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# How to solve by substitution method
One of the most important skills that students need to learn is How to solve by substitution method. Math can be a challenging subject for many students.
## How can we solve by substitution method
In addition, there are also many books that can help you How to solve by substitution method. There are a variety of online resources that can help with geometry math problems. Sites like Math is Fun and Khan Academy offer step-by-step explanations of how to solve various types of geometry problems. There are also online calculators that can be used to solve geometry problems step-by-step.
There are a few different methods that can be used to solve limits. To start, it is helpful to understand what a limit is. A limit is basically the value that a function approaches as the input gets closer and closer to a certain value. So, to solve a limit, we basically just need to find the value that the function approaches. One method of solving limits is to use direct substitution. This is where we simply plug in the value that we are approaching and see what
Square roots can be used to solve equations by taking the square root of both sides of the equation. This will isolate the variable on one side of the equation. For example, to solve the equation x^2=4, you can take the square root of both sides to get x=+/-2.
Another way to get algebra help online is to use a resource like Khan Academy. This website provides a wealth of helpful videos and articles on algebra, so you can learn at your own pace. This is a great
First, it can be helpful to break the problem down into smaller pieces and solve each piece separately. Additionally, it can be helpful to use symmetry to simplify the problem. Finally, it may be helpful to draw a diagram to visualize the problem and make it easier to identify a solution.
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# Transformation of Random Variables (Z = X-Y)
1. Dec 29, 2011
### Applejacks01
1. The problem statement, all variables and given/known data
Suppose we have a function, f(x,y) = e^-x * e^-y , 0<=x< ∞, 0<=y<∞,
where X and Y are exponential random variables with mean = 1. (For those who may not know, all this means is ∫(x*e^(-x) dx) from 0 to ∞ = 1, and the same for y)
Suppose we want to transform f(x,y) into f(z), where the transformation is Z = X-Y. Find f(z)
2. Relevant equations
f(x,y) = e^-x * e^-y , 0<=x< ∞, 0<=y<∞
Z = X-Y
3. The attempt at a solution
So I decided to transform -Y into W. So we have -Y<=W, which implies Y>=-W
∫e^-y dy from -w to ∞...
-e^-y from -w to ∞
0 + e^-(-w)
e^w
Differentiate wrt w
f(w) = e^w -∞ < w <= 0
So now we have Z = X+W
Z-W = X
We'll just let W stay as is for this problem.
The jacobian of this transformation is 1.
So we have f(z) = ∫e^-(z-w)*e^w dw from -∞ to z,
This becomes ∫e^-z*e^2w dw from -∞ to z
This becomes e^-z * (e^2w)/2 from -∞ to z
This becomes e^-z * ((e^2z)-0)/2
Which becomes (e^z)/2
The domain of z is -∞ to ∞, however, this integral does not evaluate to 1. As a matter of fact, it does not even converge.
Any help would be greatly appreciated!!
2. Dec 30, 2011
### Ray Vickson
Let f be the marginal density of X or Y. P{Z <= z} = int_{y=0..infinity} f(y)*P{X-Y <= z|Y=y} dy, and P{X-Y <= z|Y=y} = P{X <= y+z|Y=y} = P{X <= y+z} because X and Y are independent. You can get P{X <= y+z} and so do the integral. The final result is perfectly fine for all z in R and does, indeed, integrate to 1. However, you need to be careful about integration limits, since f(y) = exp(-y)*u(y) and F(y+z) = [1-exp(-y-z)]*u(y+z), where u(.) is the unit step function [u(t) = 0 for t < 0 and u(t) = 1 for t > 0]. BTW: please either use brackets (so write e^(2z)) or the "[ S U P ]" button (so write e2z) or else write "exp", as I have done. That way there is no chance of misreading what you type. Alternatively, you could use LaTeX.
RGV
3. Jan 15, 2012
### Applejacks01
So sorry, I saw your solution a while ago and it really helped me. I just remembered that I never thanked you/gave you credit for your solution. How do I give you credit for helping me?
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• 欢迎光临~
# 4. Median of Two Sorted Arrays
Given two sorted arrays `nums1` and `nums2` of size `m` and `n` respectively, return the median of the two sorted arrays.
The overall run time complexity should be `O(log (m+n))`.
Example 1:
```Input: nums1 = [1,3], nums2 = [2]
Output: 2.00000
Explanation: merged array = [1,2,3] and median is 2.
```
Example 2:
```Input: nums1 = [1,2], nums2 = [3,4]
Output: 2.50000
Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.
```
`public static double findMedian(int[] A, int[] B) { int lenA = A.length; int lenB = B.length; int totalLen = lenA + lenB; if (totalLen % 2 == 1) return helper(A, 0, lenA, B, 0, lenB, totalLen / 2 + 1); else { int var1 = helper(A, 0, lenA, B, 0, lenB, totalLen / 2); int var2 = helper(A, 0, lenA, B, 0, lenB, totalLen / 2 + 1); return (double)(var1 + var2) / 2; }}private static int helper(int[] A, int beginA, int lenA, int[] B, int beginB, int lenB, int k) { if (lenA > lenB) return helper(B, beginB, lenB, A, beginA, lenA, k); if (lenA == 0) return B[beginB + k - 1]; if (k == 1) return Math.min(A[beginA], B[beginB]); int ma = Math.min(lenA, k / 2);//把k分成两部分 int mb = k - ma; if (A[beginA + ma - 1] < B[beginB + mb - 1]) return helper(A, beginA + ma, lenA - ma, B, beginB, lenB, k - ma);//把A数组前面ma个元素去掉 else if (A[beginA + ma - 1] > B[beginB + mb - 1]) return helper(A, beginA, lenA, B, beginB + mb, lenB - mb, k - mb);//把B数组前面mb个元素去掉 else return A[beginA + ma - 1];}`
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## Density
An updated version of this page is now available containing the complete mass- volume- density module (including an online test assessment, and ideas for science projects) --
Take a look at the two boxes below. Each box has the same volume. If each ball has the same mass, which box would weigh more? Why?
The box that has more balls has more mass per unit of volume. This property of matter is called density. The density of a material helps to distinguish it from other materials. Since mass is usually expressed in grams and volume in cubic centimeters, density is expressed in grams/cubic centimeter.
We can calculate density using the formula:
Density= Mass/Volume
.
### Block I
Mass = 79.4 grams
Volume=29.8 cubic cm.
### Block II:
Mass= 25.4 grams
Volume=29.8 cubic cm.
### Now Let's calculate the density for each block.
grams/cubic cm.
If you need a calculator one is available. Click on calculator in the left frame.
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# Subtract numbers
Important: The calculated results of formulas and some Excel worksheet functions may differ slightly between a Windows PC using x86 or x86-64 architecture and a Windows RT PC using ARM architecture. Learn more about the differences.
Let's say you want to find out how many inventory items are not profitable (total inventory minus profitable items) or how many employees are approaching retirement age (total employees minus employees under 55). There are several ways to subtract numbers.
## Subtract numbers in a cell
To do this task, use the - (minus sign) arithmetic operator.
For example, if you type the following formula in a cell:
=10-5
The cell displays the following result:
5
## Subtract numbers in a range
To do this task, use the SUM function. Adding a negative number is the same as subtracting.
Note: There is no SUBTRACT function in Excel. Use the SUM function and convert any numbers that you want to subtract to their negative values. For example, SUM(100,-32,15,-6) returns 77.
### Example
1. Create a blank workbook or worksheet.
2. Select the content in cells A1-B7 in the following example.
Note: Do not select the row or column headers.
A B 1 Data 2 15000 3 9000 4 -8000 5 Formula Description (Result) 6 =A2-A3 Subtracts 9000 from 15000 (6000) 7 =SUM(A2:A4) Adds all numbers in the list, including negative numbers (16000)
3. Press CTRL+C.
4. In the worksheet, select cell A1, and then press CTRL+V.
5. To switch between viewing the results and viewing the formulas that return the results, press CTRL+` (grave accent), or on the Formulas tab, in the Formula Auditing group, click the Show Formulas button.
### How you use the SUM function
The SUM function adds all the numbers that you specify as arguments. Each argument can be a range, a cell reference, an array, a constant, a formula, or the result from another function. For example, SUM(A1:A5) adds all the numbers that are contained in cells A1 through A5 (a range). For another example, SUM(A1, A3, A5) adds the numbers that are contained in cells A1, A3, and A5 (A1, A3, and A5 are arguments).
Share
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How to Sum in Excel and All Its Formulas/Functions - Compute Expert
# How to Sum in Excel and All Its Formulas/Functions
Home >> Excel Tutorials from Compute Expert >> Excel Calculations >> How to Sum in Excel and All Its Formulas/Functions
In this tutorial, we will learn how to sum in excel completely.
There are various methods available in excel to sum depending on our needs. Excel also has special formulas for the sum process such as SUM, SUMIF, and SUMIFS.
After reading and understanding all the contents here, you should have no problem to do a sum process in excel. You will also understand what are excel sum formulas you can use to process your data. Thus, without further ado, let’s follow all the tutorial parts until the end!
Disclaimer: This post may contain affiliate links from which we earn commission from qualifying purchases/actions at no additional cost for you. Learn more
## How to Sum in Excel Using a Manual Formula Writing
The first method to sum in excel that we will discuss is by using manual sum formula writing. To do the writing, we use the plus symbol ( + ).
How to write the sum formula manually in excel? Here is the general writing form.
=number_1 + number_2 + …
Pretty simple, isn’t it? The writing form is like when you write a standard sum formula everywhere, not just in excel.
Input all numbers/cell coordinates filled with numbers you want to add with + symbols between them. Don’t forget to write an equal symbol ( = ) in front of the writing as a formula writing symbol in excel.
As an example of the implementation of this writing form to sum in excel, see below.
The numbers we want to sum in the example are spread into several cells. We input the cell coordinates of the numbers we want to sum and the + symbols to get the sum result.
After we have written the formula, press enter. You will immediately get your numbers sum result.
## How to Sum in Excel Using the SUM Formula
If you want to use an excel formula to help your numbers sum process, then you can use SUM. This formula usage is pretty easy because you only need to input all the numbers you want to sum inside it.
Generally, the SUM formula’s writing form in excel is as follows.
=SUM(number1, [number2], …)
You can input all the numbers you want to sum as needed in SUM. Usually, the input form of SUM itself is a cell range, where the numbers you want to sum are all there.
If you input more than one number/cell coordinate/cell range, don’t forget to add comma signs between your inputs.
In the example, we use the same numbers that we use in the manual sum formula writing example before. The sum results we get from the manual formula writing and the SUM methods are similar.
For the example, the SUM method is probably easier because we just need to input the cell range of the numbers. This is because the numbers we want to sum are near each other.
Input all the numbers you want to sum in SUM and enter. You will immediately get your numbers sum result!
## How to Sum in Excel Using Click + Drag
If you don’t need your sum result in a cell (you just want to know the sum result), then there is a very fast method to get it.
Just highlight all the cells which contain the numbers you want to sum. Click and drag on all the cells. Then, look at the bottom right of your excel file.
There will be the numbers sum result shows up like this (the one in the red box).
By using this method, you can get the sum result of your numbers fast!
## How to Sum Columns in Excel/to the Side
Want to sum to the side, per column, to several columns you have in your excel data table? If the columns are parallel, close, and in the same size (like most columns in a data table), then use SUM and a formula copy process.
As an example, take a look at the following data table.
We want to get the total sales quantity per month for each store. How to do this sum process fast and accurately in excel?
First, you sum the numbers in the most left column by using a formula. You can write a sum formula manually with the + symbol or use SUM.
Because the numbers we want to sum are in a cell range, it is probably easier if we use SUM.
Don’t forget to add the numbers in their cell coordinates form, not by typing them directly. This is so we can easily copy the formula later to the side to get other columns’ sum results.
After done writing the sum formula for one column, we copy the formula to the right. The way to copy it is by first highlighting the cells where your sum formula is. Then, move the pointer to the bottom right of the cell cursor (to the little box on the cell cursor) until it becomes a + symbol like this.
After your pointer becomes a +, click and drag to the right until the last column that you want to sum. When you get to the last column, release your drag. You will get the sum results of the columns in your data table!
That is how you sum columns in excel. If you understand how to copy a formula in excel, then you should get your sum results quickly.
## How to Sum Down in Excel
What if what we want to sum is numbers per row or down as per its direction? The method is pretty similar to the columns sum method earlier.
To make this clearer, look at the data table we use for the columns sum process earlier.
Now, we want to get the total sales quantity per shop branch. How to do that?
Similar to the sum process of the columns earlier, we write the sum formula for the first row numbers first.
After we get the sum result of the first row, we highlight the cell with the sum formula. Move our pointer to the bottom right of the cell cursor until it becomes a + symbol like this.
Click and drag our pointer until the last row that we want to get the numbers sum result from. Release the drag after we reach that last row. We will immediately get all the rows sum results we want!
Similar to the columns sum process before, right? You must know how to copy a formula in excel so you can get your sum results fast and accurately.
## How to Sum Between Sheets in Excel
Sometimes, maybe the numbers we want to sum are in different sheets to where we want to put our sum result. What should we do in this situation?
Fortunately, Excel makes it easy to refer to numbers in different worksheets. Just input the sheet name before we input the cell coordinate where the number is in our sum formula. We do the sum process with this approach.
We can use the sum formula we write manually or SUM for this. Generally, the writing forms for each method for this different sheets’ sum process are as follows.
Manual sum formula writing:
=‘sheet_name’number_cell_1 + ‘sheet_name’number_cell_2 + …
SUM:
=SUM(‘sheet_name’number_cell_1 , [‘sheet_name’number_cell_2] , …)
When inputting the cell containing the number, you can go to the sheet where the cell is and click the cell. By doing that, the sheet name and the cell coordinate will automatically be inputted into your formula. You don’t have to write them on your own.
Here, we want to sum numbers from the June and July sheets, which contain data like these.
How to do the sum in the “Total” sheet of the example excel file?
First, we type the = symbol in the place where we want to put the sum result as usual.
For manual sum formula writing, we can directly get the number from another sheet to sum it. We do that by going to the sheet where the number is and click the cell containing the number like this.
Do that for all the numbers we want to sum while we type our sum formula. After all the numbers are inputted, press enter. You will immediately get the sum result!
What if we want to use SUM to get the different sheets sum result? The approach is almost the same.
First, we type the SUM until the part where we need to input the numbers.
Then, we go to another sheet and input the numbers to sum by clicking the cell/dragging the cell range. Because all the numbers we want to sum in the example are in a cell range, we input like this.
After we input all the numbers we want to sum in SUM, press enter. You will immediately get the sum result!
We do all that for all the sum results we need in the “Total” sheet. The result will be like this.
Not too hard, isn’t it?
## How to Sum Time in Excel
When processing data in excel, sometimes we have time data and we need to add to them. Maybe we want to add 30 minutes or 2 hours to the time data we have.
If you add your time data with a normal number in excel, then you cannot get the result you want. To add to time data, you need to use time data too.
As an illustration of that, look at the following screenshot.
We see there how we want to add to the time data we have.
We want to add 30 minutes, 2 hours, and 600 seconds to each row of our time data. But, because we use normal numbers to add to them, we don’t get the results we want.
Let’s compare to if we use time data to add to the time data we have.
In the second screenshot, we are able to add to the time data. This is because we sum our time data with time data too. The hours, minutes, and seconds we want to add are rewritten in time forms as you can see above.
So, if you want to add to your time data, don’t forget to change all the numbers into time data first!
## How to Sum in Excel Using 1 Criterion: SUMIF
Have a criterion for the data entries which numbers you want to sum? You can use SUMIF to help you get the sum result easier.
In general, here is the writing form of SUMIF in excel:
=SUMIF(data_range, data_criterion, number_range)
In SUMIF, there are 3 inputs you must give. These are the data column/row cell range you want to evaluate, the criterion, and the numbers column/row cell range.
In the process, SUMIF will evaluate each data in the first cell range you input based on the criterion. If it passes, then SUMIF will sum the number in parallel with the data in the number cell range. With this, SUMIF can sum the numbers which data entries pass the criterion you have.
About the criterion input in SUMIF itself, you must input it correctly so you can get the result you need. You can see the writing examples of a SUMIF criterion and the meaning of them below.
Text (not case-sensitive)
Criterion ExampleExplanation
"Jim"The same as “Jim”
“<>Jim”Not the same as “Jim”
“Jim*”With “Jim” prefix
“*jim”With “jim” suffix
“J*m”“J” prefix and “m” suffix
“Jim?”“Jim” prefix with any one character suffix
“?jim”Any one character prefix with “jim” suffix
“J?m”“J” prefix, any one character, and “m” suffix
“Jim~*”The same as “Jim*”
“Jim~?”The same as “Jim?”
A bit explanation of the symbols we can use for the text criterion:
• * = represents any character with any amount
• ? = represents any one character
• ~ = use this when you want to add * or ? character for the criterion
Number
Criterion ExampleExplanation
70Equal to 70
“>70”More than 70
“<70”Less than 70
“>=70”More than or equal to 70
“<=70”Less than or equal to 70
Date
Criterion ExampleExplanation
“>”&DATE(2019,12,3)Later than 3 December 2019
We don’t recommend inputting a date criterion using direct typing (e.g. “>3-12-2019”). t can produce a wrong and unexpected result.
Cell coordinate
Criterion ExampleExplanation
“>”&B1More than the value in B1
Empty/non-empty cell
Criterion ExampleExplanation
“"Empty
“<>"Not empty
Use the correct criterion writing that fits your sum needs when using a SUMIF!
To make it clearer about this SUMIF, take a look at its implementation example in excel below.
In the example, we want to know the total sales quantity per product category (fruit and vegetable). To get them, we use SUMIF.
As explained before, the SUMIF input begins with the data cell range we want to evaluate based on our criterion. Because, here, the criterion is the product category (vegetable/fruit), we input the cell range where the product category data is. This means the cell range we input is the product category data column.
We input the criterion input based on the product category we want to sum the sales quantity of (fruit or vegetable). And for the number cell range, we input the sales quantity column cell range.
SUMIF will sum all the numbers that are in parallel with the product category we want. Because of that, we get the sum result we want!
## How to Sum in Excel Using More than 1 Criterion: SUMIFS
We can use SUMIF if we only have 1 criterion for our sum process. What if we have more than 1 criterion?
The answer is you need to use the sibling formula of SUMIF in excel, which is SUMIFS.
Generally, here is the SUMIFS writing form in excel.
=SUMIFS(number_range, data_range_1, criterion_1, …)
In SUMIFS, the data cell ranges we want to evaluate based on our criteria and the criteria are made in pairs. The criterion we input after a cell range will only evaluate the data in that cell range.
Because of that, you need to input the cell range and the criteria according to the number of criteria you need. SUMIFS will sum the numbers in parallel with the data that fulfill all the criteria you have. Just one parallel data that doesn’t pass its criterion and SUMIFS won’t involve the number in its sum process.
The criterion writing of SUMIFS itself is the same as the criterion writing of SUMIF which we have explained earlier.
To better understand the SUMIFS implementation in excel, here is the example.
In the example, we have 2 criteria for the sum process, the product category and class. Because of that, we use SUMIFS to help to add the numbers that fulfill both criteria.
We input the sales quantity column we want to sum first, followed by the data columns and their criteria. The data columns and criteria inputs begin with the category column and its category criterion. Then, we input the product class column and its criterion.
Don’t make a mistake when pairing the data cell ranges and the criteria because that will affect your sum result. Don’t forget to input parallel number and data cell ranges that have the same size in SUMIFS. This is so you can predict the result you get much easier.
Give your inputs correctly in the SUMIFS formula you use. You will get the sum result you want as seen in the example above!
## How to Sum in Excel for Filtered Data: SUBTOTAL
What if you have filtered the numbers in your data table and you want to sum those filtered numbers?
Excel provides a special formula to process filtered data like that, which is SUBTOTAL. We can also use it to sum the filtered numbers that we have!
The general writing form of SUBTOTAL to sum filtered numbers is as we can see in the following.
=SUBTOTAL(9/109, filtered_cell_range_1, …)
The 9/109 input in the SUBTOTAL is a code to make SUBTOTAL sum the numbers of your filtered cell range. Input 9 if you want to sum the numbers from the cells you hide manually too (not hidden by the filter process). If you want to ignore the numbers in those manually hidden cells, input 109.
In SUBTOTAL, the filtered cell range input containing the numbers you want to sum can be more than 1. SUBTOTAL will sum all the numbers from the cell ranges you input in it to give its result.
To make it clearer about the use of SUBTOTAL in summing filtered numbers in excel, look at this example.
In the example, we want to get the numbers sum result from the sales quantity column with SUBTOTAL. Because we haven’t filtered the table data in the screenshot above, SUBTOTAL will sum the numbers normally.
However, if we filter the data table based on, for example, the fruit category, then the SUBTOTAL result becomes like this.
Different compared to the first screenshot’s result, isn’t it? Because we use SUBTOTAL, we will get the sum result of only the numbers that pass the filter. In this example, we will get the sum result of the sales quantity from the fruit category only, which is 292.
The 9 or 109 input in SUBTOTAL in this example will give the same result. This is because there is no row that we hide manually in the data table (all the hidden rows are hidden by the excel filter process).
## How to Sum in Excel By Ignoring Errors: SUM IFERROR
Sometimes, in the number sequence we want to sum in excel, there are errors. That sometimes happens if the number sequence is a result of a data processing process.
If we want to use SUM to sum the number sequence, then the result will be an error. This is because SUM cannot add an error in the sum process that it does.
We can clean our data first before we sum it, but the process can take time. This is more so if there is much data we must check and clean one-by-one.
Is there any fast way to sum numbers in a cell range that might have errors in it? One of the ways we can implement is by combining our SUM formula with IFERROR.
Generally, here is the writing form of both formulas combined to sum a cell range with errors in it.
{=SUM(IFERROR(number_error, 0))}
We put the cell range containing numbers and errors in IFERROR before we process it with SUM. By using IFERROR like in the writing above, each error will be replaced by 0. Because of that, SUM can process the cell range that contains some errors.
In this formula combination, we use an array formula form (marked with curly braces outside the SUM and IFERROR). We need this form because we use a cell range input in IFERROR which usually processes individual data.
To change our formula writing form into an array, we press Ctrl + Shift + Enter after we write the formula (not pressing enter like when we write a normal formula).
To better understand this SUM IFERROR implementation, see the following example.
In the example’s stock quantity column, there are two errors among the numbers we want to sum. If we use a normal SUM to get the total stock quantity, we will get an error result.
Because of that, we use IFERROR in the SUM formula writing. By changing each error in the cell range we want to sum with 0, we can get our sum result
## How to Sum Much Faster in Excel: AutoSum
If you want to sum a number sequence in an excel column/row, there is a fast way to do it. The fast way is by utilizing the AutoSum feature from excel. By using AutoSum, you will immediately get a SUM writing containing the number sequence’s cell range input.
To make it clearer about this AutoSum usage in excel, take a look at the following steps.
First, highlight your number sequence or place your cell cursor one cell outside the number sequence you want to sum.
Next, activate the AutoSum feature. How? You can activate it by going to the Home or Formulas tab and click on the AutoSum button there. You can also use a shortcut by pressing the Alt and = (Windows) / Command, Shift, and T (Mac) buttons simultaneously.
Excel will automatically produce a SUM formula that sums your number sequence. The SUM writing will be placed one cell outside your number sequence!
## Exercise
After you learn how to sum in excel completely from the tutorial above, sharpen its implementation understanding by doing this exercise!
### Questions
1. How many is the number of total students from all classes?
2. How many is the number of total students for class A? Use SUMIF to answer!
3. How many is the number of total students for class 2 and 3 with code B? Use SUMIFS to answer!
Make sure the numbers you want to sum aren’t in text forms!
This text number problem sometimes shows up, especially if the numbers are from external sources. The text forms of your numbers will make them not summed when you want to sum them with other numbers!
Excel usually marks a text number with a green triangle on the top left of its cell like this.
What if you want to change a text number into a normal number? You can begin to change it by placing your cell cursor in the cell containing the text number. Usually, there will be a box with an exclamation mark triangle that shows up like this.
You just need to click on the box and choose Convert to Number from the choices that show up.
The text number in the cell will be converted to a normal number!
If there are many text numbers in the cell range you want to sum, use this method to convert them simultaneously. Highlight the cell range, go to the Data tab, and choose Text to Columns.
Then, click Finish in the dialog box that shows up.
Excel will immediately convert the text numbers in the cell range you highlight to normal numbers!
Related tutorials you should learn
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# Probability to pass the exam, if some solutions are known
There are $100$ questions, and $10$ of them are chosen randomly for an exam. For each choice $10$ questions there is the same probability. A student has managed to get $60$ of the $100$ questions and thus know only to solve them.
To pass the exam he needs to solve correctly at least $6$ of the $10$ questions.
What is the probability that the student will pass the exam?
I thinkt it's ${{6+7+8+9+10} \over 60} = {2 \over 3}$. Is that correct?
This problem is about the hypergeometric distribution. This distribution is used when you have a population of $N$ objects containing $K$ "successes", and you draw $n$ of the objects without replacement. Then the number of successes in your draw is a random variable $X$. It turns out that
$$P(X=k)=\frac{{K \choose k} {N-K \choose n-k}}{ {N \choose n}}$$
where the parentheses denote a binomial coefficient. The numerator is the number of ways to choose $k$ successes out of the total of $K$ successes times the number of ways to choose $n-k$ failures out of the total of $N-K$ failures. The denominator is the total number of ways to choose $n$ things from the whole population.
You then want $P(X \geq 6)$ which is $P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)$.
As an aside, when $k$ is much less than $K$ and $n-k$ is much less than $N-K$, the distribution of $X$ can be approximated by Binomial(n,K/N), that is, $P(X=k) \approx {n \choose k} (K/N)^k (1-K/N)^{n-k}$. This is exactly what you would get if the draws were done with replacement. In this case this approximation is pretty good.
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# Introduction to Four Addends
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## Objective
SWBAT add four two-digit numbers to find a sum.
#### Big Idea
In this lesson, students identify strategies to add four two-digit numbers together.
## Introduction/Hook
10 minutes
I put the problem of the day on the board.
James has 23 orange skittles, 39 red skittles, 26 green skittles, and 19 yellow skittles. How many skittles does he have in all?
Turn and talk: What operation should we use to solve this problem?
Students will likely indicate that they should add all four together because we want to figure out how many total skittles there are. If students are struggling to determine how to set up the problem, I give them paper or a white board to draw out the problem. Drawing the problem enables them to visualize the problem more clearly.
After students have discussed which operation to use for about one minute, I bring the class back together and ask one or two students to share which operation they chose and WHY. I use these guiding questions to prompt student discussion (1) How do you know you should add? (2) What clues does the story problem give us to tell us we should add?
Turn and Talk: Now, I want you to tell me how you think we should solve this problem? What strategies should we use?
If students simply say “we need to add the four together” push them to explain HOW they will add the four together.
After students share out their ideas have students work in groups to solve this problem—give students white boards, white board markers, cubes and place value blocks to help them solve the problem as necessary.
Some students may add all four numbers up, others may draw tens or ones, others may add two of the addends together, then add the next two, and then add the two sums. As students work, ask them guiding questions: Why did you choose this strategy? How do you know that this strategy will work?
## Introduction to New Material
10 minutes
After students have worked to solve the problem, I have two groups come forward to model how they solved it. Students may share that they used 10s and 1s to solve the problem or that they stacked all four and then regrouped. Some students may model how they used cubes to add all four together. Other students may have used an open number line to efficiently add the four numbers.
As students share their strategies, record them on an anchor chart or on the board. After students have finished sharing, post the following image on the board.
Today I am going to do a strategy spotlight where I model a strategy that I saw students working on during our problem of the day.
When my friend had to add four numbers together they started by adding two numbers together using a regrouping strategy (model adding the first two together in one of the upper circles), Then, he/she added the two other numbers together using the regrouping strategy (model adding the first two together in the other upper circle). Then he/she needed to add the two sums together. He/she added the two sums together in the lower circle.
I chose to spotlight this strategy because several of my students chose to add all four numbers together using column addition. When they added all four numbers using column addition, I noticed their accuracy was off. This strategy allows students to continue to practice their regrouping skills if they are comfortable. If one of your students does not come up with this specific strategy during the introduction to new material, you can either pick a different student strategy to spotlight or use this strategy--don't label it as "your" strategy, though, since students will likely mimic it and not take the time to develop a conceptual understanding of this skill. Read more about strategy spotlights in the linked reflection.
## Guided Practice
10 minutes
Now that we have shared our strategies as a group, you are going to work by yourself to solve a problem similar to our whole class problem. You can use any strategy that will help you get an accurate answer.
I distribute the guided practice problem as well as cubes and place value blocks for students who want to use them. As students work, I circulate and ask guiding questions:
1)What strategy are you using?
2) What steps are you taking to make sure your work is accurate?
3) What number sentence matches the story problem?
After students have had time to complete the problem, I have all of the students come back together and have students share their strategy, why they chose that strategy, and how they solved the problem. As students present, I ask them: Why did you choose that strategy? Show the class how this strategy works..., Explain how you check your work in your strategy... If students used place value blocks or cubes to solve, I have them model using these manipulatives so that their teammates can clearly see how their strategy works.
When finished, I ask students to turn and talk: How are the strategies shared similar or different from the strategy that you worked on?
This activity allows students to critique the reasoning of others (MP3) and learn new strategies for solving multiple addend problems.
## Independent Practice
10 minutes
Independent Practice is differentiated by mathematical understanding of this skill"
Group A: In need of intervention
This group will work on adding four addends using numbers 10-30 that will not require them to regroup into the hundreds. In this group, students should be focusing on using tens and ones or only adding two addends at once in order to get an accurate answer. Place value blocks and cubes will be available to this group.
Group B: Right on Track!
This group will work on adding four addends using numbers 20-60. Place value blocks and cubes will be available to this group.
NOTE: On the the numbers for group A and B are left blank so that you can fill in appropriate numbers for your groups. The numbers I've provided are suggestions based on what worked with my groups.
Group C: Extension
This group will work on adding FIVE addends using numbers 20-100. Place value blocks and cubes will be available to this group, but many students will be able to solve these problems more abstractly by regrouping or using an open number line.
## Exit Ticket
5 minutes
Now that you have worked on these kind of problems in groups and alone, it's time for you to show me what you know!
Hand out the exit ticket and allow students to take it quietly with our without manipulatives.
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# day5 - PADP 6950 Founda1ons of Policy Analysis...
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Unformatted text preview: PADP 6950: Founda1ons of Policy Analysis Intertemporal Choice & Uncertainty Angela Fer1g, PhD Special cases: Different kinds of goods to choose between Before: food and other goods, etc. Intertemporal Choice: cc now vs. cc future Uncertainty: cc if state 1 (e.g. sick) vs. cc if state 2 (e.g. healthy) 1 Intertemporal Budget Constraint c 2 = m2 + (m1 - c1 ) + r(m1 - c1 ) c 2 = m2 + (1 + r)(m1 - c1 ) c 2 + (1 + r)c1 = m2 + (1 + r)m1 c2 m2 c1 + = m1 + (1 + r) (1 + r) p1 = 1 Budget constraint in present value terms: p2 = 1 (1 + r) Intertemporal Budget Line c1 + c2 m2 = m1 + (1 + r) (1 + r) Y-intercept (c2): assume c1=0 c 2 = (1 + r)m1 + m2 c1 = m1 + m2 (1 + r) X-intercept (c1): assume c2=0 Slope: transform to form y=b+mx c 2 = (1 + r)m1 + m2 - (1 + r)c1 2 Graph How does line change when interest rate changes? If r goes from 0.05 to 0.1: slope=-(1+r) goes from -1.05 to -1.1 (steeper) c2-intercept shiYs up c1-intercept shiYs back Must pivot around endowment point. 3 Intertemporal Indifference Curve If straight line (perfect subs1tutes): consumer doesn't care whether they consume today or tomorrow If L-shaped (perfect complements): consumer wants equal amounts today and tomorrow More realis1c: convex, well-behaved indifference curve What happens to a lender when r goes up? 4 What happens to a borrower when r goes up? Discount rate Discount rate () = the rate at which we discount the value of the future We usually just let =r (interest rate) A dollar today could be invested in an interest-bearing instrument (bond) and then it would be worth \$1*(1+r) in the next period So, \$1 in the future is worth \$1/(1+r) today However, some people are more impa1ent and have a higher discount rate than r and some are very pa1ent and for them, <r Discount rates are a very important determinant of people's intertemporal choices: Risky behavior decisions Investment decisions Some say just as important as IQ in determining child's academic success 5 Net Present Value (or Discoun1ng) General formula: NPV = t =T Vt t t =1 (1 + r) Example 1: What is \$100 given to me in 1 year worth to me today (assume r=0.05)? \$100 NPV = = \$95.24 1.05 Example 2: What is \$300 given to me in 3 installments over the next 3 years worth to me today? NPV = \$100 \$100 \$100 + + = \$95.24 + \$90.70 + \$86.38 = \$272.32 1.05 1.05 2 1.05 3 Policy Applica1on: Benefit-Cost Analysis Discoun1ng is used for comparisons of different streams of benefits and/or costs over a number of periods/years. Assume discount rate=20%. Example: Investment A: returns \$20 at end of Year 1 and \$20 at end of Year 2 Investment B: returns \$12 at end of Year 1 and \$29 at the end of Year 2 A provides a total of \$40 and B provides a total \$41 so should we choose B? 6 Choice under uncertainty When deciding how much insurance to buy, really deciding between consump1on in 2 states: no loss/no illness vs. loss/illness If no insurance available, then there is no choice: (mg,mb) If insurance becomes available, then you can make a choice about how much to get. Your budget line will depend on the premium charged. Assume insurance coverage K costs K (premium): If have loss, have: If no loss, have: mb + K - K = mb + (1- )K m g - K Choice under uncertainty noins ins m g - m g + K rise c g - c g - slope = = noins = = ins run c b - c b mb - mb - (1- )K (1- ) Assume indifference curve is convex as usual. If endowment isn't at the tangent point, then buy insurance. 7 U1lity under uncertainty Another factor that must be considered (besides the premium) is: What is the probability that the bad event will occur? It is taken into considera1on through an expected u#lity func#on: U = i u(c i ) i=1 i=N Expected U1lity vs. Value If just 2 possible states: U=u(c1)+(1-)u(c2) Note this is different from expected value: EV=c1+(1-)c2 8 Risk aversion We assume that the u1lity func1on with respect to wealth/money for most is upward sloping and concave: Happier if richer, but not twice as happy if twice as rich Diminishing marginal u1lity of wealth Implica1on of this concavity: having money with certainty makes you happier than having it with some uncertainty people don't like risk Example to demonstrate the connec1on between concavity of U & risk aversion W=\$20; If sick (5% chance), then W=\$10, U=ln(W) Expected value: E(W ) = (1- )W well + W sick = 0.95 * \$20 + 0.05 * \$10 = \$19.50 Expected u#lity without risk: U(E(W )) = U(\$19.50) = ln(19.50) = 2.97 Point on the U(wealth) curve 9 Example con1nued In reality, you don't get \$19.50, you get \$10 with a 5% probability and \$20 with a 95% probability. Expected u#lity with risk: E(U) = (1- )U(W well ) + U(W sick ) = 0.95 * ln(\$20) + 0.05 * ln(\$10) = 2.965 Point below the U(wealth) curve 10 Risk premium Risk premium is the amount people are willing to pay to avoid risk: Risk premium = E(W) w/ risk that gives U1 - W w/ certainty that gives U1 = E(W)-WE(U) In this example, E(W)=\$19.50 and WE(U)=eln(W)=e2.965=\$19.39, so risk premium=11 cents 11 ...
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# What is the mean of four consecutive even integers a , b , c
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What is the mean of four consecutive even integers $$a$$ , $$b$$ , $$c$$ , $$d$$ ?
1. $$a + d = b + c$$
2. $$b + c = d - a$$
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25 Oct 2009, 11:56
Is there something missing in the Q? I didn't understand the Q
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25 Oct 2009, 12:13
Nopes, question is correct
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25 Oct 2009, 19:36
Is C the OA?
Since they are 4 even consecutive integers , b, c, d can be represented as a+2, a+4, a+6 resp
So mean is a+a+2+a+4+a+6 divided by 4 => a+3 ------> eqn 1
Now 1)
a+d = b+c
which means Mean = a+b+c+d divided by 4 => b+c divided by 2
We know nothing about a or any number. So insuff
Now from 2)
b+c = d-a
We get Mean = d/2 which yields nothing - so insuff
Together
a+d = b+c and
b+c = d-a
equating the two above we get
a+d = d-a or
2a = 0
a =0
Hence we get a
From eqn ---->1 we know Mean = a+3
Hence Mean = 0+3 = 3
Sufficient by both 1 and 2
Correct me if I am wrong. Would appreciate it
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25 Oct 2009, 19:47
IMO B
What is the mean of four consecutive even integers a , b , c , d ?
1. a + d = b + c
2. b + c = d - a
lets assume the numbers as a, a+2, a+4, a+6 since they are four consecutive integers
S1 ) given that a+d = b+c
2a +6 = 2a + 6 - no new information is given that is different from question stem - insufficient
S2) given that b+c = d-a
==>
a+2+a+4 = a+6-a
2a+6 =6 ==> a = 0, so the numbers are 0, 2,4,6 and we can calculate mean - sufficient.Hence
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Thanks, Sri
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25 Oct 2009, 19:52
srini123 wrote:
IMO B
What is the mean of four consecutive even integers a , b , c , d ?
1. a + d = b + c
2. b + c = d - a
lets assume the numbers as a, a+2, a+4, a+6 since they are four consecutive integers
S1 ) given that a+d = b+c
2a +6 = 2a + 6 - no new information is given that is different from question stem - insufficient
S2) given that b+c = d-a
==>
a+2+a+4 = a+6-a
2a+6 =6 ==> a = 0, so the numbers are 0, 2,4,6 and we can calculate mean - sufficient.Hence
Yeah I think this is the answer. I did not think about it though. Nice catch Sri
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26 Oct 2009, 02:33
Question: The stem does not specify the order of elements. it can d<a<b<c or a<b<c<d...does the solution hold true in both the cases? I am lazy to calculate today
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26 Oct 2009, 09:23
Order not known..imo c..
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26 Oct 2009, 10:48
What is the mean of four consecutive even integers a , b , c , d ?
1. a + d = b + c
2. b + c = d - a
If we don't know the order of ab,c,d i think the answer would be E
from S1 and S2 all we know is a=0
if the order was d,c,b,a then numbers are -6,-4,-2,0 and mean is -4
if the order was a,b,c,d then numbers are 0,2,4,6 and mean is 4
still both together are not sufficient. hence E
is this right ?
what is OA ?
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Re: even consecutive [#permalink] 26 Oct 2009, 10:48
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# Assumptions
## Drone assumptions
1) The maximum weight of a package will be around 5lbs / 2.5 kg, which is a little less than 10% of its own weight. This is based on the citation of Amazon, stating that the delivery drones will be able to carry up to 5 pounds. Amazon also states that the majorty of sold articles weigh less than five pounds. [1]
2) The maximum distance a delivery drone can cover is assumed to be 32 km on full battery, resulting in a flying radius of 16 km since it still has to be able to return to its deployment point. This assumption is based on research in scientific articles in which most articles proposes maximum distances between 16 – 23 km (10-15 miles) on full battery. Since 16 km appeared most the times and it is the minimum distance it can cover and therefor has the smallest chance of creating false information, we chose 16 km as our maximum radius. [2][3][4]
3) The maximum speed a fully loaded delivery drone can reach is up to 80 km/h. This is based on the maximum speed a delivery drone from amazon can reach within its battery capacity. However, to be able to use the speed within the research, it was chosen to take 60 km/h as delivery speed, since it can’t fly at maximum speed on average. [5][6]
4) The charging time of the battery is expected to be about 30 min. This charging time is used for wireless charging pads and for indoor charging stations, which are seen as one in the project. This pad deliver about 20Ah. With a battery use of 8000 - 9000 mAh this results into half an hour of charging time. [7][8]
5) It is possible for the drones to use recharge pads placed on frequent urban objects like lanterns or chimes of houses [9]
6) Each drone will cost about €3500 ($4000). According to our research an average cost for an autonomous delivery drone will be around$3000 - $5000. Taking it half way ends up at$4000, equal to €3500. [10][11]
7) It is assumed that a drone takes about 30 seconds to descend to ground level and another 30 seconds to fly up into the permitted flying height.
8) It is assumed that a drone takes about 90 seconds to enter a distribution centre, pick up a parcel and descends again towards its location.
9) Although there are areas drones are not allowed to fly (no-flight-zones), it is assumed such area's are not encountered.
10) Drones will be able to work 24 hours a day, only staying idle to recharge or to wait for an order.
## Delivery assumptions
1) To get an understanding of how many deliveries are done, an estimation has to be made on the number of packages delivered a day. This number is assumed to be around 172.000 packages a day. The amount of packages a company like Bol.com processes is approximately 200.000 orders a day [12]. However, this includes packages of all sizes. According to Amazon, 86% of the orders are packages under 2.5 kg [13].
2) For the service of parcel delivery by drone, the estimated costs are approximately €1.00 per delivery [14]
3) In the distribution centre, the entire process of organizing the parcels from the shelves to the drone will be automated, so no employees are needed. This can be applied for battery changing as well.
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What is the value of the product of x and y?
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What is the value of the product of x and y? [#permalink]
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What is the value of the product of x and y?
(1) x + y = 0
(2) The average of x^2 and y^2 is equal to the smallest positive integer multiple of two.
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Re: What is the value of the product of x and y? [#permalink]
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17 Sep 2017, 04:29
1
1
What is the value of the product of x and y? => xy ?
(1) x + y = 0
=> y=-x
=> $$xy = x * -x = -x^2$$
Insufficient
(2) The average of x^2 and y^2 is equal to the smallest positive integer multiple of two.
=> $$\frac{x^2+y^2}{2} = 2$$ ( smallest +ve integer multiple of 2 = 2)
=> $$x^2+y^2 =4$$
Insufficient
(1)+(2)
$$x+y =0$$ and $$x^2+y^2 = 4$$
=> $$x^2+y^2 =4$$
=> $$x^2+y^2+2xy =4 +2xy$$
=> $$(x+y)^2=4 +2xy$$
=> $$0 =4+2xy$$
=> xy = -2
Re: What is the value of the product of x and y? &nbs [#permalink] 17 Sep 2017, 04:29
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# A radioactive sample has a count rate of 800 counts per minute. One hour later, the count rate has fallen to 100 counts per minute. What is the half-life of the sample?
Radioactivity equation states: N = No e^(-0.693t/t1/2) --- Where t = time, t1/2 = half-life, N = count after time t, No = Initial count Using the values given; 100 = 800 e^(-0.693*1/t1/2) 100/800 = e^(-0.693t/t1/2) 0.125 = e^(-0.693/t1/2) Taking natural logs on both sides; log 0.125 = -0.693/t1/2 * log (e) -0.9031 = -0.693/t1/2 * 0.4323 t1/2 = (-0.693*0.4323)/(-0.9031) = 0.3317 hours Therefore, half-life of the sample is approximately 0.3317 hours
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1. ## Trigonometric Identities
Prove the identity tan 2x - tan x = tan x sec 2x
Thank you for your time and help
2. Originally Posted by BeckyDWood
Prove the identity tan 2x - tan x = tan x sec 2x
Thank you for your time and help
$\displaystyle \tan 2x -\tan x = \tan x \sec 2x$
take the left side
$\displaystyle \tan 2x - \tan x = \frac{\sin 2x }{\cos 2x} - \frac{\sin x }{\cos x}$
$\displaystyle \tan 2x - \tan x = \frac{\sin 2x \cos x - \sin x \cos 2x }{\cos 2x \cos x}$
$\displaystyle \tan 2x - \tan x = \frac{2 \sin x \cos ^2 x - \sin x \cos 2x }{\cos x \cos 2x }$
$\displaystyle \tan 2x - \tan x = \frac{\sin x( 2\cos ^2 x - \cos 2x }{\cos 2x \cos x}$
$\displaystyle \tan 2x - \tan x = \frac{\sin x }{\cos 2x \cos x } =\tan x \sec 2x$
3. Originally Posted by Amer
$\displaystyle \tan 2x - \tan x = \frac{\sin 2x \cos x - \sin x \cos 2x }{\cos 2x \cos x}$
note that $\displaystyle \sin 2x \cos x - \sin x \cos 2x=\sin(2x-x)=\sin x.$
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# Problem 607
Fiona kept the following records of her phone bills for 12 months:
50, 54, 57, 48, 71, 70
Find the mean and median of Fiona’s monthly phone bills.
Solution:-
First arrange the values in ascending order which is given bellow
50, 55,57,64,70,71
Mean=sum of terms /number of terms
Sum of terms=48+50+54+55+55+57+59+64+66+70+71+17=720
Number of terms=12
So mean=720/12=57
7 th term is 58
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# With total sales of less than three hundred thousand dollars
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7. With total sales of less than three hundred thousand dollars and fewer new subscribers than last year, the New England Theater Company is in danger of losing its building.
(A) of less than three hundred thousand dollars and fewer
(B) lower than three hundred thousand dollars and less
(C) lesser than three hundred thousand dollars and fewer
(D) fewer than three hundred thousand dollars and less
(E) of fewer than three hundred thousand dollars and of fewer
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04 Feb 2005, 15:20
Isn't dollar amount a countable number ? Then why use "less" ?
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04 Feb 2005, 22:31
Eliminate B D for "less subscriber", E for "sale ... of fewer subscriber", C for "lesser".
(A)
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05 Feb 2005, 00:05
OA is "A", but noone answered my ques "if less is appropriate when using count nouns" e.g. 200000 thousand ??
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05 Feb 2005, 21:10
Well we always say five is less than eight, not five is fewer than eight, don't we?
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05 Feb 2005, 21:10
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## Class 10 Maths Arithmetic Progression Notes by Toppers – Download PDF
Are you a Class 10 student looking for comprehensive arithmetic progression notes? Well, you’re in luck! In this article, we will provide you with expertly crafted Class 10 Maths Arithmetic Progression Notes prepared by top-performing students. Whether you’re struggling with understanding the concept of arithmetic progression or simply want to enhance your knowledge, these notes will serve as an invaluable resource. You can also conveniently download a PDF version of the notes for offline access. Let’s dive in!
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Class 10 Maths Arithmetic Progression – Get here the Handwritten Notes for Class 10 Arithmetic Progression. Candidates who are ambitious to qualify the Class 10 with a good score can check this article for Note. Below we provided the link to access the Notes of Class 10 Maths for the topic Arithmetic Progression. You can practice the questions and check your answers from the solutions given after the question. By practicing these resources candidates definitely get the idea of which his/her weak areas and how to prepare well for the examination.
• Class: 10th
• Subject: Math
• Topic: Arithmetic Progression
• Resource: Handwritten Notes
Maths Handwritten Notes is based on the new(reduced) syllabus by CBSE.
# CBSE Class 10 Maths Arithmetic Progression Notes
Particularly when it comes to the subject of Mathematics, students desire to have an answer key to help them in evaluating their learning and development. Refer to these solutions when practicing and solving the Mathematics exercises from NCERT Textbooks.
## 1. Introduction to Arithmetic Progression
Arithmetic Progression (AP) is a fundamental concept in mathematics that deals with a sequence of numbers in which the difference between consecutive terms remains constant. In simpler terms, it’s a sequence where each term is obtained by adding a fixed value (known as the common difference) to the previous term. AP finds applications in various fields such as physics, computer science, and economics.
## 2. Understanding the Common Difference
The common difference is a crucial element in arithmetic progression. It determines the value added or subtracted to each term to obtain the next term in the sequence. Understanding the common difference is essential for identifying the pattern and predicting subsequent terms in an AP.
## 3. Finding the nth Term of an Arithmetic Progression
To find the nth term of an arithmetic progression, we use the formula:
Tn = a + (n – 1)d
Here, Tn represents the nth term, ‘a’ is the first term, ‘n’ is the term number, and ‘d’ is the common difference. By substituting the values of ‘a,’ ‘n,’ and ‘d’ into the formula, we can easily determine any term in the AP.
## 4. Sum of an Arithmetic Progression
The sum of an arithmetic progression can be calculated using the formula:
Sn = (n/2) * [2a + (n – 1)d]
In this formula, ‘Sn’ represents the sum of the first ‘n’ terms, ‘a’ is the first term, ‘n’ is the number of terms, and ‘d’ is the common difference. By plugging in the appropriate values, we can find the sum of any given AP.
## 5. Arithmetic Mean of an Arithmetic Progression
The arithmetic mean, also known as the average, is an important concept associated with arithmetic progression. It can be calculated using the formula:
A = (T1 + Tn) / 2
Here, ‘A’ represents the arithmetic mean, ‘T1’ is the first term, and ‘Tn’ is the nth term of the AP. The arithmetic mean provides insights into the overall trend of the sequence.
## 6. Finding the Number of Terms in an Arithmetic Progression
To determine the number of terms in an arithmetic progression, we use the formula:
n = (Tn – a) / d + 1
In this formula, ‘n’ represents the number of terms, ‘Tn’ is the nth term, ‘a’ is the first term, and ‘d’ is the common difference. By rearranging the formula, we can find the value of ‘n’ when provided with other values.
## 7. Applications of Arithmetic Progression in Real Life
Arithmetic progression has numerous applications in real-life scenarios. It is used in financial calculations, such as calculating compound interest and annuities. AP is also employed in analyzing population growth, determining patterns in sports statistics, and optimizing algorithms in computer science.
## 8. Solving Problems using Arithmetic Progression
Arithmetic progression is a powerful tool for solving mathematical problems. By understanding the concepts and formulas associated with AP, you can solve a wide range of problems, including finding missing terms, determining the sum of a sequence, and predicting future terms.
## 9. Tips and Tricks for Solving Arithmetic Progression Questions
To excel in arithmetic progression problems, consider the following tips:
• Understand the given problem statement thoroughly.
• Identify the common difference and first term accurately.
• Utilize the formulas for nth term, sum, and arithmetic mean efficiently.
• Break down complex problems into simpler steps.
• Practice regularly to enhance your problem-solving skills.
## 10. Common Mistakes to Avoid in Arithmetic Progression
While dealing with arithmetic progression, students often make some common errors. These include:
• Misinterpreting the common difference.
• Incorrectly identifying the first term.
• Misusing the formulas for nth term and sum.
• Failing to recognize the patterns in the sequence.
• Rounding off intermediate calculations prematurely.
## 11. Key Formulas for Arithmetic Progression
To succeed in AP-related questions, memorize the following key formulas:
• nth term: Tn = a + (n – 1)d
• Sum of first ‘n’ terms: Sn = (n/2) * [2a + (n – 1)d]
• Arithmetic mean: A = (T1 + Tn) / 2
• Number of terms: n = (Tn – a) / d + 1
## 12. Practice Exercises for Arithmetic Progression
To strengthen your understanding of arithmetic progression, attempt the following practice exercises:
1. Find the 15th term of an AP with a common difference of 4 and a first term of 7.
2. Calculate the sum of the first 20 terms of an AP with a common difference of 3 and a first term of 2.
3. Determine the arithmetic mean of an AP with a first term of 10 and a common difference of 6.
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## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Chapter Test
These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Chapter Test
More Exercises
Question 1.
A game consists of spinning an arrow which comes to rest at one of the regions 1, 2 or 3 (shown in the given figure). Are the outcomes 1, 2 and 3 equally likely to occur? Give reasons.
Solution:
In a game,
No, the outcomes are not equally likely.
Outcome 3 is more likely to occur than the outcomes of 1 and 2.
Question 2.
In a single throw of a die, find the probability of getting
(i) a number greater than 5
(ii) an odd prime number
(iii) a number which is multiple of 3 or 4.
Solution:
In a single throw of a die
Number of total outcomes = 6 (1, 2, 3, 4, 5, 6)
(i) Numbers greater than 5 = 6 i.e., one number
Probability = $$\\ \frac { 1 }{ 6 }$$
(ii) An odd prime number 2 i.e., one number
Probability = $$\\ \frac { 1 }{ 6 }$$
(iii) A number which is a multiple of 3 or 4 which are 3, 6, 4 = 3 numbers
Probability = $$\\ \frac { 3 }{ 6 }$$ = $$\\ \frac { 1 }{ 2 }$$
Question 3.
A lot consists of 144 ball pens of which 20 are defective and the others are good. Rohana will buy a pen if it is good, but will not buy it if it is defective. The shopkeeper draws one pen at random and gives it to her. What is the probability that :
(ii) She will not buy it?
Solution:
In a lot, there are 144 ball pens in which defective ball pens are = 20
and good ball pens are = 144 – 20 = 124
Rohana buys a pen which is good only.
(i) Now the number of possible outcomes = 144
and the number of favourable outcomes = 124
Question 4.
A lot consists of 48 mobile phones of which 42 are good, 3 have only minor defects and 3 have major defects. Varnika will buy a phone if it is good but the trader will only buy a mobile if it has no major defect. One phone is selected at random from the lot. What is the probability that it is
(i) acceptable to Varnika?
Solution:
Number of total mobiles = 48
Number of good mobiles = 42
Number having minor defect = 3
Number having major defect = 3
(i) Acceptable to Varnika = 42
Probability = $$\\ \frac { 42 }{ 48 }$$ = $$\\ \frac { 7 }{ 8 }$$
(ii) Acceptable to trader = 42 + 3 = 45
Probability = $$\\ \frac { 45 }{ 48 }$$ = $$\\ \frac { 15 }{ 16 }$$
Question 5.
A bag contains 6 red, 5 black and 4 white balls. A ball is drawn from the bag at random. Find the probability that the ball drawn is
(i) white
(ii) red
(iii) not black
(iv) red or white.
Solution:
Total number of balls = 6 + 5 + 4 = 15
Number of red balls = 6
Number of black balls = 5
Number of white balls = 4
(i) Probability of a white ball will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 4 }{ 15 }$$
(ii) Probability of red ball will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 6 }{ 15 }$$ = $$\\ \frac { 2 }{ 5 }$$
(iii) Probability of not black ball will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 15-5 }{ 15 }$$
= $$\\ \frac { 10 }{ 15 }$$
= $$\\ \frac { 2 }{ 3 }$$
(iv) Probability of red or white ball will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 6+4 }{ 15 }$$
= $$\\ \frac { 10 }{ 15 }$$
= $$\\ \frac { 2 }{ 3 }$$
Question 6.
