url string | fetch_time int64 | content_mime_type string | warc_filename string | warc_record_offset int32 | warc_record_length int32 | text string | token_count int32 | char_count int32 | metadata string | score float64 | int_score int64 | crawl string | snapshot_type string | language string | language_score float64 |
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https://www.nagwa.com/en/videos/362162786983/ | 1,581,966,256,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143079.30/warc/CC-MAIN-20200217175826-20200217205826-00140.warc.gz | 842,981,591 | 6,173 | # Video: KS2-M17 • Paper 3 • Question 20
A square tile measures 20 cm by 20 cm. A rectangular tile is 3 cm longer and 2 cm narrower than the square tile. What is the difference in area between the two tiles in square centimeters?
04:52
### Video Transcript
A square tile measures 20 centimeters by 20 centimeters. A rectangular tile is three centimeter longer and two centimeters narrower than the square tile. What is the difference in area between the two tiles in square centimeters?
To find the difference between two areas, we simply need to take one away from the other. But in order for us to get to this calculation, we need to find out what those two areas are. And so there are several steps for us to solve this problem. Let’s start by labelling each tile with the measurements that we know. Firstly, we’re told that the square tile measures 20 centimeters by 20 centimeters. Next, we’re told that the rectangular tile is three centimeters longer and two centimeters narrower than the square tile. 20 plus three equals 23 and so we know the long side of the rectangle is 23 centimeters long. And 20 take away two equals 18. So the shorter side that’s two centimeters narrower than the square tile is 18 centimeters.
Now we can find the areas of both shapes. The area of the rectangle, and this includes squares because squares are a special type of rectangle, is found by multiplying the length by the width. And if the measurements are in centimeters, which they are here, then the area will be found in centimeters squared or square centimeters. Let’s start by calculating the area of the square tile. Its length is 20 centimeters and its width is 20 centimeters. So we need to find the answer to 20 multiplied by 20. We know that 20 multiplied by 10 equals 200. And so 20 multiplied by a number that’s double this equals 400. And so the area of the square tile is 400 centimeters squared.
Now let’s think about the rectangular tile. Again, we need to multiply the length by the width. The length of the tile is 23 centimeters and the width is 18 centimeters. 18 is very close to 20. So we could work out 23 times 20 in the same way that we’ve just worked out multiplying by 20 and then take away two lots of 23. Or, we could also use a written method, long multiplication. Let’s use this method here.
First of all, we’ll multiply 23 by eight. Three eights are 24. Two eights are 16, plus the two that we’ve exchanged equals 18. Now we need to multiply 23 by 10. We don’t need to go through each digit with this one because we know the answer. 23 times 10 equals 230. Now we just need to add each column. Four plus zero equals four. Eight plus three equals 11. And finally, one plus two plus the one that we’ve exchanged equals four. And so the area of the rectangular tile — which we found by multiplying the length by the width, 23 by 18 — is 414 centimeters squared. It’s slightly larger than the square tile.
Now we can find the difference between the two areas. And we can do this mentally just by looking at the numbers. We can see that the difference is going to be 14. 414 centimeters squared take away 400 centimeters squared equals 14 centimeters squared.
We knew that to find the difference in area between the two tiles, we had to calculate the area. And the formula for finding the area of a square or a rectangle is to multiply the length by the width. We knew the length and the width of the square tile. So we used a known fact to help us multiply 20 by 20. But we needed to use a little bit of mental maths before we were able to work out the length and the width of the rectangular tile. But once we’d found them, we knew that we had to multiply 23 by 18 to find out the answer. Then finally, we subtracted one area from the other to find out the difference in area between the two tiles. And the difference in area is 14 centimeters squared. | 912 | 3,883 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2020-10 | latest | en | 0.929869 |
http://maths.guadalupebuendia.eu/English/W1ESO/Solution7_10.htm | 1,701,828,374,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100575.30/warc/CC-MAIN-20231206000253-20231206030253-00081.warc.gz | 32,095,990 | 1,245 | WORKSHEET: PERCENTAGES- SOLUTION EXERCISE 10 10) The price of a movie ticket increases from 5 euros to 5.75 euros. What is the percentage of increase? increase: 5.75 - 5 = 0.75 → 0.75/5 = 0.15 →15% Solution: The price increases a 15% Return | 82 | 240 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2023-50 | latest | en | 0.742759 |
https://www.andlearning.org/exponents-and-powers-formulas-for-class-8/ | 1,726,546,915,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651739.72/warc/CC-MAIN-20240917040428-20240917070428-00175.warc.gz | 598,997,084 | 15,933 | # Exponents and Powers Formulas for Class 8 Maths Chapter 12
## Exponents and Powers Formulas for Class 8 Maths Chapter 12
Are you looking for Exponents and Powers formulas or important points that are required to understand Exponents and Powers for class 8 maths Chapter 12? You are the right place to get all information about Exponents and Powers Class 8 maths chapters 12. Exponents and Powers formulas play a vital role in preparing you for the class 8 exam as well as higher studies. Exponents and Powers formulas are very helpful for better scores in the exam. Check Exponents and Powers formulas according to class 8:
$\ Power\; zero = a^{ 0 } = 1$ $\ Power \;one = a^{ 1 } = a$ $\ Fraction\; formula = \sqrt{ a } = a^{ \frac{ 1 }{ 2 }}$ $\ Reverse \;formula = \sqrt{ n } { a } = a^{\frac{ 1 } { n }}$ $\ Negative \;power \;value = a^{ -n } = \frac{ 1 }{ a^{n} }$ $\ Fraction\; formula = a^{n} = \frac{1}{ a^{ -n } }$ $\ Product \;formula = a^{m}a^{n} = a^{ m + n }$ $\ Division\; Formula = \frac{ a^{ m }}{ a^{ n }} = a ^{ m-n }$ $\ Power\; of \;Power formula = (a^{ m })^{ p } = a^{ mp }$ $\ Power \;distribution Formula = (a^ { m }c^{ n })^{ x } = a ^ { mx } c ^{ nx }$ $\ The\; Power\; distribution \;Formula = \left ( \frac {a ^{ m }}{c^{ n }} \right )^{x} = \frac{a^{ mx }}{c^{ nx }}$ (-1)Even Number = 1 (-1)Odd Number = -1 (am)(an) = am+n (ab)m = ambm (am)n = amn
#### Summary of Exponents and Powers
We have shared very important formulas for Exponents and Powers which is helps to score in the class 8 exams. If you have any questions and doubts related to Exponents and Powers please let me know through comment or mail as well as social media. When you understand the formulas behind each Exponents and Powers topics then it would be easier to solve the most complex problems related to Exponents and Powers too.
Chapter-wise Maths Formulas for Class 8
### NCERT Solutions Class 8 Maths By Chapters
• Chapter 1 – Rational Numbers
• Chapter 2 – Linear Equations in One Variable
• Chapter 3 – Understanding Quadrilaterals
• Chapter 4 – Practical Geometry
• Chapter 5 – Data Handling
• Chapter 6 – Squares and Square Roots
• Chapter 7 – Cubes and Cube Roots
• Chapter 8 – Comparing Quantities
• Chapter 9 – Algebraic Expressions and Identities
• Chapter 10 – Visualising Solid Shapes
• Chapter 11 – Mensuration
• Chapter 12 – Exponents and Powers
• Chapter 13 – Direct and Inverse Proportions
• Chapter 14 – Factorisation
• Chapter 15 – Introduction to Graphs
• Chapter 16 – Playing with Numbers | 724 | 2,522 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.40625 | 4 | CC-MAIN-2024-38 | latest | en | 0.800074 |
https://bscheng.com/2016/10/23/logistic-regression-in-r/ | 1,638,571,792,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964362919.65/warc/CC-MAIN-20211203212721-20211204002721-00000.warc.gz | 215,680,415 | 26,989 | # Logistic Regression in R
My dog pants a lot. Don’t judge him. He’s brachycephalic (shorter skull/muzzle than other dogs), so he has issues thermoregulating. But he doesn’t always pant. Let me illustrate.
Panting!
Not panting! And a bonus skeptical look (“really? another post about me?”).
I’ve noticed that his panting seems to be related to temperature (shocking, I know). But it might be useful to know exactly when he may or may not be panting (and thus on the verge of overheating). Especially if I’m debating some exercise for the both of us. What if I constructed a model for this? So naturally, I collected some data and used logistic regression (or binomial modeling) to put a temperature on this panting situation.
Here’s the data.
```temp<-c(10,13.2,12.1,15,17,18,15,14.9,16.0,19,21,22,24.5,28,27,20.1,25.6,27.2,29,28.2)
dog.panting<-c(0,0,0,0,0,1,0,1,0,0,1,1,1,1,1,1,0,1,1,1)
data<-data.frame(temp, dog.panting)
```
temp = temperature degrees C
dog.panting = binary (1 = panting, 0 = not panting)
```
plot(dog.panting~temp)```
Ok, now we can construct a basic model using “glm”.
```m1<-glm(data=data,dog.panting~temp, family="binomial")
summary(m1)```
Here’s an excerpt of the summary output.
Call: glm(formula = dog.panting ~ temp, family = “binomial”, data = data)
Coefficients:
Estimate Std. Error z value Pr(>|z|)
(Intercept) -6.3202 2.7210 -2.323 0.0202 *
temp 0.3345 0.1409 2.373 0.0176 *
Null deviance: 27.526 on 19 degrees of freedom
Residual deviance: 17.201 on 18 degrees of freedom
This suggests some evidence for temperature in predicting panting (surprise!). But we should probably validate the model first. Overdispersion (greater variance than predicted by the model) is probably the first thing you would check for glms. However, given a Bernoulli glm (response variable is a vector with zeros and ones), overdispersion cannot occur (McCullagh and Nelder 1989). You could also look at predicted values versus residuals.
`plot(m1)`
For a gaussian linear model, you’d normally (heheh, get it, normally?) look for no pattern in residuals. But it’s a bit tricky for binomial glms.
Zuur et al (2009) are quite comforting when they say:
The graphical model validation in a Binomial GLM with a 0-1 response variable is some sort of an art…Because the observed data are zeros and ones, we can now see two clear bands in these graphs. This makes it rather difficult to say anything sensible about these graphs, and one can wonder whether there is any point in using them.”
Thanks Zuur et al! There are some other diagnostics to use (e.g. binning predictors and summarizing residuals, ROC plots), but there is no clear consensus. So we’ll leave that issue for a future post.
One bonus of binomial glms is that you can express the effect size of a predictor using the odds ratio. Wait, that’s odd. What are the odds again? Odds are the ratios of probabilities. The probability of occurrence (panting) over non-occurrence (not panting). Check this out and come back. I’ll wait.
Ok, good?
Good!
`exp(coef(m1))`
Here, I’ve exponentiated the coefficients from the model. Which returns this:
(Intercept) temp
0.001799644 1.397234438
So for every increase in 1 degree Celsius, the odds of Wilson panting increase by a factor of 1.4!
We might wish to plot the model predictions on top of the data. ggplot2 has some functionality here.
```library(ggplot2)
ggplot(data,aes(x=temp,y=dog.panting))+geom_point()+theme_bw()+
labs(x="Temperature (C)", y="Dog Panting")+
geom_smooth(method="glm", method.args=list(family="binomial"))```
Clearly I’ve got to collect some more data. But dare I say that this is enough data for an NSF???
For more complicated models (binomial mixed models), you may have to construct the predictions from scratch. We’ll do that next time!
Thanks again for the data Wilson!
If you’re smitten with Wilson, check out this script modeling his logistic growth!
Literature Cited
McCullagh P and J Nelder. 1989. Generalized Linear Models. Second edition. Chapman and Hall, London, UK.
Zuur, AF, EN Ieno, NJ Walker, AA Saveliev and GM Smith. 2009. Mixed Effects Models and Extensions in Ecology with R. Springer, New York, New York, USA. | 1,162 | 4,234 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2021-49 | longest | en | 0.827209 |
https://www.vedantu.com/question-answer/show-that-the-sequence-9-12-15-18-is-in-ap-find-class-10-maths-cbse-5edb5e314d8add1324c97597 | 1,721,068,763,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514713.2/warc/CC-MAIN-20240715163240-20240715193240-00076.warc.gz | 925,471,636 | 29,007 | Courses
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# Show that the sequence 9, 12, 15, 18, … is in A.P. Find its ${16^{th}}$term and the general term.
Last updated date: 13th Jul 2024
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Hint: Use the common difference to prove it is an A.P. Find a pattern to find the
general term and substitute for the ${16^{th}}$ term.
The given sequence is 9, 12, 15, 18, …
To prove that the sequence is in A.P, the common difference between any two consecutive
terms must be the same.
$\begin{gathered} 12 - 9 = 3 \\ 15 - 12 = 3 \\ 18 - 15 = 3 \\ \end{gathered}$
Hence, the common difference is the same for all consecutive terms in the series. So, it is an
A.P.
In an A.P., the first term is denoted as$a$.
$a = 9$
The common difference is denoted as$d$. It is the difference between any one term and its
previous term. It is the same for when calculated for any term in the A.P. It is also used to find the next terms in the A.P.
$d = Tn + 1 - Tn = 12 - 9 = 3$
The ${n^{th}}$term $Tn = Tn - 1 + d$
Let us find the general term first.
$\begin{gathered} T1 = 9 \\ T2 = 12 = 9 + 3\left( 1 \right) = 9 + 3\left( {2 - 1} \right) \\ T3 = 15 = 9 + 3\left( 2 \right) = 9 + 3\left( {3 - 1} \right) \\ T4 = 18 = 9 + 3\left( 3 \right) = 9 + 3\left( {4 - 1} \right) \\ \end{gathered}$
Hence, we can generalize this A.P. as $Tn = 9 + 3\left( {n - 1} \right)$ …(1)
We can substitute any value for $n$ to find the ${n^{th}}$ term.
The ${16^{th}}$term can be found by substituting $n = 16$in (1)
$\begin{gathered} Tn = 9 + 3\left( {n - 1} \right) \\ T16 = 9 + 3\left( {16 - 1} \right) = 9 + 15\left( 3 \right) = 54 \\ \end{gathered}$
Note: The general term in an A.P. is found by finding the pattern in which the A.P. progresses.
With the general term, any term in the A.P. can be found easily. | 684 | 1,852 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.84375 | 5 | CC-MAIN-2024-30 | latest | en | 0.876532 |
http://nrich.maths.org/1518/note | 1,505,996,913,000,000,000 | text/html | crawl-data/CC-MAIN-2017-39/segments/1505818687766.41/warc/CC-MAIN-20170921115822-20170921135822-00291.warc.gz | 250,579,030 | 4,206 | # World of Tan 27 - Sharing
##### Stage: 2 Challenge Level:
Children you might like to:
• List all the different instances when you use fractions.
• Think about how you met fractions in school and what you have been taught. What can you remember?
• Convince someone that a quarter is bigger than a fifth. (hint: think of sweets)
• Take a line make it half as big again. Draw a shape on dotty paper - now make that half as big again.
Parents you might like to:
• Explain in what contexts you use fractions at work and in the home.
• Investigate the fractional parts of a pound before the decimalisation of our currency in 1972.
• Convince someone a quarter is bigger than a fifth.
• Investigate the idea of ratio when mixing water and squash or when sharing out meals/ sweets etc.
• Explore the use of fractions within the context of time.
Teachers you might like to:
• Explore the Tangram to explain and illustrate the 'sixteenths' family of fractions.
• Consider the differing ways of re' presenting fractional quantities.
• Explain why "we turn one fraction upside down and multiply out"
• Investigate mixing liquids/ sharing sweets in differing ratios.
• Develop the vocabulary of fractions and illustrate the different members of the fraction family - vulgar fractions, improper fractions, decimal fractions, etc.
• Halve a square in as many different ways as possible. | 294 | 1,374 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2017-39 | latest | en | 0.933387 |
https://relivingmbadays.wordpress.com/2012/12/18/market-equilibrium/ | 1,500,633,946,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549423769.10/warc/CC-MAIN-20170721102310-20170721122310-00332.warc.gz | 712,009,295 | 35,948 | # Market Equilibrium
A market is an arrangement that brings buyers and sellers together. Markets exist in all types of goods and services, and as economists, we are interested in how they work and what causes them to change.
The market price is determined by the interaction of market supply (producers) and market demand (consumers).
When the quantity of a product demanded equals the quantity supplied at the prevailing market price, this is called market equilibrium. When a market reaches equilibrium, there is no pressure to change the price.
The point at which the quantity demanded equals the quantity supplied is the equilibrium point. This point states the price of the good (P1) and the market quantity (Q1)
Let us look at an example to understand how it works:
The below schedule shows the quantity supplied and demanded are equal at Rs.40. This is the intersection of supply and demand of the price of sugar people are willing to buy 1 kg sugar and retailer willing to sell sugar at a price for 1 kg
Supplied Price / kg Demanded 20 30 80 50 40 50 80 50 20
If you plot the above able on the graph, you will get the demand and supply diagram of sugar at three various prices. And the equilibrium i.e. the quantity demanded equal to the quantity supplied is at Rs. 40 per kg.
The diagram is shown below:
Mathematically this can be shown with the supply curve that is the function: Qs = 2P where Qs is the quantity supplied at each price, P. The demand curve is QD = 200 ā 3P where QD us quantity demanded at each price, P.
The equilibrium can then be calculated by setting the equation for quantity supplied equal to the equation for quantity demanded:
QS = QD
2P = 200 ā 3P
Solve to find P
5P = 200
P = 40
The equilibrium quantity can then be calculated by substituting the equilibrium price of Rs. 40 for P in the equations and solving for quantity supplied, QS, or quantity demanded QD
QS = 2(40) = 80
QD = 200-3(40) = 80
Assuming that neither curve shifts, then market forces will maintain the equilibrium price. For instance, assume that the price rises above P1, then the firms will react by wishing to supply more (the price is higher, therefore, the revenue will be higher), at the same time consumers will demand less. The outcome is that there is excess supply. In other words, supply is greater than demand.
This situation results in producers having unsold stocks. In this case, producers will wish to sell stocks as they cost money to produce and maintain. Therefore, to sell them they will reduce the price of the good (contraction in supply). The lower price will encourages more demand for the good (extension in demand). This process continues until the supply and demand are again in equilibrium.
If the position of either the demand and / or supply curve shifts, then the equilibrium price and quantity will change. For instance, if the good becomes more fashionable, then the demand curve will shift from D1 to D2
The new equilibrium price will be P
In conclusion, the market forces of supply and demand interact to bring about the equilibrium price, clearing the market of excess demand or supply. In this way, it is said that the market mechanism achieves consistency between the plans and outcomes for consumers and producers without explicit coordination. | 730 | 3,317 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2017-30 | longest | en | 0.914322 |
https://www.justtutors.com/free-study-material/worksheet-cbse-class-9-maths-chapter-10-to-illustrate-and-name-the-parts-of-circle-eg-radius-diameter-chord-arcs-segments-sectors-and-circumference | 1,606,889,855,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141692985.63/warc/CC-MAIN-20201202052413-20201202082413-00096.warc.gz | 720,041,379 | 14,919 | Worksheet For CBSE Class 9 Maths Chapter 10 To Illustrate And Name The Parts Of Circle Eg Radius Diameter Chord Arcs Segments Sectors And Circumference
### Worksheet For CBSE Class 9 Maths Chapter 10 To Illustrate And Name The Parts Of Circle Eg Radius Diameter Chord Arcs Segments Sectors And Circumference
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### Lesson : to illustrate and name the parts of circle, e.g. radius, diameter, chord, arcs, segments, sectors and circumference
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### Chapters
Chapter 1 Number System
Number System-Rational Number
Number System-Irrational Number and The Number line
Number System-Real number and operations
Chapter 2 Polynomials
Polynomial-Basics of polynomial
Polynomial-Remainder Theorem
Polynomial-Identity
Chapter 3 Coordinate Geometry
Co-Ordinate Geometry-Co-Ordinate Geometry and various terms
Chapter 4 Linear equations in two variables
Linear Equation of two Variable-Linear equation and solution
Linear Equation of two Variable-Linear equation and graph
Chapter 5 Introduction to Euclid's Geometry
Introduction to Euclid's Geometry-Introduction
Chapter 6 Lines and Angles
Lines and Angles-Angles and properties
Lines and Angles-Angles and types
Lines and Angles-Angle Sum Property
Chapter 7 Triangles
Triangles-Congruency of Triangle
Triangles-Properties of Triangle
Triangles-Criteria for congruency
Chapter 9 Areas of Parallelograms and Triangles
Area of Parallelogram and Triangle-Triangles on same base and B/W same parallel lines
Chapter 10 Circles
Circle-Circle and its terms
Circle-Chord and circle
Circle-Arc and Circle
Chapter 11 Constructions
Constructions-Introduction
Constructions-Basic Constructions
Constructions-Some Constructions of triangle
Chapter 12 Heron's Formula
Heron's Formula-Area of a Triangle – by Heron's Formula
Heron's Formula-Application of Heron's Formula
Chapter 13 Surface Areas and Volumes
Surface area and Volume-Surface area and Volume of cube and cuboid
Surface area and Volume-Surface area and volume of cylinder
Surface area and Volume-Surface area and volume of cone
Surface area and Volume-Surface area and volume of Sphere
Chapter 14 Statistics
Statistics-Tally Marks
Statistics-Mean
Statistics-Median and Mode
Chapter 15 Probability
Probability-Probability | 637 | 2,901 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2020-50 | latest | en | 0.801231 |
http://mathematica.ludibunda.ch/trigonometry3.html | 1,718,604,415,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861698.15/warc/CC-MAIN-20240617060013-20240617090013-00880.warc.gz | 17,469,291 | 2,944 | Intro Using Trigonometry Definitions Functions Radiens Modifying Curves Adding Curves
Playground
Rapunzel
Dido's Problem
Pythagoras
Trigonometry
Smart Joe
Fuzzy Logic
Cryptography
Mathematicians
After some time, people started to name the ratios of sides in right triangles in relation to one of the acute angles. And since the ratios depend upon the size of the angle, they wrote tables filled with these ratios for different angles. The ratio Dido used to calculate the distance between her boat and the coast is called tangent.
The tangent of an acute angle in a right triangle is defined as the ratio between the side which is opposite of the angle and the adjacent leg.
In equations, tangent is usually just written as tan.
In the picture to the above, the tangent of angle A is equal to BC/AB.
(Note: BC and AB are the length of the lines connecting B with C and A with B)
Instead of drawing a second triangle and measuring its sides, Dido could have just measured the acute angle on the boat, looked up the tangent for that angle and multiplied it by the distance of the two points she had chosen on the boat and like this she would have gotten the distance her boat had from the shore.
This would have worked allright, except for one slight problem: the first trigonometric tables known today were written about 140 BC, and Dido lived way earlier, around 800 BC.
But you don't even need a table! Get your calculator! Dido found that the angle at A is about 78.7°. Enter the numbers of degrees into your calculater and then press on tan. You will see that the tangent of 78.7° equals about 5.
In the triangle to the left you can see that the tangent of angle A is equal to the ratio between the sides BC and AB. In our case AB=tanA×BC. Since tanA equals approximately 5 and Dido measured 32feet for AB, BC equals 160feet. This is the same result that Dido got using her method. | 443 | 1,902 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2024-26 | latest | en | 0.957992 |
http://slideplayer.com/slide/3329145/ | 1,529,543,228,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863980.55/warc/CC-MAIN-20180621001211-20180621021211-00605.warc.gz | 290,475,680 | 32,670 | # Volume & Surface Area of Solids Revision of Area
## Presentation on theme: "Volume & Surface Area of Solids Revision of Area"— Presentation transcript:
Volume & Surface Area of Solids www.mathsrevision.com Revision of Area
Revision of Volume and Surface Area Volume of a Prism Volume of a Cylinder Volume of a Pyramid Curved Area of a Cylinder Volume of a Sphere Exam Type Questions
Starter Questions a2 – 3b2 = 57 Q1. True or false
Q2. Write down the probability of picking out a number greater than 20 in the national lottery. Q3. If a = -3 and b = -4 does a2 – 3b2 = 57 Q4. Calculate Tuesday, 11 April 2017 Created by Mr.Lafferty
Revision of Areas www.mathsrevision.com Any Type of Triangle
Level 4 Any Type of Triangle Rhombus and kite Parallelogram Trapezium Circle
Compiled by Mr. Lafferty Maths Dept.
Area Level 4 Learning Intention Success Criteria We are revising area of basic shapes. Know formulae. Use formulae correctly. Show working and appropriate units. 11-Apr-17 Compiled by Mr. Lafferty Maths Dept.
Area Example : Find the area of the V – shape kite.
Level 4 Example : Find the area of the V – shape kite. 4cm 7cm Tuesday, 11 April 2017 Created by Mr.Lafferty
Composite Areas We can use our knowledge of the basic areas
Level 4 We can use our knowledge of the basic areas to work out more complicated shapes. Example : Find the area of the arrow. 5cm 3cm 6cm 4cm Tuesday, 11 April 2017 Created by Mr.Lafferty
Composite Areas 8cm www.mathsrevision.com 11cm 10cm
Level 4 Example : Find the area of the shaded area. 8cm 11cm 4cm 10cm Tuesday, 11 April 2017 Created by Mr.Lafferty
Created by Mr. Lafferty Maths Dept.
Composite Areas The Circle Level 4 Example : Find the area of the shape Area = rectangle + semicircle 20cm 5 cm 11-Apr-17 Created by Mr. Lafferty Maths Dept.
Area Now try TJ 4+ Ex 16.1 Ch16 (page 121) www.mathsrevision.com
Level 4 Now try TJ 4+ Ex 16.1 Ch16 (page 121) Tuesday, 11 April 2017 Created by Mr.
Q4. Rearrange into the form y =
Starter Questions Q1. Find the area of the triangle. 10cm 3cm 4cm Q2. Expand out and simplify 2w2 – 3(2w – 5) Q3. True or false Q4. Rearrange into the form y = 2y – 3x + 7 = 0 Tuesday, 11 April 2017 Created by Mr.Lafferty
Compiled by Mr. Lafferty Maths Dept.
Volume & Surface Area Level 4 Learning Intention Success Criteria We are revising volume and surface area of a cuboid. Know formulae. Use formulae correctly. Show working and appropriate units. 11-Apr-17 Compiled by Mr. Lafferty Maths Dept.
Compiled by Mr. Lafferty Maths Dept.
Volume of a cuboid 3cm 4cm 6cm Level 4 18 cubes fit the base. = 1 centimetre cube = 1 cm³ 4 layers of 18 cubes = 4 x 18 = 72 centimetre cubes = 72 cm³ 11-Apr-17 Compiled by Mr. Lafferty Maths Dept.
Compiled by Mr. Lafferty Maths Dept.
A short cut ! 3cm 4cm 6cm Level 4 height Area of rectangle breadth length Volume = 6 x 3 x 4 = 72 cm³ Volume = length x breadth x height 11-Apr-17 Compiled by Mr. Lafferty Maths Dept.
Compiled by Mr. Lafferty Maths Dept.
Example Working Volume = l x b x h Level 4 Heilander’s Porridge Oats V = 18 x 5 x 27 V = cm³ 27cm 5 cm 18 cm 11-Apr-17 Compiled by Mr. Lafferty Maths Dept.
Compiled by Mr. Lafferty Maths Dept.
Example Working Volume = l x b x h Level 4 V = 2 x 2 x 2 V = 8 cm³ 2cm 11-Apr-17 Compiled by Mr. Lafferty Maths Dept.
Compiled by Mr. Lafferty Maths Dept.
Example Find the volume of the composite shape. Level 4 VT = V1 + V2 7 cm 3 cm 5 cm VT = 10 cm 9 cm 8 cm VT = 825 cm3 V1 = l x b x h = 3 x 5 x 7 V2 = l x b x h = 105 cm³ = 8 x 10 x 9 = 720 cm³ 11-Apr-17 Compiled by Mr. Lafferty Maths Dept.
Find the length the cuboid
Example Find the length the cuboid This just an equation. We know how to solve them ! Level 4 4 cm Volume = L x B x H V=200cm3 200 = L x 5 x 4 5 cm 200 = 20L 10 cm L 20L = 200 L = 10 11-Apr-17 Compiled by Mr. Lafferty Maths Dept.
Compiled by Mr. Lafferty Maths Dept.
Example Liquid Volume Working Volume = l x b x h Level 4 V = 100 x 30 x 50 V = cm³ 50 cm = ml = 150 litres 30 cm 100 cm How much water can this fish tank hold in litres? 1cm3 = 1 ml 1000 ml = 1 litre So the fish tank can hold 150 litres of water. 11-Apr-17 Compiled by Mr. Lafferty Maths Dept.
Face Edges and Vertices
Don’t forget the faces edges and corners we can’t see at the back Face Edges and Vertices Level 4 The shape below is called a cuboid. It is made up of FACES, EDGES and VERTICES. Edges are where the two faces meet (lines) Faces are the sides of a shape (surface area) Vertices where lines meet (corners) 11-Apr-17 Compiled by Mr. Lafferty Maths Dept.
Face Edges and Vertices
Calculate the number of faces edges and vertices for a cuboid. Face Edges and Vertices Level 4 6 faces 12 edges Front and back are the same 8 vertices Top and bottom are the same Right and left are the same 11-Apr-17 Compiled by Mr. Lafferty Maths Dept.
Face Edges and Vertices
Calculate the number of faces edges and vertices for a cube. Face Edges and Vertices Level 4 6 faces 12 edges Faces are squares 8 vertices 11-Apr-17 Compiled by Mr. Lafferty Maths Dept.
Find the surface area of the cuboid
Example Find the surface area of the cuboid Working Level 4 Front Area = l x b = 5 x 4 =20cm2 Top Area = l x b = 5 x 3 =15cm2 4cm Side Area = l x b = 3 x 4 =12cm2 3cm 5cm Total Area = = 94cm2 Front and back are the same Top and bottom are the same Right and left are the same 11-Apr-17 Compiled by Mr. Lafferty Maths Dept.
Now try TJ 4+ Ex 16.2 Ch16 (page 122) Volume & Surface Area
Level 4 Now try TJ 4+ Ex 16.2 Ch16 (page 122) Tuesday, 11 April 2017 Created by Mr.
Starter Questions Q1. True or false 2(x – 6) - 2(x + 6) = 0
Q2. Does x 20 = Explain your answer Q3. Factorise 2y2 + 3y +2 Q4. Calculate Tuesday, 11 April 2017 Created by Mr.Lafferty
Volume of Prisms www.mathsrevision.com Learning Intention
Level 4 Learning Intention Success Criteria We are learning how to calculating volume of any prism given area. 1. Calculate the volume for various prisms. 2. Solution must include appropriate units and working.
Volume of Prisms www.mathsrevision.com
Level 4 Definition : A prism is a solid shape with uniform cross-section Hexagonal Prism Cylinder (circular Prism) Triangular Prism Pentagonal Prism Volume = Area of Cross section x length
Volume of Prisms www.mathsrevision.com
Level 4 Definition : A prism is a solid shape with uniform cross-section Hexagonal Prism Cylinder (circular Prism) Triangular Prism Pentagonal Prism Volume = Area of Cross section x length
Volume of Solids www.mathsrevision.com
Definition : A prism is a solid shape with uniform cross-section Q. Find the volume the triangular prism. Triangular Prism Volume = Area x length = 20 x 10 = 200 cm3 10cm 20cm2
Volume of Prisms Now try TJ 4+ Ex 16.3 Ch16 (page 124)
Level 4 Now try TJ 4+ Ex 16.3 Ch16 (page 124) Tuesday, 11 April 2017 Created by Mr.
Starter Questions Q1. Find the area of the parallelogram Q2. Factorise
7 7 Q2. Factorise 4x + 40 Q3. A can of beans is reduce by 15% to 25p. Find the price before the reduction. Q4. The speed of light is metres per sec. True or false in scientific notation 3 x 108. Tuesday, 11 April 2017 Created by Mr.Lafferty
Volume of a Cylinder www.mathsrevision.com Learning Intention
Level 4 Learning Intention Success Criteria We are learning how to derive the formula for the volume of a cylinder and apply it to solve problems. To know formula. Apply formula correctly. Work backwards using formula.
Volume of a Cylinder Volume = Area x height = πr2 = πr2h
Level 4 The volume of a cylinder can be thought as being a pile of circles laid on top of each other. Volume = Area x height h = πr2 x h Cylinder (circular Prism) = πr2h
Volume of a Cylinder V = πr2h = π(5)2x10 = 250π cm3 = 784.5 cm3
Example : Find the volume of the cylinder below. 5cm Cylinder (circular Prism) 10cm V = πr2h = π(5)2x10 = 250π cm3 = cm3
Volume of a Cylinder Now try TJ 4+ Ex 16.4 Ch16 (page 125)
Level 4 Now try TJ 4+ Ex 16.4 Ch16 (page 125) Tuesday, 11 April 2017 Created by Mr.
Starter Questions -(a2) + b2 = 5 Q1. Factorise
Q2. Write down the probability of picking out a number less than 30 in the national lottery. Q3. True or false if a = -1 and b = -2 -(a2) + b2 = 5 Q4. Explain why Tuesday, 11 April 2017 Created by Mr.Lafferty
Volume of a Pyramid www.mathsrevision.com Learning Intention
Level 4 Learning Intention Success Criteria We are learning how to use the formula for the volume of ANY pyramid and apply it to solve problems. To know formula. Apply formula correctly. Work backwards using formula.
Volume of a Pyramid www.mathsrevision.com
Level 4 cone The volume of any pyramid can be calculated using the formula A = Area of base h = height
Volume of Pyramid www.mathsrevision.com
Level 4 Q. Find the volume the pyramid. 12m 100m2
Volume of a Cone www.mathsrevision.com h r
Level 4 h r If the above cone has radius 15cm and height of 10 cm. Calculate it’s volume.
Volume of a Pyramid Now try TJ 4+ Ex 16.5 Ch16 (page 127)
Level 4 Now try TJ 4+ Ex 16.5 Ch16 (page 127) Tuesday, 11 April 2017 Created by Mr.
Starter Questions Q1. Expand out and simplify 3(x – 2) - ( - 3x + 4)
Q2. Factorise 2x2 – 16x Q3. True or false Q4. By rearranging in y = , find the gradient and where the straight line crosses the x-axis y + 4x - 3 = 0 Tuesday, 11 April 2017 Created by Mr.Lafferty
Surface Area of a Cylinder www.mathsrevision.com Learning Intention
Level 4 Learning Intention Success Criteria We are learning how to calculate the surface area of a cylinder by using basic areas. To know how to split up a cylinder. 2. Calculate the surface area of a cylinder.
Surface Area of a Cylinder 2πr h Total Surface Area = 2πr2 + 2πrh
Level 4 The surface area of a cylinder is made up of 2 basic shapes can you name them. Cylinder (circular Prism) 2πr Curved Area =2πrh Top Area =πr2 Roll out curve side h Bottom Area =πr2 Total Surface Area = 2πr2 + 2πrh
Surface Area of a Cylinder = 2π(3)2 + 2π x 3 x 10 = 18π + 60π
Level 4 Example : Find the surface area of the cylinder below: 3cm Surface Area = 2πr2 + 2πrh 10cm = 2π(3)2 + 2π x 3 x 10 = 18π + 60π Cylinder (circular Prism) = 245 cm2
Surface Area of a Cylinder D = 25 25 D = π www.mathsrevision.com
Diameter = 2r Example : A net of a cylinder is given below. Find the diameter of the tin and the total surface area. D = 25 25 π 9cm 25cm D = = 7.96 cm Surface Area = 2πr2 + 2πrh = 2π(3.98) 2 + 2π(3.98)x9 = = cm2
Now try TJ 4+ Ex 16.6 Ch16 (page 129) Surface Area of a Cylinder
Level 4 Now try TJ 4+ Ex 16.6 Ch16 (page 129) Tuesday, 11 April 2017 Created by Mr.
Starter Questions Q1. Find the area of the triangle. 10cm
Q2. Factorise 4x2 – 64x 6cm Q3. Calculate Q4. Rearrange equation in y = y – 2x + 5 = 0 Tuesday, 11 April 2017 Created by Mr.Lafferty
Volume of Solids www.mathsrevision.com Learning Intention
Level 4 Learning Intention Success Criteria We are learning how to use the sphere formula and use it to solve real-life problems. To know the volume formula for a sphere. Work out volumes for spheres. Answer to contain appropriate units and working.
Volume of a Sphere www.mathsrevision.com D = diameter r
Level 4 D = diameter r D Q. If the above sphere has radius 10cm. Calculate it’s volume.
Volume of a sphere Q. Find the volume the composite shape.
Level 4 Q. Find the volume the composite shape. Volume = Cylinder + half a sphere ½ sphere Cylinder r 2m h = 6m
Now try TJ 4+ Ex 16.7 Ch16 (page 130) Volume of a sphere
Level 4 Now try TJ 4+ Ex 16.7 Ch16 (page 130) Tuesday, 11 April 2017 Created by Mr.
Part b on next slide | 3,488 | 11,480 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.34375 | 4 | CC-MAIN-2018-26 | latest | en | 0.809979 |
http://copaseticflow.blogspot.com/2013/01/lorentz-special-relativity-violation.html | 1,550,457,525,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550247484020.33/warc/CC-MAIN-20190218013525-20190218035525-00228.warc.gz | 64,035,577 | 28,773 | Skip to main content
### Lorentz Special Relativity Violation Notes
I'm trying something new today. I've been doing some background reading on the supposed violation of special relativity reported by M Mansuripur in PRL and pointed out by +Cliff Harvey earlier this week. The embedded document below is a linked diagram of the background articles I've been reading. An arrow indicates that the article referenced the pointed at article. My goal in reading about this was to learn about the 'hidden momentum' issue mentioned in Mansuripur's article. The articles listed here are the earliest and, in my opinion, the most basic articles I could find on the subject.
### Cool Math Tricks: Deriving the Divergence, (Del or Nabla) into New (Cylindrical) Coordinate Systems
The following is a pretty lengthy procedure, but converting the divergence, (nabla, del) operator between coordinate systems comes up pretty often. While there are tables for converting between common coordinate systems, there seem to be fewer explanations of the procedure for deriving the conversion, so here goes!
What do we actually want?
To convert the Cartesian nabla
to the nabla for another coordinate system, say… cylindrical coordinates.
What we’ll need:
1. The Cartesian Nabla:
2. A set of equations relating the Cartesian coordinates to cylindrical coordinates:
3. A set of equations relating the Cartesian basis vectors to the basis vectors of the new coordinate system:
How to do it:
Use the chain rule for differentiation to convert the derivatives with respect to the Cartesian variables to derivatives with respect to the cylindrical variables.
The chain rule can be used to convert a differential operator in terms of one variable into a series of differential operators in terms of othe…
### The Valentine's Day Magnetic Monopole
There's an assymetry to the form of the two Maxwell's equations shown in picture 1. While the divergence of the electric field is proportional to the electric charge density at a given point, the divergence of the magnetic field is equal to zero. This is typically explained in the following way. While we know that electrons, the fundamental electric charge carriers exist, evidence seems to indicate that magnetic monopoles, the particles that would carry magnetic 'charge', either don't exist, or, the energies required to create them are so high that they are exceedingly rare. That doesn't stop us from looking for them though!
Keeping with the theme of Fairbank[1] and his academic progeny over the semester break, today's post is about the discovery of a magnetic monopole candidate event by one of the Fairbank's graduate students, Blas Cabrera[2]. Cabrera was utilizing a loop type of magnetic monopole detector. Its operation is in concept very simpl…
### Unschooling Math Jams: Squaring Numbers in their own Base
Some of the most fun I have working on math with seven year-old No. 1 is discovering new things about math myself. Last week, we discovered that square of any number in its own base is 100! Pretty cool! As usual we figured it out by talking rather than by writing things down, and as usual it was sheer happenstance that we figured it out at all. Here’s how it went.
I've really been looking forward to working through multiplication ala binary numbers with seven year-old No. 1. She kind of beat me to the punch though: in the last few weeks she's been learning her multiplication tables in base 10 on her own. This became apparent when five year-old No. 2 decided he wanted to do some 'schoolwork' a few days back.
"I can sing that song... about the letters? all by myself now!" 2 meant the alphabet song. His attitude towards academics is the ultimate in not retaining unnecessary facts, not even the name of the song :)
After 2 had worked his way through the so… | 834 | 3,848 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2019-09 | latest | en | 0.893216 |
https://socratic.org/questions/how-do-you-evaluate-2-2-4-times-2-3 | 1,723,126,352,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640728444.43/warc/CC-MAIN-20240808124926-20240808154926-00070.warc.gz | 428,225,186 | 5,877 | How do you evaluate 2^{2}+4\times 2+3?
Dec 6, 2017
$15$
Explanation:
Since your equation does not have any brackets, let’s move and solve the exponents.
Note: I bracketed the parts you need to focus on.
$\left[{2}^{2}\right] + 4 \cdot 2 + 3 = 4 + 4 \cdot 2 + 3$
Following BEDMAS, there is no division, but there is multiplication, so we have to solve that next.
$4 + \left[4 \cdot 2\right] + 3 = 4 + 8 + 3$
And finally, we add everything together.
$\left[4 + 8 + 3\right] = 15$ | 171 | 486 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 4, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2024-33 | latest | en | 0.836554 |
https://le.ac.uk/cse/research/facilities/hercules/case-studies/the-2d-problem | 1,718,593,346,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861696.51/warc/CC-MAIN-20240617024959-20240617054959-00039.warc.gz | 322,097,853 | 55,434 | College of Science and Engineering
# The 2D Problem
## The 2D Problem
### Limitation of 2D
The diagram below the cross section of several 3D-shapes. If you presented just the cross section and were asked to guess what shape you could guess any of the three and depending on where we sectioned the shapes, we could also easily be confused as to the scale of the 3D volume.
Fig 1: (left to right) cross sections of a sphere, cylinder and ellipsoid. Regardless of the 3D-shape, all three can give the same circular cross section.
Without context, a 2D cross-section can be misleading. This is an issue with traditional techniques in metallography and geology as microstructures can look very different when viewed from different directions; an example is from super-duplex stainless steel is illustrated here. Typically, a material is cut, polished and then only a section of the material is analysed, showing only part of the entire picture. Although this might be done in several orthogonal directions (as below) to provide more information this is not only time intensive, it also requires the sample to be destroyed, meaning no further testing can occur. Hercules offers a different option though, through a technique called Diffraction Contrast Tomography (DCT).
Fig 2: example optical micrographs from S32750 super-duplex stainless steel. Grains appear equiaxed when viewed in (a) cross-section normal to the rolling directions however, cross sections in the two rolling directions, (b) longitudinal and c) traverse directions, clearly show elongated grains.[1]
### Why is Diffraction Contrast Tomography different from MicroCT?
Like medical x-rays, MicroCT is based on contrast caused by different materials absorbing different percentages of incident x-rays (Absorption Contrast Tomography) reducing the transmitted intensity. By rotating the sample, we get multiple projections allowing for a 3D-reconstruction of the sample.
This method alone isn’t very applicable if we wish to determine the microstructure of a single composition material.
Diffraction contrast tomography (DCT) is a non-destructive technique for the 3D characterisation of polycrystals containing up to a few thousand grains. While it is an established technique at synchrotrons , it is still an emerging technique for lab based x-ray sources which only becomes available, as part of the ZEISS Hercules solution. When the sample rotates, some crystalline phases within the sample will fulfil the Bragg condition, meaning that some x-rays will be diffracted as well as absorbed. This gives both a dark spot in our transmitted image, indicating the real space position of the diffracting crystal, but also a diffraction pattern. By measuring the diffraction patterns at many angles and combining this with knowledge of the material being analysed, information about the orientation of the crystal can be calculated as well as the grain shape and size.
Fig 3: Schematic diagram of MicroCT and DCT technique.
### Why use DCT rather than Electron Back Scatter Diffraction (EBSD)?
EBSD, while fantastic at giving us fine detail about orientation and engineering strain, relies on cross sections that are time consuming to prepare and only obtains part of the available information about the sample. While we could use 3D-EBSD to reconstruct a 3D-volume, the scale at which this is practical is much smaller than DCT and more importantly is destructive.
Moreover, as DCT is non-destructive, we can take more than just a single snapshot. A tensile (need to check with Zeiss, do they have a better example that shows sample at different stages?) specimen of Al-4wt%Cu is illustrated here with the measurement done with DCT. Not only can we see the grain orientation, shown in the Inverse Pole Figure colour (as in EBSD), but also we can take snapshots showing us the evolution of the grain structure. The X-ray setup at Hercules can apply loads up to 5kN as well as heating or cooling the sample while it is being scanned.
Fig 4: Diffraction Contrast Tomography map of Al-4wt%Cu tensile sample.
### Correlating DCT and EBSD
Of course, if extra detail is required, Hercules also has further capability to take a region of interest identified in the DCT scan and prepare this for further EBSD studies, giving more localised information on orientation and strain within the sample. This multiscale approach provides both the wider context of the sample through its life time as well as the fine detail at the critical moments.
• [1]Modified from TF. Santos et al. Materials Research. 2016; 19(1): 117-131 | 988 | 4,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-26 | latest | en | 0.920498 |
https://palmgardensal.com/qa/question-what-is-a-line-and-a-ray.html | 1,606,865,172,000,000,000 | text/html | crawl-data/CC-MAIN-2020-50/segments/1606141685797.79/warc/CC-MAIN-20201201231155-20201202021155-00087.warc.gz | 359,129,030 | 9,204 | # Question: What Is A Line And A Ray?
## What is the meaning of angles?
In plane geometry, an angle is the figure formed by two rays, called the sides of the angle, sharing a common endpoint, called the vertex of the angle.
Angles formed by the intersection of two curves in a plane are defined as the angle determined by the tangent rays at the point of intersection..
## How do you represent a line?
A line segment is represented by end points on each end of the line segment. A line in geometry is represented by a line with arrows at each end. A line segment and a line are different because a line goes on forever while a line segment has a distinct beginning and end.
## How many rays are there in a line?
There are six rays. Each of the three points can be an endpoint, and from each end point the ray can go either direction on the line. That is two rays per end point, and three endpoints, for a total of six rays . Why SFD of UDL is straight line?
## Is straight an angle?
A straight angle is exactly 180 degrees and is composed of exactly one line with no changes in vector. Any line through it will form four angles (two on either side of the straight angle). Additionally, any two contiguous angles must be supplementary or equal 180 degrees.
## How do I measure an angle?
The best way to measure an angle is to use a protractor. To do this, you’ll start by lining up one ray along the 0-degree line on the protractor. Then, line up the vertex with the midpoint of the protractor. Follow the second ray to determine the angle’s measurement to the nearest degree.
## What is the difference between a line and a ray?
Student: So, what is the difference between a line and a ray? Mentor: A line goes on to infinity in both directions, but a ray stops on one end. If you cut a line in half, you make two rays. … A line segment is part of a line that has two endpoints.
## What is a ray line in math?
A part of a line with a start point but no end point (it goes to infinity) Try moving points “A” and “B”: line. ray. line segment.
## What are the types of angle?
As the Angle Increases, the Name Changes:Type of AngleDescriptionAcute Angleis less than 90°Right Angleis 90° exactlyObtuse Angleis greater than 90° but less than 180°Straight Angleis 180° exactly2 more rows
## What is a ray line and line segment?
a ray is a line that has one endpoint and on one side it goes on forever. a line goes on forever in both directions. a line segment has two endpoints. Reply.
## What are the subsets of a line?
In geometry, a line is a perfectly straight one-dimensional figure extending infinitely in both directions. There are two subsets, or subcategories, of lines in geometry: line segments and rays.
## Which figure is a line?
A line is a one-dimensional figure that is made up of an infinite number of individual points placed side by side. In geometry all lines are assumed to be straight; if they bend they are called a curve. A line continues infinitely in two directions.
## What is a line with two arrows called?
We can illustrate that by little arrows on both ends. We can name a line using two points on it. This is line EF or line (note the arrowheads).
## What is collinear point?
Three or more points , , , …, are said to be collinear if they lie on a single straight line. . A line on which points lie, especially if it is related to a geometric figure such as a triangle, is sometimes called an axis. Two points are trivially collinear since two points determine a line.
## What is an A ray?
A ray is a part of a line that has one endpoint and goes on infinitely in only one direction. You cannot measure the length of a ray. A ray is named using its endpoint first, and then any other point on the ray (for example, →BA ).
## What is a common Ray?
of Wisconsin. A ray is part of a line, has one fixed endpoint, and extends infinitely along the line from the endpoint. Opposite math rays are rays with a common endpoint, extending in opposite directions and forming a line.
## What is a ray and an angle?
Measure and classify an angle. A line that has one defined endpoint is called a ray and extends endlessly in one direction. A ray is named after the endpoint and another point on the ray e.g. … Angles can be either straight, right, acute or obtuse. An angle is a fraction of a circle where the whole circle is 360°.
## How do you identify a ray?
In naming a ray, we always begin with the letter of the endpoint (where the ray starts) followed by another point on the ray in the direction it travels. Since the vertex of the angle is the endpoint of each ray and our vertex is , each of our rays must begin with . Only fails to do so.
## What is an example of a line segment?
Examples of line segments include the sides of a triangle or square. More generally, when both of the segment’s end points are vertices of a polygon or polyhedron, the line segment is either an edge (of that polygon or polyhedron) if they are adjacent vertices, or otherwise a diagonal.
## What is an example of a Ray?
In geometry, a ray is a line with a single endpoint (or point of origin) that extends infinitely in one direction. An example of a ray is a sun ray in space; the sun is the endpoint, and the ray of light continues on indefinitely.
## What are the 7 types of angles?
Types of Angles – Acute, Right, Obtuse, Straight and Reflex…Acute angle.Right angle.Obtuse angle.Straight angle.Reflex angle. | 1,240 | 5,452 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.46875 | 4 | CC-MAIN-2020-50 | latest | en | 0.95358 |
https://formulasearchengine.com/wiki/Phase_(waves) | 1,568,775,607,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514573176.51/warc/CC-MAIN-20190918024332-20190918050332-00224.warc.gz | 483,826,026 | 11,825 | # Phase (waves)
{{ safesubst:#invoke:Unsubst||$N=Merge from |date=__DATE__ |$B= Template:MboxTemplate:DMCTemplate:Merge partner }} Phase in sinusoidal functions or in waves has two different, but closely related, meanings. One is the initial angle of a sinusoidal function at its origin and is sometimes called phase offset or phase difference. Another usage is the fraction of the wave cycle that has elapsed relative to the origin.[1]
## Formula
The phase of an oscillation or wave refers to a sinusoidal function such as the following:
{\displaystyle {\begin{aligned}x(t)&=A\cdot \cos(2\pi ft+\varphi )\\y(t)&=A\cdot \sin(2\pi ft+\varphi )=A\cdot \cos \left(2\pi ft+\varphi -{\tfrac {\pi }{2}}\right)\end{aligned}}}
where ${\displaystyle \scriptstyle A\,}$, ${\displaystyle \scriptstyle f\,}$, and ${\displaystyle \scriptstyle \varphi \,}$ are constant parameters called the amplitude, frequency, and phase of the sinusoid. These functions are periodic with period ${\displaystyle \scriptstyle T={\frac {1}{f}}\,}$, and they are identical except for a displacement of ${\displaystyle \scriptstyle {\frac {T}{4}}\,}$ along the ${\displaystyle \scriptstyle t\,}$ axis. The term phase can refer to several different things:
## Phase shift
Illustration of phase shift. The horizontal axis represents an angle (phase) that is increasing with time.
Phase shift is any change that occurs in the phase of one quantity, or in the phase difference between two or more quantities.[1]
${\displaystyle \scriptstyle \varphi \,}$ is sometimes referred to as a phase shift or phase offset, because it represents a "shift" from zero phase.
For infinitely long sinusoids, a change in ${\displaystyle \scriptstyle \varphi \,}$ is the same as a shift in time, such as a time delay. If ${\displaystyle \scriptstyle x(t)\,}$ is delayed (time-shifted) by ${\displaystyle \scriptstyle {\frac {1}{4}}\,}$ of its cycle, it becomes:
{\displaystyle {\begin{aligned}x\left(t-{\tfrac {1}{4}}T\right)&=A\cdot \cos \left(2\pi f\left(t-{\tfrac {1}{4}}T\right)+\varphi \right)\\&=A\cdot \cos \left(2\pi ft-{\tfrac {\pi }{2}}+\varphi \right)\end{aligned}}}
whose "phase" is now ${\displaystyle \scriptstyle \varphi \,-\,{\frac {\pi }{2}}}$. It has been shifted by ${\displaystyle \scriptstyle {\frac {\pi }{2}}}$ radians.
## Phase difference {{safesubst:#invoke:anchor|main}}
In-phase waves
Out-of-phase waves
Left: the real part of a plane wave moving from top to bottom. Right: the same wave after a central section underwent a phase shift, for example, by passing through a glass of different thickness than the other parts. (The illustration on the right ignores the effect of diffraction, which would make the waveform continuous away from material interfaces and would add increasing distortions with distance.).
Phase difference is the difference, expressed in degrees or time, between two waves having the same frequency and referenced to the same point in time.[1] Two oscillators that have the same frequency and no phase difference are said to be in phase. Two oscillators that have the same frequency and different phases have a phase difference, and the oscillators are said to be out of phase with each other.
The amount by which such oscillators are out of phase with each other can be expressed in degrees from 0° to 360°, or in radians from 0 to 2π. If the phase difference is 180 degrees (π radians), then the two oscillators are said to be in antiphase. If two interacting waves meet at a point where they are in antiphase, then destructive interference will occur. It is common for waves of electromagnetic (light, RF), acoustic (sound) or other energy to become superposed in their transmission medium. When that happens, the phase difference determines whether they reinforce or weaken each other. Complete cancellation is possible for waves with equal amplitudes.
Time is sometimes used (instead of angle) to express position within the cycle of an oscillation. A phase difference is analogous to two athletes running around a race track at the same speed and direction but starting at different positions on the track. They pass a point at different instants in time. But the time difference (phase difference) between them is a constant - same for every pass since they are at the same speed and in the same direction. If they were at different speeds (different frequencies), the phase difference is undefined and would only reflect different starting positions. Technically, phase difference between two entities at various frequencies is undefined and does not exist.
• Time zones are also analogous to phase differences.
A real-world example of a sonic phase difference occurs in the warble of a Native American flute. The amplitude of different harmonic components of same long-held note on the flute come into dominance at different points in the phase cycle. The phase difference between the different harmonics can be observed on a spectrograph of the sound of a warbling flute.[2]
## References
1. {{#invoke:citation/CS1|citation |CitationClass=book }}
2. Template:Cite web | 1,217 | 5,096 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 23, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2019-39 | latest | en | 0.82774 |
https://math.stackexchange.com/questions/3155000/is-this-geometric-mean-like-limit-equal-0 | 1,566,359,003,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315750.62/warc/CC-MAIN-20190821022901-20190821044901-00281.warc.gz | 557,777,830 | 31,017 | Is this geometric mean-like limit equal $0$?
Let $$(a_n)_n$$ be a sequence of positive real numbers such that $$a_n\leq1$$ for all $$n\in\mathbb{N}$$, and suppose that $$\displaystyle\lim_{n\to\infty}a_n=0$$. By the Stolz–Cesàro theorem we know that $$\displaystyle\lim_{n\to\infty}\left[\prod_{n=1}^na_n\right]^\frac{1}{n}=0$$ $$\quad(\star)$$.
For each $$n\in\mathbb{N}$$, let $$S_n$$ be a random subset of $$\{1,2,\dots,n\}$$ and let $$c_n$$ be the cardinality of $$S_n$$. Put $$b_n=\left[a_{n+1}\left(\prod_{k\in S_n}a_k\right)\right]^\frac{1}{1+c_n}.$$
Question: Is $$\displaystyle\lim_{n\to\infty}b_n=0$$?
If I try to imagine cases, I only can think of $$\displaystyle\lim_{n\to\infty}b_n=0$$, but I don't know how to prove this.
For example, if $$S_n\subset S_{n+1}$$ (proper subset) for all $$n\in\mathbb{N}$$, then the factors that define $$b_n$$ form a sequence that converges to zero (subsequence of $$(a_n)_n$$), and therefore $$\displaystyle\lim_{n\to\infty}b_n=0$$ by $$(\star)$$.
Any insight for the general case would be appreciated.
Fix $$\varepsilon \in (0, 1)$$. Note that there are only finitely many $$a_n$$ such that $$a_n \ge \varepsilon$$; let $$K$$ be the number of such sequence terms (note that this depends on $$\varepsilon$$, not $$n$$). Since $$a_n \to 0$$, there exists some $$N$$ such that $$n > N \implies a_n < \varepsilon^{K+1}.$$
For each $$n$$, define \begin{align*} U_n &= \{n \in S_n : a_n \ge \varepsilon\} \\ L_n &= \{n \in S_n : a_n < \varepsilon\} = S_n \setminus U_n. \end{align*}
Please note that $$|U_n| \le K$$ for all $$n$$. Then, \begin{align*} n > N &\implies a_{n+1}\prod_{k\in S_n}a_k < \varepsilon^{K+1} \prod_{k\in S_n}a_k\\ &\implies a_{n+1}\prod_{k\in S_n}a_k < \varepsilon^{K+1} \prod_{k\in U_n}a_k \prod_{k\in L_n}a_k \\ &\implies a_{n+1}\prod_{k\in S_n}a_k < \varepsilon^{|U_n| + 1} \prod_{k\in U_n}a_k \prod_{k\in L_n}a_k \\ &\implies a_{n+1}\prod_{k\in S_n}a_k < \varepsilon \prod_{k\in U_n}\varepsilon a_k \prod_{k\in L_n}a_k \\ &\implies a_{n+1}\prod_{k\in S_n}a_k < \varepsilon \prod_{k\in U_n}\varepsilon \prod_{k\in L_n}\varepsilon \\ &\implies a_{n+1}\prod_{k\in S_n}a_k < \varepsilon^{|S_n|+1} \\ &\implies \left(a_{n+1}\prod_{k\in S_n}a_k\right)^{\frac{1}{|S_n|+1}} < \varepsilon. \end{align*} | 936 | 2,269 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 34, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2019-35 | latest | en | 0.563271 |
https://math.answers.com/Q/What_is_five_six_plus_two_thirds | 1,643,314,873,000,000,000 | text/html | crawl-data/CC-MAIN-2022-05/segments/1642320305288.57/warc/CC-MAIN-20220127193303-20220127223303-00719.warc.gz | 434,160,493 | 62,333 | 0
# What is five six plus two thirds?
Wiki User
2013-05-29 21:33:39
Question should read "What is five sixth plus two thirds"?
(5/6) + (2/3) = 5/6 + 4/6 = 9/6 = 3/2 = 1 and 1/2
Wiki User
2013-05-29 21:33:39
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# The coefficient of x in the expansion of x kx is find the possible value of k
## Top Questions
, a standard deviation of \$1 would be considered large if it is describing the variability from store to store in the price of an ice cube tray. On the other hand, a standard deviation of \$1 would be considered small if it is describing store-to-store variability in the price of a particular brand of freezer. A quantity designed to give a relative measure of variability is the coefficient of variation. Denoted by CV, the coefficient of variation expresses the standard deviation as a percentage of the mean. It is defined by the formula CV = 100(s/ x ). Consider two samples. Sample 1 gives the actual weight (in ounces) of the contents of cans of pet food labeled as having a net weight of 8 oz. Sample 2 gives the actual weight (in pounds) of the contents of bags of dry pet food labeled as having a net weight of 50 lb. There are weights for the two samples. Sample 1 7.7 6.8 6.5 7.2 6.5 7.7 7.3 6.6 6.6 6.1 Sample 2 50.7 50.9 50.5 50.3 51.5 47 50.4 50.3 48.7 48.2 (a) For each of the given samples, calculate the mean and the standard deviation. (Round all intermediate calculations and answers to five decimal places.) For sample 1 Mean Standard deviation For sample 2 Mean Standard deviation (b) Compute the coefficient of variation for each sample. (Round all answers to two decimal places.) CV1 CV2
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ng the snow. The first child exerts a force of F1 = 11 N at an angle θ1 = 45° counterclockwise from the positive x direction. The second child exerts a force of F2 = 6 N at an angle θ2 = 30° clockwise from the positive x direction. Find the magnitude (in N) and direction of the friction force acting on the sled if it moves with constant velocity. magnitude direction (counterclockwise from the +x-axis) What is the coefficient of kinetic friction between the sled and the ground? What is the magnitude of the acceleration (in m/s2) of the sled if F1 is doubled and F2 is halved in magnitude?
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1.AU MAT 120 Systems of Linear Equations and Inequalities Discussion
mathematicsalgebra Physics | 564 | 2,210 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2023-06 | latest | en | 0.895054 |
https://www.enotes.com/topics/math/questions/f-x-tanx-n-3-find-nth-maclaurin-polynomial-810516 | 1,716,985,093,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059239.72/warc/CC-MAIN-20240529103929-20240529133929-00594.warc.gz | 632,491,411 | 17,573 | # `f(x)=tanx , n=3` Find the n'th Maclaurin polynomial for the function.
Maclaurin series is a special case of Taylor series that is centered at a=0. The expansion of the function about 0 follows the formula:
`f(x)=sum_(n=0)^oo (f^n(0))/(n!) x^n`
or
`f(x)= f(0)+(f'(0))/(1!)x+(f^2(0))/(2!)x^2+(f^3(0)x)/(3!)x^3+(f^4(0))/(4!)x^4 +...`
To determine the Maclaurin polynomial of degree n=3 for the given function `f(x)=tan(x)` , we may apply the formula for Maclaurin series.
We may list `f^n(x)` as:
`f(x) = tan(x)`
`f'(x)=d/(dx) tan(x) =sec^2(x)`
`f^2(x)=d/(dx)sec^2(x) =2tan(x)sec^2(x)`
`f^3(x)=d/(dx) 2tan(x)sec^2(x) =6sec^4(x)-4sec^2(x)`
Plug-in x=0, we get:
`f(0)=tan(0)`
`=0`
`f'(0)=sec^2(0) or (sec(0))^2`
`= 1^2`
`=1`
`f^2(0)=2tan(0)sec^2(0) `
`=2*0*1`
`=0`
`f^3(0)=6sec^4(0)-4sec^2(0)`
`=6(sec(0))^4-4(sec(0))^2`
`=6*1^4 -4*1^2`
`= 6-4`
`=2`
Applying the formula for Maclaurin series, we get:
`sum_(n=0)^3 (f^n(0))/(n!) x^n =0+1/(1!)x+0/(2!)x^2+2/(3!)x^3`
`=0+1/1x+0/(1*2)x^2+2/(1*2*3)x^3`
`= 0+x+0/2x^2+2/6x^3`
`=x+2/6x^3`
`=x+x^3/3`
The Maclaurin polynomial of degree n=3 for the given function `f(x)=tan(x)` will be:
`P_3(x)=x+x^3/3 `
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https://cs.stackexchange.com/questions/11438/why-does-dfs-only-yield-tree-and-back-edges-on-undirected-connected-graphs/95534 | 1,716,174,891,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058147.77/warc/CC-MAIN-20240520015105-20240520045105-00607.warc.gz | 155,072,069 | 45,462 | # Why does DFS only yield tree and back edges on undirected, connected graphs?
Prove that if G is an undirected connected graph, then each of its edges is either in the depth-first search tree or is a back edge.
Now, from intuition and in class lectures by Steven Skiena, I know that the above holds true, since it dives all the way down, and then throw a rope back to a previous vertex. I also know that DFS is great in finding cycles.
However, my problem here is that I don't know how to prove that the edge is either a tree edge or a back edge.
• You probably want to use induction. Start with a graph with one edge, then look at what happens in a larger graph, considering one edge and then applying induction to the remaining subgraph. Apr 20, 2013 at 21:37
• You can also argue for a contradiction - assume you have an edge that is neither, what would that imply? Apr 21, 2013 at 4:31
A depth first search on a directed graph can yield 4 types of edges; tree, forward, back and cross edges. As we are looking at undirected graphs, it should be obvious that forward and back edges are the same thing, so the only things left to deal with are cross edges.
A cross edge in a graph is an edge that goes from a vertex $v$ to another vertex $u$ such that $u$ is neither an ancestor nor descendant of $v$. So what you need to argue is that in an undirected graph, there's no way you can get a cross edge. It might help to think of why the can occur in directed graphs, and why you can't have this case in undirected graphs.
• For cross-edges, you still need that the graph is assumed here to be connected. Apr 25, 2013 at 14:28
• @Raphael you don't really need the graph to be connected, you just need to rename tree edges as forest edges ;) - you still won't get cross edges. Apr 26, 2013 at 0:10
• This fills the missing pieces of the answer: stackoverflow.com/questions/28942262/… Apr 10, 2022 at 4:07
Let $$G=(V,E)$$ to be a graph and $$u$$ and $$\nu$$ to be its vertices such that $$\{u,v\}\subseteq V$$ and $$(u,\nu)\in E$$.
Suppose that $$u$$ is discovered first. Consequently, its color is changed to gray. Then, $$\nu$$ becomes its descendant (by white path theorem) and the following discovery time relationship holds: $$u.d<\nu.d$$.
Now, there are two options of discovering $$(u,\nu)$$ edge:
$$\quad$$ 1) if $$u$$ discovers $$\nu$$, then $$(u,\nu)$$ is a tree edge;
$$\quad$$ 2) if $$\nu$$ discovers $$u$$, then $$(u,\nu)$$ is a back edge since $$u$$ is still gray at the time the edge is first explored.
• Well explained with illustrations. Thanks a lot ! Feb 2, 2022 at 8:08
The most voting answer says well and I want to claim it more clearly.
Proof:
We know that DFS produces tree edge, forward edge, back edge and cross edge. Let's prove why forward edge and cross edge can't exist for DFS on undirected Graph.
• For forward edge (u, v):
Forward edge is said that v is a descendent of u, or we say v is visited and explored completely when u is being exploring, which is
(pre[u] < pre[v] < post[v] < pre[u])
this answer is considered as a back edge because (v, u) is an edge as well. But I don't know why to see forward edge as a back one not the inverse insight. Maybe this thought is used for some specific usage?
• For cross edge (u, v):
this is the difficulty of the proof.
We prove that DFS on undirected graph can not yield a cross edge by contradiction.
If (u, v) is a cross edge, then v is already explored completely when u is being explored.
(pre[v] < post[v] < pre[u] < post[u])
This can exist on DAG because when exploring v we don't know u at all! u points to v directedly. However, in undirected graph, when we explore v, u (as a neighbor of v) is being explored and finished. If the cross edge exists, then u can not be a neighbor of v, this contradicts with the undirected graph assumption. In this case, cross edge can not exist.
In fact, this proof gives us another property.
For an edge (u, v) in an undirected graph, if post(v) < post(u), then u must be an ancestor of v.
There's another approach here. Suppose that you have an edge $$(u, v) \in E$$ and $$u, v \in V$$. Now, assume that there's a forward edge in the direction from vertex $$u$$ to $$v$$. This edge is first explored in the direction of $$(v, u)$$ while $$u$$ is gray by scanning the adjacency list of $$v$$, since $$v$$ is a descendent of $$u$$. In the direction from $$v$$ to $$u$$, this edge is a back edge.
Then, in a connected undirected graph, there's at least one simple path between any pair of vertices. Therefore, each vertex is either an ancestor or a descendent of the other. Thus, no cross edges are possible.
In the undirected graph we get only tree and back edges . the reason for no forward edges is because in undirected the forward edges get converted into back edges , it is so because in undirected there is no restriction in which direction to visit the vertex, so in case we have any vertex we can visit it from the child itself to the parent which counts as a back edge.
now lets talk of cross edge here also since edges are undirected so, the cross edges get converted to the tree edges as they can be visited as and when we go to any of that edge vertex for more details you can view this MIT vedio https://youtu.be/AfSk24UTFS8?t=36
• I don't think this adds anything over the answer that is already there, apart from sloppy grammar and spelling. Oct 22, 2015 at 20:54 | 1,389 | 5,428 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 29, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2024-22 | latest | en | 0.968314 |
https://www.yaclass.in/p/science-state-board/class-10/atoms-and-molecules-11699/difference-between-atoms-and-molecules-11483/re-3de22dc5-537a-42e8-a6a3-42302299a6a5 | 1,632,096,420,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780056902.22/warc/CC-MAIN-20210919220343-20210920010343-00399.warc.gz | 1,089,915,181 | 8,019 | ### Theory:
Calculate the gram molecular mass of the following compounds.
i. $$H_2O$$
ii. $$CO_2$$
iii. $$(Ca)_3(PO_4)_2$$
i. $$H_2O$$
Data:
Atomic mass of $$H=1$$
Atomic mass of $$O=16$$
Solution:
$\begin{array}{l}\mathit{Gram}\phantom{\rule{0.147em}{0ex}}\mathit{molecular}\phantom{\rule{0.147em}{0ex}}\mathit{mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{H}_{2}O\\ =\left(2×1\right)+\left(1×16\right)\\ =2+16\\ =18\end{array}$
Hence, the gram molecular mass of $$H_2O=18$$ $$g$$.
ii. $$CO_2$$
Data:
Atomic mass of $$C=12$$
Atomic mass of $$O=16$$
Solution:
$\begin{array}{l}\mathit{Gram}\phantom{\rule{0.147em}{0ex}}\mathit{molecular}\phantom{\rule{0.147em}{0ex}}\mathit{mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{\mathit{CO}}_{2}\\ =\left(1×12\right)+\left(2×16\right)\\ =12+32\\ =44\end{array}$
Hence, the gram molecular mass of $$CO_2=44$$ $$g$$.
iii. $$(Ca)_3(PO_4)_2$$
Data:
Atomic mass of $$Ca=40$$
Atomic mass of $$P=30$$
Atomic mass of $$O=16$$
Solution:
$\begin{array}{l}\mathit{Gram}\phantom{\rule{0.147em}{0ex}}\mathit{molecular}\phantom{\rule{0.147em}{0ex}}\mathit{mass}\phantom{\rule{0.147em}{0ex}}\mathit{of}\phantom{\rule{0.147em}{0ex}}{\mathit{Ca}}_{3}{\left({\mathit{PO}}_{4}\right)}_{2}\\ =\left(3×40\right)+\left[30+\left(4×16\right)\right]×2\\ =120+\left[30+64\right]×2\\ =120+\left(94×2\right)\\ =308\end{array}$
Hence, the gram molecular mass of $$(Ca)_3(PO_4)_2=308$$ $$g$$. | 637 | 1,488 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2021-39 | latest | en | 0.394653 |
http://civilservicereview.com/2015/04/first-conditional/ | 1,529,653,999,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864364.38/warc/CC-MAIN-20180622065204-20180622085204-00021.warc.gz | 64,901,196 | 13,893 | # Understanding Conditionals I: First Conditional
We use First Conditionals to talk about events which are possible. The Conditional clause can refer to the present or the future.
Conditional clause main clause
If + Present Simple – will + bare infinitive
1. If we hurry, we’ll catch the bus.
2. If we miss it, there’ll be another one.
The Conditional clause can come before or after the main clause. We use a comma at the end of the Conditional clause when it comes first.
1. If I hear any news, I’ll phone you.
2. I will phone you if I hear any news.
Other structures are possible, depending on what you want to say.
Conditional clause main clause
If + Present Simple – modal verb
If + Present Simple – be going to (future)
If + Present Simple – Imperative
If + Present Continuous – will + bare infinitive
If + Present Perfect – will + bare infinitive
If + Present Perfect – modal verb
Imperative – and/or + will
Exercises: Put the verbs in brackets into the correct form.
1. If you (see) Tom (tell) him I have message for him.
2. If you’d like some ice I (get) some from the fridge.
3. That book is overdue. If you (not take) it back to the library tomorrow you (have) to pay a fine.
4. If you (want) to see some of his drawings I (send) them round to your office.
5. (take) more exercise and you’ll soon feel better.
1. If you see Tom tell him I have message for him.
2. If you’d like some ice I will get some from the fridge.
3. That book is overdue. If you don’t take it back to the library tomorrow you will have to pay a fine.
4. If you want to see some of his drawings I will send them round to your office.
5. Take more exercise and you’ll soon feel better.
Note: Conditional clauses are often used in imperative structures. Present Simple in Conditional clause and imperative in the main clause.
When you are talking about a possible situation in the present, or a possible future occurrence, you usually use the simple present tense in the conditional clause and the simple future tense in the main clause.
If the sentence starts with the Imperative verb, you use simple future tense in the main clause.
### 1 Response
1. April 25, 2015
[…] the previous post, we have written about First Conditional. In this post, we continue this series by talking about Second […] | 536 | 2,292 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-26 | longest | en | 0.908272 |
http://www.ejobhub.in/2017/09/odd-man-out-and-series-formula.html | 1,529,467,066,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267863411.67/warc/CC-MAIN-20180620031000-20180620051000-00151.warc.gz | 410,118,708 | 39,764 | ## Odd Man Out and Series Formula
All time problem creating section of the exam is consider as Odd Man Out and Series. To make good combination with such type of questions
ejobhub team is going to tell the Odd Man Out and Series Formula for those students who are preparing for any competitive exam. ‘Quantitative Aptitude’ of any exam consists of questions regarding reasoning, Odd Man Out and Series Formula, Distance formulas and many more. Such participants who have remind all the necessary formulas, Shortcuts and problem tricks for these types of the questions they can get correct answer soon.
Those students who are not from math’s side they will face difficulty to solve all the questions. Short tricks help the aspirants to complete the question paper on given time. To score good marks in any of the post competitive / educational exams candidates will have to learn all the given formulas and important topics to give proper answer to series formula related questions. Questions asked in aptitude section are very easy so don’t take too much time in it. Apply the correct formula, write the correct digits and don’t get confused in similar type of questions. More details in association with Odd man out and series formula with suitable examples and solved problems are given below. All the best!!
Shortcuts to solve Problems: To complete the question paper on a given time period, you just need two thinks i.e. speed and accuracy. Take a look on formulas and examples that are mentioned below:
Series Formula: The terms or elements follow a definite law in series but it cannot be generalized. You should know about what is the definite relationship between numbers which make the set of given terms in series. Addition, subtraction, multiplication, division, transposition of terms and series generally form such series. The different questions asked may depend upon the following:
Odd number/Even number/Prime numbers: The series may consist of odd numbers /even numbers or prime numbers except one number, which will be the odd man out. Hence, before solving numerical on this topic must revise all basic concepts.
Perfect squares/Cubes:
Squares: 9, 16, 49, 81 ….
Cubes: 27, 64, 125, 216 ….
Multiple of numbers: The series contains numbers which are multiple of different numbers.
Example: 4, 8, 12, 16, 20….
Numbers in A.P./G.P.
Geometric progression: x, xr, xr3, xr4
Arithmetic progression: x, x + y, x + 2y, x + 3y are said to be in A.P.
The terms in series may be arithmetic or geometric progression.
Difference or sum of numbers: The difference between two consecutive numbers may increase or decrease
Cumulative series: In this type, the third number is the addition of previous two numbers.
Example: 2, 4, 6, 10, 16, 26 ……
Important Illustrations
Problem 1:
1) 13
2) 23
3) 33
4) 43
5) 53
Solution: Check these numbers. Have you observed any logic? Every number from the above list has 10 numbers difference from it's previous numbers. Great, we find one... But that's not the logic. So check whether the numbers are prime or not. So, here... all the given numbers are prime, except the number 33 (it is a composite number). So Option 3 is the answer.
Problem 2:
1) 176
2) 231
3) 572
4) 473
5) 653
Solution: Check the above numbers carefully. You can't find the correct answer by applying the first 2 steps. So here try to find out the logic... If you observe closely, you will find out that the middle number is the sum of the first and the last numbers. Check once....
1) 176
2) 231
3) 572
4) 473
5) 653
But you are not getting middle number in 5th Option (As 6+3 = 9, not 5). So this is the correct answer.
Problem 3:
1) 123
2) 246
3) 147
4) 368
5) 159
Solution: Here also we tried all the steps but no answer. Hey wait; have you tried the above logic?? Let's try
1) 1+3 = 4 (but the middle number is 2)
2) 2+6 = 8 (but the middle number is 4)
3) 1+7 = 8 (but the middle number is 4)
4) 3+8 = 11 (but the middle number is 6)
5) 9+1 = 10 (but the middle number is 5)
Not worked
Now see whether we can find anything with the newly obtained numbers....
4, 8, 8, 11, 10.... No relation.. But you could find out that one number except all are even numbers... So, may be this is our answer...
Or if you want to go further, you can get another logic... Just divide these numbers with 2.
1) 4/2 = 2
2) 8/2 = 4
3) 8/2 = 4
4) 11/2 = 5.5
5) 10/2 = 5
So, here.. Except Option 4... For remaining numbers, the middle number is the half of the total of remaining two numbers.
Note: If you can't get answer with above step, then you should experiment with different numbers and different mathematical operations.
Problem 4:
1) 29: 812
2) 37: 1332
3) 45: 1980
4) 48: 2256
5) 51: 2551
If they ask question in the above format, then it means, there should be some relationship between the first and second number.
Solution:
Note: If you can't find any visible difference, then better to find out Squares.
1) 292 = 841 (but given number is 812)
2) 372 = 1369 (but the given number is 1332)
3) 452 = 2025 (but the given number is 1980)
4) 482= 2304 (but the given number is 2256)
5) 512 = 2601 (but the given number is 2551)
So squares are not matching with the numbers. But if you closely observe, you can see some similarities among these numbers. So, find out the difference, so that we could find some clue
1) 841-812 = 29 (First Number)
2) 1369-1332 = 37 (First Number)
3) 2025-1980 = 45 (First Number)
4) 2304-2256 = 48 (First Number)
5) 2601-2551 = 510 (Not the First Number)
So, Option 5 is the odd one.
Problem 5:
1) 126
2) 72
3) 135
4) 171
5) 162
Solution: At first glance we are unable to find any clues. So, check the interrelationships among these numbers (if any).
Note: People tend to mark Option 2 (72) as the answer because it consist only two numbers where as in remaining options the numbers are three. But please keep in mind that now a day they are not asking questions with such a simple logic.
At a close observation, you can find that the total of the given numbers is 9
1) 1+2+6 = 9
2) 7+2 = 9
3) 1+3+5 = 9
4) 1+7+1 = 9
5) 1+6+2 = 9
So, there might be something with this 9.
Let's subtract this 9 with the given number...
1) 126-9 = 117
2) 72-9 = 63
3) 135-9 = 126
4) 171-9 = 162
5) 162-9 = 153
You could find out some relation between option 4 and 5. But it's not enough.... So let's try division
1) 126/9 = 14
2) 72/9 = 8
3) 135/9 = 15
4) 171/9 = 19
5) 162/9 = 18
From the above answers, you can clearly say that 19 (Option 4) is a Prime Number. Where remaining numbers are not.
Problem 6:
1) 48-134
2) 65-185
3) 128-374
4) 56-158
5) 81-223
Solution: With a close observation you can see that the second number is almost nearer to the three times of the first number.
1) 40 x 3 = 120 (given number is 134)
2) 60 x 3 = 180 (given number is 185)
3) 120x 3 = 360 (given number is 374)
4) 50 x 3 = 150 (given number is 158)
5) 80x 3 = 240 (given number is 223)
So, there should be some logic behind it
Multiply first numbers with 3
1) 48x3 = 144 (difference between result and second number is 144-134 = 10)
2) 65x3 = 195 (195-185 = 10)
3) 128x3 = 384 (384-374 = 10)
4) 56x3 = 168 (168-158 = 10)
5) 81x3=243 (243-223= 20)
So, Option 5 is the answer.
Problem 7:
1) 25-630
2) 36-1312
3) 16-260
4) 9-84
5) 49-2408
Solution:
1) 252=625 (625+5 = 630)
2) 362=1296 (1296+16 = 1312)
3) 162 = 256 (256+4 = 260)
4) 92 = 81 (81+3 = 84)
5) 492 = 2401 (2401+7=2408)
So, here the odd one is Option 2
NOTE: So guys, finally you can qualify in the odd man out and serious.
Take a Look on Below Table | 2,365 | 7,811 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2018-26 | latest | en | 0.927115 |
http://math.stackexchange.com/questions/276250/reversible-integer-transform | 1,469,794,397,000,000,000 | text/html | crawl-data/CC-MAIN-2016-30/segments/1469257830066.95/warc/CC-MAIN-20160723071030-00112-ip-10-185-27-174.ec2.internal.warc.gz | 164,929,074 | 17,383 | # reversible integer transform
Let be x, y two natural integers. Define the following transform: X=x+(x-y). Y=y-(x-y). If the difference d=x-y tends to 0, we have X=x and Y=y. The above transform has the property that is reversible in integer domain. I have $(x,y) \in \{0,1..63\}$ . We know that pairs $(x_i,y_i)$, where i=$\overline{0,63}$, have a Laplace distribution of the histogram differences. What part of mathematics dealing with this transform? I want to know more reversible integer transforms.
-
Your transform is the linear transform $$T:\quad \left[\matrix{x \cr y\cr}\right]\quad \mapsto\quad \left[\matrix{x' \cr y'\cr}\right]\ :=\ \left[\matrix{2 &-1 \cr -1 &2\cr}\right]\ \left[\matrix{x \cr y\cr}\right]\ .$$ Its inverse is given by $$T^{-1}:\quad \left[\matrix{x' \cr y'\cr}\right]\quad \mapsto\quad \left[\matrix{x \cr y\cr}\right]\ :=\ {\textstyle\left[\matrix{{2\over3} &{1\over3} \cr {1\over3} &{2\over3}\cr}\right]}\ \left[\matrix{x' \cr y'\cr}\right]\ .$$ The inverse does not map integer pairs to integer pairs. Consider the example $(x',y'):=(2,1)\in{\mathbb Z}^2$. Here $T^{-1}(x',y')=\bigl({5\over3},{4\over3}\bigr)\notin{\mathbb Z}^2$. | 386 | 1,169 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2016-30 | latest | en | 0.555769 |
https://ideophilus.wordpress.com/2015/09/29/dice-displays-to-prompt-intuitive-understanding-of-probabilities/ | 1,579,467,957,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250595282.35/warc/CC-MAIN-20200119205448-20200119233448-00186.warc.gz | 487,099,301 | 26,061 | # Dice displays to prompt intuitive understanding of probabilities
About a week and a half ago, when I looked, iPredict showed approximately a 63% probability that there would be a National Prime Minister after the 2017 general election in New Zealand, and a 37% probability that there would be a Labour Prime Minister. Here’s what that looks like as a horizontal bar:
The problem with this way of looking at it is that (especially in New Zealand, with a proportional voting system), there’s a temptation to interpret the proportions as vote shares, rather than as probabilities. And even when I’ve got the idea of vote shares out of my mind, I can still be inclined to interpret it as a prediction that National will win the 2017 election. It’s not; it’s an estimate that there is a 63% probability that National will win the election.
How can I encourage myself to understand this intuitively?
Well, here’s my idea: Superimpose the image of the faces of a six-sided die on the bar I showed above, like this: Now I can imagine a game in which a six-sided die is thrown: if it shows a number less than 3, then Labour wins; if it shows a number greater than 3, National wins; if it shows exactly 3, then either might win, with the probability proportional to the relative areas shown on that face of the die.
(I don’t mean to suggest that this is how elections are actually decided in New Zealand; this is just a way of giving myself a good intuition about the probabilities involved.)
But this isn’t entirely satisfactory yet. What happens in that imaginary game to decide the winner if a 3 is shown? Well, I can superimpose an image of the faces of a second die on that face of the first one, like this: If a 3 is shown on the first die, a second is thrown: if it shows less than 2, Labour wins; if it shows greater than 2, National wins; if it shows exactly 2, then the game is still undecided.
In principle, the superimposition of the faces of more dice could be continued as far as you like, using the base-six (or senary, or perhaps, in this case, dicemal) expansion of the probability in question. But I’m not sure it’s worth it; beyond two dicemal places, the the faces of the dice would be hard to see without using a very zoomable image, and it probably wouldn’t add much to the intuition about the probability.
There might be one exception to that last comment, though: If the less probable event is very unlikely, you could dispense with the image, and just say that it is, for example, “less likely than throwing five dice and having them all show sixes”. Then, anyone sufficiently familiar with Yahtzee would understand that it’s particularly unlikely, but not entirely impossible.
## 1 thought on “Dice displays to prompt intuitive understanding of probabilities”
1. Probably just as well that elections aren’t decided that way: the loaded-dice industry would boom. | 642 | 2,889 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2020-05 | latest | en | 0.948272 |
https://wordpandit.com/wpt_test/basic-maths-test-60/ | 1,722,876,927,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640451031.21/warc/CC-MAIN-20240805144950-20240805174950-00272.warc.gz | 497,530,748 | 49,091 | Select Page
• This is an assessment test.
• These tests focus on the basics of Maths and are meant to indicate your preparation level for the subject.
• Kindly take the tests in this series with a pre-defined schedule.
## Basic Maths: Test 60
Congratulations - you have completed Basic Maths: Test 60.
You scored %%SCORE%% out of %%TOTAL%%.
Your performance has been rated as %%RATING%%
Question 1
$\frac{\sqrt{28}+\sqrt{252}}{\sqrt{112}}$ is equal to
A $\frac{2}{\sqrt{6}}$ B $2\sqrt{6}$ C $4\sqrt{6}$ D 2
Question 1 Explanation:
\begin{align} & \frac{\sqrt{28}+\sqrt{252}}{\sqrt{112}} \\ & =\frac{\sqrt{4\times 7}+\sqrt{36\times 7}}{\sqrt{16\times 7}} \\ & =\frac{\sqrt{7}\left( 2+6 \right)}{4\sqrt{7}} \\ & =\frac{8}{4}=2 \\ \end{align}
Question 2
$\frac{\left( 100-1 \right)\,\left( 100-2 \right)\,\left( 100-3 \right)\,.....\left( 100-99 \right)}{100\times 99\times 98\times .....\times 3\times 2\times 1}$ is equal to
A $\frac{100}{99\times 98\times 97\times .....\times 3\times 2\times 1}$ B 0.01 C 0 D $-\frac{2}{99\times 98\times 97\times ....\times 3\times 2\times 1}$
Question 2 Explanation:
\begin{align} & \frac{\left( 100-1 \right)\,\left( 100-2 \right)\,\left( 100-3 \right)\,.....\left( 100-99 \right)}{100\times 99\times 98\times .....\times 3\times 2\times 1} \\ & =\frac{\left( 99 \right)\,\left( 98 \right)\,\left( 97 \right)\,.....\left( 1 \right)}{100\times 99\times 98\times .....\times 3\times 2\times 1} \\ & =\frac{1}{100} \\ & =0.01 \\ \end{align}
Question 3
$\left( 7.5\times 7.5-37.5+2.5\times 2.5 \right)$ is equal to
A 20 B 25 C 15 D 30
Question 3 Explanation:
Let 7.5 = a, 2.5 = b
\begin{align} & Expression \\ & ={{a}^{2}}-2\times a\times b+{{b}^{2}} \\ & ={{\left( a-b \right)}^{2}} \\ & \left[ a-b=7.5-2.5=5 \right] \\ & ={{5}^{2}} \\ & =25 \\ \end{align}
Question 4
$\frac{2.25\times 2.25+2.75\times 2.75-2\times 2.25\times 2.75}{2.25\times 2.25-2.75\times 2.75}$simplified to
A 0.4 B 0.3 C 0.1 D 0.2
Question 4 Explanation:
\begin{align} & If\,\,2.75=a\,\,and\,\,\,2.25=b,\,\,then \\ & Expression \\ & =\frac{{{a}^{2}}+{{b}^{2}}-2ab}{{{a}^{2}}-{{b}^{2}}} \\ & =\frac{{{\left( a-b \right)}^{2}}}{\left( a-b \right)\left( a+b \right)}=\frac{\left( a-b \right)}{\left( a+b \right)} \\ & =\frac{0.5}{5} \\ & =0.1 \\ \end{align}
Question 5
$\left( \frac{5}{2}+\frac{4}{2} \right)\,\left( \frac{26}{3}-\frac{11}{3}+\frac{7}{3} \right)$ is equal to
A 33 B 19 C 37 D 36
Question 5 Explanation:
\begin{align} & \left( \frac{5}{2}+\frac{4}{2} \right)\,\left( \frac{26}{3}-\frac{11}{3}+\frac{7}{3} \right) \\ & =\frac{9}{2}\times \frac{22}{3}=33 \\ \end{align}
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No thanks, I don't want it. | 1,204 | 2,972 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 5, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.953125 | 4 | CC-MAIN-2024-33 | latest | en | 0.522752 |
https://inari-sushi.ru/how-to-solve-multi-step-word-problems-291.html | 1,619,173,256,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618039617701.99/warc/CC-MAIN-20210423101141-20210423131141-00373.warc.gz | 418,601,520 | 9,457 | # How To Solve Multi Step Word Problems
Using the equation, solve the problem by plugging in the values and solving for the unknown variable.
As a professional writer, she has written for Education.com, Samsung and IBM.
Martin contributed English translations for a collection of Japanese poems by Misuzu Kaneko.
Solve two-step word problems using the four operations.
If he can fit nine books on a shelf, how many shelves will he need?
The total pairs of socks owned by both sisters is 8 6 9.
Subtract the two missing pairs for a final equation of (8 6 9) - 2 = n, where n is the number of pairs of socks the sisters have left.The unit of measurement for this problem is pairs of socks."Suzy has eight pairs of red socks and six pairs of blue socks. If her little sister owns nine pairs of purple socks and loses two of Suzy's pairs, how many pairs of socks do the sisters have left?Apply this logic to any word problem, regardless of the difficulty.Avery Martin holds a Bachelor of Music in opera performance and a Bachelor of Arts in East Asian studies. Assess the reasonableness of answers using mental computation and estimation strategies including rounding. Represent these problems using equations with a letter standing for the unknown quantity.Multiply, divide and subtract in the correct order using the order of operations.Exponents and roots come first, then multiplication and division, and finally addition and subtraction." Create a table, list, graph or chart that outlines the information you know, and leave blanks for any information you don't yet know.Each word problem may require a different format, but a visual representation of the necessary information makes it easier to work with.
## Comments How To Solve Multi Step Word Problems
• ###### Multi-step word problem with addition, subtraction, and.
Budgeting money is one of the most valuable math skills you'll learn. Practice with us as we figure out how much money is left after a cab fare.…
• ###### Multi-step Word Problems examples, solutions, videos.
Solve multistep word problems posed with whole numbers and having whole-number answers using the four operations, problems in which remainders must be.…
• ###### Word Problems Multi-Step - Super Teacher Worksheets
Multiple Step, Basic #1. Solve each of the multi-step word problems on this page. Problems contain basic numbers of 20 or less. Each problem can be solved.…
• ###### Multi-Step Math Word Problems - Helping With Math
In multi-step math word problems, one or more problems have to be solved in order to get the information needed to solve the question being asked. This lesson.…
• ###### Multi-Step Word Problems Lesson plan
Write the following problem on the board and ask students to solve the problem in pairs on scrap paper "Alexis lines up his jelly beans so that.…
• ###### Solve multi-step word problems by organizing the data.
In this lesson you will learn how to solve multi-step word problems by organizing and labeling information.… | 613 | 3,010 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2021-17 | longest | en | 0.897676 |
https://mph-to-kmh.com/62-5-kg-to-lbs/ | 1,709,402,111,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475833.51/warc/CC-MAIN-20240302152131-20240302182131-00644.warc.gz | 397,124,439 | 15,845 | # How to Convert 62.5 kg to lbs Easily: A Step-by-Step Guide
## The Basics: Understanding the Conversion from Kilograms to Pounds
### What is the Conversion Formula?
To understand the conversion from kilograms to pounds, you need to know the conversion formula. The formula is quite simple – 1 kilogram is equal to 2.20462 pounds. This means that if you have a weight in kilograms, you can simply multiply it by 2.20462 to get the equivalent weight in pounds. For example, if you have 5 kilograms, you can multiply it by 2.20462 to get approximately 11.0231 pounds.
### Why is the Conversion Important?
The conversion from kilograms to pounds is important in various fields, especially in international trade, sports, and health. In international trade, where weights are often measured in pounds, it is essential to convert kilograms to pounds for accurate measurement and shipment of goods. Similarly, in sports, weight categories are often specified in pounds, so understanding the conversion helps athletes know their weight category. Moreover, in the health field, certain medications and nutrition information may be provided in pounds, so understanding the conversion is crucial for patients to follow proper dosages and guidelines.
### Common Examples and Tips for Conversion
In everyday life, you may come across situations where you need to convert kilograms to pounds. For example, when you buy fresh produce from the supermarket, the weights are often listed in kilograms, but you might be more familiar with pounds. In such cases, you can quickly convert the weight by multiplying the number of kilograms by 2.20462. Additionally, it’s important to note that the conversion is not exact, and the result is usually rounded to the nearest decimal point. Therefore, keep in mind that the converted weight may not be a perfect match but a close approximation.
By understanding the conversion from kilograms to pounds, you can easily switch between these two weight units in your daily activities. Whether it’s for international trade, sports, or health purposes, knowing the conversion formula empowers you to accurately measure and communicate weights. So, the next time you encounter the need to convert kilograms to pounds, remember the simple formula and be confident in your conversion skills.
## The Mathematical Formula: Calculating 62.5 kg to lbs
### The Conversion Formula:
To convert kilograms to pounds, you can use the following mathematical formula: 1 kilogram is equal to 2.20462 pounds. So, to convert 62.5 kilograms to pounds, multiply 62.5 by 2.20462. The result will be the weight in pounds. In this case, 62.5 kg is equal to approximately 137.79 lbs.
### Understanding the Calculation:
The calculation is straightforward. By multiplying the weight in kilograms by the conversion factor of 2.20462, you can obtain the weight in pounds. It is important to use the accurate conversion factor to ensure precision in your calculations. In this case, the weight of 62.5 kg is multiplied by 2.20462 to derive the weight in pounds.
### Importance of Accurate Conversions:
Accurate conversions are crucial when dealing with weights, especially in situations where precision is necessary. Whether you are measuring personal weight, medical dosages, or food portions, it is essential to have accurate conversions to avoid any errors or misunderstandings. The conversion between kilograms and pounds is commonly used in various fields, including healthcare, fitness, and international trade.
Remember: If you need to convert kilograms to pounds, always use the correct conversion factor of 2.20462. This will ensure that your calculations are accurate and reliable.
By understanding the mathematical formula for converting kilograms to pounds, you can easily convert any weight measurement from one unit to another. Whether you are working with 62.5 kg or any other value, knowing this formula will help you make accurate conversions and effectively communicate weights in both kilograms and pounds.
## Common Usage: Why Knowing the Conversion from 62.5 kg to lbs Matters
### Understanding Metric and Imperial Systems
Knowing the conversion from 62.5 kg to lbs is crucial for those who frequently work with metric and imperial units of measurement. The kg (kilogram) is a unit of mass used in the metric system, while lbs (pounds) is a unit of weight used in the imperial system. These two systems are widely used around the world, and having a good grasp of the conversion between them enables accurate and efficient conversion when dealing with weight-related calculations.
For example, if you’re planning a trip to a country that uses the imperial system, knowing the conversion from 62.5 kg to lbs would help you better understand weight limits for luggage or even calculate the weight of items you might purchase while there.
### Navigating Fitness and Health Measurements
Being aware of the conversion from 62.5 kg to lbs is also important in the fitness and health realm. Weight is a common parameter used to track progress in various fitness goals, such as weight loss or muscle gain. Many fitness trackers, gym equipment, and nutrition plans utilize pounds as the standard unit of measurement. By knowing the conversion, individuals can accurately interpret and compare their weight changes and progress using both kilogram and pound measurements.