A bag contains 5 red, 8 white and 7 black balls. A ball is drawn from the bag at random. Find the probability that the drawn ball is:
(i) red or white
(ii) not black
(iii) neither white nor black
Solution:
Total number of balls in a bag = 5 + 8 + 7 = 20
(i) Number of red or white balls = 5 + 8 = 13
Probability of red or white ball will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 13 }{ 20 }$$
(ii) Number of ball which are not black = 20 – 7 = 13
Probability of not black ball will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 13 }{ 20 }$$
(iii) Number of ball which are neither white nor black
= Number of ball which are only red = 5
Probability of neither white nor black ball will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$
= $$\\ \frac { 5 }{ 20 }$$
= $$\\ \frac { 1 }{ 4 }$$
Question 7.
A bag contains 5 white balls, 7 red balls, 4 black balls and 2 blue balls. One ball is drawn at random from the bag. What is the probability that the ball drawn is :
(i) white or blue
(ii) red or black
(iii) not white
(iv) neither white nor black ?
Solution:
Number of total balls = 5 + 7 + 4 + 2 = 18
Number of white balls = 5
number of red balls = 7
number of black balls = 4
and number of blue balls = 2.
(i) Number of white and blue balls = 5 + 2 = 7
Probability of white or blue balls will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 7 }{ 18 }$$
(ii) Number of red and black balls = 7 + 4 = 11
Probability of red or black balls will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 11 }{ 18 }$$
(iii) Number of ball which are not white = 7 + 4 + 2 = 13
Probability of not white balls will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 13 }{ 18 }$$
(iv) Number of balls which are neither white nor black = 18 – (5 + 4) = 18 – 9 = 9
Probability of ball which is neither white nor black will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 9 }{ 18 }$$ = $$\\ \frac { 1 }{ 2 }$$
Question 8.
A box contains 20 balls bearing numbers 1, 2, 3, 4,……, 20. A ball is drawn at random from the box. What is the probability that the number on the ball is
(i) an odd number
(ii) divisible by 2 or 3
(iii) prime number
(iv) not divisible by 10?
Solution:
In a box, there are 20 balls containing 1 to 20 number
Number of possible outcomes = 20
(i) Numbers which are odd will be,
1, 3, 5, 7, 9, 11, 13, 15, 17, 19 = 10 balls.
Probability of odd ball will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 10 }{ 20 }$$ = $$\\ \frac { 1 }{ 2 }$$
(ii) Numbers which are divisible by 2 or 3 will be
2, 3, 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, 20 = 13 balls
Probability of ball which is divisible by 2 or 3 will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 13 }{ 20 }$$
(iii) Prime numbers will be 2, 3, 5, 7, 11, 13, 17, 19 = 8
Probability of prime number will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 8 }{ 20 }$$ = $$\\ \frac { 2 }{ 5 }$$
(iv) Numbers not divisible by 10 will be
1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19 = 18
Probability of prime number will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 18 }{ 20 }$$ = $$\\ \frac { 9 }{ 10 }$$
Question 9.
Find the probability that a number selected at random from the numbers 1, 2, 3,……35 is a
(i) prime number
(ii) multiple of 7
(iii) multiple of 3 or 5.
Solution:
Numbers are 1, 2, 3, 4, 5,…..30, 31, 32, 33, 34, 35
Total = 35
(i) Prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31
which are 11
Probability of prime number will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 11 }{ 35 }$$
(ii) Multiple of 7 are 7, 14, 21, 28, 35 which are 5
Probability of multiple of 7 will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 5 }{ 35 }$$ = $$\\ \frac { 1 }{ 7 }$$
(iii) Multiple of 3 or 5 are 3, 5, 6, 9, 10, 12 ,15, 18, 20, 21, 24, 25, 27, 30, 33, 35.
Which are 16 in numbers
Probability of multiple of 3 or 5 will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 16 }{ 35 }$$
Question 10.
Cards marked with numbers 13, 14, 15,…..60 are placed in a box and mixed thoroughly. One card is drawn at random from the box. Find the probability that the number on the card is
(i) divisible by 5
(ii) a number which is a perfect square.
Solution:
Number of cards which are marked with numbers
13, 14, 15, 16, 17,….to 59, 60 are = 48
(i) Numbers which are divisible by 5 will be
15, 20, 25, 30, 35, 40, 45, 50, 55, 60 = 10
Probability of number divisible by 5 will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 10 }{ 48 }$$ = $$\\ \frac { 5 }{ 24 }$$
(ii) Numbers which is a perfect square are 16, 25, 36, 49 which are 4 in numbers.
Probability of number which is a perfect square will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 4 }{ 48 }$$ = $$\\ \frac { 1 }{ 12 }$$
Question 11.
The box has cards numbered 14 to 99. Cards are mixed thoroughly and a card is drawn at random from the box. Find the probability that the card drawn from the box has
(i) an odd number
(ii) a perfect square number.
Solution:
Cards in a box are from 14 to 99 = 86
No. of total cards = 86
One card is drawn at random
Cards bearing odd numbers are 15, 17, 19, 21, …, 97, 99
Which are 43
(i) P(E) = $$\frac { Number\quad of\quad actual\quad events }{ Number\quad of\quad total\quad events }$$
= $$\\ \frac { 43 }{ 86 }$$
= $$\\ \frac { 1 }{ 2 }$$
(ii) Cards bearing number which are a perfect square
= 16, 25, 36, 49, 64, 81
Which are 6
P(E) = $$\frac { Number\quad of\quad actual\quad events }{ Number\quad of\quad total\quad events }$$
= $$\\ \frac { 6 }{ 86 }$$
= $$\\ \frac { 3 }{ 43 }$$
Question 12.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is four times that of a red ball, find the number of balls in the bags.
Solution:
Number of red balls = 5
and let number of blue balls = x
Total balls in the bag = 5 + x
and that of red balls = $$\\ \frac { 5 }{ 5+x }$$
According to the condition,
$$\frac { x }{ 5+x } =4\times \frac { 5 }{ 5+x } =>\frac { x }{ 5+x } =\frac { 20 }{ 5+x }$$
x ≠ – 5
x = 20
Hence, number of blue balls = 20
and number of balls in the bag = 20 + 5 = 25
Question 13.
A bag contains 18 balls out of which x balls are white.
(i) If one ball is drawn at random from the bag, what is the probability that it is white ball?
(ii) If 2 more white balls are put in the bag, the probability of drawing a white ball will be $$\\ \frac { 9 }{ 8 }$$ times that of probability of white ball coming in part (i). Find the value of x.
Solution:
Total numbers of balls in a bag = 18
No. of white balls = x
(i) One ball is drawn a random
Probability of being a white ball = $$\\ \frac { x }{ 18 }$$
(ii) If 2 more white balls an put, then number of white balls = x + 2
and probability is $$\\ \frac { 9 }{ 8 }$$ times
Question 14.
A card is drawn from a well-shuffled pack of 52 cards. Find the probability that the card drawn is :
(i) a red face card
(ii) neither a club nor a spade
(iii) neither an ace nor a king of red colour
(iv) neither a red card nor a queen
(v) neither a red card nor a black king.
Solution:
Number of cards in a pack of well-shuffled cards = 52
(i) Number of a red face card = 3 + 3 = 6
Probability of red face card will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 6 }{ 52 }$$ = $$\\ \frac { 3 }{ 26 }$$
(ii) Number of cards which is neither a club nor a spade = 52 – 26 = 26
Probability of card which’ is neither a club nor a spade will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 26 }{ 52 }$$ = $$\\ \frac { 1 }{ 2 }$$
(iii) Number of cards which is neither an ace nor a king of red colour
= 52 – (4 + 2) = 52 – 6 = 46
Probability of card which is neither ace nor a king of red colour will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 46 }{ 52 }$$ = $$\\ \frac { 23 }{ 26 }$$
(iv) Number of cards which are neither a red card nor a queen are
= 52 – (26 + 2) = 52 – 28 = 24
Probability of card which is neither red nor a queen will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 24 }{ 52 }$$ = $$\\ \frac { 6 }{ 13 }$$
(v) Number of cards which are neither red card nor a black king
= 52 – (26 + 2) = 52 – 28 = 24
Probability of cards which is neither red nor a black king will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 24 }{ 52 }$$ = $$\\ \frac { 6 }{ 13 }$$
Question 15.
From pack of 52 playing cards, blackjacks, black kings and black aces are removed and then the remaining pack is well-shuffled. A card is drawn at random from the remaining pack. Find the probability of getting
(i) a red card
(ii) a face card
(iii) a diamond or a club
(iv) a queen or a spade.
Solution:
Total number of cards = 52
Black jacks, black kings and black aces are removed
Now number of cards = 52 – (2 + 2 + 2) = 52 – 6 = 46
One card is drawn
(i) No. of red cards = 13 + 13 = 26
∴Probability = $$\\ \frac { 26 }{ 46 }$$ = $$\\ \frac { 13 }{ 23 }$$
(ii) Face cards = 4 queens, 2 red jacks, 2 kings = 8
∴Probability = $$\\ \frac { 8 }{ 46 }$$ = $$\\ \frac { 4 }{ 23 }$$
(iii) a diamond on a club = 13 + 10 = 23
∴Probability = $$\\ \frac { 23 }{ 46 }$$ = $$\\ \frac { 1 }{ 2 }$$
(iv) A queen or a spade = 4 + 10 = 14
∴Probability = $$\\ \frac { 14 }{ 46 }$$ = $$\\ \frac { 7 }{ 23 }$$
Question 16.
Two different dice are thrown simultaneously. Find the probability of getting:
(i) sum 7
(ii) sum ≤ 3
(iii) sum ≤ 10
Solution:
(i) Numbers whose sum is 7 will be (1, 6), (2, 5), (4, 3), (5, 2), (6, 1), (3, 4) = 6
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 6 }{ 36 }$$ = $$\\ \frac { 1 }{ 6 }$$
(ii) Sum ≤ 3
Then numbers can be (1, 2), (2, 1), (1, 1) which are 3 in numbers
∴Probability will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 3 }{ 36 }$$ = $$\\ \frac { 1 }{ 12 }$$
(iii) Sum ≤ 10
The numbers can be,
(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),
(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, .6),
(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),
(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),
(5, 1), (5, 2), (5, 3), (5, 4), (5, 5),
(6, 1), (6, 2), (6, 3), (6, 4) = 33
Probability will be
P(E) = $$\frac { Number\quad of\quad favourable\quad outcome }{ Number\quad of\quad possible\quad outcome }$$ = $$\\ \frac { 33 }{ 36 }$$ = $$\\ \frac { 11 }{ 12 }$$
Question 17.
Two dice are thrown together. Find the probability that the product of the numbers on the top of two dice is
(i) 6
(ii) 12
(iii) 7
Solution:
Two dice are thrown together
Total number of events = 6 × 6 = 36
(i) Product 6 = (1, 6), (2, 3), (3, 2). (6, 1) = 4
Probability = $$\\ \frac { 4 }{ 36 }$$ = $$\\ \frac { 1 }{ 9 }$$
(ii) Product 12 = (2, 6), (3, 4), (4, 3), (6, 2) = 4
Probability = $$\\ \frac { 4 }{ 36 }$$ = $$\\ \frac { 1 }{ 9 }$$
(iii) Product 7 = 0 (no outcomes)
Probability = $$\\ \frac { 0 }{ 36 }$$ = 0
We hope the ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Chapter Test help you. If you have any query regarding ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 22 Probability Chapter Test, drop a comment below and we will get back to you at the earliest.
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Sale!
# ECE 275 Homework 1
\$35.00
Category:
Homework 1
Max marks: 80
Figure 1: A three-input circuit
Figure 2: A three-variable function
Problem 1 Use algebraic manipulation to find
the minimum sum-of-products expression for the
function f = x1x3+x1x¯2+ ¯x1x2x3+ ¯x1x¯2x¯3. [1,
Prob 2.12][10 marks]
Problem 2 Use algebraic manipulation to find
the minimum sum-of-products expression for the
function f = x1x¯2x¯3 + x1x2x4 + x1x¯2x3x¯4. [1,
Prob 2.13][10 marks]
Problem 3 Draw a timing diagram for the circuit in Figure 1. Show the waveforms that can
be observed on all wires (f, g, h, k, l) in the
circuit.[1, Prob 2.8][10 marks]
Problem 4 Represent the function in Figure 2
in the form of a Venn diagram and find its
minimal sum-of-products form. [1, Prob 2.17][10
marks]
Problem 5 Use algebric manipulation to prove
that (x+y)·(x+ ¯y) = x. [1, Prob 2.2] [10 marks].
Problem 6 Determine whether or not the following expressions are valid, i.e., whether the
left- and right-hand sides represent the same
function. [1, Prob 2.7][10 marks]
1. x1x¯3 + x2x3 + ¯x2x¯3 = (¯x1 + ¯x2 + x3)(x1 +
x2 + ¯x3)(¯x1 + x2 + ¯x3)
2. (x1 + x3)(¯x1 + ¯x2 + ¯x3)(¯x1 + x2) = (x1 +
x2)(x2 + x3)(¯x1 + ¯x3)
Problem 7 Design the simplest sum-ofproducts circuit that implements the function
f(x1, x2, x3) = Pm(3, 4, 6, 7). [1, Prob 2.21][10
marks]
Problem 8 Design the simplest product-ofsums circuit that implements the function
f(x1, x2, x3) = Q
M(0, 2, 5). [1, Prob 2.22][10
marks]
References
[1] S. Brown and Z. Vranesic. Fundamentals of
Digital Logic with Verilog Design: Third Edition. McGraw-Hill Higher Education, 2013.
1
ECE 275 Homework 1
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Related Articles
Repdigit Numbers
• Last Updated : 08 Jul, 2020
Repdigit Number is a number N which has all the digits in its representation in base B equal.
Some of the Repdigit number are:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 22, 33, 44, 55….
### Check if N is an Repdigit number
Given a number N, the task is to check if N is an Repdigit Number in Base B or not. If N is a Repdigit Number in Base B then print “Yes” else print “No”.
Examples:
Input: N = 2000, B = 7
Output: Yes
Explanation:
2000 in base 7 is 5555 which has all digits equal.
Input: N = 112, B = 10
Output: No
## Recommended: Please try your approach on {IDE} first, before moving on to the solution.
Approach:
• One by one find all the digits of N in base B.
• Compare every digit with its previous digit.
• If any digit is not equal to the previous digit then return false.
• Otherwise return true.
Below is the implementation of the above approach:
## C++
`// C++ implementation to check ` `// if a number is Repdigit ` ` ` `#include ` `using` `namespace` `std; ` ` ` `// Function to check if a number ` `// is a Repdigit number ` `bool` `isRepdigit(``int` `num, ``int` `b) ` `{ ` ` ``// To store previous digit (Assigning ` ` ``// initial value which is less than any ` ` ``// digit) ` ` ``int` `prev = -1; ` ` ` ` ``// Traverse all digits from right to ` ` ``// left and check if any digit is ` ` ``// smaller than previous. ` ` ``while` `(num) { ` ` ``int` `digit = num % b; ` ` ``num /= b; ` ` ``if` `(prev != -1 && digit != prev) ` ` ``return` `false``; ` ` ``prev = digit; ` ` ``} ` ` ` ` ``return` `true``; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ``int` `num = 2000, base = 7; ` ` ``isRepdigit(num, base) ? cout << ``"Yes"` ` ``: cout << ``"No"``; ` ` ``return` `0; ` `} `
## Java
`// Java implementation to check ` `// if a number is Repdigit ` `class` `GFG{ ` ` ` `// Function to check if a number ` `// is a Repdigit number ` `static` `boolean isRepdigit(``int` `num, ``int` `b) ` `{ ` ` ``// To store previous digit (Assigning ` ` ``// initial value which is less than any ` ` ``// digit) ` ` ``int` `prev = -1; ` ` ` ` ``// Traverse all digits from right to ` ` ``// left and check if any digit is ` ` ``// smaller than previous. ` ` ``while` `(num != 0) ` ` ``{ ` ` ``int` `digit = num % b; ` ` ``num /= b; ` ` ``if` `(prev != -1 && digit != prev) ` ` ``return` `false``; ` ` ``prev = digit; ` ` ``} ` ` ``return` `true``; ` `} ` ` ` `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` ` ``int` `num = 2000, base1 = 7; ` ` ``if``(isRepdigit(num, base1)) ` ` ``{ ` ` ``System.``out``.print(``"Yes"``); ` ` ``} ` ` ``else` ` ``{ ` ` ``System.``out``.print(``"No"``); ` ` ``} ` `} ` `} ` ` ` `// This code is contributed by Code_Mech `
## Python3
`# Python3 implementation to check ` `# if a number is Repdigit ` ` ` `# Function to check if a number ` `# is a Repdigit number ` `def` `isRepdigit(num, b) : ` ` ` ` ``# To store previous digit (Assigning ` ` ``# initial value which is less than any ` ` ``# digit) ` ` ``prev ``=` `-``1` ` ` ` ``# Traverse all digits from right to ` ` ``# left and check if any digit is ` ` ``# smaller than previous. ` ` ``while` `(num) : ` ` ``digit ``=` `num ``%` `b ` ` ``num ``/``/``=` `b ` ` ``if` `(prev !``=` `-``1` `and` `digit !``=` `prev) : ` ` ``return` `False` ` ``prev ``=` `digit ` ` ``return` `True` ` ` `# Driver code ` `num ``=` `2000` `base ``=` `7` `if``(isRepdigit(num, base)): ` ` ``print``(``"Yes"``) ` `else``: ` ` ``print``(``"No"``) ` ` ` `# This code is contributed by Vishal Maurya. `
## C#
`// C# implementation to check ` `// if a number is Repdigit ` `using` `System; ` `class` `GFG{ ` ` ` `// Function to check if a number ` `// is a Repdigit number ` `static` `bool` `isRepdigit(``int` `num, ``int` `b) ` `{ ` ` ``// To store previous digit (Assigning ` ` ``// initial value which is less than any ` ` ``// digit) ` ` ``int` `prev = -1; ` ` ` ` ``// Traverse all digits from right to ` ` ``// left and check if any digit is ` ` ``// smaller than previous. ` ` ``while` `(num != 0) ` ` ``{ ` ` ``int` `digit = num % b; ` ` ``num /= b; ` ` ``if` `(prev != -1 && digit != prev) ` ` ``return` `false``; ` ` ``prev = digit; ` ` ``} ` ` ``return` `true``; ` `} ` ` ` `// Driver code ` `public` `static` `void` `Main() ` `{ ` ` ``int` `num = 2000, base1 = 7; ` ` ``if``(isRepdigit(num, base1)) ` ` ``{ ` ` ``Console.Write(``"Yes"``); ` ` ``} ` ` ``else` ` ``{ ` ` ``Console.Write(``"No"``); ` ` ``} ` `} ` `} ` ` ` `// This code is contributed by Code_Mech `
Output:
```Yes
```
Time Complexity: O(n)
Reference: http://www.numbersaplenty.com/set/repdigit/
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# 11.2 Arithmetic Sequences
11.2 Arithmetic Sequences
Warm up:
1.) Find the equation of a line in slope intercept form that passes through (8, 50) and
slope is ¼.
2.) Find the equation of a line in slope intercept from that passes through the two
points (6, 10) and (21, 55)
Arithmetic
Arithmetic SSequence
equence: a sequence where the difference between consecutive terms is
constant, denoted by d.
Ex: Decide whether each sequence is arithmetic.
a) −3, 1, 5, 9, 13, …
b) 2, 5, 10, 17, 26, …
c) −10, −6, −2, 0, 2, 6, 10, …
d) 5, 11, 17, 23, 29, …
e) 1, 1, 2, 3, 5, 8, …
f) 32, 16, 8, 4, …
Ex: Write a rule for the nth term of the sequence −2, 1, 4, 7, …. Then find a20.
(Graph then make an x/y chart to help see a pattern…what do you notice?)
Rule for an Arithmetic sequence: the nth term of an arithmetic sequence can be
determined using the formula below…
an = dx + b
Ex:
Ex: Write a rule for the nth term of the sequence.
a) 50, 44, 38, 32, …. Then find a18
b) 32, 47, 62, 77, …. Then find a12
c) 17, 11, 5, −1, −7, …. Then find a15
Ex:
Ex: Finding the nth term given a term and the common difference (Hint: Think of
finding an equation using the slope and a given point.)
a) One term of an arithmetic sequence is a13 = 30. The common difference
is d =
3
.
2
b) One term of an arithmetic sequence is a8 = 50. The common difference is
d = 0.25 .
c) One term of an arithmetic sequence is a20 = −111. The common difference
is d = −6 .
d) One term of an arithmetic sequence is a2 = 18. The common difference is
d = 18 .
Ex:
Ex: Finding the nth term given two terms. (Hint: Think about how to find the
equation of a line given two points.)
a) Two terms on an arithmetic sequence are a6 = 10 and a21 = 55. Then find
n when an = 40.
b) Two terms on an arithmetic sequence are a5 = 10 and a30 = 110. Then
find n when an = −2.
c) Two terms on an arithmetic sequence are a8 = 12.5 and a15 = 30. Then
find n when an = 50
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Problem 357
# The angle formed by two tangents drawn to a circle from the same external point measures $$80^{\circ}$$. Find the measure of the minor intercepted arc.
Expert verified
The measure of the minor intercepted arc of the tangents drawn to a circle from the same external point is $$160^{\circ}$$.
See the step by step solution
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## Step 1: Draw a diagram
Draw a circle with center O and two tangents, PA and PB, from an external point P. Label the points where the tangents touch the circle as A and B. The circle, the tangents, and the intercepted arc AB are shown in the diagram below. 
## Step 2: Use the tangent property to find the measures of angles APB and AOB
One of the properties of tangents states that the tangent at a point on a circle is perpendicular to the radius through that point. Therefore, we have: Angle OAP = Angle OBP = $$90^{\circ}$$ We are given the measure of angle APB as $$80^{\circ}$$. Since the sum of angles in a triangle is $$180^{\circ}$$, we can find the measure of angle AOB as follows: Angle AOB = Angle OAP + Angle APB + Angle OBP Angle AOB = $$90^{\circ} + 80^{\circ} + 90^{\circ}$$ Angle AOB = $$260^{\circ}$$
## Step 3: Use the property of inscribed angles to find the measure of the minor intercepted arc
We know that the measure of an inscribed angle (an angle with its vertex on the circle) is half the measure of the intercepted arc. In this case, we have: Insribed Angle = Angle APB $$= 80^{\circ}$$ Using the property of inscribed angles, we can find the measure of the minor intercepted arc AB as follows: Measure of intercepted arc AB = 2 x Insribed Angle Measure of intercepted arc AB = 2 x $$80^{\circ}$$ Measure of intercepted arc AB = $$160^{\circ}$$
## Step 4: State the answer
The measure of the minor intercepted arc of the tangents drawn to a circle from the same external point is $$160^{\circ}$$.
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# Mgrl Accounting Essay
Unit I – Assignment 1
Costs can be classified into two categories, fixed and variable costs - Mgrl Accounting Essay introduction. These costs behave differently based on the level of sales volumes. Suppose we are running a restaurant and have identified certain costs along with the number of annual units sold of 1000.
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Item: Raw Materials (cost for hamburgers)
Total Annual Cost: 650
Item: Building Rent
Total Annual Cost: 9000
Identify which cost item above is fixed and variable and why? What is the cost per unit of each? Suppose we increased our sales volume to 6000 units and then to 8000 units the following year (and are still within the relevant range), what would be the total annual cost and unit cost of fixed and variable costs?
(1) Identify which cost item below is fixed and variable and why?
Raw materials: Expense for raw materials is considered as variable cost because the amount paid for raw materials increase with the amount of goods sold by the firm. The company pays \$650 for 1,000 units of hamburgers. The expense for raw materials increases proportionally with the volume of hamburgers produced or sold. The amount paid for the raw materials increases to \$3,900 for 6,000 units, and \$5,200 for 8,000 units, or at a rate of \$0.65 per unit.
Building rent: Building rest is considered a fixed cost because the total amount paid for the building rent does not change regardless of the volume of goods sold by the restaurant. Whether the firm sells 1,000 units, 6,000 units or 8,000 units, building rent remains at \$9,000.
(2) What is the cost for 1,000 units?
The variable cost per unit is: \$650/1,000 units = \$0.65 per unit.
The total variable cost is: \$650
The fixed cost per unit is: \$9,000 /1,000 units = \$9 per unit.
The total fixed cost is: \$9 per unit * 1,000 units = \$9,000
The total cost for 1,000 units is: \$650 + \$9,000 = \$9,650
The total cost per unit is: \$9 + \$0.65 = \$9.65 per unit
(3) What is the cost for 6,000 units?
The variable cost per unit is: \$650/ 1,000 units = \$0.65 per unit
The total variable cost is: \$0.65 per unit * 6,000 units = \$3,900
The fixed cost per unit is: \$9,000 / 6,000 = \$1.5 per unit
The total fixed cost is: \$1.5 per unit * 6,000 units = \$9,000
The total cost for 6,000 units is: \$3,900 + \$9,000 = \$12,900
The total cost per unit is: \$0.65 per unit + \$1.5 per unit = \$2.15 per unit.
(4) What is the cost for 8,000 units?
The variable cost per unit is:\$650/ 1,000 units = \$0.65 per unit
The total variable cost is: \$0.65 * 8,000 units = \$5,200
The fixed cost per unit is: \$9,000 /8,000 units = \$1.125per unit
The total fixed cost is: \$9,000
The total cost for 8,000 units is: \$5,200 + \$9,000 = \$14,200.
The total cost per unit is: \$0.65 + \$1.125 = \$1.775 per unit.
Unit I – Assignment 2
Discuss the role of the financial accounting and managerial accounting functions in organizations and some of their job responsibilities. What are some of the differences between the two and the type of reports they may each use?
Managerial accountants measure, interpret and report financial information for management’s use, while financial accountants provide “useful” information to people outside the organization, primarily the lenders and investors. Both managerial and financial accounting makes use of financial statements, but the degree, timing, focus and the specificity of these reports may differ.
Managerial accountants provide reports to management, on a daily, weekly, or monthly basis. Financial reports given to investors, however, are on a quarterly basis. The Securities and Exchange Commission, for example, requires public companies to submit quarterly reports, which can be accessed by anybody at its Web site, http://www.sec.gov/. Managers need access to information more quickly and more frequently in order to address any problems that would affect the firm’s goals of increasing sales, value and profitability.
Financial reports accessed by managers may contain confidential information, including the terms reached by the firm with its suppliers, which if publicly disclosed, will give advantage to competitors. Managers also need more detailed information, including how the company’s product-lines geographic segments, new products and innovation are performing; how the company’s supply chain is doing; the status of each of the company’s facilities and plants, as to capacity, volume produced, inventory, and supply. Investors and creditors are more interested in how the company’s performing as a whole — its aggregate sales, margins, net profits, liquidity, and solvency. Investors and creditors are also given disclosures as to the risks the company is facing, and material developments that would affect business operations, and the company’s reasonable financial forecast, which information would be essential for these people’s decision making.
While the the type of information provided to managers and the public may differ to some extent, both information provided are required to be accurate and factual and thus should not be contradictory. Financial information provided by the financial accountants to people outside the firm are also subject to standards set by the Financial Accounting Standards Board and monitored by agencies, including the Securities and Exchange Commission. Accountants cannot provide information to managers that says the company is substantially burning cash in majority of its segments, but disclose to investors that the company is still earning money by window dressing or manipulating the company’s books. This is an accounting fraud, which is punishable by law.
As an example, General Motors Corp., the largest producer of auto-mobiles in the United States, submits financial reports to stakeholders on a quarterly basis, and explains those results in their conference calls. Management of GM also receive reports from its managerial accountants about how each of its product lines are performing, about the company’s sales in each of its geographic segments, and the effects of the company’s recent cost cutting measures. While the degree and timing of information provided to the managers and investors may differ, both were informed that GM incurred losses of \$15.5 billion in the three months ended June 30, 2008, and the company’s outlook of lower vehicle sales for the rest of the year.
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# math
posted by .
To estimate how many more people live in Atherton than Needles, which is closer to exact answer: rounding to hundreds or thousands? Explain.
Artherto 7,127
Needles 5,346
• math -
its better to round to the hundreds because if you round to the thousands, it would be really far away from the real answer. for example, if you round 7,127 to the thousands place, it would be 7,000. but if you round it to the hundreds place, it would be 7,100 which is closer to 7,127 than to 7,000.
• math -
I am cooler then
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Real closed field
In mathematics, a real closed field is a field F that has the same first-order properties as the field of real numbers. Some examples are the field of real numbers, the field of real algebraic numbers, and the field of hyperreal numbers.
Definitions
A real closed field is a field F in which any of the following equivalent conditions are true:
1. F is elementarily equivalent to the real numbers. In other words, it has the same first-order properties as the reals: any sentence in the first-order language of fields is true in F if and only if it is true in the reals.
2. There is a total order on F making it an ordered field such that, in this ordering, every positive element of F has a square root in F and any polynomial of odd degree with coefficients in F has at least one root in F.
3. F is a formally real field such that every polynomial of odd degree with coefficients in F has at least one root in F, and for every element a of F there is b in F such that a = b2 or a = −b2.
4. F is not algebraically closed but its algebraic closure is a finite extension.
5. F is not algebraically closed but the field extension ${\displaystyle F({\sqrt {-1}})}$ is algebraically closed.
6. There is an ordering on F which does not extend to an ordering on any proper algebraic extension of F.
7. F is a formally real field such that no proper algebraic extension of F is formally real. (In other words, the field is maximal in an algebraic closure with respect to the property of being formally real.)
8. There is an ordering on F making it an ordered field such that, in this ordering, the intermediate value theorem holds for all polynomials over F with degree 0.
9. F is a real closed ring.
If F is an ordered field, the Artin–Schreier theorem states that F has an algebraic extension, called the real closure K of F, such that K is a real closed field whose ordering is an extension of the given ordering on F, and is unique up to a unique isomorphism of fields identical on F[1] (note that every ring homomorphism between real closed fields automatically is order preserving, because x ≤ y if and only if ∃z y = x + z2). For example, the real closure of the ordered field of rational numbers is the field ${\displaystyle \mathbb {R} _{\mathrm {alg} }}$ of real algebraic numbers. The theorem is named for Emil Artin and Otto Schreier, who proved it in 1926.
If (F,P) is an ordered field, and E is a Galois extension of F, then by Zorn's Lemma there is a maximal ordered field extension (M,Q) with M a subfield of E containing F and the order on M extending P. M, together with its ordering Q, is called the relative real closure of (F,P) in E. We call (F,P) real closed relative to E if M is just F. When E is the algebraic closure of F the relative real closure of F in E is actually the real closure of F described earlier.[2]
If F is a field (no ordering compatible with the field operations is assumed, nor is assumed that F is orderable) then F still has a real closure, which may not be a field anymore, but just a real closed ring. For example, the real closure of the field ${\displaystyle \mathbb {Q} ({\sqrt {2}})}$ is the ring ${\displaystyle \mathbb {R} _{\mathrm {alg} }\times \mathbb {R} _{\mathrm {alg} }}$ (the two copies correspond to the two orderings of ${\displaystyle \mathbb {Q} ({\sqrt {2}})}$ ). On the other hand, if ${\displaystyle \mathbb {Q} ({\sqrt {2}})}$ is considered as an ordered subfield of ${\displaystyle \mathbb {R} }$ , its real closure is again the field ${\displaystyle \mathbb {R} _{\mathrm {alg} }}$ .
Model theory: decidability and quantifier elimination
Although the theory of real closed fields was firstly developed by algebraists, it has received considerable attention from model theory. By adding to the ordered field axioms
• an axiom asserting that every positive number has a square root, and
• an axiom scheme asserting that all polynomials of odd degree have at least one root
one obtains a first-order theory. Alfred Tarski (1951) proved that the theory of real closed fields in the first order language of partially ordered rings (consisting of the binary predicate symbols "=" and "≤", the operations of addition, subtraction and multiplication and the constant symbols 0,1) admits elimination of quantifiers. The most important model theoretic consequences hereof: The theory of real closed fields is complete, o-minimal and decidable.
Decidability means that there exists at least one decision procedure, i.e., a well-defined algorithm for determining whether a sentence in the first order language of real closed fields is true. Euclidean geometry (without the ability to measure angles) is also a model of the real field axioms, and thus is also decidable.
The decision procedures are not necessarily practical. The algorithmic complexities of all known decision procedures for real closed fields are very high, so that practical execution times can be prohibitively high except for very simple problems.
The algorithm Tarski proposed for quantifier elimination has NONELEMENTARY complexity, meaning that no tower ${\displaystyle 2^{2^{\cdot ^{\cdot ^{\cdot ^{n}}}}}}$ can bound the execution time of the algorithm if n is the size of the problem. Davenport and Heintz (1988) proved that quantifier elimination is in fact (at least) doubly exponential: there exists a family Φn of formulas with n quantifiers, of length O(n) and constant degree such that any quantifier-free formula equivalent to Φn must involve polynomials of degree ${\displaystyle 2^{2^{\Omega (n)}}}$ and length ${\displaystyle 2^{2^{\Omega (n)}}}$ , using the Ω asymptotic notation. Ben-Or, Kozen, and Reif (1986) proved that the theory of real closed fields is decidable in exponential space, and therefore in doubly exponential time.
Basu and Roy (1996) proved that there exists a well-behaved algorithm to decide the truth of a formula ∃x1, …, ∃xk P1(x1, …, xk) ⋈ 0 ∧ … ∧ Ps(x1,…,xk) ⋈ 0 where ⋈ is <, > or =, with complexity in arithmetic operations sk+1dO(k). In fact, the existential theory of the reals can be decided in PSPACE.
Adding additional functions symbols, for example, the sine or the exponential function, can change the decidability of the theory.
Yet another important model-theoretic property of real closed fields is that they are weakly o-minimal structures. Conversely, it is known that any weakly o-minimal ordered field must be real closed.[3]
Order properties
A crucially important property of the real numbers is that it is an Archimedean field, meaning it has the Archimedean property that for any real number, there is an integer larger than it in absolute value. An equivalent statement is that for any real number, there are integers both larger and smaller. Such real closed fields that are not Archimedean, are non-Archimedean ordered fields. For example, any field of hyperreal numbers is real closed and non-Archimedean.
The Archimedean property is related to the concept of cofinality. A set X contained in an ordered set F is cofinal in F if for every y in F there is an x in X such that y < x. In other words, X is an unbounded sequence in F. The cofinality of F is the size of the smallest cofinal set, which is to say, the size of the smallest cardinality giving an unbounded sequence. For example, natural numbers are cofinal in the reals, and the cofinality of the reals is therefore ${\displaystyle \aleph _{0}}$ .
We have therefore the following invariants defining the nature of a real closed field F:
• The cardinality of F.
• The cofinality of F.
• The weight of F, which is the minimum size of a dense subset of F.
These three cardinal numbers tell us much about the order properties of any real closed field, though it may be difficult to discover what they are, especially if we are not willing to invoke the generalized continuum hypothesis. There are also particular properties which may or may not hold:
• A field F is complete if there is no ordered field K properly containing F such that F is dense in K. If the cofinality of F is κ, this is equivalent to saying Cauchy sequences indexed by κ are convergent in F.
• An ordered field F has the eta set property ηα, for the ordinal number α, if for any two subsets L and U of F of cardinality less than ${\displaystyle \aleph _{\alpha }}$ such that every element of L is less than every element of U, there is an element x in F with x larger than every element of L and smaller than every element of U. This is closely related to the model-theoretic property of being a saturated model; any two real closed fields are ηα if and only if they are ${\displaystyle \aleph _{\alpha }}$ -saturated, and moreover two ηα real closed fields both of cardinality ${\displaystyle \aleph _{\alpha }}$ are order isomorphic.
The generalized continuum hypothesis
The characteristics of real closed fields become much simpler if we are willing to assume the generalized continuum hypothesis. If the continuum hypothesis holds, all real closed fields with cardinality the continuum and having the η1 property are order isomorphic. This unique field Ϝ can be defined by means of an ultrapower, as ${\displaystyle \mathbb {R} ^{\mathbb {N} }/{\mathbf {M} }}$ , where M is a maximal ideal not leading to a field order-isomorphic to ${\displaystyle \mathbb {R} }$ . This is the most commonly used hyperreal number field in non-standard analysis, and its uniqueness is equivalent to the continuum hypothesis. (Even without the continuum hypothesis we have that if the cardinality of the continuum is ${\displaystyle \aleph _{\beta }}$ then we have a unique ηβ field of size ηβ.)
Moreover, we do not need ultrapowers to construct Ϝ, we can do so much more constructively as the subfield of series with a countable number of nonzero terms of the field ${\displaystyle \mathbb {R} ((G))}$ of formal power series on a totally ordered abelian divisible group G that is an η1 group of cardinality ${\displaystyle \aleph _{1}}$ (Alling 1962).
Ϝ however is not a complete field; if we take its completion, we end up with a field Κ of larger cardinality. Ϝ has the cardinality of the continuum which by hypothesis is ${\displaystyle \aleph _{1}}$ , Κ has cardinality ${\displaystyle \aleph _{2}}$ , and contains Ϝ as a dense subfield. It is not an ultrapower but it is a hyperreal field, and hence a suitable field for the usages of nonstandard analysis. It can be seen to be the higher-dimensional analogue of the real numbers; with cardinality ${\displaystyle \aleph _{2}}$ instead of ${\displaystyle \aleph _{1}}$ , cofinality ${\displaystyle \aleph _{1}}$ instead of ${\displaystyle \aleph _{0}}$ , and weight ${\displaystyle \aleph _{1}}$ instead of ${\displaystyle \aleph _{0}}$ , and with the η1 property in place of the η0 property (which merely means between any two real numbers we can find another).
Notes
1. ^ Rajwade (1993) pp. 222–223
2. ^ Efrat (2006) p. 177
3. ^ D. Macpherson et. al, (1998)
References
• Alling, Norman L. (1962), "On the existence of real-closed fields that are ηα-sets of power ℵα.", Trans. Amer. Math. Soc., 103: 341–352, doi:10.1090/S0002-9947-1962-0146089-X, MR 0146089
• Basu, Saugata, Richard Pollack, and Marie-Françoise Roy (2003) "Algorithms in real algebraic geometry" in Algorithms and computation in mathematics. Springer. ISBN 3-540-33098-4 (online version)
• Michael Ben-Or, Dexter Kozen, and John Reif, The complexity of elementary algebra and geometry, Journal of Computer and Systems Sciences 32 (1986), no. 2, pp. 251–264.
• Caviness, B F, and Jeremy R. Johnson, eds. (1998) Quantifier elimination and cylindrical algebraic decomposition. Springer. ISBN 3-211-82794-3
• Chen Chung Chang and Howard Jerome Keisler (1989) Model Theory. North-Holland.
• Dales, H. G., and W. Hugh Woodin (1996) Super-Real Fields. Oxford Univ. Press.
• Davenport, James H.; Heintz, Joos (1988). "Real quantifier elimination is doubly exponential". J. Symb. Comput. 5 (1–2): 29–35. doi:10.1016/s0747-7171(88)80004-x. Zbl 0663.03015.
• Efrat, Ido (2006). Valuations, orderings, and Milnor K-theory. Mathematical Surveys and Monographs. 124. Providence, RI: American Mathematical Society. ISBN 0-8218-4041-X. Zbl 1103.12002.
• Macpherson, D., Marker, D. and Steinhorn, C., Weakly o-minimal structures and real closed fields, Trans. of the American Math. Soc., Vol. 352, No. 12, 1998.
• Mishra, Bhubaneswar (1997) "Computational Real Algebraic Geometry," in Handbook of Discrete and Computational Geometry. CRC Press. 2004 edition, p. 743. ISBN 1-58488-301-4
• Rajwade, A. R. (1993). Squares. London Mathematical Society Lecture Note Series. 171. Cambridge University Press. ISBN 0-521-42668-5. Zbl 0785.11022.
• Passmore, Grant (2011). Combined Decision Procedures for Nonlinear Arithmetics, Real and Complex (PDF) (PhD). University of Edinburgh.
• Alfred Tarski (1951) A Decision Method for Elementary Algebra and Geometry. Univ. of California Press.
• Erdös, P.; Gillman, L.; Henriksen, M. (1955), "An isomorphism theorem for real-closed fields", Ann. of Math., 2, 61: 542–554, doi:10.2307/1969812, MR 0069161
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# Expression for Atmospheric Pressure with Altitude, including Tidal Forces
Is this simple to obtain? I've tried to modify the constant-gravity model to use the $GM/r^2$ form instead, but my result was shown to be wrong with some testing.
Constant Gravity Model
You can find this in many places online. Here's one. I'll repeat that work here, jumping over the calculus, and putting it in terms of Earth's radius. r0 will be the surface radius.
$$\frac{ dP }{ dr } \frac{ 1 }{ P(r) } = - \frac{ M_0 g }{ R T } \\ H \equiv \frac{ R T }{ M_0 g } \approx 7.4 \text{ km} \\ P(r) = P_0 e^{ \frac{ r_0 - r }{ H } }$$
Non-constant Gravity Model
So to generalize this (for a small moon with a diffuse atmosphere, for instance), just replace the gravity with its expression $g(r)=GM/r^2$. M0 is still the average formula mass of air, but we add M, which is the mass of the planet. This should work fine for Earth, I see no reason it shouldn't.
$$\frac{ dP }{ dr } \frac{ 1 }{ P(r) } = - \frac{ M_0 G M }{ R T } \frac{1}{r^2} \\ H' \equiv \frac{ M_0 G M }{R T } \approx 4,739 \text{ km} \\ P(r) = P_0 e^{ H' \left( \frac{1}{r} - \frac{1}{r_0} \right) }$$
It looks so nice and congruent. But I plot my non-constant gravity model side-by-side with the constant model and it's obviously wrong.
So for my question, could somebody do ANY of:
• Find something I did wrong
• Derive or find a correct model for this
• Give a clear reason it can't be obtained in a simple closed form
-
Wait, don't bother. I found the embarrassing error. It's the H' value. It should have been:
$$H' \equiv \frac{ M_0 G M }{R T } \approx 4,739,000 \text{ km}$$
I just read that output wrong. The plots now seem to fit.
Sorry to answer my own question so fast. I suppose whether any of this is right remains an open question.
-
You also have to keep in mind that the specific gas constant ($\frac{R}{M_0}$) and temperature will also vary width height. And for a dense enough atmosphere there will also be a significant increase in the gravitational parameter ($GM$) when you go farther away. – fibonatic Dec 5 '13 at 19:34
@fibonatic Yes, certainly, I shouldn't have ever implied this actually describes Earth's atmosphere. The thermosphere parameters conditions are closer to 1 amu and 1000 C, with major effects from weather, time, and geographic location. I was intending to apply this to the sci-fi concept "shell world", which it would accurately describe. But while wrong for Earth, it's still formally more accurate than the constant gravity model. – Alan Rominger Dec 5 '13 at 20:25
My solution has been challenged, and this comes from a separate source so I'm posting it as community wiki. The reference is here:
http://web.ist.utl.pt/~berberan/data/43.pdf
The proposed equation is:
$$P(z) = P(0) \exp{ \left( - \frac{ m g_0 R_0 }{ kT} \frac{ z }{ z+R_0 } \right) }$$
You could put this in terms of characteristic height if you wanted. The question in my mind is rather this solves the same equation that I was aiming for.
Any insight as to what might be wrong with the differential equation I formulated would be appreciated.
Also, their equation references equations 5 and 7 in that paper for a justification of where it comes from. I'll echo those here. Equation 5:
$$p(z) = p(H) + \frac{m}{k} \int_z^H \frac{g p(u) }{ T} du$$
Equation 7:
$$\frac{dp}{dz} = - \frac{ m g }{ k T } p$$
The idea is that you substitute in $g(z)$ in place of $g$. Honestly, I think this is what I had. Their math looks a little different, but it should come out to the same thing. I don't know what went wrong.
I think I've solved it now. The problem is that the other source uses altitude (z), whereas I have used radius (r). It also used surface gravity where I used mass and the gravitational constant. If you just re-write my equations from the OP and make these substitutions, you can match the two expressions. I have done that here:
$$z \equiv r_0 + z \\ H' \equiv \frac{ M_0 G M }{R T } \approx 4,739 \text{ km} \\ P(r) = P_0 e^{ H' \left( \frac{1}{r} - \frac{1}{r_0} \right) } \\ P(z) = P_0 e^{ H' \left( \frac{1}{r_0 + z} - \frac{1}{r_0} \right) } \\ P(z) = P_0 e^{ - \frac{ H'}{r_0} \frac{ z}{ r_0 + z} } \\ P(z) = P_0 e^{ - \frac{ M_0 G M }{R T r_0} \frac{ z}{ r_0 + z} } \\ g_0 \equiv \frac{ G M }{ r_0^2 } \\ P(z) = P_0 \exp{ \left( - \frac{ M_0 g_0 r_0 }{R T } \frac{ z}{ r_0 + z} \right) }$$
These are, in fact, the same thing.
-
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1. ## Arc Length
Find the length of an arc of degree measure 30 in a circle with an 8 meter radius. Put answer in exact form.
$\displaystyle \frac{30\;\pi\;(16)}{360}$
$\displaystyle =4.188790205$
$\displaystyle \frac{4.188790205\cdot180}{\pi}$
$\displaystyle =240\;degrees$
$\displaystyle 240\;degrees\;=\;\frac{4\;\pi}{3}\;meters$
2. Originally Posted by OzzMan
Find the length of an arc of degree measure 30 in a circle with an 8 meter radius. Put answer in exact form.
$\displaystyle \frac{30\;\pi\;(16)}{360}$
$\displaystyle =4.188790205$
$\displaystyle \frac{4.188790205\cdot180}{\pi}$
$\displaystyle =240\;degrees$
$\displaystyle 240\;degrees\;=\;\frac{4\;\pi}{3}\;meters$
Line 1: Arc length = $\displaystyle r \, \theta$ where $\displaystyle \theta$ is measured in radians.
Line 2: r = 8 and $\displaystyle \theta = \frac{\pi}{6}$.
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# Sol7 - STAT 426 HW7 1(5.24 logit π = α β 1 A β 2 S β 3 R β 4 R S where A =average number of alcoholic drink drinks consumed per day S =
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Unformatted text preview: STAT 426 HW7. 1. (5.24) logit( π ) = α + β 1 A + β 2 S + β 3 R + β 4 R * S, where A =average number of alcoholic drink drinks consumed per day. S = braceleftBigg 1 , at least one pack per day , o.w. R = braceleftBigg 1 , black , white Y = braceleftBigg 1 , yes , no a. Prediction equation: when R = 1 , logit(ˆ π ) = (ˆ α + ˆ β 3 ) + ˆ β 1 A + ( ˆ β 2 + ˆ β 4 ) S =- 6 . 7 + 0 . 1 A + 1 . 4 S when R = 0 , logit(ˆ π ) = ˆ α + ˆ β 1 A + ˆ β 2 S =- 7 + 0 . 1 A + 1 . 2 S Y-S conditional odds ratio: When R = 1 : log(odds( s = 1)) = (ˆ α + ˆ β 2 + ˆ β 3 + ˆ β 3 + ˆ β 4 ) + ˆ β 1 A ——(1) log(odds( s = 0)) = (ˆ α + ˆ β 3 ) + ˆ β 1 A ——(2) (1)-(2): log(OR(S))= ˆ β 2 + ˆ β 4 = 1 . 4 ⇒ OR(S) = exp(1 . 4) = 4 . 06 When R = 0 : log(odds( s = 1)) = (ˆ α + ˆ β 2 ) + ˆ β 1 A ——(3) log(odds( s = 0)) = ˆ α + ˆ β 1 A ——(4) (3)-(4): log(OR(S))= ˆ β 2 = 1 . 2 ⇒ OR(S) = exp(1 . 2) = 3 . 32 Prediction equation: when S = 1 , logit(ˆ...
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## This note was uploaded on 04/29/2010 for the course STAT stat 426 taught by Professor Xe during the Spring '10 term at University of Illinois at Urbana–Champaign.
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Sol7 - STAT 426 HW7 1(5.24 logit π = α β 1 A β 2 S β 3 R β 4 R S where A =average number of alcoholic drink drinks consumed per day S =
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# Taking advantage of errors in models
## What is the process error?
Bookmakers and professional bettors alike use prediction models in order to ensure profits in the betting markets. Whatever the model though, prediction is not a prophecy, as it can be influenced by error. How can bettors spot errors in prediction models and how can they exploit them? Read on to find out.
As an allegory, consider one of those shape sorting cubes toddlers play with the kind in which we need to pass the correct shape in a hole. The correct shape represents a correct prediction, but unlike the cubes toddlers have at hand; the number of possibilities are far larger.
### Choosing the wrong shape/model
The first possible error to make is to fit in the wrong shape altogether. A triangle might fit in a square if you really try hard or maybe if its much smaller, this still doesnt make it a good fit.
This is equivalent to using the wrong model for the purpose. For example, while the normal distribution seems to be a good fit to goal differences, it may not be the best predictor for home team goals scored. As demonstrated by the graph below, showing the actual and estimated number of home goals scored using the normal distributions for the 2013/14 English Premiership (using post-ante data).
With a multitude of potential models possible, the perfect fit may not be in use or, worse still, may not be available. The model is a necessary simplification of a real life scenario and is bound to be incorrect. Ways to diminish this is to apply judgement in selecting and interpreting a model as well as fitting the model to old data.
### Choosing the wrong size/parameter
Going back to the shape sorting cube analogy, the correct shape might be picked but of a different size. For example, the wrong size square is being used.
In a model building scenario, this is equivalent to using the wrong parameters. Imagine you are trying to calculate the likely number of goals scored in a certain match. The Poisson Distribution may be the correct model to use but one of the teams has recently had an 8-0 win. This, in turn, distorts the mean number of goals scored rendering this parameter useless.
In this case, a judgement needs to be used and more attention needs to be given to the standard deviation in the parameters used.
### The process error
Finally, the correct shape and size may have been picked in the shape sorting cube example, but each shape may fluctuate in size due to wear-and-tear and slight differences in production.
In a sports prediction environment, not every outcome is replicable. If this years Super Bowl final was replayed under the same conditions many times, the result would not always be the Broncos winning 24 to 10, due to natural fluctuations.
Albeit, having picked the correct model and parameters, there is always natural volatility in results (that can be measured). The best predictions are available when more relevant data is available hence why a lot of bettors find it easier to predict an English Premier League soccer match than a World Cup match.
### Conclusion
Betting companies, syndicates and private individuals all have errors in their predictions the skill is in applying judgement to take advantage of the errors posed by other parties.
In addition to searching for valuable information and advice, of course, bettors should always look for the best odds, which Pinnacle is famous for offering - read here about our unbeatable odds - and given we offer the highest limits online, a bet with us will ensure you are best placed to get the best value for your bet.
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## concerete how many kgs cement requirment
### How do I calculate volume of 50 Kg cement bag? – we civil ...
19 Sep 2018. 3 Comments. The basic things you should know to calculate the volume of 1 bag of cement are, Density of cement = 1440 kg/ m³. 1 bag = 50 kg of cement. 1 m³ = 35.3147 cubic feet. 1 m³ = 10³ liters. Lets see how can we calculate the volume of 1 cement bag by using all the values mentioned above.
### How many bags of cement are used in a 1:3:6 ratio concrete ...
Answer (1 of 2): Density of concrete =1,440kg/cubic meter Volume of 1 kg of cement =1 kg x 1 cubic meter/1440 kg=1/1440=0.000694 cubic meter Volume of 1 bag of cement (50kg)=50(0.000694)=0.035 cubic meter So for the ratio of 1:3:6, 1 for cement, 3 for sand and 6 for aggregates Volume of sand=...
### How to calculate quantity of cement,sand & aggregate in ...
For M20, 30 liters of water per 50kg cement . Total amount of water required = 30 x 403.2 / 50 = 241.92 litre. 8.064 bags of cement required for 1 cu.m of M20 grade concrete. 630 kg of Sand required for 1 cu.m of M20 grade concrete. 1260 kg of 20mm Aggregate is required for 1 cu.m of M20 grade concrete. 241.92 litre. of water is required for 1 ...
### sand for 1cum concrete m25
How much sand required for M25 concrete. We have given dry volume of Ans. :-11 bags (0.385 m3, 554 kgs) cement are required for 1 cubic metre of M25 concrete. How much sand required for M25 concrete. We have given dry volume of Ans. :-11 bags (0.385 m3, 554 kgs) cement are required for 1 cubic metre of M25 concrete.
### How to prepare Cement Concrete Road Estimate - Civilwala
Cement concrete road estimates are prepared to determine the no. of bags, the volume of aggregates, and the reinforcement required. However, for the above example, cement bags work out to be as under: Qty of Cement = 1/7 of the Dry of volume = 1/7 x 1.54 = 0.22 M 3.
### Concrete Calculator - How much concrete do you need in ...
It is also referred to "density in place". Conventional concrete has density between 1900 kg/m 3 and 2400 kg/m 3 and can endure, on day 28 of its pouring, and has a compressive strength between 2500 psi and 7000 psi (pounds per square inch) ~ 200-492 kg/cm 2. According to the Portland Cement Association concrete with strength between 7000 and ...
### How much cement sand & aggregate required for m25 …
Ans. :-554 Kgs quantity of cement required for 1 cubic metre of M25 concrete. How many cement bags required for M25 concrete step 6:-we know that 1 bag cement weight = 50 kgs,calculate cement bags required for 1 cubic metre of M25 concrete is equal to = 1/4 × 1.54 m3 × 1440 Kg/m3/50 = 11.08 = 11 bags cement. Ans. :-11 bags of cement required ...
### Concrete Calculator
Welcome to the concrete calculator. It will help you calculate the number of bags of concrete required for your path, slab or post holes.
### how many bags of cement required for 1 cubic meter of concrete
how many bags of cement required for 1 cubic meter of concrete. by amita patil (india) Q. Please tell detailed estimation of 1 cubic meter of concrete, how many bags required? A. You will need 672 pounds or 304 kilograms of cement for 1 cubic meter of concrete. We have 94 pound bags of cement …
### How many bags of cement required for 1m3 of Cement?
How to calculate No. of cement bags required for 1m 3 of cement: In order to calculate how many bags of cement required in 1m 3, you should know about the dry density of cement. In general Density = Mass/Volume. Dry Density of cement = Dry Mass/ Unit Volume. Got confused? Let me explain: In general cement consists of moisture content so Mass ...
### how much quantity of concrete required in 25 cubic meter
How much cement required for M25 concrete in Kgs step 5:-calculate cement required for 1 cubic metre of M25 concrete in Kgs ( kilogram) is equal to = 1/4 × 1.54 m3 × 1440 Kg/m3 = 554 Kgs. Ans. :-554 Kgs quantity of cement required for 1 cubic metre of M25 concrete.