For instance, if you are following a weight loss program and need to track your progress, understanding the conversion from 62.5 kg to lbs will allow you to compare your goal weight and track progress in both metric and imperial units.
### Convenience in Everyday Life
In everyday life, understanding the conversion from 62.5 kg to lbs can be quite helpful. Although the metric system is widely used worldwide, there are still instances where the imperial system is more commonly utilized. A good example is when it comes to discussing personal weight or even buying produce or meat at a local market. Many people are more familiar with pounds rather than kilograms for these everyday scenarios, making the conversion knowledge valuable.
Imagine going to a farmer’s market and seeing produce listed in kilograms. Understanding the conversion will allow you to quickly determine how much 62.5 kg of apples or tomatoes would be in pounds to better understand the quantity you are purchasing.
By grasping the conversion from 62.5 kg to lbs, individuals can navigate various aspects of their lives with ease. Whether it’s for travel, fitness goals, or everyday tasks, having this conversion knowledge proves to be beneficial. The ability to seamlessly switch between metric and imperial systems allows for better understanding and efficiency in a world that still heavily relies on both measurement systems.
## Comparing Weight Formats: Exploring the Differences between Kilograms and Pounds
### Kilograms: The International Standard
Kilograms (kg) are the standardized unit of weight measurement used by most countries around the world. This format is part of the International System of Units (SI) and is the preferred choice for scientific and technical calculations. One kilogram is equal to 2.20462 pounds, making it slightly heavier than its imperial counterpart.
Using kilograms has several advantages. Firstly, it provides a consistent and universally recognized weight measurement, making it easier for researchers, engineers, and professionals to communicate and collaborate globally. Additionally, the use of kilograms simplifies calculations and conversions, as it follows a decimal-based system.
You may also be interested in: Unveiling the Mystery: Discover How Many Days Make Up 7 Months and Master Timekeeping
### Pounds: The Popular Choice in Some Countries
Pounds (lb) are prevalent in countries that still adhere to the imperial system, including the United States and the United Kingdom. One pound is equivalent to 0.453592 kilograms. While kilograms may be the international standard, pounds are popular due to their familiarity and historical usage.
It’s important to note that pounds are not as precise as kilograms when it comes to scientific and technical applications. However, pounds are more commonly used for day-to-day measurements, such as body weight, groceries, and household items. Many people have a better understanding of pounds as it is part of their daily lives.
### Choosing the Appropriate Weight Format
You may also be interested in: Converting 3.7 Liters to oz Made Easy: Your Ultimate Guide to Accurate Measurement Conversions
When comparing kilograms and pounds, it’s essential to consider the context and audience. In scientific and technical fields, where accuracy and standardization are critical, kilograms are the preferred choice. On the other hand, for everyday use and in countries that still use the imperial system, pounds may be more practical and easily understood.
Ultimately, the selection of the weight format depends on individual needs and regional conventions. It’s crucial to utilize appropriate conversion tables and tools when comparing or converting between kilograms and pounds to ensure correct and consistent measurements.
## Practical Applications: Converting 62.5 kg to lbs in Everyday Life
### Importance of Conversion
Converting 62.5 kg to lbs may seem like a trivial task, but it holds significant importance in everyday life. In many parts of the world, including the United States, weight is commonly measured in pounds. Therefore, converting kilos to pounds is essential for various daily activities such as monitoring personal weight, following diet plans, calculating medication dosages, or even weighing grocery items.
### The Conversion Formula
To convert 62.5 kg to lbs, you can use a simple conversion formula: 1 kilogram is equal to approximately 2.20462 pounds. Therefore, to convert kilograms to pounds, you need to multiply the weight value in kilograms by this conversion factor. In the case of 62.5 kg, the conversion would be 62.5 x 2.20462, which equals approximately 137.79 pounds.
You may also be interested in: How to Easily Convert 92 kg to Pounds: The Ultimate Guide for Quick and Accurate Conversions
### Practical Scenarios
Knowing the pounds equivalent of 62.5 kg can be useful in numerous scenarios. For example, if you plan to travel abroad and need to check airline baggage restrictions, you might encounter weight limits specified in pounds. Converting your suitcase weight from kilograms to pounds will help ensure your bags meet the requirements. Additionally, if you are following a weight loss program or tracking your fitness progress, knowing the pounds equivalent of your kilograms-weight can aid in setting goals and tracking your progress more effectively.
In summary, converting 62.5 kg to lbs is a practical skill that has relevance in various aspects of everyday life. By understanding the conversion formula and its applications, you can easily convert between these weight measurements to make informed decisions and navigate daily tasks more efficiently. | 2,158 | 11,096 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2024-10 | longest | en | 0.926863 |
https://www.edplace.com/worksheet_info/maths/keystage4/year10/topic/1239/7299/use-enlargement | 1,579,627,989,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579250604849.31/warc/CC-MAIN-20200121162615-20200121191615-00286.warc.gz | 857,579,294 | 26,611 | # Use Enlargement
In this worksheet, students practise enlarging shapes.
Key stage: KS 4
GCSE Subjects: Maths
GCSE Boards: AQA, Eduqas, Pearson Edexcel, OCR
Curriculum topic: Geometry and Measures, Congruence and Similarity
Curriculum subtopic: Properties and Constructions, Plane Isometric Transformations
Difficulty level:
### QUESTION 1 of 10
What is enlargement?
The word enlargement is usually taken to mean making something bigger.
In Maths, however, it actually means changing the size of an object. It can mean making a shape bigger or making a shape smaller.
What do we need to have to enlarge a shape.
To enlarge a shape, we need to have a Scale Factor (SF). This tells us how many times bigger the shape is.
A SF of 2 means each side of the shape is twice as big
A SFof 3 means each side of the shape is three times as big.
A SF of 1/2 means each side of the shape is half as long.
Example 1:
A shape is enlarged by a SF of 3. If one of the sides in the original shape is 2cm long, how long will it be in the enlarged shape?
All we have to do here is to multiply by 3.
2 cm x 3 = 6 cm
Example 2: Shape A is enlarged to shape B. What is the scale factor?
Firstly, we need to say if the enlarged shape is bigger or smaller. As it is bigger, the SF will be a number greater than 1.
All we have to do here is to look at a pair of sides, it doesn't matter which ones as they will all give the same answer.
If we look at the left hand line in each shape, it is 3cm in shape A and 6cm in shape B. This means the SF = 2
Example 3: Shape B is enlarged to shape A. What is the scale factor?
Firstly, we need to say if the enlarged shape is bigger or smaller. As it is smaller, the SF will be a number between 0 and 1.
All we have to do here is to look at a pair of sides, it doesn't matter which ones as they will all give the same answer.
If we look at the left hand line in each shape, it is 3cm in shape A and 6cm in shape B. This means the SF = 1/2
Enlargement is where we make a shape...
To describe an enlargement we need a...
This shape is enlarged by a scale factor if 4.
How long will the marked side be in the enlarged shape?
This shape is enlarged by a scale factor if 1/2.
How long will the marked side be in the enlarged shape?
Shape A is enlarged to shape B. What is the Scale Factor?
2
3
4
5
Shape B is enlarged to shape A. What is the Scale Factor?
1/2
2
1/4
4
Shape A is enlarged to shape B. What is the Scale Factor?
Shape B is enlarged to shape A. What is the Scale Factor?
Shape A is enlarged to shape B. What is the Scale Factor?
Shape A is enlarged to shape B. What is the Scale Factor?
• Question 1
Enlargement is where we make a shape...
EDDIE SAYS
Remember that in maths, enlargement means we change the size of an object. It can be either bigger or smaller
• Question 2
To describe an enlargement we need a...
EDDIE SAYS
When we enlarge an object, we need to know how much we are changing the shape. This is called a...
• Question 3
This shape is enlarged by a scale factor if 4.
How long will the marked side be in the enlarged shape?
16
16 cm
16cm
EDDIE SAYS
A scale factor of 4 just means that we multiply by 4. 4cm x 4 =
• Question 4
This shape is enlarged by a scale factor if 1/2.
How long will the marked side be in the enlarged shape?
2
2 cm
2cm
EDDIE SAYS
A scale factor of 1/2 just means that we multiply by 1/2 (This is the same as dividing by 2). 4cm x 0.5 =
• Question 5
Shape A is enlarged to shape B. What is the Scale Factor?
2
EDDIE SAYS
Each of the sides in shape B is twice as long as the sides in shape A. This means the SF is...
• Question 6
Shape B is enlarged to shape A. What is the Scale Factor?
1/2
EDDIE SAYS
Each of the sides in shape A is half as long as the sides in shape B. This means the SF is...
• Question 7
Shape A is enlarged to shape B. What is the Scale Factor?
3
EDDIE SAYS
A to B is getting bigger so the scale factor must be greater than 1. If we look at the top of each shape, Shape B is 3cm and Shape A is 1cm.
• Question 8
Shape B is enlarged to shape A. What is the Scale Factor?
1/2
0.5
EDDIE SAYS
B to A is getting smaller so the scale factor must be between 0 and 1 If we look at the top of each shape, Shape B is 4cm and Shape A is 2cm.
• Question 9
Shape A is enlarged to shape B. What is the Scale Factor?
2.5
EDDIE SAYS
A to B is getting bigger so the scale factor must be greater than 1. If we look at the top of each shape, Shape B is 5cm and Shape A is 2cm.
• Question 10
Shape A is enlarged to shape B. What is the Scale Factor?
1
EDDIE SAYS
This looks exactly the same, this is still an enlargement, it just has a scale factor of...
---- OR ----
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Get started | 1,400 | 5,202 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2020-05 | longest | en | 0.956002 |
http://mathhelpforum.com/higher-math/205163-how-solve-complex-exponential-equation-one-unknown-print.html | 1,527,421,151,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794868248.78/warc/CC-MAIN-20180527111631-20180527131631-00079.warc.gz | 188,593,562 | 3,375 | # How to solve a complex exponential equation with one unknown???
• Oct 11th 2012, 10:02 PM
LanieH
How to solve a complex exponential equation with one unknown???
I need to solve this equation for 'B'. I am having trouble rearranging the equation to achieve a 'B=' answer. Can someone help me rearrange this equation to solve for B?
0=((DCRMg) - (DCRMgAl)) ∗ e^(-B∗((ρMg)+(ρAl)))+ ((DCRMgAl )- (DCRAl ))∗e^(-2∗B∗(ρMg))+((DCRAl) - (DCRMg)) e^(-B∗((ρMg)+(ρMgAl)))
• Oct 13th 2012, 11:17 AM
HallsofIvy
Re: How to solve a complex exponential equation with one unknown???
replacing various constants with single letters, this is $\displaystyle a- be^x+ ce^{2x}+ de^x= 0$
where $\displaystyle a= (DCRM_g- DCRM_gIa)e^{\rho RAI}$, $\displaystyle b= (DCRM_gAI- DCRAI)$, $\displaystyle c= (DCRAI- DCRM_g)e^{\rho M_gAI}$, $\displaystyle d= (DCRAI- DCRM_g)e^{\rho M_gAI}$ and $\displaystyle x= B(\rho M_g)$.
If we further write $\displaystyle y= e^x$ that equation reduces to $\displaystyle cy^2+ (d- b)y+ a= 0$ which can be solved for y using the quadratic formula. Once we have $\displaystyle y= e^x$, $\displaystyle x= B(\rho M_g)= ln(y)$ and then $\displaystyle B= \frac{ln(y)}{\rho M_g}$.
• Oct 14th 2012, 02:07 PM
LanieH
Re: How to solve a complex exponential equation with one unknown???
Thank you for your help HallsofIvy. Could you possibly re-write the equation as 'B=.......' I am struggling with this....
'
• Oct 14th 2012, 07:44 PM
LanieH
Re: How to solve a complex exponential equation with one unknown???
I have been recently told that this equation must be solved using Excel Solver. I have no idea how to use this program in order to calculate B.
I have created an excel formula relative to the above equation, but I am stuck at this point. Are you able to show me how you would solve the equation for B in Excel Solver?
=(B50-C50)*EXP(-B*($H$44+$I$44))+(C50-D50)*EXP((-2*B($H$44)))+(D50-B50)*EXP((-B($H$44+$J$44)))
• Oct 14th 2012, 08:31 PM
MaxJasper
Re: How to solve a complex exponential equation with one unknown???
You can use exact solutions given here in your Excel module.
• Oct 14th 2012, 08:32 PM
LanieH
Re: How to solve a complex exponential equation with one unknown???
The link didn't work... | 723 | 2,215 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.8125 | 4 | CC-MAIN-2018-22 | latest | en | 0.844321 |
https://www.elfsong.cn/tag/linked-list/ | 1,701,812,386,000,000,000 | text/html | crawl-data/CC-MAIN-2023-50/segments/1700679100568.68/warc/CC-MAIN-20231205204654-20231205234654-00457.warc.gz | 836,029,175 | 9,344 | # Flatten a Multilevel Doubly Linked List
You are given a doubly linked list which in addition to the next and previous pointers, it could have a child pointer, which may or may not point to a separate doubly linked list. These child lists may have one or more children of their own, and so on, to produce a multilevel data structure, as shown in the example below.
Flatten the list so that all the nodes appear in a single-level, doubly linked list. You are given the head of the first level of the list.
Example:
Input: 1---2---3---4---5---6--NULL
|
7---8---9---10--NULL
|
11--12--NULL
Output: 1-2-3-7-8-11-12-9-10-4-5-6-NULL
"""
# Definition for a Node.
class Node:
def __init__(self, val, prev, next, child):
self.val = val
self.prev = prev
self.next = next
self.child = child
"""
class Solution:
def flatten(self, head: 'Node') -> 'Node':
return None
stack = list()
while cur:
if cur.child:
if cur.next:
stack += [cur.next]
cur.next = cur.child
cur.next.prev = cur
cur.child = None
if not cur.next and stack:
temp = stack.pop()
cur.next = temp
temp.prev = cur
cur = cur.next
# Flatten Binary Tree to Linked List
Given a binary tree, flatten it to a linked list in-place.
For example, given the following tree:
1
/ \
2 5
/ \ \
3 4 6
The flattened tree should look like:
1
\
2
\
3
\
4
\
5
\
6
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def flatten(self, root: TreeNode) -> None:
"""
Do not return anything, modify root in-place instead.
"""
node, stack = root, []
while node:
if node.right:
stack.append(node.right)
node.right = node.left
node.left = None
if not node.right and stack:
temp = stack.pop()
node.right = temp
node = node.right | 502 | 1,797 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2023-50 | longest | en | 0.736818 |
https://www.coursehero.com/file/14070674/Exam-2-solutions/ | 1,544,690,987,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376824601.32/warc/CC-MAIN-20181213080138-20181213101638-00239.warc.gz | 861,538,220 | 113,847 | Exam 2-solutions
# Exam 2-solutions - Version 041 Exam 2 markert(55325 This...
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Version 041 – Exam 2 – markert – (55325) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points The current in a wire decreases with time according to the relationship I = (3 . 03 mA) e a t where a = 0 . 13328 s 1 . Determine the total charge that passes through the wire from t = 0 to the time the current has diminished to zero. 1. 0.0222839 2. 0.00877851 3. 0.0234094 4. 0.0282113 5. 0.0181573 6. 0.0164316 7. 0.0143307 8. 0.0252101 9. 0.0208583 10. 0.0227341 Correct answer: 0 . 0227341 C. Explanation: I = d q dt q = integraldisplay t t =0 I dt = integraldisplay t =0 (0 . 00303 A) e ( 0 . 13328 s - 1 ) t dt = (0 . 00303 A) e (0 . 13328 s - 1 ) t - 0 . 13328 s 1 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle 0 = 0 . 0227341 C . 002 10.0points Two parallel conducting plates separated by a distance d are connected to a battery of voltage E . What is correct if the plate separation is doubled while the battery remains connected? 1. The potential difference between the plates is doubled. 2. The capacitance is unchanged. 3. The electric charge on the plates is halved. correct 4. The potential difference between the plates is halved. 5. The electric charge on the plates is dou- bled. Explanation: The capacitance of the two parallel con- ducting plates is given by C = ǫ 0 A d , so C = ǫ 0 A d = ǫ 0 A 2 d = 1 2 ǫ 0 A d = 1 2 C and when the separation d is doubled, the capacitance is halved. The battery remains connected during the whole process, so the potential difference re- mains the same throughout. Q = C V, so the electric charge on the plates is also halved. 003 10.0points Consider two coaxial rings of 17 . 3 cm radius and separated by 12 . 2 cm . Find the electric potential at a point on their common axis midway between the two rings, assuming that each ring carries a uniformly distributed charge of 6 . 97 μ C . The value of the Coulomb constant is 8 . 98755 × 10 9 N · m 2 / C 2 . 1. 325817.0 2. 272568.0 3. 342272.0 4. 346022.0 5. 372068.0 6. 378302.0 7. 289750.0 8. 321695.0 9. 682986.0 10. 257074.0
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Version 041 – Exam 2 – markert – (55325) 2 Correct answer: 6 . 82986 × 10 5 V. Explanation: Let : d = 12 . 2 cm = 0 . 122 m , q = 6 . 97 μ C = 6 . 97 × 10 6 C , r = 17 . 3 cm = 0 . 173 m , and k e = 8 . 98755 × 10 9 N · m 2 / C 2 . With d the distance between the rings, con- sider the potential due only to one ring of radius r and charge Q . The point of measure- ment is a distance L = radicalBigg r 2 + parenleftbigg d 2 parenrightbigg 2 = radicalBigg (0 . 173 m) 2 + parenleftbigg 0 . 122 m 2 parenrightbigg 2 = 0 . 183439 m from any point on the ring. If the charge in an infinitesimal section of the ring is dq , then the potential due to this section is dV = k e dq L , and the potential due to the entire ring is V = integraldisplay dV = k e Q L . Note: All infinitesimal sections are the same distance from the point of measurement and contribute equally. For the two rings, V = 2 k e q L = 2 ( 8 . 98755 × 10 9 N · m 2 / C 2 ) × 6 . 97 × 10 6 C 0 . 183439 m = 6 . 82986 × 10 5 V . 004 10.0points A 0 . 8 V potential difference is maintained across a 2 . 3 m length of tungsten wire that has a cross-sectional area of 0 . 9 mm 2 .
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Jill Tulane University ‘16, Course Hero Intern | 1,428 | 4,644 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2018-51 | latest | en | 0.841992 |
https://kristinbell.org/category/mathematics/ | 1,680,175,513,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296949181.44/warc/CC-MAIN-20230330101355-20230330131355-00108.warc.gz | 412,241,479 | 39,477 | # TI-89 Calculator and Batteries Cozy
Well, since I was making storage pocket things, I decided to make a calculator cozy for my TI-89. I also needed a place to hold extra batteries, because I’d rather not be in the middle of a test and have my calculator die with no back up batteries available. Yes! This does happen! I’m still debating about adding the extra ribbons on top to secure the calculator, but I don’t know if it really needs it, because it fits pretty snug. I made this out of the Mars Martian fabric that I created. :) UPDATE: I decided to add the bows for closure. :)
# Cats with Parachutes in Math Books!
I just posted this on my FB page, but I had to post here. I think this is so funny. This pic comes from my Differential Equations textbook. They were talking about modeling using differential equations, and in the process talked about cats falling from high rises, so they included this pic of a cat with a parachute! Math nerds rock! :) lol
# Having Fun with Ellipsoids!
We are currently working on understanding strange things like ellipsoids and hyperbolic paraboloids and all sorts of strange things like that in my Calc 3 class. It seems like it wouldn’t be that hard, but it is pretty hard to draw these things correctly! This week we got a take home quiz! Thank goodness! I’m posting one example where we are supposed to name the shape and then draw it and talk about the plane sections that intersect it. I don’t know if I’m doing it correctly or not. Guess I will find out!
Basically, how you know it is an equation for an ellipsoid is that an ellipsoid has a standard form for the equation which is basically (x/a)^2 + (y/b)^2 + (z/c)^2 = 1. Now there are many variations on that, so that it might look a bit different, but if it is too different it will be a different shape, like a hyperboloid of one sheet or something else like that. The a’s give you the ellipse size along the x-axis, the b’s give you the ellipse parts for the y-axis and the c’s give you the ellipse information for the z-axis. It is kind of hard to explain in words! lol I guess what you first need to know is that a regular old ellipse is drawn on the x and y axis where the a number gives you information on the number of steps away from the origin along the x-axis and the b number gives you the number of steps away from the origin along the y-axis. I guess I better post an example.
In the example below #1b, you look at the bottom numbers to give you the size of the ellipse. The x-axis number is 5, so it is five units wide on each side. The y-axis number is 10, so it is 10 units wide on each side of the origin. The x-3 and y-3 has to do with shifting the graph away from the origin up and over 3 units each.
Anyway, from what I can gather, to draw the ellipsoid (not ellipse) like in example 2a, you need to draw the various ellipses that make up the ellipsoid in order to get the final shape, so you set x, y and z equal to zero for three different cases to get your ellipse equations. Blah blah blah. This probably makes no sense to anyone, but I thought I’d share! lol
Hyperbola (top) and Ellipse (bottom)
Ellipsoid Example
# New Calculus Fabric Arrived Today!
More Calc2 Fabric
Orange Chevrons with Balloons
I made another Calculus fabric through spoonflower (http://www.spoonflower.com/fabric/1187406), because I was so excited about my first Calculus fabric. Well, the material arrived in the mail today! How exciting!!! I also got my design of balloons on orange chevrons, but I have to work on perfecting the chevrons and getting it how I like it. So fun to get fabric in the mail! :) Click the photos for a larger view. :)
# Yellow-Backed Calculus Pillow
Calculus Pillow
I just finished making my yellow-backed Calculus pillow. It has soft fleece on the back and my Calculus fabric that I made on the front. Here is the link for the Calc Fabric: http://www.spoonflower.com/fabric/1147427. It took a lot of stuffing to fill it, and I probably should have made it with an insert instead of filling, but meh! I love it anyway! Should help me get comfy when I need to do my Calc homework! haha. And yes, it is made from actual Calculus homework! :) Yay!!!
# It Is About Math and Cats People! LOL
Yes, my life is all about math and cats. Here is another cat I’ve finished making. This one is for a friend of mine who likes green. :)
I have one cat for sale in my Etsy shop here: http://www.etsy.com/shop/ArtByKristinBell.
And I’ve made a new Calculus fabric out of my math homework and uploaded it to spoonflower.
Here’s my shop on spoonflower: http://www.spoonflower.com/profiles/kristinbell
OMG I was so excited when someone bought one of my fabrics! You have no idea how thrilling it is!!!
I’m quite a boring person…obsessed with making cats and doing math homework! hehe. Thank god the anxiety has calmed down quite a lot, so I can go about my business and get stuff done!!! What a relief! :) :) :)
# More Fabric From Spoonflower! Yay! Nerd Alert!
Okay, watch out! I’m totally going to nerd out now! I took my calculus homework and created a fabric!!! hahaha Other fabrics include a larger version of the playing kitty, red-white-blue kitty, a neuron, and then I wanted to show you guys what my martiannaut and Judie the Utie look like by the yard! On some I have swatches and fat quarters, because I wanted to see what the repeating pattern was going to look like. The math one is a fat quarter and just a close up of a section of the fabric. It is all so exciting!!! I LOVE FABRIC! If only it were free!!! One of my favorites really is the math one I have to admit! lol Here’s my spoonflower shop address should you feel motivated to get any of my fabrics or if you want to make your own: http://www.spoonflower.com/profiles/kristinbell JUST BEWARE THE ADDICTION!!!! LOL
The Calc2 Fat Quarter with REAL Calc2 Homework!!! hahaha
Calc2 section
Neuron Swatch
Neuron Fat Quarter
Playing Kitty Large Swatch
Playing Kitty Large Fat Quarter
Red, White & Blue Kitty Swatch
Red, White & Blue Kitty Fat Quarter
A yard of Martiannaut
A yard of Judie the Utie
# Calc Example: For When You Hit sqrt(2-2cos(t)) And Need To Integrate!
This example almost made my brain explode! I didn’t know about the Weierstrass substitution method (see this link for more info.: http://en.wikipedia.org/wiki/Weierstrass_substitution) when I began this problem…which is what I needed to complete the problem. I’m still unsure if I did it correctly, because we were supposed to find the integral over 0 to 2*pi, but the integration method only works over 0 to pi, so I ended up multiplying the answer by 2 (because 2 times pi is 2*pi) which got me the answer the book was asking for, but it seemed odd to do it that way. The question is number 18, section 13.3 from the Jon Rogawski Calc book in case you are interested. I think it is a pretty great textbook by the way…I mean as far as math textbooks go! We were also supposed to find when the speed was at its max, but I was so tired I just figured it was at pi and gave up! So, here’s the problem!
# Calc 4 Vector Integration Problem
Let me just say: Calc 4 is confusing!!! We are working with vectors and all sorts of weird 3D shapes and it is strange! Here is an example homework problem. I hope I survive Calc 4!!! | 1,788 | 7,306 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2023-14 | latest | en | 0.95106 |
http://www.math-only-math.com/worksheet-to-find-profit-and-loss.html | 1,502,941,951,000,000,000 | text/html | crawl-data/CC-MAIN-2017-34/segments/1502886102891.30/warc/CC-MAIN-20170817032523-20170817052523-00232.warc.gz | 603,754,772 | 9,426 | # Worksheet to Find Profit and Loss
In practice math worksheet to find profit and loss we will solve numerous problems on calculating profit percent and loss percent. Before solving the worksheet to find profit and loss we can review the word problems on profit and loss.
1. Find the gain or loss per cent when:
(i) CP = $620 and SP =$ 713
(ii) CP = $675 and SP =$ 630
(iii) CP = $345 and SP =$ 372.60
(iv) CP = $80 and SP =$ 76.80
2. Find the selling price when:
(i) CP = $1 650 and gain% = 4% (ii) CP =$ 915 and gain% = 6²/₃%
(iii) CP = $875 and loss% = 12% (iv) CP =$ 645 and loss% = 13¹/₃%
3. Find the cost price when:
(i) SP = $1596 and gain% = 12% (ii) SP =$ 2431 and loss% = 6¹/₂%
(iii) SP = $657.60 and loss% = 4% (iv) SP =$ 34.40 and gain% = 7¹/₂%
Try to practice and solve the questions given in the worksheet to find profit and loss word problems:
4. Mack bought an iron safe for $5580 and paid$ 170 for its transportation. Then, he sold it for $6440. Find the gain per cent earned by him. 5. Rob purchased an old car for$ 73500. He spent $10300 on repairs and paid$ 2600 for its insurance. Then he sold it to a mechanic for $84240. What was his percentage gain or loss? Hint: Overheads =$ (10300 + 2600) = $12900. 6. Harry bought 20 kg of rice at$18 per kg and 25 kg of rice at $16 per kg. He mixed the two varieties and sold the mixture at$ 19 per kg. Find his gain percent in the whole transaction.
7. Coffee costing $250 per kg was mixed with chicory costing$ 75 per kg in the ratio 5 : 2 for a certain blend. If the mixture was sold at $230 per kg, find the gain or loss percent. Hint: Let 5 kg of coffee be mixed with 2 kg of chicory. 8. If the selling price of 16 hats is equal to the cost price of 17 hats, find the gain per cent earned by the dealer. To calculate the profit or loss % lets practice the questions given below in worksheet to find profit and loss. 9. The cost price of 12 candles is equal to the selling price of 15 candles. Find the loss percent. 10. Kelvin gains an amount equal to the selling price of 5 cassettes, by selling 125 cassettes. Find the gain percent. 11. By selling 45 oranges, a man loses a sum equal to the selling price of 3 oranges. Find his loss percent. 12. Apples are bought at 6 for$ 10 and sold at 4 for $9. Find the gain or loss per cent. 13. Vendor purchased bananas at$ 16 per dozen and sold them at 10 for $18. Find his gain or loss per cent. 14. A man bought apples at 10 for$ 25 and sold them at $25 per dozen. Find his gain or loss percent. Answers for worksheet to find profit and loss are given below to check the exact answers of the questions. ### Answers: 1. (i) Gain = 15% (ii) Loss = 6²/₃% (iii) Gain = 8% (iv) Loss = 4% 2. (i)$ 1716
(ii) $976 (iii)$ 770
(iv) $559 3. (i)$ 1425
(ii) $2600 (iii)$ 685
(iv) \$ 32
4. 12%
5. Loss = 2¹/₂%
6. 12¹/₂%
7. Gain = 15%
8. 6¹/₄%
9. 20%
10. 4¹/₆%
11. 6¹/₄%
12. Gain = 35%
13. Gain = 35%
14. Loss = 16²/₃%
Profit, Loss and Discount
Calculating Profit Percent and Loss Percent
Word Problems on Profit and Loss
Examples on Calculating Profit or Loss
Practice Test on Profit and Loss
Discount
Practice Test on Profit Loss and Discount
Profit, Loss and Discount - Worksheets
Worksheet to Find Profit and Loss
Worksheets on Profit and Loss Percentage
Worksheet on Gain and Loss Percentage
Worksheet on Discounts | 1,016 | 3,358 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.625 | 5 | CC-MAIN-2017-34 | longest | en | 0.929944 |
https://math.stackexchange.com/questions/2893941/understanding-a-particular-method-of-solving-generalized-version-of-pells-equat | 1,563,545,199,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195526254.26/warc/CC-MAIN-20190719140355-20190719162355-00365.warc.gz | 472,976,561 | 36,592 | # Understanding a particular method of solving generalized version of Pell's equation
So, I have understood how to solve Diophantine equations of the form $$x^2-Dy^2=1.$$ However, when I was reading the solution of the generalized Pell's equation $$x^2-Dy^2=c,$$ I got stuck. I knew how to solve it using continued fraction, but the book used an intriguing method, but skips over so much detail that I can't understand it. It seems to be using a method similar to Euclid's algorithm, which reduces the $c$ step by step, finally reaching $c=1$.
Here is the translated text:
Example 5. Find all integer solutions of the equation $x^2-15y^2=61$.
Solution: This is not a standard Pell's equation, therefore we need to transform it.
First, solve for the equation $l^2\equiv15(\mathrm{mod}\;61)\;(0\le l\le\frac{61}2)$, which is equivalent to $l^2=15+61h\;(l^2\le900)$. Since $0\le h\le \left[\frac{900}{61}\right]=14$, trying $h=0,1,\dots,14$ gives us $h=10, l=25$.
Next, find the solutions of $x^2_1-15y^2_2=10$, which reduces to to solving $l^2=15+10h\;(l\le\frac{10}{2}=5)$. We can see that $l=5,h=1$. Therefore $x_1=5, y_1=1$.
Next, solve the equation $x^2_2-15y^2_2=1$. This is a Pell's equation, whose fundamental solution is $x_2=4, y_2=1$.
Therefore, $x_2+\sqrt{15}y_2=\pm(4+\sqrt{15})^n$,
$x_1+\sqrt{15}y_1=\pm(4+\sqrt{15})^n(5\pm\sqrt{15})$,
$x+\sqrt{15}y=\pm(4+\sqrt{15})^n(5\pm\sqrt{15})(\frac{25\pm\sqrt{15}}{10})$.
As the three $\pm$ signs are mutually independent, the solution to the equation is therefore $$x+\sqrt{15}y=\pm(4+\sqrt{15})^n(14+3\sqrt{15})$$ and $$x+\sqrt{15}y=\pm(4+\sqrt{15})^n(11+2\sqrt{15})$$
• Where did you find this text? It is absolutely abominable. – Servaes Aug 25 '18 at 9:52
• It's from a Chinese booklet on number theory, and is the only source that has this particular method mentioned. @Servaes – Trebor Aug 25 '18 at 10:03
• math.stackexchange.com/questions/1719280/… – individ Aug 25 '18 at 10:16
The point is to first find a single solution to $x^2-15y^2=61$. Then every other solution is obtained by multiplying by units. I won't prove this here.
First, solving $l^2\equiv15\pmod{61}$ yields the identity $$25^2=15+61\cdot10\qquad\text{ so }\qquad \frac{(25+\sqrt{15})(25-\sqrt{15})}{10}=61.$$ Now suppose we have integers $z,w\in\Bbb{Z}$ such that $z^2-15w^2=10$. Then $x,y\in\Bbb{Z}$ defined by $$x+y\sqrt{15}:=(25\pm\sqrt{15})(z\pm w\sqrt{15}),$$ will satisfy $x^2-15y^2=61$. So we've reduced the problem to finding a solution to $z^2-15w^2=10$.
By the exact same method, solving $k^2\equiv15\pmod{10}$ yields the identity $$5^2=15+10\qquad\text{ so }\qquad(5+\sqrt{15})(5-\sqrt{15})=10.$$ Putting these together shows that $$61=\frac{(25+\sqrt{15})(25-\sqrt{15})}{(5+\sqrt{15})(5-\sqrt{15})} =\frac{1}{10^2}(5+\sqrt{15})(5-\sqrt{15})(25+\sqrt{15})(25-\sqrt{15}).$$ So all $x,y\in\Bbb{Z}$ with $x^2-15y^2=61$ are of the form $$x+y\sqrt{15}=u\cdot\frac{1}{10}(5\mp\sqrt{15})(25\pm\sqrt{15}),$$ where $u\in\Bbb{Z}[\sqrt{15}]^{\times}$ is any unit.
For values other than $61$ the process might be longer or shorter; here we were reduce to the classical Pell equation in two steps, consecutively solving the diophantine equations $$x^2-15y^2=61,\qquad x^2-15y^2=10,\qquad x^2-15y^2=1.$$ What can be said is that the right hand side decreases with every step, so this process does eventually terminate. | 1,212 | 3,356 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.375 | 4 | CC-MAIN-2019-30 | longest | en | 0.84549 |
https://www.mathcelebrity.com/search.php?searchInput=margin%20of%20error | 1,716,210,700,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058278.91/warc/CC-MAIN-20240520111411-20240520141411-00485.warc.gz | 772,708,731 | 10,961 | # margin of error 19 results
margin of error - a statistic expressing the amount of random sampling error in the results of a survey.
Formula: σp + z-score
a confidence interval for a population mean has a margin of error of 0.081
a confidence interval for a population mean has a margin of error of 0.081
a confidence interval for a population mean has a margin of error of 0.081
Margin of error = Interval Size/2 0.081 = Interval Size/2 Cross Multiply: Interval Size = 0.081 * 2 Interval Size = [B]0.162[/B]
A random sample of 149 students has a test score average of 61 with a standard deviation of 10. Find
A random sample of 149 students has a test score average of 61 with a standard deviation of 10. Find the margin of error if the confidence level is 0.99. (Round answer to two decimal places) Using our [URL='https://www.mathcelebrity.com/normconf.php?n=149&xbar=61&stdev=10&conf=99&rdig=4&pl=Large+Sample']confidence interval of the mean calculator[/URL], we get [B]58.89 < u < 63.11[/B]
A sample of 71 college students yields a mean annual income of \$3595. Assuming that ? = \$898, find t
A sample of 71 college students yields a mean annual income of \$3595. Assuming that ? = \$898, find the margin of error in estimating µ at the 99% level of confidence
A sample of 71 college students yields a mean annual income of \$3595. Assuming that ? = \$898, find t
A sample of 71 college students yields a mean annual income of \$3595. Assuming that ? = \$898, find the margin of error in estimating µ at the 99% level of confidence
based on a sample of size 41, a 95% confidence interval for the mean score of all students,µ, on an
based on a sample of size 41, a 95% confidence interval for the mean score of all students,µ, on an aptitude test is from 60 to 66. Find the margin of error
Confidence Interval of a Proportion
Free Confidence Interval of a Proportion Calculator - Given N, n, and a confidence percentage, this will calculate the estimation of confidence interval for the population proportion π including the margin of error. confidence interval of the population proportion
Find Necessary Sample Size
The monthly earnings of a group of business students are are normally distributed with a standard deviation of 545 dollars. A researcher wants to estimate the mean monthly earnings of all business students. Find the sample size needed to have a confidence level of 95% and a margin of error of 128 dollars.
Margin of Error from Confidence Interval
Free Margin of Error from Confidence Interval Calculator - Given a confidence interval, this determines the margin of error and sample mean.
Point Estimate and Margin of Error
Free Point Estimate and Margin of Error Calculator - Given an upper bound and a lower bound and a sample size, this calculate the point estimate, margin of error.
Solve Problem
based on a sample of size 41, a 95% confidence interval for the mean score of all students,µ, on an aptitude test is from 60 to 66. Find the margin of error
Solve Problem
A sample of 71 college students yields a mean annual income of \$3595. Assuming that ? = \$898, find the margin of error in estimating µ at the 99% level of confidence
Solve Problem
[URL]http://www.mathcelebrity.com/marginoferror.php?num=60%2C66&pl=Calculate+Margin+of+Error+and+Sample+Mean[/URL]
Solve the problem
a confidence interval for a population mean has a margin of error of 0.081. Determine the length of the confidence interval
standard deviation of 545 dollars. Find the sample size needed to have a confidence level of 95% and
standard deviation of 545 dollars. Find the sample size needed to have a confidence level of 95% and margin of error 128
standard deviation of 545 dollars. Find the sample size needed to have a confidence level of 95% and
Standard Error (margin of Error) = Standard Deviation / sqrt(n) 128 = 545/sqrt(n) Cross multiply: 128sqrt(n) = 545 Divide by 128 sqrt(n) = 4.2578125 Square both sides: [B]n = 18.1289672852 But we need an integer, so the answer is 19[/B]
The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognit
The brand manager for a brand of toothpaste must plan a campaign designed to increase brand recognition. He wants to first determine the percentage of adults who have heard of the bran. How many adults must he survey in order to be 90% confident that his estimate is within seven percentage points of the true population percentage? [IMG]https://ci5.googleusercontent.com/proxy/kc6cjrLvUq64guMaArhfiSR0mOnTrBwB9iFM9u9VaZ5YYn86CSDWXr1FNyqxylwytHdbQ3iYsUDnavt-zvt-OK0=s0-d-e1-ft#http://latex.codecogs.com/gif.latex?%5Chat%20p[/IMG] = 0.5 1 - [IMG]https://ci5.googleusercontent.com/proxy/kc6cjrLvUq64guMaArhfiSR0mOnTrBwB9iFM9u9VaZ5YYn86CSDWXr1FNyqxylwytHdbQ3iYsUDnavt-zvt-OK0=s0-d-e1-ft#http://latex.codecogs.com/gif.latex?%5Chat%20p[/IMG] = 0.5 margin of error (E) = 0.07 At 90% confidence level the t is, alpha = 1 - 90% alpha = 1 - 0.90 alpha = 0.10 alpha / 2 = 0.10 / 2 = 0.05 Zalpha/2 = Z0.05 = 1.645 sample size = n = (Z[IMG]https://ci4.googleusercontent.com/proxy/mwhpkw3aM19oMNA4tbO_0OdMXEHt9juW214BnNpz4kjXubiVJgwolO7CLbmWXXoSVjDPE_T0CGeUxNungBjN=s0-d-e1-ft#http://latex.codecogs.com/gif.latex?%5Calpha[/IMG] / 2 / E )2 * [IMG]https://ci5.googleusercontent.com/proxy/kc6cjrLvUq64guMaArhfiSR0mOnTrBwB9iFM9u9VaZ5YYn86CSDWXr1FNyqxylwytHdbQ3iYsUDnavt-zvt-OK0=s0-d-e1-ft#http://latex.codecogs.com/gif.latex?%5Chat%20p[/IMG] * (1 - [IMG]https://ci5.googleusercontent.com/proxy/kc6cjrLvUq64guMaArhfiSR0mOnTrBwB9iFM9u9VaZ5YYn86CSDWXr1FNyqxylwytHdbQ3iYsUDnavt-zvt-OK0=s0-d-e1-ft#http://latex.codecogs.com/gif.latex?%5Chat%20p[/IMG] ) = (1.645 / 0.07)^2 *0.5*0.5 23.5^2 * 0.5 * 0.5 552.25 * 0.5 * 0.5 = 138.06 [B]sample size = 138[/B] [I]He must survey 138 adults in order to be 90% confident that his estimate is within seven percentage points of the true population percentage.[/I]
The monthly earnings of a group of business students are are normally distributed with a standard de
The monthly earnings of a group of business students are are normally distributed with a standard deviation of 545 dollars. A researcher wants to estimate the mean monthly earnings of all business students. Find the sample size needed to have a confidence level of 95% and a margin of error of 128 dollars.
The monthly earnings of a group of business students are are normally distributed with a standard de
The monthly earnings of a group of business students are are normally distributed with a standard deviation of 545 dollars. A researcher wants to estimate the mean monthly earnings of all business students. Find the sample size needed to have a confidence level of 95% and a margin of error of 128 dollars. | 1,933 | 6,713 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2024-22 | longest | en | 0.816116 |
http://www.physicsforums.com/showthread.php?t=274302&page=4 | 1,387,650,861,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1387345776439/warc/CC-MAIN-20131218054936-00054-ip-10-33-133-15.ec2.internal.warc.gz | 656,499,155 | 10,214 | Help with Laplace Transformations and 2nd order ODEs
by TFM
Tags: laplace, odes, order, transformations
P: 774 For 1/(s+1)^3, let's just look at 1/s^3 ; we know that t^2 is 2/s^3 so 1/s^3 will be the inverse of (1/2)(t^2) right? but, we dont want 1/s^3, we want 1/(s+1)^3, so what do we do (like we did before)?
P: 1,031
First Shft Theorem
Ok, the first shift theorem says that: if you have a function that looks like: e^(at)F(t); the laplace of that is the laplace of F(t) but instead of putting p, you put p-a eg: L(e^3t(t)) we know that laplace of t is 1/p^2 (you know that from the tables right?) and so here a=3 so the laplace of e^(3t)(t)=1/(p-3)^2 so, where ever you have p, you replace with p-a in this case, p-3 Do you agree??
$$t^2 = 2/s^3$$
$$\frac{1}{2}t^2 = \frac{1}{s^3}$$
using FST
$$\frac{1}{2}t^2 = \frac{1}{b^3}$$, where b = s + 1
would this make it:
\frac{1}{2}(t+1)^2 = \frac{1}{b^3} [/tex]
???
TFM
P: 774 I dont see any e's in your answer (there should be). now as we did before, the second fraction is the laplace transform of: 4te^(-2t) because if you find the laplace of 4t, this will give you 4/s^2 but we dont want s at the bottom, we want (s+2) so thats why we use the first shift theorem and times the 4t by e^(-2t). now, can you do the same for the 1/(s+1)^3
P: 1,031
Okay so:
4t will give you 4/s^2,but we don't want s at the bottom, we want (s+2) so thats why we use the first shift theorem and times the 4t by e^(-2t).
so would this mean:
$$t^2 would give: \frac{2}{s^3}$$
thus would:
$$\frac{2}{(s + 1)^3} be \frac{1}{2}t^2 e^{-t}$$
???
TFM
P: 774 very good!!
P: 1,031 Great So now: $$Y = \frac{1}{(s + 2)^3} + \frac{4}{(s + 2)^2}$$ this goes to: $$Y = \frac{1}{2}t^2 e^{-t} + 4te^{-2t}$$ So now we put this value of Y back into: $$s^2(Y) - 4 + 4(sY) + 4Y = \frac{1}{s + 2}$$ ??? TFM
P: 774 y(t)=(1/2)t^2(e^-t) + 4t(e^-2t) This is the answer. dont substitute anything into anything.
P: 1,031 Excellent. So $$y(t) = \frac{1}{2}t^2 e^{-t} + 4te^{-2t}$$ Tomorrow, I'll start (c), if thats okay? TFM
P: 774 yep thats fine
P: 1,031 Okay, so: (c): $$y'' + y = sin(t), y_0 = 1, y_0' = 0$$ L(y(t)) = Y L(y'(t)) = sL(y(t)) - y(0) L(y''(t)) = sL(y'(t))-y'(0) = s^2(L(y(t))-sy(0)-y'(0) there is no y', $$L(y(t)) = Y$$ $$L(y'(t)) = sL(y(t)) - 1$$ $$L(y''(t)) = sL(y'(t))-0 = s^2(L(y(t))-sy(0)-y'(0)$$ and for the sin(t) $$sin(\alpha t) = \frac{\alpha}{p^2 + \alpha^2}$$ since alpha = 1: $$sin(t) = \frac{1}{p^2 + 1}$$ Is this okay, I am not quite sure because the lack of y' TFM
P: 774 That is absolutely right. what next?
P: 1,031 Okay so: $$L(y(t)) = Y$$ $$L(y'(t)) = sL(y(t)) - 1$$ $$L(y''(t)) = s^2(L(y(t))-sy(0)$$ So now I substitute these into the original equation: $$y'' + y = sin(t)$$ $$s^2(L(y(t))-sy(0) + Y = \frac{1}{p^2 + 1}$$ Is this okay? Do I substitute y(0) into sy(0), or is there another value/leave it alone? TFM
P: 774 That's fine substitute y(0) and also substitue the value L(y(t))
P: 1,031 Okay, so $$s^2(L(y(t))-sy(0) + Y = \frac{1}{p^2 + 1}$$ $$s^2(Y)-1 + Y = \frac{1}{p^2 + 1}$$ $$Ys^2 + Y - 1 = \frac{1}{p^2 + 1}$$ Now I need to make the Y the subject: $$Ys^2 + Y = \frac{1}{p^2 + 1} + 1$$ factorise out $$Y(s^2 + 1) = \frac{1}{p^2 + 1} + 1$$ divide through: $$Y = \frac{\frac{1}{p^2 + 1} + 1}{s^2 + 1}$$ Now I need to find the inverse. Firstly, split it up into two fractions: $$Y = \frac{\frac{1}{p^2 + 1}}{s^2 + 1} + \frac{1}{s^2 + 1}$$ Does this look okay? TFM
P: 774 in ur second line, you have: s^y-1+y=1/(s^2+1) the -1 on the left hand side should be -s (since you have:L(y'')=s(sy-1) and remember: s is p so dont put both...choose one of them :)
P: 1,031 Okay so: $$s^2(Y)-1 + Y = \frac{1}{p^2 + 1}$$ $$Ys^2 + sY - 1 = \frac{1}{s^2 + 1}$$ Does this look okay now? TFM
P: 774 you should have: Ys^2 - s + Y=1/(s^2+1) i think you made a mistake while substituting.
P: 1,031 Okay, back a bit originally: $$s^2(L(y(t))-sy(0) + Y = \frac{1}{s^2 + 1}$$ y(0) = 1 L(y(t)) = Y put these in: $$s^2(Y) - 1 + Y = \frac{1}{s^2 + 1}$$ goes to: $$Ys^2 - 1 + Y = \frac{1}{s^2 + 1}$$ I still have a minus one where there should be a minus s.???? TFM
Related Discussions Calculus & Beyond Homework 3 Differential Equations 4 Differential Equations 6 Calculus & Beyond Homework 0 Calculus & Beyond Homework 2 | 1,794 | 4,280 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2013-48 | longest | en | 0.861005 |
https://www.obcina-ig.si/uncommon-article-gives-you-the-facts-on-mass-in-physics-that-only-a-few-people-know-exist/ | 1,600,760,092,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600400204410.37/warc/CC-MAIN-20200922063158-20200922093158-00227.warc.gz | 989,526,613 | 14,437 | # Uncommon Article Gives You the Facts on Mass in Physics That Only a Few People Know Exist
## The Death of Mass in Physics
Therefore the pressure will in fact be a tiny bit bigger. In the majority of common instances, mass is decided by weighing the object and employing the force of gravity to compute the value automatically. Sunday Mass is a significant sociocultural component in Roman Catholic life. research paper writer He additionally investigated gravity. Gravity is extremely anomalous.
## Mass in Physics for Dummies
In studying density, it can be useful to work a sample problem working with the formula for density, as stated in the past section. These are lollapalooza outcomes. In the instance of non-compact materials, an individual must also be mindful in shaping the mass of the material sample. https://www.ucdavis.edu/ Generally speaking, an operation performed on two numbers will lead to a new number. Stick to the steps and you may establish the speed for your mathematical issue.
Scales are utilized to measure. The V rishon doesn’t have enough mass to get this done. You have to use radians to assess the angle. Equations involving physical quantities should have precisely the same dimensions on either side, and the dimensions have to be the appropriate ones for the quantity calculated.
The debut of Newton’s laws of motion and the growth of Newton’s law of universal gravitation resulted in considerable further evolution of the idea of weight. Plasma won’t be discussed in depth within this chapter because plasma has quite different properties from the 3 other common phases of matter, discussed within this chapter, as a result of strong electrical forces between the charges. By way of example, time doesn’t seem to be divided up into discrete quanta as are most other facets of reality.
This is due to binding energy. same day essays eu In reality, sometimes I believe the universe is daring us to try and work out how it works. The planet Earth is only a small portion of the universe, but it’s the house of human beings and all known life within it.
## The Fundamentals of Mass in Physics You Can Benefit From Starting Right Away
Fluid dynamics is a special area of physics in that it’s the study of fluids and their physical properties. To help better understand this, I developed a whole lot of examples. This has resulted in lots of specialization in physics, and the introduction, development and application of tools from different fields. Here are a few practice questions that you’re able to try.
However, it isn’t only for rocket scientists! The majority of the predictions from these types of theories are numerical. Newton’s 2nd law is utilized in the sphere of Rocket Science. Conservation of momentum is very helpful for understanding the area of particle physics.
## A Secret Weapon for Mass in Physics
Suppose you should push with 5-Newton of force on a sizable box to move it across the ground. This label is known as inertial mass. Beneath this condition, the individual will rise from the ground of the lift and stick with the ceiling of the lift. Thus, though the pillow cases are the exact same dimensions, and both are filled to the exact level, one has much increased mass than the other. She acquires an intense urge to know which of the 2 bricks are most massive.
Like solids, liquids are composed of distinct quantities of atoms and molecules. If one isn’t careful, mass can be transformed into energy. The mass lets you know how many particles you’ve got, not what they weigh.
Momentum has become the most important quantity once it comes to handling collisions in physics. The more inertia an object has, the more mass it has. The photon and vector bosons have various masses since they are created of distinct rishons.
To learn, you will need to convert the miles into kilometers. So that your mass doesn’t change based on where you’re. However, it is not the same as mass. But my weight isn’t 70 kilograms. The weight of the human body differs by place. Demonstrate how to figure your weight on the moon and compose the solution on the board.
They’ve discovered the very first ever stable compound from helium with sodium, at quite significant pressures. The main reason for producing distinction between mass and weight is because gravity isn’t the exact same everywhere. For instance, a massive balloon can have very little mass, and a lead bullet can be quite small but have a lot of mass. COM seems to carry the entire mass of the human body. In doing this, we’ve essentially considered a rigid body for a point particle.
Steel comes in many unique forms. Theories which try to explain these data are made. Measurement is an integral component of human race, without it there is not going to be a trade, zero statistics. He is a measure of the quantity of matter contained in an object.
## The 5-Minute Rule for Mass in Physics
Rocks are from time to time used along coasts to stop erosion. From an experimental perspective, we just wish to create a whole lot of them and see what the results are. Reality on such level is confusing.
That’s 1 reason she and a lot of her colleagues wish to create a bigger, better machine. They’re two things. Again this depends upon your requirements! Because the quantity of stuff which makes you up is always the exact same. Basically what they’re saying is that there’s a difference between how things are theoretically defined and the way you’d measure them in practice.
## The Good, the Bad and Mass in Physics
It can likewise be determined empirically. But if then requested to explain just what it is and the way it works you’d be stumped. We are supposed to leave it. Or when you haven’t, don’t hesitate to watch it. | 1,178 | 5,749 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2020-40 | latest | en | 0.942574 |
https://www.iitianacademy.com/ib-dp-physics-c-4-standing-waves-and-resonance-ib-style-question-bank-sl-paper-2/ | 1,726,392,982,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651622.79/warc/CC-MAIN-20240915084859-20240915114859-00618.warc.gz | 756,064,133 | 38,213 | Home / C.4 Standing waves and resonance Paper 2- IBDP Physics 2025- Exam Style Questions
C.4 Standing waves and resonance Paper 2- IBDP Physics 2025- Exam Style Questions
IBDP Physics SL 2025 – C.4 Standing waves and resonance SL Paper 2 Exam Style Questions
Topic: C.4 Standing waves and resonance SL Paper 2
Standing Waves, Vibrating Strings, Vibrating Air Columns, Resonance, Damping
Question-C.4 Standing waves and resonance
On a guitar, the strings played vibrate between two fixed points. The frequency of vibration is modified by changing the string length using a finger. The different strings have different wave speeds. When a string is plucked, a standing wave forms between the bridge and the finger.
(a) Outline how a standing wave is produced on the string. [2]
(b) The string is displaced 0.4 cm at point P to sound the guitar. Point P on the string vibrates with simple harmonic motion (shm) in its first harmonic with a frequency of 195 Hz. The sounding length of the string is 62 cm.
(i) Show that the speed of the wave on the string is about 240 m . [2]
(ii) Sketch a graph to show how the acceleration of point P varies with its displacement from the rest position. [1]
(iii) Calculate, in m , the maximum velocity of vibration of point P when it is vibrating with a frequency of 195 Hz. [2]
(iv) Calculate, in terms of g, the maximum acceleration of P. [2]
(v) Estimate the displacement needed to double the energy of the string. [2]
(c) The string is made to vibrate in its third harmonic. State the distance between consecutive nodes. [1]
Ans:
a «travelling» wave moves along the length of the string and reflects «at fixed end» superposition/interference of incident and reflected waves the superposition of the reflections is reinforced only for certain wavelengths
b i λ = 2 l = 2 × 0.62 = «1.24 m» v = f λ = 195 ×1.24 = 242 «m s-1» Answer must be to 3 or more sf or working shown for MP2
b ii straight line through origin with negative gradient
b iii max velocity occurs at x 0 = V= «(2π)(195) $$\sqrt{0.004^{2}}$$ » = 4.9«ms-1 »
b iv a = (2π 195)2 × 0.004 = 6005 «m s−2 » = 600g v use of E =A 2 RO Xo2 A = 0.4 = 0.57 «cm» ≅ 0.6 «cm»
c $$\frac{62}{3}$$ = 21«cm»
Question
A large cube is formed from ice. A light ray is incident from a vacuum at an angle of 46˚ to the normal on one surface of the cube. The light ray is parallel to the plane of one of the sides of the cube. The angle of refraction inside the cube is 33˚.
Each side of the ice cube is 0.75 m in length. The initial temperature of the ice cube is –20 °C.
a.i. Calculate the speed of light inside the ice cube.[2]
a.ii. Show that no light emerges from side AB.[3]
a.iiii. Sketch, on the diagram, the subsequent path of the light ray.[2]
b.i. Determine the energy required to melt all of the ice from –20 °C to water at a temperature of 0 °C.
Specific latent heat of fusion of ice = 330 kJ kg–1
Specific heat capacity of ice = 2.1 kJ kg–1 k–1
Density of ice = 920 kg m–3[4]
b.ii. Outline the difference between the molecular structure of a solid and a liquid.[1]
Markscheme
a.i.
«v = c$$\frac{{{\text{sin }}i}}{{{\text{sin }}r}}$$ =» $$\frac{{3 \times {{10}^8} \times {\text{sin}}\left( {33} \right)}}{{{\text{sin}}\left( {46} \right)}}$$
2.3 x 108 «m s–1»
a.ii.
light strikes AB at an angle of 57°
critical angle is «sin–1$$\left( {\frac{{2.3}}{3}} \right)$$ =» 50.1°
49.2° from unrounded value
angle of incidence is greater than critical angle so total internal reflection
OR
light strikes AB at an angle of 57°
calculation showing sin of “refracted angle” = 1.1
statement that since 1.1>1 the angle does not exist and the light does not emerge[Max 3 marks]
a.iiii.
total internal reflection shown
ray emerges at opposite face to incidence
Judge angle of incidence=angle of reflection by eye or accept correctly labelled angles
With sensible refraction in correct direction
b.i.
mass = «volume x density» (0.75)3 x 920 «= 388 kg»
energy required to raise temperature = 388 x 2100 x 20 «= 1.63 x 107
energy required to melt = 388 x 330 x 103 «= 1.28 x 108
1.4 x 108 «J» OR 1.4 x 105 «kJ»
Accept any consistent units
Award [3 max] for answer which uses density as 1000 kg–3 (1.5× 108 «J»)
b.ii.
in solid state, nearest neighbour molecules cannot exchange places/have fixed positions/are closer to each other/have regular pattern/have stronger forces of attraction
in liquid, bonds between molecules can be broken and re-form
OWTTE
Accept converse argument for liquids[Max 1 Mark]
Scroll to Top | 1,337 | 4,616 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2024-38 | latest | en | 0.773524 |
https://socratic.org/questions/how-do-you-solve-and-graph-a-7-10 | 1,590,435,864,000,000,000 | text/html | crawl-data/CC-MAIN-2020-24/segments/1590347389355.2/warc/CC-MAIN-20200525192537-20200525222537-00444.warc.gz | 543,070,508 | 6,165 | # How do you solve and graph a+7<10?
##### 1 Answer
Apr 27, 2017
$a < 3$
#### Explanation:
Subtract 7 from both sides to get
$a < 3$
Place an open dot on $a = 3$ because the point does not satisfy the inequality. Then draw an arrow in the $- \setminus \infty$ direction (left) because $a$ is going to be less than 3. | 102 | 321 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 5, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2020-24 | longest | en | 0.886511 |
https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_29 | 1,708,579,835,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947473690.28/warc/CC-MAIN-20240222030017-20240222060017-00838.warc.gz | 109,166,199 | 10,922 | # 1963 AHSME Problems/Problem 29
## Problem
A particle projected vertically upward reaches, at the end of $t$ seconds, an elevation of $s$ feet where $s = 160 t - 16t^2$. The highest elevation is:
$\textbf{(A)}\ 800 \qquad \textbf{(B)}\ 640\qquad \textbf{(C)}\ 400 \qquad \textbf{(D)}\ 320 \qquad \textbf{(E)}\ 160$
## Solution
The highest elevation a particle can reach is the vertex of the quadratic. The x-value that can get the maximum is $\frac{-160}{-2 \cdot 16} = 5$, so the highest elevation is $160(5) - 16(5^2) = 400$ feet, which is answer choice $\boxed{\textbf{(C)}}$. | 196 | 585 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 7, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2024-10 | latest | en | 0.688879 |
http://joyceh1.blogspot.com/2014/11/common-core-8-using-ti-83-graphing.html | 1,527,322,181,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867374.98/warc/CC-MAIN-20180526073410-20180526093410-00392.warc.gz | 163,666,166 | 14,736 | Friday, November 7, 2014
Common Core 8: Using TI 83+ Graphing Calculator Technology
The students worked on a problem from the textbook called "The Big C's". They discovered that the rule was y=6x+3 where x is the figure number and y is the number of tiles in that figure.
They discovered this after making a table of values. We used graphing calculators to graph the table of values and then graph the rule. I'm writing the steps up for my own reference and for students to access later.
First off, press STAT, then press Edit...
Here you will see a screen with 3 columns labeled L1, L2, and L3.
L1 represents your x values, and L2 represents your y values. Enter in a list of points, in this case it was (1,9) (2, 15) (3, 21)
When you press GRAPH in the top right, you'll get 3 points.
For some students, no points showed up. I troubleshooted this problem and found out that you need to press "2nd" "STAT PLOT" (where the Y= button is) and make sure that your Plot 1 is on so that the L1 and L2 values show up.
They seem to form a straight line if they were connected. Since there is no such thing as a Figure 1.5, this is called a discrete graph. The points are separate and are not meant to be connected.
We then pressed the Y= button in the top right and entered in the rule, 6x+3. When you graph that, it connects those previously unconnected points, forming a continuous graph.
I will add a picture of student work to give some context to the problem. | 373 | 1,468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5 | 4 | CC-MAIN-2018-22 | latest | en | 0.955147 |
http://jacara.info/working-capital-per-dollar-357299 | 1,569,102,054,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574665.79/warc/CC-MAIN-20190921211246-20190921233246-00389.warc.gz | 103,621,051 | 30,889 | # Working Capital Per Dollar of Sales and Turnover Ratio
Once you know what working capital is, the question becomes, "How much working capital should a company have on its balance sheet?"
The answer, and metric, is often expressed as a financial ratio known as working capital per dollar of sales. The calculation itself is straightforward. You take total revenue and divide it into the working capital of a firm.
Precisely how much working capital per dollar of sales is optimal for a given firm will depend on multiple things, including the industry and sector in which the business operates. A business that sells a lot of low-cost items and cycles through its inventory rapidly (such as a convenience store, grocery store, or discount retailer) may only need 10-15% of working capital per dollar of sales.
A manufacturer of heavy machinery and high-priced items with a slower inventory turn may require 20-25% working capital per dollar of sales. A company such as Coca-Cola would probably fall somewhere between the two.
Generally, the best way to determine the range in which a particular company should fall is to compare it to its competitors. If you're looking at the candy business and nearly all candy manufacturers are between [x] and [y] working capital per dollar of sales and you come upon a potential investment that has 1/3rd or 4x that amount, it should raise your eyebrows and warrant further investigation, because the odds are high that there is something going on you need to understand.
### How to Calculate
Here's the formula for Working Capital per Dollar of Sales:
Working Capital ÷ Total Sales (Found on the income statement)
### Practice Sample Calculation I
The first calculation we'll do comes from an old annual report of Goodrich.
First, we calculate working capital, which came to \$933 million. Next, we looked for total sales on the income statement, which was \$4.3 billion. Finally, we plug them into the working capital as a percentage of revenue formula:
Working Capital of \$933 million ÷ Total Sales of \$4.3 billion = .2138, or 21.38%
As a manufacturer of heavy-duty machinery, Goodrich fell within the 20-25% working capital per dollar of sales range that was typical for its peers; a good showing that cast it in a favorable light.
### Practice Sample Calculation II
Now, let's look at a 2014 annual report for another business, Johnson & Johnson, and calculate the working capital per dollar of sales for it.
First, we calculate working capital. We pull the firm's 10-K filing and see that current assets were \$59.3 billion and current liabilities \$25.1 billion. This leaves working capital of \$34.2 billion.
Next, we look at the income statement and find total sales came to \$74.3 billion for the year.
Now, we combine the two into the working capital per dollar of sales calculation:
\$34.2 billion working capital ÷ \$74.3 billion sales = .46, or 46% working capital per dollar of sales.
### Working Capital Turnover
Some analysts prefer to invert working capital per dollar of sales into a financial metric known as working capital turnover. To calculate working capital turnover, you take the working capital per dollar of sales and divide it into one. For example, in the case of Johnson & Johnson, you'd take 1 ÷ .46 to arrive at 2.17.
You can reverse engineer the working capital turnover ratio back into working capital per dollar of sales by taking it and divide it into 1. For example, in this case, you'd take 1 ÷ 2.17 and get .46, or 46%. | 765 | 3,523 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2019-39 | latest | en | 0.952538 |
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# Newton’s third law of motion
Module by: Sunil Kumar Singh. E-mail the author
Summary: Newton’s third law of motion characterizes existence of force in pair only.
Third law of motion is different to other two laws of motion in what it describes. This law states about an important characteristic of force rather than the relation between force and motion as described by the first two laws.
Definition 1: Newton’s third law of motion
One body interacts with other body exerting force on each other, which are equal in magnitude, but opposite in direction.
The action and reaction pair acts along the same line. Their points of application are different as they act on different bodies. This is a distinguishing aspect of third law with respect first two laws, which consider application of force on a single entity.
The law underlines the basic manner in which force comes into existence. Force results from interaction of two bodies, always appearing in pair. In other words, the existence of single force is impossible. In the figure below, we consider a block at rest on a table. The block presses the table down with a force equal to its weight (mg). The horizontal table surface, in turn, pushes the block up with an equal normal force (N), acting upwards.
N = mg N = mg
In this case, the net force on the block and table is zero. The force applied by the table on the block is equal and opposite to the force due to gravity acting on it. As such, there is no change in the state of block. Similarly, net force on the table is zero as ground applies upward reaction force on table to counterbalance the force applied by the block. We should, however, be very clear that these action and reaction force arising from the contact are capable to change the state of motion of individual bodies, provided they are free to move. Consider collision of two billiard balls. The action and reaction forces during collision change the course of motion (acceleration of each ball).
The "action" and "reaction" forces are external forces on individual bodies. Depending on the state of a body (i.e. the state of other forces on the body), the individual "action" or "reaction" will cause acceleration in the particular body. For this reason, book and table do not move on contact, but balls after collision actually moves with certain acceleration.
The scope of this force is not limited to interactions involving physical contact. This law appears to apply only when two bodies come in contact. But, in reality, the characterization of force by third law is applicable to all force types. This requirement of pair existence is equally applicable to forces like electrostatic or gravitational force, which act at a distance without coming in contact.
Let us consider the force between two charges q 1 q 1 and - q 2 - q 2 placed at a distance "r" apart. The magnitude of electrostatic force is given by :
F = q 1 q 2 4 π ε 0 r 2 F = q 1 q 2 4 π ε 0 r 2
The charge q 1 q 1 applies a force F 21 F 21 on the charge q 2 q 2 and charge q 2 q 2 applies a force F 12 F 12 on q 1 q 1 . The two forces are equal in magnitude, but opposite in direction such that :
F 12 = - F 21 F 12 + F 21 = 0 F 12 = - F 21 F 12 + F 21 = 0
(1)
## Note:
Here, we read the subscripted symbol like this : F 12 F 12 means that it is a force, which is applied on body 1 by body 2.
It should be emphasized that though vector sum of two forces is zero, but this condition does not indicate a state of equilibrium. This is so because two forces, often called as action and reaction pair, are acting on different bodies. Equilibrium of a body, on the other hand, involves consideration of external forces on the particular body.
## Deduction of Third law from Newton’s Second law
We have pointed out that “action” and “reaction” forces are external forces on the individual bodies. However, if we consider two bodies forming a “system of two bodies”, then action and reaction pairs are internal to the system of two bodies. The forces on the system of bodies are :
F = F int + F ext F = F int + F ext
(2)
If no external forces act on the system of bodies, then :
F ext = 0 F ext = 0
From second law of motion, we know that only external force causes acceleration to the body system under consideration. As such, acceleration of the “system of two bodies” due to net internal forces should be zero. Hence,
F int = 0 F int = 0
This is possible when internal forces are pair forces of equal magnitude, which are directed in opposite directions.
The internal forces are incapable to produce acceleration of the system of bodies. The term “system of bodies” is important (we shall discuss the concept of system of bodies and their motion in separate module). The acceleration of the system of bodies is identified with a point known as center of mass. When we say that no acceleration is caused by the pair of third law forces, we mean that the “center of mass” has no acceleration. Even though individual body of the system is accelerated, but “center of mass” is not accelerated and hence, we say that "system of bodies" is not accelerated.
Referring to two charged body system that we referred earlier, we can consider forces F 12 F 12 and F 21 F 21 being the internal forces with respect to the system of two charged bodies. As such, applying Newton’s second law,
F i n t = F 12 + F 21 = 0 F i n t = F 12 + F 21 = 0
F 12 = F 21 F 12 = F 21
The “action” and “reaction” forces are, therefore, equal in magnitude, but opposite in direction. Clearly, third law is deducible from second law of motion.
The important point to realize here is that “action” and “reaction” forces are external forces, when considered in relation to individual bodies. Each of the two forces is capable to produce acceleration in individual bodies. The same forces constitute a pair of equal and opposite internal forces, when considered in relation to the system of bodies. In this consideration, the center of mass of the body system has no acceleration and net internal force is zero.
## Summary
Here, we summarize the interpretation of third law of motion as discussed above :
• This law does not address the issue of force and acceleration as applied to a body like first two laws.
• The law characterizes the nature of force irrespective of its class and genesis that they exist in pair of two equal but opposite forces. The existence of a single force is impossible.
• The pair of forces acts on two different bodies.
• The two forces do not neutralize each other, because they operate on different bodies. Each of the bodies will accelerate, if free to do so.
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https://www.stechies.com/pi-python/ | 1,652,679,056,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662509990.19/warc/CC-MAIN-20220516041337-20220516071337-00422.warc.gz | 1,244,308,496 | 14,125 | # How to use pi in Python
Every student in their life has come across pi at least once. It's often used in mathematical formulas to calculate areas, perimeters, volume, circumference, etc.
Pi has an irreplaceable position in the following domains:
1. quantum physics
2. space flights
3. medical procedures, and others.
In this article, we will learn how to use pi in Python and applications of pi in Python.
## What is pi?
Pi is nothing but a mathematical constant. The approximate value of pi is 3.141592653589793. In mathematics, the Greek letter π represents pi. The pi (π ) is a constant of the math library in Python that returns the value 3.141592653589793. Pi helps calculate the area and circumference of the circle or other mathematical figures. The symbol π (pronounced out as "pi" or "pie") is a mathematical constant that represents a number equal to 3.14159 (for short). It is used in Euclidean geometry to define the ratio of a circle's circumference with respect to its diameter. Also, it has various equivalent illustrations. This unique number emerges in many formulas of the different verticals of mathematics and physics.
## Use pi (π) in Python.
Let's take some examples to understand the application of pi (π) in Python:
## Using math.pi
Math is a common module that comes with a lot of mathematical, geometric and trigometric built-in functions.
The first example will be using the math module. In Python, the math module allows access to several mathematical functions. The math module gets imported first to have access to the pi value.
Program:
``````# Importing math Library
import math
# This will print the value of pi in the output
print (math.pi)``````
OUTPUT:
The above example is the most standard way to call/get/use the value of pi (π)
## Using NumPy for Pi
NumPy (Numeric Python) is a popular data science library that also comes with the Pi constant. One can also access the value of pi (π) using numpy.pi. The numpy module also allows access to several constants and mathematical functions and performs several mathematical operations on arrays. To access the pi (π) value using numpy, one must import the numpy module first.
Program:
``````import numpy
print(numpy.pi)``````
Output:
## Using SciPy for Pi
SciPy (Scientific Python) is also a Python module that allows accessing the value of pi(π). The scipy module comes into use to solve mathematical, engineering, and technical problems. To access the pi (π) value using scipy, one must import the scipy module first.
Program:
``````import scipy
print(scipy.pi)
OUTPUT:``````
Output:
Even though they return the exact value of pi, Why Python has three modules to access pi (π)
The reason behind having three modules is that:
While using a module, say math or NumPy module, there is no need to import another module to access pi(π)
## Access pi (π) using math.radians
Using math.radians is the most unusual way to access the value of pi (π). The function radians() converts degrees to radians. To access the pi (π) value using math.radian, one must import the math module first and enter the code as shown below:
Program:
``````import math
math.radians(180) # Here pi defines the 180 degree, passing the value 180 will print the value of pi (π)``````
Output:
## Programs using Pi (π)
Below are some programs in Python that demonstrate the use of pi (π):
Area of a circle:
Code:
``````pi = 3.14 # The value of pi has been initialized in this code as 3.14
ra = float(input("Circle's radius:"))
Area = pi * ra * ra
print("Circles area: = %.2f" %Area)``````
Output:
## Calculating the circumference of a circle:
Program:
``````pi = 3.14 #The value of pi taken in this code is 3.14
radius = float(input("Circle's radius:"))
Circumf =2 * pi * radius
print("Circumference of the circle is: = %.2f" %Circumf)``````
Output:
Wrapping up:
A lot of Python application requires mathematical operations and values to define the various formulae and engineering works such as civil and mechanical engineering. Such software often use the Pi (π) to calculate the area and circumference of the circle or other figures. In Python, pi is a math library constant that returns the value 3.141592653589793.
There are three modules available in Python for accessing the pi (π) value. The reason behind having three modules is to eliminate the need to import a separate module to access the value of pi (π). An unusual or unconventional method of accessing pi (π) is using math.radians.
The radians() convert degree to radians. Among all the above methods, the most efficient way of using the pi is to either use the math.pi or if you are using numpy or scipy module, then do not mention the math.pi separately for using pi constant.
Recommended Posts: | 1,114 | 4,764 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.71875 | 4 | CC-MAIN-2022-21 | latest | en | 0.845632 |
https://mathspace.co/textbooks/syllabuses/Syllabus-914/topics/Topic-19401/subtopics/Subtopic-259815/ | 1,709,288,622,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947475238.84/warc/CC-MAIN-20240301093751-20240301123751-00185.warc.gz | 379,983,481 | 79,146 | # 2.04 Review: Graphs and characteristics of quadratic functions
Lesson
## Identifying key features
In Algebra I, we explored a few key features of quadratic functions. In this lesson, we are going to continue to distribute our understanding of quadratic functions.
Remember!
A quadratic function is a nonlinear function that can be written in general form:
$y=ax^2+bx+c$y=ax2+bx+c, where $a$a$\ne$$00 and vertex occurs at x=\frac{-b}{2a}x=b2a and can be written in vertex form (or standard form): y=a\left(x-h\right)^2+ky=a(xh)2+k, where aa\ne$$0$0 and vertex is $\left(h,k\right)$(h,k)
The U-shaped graph of a quadratic function is called a parabola.
The quadratic function produces a continuous function.
### Transformational graphing
A parent function is the most basic function of a family of functions. It preserves the shape of the entire family. The graph below models the function $f\left(x\right)=x^2$f(x)=x2, which is most basic quadratic function.
• The $x$x- and $y$y-intercepts are at the origin $\left(0,0\right)$(0,0).
• The extrema is an absolute minimum of $0$0.
• The domain of the function is all real numbers, $\left(-\infty,\infty\right)$(,)
• The range of the function is all real numbers greater than $0$0, i.e. $\left[0,\infty\right)$[0,)
• The vertex, or turning point, is at the origin $\left(0,0\right)$(0,0). The axis of symmetry is the vertical line $x=0$x=0.
### Translation
When a graph is shifted either horizontally or vertically, it is called a translation.
Horizontally: $g(x)=f(x-h)$g(x)=f(xh) is $f\left(x\right)$f(x) translated $h$h units right when $h>0$h>0 or $h$h units left when $h<0$h<0
Vertically: $g\left(x\right)=f\left(x\right)+k$g(x)=f(x)+k is $f\left(x\right)$f(x) translated $k$k units up when $k>0$k>0 or $k$k units down when $k<0$k<0
#### Exploration
Let's say we want to move our parent graph of $f\left(x\right)=x^2$f(x)=x2 to the left two units. To do this, we subtract negative two from the $x$x-value inside parentheses: $f\left(x\right)=\left(x-\left(-2\right)\right)^2$f(x)=(x(2))2, which is the same as $f\left(x\right)=\left(x+2\right)^2$f(x)=(x+2)2. So shifts to the right are represented by subtracting a number inside the parentheses, while shifts to the left are represented by adding a number inside the parentheses.
What is the impact of the shift on the key features of the graph?
• The $x$x-intercept is shifted to $\left(-2,0\right)$(2,0). The $y$y-intercept has also moved, and is at the point $\left(0,4\right)$(0,4).
• The extrema is an absolute minimum of $0$0.
• The domain of the function is all real numbers, $\left(-\infty,\infty\right)$(,)
• The range of the function is all real numbers greater than $0$0, i.e. $\left[0,\infty\right)$[0,)
• The vertex, or turning point, is at the $x$x-intercept of $\left(-2,0\right)$(2,0). The axis of symmetry is the vertical line $x=-2$x=2.
Now, let's explore the what happens when we shift the parent function up one unit. We will add one to the parent function, $x^2$x2: $f\left(x\right)=x^2+5$f(x)=x2+5. Vertical shifts up are represented by adding a number outside of the parentheses, while vertical shifts down are represented by subtracting a number outside of the parentheses.
What is the impact of the shift on the key features of the graph?
• The function no longer has any $x$x-intercepts. The y-intercept has been shifted to the point $\left(0,1\right)$(0,1).
• The extrema is an absolute minimum of $1$1.
• The domain of the function is all real numbers, $\left(-\infty,\infty\right)$(,)
• The range of the function is all real numbers greater than or equal to $1$1, i.e. $\left[1,\infty\right)$[1,)
• The vertex, or turning point, is at the $y$y-intercept $\left(0,1\right)$(0,1). The axis of symmetry is the vertical line $x=0$x=0.
### Dilation and reflection
Dilation has the effect of stretching or compressing the parent function graph while the reflection flips the graph over the $x$x-axis.
Dilation: $g\left(x\right)=af\left(x\right)$g(x)=af(x) if $a>1$a>1 stretches away from $x$x-axis; if $00<a<1 compresses towards$x$x-axis Reflection:$g\left(x\right)=-f\left(x\right)$g(x)=f(x) reflected over the$x$x-axis Note: When sketching the graph of a reflection, the ordered pair changes from$\left(x,y\right)$(x,y)$\longrightarrow\left(x,-y\right)$(x,y) for a reflection in the$x$x-axis. ### Exploration When given the parent function,$y=x^2$y=x2, is multiplied by a real number greater than$0$0, the parabola changes its shape. It remains a U-shape; however the U-shape becomes either compressed or stretched. The graph below shows the parent function,$y=x^2$y=x2, and two other functions. What do we notice about the change in shape of the parabola: • The parabola$y=3x^2$y=3x2 is steeper (compressed) than its parent function. • The$y$y-value of each point on$y=3x^2$y=3x2 parabola is three times the$y$y-value of the point on its parent function with the same$x$x-value. Hence, the graph has been compressed by a factor of$3$3 • The parabola$y=0.5x^2$y=0.5x2 is not as steep as its parent function. • The$y$y-value of each point on$y=0.5x^2$y=0.5x2 parabola is half the y-value of the point on its parent function with the same$x$x-value. Hence, the graph has been stretched by a factor of$0.5$0.5 • The following is unchanged from the parent function: • The$x$x- and$y$y-intercepts are at the origin$\left(0,0\right)$(0,0). • The extrema is an absolute minimum of$0$0. • The domain of the function is all real numbers,$\left(-\infty,\infty\right)$(,) • The range of the function is all real numbers greater than$0$0,$\left[0,\infty\right)$[0,) • The vertex, or turning point, is at the origin$\left(0,0\right)$(0,0). The axis of symmetry is the vertical line$x=0$x=0 Now, what happens when we multiply the parent parent function by a negative$1$1? What do we notice about the change in shape of the parabola: • The parabola$y=-x^2$y=x2 maintains the same shape as its parent function. • The ordered pairs on$y=3x^2$y=3x2 parabola are the opposite of those on its parent function:$\left(x,y\right)$(x,y)$\longrightarrow$$\left(-x,-y\right)(x,y). Hence, the graph has been flipped across the x-axis • The parabola does not have an absolute minimum. It does have an absolute maximum at 00 • The range of the graph y=-x^2y=x2 is all real numbers less than or equal to 00, \left(-\infty,0\right](,0] • The following is unchanged from the parent function: • The xx- and yy-intercepts are at the origin \left(0,0\right)(0,0). • The domain of the function is all real numbers, \left(-\infty,\infty\right)(,) • The vertex, or turning point, is at the origin \left(0,0\right)(0,0). The axis of symmetry is the vertical line x=0x=0. ### Increasing and decreasing intervals A function can be described as increasing, decreasing, or constant over a specified interval or the entire domain. Since a parabola is U-shaped, the graph has specific intervals where it increases and decreases. Because the vertex is the greatest or least point on a parabola, its y-coordinate is the maximum value or minimum value of the function. The vertex of a parabola lies on its line of symmetry. So, the graph of the function is increasing on one side of the line of symmetry and decreasing on the other side. Notice, that the intervals do not include 00. The vertex is the point where the graph changes direction; therefore this point cannot be increasing or decreasing. #### Practice questions ##### question 1 Consider the graph of the function y=f\left(x\right)y=f(x) and answer the following questions. 1. What is the absolute minimum of the graph? 2. Hence determine the range of the function. y\ge\editable{}y 3. Over what interval of the domain is the function increasing? x>\editable{}x> ##### question 2 This is a graph of y=x^2y=x2. Loading Graph... 1. How do we shift the graph of y=x^2y=x2 to get the graph of y=\left(x-2\right)^2y=(x2)2? Move the graph downwards by 22 units. A Move the graph to the right by 22 units. B Move the graph to the left by 22 units. C Move the graph upwards by 22 units. D 2. Hence plot y=\left(x-2\right)^2y=(x2)2 on the same graph as y=x^2y=x2. Loading Graph... ##### question 3 Consider the equation y=x^2+5y=x2+5. 1. Complete the set of solutions for the given equation. AA((-33, \editable{}$$)$),$B$B$($($-2$2,$\editable{}$$)), CC((-11, \editable{}$$)$),$D$D$($($\editable{}$,$5$5$)$),$E$E$($($1$1,$\editable{}$$)), FF((22, \editable{}$$)$),$G$G$($($3$3,$\editable{})$) 2. Plot the points$C$C,$D$D and$E\$E on the coordinate axes.
3. Plot the curve that results from the entire set of solutions for the equation being graphed.
### Outcomes
#### MGSE9-12.F.IF.7
Graph functions expressed algebraically and show key features of the graph both by hand and by using technology.
#### MGSE9-12.F.IF.7c
Graph polynomial functions, identifying zeros when suitable factorizations are available, and showing end behavior. | 2,617 | 8,898 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.90625 | 5 | CC-MAIN-2024-10 | latest | en | 0.766594 |
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## Q. 2.35
The specific internal energy of a gas is given by the relation:
u = 3.65 pν + 100 kJ/kg
Where p is in kPa and ν in m³/kg.
3 kg and 0.30 m³ of this gas expands from an initial pressure of 600 kPa to 100 kPa according to the law given by : $p V^{1 .25}$ = C. If the expansion is quasi – static, calculate (a) heat transferred, (b) change in internal energy and (c) work transferred.
## Verified Solution
$V_2=V_1\left\lgroup \frac{p_1}{p_2}\right\rgroup^{\frac{1}{n}}=0.3\left(\frac{600}{100}\right)^{\frac{1}{1.25}}=1.2579 \mathrm{~m}^3$
(b)
\begin{aligned}d U &=U_2-U_1=3.65\left(p_2 V_2-p_1 V_1\right) \\&=3.65(100 \times 1.2579-600 \times 0.3)=-197.87 \mathrm{~kJ}\end{aligned}
(c)
\begin{aligned}W &=\int p \cdot d V=\frac{p_1 V_1-p_2 V_2}{n-1} \\&=\frac{600 \times 0.3-100 \times 1.2579}{1.25-1}=216.84 \mathrm{~kJ}\end{aligned}
(d) Q = dU + W = – 197.87 + 216.84 = 18.97 kJ | 580 | 1,715 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 4, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2023-06 | latest | en | 0.700152 |
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# geometry
+1
136
2
what is the equation of the line that passes through(1,2) and is parallel to the line that passes through the(0,5) and (-4,8)
Mar 4, 2019
#1
+1
The line that passes through (0,5) and (-4,8) has the slope (y2-y1)/(x2-x1) = (8-5)/(-4-0) = 3/-4 = -3/4
y = mx + b, plug in the point (1,2):
2 = (-3/4)(1) + b
2 = -3/4 + b
b = 2.75
y = (-3/4)x + 2.75
Mar 4, 2019
#2
+106532
+1
The slope of the line passing through (0, 5) and (-4, 8) is given by
[ 8 - 5 ] / [ -4 - 0 ] = 3/-4 = -3/4
A line parallel to this has the same slope so we have
y = (-3/4) ( x - 1) + 2 simplify
y = (-3/4)x + 3/4 + 2
y = (-3/4)x + 3/4 + 8/4
y = (-3/4)x + 11/4
Mar 4, 2019 | 334 | 696 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2020-05 | latest | en | 0.843909 |
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# Fun with Logic!
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## Michelle Hirschboeck
on 26 August 2017
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#### Transcript of Fun with Logic!
Fun with Logic!
Implication means that if the antecedent (first proposition) of the conditional is true, then the consequent (last proposition) must also be true.
A conditional is true when this relationship is so.
AND
Connects two or more atomic propositions to make a conjunction
&
IMPLIES
aka 'if/then'
Connects two or more propositions to make a conditional.
v
Any number of atomic propositions may be connected by 'if/then' to make a conditional,
AS LONG AS the antecedent always implies the consequent.
Any number of atomic propositions may be connected in any order by 'or' to make a disjunction.
A disjunction is true if at least one part of it is true.
When 'or' is used in an inclusive sense, more than one part may be true.
When 'or' is used in an exclusive sense, only one part may be true.
Any number of atomic propositions may be connected in any order by '&' to make a conjunction.
A conjunction is true only if ALL of its parts are true.
If one part is false, then the whole conjunction is false.
P,Q...
Variables which stand in for any atomic proposition:
sentence or statement of fact
OR
Connects two or more propositions to make a disjunction
NOT BOTH
Connects two propositions to make a disjunction in which at least one part must be false
~
NOT
Negates a proposition; gives it the opposite truth value
THEREFORE
signifies a conclusion
IFF aka
'if and only if'
Connects two propositions to make a bi-conditional
And when this relationship is so, and the consequent is FALSE, then the antecedent must also be false.
If the antecedent and consequent are so related that they are always either both true or both false,
then their relationship is called a bi-conditional.
Full transcript | 528 | 2,317 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2019-04 | latest | en | 0.88078 |
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## Circumradius of Hexagon given area Solution
STEP 0: Pre-Calculation Summary
Formula Used
rc = sqrt((2*A)/(3*sqrt(3)))
This formula uses 1 Functions, 1 Variables
Functions Used
sqrt - Squre root function, sqrt(Number)
Variables Used
Area - The area is the amount of two-dimensional space taken up by an object. (Measured in Square Meter)
STEP 1: Convert Input(s) to Base Unit
Area: 50 Square Meter --> 50 Square Meter No Conversion Required
STEP 2: Evaluate Formula
Substituting Input Values in Formula
rc = sqrt((2*A)/(3*sqrt(3))) --> sqrt((2*50)/(3*sqrt(3)))
Evaluating ... ...
rc = 4.38691337650831
STEP 3: Convert Result to Output's Unit
4.38691337650831 Meter --> No Conversion Required
4.38691337650831 Meter <-- Circumradius
(Calculation completed in 00.001 seconds)
## < 8 Circumradius of Hexagon Calculators
Circumradius of Hexagon given height and central angle
circumradius = Height/(2*tan((Central Angle*pi/180))) Go
Circumradius of Hexagon given area
circumradius = sqrt((2*Area)/(3*sqrt(3))) Go
Circumradius of hexagon given side and central angle
circumradius = Side/(2*sin(Angle A/2)) Go
Circumradius of Hexagon given short diagonal
circumradius = Diagonal/sqrt(3) Go
Circumradius of Hexagon given long diagonal
circumradius = Long diagonal/2 Go
Circumradius of Hexagon given perimeter
circumradius = Perimeter/6 Go
Circumradius of Hexagon given width
circumradius = Width/2 Go
circumradius = Side Go
### Circumradius of Hexagon given area Formula
rc = sqrt((2*A)/(3*sqrt(3)))
## What is a hexagon?
A regular hexagon is defined as a hexagon that is both equilateral and equiangular. It is bicentric, meaning that it is both cyclic (has a circumscribed circle) and tangential (has an inscribed circle). The common length of the sides equals the radius of the circumscribed circle or circumcircle, which equals 2/sqrt(3) times the apothem (radius of the inscribed circle). All internal angles are 120 degrees. A regular hexagon has six rotational symmetries
## How to Calculate Circumradius of Hexagon given area?
Circumradius of Hexagon given area calculator uses circumradius = sqrt((2*Area)/(3*sqrt(3))) to calculate the Circumradius, The Circumradius of hexagon given area formula is defined as straight line from the centre to the circumference of hexagon, where radius= radius of hexagon , where a is side and Rc is circumradius of hexagon. Circumradius and is denoted by rc symbol.
How to calculate Circumradius of Hexagon given area using this online calculator? To use this online calculator for Circumradius of Hexagon given area, enter Area (A) and hit the calculate button. Here is how the Circumradius of Hexagon given area calculation can be explained with given input values -> 4.386913 = sqrt((2*50)/(3*sqrt(3))).
### FAQ
What is Circumradius of Hexagon given area?
The Circumradius of hexagon given area formula is defined as straight line from the centre to the circumference of hexagon, where radius= radius of hexagon , where a is side and Rc is circumradius of hexagon and is represented as rc = sqrt((2*A)/(3*sqrt(3))) or circumradius = sqrt((2*Area)/(3*sqrt(3))). The area is the amount of two-dimensional space taken up by an object.
How to calculate Circumradius of Hexagon given area?
The Circumradius of hexagon given area formula is defined as straight line from the centre to the circumference of hexagon, where radius= radius of hexagon , where a is side and Rc is circumradius of hexagon is calculated using circumradius = sqrt((2*Area)/(3*sqrt(3))). To calculate Circumradius of Hexagon given area, you need Area (A). With our tool, you need to enter the respective value for Area and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
How many ways are there to calculate Circumradius?
In this formula, Circumradius uses Area. We can use 8 other way(s) to calculate the same, which is/are as follows -
• circumradius = Side
• circumradius = sqrt((2*Area)/(3*sqrt(3)))
• circumradius = Height/(2*tan((Central Angle*pi/180)))
• circumradius = Long diagonal/2
• circumradius = Perimeter/6
• circumradius = Diagonal/sqrt(3)
• circumradius = Width/2
• circumradius = Side/(2*sin(Angle A/2))
Let Others Know | 1,171 | 4,447 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2021-49 | latest | en | 0.75388 |
https://kmmiles.com/70315-km-in-miles | 1,660,609,206,000,000,000 | text/html | crawl-data/CC-MAIN-2022-33/segments/1659882572215.27/warc/CC-MAIN-20220815235954-20220816025954-00034.warc.gz | 344,010,042 | 6,678 | kmmiles.com
# 70315 km in miles
## Result
70315 km equals 43665.615 miles
You can also convert 70315 miles to km.
## Conversion formula
Multiply the amount of km by the conversion factor to get the result in miles:
70315 km × 0.621 = 43665.615 mi
## How to convert 70315 km to miles?
The conversion factor from km to miles is 0.621, which means that 1 km is equal to 0.621 miles:
1 km = 0.621 mi
To convert 70315 km into miles we have to multiply 70315 by the conversion factor in order to get the amount from km to miles. We can also form a proportion to calculate the result:
1 km → 0.621 mi
70315 km → L(mi)
Solve the above proportion to obtain the length L in miles:
L(mi) = 70315 km × 0.621 mi
L(mi) = 43665.615 mi
The final result is:
70315 km → 43665.615 mi
We conclude that 70315 km is equivalent to 43665.615 miles:
70315 km = 43665.615 miles
## Result approximation
For practical purposes we can round our final result to an approximate numerical value. In this case seventy thousand three hundred fifteen km is approximately forty-three thousand six hundred sixty-five point six one five miles:
70315 km ≅ 43665.615 miles
## Conversion table
For quick reference purposes, below is the kilometers to miles conversion table:
kilometers (km) miles (mi)
70316 km 43666.236 miles
70317 km 43666.857 miles
70318 km 43667.478 miles
70319 km 43668.099 miles
70320 km 43668.72 miles
70321 km 43669.341 miles
70322 km 43669.962 miles
70323 km 43670.583 miles
70324 km 43671.204 miles
70325 km 43671.825 miles
## Units definitions
The units involved in this conversion are kilometers and miles. This is how they are defined:
### Kilometers
The kilometer (symbol: km) is a unit of length in the metric system, equal to 1000m (also written as 1E+3m). It is commonly used officially for expressing distances between geographical places on land in most of the world.
### Miles
A mile is a most popular measurement unit of length, equal to most commonly 5,280 feet (1,760 yards, or about 1,609 meters). The mile of 5,280 feet is called land mile or the statute mile to distinguish it from the nautical mile (1,852 meters, about 6,076.1 feet). Use of the mile as a unit of measurement is now largely confined to the United Kingdom, the United States, and Canada. | 626 | 2,288 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.96875 | 4 | CC-MAIN-2022-33 | latest | en | 0.817873 |
http://clay6.com/qa/22874/let-f-x-y-f-x-f-y-forall-x-y-in-r-suppose-that-f-3-3-and-f-0-11-then-f-3-is | 1,480,868,498,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541324.73/warc/CC-MAIN-20161202170901-00028-ip-10-31-129-80.ec2.internal.warc.gz | 52,441,227 | 27,247 | Browse Questions
# Let $f(x+y)=f(x)f(y)$ $\forall x,y\in R$. Suppose that $f(3)=3$ and $f'(0)=11$, then $f'(3)$ is given by
$(a)\;22\qquad(b)\;44\qquad(c)\;28\qquad(d)\;33$
$f'(3)=\lim\limits_{h\to 0}\large\frac{f(3+h)-f(3)}{h}$
$\qquad=\lim\limits_{h\to 0}\large\frac{f(3)f(h)-f(3)}{h}$
$\Rightarrow f'(3)=3\lim\limits_{\large h\to 0}\large\frac{f(h)-1}{h}$-----(1)
$f(3)=3$
Now $f(x+0)=f(x)f(0)$
$\Rightarrow f(x)(f(0)-1)=0$
$\Rightarrow$ either $f(x)=0$ or $f(0)=1$
But $f(3)=3\neq 0$
$\therefore f(0)=1$
From (1),$f'(3)=3f'(0)$
$\qquad\qquad\quad\;\;=3\times 11$
$\qquad\qquad\quad\;\;=33$
Hence (d) is the correct option. | 306 | 629 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2016-50 | longest | en | 0.274152 |
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Topic: Just kidding! Here's the real proof of FLT.
Replies: 0
James Harris Posts: 16 Registered: 12/12/04
Just kidding! Here's the real proof of FLT.
Posted: Sep 7, 1996 3:40 PM
This is a strange one but it works, so what can I say?
Prove x^n + y^n = z^n false when x,y,z relatively prime integers and n a
Natural prime greater than 2.
use x - delta x = my, y - delta y = mx, z + delta y = mz
this requires that m = (z+y)/(z+x), delta x = -(y-x)(x+y+z)/(z+x)
delta y = z(y-x)/(z+x)
substituting into the first equation gives
(x - delta x)^n + (y - delta y)^n = (z + delta y)^n
Multiplying out and subtracting off x^n + y^n = z^n , you
have all terms multiplied by n except
(delta x)^n + 2(delta y)^n
(I added (delta x)^n to both sides)
But that proves that (x+y+z)^n + 2z^n must be divisible by n
so by F. Little T. x+y+z + 2z must be divisible by n, and therefore
z [x+y-z+x+y+z+2z=2(x+y+z)]must be divisible by n. (yes it's ugly but it
works)
Note that z+x and (y-x) can't be divisible by n, which is obvious.
So, just in case you think that I've just managed to prove that z is
divisible by n take the following
x - delta x = -mz, y - delta y = my, z = delta y = -mx
do the same thing and you prove that y must be divisible by n.
Well, that does it and I just have to say that the above is ugly. I can
understand why a certain someone didn't publish it before. Personally,
I'm feeling a bit nauseous.
Signing out.
James S.
do the rest of the above and you prove that y must be divisible by n
Work out everything the same as above and you get that y must be
divisible by n. | 535 | 1,806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2018-17 | latest | en | 0.93045 |
http://www.ask.com/web?qsrc=3053&o=102140&oo=102140&l=dir&gc=1&q=What+Is+the+Formula+for+Area+of+a+Triangle | 1,495,989,728,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463610342.84/warc/CC-MAIN-20170528162023-20170528182023-00037.warc.gz | 530,584,004 | 16,228 | Web Results
The formula for finding the area of a triangle is A=(1/2)BxH.
B is the base of the circle and H is the height.
www.virtualnerd.com/pre-algebra/perimeter-area-volume/perimeter-and-area/area/triangle-area-formula
Did you know that the formula for the area of a triangle can be found by using the formula for the ... area; triangle; base; height; formula; half of parallelogram.
www.virtualnerd.com/pre-algebra/perimeter-area-volume/perimeter-and-area/area-formulas-examples/triangle-area-example
Then just take those values and plug them into the formula for the area of a triangle ... area; triangle; base; height; solve for area; formula; plug in; triangle area.
www.mathwarehouse.com/geometry/triangles/area
Area of a Triangle tutorial. Pictures, examples and many practice problems on how to find the area of a triangle from its base and its height.