### Portland Cement Volume Calculator - findnchoose
Portland Cement Volume Calculator, Calculate the Number of Bags of Portland Cement that are Required for a Project. Portland Cement Volume Calculator Calculate the quantity of concrete you can produce ... cement or sand as needed. Concrete must be used within 2 hours of mixing, the sooner the better. Avoid over-working any poured concrete, as this
### How much concrete does a 40kg bag of cement make?
How much concrete does a 40kg bag of cement make? 5. YIELD. Using proportions suggested, one 40 kg (88 lb) bag of cement will produce about 3.5 ft³ (0.1 m³) of concrete mix approximately. Click to see full answer. Also know, how much concrete does a 40kg bag make? The addition of 4 litres of water to a 40kg bag will produce around 18.5 litres ...
### How many 20kg bags concrete per cubic meter? - Answers
FOR 1cubic meter of concrete block,firstly we must know the mix ratio but if e.g we have the ratio of 1:2:4, 1 is cement 2 is sand 4 is aggregate than we …
### How to Calculate Quantities of Cement, Sand and Aggregate ...
Considering concrete density = 2400 kg/cum, One bag of cement and other ingredients can produce = 400/2400 = 0.167 Cum of concrete (1:2:4) 01 bag cement yield = 0.167 cum concrete with a proportion of 1:2:4. 01 cum of concrete will require. Cement required = 1/0.167 = 5.98 Bags ~ 6 Bags. Sand required = 115/0.167 = 688 Kgs or 14.98 cft
### How To Calculate Cement, Sand & Aggregates Quantity in ...
Now we start calculation to find Cement, Sand and Aggregate quality in 1 cubic meter concrete. CALCULATION FOR CEMENT QUANTITY; Cement= (1/5.5) x 1.54 = 0.28 m 3 ∴ 1 is a part of cement, 5.5 is sum of ratio Density of Cement is 1440/m 3 = 0.28 x 1440 = 403.2 kg We know each bag of cement is 50 kg
### Calculate Quantities of Materials for Concrete -Cement ...
Thus, the quantity of cement required for 1 cubic meter of concrete = 0.98/0.1345 = 7.29 bags of cement. The quantities of materials for 1 m3 of concrete production can be calculated as follows: The weight of cement required = 7.29 x 50 = 364.5 kg. Weight of fine aggregate (sand) = 1.5 x 364.5 = 546.75 kg.
### How many 20 kg bags of cement per cubic meter of 25Mpa ...
How many 20 kg bags of cement per cubic meter of 25Mpa concrete? Asked by Wiki User. See Answer. Top Answer. Wiki User Answered 02:23:54. 108. 0 0 1. ... 😮. 0. 😂. 0. Add a Comment. Your Answer. Related Questions. How many cement bags per 1 cubic meter of concrete? ...
### Question: How Much Cement Is Required To Plaster A Wall ...
So, One bag of cement (50 Kgs) has to be mixed with 115 kgs of Sand, 209 Kgs of aggregate and 27.5 kgs of water to produce M20 grade concrete. How many 50kg bags of cement are in a cubic meter? Procedure To Calculate Cement Bags In 1 Cubic Meter: Weight of 1 bag cement = 50 kg…
### 1cum concrete how much cement? - Quora
Answer (1 of 6): The cement content in 1 Cum of concrete can vary. It depends on the Grade of concrete you are referring to. for instance, as per IS456, RCC code, The minimum content of cementitious (Cement + Flyash or PFA) in different grade of concrete is: 1. M20: 300 kg 2. M25: 300 kg …
### How to Calculate Cement, Sand and Aggregate required …
Cement : Sand : Aggregate (in Kgs) is 50 kgs : 115 kgs : 209 kgs (by weight) Water required for the mixture = 27.5 kgs. Total weight of concrete ingredients = 50+115+209+27.5 = 401.5 say 400 kg. Density of concrete = 2400 kg/cum. So, 1 bag of cement produces = 400/2400 = 0.167 cum
### How much cement sand & aggregate required for M10 concrete ...
Ans. :- 222 Kgs quantity of cement required for 1 cubic metre of M10 concrete. How many cement bags required for M10 concrete step 6:- we know that 1 bag cement weight = 50 kgs,calculate cement bags required for 1 cubic metre of M10 concrete is equal to = 1/10 × 1.54 m3 × 1440 Kg/m3/50 = 4.44 = 4.5 bags cement. Ans. :- 4.5 bags of cement ...
### Water-Cement Ratio (W/C) of Concrete | online civil ...
The quantity of water is usually expressed in litres per bag of cement and hence the water-cement ratio reduces to the quantity of water required in litres per kg of cement as 1 litre of water weighs 1 kg. For instance, if water required for, 1 bag of cement is 30 litres, the water-cement ratio is equal to 30/50= 0.60. The important points to ...
### Concrete 1 cubic meter volume to kilograms converter
How heavy is concrete? Calculate how many kilograms ( kg - kilo ) of concrete are in 1 cubic meter ( 1 m3 ). Specific unit weight of concrete - amount properties converter for conversion factor exchange from 1 cubic meter m3 equals = 2,406.53 kilograms kg - kilo exactly for the masonry material type. To convert concrete measuring units can be useful when building with concrete and where ...
### CONCRETE CALCULATOR - how many bags of concrete do I need?
For concrete, the procedure for volume is as follows: length x width x thickness. You need to divide the total cubic yards that the yield needs to determine the number of concrete bags you need. Use the following yields per each bag size: 40-pound bag yields .011 …
### EXAMPLE BELOW - BGC Cement – BGC Cement
(required to make up 1m3 of concrete) Cement Sand Coarse Aggregate Water Cement (20kg Bags) Sand (m3) Coarse Aggregate (m3) High Structural Strength Grade concrete for thin reinforced walls, slender reinforced columns, fence columns, heavy duty floors 1 1.5 3 0.66 40 MPa 21 (=420kg/m3) 0.5 1 Commonly adopted mixture for reinforced concrete
### How To Calculate Cement Bags Per Cubic Meter Concrete ...
Therefore required cement = 1440 x 0.221 m 3 = 318.24 Kg or (318.24/50) = 6.37 Bags. Cement Bag Per Cubic Metre Concrete Table – Nominal Mix
### How many bags of cement do I need for 1 m3 of m25 concrete ...
How much cement sand and aggregate required for m25 concrete. quantity of cement in m25 concrete. 11.088 no (554.4 kg) of cement bags are used in m25 grade of 1m3 concrete. volume of sand in cft for m25 concrete. 1m3 = 35.32 cft. Volume =( 1/4) × 1.54×35.32. volume of aggregates in cft for m25 concrete. How many bags of cement make 1m3 of ...
### How Much Concrete Does A 40kg Bag Of Cement Make?
Using proportions suggested, one 40 kg (88 lb) bag of cement will produce about 3.5 ft³ (0.1 m³) of concrete mix approximately. People Also Asked, How much concrete does a 40kg bag make? The addition of 4 litres of water to a 40kg bag will produce around 18.5 litres of concrete, enough for a rectangular slab measuring 1000 x 480 x 40mm or ...
### Question: How Do You Calculate Sand And Cement For Mortar ...
How many bags of cement do I need for 1 cubic meter? Thus, the quantity of cement required for 1 cubic meter of concrete = 0.98/0.1345 = 7.29 bags of cement. Do I need plasticiser in mortar? The longer the mixing (in a mixer) the more a poor sand will become workable.
### Quick Answer: How To Calculate Unit Weight Of Concrete ...
Thus, the quantity of cement required for 1 cubic meter of concrete = 0.98/0.1345 = 7.29 bags of cement. The quantities of materials for 1 m3 of concrete production can be calculated as follows: The weight of cement required = 7.29 x 50 = 364.5 kg. Weight of fine aggregate (sand) = 1.5 x 364.5 = 546.75 kg.
### Often asked: How Much Concrete Does A Concrete Truck Carry ...
How many bags of cement make 1m3 of concrete? Thus, the quantity of cement required for 1 cubic meter of concrete = 0.98/0.1345 = 7.29 bags of cement . The quantities of materials for 1 m3 of concrete production can be calculated as follows: The weight of cement required = 7.29 x 50 = 364.5 kg.
### For 1 cubic meter of concrete how many kgs cement required
For 1 cubic meter of concrete how many kgs cement required Products. As a leading global manufacturer of crushing, grinding and mining equipments, we offer advanced, reasonable solutions for any size-reduction requirements including, For 1 cubic meter of concrete how many kgs cement required, quarry, aggregate, and different kinds of minerals.
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# Probability and Statistics Help Plz?
• Feb 1st 2009, 06:00 PM
mathwiz2006
Probability and Statistics Help Plz?
An urn contains five balls, one marked WIN and four marked LOSE. You and another player take turns selecting a ball from the urn, one at a time. The first person to select the WIN ball is the winner. If you draw first, find the probability that you will win if the sampling is done
(a) with replacement - answer = 5/9
(b) without replacement - answer = 3/5
(a) 1/5 + [(4/5)(4/5)(1/5)] + [(4/5)(4/5)(4/5)(4/5)(1/5)] = 1281/3125
(b) 1/5 + [(4/5)(3/4)(1/3)] + [(4/5)(3/4)(2/3)(1/2)(1/1)] = 3/5
I was able to solve (b) with no problem . For (a) I did the same thing except took replacement into consideration, but keep getting 1281/3125
• Feb 1st 2009, 06:40 PM
awkward
Quote:
Originally Posted by mathwiz2006
An urn contains five balls, one marked WIN and four marked LOSE. You and another player take turns selecting a ball from the urn, one at a time. The first person to select the WIN ball is the winner. If you draw first, find the probability that you will win if the sampling is done
(a) with replacement - answer = 5/9
(b) without replacement - answer = 3/5
(a) 1/5 + [(4/5)(4/5)(1/5)] + [(4/5)(4/5)(4/5)(4/5)(1/5)] = 1281/3125
(b) 1/5 + [(4/5)(3/4)(1/3)] + [(4/5)(3/4)(2/3)(1/2)(1/1)] = 3/5
I was able to solve (b) with no problem . For (a) I did the same thing except took replacement into consideration, but keep getting 1281/3125
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# Looking at Data
All Classroom Lessons
A Lesson with Third, Fourth, and Fifth Graders
by Rusty Bresser and Caren Holtzman
This activity, Looking at Data, is excerpted from Mini-lessons for Math Practice, Grades 3–5, by Rusty Bresser and Caren Holtzman (Math Solutions Publications, 2006). The book presents ideas for providing opportunities for students to practice the things they have learned, with practice defined broadly to include understanding as well as skill. In this instance, students practice what they’ve learned about identifying the median, mode, and range from data, which also provides them practice with computing and numerical reasoning.
I posed a question to the class in order to generate estimates that would become the data for our lesson. I had a mixed group of third through fifth graders, so I chose a question that would be accessible to the younger students while delivering some intrigue to the older ones.
“If I started at the front door and walked all the way around the perimeter of the school, about how many cars would I find parked on the street?” I asked.
Two sides of the school were bordered by streets. From the classroom window, students could see one side of the school’s perimeter lined with cars. Many started from there, counting the visible cars and using that number as a starting place for estimating. As individuals shared their estimates I used a thick, dark marker to record each on a sticky note. I arbitrarily slapped each sticky note on the board before taking the next estimate. (I chose to write the numbers myself in order to save time, but I have also done this activity by giving each student a sticky note and marker and having students record and post their own estimates.)
When we finished recording and posting the estimates, we had a board haphazardly filled with sticky notes. This “mess of numbers” became the source of a rich discussion about data.
“Hmm,” I said, “there are a lot of numbers up here. Look at them carefully and see if you can identify the lowest number and the highest estimate on the board.”
I gave the students a few moments to peruse the board and waited until many hands were raised. I called on Kellen, who told us 16 and 210. I checked in with the rest of the students to see if they agreed. After establishing consensus, I pulled the 16 sticky off the board and put it at the far left end near the bottom. Then I put the 210 sticky at the right side of the board.
“OK,” I said, “the rest of your estimates are between sixteen and two hundred ten. Those two numbers help us figure out the range. What is the range?”
I saw that some of the students were counting how many sticky notes were on the board instead of calculating the difference between the highest and lowest numbers, so I clarified my question. “I’m noticing that some of you are trying to figure out how many sticky notes are on the board. That will tell us how many pieces of data we have, how many estimates you made. But to find the range, you need to figure out the difference between our lowest and highest estimates, between sixteen and two hundred ten. The range tells us something about how spread apart our estimates are.” I recorded on the board the problem they needed to solve to find the range: 210 – 16.
This explanation helped the students get on track with what they had to do, but now I noticed that some students were struggling to do the arithmetic mentally. I told the students that they could work on the problem in their math journals if they liked. After a couple of minutes I called for their attention and asked who would like to tell us how he or she found the range. Here was a nice opportunity to have a brief number talk.
“It’s one hundred ninety-four,” D’andre announced. The class agreed with the answer.
“How do you know?” I asked.
“Because I just did two ten minus sixteen and that’s one ninety-four,” he explained. He held up his math journal to show that he had used the standard subtraction algorithm.
“Did anyone figure in a different way?” I asked.
“Tell us how you did that,” I encouraged.
“Well, I added four to sixteen to get to twenty. Then I added a hundred, so I was at one hundred twenty. Then I added ninety more to get to two hundred ten.”
“So how did you know the range was one hundred ninety-four?” I asked.
“Because I added four plus one hundred plus ninety, and that’s one ninety-four,” Marsha explained.
Since Marsha’s strategy might not make complete sense to students who had not used it or been exposed to it before, I decided to record it on the board to demonstrate her thinking for other students. I wrote:
16 + 4 = 20
20 + 100 = 120
120 + 90 = 210
4 + 100 + 90 = 194
Tina shared next. “I added and subtracted,” she said.
“How did you do that?” I queried.
Tina explained, “Well I started at sixteen and added one hundred. Then I was at one hundred sixteen. I added another one hundred and that made two hundred sixteen. I was past two hundred ten, so I had to subtract six.”
I realized this method called for a visual model as well. I wrote on the board:
16 + 100 = 116
116 + 100 = 216
216 – 6 = 210
100 + 100 = 200, 200 – 6 = 194
“Great,” I said. “The range is one hundred ninety-four, which helps us know about how far apart your lowest and highest estimates are. There are some other things to look at that also help you analyze data and figure out what it all might mean. But as I’m looking at the board, I still see a big mess. Let’s put the numbers in order. We already have the sixteen at one end and the two hundred ten at the other. Who would like to help arrange the rest of the numbers?”
Many hands flew into the air. I called on Esteban and Paulina to come forward and rearrange the sticky notes. They got a lot of help from the rest of the class. I helped them by asking questions such as, “Which is the next higher number?” and “Are there any numbers between these two numbers?” I also showed them that when two sticky notes had the same number, they should put one above the other. It took them only about a minute to get the sticky notes in order. They did not place the numbers proportionally, with spaces left for missing numbers, but merely from least to greatest. Still, we were now able to think about the median and the mode.
“Now let’s find the mode. How can we tell what the mode is?” I asked the class.
“It’s the number you see the most,” Tina reminded us.
“So what is the mode?” I asked. Because sticky notes with the same number were placed one above the other, it was easy to see that one hundred was the mode because it had the tallest column.
“Now let’s see if we can find the median. How can we do that?” I asked.
“The median is the one in the middle,” Chip explained.
“So how can we tell which one is in the middle?” I wondered aloud.
“Start at each end and move into the middle,” Nicole offered.
“Let’s try that,” I agreed.
I asked for two volunteers to help. Michael and Tanisa obliged. Michael started at 16 and Tanisa started at 210. They each put a finger on the first sticky note on their end and then simultaneously moved their finger toward the middle, one sticky note at a time. They met when Michael had a finger on 60 and Tanisa was touching 75.
“Hmm,” I observed, “there seem to be two sticky notes in the middle, not just one. I wonder what that’s about.”
“It’s because there’s an even number of estimates,” Miguel explained.
“Yes,” I agreed. “Does anybody know how we determine the median if there are two numbers in the middle instead of just one?”
There were universal looks of uncertainty, so I explained the mathematical convention to the students. “If you have an odd number of data, you can find the one that’s exactly in the middle, and that’s the median. If you have an even number, you take the two numbers in the middle and find the number that’s halfway between those two.”
Having told them the convention, I gave the students a few moments to determine that the median was 67.5.
“OK,” I said, “let’s review what we know about the data.” I wrote on the board:
Range: 194
Mode: 100
Median: 67.5
“This information tells us more about your estimates than when we just had a messy bunch of numbers on the board,” I said. “If we went outside and counted the cars parked around the school, we could then compare the actual number to this information. Then, if we collected estimates on another day, when a different number of cars would most likely be parked, or on a weekend day when no one was in school, we’d have different estimates to use to figure our range, mode, and median.”
Related Publication:
Minilessons for Math Practice, Grades 3–5
by Rusty Bresser and Caren Holtzman
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# SOLUTION: How do I go from knowing the rational roots (2, -3, -6) to the equation
Algebra -> Algebra -> Real-numbers -> SOLUTION: How do I go from knowing the rational roots (2, -3, -6) to the equation Log On
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Algebra: Real numbers, Irrational numbers, etc Solvers Lessons Answers archive Quiz In Depth
Question 284075: How do I go from knowing the rational roots (2, -3, -6) to the equation Answer by richwmiller(9144) (Show Source): You can put this solution on YOUR website!set up factors (x-2)*(x+3)*(x+6)=0 notice that when each factor is set to 0 they work out to the roots x-2=0 x=2 etc now multiply out the factors (x-2)*(x+3)*(x+6) use foil to multiply (x+3)*(x+6) x^2+9 x+18 now multiply each term by x and then do the same for -2 combine and add them together to get x^3+7x^2-36 = 0
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http://clay6.com/qa/18263/a-determinant-is-chosen-at-random-from-the-set-of-all-determinants-of-order
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# A determinant is chosen at random from the set of all determinants of order 2 with elements 0 or 1 only. The probability that the value of determinant chosen is +ve is
$\begin{array}{1 1} \frac{3}{16} \\ 3 \\ 4 \\ 5 \end{array}$
With 0 and 1 as elements there are $2\times 2\times 2\times 2=16$ determinants of order $2\times 2$ out of which only.
$\begin{vmatrix}1 &0\\0&1\end{vmatrix}1\begin{vmatrix}1 &1\\0&1\end{vmatrix}1\begin{vmatrix}1 &0\\1&1\end{vmatrix}$ are the three determinant whose value is +ve.
$\therefore$ Required probability =$\large\frac{3}{16}$
Hence (a) is the correct answer.
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# Thread: Help with basic explanation
1. ## Help with basic explanation
Hey guys, I'm really bad at maths and have to take this bridging course as I need to do one subject for my degree for the maths aspect of it. Was just wondering if anyone can help me or show me where I can get some help..
I know the answers the these, I just can't articulate it and provide proper examples, so frustrating! Here they are...
Explain in your own words why brackets need to have the highest precedence amongst all mathematical operators
Explain in your own words using everyday examples and diagrams:
a. Why adding a negative number is the same as subtracting a positive number of the same size.
b. Why subtracting a negative number is the same as adding a positive number of the same size.
c. Why multiplying a negative number by a positive number gives you a positive answer.
d. Why multiplying a negative number by a negative number gives a positive answer.
e. Why it is not possible to take the square root of a negative number.
2. Originally Posted by mathsilliterate
e. Why it is not possible to take the square root of a negative number.
Well consider $\sqrt{36} = -6$ and $6$
Why? because $-6\times -6 = 36$ and $6\times 6 = 36$ . As you need to find a number that multiplies by itself to make $36$
So for $\sqrt{-36}$ We need two numbers that are the same that multiply to give $-36$ . But because $-36$ is negative we can never find this number, as two numbers that mulitply to get a negative answer, one must be positive and the other neagtive, therefore never the same.
3. Originally Posted by pickslides
post
Awesome man! Thanks a lot and that actually makes sense to me. This is what I got for the others:
Explain in your own words why brackets need to have the highest precedence amongst all mathematical operators.
According to the BIDMAS rule, brackets must be done first in the instance that a certain combination of numbers need to be calculated before others - in order to get the desired answer. For example:
5 + 3 x 8 = 29
Brackets are not needed on that occasion. But when an answer needs to be calculated against this:
8 x (3 + 5) = 64
Numbers are being calculated in different orders to get the "desired" answer.
Do you guys think that's an ok answer?
4. Originally Posted by mathsilliterate
Explain in your own words why brackets need to have the highest precedence amongst all mathematical operators
Hmm, how would you explain this in words? To avoid confusion, perhaps? The PEMDAS order of operations is more of a convention than anything else, it is just a way to ensure that when faced with some sort of arithmetical computation, everyone will work it out the same way and get to the same answer. Brackets indicate what should be computed first, and so that takes precedence.
d. Why multiplying a negative number by a negative number gives a positive answer.
Based on the way the questions are phrased, it doesn't seem like rigorous proofs are required. Maybe example proofs can work though. So you can provide an example and give your reasoning in one shot. For example, for the above problem, see: Multiplying integers
They show that negative * negative = positive by using an example, -2 * -5 = 10.
A similar technique can be used to answer part (c)
5. Originally Posted by mathsilliterate
Awesome man! Thanks a lot and that actually makes sense to me. This is what I got for the others:
Explain in your own words why brackets need to have the highest precedence amongst all mathematical operators.
According to the BIDMAS rule, brackets must be done first in the instance that a certain combination of numbers need to be calculated before others - in order to get the desired answer. For example:
5 + 3 x 8 = 29
Brackets are not needed on that occasion. But when an answer needs to be calculated against this:
8 x (3 + 5) = 64
Numbers are being calculated in different orders to get the "desired" answer.
Do you guys think that's an ok answer?
it seems to be an ok answer here.
6. Originally Posted by mathsilliterate
Explain in your own words using everyday examples and diagrams:
a. Why adding a negative number is the same as subtracting a positive number of the same size.
b. Why subtracting a negative number is the same as adding a positive number of the same size.
for these, along with examples, i think you can use the number line to help you explain. remember, we think of adding positive numbers as moving to the right on the number line, subtracting positive numbers amounts to moving to the left.
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Teachers Pay Teachers
# Measurement and Data All Standards 2MD Second Grade Common Core Math Worksheets
Subjects
Resource Types
Product Rating
4.0
File Type
Word Document File
2.17 MB | 115 pages
### PRODUCT DESCRIPTION
Measurement and Data - 2MD Second Grade Common Core Math Worksheets
86 worksheets, some with answer keys covering every second grade common core measurement and data standard.
Measure and estimate lengths in standard units.
CCSS.MATH.CONTENT.2.MD.A.1
Measure the length of an object by selecting and using appropriate tools such as rulers, yardsticks, meter sticks, and measuring tapes.
CCSS.MATH.CONTENT.2.MD.A.2
Measure the length of an object twice, using length units of different lengths for the two measurements; describe how the two measurements relate to the size of the unit chosen.
CCSS.MATH.CONTENT.2.MD.A.3
Estimate lengths using units of inches, feet, centimeters, and meters.
CCSS.MATH.CONTENT.2.MD.A.4
Measure to determine how much longer one object is than another, expressing the length difference in terms of a standard length unit.
Relate addition and subtraction to length.
CCSS.MATH.CONTENT.2.MD.B.5
Use addition and subtraction within 100 to solve word problems involving lengths that are given in the same units, e.g., by using drawings (such as drawings of rulers) and equations with a symbol for the unknown number to represent the problem.
CCSS.MATH.CONTENT.2.MD.B.6
Represent whole numbers as lengths from 0 on a number line diagram with equally spaced points corresponding to the numbers 0, 1, 2, ..., and represent whole-number sums and differences within 100 on a number line diagram.
Work with time and money.
CCSS.MATH.CONTENT.2.MD.C.7
Tell and write time from analog and digital clocks to the nearest five minutes, using a.m. and p.m.
CCSS.MATH.CONTENT.2.MD.C.8
Solve word problems involving dollar bills, quarters, dimes, nickels, and pennies, using \$ and ¢ symbols appropriately. Example: If you have 2 dimes and 3 pennies, how many cents do you have?
Represent and interpret data.
CCSS.MATH.CONTENT.2.MD.D.9
Generate measurement data by measuring lengths of several objects to the nearest whole unit, or by making repeated measurements of the same object. Show the measurements by making a line plot, where the horizontal scale is marked off in whole-number units.
CCSS.MATH.CONTENT.2.MD.D.10
Draw a picture graph and a bar graph (with single-unit scale) to represent a data set with up to four categories. Solve simple put-together, take-apart, and compare problems1 using information presented in a bar graph.
Total Pages
115
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4.0
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# Search by Topic
#### Resources tagged with smartphone similar to Dotty Circle:
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### There are 66 results
Broad Topics > Information and Communications Technology > smartphone
### Weights
##### Stage: 3 Challenge Level:
Different combinations of the weights available allow you to make different totals. Which totals can you make?
### Make 37
##### Stage: 2 and 3 Challenge Level:
Four bags contain a large number of 1s, 3s, 5s and 7s. Pick any ten numbers from the bags above so that their total is 37.
### Two and Two
##### Stage: 3 Challenge Level:
How many solutions can you find to this sum? Each of the different letters stands for a different number.
### Consecutive Numbers
##### Stage: 2 and 3 Challenge Level:
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
### Number Daisy
##### Stage: 3 Challenge Level:
Can you find six numbers to go in the Daisy from which you can make all the numbers from 1 to a number bigger than 25?
### Consecutive Negative Numbers
##### Stage: 3 Challenge Level:
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
### Ben's Game
##### Stage: 3 Challenge Level:
Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. After this they all had the same number of counters.
### Summing Consecutive Numbers
##### Stage: 3 Challenge Level:
Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way?
### Squares in Rectangles
##### Stage: 3 Challenge Level:
A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all?
##### Stage: 3 Challenge Level:
A game for 2 or more people, based on the traditional card game Rummy. Players aim to make two `tricks', where each trick has to consist of a picture of a shape, a name that describes that shape, and. . . .
### Fence It
##### Stage: 3 Challenge Level:
If you have only 40 metres of fencing available, what is the maximum area of land you can fence off?
### Mirror, Mirror...
##### Stage: 3 Challenge Level:
Explore the effect of reflecting in two parallel mirror lines.
##### Stage: 3 Challenge Level:
How many different symmetrical shapes can you make by shading triangles or squares?
### Eight Hidden Squares
##### Stage: 2 and 3 Challenge Level:
On the graph there are 28 marked points. These points all mark the vertices (corners) of eight hidden squares. Can you find the eight hidden squares?
### Marbles in a Box
##### Stage: 3 Challenge Level:
How many winning lines can you make in a three-dimensional version of noughts and crosses?
### Special Numbers
##### Stage: 3 Challenge Level:
My two digit number is special because adding the sum of its digits to the product of its digits gives me my original number. What could my number be?
### Product Sudoku
##### Stage: 3 Challenge Level:
The clues for this Sudoku are the product of the numbers in adjacent squares.
### Who Is the Fairest of Them All ?
##### Stage: 3 Challenge Level:
Explore the effect of combining enlargements.
### Handshakes
##### Stage: 3 Challenge Level:
Can you find an efficient method to work out how many handshakes there would be if hundreds of people met?
### Children at Large
##### Stage: 3 Challenge Level:
There are four children in a family, two girls, Kate and Sally, and two boys, Tom and Ben. How old are the children?
### Route to Infinity
##### Stage: 3 Challenge Level:
Can you describe this route to infinity? Where will the arrows take you next?
### On the Edge
##### Stage: 3 Challenge Level:
If you move the tiles around, can you make squares with different coloured edges?
### Cuboid Challenge
##### Stage: 3 Challenge Level:
What size square corners should be cut from a square piece of paper to make a box with the largest possible volume?
### Painted Cube
##### Stage: 3 Challenge Level:
Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces?
### Cuboids
##### Stage: 3 Challenge Level:
Find a cuboid (with edges of integer values) that has a surface area of exactly 100 square units. Is there more than one? Can you find them all?
### An Unusual Shape
##### Stage: 3 Challenge Level:
Can you maximise the area available to a grazing goat?
### Sweet Shop
##### Stage: 3 Challenge Level:
Five children went into the sweet shop after school. There were choco bars, chews, mini eggs and lollypops, all costing under 50p. Suggest a way in which Nathan could spend all his money.
### Picturing Square Numbers
##### Stage: 3 Challenge Level:
Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153?
### American Billions
##### Stage: 3 Challenge Level:
Play the divisibility game to create numbers in which the first two digits make a number divisible by 2, the first three digits make a number divisible by 3...
### Make 100
##### Stage: 2 Challenge Level:
Find at least one way to put in some operation signs (+ - x ÷) to make these digits come to 100.
### Searching for Mean(ing)
##### Stage: 3 Challenge Level:
Imagine you have a large supply of 3kg and 8kg weights. How many of each weight would you need for the average (mean) of the weights to be 6kg? What other averages could you have?
### Days and Dates
##### Stage: 3 Challenge Level:
Investigate how you can work out what day of the week your birthday will be on next year, and the year after...
### Litov's Mean Value Theorem
##### Stage: 3 Challenge Level:
Start with two numbers and generate a sequence where the next number is the mean of the last two numbers...
### Largest Product
##### Stage: 3 Challenge Level:
Which set of numbers that add to 10 have the largest product?
### Counting Factors
##### Stage: 3 Challenge Level:
Is there an efficient way to work out how many factors a large number has?
### Attractive Rotations
##### Stage: 3 Challenge Level:
Here is a chance to create some attractive images by rotating shapes through multiples of 90 degrees, or 30 degrees, or 72 degrees or...
### Consecutive Seven
##### Stage: 3 Challenge Level:
Can you arrange these numbers into 7 subsets, each of three numbers, so that when the numbers in each are added together, they make seven consecutive numbers?
### How Much Can We Spend?
##### Stage: 3 Challenge Level:
A country has decided to have just two different coins, 3z and 5z coins. Which totals can be made? Is there a largest total that cannot be made? How do you know?
### Keep it Simple
##### Stage: 3 Challenge Level:
Can all unit fractions be written as the sum of two unit fractions?
### M, M and M
##### Stage: 3 Challenge Level:
If you are given the mean, median and mode of five positive whole numbers, can you find the numbers?
### Sissa's Reward
##### Stage: 3 Challenge Level:
Sissa cleverly asked the King for a reward that sounded quite modest but turned out to be rather large...
### Mixing More Paints
##### Stage: 3 Challenge Level:
Is it always possible to combine two paints made up in the ratios 1:x and 1:y and turn them into paint made up in the ratio a:b ? Can you find an efficent way of doing this?
### In the Playground
##### Stage: 1 and 2 Challenge Level:
What can you say about the child who will be first on the playground tomorrow morning at breaktime in your school?
### Rule of Three
##### Stage: 3 Challenge Level:
If it takes four men one day to build a wall, how long does it take 60,000 men to build a similar wall?
### How Many Miles to Go?
##### Stage: 3 Challenge Level:
How many more miles must the car travel before the numbers on the milometer and the trip meter contain the same digits in the same order?
### Cola Can
##### Stage: 3 Challenge Level:
An aluminium can contains 330 ml of cola. If the can's diameter is 6 cm what is the can's height?
### Sending a Parcel
##### Stage: 3 Challenge Level:
What is the greatest volume you can get for a rectangular (cuboid) parcel if the maximum combined length and girth are 2 metres?
### Fractions Jigsaw
##### Stage: 3 Challenge Level:
A jigsaw where pieces only go together if the fractions are equivalent.
### Mixing Paints
##### Stage: 3 Challenge Level:
A decorator can buy pink paint from two manufacturers. What is the least number he would need of each type in order to produce different shades of pink.
### 1 Step 2 Step
##### Stage: 3 Challenge Level:
Liam's house has a staircase with 12 steps. He can go down the steps one at a time or two at time. In how many different ways can Liam go down the 12 steps?
| 2,085 | 8,998 |
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## Section51.3Diffraction Through a Single Slit
In this section we study the diffraction phenomenon in more detail. As our first example, we consider a light source in front of a narrow slit in an opaque material as shown in Figure 51.3.1. Suppose the light source is either far away or at the focal plane of a lens so that the waves are planar at the slit.
We place a screen behind the slit to observe the diffraction pattern there when it is placed at different distances from the slit. If the screen is placed immediately behind the slit, we find a shadow of the slit on the screen. When we move the screen further out, the shadow develops fringes whose pattern changes as you move the screen further out. These patterns are called Fresnel or near-field diffraction}. Moving the screen further away from the slit you reach a region where the patterns stabilize and although they spread out more as you move the screen further away, the pattern remains the same. We call this pattern the Fraunhofer or far-field diffraction after Joseph von Fraunhofer, who built the first diffraction grating consisting of 260 closely, spaced parallel wires in 1821, and measured wavelengths of colored lights.
In the following, we will study a trick that lets us predict the directions for diffraction minima. In the next section we will study variation of intensity in a diffraction pattern. We can use the formula of intensity to find the direction of diffraction maxima as well as minima and the width of the peaks corresponding to the maxima.
### Subsection51.3.1Minima in a Single-slit Diffraction
How can we predict the locations of bright and dark circles on the screen of a single-slit Fraunhofer diffraction? It turns out that we can find the minima of the diffraction on the screen by a clever procedure rather than looking at how intensity of the wave varies with the direction. The trick is to imagine the wavefront at the slit as consisting of infinite number of wave sources and try to pair up two points whose waves will produce a destructive interference on the screen.
For instance, say we divide up the slit into two equal zones, we obtain one set of pairs of sources which consist of a point at the top of the upper zone and a point at the top of the lower zone, the next point of the top zone is paired with the corresponding point of the lower zone, etc., as illustrated in Figure 51.3.2.
Since, the screen is far away, all the rays will be approximately parallel there and the path differences between the members of every pair will be same, regardless of their location in the zones. Thus if the path difference of one of these pairs is equal to half a wavelength, they will all interfere destructively at the screen.
We can work out the path difference from the triangle in Figure 51.3.2. With the slit width $b\text{,}$ the path difference between rays in the direction $\theta$ from the top of one zone and and the top of the zone below it will be
\begin{equation*} \frac{b}{2}\, \sin\theta. \end{equation*}
To be destructive, this will have to equal to $\lambda_2\text{.}$ You might say that we can capure this by making this equal to an odd multiple of $\lambda/2\text{,}$ but this trick does not work that way. To find the other diffraction minima, the trick is to divide th wavefront at the slit into 4 parts, 8 parts, 16 parts, etc., and make sure that the path difference between top of one zone and the top of the immediate next zone is $\pm \lambda/2\text{.}$
\begin{equation*} \frac{b}{2}\, \sin\theta = \pm \frac{\lambda}{2}. \end{equation*}
Simplifying this formula we get, as we will recognize as $m^\prime=\pm 1$ diffraction minimum when we study the intensity below,
$$b\sin\theta = \pm \lambda.\tag{51.3.1}$$
Similarly, if you divide up the wave at the slit into four equal zones as shown in Figure 51.3.3.
Let us label the zones 1, 2, 3 and 4,. We ensure that (1,2) cancel each other and (3,4) cancel each other. Since, the screen is supposed to be far away, we will get the same condition for the two pairs.
\begin{equation*} \frac{b}{4}\, \sin\theta = \pm \frac{\lambda}{2}. \end{equation*}
Simplifying this formula we get, as we will recognize as $m^\prime=\pm 2$ diffraction minimum when we study the intensity below,
$$b\sin\theta = \pm2\lambda.\tag{51.3.2}$$
You can generalize your results to
\begin{equation*} b\sin\theta = m^\prime \lambda, \ \ \ (m^\prime = \pm 1, \pm 2, \pm 3, \cdots.) \end{equation*}
Light of wavelength $550\text{ nm}$ passes through a slit of width $2.0\:\mu\text{m}\text{.}$ Find the location of $m^\prime=-2,\ -1,\ 1,\ 2$ minima on a screen $50\text{ cm}$ away in terms of (a) angles subtendend with the horizontal direction from the center of the slit, and (b) $y$ coordinates of the positions of the dark bands using a $y$ axis pointed up with origin at the center of the central maximum.
Hint
(a Use the condition for minima. (b) Use $y=L\tan\theta\text{.}$
(a) $16^{\circ},\ 33.4^{\circ}\text{,}$ (b) $\pm 14\ \textrm{cm},\ \pm 33\ \textrm{cm}$ .
Solution 1 (a)
(a) The minima are given by the following condition
\begin{equation*} b\sin\theta = m^\prime \lambda \ \ (m^\prime = \pm 1, \pm 2, \pm 3, \cdots)\ \ \text{(destructive)}. \end{equation*}
Hence, the four minima around the central maxima will have $m^\prime = \pm 1, \pm 2\text{.}$ The corresponding directions from the slit are
\begin{align*} \amp \theta_{\pm 1} = \pm\sin\left( \frac{\lambda}{b}\right) = \pm 16^{\circ}\\ \amp \theta_{\pm 2} = \pm\sin\left( \frac{2\lambda}{b}\right) = \pm 33.4^{\circ} \end{align*}
Pictorially, the directions from the slit for the minima are illustrated in Figure 51.3.5.
Solution 2 (b)
The location of the diffraction minima on the screen can be deduced from the right-angled triangles. Let the $y$-axis be pointed up on the screen, then we will have
\begin{equation*} y = (50\ \textrm{cm})\ \tan\theta. \end{equation*}
Denote the positions of the four minima by $y_{+1}\text{,}$ $y_{-1}\text{,}$ $y_{+2}$ and $y_{-2}\text{.}$
\begin{align*} \amp y_{\pm 1} = (50 \textrm{cm})\ \tan\theta_{\pm 1} = \pm 14\ \textrm{cm}\\ \amp y_{\pm 2} =(50 \textrm{cm})\ \tan\theta_{\pm 2} = \pm 33\ \textrm{cm} \end{align*}
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# Yahoo Web Search
1. ### True or false: if f'( r ) exists, then lim x -> r f(x) = f( r )?
assume r is any number. { r ∈ ℝ } as x becomes sufficiently close to that number, f(x) becomes arbitrarily close to f( r ). for example: say r = 7 which means: f( r ) = f(7) as x approaches...
1 Answers · Science & Mathematics · 14/10/2013
2. ### Make a the subject of R = ?
R =(ab)/(a+b) R (a+b) = ab Ra + Rb = ab Rb = ab - Ra Rb = a(b- R ) a = (Rb)/(b- R )
1 Answers · Education & Reference · 01/11/2008
3. ### how to understand r = 4csc(theta) ?
r = 4 cscθ then r sinθ = 4. now, r cosθ = x, while r sinθ = y.... these are the equations needed to transform cartesian to polar ... thus you have y = 4. (a horizontal line) §
1 Answers · Science & Mathematics · 14/04/2008
4. ### Prove that R is commutative if and only if R ' is commutative?
Let f: R -> R ' be an isomorphism. If R is commutative, and a... of R ', then a' = f(a) and b' = f(b) for some a,b in R . Then we have: a'b' = f(a)f(b) = f(ab) = f(ba) = f(b)f(a) = b'...
2 Answers · Science & Mathematics · 05/12/2012
5. ### help with an easy math problem making ' r ' the subject?
r ^2+ r =A/(2ph)? If so, then r ^2+ r -A/(2ph)=0=> r =[-1+/-sqrt(1+2A/(ph)]/2 => r =[-1+sqrt(1+2A/(ph))]/2 or r =[-1-sqrt(1+2A/(ph))]/2
2 Answers · Science & Mathematics · 20/09/2013
6. ### What is the rectangular equation of r =sin3theta?
Use the facts that sin(θ) = y/ r and r = √(x^2 + y^2). First, for r = 3sin(θ): r = 3sin(θ) r = 3(y/ r ... is a circle. In the subject line, I'll assume you mean r = sin(3θ) and not r = sin^3 (θ). You'll have to use a trig identity for...
1 Answers · Science & Mathematics · 19/05/2007
7. ### Math: Solve for c if r = 7(c-x)?
r = 7(c - x) r /7 = 7(c-x)/7 r /7 = c - x r /7 + x = c
2 Answers · Science & Mathematics · 16/02/2012
8. ### Make R subject of formula: please?
( R + r )( R - r ) = A/pi R²-r² = A/pi <--- this is the difference of two squares method R² = A/pi + r² R = √(A/pi+r²)
4 Answers · Science & Mathematics · 26/04/2012
9. ### Help! Algebra! | R - 3 | = 10, {__ , __}?
- ( R -3 ) = 10 - r +3 = 10 -3 -3 --------------- - r =7 ----------- -1 -1 r = -7 ( r -3)=10 +3 +3 -------------- r =13 {-7,13}
2 Answers · Science & Mathematics · 02/11/2011
10. ### How do I convert r =2/sin(theta) to rectangular?
r = 2 / sin(t) r * sin(t) = 2 y = 2 x = r * cos(t) y = r * sin(t) x^2 + y^2 = r ^2
2 Answers · Science & Mathematics · 03/12/2017
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https://web2.0calc.com/questions/if-terry-took-1-day-of-rest-every-9-5-days-out-of-142-how-many-days-did-he-actually-run
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+0
# if terry took 1 day of rest every 9.5 days out of 142 how many days did he actually run
+3
535
5
if terry took 1 day of rest every 9.5 days out of 142 how many days did he actually run
Guest Sep 26, 2014
#3
+92429
+8
Dviding 142/9.5 we have that there are about 14.9 "blocks" of 9.5 days in 142 days.Now, assuming that the day of rest always occurs at the end of the 9.5 day period, then he rested about 14 days. Notice that 14 * 9.5 = 133 days. And 8.5 more days = 141.5 days......so, to be technically correct, he still has 1/2 of day of rest at the end. So he actually rests 14.5 days !!!!
CPhill Sep 26, 2014
#1
+3
Guest Sep 26, 2014
#2
+94088
+8
8.5/9.5*142
Melody Sep 26, 2014
#3
+92429
+8
Dviding 142/9.5 we have that there are about 14.9 "blocks" of 9.5 days in 142 days.Now, assuming that the day of rest always occurs at the end of the 9.5 day period, then he rested about 14 days. Notice that 14 * 9.5 = 133 days. And 8.5 more days = 141.5 days......so, to be technically correct, he still has 1/2 of day of rest at the end. So he actually rests 14.5 days !!!!
CPhill Sep 26, 2014
#4
+94088
+3
So mine is wrong? I will admit I did not think about it very hard but i thought it would be ok.
Melody Sep 26, 2014
#5
+8263
0
??? What do you mean?
😧😧😧
DragonSlayer554 Sep 26, 2014
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Indetifying a vector space
I'm trying to identify which sets are vector spaces. I know that a set $V$ is a vector space if its elements (vectors) have addition and scalar multiplication so that the result is also within the set $V$. I'm trying to identify the following sets.
1. Set of real $m\times n$ matrices $\mathbb{R}^{m\times n}$
If we do operations for two matrices it always yields and matrix within $\mathbb{R}^{m\times n}$. So this is a vector space.
1. Line $\{(x,4x) \in \mathbb{R}^2 \: | \: x\in\mathbb{R}\}$
EDIT: Is a vector subspace of $\mathbb{R}^2$.
1. $\mathbb{R}^n = \{(x_1, \ldots, x_n) \ \vert \ x_i \in \mathbb{R}\}$
Clearly a vector space because for all $n$-vectors in it the operations hold.
1. Curve $\{(x,x^2) \in \mathbb{R}^2 \: | \: x\in\mathbb{R}\}$
Addition does not hold on this set $(1,1) + (2,4) = (3,5) \notin V$ so it is not a vector space.
1. Line $\{(x,3x-2) \in \mathbb{R}^2 \: | \: x\in\mathbb{R}\}$. Same as the other line and not an vector space.
EDIT: Under the vector space axioms addition $\mathbf a + \mathbf b \in V$ when $\mathbf a, \mathbf b \in V$ and $\lambda \mathbf a \in V$ thus if $\mathbf b = -1\cdot \mathbf a$ then $\mathbf a + \mathbf b = \mathbf 0 \notin V$. So additive indentity does not hold and the set is not a vector space.
1. Set of matrices in $\{ M\in\mathbb{R}^{n\times n}\: | \: \det(M)=1\}$.
\begin{align*} \text{det}\left(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}\right) = 4 \neq 1\end{align*}
OR
\begin{align*} \text{det}\left(\lambda \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \right) = \lambda \neq 1 \text{ when } \lambda \neq 1\end{align*}
EDIT: So addition from said set yields a matrix with non 1 determinant. Not a vector space.
• Your reasoning that $\tt 3.$ isn't a vector space isn't correct, $\mathbf{x}+\mathbf{y}=(3,12)$ has the form $(\alpha,4\alpha)$, so it is certainly in that set. Commented Feb 21, 2016 at 12:40
• Number 2. seems a vector subspace of $\mathbb{R}^2$ to me. Take two elements in the set, $(\lambda, 4\lambda)$ and $(\mu, 4\mu)$, then $(\lambda, 4\lambda) + (\mu, 4\mu) = (\lambda+\mu , 4(\lambda+\mu))$ which is of the required form. Also if $k\in \mathbb{R}$ then $k(\lambda, 4\lambda)= (k\lambda, 4k\lambda)$ is of the required form. An easy intuition is that this set is a straight line that contains the 0-vector, so it should indeed be a vector space. Commented Feb 21, 2016 at 12:40
• Reasoning for 5 is wrong although deduction is correct. 5-is not a vector space as the additive identity is not present in the set. Commented Feb 21, 2016 at 12:42
• 6 is clearly not a vector space. Take 2 identity matrices, add them and take their determinant Commented Feb 21, 2016 at 12:44
• To qualify as a vector space , don't miss the fact that the additive identity should also be a part of the space Commented Feb 21, 2016 at 12:47
Hints:
1. The deduction is correct, although you should actually check all the vector space axioms to conclude that it is one.
2. Your reasoning that $\tt 3.$ isn't a vector space isn't correct, $\mathbf{x}+\mathbf{y}=(3,12)$ has the form $(\alpha,4\alpha)$, so it is certainly in that set. $-$ Edit 1: That's correct, assuming that you already know that $\mathbf{R}^2$ is a vector space.
3. Same comment as in 1.
6. I assume that the field you're working over is $\mathbf{R}$. So take an element $c\in\mathbf{R}$, and suppose that $M$ is in that set, does it follow that $\det(cM)=1$? (i.e. does it follow that it is closed under scalar multiplication?) $-$ Edit 1: You showed that it isn't closed under addition, hence not a vector space, and that's correct.
Number (2) seems a vector subspace of $\mathbb{R}^2$ to me. Take two elements in the set, $(\lambda, 4\lambda)$ and $(\mu, 4\mu)$, then $(\lambda, 4\lambda) + (\mu, 4\mu) = (\lambda+\mu , 4(\lambda+\mu))$ which is of the required form. Also if $k\in \mathbb{R}$ then $k(\lambda, 4\lambda)= (k\lambda, 4k\lambda)$ is of the required form. An easy intuition is that this set is a straight line that contains the 0-vector, so it should indeed be a vector space.
For the last one (6) just use the following fact about determinants of matrices. If $A$ i an $n\times n$-matrix and $\lambda\in \mathbb{R}$ a real scalar, then $$\det (\lambda A) =\lambda^n \det(A) = \lambda^n,$$ where the latter follows if $A$ is in the set you defined.
Now, if $\lambda \neq 1$ then $\det(\lambda A) = \lambda^n \neq 1$ and hence $\lambda A$ is not in the set you defined, and hence it cannot be a vector space.
A last comment, since $a,b\in V$ then $a+b\in V$ and $a\in V$, $\lambda \in \mathbb{R}$ then $\lambda a\in V$, this in particular implies that $0\in V$ (the zero vector should always belong to the space, take $a\in V$ then $(-1)*a =-a\in V$ and hence $a-a=0\in V$). Then, since checking this is fairly quick, you may do this for number (5), where $(0,0)\notin \{(x,3x-2)\in \mathbb{R}^2|x\in \mathbb{R}\}$.
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# Thread: Find the domain of this function?
1. ## Re: Find the domain of this function?
Originally Posted by explodingtoenails
So when would abs bars actually affect the value of a binomial?
using the definition of absolute value (which I recommend you take another look at in your textbook or other resource)
$|x^2 - 4| = x^2 - 4$ if ( $x^2-4) \ge 0$
$|x^2 - 4| = -(x^2 - 4) = 4-x^2$ if $(x^2-4) < 0$
now, which case should be used if $|x^2-4| = 0$ ?
2. ## Re: Find the domain of this function?
Originally Posted by skeeter
using the definition of absolute value (which I recommend you take another look at in your textbook or other resource)
$|x^2 - 4| = x^2 - 4$ if ( $x^2-4) \ge 0$
$|x^2 - 4| = -(x^2 - 4) = 4-x^2$ if $(x^2-4) < 0$
now, which case should be used if $|x^2-4| = 0$ ?
The first case?
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+0
# Algebra... Help~
+1
38
1
+476
Find all \$p\$ which satisfy both the inequalities \$0\ge 54p-144\$ and \$0>12-20p\$. Express your answer in interval notation, reducing any fractions in your answer.
gueesstt Apr 14, 2018
Sort:
#1
+6943
+6
Find all \(p\) which satisfy both the inequalities \(0\ge 54p-144\) and \(0>12-20p\) .
0 ≥ 54p - 144
Add 144 to both sides of the inequality.
144 ≥ 54p
Divide both sides by 54 .
144/54 ≥ p
8/3 ≥ p
0 > 12 - 20p
Add 20p to both sides.
20p > 12
Divide both sides by 20 .
p > 12/20
p > 3/5
We can combine 8/3 ≥ p and p > 3/5 to get...
8/3 ≥ p > 3/5 which is the same as...
3/5 < p ≤ 8/3 And 8/3 is greater than 3/5 so this is valid.
In interval notation, it is (3/5, 8/3]
hectictar Apr 14, 2018
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## Friday, March 27, 2015
### Algebraic side effects
Haskell differentiates itself from most other functional languages by letting you reason mathematically about programs with side effects. This post begins with a pure Haskell example that obeys algebraic equations and then generalizes the example to impure code that still obeys the same equations.
#### Algebra
In school, you probably learned algebraic rules like this one:
``f * (xs + ys) = (f * xs) + (f * ys)``
Now let's make the following substitutions:
• Replace the mathematical multiplication with Haskell's `map` function
• Replace the mathematical addition with Haskell's `++` operator
These two substitutions produce the following Haskell equation:
``map f (xs ++ ys) = (map f xs) ++ (map f ys)``
In other words, if you concatenate the list `xs` with the list `ys` and then `map` a function named `f` over the combined list, the result is indistinguishable from `map`ping `f` over each list individually and then concatenating them.
Let's test this equation out using the Haskell REPL:
``````>>> map (+ 1) ([2, 3] ++ [4, 5])
[3,4,5,6]
>>> (map (+ 1) [2, 3]) ++ (map (+ 1) [4, 5])
[3,4,5,6]``````
#### Evaluation order
However, the above equation does not hold in most other languages. These other languages use function evaluation to trigger side effects, and therefore if you change the order of evaluation you change the order of side effects.
Let's use Scala to illustrate this. Given the following definitions:
``````>>> def xs() = { print("!"); Seq(1, 2) }
>>> def ys() = { print("?"); Seq(3, 4) }
>>> def f(x : Int) = { print("*"); x + 1 }``````
.. the order of side effects differs depending on whether I concatenate or map first:
``````>>> (xs() ++ ys()).map(f)
!?****res0: Seq[Int] = List(2, 3, 4, 5)
>>> (xs().map(f)) ++ (ys().map(f))
!**?**res1: Seq[Int] = List(2, 3, 4, 5)``````
One line #1, the two lists are evaluated first, printing `"!"` and `"?"`, followed by evaluating the function `f` on all four elements, printing `"*"` four times. On line #2, we call `f` on each element of `xs` before beginning to evaluate `ys`. Since evaluation order matters in Scala we get two different programs which print the punctuation characters in different order.
#### The solution
Haskell, on the other hand, strictly separates evaluation order from side effect order using a two-phase system. In the first phase you evaluate your program to build an abstract syntax tree of side effects. In the second phase the Haskell runtime executes the tree by interpreting it for its side effects. This phase distinction ensures that algebraic laws continue to behave even in the presence of side effects.
To illustrate this, we'll generalize our original Haskell code to interleave side effects with list elements and show that it still obeys the same algebraic properties as before. The only difference from before is that we will:
• Generalize pure lists to their impure analog, `ListT`
• Generalize functions to impure functions that wrap side effects with `lift`
• Generalize `(++)` (list concatenation) to `(<|>)` (`ListT` concatenation)
• Generalize `map` to `(=<<)`, which streams elements through an impure function
This means that our new equation becomes:
``````-- f * (xs + ys) = (f * xs) + (f * ys)
f =<< (xs <|> ys) = (f =<< xs) <|> (f
=<< ys)``````
You can read this as saying: if we concatenate `xs` and `ys` and then stream their values through the impure function `f`, the behavior is indistinguishable from streaming each individual list through `f` first and then concatenating them.
Let's test this equation out with some sample definitions for `xs`, `ys`, and `f` that mirror their Scala analogs:
``````>>> import Control.Applicative
>>> import Pipes
>>> let xs = do { lift (putChar '!'); return 1
<|> return 2 }
>>> let ys = do { lift (putChar '?'); return 3
<|> return 4 }
>>> let f x = do { lift (putChar '*'); return (x + 1) }
>>> runListT (f =<< (xs <|> ys)) -- Note:
!**?**>>> runListT ((f =<< xs) <|> (f =<< ys))
!**?**>>>``````
The resulting punctuation order is identical. Many people mistakenly believe that Haskell's mathematical elegance breaks down when confronted with side effects, but nothing could be further from the truth.
#### Conclusion
Haskell preserves algebraic equations even in the presence of side effects, which simplifies reasoning about impure code. Haskell separates evaluation order from side effect order so that you spend less time reasoning about evaluation order and more time reasoning about your program logic.
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# HELP - I FORGOT ALGEBRA
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HELP - I FORGOT ALGEBRA [#permalink]
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04 Aug 2010, 18:28
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I can set up the equation for GMAT CLUB TEST 15 problem 20 but cannot seem to solve. Is there anyone that can break this down for me? I am banging my head up against the wall in frustration. (see below).
Tom read a book containing 480 pages by reading the same number of pages each day. If he would have finished the book 5 days earlier by reading 16 pages a day more, how many days did Tom spend reading the book?
(C) 2008 GMAT Club - m15#20
10
12
15
16
18
Any help is much appreciated!
[Reveal] Spoiler: OA
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Re: HELP - I FORGOT ALGEBRA [#permalink]
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04 Aug 2010, 21:23
I would work backwards.