Understand why the formula for the area of a triangle is one half base times height, which is half of the area of a parallelogram.
Intuition for why the area of a triangle is one half of base times height. ... To see why the formula makes sense, drag the dot all the way to the right: Created with ...
www.mathgoodies.com/lessons/vol1/area_triangle.html
Since the area of a parallelogram is A = b x h , the area of a triangle must be one- half the area of a parallelogram. Thus, the formula for the area of a triangle is: ...
www.wikihow.com/Calculate-the-Area-of-a-Triangle
The most common way to find the area of a triangle is to take half of the base times the height. Numerous other formulas exist, however, for finding the area of a ... | 389 | 1,617 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2017-22 | longest | en | 0.851117 |
https://convertoctopus.com/76-8-cubic-inches-to-pints | 1,632,550,666,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057598.98/warc/CC-MAIN-20210925052020-20210925082020-00545.warc.gz | 230,646,529 | 7,949 | ## Conversion formula
The conversion factor from cubic inches to pints is 0.034632034632111, which means that 1 cubic inch is equal to 0.034632034632111 pints:
1 in3 = 0.034632034632111 pt
To convert 76.8 cubic inches into pints we have to multiply 76.8 by the conversion factor in order to get the volume amount from cubic inches to pints. We can also form a simple proportion to calculate the result:
1 in3 → 0.034632034632111 pt
76.8 in3 → V(pt)
Solve the above proportion to obtain the volume V in pints:
V(pt) = 76.8 in3 × 0.034632034632111 pt
V(pt) = 2.6597402597461 pt
The final result is:
76.8 in3 → 2.6597402597461 pt
We conclude that 76.8 cubic inches is equivalent to 2.6597402597461 pints:
76.8 cubic inches = 2.6597402597461 pints
## Alternative conversion
We can also convert by utilizing the inverse value of the conversion factor. In this case 1 pint is equal to 0.37597656249918 × 76.8 cubic inches.
Another way is saying that 76.8 cubic inches is equal to 1 ÷ 0.37597656249918 pints.
## Approximate result
For practical purposes we can round our final result to an approximate numerical value. We can say that seventy-six point eight cubic inches is approximately two point six six pints:
76.8 in3 ≅ 2.66 pt
An alternative is also that one pint is approximately zero point three seven six times seventy-six point eight cubic inches.
## Conversion table
### cubic inches to pints chart
For quick reference purposes, below is the conversion table you can use to convert from cubic inches to pints
cubic inches (in3) pints (pt)
77.8 cubic inches 2.694 pints
78.8 cubic inches 2.729 pints
79.8 cubic inches 2.764 pints
80.8 cubic inches 2.798 pints
81.8 cubic inches 2.833 pints
82.8 cubic inches 2.868 pints
83.8 cubic inches 2.902 pints
84.8 cubic inches 2.937 pints
85.8 cubic inches 2.971 pints
86.8 cubic inches 3.006 pints | 552 | 1,865 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.09375 | 4 | CC-MAIN-2021-39 | latest | en | 0.738084 |
https://metanumbers.com/31242 | 1,685,383,706,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224644907.31/warc/CC-MAIN-20230529173312-20230529203312-00139.warc.gz | 443,873,028 | 7,694 | # 31242 (number)
31,242 (thirty-one thousand two hundred forty-two) is an even five-digits composite number following 31241 and preceding 31243. In scientific notation, it is written as 3.1242 × 104. The sum of its digits is 12. It has a total of 4 prime factors and 16 positive divisors. There are 10,080 positive integers (up to 31242) that are relatively prime to 31242.
## Basic properties
• Is Prime? No
• Number parity Even
• Number length 5
• Sum of Digits 12
• Digital Root 3
## Name
Short name 31 thousand 242 thirty-one thousand two hundred forty-two
## Notation
Scientific notation 3.1242 × 104 31.242 × 103
## Prime Factorization of 31242
Prime Factorization 2 × 3 × 41 × 127
Composite number
Distinct Factors Total Factors Radical ω(n) 4 Total number of distinct prime factors Ω(n) 4 Total number of prime factors rad(n) 31242 Product of the distinct prime numbers λ(n) 1 Returns the parity of Ω(n), such that λ(n) = (-1)Ω(n) μ(n) 1 Returns: 1, if n has an even number of prime factors (and is square free) −1, if n has an odd number of prime factors (and is square free) 0, if n has a squared prime factor Λ(n) 0 Returns log(p) if n is a power pk of any prime p (for any k >= 1), else returns 0
The prime factorization of 31,242 is 2 × 3 × 41 × 127. Since it has a total of 4 prime factors, 31,242 is a composite number.
## Divisors of 31242
1, 2, 3, 6, 41, 82, 123, 127, 246, 254, 381, 762, 5207, 10414, 15621, 31242
16 divisors
Even divisors 8 8 4 4
Total Divisors Sum of Divisors Aliquot Sum τ(n) 16 Total number of the positive divisors of n σ(n) 64512 Sum of all the positive divisors of n s(n) 33270 Sum of the proper positive divisors of n A(n) 4032 Returns the sum of divisors (σ(n)) divided by the total number of divisors (τ(n)) G(n) 176.754 Returns the nth root of the product of n divisors H(n) 7.74851 Returns the total number of divisors (τ(n)) divided by the sum of the reciprocal of each divisors
The number 31,242 can be divided by 16 positive divisors (out of which 8 are even, and 8 are odd). The sum of these divisors (counting 31,242) is 64,512, the average is 4,032.
## Other Arithmetic Functions (n = 31242)
1 φ(n) n
Euler Totient Carmichael Lambda Prime Pi φ(n) 10080 Total number of positive integers not greater than n that are coprime to n λ(n) 2520 Smallest positive number such that aλ(n) ≡ 1 (mod n) for all a coprime to n π(n) ≈ 3369 Total number of primes less than or equal to n r2(n) 0 The number of ways n can be represented as the sum of 2 squares
There are 10,080 positive integers (less than 31,242) that are coprime with 31,242. And there are approximately 3,369 prime numbers less than or equal to 31,242.
## Divisibility of 31242
m n mod m 2 3 4 5 6 7 8 9 0 0 2 2 0 1 2 3
The number 31,242 is divisible by 2, 3 and 6.
• Arithmetic
• Abundant
• Polite
• Square Free
## Base conversion (31242)
Base System Value
2 Binary 111101000001010
3 Ternary 1120212010
4 Quaternary 13220022
5 Quinary 1444432
6 Senary 400350
8 Octal 75012
10 Decimal 31242
12 Duodecimal 160b6
20 Vigesimal 3i22
36 Base36 o3u
## Basic calculations (n = 31242)
### Multiplication
n×y
n×2 62484 93726 124968 156210
### Division
n÷y
n÷2 15621 10414 7810.5 6248.4
### Exponentiation
ny
n2 976062564 30494146624488 952698128842254096 29764194941289702467232
### Nth Root
y√n
2√n 176.754 31.4953 13.2949 7.92406
## 31242 as geometric shapes
### Circle
Diameter 62484 196299 3.06639e+09
### Sphere
Volume 1.27734e+14 1.22656e+10 196299
### Square
Length = n
Perimeter 124968 9.76063e+08 44182.9
### Cube
Length = n
Surface area 5.85638e+09 3.04941e+13 54112.7
### Equilateral Triangle
Length = n
Perimeter 93726 4.22647e+08 27056.4
### Triangular Pyramid
Length = n
Surface area 1.69059e+09 3.59377e+12 25509
## Cryptographic Hash Functions
md5 19ddc61af8f213d2c43c17204efab297 d644e1046d4140f2dd3ac6361b68a11fd333bf30 cb94679584197c2e2c89e1cf9070e88c4651429443d7711f64e90ffd3aedcb64 376b376a77df471911c1f86aab526a1b8c70598e81eec1912418980b3e58e404bcc72d4de86f75dd5338a728886ab774ffe0eaff79443eecd70326030c455680 4c956475eb1e166fe8b53ec2556e99d9cd7d6e16 | 1,471 | 4,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2023-23 | latest | en | 0.800377 |
https://www.weegy.com/?ConversationId=5ZE8D9HF | 1,631,896,415,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780055684.76/warc/CC-MAIN-20210917151054-20210917181054-00453.warc.gz | 1,088,119,948 | 12,150 | Which statement supports the conclusion that dominant genes mask recessive genes? Genes can be paired in three different combinations. Genetic information is passed on during reproduction. A purple flower bred with a white flower produces a purple flower. Punnett squares are used to visually represent genetic inheritance.
s
A purple flower bred with a white flower produces a purple flower supports the conclusion that dominant genes mask recessive genes.
Question
Updated 12/27/2017 1:15:09 PM
Rating
8
A purple flower bred with a white flower produces a purple flower supports the conclusion that dominant genes mask recessive genes.
Questions asked by the same visitor
7(x + 2) = 6(x + 5)
Question
Updated 11/19/2014 9:41:07 AM
7(x + 2) = 6(x + 5)
7x + 14 = 6x + 30;
7x - 6x = 30 - 14;
x = 16
Confirmed by yumdrea [11/19/2014 9:48:27 AM]
Solve for x. 13(x - 3) = 39
Weegy: 13(x - 3) = 39 13x - 39 = 39 13x = 39 + 39 13x = 78 x = 78/13 x = 6 User: Simplify the given equation. 5x + 2(x - 3) = -2(x - 1) (More)
Question
Updated 11/19/2014 9:07:22 AM
5x + 2(x - 3) = -2(x - 1);
5x + 2x - 6 = -2x + 2;
7x - 6 = -2x + 2;
7x + 2x = 2 + 6;
9x = 8;
x = 8/9
Solve for x. 5(x - 3) + 4(x + 3) = 3(x - 1)
Weegy: 5(x - 3) + 4(x + 3) = 3(x - 1) 5x - 15 + 4x + 12 = 3x - 3; 5x + 4x - 3x = -3 - 12 + 15; 6x = 0; x = 0 User: Solve for x. 7(x - 3) + 3(4 - x) = -8 x = 1/4 x = 25/10 x = -16/19 x = -41/10 Weegy: 7(x - 3) + 3(4 - x) = -8; 7x - 21 + 12 - 3x = -8; 4x - 9 = -8; 4x = -8 + 9; 4x = 1; x = 1/4 User: Solve for x. 3(x + 2) + 4(x - 5) = 10 x = 36/7 x = 24/7 x = 16/7 x = 9/7 Weegy: 3(x + 2) + 4(x - 5) = 10; 3x + 6 + 4x - 20 = 10; 7x - 14 = 10; 7x = 10 + 14; 7x = 24; x = 24/7 User: Simplify the given equation. 6 - (3x + 10) + 4(2 - x) = 15 4 - 7x = 15 4 - 4x = 15 12 - 7x = 15 Weegy: 6 - (3x + 10) + 4(2 - x) = 15; 6 - 3x - 10 + 8 - 4x = 15; 6 - 7x - 2 = 15; -7x + 4 = 15; -7x = 15 - 4; -7x = 11; x = -11/7 User: Solve for x. 8(x + 1) - 3(x + 4) = 7(2 - x) (More)
Question
Updated 11/19/2014 10:32:51 AM
8(x + 1) - 3(x + 4) = 7(2 - x);
8x + 8 - 3x - 12 = 14 - 7x;
5x - 4 = 14 - 7x;
5x + 7x = 14 + 4;
12x = 18;
x = 18/12;
x = 3/2
Solve for x. 7(x + 2) = 6(x + 5)
Weegy: Solution: x/2 = 6 , x = 2(6) , x = 12 (answer) User: Solve for x. 3(x + 2) + 4(x - 5) = 10 Weegy: 5(x - 3) + 4(x + 3) = 3(x - 1) 5x - 15 + 4x + 12 = 3x - 3; 5x + 4x - 3x = -3 - 12 + 15; 6x = 0; x = 0 User: Solve for x. 3(x + 2) + 4(x - 5) = 10 x = 36/7 x = 24/7 x = 16/7 x = 9/7 (More)
Question
Updated 11/19/2014 9:21:37 AM
7(x + 2) = 6(x + 5);
7x + 14 = 6x + 30;
7x - 6x = 30 - 14;
x = 16
3(x + 2) + 4(x - 5) = 10;
3x + 6 + 4x - 20 = 10;
7x - 14 = 10;
7x = 10 + 14;
7x = 24;
Simplify 10(2x + 3) - 20. 20x + 10 20x - 17 20x - 10
Question
Updated 11/15/2014 2:53:13 AM
10(2x + 3) - 20
= 10*2x + 10*3 - 20
= 20x + 30 - 20
= 20x + 10
Confirmed by jeifunk [11/15/2014 2:52:58 AM]
34,443,592
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Home | Contact | Blog | About | Terms | Privacy | © Purple Inc. | 2,117 | 4,854 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2021-39 | latest | en | 0.774387 |
https://math.stackexchange.com/questions/551361/rules-for-product-and-summation-notation/551391 | 1,696,318,807,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511055.59/warc/CC-MAIN-20231003060619-20231003090619-00254.warc.gz | 415,700,314 | 38,760 | Rules for Product and Summation Notation
When we deal with summation notation, there are some useful computational shortcuts, e.g.: $$\sum\limits_{i=1}^{n} (2 + 3i) = \sum\limits_{i=1}^{n} 2 + \sum\limits_{i=1}^{n} 3i = 2n + \sum\limits_{i=1}^{n}3i$$
However, I don't think I know all the useful shortcuts here. Are there other computational tricks one should be aware of? What's a good way for thinking about this?
More importantly, consider product notation: $$\prod\limits_{i=1}^{n} (\sqrt{2} - \sqrt[n]{2})$$
I don't know what the shortcuts here are. What are some of the more effective ways of attacking such computations?
• Since the product notation you expressed does not contain any $i$, you are multiplying the same thing $n$ times and thus raising it to the $n$th power. You would have $(\sqrt{2}-\root{n}\of{2})^n$. Not sure how to go from there, though. Apr 14, 2020 at 14:10
Some sum identities:
$$\sum_{n=s}^t C\cdot f(n) = C\cdot \sum_{n=s}^t f(n)$$
$$\sum_{n=s}^t f(n) + \sum_{n=s}^{t} g(n) = \sum_{n=s}^t \left[f(n) + g(n)\right]$$
$$\sum_{n=s}^t f(n) - \sum_{n=s}^{t} g(n) = \sum_{n=s}^t \left[f(n) - g(n)\right]$$
$$\sum_{n=s}^t f(n) = \sum_{n=s+p}^{t+p} f(n-p)$$
$$\sum_{n=s}^j f(n) + \sum_{n=j+1}^t f(n) = \sum_{n=s}^t f(n)$$
$$\sum_{n\in A} f(n) = \sum_{n\in \sigma(A)} f(n)$$
$$\sum_{i=k_0}^{k_1}\sum_{j=l_0}^{l_1} a_{i,j} = \sum_{j=l_0}^{l_1}\sum_{i=k_0}^{k_1} a_{i,j}$$
$$\sum_{n=0}^t f(2n) + \sum_{n=0}^t f(2n+1) = \sum_{n=0}^{2t+1} f(n)$$
$$\sum_{n=0}^t \sum_{i=0}^{z-1} f(z\cdot n+i) = \sum_{n=0}^{z\cdot t+z-1} f(n)$$
$$\sum_{n=s}^t \ln f(n) = \ln \prod_{n=s}^t f(n)$$
$$c^{\left[\sum_{n=s}^t f(n) \right]} = \prod_{n=s}^t c^{f(n)}$$
$$\sum_{i=m}^n 1 = n+1-m$$
$$\sum_{i=1}^n \frac{1}{i} = H_n$$
$$\sum_{i=1}^n \frac{1}{i^k} = H^k_n$$
$$\sum_{i=m}^n i = \frac{n(n+1)}{2} - \frac{m(m-1)}{2} = \frac{(n+1-m)(n+m)}{2}$$
$$\sum_{i=0}^n i = \sum_{i=1}^n i = \frac{n(n+1)}{2}$$
$$\sum_{i=0}^n i^2 = \frac{n(n+1)(2n+1)}{6} = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}$$
$$\sum_{i=0}^n i^3 = \left(\frac{n(n+1)}{2}\right)^2 = \frac{n^4}{4} + \frac{n^3}{2} + \frac{n^2}{4} = \left[\sum_{i=1}^n i\right]^2$$
$$\sum_{i=0}^n i^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} = \frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} - \frac{n}{30}$$
$$\sum_{i=0}^n i^p = \frac{(n+1)^{p+1}}{p+1} + \sum_{k=1}^p\frac{B_k}{p-k+1}{p\choose k}(n+1)^{p-k+1}$$
$$\left(\sum_{i=m}^n i\right)^2 = \sum_{i=m}^n ( i^3 - im(m-1) )$$
$$\sum_{i=m}^n i^3 = \left(\sum_{i=m}^n i\right)^2 + m(m-1)\sum_{i=m}^n i$$
$$\sum_{i=m}^{n-1} a^i = \frac{a^m-a^n}{1-a}$$
$$\sum_{i=0}^{n-1} a^i = \frac{1-a^n}{1-a}$$
$$\sum_{i=0}^{n-1} i a^i = \frac{a-na^n+(n-1)a^{n+1}}{(1-a)^2}$$
$$\sum_{i=0}^{n-1} i 2^i = 2+(n-2)2^{n}$$
$$\sum_{i=0}^{n-1} \frac{i}{2^i} = 2-\frac{n+1}{2^{n-1}}$$
$$\sum_{i=0}^n {n \choose i} = 2^n$$
$$\sum_{i=1}^{n} i{n \choose i} = n2^{n-1}$$
$$\sum_{i=0}^{n} i!\cdot{n \choose i} = \sum_{i=0}^{n} {}_{n}P_{i} = \lfloor n!\cdot e \rfloor$$
$$\sum_{i=0}^{n-1} {i \choose k} = {n \choose k+1}$$
$$\sum_{i=0}^n {n \choose i}a^{(n-i)} b^i=(a + b)^n$$
$$\sum_{i=0}^n i\cdot i! = (n+1)! - 1$$
$$\sum_{i=1}^n {}_{i+k}P_{k+1} = \sum_{i=1}^n \prod_{j=0}^k (i+j) = \frac{(n+k+1)!}{(n-1)!(k+2)}$$
$$\sum_{i=0}^n {m+i-1 \choose i} = {m+n \choose n}$$
• Oh wow. That's very thorough. Thank You.
– Newb
Nov 4, 2013 at 10:48
• Did you copy those from somewhere? What about product identities? Dec 20, 2013 at 3:04
• Nah - it's obvious that LTS just entered them in the spur of the moment off the top of her/his head. However, I suggest you save these, because they will prove very useful. Dec 11, 2014 at 6:21
Summation of product of two functions.
$$\sum_{i=1}^{n} f(x) g(x)$$ = $$\sum_{i=1}^{n}f(x)∑g(x) - [ f(x)∑g(x-1) + f(x-1)∑g(x-2) + f(x-2)∑g(x-3) + .... + f(2)∑g(1) + g(x)∑f(x-1) + g(x-1)∑f(x-2) + g(x-2)∑f(x-3) + .... + g(2)∑f(1)]$$
• Here is some MathJax tutorial
– ASB
Feb 25, 2015 at 8:30
• you want to write the Abel summation formula ? Feb 25, 2015 at 8:32
Here are the formula for the sum of the first $n$ natural numbers and the first $n$ squares. There are similar formula for the sum of the first $n$ cubes etc...
$$\sum_{i=1}^{n} i = \frac{n(n+1)}{2}$$ $$\sum_{i=1}^{n} i^{2} = \frac{n(n+1)(2n+1)}{6}$$
The formula for the sum of an arithmetic series is also useful:
if we know the first term $a_1$ and the last term $a_n$, and the series has $n$ terms, then the sum will be $$\frac{n(a_1+a_n)}{2}$$
I suppose $$\prod\limits_{i=1}^{n}(x)$$ means multiplying from 1 to n, which is n factorial. So maybe it will help. Also $$(\sqrt{2} - \sqrt[n]{2})$$ does not contain any i. | 2,221 | 4,620 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.03125 | 4 | CC-MAIN-2023-40 | latest | en | 0.647428 |
https://nabrah.com/092bk8c/b17299-area-of-parallelogram-inside-a-rectangle | 1,624,042,868,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487640324.35/warc/CC-MAIN-20210618165643-20210618195643-00421.warc.gz | 388,479,153 | 11,694 | # area of parallelogram inside a rectangle
Examples: Find the area and perimeter of a rectangle … Area = b×h = 4×5 = 20 Area of a Parallelogram: Quick Interactive Discovery. Having made a simple movement of the triangle on the other side of the trapezoid, we get a rectangle equal in area to our parallelogram. Perimeter of Rectangle = 2(b) + 2(h), Area of Parallelogram = b×h The area of a parallelogram is the $$base \times perpendicular~height~(b \times h)$$. D. 33 units. A rectangle is defined as a quadrilateral where all four of its angles are right angles. A. a) Create a rectangle by cutting off the right-angled triangle and moving it. What is the area of the parallelogram? The area is $${7~cm}\times{3~cm}={21~cm}^{2}$$. But an interesting fact is that the area can also be calculated. Level game: 5th, 6th and 7th grade Angles in a parallelogram Area of a triangle Area of a square Area of a rectangle Area of a rhombus Area of a trapezoid Area of a Parallelogram: Quick Interactive Discovery. Here is a summary of the steps we followed to show a proof of the area of a parallelogram. The figure shows a parallelogram inside a rectangle outline: -5 . Rectangle vs Parallelogram. Area of A = a 2 = 20m × 20m = 400m 2 Part B is a triangle. Be sure to include partial squares! We can see that by taking a triangle out of one side of the rectangle and moving it to the other side, we get a parallelogram. The diagonals bisect each other. The area of any rectangular place is or surface is its length multiplied by its width. A parallelogram is a quadrilateral, or four-sided shape, with two sets of parallel sides. Area of B = ½b × h = ½ × 20m × 14m = 140m 2 Every angle of rectangle is a right angle (900). The diagonals are congruent. 130 units. It is a special case of the quadrilateral. The area of a parallelogram is also equal to its base multiplied by its height. Therefore, area of the parallelogram can be stated as Area of parallelogram = base × height = AB×h The area of the parallelogram is independent of … The diagonals are congruent. For example, if you were trying to find the area of a parallelogram that has a length of 10 and a height of 5, you'd multiply 10 by 5 and get 50. The area of the parallelogram is squares. Area of a Parallelogram. After you set the filling slider, try to count the number of squares inside this parallelogram. Use the perpendicular height of the parallelogram, not the sloping height. Finding the area of a rectangle, for example, is easy: length x width, or base x height. C. 46 units. You can see that this is true by rearranging the parallelogram to make a rectangle. Parallelogram inscribed in a quadrilateral Try this Drag any orange dot and note that the red lines always form a parallelogram. Area of a parallelogram is a region covered by a parallelogram in a two-dimensional plane. A rectangle is a four-sided closed figure with each side making a right angle with another. E. None of the above. 2 Calculate the areas of the parallelograms.) The figure shows a parallelogram inside a rectangle outline: What is the area of the parallelogram? Therefore, the area of the parallelogram is 50. A rectangle is also a parallelogram.. Whose parallelogram has the largest area? What is the area of a parallelogram with base 30 ft and height 25 ft? Parallelograms - area. You can see that this is true by rearranging the parallelogram to make a rectangle. Whose parallelogram has the largest area? Area of a parallelogram: Practice finding area of a parallelogram using its side lenghts and height. The length of any linear geometric shape is … The length of the rectangle is 3 over 5 foot, and the width of the rectangle is 2 over 5 foot. Meracalculator is a free online calculator’s website. Examples: 55 sq ft. B. The area of the parallelogram is 72 cm². Our tips from experts and exam survivors will help you through. The opposite sides are parallel and congruent. How to Find the Area of a Parallelogram? 3. Area of a parallelogram = base × height = 9 × 8 = 72. An easy to use, free area calculator you can use to calculate the area of shapes like square, rectangle, triangle, circle, parallelogram, trapezoid, ellipse, octagon, and sector of a circle. Perimeter = 2(b) + 2(h) = 2(4) + 2(5) = 8 + 10 = 18. The area of the quadrilaterals can be calculated by the formula (base)x(height). What is the area of the parallelogram? Area of a triangle calculation using all different rules, side and height, SSS, ASA, SAS, SSA, etc. A Rectangle is a plane figure with four straight sides and four right angles, one with unequal adjacent sides, in contrast to a square. The diagonals bisect each other. A) 1/25 square feet B) 11/25 square feet C) 3/25 square feet D) 4/25 square feet The figure shows a parallelogram inside a rectangle outline: A parallelogram is shown within a rectangle. Area of a Parallelogram: Quick Interactive Discovery. The opposite sides are parallel and congruent. The given figure shows a parallelogram inside a rectangle. Examples: Find the area and perimeter of a rectangle for the breadth 4 … In order to find the area of the shaded region, we must first find the areas of the rectangle and parallelogram. The figure shows a parallelogram inside a rectangle outline What is the area of the parallelogram? The sum of the interior angles in a quadrilateral is 360 degrees. Remember to always include units with your answer. This tutorial shows how to find the area of a triangle located inside of a rectangle. Read about our approach to external linking. Find the area and perimeter of a rectangle for the breadth 4 and the height 5. Area of a parallelogram 1 On a piece of squared paper, copy this parallelogram and cut it out. 5/16 square foot C.9/32 square foot D. 6/16 square foot The two equal bases of the triangles outside the parallelogram are labeled 1 over 5 foot. Next, find the area of the parallelogram. The measure of the base and height are expressed as integers ≤ 20 in level 1 and ≥ 10 in level 2; plug these values into the formula, Area = base * height, to solve for the area of parallelograms in this set of printable worksheets for 5th grade and 6th grade children. Recall how to find the area of a parallelogram: We know the formula for finding the rectangle’s area is base • height, so the area of this shape is 8 • 5, which is 40 square units. In the app below, use the filling slider to make the parallelogram light enough so that you can see the white gridlines through it.But don't touch anything else yet! The area of a rectangle is length × width, which can also be written as base × height. The area of a parallelogram is the $$base \times perpendicular~height~(b \times h)$$.. You can see that this is true by rearranging the parallelogram to make a rectangle. To find the area of a parallelogram, multiply the base by the height. In order to find the area of the shaded region, we must first find the areas of the rectangle and parallelogram. A. Area of a Parallelogram : The Area is the base times the height: Area = b × h (h is at right angles to b) Example: A parallelogram has a base of 6 m and is 3 m high, what is its Area? The height is already provided it's $$\frac{1}{3}$$ In rectangle, EDCF, segment ED = segment FC because in a rectangle opposite sides are equal We have found two sides that are equal! A rectangle is also a parallelogram, the only difference lies in the angles constituting the figure. A parallelogram is simply a rearranged rectangle. Its sides are generally called Lengths and Breadths. Be sure to include partial squares! B. Determine the area inside the rectangle that is not occupied by the parallelogram. Also, the area of a parallelogram is twice the area of a triangle provided the triangle and the parallelogram are on the same base and between the same parallels. A. Highlighted section is moved to form a rectangle. After you set the filling slider, try to count the number of squares inside this parallelogram. Be sure to include partial squares! Q5: The table shows the dimensions of parallelograms drawn by three students. In Geometry, a parallelogram is a two-dimensional figure with four sides. Also, explore the surface area or volume calculators, as well as hundreds of other math, finance, fitness, and health calculators. The diagonals are congruent. For example, a garden shaped as a rectangle with a length of 10 yards and width of 3 yards has an area of 10 x 3 = 30 square yards. Recall how to find the area of a parallelogram: What is the perimeter of a rectangle with length 13 and width 10? The diagonals bisect each other. Area of a Parallelogram: Quick Interactive Discovery. The area of a shape is a measure of the two dimensional space that it covers. The first way to calculate the area of a parallelogram is associated with one of the simplest shapes - a rectangle. A rectangle is defined as a quadrilateral where all four of its angles are right angles.. In the app below, use the filling slider to make the parallelogram light enough so that you can see the white gridlines through it.But don't touch anything else yet! Inside Any Quadrilateral . Perimeter of Parallelogram = 2(b) + 2(h). 187.5 sq ft. C. 375 sq ft. D. 750 sq ft. 4. Calculate the area, perimeter of Rectangle(Rectangle Calculator), Area of Rectangle = b*h Radio 4 podcast showing maths is the driving force behind modern science. After you set the filling slider, try to count the number of squares inside this parallelogram. In Geometry, a parallelogram is a two-dimensional figure with four sides. A parallelogram where all angles are right angles is a rectangle! calculus. Viewed sideways it has a base of 20m and a height of 14m. A parallelogram is inscribed in a rectangle such that its vertices are the midpoints of the sides of the rectangle. Remember to use the perpendicular height of the parallelogram. Recall how to find the area of a rectangle: Substitute in the given length and height to find the area of the rectangle. The difference between the Rectangle and Parallelogram is that even though the opposite sides of both are parallel and equal, all the angles of a rectangle are 90 degrees. It is a special case of the quadrilateral. The area of the rectangle is squares. The length of the rectangle is 3 over 5 foot, and the width of the rectangle is 2 over 5 foot. Squares, rectangles, and rhombuses are special types of parallelograms, though most people think of a "slanted" rectangle, with two diagonal sides and two flat sides, when they think of the parallelogram. Determine the area inside the rectangle that is not occupied by the parallelogram. Draw a parallelogram. This free area calculator determines the area of a number of common shapes using both metric units and US customary units of length, including rectangle, triangle, trapezoid, circle, sector, ellipse, and parallelogram. Area of the parallelogram can be calculated by the product of the length of one side and the height to the opposite side. A rectangle is defined as a quadrilateral where all four of its angles are right angles.. Be sure to include partial squares! 23 units. Cut a right triangle from the parallelogram. Step 2: Find the perimeter. The two equal bases of the triangles outside the parallelogram are labeled 1 over 5 foot. A rectangle is also a parallelogram.. In the app below, use the filling slider to make the parallelogram light enough so that you can see the white gridlines through it.But don't touch anything else yet! $$base \times perpendicular~height~(b \times h)$$. The width (or height) of the crate -- the distance straight across from the base to the other side -- could vary depending on the inside angles of vertices A, B, C and D. We need to find the width (or height) h of the parallelogram; that is, the distance of a perpendicular line drawn from base C D to A B. Step 1: Find the area. Example: find the area of a rectangle. Proof: if we draw a height h in a parallelogram, then we will divide it into two figures - a triangle and a trapezoid. Area of a parallelogram is a region covered by a parallelogram in a two-dimensional plane. The area of a parallelogram is the $$base \times perpendicular~height~(b \times h)$$. b) Complete the sentences. Parallelograms - area. The sum of the interior angles in a quadrilateral is 360 degrees. Area of Parallelograms | Integers - Type 1. All of the area formulas for general convex quadrilaterals apply to parallelograms. If you noticed the three special parallelograms in the list above, you already have a sense of how to find area. The opposite sides are parallel and congruent. Area of a triangle constructed on the base of parallelogram and touching at any point on the opposite parallel side of the parallelogram can be given as = 0.5 * base * height Hence, Area of ▲ABM = 0.5 * b * h Below is the implementation of the above approach: Regular parallelogram showing base and height. A Rectangle is a quadrilateral whose opposite sides are parallel and are equal and whose angles are equal. Next, find the area of the parallelogram. In the app below, use the filling slider to make the parallelogram light enough so that you can see the white gridlines through it.But don't touch anything else yet! The figure shows a parallelogram inside a rectangle outline: A parallelogram is shown within a rectangle. Determine the area inside the rectangle that is not occupied by the parallelogram. 1/32 square foot B. Formulas, explanations, and graphs for each calculation. In order to answer this question, we recall that the area of any rectangle is equal to its base multiplied by its height. Q5: The table shows the dimensions of parallelograms drawn by three students. The given figure shows a parallelogram inside a rectangle. Recall how to find the area of a rectangle: Substitute in the given length and height to find the area of the rectangle. b) a area = cm2 area = cm2 ... Maximal area of equilateral triangle inside rectangle… If you find the midpoints of each side of any quadrilateral , then link them sequentially with lines, the result is always a parallelogram . To make calculations easier meracalculator has developed 100+ calculators in math, physics, chemistry and health category. Here you can calculate the area and perimeter of Rectangle/parallelogram. Area of parallelograms In this parallelogram, you can cut the triangle from the left of the parallelogram and move it to the right side of the parallelogram, making a rectangle. To find the area of a parallelogram, use the formula area = bh, where b is the length of the parallelogram and h is the height. If the area of the parallelogram is 60, and one of the dimensions of the rectangle is 12, what is the other dimension of the rectangle? ... Well to find the Area of a Parallelogram you must use the formula: A = B * H. But this image doesn't give you the absolute base. Area of a Parallelogram Formula A shape's area is measured in squares, eg square centimetres, square metres and square kilometres. Like other parallelograms, opposite sides in a rectangle are parallel to each other. After you set the filling slider, try to count the number of squares inside this parallelogram. The given figure shows a parallelogram inside a rectangle. 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( b \times h ) \ ) of Rectangle/parallelogram of parallelograms drawn by three students the and!, SAS, SSA, etc tutorial shows how to find the perimeter sets of parallel sides with. × height = 9 × 8 = 72 is a bit difficult but, solvable 2. A region covered by a parallelogram is a rectangle by cutting off the right-angled triangle moving. Any rectangular place is or surface is its length multiplied by its width shown within a rectangle each.!: length x width, or four-sided shape, with two sets of parallel sides:! Two-Dimensional figure with each side making a right angle ( 900 ) a sense of how to the! S website a shape 's area is measured in squares, eg square centimetres, square metres square. Interactive Discovery quadrilateral is 360 degrees, etc after you set the filling slider, try to count number. 20M = 400m 2 Part b is a four-sided closed figure with each side making a right (! The only difference lies in the list above, you already have sense... B×H = 4×5 = 20 Step 2: find the perimeter ) x ( height ) note. Shows a parallelogram inside a rectangle outline: a parallelogram is a rectangle its! Of a rectangle h ) \ ) from experts and exam survivors will help you through x... With base 30 ft and height to find the area of the triangles outside the.... It covers base × height = 9 × 8 = 72 by the formula ( base perpendicular~height~. The length of the rectangle is also a parallelogram 1 On a piece of squared,... But an interesting fact is that the red lines always form a parallelogram = base height! The sum of the area of a parallelogram inside a rectangle, for example, is easy length! 5 foot using all different rules, side and height 25 ft general convex quadrilaterals apply parallelograms. List above, you already have a sense of how to find the area perimeter. Form a parallelogram is 50, the area of the steps we followed to show a proof the... Viewed sideways it has a base of 20m and a height of the steps we to... This parallelogram recall how to find the areas of the rectangle is a of... Q5: the table shows the dimensions of parallelograms drawn by three students ) Create a is! 375 sq ft. D. 750 sq ft. C. 375 sq ft. C. 375 sq ft..! Are right angles is a rectangle: Substitute in the list above, you already have a sense of to! Parallel to each other find the areas of the area of the interior angles a. Areas of the shaded region, we must first find the area of a parallelogram = base × height 9. 4 and the width of the interior angles in a quadrilateral where all of. The driving force behind modern science angles constituting the figure shows a parallelogram inside a rectangle is a of... Other parallelograms, opposite sides in a quadrilateral, or base x height parallelogram inside a rectangle the... Filling slider, try to count the number of squares inside this parallelogram and it. By three students x ( height ) base 30 ft and height to find the of... 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Four-Sided shape, with two sets of parallel sides all four of its angles right... 360 degrees by the parallelogram the table shows the dimensions of parallelograms drawn by three students with 30... Parallel to each other can calculate the area of a rectangle whose opposite sides parallel... Be calculated by the parallelogram to make a rectangle and the width the! A base of 20m and a height of 14m can be calculated to use the perpendicular of. Outside the parallelogram are labeled 1 over 5 foot by a parallelogram is a quadrilateral where all of! Difficult but, solvable you set the filling slider, try to the! All four of its angles are equal and whose angles are equal whose! Figure shows a parallelogram = base × height = 9 × 8 = 72 a online. Base multiplied by its height difference lies in the angles constituting the figure the sum the! 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The area of a parallelogram is equal to its base multiplied by its.... All different rules, side and height to find the area of the rectangle that not... The breadth 4 and the width of the area of a rectangle: Substitute the! Its angles are right angles surface is its length multiplied by its height examples: find the area a. Followed to show a proof of the rectangle and parallelogram angles are right angles a four-sided closed with! Given figure shows a parallelogram with base 30 ft and height force behind modern science height! } = { 21~cm } ^ { 2 } \ ) general convex quadrilaterals to! Of the area can also be calculated by the parallelogram to make rectangle! | 5,819 | 24,361 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.59375 | 5 | CC-MAIN-2021-25 | latest | en | 0.88938 |
https://www.onlinemath4all.com/bisectors-of-a-triangle.html | 1,696,051,436,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510603.89/warc/CC-MAIN-20230930050118-20230930080118-00559.warc.gz | 1,014,130,131 | 7,789 | # BISECTORS OF A TRIANGLE
In a triangle, bisector is the line which divides a side of the triangle into two equal halves. In other words, the bisector will always intersect at the mid points of a side of the triangle.
Perpendicular Bisector of a Triangle :
In a triangle, perpendicular bisector is a line (or ray or segment) that is perpendicular to a side of the triangle at the midpoint of the side.
When three or more lines (or rays or segments) intersect in the same point, they are called concurrent lines (or rays or segments).
The point of intersection of the lines is called the point of concurrency. In any triangle, the three perpendicular bisectors are concurrent. The point of concurrency can be inside the triangle, on the triangle, or outside the triangle as shown in the diagram below.
The point of concurrency of the perpendicular bisectors in a triangle is called the circumcenter of the triangle.
Angle Bisector of a Triangle :
In a triangle, an angle bisector is a line which bisects an angle of the triangle.
The three angle bisectors are concurrent. The point of concurrency of the angle bisectors is called the incenter of the triangle and it always lies inside the triangle.
It has been illustrated in the diagram shown below.
## Concurrency of Perpendicular Bisectors of a Triangle
In a triangle, the perpendicular bisectors intersect at a point that is equidistant from the vertices of the triangle.
PA = PB = PC
## Concurrency of Angle Bisectors of a Triangle
In a triangle, the angle bisectors intersect at a point that is equidistant from the sides of the triangle.
PD = PE = PF
## Solved Examples
Example 1 :
Construct a perpendicular bisector to a line segment.
Solution :
Step 1 :
Draw the line segment AB.
Step 2 :
With the two end points A and B of the line segment as centers and more than half the length of the line segment as radius draw arcs to intersect on both sides of the line segment at C and D.
Step 3 :
Join C and D to get the perpendicular bisector of the given line segment AB.
In the diagram above, CD is the perpendicular bisector of the line segment AB.
Example 2 :
Construct the circumcenter of the triangle ABC with AB = 5 cm, A = 70° and B = 70°.
Solution :
Step 1 :
Draw triangle ABC with the given measurements.
Step 2 :
Construct the perpendicular bisectors of any two sides (AC and BC) and let them meet at S which is the circumcenter.
Example 3 :
A company plans to build a distribution center that is convenient to three of its major clients as shown in the diagram below.
The planners start by roughly locating the three clients on a sketch and finding the circumcenter of the triangle formed.
(i) Explain why using the circumcenter as the location of a distribution center would be convenient for all the clients.
(ii) Make a sketch of the triangle formed by the clients. Locate the circumcenter of the triangle. Tell what segments are congruent.
Solution (i) :
Because the circumcenter is equidistant from the three vertices, each client would be equally close to the distribution center.
Solution (ii) :
Label the vertices of the triangle as E, F, and G. Draw the perpendicular bisectors. Label their intersection as D.
By theorem 1 given above, in a triangle, the perpendicular bisectors intersect at a point that is equidistant from the vertices of the triangle.
So,
DE = DF = DG
Example 4 :
In the diagram shown below, the angle bisectors of ΔMNP meet at point L.
(i) What segments are congruent?
(ii) Find LQ and LR
Solution (i) :
By theorem "Concurrency of Angle Bisectors of a Triangle", the three angle bisectors of a triangle intersect at a point that is equidistant from the sides of the triangle.
So, we have
LR ≅ LQ ≅ LS
Solution (ii) :
By theorem "Concurrency of Angle Bisectors of a Triangle", the three angle bisectors of a triangle intersect at a point that is equidistant from the sides of the triangle.
Use the Pythagorean Theorem to find LQ in ΔLQM.
LQ2 + MQ2 = LM2
Substitute MQ = 15 and LM = 17.
LQ2 + 152 = 172
Simplify.
LQ2 + 225 = 289
Subtract 225 from both sides.
LQ2 = 64
LQ2 = 82
LQ = 8 units
Because LR ≅ LQ,
LR = 8 units
Kindly mail your feedback to v4formath@gmail.com
## Recent Articles
1. ### Finding Two Numbers with the Given Difference and Product
Sep 29, 23 10:55 PM
Finding Two Numbers with the Given Difference and Product
2. ### Finding Two Numbers with the Given Sum and Product
Sep 29, 23 10:49 PM
Finding Two Numbers with the Given Sum and Product | 1,085 | 4,527 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2023-40 | latest | en | 0.869081 |
https://math.answers.com/math-and-arithmetic/All_rectangles_are_also_. | 1,722,680,104,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640365107.3/warc/CC-MAIN-20240803091113-20240803121113-00463.warc.gz | 306,635,303 | 47,192 | 0
# All rectangles are also .
Updated: 12/17/2022
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7y ago
parallelograms.
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Related questions
### Are all square is also rectangles?
No, but all squares are all rectangles.
### Is every property of parallelograms also a property of all rectangles?
Yes. All rectangles are also parallelograms, but not all parallelograms are rectangles.
### Why are some rectangles squares?
All squares are rectangles also, but not all rectangles are squares, only equilateral rectangles are considered square.
### Can square can be a rectangle?
A square is a specialized type of a rectangle. All squares are also rectangles, but only some rectangles are squares.
### Are all non squares also non rectangles?
All non rectangles are non squares.
### Are all rhombuses also rectangles?
No. Rhombuses that are also rectangles are called squares. Rhombuses are parallelograms with 4 equal sides, while rectangles are parallelograms where all the angles are right.
### Can a parallelogram also be a rectangle?
Some parallelograms are rectangles; all rectangles are parallelograms.
### Are all square rectangle?
Yes, all squares are also rectangles. But not all rectangles are also squares. Squares are a specific type of rectangle that has all equal length sides and all 90 degree vertices.
### Are all parallelograms also rectangles?
No because rectangles have to have right (90 degree) angles.
### Are all square also rectangles?
All squares and rectangles belong to the class of 4 sided quadrilaterals
### Why is a square also a rectangle?
Because there opposite sides are congruent. And all squares are rectangles, but not all rectangles are squares!
### Are all rectangles parallel?
Yes and also all squares. | 387 | 1,817 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-33 | latest | en | 0.906088 |
https://digitdogchallenges.com/2018/04/27/mouse-count-4/ | 1,675,606,478,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500255.78/warc/CC-MAIN-20230205130241-20230205160241-00461.warc.gz | 231,198,918 | 27,839 | Posted in Mouse Count
# Mouse Count 4
#### More ideas for reasoning and problem-solving using the book Mouse Count by Ellen Stoll Walsh
The aim of the activity is to encourage learners to think and talk mathematically. Ask learners to discuss and find ways to solve the problem. Ask learners to explain their thinking and show, in any way they want, how the problem has been solved.
The problem of The Missing Mice
The snake put ten mice in his jar.
“Ten mice are enough. Now I am going to eat you up, little, warm and tasty, ” said the snake.
“Wait”, said one of the mice. “The jar isn’t full yet. And look at the big mouse over there.”
The snake was very greedy. He hurried off to get the big mouse.
When he got back…….what did he see?
Can you work out how many mice escaped while the snake was away?
How does the snake know that some mice have escaped?
What do you need to do first to solve the problem?
How many mice are still in the jar? How many have got away?
Explain how you can find out how many escaped. You might want to use a ten frame or Numicon shape to help you. You might want to draw a picture.
Can you write a number sentence for this problem?
Convince me that you have worked out the problem.
Encourage learners to predict and estimate by asking questions such as:
I wonder if more than 1 escaped……..what do you think?
I wonder if more than 5 escaped………
What if……………
………there was a different number of mice in the jar when the snake got back?
……..the snake started with a different number?
Encourage learners to make up some problems of their own using different numbers and different contexts. | 377 | 1,640 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2023-06 | latest | en | 0.968029 |
https://scicomp.stackexchange.com/questions/33028/fast-algorithm-for-computing-cofactor-matrix/33033 | 1,632,223,109,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057202.68/warc/CC-MAIN-20210921101319-20210921131319-00314.warc.gz | 567,229,383 | 40,607 | # Fast algorithm for computing cofactor matrix
I wonder if there is a fast algorithm, say ($$\mathcal O(n^3)$$) for computing the cofactor matrix (or conjugate matrix) of an $$N\times N$$ square matrix. And yes, one could first compute its determinant and inverse separately and then multiply them together. But how about this square matrix is non-invertible?
What would it mean by "This probably means that also for non-invertible matrixes, there is some clever way to calculate the cofactor (i.e., not use the mathematical formula that you use above, but some other equivalent definition)."?
• welcome to scicomp! I'm curious, too: why do you want to numerically compute the co-factors of a singular matrix (i.e. what's your target application)? Jul 9 '19 at 11:58
• I am implementing an algorithm on a large-scale setting that computes conjugate matrix in iterative steps. And I find it is the bottleneck. Maybe there is some mathematical foundation for why it has to be conjugate (instead of others, like pseudoinverse * pseudo-determinant, which all led to divergence on small-scale instance). I am just curious if some one have developed an efficient way to deal with this computational issue. Jul 9 '19 at 14:36
• If I understand the goal, you want to compute all $n^2$ cofactors with $O (n^3)$ effort. The structure of the underlying matrix will be needed to evaluate the stability of such algorithms. Jul 10 '19 at 14:16
So, a cofactor matrix is a transpose of an adjugate matrix. I know of the following paper:
There, the author works on an algorithm of computing an adjugate matrix $$\text{adj}(A)$$ when $$A$$ is nearly singular or singular. For such matrices, one can make use of the factorizations. Suppose, we have found:
$$A=XDY \tag{1} \label{eq1}$$ where $$X$$ and $$Y$$ are well-conditioned, and $$D$$ is a diagonal matrix. Now, we can write the adjugate matrix, as follows:
$$\text{adj}(A)=\text{det}(X)\text{det}(D)\text{det}(Y)\left(Y^{-1}D^{-1}X^{-1}\right) \label{eq2} \tag{2}$$
There are several standard decompositions that satisfy (with various guarantees) $$\eqref{eq1}$$: SVD, LU with full pivoting, pivoted QR, and pivoted QLP. Now, the matrix $$D$$ enters the $$\eqref{eq2}$$ twice: as a $$D^{-1}$$ and $$\text{det}(D)$$ which seems like a problem in case the matrix is truly singular. The author of the paper argues (and justifies by the perturbation series analysis) that
1. In floating-point arithmetic a true zero is unlikely
2. If it really happens, a small perturbation should be applied to that and algorithm proceeds as per $$\eqref{eq2}$$ with perturbed zero entries of $$D$$.
The perturbation theory is unusual because although $$\text{adj}(A)$$ and $$A^{-1}$$ differ only by a scalar factor, the matrix $$A^{-1}$$ has singularities while $$\text{adj}(A)$$ is analytic - in fact, it is a multinomial in the elements of $$A$$. It turns out that multiplying by the determinant smooths out the singularities to give an elegant perturbation expansion.
...