1) You are given number of days, so divide 480 by one of the choices to get the number of pages read per day.
3) Divide 480 by that total, see if that gets to the 5 less than number of days you used.
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Re: HELP - I FORGOT ALGEBRA [#permalink]
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05 Aug 2010, 07:24
If you wish to formally set it up, you could write two equations with two unknowns. In this case, we do not know how many pages per day he was reading (I will call this rate r), and we don't know how many days it took him to read the book (I will call this d).
Using dimensional analysis we know that
total # of pages * # of days per page = # of days spent
(pgs) * (days/pg) = (days)
$$480 * \frac{1}{pgs/day} = d$$
$$480 * \frac{1}{r} = d$$
Similarly, we can write a second equation for the additional piece of given information (that we could have finished 5 days sooner reading at a rate of 16 more pages a day).
$$480 * \frac{1}{(r+16)} = d-5$$
Now we have two equations with two unknowns. I would solve one of the two for r and then substitute into the other equation for r. This will give you a single equation that you can solve for d, the desired quantity.
In this case the final equation is a quadratic that can be factored to yield two solutions, one of which is negative and thus can be ignored (it didn't take him negative days to complete the book).
$$(d-15)(d+10) = 0$$
Hence d must be 15.
As you can see, this process is likely more time consuming than using the multiple choices to work backwards.
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Re: HELP - I FORGOT ALGEBRA [#permalink]
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05 Aug 2010, 08:23
My method is a little time intensive:
let n be the days taken to read the book initially
now the with increased number of pages being read everyday (+16) days taken = n-5
We can get the number of pages read per day by
number of pages in the book / days taken to read the book
so with slow speed = 480/n
with fast speed = 480/(n-5)
We also know that the difference between the number of pages per day between the two different speeds is 16
480/(n-5) - 480/n = 16
approach 1 : simplify it make it a quadratic equation and you will get the answer as 15
approach 2 : plug int he values
10 = > (96-48) ! = 16
12 = > 480/7 - 40 != 16
15 = > 48 - 32 = 16 <Bingo>
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Re: HELP - I FORGOT ALGEBRA [#permalink]
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05 Aug 2010, 11:04
This can be solved by setting up two equations concerning the total number of pages.
Let him read p pages per day initially and let it take d days for him to finish reading. So the total number of pages is pd.
$$pd = 480$$ (1)
In the other condition stated, it says that he could have finished the book 5 days earlier if he had read 16 more pages per day.
This means that he reads (p+16) pages per day and it takes him (d-5) days to finish the book. The total number of pages in this case is (p+16)(d-5).
But note that this is the same as the total number of pages hasn't changed.
$$(p+16)(d-5) = 480$$
Expanding this we get: $$pd - 5p + 16d - 80 = 480$$ (2)
Substituting (1) into (2) in place of pd, we get:
$$480 - 5p + 16d - 80 = 480$$
$$16d - 5p = 80$$
Now we can substitute $$p = \frac{480}{d}$$ into this equation to get the following quadratic:
$$16d - \frac{2400}{d} = 80$$
$$16d^2 - 80d - 2400 = 0$$
Dividing by 16 throughout
$$d^2 - 5d - 150 = 0$$
$$d^2 - 15d + 10d - 150 = 0$$
$$(d-15)(d+10) = 0$$
$$d = 15$$ or $$-10$$
But d, being the number of days, cannot be negative. Hence d = 15 is the answer. Answer choice C.
Hope this helps.
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Re: HELP - I FORGOT ALGEBRA [#permalink]
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05 Aug 2010, 22:09
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toddrud wrote:
I can set up the equation for GMAT CLUB TEST 15 problem 20 but cannot seem to solve. Is there anyone that can break this down for me? I am banging my head up against the wall in frustration. (see below).
Tom read a book containing 480 pages by reading the same number of pages each day. If he would have finished the book 5 days earlier by reading 16 pages a day more, how many days did Tom spend reading the book?
(C) 2008 GMAT Club - m15#20
10
12
15
16
18
Any help is much appreciated!
Better work with choices....
E is out straightaway because 480/18 is not an integer....Tom can not read pages in fractions....
Now With choices...
A
480/10 = 48 pages add 16 pages 64 480/64 doesnt gives 5 days (not an integer)
B
480/12 = 40 pages add 16 pages 56 480/56 doesnt gives 7 days (not an integer)
C
480/15 = 32 pages add 16 pages 48 gives 480/48 as 10 days......Answer...
No need to check D
Hope this helps ......( I would better avoid forming equations in this case)
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Re: HELP - I FORGOT ALGEBRA [#permalink] 05 Aug 2010, 22:09
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Home » College Entrance ExamLETPMA Entrance ExamReviewersUPCAT » Classification of Triangles (Plus, Theorems and Postulates)
Classification of Triangles (Plus, Theorems and Postulates)
Tasty pizza, magnificent roofs, and timeless pyramids: What’s common among these three?
Yes, they are all triangles!
With their simple form consisting of three sides, triangles are one of the essential elements in geometry. Mathematicians’ fascination with these figures paved the way for numerous engineering, architecture, and navigation developments.
Using this reviewer, you can refresh your knowledge of triangles, their classification, properties, and valuable theorems.
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Classification of Triangles
We all know that triangles are three-sided polygons with three angles. Let us start our discussion of triangles with their classification. We can classify triangles according to their interior angles or sides.
1. Classification of Triangles According to Angles
We can classify triangles according to the measurement of their interior angles.
a. Acute Triangle
A triangle is acute if all of its interior angles are acute. This means that all of the interior angles of an acute triangle have a measurement between 0 degrees to 90 degrees.
b. Right Triangle
A right triangle has precisely one interior angle, a right angle (90° angle), and two interior angles are acute angles.
Right triangles are of great importance in geometry (and trigonometry). It is the subject of the famous mathematical theorem called the Pythagorean theorem. That’s why we must dedicate a separate discussion to this particular type of triangle.
c. Obtuse Triangle
An obtuse triangle has precisely one obtuse interior angle, and two interior angles are acute.
2. Classification of Triangles According to Sides
If we classify triangles according to their sides, we will have three types of triangles: scalene, isosceles, and equilateral.
a. Scalene Triangle
A scalene triangle has all of its sides non-congruent. This means that all sides of a scalene triangle have different measurements.
b. Isosceles Triangle
An isosceles triangle has two congruent sides. This means that two sides of an isosceles triangle have equal measurement or length.
c. Equilateral Triangle
An equilateral triangle has congruent sides. This is the opposite of a scalene triangle.
Congruent Triangles
Two triangles are congruent if all their corresponding sides and angles are equal. For instance, take a look at the given image below:
AB is congruent to its corresponding side DE; AC is congruent to its corresponding side DF; and BC is congruent to its corresponding side EF. Furthermore, all the corresponding angles of the triangles are equal in degree measurement. Therefore, triangles ABC and DEF are congruent triangles.
1. Corresponding Sides of Triangles
Look at the given triangles below. These triangles are named Triangle ABC and Triangle DEF.
Note that side AB of triangle ABC is in the same position as side DE of triangle DEF. Thus, we call them corresponding sides
Therefore, the corresponding sides of triangles are the sides of triangles located in the same location.
2. Corresponding Angles of Triangles
The corresponding angles of triangles are similar to the concept of corresponding sides. Corresponding angles of triangles are the angles of triangles located in the same location.
In the figure below, angles 1 and 4 are corresponding angles. Similarly, angles 2 and 5 are also corresponding angles.
Why do we have to talk about corresponding sides and angles? Well, an important theorem relates congruent triangles to their corresponding parts (the sides and angles). We will discuss this theorem in-depth in the next section.
3. Corresponding Parts of a Triangle are Congruent (CPCTC) Theorem
CPCTC theorem states that if two triangles are congruent, their corresponding parts (sides and angles) are also congruent.
Suppose that triangles ABC and DEF below are congruent triangles:
According to the CPCTC theorem, these triangles’ corresponding sides and angles are also congruent. Thus, side AB is congruent to side DE, side BC is congruent to side EF, and side AC is congruent to side DF.
Similarly, angle 1 is congruent to angle 4, angle 2 is congruent to angle 5, and angle 3 is congruent to angle 6.
The tick marks above show which sides and angles are congruent.
Sample Problem: Suppose that triangles PRQ and PTS are congruent. Determine the measurement of angle y.
Solution:
Based on the figure above, angle PRQ, which measures 45 degrees, and angle PTS (or angle y) are corresponding angles. Note that triangles PRQ and PTS are congruent. Hence, we can conclude that angles PRQ and PTS are congruent using the CPCTC theorem. Therefore, angle y (or angle PTS) is also 45 degrees.
Triangle Congruence Postulates and Theorems
This section discusses other criteria for determining whether two triangles are congruent. These criteria are some of the geometric postulates that allow us to conclude that two triangles are congruent even without knowing the measurements of every side and angle of the triangles.
1. Side-Angle-Side (SAS) Postulate
SAS Postulate states that if two corresponding sides and their included angle (the angle located between them) of a triangle are congruent, the triangles are congruent.
In the example below, note that sides AB and DE are congruent (15 cm). Meanwhile, sides AC and DF are also congruent (15 cm). Their included angles are also congruent (both 40 degrees). By SAS Postulate, we can conclude that triangles ABC and DEF are congruent.
2. Angle-Side-Angle (ASA) Postulate
ASA Postulate states that if two corresponding angles and their included side (the side located between them) of a triangle are congruent, the triangles are congruent.
In the figure below, angles LJK and OMN are congruent (40 degrees). Meanwhile, angles JLK and MON are also congruent (60 degrees). Their included sides (JL and MO, respectively) are also congruent (15 cm). As per the ASA postulate, we can conclude that triangles JKL and MNO are congruent.
3. Side-Side-Side (SSS) Postulate
The SSS postulate is the most intuitive of the three. It states that if all three sides of two triangles are congruent, they are congruent.
In the figure below, we can conclude using the SSS postulate that the triangles PQR and XYZ are congruent since all of their corresponding sides are congruent.
More Useful Theorems on Triangles
This section will discuss two useful theorems about triangles that you must learn in geometry.
1. Isosceles Triangle Theorem
“If two sides of a triangle are congruent (i.e., that triangle is an isosceles triangle), then the angles opposite those sides are congruent.”
As we have mentioned earlier, an isosceles triangle is one whose two sides are congruent (the remaining side is not congruent to either).
The isosceles triangle theorem tells us that in an isosceles triangle, the angles opposite to the congruent or equal sides are also congruent. These angles are also called base angles.
In the figure above, the triangle PQR is isosceles. Side PQ is congruent to side QR. The angles opposite to these congruent sides are angles ∠QPR and ∠PRQ. Per the isosceles triangle theorem, we can state that angles ∠QPR and ∠PRQ are congruent. If m∠QPR = 40°, then m∠PRQ = 40°.
Sample Problem: Using the exact figure above, determine the measure of ∠PQR if m∠QPR and m∠PRQ = 40°
By the isosceles triangle theorem, ∠QPR and ∠PRQ are congruent angles. Hence, m∠QPR = 40°.
Now, how do we find the measure of ∠PQR?
If you remember, the sum of the interior angles of a triangle is always 180°. Therefore, if we add the measurements of angles ∠PQR, ∠QPR, and ∠PRQ, then the sum must be 180°:
m∠PQR + m∠QPR+ m∠PRQ = 180°
We know that both m∠QPR and m∠PRQ are 40°:
m∠PQR + 40 + 40 = 180°
Let us solve for m∠QRP:
m∠PQR + 40° + 40° = 180°
m∠PQR + 80° = 180° Combining like terms
m∠PQR = – 80° + 180° Transposition method
m∠PQR = 100°
Therefore, the measurement of ∠PQR is 100°.
2. Triangle Inequality Theorem
The sum of any two sides of a triangle is always greater than or equal to the third side.”
Let us understand this theorem by analyzing the image above. We have assigned variables to the lengths of the triangle’s sides in the figure. The side lengths are a, b, and c.
The triangle inequality theorem states that in any triangle, when you add two sides, the result will always be larger than or equal to the side that you didn’t include in the addition.
Suppose we add sides a and b. By the triangle inequality theorem, a+ b must always be greater than or equal to the side we didn’t include in the addition process (c). Hence, a + b ≥ c.
However, we can add any two sides of the given triangle. For instance, if we add sides a and c instead, the triangle inequality theorem states that a + c must be larger than or equal to b or a + c ≥ b.
Suppose that two sides of a triangle were given, and we want to determine the possible value of the third side. As a consequence of the triangle inequality theorem, the possible length of the third side can be any real number within this range:
First side – second side < Third side < First side + second side
Don’t fret if you cannot immediately grasp what we are discussing above. Through our examples below, you’ll get a better understanding of the concept mentioned above:
Sample Problem 1: Suppose that two sides of a triangle have measures of 10 cm and 19 cm. Determine the possible measurement of the third side.
Solution:
We are given two sides of a triangle which are 10 cm and 19 cm. The measurement of the third side is unknown, and we must determine the possible values for its length.
We cannot just guess the length of the third side since if our guess is too small, the three sides might not form a triangle. The same applies when our guess is too large; the three sides may not form a triangle. We have to be precise in determining the possible values of the third side to ensure that these three sides will form a triangle.
To find the possible lengths of the third side, we use the following:
First side – second side < Third side < First side + second side
Our first and second sides are 19 cm and 10 cm, respectively (to avoid negative values):
19 – 10 < third side < 19 + 10
9 < third side < 29
The result states that the third side’s length can be any real number between 9 and 29 (including 9 and 29). Hence, the possible measurements of the third side can be any number between 9 and 29 inclusively.
This means that if we let the measurement of the third side be any value from 9 to 29, the three sides will form a triangle. For instance, if we let the third side = 15, line segments with 10, 19, and 15 cm measurements will form a triangle. However, if we pick a value outside the range of 9 to 29 (for instance, 8), the sides or line segments will not form a triangle.
Hence, the possible measurement of the third side is any real number between 9 and 29 inclusively.
Sample Problem: What is the largest possible length of the third side of a triangle if its first and second sides measure 12 cm and 9 cm, respectively?
Solution:
Using the formula we have used in the previous example:
First side – second side < Third side < First side + second side
According to the given problem, the first side is 12 cm long, while the second is 9 cm long. By substitution, we have:
12 – 9 < Third Side < 12 + 9
3 < Third Side < 21
Based on our solution above, the length of the third side is the set of all real numbers between 3 and 21 inclusively. Note that the largest value within this set is 21. Since we are looking for the largest possible measurement of the third side of the triangle, the answer is 21 cm.
Next topic: Conversion of Units of Measurement
Previous topic: Angles
Test Yourself!
3. Math Mock Exam + Answer Key
Written by Jewel Kyle Fabula
Jewel Kyle Fabula
Jewel Kyle Fabula is a Bachelor of Science in Economics student at the University of the Philippines Diliman. His passion for learning mathematics developed as he competed in some mathematics competitions during his Junior High School years. He loves cats, playing video games, and listening to music.
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# Coordinate transformation in ODE's with a unit step function
Consider the following general set of 2 ODE's
$$\dot{x}=\Theta(\dot{x} )f_1(x,y)+(1-\Theta(\dot{x}))f_2(x,y)$$ $$\dot{y}=(1-\Theta(\dot{y}))g_1(x,y)+\Theta(\dot{y})g_2(x,y)$$
where $$\Theta(x)$$ is the unit step function.
what it means is when $$\dot{x}>0$$ then the dynamics are govern by $$f_1$$ and when $$\dot{x}<0$$ then the dynamics are govern by $$f_2$$. With $$\dot{y}$$ it is the other way around.
Now let's consider that I can make in $$f_1,f_2,g_1,g_2$$ the transformations $$x-y\rightarrow d$$ and $$y-y\rightarrow 0$$, which leaves me with
$$\dot{x}=\Theta(\dot{x} )f_1(d)+(1-\Theta(\dot{x}))f_2(d)$$ $$\dot{y}=(1-\Theta(\dot{y}))g_1(d)+\Theta(\dot{y})g_2(d)$$
I would like to apply the transformation over the derivatives, such that I get an equation for $$\dot{d}$$ as well
$$\dot{d}=\dot{x}-\dot{y}=\Theta(\dot{x} )f_1(d)+(1-\Theta(\dot{x}))f_2(d)-(1-\Theta(\dot{y}))g_1(d)-\Theta(\dot{y})g_2(d)$$
If I apply the transformation on the unit step function as well, it seems that I lose information about the system
$$\dot{d}=\Theta(\dot{d} )f_1(d)+(1-\Theta(\dot{d}))f_2(d)-(1-\Theta(0))g_1(d)-\Theta(0)g_2(d)$$
The unit step can be defined when $$\Theta(0)=\frac{1}{2}$$ or $$\Theta(0)=1$$. In both cases I lose the information about $$\dot{y}$$.
Is there any way to make a transformation of such a system to $$\dot{d}$$?
• You lost information when you redefined $f_1(x,y) \to f_1(x-y)$, which isn't always true. – Dylan May 20 at 8:43
• @Dylan, I assume here that the functional form of $f_1$ can do so, for example, $f_1(x,y)=4(x-y)+e^{x-y}$. – jarhead May 20 at 9:20
• Then I'm not sure what you're asking? Solve the reduced system for $d$, then plug it back into the original equations of $\dot x$ and $\dot y$ – Dylan May 20 at 9:43
• @Dylan, yes but when you substitute $x\rightarrow x-y=d$ and $y\rightarrow y-y=0$ you end up with $\Theta(\dot{y}) \rightarrow \Theta(0)$ and lose the sign of the derivative which is also dynamic. – jarhead May 20 at 9:50
• That should tell you that the substitution isn't valid. You can't reduce one variable to $0$ without justifying it. – Dylan May 20 at 15:45
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# Convert Seconds to Months (s to mo Conversion)
1 s = 3.8051333379604E-7 mo
## How to convert seconds to months?
To convert seconds to months, multiply the value in seconds by 0.00000038051333379604.
You can use the conversion formula :
months = seconds × 0.00000038051333379604
To calculate, you can also use our seconds to months converter, which is a much faster and easier option as compared to calculating manually.
## How many months are in a second?
There are 3.8051333379604E-7 months in a second.
1 second is equal to 3.8051333379604E-7 months.
• 1 second = 3.8051333379604E-7 months
• 2 seconds = 7.6102666759208E-7 months
• 3 seconds = 1.1415400013881E-6 months
• 4 seconds = 1.5220533351842E-6 months
• 5 seconds = 1.9025666689802E-6 months
• 10 seconds = 3.8051333379604E-6 months
• 100 seconds = 3.8051333379604E-5 months
## Examples to convert s to mo
Example 1:
Convert 50 s to mo.
Solution:
Converting from seconds to months is very easy.
We know that 1 s = 3.8051333379604E-7 mo.
So, to convert 50 s to mo, multiply 50 s by 3.8051333379604E-7 mo.
50 s = 50 × 3.8051333379604E-7 mo
50 s = 1.9025666689802E-5 mo
Therefore, 50 seconds converted to months is equal to 1.9025666689802E-5 mo.
Example 2:
Convert 125 s to mo.
Solution:
1 s = 3.8051333379604E-7 mo
So, 125 s = 125 × 3.8051333379604E-7 mo
125 s = 4.7564166724505E-5 mo
Therefore, 125 s converted to mo is equal to 4.7564166724505E-5 mo.
For faster calculations, you can simply use our s to mo converter.
## Seconds to months conversion table
Seconds Months
0.001 s 3.8051333379604E-10 mo
0.01 s 3.8051333379604E-9 mo
0.1 s 3.8051333379604E-8 mo
1 s 3.8051333379604E-7 mo
2 s 7.6102666759208E-7 mo
3 s 1.1415400013881E-6 mo
4 s 1.5220533351842E-6 mo
5 s 1.9025666689802E-6 mo
6 s 2.2830800027762E-6 mo
7 s 2.6635933365723E-6 mo
8 s 3.0441066703683E-6 mo
9 s 3.4246200041644E-6 mo
10 s 3.8051333379604E-6 mo
20 s 7.6102666759208E-6 mo
30 s 1.1415400013881E-5 mo
40 s 1.5220533351842E-5 mo
50 s 1.9025666689802E-5 mo
60 s 2.2830800027762E-5 mo
70 s 2.6635933365723E-5 mo
80 s 3.0441066703683E-5 mo
90 s 3.4246200041644E-5 mo
100 s 3.8051333379604E-5 mo
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Ch. 13 Frequency analysis
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# Ch. 13 Frequency analysis - PowerPoint PPT Presentation
Ch. 13 Frequency analysis. TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: A A A A A A A A A A. Force linear system with input x(t) = A sin t . Here is the output y(t):. Chapter 14. u(t) = A sin( ! t). u. y. As t ! 1 :
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### Ch. 13 Frequency analysis
TexPoint fonts used in EMF.
Read the TexPoint manual before you delete this box.: AAAAAAAAAA
Force linear system with input x(t) = A sin t .
Here is the output y(t):
Chapter 14
u(t) = A sin(!t)
u
y
As t!1:
y(t) = AR¢ A sin(!t + Á)
General (VERY SIMPLE):
AR = |G(j!)|
Á = Å G(j!)
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SINUSOIDAL RESPONSE OF FIRST-ORDER SYSTEM
k = 1, ¿ = 1 s
y(t) = AR¢ sin(!t + Á)
u(t) = sin(!t)
Plots: VARY ! from 0.1 to 30 rad/s
w=0.3; tau=1; t = linspace(0,20,1000);
u = sin(w*t);
AR = 1/sqrt((w*tau)^2+1)
phi = - atan(w*tau), phig=phi*180/pi, dt=-phi/w
y = AR*sin(w*t+phi);
plot(t,y,t,u)
Mathematics. Complex numbers, j2=-1
Im(G)
G(j!)=R+jI
I
Á=Å G
R
Re(G)
Chapter 14
Polar form
Multiply complex numbers:
Chapter 14
Simple method to find sinusoidal response of system G(s)
• Input signal to linear system: u = u0 sin(! t)
• Steady-state (“persistent”, t!1) output signal: y = y0 sin(! t + Á)
• What is AR = y0/u0 and Á?
• Solution (extremely simple!)
• Find system transfer function, G(s)
• Let s=j! (imaginary number, j2=-1) and evaluate G(j!) = R + jI (complex number)
• Then (“believe it or not!”)
• AR = |G(j!)| (magnitude of the complex number)
• Á = Å G(j!) (phase of the complex number)
Im(G)
G(j!)=R+jI
I
R
Re(G)
|G|
Å G
Example 13.1:
1.
2.
Chapter 14
R I
3.
Gain and phase shift
of sinusoidal response!
SIMPLER: Polar form; see board
Chapter 14
-1 rad = -57o at !µ = 1
=-!µ
Figure 14.4 Bode diagram for a time delay, e-qs.
Peak goes to infinity when ³! 0
Phase increases for LHP zero
Oops! Phase drops for RHP zero
Bode Diagram
40
30
Magnitude (dB)
20
10
0
90
45
Phase (deg)
0
-45
-90
-4
-3
-2
-1
0
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Electrical engineers (and Matlab) use decibel for gain
• |G| (dB) = 20 log10|G|
s=tf('s')
g = 10*(100*s+1)/[(10*s+1)*(s+1)]
bode(g) % gives AR in dB
*
*To change magnitude from dB to abs: Right click + properties + units
Other way: |G| = 10|G|(dB)/20
GM=2 is same as GM = 6dB
ASYMPTOTES
Frequency response of term (Ts+1): set s=j!.
Asymptotes:
(j!T + 1) ¼ 1 for !T ¿ 1
(j!T+ 1) ¼j!Tfor !T À1
Note: Integrator (1/s) has one “asymptote”:
=1/(j!)= -j!-1 at all frequencies
Rule for asymptotic Bode-plot, L = k(Ts+1)/(¿s+1)….. :
(a) If constant (L=k): Gain=k and Phase=0o
(b) If integrator (L=k/s):
Phase: -90o.
Gain: Has slope -1 (on log-log plot) and gain=1 at !=k
2. Break frequencies:
Figure 14.6 Bode plots of ideal parallel PID controller and series PID controller with derivative filter (α = 0.1).
Ideal parallel:
Series with Derivative Filter:
Chapter 14
CLOSED-LOOP STABILITY
• L = gcgm = loop transfer function with negative feedback
• Bode’s stability condition: |L(!180)|<1|
• Limitations
• Open-loop stable (L(s) stable)
• Phase of L crosses -180o only once
• More general: Nyquist stability condition:
Locus of L(j!) should encircle the
(-1)-point P times in the anti-clockwise
direction (where P = no. of unstable
poles in L).
Stable plant (P=0): Closed-loop stable if L has no encirclements of -1
(=Bode’s stability condition)
|L(j!)| =
c
180
Chapter 14
Å L(j!)=
c
180
Sigurd’s preferred notation in red
Time delay margin, ¢µ = PM[rad]/!c
Slope=-1
-2
-1
GM=1/0 = 1
-2
Slope
Help lines
-2
-1
!=0.5
!=0.01
!=0.05
-90o
-90o
PM=57o
-180o
-180o
L(s): SIMC PI-control with ¿c=4 for g(s) = 1/(100s+1)(2s+1)
!180 = 1
TexPoint fonts used in EMF.
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Slope=-1
With added delay, e-µs with µ=2
No change in gain
-2
-1
GM=1/0,4=2.5
-2
!=0.5
!=0.01
!=0.05
-90o
-90o
With added delay, e-µs with µ=2.
Contribution to phase is:
-5.7o at !=0.1/µ = 0.05
-57o at !=1/µ = 0.5
PM=35o
-180o
-180o
!180 = 0.4
!c = 0.19
Example. PI-control of integrating process with delay
• g(s) =k’e-µs/s
• Derive: Pu = 4µ and Ku = (¼/2)/(k’µ)
• PI-controller, c(s) = Kc (1+1/¿Is)
Task: Compare Bode-plot (L=gc), robustness and simulations (use k’=1, µ=1).
Bode Diagram
Bode Diagram
Gm = 2.96 (at 1.49 rad/s) , Pm = 46.9 deg (at 0.515 rad/s)
Gm = 1.87 (at 1.35 rad/s) , Pm = 24.9 deg (at 0.76 rad/s)
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Magnitude (abs)
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-180
Phase (deg)
Phase (deg)
-360
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-540
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-720
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SIMC-PI
Ziegler-Nichols PI
GM
GM
PM
PM
SIMC is a lot more robust:
ZN: ¢µ = 24.9*(3.14/180)/0.76 = 0.572s
SIMC: ¢µ = 46.9*(3.14/180)/0.515 = 1.882s
s=tf('s')
g = exp(-s)/s
Kc=0.707, taui=3.33
c = Kc*(1+1/(taui*s))
L1 = g*c
figure(1), margin(L1) % Bode-plot with margins
% To change magnitude from dB to abs: Right click + properties + units
Kc=0.5, taui=8
c = Kc*(1+1/(taui*s))
L2 = g*c
figure(2), margin(L2)
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SIMC
OUTPUT, y
ZN
s=tf('s')
g = exp(-s)/s
Kc=0.707, taui=3.33, taud=0 % ZN
sim('tunepid4')
plot(Tid,y,'red',Tid,u,'red')
Kc=0.5, taui=8, taud=0 % SIMC
sim('tunepid4')
plot(Tid,y,'blue',Tid,u,'blue')
INPUT, u
t=0: setpoint change, t=20: input disturbance
• Conclusion: Ziegler-Nichols (ZN) responds faster to the input disturbance,
• but is much less robust.
• ZN goes unstable if we increase delay from 1s to 1.57s.
• SIMC goes unstable if we increase delay from 1s to 2.88s.
4
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SIMC
OUTPUT, y
ZN
INPUT, u
t=0: setpoint change, t=20: input disturbance
ZN is almost unstable when the delay is increased from 1s to 1.5s.
SIMC does not change very much
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Closed-loop frequency response
e
SIMC: Ms=1.70
ZN: Ms = 2.93
SIMC
ZN
w = logspace(-2,1,1000);
[mag1,phase]=bode(1/(1+L1),w);
[mag2,phase]=bode(1/(1+L2),w);
figure(1), loglog(w,mag1(:),'red',w,mag2(:),'blue',w,1,'-.')
axis([0.01,10,0.001,10])
NO EFFECT
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# Graphing a Quadratic Function Given in General Form
• Mar 31st 2012, 06:58 AM
Trypanosoma
Graphing a Quadratic Function Given in General Form
If I understand correctly, the formula is:
$-\frac{b}{2a}, \frac{4ac-b^2}{4a}$
Here's what I have:
$33x^2 -2x +15$
First:
$-\frac{2}{2(33)}$ = $\frac{1}{33}$
Easy enough. Now though...
$\frac{4(33)(15)}\frac{4(33)}$
Doing it that way comes up with wildly incorrect answers. I'd type it out but it's just all wrong. The answer in the book is...
$\frac{494}{33}$
However, doing...
$15-\frac{-2^2}{4(33)}$ does get me
$\frac{494}{33}$
So that's really $C-\frac{(-b)^2}{4(a)}$
So my question is: Why? I'm missing something in the translation and I'm hoping to get some kind of answer. Thanks for any and all help.
• Mar 31st 2012, 10:55 AM
emakarov
Re: Graphing a Quadratic Function Given in General Form
Quote:
Originally Posted by Trypanosoma
If I understand correctly, the formula is:
$-\frac{b}{2a}, \frac{4ac-b^2}{4a}$
Being psychic, I know that you meant the formula for the vertex coordinates of the parabola ax^2+bx+c (Smile).
Quote:
Originally Posted by Trypanosoma
$33x^2 -2x +15$
First:
$-\frac{2}{2(33)}$ = $\frac{1}{33}$
Should be $-\frac{-2}{2(33)}$
Quote:
Originally Posted by Trypanosoma
$\frac{4(33)(15)}\frac{4(33)}$
Should be $\frac{4(33)(15)-4}\frac{4(33)}$
Quote:
Originally Posted by Trypanosoma
Doing it that way comes up with wildly incorrect answers.
Not wildly incorrect; the difference between this and the correct answer is only 1/33.
Quote:
Originally Posted by Trypanosoma
So that's really $C-\frac{(-b)^2}{4(a)}$
So my question is: Why?
The y-coordinate of the vertex is $\frac{4ac-b^2}{4a}=c-\frac{b^2}{4a}$.
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## Video on Simplices and Simplicial Complexes
Professor Wildberger is extremely kind to upload his videos which would be very useful to any Math student studying Topology. Simplices / Simplicial Complexes are usually the first chapter in a Algebraic Topology book.
Check out also Professor Wildberger’s book on Rational Trigonometry, something that is quite novel and a new approach to the subject of Trigonometry. For instance, it can be used for rational parametrisation of a circle.
## Pick’s Theorem Proof (Video)
This is an excellent video I found on Youtube by Professor Wildberger on the proof of Pick’s Theorem. It is easy enough for a high school student to understand!
Pick’s Theorem is a formula $A=I+\frac{B}{2}-1$ which gives the area of a simple polygon whose vertices lie on points with integer coordinates. Surprisingly, it is a relatively modern theorem, the result was first described by Georg Alexander Pick in 1899.
Using Pick’s Formula, the area of the above polygon is $A=I+\frac{B}{2}-1=7+\frac{8}{2}-1=10$. We can also see that it is the sum of two triangles $A=\frac{1}{2}(4)(2)+\frac{1}{2}(4)(3)=10$.
Amazing Formula!
Source: http://en.wikipedia.org/wiki/Pick’s_theorem
A recent visitor to my website bought this book. Highly interesting and suitable for parents of young children. Three to seven is a critical period where the brain develops, hence learning about how to teach math to preschoolers is of great significance for young parents.
This book is a captivating account of a professional mathematician’s experiences conducting a math circle for preschoolers in his apartment in Moscow in the 1980s. As anyone who has taught or raised young children knows, mathematical education for little kids is a real mystery. What are they capable of? What should they learn first? How hard should they work? Should they even “work” at all? Should we push them, or just let them be? There are no correct answers to these questions, and the author deals with them in classic math-circle style: he doesn’t ask and then answer a question, but shows us a problem–be it mathematical or pedagogical–and describes to us what happened. His book is a narrative about what he did, what he tried, what worked, what failed, but most important, what the kids experienced. This book does not purport to show you how to create precocious high achievers. It is just one person’s story about things he tried with a half-dozen young children. Mathematicians, psychologists, educators, parents, and everybody interested in the intellectual development in young children will find this book to be an invaluable, inspiring resource. Titles in this series are co-published with the Mathematical Sciences Research Institute (MSRI).
## AlgTop1: One-dimensional objects
This is a continuation of the series of Algebraic Topology videos. Previous post was AlgTop 0.
Professor Wildberger is an interesting speaker. He holds some unorthodox views, for instance he doesn’t believe in “real numbers” or “infinite sets”. Nevertheless, his videos are excellent and educational. Highly recommended to watch!
The basic topological objects, the line and the circle are viewed in a new light. This is the full first lecture of this beginner’s course in Algebraic Topology, given by N J Wildberger at UNSW. Here we begin to introduce basic one dimensional objects, namely the line and the circle. However each can appear in rather a remarkable variety of different ways.
Author: NJ Wildberger
This revolutionary book establishes new foundations for trigonometry and Euclidean geometry. It shows how to replace transcendental trig functions with high school arithmetic and algebra to dramatically simplify the subject, increase accuracy in practical problems, and allow metrical geometry to be systematically developed over a general field. This new theory brings together geometry, algebra and number theory and sets out new directions for algebraic geometry, combinatorics, special functions and computer graphics. The treatment is careful and precise, with over one hundred theorems and 170 diagrams, and is meant for a mathematically mature audience. Gifted high school students will find most of the material accessible, although a few chapters require calculus. Applications include surveying and engineering problems, Platonic solids, spherical and cylindrical coordinate systems, and selected physics problems, such as projectile motion and Snell’s law. Examples over finite fields are also included.
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# Electrical Engineering - Ohm's Law - Discussion
### Discussion :: Ohm's Law - General Questions (Q.No.15)
15.
If 24 V are applied across a resistor and there are 10.9 mA of current, the resistance is
[A]. 220 k [B]. 22 k [C]. 2.2 k [D]. 220
Explanation:
No answer description available for this question.
Raghu said: (Mar 3, 2011) V=RI 24/0.0109=2201 means 2.2KV
Varatharajan said: (Jan 7, 2012) Hi raghu, are you understand for the sum. The sum is required to resistance value. So the answer is R=V/I R=24/10.9*10^-3 R=2.2*10^-3 R=2.2KOhm
Sowjanya said: (Mar 12, 2014) R= V/I. R= 24/10.9*10^-3. R= 2.2 kohm.
Ali Shan said: (Dec 28, 2014) I am not satisfied to all explanation please tell me about your question.
Arun said: (Mar 12, 2016) R = V/I.
Ramalingam said: (Jul 29, 2016) 24 / 10.90000 = 2.20183481.
Ashok said: (Jun 4, 2017) Thank you everyone for the given explanation.
Gowri said: (Jul 15, 2019) v = IR. R = V/I. R = 24/10.9*10^-3 =2.2*10^3. R = 2.2 Kiloohm.
Koushik said: (Jul 21, 2020) By using Ohms Law. R=V/I = 24รท10.9 =2.2kilo Ohm.
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# Name: Date: Period: ______ A2 Exponential Functions Write an
```Name: ______________________________________________________ Date: ______________ Period: _______
A2 Exponential Functions
Write an exponential function to model each situation. Find the value of the function after 8 years.
1. A \$12,000 car depreciates 25% each year.
3. A population of 2785 brown bears increases 3%
each year.
2. A \$22,000 truck depreciates 12% per year.
4. A \$45,000 investment increases at a rate of
9.8% per year.
5. The population of the animals you are studying is decreasing by 1.5% each year. There were about 2,000,000 of
them world-wide this year.
Identify each function as modeling either exponential growth or exponential decay. What is the percent of increase or
decrease for each function?
6. 𝑦 = 72(1.6)𝑥
8. 𝑦 = 2.02(2.01)𝑥
7. 𝑦 = 24(0.8)𝑥
9. 𝑦 = 0.9(0.92)𝑥
Solve:
10. Suppose you have 1 g of a mixture that is losing 1% of mass each hour by evaporation.
a. What is the decay factor of the mixture?
b. Write an equation to model the mass of the mixture.
11. In 1985, there were 285 cell phone subscribers in the small town of Centerville. The number of subscribers
increased by 75% per year after 1985. How many cell phone subscribers were in Centerville in 1994?
12. The population of Winnemucca, Nevada, can be modeled by 𝑃 = 6191(1.04)𝑡 where t is the number of years
since 1990. What was the population in 1990? By what percent did the population increase by each year?
13. You have inherited land that was purchased for \$30,000 in 1960. The value of the land increased by
approximately 5% per year. What is the approximate value of the land in the year 2014?
14. An adult takes 400 mg of ibuprofen. Each hour, the amount of ibuprofen in the person’s system decreases by
about 29%. How much ibuprofen is left after 6 hours.
```
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Education.com
Try
Brainzy
Try
Plus
# Tip #18 to Get a Top ACT Math Score (page 3)
By McGraw-Hill Professional
Updated on Sep 7, 2011
1. B This question intimidates some students; it seems too theoretical. But "Make It Real" makes it so easy. Just choose a real number for p (remember that the question says that the number that you choose has to be odd). Let's say p = 5. Then try p = 5 in all the answer choices to see which one does what the question asks, which one gives an odd number. The answer is B since 2(5) – 1 = 9.
2. J First, "product" means "multiply." We will review this more in Skill 27. "Make It Real," let's say, m = –3 and n = 3. Now it's easy, the product (multiplication) of –3 and 3 is –9. Choice J looks good, but to be safe, try a few more possibilities. After you've done three, you can be sure that your answer is correct. And by then you might realize that "Of course, a positive times a negative is always a negative."
3. D Choose a number for b, let's say b = 5 (notice the question states that b can be any real number besides 3 and 7). Now it's easy, solve:
4. Skill 12 review: the value of b cannot equal 3 or 7 because either number would give us a 0 on the bottom of the fraction, which is illegal in math and would cause the equation to be "undefined" and would cause lots of important folks to get very upset.
5. H We are into the "hards" now. But "Make It Real" will knock these down a notch to mediums or even easies. Let's say k = 100 and q = 20. So, of the 100 kids in a class, 20 have NOT seen Monty Python and the Holy Grail. Once we've "Made It Real," it becomes obvious that if 20 have NOT seen it, then 80 have seen it. (When you are using "Make It Real" with percents, use 100 when you can, it makes questions really easy.) Now, based on k = 100 and q = 20, we want an answer of 80. So try each answer choice, using k = 100 and q = 20 and find the one answer choice that equals 80.
6. Choice H is correct because
If two choices work, just choose new "Make It Real" numbers and have a tiebreaker.
People think that this might take forever to do, but try it; it takes like 30 seconds.
7. E Let's say m = 3 and n = 5. Then –3m – 2n – 4 = –3(3) – 2(5) – 4 = –23. The question asks what happens when m increases by 2 and n decreases by 1, so let's solve for m = 5 and n = 4. Now, –3m – 2n – 4 = –3(5) – 2(4) – 4 = –27. So the result changed from –23 to –27; it decreased by 4.
Go to: Tip #19
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# Chapter 1 Notes
```1.1 Points, Lines and Planes
Undefined Terms – Words/items that are only explained using
examples or descriptions.
Collinear – Points that lie on the same line.
Coplanar – Points that lie on the same plane.
Intersection – The set of points that two or more geometric
figures have in common.
Space – A boundless, three-dimensional set of all points. Can
contain lines and planes.
1.2 Linear Measure
Line Segment – A measurable part of a line that consists of two
points, called endpoints, and all of the points between them.
Congruent Segments – Segments that have the same measure.
Constructions – A method of creating geometric figures without the
benefit of measuring tools.
Generally a compass and straightedge are used.
Precision – A clustering of a group of measurements.
Depends on the smallest unit of measure available on
measuring tool.
Accuracy – How close a measured value comes to the actual or
desired value.
1.3 Distance and Midpoints
Distance – The length of the segment between two points.
Distance Formula
Midpoint – The point halfway between the endpoints of a
segment.
Midpoint Formula
Segment Bisector – Any segment, line, or plane that
intersects a segment at its midpoint.
1.4 Angle Measure
Ray – A part of a line that has one endpoint and extends
indefinitely in one direction.
Opposite Rays – Two rays on the same line that share a
common endpoint.
Angle – An object formed by two noncollinear rays that have
a common endpoint.
Sides – The rays that make up the angle.
Vertex – The common endpoint of an angle.
Degree – The unit used to measure angles. Equal to 1/360th
of a turn around a circle.
Angle Bisector – A ray that divides an angle into two
congruent angles.
1.5 Angle Relationships
Perpendicular – Lines, segments, or rays that form right
angles.
1.6 Two Dimensional Figures
Concave – If any of the lines that contain each side is extended and
goes through the inside.
Convex – If none of the lines that contain each side is extended and
never pass through the interior.
Names of Polygons
Number of Sides
3
4
5
6
7
8
9
10
12
N
Type (Name) of Polygon
Triangle
Pentagon
Hexagon
Heptagon
Octagon
Nonagon
Decagon
Dodecagon
n-gon
Equilateral, Equiangular, and Regular
Equilateral – All sides are equal.
Equiangular – All angles are equal.
Regular – All sides and angles are equal. It would be both
Equilateral and Equiangular.
Perimeter – The sum of the lengths of the sides of a polygon.
Circumference – The distance around a circle.
Area – The number of square units needed to cover a surface.
1.7 Three-Dimensional Figures
Polyhedron – A solid with all flat surfaces that enclose a single
region of space.
Face – The polygons that make up the flat surface.
Edge – The line segments where the faces intersect.
Vertices – The point where three or more edges intersect.
Regular Polyhedron – A polyhedron with all faces being regular
congruent polygons.
Surface Area – A two-dimensional measurement of the surface of a
solid figure.
Volume – The measure of the amount of space enclosed by a solid
figure.
```
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Enter the initial points before transfer and the transfer rate into the calculator to determine the total points after transfer. This calculator can also evaluate any of the variables given the others are known.
## Point Transfer Formula
The following formula is used to calculate the point transfer between different reward programs.
TP = P * R
Variables:
• TP is the total points after transfer P is the initial points before transfer R is the transfer rate (decimal)
To calculate the total points after transfer, multiply the initial points before transfer by the transfer rate. The result will be the total points after transfer.
## What is a Point Transfer?
A point transfer refers to the process of moving points or rewards from one account to another, often within the same rewards program or between partnered programs. This is commonly seen in credit card rewards programs, frequent flyer programs, or hotel loyalty programs. The ability to transfer points allows users to maximize their rewards by consolidating points in one place or using them for the most valuable redemptions. The rules and rates for point transfers vary by program.
## How to Calculate Point Transfer?
The following steps outline how to calculate the Point Transfer using the formula TP = P * R.
1. First, determine the initial points before transfer (P).
2. Next, determine the transfer rate (R) as a decimal.
3. Next, gather the formula from above = TP = P * R.
4. Finally, calculate the total points after transfer (TP).
5. After inserting the variables and calculating the result, check your answer with the calculator above.
Example Problem :
Use the following variables as an example problem to test your knowledge.
initial points before transfer (P) = 150
transfer rate (R) = 0.75
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# Difference between revisions of "2009 AMC 10A Problems/Problem 1"
## Problem
One can, can hold $12$ ounces of soda, what is the minimum number of cans needed to provide a gallon ($128$ ounces) of soda?
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 9\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 11$
## Solution 1
$10$ cans would hold $120$ ounces, but $128>120$, so $11$ cans are required. Thus, the answer is $\mathrm{(E)}$.
## Solution 2
We want to find $\left\lceil\frac{128}{12}\right\rceil$ because there are a whole number of cans.
$\frac{128}{12} = 10R8\longrightarrow 11\longrightarrow \fbox{(E)}.$
2009 AMC 10A (Problems • Answer Key • Resources) Preceded byFirst Question Followed byProblem 2 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
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7 July, 12:25
A factory adds 3 red drops and 2 blue drops of coloring to white paint to make each pint of purple paint. The factory will make 50 gallons of this purple paint. How many drops of red and blue coloring will the factory need in the 50-gallon batch of purple paint?
+3
1. 7 July, 12:38
0
1200 red drops and 800 blue drops
Step-by-step explanation:
1 gallon = 8 pints
50 gallons = 50*8 = 400 pints
Red drops:
400 * 3 = 1200 drop
Blue drops:
400 * 2 = 800 drops
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# Ring-Planet
Planetary gear set of carrier, planet, and ring wheels with adjustable gear ratio and friction losses
• Library:
• Simscape / Driveline / Gears / Planetary Subcomponents
## Description
The Ring-Planet block represents a carrier, a ring gear, and a set of planet gears. The planet gears are connected to and rotate with respect to the carrier. The planet and ring gears corotate with a fixed gear ratio that you specify. A ring-planet and a sun-planet gear are basic elements of a planetary gear set. For model details, see Equations.
### Thermal Model
You can model the effects of heat flow and temperature change by enabling the optional thermal port. To enable the port, set Friction model to ```Temperature-dependent efficiency```.
### Equations
Ideal Gear Constraints and Gear Ratios
The Ring-Planet block imposes one kinematic and one geometric constraint on the three connected axes:
`${r}_{\text{R}}{\omega }_{\text{R}}={r}_{\text{C}}{\omega }_{\text{C}}+{r}_{\text{P}}{\omega }_{\text{P}}$`
`${r}_{\text{R}}={r}_{\text{C}}+{r}_{\text{P}}$`
The ring-planet gear ratio is
`${g}_{\text{RP}}={r}_{\text{R}}/{r}_{\text{P}}={N}_{\text{R}}/{N}_{\text{P}},$`
where N is the number of teeth on each gear. In terms of this ratio, the key kinematic constraint is
The three degrees of freedom reduce to two independent degrees of freedom. The gear pair is (1, 2) = (P, R).
Warning
The ring-planet gear ratio gRP must be strictly greater than one.
The torque transfer is:
In the ideal case where there is no torque loss, τloss = 0.
Nonideal Gear Constraints and Losses
In the nonideal case, τloss ≠ 0. For more information, see Model Gears with Losses.
### Variables
Use the Variables settings to set the priority and initial target values for the block variables before simulating. For more information, see Set Priority and Initial Target for Block Variables.
### Assumptions and Limitations
• Gear inertia is assumed to be negligible.
• Gears are treated as rigid components.
## Ports
### Conserving
expand all
Rotational mechanical conserving port associated with the planet gear carrier.
Rotational mechanical conserving port associated with the ring gear.
Rotational mechanical conserving port associated with the planet gear.
Thermal conserving port associated with heat flow. Heat flow affects the power transmission efficiency by altering the gear temperatures.
#### Dependencies
To enable this port, set Friction model to `Temperature-dependent efficiency`.
## Parameters
expand all
### Main
Ratio, gRP, of the ring gear to planet gear rotations as defined by the number of ring gear teeth divided by the number of planet gear teeth. This gear ratio must be strictly greater than 1.
### Meshing Losses
Friction model for the block:
• `No meshing losses - Suitable for HIL simulation` — Gear meshing is ideal.
• `Constant efficiency` — Transfer of torque between the gear wheel pairs is reduced by a constant efficiency, η, such that 0 < η ≤ 1.
• `Temperature-dependent efficiency` — Transfer of torque between the gear wheel pairs is defined by the table lookup based on the temperature.
Torque transfer efficiency, ηRP, for the outer and inner planet gear wheel pair meshing. This value must be in the range (0,1].
#### Dependencies
To enable this parameter, set Friction model to `Constant efficiency`.
Vector of temperatures used to construct a 1-D temperature-efficiency lookup table. The vector elements must increase from left to right.
#### Dependencies
To enable this parameter, set Friction model to `Temperature-dependent efficiency`.
Vector of output-to-input power ratios that describe the power flow from the ring gear to the planet gear, ηRP. The block uses the values to construct a 1-D temperature-efficiency lookup table.
Each element is an efficiency that relates to a temperature in the Temperature vector. The length of the vector must be equal to the length of the Temperature vector. Each element in the vector must be in the range (0,1].
#### Dependencies
To enable this parameter, set Friction model to ```Temperature-dependent efficiency```.
Power threshold, pth, above which full efficiency is in effect. Below this value, a hyperbolic tangent function smooths the efficiency factor.
When you set Friction model to `Constant efficiency`, the block lowers the efficiency losses to zero when no power is transmitted. When you set Friction model to `Temperature-dependent efficiency`, the block smooths the efficiency factors between zero when at rest and the values provided by the temperature-efficiency lookup tables at the power thresholds.
#### Dependencies
To enable this parameter, set Friction model to `Constant efficiency` or ```Temperature-dependent efficiency```.
### Viscous Losses
Viscous friction coefficient, μP, for the planet-carrier gear motion.
### Thermal Port
To enable this parameter, set Friction model to `Temperature-dependent efficiency`.
Thermal energy required to change the component temperature by a single temperature unit. The greater the thermal mass, the more resistant the component is to temperature change.
#### Dependencies
To enable this parameter, set Friction model to ```Temperature-dependent efficiency```.
expand all
## Version History
Introduced in R2011a
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## How Computers Generate Random Numbers - Chapter 11
Chapter 10 Conclusions
### 11.0 THE RANDOM SEQUENCE {1, 1, 1, 1,... ,1}
Let us come full circle and reconsider the original problem posed at the beginning of this paper.
Consider the case of rolling a single die. A theoretician picks up the die, examines it, and makes the following statement: "The die has six sides, each side is equally likely to turn up, therefore the probability of any one particular side turning up is 1 out of 6 or 1/6. Furthermore, each throw of the die is completely independent of all previous throws." With that the theoretician puts the die down and leaves the room without ever having thrown it once.
An experimentalist, never content to accept what a cocky theoretician states as fact, picks up the die and starts to roll it. He rolls the die twelve times and records the outcome of each roll. His results are:
{ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 }
We now ask the question, "What is the probability of throwing a 1 on the next roll?"
There are three answers to this question. Each answer depends on what we believe to be the true properties of the chance experiment.
The theoretician claims the above result is a legitimate possible outcome according to his model of the chance experiment, as equally likely to occur as any of the other 612 = 2,176,782,336 possible outcomes of rolling a six sided die twelve times. We may choose to believe the theoretician's proclamation that each side of the die is still equally likely to show up, and each throw of the die is still completely independent of all previous throws. In this case the results of the previous twelve throws tell us nothing about what the next throw might be, and the probability of the next throw being a 1 is still 1/6 by definition.
The theoretician may support his argument by citing the physics of the experiment and the great uncertainty of the initial condition of the die while being shaken in the hand.
The experimentalist, never content to accept what a cocky theoretician states as fact, feels that we should take into account the previous results of rolling the die and reconsider the theoretician's proposed model of the chance experiment. Perhaps the die is actually unfair, heavily weighted in a lopsided way as to render the throwing of a 1 quite likely, and the throwing of any other number quite unlikely. The experimentalist comes up with a new model of the chance experiment, claiming the probability of throwing a 1 is near 100%, and the probability of throwing any other number is extremely low. We may then choose to abandon the theoretician's initial model of the chance experiment and refer to the new updated model of the chance experiment based on recent past experience. In this case the new model declares the probability of throwing a 1 on the next roll to be near 100% certain.
The experimentalist may support his argument that a new model of the chance experiment is justified by performing the analysis presented in Chapter 2.
At this point a computer programmer enters the room and declares the die isn't a real die after all but merely a computer simulation of a die. The computer uses a psuedo-random number generator algorithm which in the long run simulates well a sequence of random numbers—and that is the catch right there which throws off the probability of throwing a 1 on the next roll. If we believe beforehand that our chance experiment is one which in the long run will appear to be random, then we can claim the probability of the next simulated throw of the die being a 1 is less than 1/6, because it's high time some other number showed up. This is equivalent to the die having a memory of previous results and adjusting it's next outcome accordingly to maintain the appearance of randomness.
So we have three answers to the question, "What is the probability of throwing a 1 on the next throw."
1. The probability is 1/6 because we believe the theoretician.
2. The probability is greater than 1/6 because we believe the experimentalist.
3. The probability is less than 1/6 because we believe the computer programmer.
Much confusion about probability theory results from failing to comprehend and understand the above three possibilities. It all has to do with what we choose to believe are the properties of the chance experiment.
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# how to find angle of isosceles triangle with 3 sides
Input: A = 2, B = 2, C = 2 Output: Acute-angled Triangle Isosceles triangle Calculate the perimeter of isosceles triangle with arm length 73 cm and base length of 48 cm. the third side. Step 3 Calculate Adjacent / Hypotenuse = 6,750/8,100 = 0.8333; Step 4 Find the angle from your calculator using cos-1 of 0.8333: Double the result of the previous step. An isosceles triangle is a triangle with two sides of the same length. If you know the side lengths a, b and c you can find cos(C) and hence the measure of the angle C. Harley . So the two base angles are going to be congruent. Answer. In this triangle, the bisector of an angle lies perpendicular to the opposite side. Step 2: To find The value of ∠F. For an isosceles triangle, along with two sides, two angles are also equal in measure. Hi Lucy. The interior angles of a triangle add up to 180 degrees. Square the length of the side. Step 1 The two sides we know are Adjacent (6,750) and Hypotenuse (8,100). The formula is derived from Pythagorean theorem Angles. Ans: An isosceles triangle can be defined as a special type of triangle whose at least 2 sides are equal in measure. But in every isosceles right triangle, the sides are in the ratio 1 : 1 : , as shown on the right. Properties of Isosceles triangle. If all three sides are the same length it is called an equilateral triangle.Obviously all equilateral triangles also have all the properties of an isosceles triangle. #c# is the hypotenuse of a right triangle. Given three integers A, B, and C which denotes the sides of a triangle, the task is to check that the triangle is a right-angled, acute-angled or obtuse-angled triangle.. Calculate Angle Bisector of Two Sides using this online Angle Bisector of Isosceles Triangle Calculator. In an isosceles triangle, knowing the side and angle α, you can calculate the height, since the side is hypotenuse and the height is the leg, then the height will be equal to the product of the sine of the angle … The angle which is opposite to the base is called the vertex angle and the point associated with that angle is called as apex. And so the third angle needs to be the same. These two equal sides always join at the same angle to the base (the third side), and meet directly above the midpoint of the base. They giving you almost 90 is coincidental. Examples: Input: A = 1, B = 4, C = 3 Output: Obtuse-angled Triangle Explanation: Triangle with the sides 1, 2 and 3 is an obtuse-angled triangle. math.degrees takes radians and gives degrees. In this tutorial, see how identifying your triangle first can be very helpful in solving for that missing measurement. If you are, that knowledge can help you. In geometry, an isosceles triangle is a triangle that has two sides of equal length. The figure shown below will be used for sides and angle notations. The base of this isosceles triangle is given as 12cm. The base is 650 and the hypotenuse (the opposite side of the right angle) is 9000. So … The Morley triangle is a special equilateral (and thus acute) triangle that is formed from any triangle where the vertices are the intersections of the adjacent angle trisectors. To solve a triangle means to know all three sides and all three angles. Every triangle with two angle bisectors is called Isosceles triangle. In our calculations for a right triangle we only consider 2 known sides … Exercise worksheet on 'How to find a missing angle in an isosceles triangle.' In an Isosceles Triangle, the median drawn to the base is the angle bisector. Hey everyone, Just another question. Property 1: The area of an isosceles triangle is the amount of region enclosed by it in a two-dimensional space. So first of all, we see that triangle ABC is isosceles. The following two theorems — If sides, then angles and If angles, then sides — are based on a simple idea about isosceles triangles that happens to work in both directions: If sides, then angles: If two sides of a triangle are congruent, then the angles opposite those sides are congruent. X is the angle. – Teepeemm Sep 3 '13 at 3:26 The area of an isosceles triangle is the amount of region enclosed by it in a two-dimensional space. (a) Find the area, A, of the triangle. Imagine you have half of the triangle, a right triangle. Find the length of a side (base) if given equal sides and angle How to find the missing side of an isosceles triangle - Calculator Online Home List of all formulas of the site Calculators to solve isosceles triangle problems depending on which sides and angles you given. If the rate of the sides an isosceles triangle is 7:6:7, find the base angle correct to the nearest degree. An Isosceles Triangle can have an obtuse angle, a right angle, or three acute angles. Solved: The included angle of the two sides of constant equal length s of an isosceles triangle is \\theta. Thus in an isosceles triangle to find altitude we have to draw a perpendicular from the vertex which is common to the equal sides. In an Isosceles triangle, the two equal sides are called legs and the third side is called as base. This is the other leg right over there. Apex angle of Isosceles triangle is the angle between the lines that joins the pointed end. Let us take a triangle ABC, whose vertex angles are ∠A, ∠B, and ∠C, and sides are a,b and c, as shown in the figure below. Take a look! The word isosceles is pronounced "eye-sos-ell-ease" with the emphasis on the 'sos'.It is any triangle that has two sides the same length. This is the Law of Cosines. Isosceles Triangle Calculator and Solver. By their interior angles, triangles have other classifications: Right-- One right angle (90 °) and two acute angles; Oblique-- No right angles; Oblique Triangles There are two different heights of an isosceles triangle; the formula for the one from the apex is: hᵇ = √(a² - (0.5 * b)²), where a is a leg of the triangle and b a base. Two of the angles in an isosceles triangle are equal. The isosceles triangle is an important triangle within the classification of triangles, so we will see the most used properties that apply in this geometric figure. In this type of right triangle, the sides corresponding to the angles 30°-60°-90° follow a ratio of 1:√ 3:2. This angle, is the same as that angle. Or we could also call that DC. All I know is the ratios between the sides and of course the value of the angles (as you can see in the image). Let's say I am told the sides lengths are 10, 4 and 7. Solution Step 1: Giveno ∆DEF is an isosceles triangle ∠D = 1000. This is one leg. First I should assign reference letters. Solve the isosceles right triangle whose side is 6.5 cm. Ignore the other side. In an isosceles triangle DEF, if an interior angle ∠D = 1000 then find the value of ∠F? So sine x = 650/9000. How to find the height of an isosceles triangle. Because it's an isosceles triangle, this 90 degrees is the same as that 90 degrees. Find the length of one of the non-hypotenuse sides. Trying to find a missing interior angle measurement in a triangle? Isosceles triangle Calculate the area of an isosceles triangle, the base of which measures 16 cm and the arms 10 cm. The Pythagorean Theorem states: #a^2 + b^2 = c^2# Where: #a# and #b# are sides of a right triangle. How do I find the hypotenuse of isosceles right triangle? Square root the result of step 3. The sides of an isosceles triangle are 3, 3, and 4. Given an isosceles triangle with equal sides of length b, base angle α < 4 π and R, r the radii and O, I the centres of the circumcircle and incircle, respectively.Then This question has multiple correct options The golden triangle is an acute isosceles triangle where the ratio of twice the the side to the base side is the golden ratio. Also, in an isosceles triangle, two equal sides will join at the same angle to the base i.e. Most mathematicians agree that the classic equilateral triangle can also be considered an isosceles triangle, because an equilateral triangle has two congruent sides. You can test this yourself with a ruler and two pencils of equal length: if you try to tilt the triangle to one direction or the other, you cannot get the tips of the pencils to meet. An isosceles triangle is a triangle with two sides of equal length. Step 3: Approach We know that the sum of all interior angles in a triangle = 1800 Hence, ∠D + ∠E + … See if you're working with a special type of triangle such as an equilateral or isosceles triangle. I’m trying to figure out how to find the shaded angle of an isoceles triangle. math.tan takes radians and gives a ratio. Thus, in this type of triangle, if the length of one side and the side's corresponding angle is known, the length of the other sides can be determined using the above ratio. In the triangle below sides … The general formula for the area of triangle is equal to half the product of the base and height of the triangle. Step 2 SOHCAHTOA tells us we must use Cosine. An isosceles triangle is a special case of a triangle where 2 sides, a and c, are equal and 2 angles, A and C, are equal. A single trigonometric function will only work if you already have a right angle in the triangle, and are trying to find a different angle. Since the sum of the angles of a triangle is always 180 degrees, we can figure out the measure of the angles of an equilateral triangle: The isosceles triangle : The isosceles triangle (I can NEVER remember how to spell isosceles) has two sides that are the same length (congruent) and two angles that are the same size (congruent). An isosceles triangle has two sides that are equal called legs. Triangle area for Two Sides and the Included Angle Now, the question comes, when we know the two sides of a triangle and an angle included between them, then how to find its area. So we know that this angle … You can figure out the top angle of the half triangle using Sine: sin x = opposite / hypotenuse. Find the measure of the vertex angle to the nearest degree. Here's an example of how I would use Harley's "Law of Cosines" to find the angles of a triangle. And because it's isosceles, the two base angles are going to be congruent. Because the line which bisects and isosceles triangles is at a right angle to the base we can use the Pythagorean Theorem to find the height. The above figure shows […] Since this is an isosceles right triangle, the only problem is to find the unknown hypotenuse. This is the length of the hypotenuse. This page includes a lesson covering 'how to find a missing angle in an isosceles triangle' as well as a 15-question worksheet, which is printable, editable, and sendable. Going to be congruent triangle with two sides using this online angle bisector of an angle perpendicular. The hypotenuse of isosceles triangle Calculate the perimeter of isosceles triangle Calculate the perimeter of triangle. 'S an isosceles triangle is a triangle add up to 180 degrees the angle... Calculators to solve a triangle ) find the angles of a right triangle by it a! Is opposite to the opposite side of the angles of a right angle ) is 9000 opposite to nearest! An example of how I would use Harley 's Law of Cosines to... Tells us we must use Cosine side to the base of which measures 16 cm the! Formula for the area of an isosceles triangle, the median drawn the! Of how I would use Harley 's Law of Cosines '' to find the unknown.. Joins the pointed end this tutorial, see how identifying your triangle first can be very helpful solving... Must use Cosine same angle to the nearest degree would use Harley 's of! Sides are equal in measure Law of Cosines '' to find the measure of the angles of a angle! An isosceles triangle ∠D = 1000 defined as a special type of triangle is given as 12cm 16 and! Are equal called legs and the third angle needs to be congruent a type! Side is the same as that 90 degrees is the golden triangle is 7:6:7, find the of... Property 1:, as shown on the 'sos'.It how to find angle of isosceles triangle with 3 sides any triangle that has two sides know! 90 degrees the nearest degree measure of the triangle in every isosceles right triangle out how to find measure. Of constant equal length s of an isosceles triangle with two sides we know that angle. First of all, we see that triangle ABC is isosceles side of the non-hypotenuse.! Know are Adjacent ( 6,750 ) and hypotenuse ( 8,100 ) the right,... Of 48 cm bisectors is called isosceles triangle Calculate the perimeter of isosceles triangle is a triangle two. Using Sine: sin x = opposite / hypotenuse height of an isosceles triangle. Will join at the same that knowledge can help you third angle needs to be the same to. Two-Dimensional space to solve a triangle bisectors is called as apex angles of a triangle has! 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# How do I interpret a negative Z-test (two tailed for 2 proportions)?
I used my calculator to do a two tailed Z test for two proportions. Here's the data I input:
X1 is 834
n1 is 3610
X2 is 145
n2 is 495
I chose p1 not equal to p2
I set alpha to 0.05
My hypotheses are:
$H_0: \pi_1= \pi_2$
$H_a: \pi_1 \neq \pi_2$
The calculated Z score is $-3.03$, which gives a $p = .002$, so I am rejecting the null.
Here is my question: what does the negative Z score mean:
a) that the $X_1/n_1$ proportion is > than the $X_2/n_2$ proportion? OR
b) that the $X_2/n_2$ proportion is > than the $x_1/n_1$ proportion?
• You can calculate which of the two sample proportions is larger. It's not going to lead you to conclude the population proportions differ in the opposite direction. – Glen_b Jul 20 '17 at 3:16
• If the one sided p-value is 0.002 the two-sided p-value is 0.004 approximately. – Michael Chernick Jul 20 '17 at 5:13
• Thank you but I still don't understand. Since the difference is significant based on the score, can I say that the x1/n1 proportion is greater than the X2/n2 proportion? I am just trying to understand what it means to get a negative Z score. Thanks. I am just 17 and trying to learn. – JHB Jul 20 '17 at 12:24
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Consider the polynomial below:
((x^2 - 85)^2 - 4176)^2 - 2880^2 = (x^2 - 1)*(x^2 - 11^2)*(x^2 - 13^2)*(x^2 - 7^2)
This polynomial could be considered a ''type 3'' polynomial,since it can be generated by only 3 squaring operations (ignoring the square on 2880 of course), and it has 2^3 distinct integer roots.
In general, if the integer ''2A'' can be written as a sum of two squares (ie x^2 + y^2) in at least two different ways, then there exists some ''C'' and ''D'' such that
((x^2 - A)^2 - B)^2 - C^2 = (x^2 - a^2)*(x^2 - b^2)*(x^2 - c^2)*(x^2 - d^2)
where ''a,b,c,d'' are all integers
Question:
Do there exist any ''type 4'' polynomials that have this property? That is a polynomial that can be generated in four squaring operations, that has 2^4 integer roots.
I'm having trouble following what exactly the theorem is that you're citing. Here's what I am reading: "Given a degree-8 polynomial $\displaystyle P_8$ with roots $\displaystyle \pm a$, $\displaystyle \pm b$, $\displaystyle \pm c$, and $\displaystyle \pm d$, and given a number A, there exists a B and C such that $\displaystyle P_8=((x^2 - A)^2 - B)^2 - C^2$ if 2A is expressible as the sum of two squares in at least two unique ways." Is this correct?
This seems inconsistent. By expanding both sides of the equation, it is clear that given a,b,c,d, there exists one and only one combination A,B,C making this equation work, found by:
$\displaystyle A=\frac14(a^2+b^2+c^2+d^2)$
$\displaystyle B=3A^2-\frac12(a^2b^2+a^2c^2+a^2d^2+b^2c^2+b^2d^2+c^2d^2)$
$\displaystyle C^2=(A^2-B)^2-a^2b^2c^2d^2$
The algebra is long but elementary, and I have checked this with the raw data you provided, a=1,b=11,c=13,d=7, and indeed A=85,B=4176,C=2880. In other words, you cannot choose A - it is dependent on a,b,c,d. Is the proper wording of the theorem, "If 2A is expressible as the sum of two squares in two unique ways, then there exist integers B and C..."
I think your overall question can be answered by analyzing the proof of the theorem you are citing, and trying to generalize it.
3. Hi Media_Man
I completely forgot about this post. I posted it a little while before your response and then left for the cottage (where theres no internet connection) for over a week. Anyway I've finished much of my summer job responsibilities, so I've got time now to talk about mathematics if your up to it.
This question comes from one of Pomerance's and Crandall's book titled ''Prime Numbers: A computational perspective''. It is the first ''research type question'' found in Chapter 6.
The 2005 edition can be downloaded at the bottom of this page: Crandall R., Pomerance C. Prime numbers. A computational perspective (2ed., Springer, 2005)(604s).pdf: eKnigu
The purpose here is to try and develop a powerful integer factoring algorithm, by creating polynomials of the form
$\displaystyle P_k(x) = (x^2 - a_1)^2 - a_2)^2....)^2 - a_k^2$, which have $\displaystyle 2^k$ distinct integer roots. The idea is that for any composite integer n, if we calculate $\displaystyle P_k(x) mod(n)$ for various integers $\displaystyle x$, then eventually, were bound to find a value of $\displaystyle x$, in which $\displaystyle gcd(P_k(x) mod(n),n) = p$ for some prime factor p of n.
For example, suppose we wanted to factor the integer $\displaystyle n$ with the polynomial $\displaystyle P_8(x)=((x^2 - 85)^2 - 4176)^2 - 2880^2 = (x^2 - 1)(x^2 - 7^2)(x^2 - 11^2)(x^2 - 13^2)$.
It can be shown that if $\displaystyle x \equiv$ $\displaystyle \pm 1$, $\displaystyle \pm 7$, $\displaystyle \pm 11$, and $\displaystyle \pm 13$ $\displaystyle \mod p$, then $\displaystyle gcd(P_8(x),n) = p$. Hence if one takes the product $\displaystyle P_8(x_1)P_8(x_2)....P_8(x_i) \mod n = s$, then it becomes increasing likely $\displaystyle gcd(s,n)$ will give a factorization of $\displaystyle n$ (unless of course $\displaystyle gcd(s,n) = n$ in which case we need only do some backtracking).
The above algorithm becomes more efficient (ie factors $\displaystyle n$ more quickly), when the polynomial $\displaystyle P_k(x)$ has more distinct integer roots. As such, it becomes desirable to find polynomials of this form for larger $\displaystyle k$.
To date, no one has found such polynomials for $\displaystyle k$ larger than 4 and such polynomials would have to satisfy some very strict conditions in order to have $\displaystyle 2^k$ distinct integer roots. Many do not believe such polynomials exist. This probably isn't too surprising since if one could find such polynomials for arbitrary $\displaystyle k$, one could create a polynomial time factoring algorithm, which would be very bad for things like internet security!
With all that said, there are other types of polynomials one can consider besides the above. I'll post on one of these more fruitful polynomials shortly.
4. ## Here's what I have so far...
Given $\displaystyle a,b,c,d$, let $\displaystyle P_8(x)=(x^2-a^2)(x^2-b^2)(x^2-c^2)(x^2-d^2)$ --(1). We seek integers A,B,C such that $\displaystyle P_8(x)=((x^2-A)^2-B)^2-C^2$ --(2). By expanding (1) we can easily write $\displaystyle P_8(x)=x^8+R_6x^6+R_4x^4+R_2x^2+R_0$ and find the coefficients $\displaystyle R_i$. By expanding (2) we can find equations relating $\displaystyle A,B,C$ to $\displaystyle R_6,R_4,R_2,R_0$. Solving,
$\displaystyle A=\frac{R_6}{4}$
$\displaystyle B=3A^2-\frac{R_2}{2}$
$\displaystyle C^2=(A^2-B)^2-R_0$
There are actually four equations and three unknowns, but the "extra" equation always holds. In linear algebra terms, this system has rank 3. I put together a quick java app to crank out some examples, and uncovered just these two so far:
$\displaystyle (x^2-1^2)(x^2-7^2)(x^2-11^2)(x^2-13^2)=((x^2-85)^2-4176)^2-2880^2$ $\displaystyle \to 2A=170=1^2+13^2=7^2+11^2$
$\displaystyle (x^2-1^2)(x^2-11^2)(x^2-13^2)(x^2-17^2)=((x^2-145)^2-10656)^2-10080^2$ $\displaystyle \to 2A=290=1^2+17^2=11^2+13^2$
NOTICE A PATTERN?
$\displaystyle 290=1^2+17^2=11^2+13^2$ where the roots are $\displaystyle 1,17,11,13$. This actually follows from the algebra pretty simply, if you want to chase down a proof. The important thing is we can find numbers like 290 extremely easily. Example, pick and factor a random number like 200: $\displaystyle 15^2-5^2=(15+5)(15-5)=20*10=200=2*100=(51+49)(51-49)=51^2-49^2$. Therefore, $\displaystyle 15^2+49^2=5^2+51^2=2626$. Using $\displaystyle 5,15,49,51$ as roots, we have:
$\displaystyle (x^2-5^2)(x^2-15^2)(x^2-49^2)(x^2-51^2)=((x^2-1313)^2-1421344)^2-237600^2$
Your original post can be rephrased formally as such: Given positive integers a,b,c,d, define A and B as shown above (it can be shown trivially they will always be integers). If 2A is expressible as the sum of two squares in two unique ways, then C, as constructed above, will be an integer.
*I am working now on extending all of this from the 3 case to the 4 case, which incidentally, doubles the complexity. See if you can verify what I have so far in this post. Updates soon.
5. Hi Media_Man
I knew most of the stuff regarding the k = 3 you mentioned. I'm curious though by what you mean by ''doubling complexity'' regarding the k = 4 case?
I had a theorem awhile back on a necessary condition needed for k = 4, but I can't remember what it was right now. On a positive note though, k = 4 is definitely doable. Furthermore, if $\displaystyle (a_1,a_2,a_3,a_4)$ generates a k = 4 ladder, then $\displaystyle (x^2a_1,x^4a_2,x^8a_3,x^8a_4)$ also generates a k = 4 ladder. Using the one ''k = 4'' polynomial generated by $\displaystyle (67405,3525798096,533470702551552000,4692082091913 21600)$ from Pomerance's book, one can generate an infinite number of k = 4 such polynomials!
A similar statement can also be made for larger k values as well, but unfortunately no one has been able to find even a single polynomial beyond k = 4.
6. ## K=4
Here is what you get when expanding the relation $\displaystyle (((x^2-A)^2-B)^2-C)^2-D^2=$ $\displaystyle (x^2-a^2)(x^2-b^2)(x^2-c^2)(x^2-d^2)(x^2-e^2)(x^2-f^2)(x^2-g^2)(x^2-h^2)$ $\displaystyle =x^{16}+R_1x^{14}+R_2x^{12}+R_3x^{10}+R_4x^8+R_5x^ 6+R_6x^4+R_7x^2+R_8$ :
$\displaystyle A=\frac{R_1}8$
$\displaystyle B=7A^2-\frac{R_2}4$
$\displaystyle C=35A^4-30A^2B+3B^2-\frac{R_4}2$
$\displaystyle D^2=A^8-4A^6B+6A^4B^2-2A^4C-4A^2BC-2B^2C+C^2-R_8$
Conditions:
$\displaystyle 8A(5A^2+2A-3B)=R_3$
$\displaystyle 8A(7A^4+3B^2-9A^2B+8C)=R_5$
$\displaystyle 4A^2(7A^4-15A^2B+9B^2+12C)-4B(B^2-C)=R_6$
$\displaystyle 8A(A^6-3A^4B+2A^2B^2-A^2C-B^3+BC)=R_7$
These four "conditions" are extra equations that are undoubtedly always true. The only way I can imagine proving this is substituting $\displaystyle A,B,C,D,R_1,R_2,...,R_8$ all in terms of $\displaystyle a,b,c,d,e,f,g,h$ and simplifying. Obviously, I am trying to find a simple shortcut around this.
This is a monumental task to be done by hand, but perhaps you have access to Maple or Mathematica? Also, I am pleased you have an actual example in the k=4 case that can be tested, though again, I have no resources for handling numbers that big. Once these verifications are made, the real meat of your quest is finding criteria for which D is a nonzero integer.
(You see what I mean by "doubling complexity"? For k=5, we will have 16 equations with 5 unknowns, leaving 11 "conditions" and undoubtedly a very long definition of $\displaystyle E^2$ for which we seek E as an integer.)
7. ## A Simpler Way = A Proof = Pythagorean Triples?
Consider: $\displaystyle ((x^2-A)^2-B)^2-C^2=(x^2-r_0^2)(x^2-r_1^2)(x^2-r_2^2)(x^2-r_3^2)$ means that $\displaystyle ((r_i^2-A)^2-B)^2-C^2=0$ for any $\displaystyle i$. So $\displaystyle r_i^2=A\pm\sqrt{B\pm C}$. Naming these roots as follows...
$\displaystyle r_0^2=A-\sqrt{B-C}$
$\displaystyle r_1^2=A-\sqrt{B+C}$
$\displaystyle r_2^2=A+\sqrt{B-C}$
$\displaystyle r_3^2=A+\sqrt{B+C}$
... it is clear that $\displaystyle r_0^2+r_2^2=r_1^2+r_3^2=2A$. This proves your original observation. It also provides a means of construction.
All pythagorean triples can be generated using this relation: Choose positive integers $\displaystyle m>n$, and $\displaystyle (m^2-n^2)^2+(2mn)^2=(m^2+n^2)^2$, so $\displaystyle T(m,n)=(a,b,c)=(m^2-n^2,2mn,m^2+n^2)$. Notice that by construction, $\displaystyle c\pm b$ is always a perfect square. Here is your recipe for constructing a ladder:
1. Pick two triples with the same hypotenuse: $\displaystyle T(7,4)=(33,56,65)$ and $\displaystyle T(8,1)=(63,16,65)$
2. Calculate B and C from the middle terms of the triples as such: $\displaystyle B=\frac12(56^2+16^2)=1696$ and $\displaystyle C=\frac12(56^2-16^2)=1440$
3. Let A be the length of the hypotenuse, $\displaystyle A=65$.
4. Your four roots are $\displaystyle r_i^2=A\pm\sqrt{B\pm C}=65\pm\sqrt{1696\pm 1440}=(7^2,9^2,3^2,11^2)$. These also can be gotten by taking the values $\displaystyle m\pm n$ in your choice of triples, $\displaystyle 7\pm 4$, $\displaystyle 8\pm 1$
5. Your final ladder is $\displaystyle ((x^2-65)^2-1696)^2-1440^2=(x^2-7^2)(x^2-9^2)(x^2-3^2)(x^2-11^2)$
8. I've read through your posts (including your new thread). It's been difficult and I'm still trying to gather my thoughts on what to do for the k = 4 case. Anyway keep up the good work. Hopefully I'll have something to add tonight.
9. ## Finding Special Triples - Pythagoras Lives!
Turns out this problem is intimately connected with Pythagorean Triples. Consider the k=4 case: $\displaystyle r_i^2=a_1\pm\sqrt{a_2\pm\sqrt{a_3\pm a_4}}$.
Using a binary counting system for convenience,
$\displaystyle r_{---}=r_{000}=r_0 = 101$
$\displaystyle r_{--+}=r_{001}=r_1 = 131$
$\displaystyle r_{-+-}=r_{010}=r_2 = 77$
$\displaystyle r_{-++}=r_{011}=r_3 = 11$
$\displaystyle r_{+--}=r_{100}=r_4 = 353$
$\displaystyle r_{+-+}=r_{101}=r_5 = 343$
$\displaystyle r_{++-}=r_{110}=r_6 = 359$
$\displaystyle r_{+++}=r_{111}=r_7 = 367$
(1) -- $\displaystyle r_0^2+r_4^2=r_1^2+r_5^2=r_2^2+r_6^2=r_3^2+r_7^2=2a _1$ | $\displaystyle a_1=67405$
(2) -- $\displaystyle (r_0r_4)^2+(r_2r_6)^2=(r_1r_5)^2+(r_3r_7)^2=2a_1^2-2a_2$ | $\displaystyle a_2=3525798096$
(3) -- $\displaystyle a_2^2-(a_1^2-r_0^2r_4^2)(a_1^2-r_2^2r_6^2)=a_3+a_4$ | $\displaystyle a_3=533470702551552000$
(4) -- $\displaystyle a_2^2-(a_1^2-r_1^2r_5^2)(a_1^2-r_3^2r_7^2)=a_3-a_4$ | $\displaystyle a_4=469208209191321600$
Above is a run-through of how to calculate $\displaystyle a_i$'s given the roots $\displaystyle r_i$'s. Below is how to find a suitable set of roots:
ENTER PYTHAGORAS: $\displaystyle T(m,n)=(n^2-m^2,2mn,m^2+n^2)$
$\displaystyle T_0(r_0,r_4)=(r_4^2-r_0^2,2r_0r_4,r_0^2+r_4^2)$
$\displaystyle T_1(r_1,r_5)=(r_5^2-r_1^2,2r_1r_5,r_1^2+r_5^2)$
$\displaystyle T_2(r_2,r_6)=(r_6^2-r_2^2,2r_2r_6,r_2^2+r_6^2)$
$\displaystyle T_3(r_3,r_7)=(r_7^2-r_3^2,2r_3r_7,r_3^2+r_7^2)$
Notice that by (1), the four hypotenuses of these four triples are all equal. Notice that by (2), $\displaystyle T_{0b}^2+T_{2b}^2=T_{1b}^2+T_{3b}^2$, where the subscript b indicates the second leg of the triple, or the middle term of the above relation. SO, FIND FOUR PYTHAGOREAN TRIPLES MEETING THESE TWO SIMPLE CRITERIA, AND THE M,N VALUES GENERATING THEM GIVE YOU YOUR EIGHT ROOTS!
$\displaystyle T_0(101,353)=(114408,71306,134810)$
$\displaystyle T_1(131,343)=(100488,86866,134810)$
$\displaystyle T_2(77,359)=(122952,55286,134810)$
$\displaystyle T_3(11,367)=(134568,8074,134810)$
$\displaystyle 71306^2+55286^2=86866^2+8074^2$
$\displaystyle 114408^2+122952^2=100488^2+134568^2$ (This is incidental, and follows easily. It is not an another condition.)
This is about as far as I've gotten. I am trying to derive a method of finding these "triplet quadruples" explicitly, as opposed to searching by trial. This, I believe, is where the fundamental roadblock lies.
10. ## Some conditions on k = 4
Your above post seems similar to a derivation I tried awhile ago. Let me know if the following below is equivalent to what you have.
Originally Posted by Media_Man
(1) -- $\displaystyle r_0^2+r_4^2=r_1^2+r_5^2=r_2^2+r_6^2=r_3^2+r_7^2=2a _1$ | $\displaystyle a_1=67405$
(2) -- $\displaystyle (r_0r_4)^2+(r_2r_6)^2=(r_1r_5)^2+(r_3r_7)^2=2a_1^2-2a_2$ | $\displaystyle a_2=3525798096$
(3) -- $\displaystyle a_2^2-(a_1^2-r_0^2r_4^2)(a_1^2-r_2^2r_6^2)=a_3+a_4$ | $\displaystyle a_3=533470702551552000$
(4) -- $\displaystyle a_2^2-(a_1^2-r_1^2r_5^2)(a_1^2-r_3^2r_7^2)=a_3-a_4$ | $\displaystyle a_4=469208209191321600$
Above is a run-through of how to calculate $\displaystyle a_i$'s given the roots $\displaystyle r_i$'s. Below is how to find a suitable set of roots:
ENTER PYTHAGORAS: $\displaystyle T(m,n)=(n^2-m^2,2mn,m^2+n^2)$
$\displaystyle T_0(r_0,r_4)=(r_4^2-r_0^2,2r_0r_4,r_0^2+r_4^2)$
$\displaystyle T_1(r_1,r_5)=(r_5^2-r_1^2,2r_1r_5,r_1^2+r_5^2)$
$\displaystyle T_2(r_2,r_6)=(r_6^2-r_2^2,2r_2r_6,r_2^2+r_6^2)$
$\displaystyle T_3(r_3,r_7)=(r_7^2-r_3^2,2r_3r_7,r_3^2+r_7^2)$
Notice that by (1), the four hypotenuses of these four triples are all equal.
I've noticed that the term $\displaystyle a_i$ which needs the most conditioning, is $\displaystyle a_1$. As you've mentioned, for k = 4, it must be the case that $\displaystyle 2a_1$ can be written as a sum of two squares in at least 4 ways.
Recall now that an integer $\displaystyle n$ can be written as a sum of two squares, iff there are no primes $\displaystyle p \equiv 3mod4$ that divide $\displaystyle n$ to an odd exponent. This eliminates a ton of candidates for what $\displaystyle a_1$ can be.
Furthermore (and this is just an observation), I've noticed that many integers that can be expressed as a sum of two squares in many distinct ways (or as you've put it, the hypotenus of many different triangles), is a product of many primes of the form $\displaystyle p \equiv 1mod4$.
Notice that any k = 4 ladder generated by $\displaystyle (a_1,a_2,a_3,a_4)$ can be written as a product of two k = 3 ladders $\displaystyle (b_1,b_2,b_3)$ and $\displaystyle (c_1,c_2,c_3)$ where $\displaystyle b_1 = c_1$ and $\displaystyle b_2 = c_2$.
The above uses the idea that if $\displaystyle p(x) - q(x) = c$ for some constant $\displaystyle c$, then $\displaystyle p(x)q(x) = (q(x) + 0.5c)^2 - (0.5c)^2$.
If we have two k = 3 polynomials, where $\displaystyle b_1 = c_1$ and $\displaystyle b_2 = c_2$, then the difference between them is $\displaystyle b_3^2 - c_3^2$. So long as this integer is even, then $\displaystyle p(x)q(x)$ will be a k = 4 ladder.
A motivating question from the above is how can we find two k = 3 polynomials, where $\displaystyle b_1 = c_1$ and $\displaystyle b_2 = c_2$?
I believe that the main necessary condition, is the following:
Let $\displaystyle (k_1,k_2,k_3,k_4)$ be the integers such that
$\displaystyle 2a_1 = (a_1 - k_i) + (a_1 + k_i) = r_i^2 + r_j^2$
(note that in the above, only 4 specific $\displaystyle (i,j)$ are used to define each $\displaystyle k_i$).
Relabeling the indices if necessary, if these $\displaystyle k_i$ satisfy the condition that $\displaystyle k_1^2 + k_2^2 = k_3^2 + k_4^2$, then it follows that $\displaystyle a_1$ is the first term in some sequence of $\displaystyle (a_1,a_2,a_3,a_4)$ that generates a k = 4 ladder.
I haven't found the time or energy to work out a proof to this yet. However I wonder if this is the same thing which you have in your previous post
For $\displaystyle a_1 = 67405$, I know the set $\displaystyle (k_1,k_2,k_3,k_4)$ would have to come from members of the following list:
$\displaystyle 9324$
$\displaystyle 14964$
$\displaystyle 40836$
$\displaystyle 50244$
$\displaystyle 57244$
$\displaystyle 61476$
$\displaystyle 62916$
$\displaystyle 67284$
The reason why the four $\displaystyle k_i$ must come from the above, is that these are the only integers $\displaystyle k$, such that the pair $\displaystyle (67405 + k,67405 - k)$ will be squares.
11. Most importantly, all squaring ladders of "degree" k MUST be the product of two ladders of "degree" k-1 where $\displaystyle a_1,a_2,...a_{k-1}$ are the same. This is an interesting observation, but a bit anticlimactic. Consider...
$\displaystyle P_1(x)=((x^2-67405)^2-3525798096)^2-1001338560^2$
$\displaystyle P_2(x)=((x^2-67405)^2-3525798096)^2-253500480^2$
$\displaystyle P_1(x)P_2(x)=(((x^2-67405)^2-3525798096)^2-$ $\displaystyle 533470702551552000 )^2-469208209191322000^2$
Now, recall our method of constructing k=3 ladders:
1. Find a hypotenuse shared by two triples...
$\displaystyle T_0(106,237)=(44933,50244,67405)$
$\displaystyle T_2(178,189)=(4037,67284,67405)$ -- $\displaystyle P_1$
$\displaystyle T_1(126,227)=(35653,57204,67405)$
$\displaystyle T_3(141,218)=(27643,61476,67405)$ -- $\displaystyle P_2$
2. B is the sum of the squares of the middle legs.
$\displaystyle B=a_2=\frac12(50244^2+67284^2)=3525798096$ -- $\displaystyle P_1$
$\displaystyle B=a_2=\frac12(57204^2+61476^2)=3525798096$ -- $\displaystyle P_2$
3. Calculate C by the difference instead of the sum from above.
$\displaystyle C=a_3=\frac18(67284^2-50244^2)=1001338560$ -- $\displaystyle P_1$
$\displaystyle C=a_3=\frac18(61476^2-57204^2)=253500480$ -- $\displaystyle P_2$
So as you can see, in order to find two k=3 ladders with A and B in common, we are bound by the exact same conditions to search for as we have been using already. However, this will be useful for the k=5 case. Consider: If a k=5 case exists, then there exists two k=4 cases with a common A,B,C. If we can find a formula to generate ALL k=4 cases and prove that no two share an A,B,C, we will successfully prove that no k=5 cases exist (as is probably the case).
Alunw is doing a great job spinning out k=4 examples in the other thread (why do we have two??? ). Let's keep collecting solutions and see if we can find a pattern and a formula!
12. ## Two
Ah okay I have a better understanding now, thanks.
We can try and find more than two cases if you'd like. Perhaps I'll ask my comp sci friend to see if he can arrange to have alumw's code run on several terminals at school. This might be quicker and might not, depending on whether or not the guys in charge will let me download something like Pari onto the laboratory computers.
Have you read the research question for this (6.17) in Crandall and Pomerance's book?
I've generally noticed that when working on these questions is that the questions themselves are typically unfruitful. Indeed, if k = 5 or higher were possible (or findable), there are many professionals who would have located them by now and this wouldn't be a research question. A general rule of thumb here is that usually one needs to alter the question into a new one, in order to find something fruitful. One idea is to loosen the restrictions on the definition of a squaring ladder, and instead find polynomials of the form $\displaystyle P_k(x) = ((..(x^2 - a_1(x))^2 - a_2(x))^2 - .......)^2 - a_k(x)^2$ where $\displaystyle a_i(x)$ are polynomials of either small degree, or themselves squaring ladders. Furthermore, these polynomials don't necessarily have to be a product of only linear terms. If we decide thats its okay to settle for something less than perfection, we could instead just find polynomial ladders that have a maximal number of integer roots. For instance, is there a k = 5 ladder that has less than $\displaystyle 2^5$ distinct integer roots, but more than $\displaystyle 2^4$.
13. ## Unifying Patterns, k=5
I've finally looked at this question enough to see the "beauty" in the structure and the patterns Below is a rough sketch on how to find a k=5 ladder, if it exists, based on the fact that it must be the product of 8 k=2 ladders...
Suppose $\displaystyle A=a^2+b^2=c^2+d^2=e^2+f^2=g^2+h^2=i^2+j^2=k^2+l^2= m^2+n^2=o^2+p^2$. This is Test 1, finding a large number expressible as the sum of two squares in eight unique ways. (Notice that A must have at least 8 prime factors of the form 4k+1.)
Define $\displaystyle B_1=2ab, B_2=2cd, B_3=2ef, B_4=2gh, B_5=2ij, B_6=2kl, B_7=2mn, B_8=2op$. We have here 8 k=2 ladders: $\displaystyle (x^2-A)^2-B_i^2$
Multiplying them in pairs to get a k=3 ladder, $\displaystyle B=\frac12(B_1^2+B_2^2)=\frac12(B_3^2+B_4^2)=\frac1 2(B_5^2+B_6^2)=\frac12(B_7^2+B_8^2)$
Or $\displaystyle B=2(a^2b^2+c^2d^2)=2(e^2f^2+g^2h^2)=2(i^2j^2+k^2l^ 2)=2(m^2n^2+o^2p^2)$. This is Test 2, that our variables a-p satisfy this constraint to produce the same $\displaystyle B$.
Now define $\displaystyle C_1=\frac12|B_1^2-B_2^2|, C_2=\frac12|B_3^2-B_4^2|, C_3=\frac12|B_5^2-B_6^2|, C_4=\frac12|B_7^2-B_8^2|$. We have here 4 k=3 ladders: $\displaystyle ((x^2-A)^2-B)^2-C_i^2$
Multiplying these in pairs to get a k=4 ladder, $\displaystyle C=\frac12(C_1^2+C_2^2)=\frac12(C_3^2+C_4^2)$
Or $\displaystyle C=2(a^2b^2-c^2d^2)^2+2(e^2f^2-g^2h^2)^2=2(i^2j^2-k^2l^2)^2+2(m^2n^2-o^2p^2)^2$. This is Test 3, that using a-h produces the same value for C as using i-p. This is also the last constraint.
Now define $\displaystyle D_1=\frac12|C_1^2-C_2^2|, D_2=\frac12|C_3^2-C_4^2|$
Or $\displaystyle D_1=2\left|(a^2b^2-c^2d^2)^2-(e^2f^2-g^2h^2)^2\right|, D_2=2\left|(i^2j^2-k^2l^2)^2-(m^2n^2-o^2p^2)^2\right|$. This gives us 2 k=4 ladders: $\displaystyle (((x^2-A)^2-B)^2-C)^2-D_i^2$
Multiplying to give our ultimate k=5 ladder, $\displaystyle D=\frac12(D_1^2+D_2^2)$ and $\displaystyle E=\frac12|D_1^2-D_2^2|$
Or $\displaystyle D=2\left((a^2b^2-c^2d^2)^2-(e^2f^2-g^2h^2)^2\right)^2+2\left((i^2j^2-k^2l^2)^2-(m^2n^2-o^2p^2)^2\right)^2$
And $\displaystyle E=2\left|\left((a^2b^2-c^2d^2)^2-(e^2f^2-g^2h^2)^2\right)^2-\left((i^2j^2-k^2l^2)^2-(m^2n^2-o^2p^2)^2\right)^2\right|$
And voila, we have our k=5 ladder, $\displaystyle ((((x^2-A)^2-B)^2-C)^2-D)^2-E^2$. The challenge then, is finding 16 positive integers a-p that satisfy all three constraints. Notice that, all told, there are eleven equations that must be met. So if we really want to brute force it, we can solve the system of 11 equations and 16 unknowns down to a 5-variable Diophantine equation. It should be unfathomably complicated, but it still boils the k=5 case down to finding an integer solution to a 5-variable equation.
This post is not at all a practical way to go about the problem, but perhaps we can all benefit from recognizing the beautifully nested patterns that finally emerge.
14. ## Nevermind
On second thought, the system is not that hard to reduce, but still yields no easy results.
$\displaystyle A=a^2+b^2=c^2+d^2=e^2+f^2=g^2+h^2=i^2+j^2=k^2+l^2= m^2+n^2=o^2+p^2$ -- (1)
$\displaystyle \frac12B=a^2b^2+c^2d^2=e^2f^2+g^2h^2=i^2j^2+k^2l^2 =m^2n^2+o^2p^2$ -- (2)
$\displaystyle (a^2b^2-c^2d^2)^2+(e^2f^2-g^2h^2)^2=(i^2j^2-k^2l^2)^2+(m^2n^2-o^2p^2)^2$ -- (3)
The above system simplifies as follows: Given independent variables $\displaystyle a,e,i,m,A$...
$\displaystyle B=4\frac{(a^4+e^4-i^4-m^4)A^2+(a^6+e^6-i^6-m^6)A+(a^8+e^8-i^8-m^8)}{(a^2+e^2-i^2-m^2)A+(a^4+e^4-i^4-m^4)}$
$\displaystyle c^2,d^2=A\pm\sqrt{A^2+4a^2A-4a^4-2B}$
$\displaystyle g^2,h^2=A\pm\sqrt{A^2+4e^2A-4e^4-2B}$
$\displaystyle k^2,l^2=A\pm\sqrt{A^2+4i^2A-4i^4-2B}$
$\displaystyle o^2,p^2=A\pm\sqrt{A^2+4m^2A-4m^4-2B}$
$\displaystyle b^2=A-a^2, f^2=A-e^2, j^2=A-i^2, n^2=A-m^2$
If anyone wants to take a crack at it, be my guest. Find positive integers A,B,a-p satisfying the above or prove none exist.
15. Clever stuff. Perhaps you should be an Algebraist. Just looking at all that algebra makes me cringe though, which is probably why I never had the stomach to even attempt k = 5 (in addition to the fact that I didn't think it was doable). I probably should have originally put up a different question back a month ago, however at that time I was fairly enthusiastic about this problem. There is another Integer Factoring question from Chapter 5 I'm going to put up (however I'm gonna modify it) some time in the near future.
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# Show f(x) >= g(x) given integral of f is >= g on any subinterval of [a,b]
afkguy
## Homework Statement
f,g are cont. fcns on [a,b] and $$\int f$$ $$\geq$$ $$\int g$$ for any subinterval [i,j] of [a,b]. Show f(x) $$\geq$$ g(x) on [a,b].
## Homework Equations
Don't know if I should have used something like the fundamental theorem or mean value of integrals or something. I was using theorems like those for other problems in this section but wasn't sure about this one so I took a different direction.
## The Attempt at a Solution
Assume that this isn't true, i.e., there exists some c in [a,b] s.t. g(c) > f(c).
f and g are continuous so let's choose our $$\epsilon$$ = g(c) - f(c), so there should exist some $$\delta$$ > 0 where:
|f(x) - f(c)| < $$\epsilon$$ whenever |x - c| < $$\delta$$
Thus 2f(c) - g(c) < f(x) < g(c) when x is in (c - $$\delta$$, c + $$\delta$$)
So f(x) < g(c) whenever x is in (c - $$\delta$$, c + $$\delta$$).
So consider any partition p of any subinterval [i,j] of [a,b] where the one of the subintervals of [i,j] is (c - $$\delta$$, c + $$\delta$$).
I stopped here because I'm pretty sure I chose the wrong epsilon and I wasn't really sure if I was even in the right direction.
Asphyxiated
well i am not sure how you would go about 'proving' this but knowing that the area of f is always greater than or equal to g on any sub interval [i,j] in the interval [a,b] implies that f never intersects g, they may touch and be equal but f will never be less than g because the area would be less on the sub interval [i,j] where f<g. The fact that this is true with any sub interval (implying infinitesimal intervals as well) means that f>=g.
Homework Helper
Proof by contradiction is a good strategy here and you had good ideas, but you are muddling yourself up by dealing with details that we don't need, such as finding epsilons.
1) Assume g(c) > f(c) for some c in [a,b]
2) Use the continuity property to deduce there is some neighborhood (c - e, c + e) ( e>0 ) where g > f in that neighborhood.
3) Consider the integral over this interval to find a contradiction.
Homework Helper
Gold Member
To prove 2) stated above use the function g(x)-f(x) which is also continuous and you know it has a positive limit at x=c.
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### Work Sample : Taylor
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## What's the problem? - Taylor : Grade Commentary
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Taylor has demonstrated extensive knowledge and understanding of fractions and decimals. Complex questions involving more than one operation, improper fractions, brackets and large numbers have been used in mathematically creative, multi-step questions with a high degree of consistency and accuracy. This work sample demonstrates characteristics of work typically produced by a student performing at grade A standard at the end of Stage 3.
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## Foundation Statement strands
The following strands are covered in this activity:
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Students ask questions and undertake investigations, selecting appropriate technological applications and problem-solving strategies. They use mathematical terminology and some conventions and they give valid reasons when comparing and selecting from possible solutions, making connections with existing knowledge and understanding.
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Students record and describe geometric and number patterns using tables and words. They construct, verify and complete number sentences involving the four operations.
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# A Complete Course in Algebra
Leach, Shewell, and Sanborn, 1885
### Hva folk mener -Skriv en omtale
Vi har ikke funnet noen omtaler på noen av de vanlige stedene.
### Innhold
DEFINITIONS AND NOTATION 1 ADDITION 15 USE OF PARENTHESES 24 DIVISION 36 FORMULÆ 43 FACTORING 51 LOWEST COMMON MULTIPLE 75 SIMPLE EQUATIONS 106
THE THEORY OF EXPONENTS 180 RADICALS 190 QUADRA RATIC EQUATIONS 211 PROBLEMS INVOLVING QUADRATIC EQUATIONS 223 EQUATIONS IN THE QUADRATIC FORM 228 THEORY OF QUADRATIC EQUATIONS 246 RATIO AND PROPORTION 255 ARITHMETICAL PROGRESSION 264
SIMPLE EQUATIONS CONTAINING Two UNKNOWN 132 MORE THAN 144 TITY 158 EvoluTJON 165
GEOMETRICAL PROGRESSION 273 BINOMIAL THEOREM 282 ANSWERS 307 Opphavsrett
### Populære avsnitt
Side 166 - Arts. 200 and 201 we derive the following rule : Extract the required root of the numerical coefficient, and divide the exponent of each letter by the index of the root.
Side 213 - In any trinomial square (Art. 108), the middle term is twice the product of the square roots of the first and third terms...
Side 44 - ... the product of the two, plus the square of the second. In the third case, we have (a + b) (a — 6) = a2 — b2. (3) That is, the product of the sum and difference of two quantities is equal to the difference of their squares.
Side 49 - The exponent of b in the second term is 1, and increases by 1 in each succeeding term.
Side 255 - The first and fourth terms of a proportion are called the extremes; and the second and third terms the means. Thus, in the proportion a : b = с : d, a and d are the extremes, and b and с the means.
Side 258 - In a series of equal ratios, any antecedent is to its consequent, as the sum of all the antecedents is to the sum of all the consequents. Let a: b = c: d = e:f.
Side 5 - If equal quantities be divided by the same quantity, or equal quantities, the quotients will be equal. 5. If the same quantity be both added to and subtracted from another, the value of the latter will not be changed.
Side 44 - The square of the sum of two quantities is equal to the SQuare of the first, plus twice the product of the first by the second, plus the square of the second.
Side 107 - Any term may be transposed from one side of an equation to the other by changing its sign. For, consider the equation x + a = b.
Side 227 - A' courier proceeds from P to Q in 14 hours. A second courier starts at the same time from a place 10 miles behind P, and arrives at Q at the same time as the first courier. The second courier finds that he takes half an hour less than the first to accomplish 20 miles. Find the distance from P to Q.
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# On the centralizers of $n$-cycles and conjugacy in $A_n$
I'd appreciate comments on the validity of these attempted proofs. Thanks.
Let $a$ be an $n$-cycle in $S_n$.
a) Show that the centralizer of $a$ in $S_n$ is $\langle a \rangle$.
b) Assume that $n$ is odd. Show that $A_n$ contains an $n$-cycle $b$ that is not conjugate to $a$ in $A_n$.
a) Since the number of $k$-cycles in $S_n$ is given by $\frac{n!}{(n-k)!k}$, the number of $n$-cycles in $S_n$ is $(n-1)!$. Since two elements of $S_n$ are conjugate iff they have the same cycle type, $|\mathrm{cl}(a)|=(n-1)!$ (where $\mathrm{cl}(a)$ is the conjugacy class of $a$). But $|\mathrm{cl}(a)|=|S_n:C_{S_n}(a)|$, or $|C_{S_n}(a)|=\frac{|S_n|}{|\mathrm{cl}(a)|}=\frac{n!}{(n-1)!}=n$. Since $\langle a \rangle \le C_{S_n}(a)$, it follows that $\langle a \rangle = C_{S_n}(a)$.
b) since $n$ is odd, all $n$-cycles are contained in $A_n$, so $A_n$ contains $(n-1)!\,\,$ $n$-cycles. Since, for $a \in S_n$, $C_{A_n}(a)=C_{S_n}(a) \cap A_n = \langle a \rangle \cap A_n$, we have $|\mathrm{cl}_{A_n}(a)|=|A_n:C_{A_n}(a)|=\frac{\frac{1}{2}n!}{n}=\frac{n!}{2n}=\frac{1}{2}(n-1)!$, and so $a$ cannot be conjugate to all $(n-1)!$ $n$-cycles in $A_n$ and there exists such an element $b$.
• Yes, those are both correct. And clearly written, too. – Christopher Feb 5 '14 at 17:22
• Great, thanks.$\,$ – Alex Petzke Feb 6 '14 at 1:46
• Perhaps you made a little mistake when you wrote: $C_{A_n}(a) =C_{A_n} \cap A_n$. Did you mean $C_{S_n}(a) =C_{A_n} \cap A_n$? – Jonas Gomes Mar 28 '14 at 14:29
• Yes, this should really be a comment. But I did make a correction, though it is not what you suggested. It should be $C_{A_n}(a)=C_{S_n}(a) \cap A_n$, not $C_{A_n}(a)=C_{A_n}(a) \cap A_n$. – Alex Petzke Mar 28 '14 at 15:20
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Vertical Angles
## Two congruent, non-adjacent angles formed by intersecting lines.
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Vertical Angles
What if you were given two angles of unknown size and were told they are vertical angles? How would you determine their angle measures? After completing this Concept, you'll be able to use the definition of vertical angles to solve problems like this one.
### Watch This
Watch the first part of this video.
Then watch the third part of this video.
### Guidance
Vertical angles are two non-adjacent angles formed by intersecting lines. $\angle 1$ and $\angle 3$ are vertical angles and $\angle 2$ and $\angle 4$ are vertical angles.
The Vertical Angles Theorem states that if two angles are vertical angles, then they are congruent.
#### Example A
Find $m\angle 1$ .
$\angle 1$ is vertical angles with $18^\circ$ , so $m\angle 1 = 18^\circ$ .
#### Example B
If $\angle ABC$ and $\angle DEF$ are vertical angles and $m\angle ABC =(4x+10)^\circ$ and $m \angle DEF=(5x+2)^\circ$ , what is the measure of each angle?
Vertical angles are congruent, so set the angles equal to each other and solve for $x$ . Then go back to find the measure of each angle.
$4x+10&=5x+2\\ x&=8$
So, $m\angle ABC = m\angle DEF=(4(8)+10)^\circ =42^\circ$
#### Example C
True or false: vertical angles are always less than $90^\circ$ .
This is false, you can have vertical angles that are more than $90^\circ$ . Vertical angles are less than $180^\circ$ .
### Guided Practice
Find the value of $x$ or $y$ .
1.
2.
3.
1. Vertical angles are congruent, so set the angles equal to each other and solve for $x$ .
$x+16&=4x-5\\3x&=21\\ x&=7^\circ$
2. Vertical angles are congruent, so set the angles equal to each other and solve for $y$ .
$9y+7&=2y+98\\7y&=91\\y&=13^\circ$
3. Vertical angles are congruent, so set the angles equal to each other and solve for $y$ .
$11y-36&=63\\11y&=99\\y&=9^\circ$
### Practice
Use the diagram below for exercises 1-2. Note that $\overline{NK} \perp \overleftrightarrow{IL}$ .
1. Name one pair of vertical angles.
1. If $m\angle INJ = 63^\circ$ , find $m\angle MNL$ .
For exercise 3, determine if the statement is true or false.
1. Vertical angles have the same vertex.
1. If $\angle ABC$ and $\angle DEF$ are vertical angles and $m\angle ABC =(9x+1)^\circ$ and $m \angle DEF=(5x+29)^\circ$ , what is the measure of each angle?
2. If $\angle ABC$ and $\angle DEF$ are vertical angles and $m\angle ABC =(8x+2)^\circ$ and $m \angle DEF=(2x+32)^\circ$ , what is the measure of each angle?
3. If $\angle ABC$ and $\angle DEF$ are vertical angles and $m\angle ABC =(x+22)^\circ$ and $m \angle DEF=(5x+2)^\circ$ , what is the measure of each angle?
4. If $\angle ABC$ and $\angle DEF$ are vertical angles and $m\angle ABC =(3x+12)^\circ$ and $m \angle DEF=(7x)^\circ$ , what is the measure of each angle?
5. If $\angle ABC$ and $\angle DEF$ are vertical angles and $m\angle ABC =(5x+2)^\circ$ and $m \angle DEF=(x+26)^\circ$ , what is the measure of each angle?
6. If $\angle ABC$ and $\angle DEF$ are vertical angles and $m\angle ABC =(3x+1)^\circ$ and $m \angle DEF=(2x+2)^\circ$ , what is the measure of each angle?
7. If $\angle ABC$ and $\angle DEF$ are vertical angles and $m\angle ABC =(6x-3)^\circ$ and $m \angle DEF=(5x+1)^\circ$ , what is the measure of each angle?
### Vocabulary Language: English
Vertical Angles
Vertical Angles
Vertical angles are a pair of opposite angles created by intersecting lines.
Vertical Angles Theorem
Vertical Angles Theorem
The Vertical Angles Theorem states that if two angles are vertical, then they are congruent.
### Explore More
Sign in to explore more, including practice questions and solutions for Vertical Angles.
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# Statistics Problems
1. Indicate whether the following variables are categorical or quantitative:
1 Favorite food.
2 Favorite profession.
3 Number of goals scored by your favorite team last season.
4 Number of students at your school.
5 The eye color of your classmates.
2. Indicate whether the following variables are discrete or continuous:
1 Number stocks sold every day in the stock exchange.
2Hourly temperatures recorded at an observatory.
4 The diameter of the wheels of several cars.
5 Number of children from 50 families.
6 Annual Census of Americans.
3. Classify the following variables as categorical, quantitative discrete or continuous.
1 The nationality of a person.
2 Number of liters of water contained in a tank.
3 Number of books on a library shelf.
4 Sum of points tallied from a set of dice.
5 The profession of a person.
6 The area of the different tiles on a building.
4. The marks obtained by a group of students in a test are:
15, 20, 15, 18, 22, 13, 13, 16, 15, 19, 18, 15, 16, 20, 16, 15, 18, 16, 14, 13.
Construct a frequency distribution table for the data and draw the corresponding frequency polygon.
5. Given the following series:
3, 3, 4, 3, 4, 3, 1, 3, 4, 3, 3, 3, 2, 1, 3, 3, 3, 2, 3, 2, 2, 3, 3, 3, 2, 2, 2, 2, 2, 3, 2, 1, 1, 1, 2, 2, 4, 1.
Construct a frequency distribution table for the data and draw the corresponding bar chart.
6. Given the following series:
5, 2, 4, 9, 7, 4, 5, 6, 5, 7, 7, 5, 5, 2, 10, 5, 6, 5, 4, 5, 8, 8, 4, 0, 8, 4, 8, 6, 6, 3, 6, 7, 6, 6, 7, 6, 7, 3, 5, 6, 9, 6, 1, 4, 6, 3, 5, 5, 6, 7.
Construct a frequency distribution table for the data and draw the corresponding bar chart.
7. The weights of 65 children are represented by the following table:
Weight [50, 60) [60, 70) [70, 80) [80,90) [90, 100) [100, 110) [110, 120) fi 8 10 16 14 10 5 2
1 Construct the frequency table.
2 Plot the histogram and frequency polygon.
8. 40 students in a class have obtained the following test scores out of 50.