However, if $$A$$ is ill-conditioned - that is, if $$A$$ is nearly singular - the inverse will be inaccurately computed. Nonetheless, we will show that this method, properly implemented, can give an accurate adjugate, even when the inverse has been computed inaccurately.
Take a look at the detailed discussion in the paper on the advantages and disadvantages of the proposed factorizations.
• Going via the SVD route guarantees the well-conditioning of $$X$$ and $$Y$$; however, finding their determinants might be tricky (even though, they are just signs: $$\text{det}(X,Y)=\pm 1$$.
• On the contrary, both full-pivoted LU and pivoted QR should lead to an easy $$\mathcal O(N^3)$$ algorithm. For example full-pivoted LU:
$$A=\Pi_\text{R} LDU\Pi_\text{C}$$
results in
$$\text{adj}(A)=\text{det}(\Pi_\text{R})\text{det}(D)\text{det}(\Pi_\text{C})\left( \Pi_\text{C}^{T}U^{-1}D^{-1}L^{-1}\Pi_\text{R}^{T}\right)$$ where $$\text{det}(\Pi_\text{R})=(-1)^{\text{number of row interchanges}}$$ and all computations are straightforward.
So, that gives an $$\mathcal O(N^3)$$ algorithm to compute the adjugate matrix since all the components are at most $$\mathcal O(N^3)$$: finding the inverse of well-conditioned matrices, LU-decomposition, matrix-matrix multiplication, calculation of easy determinants. However, as opposed to SVD, the $$X$$ and $$Y$$ tend to be well-conditioned, but might not be (see the detailed discussion in the paper). In practice, I don't think it would be an issue. And worst comes to worst, you might just have to use both methods in such special cases.
• Since $\det(D) = \prod_{i=1}^n d_{ii}$, it seems to me that one can use the identity $\det(D) D^{-1} = \operatorname{diag}(e_i)$, where $e_i = \prod_{j\neq i} d_{jj}$. This gives an algorithm that works without divisions by zero even if $D$ is exactly singular: no need to introduce "small perturbations". Feb 23 '20 at 20:09
• @FedericoPoloni interesting, I did not think of it while reading that paper. That seems like a small but valuable improvement over the 1998 paper to me. Feb 23 '20 at 20:57
• Actually that formula is literally in the Stewart paper (Eqns 1.4 and 1.5) for the special case of the SVD, but surprisingly the author does not use it later and resorts to that perturbation argument. Maybe I am missing something here, I did not read the paper carefully. Feb 23 '20 at 21:31 | 1,449 | 5,384 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 34, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2021-39 | latest | en | 0.895776 |
https://oeis.org/A254669 | 1,653,554,330,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662604495.84/warc/CC-MAIN-20220526065603-20220526095603-00544.warc.gz | 487,969,173 | 4,333 | The OEIS is supported by the many generous donors to the OEIS Foundation.
Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
A254669 Consider A098550(x_n)=2*prime(n). Let a(n) be number of even numbers<=2*prime(n) in A098550 in positions <=x_n. 3
2, 3, 5, 4, 10, 11, 16, 19, 23, 28, 28, 36, 40, 40, 46, 52, 58, 60, 67, 70, 72, 76, 83, 89, 97, 99, 102, 107, 108, 113, 126, 129, 136, 137, 149, 151, 155, 163, 167, 172, 177, 180, 190, 193, 197, 199, 207, 220, 225, 227, 231, 239, 241, 247, 252, 262, 267, 270 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 COMMENTS Conjecture. lim a(n)/prime(n)=1, as n goes to infinity. LINKS Peter J. C. Moses, Table of n, a(n) for n = 1..1000 David L. Applegate, Hans Havermann, Bob Selcoe, Vladimir Shevelev, N. J. A. Sloane, and Reinhard Zumkeller, The Yellowstone Permutation, arXiv preprint arXiv:1501.01669, 2015. EXAMPLE Let n=10, p_n=29, 2*p_n=58, 58=A098550(63). We consider all even terms <= 58 up to the position 63. They are 2,4,8,14,6,12,16,10,20,22,26,28,32,18,24,34,36,30,38,42,44,40,50,48,52,46,56,58. We have 28 such numbers. Thus a(10)=28. MATHEMATICA terms = 58; f[lst_] := Block[{k = 4}, While[GCD[lst[[-2]], k] == 1 || GCD[lst[[-1]], k]>1 || MemberQ[lst, k], k++]; Append[lst, k]]; A098550 = Nest[f, {1, 2, 3}, 12 terms]; a[n_] := Module[{p, pos}, p = Prime[n]; pos = FirstPosition[A098550, 2 p][[1]]; Count[A098550[[1 ;; pos]], k_ /; EvenQ[k] && k <= 2 p]]; Array[a, terms] (* Jean-François Alcover, Dec 12 2018, after Robert G. Wilson v in A098550 *) CROSSREFS Cf. A098550. Sequence in context: A246841 A066417 A347348 * A227913 A079521 A325549 Adjacent sequences: A254666 A254667 A254668 * A254670 A254671 A254672 KEYWORD nonn AUTHOR Vladimir Shevelev, Feb 04 2015 EXTENSIONS More terms from Peter J. C. Moses, Feb 04 2015 STATUS approved
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Last modified May 26 04:26 EDT 2022. Contains 354074 sequences. (Running on oeis4.) | 837 | 2,173 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2022-21 | latest | en | 0.5846 |
https://socratic.org/questions/how-do-you-find-all-the-real-and-complex-roots-of-125x-3-343-0 | 1,721,456,645,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763515020.57/warc/CC-MAIN-20240720052626-20240720082626-00418.warc.gz | 460,455,933 | 6,390 | How do you find all the real and complex roots of 125x^3 + 343 = 0?
Mar 2, 2016
$- \frac{7}{5}$ and $\frac{7 \pm 7 \sqrt{3} i}{10}$
Explanation:
You can determine the real root of
$125 {x}^{3} + 343 = 0$
like this
$x = \sqrt[3]{- \frac{343}{125}}$
The real solution is $- \frac{7}{5}$
But you can determine all the roots by solving the equation in complex form (remember that e^(pii)=-1) :
$x = \sqrt[3]{\left(\frac{343}{125}\right) {e}^{\pi i}}$
So we would have the solutions:
$\sqrt[n]{\rho {e}^{\theta \pi i}} = \sqrt[n]{\rho} \times {e}^{\left(\frac{2 k + \theta \pi}{n}\right) i} , k \in \mathbb{Z}$
$x = \frac{7}{5} \times {e}^{\left(\frac{\pi + 2 k \pi}{3}\right) i}$
When k is 1, we obtain the real solution -7/5.
When k is 0 or -1, we obtain
$x = \frac{7}{5} \times {e}^{\frac{\pm \pi i}{3}} = \frac{7}{5} \times \left[\cos \left(\pm \pi \frac{i}{3}\right) + i \sin \left(\pm \pi \frac{i}{3}\right)\right]$
which is
$= \frac{7}{5} \times \frac{1 \pm \sqrt{3}}{2} = \frac{7 \pm 7 \sqrt{3} i}{10}$
Mar 3, 2016
$x = - \frac{7}{5} , \frac{7}{10} \pm \frac{7 \sqrt{3}}{10} i$
Explanation:
Factor as a sum of cubes, since
$125 {x}^{3} = {\left(5 x\right)}^{3}$
$343 = {7}^{3}$
Sums of cubes factor as follows:
${a}^{3} + {b}^{3} = \left(a + b\right) \left({a}^{2} - a b + {b}^{2}\right)$
Thus,
${\left(5 x\right)}^{3} + {7}^{3} = \left(5 x + 7\right) \left({\left(5 x\right)}^{2} - 5 x \left(7\right) + {7}^{2}\right)$
$0 = \left(5 x + 7\right) \left(25 {x}^{2} - 35 x + 49\right)$
This can be split up into two equations:
$5 x + 7 = 0 \text{ "=>" } x = - \frac{7}{5}$
And
$25 {x}^{2} - 35 x + 49 = 0$
$x = \frac{35 \pm \sqrt{{\left(- 35\right)}^{2} - 4 \left(25\right) \left(49\right)}}{2 \left(25\right)}$
$x = \frac{35 \pm \sqrt{1225 - 4900}}{50}$
$x = \frac{35 \pm \sqrt{- 3675}}{50}$
$x = \frac{35 \pm \sqrt{3 \cdot {5}^{2} \cdot {7}^{2}} i}{50}$
$x = \frac{35 \pm 35 \sqrt{3} i}{50}$
$x = \frac{7}{10} \pm \frac{7 \sqrt{3}}{10} i$ | 883 | 1,978 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 25, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-30 | latest | en | 0.493643 |
https://sciforums.com/threads/displaying-equations-using-tex.61223/#post-1240085 | 1,712,990,359,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816586.79/warc/CC-MAIN-20240413051941-20240413081941-00731.warc.gz | 487,949,673 | 13,715 | # Displaying equations using Tex
Discussion in 'Physics & Math' started by Pete, Dec 20, 2006.
1. ### PeteIt's not rocket surgeryRegistered Senior Member
Messages:
10,167
Good news, everyone!
Plazma has told me that we can now use $$tags. Example: [tex]\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Displays as:
$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
I wonder if someone who is good at LaTex could write a short tutorial of the basics, with some common SciForums examples? I'll write one up myself, but I'm a complete beginner at LaTex so it will take a little while.
In the meantime, here's an external tutorial site: LaTex Math Tutorial for mimeTeX
Pete
Last edited by a moderator: Sep 26, 2021
danshawen likes this.
3. ### James RJust this guy, you know?Staff Member
Messages:
39,421
Wow! This is a great idea. Thanks, Plazma!
Ok, basics of LaTeX, very briefly:
Grouping, where necessary, is done by enclosing expressions in curly brackets: {}
Superscripts and subscripts:
a^2 gives a superscript:
$a^2$
a_2 gives a subscript:
$a_2$
a^b_c does both:
$a^b_c$
Sometimes, grouping is necessary, as in a^{b+c}:
$a^{b+c}$
LaTeX functions are indicated by a backslash followed by the function name. For example \sin \int \pi gives:
$\sin \qquad \int \qquad \pi$
Greek letters are produced with a backslash and the letter name:
\alpha, \beta, \gamma, \phi
$\alpha, \beta, \gamma, \phi$
Fractions use the \frac function, with the numerator and denominator enclosed in curly brackets:
\frac{mc^2}{a + b + c^3}
$\frac{mc^2}{a + b + c^3}$
Sums and integrals can be produced as follows:
\int \limits_a^b 4x^2~dx
\sum \limits_{n=1}^4 n = 10
$\int \limits_a^b 4x^2~dx$
$\sum \limits_{n=1}^4 n = 10$
Note that a space is inserted in the integral above using a tilde (~).
Last edited: Dec 21, 2006
Pinball1970 and danshawen like this.
5. ### AbsaneRocket SurgeonValued Senior Member
Messages:
8,989
Tex is short for LaTeX?
FREAKIN' SWEET! Now I just have to learn how to use it
7. ### quadraphonicsBloodthirsty BarbarianValued Senior Member
Messages:
9,391
Nice, but what's up with the automatic "x = " at the starting of every tex input?
$\frac{\partial}{\partial z} \int_{a(z)}^{b(z)} f(x,z) \, dx = \int_{a(z)}^{b(z)} \frac{\partial f}{\partial z} \, dx + f(b(z),z) \frac{\partial b}{\partial z} - f(a(z),z) \frac{\partial a}{\partial z}$
danshawen likes this.
8. ### James RJust this guy, you know?Staff Member
Messages:
39,421
Houston, we have a problem!
danshawen likes this.
9. ### quadraphonicsBloodthirsty BarbarianValued Senior Member
Messages:
9,391
Actually, Tex is a lower-level system that LaTex is built on top of, although it should be said that most people only use Tex in the context of LaTex.
10. ### D HSome other guyValued Senior Member
Messages:
2,257
TeX and LaTeX are not the same. TeX is a macro language designed by Donald Knuth back in the 1960s for writing technical papers with lots of math. LaTeX is a macro package for TeX and was written by Leslie Lamport back in the 1980s. Writing in plain TeX is a bear, plain and simple. LaTex makes writing a math paper a breeze.
Last edited: Dec 21, 2006
11. ### PeteIt's not rocket surgeryRegistered Senior Member
Messages:
10,167
Looks like a problem with the parsing of the $tag. The mimetex.cgi renderer is working, but "x=" is added to the front of the [tex] element content before being sent to the renderer. Plazma is notified! Pete$
12. ### cosmodelRegistered Senior Member
Messages:
62
test math
$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
13. ### AbsaneRocket SurgeonValued Senior Member
Messages:
8,989
$\gamma = \lim_{n \rightarrow \infty } \left( \left( \sum_{k=1}^n \frac{1}{k} \right) - \ln (n) \right)=\int_1^\infty\left({1\over\lfloor x\rfloor}-{1\over x}\right)\,dx$
My favorite constant.
14. ### James RJust this guy, you know?Staff Member
Messages:
39,421
Ok, looks like the problem is fixed now.
I'm excited, people! Ooh, maths. It makes me go all gooey inside.
15. ### AbsaneRocket SurgeonValued Senior Member
Messages:
8,989
Hrm. Does the TeX display incorrectly for IE or is it just this machine I am on (computer at school)? Some of the commands don't show right here (like in my Gamma constant equation, I see 1/[\overx?] instead of 1/x in Tex.
16. ### James RJust this guy, you know?Staff Member
Messages:
39,421
The space is important after the \over command.
It should be
1 \over x
which comes out as:
$1 \over x$
Or, you can use \frac:
\frac{1}{x}
is displayed as
$\frac{1}{x}$
17. ### PeteIt's not rocket surgeryRegistered Senior Member
Messages:
10,167
Thanks to Plazma for having the bug corrected.
Should we leave this thread stickied, or just link to it in the FAQ?
18. ### James RJust this guy, you know?Staff Member
Messages:
39,421
Leave it sticky for now, I think.
19. ### AbsaneRocket SurgeonValued Senior Member
Messages:
8,989
Well, the code produces the correct equations from my computer. However, it does not output everything from my school's computer. Maybe they are just too old.
20. ### invert_nexusZe do caixaoValued Senior Member
Messages:
9,686
Makes me wish I was a mathematician...
21. ### Blue_UKDrifting MindValued Senior Member
Messages:
1,449
I haven't returned to sciforums in ages - I'm glad to see that James and Pete are still about and more glad to see that someone has made an equation editor for bb code.
Absane, could you explain that constant? I can't grasp it! Surely as x approaches inf. then ln(x) also approaches something high too?
$\int{\frac{2x}{x^2}} = ln(x^2)$
22. ### James RJust this guy, you know?Staff Member
Messages:
39,421
Yes, but infinity minus infinity can be finite.
23. ### AbsaneRocket SurgeonValued Senior Member
Messages:
8,989
I don't have a proof that the series converges, but I am sure that it is not too hard.
It goes something like this:
1 - ln(1) = 1
1 + 1/2 - ln(2) = 0.806852819
1 + 1/2 + 1/3 - ln(3) = 0.734721045
1 + 1/2 + 1/3 + ... + 1/100 - ln(100) = 0.582207331651
The series converges to 0.5772156649...
No one knows if this number is irrational. According to Mathworld, "The famous English mathematician G. H. Hardy is alleged to have offered to give up his Savilian Chair at Oxford to anyone who proved gamma to be irrational"
http://mathworld.wolfram.com/Euler-MascheroniConstant.html
But this series is a perfect example as to why we cannot assume infinity - infinity = 0 or any other number for that matter. | 1,910 | 6,423 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2024-18 | latest | en | 0.820134 |
https://www.printablemultiplication.com/free-printable-multiplication-table-pdf/ | 1,620,874,217,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243992721.31/warc/CC-MAIN-20210513014954-20210513044954-00434.warc.gz | 987,751,158 | 78,280 | # Free Printable Multiplication Table Pdf
Discovering multiplication soon after counting, addition, as well as subtraction is good. Kids find out arithmetic by way of a all-natural progression. This progress of studying arithmetic is generally the adhering to: counting, addition, subtraction, multiplication, and lastly department. This document contributes to the issue why find out arithmetic with this sequence? Furthermore, why find out multiplication soon after counting, addition, and subtraction before section?
## The subsequent facts answer these concerns:
1. Children discover counting very first by associating aesthetic objects using their hands. A concrete case in point: The amount of apples exist inside the basket? More abstract case in point is how older are you?
2. From counting phone numbers, the subsequent plausible step is addition combined with subtraction. Addition and subtraction tables can be quite beneficial training helps for the kids because they are aesthetic equipment creating the move from counting easier.
3. Which will be acquired up coming, multiplication or division? Multiplication is shorthand for addition. At this time, youngsters use a organization grasp of addition. Consequently, multiplication will be the after that logical kind of arithmetic to discover.
## Overview basic principles of multiplication. Also, assess the essentials the way you use a multiplication table.
Let us review a multiplication illustration. By using a Multiplication Table, grow a number of times about three and have an answer 12: 4 x 3 = 12. The intersection of row a few and column a number of of a Multiplication Table is twelve; twelve is the response. For the kids beginning to find out multiplication, this can be straightforward. They are able to use addition to resolve the issue therefore affirming that multiplication is shorthand for addition. Example: 4 x 3 = 4 4 4 = 12. It is really an superb overview of the Multiplication Table. The added benefit, the Multiplication Table is visible and demonstrates back to understanding addition.
## Where will we start discovering multiplication using the Multiplication Table?
1. Initial, get informed about the table.
2. Start with multiplying by 1. Commence at row number 1. Go on to line # 1. The intersection of row a single and column the first is the best solution: one particular.
3. Perform repeatedly these methods for multiplying by one. Flourish row one by columns a single through a dozen. The answers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 correspondingly.
4. Recurring these methods for multiplying by two. Multiply row two by posts one via five. The answers are 2, 4, 6, 8, and 10 correspondingly.
5. Allow us to leap ahead. Recurring these actions for multiplying by 5. Increase row 5 by columns one particular by means of twelve. The answers are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 respectively.
6. Now let us raise the level of trouble. Perform repeatedly these actions for multiplying by three. Multiply row three by posts a single by means of a dozen. The responses are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 correspondingly.
7. In case you are confident with multiplication up to now, try a examination. Solve the next multiplication issues in your mind after which examine your answers for the Multiplication Table: grow half a dozen as well as 2, flourish 9 and 3, multiply a single and 11, increase four and 4, and flourish several and 2. The problem solutions are 12, 27, 11, 16, and 14 correspondingly.
In the event you got four away from 5 various troubles right, design your very own multiplication assessments. Determine the responses in your mind, and appearance them utilizing the Multiplication Table. | 846 | 3,752 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2021-21 | longest | en | 0.929568 |
https://fr.mathworks.com/matlabcentral/cody/problems/44309-pi-digit-probability/solutions/1908290 | 1,601,596,130,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600402132335.99/warc/CC-MAIN-20201001210429-20201002000429-00591.warc.gz | 406,544,265 | 16,832 | Cody
# Problem 44309. Pi Digit Probability
Solution 1908290
Submitted on 21 Aug 2019
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Fail
N = 101; n = 3; y_correct = 0.1200; assert(abs(pidigit(N,n)-y_correct)<0.0001) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[101,201,202,203,1001]),regexp(fileread('pidigit.m'),'[\d\.\+\-\*\/]+','match')))) % modified from the comment of Alfonso on https://www.mathworks.com/matlabcentral/cody/problems/44343
Error in solution: Line: 4 Column: 24 Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.
2 Fail
N = 201; n = 6; y_correct = 0.0750; assert(abs(pidigit(N,n)-y_correct)<0.0001) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[101,201,202,203,1001]),regexp(fileread('pidigit.m'),'[\d\.\+\-\*\/]+','match'))))
Error in solution: Line: 4 Column: 24 Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.
3 Fail
N = 202; n = 6; y_correct = 0.0796; assert(abs(pidigit(N,n)-y_correct)<0.0001) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[101,201,202,203,1001]),regexp(fileread('pidigit.m'),'[\d\.\+\-\*\/]+','match'))))
Error in solution: Line: 4 Column: 24 Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.
4 Fail
N = 203; n = 6; y_correct = 0.0792; assert(abs(pidigit(N,n)-y_correct)<0.0001) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[101,201,202,203,1001]),regexp(fileread('pidigit.m'),'[\d\.\+\-\*\/]+','match'))))
Error in solution: Line: 4 Column: 24 Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters.
5 Fail
N = 1001; n = 9; y_correct = 0.1050; assert(abs(pidigit(N,n)-y_correct)<0.0001) assert(~any(cellfun(@(x)ismember(max([0,str2num(x)]),[101,201,202,203,1001]),regexp(fileread('pidigit.m'),'[\d\.\+\-\*\/]+','match'))))
Error in solution: Line: 4 Column: 24 Invalid expression. When calling a function or indexing a variable, use parentheses. Otherwise, check for mismatched delimiters. | 725 | 2,313 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2020-40 | latest | en | 0.389851 |
https://www.termpaperwarehouse.com/essay-on/Qnt-561-Week-2-Weekly-Learning/486534 | 1,696,438,538,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233511386.54/warc/CC-MAIN-20231004152134-20231004182134-00440.warc.gz | 1,108,856,906 | 10,637 | # Qnt 561 Week 2 Weekly Learning Assessment – Assignment
In: Other Topics
Submitted By uopstudents
Words 1187
Pages 5
Chapter 5 Exercise 4
A large company must hire a new president. The Board of Directors prepares a list of five candidates, all of whom are equally qualified. Two of these candidates are members of a minority group. To avoid bias in the selection of the candidate, the company decides to select the president by lottery.
a. What is the probability one of the minority candidates is hired? (Round your answer to 1 decimal place.)
b. Which concept of probability did you use to make this estimate?
Chapter 5 Exercise 14
The chair of the board of directors says, "There is a 50% chance this company will earn a profit, a 30% chance it will break even, and a 20% chance it will lose money next quarter." a. Use an addition rule to find the probability the company will not lose money next quarter. (Round your answer to 2 decimal places.)
b. Use the complement rule to find the probability it will not lose money next quarter. (Round your answer to 2 decimal places.)
Find the Weekly Learning Assessment answers here QNT 561 Week 2 Weekly Learning Assessments Chapter 5 Exercise 22
A National Park Service survey of visitors to the Rocky Mountain region revealed that 50% visit Yellowstone Park, 40% visit the Tetons, and 35% visit both.
a. What is the probability a vacationer will visit at least one of these attractions? (Round your answer to 2 decimal places.)
b. What is the probability .35 called?
c. Are the events mutually exclusive?
Chapter 5 Exercise 34
P(A1) = .20, P(A2) = .40, and P(A3) = .40. P(B1|A1) = .25. P(B1|A2) = .05, and P(B1|A3) = .10.
Use Bayes' theorem to determine P(A3|B1). (Round your answer to 4 decimal places.)
Want to download the Complete Weekly Assignment of QNT/561 Class..?? Click QNT 561 Weekly Learning Assessments
Chapter 5 Exercise 40
Solve the following:
a. 20! 17!
b. 9P3
c. 7C2
Chapter 6 Exercise 4
Which of...
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Words: 1756 - Pages: 8 | 910 | 3,510 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2023-40 | latest | en | 0.875852 |
https://homework.cpm.org/category/CON_FOUND/textbook/mc2/chapter/11/lesson/11.3.3/problem/11-113 | 1,571,654,107,000,000,000 | text/html | crawl-data/CC-MAIN-2019-43/segments/1570987769323.92/warc/CC-MAIN-20191021093533-20191021121033-00153.warc.gz | 529,014,414 | 15,367 | ### Home > MC2 > Chapter 11 > Lesson 11.3.3 > Problem11-113
11-113.
1. Simplify each expression. Homework Help ✎
Find a common denominator.
$- \frac{44}{60} + \frac{45}{60}$
$\frac{1}{60}$
Use the associative property of addition.
$(5-2) + ( \frac{2}{9} - \frac{1}{10} )$
$3 \frac{11}{90}$
See (a). | 116 | 306 | {"found_math": true, "script_math_tex": 4, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2019-43 | longest | en | 0.492588 |
https://www.smartstudy.com/gre/article/72425.html | 1,623,540,084,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487586465.3/warc/CC-MAIN-20210612222407-20210613012407-00416.warc.gz | 891,665,431 | 16,721 | GRE数学备考时，首先同学们要打好基础，包括一些基本的知识点、词汇、公式、定理等等，都要大家能够熟悉掌握，下面智课网给大家提供的内容就是GRE数学备考重点手册【主要符号】，一起来熟记下来吧!
主要符号：
＋ plus ;positive － minus ;negative × multiplied by ;times ÷ divided by ＝ equals ≈ approximately equals ≠ not equal to ＜ less than ＞ greater than ≤ equal to or less than ≥ equal to or greater than o ( ) round brackets ;parentheses [ ] square brackets { } braces ∈ is a member of the set ⊂ is a subset of ∽ similar to ≌ congruent to * denotes an operation ∴ therefore ∵ because ∶ ratio sign, divided by, is to ∷ equals, as(proportion) square root of cube root of ∥ parallel to ⊥ perpendicular to, at right angles with ∠ angle ∟ right angle º degree ′ minute ″ second ⊙ circle A⁀B arc AB e the base of natural logarithms,approx.2.71828 x! factorial x, x(x-1)(x-2)---1 lognx log x to the base n π pi lnx log x to the base e(natural logarithm) lgx log x to the base 10(common logarithm) |x| the absolute value of x
本文是智课网给大家带来的GRE数学备考重点手册【主要符号】，熟练掌握下来吧，对大家以后考试会有很大帮助的! | 353 | 975 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2021-25 | latest | en | 0.577902 |
www.1900s.org.uk | 1,716,478,465,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058642.50/warc/CC-MAIN-20240523142446-20240523172446-00062.warc.gz | 538,541,374 | 7,486 | # Log tables for multiplying and dividing
Log tables and antilog tables were the only way to multiply and divide large numbers quickly, easily and accurately before today's electronic calculators. The only alternatives were slide rules which were easy to use but not reliably accurate; mechanical calculating machines which were slow and fiddly; and what was called long multiplication and division which was based on slow and clumsy mental arithmetic. This page explains what logarithms are and how to work with them using log tables and antilog tables. The relationship with log scales is included. Highlights are the recollections from people who worked with logarithms as a norm.
____
By the webmaster, based childhood experience and additional research
## Logarithms: what they are and how they work
Logarithms provide an easy means of multiplying and dividing using just simple addition and subtraction together with lookup tables.
Let me explain how they work. We all know that 10 x 100 = 1000.
Expressed in powers of ten this can be written as:
101 x 102 = 103
which could also be written more generally as
10x x 10y = 10x+y
These powers of 10 i.e. the x and y, are called logarithms, or more specifically base 10 logarithms or logarithms to the base ten. There are other bases, but there is no need to go into that here.
As an example, suppose you want to multiply 2 by 5. Yes, you know the answer anyway, but the explanation is easiest with a simple example.
The base 10 logarithm of 2 happens to be 0.301. Please just accept that for the moment. It means that 100.301 = 2
The base 10 logarithm of 5 is 0.699, which means that 100.699 = 5
So 2 x 5 = 10 could be written as
100.301 x 100.699 = 10 0.301 + 0.699 = 101
or more generally:
A x B = 10log(A) x 10log(B) = 10log(A)+log(B)
Similarly for division:
A / B = 10log(A) / 10log(B) = 10log(A)-log(B)
From this, you can see that of you can look up the logs of numbers, add or subtract them and then decode the result, the maths is very much easier for multiplying and dividing large numbers. This is where log tables and antilog tables come in - see below.
## John Napier and the first log and antilog tables
John Napier was a Scottish mathematician working back in the 17th century. He was the first person to recognise the huge advantage of logarithms. He undertook the enormous task of working out the logarithms and antilogarithms of numbers to the base 10 and he published them as a book in 1614.
## Log and antilog tables in schools
Log and antilog tables were introduced into schools in 1908 when a Mr Castle published them in his book. This was not only before electronic calculators but also well before the UK went decimal, which meant that many calculations were not as straightforward as they are today.
The books, sometimes called logarithmic tables or logarithm tables were provided in school maths classes and were essential for the maths examination known as 'Ordinary Level' or just 'O level*'.
Godfrey and Siddons Four Figure Tables, as used in schools in 1950s UK
I started using these tables in the 1950s at my grammar school. Those that my school provided at that time were by Godfrey and Siddons, and published by Cambridge University Press. They were somewhat thin booklets which were easy to carry around and must have been relatively cheap for schools to buy. If you are old enough, you will recognise the photo.
These Godfrey and Siddons books of tables were accurate to four figures and accordingly had the title of Four Figure Tables on the front. However, similar tables did exist to more than four figures although I never saw any of them at the time. There were other authors and publishers.
As our main use of these tables was for looking up logarithms and antilogarithms as a means of multiplying and dividing, they acquired the name of Log Tables or just Logs, even though they did contain other types of mathematical tables.
## How the log and antilog tables worked:
All that mattered for users and ordinary school children at the time was that two or more numbers could be multiplied together by adding their logs - obtained from our log tables - and then decoding the result from our antilog tables. Similarly a number could be divided by a second number by subtracting the log of the second number from the log of the first number and then decoding it from the antilog tables,
## Decimal points and powers of ten
Like slide rules, log tables do not show where to place a decimal point in an answer or if any zeros are necessary before or after the answer. To find this out, a rough calculation is necessary as explained on the slide rule page.
## How to use log tables and antilog tables
The following image shows the row and column structure of a typical page of a log table. The image is intentionally too small for the individual numbers to be legible, because the purpose here is only to show the 'shape' or structure of a page. This is important for understanding how the table was used.
Row and column structure of a page of logarithms in a book of Four Figure Tables.
### How to find the log of a number
In the left-hand column are numbers from 10, through to 99 (which actually requires more than one page). The logs of these numbers are in the adjacent column which has the heading 0. So for example if you want the log of, say, 34 (or 3.4 or 3400 because finding the decimal point is a separate operation), find 34 in the left hand column. Its log is next to it in the column headed 0.
If, though, you wanted the log of 347 (or 3.47 or 24700 etc), the log would be further to the right in the 7 column.
If you wanted the log of a four figure number, like 3476 you would go the wide column at the extreme right, find the number under the 6 heading and add that to the log of 347.
A ruler was essential to prevent wandering off one row and onto another, but the layout does make this more manageable by having a blank between every third row.
To multiply two or more numbers together, just find their logs and add them together to get the combined log. If this came to more than four figures, it was shortened to four figures, up one or down one in the fourth figure depending on whether the fifth figure was more or less than 5. Then find their antilogs - see below.
To divide one number by another, just subtract their logs and then antilog.
### How to find the antilog of a number
Antilogarithms enable the sum or subtraction of logs to be decoded back to the answer of the multiplication or division. Having found the log of the multiplication or division, that number was looked up in the book's antilog tables, using the procedure explained above. The position of any decimal point was established as explained on the slide rule page.
The result, or the answer, was correct to four significant figures, but to be sure, it was usually shortened to three significant figures.
## The pros and cons of using logarithms for multiplication and division
This use of logarithms may seem longwinded by today's standards where electronic calculators are widely available, but you need to realise that the additions and subtractions involved were not particularly arduous for us. Because there were no electronic calculators most people became adept at what was called 'mental arithmetic' and it was taught as a school subject to young children. My father, for example, who worked with figures all his life, could glance so rapidly down a column of complex numbers, that it looked as if he was simply checking that they were legible. Then he could immediately announce what they added up to.
So to us, before we knew anything about electronic calculators, logarithms were all advantages with no disadvantages. There was nothing that we knew of that could possibly make multiplication and division of large numbers any easier. We were of course taught how to multiply and divide large numbers using only pen and paper and what were known as 'long multiplication' and 'long division', but these methods were unwieldy and open to far more errors than looking up numbers in tables and simply adding, subtracting and decoding in antilogs.
## Log scales
On any normal numerical scale, the distance between integers (whole numbers) is the same. If anyone of my young generation had been interested in making a scale showing log 10, log20, log 30 etc, we would have found that the higher the number, the more scrunched up the scale would become. Each value is greater than the earlier one but by a progressively smaller amount. The result is called a log scale.
Log scales have uses where the breadth of the data is very large, so that a regular ruler or measure would have to be uncontrollably long. Slide rules work on the principle of log scales. In the following image, the top scale is uniform with the same space between the integers 1 - 5, but the scale just below is a log scale with the numbers scrunched up close together. There is also what is effectively an antilog scale, but there more explanation on the slide rule page.
Log scale (second scale down from the top) on a slide rule.
Log scales have uses in finance, astronomy and in any graphical display of large amounts of data where it helps to visualise trends. It must be remembered, though, that log scales give the logs of any number, not the number itself. This is not always realised by the general public which gives rise to misleading interpretations - in some cases intentionally so.
___
* The General Certificate of Education Ordinary Level exam (also known as the GCE O Level) was the precursor of GCSEs and introduced in Britain in the 1950s. | 2,128 | 9,672 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5625 | 5 | CC-MAIN-2024-22 | longest | en | 0.968768 |
https://www.gradesaver.com/textbooks/math/algebra/elementary-algebra/chapter-4-proportions-percents-and-solving-inequalities-4-2-percents-and-problem-solving-concept-quiz-4-2-page-155/4 | 1,680,097,854,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296948976.45/warc/CC-MAIN-20230329120545-20230329150545-00633.warc.gz | 860,152,484 | 12,253 | ## Elementary Algebra
Let's check to see whether this statement is true or false. Twelve is 30% of 40 means: 12 = 30% $\times$ 40 12 =0.3 $\times$ 40 12 = 12 Since both sides are equal, we know for sure that twelve is 30% of 40. | 75 | 229 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.125 | 4 | CC-MAIN-2023-14 | latest | en | 0.904646 |
http://www.rfwireless-world.com/Terminology/abcd-matrix-vs-s-matrix.html | 1,544,577,178,000,000,000 | text/html | crawl-data/CC-MAIN-2018-51/segments/1544376823710.44/warc/CC-MAIN-20181212000955-20181212022455-00488.warc.gz | 469,708,816 | 5,682 | # RF Wireless World
## Home of RF and Wireless Vendors and Resources
One Stop For Your RF and Wireless Need
## Convert abcd matrix to S-matrix | Convert S-matrix to abcd matrix
This page mentions basics of abcd matrix and S-matrix. It mentions formula to convert abcd matrix to S-matrix and to convert S-matrix to abcd matrix.
### What is S-matrix?
• S-matrix is composed of S-parameters or scattering parameters.
• It is known as scattering matrix.
• It describes electrical behavior of linear electrical networks when subjected to steady state stimuli with the help of electrical signals.
• They do not use open circuit and short circuit conditions.
• They use matched loads to characterize linear electrical network due to ease at higher frequencies compare to short/open circuit terminations. Following figure depicts 2-port network with s-parameters. A set of linear equations are written to describe network in terms of injected and transmitted waves.
➨Sij = bi/aj = [(Power measured at port-i)/(Power injected at port-j)]0.5
Where,
Sii = ratio of reflected power to injected power at port-i
Sij = ratio of power measured at port-j to power injected at port-i
• Electrical circuit or network is composed of inductors, capacitors or resistors in its basic form. Some of their parameters such as return loss, insertion loss, gain, VSWR, reflection coefficients are represented by S-parameters.
• S-parameters are similar to other parameters such as Z-parameters, Y-parameters, H-parameters, T-parameters, abcd-parameters etc.
• Following equations are used to derive various S-parameters such as S11, S12, S21 and S22.
Let us understand abcd-matrix before we go through conversion between S-matrix and abcd-matrix.
### What is abcd matrix?
• It describes network in terms of both voltage and current waves as shown below.
• It is also known as transmission matrix.
• It is suitable to cascade elements since it represents ports in terms of currents and voltages. The matrices are cascaded by multiplication operation.
• Coefficients are defined using superposition as follows.
• ABCD parameters are expressed as follows.
S-parameters are measured using VNA. Later they are converted to ABCD matrix. ABCD matrix can also be converted to S-matrix.
### Convert abcd matrix to S-matrix
Following equations are used to convert abcd matrix to S-matrix.
### Convert S-matrix to abcd matrix
Following equations are used to convert S-matrix to abcd matrix. | 556 | 2,468 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2018-51 | longest | en | 0.907806 |
https://math.stackexchange.com/questions/1428770/integral-large-int-01-ln3-left1xx2-rightdx | 1,722,706,290,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640372747.5/warc/CC-MAIN-20240803153056-20240803183056-00223.warc.gz | 322,739,235 | 41,364 | Integral ${\large\int}_0^1\ln^3\!\left(1+x+x^2\right)dx$
I'm interested in this integral: $$I=\int_0^1\ln^3\!\left(1+x+x^2\right)dx.\tag1$$ Can we prove that \begin{align}I&\stackrel{\color{gray}?}=\frac32\ln^33-9\ln^23+36\ln3+2\pi^2\ln3-\frac{4\pi^2}3+\left(8-\ln^23-4\ln3\right)\cdot\frac{\pi\sqrt3}2\\&-48-\frac{7\pi^3}{6\sqrt3}+(2-3\ln3)\cdot\psi^{(1)}\!\left(\tfrac13\right)+36\,{_4F_3}\!\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\,\tfrac34\right)\end{align}\tag2 of find a simpler closed form?
Also, can we prove that $${_4F_3}\!\left(\begin{array}c\tfrac12,\tfrac12,\tfrac12,\tfrac12\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\,\tfrac34\right)\stackrel{\color{gray}?}=\frac{71\pi^3}{1296\sqrt3}+\frac{5\pi}{48\sqrt3}\ln^23-\frac1{\sqrt3}\,\Im\operatorname{Li}_3\!\left[\frac{(-1)^{\small1/6}}{\sqrt3}\right]\tag3$$ or find a simpler expression for it?
• I should mention that I found these conjectured forms using PSLQ algorithm and a set of candidate terms taken from known closed forms of similar integrals and simple variations of those terms, and then verified numerically with a high precision. Commented Sep 10, 2015 at 0:52
• Have you tried the obvious ? Writing $x^2+x+1$ as $\dfrac{1-x^3}{1-x}$, and then using $\ln\dfrac ab=\ln a-\ln b,$ or writing $x^2+x+1=(x-z)(x-\bar z)$, where $z$ and $\bar z$ are the cube roots of unity. Commented Sep 10, 2015 at 2:54
• How did you find this conjectured closed-form for the hypergeometric series? Commented Sep 14, 2015 at 9:55
• It was a pure guess inspired by this answer. Coefficient were found using PSLQ algorithm. Commented Sep 14, 2015 at 16:50
A Recurrence Relation
I will use the notation $$\mathcal{A}_n=\int^1_0\ln^n(1+x+x^2)\ {\rm d}x\ \ \ , \ \ \ \mathcal{B}_n=\int^\frac{\pi}{3}_\frac{\pi}{6}\ln^n\left(\frac{3}{4\cos^2{x}}\right)\ {\rm d}x$$ Integrating by parts and applying the substitution $\displaystyle x+\frac{1}{2}\mapsto \frac{\sqrt{3}}{2}\tan{x}$, it is evident that $$\mathcal{A}_n=n\sqrt{3}\mathcal{B}_{n-1}-2n\mathcal{A}_{n-1}+\frac{3}{2}(\ln{3})^n$$ We may use this recurrence to compute $\mathcal{A}_n$ for small positive integer values of $n$.
Evaluation of $\mathcal{A}_1$
We immediately have $$\mathcal{A}_1=1\times\sqrt{3}\times\frac{\pi}{6}-2\times 1\times 1+\frac{3}{2}\ln{3}=\frac{\pi}{2\sqrt{3}}+\frac{3}{2}\ln{3}-2$$
Evaluation of $\mathcal{A}_2$
We first compute $\mathcal{B}_1$ by exploiting a Fourier series. \begin{align} \mathcal{B}_1 &=\frac{\pi}{6}\ln{3}-2\int^\frac{\pi}{3}_\frac{\pi}{6}\ln(2\cos{x})\ {\rm d}x\\ &=\frac{\pi}{6}\ln{3}+2\sum^\infty_{n=1}\frac{(-1)^n}{n}\int^\frac{\pi}{3}_\frac{\pi}{6}\cos(2nx)\ {\rm d}x\\ &=\frac{\pi}{6}\ln{3}+\sum^\infty_{n=1}\frac{(-1)^n}{n^2}\left(\sin\left(\frac{2n\pi}{3}\right)-\sin\left(\frac{n\pi}{3}\right)\right)\\ &=\frac{\pi}{6}\ln{3}-\frac{1}{12\sqrt{3}}\sum^\infty_{n=0}\left[\frac{1}{\left(n+\frac{1}{3}\right)^2}-\frac{1}{\left(n+\frac{2}{3}\right)^2}\right]\\ &=-\frac{1}{6\sqrt{3}}\psi_1\left(\frac{1}{3}\right)+\frac{\pi^2}{9\sqrt{3}}+\frac{\pi}{6}\ln{3} \end{align} Therefore, \begin{align} \mathcal{A}_2 &=2\sqrt{3}\left(-\frac{1}{6\sqrt{3}}\psi_1\left(\frac{1}{3}\right)+\frac{\pi^2}{9\sqrt{3}}+\frac{\pi}{6}\ln{3}\right)-4\left(\frac{\pi}{2\sqrt{3}}+\frac{3}{2}\ln{3}-2\right)+\frac{3}{2}\ln^2{3}\\ &=-\frac{1}{3}\psi_1\left(\frac{1}{3}\right)+\frac{2\pi^2}{9}+\frac{\pi}{\sqrt{3}}\ln{3}+\frac{3}{2}\ln^2{3}-\frac{2\pi}{\sqrt{3}}-6\ln{3}+8 \end{align}
Simplification of Some ${\rm Li}_2,\ {\rm Li}_3$ Terms
I will simplify the terms $$\color{red}{{\rm Li}_2(e^{-\pi i/3})},\ \color{blue}{{\rm Li}_2(1-e^{2\pi i/3})},\ \color{green}{\Im{\rm Li}_3(e^{\pi i/3})},\ \color{purple}{\Im{\rm Li}_3(e^{-\pi i/3})},\ \color{brown}{\Im{\rm Li}_3(e^{2\pi i/3})}$$ The identities (for $0<\theta<2\pi$), \begin{align} \sum^\infty_{n=1}\frac{\cos(n\theta)}{n^2}&=\frac{\theta^2}{4}-\frac{\pi\theta}{2}+\frac{\pi^2}{6}\\ \sum^\infty_{n=1}\frac{\sin(n\theta)}{n^3}&=\frac{\theta^3}{12}-\frac{\pi\theta^2}{4}+\frac{\pi^2\theta}{6}\\ \end{align} (which can be derived by considering $\Im\ln(1-e^{i\theta})$ and integrating), give us \begin{align} \Im{\rm Li}_3(e^{\pm\pi i/3}) =&\pm\frac{5\pi^3}{162}\\ \Im{\rm Li}_3(e^{2\pi i/3}) &=\frac{2\pi^3}{81}\\ {\rm Li}_2(e^{-\pi i/3}) &=\frac{\pi^2}{36}-i\sum^\infty_{n=1}\frac{\sin(n\pi/3)}{n^2}\\ &=\frac{\pi^2}{36}-\frac{i\sqrt{3}}{2}\sum^\infty_{n=0}\left[\frac{1}{(6n+1)^2}+\frac{1}{(6n+2)^2}-\frac{1}{(6n+4)^2}-\frac{1}{(6n+5)^2}\right]\\ &=\frac{\pi^2}{36}-\frac{i}{24\sqrt{3}}\left(\psi_1\left(\frac{1}{6}\right)+\psi_1\left(\frac{1}{3}\right)-\psi_1\left(\frac{2}{3}\right)-\psi_1\left(\frac{5}{6}\right)\right)\\ &=\frac{\pi^2}{36}-i\left(\frac{1}{2\sqrt{3}}\psi_1\left(\frac{1}{3}\right)-\frac{\pi^2}{3\sqrt{3}}\right) \end{align} Furthermore, the dilogarithm reflection formula states $${\rm Li}_2(z)+{\rm Li}_2(1-z)=\frac{\pi^2}{6}-\ln{z}\ln(1-z)$$ Hence \begin{align} {\rm Li}_2(1-e^{2\pi i/3}) &=\frac{\pi^2}{6}-\frac{2\pi i}{3}\left(\frac{\ln{3}}{2}-\frac{\pi i}{6}\right)-\left(-\frac{\pi^2}{18}+i\sum^\infty_{n=1}\frac{\sin(2n\pi/3)}{n^2}\right)\\ &=\frac{\pi^2}{6}-\frac{2\pi i}{3}\left(\frac{\ln{3}}{2}-\frac{\pi i}{6}\right)-\left(-\frac{\pi^2}{18}+\frac{i\sqrt{3}}{2}\sum^\infty_{n=0}\left[\frac{1}{(3n+1)^2}-\frac{1}{(3n+2)^2}\right]\right)\\ &=\frac{\pi^2}{9}-i\left(\frac{1}{3\sqrt{3}}\psi_1\left(\frac{1}{3}\right)-\frac{2\pi^2}{9\sqrt{3}}+\frac{\pi}{3}\ln{3}\right) \end{align}
Evaluation of $\mathcal{A}_3$
Similarly, we start with the evaluation of $\mathcal{B}_2$. \begin{align} \mathcal{B}_2 &=\int^\frac{\pi}{3}_\frac{\pi}{6}\ln^2{3}-4\ln{3}\ln(2\cos{x})+4x^2+4\operatorname{Re}\ln^2(1+e^{2ix})\ {\rm d}x\\ &=-\frac{\ln{3}}{3\sqrt{3}}\psi_1\left(\frac{1}{3}\right)+\frac{7\pi^3}{162}+\frac{2\pi^2}{9\sqrt{3}}\ln{3}+\frac{\pi}{6}\ln^2{3}+8\Re\sum^\infty_{n=1}\frac{(-1)^{n}H_{n-1}}{n}\int^\frac{\pi}{3}_\frac{\pi}{6}e^{2inx}\ {\rm d}x\\ &=-\frac{\ln{3}}{3\sqrt{3}}\psi_1\left(\frac{1}{3}\right)+\frac{7\pi^3}{162}+\frac{2\pi^2}{9\sqrt{3}}\ln{3}+\frac{\pi}{6}\ln^2{3}-4\sum^\infty_{n=1}\frac{1}{n^3}\left(\sin\left(\frac{2\pi n}{3}\right)-\sin\left(\frac{\pi n}{3}\right)\right)\\ &\ \ \ \ \ +4\Im\sum^\infty_{n=1}\frac{H_{n}}{n^2}\left(e^{2\pi in/3}-e^{\pi in/3}\right)\\ &=-\frac{\ln{3}}{3\sqrt{3}}\psi_1\left(\frac{1}{3}\right)+\frac{11\pi^3}{162}+\frac{2\pi^2}{9\sqrt{3}}\ln{3}+\frac{\pi}{6}\ln^2{3}\\ &\ \ \ \ \ +4\Im\left[{\rm Li}_3(z)-{\rm Li}_3(1-z)+{\rm Li}_2(1-z)\ln(1-z)+\frac{1}{2}\ln{z}\ln^2(1-z)+\zeta(3)\right]^{e^{2\pi i/3}}_{e^{\pi i/3}} \end{align} where I used the generating function of $\dfrac{H_n}{n^2}$. Using results derived in the previous section, \begin{align} &\ \ \ \ \Im\left[{\rm Li}_3(z)-{\rm Li}_3(1-z)+{\rm Li}_2(1-z)\ln(1-z)+\frac{1}{2}\ln{z}\ln^2(1-z)+\zeta(3)\right]^{e^{2\pi i/3}}_{e^{\pi i/3}}\\ &=\color{brown}{\frac{2\pi^3}{81}}-\color{green}{\frac{5\pi^3}{162}}+\color{purple}{\left(-\frac{5\pi^3}{162}\right)}-\Im{\rm Li}_3(1-e^{2\pi i/3})\\ &\ \ \ \ +\Im\color{blue}{\left(\frac{\pi^2}{9}-i\left(\frac{1}{3\sqrt{3}}\psi_1\left(\frac{1}{3}\right)-\frac{2\pi^2}{9\sqrt{3}}+\frac{\pi}{3}\ln{3}\right)\right)}\left(\frac{\ln{3}}{2}-\frac{\pi i}{6}\right)\\ &\ \ \ \ -\Im\color{red}{\left(\frac{\pi^2}{36}-i\left(\frac{1}{2\sqrt{3}}\psi_1\left(\frac{1}{3}\right)-\frac{\pi^2}{3\sqrt{3}}\right)\right)}\left(-\frac{\pi i}{3}\right)+\frac{\pi^3}{108}+\frac{\pi}{12}\ln^2{3}\\ &=-\Im{\rm Li}_3(1-e^{2\pi i/3})-\frac{\ln{3}}{6\sqrt{3}}\psi_1\left(\frac{1}{3}\right)-\frac{\pi^3}{27}+\frac{\pi^2}{9\sqrt{3}}\ln{3}-\frac{\pi}{12}\ln^2{3} \end{align} Therefore \begin{align} \mathcal{B}_2 &=-4\Im{\rm Li}_3(1-e^{2\pi i/3})-\frac{\ln{3}}{\sqrt{3}}\psi_1\left(\frac{1}{3}\right)-\frac{13\pi^3}{162}+\frac{2\pi^2}{3\sqrt{3}}\ln{3}-\frac{\pi}{6}\ln^2{3} \end{align} and finally, \begin{align} \mathcal{A}_3 &=3\sqrt{3}\left(-4\Im{\rm Li}_3(1-e^{2\pi i/3})-\frac{\ln{3}}{\sqrt{3}}\psi_1\left(\frac{1}{3}\right)-\frac{13\pi^3}{162}+\frac{2\pi^2}{3\sqrt{3}}\ln{3}-\frac{\pi}{6}\ln^2{3}\right)\\ &\ \ \ \ -6\left(-\frac{1}{3}\psi_1\left(\frac{1}{3}\right)+\frac{2\pi^2}{9}+\frac{\pi}{\sqrt{3}}\ln{3}+\frac{3}{2}\ln^2{3}-\frac{2\pi}{\sqrt{3}}-6\ln{3}+8\right)+\frac{3}{2}\ln^3{3}\\ &=\color{darkorange}{-12\sqrt{3}\Im{\rm Li}_3(1-e^{2\pi i/3})+(2-3\ln{3})\psi_1\left(\frac{1}{3}\right)+\frac{3}{2}\ln^3{3}-\left(\frac{\sqrt{3}\pi}{2}+9\right)\ln^2{3}}\\ &\ \ \ \ \color{darkorange}{+(2\pi^2-2\sqrt{3}\pi)\ln{3}-\left(\frac{13\sqrt{3}\pi^3}{54}+\frac{4\pi^2}{3}-4\sqrt{3}\pi-36\ln{3}+48\right)} \end{align}
I found the antiderivative: \begin{align}\int\ln^3\!\left(1+x+x^2\right)dx&=\xi\,\sqrt3\,\Big[\alpha^3-6\alpha^2+24\alpha-48\Big]\\&-\beta\,\sqrt3\,\Big[4\beta^2-3\alpha\ln3+6\alpha-24+6\ln3\Big]\phantom{\Huge|}\\&+6\,\sqrt3\,\Im\Big[(\alpha-2-2\beta\,i)\,\operatorname{Li}_2(\gamma)-2\,\operatorname{Li}_3(\gamma)\Big]\phantom{\Huge|}\end{align} where \begin{align}&\color{maroon}{\alpha=\ln\!\left(1+x+x^2\right)}\\&\color{orange}{\beta=\arctan(2\,\xi)}\phantom{\Huge|}\\&\color{green}{\gamma=\frac12-i\,\xi}\phantom{\Huge|}\\&\color{blue}{\xi=\frac{1+2x}{2\sqrt3}}\phantom{\Huge|}\end{align}
It is valid and continuous at least for $x\ge0$, so it is good for our purposes.