3, 15, 24, 28, 33, 35, 38, 42, 23, 38, 36, 34, 29, 25, 17, 7, 34, 36, 39, 44, 31, 26, 20, 11, 13, 22, 27, 47, 39, 37, 34, 32, 35, 28, 38, 41, 48, 15, 32, 13.
1 Construct the frequency table.
2 Draw the histogram and frequency polygon.
9. Given the statistical distribution of the table.
xi 61 64 67 70 73 fi 5 18 42 27 8
Calculate:
1 The mode, median and mean.
2 The range, average deviation, variance and standard deviation.
10.Calculate the mean, median and mode for the following set of numbers: 5, 3, 6, 5, 4, 5, 2, 8, 6, 5, 4, 8, 3, 4, 5, 4, 8, 2, 5, 4.
11 Find the variance and standard deviation for the following data series:
12, 6, 7, 3, 15, 10, 18, 5.
12 Find the mean, median and mode for the following set of numbers:
3, 5, 2, 6, 5, 9, 5, 2, 8, 6.
13. Find the average deviation, variance and standard deviation for the following series of numbers:
2, 3, 6, 8, 11.
12, 6, 7, 3, 15, 10, 18, 5.
14 The test results from a group of employees from a factory are represented in the following table:
fi [38, 44) 7 [44, 50) 8 [50, 56) 15 [56, 62) 25 [62, 68) 18 [68, 74) 9 [74, 80) 6
Draw the histogram and the cumulative frequency polygon.
15. Given the series:
3, 5, 2, 7, 6, 4, 9.
3, 5, 2, 7, 6, 4, 9, 1.
Calculate:
The mode, median and mean.
The average deviation, variance and standard deviation.
The quartiles 1 and 3.
The deciles 2 and 7.
The percentiles 32 and 85.
16. A statistical distribution is given by the following table:
[10, 15) [15, 20) [20, 25) [25, 30) [30, 35) fi 3 5 7 4 2
Calculate:
The mode, median and mean.
The range, average deviation and variance.
The quartiles 1 and 3.
The deciles 3 and 6.
The percentiles 30 and 70.
17. Given the statistical distribution:
[0, 5) [5, 10) [10, 15) [15, 20) [20, 25) [25, ∞) fi 3 5 7 8 2 6
Calculate the mode.
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# From the dam of the hydroelectric power station, water falls every minute with a volume of 600 m
From the dam of the hydroelectric power station, water falls every minute with a volume of 600 m (3 degrees). The height of the dam is 8 m. Determine the power developed by the flow of falling water. The density of water is 1000 kg / m (3 degrees)
t = 1 min = 60 s.
V = 600 m ^ 3.
h = 8 m.
ρ = 1000 kg / m ^ 3.
g = 9.8 m / s ^ 2.
N -?
Power N is a physical quantity that characterizes the speed of performing work A and is determined by the formula: N = A / t, where t is the time of performing work A.
We will find work A by the change in the potential energy of water when it falls: A = ΔEp.
The change in potential energy ΔЕп is found by the formula: ΔЕп = m * g * h, m is the mass of the fallen water, g is the acceleration of gravity, h is the height of the fall of water.
We find the mass of water m by the formula: m = ρ * V.
The formula for determining the power will take the form: N = ρ * V * g * h / t.
N = 1000 kg / m ^ 3 * 600 m ^ 3 * 9.8 m / s ^ 2 * 8 m / 60 s = 784000 W.
Answer: the power of the dam is N = 784000 W.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.
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# What is the probability of a random string of numbers having a consistent pattern?
I saw a video by Michael Shermer today, where he said people who believed in ESP etc. were more likely to see patterns in selected pictures than the average person, whether or not the picture held a pattern.
This got me thinking, what is the probability of a random data set having a pattern? Of course with pictures it would be astronomically high without a lot of approximation, but what about a simple string of numbers? Is there a rule that can describe this?
Here is a random string I have generated as an example:
8177322396
Within this, if I had only produced 3223, it could be an inflection, or a repeated pattern 2233… If I only had 396, it could be 3,9,6,12,9,15… If I only had 732, it could be the end of the pattern 32,16,7,3,2. As a whole though, the string has no pattern that I can see. The probability of a pattern therefore decreases with string length, but is there some sort of relationship that we can describe empirically?
Observing members: 0 Composing members: 0
I think if it’s truely random then it is stochastic, but true randomness may not even be possible, so who knows?
dpworkin (27035)
@dpworkin The point is the ability to read patterns that aren’t really there into a stochastic data set. For example, there is no pattern that can be used to predict the decimal places of e, but most people recognise the “18281828”, and wonder how long the pattern lasts. Later on someone might recognise the pattern “996996”, but of course the pattern is not intended.
Isn’t it the nature of randomness to be random? That would mean that are times when you have distinct patterns, and other times not. I heard that when people try and make a random line of numbers, it is usually obvious because it has so few patterns, when they do occur in nature and randomly.
skfinkel (13478)
@skfinkel That would mean that are times when you have distinct patterns, and other times not.
Of course, but in what proportions?
I thought that when we study math in school, its broken to discrete and statistical mathematics. When we observe random patterns, that’s when statistical maths is used to deduce the pattern in that randomness. I know some people have the ability to understand pattern easier than others, there is a movie about John Nash, a great mathematician who decifers code for the military in the war, interesting movie to watch.
mea05key (1802)
I understood the question, I just didn’t know how to predict the answer.
dpworkin (27035)
In a truly random sequence of digits, any given finite pattern is not only possible but inevitable. In other words, as the length of the sequence approaches infinity, the probability of any specified finite sequence approaches 1, which is certainty.
In your example of “8177322396”, which is ten digits long, it’s frequency of occurrence among all 10-digit strings should be, on average one in 10^10 (once per ten billion). So if you looked at, say, 10^15 such 10-digit strings you should find yours many times over.
The same is true for any finite sequence no matter how large—say a trillion 3’s in a row. the “law of large numbers” can seem counter-intuitive.
By the way, the digits of irrational numbers such as pi, the square root of 2, etc., exhibit the same statistical properties as random sequences. 8177322396 ought to occur sooner or later, as would a trillion 3’s in a row—though we might not be able to actually calculate where in any reasonable amount of time such as the lifetime of the universe.
gasman (11296)
Part of the problem is determining what is meant by pattern. Suppose that I generated a random digit sequence using nuclear decay, and I modified it so that every time there were 7 1’s in a row, the seventh 1 is replaced by randomly choosing one of the other digits. This sequence is no longer random, and there are computer programs that test for randomness that would recognize that it is not random. But would you say that it has a pattern?
Clearly the 9th digit is nine… which happens in 10% of random digit sequences with 9 digits in them.
Patterns can be invented to fit a random sample, once the sample is available. The chance that that sample could have been guessed and would show up, depends on the pattern definition, and the random number generation algorithm. Sometimes the people generating the randomness unintentionally build in some patterns, too. And sometimes, subconscious figuring can look like something else. For example, horses that seem to be able to do math, but are really picking up on the way the asker asks the question, and learns how many times to stomp their hoof to get a biscuit, based on the asker’s expectation and clues they didn’t realize they were giving.
Zaku (22533)
It depends on how long the string is. If it is infinitely long, the chance is zero.
If you want relatively accurate proportions, your “patterns” need to be formally defined within the possibility domain – and thus you can get to ratios. Truly random assigning will result in a value close to that ratio(s). Testing to see if someone is psychic you can use their test results to the ratio metrics to see if they do significantly better than random.
ipso (4476)
@ipsoTesting to see if someone is psychic you can use their test results to the ratio metrics to see if they do significantly better than random.” Yes, this is one of the basic tests that paranormal claims must pass to demonstrate scientific validity. Generally statistical significance (low p-value) is lacking in tests of psychic ability. There’s a vast skeptical literature on the long and sad history of junk science, fake science, and hoaxes on which evidence for psychic abilities rests.
@Zaku: Read about Clever Hans: ”Though the experiment strongly indicated that the horse probably had no real grasp of math, it did uncover an extraordinary insight…There is evidence to indicate that horses may possess an enhanced sensitivity to inconspicuous body language, perhaps as a key part of their social interactions with other horses” (the point being the horse trainer was subtly cuing the horse to stop tapping while probably unaware of it himself). Which calls for a Feynman quote: Science is a way of trying not to fool yourself. The first principle is that you must not fool yourself, and you are the easiest person to fool. [1964 lecture]
Back to the original question: ”What is the probability of a random data set having a pattern” As noted by others above, pattern is a vague term. A small number of simple rules can generate non-trivial infinite patterns—think computer programs. The results may be pseudo-random or chaotic—but deterministic nonetheless. (True randomness occurs in nature, e.g. in the form of radioactive decay, though the probabilities are precisely measured and explained by theory.)
But then if an underlying pattern exists which is generating the sequence, then it isn’t really “random” after all, is it? At some level, being random and having a pattern are mutually exclusive properties. In fact randomness is even more slippery a concept than pattern. Making the question a bit ambiguous.
In Carl Sagan’s sci-fi novel Contact the aliens send blocks of data whose length is the product of two prime numbers, suggesting rearrangement as a 2-dimensional raster-scan bit image (which indeed pans out in the story). Many seemingly random sequences can turn out to be highly patterned, organized, and intelligent.
But wait: Pattern recognition is what our Home sapiens brains have been honed to do over millions of years. As Michael Shermer put it in Why People Believe Weird Things, ”We evolved to be skilled, pattern-seeking, cause-finding creatures. Those who were best at finding patterns (e.g., standing upwind of game animals is bad for the hunt, cow manure is good for the crops, etc.) left behind the most offspring. We are their descendants.
Again there’s a huge amount written about the human tendency to find signals in noise (seeing the Virigin Mary in fried tortillas, for instance). Eerie-seeming coincidences in numerical sequences are no different, and can give the illusion of “pattern” where none actually exists, such as the first few decimal digits of e, mentioned in somebody’s answer above.
Bottom line: You can always find patterns in random, or seemingly random, data. Whether such patterns are actually meaningful is usually in doubt. It’s easy to fool yourself.
gasman (11296)
Isn’t randomness in and of itself a pattern?
@thekoukoureport You’ve uncovered another ambiguity of the word ‘pattern’. Randomness has certain definite mathematical statistical properties. If you label that ‘pattern’ then the answer is definitely yes.
But randomness is inherently non-deterministic—following no rules to predict future behavior. If you label that ‘unpatterned’ then the answer is definitely no.
You can always find some kind of rule or pattern to account—after the fact—for some finite list of numbers. We do it all the time to devise a mnemonic rule for remembering phone numbers, etc. Some people use geometric patterns on a phone keypad. None of this has much bearing on underlying randomness.
gasman (11296)
I’d say the probability is 100% for real numbers like e or pi or sqrt(2). All you need is plenty of computing time to find plenty of patterns. For example
3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587006606315588174881520920962829254091715364367892590360011330530548820466521384146951941511609
I found a pattern: 74
It occurs three times.
mattbrowne (31595)
One definition of randomness, Kolmogorov complexity , talks about the number of lines of code needed to reproduce a pattern. If the program is no better than one required to individually name each item then the pattern is random. I am not that familiar with how this is used, but I would think that it would have to be modified, since a data compaction algorithm could do at least some compaction on random data. For example, random data will have sequences where a single character is repeated several times and the compaction algorithm can take advantage of this by using an escape sequence to indicate the repetition.
@LostInParadise I can’t say I follow you completely, having no math, but I was still fascinated by your answer and found a lot in it to think about. Thanks.
dpworkin (27035)
@LostInParadise Thanks for your answer. As far as I understand it, compression algorithms only work well when there is a lot of homogeneous data. If you only have three repeated digits in a thread of three hundred, then the extra two bytes (or however many) to define how many digits and where in the look-up table to find what digit to repeat would take up more room than that saved. For example, saying “two sixes” is longer than 66, but “ten nines” is shorter than 9999999999.
But I think the point @LostInParadise was making (and it’s an excellent one: G.A.) is that repetition is not the only hallmark of patterns that can be compressed. For example (as one of the related links points out) you could have a full-color, hi-res image of the Mandelbrot set containing millions or billions of pixels, yet it can be easily described by a simple equation consisting of only a handful of symbols & a few other bits of data (starting seed, zoom scale, etc) even though no explicit repeating groups of data are involved.
With truly random data, on the other hand, any computer algorithm you might devise to describe it would itself would be longer or more complex than the data it’s describing. I was only dimly aware of Kolmogorov theory but I see now that it’s highly relevant to the original question.
gasman (11296)
@gasman I understand that, I was simply trying to illustrate that compression only works well for strong patterns, not for data sets that are a distant approximation to a pattern. ‘Homogeneous’ wasn’t the best choice of words though.
What I was trying to say is that in a long enough sequence of random digits there are going to be some long sequences of digits and other repeating patterns, allowing for some compaction, although I imagine the space saving will not be very significant.
By definition there can be no meaningful pattern in randomness. It is true that a billion monkeys typing for eternity will produce the works of Shakespeare but any literary gems they produce will be hopelessly lost in a chaos of letters. If you tried to prevent the monkeys from typing out Macbeth the sequence would cease to be random.
flutherother (29399)
Depending on what you want from your pattern.
If you want “999” then the longer the string, the higher the probability.
But people will find patterns wherever they want. The whole problem skeptics have with ESP is that the “pattern” is not defined ahead of time.
roundsquare (5517)
or
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Cultivating Teamwork in Math Class
Establishing community in the classroom can be a challenge. Here's an activity my students participated in on the first day of school. I learned about this activity while participating in the Advanced Educator International Space Camp in Huntsville, Alabama. The objective is for two crews of astronauts to exchange positions in cramped quarters when a new crew shows up to relieve the old crew at the International Space Station.
How the Game is Played
1. Only one person can move at a time.
2. Only movement forward (in the direction a person faces) is allowed. In the above diagram, orange players can only move right, while blue players can only move left.
3. A person can move to an empty space in front of them.
4. A person can jump an opposing team member in front of them.
What Does It Mean to Win the Game?
The teams win the challenge when they have exchanged their original positions. See the ending position diagram for an example of what this looks like.
What If...?
A teacher could use an agility (speed) ladder for this activity. Or if a ladder isn't handy, use tape. This is a shot of my classroom the first day. It's pretty unlikely a teacher would have only eight students. I put down three tape ladders in my classroom on the floor. If the number of students is not a multiple of eight (like my class was), the teacher could place the extra students on the side as coaches. To up the responsibility of the coach, add the rule that no one inside the ladder can talk to anyone else.
Teacher Moves
While this activity was going on, I floated between the groups and listened very carefully. I wanted to learn about which of my students would step up and take initiative; which would be a leader; which would be concerned about the frustration of others and take action to minimize other students' discomfort/anxiety. This activity helped me better understand how to assign groups for course work in a meaningful way.
Examples of Student Moves
Below is an example of what some students might do.
If the third blue player from the right jumps the lone orange player, the blue team has a problem. With two blue players in adjacent cells, the game is gridlocked and ends.
Computational Thinking
Once the students came up with the solution, I gave them the sequence "1-2-3-4-4-3-2-1" and asked them how it relates to this situation. Think of this sequence as the answer key.
1: Orange moves first
2: Blue moves next - twice
3: Orange moves three times
4: Blue moves four times
4: Orange moves four times
3: Blue moves three times
2: Orange moves two times
1: Blue moves one time
Low Entry, High Ceiling (Extending the Task)
• Ask the students to come up with some pseudo-code to describe how they would build this game on a computer using programming applications.
• Ask the students whether the strategy remains the same if there are teams of five? Or if there are two empty middle squares? Three empty middle squares?
• Ask the students to write a program that allows the user to watch the game. Then ask the students to write a program that allows the user to play the game.
• Example from my classroom I had two students come up with different lines of thinking for coding this game on a computer. One student thought of a number line to label each cell, using the values -4, -3, -2, -1, 0, 1, 2, 3, 4. Another student thought of simply have numbers represent each student. The starting configuration would be
1 2 3 4 _ 5 6 7 8. Then, each move would be a shuffling of the sequence. The second row would be 1 2 3 _ 4 5 6 7 8. The third row would be 1 2 3 5 4 _ 6 7 8. The fourth row would be 1 2 3 5 _ 4 6 7 8. We had a really spirited discussion of the issues that could arise from each organizational coding strategy.
PAEMST Ceremony
Below is a photo from our photo session with President Obama on Friday 7/31. I am on the far left!
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# How can I make a guess sidel function with the given variables: GS(Am,bm,initial_guess,10^-4) and the range is 𝑁 = [20, 40, 80, 160, 320, 640, 1280]
1 Ansicht (letzte 30 Tage)
poppy am 27 Nov. 2022
Beantwortet: Ishan am 29 Nov. 2022
How can I make a guess sidel function with the given variables: GS(Am,bm,initial_guess,10^-4) and the range is 𝑁 = [20, 40, 80, 160, 320, 640, 1280]?
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### Antworten (1)
Ishan am 29 Nov. 2022
Hi Anujan,
If you want to solve a linear equation by Gauss-Seidel method, you can use the below function to do so:
%A = input('Enter a Co-effecient Matrix A: ');
A = [10 3 1;3 10 2;1 2 10];
%B = input('Enter Source Vector B: ');
B = [19;29;35];
%P = input('Enter initial Guess Vector: ');
P = [0;0;0];
%n = input('Enter no. of iterations: ');
n = 10;
%e = input('Enter your tollerance: ');
e = 0.0001
e = 1.0000e-04
N = length(B);
X = zeros(N,1);
Y = zeros(N,1);
for j=1:n
for i = 1:N
X(i) = (B(i)/A(i,i))-(A(i,[1:i-1,i+1:N])*P([1:i-1,i+1:N]))/A(i,i);
P(i) = X(i);
end
fprintf('Iteration no. %d\n', j)
X
if abs(Y-X)<e
break
end
Y=X;
end
Iteration no. 1
X = 3×1
1.9000 2.3300 2.8440
Iteration no. 2
X = 3×1
0.9166 2.0562 2.9971
Iteration no. 3
X = 3×1
0.9834 2.0056 3.0005
Iteration no. 4
X = 3×1
0.9983 2.0004 3.0001
Iteration no. 5
X = 3×1
0.9999 2.0000 3.0000
Iteration no. 6
X = 3×1
1.0000 2.0000 3.0000
Iteration no. 7
X = 3×1
1.0000 2.0000 3.0000
Hope this helps you solve your problem!
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# Statistics
posted by .
Two cards are drawn in succession from a standard deck of 52 cards without replacing the first one before the second.
a) Are these outcomes are independent? Why?
b) Find P ( 3 on 1st card and 10 on the 2nd card)
c) Find P( 10 on the 1st card and 3 on 2nd card)
d) Find the Probability of drawing 10 and a 3 in either order
• Statistics -
a) The choice of the first card without replacement changes the probability of the second card.
b) 4/52 and 4/51. To find probability that all events would occur, multiply the probability of the individual events.
c) Same
## Similar Questions
1. ### Statistics
you draw two cards from a standard deck of 52 cards, but before you draw the second card, you put the first one back and reshuffle the deck. a) are the outcomes on the two cards independent?
2. ### Intro to Statistics
Two cards are drawn at random from a standard 52-card deck of playing cards without replacement. Find the probability that the first card is a heart and the second is red.
3. ### math
Two cards are drawn in succession without replacement from a standard deck of 52 cards. What is the probability that the first card is a spade given that the second card is a club?
4. ### Math
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6. ### Math
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7. ### Math
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8. ### math - help please
Assume that 2 cards are drawn in succession and without replacement from a standard deck of 52 cards. Find the probability that the following occurs: the second card is a 9, given that the first card was a 9.
9. ### Statistics
. Two cards are drawn at random, without replacement, from a standard 52-card deck. Find the probability that: (a) both cards are the same color (b) the first card is a face card and the second card is black
10. ### math
Two cards are drawn in succession without replacement from a standard deck of 52 cards. What is the probability that the first card is a spade given that the second card is a diamond?
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# Matrix Eigenvectors
• Apr 25th 2010, 08:04 PM
Laydieofsorrows
Matrix Eigenvectors
A is a 3 x 3 matrix with eigenvectors v1 = [1 0 0], v2 = [1 1 0], v3 = [1 1 1] (all vertical) corresponding to eigenvalues x1 = -1/3, x2 = 1/3, x3 = 1, respectively and x = [2 1 2].
Find (A^k)x. What happens as k approaches infinity?
I don't even know where to begin on this one.
Thanks for the help!
• Apr 25th 2010, 10:09 PM
eigenvex
First use the eigenvalue equation (Av = lambda*v) to backsolve for the values in A. You'll also have to use the property that the trace of a matrix (sum of the diagonal values) is also equal to the sum of that matrix's eigenvalues.
Once you have A, you'll see it is an upper-triangular matrix with values <= 1....and you can see what will happen if you raise this matrix to a high power. (if you can't see it, just type it into matlab and do A^5, A^6, ... until you see what is happening)
• Apr 26th 2010, 01:07 AM
Defunkt
Quote:
Originally Posted by Laydieofsorrows
A is a 3 x 3 matrix with eigenvectors v1 = [1 0 0], v2 = [1 1 0], v3 = [1 1 1] (all vertical) corresponding to eigenvalues x1 = -1/3, x2 = 1/3, x3 = 1, respectively and x = [2 1 2].
Find (A^k)x. What happens as k approaches infinity?
I don't even know where to begin on this one.
Thanks for the help!
First, note that $x=2v_3-v_2+v_1$. Now,
$A^kx = A^k(2v_3-v_2+v_1) = 2(A^kv_3) - A^kv_2 + A^kv_1$.
Use the fact that if $\lambda$ is an eigenvalue of a matrix C with eigenvector $v$ then $\lambda ^k$ is en eigenvalue of $C^k$ with eigenvector $v$ (prove this if you haven't already) to get:
$A^kx = 2(x_3^kv_3) - x_2^kv_2 + x_1^kv_1$
and finish.
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10.1 – Finding Who is X & who is Y?
I hope you have gained a good understanding of linear regression. Also, how to conduct linear regression operations on two sets data using MS Excel. We are referring to two variables X and Y.
X is the variable (independent one), and Y the variable (dependent one). You'd know X and eventually Y if you spent some time considering this.
Let's just run a linear regression on 2 stocks - HDFC Bank or ICICI Bank - and see what we get.
I am setting ICICI Bank to be X and HDFC Bank to be Y.Before we proceed,let's take a quick note-
1. You must ensure that your data is accurate - adjust for bonuses, splits, and other corporate actions
2. Check that the dates are correct. For example, I have data for both stocks from the 4th of December 2015 to the 4th of December 2017.
Here's how the data looks (IMAGE 1)
I will run linear regression on these stocks (I have explained how to do it in the previous chapter). Also, please note that I am only focusing on stock prices and not stock returns.
(IMage 2
The linear regression results are as follows:
(Image 3).
The equation is -ICICI is independent, HDFC is dependent.
HDFC =
Price of ICICI * 7.613 – 663.677
Assuming you know the equation, I assume that you do. If you are unfamiliar with the equation, I recommend that you read the preceding chapters. Here's a quick summary: The equation attempts to predict HDFC's price using ICICI's price.
We are trying to 'express the HDFC price in terms of ICICI.
Let's reverse this: I will make ICICI dependent and HDFC independent.
These are the results.
(IMAGE 4).
The equation is -
ICICI = HDFC + 0.09 + 142.4677
You can therefore regress in two different ways for the two stocks you have just mentioned by reordering which stock has the dependent variable and which is the independent variable.
The question is: How do you decide which should be considered dependent and which independent? Also, it is up to you to decide which order makes sense.
Three things are required to answer this question:
1. Standard Error
2. Standard Error in intercept
3. The sum of the two variables above.
The linear equation above essentially expresses the price variation of ICICI in terms HDFC. Refer to the equation. The expression of price variation in one stock using the price of another stock as a reference cannot be 100%. It would be 100% if there was no play.
The equation must be strong enough to explain as much variation as possible in the price of the dependent variable, while keeping the independent variable in context. This equation is stronger than the one before it.
We now have to ask the obvious question: How do we determine how strong the linear regression equation? Here is the ratio.
Standard Error of Incept / Standard Error comes into play. Before we can talk about the actual ratio, it is important to first understand the numerator as well as the denominator.
Below is the linear regression equation for ICICI (as independent) and HDFC (as dependent).
HDFC =
Price of ICICI * 7.613 – 663.677
This basically means that if I know the price for ICICI, I should also be able predict the price for HDFC. In reality, however, the actual HDFC price is different from what was predicted. This is known as the "Residuals".
following a snapshot is given,for the residual explaining the price of HDFC keeping ICICI as a variable which is independent.
I often get asked a common question when I discuss the regression equation and residuals. Is it possible to use regression if there are residuals every time? Also, we cannot rely on an equation that fails to accurately predict anything, even once.
It is a valid question. It is a fair question.
This was not about predicting the stock's price based on the independent stock. It was always about residuals!
Let me tell you, the residuals exhibit a certain behavior. Once we understand the pattern and can identify it, we can then work backwards to create a trade. This trade is a pair trade because it involves simultaneously buying and selling both stocks.
We will delve into this more in the next chapters. For now, however, let's focus on the 'Standard Error,' the denominator of the Standard Error of Incept / Standard Error equation
When you perform a linear regression operation, the standard error is one variable that gets reported. The snapshot below shows the same.
(Image 6).
The standard error is the standard deviation of residuals. The residuals are a time series array. If you calculate the standard deviation from the residuals, you will get the standard error.
Let me actually manually calculate the standard error for the residuals. I'm doing this with X = ICICI, and y = HFC
(IMAGE 7).
Excel tells me that the standard deviation of 152.665. The 152.819 standard error is reported in the summary output. It is possible to ignore the minor differences.
It is not easy to calculate the 'Standard error of the Intercept'. It is reported in the regression report. Here is the standard error for the intercept, with x = ICICI y = HDFC
(IMAGe 8).
Remember, the regression equation is -
y=M*x+
Where?
M = Slope
C = Intercept
You will see that both M and C are estimates. How are they calculated? These are based on historical data that has been provided to the regression algorithm. There may be noise components in the data, but few outliers. This means that estimates can be wrong.
The Standard Error of the Intercept measures the variance of the estimated intercept. This helps to understand how much the intercept can vary. This can be taken as likely to the 'Standard error' itself. To summarize -
• Standard Error of Incept - The variance from the intercept
• Standard Error is the variance of residuals.
Let's now bring back the "Error Ratio" after we have defined these variables. The term "Error Ratio" is not a standard term. I created it to make it easier to understand.
However, we do know that the error ratio is -
Error Ratio = The Standard Error Of Intercept / The Standard Error
I calculated the same as -
1. ICICI as X, HDFC as y = 0.401
2. HDFC as X, ICICI at y = 0.227
The error ratio is a key factor in deciding whether to assign X or Y to stock. The lower the error ratio, the better. We will assign HDFC (X) to ICICI (Y), as ICICI (Y).
Although I would love to give the reasons behind why we use the error ratio to designate X and Y I won't. This will be a topic I return to when I am able to take up pair trading.
Now, calculate the error ratio to determine which stock is dependent and which independent.
keypoints
1. X is the independent stock, Y the dependent stock
2. To figure out between the stock that has to be X and Y depends upon the 'Error Ratio' .
3. Estimates of both the slope and intercept are made from the linear regression equation
4. Error Ratio = Standard error of the Intercept/Standard Error
5. The standard error is the standard deviation from the residuals
6. The standard error of intercept provides an indication of the variance of intercept
7. Regress Stock 1 and Stock 2, or Stock 2 and Stock 1, depending on which has the lowest error ratio, to determine which stock is dependent and which is independent
8. Certain properties of residuals can be used to identify pairs trading patterns.
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# Pre-Calculus
posted by on .
The paddle wheel of a boat measures 16 feet in diameter and is revolving at a rate of 20 rpm. The maximum depth of the paddle wheel under water is 1 foot. Suppose a point is located at the lowest point of the wheel at t=0.
1) Write a cosine function with phase shift 0 for the height above water at the initial point after t seconds
2) use your function to find the height of the initial point after 5.5 seconds
3) and find the values of x for which the equation sin x= -1 is true.
Thanks for any assistance
• Pre-Calculus - ,
w = angular rate = 20*2pi/60 = 2pi/3 radians/sec
r = 16/2 = 8 feet
height of point above axle of wheel = -8cos wt
height of point above water = h = 7 - 8 cos wt
since w = 2 pi/3
h = 7 - 8 cos (2 pi t /3)
if t = 5.5
h = 7 - 8 cos (11 pi/3)
= -.839 ft
I do not understand what 3) has to do with this.
sin anything is -1 for 3pi/2, 3pi/2 + n*2pi
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# What number is 5 less than half of 32?
$5$ less than half of $32$ would be $11$.
Half of $32 = \left(\frac{1}{2}\right) \cdot 32 = 16$
$5$ less than half of $32$ is the same as $5$ less than $16$.
$16 - 5 = 11$
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# Fourier Analysis. u m, a n u n = am um, u m
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## Transcription
1 Fourier Analysis Fourier series allow you to expand a function on a finite interval as an infinite series of trigonometric functions. What if the interval is infinite? That s the subject of this chapter. Instead of a sum over frequencies, you will have an integral Fourier Transform For the finite interval you have to specify the boundary conditions in order to determine the particular basis that you re going to use. On the infinite interval you don t have this large set of choices. After all, if the boundary is infinitely far away, how can it affect what you re doing over a finite distance? But see section In section 5.3 you have several boundary condition listed that you can use on the differential equation u λu and that will lead to orthogonal functions on your interval. For the purposes here the easiest approach is to assume periodic boundary conditions on the finite interval and then to take the limit as the length of the interval approaches infinity. On L < x < +L, the conditions on the solutions of u λu are then u( L) u(+l) and u ( L) u (+L). The solution to this is most conveniently expressed as a complex exponential, Eq. (5.19) u(x) e ikx, where u( L) e ikl u(l) e ikl This implies e 2ikL 1, or 2kL 2nπ, for integer n, ±1, ±2,.... With these solutions, the other condition, u ( L) u (+L) is already satisfied. The basis functions are then u n (x) e iknx e nπix/l, for n, ±1, ±2, etc. (15.1) On this interval you have the Fourier series expansion f(x) a n u n (x), and um, f u m, a n u n am um, u m In the basis of Eq. (15.1) this normalization is u m, u m 2L. Insert this into the series for f. un, f f(x) u n (x) 1 un, u n 2L n n un, f u n (x) (15.2) Now I have to express this in terms of the explicit basis functions in order to manipulate it. When you use the explicit form you have to be careful not to use the same symbol (x) for two different things in the same expression. Inside the u n, f there is no x left over it s the dummy variable of integration and it is not the same x that is in the u n (x) at the end. Denote k n πn/l. f(x) 1 2L n L L dx u n (x ) * f(x ) u n (x) 1 2L L dx e iknx f(x ) e iknx n L Now for some manipulation: As n changes by 1, k n changes by k n π/l. f(x) 1 2π 1 2π n n π L L dx e iknx f(x ) e iknx L James Nearing, University of Miami 1 L e iknx k n dx e iknx f(x ) (15.3) L
2 For a given value of k, define the integral L g L (k) dx e ikx f(x ) L 15 Fourier Analysis 2 If the function f vanishes sufficiently fast as x, this integral will have a limit as L. Call that limit g(k). Look back at Eq. (15.3) and you see that for large L the last factor will be approximately g(k n ), where the approximation becomes exact as L. Rewrite that expression as f(x) 1 2π n As L, you have k n, and that turns Eq. (15.4) into an integral. f(x) dk 2π eikx g(k), where g(k) e iknx k n g(k n ) (15.4) dx e ikx f(x) (15.5) The function g is called* the Fourier transform of f, and f is the inverse Fourier transform of g. Examples For an example, take the step function { 1 ( a < x < a) f(x) (elsewhere) a g(k) dx e ikx 1 a 1 [e ika e +ika] ik 2 sin ka k then (15.6) The first observation is of course that the dimensions check: If dx is a length then so is 1/k. After that, there is only one parameter that you can vary, and that s a. As a increases, obviously the width of the function f increases, but now look at g. The first place where g(k) is at ka π. This value, π/a decreases as a increases. As f gets broader, g gets narrower (and taller). This is a general property of these Fourier transform pairs. Can you invert this Fourier transform, evaluating the integral of g to get back to f? Yes, using the method of contour integration this is very easy. Without contour integration it would be extremely difficult, and that is typically the case with these transforms; complex variable methods are essential to get anywhere with them. The same statement holds with many other transforms (Laplace, Radon, Mellin, Hilbert, etc. ) The inverse transform is dk 2 sin ka dk e ika e ika eikx 2π k C 1 2π eikx ik i C 2 dk 2π 1 k [e ik(x+a) e ik(x a)] C 1 C 2 * Another common notation is to define g with an integral dx/ 2π. That will require a corresponding dk/ 2π in the inverse relation. It s more symmetric that way, but I prefer the other convention.
3 15 Fourier Analysis 3 1. If x > +a then both x+a and x a are positive, which implies that both exponentials vanish rapidly as k +i. Push the contour C 2 toward this direction and the integrand vanishes exponentially, making the integral zero. 2. If a < x < +a, then only x + a is positive. The integral of the first term is then zero by exactly the preceding reasoning, but the other term has an exponential that vanishes as k i instead, implying that you must push the contour down toward i. dk 1 i C 3 2π k eik(x a) C 4 +i 1 ( 1)2πi Res 2π k e ik(x a) k i 1 2π. 2πi 1 C 3 C 4 The extra ( 1) factor comes because the contour is clockwise. 3. In the third domain, x < a, both exponentials have the form e ik, requiring you to push the contour toward i. The integrand now has both exponentials, so it is analytic at zero and there is zero residue. The integral vanishes and the whole analysis takes you back to the original function, Eq. (15.6). Another example of a Fourier transform, one that shows up often in quantum mechanics f(x) e x2 /σ 2, so g(k) dx e ikx e x2 /σ 2 The trick to doing this integral is to complete the square inside the exponent. dx e ikx x2 /σ 2 ikx x 2 /σ 2 1 σ 2 [ x 2 + σ 2 ikx σ 4 k 2 /4 + σ 4 k 2 /4 ] 1 σ 2 [ (x + ikσ 2 /2) 2 + σ 4 k 2 /4 ] The integral of f is now g(k) e σ2 k 2 /4 dx e x 2 /σ 2 where x x + ikσ/2 The change of variables makes this a standard integral, Eq. (1.1), and the other factor, with the exponential of k 2, comes outside the integral. The result is g(k) σ π e σ2 k 2 /4 (15.7) This has the curious result that the Fourier transform of a Gaussian is* a Gaussian Convolution Theorem What is the Fourier transform of the product of two functions? It is a convolution of the individual transforms. What that means will come out of the computation. Take two functions f 1 and f 2 with Fourier transforms g 1 and g 2. dx f 1 (x)f 2 (x)e ikx dk dx 2π g 1(k )e ik x f 2 (x)e ikx * Another function has this property: the hyperbolic secant. Look up the quantum mechanical harmonic oscillator solution for an infinite number of others.
4 15 Fourier Analysis 4 dk 2π g 1(k ) dx e ik x f 2 (x)e ikx dk 2π g 1(k ) dx f 2 (x)e i(k k )x dk 2π g 1(k )g 2 (k k ) (15.8) The last expression (except for the 2π) is called the convolution of g 1 and g 2. dx f 1 (x)f 2 (x)e ikx 1 2π (g 1 g 2 )(k) (15.9) The last line shows a common notation for the convolution of g 1 and g 2. What is the integral of f 2 over the whole line? dk dx f * (x)f(x) dx f * (x) 2π g(k)eikx dk 2π g(k) dx f * (x)e ikx [ ] dk 2π g(k) dx f(x)e ikx * dk 2π g(k)g* (k) (15.1) This is Parseval s identity for Fourier transforms. There is an extension to it in problem Time-Series Analysis Fourier analysis isn t restricted to functions of x, sort of implying position. They re probably more often used in analyzing functions of time. If you re presented with a complicated function of time, how do you analyze it? What information is present in it? If that function of time is a sound wave you may choose to analyze it with your ears, and if it is music, the frequency content is just what you will be listening for. That s Fourier analysis. The Fourier transform of the signal tells you its frequency content, and sometimes subtle periodicities will show up in the transformed function even though they aren t apparent in the original signal. A function of time is f(t) and its Fourier transform is g(ω) dt f(t) e iωt with f(t) g(ω) e iωt 2π The sign convention in these equations appear backwards from the one in Eq. (15.5), and it is. One convention is as good as the other, but in the physics literature you ll find this pairing more common because of the importance of waves. A function e i(kx ωt) represents a wave with (phase) velocity ω/k, and so moving to the right. You form a general wave by taking linear combinations of these waves, usually an integral. Example When you hear a musical note you will perceive it as having a particular frequency. It doesn t, and if the note has a very short duration it becomes hard to tell its* pitch. Only if its duration is long * Think of a hemisemidemiquaver played at tempo prestissimo.
5 15 Fourier Analysis 5 enough do you have a real chance to discern what note you re hearing. This is a reflection of the facts of Fourier transforms. If you hear what you think of as a single note, it will not last forever. It starts and it ends. Say it lasts from t T to t +T, and in that interval it maintains the frequency ω. f(t) Ae iω t ( T < t < T ) (15.11) The frequency analysis comes from the Fourier transform. T g(ω) T dt e iωt Ae iω t A ei(ω ω )T e i(ω ω )T i(ω ω ) 2A sin(ω ω )T (ω ω ) This is like the function of Eq. (15.6) except that its center is shifted. It has a peak at ω ω instead of at the origin as in that case. The width of the function g is determined by the time interval T. As T is large, g is narrow and high, with a sharp localization near ω. In the reverse case of a short pulse, the range of frequencies that constitute the note is spread over a wide range of frequencies, and you will find it difficult to tell by listening to it just what the main pitch is supposed to be. This figure shows the frequency spectrum for two notes having the same nominal pitch, but one of them lasts three times as long as the other before being cut off. It therefore has a narrower spread of frequencies. T larger T smaller ω Example Though you can do these integrals numerically, and when you are dealing with real data you will have to, it s nice to have some analytic examples to play with. I ve already shown, Eq. (15.7), how the Fourier transform of a Gaussian is simple, so start from there. If g(ω) e (ω ω ) 2 /σ 2 then f(t) σ 2 π e iω t e σ2 t 2 /4 If there are several frequencies, the result is a sum. g(ω) n A n e (ω ωn)2 /σ 2 n f(t) n A n σ n 2 π e iωnt e σ2 nt 2 /4 In a more common circumstance you will have the time series, f(t), and will want to obtain the frequency decomposition, g(ω), though for this example I worked backwards. The function of time is real, but the transformed function g is complex. Because f is real, it follows that g satisfies g( ω) g * (ω). See problem Real Imag f g
6 15 Fourier Analysis 6 This example has four main peaks in the frequency spectrum. The real part of g is an even function and the imaginary part is odd. Real f g Imag This is another example with four main peaks. In either case, if you simply look at the function of time on the left it isn t obvious what sort of frequencies are present. That s why there are standard, well-developed computer programs to do the Fourier analysis Derivatives There are a few simple, but important relations involving differentiation. What is the Fourier transform of the derivative of a function? Do some partial integration. F( f) dt e iωt df dt eiωt f(t) iω dt e iωt f(t) iωf(f) (15.12) Here I ve introduced the occasionally useful notation that F(f) is the Fourier transform of f. The boundary terms in the partial integration will go to zero if you assume that the function f approaches zero at infinity. The n th time derivative simply give you more factors: ( iω) n on the transformed function Green s Functions This technique showed up in the chapter on ordinary differential equations, section 4.6, as a method to solve the forced harmonic oscillator. In that instance I said that you can look at a force as a succession of impulses, as if you re looking at the atomic level and visualizing a force as many tiny collisions by atoms. Here I ll get to the same sort of result as an application of transform methods. The basic technique is to Fourier transform everything in sight. The damped, forced harmonic oscillator differential equation is Multiply by e iωt and integrate over all time. integration as in Eq. (15.12). m d2 x dt 2 + bdx dt + kx F (t) (15.13) You do the transforms of the derivatives by partial dt e iωt [Eq. (15.13)] mω 2 x ibω x + k x F, where x(ω) dt e iωt x(t) This is an algebraic equation that is easy to solve for the function x(ω). F (ω) x(ω) mω 2 ibω + k Now use the inverse transform to recover the function x(t). F x(t) 2π e iωt x(ω) (ω) 2π e iωt mω 2 ibω + k e iωt dt F 2π mω 2 ibω + k (t )e iωt dt F (t e iωt ) 2π mω 2 ibω + k eiωt (15.14)
7 15 Fourier Analysis 7 In the last line I interchanged the order of integration, and in the preceding line I had to be sure to use another symbol t in the second integral, not t. Now do the ω integral. 2π e iωt mω 2 ibω + k eiωt 2π e iω(t t ) mω 2 ibω + k (15.15) To do this, use contour integration. The singularities of the integrand are at the roots of the denominator, mω 2 ibω + k. They are C 1 ω ib ± b 2 + 4km 2m ω ± C 2 Both of these poles are in the lower half complex plane. The contour integral C 1 is along the real axis, and now I have to decide where to push the contour in order to use the residue theorem. This will be governed by the exponential, e iω(t t ). First take the case t < t, then e iω(t t ) is of the form e +iω, so in the complex ω-plane its behavior in the ±i directions is as a decaying exponential toward +i ( e ω ). It is a rising exponential toward i ( e + ω ). This means that pushing the contour C 1 up toward C 2 and beyond will make this integral go to zero. I ve crossed no singularities, so that means that Eq. (15.15) is zero for t < t. Next, the case that t > t. Now e iω(t t ) is of the form e iω, so its behavior is reversed from that of the preceding paragraph. It dies off rapidly toward i and rises in the opposite direction. That means that I must push the contour in the opposite direction, down to C 3 and to C 4. Because of the decaying exponential, the large arc of the contour that is pushed down to i gives zero for its integral; the two lines that parallel the i-axis cancel each other; only the two residues remain. C 3 e iω(t t ) 2π mω 2 ibω + k 2πi Res ω ± C 4 (15.16) The denominator in Eq. (15.15) is m(ω ω + )(ω ω ). Use this form to compute the residues. Leave the 1/2π aside for the moment and you have e iω(t t ) mω 2 ibω + k e iω(t t ) m(ω ω + )(ω ω ) The residues of this at ω ± are the coefficients of these first order poles. at ω + : e iω +(t t ) m(ω + ω ) and at ω : e iω (t t ) m(ω ω + )
8 The explicit values of ω ± are ω + ib + b 2 + 4km and 2m b Let ω 2 + 4km 2m The difference that appears in the preceding equation is then 15 Fourier Analysis 8 ω ib b 2 + 4km 2m and γ b 2m ω + ω (ω iγ) ( ω iγ) 2ω Eq. (15.16) is then [ ] 2π. e iω(t t ) e mω 2 ibω + k i i(ω iγ)(t t ) 2mω + e i( ω iγ)(t t ) +2mω i ) [ 2mω e γ(t t e iω (t t ) + e +iω (t t ) ] 1 mω e γ(t t ) sin ( ω (t t ) ) Put this back into Eq. (15.14) and you have t x(t) dt F (t )G(t t ), where G(t t ) 1 mω e γ(t t ) sin ( ω (t t ) ) (15.17) If you eliminate the damping term, setting b, this is exactly the same as Eq. (4.34). The integral stops at t t because the Green s function vanishes beyond there. The motion at time t is determined by the force that was applied in the past, not the future. Example Apply a constant external force to a damped harmonic oscillator, starting it at time t and keeping it on. What is the resulting motion? { (t < ) F (t) F 1 (t > ) where F 1 is a constant. The equation (15.17) says that the solution is (t > ) t t x(t) dt F (t )G(t t ) dt F 1 G(t t ) t F 1 dt 1 mω e γ(t t ) sin ( ω (t t ) ) F 1 2imω F 1 2imω F 1 2imω t F 1 m(γ 2 + ω 2 ) dt e γ(t t ) [ e iω (t t ) e iω (t t ) ] [ ] 1 )(t t ) 1 t t )(t t γ iω e( γ+iω ) γ + iω e( γ iω t [ 2iω γ 2 + ω 2 e γt [ 2iγ sin ω t γ 2 + ω 2 + 2iω cos ω t ] ] [1 e γt[ cos ω t + γ ω sin ω t ]] Check the answer. If t it is correct; x() as it should. If t, x(t) goes to F 1 / ( m(γ 2 +ω 2 ) ) ; is this correct? Check it out! And maybe simplify the result in the process. Is the small time behavior correct?
9 15 Fourier Analysis Sine and Cosine Transforms Return to the first section of this chapter and look again at the derivation of the Fourier transform. It started with the Fourier series on the interval L < x < L and used periodic boundary conditions to define which series to use. Then the limit as L led to the transform. What if you know the function only for positive values of its argument? If you want to write f(x) as a series when you know it only for < x < L, it doesn t make much sense to start the way I did in section Instead, pick the boundary condition at x carefully because this time the boundary won t go away in the limit that L. The two common choices to define the basis are u() u(l), and u () u (L) (15.18) Start with the first, then u n (x) sin(nπx/l) for positive n. The equation (15.2) is unchanged, save for the limits. f(x) a n u n (x), 1 and um, f u m, a n u n am um, u m n1 In this basis, u m, u m L/2, so f(x) un, f u n (x) 2 un, u n L n1 un, f u n (x) Now explicitly use the sine functions to finish the manipulation, and as in the work leading up to Eq. (15.3), denote k n πn/l, and the difference k n π/l. n1 f(x) 2 L 2 π L dx f(x ) sin nπx nπx sin 1 L L sin nπx L L k n dx f(x ) sin nπx /L (15.19) 1 For a given value of k, define the integral L g L (k) dx sin(kx )f(x ) If the function f vanishes sufficiently fast as x, this integral will have a limit as L. Call that limit g(k). Look back at Eq. (15.19) and you see that for large L the last factor will be approximately g(k n ), where the approximation becomes exact as L. Rewrite that expression as f(x) 2 π sin(k n x) k n g(k n ) (15.2) 1 As L, you have k n, and that turns Eq. (15.2) into an integral. f(x) 2 π dk sin kx g(k), where g(k) dx sin kx f(x) (15.21)
10 15 Fourier Analysis 1 This is the Fourier Sine transform. For a parallel calculation leading to the Cosine transform, see problem 15.22, where you will find that the equations are the same except for changing sine to cosine. f(x) 2 π dk cos kx g(k), where g(k) dx cos kx f(x) (15.22) What is the sine transform of a derivative? Integrate by parts, remembering that f has to approach zero at infinity for any of this to make sense. dx sin kx f (x) sin kxf(x) For the second derivative, repeat the process. k dx cos kx f(x) k dx cos kx f(x) dx sin kx f (x) kf() k 2 dx sin kx f(x) (15.23) 15.7 Wiener-Khinchine Theorem If a function of time represents the pressure amplitude of a sound wave or the electric field of an electromagnetic wave the power received is proportional to the amplitude squared. By Parseval s identity, the absolute square of the Fourier transform has an integral proportional to the integral of this power. This leads to the interpretation of the transform squared as some sort of power density in frequency. g(ω) 2 is then a power received in this frequency interval. When this energy interpretation isn t appropriate, g(ω) 2 is called the spectral density. A useful result appears by looking at the Fourier transform of this function. 2π g(ω) 2 e iωt 2π g* (ω)e iωt dt f(t )e iωt dt f(t ) 2π g* (ω)e iωt e iωt [ ] * dt f(t t) ) 2π g(ω)e iω(t dt f(t )f(t t) * (15.24) When you re dealing with a real f, this last integral is called the autocorrelation function. It tells you in some average way how closely related a signal is to the same signal at some other time. If the signal that you are examining is just noise then what happens now will be unrelated to what happened a few milliseconds ago and this autocorrelation function will be close to zero. If there is structure in the signal then this function gives a lot of information about it. The left side of this whole equation involves two Fourier transforms ( f g, then g 2 to it s transform). The right side of this theorem seems to be easier and more direct to compute than the left, so why is this relation useful? It is because of the existence of the FFT, the Fast Fourier Transform, an algorithm that makes the process of Fourier transforming a set of data far more efficient than doing it by straight-forward numerical integration methods faster by factors that reach into the thousands for large data sets.
11 15 Fourier Analysis 11 Problems 15.1 Invert the Fourier transform, g, in Eq. (15.7) What is the Fourier transform of e ik x x 2 /σ 2? Ans: A translation of the k case 15.3 What is the Fourier transform of xe x2 /σ 2? 15.4 What is the square of the Fourier transform operator? That is, what is the Fourier transform of the Fourier transform? 15.5 A function is defined to be f(x) { 1 ( a < x < a) (elsewhere) What is the convolution of f with itself? (f f)(x) And graph it of course. Start by graphing both f(x ) and the other factor that goes into the convolution integral. Ans: (2a x ) for ( 2a < x < +2a), and zero elsewhere Two functions are f 1 (x) { 1 (a < x < b) (elsewhere) and f 2 (x) { 1 (A < x < B) (elsewhere) What is the convolution of f 1 with f 2? And graph it Derive these properties of the convolution: (a) f g g f (b) f (g h) (f g) h (c) δ(f g) f δg+g δf where δf(t) tf(t), δg(t) tg(t), etc. (d) What are δ 2 (f g) and δ 3 (f g)? 15.8 Show that you can rewrite Eq. (15.9) as F(f g) F(f). F(g) where the shorthand notation F(f) is the Fourier transform of f Derive Eq. (15.1) from Eq. (15.9) What is the analog of Eq. (15.1) for two different functions? That is, relate the scalar product of two functions, f1, f 2 f 1 *(x)f 2(x) to their Fourier transforms. Ans: g * 1 (k)g 2(k) dk/2π In the derivation of the harmonic oscillator Green s function, and starting with Eq. (15.15), I assumed that the oscillator is underdamped: that b 2 < 4km. Now assume the reverse, the overdamped case, and repeat the calculation Repeat the preceding problem, but now do the critically damped case, for which b 2 4km. Compare your result to the result that you get by taking the limit of critical damping in the preceding problem and in Eq. (15.17).
12 15 Fourier Analysis Show that if f(t) is real then the Fourier transform satisfies g( ω) g * (ω). What are the properties of g if f is respectively even or odd? Evaluate the Fourier transform of { ( ) A a x ( a < x < a) f(x) (otherwise) How do the properties of the transform vary as the parameter a varies? Ans: 2A(1 cos ka) / k Evaluate the Fourier transform of Ae α x. Invert the transform to verify that it takes you back to the original function. Ans: 2α/(α 2 + k 2 ) Given that the Fourier transform of f(x) is g(k), what is the Fourier transform of the function translated a distance a to the right, f 1 (x) f(x a)? Schroedinger s equation is i h ψ t h2 2 ψ 2m x 2 + V (x)ψ Fourier transform the whole equation with respect to x, and find the equation for Φ(k, t), the Fourier transform of ψ(x, t). The result will not be a differential equation. Ans: i h Φ(k, t)/ t ( h 2 k 2 /2m)Φ + (v Φ)/2π Take the Green s function solution to Eq. (15.13) as found in Eq. (15.17) and take the limit as both k and b go to zero. Verify that the resulting single integral satisfies the original second order differential equation (a) In problem you have the result that a double integral (undoing two derivatives) can be written as a single integral. Now solve the equation d 3 x dt 3 F (t) C directly, using the same method as for Eq. (15.13). You will get a pole at the origin and how do you handle this, where the contour of integration goes straight through the origin? Answer: Push the contour up as in the figure. Why? This is what s called the retarded solution for which the value of x(t) depends on only those values of F (t ) in the past. If you try any other contour to define the integral you will not get this property. (And sometimes there s a reason to make another choice.) (b) Pick a fairly simple F and verify that this gives the right answer. Ans: 1 t 2 dt F (t )(t t ) Repeat the preceding problem for the fourth derivative. Would you care to conjecture what 3 1 / 2 integrals might be? Then perhaps an arbitrary non-integer order? Ans: 1 t 6 dt F (t )(t t ) What is the Fourier transform of xf(x)? Ans: ig (k)
13 15 Fourier Analysis Repeat the calculations leading to Eq. (15.21), but for the boundary conditions u () u (L), leading to the Fourier cosine transform For both the sine and cosine transforms, the original function f(x) was defined for positive x only. Each of these transforms define an extension of f to negative x. This happens because you compute g(k) and from it get an inverse transform. Nothing stops you from putting a negative value of x into the answer. What are the results? What are the sine and cosine transforms of e αx. In each case evaluate the inverse transform What is the sine transform of f(x) 1 for < x < L and f(x) otherwise. Evaluate the inverse transform Repeat the preceding calculation for the cosine transform. Graph the two transforms and compare them, including their dependence on L Choose any different way around the pole in problem 15.19, and compute the difference between the result with your new contour and the result with the old one. Note: Plan ahead before you start computing.
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# Now lets move on to factorizations that may | Ch 6.5 - 68
ISBN: 9780321758941 177
## Solution for problem 6.1.603 Chapter 6.5
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Problem 6.1.603
Now lets move on to factorizations that may require two or more techniques. In Exercises 1780, factor completely, or state that the polynomial is prime. Check factorizations using multiplication or a graphing utility. 100y2 49
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BIOL 1030 Lecture 2/25 Notes (Folkerts) Flowers Vary In… Ovary position - Hypogynous: below the ovary (superior ovary) (1) - Epigynous: when everything is above the ovary (inferior ovary) (2) - Perigynous: inbetween ovary position (3) o Hypanthium: a cup like structure in which the ovary sits if it is a perigynous case. Fusion of Parts - Connation - Adnation Presence/Absence of Parts - Complete: if all 4 whorrols are prese
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##### ISBN: 9780321758941
The full step-by-step solution to problem: 6.1.603 from chapter: 6.5 was answered by , our top Math solution expert on 12/23/17, 04:54PM. The answer to “Now lets move on to factorizations that may require two or more techniques. In Exercises 1780, factor completely, or state that the polynomial is prime. Check factorizations using multiplication or a graphing utility. 100y2 49” is broken down into a number of easy to follow steps, and 35 words. Since the solution to 6.1.603 from 6.5 chapter was answered, more than 240 students have viewed the full step-by-step answer. This full solution covers the following key subjects: . This expansive textbook survival guide covers 119 chapters, and 11220 solutions. This textbook survival guide was created for the textbook: Introductory & Intermediate Algebra for College Students, edition: 4. Introductory & Intermediate Algebra for College Students was written by and is associated to the ISBN: 9780321758941.