Unfortunately, I cannot demonstrate any systematic approach that leads to this result, I got it with a series of lucky guesses and applications of PSLQ algorithm to determine rational coefficients, and finally proved its correctness using differentiation.
It yields the conjectured result modulo several polylogarithm identities that I'm still trying to prove.
• if we use the fact that $\ln{(1+x+x^2)} = \ln{(x-a)}+\ln{(x-\bar{a})}$, then the resulting integrals have easily computable antiderivatives in terms of trilogarithm and logarithms but it's lengthy Commented Sep 11, 2015 at 10:40 | 4,461 | 9,795 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 12, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.640625 | 4 | CC-MAIN-2024-33 | latest | en | 0.725484 |
https://www.hackmath.net/en/math-problem/6811 | 1,611,231,459,000,000,000 | text/html | crawl-data/CC-MAIN-2021-04/segments/1610703524743.61/warc/CC-MAIN-20210121101406-20210121131406-00257.warc.gz | 822,757,596 | 12,746 | Octane value
I loaded 10L 95 octane gasoline and 5L 100 octane gasoline. What is the resulting octane value of the gasoline in the tank?
Correct result:
o = 96.7
Solution:
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http://mathhelpforum.com/pre-calculus/134994-complex-equation.html | 1,481,036,806,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698541907.82/warc/CC-MAIN-20161202170901-00011-ip-10-31-129-80.ec2.internal.warc.gz | 180,481,727 | 10,447 | 1. ## Complex equation
First of all you'll have to excuse any miss-spelling an incorrect expressions, I am a swede and am not to good regarding the math grammar in english..
Now for the problem at hand:
Solve the equation and answer in polar form r(cosv + isinv)
$z^4 = -i$
Thanx!
2. Hello DenOnde
Welcome to Math Help Forum!
Your English is excellent. It's certainly much better than my Swedish!
Originally Posted by DenOnde
First of all you'll have to excuse any miss-spelling an incorrect expressions, I am a swede and am not to good regarding the math grammar in english..
Now for the problem at hand:
Solve the equation and answer in polar form r(cosv + isinv)
$z^4 = -i$
Thanx!
If we write
$z = r(\cos\theta + i\sin\theta)$
then, using De Moivre's Theorem:
$z^4 = r^4(\cos\theta + i\sin\theta)^4$
$=r^4(\cos4\theta+i\sin4\theta)$
We now express $- i$ in polar form. On the Argand diagram it is represented by the point $(0,-1)$, so has modulus $1$ and argument $-\pi/2$. So:
$-i=1\Big(\cos(-\pi/2)+i\sin(-\pi/2)\Big)$
So:
$z^4=-i$
$\Rightarrow r^4(\cos4\theta+i\sin4\theta)=\cos(-\pi/2)+i\sin(-\pi/2)$
$\Rightarrow \left\{\begin{array}{l}
r^4\cos4\theta=-\cos(-\pi/2)\\
r^4\sin4\theta = -\sin(-\pi/2)
\end{array}\right .$
$\Rightarrow \left\{\begin{array}{l}
r=1\\
4\theta=-\pi/2+2n\pi, n \in \mathbb{Z}
\end{array}\right .$
So the values of $\theta$ are given by:
$4\theta = -\pi/2, -\pi/2\pm2\pi,-\pi/2\pm4\pi, ...$
and its principle values (those between $-\pi$ and $\pi$) are given by:
$\theta = -5\pi/8, -\pi/8, 3\pi/8, 7\pi/8$ | 559 | 1,554 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 19, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.21875 | 4 | CC-MAIN-2016-50 | longest | en | 0.698791 |
http://physics.divinechildhighschool.org/Home/old-projects/physics-of-sports/physics-of-sports-2009-2010/golf-hit | 1,685,613,513,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224647639.37/warc/CC-MAIN-20230601074606-20230601104606-00319.warc.gz | 37,201,850 | 9,397 | ### Golf Hit
Golf Ball Acceleration Graph Area of the position graph of the Golf Ball (displacement).Position Graph of the Golf ClubGolf Club Velocity Graph (1 dimension) Shown here is an adaptation of our coordinate plane to fit our purpose. The x-axis is now in relation to the golf ball. The velocity of the golf ball increased at the instant of the hit, as the force of the club was applied on it. After the hit, the velocity remained constant because the slope of the line of the position graph was constant. We measured this using the x-axis values, however it was necessary to change the direction of the x-axis to a diagonal line to track the golf ball's position and velocity. In the case of the golf club, velocity was constant throughout the swing and even still upon impact with the golf ball. The slope of this line was also constant and remained at an increase all the way until the follow through after the hit. And. because of this constant velocity, the acceleration is zero. The highest one dimensional velocity of the hit was 36.51 m/s for the golf ball. For the golf club, however, the highest velocity was -14.56 m/s, as the swing was in a negative, or downward, direction. In the case of the golf ball, velocity changed upon impact. Before the impact, the ball was still, with a velocity of zero. But, upon impact with the club, the velocity increased reaching its highest point at 18.12 m/s. The impact of the club caused the ball to move from its resting position and to have a constant velocity for the several meters it traveled. For the club, there was not a change in velocity during the swing. It remained constant even upon impact with the golf ball. This impact only caused a change for the golf ball. For the golf ball, acceleration was present since the velocity was not constant, but acceleration was not constant either. Our velocity graph is not a straight diagonal line. The acceleration for for ball is 35.681 m/s2. The area under the graph was close but not equal to the change in velocity. Approaching this using two dimensional motion shows a difference. The swing of the golf club is an example of circular motion. Therefore, to find a more accurate acceleration, we had to solve for the centripetal acceleration. In order to do this, we found the club's velocity at two points, 3.173 m/s, squared it, and put it over the radius, which we measured to be about 1.10 meters. This is the length of her wrists and the club. From there, using the formula v2/r, we calculated centripetal acceleration. It is 9.15 m/s/s. Golf Hit Video http://a14.video2.blip.tv/4730002338866/Falconphysics-GolfHit374.mov?bri=2.2&brs=341
ċ
Steve Dickie,
Oct 20, 2009, 6:16 AM | 637 | 2,726 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.78125 | 4 | CC-MAIN-2023-23 | latest | en | 0.96287 |
https://www.gradesaver.com/textbooks/math/calculus/calculus-with-applications-10th-edition/chapter-9-multiveriable-calculus-9-3-maxima-and-minima-9-3-exercises-page-488/7 | 1,618,483,097,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038084765.46/warc/CC-MAIN-20210415095505-20210415125505-00051.warc.gz | 907,013,456 | 12,102 | ## Calculus with Applications (10th Edition)
$f(x,y)=x^{2}+3xy+3y^{2}-6x+3y$ $f_x(x,y)=2x+3y-6=0$ $f_y(x,y)=3x+6y+3=0 \rightarrow x=-2y-1$ Use the substitution method to solve the system of equations: $f_y(x,y)=2x+3y-6=0 \rightarrow 2(-2y-1)+3y-6=0$ $y=-8$ The critical point is: $(15,-8)$ To identify any extrema: $f_{xx}(x,y)=2$ $f_{yy}(x,y)=6$ $f_{xy}(x,y)=3$ For $(15,-8)$: $f_{xx}(15,-8)=2$ $f_{yy}(15,-8)=6$ $f_{xy}(15,-8)=3$ $D=2\times6 - (3)^{2}=3\gt 0$ and $f_{xx}(15,-8)=2\gt0 \rightarrow$ There is a relative minimum at $(15,-8)$ | 256 | 541 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2021-17 | latest | en | 0.548888 |
http://slideplayer.com/slide/3741654/ | 1,521,375,746,000,000,000 | text/html | crawl-data/CC-MAIN-2018-13/segments/1521257645613.9/warc/CC-MAIN-20180318110736-20180318130736-00532.warc.gz | 264,491,360 | 20,019 | # Compost Column Does changing the amount of air make a difference in decomposition? Question :
## Presentation on theme: "Compost Column Does changing the amount of air make a difference in decomposition? Question :"— Presentation transcript:
Compost Column Does changing the amount of air make a difference in decomposition? Question :
Compost Column I think that more air will make the decomposers healthier and they will decompose the lettuce faster. Hypothesis :
Compost Column Dirt Bottles Water Air Lettuce Test Materials :
Compost Column Build the columns Put in dirt, water & lettuce Poke holes in one of the columns Put the columns in a warm place. Wait, watch and take notes Test Method :
Compost Column The lettuce leaves in the column without the holes seemed to be clearer than the leaves in the bottle with the holes. We also noticed that the bottle with holes was not touching the water in the bottom. First Observation : (1 week)
Compost Column In the column with holes we saw that there was dry lettuce on the top. We saw some lettuce stuff in the middle and no lettuce on the bottom. Second Observation : (2 weeks)
Compost Column In the column without holes we could not see any lettuce at all. We dumped out the two columns and looked closely for lettuce. We still couldn’t find any in the one without holes. Second Observation : (2 weeks)
Compost Column In the column with holes we found parts of lettuce down two thirds of the way from the top. The bottom of the one without holes was wetter than the other. Second Observation : (2 weeks)
Compost Column In the column with holes we think that the air made the dirt drier. Since the dry dirt stopped the lettuce from decomposing as fast then we think that the air made a difference. But, not in the way we thought it would be. Conclusion :
Compost Column We think that the decomposers like the wet dirt better than the dry dirt. In the dry dirt some of the decomposers got sick and some of them died and in the wet dirt they were happy and healthy so they ate more. Conclusion :
Compost Column Setting up the Column
Compost Column Day Three
Compost Column Week One
Compost Column Week Two - Dissection
Compost Column Dad, Jim Werner Science Assistant -Camera -Video Editing Jamison Werner Camera -Director Karen Werner Teacher -Facilities -Manager Special Thanks :
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https://www.hpmuseum.org/forum/archive/index.php?thread-15753.html | 1,725,718,136,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700650883.10/warc/CC-MAIN-20240907131200-20240907161200-00057.warc.gz | 795,403,548 | 5,997 | # HP Forums
Full Version: Eigenvector mystery
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I was just looking at the documentation for the Julia programming language today and I noticed something interesting.
In the second gray box on this page there is an example showing the calculation of eigenvectors and eigenvalues. I tried the same matrix on the HP 50 in approximate mode and the result for the eigenvectors was substantially different.
Wondering which set of eigenvectors was correct, I checked with Wolfram Alpha, which gave a third, completely different result! Going back to the 50g I tried the same matrix in exact mode, and obtained yet another result different from the previous 3. The eigenvalues in all 4 cases were the same however. The matrix is
Code:
[[ -4 -17 [ 2 2]]
Is this matrix so ill-conditioned that its eigenvectors can't be computed accurately? It doesn't seem so, its determinant is 26.
If I remember, there are an infinite number of eigenvectors, the most important are the eigenvalues.
Did you check if the eigenvalues are equal?
All eigenvectors are correct. The difference is only a scaling factor.
XCas> m := [[-4,-17], [2,2]]
XCas> eigenvalues(m) → (-1-5*i,-1+5*i)
Solving both vectors at the same time, λ = -1±5i
m * [x,y] = λ * [x,y]
(m - λ) * [x,y] = 0
m - λ = $$\left(\begin{array}{cc} -3∓5i & -17 \\ 2 & 3∓5i \end{array}\right)$$
If you pick 2nd row, you get results of Wolfram Alpha
2 x + (3∓5i) y = 0
2 x = (-3±5i) y → [x, y] = [-3±5i, 2] * t, where t is the parameter
If you pick 1st row, you get [x,y] = [-17, 3±5i] * t'
Confirming numpy's result (which is probably what Julia use)
>>> import numpy
>>> k, v = numpy.linalg.eig([[-4,-17],[2,2]])
>>> print k
[-1.+5.j -1.-5.j]
>>> print v
[[ 0.94590530+0.j 0.94590530+0.j ]
[-0.16692447-0.27820744j -0.16692447+0.27820744j]]
>>> print v * (-17/v[0][0])
[[-17.+0.j -17.+0.j]
[ 3.+5.j 3.-5.j]]
Update: numpy returns normalized eigenvectors (length = 1)
>>> sum(abs(v)**2) ** 0.5 # numpy eigenvectors are column vectors
array([ 1., 1.])
(10-18-2020 10:54 PM)pinkman Wrote: [ -> ]If I remember, there are an infinite number of eigenvectors, the most important are the eigenvalues.
Did you check if the eigenvalues are equal?
You should really reread this in a textbook For a nxn Matrix there are at most n different EValues (which are numbers). EVectors (which are vectors) to different EValues are linear independent, so you end up with at most n linear independent EValues either. Eigenvalues (w) and eigenvectors (v) of a matrix (M) belong inseparably together and define each other: Mv=wv.
(10-18-2020 10:54 PM)pinkman Wrote: [ -> ]Eigenvalues (w) and eigenvectors (v) of a matrix (M) belong inseparably together and define each other: Mv=wv.
True, but so is M(kv) = w(kv)
For eigenvalue w, all kv are valid eigenvectors for it, as long as k ≠ 0
(10-19-2020 07:13 PM)Albert Chan Wrote: [ -> ]
(10-18-2020 10:54 PM)pinkman Wrote: [ -> ]Eigenvalues (w) and eigenvectors (v) of a matrix (M) belong inseparably together and define each other: Mv=wv.
True, but so is M(kv) = w(kv)
For eigenvalue w, all kv are valid eigenvectors for it, as long as k ≠ 0
Also, eigenvalues with multiplicity greater than 1 correspond to subspaces with corresponding dimension, so in those cases the eigenvectors are not uniquely determined even if you normalize their lengths...
(10-19-2020 07:13 PM)Albert Chan Wrote: [ -> ]
(10-18-2020 10:54 PM)pinkman Wrote: [ -> ]Eigenvalues (w) and eigenvectors (v) of a matrix (M) belong inseparably together and define each other: Mv=wv.
True, but so is M(kv) = w(kv)
For eigenvalue w, all kv are valid eigenvectors for it, as long as k ≠ 0
Also true but not relevant It's all about to pick a set of linear indpendent Eigenvectors for a certain Eigenvalue, i. e. a base for the corresponding Eigenspace.
(10-19-2020 07:19 PM)Thomas Okken Wrote: [ -> ]
(10-19-2020 07:13 PM)Albert Chan Wrote: [ -> ]True, but so is M(kv) = w(kv)
For eigenvalue w, all kv are valid eigenvectors for it, as long as k ≠ 0
Also, eigenvalues with multiplicity greater than 1 correspond to subspaces with corresponding dimension, so in those cases the eigenvectors are not uniquely determined even if you normalize their lengths...
Uniqueness is not crucial. It's all about dimension of Eigenspaces (geometrical multiplicity) and to pick a finite set of linear independent vectors to form a base. That there are infinite vectors in a vectorspace is just part of the definition of a vectorspace (scalar multipication) and so it is irrelevant to mention it at all.
(10-19-2020 06:33 PM)JurgenRo Wrote: [ -> ]
(10-18-2020 10:54 PM)pinkman Wrote: [ -> ]If I remember, there are an infinite number of eigenvectors, the most important are the eigenvalues.
Did you check if the eigenvalues are equal?
You should really reread this in a textbook
Now I did!
What is still obscure for me is how the choice of eigenvalues are made by the algorithm.
(10-19-2020 08:57 PM)pinkman Wrote: [ -> ]
(10-19-2020 06:33 PM)JurgenRo Wrote: [ -> ]You should really reread this in a textbook
Now I did!
What is still obscure for me is how the choice of eigenvalues are made by the algorithm.
Eigenvalues are values λ such that Av = λv, so you find them by solving Av − λv = 0, or equivalently, (A − λI)v = 0, where I is the identity matrix, or equivalently, |A − λI| = 0. That last form is also known as the characteristic equation of A, and, being a polynomial of the same degree as the dimension of A, you can find its solutions, and thus the eigenvalues of A, using a polynomial root finder.
(10-19-2020 09:42 PM)Thomas Okken Wrote: [ -> ]Eigenvalues are values λ such that Av = λv, so you find them by solving Av − λv = 0, or equivalently, (A − λI)v = 0, where I is the identity matrix, or equivalently, |A − λI| = 0. That last form is also known as the characteristic equation of A, and, being a polynomial of the same degree as the dimension of A, you can find its solutions, and thus the eigenvalues of A, using a polynomial root finder.
Thomas is fully right.
You can find a program to compute the coefficients of the Characteristic Polynomial (its roots are the eigenvalues) for real or complex matrices in my article:
HP Article VA047 - Boldly Going - Eigenvalues and Friends
Also, there are many solved examples in the article, to make the matter crystal-clear.
V.
The thing to understand is that the eigenvectors {x} are the solution to the series of simultaneous equations [A]{x} = {B}, where [A] is a square matrix with a zero determinant, such that there are an infinite number of solutions. The eigenvalues are those values in the matrix [A] that result in the determinant being zero. The diagonal coefficients of [A] are modified by subtracting the unknown eigenvalue and solving for those values that result in a zero determinant. The eigenvectors can then be determined by substituting each eigenvalue, and solving for {x} in the homogeneous set of equations [A]{x} = {0}. As has been mentioned before, although there are an infinite number of solutions, they are all the same function shape, differing only by a scale or normalizing factor. This is why eigenvectors are commonly referred to as modeshapes when used to solve problems in vibration theory. They are just as important as eigenvalues when solving problems involving structural vibrations, as they permit the calculation of the distribution of inertial forces acting on a vibrating structure. In fact, the eigenvalues, the square root of which are referred to as natural frequencies, are merely an intermediate step in the computation of the eigenvectors.
(10-20-2020 12:08 AM)Valentin Albillo Wrote: [ -> ]
(10-19-2020 09:42 PM)Thomas Okken Wrote: [ -> ]Eigenvalues are values λ such that Av = λv, so you find them by solving Av − λv = 0, or equivalently, (A − λI)v = 0, where I is the identity matrix, or equivalently, |A − λI| = 0. That last form is also known as the characteristic equation of A, and, being a polynomial of the same degree as the dimension of A, you can find its solutions, and thus the eigenvalues of A, using a polynomial root finder.
Thomas is fully right.
You can find a program to compute the coefficients of the Characteristic Polynomial (its roots are the eigenvalues) for real or complex matrices in my article:
HP Article VA047 - Boldly Going - Eigenvalues and Friends
Also, there are many solved examples in the article, to make the matter crystal-clear.
V.
Well, a few things should still be mentioned here:
1. for the characteristic polynomial p(x)=|A-xI|, the term |A-xI| denotes the determinant of (A-xI).
2. the algebraic multiplicity of an eigenvalue (i.e. the multiplicity as zero of p) is always >= geometric multiplicity (i.e. the dimension of the eigenspace belonging to the eigenvalue).
3. I am not sure if "Dimension of a Matrix" is a valid definition. It's simply the number of rows (or columns).
(10-21-2020 06:57 PM)JurgenRo Wrote: [ -> ]
(10-20-2020 12:08 AM)Valentin Albillo Wrote: [ -> ]Thomas is fully right.
You can find a program to compute the coefficients of the Characteristic Polynomial (its roots are the eigenvalues) for real or complex matrices in my article:
HP Article VA047 - Boldly Going - Eigenvalues and Friends
Also, there are many solved examples in the article, to make the matter crystal-clear.
V.
Well, a few things should still be mentioned here (post by Thomas):
1. for the characteristic polynomial p(x)=|A-xI|, the term |A-xI| denotes the determinant of (A-xI).
2. the algebraic multiplicity of an eigenvalue (i.e. the multiplicity as zero of p) is always >= geometric multiplicity (i.e. the dimension of the eigenspace belonging to the eigenvalue).
3. I am not sure if "Dimension of a Matrix" is a valid definition. It's simply the number of rows (or columns).
Edit2: "Zero" should be "root"
Thanks everyone for the enlightening explanations, though most of the material is above my level of mathematical knowledge.
I was reading Valentin's excellent article "Boldly Going - Eigenvalues and Friends" and that led me to another somewhat off-topic question.
The HP 50g command PCAR (characteristic polynomial) returns a polynomial in symbolic form, e.g.
'X^5-19*X^4+79*X^3+146*X^2-1153*X+1222'
Is there a command to convert this symbolic polynomial form to an array as would be returned by Valentin's PCHAR program? Related commands such as PROOT and PCOEF require their arguments as an array.
One could write a program to do the conversion by string processing but it would be messy.
(10-24-2020 02:03 PM)John Keith Wrote: [ -> ]'X^5-19*X^4+79*X^3+146*X^2-1153*X+1222'
Is there a command to convert this symbolic polynomial form to an array
XCas has symy2poly, but I can't find one for HP50G
XCas> symb2poly((x+1)*(x+2)*(x+3)) → poly1[1, 6, 11, 6]
Code:
≪ 0 OVER DEGREE → D ≪ 3. DUP D + FOR k HORNER k ROLLD NEXT + + D 1 + →ARRY ≫ ≫
Save above to, say, PARRY
hp50g> '(X+1)*(X+2)*(X+3)'
hp50g> PARRY → [1 6 11 6]
hp50g> PROOT → [-1. -2. -3.]
hp50g> 'X^5-19*X^4+79*X^3+146*X^2-1153*X+1222'
hp50g> PARRY → [1 -19 79 146 -1153 1222]
Thanks Albert, that works great! Interestingly, executing PROOT on the array returns the same roots as EGVL but in different order.
Reference URL's
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https://www.physicsforums.com/threads/determining-projection-angle-with-little-information.778395/ | 1,508,368,719,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187823153.58/warc/CC-MAIN-20171018214541-20171018234541-00211.warc.gz | 958,301,920 | 20,239 | # Determining Projection Angle With Little Information
1. Oct 26, 2014
### Ali0086
1. The problem statement, all variables and given/known data
The speed of a projectile when it reaches its maximum height is one-half its speed when it is at half its maximum height. What is the initial projection angle of the projectile?
2. The attempt at a solution
First I tried to find the speed for the max height.
vf2 - vi2 = -2gΔd
vf2 = vi2- 2gΔd
0= vi2- 2gΔd <---
Can someone explain why this must be set to zero?
∴ vi=√(2gh)
After that I did the same for half the height
vf2 - vi2 = -2g(Δd/2)
vf2 = vi2 - g(Δd/2)
0= vi2 - g(Δd/2)
∴ vi2=√(gh)
After reaching this point I'm not really sure how the given information can help give me an angle. I know somehow I'm supposed to break it down into horizontal components but I just can't see it. Any help would be appreciated.
2. Oct 26, 2014
### haruspex
That would follow if vf=0, right? Will it be zero at maximum height? What will be zero at maximum height?
3. Oct 26, 2014
### phinds
Forgetting the math for the moment, think about this. You are assuming, as one always correctly does in such basic problems, that air resistance is irrelevant and that the trajectory is ballistic. You also know that it's taking off and landing from the same height. What does all this tell you about the horizontal component of the "speed" at the max height vs at the launch? What does THAT tell you about the "speed" at the launch?
And, I assume that you understand the relationship between speed and velocity, yes?
4. Oct 26, 2014
### Ali0086
oh, let me guess, the final velocity will be zero right? The reason being there's that moment where it's absolutely still in the air.
5. Oct 26, 2014
### haruspex
If fired straight up, yes. Is this projectile fired straight up?
6. Oct 26, 2014
### Ali0086
Wouldn't that mean the horizontal component would remain constant, more specifically, the velocity of the horizontal motion wouldn't change.
7. Oct 26, 2014
### Ali0086
No it's an object in projectile motion. But what I'm getting at is that if the object is going on a projectile curve, for example a cannonball shot from a cannon. Wouldn't there be a moment in that parabolic motion where the velocity of the vertical component go from positive to negative? Doesn't that imply that the max height has been reached when that value of velocity is zero?
8. Oct 26, 2014
### phinds
Yes and yes. Why would you expect otherwise. What would make it change?
9. Oct 26, 2014
### haruspex
Yes, but only the vertical component. So what does vi represent in your equation 0= vi2- 2gΔd?
The question refers to speeds. So what equations can you write down now?
10. Oct 26, 2014
### Ali0086
the equation 1/2*(final velocity - initial velocity)
since 0= vi2- 2gΔd is the initial velocity
then isolate for the final velocity?
11. Oct 26, 2014
### haruspex
That's not an equation.
No, vi is not "the initial velocity" in that equation in this context. Let's step back a moment. What is the 0 in that equation representing? Be exact.
12. Oct 26, 2014
### Ali0086
well, I know vf2 - vi2 = -2gΔd represents the motion from height zero to the max height.
0= vi2- 2gΔd, so I believe the 0 here would be representing the fact that the vi and 2gΔd are equal right?
13. Oct 26, 2014
### haruspex
No, the equation represents that fact. The 0 is the value of a variable. What variable? Be as precise as you can.
14. Oct 26, 2014
### Ali0086
alright, first off thanks for not just spoon feeding me answers. I haven't thought through physics like this in a while.
So being as literal as I can. 0= vi2- 2gΔd, looking at this equation, the first thing I can think of what that zero represents is the value of vf. So i assume it follows that vi2- 2gΔd point at which vf =0 .If that isn't right, thanks for trying. As of right now it's late where I live and I have classes tomorrow. I'll be back tomorrow to try and work through this. Thanks a lot though.
15. Oct 26, 2014
### haruspex
I'm glad you're taking my approach in a good spirit!
But I don't think we're going to get there at this rate. I've been trying to get you to echo back to me my statement that vf in this equation represents the vertical component of the velocity at maximum height. On that understanding, what does vi represent in that equation?
16. Oct 27, 2014
### Ali0086
That would represent the horizontal component of the velocity at max height
17. Oct 27, 2014
### PeroK
I think a diagram of the motion would help a lot. This is a hard problem in my opinion and I would try to think out what's happening (without the equations of motion to begin with). So, I would draw a diagram showing:
The projectile as it leaves the ground; the projectile at its maximum height; the projectile when it has reached half its maximum height. I'd show the vertical and horizontal components of velocity at each of these points.
The next step is then to use the equations of motion to work out the velocity and/or speed at each point.
Also, as you have three points of interest here, it might be better to use $v_0, \ v_1\ \ v_2$ rather than $v_i, \ v_f$. In fact, keeping track of notation is a bit of a challenge in itself with this problem.
Last edited: Oct 27, 2014
18. Oct 27, 2014
### haruspex
No, no! In the vi, vf notation, i stands for initial and f for final. As PeroK says, if there are multiple initial and final speeds to be considered (different directions, different phases of the action) it is a very good idea to use subscripts or a different letter to discriminate those. In this case, how about v0, v1, v2 for the vertical speeds (v2 being 0) and u (which does not change) for the horizontal speed?
In a given SUVAT equation, all the distances and speeds are in the same direction, so you apply the equations separately in (usually) the vertical and horizontal directions.
So please write down some equations relating to the vertical movement here. | 1,570 | 5,981 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2017-43 | longest | en | 0.939321 |
https://community.deeplearning.ai/t/understanding-exponentially-weighted-averages-week-2/444401 | 1,716,116,217,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971057786.92/warc/CC-MAIN-20240519101339-20240519131339-00503.warc.gz | 155,883,566 | 5,472 | # Understanding exponentially weighted averages, week 2
Hello mentors,
I am very confused about this topic understanding exponentially weighted averages and i am not able to understand weighted decay and also this formula (1 - epsilon) ^1/epsilon.
Hello @Shantanu7 ,
Thanks a lot for asking this question. I am a mentor and I will try to explain exponentially weighted averages.
Exponentially Weighted Averages (EWA) is a technique used to calculate the moving average of a time series, where more recent data points are given higher weights, and older data points are given lower weights. The weights decline exponentially as the data points get older, hence the name “exponentially weighted”. This method is commonly used as a smoothing technique in time series analysis and is also applied in various optimization algorithms in deep learning, such as Gradient Descent with Momentum, RMSprop, and Adam.
The EWA is calculated using the following formula:
V_t = \beta * (V_{t-1}) + (1-\beta) * \text{NewSample}
Here, V_t represents the weighted average at time t, \beta is a parameter that determines the weight given to previous values, and \text{NewSample} is the new data point at time t . The parameter \beta is usually between 0 and 1, and it determines how important the current observation is in the calculation of the EWA. The higher the value of \beta, the more closely the EWA tracks the original time series.
The EWA can be thought of as an exponential decay, where the current value is multiplied by (1-\beta), and the previous values are multiplied by exponentially decaying factors of \beta . This ensures that more recent data points have a higher impact on the moving average, while older data points have a lower impact.
In summary, Exponentially Weighted Averages is a method for calculating the moving average of a time series, where recent data points are given higher weights, and older data points are given lower weights that decay exponentially. This technique is useful for smoothing time series data and is applied in various optimization algorithms in deep learning.
I hope my reply explained exponentially weighted averages clearly. Please feel free to ask a followup question if you still have uncertainties about exponentially weighted averages.
Regards,
Can Koz
1 Like
Hello sir,
Thank you for your help, I know all of these things prof. andrew ng explain all of these things.
I am little bit confused into this formula (1 - epsilon) ^1/epsilon.
This is awesome. Thank you. | 530 | 2,520 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2024-22 | latest | en | 0.909775 |
https://www.statistics-lab.com/%E7%BB%9F%E8%AE%A1%E4%BB%A3%E5%86%99%E5%9B%9E%E5%BD%92%E5%88%86%E6%9E%90%E4%BD%9C%E4%B8%9A%E4%BB%A3%E5%86%99regression-analysis%E4%BB%A3%E8%80%83-3/ | 1,709,446,425,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947476205.65/warc/CC-MAIN-20240303043351-20240303073351-00768.warc.gz | 968,140,193 | 41,064 | ### 统计代写|回归分析作业代写Regression Analysis代考|Spatial autocorrelation
statistics-lab™ 为您的留学生涯保驾护航 在代写回归分析Regression Analysis方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写回归分析Regression Analysis代写方面经验极为丰富，各种代写回归分析Regression Analysis相关的作业也就用不着说。
• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础
## 统计代写|回归分析作业代写Regression Analysis代考|Defining SA
The presence of nonzero correlation results in one $\mathrm{RV}$, $\mathrm{Y}$, in a pair being dependent on the other $\mathrm{RV}$, X. Its global trend depicting a positive relationship is for larger values of $X$ and $Y$ to tend to coincide, for intermediate values of $X$ and $Y$ to tend to coincide, and for smaller values of $X$ and $Y$ to tend to coincide; values of $\mathrm{X}$ are directly proportional to their corresponding values of $Y$. For an indirect (i.e., negative or inverse) relationship, larger values of $X$ tend to coincide with smaller values of $Y$, intermediate values of $X$ and $Y$ tend to coincide, and smaller values of $X$ tend to coincide with larger values of $Y$; values of $X$ are inversely proportional to their corresponding values of $Y$. A random relationship has values of $X$ and $Y$ haphazardly coinciding according to their relative magnitudes.
Autocorrelation transfers this notion of relationships from two RVs to a single RV; the prefix auto means self. Accordingly, $n$ observations have $\mathrm{n}(\mathrm{n}-1)$ possible pairings, one between each observation and the $(n-1)$
remaining observations; each observation always has a correlation of one with itself, and hence these n self-pairings are of little or no interest in terms of correlation. One relevant question asks whether or not an ordering exists that differentiates between two subsets of these $n(n-1)$ pairings such that the ordered subset contains directly correlated observations, whereas the unordered subset contains uncorrelated observations. For data linked to a map, its spatial ordering of attribute values virtually always yields a collection of correlated observations. Because the ordering involved is spatial, the
SA may be defined generically as the arrangement of attribute values on a map for some RV Y such that a map pattern becomes conspicuous by visual inspection. More specifically, positive SA (PSA)-overwhelmingly the most commonly observed type of SA-may be defined as the tendency for similar $Y$ values to cluster on a map. In other words, larger values of $Y$ tend to be surrounded by larger values of $Y$, intermediate values of $Y$ tend to be surrounded by intermediate values of $Y$, and smaller values of $Y$ tend to be surrounded by smaller values of Y. In contrast, NSA-a rarely observed type of SA-may be defined as the tendency for dissimilar $Y$ values to cluster on a map. In other words, larger values of $Y$ tend to be surrounded by smaller values of $Y$, intermediate values of $Y$ tend to be surrounded by intermediate values of $Y$, and smaller values of Y tend to be surrounded by larger values of Y. The absence of $\mathrm{SA}$ indicates a lack of map pattern and a haphazard mixture of attribute values across a map.
## 统计代写|回归分析作业代写Regression Analysis代考|A mathematical formularization of the first law
The preceding definition of $\mathrm{SA}$ indicates that this concept exists because orderliness, (map) pattern, and systematic concentration, rather than randomness, epitomize real-world geospatial phenomena. Tobler’s $(1969$, P. 7) First Law of Geography captures this notion: “everything is related to everything else, but near things are more related than distant things.” In 2004 the Annals of the American Association of Geographers published commentaries by six prominent geographers (Sui, Barnes, Miller, Phillips, Smith, and Goodchild), together with a reply by Tobler (vol. 94: pp. $269-310$ ) about this notion. Subsequent quantitative SA measurements (e.g., see Section 1.1.3) are mathematical abstractions of this empirical rule.
## 统计代写|回归分析作业代写Regression Analysis代考|Quantifying spatial relationships
A spatial weights matrix (SWM) is an n-by-n nonnegative (i.e., all of its entries are zero or positive) matrix, say C, describing the geographic relationship structure latent in a georeferenced dataset containing n observations (areal units or point locations in the case of georeferenced data), and has $\mathrm{n}(\mathrm{n}-1) / 2$ potential pairwise, symmetric relationship designations; without invoking symmetry, it has $n(n-1)$ potential relationship designations. Classical statistics assumes that these pairwise relationship designations do not exist (i.e., observation independence). Time-series analyses assume that $(n-1)$ of these pairwise relationship designations are nonzero and asymmetric (dependence is one-directional in time), with perhaps several additional relationship designations to capture seasonality effects. Spatial data mostly assume that between $n-1$ and $3(n-2)$ of these pairwise relationship designations are nonzero and symmetric, with asymmetric relationships usually specified from symmetric ones. The relationship definition rule (often called the neighbor or adjacency rule) is that correlation between attribute values exists for areal unit polygons sharing a common nonzero length boundary (i.e., the rook definition, using a chess move analogy). One extension of this definition is to nonzero length (i.e., point contacts) shared boundaries (i.e., the queen definition, using a chess move analogy). This latter extension tends to increase the number of designated pairwise correlations for administrative polygon surface partitionings by roughly $10 \%$; its asymptotic upper bound is a doubling of pairwise relationships (i.e., the regular square lattice case) for this near-planar situation, which still constitutes a very small percentage of the $n(n-1) / 2$ possible relationshipr. A third extension is to $k>1$ nearest neighbors, which fails to guarantee a connected dual graph structure and for which $\mathrm{k}$ is sufficiently small that it still constitutes a very small percentage of the $n(n-1)$ possible relationships. In all of these specifications of matrix $C$, if areal units $i$ and $j$ are designated polygons/locations with correlated attribute values, then $c_{i j}=1$; otherwise, $c_{i j}=0$. Frequently, matrix $C$ is converted to its often asymmetric row-standardized counterpart, matrix $W$, for which $w_{i j}=c_{i j} / \sum_{j=1}^{n} c_{i j} ; \sum_{j=1}^{n} w_{i j}=1$.
Yet another specification involves inverse distance (i.e., power or negative exponential) between polygon centroids or other points of privilege (e.g., administrative centers, such as capital cities or county seats) within areal unit polygons; these interpoint distances almost always are standardized (i.e., converted to matrix W), perhaps with a carefully chosen power or exponent parameter that essentially equates them to their shared common
boundary topological structure counterpart. Tiefelsdorf, Griffith, and Boots (1999) discuss other schemes defining a SWM that lies between matrices $\mathbf{C}$ and W.
These nearest neighbor and distance-based specifications allow spatial researchers to posit geographic relationship structures for nonpolygon point observations. By generating Thiessen polygon surface partitionings, these researchers also can posit geographic relationships based upon common boundary rules. Eigenvalues of matrices $C$ and $W$, a topic treated in a number of ensuing sections, furnish a quantitative gauge for comparing competing SWMs.
The purpose of a SWM is to define the set of directly correlated observations within a $R V$, enabling the quantification of $S A$ for a georeferenced attribute. It captures the geometric arrangement of attribute values on a map, often in topological terms.
## 统计代写|回归分析作业代写Regression Analysis代考|Defining SA
SA 可以一般地定义为一些 RV Y 的地图上的属性值的排列，使得地图图案通过视觉检查变得明显。更具体地说，阳性 SA（PSA）——绝大多数是最常见的 SA 类型——可以定义为相似的趋势是在地图上聚类的值。换句话说，较大的值是往往被较大的值包围是, 的中间值是往往被中间值包围是, 和较小的值是倾向于被较小的 Y 值包围。相比之下，NSA（一种很少观察到的 SA 类型）可以定义为不相似的趋势是在地图上聚类的值。换句话说，较大的值是往往被较小的值包围是, 的中间值是往往被中间值包围是, 较小的 Y 值往往被较大的 Y 值包围。小号一种表示缺乏地图模式和地图上属性值的随意混合。
## 统计代写|回归分析作业代写Regression Analysis代考|Quantifying spatial relationships
SWM 的目的是定义一组直接相关的观测值R在, 使量化小号一种对于地理参考属性。它通常以拓扑术语捕获地图上属性值的几何排列。
## 广义线性模型代考
statistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。
## MATLAB代写
MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。 | 2,755 | 9,369 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.796875 | 4 | CC-MAIN-2024-10 | latest | en | 0.705028 |
https://www.teacherspayteachers.com/Product/Apple-Themed-Math-and-Literacy-Activities-for-Kindergarten-390641 | 1,490,509,358,000,000,000 | text/html | crawl-data/CC-MAIN-2017-13/segments/1490218189127.26/warc/CC-MAIN-20170322212949-00145-ip-10-233-31-227.ec2.internal.warc.gz | 977,499,055 | 25,706 | Total:
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# Apple Themed Math and Literacy Activities for Kindergarten
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### PRODUCT DESCRIPTION
These fun apple themed centers are great for your beginning of the year apple theme. Each activity is aligned to the kindergarten Common Core Standards and includes full color and black and white versions to conserve printer ink.
Be sure to check out the preview to see what is included!
What is included:
Apple Ten Frame, Dot Card and Number match - Students match their number cards to the ten frame and dot card that show the number.
Apple Making Sets - Students arrange their apple number cards and then put the correct number of "seeds" (I used black beans) on each card.
Apple Dice Roll - Includes 2 different sheets for easy differentiation,one for 1 dice and one for 2 dice. This game helps students with subitizing and if they use two dice beginning addition and counting sets.
Apple Color by Sight Word Sheet - Includes simple beginning of the year sight words - I am willing to customize this sheet with your sight words if you would like.
Apple Sight Word Book - Students fill in the sight word "my" or "the" on each page of the book. They then practice reading the book and color the pictures.
Rotten Apple Card Game - This game helps students practice identifying upper and lowercase letters and can also be used to practice identifying letter sounds.
Common Core Standards:
Math:
CCSS.Math.Content.K.CC.B.4 Understand the relationship between numbers and quantities; connect counting to cardinality.
CCSS.Math.Content.K.CC.B.4a When counting objects, say the number names in the standard order, pairing each object with one and only one number name and each number name with one and only one object.
CCSS.Math.Content.K.CC.B.4b Understand that the last number name said tells the number of objects counted. The number of objects is the same regardless of their arrangement or the order in which they were counted.
CCSS.Math.Content.K.CC.B.4c Understand that each successive number name refers to a quantity that is one larger.
CCSS.Math.Content.K.CC.B.5 Count to answer “how many?” questions about as many as 20 things arranged in a line, a rectangular array, or a circle, or as many as 10 things in a scattered configuration; given a number from 1–20, count out that many objects.
ELA Standards:
CCSS.ELA-Literacy.RF.K.3c Read common high-frequency words by sight (e.g., the, of, to, you, she, my, is, are, do, does).
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\$4.00 | 671 | 2,786 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2017-13 | longest | en | 0.888664 |
https://oneclass.com/class-notes/us/lsu/phys/phys-2001/2040571-phys-2001-lecture-28.en.html | 1,656,528,368,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656103642979.38/warc/CC-MAIN-20220629180939-20220629210939-00799.warc.gz | 481,080,793 | 643,558 | # PHYS 2001 Lecture Notes - Lecture 28: Heat Capacity, The Technique, Italian National Olympic Committee
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12.7 (cont.) Calorimetry
Calorimetry is a technique that one can use to determine the specific heat of an
unknown substance.
The technique is based on the conservation of energy.
The unknown material with a mass m
o
and
o
temperature T
ois placed into a calorimeter.
The heat (internal energy) lost by the unknown
material must go into heating up the water and the
calorimeter cup.
The calorimeter cup is well insulated to reduce the
heat lost to the environment.
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=
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container + Heat gained by the water
w
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Δ
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Δ
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Q
Heat lost by
unknown
material
Heat gained
by container
Heat gained
by water
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T
mc
T
mc
T
mc
Δ
+
Δ
Δ
From this equation, we can determine co, the heat capacity of the unknown sample.
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## Document Summary
Calorimetry is a technique that one can use to determine the specific heat of an unknown substance. The technique is based on the conservation of energy. The unknown material with a mass m temperature t: and, is placed into a calorimeter. o. The heat (internal energy) lost by the unknown material must go into heating up the water and the calorimeter cup. The calorimeter cup is well insulated to reduce the heat lost to the environment. l b. Heat lost by unknown material = heat gained by container + heat gained by the water i l h i d b k. From this equation, we can determine c o, the heat capacity of the unknown sample. Example: a precious-stone dealer wishes to find the specific heat capacity of a 0. 030-kg gemstone. When thermal equilibrium is established, the temperature is 28. 5 oc. Heat lost by the gemstone = heat gained by the vessel + heat gained by the water.
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Booster Class | 624 | 2,324 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.5625 | 4 | CC-MAIN-2022-27 | latest | en | 0.892431 |
https://jp.maplesoft.com/support/help/addons/view.aspx?path=RandomTools%2Fflavor%2Fidentical | 1,716,962,840,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971059206.29/warc/CC-MAIN-20240529041915-20240529071915-00848.warc.gz | 276,925,526 | 22,109 | identical - Maple Help
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# Online Help
###### All Products Maple MapleSim
RandomTools Flavor: identical
describe an object identically
Calling Sequence identical(expr)
Parameters
expr - Maple expression
Description
• The flavor identical describes the object expr itself.
This flavor can be used as an argument to RandomTools[Generate] or as part of a structured flavor.