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Practice 8-1 Exploring Exponential Models Without graphing, determine whether each equation represents exponential growth or exponential decay. ... =4 48. log 4x =-1 49. 2 log -log 3 =1 Solve by graphing. Round answers to the nearest hundredth. 50. 10n =3 51. 3y =5 52. 10k-2 =20 53. 5x =4 54 ...
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Exploring Exponential Models - BakerMath.org
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Algebra 2 Benchmark 4 Practice - Lavergne High School
Identify the letter of the choice that best completes the statement or answers the question. ____ 1. EQ1 Let f(x) =3x +2 ... 1 Algebra 2 Benchmark 4 Practice Answer Section MULTIPLE ... L1 REF: 8-1 Exploring Exponential Models 8. ANS: G DIF: L1 REF: 8-1 Exploring Exponential Models 9 ...
http://www.lhs.rcs.k12.tn.us/teachers/frydm/frydm_files/ExamView%20Pro%20-%20Algebra%202%20Benchmark%204%20w%20EQs%20Practice.pdf
Date added: August 8, 2013 - Views: 5
ANSWERS - Mercer Island School District
(continued) 7-1 Form G Exploring Exponential Models For each annual rate of change, fi nd the corresponding growth or decay factor. 17. 145% 18. 210% 19. 40% 20. 1200% 21. ... ANSWERS 7-1 Practice (continued) Form K Exploring Exponential Models
http://www.mercerislandschools.org/cms/lib3/WA01001855/Centricity/Domain/686/solutions%20wksts%20ch7.pdf
Date added: March 18, 2014 - Views: 2
Exploring Function Graphs – Grade 10 1 Ohio Standards Connection Patterns, Functions, and ... exponential functions and inverse variation from given data. ... Possible Answers Function Descriptions Exponential Functions Is a single curve
http://ims.ode.state.oh.us/ODE/IMS/Lessons/Web_Content/CMA_LP_S04_BA_L10_I01_01.pdf
Date added: February 8, 2012 - Views: 30
The High School Math Project Rhinos and M&M’s
exploring exponential models more accessible. It is important to emphasize the patterns in the tables, graphs, and algebraic rules for ... 1. Answers will vary. Number of M&M’s Remaining Trial Number Number of M&M’s remaining 0 140 1 76 2 39 3 22 4 12 5 8 6 3
http://www-tc.pbs.org/teachers/mathline/lessonplans/pdf/hsmp/rhinos.pdf
Date added: August 8, 2013 - Views: 5
Exploring Exponential Models For each annual rate of change, nd the corresponding growth or decay factor. 17. 1 45% 18. 2 10% 19. 2 40% 20. 1 200% 21. 1 ... Practice Form G Exploring Exponential Models Graph each function. 1. y 5
Date added: September 16, 2013 - Views: 7
Unit 5 Exploring Exponentia (Exponents and Logarithms)
e. Understand and use basic exponential functions as models of real phenomena. f ... reviewed and students practice working with rational exponents. ... The Planet of Exponentia Learning Task
https://www.georgiastandards.org/Frameworks/GSO%20Frameworks/Acc-Math-II-Unit-5-SE-Exponential-Logarithmic-Functions.pdf
Date added: April 28, 2012 - Views: 17
PLAN AND STATISTICS Comparing Linear, Exponential, 9.8 and ...
Practice Level A (p. 116) Practice Level B (p. 117) Practice Level C ... Comparing Linear, Exponential, and Quadratic Models ... 9.8 R E A L L I F E L EXPLORING DATA AND STATISTICS LINEAR MODEL EXPONENTIAL MODEL QUADRATIC MODEL
Date added: September 12, 2014 - Views: 1
Equations and InequalitiesEquations and Inequalities - ClassZone
1.5 Problem Solving Using Algebraic Models: EXPLORING DATA AND STATISTICS 33 ... 1.7 Solving Absolute Value Equations and Inequalities 50 CONCEPT ACTIVITY: Absolute Value Equations and Inequalities, 49 QUIZ 3, 56 ASSESSMENT Skill Review, 2 Quizzes, 17, 40, 56 ... 8.1 Exponential Growth 465
Date added: November 16, 2012 - Views: 63
Course: Algebra 1 Unit #5: Quadratic Functions
Algebra 1 Quadratic Functions This document is the property of MAISA. Page 1 of 7 8 ... different classes of polynomial functions by exploring the graphs of the functions. The ... parabolas, inverse models, and exponential models. Students investigate patterns and various models using the ...
Date added: March 24, 2013 - Views: 42
Prentice Hall Mathematics: Algebra 2 © 2004 Correlated to ...
Radical Functions, 408-413; 8-1: Exploring Exponential Models, 422-429; 8-2: Properties of Exponential Functions, 431-436; 8-3: Logarithmic Functions as Inverses, 438-444; 445; 9-2: Graphing Inverse Variations, 485-490; 9-3: Rational Functions and Their
http://k12pearson.com/statepage/correlation/ph_corr/al/AL_PH_Math_Algebra_2_with_Trig_2004.pdf
Date added: January 21, 2013 - Views: 3
HSM12CC A2 07 AO - Stacie Kyhn - AAEC Math Department
7-1 Practice Form K Exploring Exponential Models ... Exploring Exponential Models For each annual rate of change, • nd the corresponding growth or decay factor. 14. ... Answers may vary. Sample: Someone deposits \$200 into a bank account that
Date added: July 3, 2014 - Views: 2
Pearson Georgia High School Mathematics Advanced Algebra
Advanced Algebra ©2014 to the Gwinnett County ... to practice problem-solving strategies: ACTIVITY LAB 1-4a: ... Logarithms for Exponential Models, 8-1 Exploring Periodic Data, TECHNOLOGY LAB 8-4: Graphing Trigonometric Functions, TECHNOLOGY
Date added: December 1, 2013 - Views: 7
Algebra 1 Course Outline
1-1 Properties of Real Numbers . 1-2 Algebraic Expressions . ... 2-4 Using Linear Models . 2-5 Absolute Value Functions and Graphs . Technology: Use TI-83 to find ... (Use 3-6 for additional practice with matrix equations, if desired)
http://www.lcsedu.net/sites/default/files/pdfs/curriculum/timelines/algebra-ii-seqence-08-09.pdf
Date added: September 8, 2014 - Views: 1
Mathematical Models With Applications Texas Edition Answers
Thinking With Mathematical Models Answers Investigation 1 Additional Practice. 1. a. 16; 19 b. t. 3n. 1 . ... Mathematics Thinking With Mathematical Models Grade 8 INVESTIGATION 1: EXPLORING DATA PATTERNS. ... Math 1314 Exponential Functions as Mathematical Models.
Date added: September 17, 2014 - Views: 1
... Thinking With Mathematical Models Investigation 1: Exploring ... Growing Investigation 1: Exponential Growth (15), Investigation 2: Examining Growth Patterns (29); Frogs, Fleas, and Painted Cubes Investigation 2: Quadratic Expressions (35); Say It With Symbols Investigation 2: Combining ...
Date added: February 12, 2012 - Views: 27
laschools.net
Practice Exploring Exponential Models Graph each function. Date Form G ... Exploring Exponential Models Date Form G For each annual rate of change, ... Answers 000000 000000 000000 000000 000000 000000 000000 000000 000000
Date added: April 8, 2014 - Views: 2
Algebra 2B: Full Course Summary - National Connections Academy
• This course is a core course at the “Basic” level in Connections Academy’s system, which titles courses as Basic, ... Exploring Exponential Models: 1 . 2. Exploring Exponential Models: 2 ... solve rational equations and check the solutions for extraneous answers. Finally, ...
Date added: February 14, 2012 - Views: 7
HRW Algebra One Interactions, Course 2 - ThinkCentral
HRW Algebra One Interactions, Course 2 Table of Contents Chapter 1 Functions, Equations, ... 6.5 Exponential Functions ... 7.1 Exploring Polynomial Functions 7.2 Adding and Subtracting Polynomials 7.3 Exploring Multiplication Models 7.4 Exploring Multiplication of Binomials 7.5 Common Factors
http://go.hrw.com/resources/go_mk/ma/maaoi/AOIC2TOC.PDF
Date added: January 29, 2012 - Views: 18
Pearson Georgia High School Mathematics Analytic Geometry
Integrated Geometry 1 Gwinnett County Academic Knowledge and Skills (AKS) ... opportunities to practice problem-solving strategies: ACTIVITY LAB 3-5: ... 7-1 Exploring Exponential Models, TECHNOLOGY LAB 7-5: Using Logarithms for Exponential Models,
http://www.pearsonschool.com/correlations/Gwinnett_Int_GEO_GA_HS_Analytic_Geometry_2014_final.pdf
Date added: September 28, 2013 - Views: 4
Exploring the Exponential Function
Exploring the Exponential Function When the value of a variable increases or decreases by a constant ... these activities, you will be able to identify exponential growth or decay from equations. In these activities, you will practice developing "eyeball" models for situations involving ...
http://education.ti.com/~/media/72689ABA2BA9442888068BF8DC9F2B19
Date added: September 17, 2014 - Views: 1
Exploring Exponential Growth and Decay Functions
Title: Exponential Growth And Decay Brief Overview: ... when a > 0 and b > 1 when a > 0 and 0 < b < 1 This is called an exponential regression model. You will need to plot data points using the STAT menu. 1) Plot Data Points as described on previous worksheet.
Date added: November 5, 2011 - Views: 139
Answer Key © Glencoe/McGraw ... Sample answers are given. 1. Kelton 2. 3 out of 4 3. Steve 4. 2.5 5. Jack 6. \$6.75 7. Monique 8. 2 out of 5 9. Kelton 10. Kelton ... models the data because it organizes the data so you can easily see the range of ages
Date added: August 10, 2014 - Views: 7
Mathematical Models With Applications Texas Edition
Thinking With Mathematical Models Answers. Investigation 1 Additional Practice. 1. a. 16; 19 b. t. 3n. 1 . 19. 9. Thinking With Mathematical Models Answers. Figure number. 1 2. 3. 4. 5. ... ACE ANSWERS. 1. Investigation 1 Exploring Data Patterns. 31.
http://www.ursbook.com/pdf/mathematical-models-with-applications-texas-edition.pdf
Date added: September 8, 2014 - Views: 2
Unit 3 Exploring Exponentia (Exponents and Logarithms)
reviewed and students practice working with rational exponents. ... The unit emphasizes the reasonableness of results and models. ... Write each exponential as a logarithm or logarithm as an exponential. 1. log 10 (1/100) = -2 2. 53 = 125 3. log 2
https://www.georgiastandards.org/Frameworks/GSO%20Frameworks/Math%20III%20Unit%203%20SE.pdf
Date added: October 23, 2011 - Views: 16
To download free 2.1 quadratic functions and models academics you need to ... Graph exponential functions. 1, 2, 3, 6, 8, 9. (488493). Investigation. Explore geometric . Standardized Test Practice, . Answer Key Transparencies. This PDF book contain 11 2 practice exponential functions answer key ...
Date added: June 12, 2014 - Views: 1
Tesccc Hs Mathematical Models
ACE ANSWERS. 1. Investigation 1 Exploring Data Patterns. 31. ... Models Answers. Thinking With Mathematical Models Answers. Investigation 1 Additional Practice. 1. a. 16; 19 b. t. 3n. 1 . 19. 9.
http://www.isohd.com/pdf/tesccc-hs-mathematical-models.pdf
Date added: September 17, 2014 - Views: 1
Grade 8 Math Curriculum Alignment with State Standards
... Growing, Growing Investigation 1: Exponential Growth (7–10, 17–18) TG: Growing, Growing, Growing Investigation 1: ... Thinking With Mathematical Models Investigation 1: Exploring Data Patterns (19), Investigation 2: ... 8.N.3.9 Estimate answers and use
Date added: May 19, 2013 - Views: 8
Deliver instruction (Block 4) - Piedra Vista High School
Describe the scenario from page 1 of the Exploring "Geometric sequences and series." ... What is the constant multiplier of the exponential function that models this situation? What is its domain? ... Have them practice writing the sum in sigma notation. [SAS 3, ...
http://pvhs.fms.k12.nm.us/teachers/kevers/Agile%20mind/Unit%201%20Block%204.pdf
Date added: September 16, 2013 - Views: 3
Mathematics Pacing Guide High School Algebra
Mathematics Pacing Guide: High School Algebra Course Math_Pacing Guide Algebra_07.27.11_v1 Page | 1 This Pacing Guide is to be used in conjunction with the Curriculum Guide and the Instructional Design Lesson Plan.
http://www4.uwm.edu/Org/mmp/PDFs/Yr9_PDFs/MathPacingGuide_HS_Algebra.pdf
Date added: February 29, 2012 - Views: 30
Unit 5 Exploring Exponentia (Exponents and Logarithms
and graphs of logarithms. Exponential and logarithmic functions behave the same as other functions with respect ... Understand and use basic exponential functions as models of real phenomena. ... validity of their answers.
Date added: December 1, 2013 - Views: 4
View Park Prep High School
2.8 Exploring Data: Quadratic Models ... 3.1 Exponential Functions and Their Graphs [A2 11.0, A2 12.0] ... practice the attention to detail and depth that we require on tests. Finally, you’ll be interacting with problems more deeply.
http://p2cdn3static.sharpschool.com/UserFiles/Servers/Server_29660/File/Pre-Calculus%20Syllabus%20A%20_Slayton_.pdf
Date added: September 17, 2014 - Views: 1
Unit 3 Exploring Exponentia (Exponents and Logarithms)
reviewed and students practice working with rational exponents. ... y = 1, not an exponential function. ii. ... could be written as t = r 1/.53 = r 1.89. 15/8 = 1.875, so the exponent values are quite close. 6.
http://schools.liberty.k12.ga.us/subsites/curriculum/securedocs/Teacher%20Tools/Curriculum%20Maps%20and%20Units/Mathematics/High%20School%20Math/Math%20III/math%20III%20units/math%20III%20unit%203%20exponents%20and%20logarithms.pdf
Date added: March 5, 2014 - Views: 1
discussion: have groups present solutions. Practice. 1. This PDF book provide practice a exploring ... and solving exponential equations. 1 transformations previously seen to the exponential parent functions. Horizontal€. ... 2.1 QUADRATIC FUNCTIONS AND MODELS Academics.
Date added: June 13, 2014 - Views: 1
5.5 Making Connections: Exponential Models ... Practice Exam ... Answers ..... Overview Calculus and vectors play an important role in many activities, from business and economics to the social, medical ...
http://highered.mheducation.com/sites/dl/free/0070735824/582661/mhr_calculus12_studyguide_toc.pdf
Date added: July 10, 2014 - Views: 2
STANDARDS FOR M PRACTICE - Michael Serra
the Koch curve as the basis for exploring exponential growth. ... Take Another Look 1 Discovering Advanced Algebra Lesson: Lesson 7.5: ... Lesson 6.8: Decreasing Exponential Models and Half-Life Discovering Advanced Algebra Lessons:
http://www.michaelserra.net/files/commoncore_correlation_dm.pdf
Date added: January 28, 2012 - Views: 61
Common Core State Standards for Mathematics
Common Core State StandardS for matHematICS table of Contents Introduction 3 ... 3/8 = 1/8 + 2/8 ; ... and evaluate probability models. mathematical Practices 1. Make sense of problems and persevere in solving them.
http://www.k12.wa.us/CoreStandards/Mathematics/pubdocs/CCSSI_MathStandards.pdf
Date added: November 6, 2012 - Views: 58
Integrated Math Course 1 - Escondido Union High School District
Students who complete Integrated Math 1 & 2 satisfy the California Educational Code ... Linear, Quadratic, and Exponential Models Construct and compare linear, quadratic, and exponential ... 8 CCSS.Math.Practice.MP8 Look for and express regularity in repeated reasoning
http://www.euhsd.k12.ca.us/cms/lib07/CA01001539/Centricity/Domain/48/Integrated%20Math%201%20_DLA.pdf
Date added: September 28, 2013 - Views: 6
Date added: September 17, 2014 - Views: 1
CCGPS Coordinate Algebra - Bibb County Public School District
CCGPS Coordinate Algebra Unpacked Standards Page 1 of 32 Bibb County School System May 2012 CCGPS ... models is a Standard for Mathematical Practice, ... Linear, Quadratic, and Exponential Models (F-LE) Geometry Congruence (G-CO)
http://schools.bibb.k12.ga.us/cms/lib01/GA01000598/Centricity/Domain/94/CCGPS%20Coordinate%20Algebra%20Unpacked%20Standards.pdf
Date added: December 25, 2013 - Views: 2
Syllabus
Section 3.1 Exponential Functions and Their Graphs ... (work) must accompany the answers for full credit. ... Section 2.8 Exploring Data: Quadratic Models Classify scatter plots Find a quadratic model 165: 1-10,[16] 18 Review Quiz 19
http://www.cbennett.nuames.org/Pre_Calculus/Syllabus_PreCalculus.pdf
Date added: May 6, 2013 - Views: 3
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This content is taken from the Davidson Institute of Science Education at the Weizmann Institute of Science's online course, An Introduction to Recreational Math: Fun, Games, and Puzzles. Join the course to learn more.
3.3
How to win at Nim
Nim is a really fascinating game. Now for some of the math behind it…
The solution to the 1,2,3 one pile Nim game
Here is the answer to the 1,2,3 ten stone, one pile Nim game.
What we have to realise is that the player who is left with a pile of 3,2 or 1 stones is going to be the winner, but the player who is left with 4 stones in the pile is definitely a loser.
So, the winning strategy of the first player is to try and leave the second player with a four stone pile.
If we had 5,6 or 7 stones in the original pile, the first player could easily do this, by taking 1,2 or 3 stones to leave the second player with 4 stones and a losing position.
Unfortunately, the initial pile is bigger and has ten stones.
The question now becomes how can the first player force the second player to leave him with 5,6 or 7 stones - positions from which he (the first player) can win.
Luckily, the first player can do this by taking two stones from the initial ten stone pile, leaving 8 stones in the pile for the second player. However many stones the second player now takes, 1,2 or 3, he leaves 5, 6 or 7 stones in the pile from which the first player can take the necessary number of stones to leave the second player with the losing four stone position.
In fact, you may have realised by now, that any multiple of 4 is a losing position for the second player. This means that if you play Nim against someone using the 1,2 or 3 rule, make sure you don’t have a multiple of four number of stones in your initial pile…
A friend of mine, the stand up mathematician, Matt Parker, has a beautiful explanation of the 1,2,3 Nim game. You can see it in the following YouTube video.
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1. Strange angle differentiation
Ok first time being here but I could use some help. The problem is this.
-There are 2 blood vessels seperated by angle X
-These two blood vessels have radii of R and r respectively
-The length "a" is the length of the primary blood vessel while the height "b" is the length from the end of the primary blood vessel to the branching vessel.
-The resistance to blood flow is as follows:
T= ((a-bcot(x))/R^4)-((bcsc(x))/r^4
-R,r,a, and b are all constants.
-Which angle "x" would provide the least resistance of blood?
I worked out the derivative to look like
T'=((bcsc(x)^2/R^4)+((bcsc(x)cot(x)/r^4)
But I have no idea how to get a minimum of resistance since the only angles to make this 0 or undefined are angles such as 0 90 or 180 which are not the answers. Have I done something wrong in my derivative? Is there a specific answer?
2. Originally Posted by Blackstar347
Ok first time being here but I could use some help. The problem is this.
-There are 2 blood vessels seperated by angle X
-These two blood vessels have radii of R and r respectively
-The length "a" is the length of the primary blood vessel while the height "b" is the length from the end of the primary blood vessel to the branching vessel.
-The resistance to blood flow is as follows:
T= ((a-bcot(x))/R^4)+((bcsc(x))/r^4
-R,r,a, and b are all constants.
-Which angle "x" would provide the least resistance of blood?
I worked out the derivative to look like
T'=((bcsc(x)^2/R^4)-((bcsc(x)cot(x)/r^4)
But I have no idea how to get a minimum of resistance since the only angles to make this 0 or undefined are angles such as 0 90 or 180 which are not the answers. Have I done something wrong in my derivative? Is there a specific answer?
First of all, I'm pretty sure there's meant to be a plus between the two terms in T since the first and second terms represent resistance to blood flow in the primary and branching blood vessel respectively. I've made the necessary corrections in red.
The derivative can be re-written as $\frac{dT}{dx} = \frac{b}{R^4 \sin^2 x} - \frac{b \cos x}{r^4 \sin^2 x} = \frac{b(r^4 - R^4 \cos x)}{R^4 r^4 \sin^2 x}$.
$\frac{dT}{dx} = 0 \Rightarrow r^4 - R^4 \cos x = 0 \Rightarrow \cos x = \left(\frac{r}{R}\right)^4$.
There will be a solution since r < R (why?).
Now you must test the nature of the solution. Given the model, you're only interested in solutions lying in the domain $\left(\alpha, \, \frac{\pi}{2} \right)$ where $\alpha > 0$ will depend on the value of a and b (I'll let you work out the relationship).
3. Well the branching blood vessel has a smaller radius than the main blood vessel which is why r<R...hmm I feel like I'm missing a value somewhere. But thanks for the tips the derivative there seems alot easier to understand.
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Pre-Calculus Homework Solutions 195
# Pre-Calculus Homework Solutions 195 - 26 Since the measure...
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26. Since the measure of the longest side is } Ï 3 7 5 4 w } , } Ï 3 7 5 4 w } must be c , and a or b are } 1 5 } and } 1 7 } . a 2 1 b 2 5 ¬ c 2 1 } 1 5 } 2 2 1 1 } 1 7 } 2 2 0 ¬ 1 } Ï 3 7 5 4 w } 2 2 } 2 1 5 } 1 } 4 1 9 } 0 ¬ } 1 7 2 4 25 } } 1 7 2 4 25 } 5 ¬ } 1 7 2 4 25 } These segments form the sides of a right triangle since they satisfy the Pythagorean Theorem. However, the three numbers are not whole numbers. Therefore, they do not form a Pythagorean triple. 27. Since the measure of the longest side is } 3 3 5 6 } , } 3 3 5 6 } must be c , and a or b are } Ï 2 3 w } and } Ï 3 2 w } . a 2 1 b 2 5 ¬ c 2 1 } Ï 2 3 w } 2 2 1 1 } Ï 3 2 w } 2 2 0 ¬ 1 } 3 3 5 6 } 2 2 } 3 4 } 1 } 2 9 } 0 ¬ } 1 1 2 2 2 9 5 6 } } 3 3 5 6 } Þ ¬ } 1 1 2 2 2 9 5 6 } Since } 3 3 5 6 } Þ } 1 1 2 2 2 9 5 6 } , segments with these measures cannot form a right triangle. Therefore, they do not form a Pythagorean triple. 28.
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A291638 Numbers n such that psi(n) and sigma(n) both are perfect squares and n is not a squarefree number. 0
370440, 1704024, 3926664, 11039112, 13854456, 21707784, 25264008, 28375704, 40822488, 44378712, 57862728, 59196312, 63937944, 75051144, 79051896, 79940952, 103500936, 107946216, 128394504, 134766072, 162178632, 169735608, 177737112, 191517480, 193530168, 195221880, 196407288, 215077464 (list; graph; refs; listen; history; text; internal format)
OFFSET 1,1 LINKS Table of n, a(n) for n=1..28. EXAMPLE 370440 = 2^3*3^3*5*7^3 is a term because psi(370440) = 2^2*3^2*7^2*(1+2)*(1+3)*(1+5)*(1+7) = 2^8*3^4*7^2 and sigma(370440) = (1+2+2^2+2^3)*(1+3+3^2+3^3)*(1+5)*(1+7+7^2+7^3) = 2^8*3^2*5^4. PROG (PARI) a001615(n) = my(f=factor(n)); prod(i=1, #f~, f[i, 1]^f[i, 2] + f[i, 1]^(f[i, 2]-1)); isok(n) = !issquarefree(n) && issquare(a001615(n)) && issquare(sigma(n)) \\ after Charles R Greathouse IV at A001615 CROSSREFS Cf. A000203, A001615, A006532, A291167. Sequence in context: A172666 A251209 A252800 * A219284 A017537 A234792 Adjacent sequences: A291635 A291636 A291637 * A291639 A291640 A291641 KEYWORD nonn AUTHOR Altug Alkan, Aug 28 2017 STATUS approved
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# Cochran's Q test - overview
This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
Cochran's Q test
Pearson correlation
Spearman's rho
Independent/grouping variableVariable 1Variable 1
One within subject factor ($\geq 2$ related groups)One quantitative of interval or ratio levelOne of ordinal level
Dependent variableVariable 2Variable 2
One categorical with 2 independent groupsOne quantitative of interval or ratio levelOne of ordinal level
Null hypothesisNull hypothesisNull hypothesis
H0: $\pi_1 = \pi_2 = \ldots = \pi_I$
Here $\pi_1$ is the population proportion of 'successes' for group 1, $\pi_2$ is the population proportion of 'successes' for group 2, and $\pi_I$ is the population proportion of 'successes' for group $I.$
H0: $\rho = \rho_0$
Here $\rho$ is the Pearson correlation in the population, and $\rho_0$ is the Pearson correlation in the population according to the null hypothesis (usually 0). The Pearson correlation is a measure for the strength and direction of the linear relationship between two variables of at least interval measurement level.
H0: $\rho_s = 0$
Here $\rho_s$ is the Spearman correlation in the population. The Spearman correlation is a measure for the strength and direction of the monotonic relationship between two variables of at least ordinal measurement level.
In words, the null hypothesis would be:
H0: there is no monotonic relationship between the two variables in the population.
Alternative hypothesisAlternative hypothesisAlternative hypothesis
H1: not all population proportions are equalH1 two sided: $\rho \neq \rho_0$
H1 right sided: $\rho > \rho_0$
H1 left sided: $\rho < \rho_0$
H1 two sided: $\rho_s \neq 0$
H1 right sided: $\rho_s > 0$
H1 left sided: $\rho_s < 0$
AssumptionsAssumptions of test for correlationAssumptions
• Sample of 'blocks' (usually the subjects) is a simple random sample from the population. That is, blocks are independent of one another
• In the population, the two variables are jointly normally distributed (this covers the normality, homoscedasticity, and linearity assumptions)
• Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
Note: these assumptions are only important for the significance test and confidence interval, not for the correlation coefficient itself. The correlation coefficient just measures the strength of the linear relationship between two variables.
• Sample of pairs is a simple random sample from the population of pairs. That is, pairs are independent of one another
Note: this assumption is only important for the significance test, not for the correlation coefficient itself. The correlation coefficient itself just measures the strength of the monotonic relationship between two variables.
Test statisticTest statisticTest statistic
If a failure is scored as 0 and a success is scored as 1:
$Q = k(k - 1) \dfrac{\sum_{groups} \Big (\mbox{group total} - \frac{\mbox{grand total}}{k} \Big)^2}{\sum_{blocks} \mbox{block total} \times (k - \mbox{block total})}$
Here $k$ is the number of related groups (usually the number of repeated measurements), a group total is the sum of the scores in a group, a block total is the sum of the scores in a block (usually a subject), and the grand total is the sum of all the scores.
Before computing $Q$, first exclude blocks with equal scores in all $k$ groups.
Test statistic for testing H0: $\rho = 0$:
• $t = \dfrac{r \times \sqrt{N - 2}}{\sqrt{1 - r^2}}$
where $r$ is the sample correlation $r = \frac{1}{N - 1} \sum_{j}\Big(\frac{x_{j} - \bar{x}}{s_x} \Big) \Big(\frac{y_{j} - \bar{y}}{s_y} \Big)$ and $N$ is the sample size
Test statistic for testing values for $\rho$ other than $\rho = 0$:
• $z = \dfrac{r_{Fisher} - \rho_{0_{Fisher}}}{\sqrt{\dfrac{1}{N - 3}}}$
• $r_{Fisher} = \dfrac{1}{2} \times \log\Bigg(\dfrac{1 + r}{1 - r} \Bigg )$, where $r$ is the sample correlation
• $\rho_{0_{Fisher}} = \dfrac{1}{2} \times \log\Bigg( \dfrac{1 + \rho_0}{1 - \rho_0} \Bigg )$, where $\rho_0$ is the population correlation according to H0
$t = \dfrac{r_s \times \sqrt{N - 2}}{\sqrt{1 - r_s^2}}$
Here $r_s$ is the sample Spearman correlation and $N$ is the sample size. The sample Spearman correlation $r_s$ is equal to the Pearson correlation applied to the rank scores.
Sampling distribution of $Q$ if H0 were trueSampling distribution of $t$ and of $z$ if H0 were trueSampling distribution of $t$ if H0 were true
If the number of blocks (usually the number of subjects) is large, approximately the chi-squared distribution with $k - 1$ degrees of freedomSampling distribution of $t$:
• $t$ distribution with $N - 2$ degrees of freedom
Sampling distribution of $z$:
• Approximately the standard normal distribution
Approximately the $t$ distribution with $N - 2$ degrees of freedom
Significant?Significant?Significant?
If the number of blocks is large, the table with critical $X^2$ values can be used. If we denote $X^2 = Q$:
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
$t$ Test two sided:
$t$ Test right sided:
$t$ Test left sided:
$z$ Test two sided:
$z$ Test right sided:
$z$ Test left sided:
Two sided:
Right sided:
Left sided:
n.a.Approximate $C$% confidence interval for $\rho$n.a.
-First compute the approximate $C$% confidence interval for $\rho_{Fisher}$:
• $lower_{Fisher} = r_{Fisher} - z^* \times \sqrt{\dfrac{1}{N - 3}}$
• $upper_{Fisher} = r_{Fisher} + z^* \times \sqrt{\dfrac{1}{N - 3}}$
where $r_{Fisher} = \frac{1}{2} \times \log\Bigg(\dfrac{1 + r}{1 - r} \Bigg )$ and the critical value $z^*$ is the value under the normal curve with the area $C / 100$ between $-z^*$ and $z^*$ (e.g. $z^*$ = 1.96 for a 95% confidence interval).
Then transform back to get the approximate $C$% confidence interval for $\rho$:
• lower bound = $\dfrac{e^{2 \times lower_{Fisher}} - 1}{e^{2 \times lower_{Fisher}} + 1}$
• upper bound = $\dfrac{e^{2 \times upper_{Fisher}} - 1}{e^{2 \times upper_{Fisher}} + 1}$
-
n.a.Properties of the Pearson correlation coefficientn.a.
-
• The Pearson correlation coefficient is a measure for the linear relationship between two quantitative variables.
• The Pearson correlation coefficient squared reflects the proportion of variance explained in one variable by the other variable.
• The Pearson correlation coefficient can take on values between -1 (perfect negative relationship) and 1 (perfect positive relationship). A value of 0 means no linear relationship.
• The absolute size of the Pearson correlation coefficient is not affected by any linear transformation of the variables. However, the sign of the Pearson correlation will flip when the scores on one of the two variables are multiplied by a negative number (reversing the direction of measurement of that variable).
For example:
• the correlation between $x$ and $y$ is equivalent to the correlation between $3x + 5$ and $2y - 6$.
• the absolute value of the correlation between $x$ and $y$ is equivalent to the absolute value of the correlation between $-3x + 5$ and $2y - 6$. However, the signs of the two correlation coefficients will be in opposite directions, due to the multiplication of $x$ by $-3$.
• The Pearson correlation coefficient does not say anything about causality.
• The Pearson correlation coefficient is sensitive to outliers.
-
Equivalent toEquivalent ton.a.
Friedman test, with a categorical dependent variable consisting of two independent groups.OLS regression with one independent variable:
• $b_1 = r \times \frac{s_y}{s_x}$
• Results significance test ($t$ and $p$ value) testing $H_0$: $\beta_1 = 0$ are equivalent to results significance test testing $H_0$: $\rho = 0$
-
Example contextExample contextExample context
Subjects perform three different tasks, which they can either perform correctly or incorrectly. Is there a difference in task performance between the three different tasks?Is there a linear relationship between physical health and mental health?Is there a monotonic relationship between physical health and mental health?
SPSSSPSSSPSS
Analyze > Nonparametric Tests > Legacy Dialogs > K Related Samples...
• Put the $k$ variables containing the scores for the $k$ related groups in the white box below Test Variables
• Under Test Type, select Cochran's Q test
Analyze > Correlate > Bivariate...
• Put your two variables in the box below Variables
Analyze > Correlate > Bivariate...
• Put your two variables in the box below Variables
• Under Correlation Coefficients, select Spearman
JamoviJamoviJamovi
Jamovi does not have a specific option for the Cochran's Q test. However, you can do the Friedman test instead. The $p$ value resulting from this Friedman test is equivalent to the $p$ value that would have resulted from the Cochran's Q test. Go to:
ANOVA > Repeated Measures ANOVA - Friedman
• Put the $k$ variables containing the scores for the $k$ related groups in the box below Measures
Regression > Correlation Matrix
• Put your two variables in the white box at the right
• Under Correlation Coefficients, select Pearson (selected by default)
• Under Hypothesis, select your alternative hypothesis
Regression > Correlation Matrix
• Put your two variables in the white box at the right
• Under Correlation Coefficients, select Spearman
• Under Hypothesis, select your alternative hypothesis
Practice questionsPractice questionsPractice questions
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Mensuration Practice Set 3
5 Steps - 3 Clicks
Mensuration Practice Set 3
Introduction
Mensuration Practice set 3 is a topic in Geometry which is a branch of mathematics. Mensuration deals with length, area, and volume of different kinds of shape- both 2D and 3D. The article Mensuration Practice Quiz provides information about Mensuration, an important topic of Mathematics Consists of different types of Mensuration questions with solutions useful for candidates preparing for different competitive examinations like RRB, RRB ALP/Technical Exams/Junior Engineer Recruitment Exams, SSC CGL, SSC CHSL, IBPS, SBI PO, SBI Clerks, CAT and etc.
Quiz
Q1. In a rectangle the ratio of the length and breadth is 3:2. If each of the length and breadth is increased by 4m their ratio becomes 10:7. The area of the original rectangle in m² is?
A. 846
B. 864
C. 840
D. 876
Explanation:
$$[3x + \frac {4}{2x + 4}] = \frac {10}{7}$$
$$x = 12$$
Area of the original rectangle = $$3x \times 2x = 6x²$$
Area of the original rectangle = $$6 \times 144 = 864 m²$$
Q2. The perimeter of a rectangle and a square is 160 cm each. If the difference between their areas is 500 cm. If the area of the rectangle is less than that of a Square then find the area of the rectangle?
A. 1800 cm²
B. 1500 cm²
C. 1100 cm²
D. 1600 cm²
Explanation:
Perimeter of rectangle = Perimeter of Square = 160
4a = $$160 \Rightarrow a$$ = 40
Area of square = 1600
1600 – lb = 500
lb = 1100 cm²
Q3. Circumference of a circle A is $$\frac {22}{7}$$ times perimeter of a square. Area of the square is 441 cm². What is the area of another circle B whose diameter is half the radius of the circle A(in cm²)?
A. 354.5
B. 346.5
C. 316.5
D. 312.5
Explanation:
Area = 441 cm²
a = 21 cm
Perimeter of Square = $$4 \times 21$$
Circumference of a Circle = $$4 \times 21 \times \frac {22}{7}$$
2 $$\pi r = 4 \times 3 \times 22$$
r = $$\frac {12 \times 22 \times 7}{2} \times 22$$ = 42 cm
Radius of Circle B = $$\frac {42}{4}$$ = 10.5 cm
Area of Circle = $$\pi r² = \frac {22}{7} \times 10.5 \times 10.5$$ = 346.5 cm²
Q4. The area of a rectangle is equal to the area of a square whose diagonal is $$12 \sqrt {2}$$ meter. The difference between the length and the breadth of the rectangle is 7 meter. What is the perimeter of a rectangle ? (in meter).
A. 68 meter
B. 50 meter
C. 62 metre
D. 64 metre
Explanation:
d = $$a \sqrt {2}$$
$$12 \sqrt {2} = a \sqrt {2}$$
a = 12
$$l \times b$$ = a² = (12²) = 144
l – b = 7
l = b + 7
$$(b + 7) \times (b)$$= 144
b² + 7b – 144 = 0
b = 9
l = 16
2(l + b) = 2(16 + 9) = 50 m
Q5. The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and breadth by 2 units, then the area is increased by 67 square units. Find the area of the rectangle?
A. 159 m
B. 179 m
C. 147 m
D. 153 m
Explanation:
Length = $$x$$
Breadth = $$y$$
$$xy – (x-5)(y+3)$$ = 9
$$3x – 5y – 6$$= 0 …………….(i)
$$(x+3)(y+2) – xy$$ = 67
$$2x + 3y -61$$ = 0 …………….(ii)
solving (i) and (ii)
$$x = 17 m$$
$$y = 9 m$$
Area of the Rectangle = 153 m
Q1. The length of a plot is four times its breath. A playground measuring 900 square meters occupies one fourth of the total area of a plot. What is the length of the plot in meter.?
A. 150
B. 130
C. 160
D. 120
Explanation:
Area of the plot = $$(4 \times 900) m²$$
= $$3600 m²$$
Breadth = $$y$$ meter
Length = $$4y$$ meter
Now area = $$4y \times y = 3600 m²$$
$$\Rightarrow y² = 900 m²$$
$$\Rightarrow y = 30 m$$
∴ Length of plot = $$4y = 120 m$$
Q2. The sum of the radius and height of a cylinder is 19 m. The total surface area of the cylinder is 1672 m², what is the radius and height of the cylinder? (in m)
A. 12, 7
B. 11, 8
C. 13, 6
D. 14, 5
Explanation:
r + h = 19 m
2 $$\pi r(r + h)$$ = 1672
r = $$1672 \times \frac {7} {2} \times 22 \times 19$$ = 14
r = 14
h = 5
Q3. If each side pair of opposite sides of a square is increased by 20 m, the ratio of the length and breadth of the rectangular so formed becomes 5:3. The area of the old square is?
A. 990 m²
B. 900 m²
C. 930 m²
D. 945 m²
Explanation:
$$\frac {(x+20)} {x} = \frac {5} {3}$$
$$3x + 60 = 5x$$
$$x = 30 m$$
Area = $$900 m²$$
Q4. The length of a park is four times its breadth. A playground whose area is 1024 m² covers 1/4th part of the park. The length of the park is?
A. 118 m
B. 142 m
C. 124 m
D. 128 m
Explanation:
l = 4b
Area of the park = $$4 \times 1024 m²$$
$$l \times b = 4 \times 1024$$
$$l times \frac {l}{4} = 4 \times 4 \times 1024$$
$$l² = 1024 \times 16$$
$$l = 32 \times 4 = 128 m$$
Q5. The width of a rectangular piece of land is $$\frac {1}{{4}{th}}$$ of its length. If the perimeter of the piece of land is 320m its length is?
A. 140 m
B. 128 m
C. 120 m
D. 156 m
Explanation:
length = $$l$$
breadth = $$\frac {l}{4}$$
$$2(l + b)$$ = 320
$$2(l + \frac {l}{4})$$ = 320
$$l = 320 \times \frac {4}{10}$$ = 128m
Q1. A deer and a rabbit can complete a full round on a circular track in 9 minutes and 5 minutes respectively. P, Q, R, and S are the four consecutive points on the circular track which are equidistant from each other. P is opposite to R and Q is opposite to S. After how many minutes will they meet together for the first time at the starting point when both have started simultaneously from the same point in the same direction?
A. 15 minutes
B. 25 minutes
C. 35 minutes
D. 45 minutes
Explanation:
Time is taken by a deer to complete one round = 9 minutes
Time is taken by a rabbit to complete one round = 5 minutes
They meet together for the first time at the starting point = LCM of 9 and 5 = 45 minutes.
Q2. A deer and a rabbit can complete a full round on a circular track in 9 minutes and 5 minutes respectively. P, Q, R, and S are the four consecutive points on the circular track which are equidistant from each other. P is opposite to R and Q is opposite to S. After how many minutes will they meet together for the first time when both have started simultaneously from the same point in the same direction(in min)?
A. $$\frac {15}{4}$$
B. $$\frac {45}{4}$$
C. $$\frac {35}{4}$$
D. $$\frac {25}{4}$$
Explanation:
Circumference of the track = LCM of 9 and 5 = 45 m.
The ratio of time of deer and rabbit = 9 : 5
The ratio of the speed of deer and rabbit = 5 : 9
Relative Speed = 4 m/min
They meet together for the first time at the starting point = $$\frac {45}{4}$$min
Q3. A cylindrical cistern whose diameter is 14 cm is partly filled with water. If a rectangular block of iron 22 cm in length, 14 cm in breadth and 7 cm in thickness is wholly immersed in water, by how many centimeters will the water level rise?
A. 10 cm
B. 14 cm
C. 12 cm
D. 15 cm
Explanation:
Volume of the block = $$22 \times 14 \times 7$$
Radius of the cistern = $$\frac {14}{2} = 7$$
Volume of the Cylinder = $$\frac {22}{7} \times R2 \times h$$
$$\frac {22}{7} \times R2 \times h = \frac {22}{7} \times 7 \times 7 \times h$$
$$\frac {22}{7} \times 7 \times 7 \times h = 22 \times 14 \times 7 \Rightarrow h$$= 14
Q4. A well with 28 m inside diameter is dug out 18 m deep. The earth taken out of it has been evenly spread all around it to a width of 21 m to form an embankment. Find the height of the embankment.
A. 3 m
B. 8 m
C. 9 m
D. 6 m
Explanation:
$$\frac {22}{7[(R2) – (r2)]} \times h = \frac {22}{7(7 \times 7 \times 18)}$$
[(352) – (72)] h = $$14 \times 14 \times 18$$
$$(42 \times 28) h = 14 \times 14 \times 18$$
h = 3 m
Q5. The radii of two cylinders are in the ratio 4:5 and their heights are in the ratio 5 : 7, What is the ratio of their curved surface areas?
A. 2 : 5
B. 4 : 7
C. 7 : 4
D. 2 : 3
Explanation:
$$\frac {2 \pi r1 h1}{2 \pi r2 h2} = [\frac {4}{5} \times {5}{7}]$$ = 4 : 7
Exams
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Enter the Young’s Modulus, temperature expansion coefficient, change in temperature, and the cross-sectional area into the calculator to determine the thermal expansion force on a restricted pipe.
## Thermal Expansion Force Formula
The following equation is used to calculate the Thermal Expansion Force.
F = E * a * T * A
• Where F is the force (lbf)
• E is the Young’s modulus of the material (PSI)
• a is the thermal expansion coefficient (in/in * F)
• T is the change in temperature (F)
• A is the cross-sectional area (in^2)
## What is a Thermal Expansion Force?
Definition:
A thermal expansion force measures the total force acting on an object in a constricted space due to a temperature change, causing an expansion of the object.
## How to Calculate Thermal Expansion Force?
Example Problem:
The following example outlines the steps and information needed to calculate the Thermal Expansion Force.
First, determine the Young’s modulus of the material . In this example, this is found to be 6000 PSI.
Next, determine the thermal expansion coefficient. For this problem, this is found to be 4 in/in*F.
Next, determine the change in temperature. In this case, the change in temperature is measured to be 8 F.
Next, determine the cross-sectional area. This cross-sectional area is 11 in^2.
Finally, calculate the thermal expansion force using the formula above:
F = E * a * T * A
F = 6000*4*8*11
F = 2,112,000 lbf
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# The Arithmetic Derivative
The derivative of a function is a cornerstone of mathematics, engineering, physics, biology, chemistry, and a large number of other sciences as well. Today we're going to be calculating something only tangentially related: the arithmetic derivative.
# Definition
The arithmetic derivative a(n) or n' is defined here (A003415) by a number of properties that are similar to the derivative of a function.
• a(0) = a(1) = 0,
• a(p) = 1, where p is any prime, and
• a(mn) = m*a(n) + n*a(m).
The third rule is based on the product rule for differentiation of functions: for functions f(x) and g(x), (fg)' = f'g + fg'. So with numbers, (ab)' = a'b + ab'.
Also of note, since the arithmetic derivative can be extended to the negative numbers via this simple relation, a(-n) = -a(n), the input may be negative.
# Rules
• Write a program or function that, given any integer n, returns the arithmetic derivative of n.
• Inputs will be -230 < n < 230, to avoid problems with integer sizes and numbers too large to factor in a reasonable amount of time. Your algorithm should still be able to theoretically calculate the arithmetic derivative of numbers outside this range.
• Built-ins for symbolic math, prime factorization and differentiation are allowed.
# Examples
> a(1)
0
> a(7)
1
> a(14) # a(7)*2 + a(2)*7 = 1*2 + 1*7 = 9
9
> a(-5) # a(-5) = -a(5) = -1
-1
> a(8) # a(8) = a(2**3) = 3*2**2 = 12
12
> a(225) # a(225) = a(9)*25 + a(25)*9 = 6*25 + 10*9 = 150 + 90 = 240
240
> a(299792458) # a(299792458) = a(2)*149896229 + a(7)*42827494 + a(73)*4106746 + a(293339)*1022 = 1*149896229 + 1*42827494 + 1*4106746 + 1*1022 = 149896229 + 42827494 + 4106746 + 1022 = 196831491
196831491
As always, if the problem is unclear, please let me know. Good luck and good golfing!
• What, exactly, is prime in a(prime)? Is it just a prime number? Commented Apr 17, 2019 at 20:03
• Also, I don't get how you decomposed the last example. Commented Apr 17, 2019 at 20:20
• @Stackstuck Yes, it's any prime. I've edited for clarity. Also, I added to the last example to hopefully make it clearer. Commented Apr 18, 2019 at 6:31
# MATL, 12 bytes
|1>?GtYf/s}0
Try it online!
### Explanation
Consider an integer a with |a|>1, and let the (possibly repeated) prime factors of |a| be f1, ..., fn. Then the desired result is a·(1/f1 + ... + 1/fn).
|1> % take input's absolute value. Is it greater than 1?
? % if so:
Gt % push input twice
Yf % prime factors. For negative input uses its absolute value
/ % divide element-wise
s % sum of the array
} % else:
0 % push 0
• Isn't the sum of the prime factors of 1 equal to 0? Or doesn't that work in MATL? Commented Apr 3, 2016 at 14:18
• @wythagoras Actually 1 gives 1 as its "prime" number decomposition. It's a strange result (an empty array would be more meaningful). But that's how Matlab works. And also CJam. So I guess there must be good reason to output 1 in that case? What do you think? I've been tempted to redefine the Yf function to output an empty array for 1, but I wasn't sure Commented Apr 3, 2016 at 15:23
• Pyth gives an empty array, fwiw. Commented Apr 5, 2016 at 4:34
• @isaacg Thanks! Maybe I'll change that Commented Apr 5, 2016 at 8:35
• Same in Mathematica (was almost a problem once) Commented Apr 8, 2016 at 4:28
## Python, 59 bytes
f=lambda n,p=2:+(n*n>1)and(n%p and f(n,p+1)or p*f(n/p)+n/p)
A recursive function. On large inputs, it runs out of stack depth on typical systems unless you run it with something like Stackless Python.
The recursive definition is implemented directly, counting up to search for candidate prime factors. Since f(prime)=1, if n has a prime p as a factor, we have f(n) == p*f(n/p)+n/p.
• Don't you need input and print? At least when I run this (Python 2), I get no result. Commented Apr 3, 2016 at 18:05
• @wythagoras By default, functions are allowed as an alternative to programs. Also, this challenge says "program or function".
– xnor
Commented Apr 3, 2016 at 18:14
# Jelly, 8 7 bytes
-1 byte by @Dennis
ÆfḟṠ³:S
Uses the same formula everyone else does. However, there's a little trick to deal with 0.
o¬AÆfİS× Main link. Inputs: n
o¬ Logical OR of n with its logical NOT
That is, 0 goes to 1 and everything else goes to itself.
A Then take the absolute value
Æf get its list of prime factors
İ divide 1 by those
S sum
× and multiply by the input.
Try it here.
• Can you please add an explanation? I like answers to have explanations before I upvote them. Commented Apr 3, 2016 at 19:10
• @Sherlock9 Done. Commented Apr 3, 2016 at 21:51
• I see that your answer has been golfed and the explanation is now out-of-date. Would you please fix that? Thanks :D Commented Apr 12, 2016 at 6:01
## Python 2, 877876 74 bytes
a=b=input()
d=2
s=0
while d<=abs(b):
if a%d==0:
a=a/d
s+=b/d
else:
d+=1
print s
Improvements thanks to @Maltysen:
a=b=input()
d=2
s=0
while d<=abs(b):
if a%d==0:a/=d;s+=b/d
else:d+=1
print s
Further improvement by two bytes:
a=b=input()
d=2
s=0
while abs(a)>1:
if a%d<1:a/=d;s+=b/d
else:d+=1
print s
Further improvement thanks to @xnor:
a=b=input()
d=2
s=0
while a*a>1:
if a%d<1:a/=d;s+=b/d
else:d+=1
print s
### Explanation
The arithmetic derivative of a is equal to a times the sum of the reciprocals of the prime factors of a. No exception for 1 is needed since the sum of the reciprocals of the prime factors of 1 is zero.
• abs(a)>1 can be a*a>1.
– xnor
Commented Apr 3, 2016 at 17:22
• @xnor Yes, thank you. Commented Apr 3, 2016 at 17:27
• Replace line 2 with d,s = 2,0 Commented Apr 4, 2016 at 12:42
• @AgnishomChattopadhyay Both are 8 bytes in total. Commented Apr 4, 2016 at 15:44
Thanks @nimi!
I still have no idea when what indentations cause what interpretation, this is the shortest I managed so far, and as always, I'm sure it can be golfed a lot more. I'm going to try again in the evening.
n#(x:_)|y<-div n x=x*a y+y*a x;_#_=1
a n|n<0= -a(-n)|n<2=0|1<2=n#[i|i<-[2..n-1],mod n i<1]
• Thank you very much, teacher=) I can always learn so much whenever you help me out here! Feel free to add your version as your own answer! Commented Apr 3, 2016 at 15:35
# J, 3027 19 chars
Thanks to @Dennis for chopping off 3 characters.
Thanks to @Zgarb for chopping off 8 characters.
0:(*[:+/%@q:@|)@.*
Try it online!
Sample input:
0:(*[:+/%@q:@|)@.* _8
_12
0:(*[:+/%@q:@|)@.* 0
0
0:(*[:+/%@q:@|)@.* 8
12
How it works:
0:(*[:+/%@q:@|)@.* N
XX[email protected] if Z then Y else X end
0: X: return 0
Z Z: signum(N)
N
* *
%@ reciprocal_all( )
q:@ all_prime_factors( )
| abs( )
N
# Pyth - 10 8 bytes
Lovin' the implicit input! Should bring it on par with Jelly for most things (Except Dennis' golfing skills).
*scL1P.a
* Times the input, implicitly (This also adds the sign back in)
s Sum
cL1 Reciprocal mapped over lit
P Prime factorization
.a Absolute value of input, implicitly
n%p|n*n<2=0|mod n p>0=n%(p+1)|r<-div n p=r+p*r%2
(%2)
Implements the recursive definition directly, with an auxiliary variable p that counts up to search for potential prime factors, starting from 2. The last line is the main function, which plugs p=2 to the binary function defined in the first line.
The function checks each case in turn:
• If n*n<2, then n is one of -1,0,1, and the result is 0.
• If n is not a multiple of p, then increment p and continue.
• Otherwise, express n=p*r, and by the "derivative" property, the result is r*a(p)+p*a(r), which simplifies to r+p*a(r) because p is prime.
The last case saves bytes by binding r in a guard, which also avoids the 1>0 for the boilerplate otherwise. If r could be bound earlier, the second condition mod n p>0 could be checked as r*p==n, which is 3 bytes shorter, but I don't see how to do that.
# Japt-x, 1613 10 bytes
ÒU©a k £/X
- 6 bytes thanks to @Shaggy
Try it online!
• Both fail for negative numbers because, for some reason, N.k() doesn't work on them. Commented Feb 28, 2019 at 16:54
• Here's a fix, with some golfing. Commented Feb 28, 2019 at 16:58
• Or -2 more byes with the -x flag. Commented Feb 28, 2019 at 16:59
• Knocked another byte off Commented Feb 28, 2019 at 18:10
• @Shaggy Thanks, nice Commented Feb 28, 2019 at 19:26
# APL (Dyalog Extended), 13 9 bytes
A simple solution. The Dyalog Unicode version was simply a longer version of this so it has been omitted.
Edit: Saved 4 bytes by adopting the method in lirtosiast's Jelly solution.
{+/⍵÷⍭|⍵}
Try it online!
Ungolfing
{+/⍵÷⍭|⍵}
{ } A dfn, a function in {} brackets.
⍭|⍵ The prime factors of the absolute value of our input.
⍵÷ Then divide our input by the above array,
giving us a list of products for the product rule.
+/ We sum the above numbers, giving us our arithmetic derivative.
# Ruby, 8766807570 68 bytes
This answer is based on Luis Mendo's MATL answer, wythagoras's Python answer, and the idea that the arithmetic derivative of a number m is equal to m·(1/p1 + 1/p2 + ... + 1/pn) where p1...pn is every prime factor of n to multiplicity.
->n{s=0;(2...m=n.abs).map{|d|(m/=d;s+=n/d)while m%d<1};m<2?0:s+0**s}
This function is called in the following way:
> a=->n{s=0;(2...m=n.abs).map{|d|(m/=d;s+=n/d)while m%d<1};m<2?0:s+0**s}
> a[299792458]
196831491
Ungolfing:
def a(n)
s = 0
m = n.abs
(2...m).each do |z|
while m%d == 0
m /= d
s += n / d
end
end
if s == 0
if n > 1
s += 1 # if s is 0, either n is prime and the while loop added nothing, so add 1
# or n.abs < 2, so return 0 anyway
# 0**s is used in the code because it returns 1 if s == 0 and 0 for all other s
end
end
return s
end
# Julia, 72 43 bytes
n->n^2>1?sum(p->n÷/(p...),factor(n^2))/2:0
This is an anonymous function that accepts an integer and returns a float. To call it, assign it to a variable.
For an input integer n, if n2 ≤ 1 return 0. Otherwise obtain the prime factorization of n2 as a Dict, then for each prime/exponent pair, divide the prime by its exponent, then divide n by the result. This is just computing n x / p, where p is the prime factor and x is its exponent, which is the same as summing n / p, x times. We sum the resulting array and divide that by 2, since we've summed twice as much as we need. That's due to the fact that we're factoring n2 rather than n. (Doing that is a byte shorter than factoring |n|.)
Saved 29 bytes thanks to Dennis!