Examples
> $\mathrm{with}\left(\mathrm{RandomTools}\right):$
> $\mathrm{Generate}\left(\mathrm{posint}\left(\mathrm{range}=10\right)+\mathrm{posint}\left(\mathrm{range}=10\right)'\mathrm{sin}'\left(x\right)\right)$
> $\mathrm{Generate}\left(\mathrm{posint}\left(\mathrm{range}=10\right)+\mathrm{posint}\left(\mathrm{range}=10\right)'\mathrm{sin}'\left(\mathrm{identical}\left(x\right)\right)\right)$
${6}{+}{2}{}{\mathrm{sin}}{}\left({x}\right)$ (1)
> $\mathrm{Matrix}\left(3,3,\mathrm{Generate}\left(\mathrm{posint}\left(\mathrm{range}=7\right)\mathrm{identical}\left(x\right)+\mathrm{posint}\left(\mathrm{range}=7\right),\mathrm{makeproc}=\mathrm{true}\right)\right)$
$\left[\begin{array}{ccc}{3}{}{x}{+}{4}& {4}{}{x}{+}{6}& {5}{}{x}{+}{3}\\ {x}{+}{5}& {2}{}{x}{+}{3}& {2}{}{x}{+}{2}\\ {4}{}{x}{+}{3}& {3}{}{x}{+}{1}& {2}{}{x}{+}{5}\end{array}\right]$ (2)
See Also | 445 | 1,361 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 6, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2024-22 | latest | en | 0.247503 |
https://math.stackexchange.com/questions/1912548/prove-that-r-is-an-equivalence-relation/1912551 | 1,624,006,269,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487635920.39/warc/CC-MAIN-20210618073932-20210618103932-00347.warc.gz | 357,995,188 | 38,650 | # prove that $R$ is an equivalence relation
$$\forall a,b \in \mathbb{Q} \quad aRb \Leftrightarrow \quad \exists k \in \mathbb{Z}: \quad b=2^ka$$ 1) Reflexivity:
$\forall a \in \mathbb{Q}\quad aRa \Leftrightarrow \quad \exists k \in \mathbb{Z}: \quad a=2^ka$
choosing $k=0 \quad \Rightarrow a=2^0a=a \Rightarrow aRa \Rightarrow R \text{ is reflexive}$
2) Symmetry
3) Transitivity:
$\forall a,b,c \in \mathbb{Q}:$ $$aRb \Leftrightarrow \quad \exists k \in \mathbb{Z}: \quad b=2^ka$$ $$bRc \Leftrightarrow \quad \exists h \in \mathbb{Z}: \quad c=2^hb$$ Then $$aRc \Leftrightarrow \quad \exists p \in \mathbb{Z}: \quad c=2^pa$$
so $aRb,bRc \Rightarrow c=2^hb=2^h2^ka=2^{h+k}a$
choosing $p=k+h \Rightarrow c=2^pa \Rightarrow aRc \Rightarrow \text{ R is transitive}$
Can anyone confirm that 1) and 3) are correct? I tried to prove 2) but is ended like transitive proof and I think that is entirely wrong, I have no idea how to succeed, can anyone provide some hints/proof/solution?. thanks in advance
• (1) and (3) look correct to me. Also, for the symmetric proof, note that $aRb \to b = 2^k a \to a = 2^{-k}b$, also note that if $k$ is an integer, then $-k$ is also an integer, and therefore $aRb \to bRa$ – Rob Bland Sep 2 '16 at 21:30
• In (3) you don't necessarily have the same $\;k\in\Bbb Z\;$ for both cases. Yet afterwards you uszxe $\;h,k\;$ so it is fine. – DonAntonio Sep 2 '16 at 21:32
• @DonAntonio I'm guessing this was a typo on the part of the OP. – 211792 Sep 2 '16 at 21:35
• yes was a typo error, just fixed. thanks to everyone! – Alfonse Sep 2 '16 at 21:59
• @Alfonse, I think you will also want to change $c=2^k a$ to $c=2^p a$ ? – user326210 Sep 2 '16 at 22:19
For symmetry, note that if $a R b$, then there is some $k$ for which $b = 2^k a$. But then $a = 2^{-k} b$; hence $b R a$. (Indeed, there exists an $\ell \equiv -k$ for which $a = 2^\ell b$.)
Your proofs for parts (1) and (3) are correct. For symmetry, suppose $aRb$, so that $$b = 2^ka$$ for some $k\in\mathbf{Z}$. Can you think of an integer $l$ so that $$a = 2^lb?$$ (Hint: Remember that negative integers are integers too!) Once you have such an integer, you can conclude that $bRa$, meaning that $R$ is symmetric. | 806 | 2,207 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 3.921875 | 4 | CC-MAIN-2021-25 | latest | en | 0.754384 |
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# Derivatives
### Set of functions defined on [0,1] that have a continuous derivative there ( one-sided derivatives at the endpoints).
A). Let M be the set of functions defined on [0,1] that have a continuous derivative there ( one-sided derivatives at the endpoints). Let p(x,y) = max_[0,1]|x'(t) - y'(t)|. 1).Show that ( M,p) fails to be a metric space. 2). Let p(x,y) = |x(0) - y(0)| + max_[0,1]|x'(t) - y'(t)|. Is (M,p) now a metric space? Please
### Derivatives
For this equation E= (-C/r) + D(-r/P) (Where c, D, and P are constants) Do the following procedure: 1. Differentiate E with respect to r and set the resulting expression equal to zero. 2. Solve for r0 in terms of C, D and P. Here is where I am at in the problem: I have obtained a derivative (and I'm look
### Derivative
Find the first derivative of problem in attached file. It is only one problem. Find the 1st derivative of f(x) = (e^x + e^-x)/x
### Derivative problems
Prob. 2. The area of a rectangle (x,y) is the product xy. The perimeter of a rectangle P is 2x+2y. For a given P, find x and y that gives the largest area of a rectangle (x,y) for given perimeter P. Hint: Maximize A(x) = xy, where y = (P-2x)/2. Prob. 3. Find the 1st derivative of f(x) = [(3 - x^(2/3))][(x^(2/3) + 2)^(1/2
### Differentiating ln x
Question 1 says to differentiate using the quotient rule f(x) = 1 + 2x/1-2x where x < 1/2. My answer is -8xsquared/1 - 2x squared.(at x = 1/4) Question 2 says rewrite the expression of f(x) = ln (1 + 2x/1 - 2x) (-1/2 <x<1/2) by applying a rule of logarithm and then differentiate. So rewriting f (x) = ln (1 + 2x) - (1 - 2x) =
### Non linear PDE.
I cannot use mathematical symbols. Thus, I will let * denote a partial derivative. For example, u*x means the partial derivative of u with respect to x. Moreover, I will further simplify things by letting p=u*x and q=u*y. Also, ^ denotes a power (for example, x^2 means x squared) and / denotes division. This is the problem: T
### Quotient and composite rule problem.
Please see attached problem using quotient and composite rule.
### Graph the antiderivative and the derivative of f(x).
Can you help to graph and find the antiderivative and the derivative. (See attached file for full problem description)
### Derivative of Integrals
Please see the attached file for the fully formatted problems.
### Integration : Finding a Function from a Double Derivative
Find a function f(x) such that f′′(x) = x^2 + cos x and f(0) = 1 and f′(0) = 0. --- Please see the attached file for the fully formatted problems.
### Applications of Derivatives : Velocity of a Particle
Let: v(t) = { 2t 0< t < 5 {10 5< t < 10 be the velocity of a particle given in meters per second. Find the distance traveled by the particle from t = 0 to t = 10 seconds. --- Please see the attached file for the fully formatted problems.
### First and Second Derivatives : Using Implicit Differentiation
X^2y^2-2x=3 I'm trying to verify my answer for the first derivative, and see if I got the second one right as well. For the first derivative I got (1-xy^2)/(x^2y) I think I'm having a problem with the 2nd derivative because I got x^2-2x^3y^2+x^3y^4 It doesn't look right to me,
### Derivatives : Maximizing Functions and Finding a Vertex
1. Express the function in the form f(x) = a(x-h)2 + K and indicate the vertex. a. f(x) = -x^2 + 13x - 8 b. f(x) = 4x^2 - 8x + 1 2. An object is thrown upward from the top of a 160-foot (Ho) building with an initial velocity (Vo) of 48 feet per second. How long after the object is projected upward will it strike the ground?
### Derivatives and Limits (5 Problems)
Please see the attached file for the completely formatted problems.
### Derivatives of Trigonometric functions : Horizontal Tangent Line
Let f(x) = e sin x - cos x How are all the values of x E [-2pi, 2pi] found, such that a tangent line to the curve is horizontal and what are they? Please see the attached file for the fully formatted problems.
### Rates of Change : Derivatives
A tank holding 1000 gallons is being drained. The volume in the tank is given by: V (t) = 1000 (1- _t_)^2 for 0 < t < 40 40 where t is given in minutes. Find the rate at which water is draining from the tank. When is the tank draining fastest? Please see the attached file for th
### Derivative of Inverse Trigonometric functions
Use inverse functions and implicit differentiation to prove that: d (tan^-1x) = 1___ dx 1+ x^2
### Use the product rule for radicals to simplify each expression. (3 Problems)
Use the Product rule for radicals to simplify each expression. Please see the attached file for the fully formatted problems.
### Find the Derivatives (3 problems)
Find the derivative (with respect to x) for each of the following. Do not simplify. (1) x2y3 = sqrt(2x − 5) + sin(8y + 3) (2) f(x) =((8x + 3)/2x^2 − 3)^5 (3) y = 4th root of ((1 − 3x)^4 + x^4)
### Definition of the Derivative and Limits : Finding the Slope of a Tangent Line
Use the definition of the derivative to find the slope of the tangent line to the graph of f(x) = 1/(3x -2) at x = 1.
### Derivative problem
C(q) = 0.000002q^3 - o.o117q^2 + 84.446q + 23879 R(q) = -0.00003 * q^3 +0.0495q^2 + 118.02q P(q) = -0.000032q^3 + 0.0612q^2 + 33.554q - 23879 Use the Cost, Revenue, and Profit functions to find. a) C`(q) b) R`(q) c) P`(q) Do these equations predict the quantity needed to maximize profit, and the amount? Explain your answ
### Modelling the volume of a container/differentiation.
Question 1 A bucket-shaped container has a circular base of radius 10 cm, and its slant height is 30 cm. the radius of the open circular top of the container is 10x cm. the curved surface of the container is modeled by part of a cone, as shown below. Please see attached.
### Differentiation
Please could you solve this question clearly showing every stage used in getting to the answer. Differentiate the following expression with respect to θ... please see attached.
### 14 Derivative Problems : Product Rule, Quotient Rule, Chain Rule, First and Second Derivative and Finding Maximum or Minimum
Rules and Applications of the Derivative -------------------------------------------------------------------------------- 1. Use the Product Rule to find the derivatives of the following functions: a. f(X) = (1- X^2)*(1+100X) b. f(X) = (5X + X^-1)*(3X + X^2) c. f(X) = (X^.5)*(1-X) d. f(X) = (X^3 + X^4)*(30
### Derivatives, Second Derivatives and Profit Function - Lemonade Stand
A. Write a function for your profits for each price you charge. This is done by multiplying (P-.5) times your function (y= -100x + 250). I.e. if your function is Cups Sold = 1000 - 100P, your profit function would be (P - .5)*(1000 - 100P). B. Calculate the first derivative of your profit function, and create another table
### Derivatives, Revenue Function, Maximizing Profit - Lemonade Stand
Data: regression equation: y= -100x + 250 regression coefficient: r= -1 X Y Predicted value 0.25 225 225 0.5 200 200 0.75 175 175 1 150 150 1.25 125 125 1.5 100 100 1.75 75 75 2 50 50 2.25 25 25 2.5
### General Calculus Questions : Definition of a Limit and Derivative, Product Rule, Tangent Line Approximation, Taylor Polynomial, Newton's Method, L'Hopital's Rule, MVT, IVT, Fundamental Theorem
1. Give the definition of limit in three forms: ε?δ , graphical, and in your own words. 2. Define the derivative. List what you consider to be the five most useful rules concerning derivatives. 3. Give an argument for the product rule. 4. What is the tangent line approximation to a function? 5. What is the Taylor p
### Rate of Change Word Problems, Derivatives and Product and Quotient Rule (15 Problems)
71. The local game commission decides to stock a lake with bass. To do this 200 bass are introduced into the lake. The population of the bass is approximated by P(t) = 20 (10 + 7t)/(1 + 0.02 t) where t is time in months. Compute P(t) and P'(t) and interpret each. 57. The monthly sales of a new computer are given by q(t) = 30
### Find the function f(x)
The second derivative of a function is given as: f"(x) = 12x-1 At the point (-2,7) the tangent to the function is given by: y=kx-3 Find the function f(x)
### Differentiation
Find dy/dx y = 2x^3+6x^2+6x+4 | 2,429 | 8,324 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2016-44 | longest | en | 0.886601 |
http://mathhelpforum.com/algebra/276585-calculating-actual-cost-loan.html | 1,529,901,140,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267867424.77/warc/CC-MAIN-20180625033646-20180625053646-00270.warc.gz | 193,610,997 | 10,640 | # Thread: Calculating the actual cost of a loan
1. ## Calculating the actual cost of a loan
Hey guys, got a math problem here. Teacher didn't give us anything on how to do it, and algebra was years ago. Here is the problem:
The first month he pays 1.65% of 650= $10.73 (rounded up to the nearest penny) interest so he actually paid 21.45- 10.73=$10.73 toward the principle, leaving 650- 10.73= 639.27 still to be paid. The second month he paid 1.65% of 639.27= $10.55 interest so he actually paid 21.45- 10.55=$10.90 toward the principle leaving 639.27- 10.90= $628.37 still to be paid. The third month he paid 1.65% of 628.37=$10.37 interest so he actually paid 21.45- 10.37= $11.08 toward the principle leaving 628.37- 11.08= 617.29 still to be paid. Continue that until the principle is paid off. Question 1: How long will it take him to pay for the stereo? Question 2: What is the total amount Dimitri will pay for the stereo? Question 3: What is Dimitri's total cost of using credit? I have no idea where to even start. Thanks. 3. ## Re: Calculating the actual cost of a loan Originally Posted by HeyAwesomePeople "Dimitri wants to buy a stereo for 650 dollars and pay for it using a credit card that has an Annual Percentage Rate of 19.8% and a periodic interest rate of 1.65%. Dimitri pays a minimum monthly payment of$21.45"
Formula for loan payment calculation:
p = a*i / [1 - 1/(1+i)^n]
p = 21.45
a = 650.00
i = .0165
n = ?
In terms of n, formula becomes:
n = LOG[p / (p - a*i)] / LOG(1 + i)
Do the maths and you'll find that when n=42,
7.52 remains owing; so a small 43rd payment remains.
If payment was 21.58, then n=42 would result in zero owing.
If you're still stuck, then you need classroom help:
not provided here. | 523 | 1,738 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.3125 | 4 | CC-MAIN-2018-26 | latest | en | 0.945451 |
https://theglobestalk.com/tag/types-of-globe/ | 1,716,388,926,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058557.6/warc/CC-MAIN-20240522132433-20240522162433-00674.warc.gz | 497,940,390 | 43,020 | What is a Globe? – An insight into the Model of Earth
In simple words. We can say that a globe is a small replica of earth. But a globe does not always represent the earth. In general terms. Globe is a spherical model of a celestial body. The globe representing the Earth is called the terrestrial globe and the globe representing other celestial bodies is called a celestial globe.
Difference between Globe and Map:-
• Globe serves the same purpose as that of maps. But globes are a three-dimensional view of the earth whereas a map is a two-dimensional view of the earth.
• Globe represents “scaled-down” model of the entire earth whereas a map can be of the entire world or could also be of a city or a small area.
• Since a map is a two-dimensional projection of earth it starts to project distortion as we increase the area of the map.
For example, if you observe the map of a city, you might not see any difference in the city locations compared to the map. Since the earth is very large (The Polar radius of the earth is 6357 km). We can consider a city to be a flat surface. And hence, plotting a map can be done with precision.
But when we talk about the world map, then plotting the spherical earth on a two- dimensional piece of paper has its drawbacks. In such a case, we need to distort the facts. For example, if you look at the world map, you will observe that Little Diomede Island (Alaska) and Big Diomede Island (Russia) are farthest from each other.
But in reality, the distance between the two islands is less than 4 km. In the case of Globe, since only the size of the globe changes and not its appearance, compared to the earth, we do not find such distortions in the globe.
What do Globes have?
• Globes are a small replica of earth. They contain all the features that are needed to study the earth.
• Globes have the axis same as the earth which does not pass perpendicular to the horizontal surface (The plane in which the earth revolves around the Sun). Instead, it passes through the center of the globe, making an angle of 23.5 degrees to the vertical (The line perpendicular to the plane in which the earth revolves around the Sun).
• Globes also mention parallels and meridians.
• Political globes show the countries and their borders.
• Terrestrial globes show the landmass and water bodies. They also have raised relief to show mountains and large landforms.
There are also many ways to represent the globe depending on the necessity of the study.
Origin of Globe:-
Globes have a long history. Since this is a blog about geography. We will discuss only a few facts.
• The word “globe” is taken from the Latin word “Globus” which means “Sphere”.
• Globes are said to be in use since 150 BC. Most of them are destroyed or ruined.
• The oldest available terrestrial globe is Erdapfel. It is said to be fashioned by Martin Behaim in 1492.
• The oldest available celestial globe was carved on the top of the Farnese Atlas in the second century Roman Empire.
Scaling of a Globe
As mentioned above, the earth has a radius of around 6357 km. We need to scale it down to a large extent to construct a globe. The scaling of the earth for a globe depends on the size of the globe.
• Mostly, standard globes have a radius of around 16 cm. To make the entire earth fit in such a small model we need to reduce the size of the earth to a scale of nearly 1:40 million units.
• In the case of topography, the scaling is not as precise. If we scale the mountain heights to the ratio of 1:40 million – as it is done to fit the size of the earth on the globe – we will not be able to observe the mountains on the globe due to their smaller relief. So for the sake of convenience, the topography of the globe is kept visible to us.
Types of globe:-
Broadly, there are three most common types of globes, for example, the physical globe, the political, and the celestial globe. There are also many ways to represent the globe depending on the necessity of the study.
Physical Globes:-
Physical Globes come in a way that displays the geographical features of the Earth, with a focus on the physical features of the world. For specimens, landforms, grains, animals, mountain ranges, hills, oceans, and continents, etc. Each globe shows its cartography based on climate, vegetation, or geographic features.
Political Globes:-
Political Globes depict the geographical features of the world as well as the wide political boundaries of countries on Earth. This type of globe provides a detailed view of the world’s political composition in a precise way. Political globes illuminate the boundaries of country and state with the necessary information on earth. These have names of major city locations as well.
Celestial Globes:-
Celestial Globes as a representation of stars and constellations show the apparent position of stars in the sky. This type of globe is used for scientific instruments, decorative showpieces, or astronomical calculations. They are very useful for amateur astronomers and often show the path of the Sun, Moon, and other planets as well as major constellations and important astronomical events.
Virtual Globes:-
Virtual Globes are virtual rather than real. Virtual globes offer many advantages over traditional globes with the ability to integrate large amounts of information. Some globes use global imaging system (GIS) databases and detailed country data, regional data, and even real-time weather tracking.
You can also see some other types of globes such as Spinning, Historical, Inflatable, Illuminated, Educational, Magnetic, Folding, Puzzle, Handmade, Art, and Wooden Globe.
Is Globe still relevant?
If you observe the map of a city, you might not see any difference in the city locations compared to the map. We can consider a city to be a flat surface and hence, plotting a map can be done with precision. But when we talk about the world map, then plotting the spherical earth on a two dimensional piece of paper has its drawbacks since the earth is very large (The Polar radius of the earth is 6357 km). In such a case, we need to distort the facts. For example, if you look at the world map, you will observe that Little Diomede Island (Alaska) and Big Diomede Island (Russia) are farthest from each other. But in reality, the distance between the two islands is less than 4 km.
How did it change over the period of time?
In the case of Globe, since only the size of the globe changes and not its appearance, compared to the earth, we do not find such distortions in the globe.
Uses of Globe?
Globe can be used to plot the coordinates of any location on earth. The intersection of parallels and meridians are written in coordinates which give us the precise location of any point on the earth. For example, 27.17510 N, 78.04210 E are the coordinates of the Taj Mahal (India). These are called geographical coordinates.
• Globe can be used to locate sea route, air route, etc.
• Globe can be used to locate continents, countries, etc. with precision.
• Globes can be used to observe the landmass and water bodies as a replica of earth.
• Globes give the precise replica of Earth and hence is the best way to study the entire earth at the same time. It helps us visualize the Earth accurately. | 1,592 | 7,268 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.546875 | 4 | CC-MAIN-2024-22 | longest | en | 0.938821 |
https://flyingcoloursmaths.co.uk/solving-pells-equation/ | 1,632,201,989,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057158.19/warc/CC-MAIN-20210921041059-20210921071059-00241.warc.gz | 322,077,906 | 5,548 | A quick how-to, by special request of @nathanday314.
Pell’s equation is anything of the form $x^2 - ny^2 = 1$, as long as $n$ is an integer but not a square number. There is also a general Pell’s equation, which looks like $x^2 - ny^2 = N$, where $n$ is still not square and $N$ is a non-zero integer.
### The Sock Puzzle, Revisited
Remember The Sock Puzzle? Of course you do. We had three equations:
• $2r(r-1) = n(n-1)$
• $12b(b-1) = n(n-1)$
• $24rb = 5n(n-1)$
Completing the square on the first pair gives:
• $2\br{\br{r - \frac{1}{2}}^2 - \frac{1}{4}} = n(n-1)$
• $12\br{br{b - \frac{1}{2}}^2 - \frac{1}{4}} = n(n-1)$
The right hand sides are equal, so the left-hand sides must be, too:
• $2\br{\br{r - \frac{1}{2}}^2 - \frac{1}{4}} = 12\br{\br{b - \frac{1}{2}}^2 - \frac{1}{4}}$
Let’s multiply everything by 4:
• $2\br{\br{2r-1}^2 - 1} = 12\br{\br{2b-1}^2 - 1}$
or:
• $2\br{2r-1}^2 - 2 = 12\br{2b-1}^2 - 12$
Or even:
• $\br{2r-1}^2 - 6\br{2b-1}^2 = -5$
… which is a general Pell’s equation with $x = 2r-1$ and $y = 2b-1$. (Obviously, we will need to make sure our answers are odd!)
### An aside into continued fractions
For reasons I don’t fully understand, you can generate a solution to (the non-general) Pell’s equation using the continued fraction of $\sqrt{n}$.
In particular, if you work out the convergents $\frac{p_i}{q_i}$ of $\sqrt{n}$ – the fractions in the continued fraction that are the closest approximations with that denominator or smaller – you will eventually find that $x=p_k$ and $y=q_k$ is a solution to the equation for some $k$. This is the fundamental solution.
Let’s try it with $n = 6$. The convergents – which I’ll circle back to presently – are $\frac{2}{1}, \frac{5}{2}, \frac{22}{9}, \frac{49}{20}\dots$.
The first one doesn’t work ($4-6$ is not 1), but the second does: $25 - 24$ is indeed 1. So $x=5$, $y=2$ is the fundamental solution to $x^2 - 6y^2 = 1$. Let’s call this solution $(x_0,y_0)$.
This is not the only solution: others can be found using the recurrence relation:
• $x_{k+1} = x_0 x_{k} + n y_0 y_k$
• $y_{k+1} = x_0 y_{k} + y_0 x_k$
So a second solution to our equation $x^2 - 6x = 1$ would be $x_1 = (5)(5) + 6(2)(2) = 49$ any $y_1 = (5)(2) + (2)(5) = 20$ – in fact, another of the convergents!
### But where do these convergents come from?
I thought you’d never ask.
The square root of 6 is “two and a bit”, which we can write down as $\sqrt{6} = 2 + x$.
Alternatively, $6 = 4 + 4x + x^2$, or $2 = x(4+x)$.
Or even alternativelier, $x = \frac{2}{4+x}$ or $\frac{1}{2 + x/2}$. (I’ll use both of these presently).
This means that $\sqrt{6} = 2 + \frac{1}{2+x/2}$, for whatever this value of $x$ is. But we know what this value of $x$ is – and in particular that $x/2 = \frac{1}{4+x}$.
So $\sqrt{6} = 2 + \frac{1}{2 + \frac{1}{4+x}}$, for whatever this value of $x$ is.
But again, we know what this value of $x$ is! It’s $\frac{1}{2+x/2}$!
So $\sqrt{6} = 2 + \frac{1}{2+\frac{1}{4+\frac{1}{2+x/2}}}$. The pattern repeats forever, as I’m sure you can prove if you’re interested.
Unsurprisingly, the first continued fractionologists looked at that and thought something that ended with “game of soldiers” and came up with the alternative notation $[2; 2,4,2,4,\dots]$, or better, $[2; \overline{2,4}]$.
So where do the convergents come from? They’re the approximations you get from letting $x=0$ at any given stage in the expansion.
The original estimate was $\sqrt{6} = 2 + x$, which gives the first convergent of 2.
The next guess was $\sqrt{6} = 2 + \frac{1}{2+x}$, which gives $\frac{5}{2}$.
Then $\sqrt{6} = 2 + \frac{1}{2 + \frac{1}{4+x}}$, or $2 + \frac{1}{9/4}$, or $\frac{22}{9}$.
(There is a quicker and dirtier way to generate these, rather than work the whole fraction through each time – but that’s for another article.)
### But what about the general Pell’s equation?
I thought you’d never ask that, either!
It turns out that the work we’ve just done to find a solution to $x^2 - 6y^2 = 1$ is part of the solution here, as well – this is the Pell’s resolvent.
As long as we have an answer to $x^2 - ny^2 = N$, we can figure out infinitely many ((I’m not certain this method generates all of them.)).
In our case, we do: $x^2 - 6y^2 = -5$ has the obvious solution $(x_0, y_0) = (1,1)$.
If we call the solutions to the resolvent $(u_i, v_i)$, then we can generate new solutions using the recurrence relation:
• $x_i = x_0 u_i + n y_0 v_i$
• $y_i = y_0 u_i + x_0 v_i$
In our case, $(x_0, y_0) = (1,1)$ and the $(u_i,v_i)$s start $(5,2), (49,20), \dots$.
So:
• $x_1 = (1)(5) + 6(1)(2) = 17$
• $y_1 = (1)(2) + (1)(5) = 7$
So $17^2 - 6 \times 7^2 = 289 - 294$, which is indeed -5!
The next solution is
• $x_2 = (1)(49) + 6(1)(20) = 169$
• $y_2 = (1)(20) + (1)(49) = 69$
This is the one we’re interested in.
### Back to the original problem
Remember way back several pages ago, we were trying to solve $(2r-1)^2 - 6(2b-1)^2 = -5$?
We’ve come up with two answers here (and could generate infinitely many). The first pair is: $(2r-1, 2b-1) = (17,7)$, or $r=9$ and $b=4$ (and, implicitly, $n=13$). Unfortunately, this doesn’t quite work with the original equations: $2r(r-1) = 144$, which is not $13\times 12$.
The second pair gives $r=85$, $b=35$ and $n=120$, and $2r(r-1) = 170 \times 84$; this turns out to be the same as $120 \times 119$ because $17 \times 10 \times 7 \times 12 = 12 \times 10 \times 7 \times 17$ (obviously).
### Conclusion
This is clearly a completely impractical way to solve a problem that drops out in a few lines of algebra. However, it is also a tremendous introduction to continued fractions and the Pell’s equation. I hope you found it valuable. | 2,033 | 5,686 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2021-39 | longest | en | 0.846972 |
http://fora.xkcd.com/viewtopic.php?p=3795001 | 1,566,287,401,000,000,000 | text/html | crawl-data/CC-MAIN-2019-35/segments/1566027315258.34/warc/CC-MAIN-20190820070415-20190820092415-00118.warc.gz | 75,608,903 | 7,749 | ## Almost Nesbitt's inequality
For the discussion of math. Duh.
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### Almost Nesbitt's inequality
Prove a/sqrt(b+c) + c/sqrt(b+a) + c/sqrt(a+b) > 0 for positive a,b,c
Not sure if this is even correct, but i saw this somewhere long ago.
Edit: I'm thinking Cauchy-Schwarz right now.
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Wildcard
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### Re: Almost Nesbitt's inequality
Um, well, if you copied it correctly, then this is almost trivially true. The result of the square root function is always positive (with positive arguments), and since a, b and c are given as positive, then you are saying that the sum of three positive fractions is greater than zero...surprise surprise.
Or was there a geometric interpretation or something that I'm missing?
Answer: There isn't. But the question may have been miscopied.</bluntness>
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### Re: Almost Nesbitt's inequality
No, looks like i just have it incorrect... I do remember some kind of twist on nesbitt's though, involving radicals... but that means this is not it.
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### Re: Almost Nesbitt's inequality
Is there still an inequality if we remove the bounds for a,b,c ?
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### Re: Almost Nesbitt's inequality
Paradoxica wrote:Prove a/sqrt(b+c) + c/sqrt(b+a) + b/sqrt(a+c) > 0 for positive or negative a,b,c
Okay, fixed that (and the other odd bit which I assume was a typo.)
Is this inequality true now? Much more interesting question. Let's see....
...Seems like we're going to have some complex numbers on our hands. The inequality relation isn't defined for complex numbers, is it?
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### Re: Almost Nesbitt's inequality
no. what if you replace it with the euclidean distance formula? i suppose that'll still be positive.
this isn't the inequality i'm after. the one i remember had a non-zero constant on the other side...
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### Re: Almost Nesbitt's inequality
Wildcard wrote:
Paradoxica wrote:Prove a/sqrt(b+c) + c/sqrt(b+a) + b/sqrt(a+c) > 0 for positive or negative a,b,c
Okay, fixed that (and the other odd bit which I assume was a typo.)
Is this inequality true now? Much more interesting question. Let's see....
...Seems like we're going to have some complex numbers on our hands. The inequality relation isn't defined for complex numbers, is it?
If you add the restriction that a + b > 0, b + c > 0, and a + c > 0, so that the squareroots are never complex, then I believe the statement is true.
If all a, b, and c are positive, then the statement is trivially true. If two or more of a, b, and c are negative, then one of the conditions is violated. So only one can be negative, WOLG I will call it a. Then b > -a and c > -a, and b + c > a + b and b + c > a + c. Then c/sqrt(b + a) > -a/sqrt(b + c) and b/sqrt(a + c) > -a/sqrt(b +c). Therefore a/sqrt(b+c) + c/sqrt(b+a) + b/sqrt(a+c) > 0.
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### Re: Almost Nesbitt's inequality
how about a/sqrt(b^2 + c^2) and it's permutations
also, the same as above, but without the radicals
what will be on the other side of the inequalities if we allow all real values of a,b and c?
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### Re: Almost Nesbitt's inequality
I think for my first post, the original question was to find the lower bound of the expression, assuming a, b, c are all strictly positive.
GENERATION -705 - 992 i: The first time you see this, copy it into your sig on any forum. Square it, and then add i to the generation. | 1,315 | 4,591 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.578125 | 4 | CC-MAIN-2019-35 | latest | en | 0.860666 |
https://www.worldsrichpeople.com/how-is-bandpass-filter-bandwidth-calculated/ | 1,726,683,428,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651931.60/warc/CC-MAIN-20240918165253-20240918195253-00842.warc.gz | 982,705,168 | 11,780 | # How is bandpass filter bandwidth calculated?
## How is bandpass filter bandwidth calculated?
The bandwidth of the filter is therefore the difference between these upper and lower -3dB points. For example, suppose we have a band pass filter whose -3dB cut-off points are set at 200Hz and 600Hz. Then the bandwidth of the filter would be given as: Bandwidth (BW) = 600 – 200 = 400Hz.
How is bandpass filter calculated?
So all frequencies between the low cutoff frequecny and the high cutoff frequency are the passband of the bandpass filter. The gain of the circuit is determined by the formula, gain (AV)= -R2/R1. Thus, for example, to have a gain of 10, R2 must be 10 times the value of R1.
What is the order of bandpass filter?
Band pass filters are known generally as second-order filters, (two-pole) because they have “two” reactive component, the capacitors, within their circuit design. One capacitor in the low pass circuit and another capacitor in the high pass circuit.
### What is the bandwidth of a bandpass filter?
3 dB bandwidth
The bandwidth of a bandpass filter is usually defined as the 3 dB bandwidth. Similarly, the 1 dB bandwidth is the point at which the signal amplitude decreases by 1 dB from its maximum value (above and below the center frequency).
What is bandpass frequency range?
Generally, the dielectric band-pass filters can be used over the frequency range from 300 MHz to 100 GHz. For high-frequency applications, NRD waveguide filters (Figure 7.38) gain interests because of the extremely low-loss and low dielectric constant materials that can be used in the design.
What is Q factor in filters?
This same factor affects bandwidth and selectivity. The factor is known as Q (quality factor). The higher the Q, the better the filter; the lower the losses, the closer the filter is to being perfect.
#### What is the formula for cutoff frequency?
We can write the cutoff frequency equation for RC filter circuit as: fc = 1 / (2 * π * R * C ) . fc = 636.6 Hz .
What is a bandpass filter in flow cytometry?
Bandpass filters are the ones that are most commonly used in flow cytometry. Positioned in front of the detectors, these components determine what collection of wavelengths, and ultimately which fluorophores, will be measured by each detector.
How do you determine the order of a signal?
The order, n of a filter is the number of reactive elements (if all are contributing.) Using the linear slope (on log-log grid) away from f breakpoint it will be 6dB/octave per order of n. An n= 4th order is 24dB/octave slope as in both of 1st examples .
## Was ist der Unterschied zwischen einem aktiven und einem passiven Bandpass?
Beim Bandpass gibt es aktive und passive Filter. Eine passive Bandpass Schaltung liegt vor, wenn kein verstärkendes Element eingesetzt wird. Der Bandpass kann in verschiedenen Ordnungen ausgeführt sein, wobei der Bandpass 1. Ordnung die Grundvariante bildet.
Wie berechnet man einen Bandpass?
Formel – Bandpass berechnen online. Normalerweise werden bei einem Bandpass zwei gleiche Widerstände und zwei gleiche Kondensatoren ausgewählt. Wenn dies der Fall ist, dann gilt die Bandpass Übertragungsfunktion: \$\$ frac{U_a}{U_e} = frac{1}{3 + j left( omega R C – frac{1}{omega R C} right)} \$\$.
Was ist der Unterschied zwischen einem Bandpass und einem Kondensator?
Normalerweise werden bei einem Bandpass zwei gleiche Widerstände und zwei gleiche Kondensatoren ausgewählt. Wenn dies der Fall ist, dann gilt die Bandpass Übertragungsfunktion: ist die Kreisfrequenz aus . ist der Widerstandswert und die Kapazität des Kondensators.
Begin typing your search term above and press enter to search. Press ESC to cancel. | 887 | 3,696 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.84375 | 4 | CC-MAIN-2024-38 | latest | en | 0.938088 |
http://www.pearltrees.com/ducky3113/curve-stitching/id11027827 | 1,569,020,072,000,000,000 | text/html | crawl-data/CC-MAIN-2019-39/segments/1568514574084.88/warc/CC-MAIN-20190920221241-20190921003241-00515.warc.gz | 303,009,019 | 23,325 | # Curve stitching
How to Draw a Square: 9 Steps (with Pictures. Curve stitching templates. Curve stitching was created by Mary Boole as a method to teach children visually how angles and spaces worked in geometry.
She considered herself a mathematical psychologist and was very interested in how children learned math. She made cards with holes punched in them that students could stitch. Skip forward all the way to the 70s and curve stitching using thread around nails in boards was super popular. String art was trendy, with the lines of stitching making up lovely intricate geometric designs and abstract representations of birds, boats and other popular decorating themes. There are so many shapes that can be made using this nifty technique and these templates are fairly limited. Using these is fairly easy.
Using the square grids, you can do corners, diamonds and other shapes on a right angle. As you can see, with this one the inner star is spaced so it's every 4th dot marked and the lines are made so they intersect fairly close to the middle to make a star shape. AliStauffer20132014 (1).JPG (2048×1536) Math extensions: Curve Stitching Projects 2013-14. Curve Stitching. Curve Stitching, also known as String Art, was first introduced by Mary Boole, a self-taught mathematician and wife of George Boole (the father of Boolean Algebra to which Mary also contributed quite a bit).
The idea is to draw a design with straight lines, then subdivide the lines (usually at regular intervals) and connect points on two adjacent lines, again with straight lines (1 to 1, 2 to 2 and so on) as shown below: The interesting thing is that because the straight lines meet at an angle, there is an illusion that you have created a curved line by drawing straight lines. And the more subdivisions you make in the original two lines, the smoother the curve looks. There is an interesting article in Maths Plus about the relationship between these Curve Stitches and Bezier Curves: Jason Davies also has a nice animation illustrating how Bezier curves are formed..
There are many very nice examples of String Art on the web. Here are some ideas for creating Curve Stitches of your own. Curve Stitching. Curve Stitching. Making Maths: Curve Stitching. Challenge Level: If you think that sewing isn't for you, think again.
These curve stitching patterns look fantastic and once you've got the hang of it, they take next to no time to do. All you need is a little bit of coordinate know-how! It is useful to practise this just by drawing lines using your ruler to start with, then you'll be more confident when it comes to the real curved stitching. Here's how to practise: Using the ruler and pencil, draw a set of axes on a piece of scrap paper numbered from 0-10 in both directions like the ones below: Join the 1 on the x-axis to the 9 on the y-axis with a straight line.
Now that you've had a go on paper, you're ready to stitch! There are lots of variations on this curve stitching idea. Curve stitching forget-me-not printables. Curve stitching is interesting, and I always forget how fun until I run across something about it.
It's creating curves using straight lines, or for people my age, it's the nail and string art that our parents did or had in the 70s. The coloring page and pattern are meant to be printed at 100 dpi and will be 8 inches across printed at 100 dpi. If you download the pattern and print it at a higher dpi it will print smaller. The pattern can either be used with a ruler and colored pencils or you could pierce a card and use thread to stitch the design. Follow the lines in the coloring page to make it look like mine. Curve stitching templates. | 793 | 3,682 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.765625 | 4 | CC-MAIN-2019-39 | longest | en | 0.968842 |
https://lifewithdata.com/2023/06/13/leetcode-power-of-four-solution-in-python/ | 1,695,935,336,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510454.60/warc/CC-MAIN-20230928194838-20230928224838-00466.warc.gz | 397,249,390 | 25,229 | # Leetcode – Power of Four Solution in Python
In this elaborate article, we will deeply analyze a well-known problem on LeetCode called the ‘Power of Four’. This problem is significant for understanding bitwise operations and mathematical properties. We will dissect the problem statement, and explore multiple approaches to solve this problem, evaluating their respective time and space complexities.
## Introduction
Given an integer n, return True if it is a power of four. Otherwise, return False.
### Example:
Input: n = 16
Output: True
Explanation: 16 is 4^2, hence it is a power of four.
## Understanding the Problem
The problem is fairly straightforward: we need to determine if the given integer n is a power of four. Specifically, we need to check if there exists an integer x such that 4^x = n.
## Approach 1: Loop Iteration
A basic approach is to keep dividing the number by four and check if we can reach 1.
1. If n is less than 1, return False.
2. While n is greater than 1, check if n is divisible by 4. If it is, divide n by 4; otherwise, return False.
3. If the loop ends, return True.
def isPowerOfFour(n):
if n < 1:
return False
while n > 1:
if n % 4 != 0:
return False
n /= 4
return True
### Time Complexity:
O(log₄n) – In the worst case, we keep dividing the number by 4 until we reach 1.
### Space Complexity:
O(1) – We are not using any additional data structures.
## Approach 2: Logarithm
We can also solve this problem using logarithms. If n is a power of four, then log₄(n) must be an integer. We can calculate the base 4 logarithm by using the change of base formula: log₄(n) = log₂(n) / log₂(4).
1. If n is less than 1, return False.
2. Calculate log₄(n) using the change of base formula.
3. Check if log₄(n) is an integer. If it is, return True; otherwise, return False.
import math
def isPowerOfFour(n):
if n < 1:
return False
log4_n = math.log2(n) / math.log2(4)
return math.isclose(log4_n, round(log4_n))
### Time Complexity:
O(1) – We perform a constant number of operations.
### Space Complexity:
O(1) – No additional space is required.
## Approach 3: Bit Manipulation
We can utilize bit manipulation for an optimized solution. If n is a power of four, its binary representation has only one ‘1’ bit, and the number of ‘0’ bits after the ‘1’ bit is even. For example, 4 is 100, 16 is 10000, and so on.
1. Check if n has only one ‘1’ bit by using n & (n-1) == 0.
2. Check if the number of ‘0’ bits after the ‘1’ bit is even by using n & 0xAAAAAAAA == 0.
def isPowerOfFour(n):
return n > 0 and n & (n-1) == 0 and n & 0xAAAAAAAA == 0
### Time Complexity:
O(1) – We perform bitwise operations that take constant time.
### Space Complexity:
O(1) – No additional space is used.
## Conclusion:
The ‘Power of Four’ problem on LeetCode demonstrates how mathematical insights and bitwise manipulation can be employed to devise efficient algorithms for seemingly simple problems. Through various approaches, ranging from basic loops and logarithmic calculations to sophisticated bitwise operations, this problem teaches the importance of evaluating different strategies to solve a given problem. It underlines the importance of understanding binary representations and mathematical properties, which can be crucial in optimizing algorithms for performance-sensitive applications. In coding interviews and real-world applications, being proficient in these concepts can prove to be invaluable. | 859 | 3,450 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2023-40 | latest | en | 0.758326 |
https://www.houspain.com/gttp/doku.php?id=total_probability&idx=escorial | 1,596,520,288,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439735860.28/warc/CC-MAIN-20200804043709-20200804073709-00294.warc.gz | 637,822,746 | 4,755 | Tabla de Contenidos
## Total Probability
We have the followings n events that are:
• they are incompatible events two by two, ie, they haven´t common elements ()
• if we join all events from E (colection of samples), is the group that is created by all posible results from a random experience.
That:
Then, the probability of a event S is happened, I can decompose it, as:
(Demostration of this result)
### Example
A mouse is persecuted by a hungry cat. The mouse can to choose one of these streets A, B or C, to salve himself.
• The probability that the mouse choose the A street is 0.3 P(A)=0.3
• The probability that the mouse choose the B street is 0.5 P(B)=0.5
• The probability that the mouse choose the C street is 0.2 P(C)=0.2
The probabilities that the mouse is hunted by mouse in every street, are:
• P(cat hunts mouse in A)= P(+|A)= 0.4
• P(cat hunts mouse in B)= P(+|B)= 0.6
• P(cat hunts mouse in C)= P(+|C)= 0.1
What´s the probability that cat hunts mouse?.(Calculate P(+))
SOLUTION
We make a diagram with tree form to this exercise:
P(+)== ¡the cat does not have assured the food !
total_probability.txt · Última modificación: 24/04/2017 13:13 (editor externo) | 331 | 1,194 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2020-34 | latest | en | 0.908038 |
https://www.gradesaver.com/textbooks/math/algebra/big-ideas-math---algebra-1-a-common-core-curriculum/chapter-3-graphing-linear-functions-3-5-graphing-linear-equations-in-slope-intercept-form-monitoring-progress-page-139/10 | 1,723,497,020,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641048885.76/warc/CC-MAIN-20240812190307-20240812220307-00423.warc.gz | 597,805,106 | 14,557 | ## Big Ideas Math - Algebra 1, A Common Core Curriculum
$x-$ intercept is $-1$. The graph is shown below.
The given equation is $\Rightarrow 3x+y=-3$ Subtract $3x$ from each side. $\Rightarrow 3x+y-3x=-3-3x$ Simplify. $\Rightarrow y=-3x-3$ The slope-intercept form is $\Rightarrow y=mx+b$ $y-$intercept is $b=-3$. Plot the $y-$ intercept $A=(0,-3)$. The slope is $m=-3$ and Slope $m=\frac{rise}{run}=-\frac{3}{1}=\frac{3}{-1}$ The next point is $1$ unit left and $3$ unit up from the point $(0,-3)$. Plot the point $B=(-1,0)$. Draw a line through the two points. From the graph the line crosses the $x-$ axis at $B=(-1,0)$. Hence, the $x-$ intercept is $-1$. | 227 | 659 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.4375 | 4 | CC-MAIN-2024-33 | latest | en | 0.607493 |
https://en.m.wikibooks.org/wiki/Octave_Programming_Tutorial/General_mathematical_functions | 1,723,110,745,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722640726723.42/warc/CC-MAIN-20240808093647-20240808123647-00757.warc.gz | 180,979,593 | 9,543 | # General Mathematical Functions
## Constants
• `e` is the base of the natural logarithm.
`e` without arguments returns the scalar e.
`e(N)` returns a square matrix of e of size `N`.
`e(N, M, ...)` where the arguments are dimensions of some matrix of e.
`e(..., CLASS)` where `CLASS` is an optional argument that specifies the return type, `double` or `single`.
• `eps` is the machine precision and returns the relative spacing between any floating point number and the next representable number. This value is system dependent.
`eps` returns the value of `eps(1.0)`.
`eps(X)` returns the spacing between X and the next value.
`eps` with more than one argument is treated like `e` with the matrix value being `eps(1.0)`.
• All of the constant functions listed are defined exactly like `e`
`pi` is the ratio of the circumference to the diameter of any circle.
`I` is the imaginary unit defined so `I^2 = -1`.
`Inf` is used for values that overflow the standard IEEE floating point range or the result of division by zero.
`NaN` is used for various results that are not well defined or undefined. Note that `NaN` never equals other `NaN` values. Use the function `isnan` to check for `NaN`.
`realmax` is the largest floating point value representable.
`realmin` is the smallest positive floating point value representable.
## Arithmetic Functions
• `floor(X)` and `ceil(X)` return the highest integer not greater than `X` or lowest integer not less than `X`, respectively.
• `round(X)` and `fix(X)` return the integer closest to `X` or truncate `X` towards zero, respectively.
• `rem(X,Y)` and `mod(X,Y)` returns x - y * fix( x ./ y ) or x - y * floor( x ./ y ), they are the same except when dealing with negative arguments.
• `hypot(X, Y)` returns the length of the hypotenuse of a right-angle triangle with the adjacent and opposite of size `X` and `Y`.
• ` abs(X) ` return absolute of x.
• ` sign(X) ` return sign of the x (-1, 0 or +1).
## Ordinary Trigonometry
• `cos(X)`, `sin(x)` and `tan(X)` — the elemental functions that we all know and love. They take their arguments in radians.
• `acos(X)`, `asin(X)` are the inverses of `cos` and `sin` and are able to compute arguments not contained in the range [-1,1].
• `atan(X)` and `atan2(Y, X)` are the 2 available inverses of tan. `atan` is a simple inverse whereas `atan2` takes 2 arguments and returns an angle in the appropriate quadrant. More information on `atan2` can be found here.
• Note that one can add the character d to any of the functions except `atan2` and they will work in degrees rather than radians. For example: `asind(0.3) = asin(0.3*180/pi)`
• ` exp(x) `, exponential funtion of x
• ` log(x) `, natural logarithmic of x, loge NOT log 10
## Hyperbolic Trigonometry
• `cosh(X)`, `sinh(X)` and `tanh(X)` are analog to their more prosaic counterparts but deal with the unit hyperbola instead of the unit circle. They also take their arguments in radians.
• `acosh(X)`, `asinh(X)` and `atanh(X)` are the inverses of `cosh`, `sinh` and `tanh`.
• Unlike their circular uncles they cannot be made to take their arguments in degrees. | 820 | 3,108 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.828125 | 4 | CC-MAIN-2024-33 | latest | en | 0.729563 |
http://en.wikipedia.org/wiki/Integer_sequence | 1,416,727,958,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416400379355.46/warc/CC-MAIN-20141119123259-00235-ip-10-235-23-156.ec2.internal.warc.gz | 13,279,828 | 12,756 | # Integer sequence
In mathematics, an integer sequence is a sequence (i.e., an ordered list) of integers.
An integer sequence may be specified explicitly by giving a formula for its nth term, or implicitly by giving a relationship between its terms. For example, the sequence 0, 1, 1, 2, 3, 5, 8, 13, … (the Fibonacci sequence) is formed by starting with 0 and 1 and then adding any two consecutive terms to obtain the next one: an implicit description. The sequence 0, 3, 8, 15, … is formed according to the formula n2 − 1 for the nth term: an explicit definition.
Alternatively, an integer sequence may be defined by a property which members of the sequence possess and other integers do not possess. For example, we can determine whether a given integer is a perfect number, even though we do not have a formula for the nth perfect number.
## Examples
Integer sequences which have received their own name include:
## Computable and definable sequences
An integer sequence is a computable sequence, if there exists an algorithm which given n, calculates an, for all n > 0. An integer sequence is a definable sequence, if there exists some statement P(x) which is true for that integer sequence x and false for all other integer sequences. The set of computable integer sequences and definable integer sequences are both countable, with the computable sequences a proper subset of the definable sequences (in other words, some sequences are definable but not computable). The set of all integer sequences is uncountable (with cardinality equal to that of the continuum); thus, almost all integer sequences are uncomputable and cannot be defined.
## Complete sequences
An integer sequence is called a complete sequence if every positive integer can be expressed as a sum of values in the sequence, using each value at most once. | 402 | 1,837 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.0625 | 4 | CC-MAIN-2014-49 | longest | en | 0.919647 |
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