# Seriously, 171411 12 bytes
My first ever Seriously answer. This answer is based on Luis Mendo's MATL answer and the idea that the arithmetic derivative of a number m is equal to m·(1/p1 + 1/p2 + ... + 1/pn) where p1...pn is every prime factor of n to multiplicity. My addition is to note that, if m = p1e1·p2e2·...·pnen, then a(m) = m·(e1/p1 + e2/p2 + ... + en/pn). Thanks to Mego for golfing and bug fixing help. Try it online!
,;wi@/MΣ*l
Ungolfing:
, get a single input
;w duplicate input and get prime factorization, p_f
for input [-1..1], this returns [] and is dealt with at the end
M map the function inside to p_f
i pop all elements of p_f[i], the prime and the exponent, to the stack
@ rotate so that the exponent is at the top of the stack
/ divide the exponent by the prime
Σ sum it all together
* multiply this sum with the input
l map and multiply do not affect an empty list, so we just take the length, 0
l is a no-op for a number, so the result is unchanged for all other inputs
# Jolf, 13 bytes
*jmauΜm)jd/1H
Kudos to the MATL answer for the algorithm! Try it here, or test them all at once. (Outputs [key,out] in an array.)
## Explanation
*jmauΜm)jd/1H
*j input times
m)j p.f. of input
Μ d/1H mapped to inverse
u sum of
ma abs of
# Mathematica 10.0, 39 bytes
Tr[If[#>1,#2/#,0]&@@@FactorInteger@#]#&
• Can you please add an explanation? I like answers to have explanations before I upvote them. Commented Apr 4, 2016 at 9:31
• @Sherlock9 This is a quite uninteresting answer so I don't plan to add one. It's OK if no one upvotes it. Commented Apr 4, 2016 at 10:24
• Alright, then. Have a good day :) Commented Apr 4, 2016 at 10:25
• In the current Mathematica version, FactorInteger@1 yields {1,1}, so the If function is no longer necessary, saving 10 bytes. Commented Aug 26, 2016 at 5:52
• @GregMartin Seriously? That's even more inconsistent than the value, {{1,1}}, returned by my version ({} is the expected result to me). Commented Aug 26, 2016 at 6:31
# APL(NARS), 35 char, 70 bytes
{1≥a←∣⍵:0⋄1=≢k←πa:×⍵⋄c+m×∇c←⍵÷m←↑k}
test and how to use:
f←{1≥a←∣⍵:0⋄1=≢k←πa:×⍵⋄c+m×∇c←⍵÷m←↑k}
f 14
9
f 8
12
f 225
240
f ¯5
¯1
f 299792458
196831491
I thought that it would not be ok because I don't know if c variable is composed (and not a prime)... But seems ok for the test...
# 05AB1E, 7 4 bytes
ÄÒ÷O
Explanation:
Ä # Take the absolute value of the (implicit) input
Ò # Get all its prime factors (with duplicates)
÷ # Integer divide the (implicit) input by each of these prime factors
O # And take the sum (which is output implicitly)
# Vyxals, 3 bytes
ȧǐ/
Try it Online!
ȧ # Abs. val
ǐ # All prime factors
# (Implicit input)
/ # Divided by each factor.
# (s flag) sum of stack
-1 thx to Underslash
• 3 bytes -s If you remove the -r flag, you can get implicit input for the dividing. Commented May 30, 2021 at 4:21
• @Underslash Thx! Commented May 30, 2021 at 5:38
# Pari/GP, 44 bytes
n->n*vecsum([i[2]/p|i<-factor(n)~,0<p=i[1]])
Try it online!
# Perl 5, 62 bytes
perl -MMath::Prime::Util=:all -E"map$i+=1/$_,factor abs($j=<>);say$i*$j" Uses the formula (from OEIS): If n = Product p_i^e_i, a(n) = n * Sum (e_i/p_i). ## Perl 6, 90 sub A(\n) {0>n??-A(-n)!!(n>1)*{$_??n/$_*A($_)+$_*A n/$_!!1}(first n%%*,2..^n)};say A slurp
This might be a bit slow for large numbers. Replace 2..^n with 2..n.sqrt for longer code but faster computation.
# Ink, 183 bytes
==function a(n)
{n<0:
~return-a(-n)
}
{n<2:
~return 0
}
~temp f=t(n,2)
{f:
~return a(n/f)*f+n/f
}
~return 1
==function t(n,i)
{n>1&&n-i:
{n%i:
~return t(n,i+1)
}
~return i
}
~return 0
Try it online!
I refuse to believe this is a good solution, but I can't see a way to improve it either.
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latest
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en
| 0.851496 |
http://stackoverflow.com/questions/11301924/how-to-find-4-numbers-whose-xor-is-equal-to-zero-in-an-array?answertab=active
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# How to find 4 numbers whose XOR is equal to zero in an array? [closed]
Given an array `A[]` of size `n<=100000` and `0<=A[i]<=2^64`, and every two adjacent element of `A[]` have exactly one different digit in their binary representation. Now we have to check if there exist any `4` elements `A[i1],A[i2],A[i3] and A[i4]` in array such that `1<=i1<i2<i3<i4<=n` and `A[i1] xor A[i2] xor A[i3] xor A[i4] = 0`.
-
## closed as too localized by csgillespie, Raghav Sood, angainor, mah, RichardTheKiwiOct 8 '12 at 19:48
This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.
It would help if you told us what you had done so far. – zellio Jul 2 '12 at 22:23
I have no idea how to start. It would be a great help if you can suggest me some topic to go through before giving this problem a shot. – hell coder Jul 2 '12 at 22:28
The solution should be apparent if you write our a 4 digit binary number, then a number following it that obeys the rule that only one digit changes (that rule is key). Then consider, how could you xor your way to 0 with two more numbers where each subsequent number can only change by one digit. You should see the pattern that you'll be looking for. – hatchet Jul 2 '12 at 22:30
@btilly I posted an answer which will not work so it has been deleted. The failing assumption was that if (u,v) are separated by j steps, then (x,y) will also differ by j steps. This will not work since the bits toggled during the j steps need not be unique – VSOverFlow Jul 3 '12 at 2:30
This is a question from a currently running algorithmic competition at CodeChef. The link the the question being http://www.codechef.com/JULY12/problems/GRAYSC .So it is the usage of unfair means and the question should be removed.
-
Since two adjacent numbers differ by a single binary bit, the result of XOR'ing two adjacent numbers together is very predictable:
`x` represents a bit whose value doesn't matter, but is the same as the corresponding `x` below/above it.
`````` xxxxxxxxxxxxx1xxxxxxxxxxxx
xor: xxxxxxxxxxxxx0xxxxxxxxxxxx
__________________________________
00000000000001000000000000
``````
This is true for ANY two adjacent numbers in the set you described.
If you can find another pair of adjacent numbers, that differ by the same bit, they will ALSO XOR-out to the same value: `00000000000001000000000000`
And finally, XOR'ing those two values together will DEFINITELY give you a final answer of zero.
So the algorithm you want looks like:
• Identify which bit is different between a number and its successor (bit 0-63)
• Find another pair of numbers that also differ by the same bit identified in step 1.
Note that if N > 64, then by the pigeonhole priniciple, there MUST be a solution like this.
If `N <= 64`, there may be a solution of this form, or a solution of another form, I didn't describe here.
There are other constraints that could also find 4 numbers that xor together to 0.
But this is probably the simplest way to search for a solution.
Failing to find a solution this way does not guarantee that there is no solution, however.
-
This method would fail, as finding two pairs of numbers that differ by the same bit (as in a single bit, as you described) violates the original constraint that (A[a1] < A[a2] < A[a3] < A[a4]). Finding two pairs of numbers that differ by the same group of bits, on the other hand, yields a range of numbers that can be ordered, allowing the constraint (A[a1] < A[a2] < A[a3] < A[a4]) to be upheld. – aps2012 Jul 4 '12 at 6:49
@aps: 10101 and 10001 differ by the middle bit. Numbers 10100 and 10000 also differ by the middle bit. 10000 < 10001 < 10100 < 10101. – abelenky Jul 6 '12 at 21:03
@albenky So they do. I stand corrected. +1 for technical accuracy. – aps2012 Jul 7 '12 at 1:05
Hint: if you are completely stuck, then try the O(n^4) solution:
``````for i = 0 to N-3
for j = i+1 to N-2
for k = j+1 to N-1
for l = k+1 to N
if a[i] XOR a[j] XOR a[k] XOR a[l] is equal to 0 then
print a[i], a[j], a[k], a[l]
``````
-
... and for N == 100000, you could examine all the possibilities by hand over a nice cup of tea and come up with the answer in less time that it would take this algorithm to complete... :-) – aps2012 Jul 4 '12 at 6:57
Make Some careful observations about all possible values of xor of two adjacent numbers, in total how many distinct values are possible? Once you figure this out you will have your answer.
-
Hint: Look at the truth table for XOR:
``````A | B | Output
- + - + - - -
0 | 0 | 0
0 | 1 | 1
1 | 0 | 1
1 | 1 | 0
``````
Now, what binary number XORs with 10101 to be 0?
Those are the types of numbers to look for.
Edit: Once you find out what the relationship between those numbers are, which you seem to have from your comment on @abelenky 's answer, you have to look at the other parts of the problem. Since each adjacent element has exactly one different digit in their binary representation, what does that mean about the likelihood that two adjacent elements are what we are looking for?
-
If I am getting it right then we have to check if there is a pair of adjacent numbers (a,b) and (c,d) following given constraints and (a^b==c^d). Am i correct? – hell coder Jul 2 '12 at 22:41
Hint: Under these rules, could two numbers XOR to be zero? Why or why not? What would those two numbers look like with regard to each other?
-
they must be equal i guess. no? – hell coder Jul 2 '12 at 22:29
No. I do not respond to queries using terms like "plz", "intr8sting", "knw". – abelenky Jul 3 '12 at 20:03
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| 3.6875 | 4 |
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latest
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https://www.physicsforums.com/threads/is-the-integral-zero-for-closed-paths-in-complex-analysis.958405/
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Is the Integral Zero for Closed Paths in Complex Analysis?
• Gwinterz
In summary, the conversation discusses how to show that an integral is equal to zero and whether the condition in the premise is on the absolute value of a function or the function itself. It is mentioned that the integral must be equal to zero if it is less than or equal to zero, and that being a closed path may also contribute to the integral being equal to zero.
Gwinterz
Hey, I have been stuck on this question for a while:
I have tried to follow the hint, but I am not sure where to go next to get the result.
Have I started correctly? I am not sure how to show that the integral is zero.
If I can show it is less than zero, I also don't see how that shows it is always zero.
Thanks in advance for any help.
Attachments
• WClcaAN.jpg
17.6 KB · Views: 576
The condition in the premise is on ##\left|f(z)\right|##, not ##f(z)##. If you have in fact shown that ##\left|f(z)\right| \leq 0##, then it must be equal to 0 since it can't be negative.
Gwinterz said:
Hey, I have been stuck on this question for a while:
View attachment 232707
I have tried to follow the hint, but I am not sure where to go next to get the result.
Have I started correctly? I am not sure how to show that the integral is zero.
If I can show it is less than zero, I also don't see how that shows it is always zero.
Thanks in advance for any help.
Something cannot be less than zero and equal to zero at the same time. However, since you have non-strict inequalities "##\leq##" there is a chance you can show the thing is ##\leq 0##. Then (being a norm in the complex plane) it must also be ##\geq 0##, hence must ##= 0.##
Thanks guys that makes sense.
Is the integral equal to zero because its a closed path?
1. What is complex analysis?
Complex analysis is a branch of mathematics that deals with the study of complex numbers and functions. It involves the analysis of functions with complex variables and the properties and behavior of these functions.
2. What are complex numbers?
Complex numbers are numbers that have both a real and imaginary part. They are written in the form a + bi, where a and b are real numbers and i is the imaginary unit (√-1).
3. What are the applications of complex analysis?
Complex analysis has many applications in mathematics, physics, engineering, and other fields. It is used in the study of differential equations, signal processing, fluid dynamics, and quantum mechanics, among others.
4. What are some key concepts in complex analysis?
Some key concepts in complex analysis include analytic functions, Cauchy-Riemann equations, contour integration, and the Cauchy integral theorem. Other important concepts include Laurent series, residue theory, and conformal mapping.
5. How is complex analysis different from real analysis?
Complex analysis deals with functions with complex variables, while real analysis deals with functions with real variables. In complex analysis, functions are defined in terms of complex numbers and their properties, while in real analysis, functions are defined in terms of real numbers. Additionally, complex analysis involves studying functions in the complex plane, while real analysis involves studying functions on the real number line.
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https://www.univerkov.com/the-length-of-the-first-rectangle-is-36-cm-and-the-length-of-the-second-rectangle-is-6-cm-less/
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The length of the first rectangle is 36 cm, and the length of the second rectangle is 6 cm less.
The length of the first rectangle is 36 cm, and the length of the second rectangle is 6 cm less. Both rectangles have the same area. Find the width of the second rectangle if it is known that the width of the first rectangle is 15cm.
1) Let’s find out what area the first rectangle has, if it is known that its length is 36 cm, and its width is 15 cm.
36 * 15 = 540 m².
2) Let’s find out how long the second rectangle has, if it is known that it is 6 cm less than the length of the first rectangle, and the first rectangle has a length of 36 cm.
36 – 6 = 30 cm.
3) Find out what width the second rectangle has, if it is known that its area is 540 m², its length is 30 cm.
540/30 = 18 cm.
Answer: The width of the second rectangle is 18 cm.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.
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Name: ___________________Date:___________________
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### Middle/High School Algebra, Geometry, and Statistics (AGS)3.8 Factorization: (2a+b)2 - c2
Example: Factorize, 4a2+4ab+b2-c2. Solution: Given that, 4a2+4ab+b2-c2. We can write this as, [(2a)2+2.2a.b+(b)2]-c2 This is in the form of (a+b)2 = a2+2ab+b2. Here a = 2a and b = b, substitute these values in the formula, we get, = (2a+b)2-c2 This is in the form of a2-b2 = (a+b)(a-b) Here a = 2a+b and b = c, Substitute these values in the formula, we get, = (2a+b+c)(2a+b-c) Directions: Solve the following problems. Also write at least 10 examples of your own.
Name: ___________________Date:___________________
### Middle/High School Algebra, Geometry, and Statistics (AGS)3.8 Factorization: (2a+b)2 - c2
Q 1: Factorize, x2+4xy+4y2-4.None of these(x+2y+2)(x+2y+2)(x-2y+2)(x+2y-2)(x+2y+2)(x+2y-2) Q 2: Factorize,4+12x+9x2-4y2.(2+3x-2y)(2+3x-2y)None of these(2+3x+2y)(2+3x-2y)(2+3x+2y)(2-3x-2y) Q 3: Factorize, x2+6x+9-16y2. None of these(x+3+4y)(x+3-4y)(x-3+4y)(x+3-4y)(x+3+4y)(x+3+4y) Q 4: Factorize, 9x2+12x+4-36y2. (3x-2+6y)(3x+2-6y)(3x+2+6y)(3x+2-6y)(3x+2+6y)(3x-2-6y)None of these Question 5: This question is available to subscribers only! Question 6: This question is available to subscribers only!
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# Multiplying 2 and 3-Digit by 1-Digit Numbers
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## Interactive Math Practice - Multiplying 2 and 3 Digit Numbers by 1 Digit Numbers
It's time to practice basic multiplication skills! In this fifth grade-level multiplication lesson, students will practice multiplying 2- and 3-digit numbers by 1-digit numbers. The multiplication problems are presented in several formats: vertical multiplication problems, horizontal multiplication problems, and multiplication word problems. Here are some examples of math problems students may be asked in this online math practice activity: "22 x 7 =", "876 x 3 =", and "Discovery Charter School's PTA was hosting an ice cream social. They ordered four boxes of 480 plastic spoons for the event. How many spoons did the PTA order in all?"
If students need a little extra help answering their multiplication problems, they can click on the "Hint" button on their practice screen. They will be prompted to take the first step in solving the problem, multiplying the ones. If students answer a question incorrectly, they will be showed a detailed explanation page with the correct answer, including a custom graphic that provides step-by-step instructions for how to solve the multiplication problem correctly, so students can learn from their mistakes and improve their skills.
## How does iKnowIt.com work?
As with all the interactive math practice activities on "I Know It," this basic multiplication lesson includes several features that help students make the most out of their math practice time. A progress-tracker in the upper-right corner of the screen shows students how many math problems they've solved in the lesson so far, and how many are left to go. A score-tracker beneath that shows them how many math problems they have solved correctly so far. A speaker icon in the upper-left corner of the screen gives students the option of having the question read aloud to them in a clear voice. This option is great for ESL/ELL students or students who prefer auditory learning.
We hope when you try out this fifth grade multiplication lesson with your class, you and your students will enjoy the bright and colorful design of our interactive math practice program, including its kid-friendly features, whimsical animated characters, and motivating reward system. Be sure to explore the hundreds of other math topics available on "I Know It."
## Free Trial and Membership Options
Sign up for a free sixty-day trial of iKnowIt.com so your class can try out this basic multiplication lesson at no cost! In the free trial mode, your students will be able to play any math game on our website for free, but they will be limited to a total of twenty-five questions per day across all “I Know It” math lessons. For unlimited access to the website, you will need to become a paying member.
Here are some of the benefits you will enjoy as a member of iKnowIt.com: You can create a class roster and add your students to it; you can assign individual user names and passwords to each student; you can give different students different lessons to complete based on their individual needs and skills; you can change lesson settings at any time; and you can track your students' progress through their lessons, including downloading, emailing, and printing student progress reports.
We think your students will find it easy and fun to use the student interface of iKnowIt.com! When they log into the site with their unique user name and password, they will be shown a kid-friendly homepage, where they will find their lessons to complete under the "My Assignments" section. They will also have the option to explore other lessons at their grade level, and if you choose to let them, they can explore lessons at other grade levels too. (Grade levels appear as "Level A" through "Level E" in the student mode.)
## Level
This Level E multiplication lesson may be ideal for fifth grade students.
## Common Core Standard
5.NBT.5, MA.5.NSO.2.1, MA.5.AR.1.1, 5.3B
Number And Operations In Base Ten
Perform Operations With Multi-Digit Whole Numbers And With Decimals To Hundredths.
Fluently multiply multi-digit whole numbers using the standard algorithm.
## You might also be interested in...
Multiplying 4-Digit by 1-Digit Numbers (Level E)
In this fifth grade-level math lesson, students will practice multiplying 4-digit by 1-digit numbers. Students will solve horizontal multiplication problems, vertical multiplication problems, and multiplication word problems.
Multiplying 2-Digit by 2-Digit Numbers (Level E)
Students will practice multiplication skills in this fifth grade math lesson. Math problems include vertical multiplication problems, horizontal multiplication problems, and multiplication word problems.
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## Saturday 25 May 2024
### On This Day in Math - May 25
Some people are always critical of vague statements.
I tend rather to be critical of precise statements;
they are the only ones which can correctly be labeled wrong.
~ Raymond Smullyan on criticism
The 145th day of the year; 145= 1! + 4! + 5!. There are only four such numbers in base ten. 1, 2 and 145 are three of them, what's the fourth? (answer at bottom of post) Such numbers are called factorions, a term created by Cliff Pickover in 1995
145 is the result of 34 + 43, making it a Leyland number. a number of the form xy + yx where x and y are integers greater than 1. They are named after the British number theorist, Paul Leyland. (There are ten days of the year that are Leyland numbers)
Prime Curios points out several curiosities related to 145, The 145th prime number is 829 and their concatenation, 145829 is prime. And the largest prime factor of 145, is 1+4+5+8+2+9. and 149 is congruent to 1 in mod 8, mod 2, and mod 9
The process of summing the squares of the digits of a decimal number has two results, one is the eventual decent to 1, and being called a Happy number. 145 is the largest Unhappy number. Unhappy numbers eventually land on one of the numbers in the eight cycle, 4 → 16 → 37 → 58 → 89 → 145 → 42 → 20 → 4 ... (any three digit number, for example, produces a sum of squares less than or equal to 243. Any of the numbers you land on that are greater than 145 and less than 243 has a sum of squares of its digits that is less than itself, and eventually they land on one of the chains that lead to the eight cycle shown. Some numbers (like 99) iterate to a number greater than 145, but they then recede back into the inexorable "cycle of unhappiness" above. (A great exploration for students to create the trees of all numbers less than 200 that go to either 1, or the unhappy cycle)
The 145th prime number is 829, and 145829 is prime. Notice also, that the largest prime factor of 145 = 1+4+5+8+2+9, and that 145 is congruent to 1, mod 8, mod 2, and mod 9. *Prime Curios
145 is a palindrome in base 12, (101)
145 can be written as the sum of two squares in two different ways, 122+12 = 82+92
EVENTS
1581 John Dee, mathematician and mystic, first saw spirits in his crystal globe. [Daniel Cohen, Masters of the Occult , Dodd and Mead, 1971, p. 28] *VFR
It is believed that Dee used his Claude glass as a crystal ball to look into the future, a practice known as scrying and a form of divination.
*Wik
1694 Isaac Newton to Nathaniel Hawes: “A Vulgar Mechanick can practice what he has been taught or seen done, but if he is in an error he knows not how to find it out and correct it, and if you put him out of his road, he is at a stand; Whereas he that is able to reason nimbly and judiciously about figure, force and motion, is never at rest till he gets over every rub.” Westfall takes this as the title of his fabulous biography of Newton, Never at Rest. Reportedly he received one of the early Golden Fleece Awards of Senator Proxmire for a grant to write this book. [But we have been unable to find a reference for this last statement.] *VFR
1721 John Copson of High Street in Philadelphia posted an offering in the American Weekly Mercury on this date announcing that he would open an office to provide fire insurance for "vessels, goods, and merchandise." With that, he became the first Insurance agent in continental America. *Kane, First Famous Facts He opened his office near Penn’s Landing in Philadelphia on June 2 of that year.
The actual complete ad read:
"Assurances from Losses happening at Sea ect. [sic] being found to be very much for the Ease and benefit of the Merchants and Traders in general, and whereas the merchants of this city of Philadelphia and other parts, have been obliged to send to London for such Assurance, which has not only been tedious and troublesome, but even very precarious. For remedying of which, An Office of Publick Insurance on Vessels, Goods and Merchandizes, will, on Monday next, be Opened, and Books kept by John Copson of this City, at this House in the High Street, where all Persons willing to be Insured may apply: And Care shall be taken by the said J. Copson That the Assurors or Underwriters be Persons of undoubted Worth and reputation and of considerable Interest in this City and Province."
1747 Benjamin Franklin describes his electrical experiments in a letter to Peter Collinson. " Hence have arisen some new Terms among us. .... Or rather B is electrised plus and A minus. And we daily in our Experiments electrise Bodies plus or minus as we think proper. These Terms we may use till your Philosophers give us better. To electrise plus or minus, no more needs to be known than this; that the Parts of the Tube or Sphere, that are rub'd, do, in the Instant of the Friction, attract the Electrical Fire, and therefore take it from the Thing rubbing: the same Parts immediately, as the Friction upon them ceases, are disposed to give the Fire they have received, to any Body that has less." *Collinson, "Experiments and Observations on Electricity Made at Philadelphia in America."
1773 After more than a decade as master of his school in Newcastle, author, and shameless self promoter, Charles Hutton was appointed professor of mathematics of the Royal Military Academy at Woolwich. In 18 years he had gone from poorly educated coal miner, to Professor. *Gunpowder and Geometry, Benjamin Wardhaugh
He was professor of mathematics at the Royal Military Academy, Woolwich from 1773 to 1807. He is remembered for his calculation of the density of the earth from Nevil Maskelyne's measurements collected during the Schiehallion experiment.
1832 Galois writes to his friend, Auguste Chevalier, about his broken romance for Stephanie. The fatal duel looms only five days away. Mario Livio tells the story wonderfully in "The Equation That Couldn't be Solved".
1842 Johann Christian Doppler (1803–1853) presented a lecture on the Doppler effect. It was first experimentally verified in 1845 using a locomotive drawing an open car with several trumpeters. *VFR
The hypothesis was tested for sound waves by Buys Ballot in 1845.
Buys Ballot was a Dutch chemist and meteorologist after whom Buys Ballot's law and the Buys Ballot table are named. He was first chairman of the International Meteorological Organization, the organization that would become the World Meteorological Organization.
*Wik
1844 the first news communicated by telegraph in the U.S. was sent 80 miles to the Baltimore Patriot, Maryland, from Washington, D.C. giving the information that "One o'clock. There has just been made a motion in the House to go into committee of the whole on the Oregon question. Rejected. Ayes 79 - Nays 86." This was just one day after Samuel Morse transmitted his famous "What hath God wrought!" message from the U.S. Supreme Court room and opened America's first telegraph line linking Washington and Baltimore.*TIS
1862 Caricature published to celebrate first aerial photographer, Nadar, was Published in Le Boulevard. Nadar "elevating photography to the condition of art", caricature by Honoré Daunier. The first known aerial photograph was taken in 1858 by French photographer and balloonist, Gaspar Felix Tournachon, known as "Nadar". In 1855 he had patented the idea of using aerial photographs in mapmaking and surveying, but it took him 3 years of experimenting before he successfully produced the very first aerial photograph. It was a view of the French village of Petit-Becetre taken from a tethered hot-air balloon, 80 meters above the ground. This was no mean feat, given the complexity of the early collodion photographic process, which required a complete darkroom to be carried in the basket of the balloon! Unfortunately, Nadar's earliest photographs no longer survive, and the oldest aerial photograph known to be still in existence is James Wallace Black's image of Boston from a hot-air balloon, taken in 1860. Following the development of the dry-plate process, it was no longer necessary carry so much equipment, and the first free flight balloon photo mission was carried out by Triboulet over Paris in 1879. *History of Aerial Photography
1946 The Soviet Union issued two stamps to celebrate the 125th anniversary of Pafnuti Lvovich Chebyshev (1821–1894). [Scott #1050-1].*VFR (see May 16th Births)
1961 the formal announcement of an American lunar landing was made by President John F. Kennedy speaking to the Congress: "I believe that this nation should commit itself to achieving the goal, before this decade is out, of landing a man on the Moon and returning him safely to the Earth. No single space program in this period will be more impressive to mankind, or more important in the long-range exploration of space; and none will be so difficult or expensive to accomplish." *TIS
1968
The Gateway Arch in St Louis, Missouri was inaugurated on this date by U S Vice President, Hubert Humphrey. Although it is often mistaken for a parabola, the arch is built in the form of an inverted, weighted catenary arch. It is the world's tallest arch, the tallest man-made monument in the Western Hemisphere, and Missouri's tallest accessible building. Built as a monument to the westward expansion of the United States,the centerpiece of the Jefferson National Expansion Memorial and has become an internationally famous symbol of St. Louis. Appropriately, Thomas Jefferson is credited for the first use of the term catenary in English in correspondence with Thomas Paine on bridge construction.. *Wik *@HistoryTime_
2142 The next total solar eclipse in Ostend, Belgium. The last total solar eclipse took place more than 11 centuries ago, 29 September 878. But only 9 years later, on 14 June 2151, there will be another one. *NASA Solar Eclipse Catalog (I know I can hardly wait!)
BIRTHS
1828 Karl Peterson (25 May 1828 in Riga, Russia (now Latvia) - 19 April 1881 in Moscow, Russia)
was a Latvian mathematician worked in differential geometry and partial differential equations. .. by means of a uniform general method, he deduced nearly all the devices known at that time for finding general solutions of different classes of equations.
Largely because he was not at a university his results were not well known but they did influence Egorov in Moscow, but Peterson gained an international reputation only when Darboux and Bianchi used his results.
*SAU
1873 Michele Angelo Besso (25 May 1873 Riesbach – 15 March 1955 Genoa) was a Swiss/Italian engineer of Jewish Italian (Sephardi) descent. He was a close friend of Albert Einstein during his years at the Federal Polytechnic Institute in Zurich, today the ETH Zurich, and then at the patent office in Bern. Besso is credited with introducing Einstein to the works of Ernst Mach, the sceptical critic of physics who influenced Einstein's approach to the discipline. Einstein called Besso "the best sounding board in Europe" for scientific ideas.
In a letter of condolence to the Besso family Albert Einstein wrote his now famous quote "Now Besso has departed from this strange world a little ahead of me. That means nothing. People like us, who believe in physics, know that the distinction between past, present and future is only a stubbornly persistent illusion" *Wik
1919 Raymond Merrill Smullyan ( May 25, 1919 -February 6, 2017) is an American mathematician, concert pianist, logician, Taoist philosopher, and magician. His first career (like Persi Diaconis a generation later) was stage magic. He then earned a BSc from the University of Chicago in 1955 and his Ph.D. from Princeton University in 1959. He is one of many logicians to have studied under Alonzo Church. Smullyan is the author of many books on recreational mathematics, recreational logic, etc. Most notably, one is titled "What Is the Name of This Book?". *Wik For example the book is described on the cover as follows:"Beginning with fun-filled monkey tricks and classic brain-teasers with devilish new twists, Professor Smullyan spins a logical labyrinth of even more complex and challenging problems as he delves into some of the deepest paradoxes of logic and set theory, including Gödel's revolutionary theorem of undecidability."
Martin Gardner described this book in Scientific American as:"The most original, most profound and most humorous collection of recreational logic and mathematics problems ever written."
1921 Jack Steinberger (born Hans Jakob Steinberger; May 25, 1921 – December 12, 2020) was a German-born American physicist noted for his work with neutrinos, the subatomic particles considered to be elementary constituents of matter. He was a recipient of the 1988 Nobel Prize in Physics, along with Leon M. Lederman and Melvin Schwartz, for the discovery of the muon neutrino. Through his career as an experimental particle physicist, he held positions at the University of California, Berkeley, Columbia University (1950–68), and the CERN (1968–86). He was also a recipient of the United States National Medal of Science in 1988, and the Matteucci Medal from the Italian Academy of Sciences in 1990.
1954 Clyde P. Kruskal (born May 25, 1954) is an American computer scientist, working on parallel computing architectures, models, and algorithms. He got his A.B. degree in mathematics and computer science from Brandeis University, M.Sc. (1978) and Ph.D. (1981) from New York University under Jack Schwartz. Since then he has worked as assistant professor at University of Illinois (1981–85) and University of Maryland, College Park (1985–88), as associate professor (1988–). He has published extensively, becoming an ISI highly cited researcher. His father was the world-renowned mathematician Martin Kruskal. *Wik
DEATHS
1555 Gemma Frisius ; (December 9, 1508–May 25, 1555) was a Dutch mathematician who applied his mathematical expertise to geography, astronomy and map making. He became the leading theoretical mathematician in the Low Countries.*SAU Frisius created or improved many instruments, including the cross-staff, the astrolabe and the astronomical rings. His students included Gerardus Mercator (who became his collaborator), Johannes Stadius, John Dee, Andreas Vesalius and Rembert Dodoens.*Wik
It was Frisius who created the modern symbol for degrees, o , in the 1569 edition of Arithmeticae practicae moethodus facilis. *Cajori
Math Historian Thony Christie informed me that "Regnier Gemma Frisius", a name sometimes used for him (and I had) was not his name. "His birth name was Jemme Reinerzoon, i.e. Jemma son of Reiner, his humanist name simply Gemma Frisius."
1676 Johann Rahn (10 March 1622 in Zürich, Switzerland - 25 May 1676 in Zürich, Switzerland)
was a Swiss mathematician who was the first to use the symbol "÷",called an obelus, for a division symbol in Teutsche Algebra (1659). The invention is also sometimes credited to British Mathematician John Pell who was Rahn's tutor for a while. The book, written in German, contains an example of Pell's equation also.
1818 Caspar Wessel (June 8, 1745 – March 25, 1818) was a Danish-Norwegian mathematician.
He was born in Jonsrud, Vestby, Akershus, Norway. In 1763, having completed secondary school, he went to Denmark for further studies (since Norway didn't have any university in 1763). In 1778 he got the degree of candidatus juris, which is one law degree. In 1794 he was hired as a surveyor; in 1798, a Royal inspector of Surveying.
As surveying is related to mathematics, he later studied the geometrical importance of complex numbers. His most important paper, Om directionens analytiske betegning, (On the Analytical Representation of Direction) was published in 1799 by the Royal Danish Academy of Sciences and Letters. Since it was in Danish, it was not noticed by many people. Later, Jean-Robert Argand and Carl Friedrich Gauss's paper showed the same results.
One of the more important, but missed ideas shown in Wessel's Om directionens analytiske betegning was vectors. Wessel's main thing he wanted to show in the paper was not this, but he felt that the concept of numbers, with length and direction would be needed. Wessel's thoughts on addition was: "Two straight lines are added if we unite them in such a way that the second line begins where the first one ends and then pass a straight line from the first to the last point of the united lines. This line is the sum of the united lines". Today, the same idea is used when adding vectors.
His paper was printed in a French translation in 1899. It was released in English in 1999 as "On the analytic representation of direction" (ed. J. Lützen et al.).
1956 Johann Radon (16 December 1887 – 25 May 1956) worked on the calculus of variations, differential geometry and measure theory.
Radon is known for a number of lasting contributions, including:
• his part in the Radon–Nikodym theorem;
• the Radon measure concept of measure as linear functional;
• the Radon transform, in integral geometry, based on integration over hyperplanes — with application to tomography for scanners (see tomographic reconstruction);
• Radon's theorem, that d + 2 points in d dimensions may always be partitioned into two subsets with intersecting convex hulls
• the Radon-Hurwitz numbers.
• He is possibly the first to make use of the so called Radon-Riesz property. *Wik
1989 Ruby Violet Payne-Scott, (28 May 1912 – 25 May 1981) was an Australian pioneer in radiophysics and radio astronomy, and was the first female radio astronomer.
One of the more outstanding physicists that Australia has ever produced and one of the first people in the world to consider the possibility of radio astronomy, and thereby responsible for what is now a fundamental part of the modern lexicon of science, she was often the only woman in her classes at the University of Sydney.
Her career arguably reached its zenith while working for the Australian government's Commonwealth Scientific and Industrial Research Organisation (then called CSIR, now known as CSIRO) at Dover Heights, Hornsby and especially Potts Hill in Sydney. Some of her fundamental contributions to solar radio astronomy came at the end of this period. She is the discoverer of Type I and Type III bursts and participated in the recognition of Type II and IV bursts.
She played a major role in the first-ever radio astronomical interferometer observation from 26 January 1946, when the sea-cliff interferometer was used to determine the position and angular size of a solar burst. This observation occurred at either Dover Heights (ex Army shore defence radar) or at Beacon Hill, near Collaroy on Sydney's north shore (ex Royal Australian Air Force surveillance radar establishment - however this radar did not become active until early 1950).
During World War II, she was engaged in top secret work investigating radar. She was the expert on the detection of aircraft using PPI (Plan Position Indicator) displays. She was also at the time a member of the Communist Party and an early advocate for women's rights. The Australian Security Intelligence Organisation (ASIO) was interested in Payne-Scott and had a substantial file on her activities, with some distortions.
*Wik
*Wik
**The fourth factorion is 40585 = 4! + 0! + 5! + 8! + 5!
Credits
*CHM=Computer History Museum
*FFF=Kane, Famous First Facts
*NSEC= NASA Solar Eclipse Calendar
*RMAT= The Renaissance Mathematicus, Thony Christie
*SAU=St Andrews Univ. Math History
*TIA = Today in Astronomy
*TIS= Today in Science History
*VFR = V Frederick Rickey, USMA
*Wik = Wikipedia
*WM = Women of Mathematics, Grinstein & Campbell
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# Lesson 6: Neural Networks, In Practice
## A.K.A. Neural Networks (Part 2), Part 2
by CAIS++
In this section, we’ll take care of a few loose ends that we may have skimmed over in the previous lessons. These topics will become especially relevant for when you try to implement neural networks in your own projects. Be sure to go through Lessons 4 and 5 before continuing on to this lesson!
## Loss Functions
In our equations of backpropagation, we have used a generalized loss function $$J$$. This cost function is what we aim to through stochastic gradient descent (SGD) and tells us how well the model is doing given the model parameters.
In earlier lessons, we saw the mean squared error used as a loss function. However, the mean squared error may not always be the best cost function to use. In fact, a more popular loss function is the cross entropy cost function. Before we get more into the cross-entropy cost function, let's look into the softmax classification function.
### Softmax Classifier
Let's say you are building a neural network to classify between two classes. Our neural network will look something like the following image. Notice that there are two outputs $$y_1$$ and $$y_2$$ representing class one and two respectively.
We are given a set of data points $$\textbf{X}$$ and their corresponding labels $$\textbf{Y}$$. How might we represent the labels? A given point is either class one or class two. The boundary is distinct. If you remember the linear classification boundary from earlier we said that any output greater than 0 was class one and any output less than 0 was class two. However, that does not really work here. A given data point $$\textbf{x}_i$$ is simply class one or class two. We should not have data points be more class one than other data points.
We will use one hot encoding to provide labels for these points. If a data point has the label of class one simply assign it the label vector $$\textbf{y}_i = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$ and for a data point of class two assign it the label vector $$\textbf{y}_i = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$
Say our network outputs the value $$\begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$$ where $$c_1, c_2$$ are just constants. We can say the network classified the input as class one if $$c_1 > c_2$$ or classified as class two if $$c_2 > c_1$$. Let's use the softmax function to interpret these results in a more probabilistic manner.
The softmax function is defined as the following $$q(\textbf{c}) = \frac{e^{c_i}}{\sum_j e^{c_j}}$$ Where $$c_i$$ is the scalar output of the $$ith$$ element of the output vector. Think of the numerator as converting the output to an un-normalized probability. Think of the denominator as normalizing the probability. This means that for every output $$i$$ the loss function will have an output between 0 and 1. Another convenient property of the softmax function is that the sum of all the output probabilties $$i$$ will sum to $$1$$, making this activation function more suitable to working with probability distributions.
### Entropy
We need to take one more step before we can use the softmax function as a loss function. This requires some knowledge of what entropy is. Think about this example. Say you were having a meal at EVK, one of the USC dining halls. If your meal is bad, this event does not carry much new information, as the meals are almost guaranteed to be bad at EVK. However, if the meal is good, this event carries a lot of information, since it is out of the ordinary. You would not tell anyone about the bad meal (since a bad meal is pretty much expected), but you would tell everyone about the good meal. Entropy deals with this measure of information. If we know an underlying distribution $$y$$ to some system, we can define how much information is encoded in each event. We can write this mathematically as: $$H(y) = \sum_i y_i \log \left( \frac{1}{y_i} \right) = - \sum_i y_i \log ( y_i )$$
### Cross Entropy
This definition assumes that we are operating under the correct underlying probability distribution. Let's say a new student at USC has no idea what the dining hall food is like and thinks EVK normally serves great food. This freshman has not been around long enough to know the true probability distribution of EVK food, and instead assumes the probability distribution $$y'_i$$. Now, this freshman incorrectly thinks that bad meals are uncommon. If the freshman were to tell a sophomore (who knows the real distribution) that his meal at EVK was bad, this information would mean little to the sophomore because the sophomore already knows that EVK food is almost always bad. We can say that the cross entropy is the encoding of events in $$y$$ using the wrong probability distribution $$y'$$. This gives $$H(y, y') = - \sum_i y_i \log y'_i$$
Now let's go back to our neural network classification problem. We know the true probability distribution for any sample should be just the one hot encoded label of the sample. We also know that our generated probability distribution is the softmax function. This gives the final form of our cross entropy loss. $$L_i = -\log \left( \frac{e^{c_i}}{\sum_j e^{c_j}} \right)$$ Where $$y_i = 1$$ for the correct label and $$y'$$ is the softmax function. This loss function is often called the categorical cross entropy loss function because it works with categorical data (i.e. data that can be classified into distinct classes).
And while we will not go over it here, know that this function has calculable derivatives as well. This allows it to be used just the same as the mean squared error loss function. However, the cross entropy loss function has many desirable properties that the mean squared error does not have when it comes to classification.
Let's say you are trying to predict the classes cat or dog. Your neural network has a softmax function on the output layer (as it should because this is a classification problem). Let's say for two inputs $$\textbf{x}_1,\textbf{x}_2$$ the network respectively outputs $$\textbf{a}_1 = \begin{bmatrix} 0.55 \\ 0.45 \end{bmatrix}, \textbf{a}_2 = \begin{bmatrix} 0.44 \\ 0.56 \end{bmatrix}$$ where the corresponding labels are $$\textbf{y}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \textbf{y}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$ As you can see, the network only barely classified each result as correct. But by only looking at the classification error, the accuracy would have been 100%.
Take a similar example where the output of the network is just slightly off. $$\textbf{a}_1 = \begin{bmatrix} 0.51 \\ 0.49 \end{bmatrix}, \textbf{a}_2 = \begin{bmatrix} 0.41 \\ 0.59 \end{bmatrix}, \textbf{y}_1 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \textbf{y}_2 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$ Now in this case, we would have a 0% classification accuracy.
Let's see what our cross entropy function would have given us in each situation when averaged across the two samples. In the first situation: $$-(\log(0.55) + \log(0.56)) / 2 = 0.59$$ In the second situation: $$-(\log(0.49) + \log(0.59)) / 2 = 0.62$$ Clearly, this result makes a lot more sense for our situation than just having a cost value of $$0$$ for a barely-correct classification.
Overall, the choice of the correct loss function is dependent on the problem, and is a decision you must make in designing your neural network. Always keep in mind the general equations for stochastic gradient descent will have the form: $$\mathbf{W} (k) = \textbf{W}(k-1) - \alpha \nabla J(\textbf{x}_k, \textbf{W}(k-1))$$ $$\mathbf{b} (k) = \textbf{b}(k-1) - \alpha \nabla J(\textbf{x}_k, \textbf{b}(k-1))$$ Where $$J$$ is the loss function. Furthermore, the same form of backpropagation equations will still apply with backpropagating the error terms through the network.
## Optimization In Practice (Continued)
### Mini-batch Algorithm
First let's review our SGD algorithm shown below. Note that $$\theta$$ is commonly used to refer to all the parameters of our network (including weights and biases). $$\mathbf{\theta} (k) = \mathbf{\theta}(k-1) - \alpha \nabla J(\textbf{x}_k, \mathbf{\theta}(k-1))$$ As this algorithm is stochastic gradient descent, it operates on one input example at a time. This is also referred to as online training. However, this is not an accurate representation of the gradient, as it is only over a single training input, and is not necessarily reflective of the gradient over the entire input space. A more accurate representation of the gradient could be given by the following. $$\mathbf{\theta} (k) = \mathbf{\theta}(k-1) - \alpha \nabla J(\textbf{x}, \mathbf{\theta}(k-1))$$ The gradient at each iteration is now being computed across the entire input space. This is referred to as batch gradient descent, which actually turns out to be a kind of confusing name in practice (as we'll see in a bit).
In practice, neither of these approaches are desirable. The first does not give a good enough of an approximation of the gradient -- the second is computationally infeasible, since for each iteration, the gradient of the cost function for the entire dataset has to be computed. Mini-batch methods are the solution to this problem.
In mini-batch training, a sample set of all the training examples are used to compute the cost gradient. The average of these gradients for each sample is then used. This approach offers a good trade off between speed and accuracy. The equation for this method is given below, where $$Q$$ is the number of samples in the mini-batch and $$\alpha$$ is the learning rate. $$\mathbf{\theta} (k) = \mathbf{\theta}(k-1) - \frac{\alpha}{Q} \sum_{q=1}^{Q}\nabla J(\textbf{x}_q, \mathbf{\theta}(k-1))$$ Remember that batch gradient descent is over the whole input space, while mini-batch is just over a smaller subset at a time.
Of course, it would make sense that the samples have to be randomly drawn from the input space, as sequential samples will likely have some correlation. The typical mini-batch sampling procedure is to randomly shuffle the input space, and then to sample sequentially from the scrambled inputs.
### Initializations
At this point, you may be wondering how the parameters of a neural network are typically initialized. So far, the learning procedure has been described, but the actual initial state of the network has not been discussed.
You may think that how a network is initialized does not necessarily matter. After all, the network should eventually converge to the correct parameters right? Unfortunately, this is not the case with neural networks; as it turns out, the initialization of the parameters matters greatly. Initializing to small random weights typically works. But typically, the standard for weight initialization is the normalized initialization method.
In the normalized initialization method, weights are randomly drawn from the following uniform distribution: $$\textbf{W} \sim U \left( -\frac{6}{\sqrt{m+n}}, \frac{6}{m+n} \right)$$ Where $$m$$ is the number of inputs into the layer and $$n$$ is the number of outputs from the layer.
As for the biases, typically just assigning them to a value of 0 works.
### Challenges in Optimization
In mathematical optimization, we optimize (i.e. find the minimum/maximum of) some function $$f$$. In machine learning, we can think of training a neural network as a specific type of optimization (gradient descent) applied to a specific cost function.
As you may have expected, there are some concerns that can arise in this training phase: for example, local minima. Any deep neural network is guaranteed to have a very large number of local minima. Take a look at the below surface. This surface has two minima: one local, and one global. If you look at the contour map below, you can see that the algorithm converges to the local minimum instead of the global minimum.
Should we take measures to stop our neural network from converging at a local (rather than a global) minimum? Local minima would be a concern if the cost function evaluated at the local minima was far greater than the cost function evaluated at the global minima. It turns out that in practice, this difference is often negligible. Most of the time, simply finding any minima is sufficient in the case of deep neural networks.
Some other potential problem points are saddle points, plateaus, and valleys. In practice, neural networks can often escape valleys or saddle points. However, they can still pose a serious threat to neural networks, since they can have cost values much greater than at the global minimum. Even more dangerous to the training process are flat regions on the cost surface. Small initial weights are chosen in part to avoid these flat regions.
In general, more flat areas are problematic for the rate of convergence. It takes a lot of iterations for the gradient descent algorithm to get over flatter regions. One's first thought may be to increase the learning rate of the algorithm, but too high of a learning rate will result in divergence at steeper areas of the performance surface. When this algorithm with a high learning rate goes across something like a valley, it will oscillate out of control and diverge. An example of this is shown below.
At this point, it should be clear that several modifications to backpropagation need to be made to allow solve this oscillation problem and to fix the learning rate issue.
### Momentum
For this concept, it is useful to think of the progress of the algorithm as a point traveling over the cost surface. Momentum in neural networks is very much like momentum in physics. And since our 'particle' traveling over the cost surface has unit mass, momentum is just the velocity of our motion. The equation of backprop with momentum is given by the following. $$\textbf{v}(k) = \lambda \textbf{v}(k-1) - \alpha \nabla J(\textbf{x}, \mathbf{\theta}(k-1))$$ $$\mathbf{\theta} (k) = \mathbf{\theta}(k-1) + \textbf{v}(k)$$ The effect of applying this can be seen in the image below. Momentum dampens the oscillations and tends to make the trajectory continue in the same direction. Values of $$\lambda$$ closer to 1 give the trajectory more momentum. Keep in mind that $$\lambda$$ itself does not actually represent the magnitude of the particle's momentum; instead, it is more like a force of friction for the particle's trajectory. Typical values for $$\lambda$$ are 0.5, 0.9, 0.95 and 0.99.
Nesterov momentum is an improvement on the standard momentum algorithm. With Nesterov momentum, the gradient of the cost function is considered after the momentum has been applied to the network parameters at that iteration. So now we have: $$\textbf{v}(k) = \lambda \textbf{v}(k-1) - \alpha \nabla J(\textbf{x}, \mathbf{\theta}(k-1) + \lambda \textbf{v}(k-1))$$ $$\mathbf{\theta} (k) = \mathbf{\theta}(k-1) + \textbf{v}(k)$$ In general, Nesterov momentum outperforms standard momentum.
One of the most difficult hyper-parameters to adjust in neural networks is the learning rate. Take a look at the image below to see the effect of learning different learning rates on the minimization of the loss function.
As from above, we know that the trajectory of the algorithm over flat sections of the cost surface can be very slow. It would be nice if the algorithm could have a fast learning rate over these sections, but a slower learning rate over steeper and more sensitive sections. Furthermore, the direction of the trajectory is more sensitive in some directions as opposed to others. The following algorithms will address all of these issues with adaptive learning rates.
The Adaptive Gradient algorithm (AdaGrad) adjusts the learning rate of each network parameter according to the history of the gradient with respect to that network parameter. This is an inverse relationship, so if a given network parameter has had large gradients (i.e. steep slopes) in the recent past, the learning rate will scale down significantly.
Whereas before there was just one global learning rate, there is now a per parameter learning rate. We set the vector $$\textbf{r}$$ to be the accumulation of the parameter's past gradients, squared. We initialize this term to zero. $$\textbf{r} = 0$$ Next we compute the gradient as usual $$\textbf{g} = \frac{1}{Q} \sum_{q=1}^{Q}\nabla J(\textbf{x}_q, \mathbf{\theta}(k-1))$$ And then accumulate this gradient in $$r$$ to represent the history of the gradient. $$\textbf{r} = \textbf{r} + \textbf{g}^2$$ And finally, we compute the parameter update $$\mathbf{\theta} (k) = \mathbf{\theta}(k-1) - \frac{\alpha}{\delta + \sqrt{\textbf{r}}} \odot g$$ Where $$\alpha$$ is the global learning rate, and $$\delta$$ is an extremely small constant ( $$10^{-7}$$ ). Notice that an element wise vector multiplication is being performed (by the $$\odot$$ operator). Remember that each element of the gradient represents the partial derivative of the function with respect to a given parameter. The element wise multiplication will then scale the gradient with respect to a given parameter appropriately. The global learning rate is usually not difficult to choose, and normally works as just 0.01.
However, a problem with this algorithm is that it considers the whole sum of the squared gradient since the beginning of training. In practice, This results in the learning rate decreasing too much too early.
### RMSProp
RMSProp is regarded as the go-to optimization algorithm for deep neural networks. It is similar to AdaGrad, but includes a decay over the accumulation of the past gradient-squared, so the algorithm "forgets" gradients far in the past.
As normal, compute the gradient. $$\textbf{g} = \frac{1}{Q} \sum_{q=1}^{Q}\nabla J(\textbf{x}_q, \mathbf{\theta}(k-1))$$ Now, this is where the algorithm changes with the introduction of the decay term $$\rho$$, which is set somewhere between 0 and 1. $$\textbf{r} = \rho \textbf{r} + (1 - \rho) \textbf{g}^2$$ And the parameter update is the same. $$\mathbf{\theta} (k) = \mathbf{\theta}(k-1) - \frac{\alpha}{\delta + \sqrt{\textbf{r}}} \odot g$$
### Second Order Algorithms
Second order algorithms make use of the second derivative to "jump" to the critical points of the cost function. Further discussion of these algorithms is outside the scope of this tutorial. However, these algorithms do not work very well in practice. First of all, it is computationally infeasible to compute the second order derivatives. Second of all, for a complex performance surface with many critical points, it is very likely the second order method would go in the completely wrong direction. Overall, gradient descent first order methods have been shown to perform better, so I would not worry about knowing what second order algorithms are all about. But know that they exist and are an active area of research.
## Regularization (Continued)
Recall from our previous lesson that regularization is a technique that we can use to reduce overfitting by producing more robust models that are more capable of making general predictions in the real world. In this section, we’ll discuss a variety of specific methods that we can use to address this problem of overfitting.
(If you need a refresher on the general idea behind regularization, take a look back at our previous lesson here.)
### $$L^2$$ Parameter Regularization
Parameter regularization aims to reduce overfitting by penalizing for overly complex models. We can do this by adding an extra term to our cost function that penalizes large weights (which usually indicate overfitting), which helps result in smoother model predictions. This type of regularization defines the parameter norm penalty as the following. $$\Omega(\theta) = \frac{1}{2} \lVert \textbf{w} \rVert_2^2$$ and the total objective function: $$\hat{J}(\theta, \textbf{X}, \textbf{y}) = J(\theta, \textbf{X}, \textbf{y}) + \frac{\alpha}{2} \textbf{w}^T \textbf{w}$$ Evidently, this regularization will penalize larger weights. In theory, this should help prevent the model from overfitting. It is common to employ $$L^2$$ regularization when the number of observations is less than the number of features. Similar to $$L^2$$ regularization is $$L^1$$, which you can probably expect is just $$\Omega(\theta) = \frac{1}{2} \lVert \textbf{w} \rVert_1$$. In almost all cases, $$L^2$$ regularization outperforms $$L^1$$ regularization.
### Early Stopping
When we are working with a dataset, we split that dataset up into testing and training datasets. The training dataset is used to adjust the weights of the network. The test dataset is used to check the accuracy of the model on data that has never been seen before.
However, the training dataset can be divided again into the training data and a small subset of data called the validation set. The validation set is used during training to ensure that the model is not overfitting. (You can think of the validation set as lying somewhere between the training set, which is used at every step of the training process, and the test set, which is only used at the very end of the training process to evaluate the model's generalizability.) This validation data is not ever used to train the model. The validation accuracy refers to the model's accuracy over the validation set. The goal is to minimize the validation accuracy through tuning hyperparameters of the network. The network is only evaluated on the test dataset with the fully tuned model.
Take a look at the below graph showing validation loss versus training loss. It should be clear that at a certain point, the model overfits on the training data and begins to suffer in validation accuracy despite this not being reflected in the training accuracy.
The solution to this is to simply stop training once the validation set loss has not improved for some time. Just like $$L^1$$ and $$L^2$$ regularization, this is a method of decreasing overfitting on the training dataset.
### Ensemble Methods
Bagging (short for bootstrap aggregation, a term in statistics) is the technique of making a model generalize better by combining multiple weaker learners into a stronger learner. Using this technique, several models are trained separately and their results are averaged for the final result. This ideal of one model being composed of several independent models is called an ensemble method. Ensemble methods are a great way to fine tune your model to make it generalize better on test data. Ensemble methods apply to more than just neural networks, and can be used on any machine learning technique. Almost all machine learning competitions are won using ensemble methods. Often times, these ensembles can be comprised of dozens and dozens of learners.
The idea is that if each model is trained independently of each other, they will have their own errors on the test set. However, when the results of the ensemble learners are averaged the error should approach zero.
Using bagging, we can even train a multiple models on the same dataset but be sure that the models were trained independently. With bagging, $$k$$ different datasets of the same size are constructed from the original dataset for $$k$$ learners. Each dataset is constructed by sampling from the original dataset with some probability with replacement. So there will be duplicate and missing values in the constructed dataset.
Furthermore, differences in model initialization and hyperparameter tuning can make ensembles of neural networks particularly favorable.
### Dropout
Dropout is a very useful form of regularization when used on deep neural networks. At a high level, dropout can be thought of randomly removing neurons from some layer of the network with a probability $$p$$. Removing certain neurons helps prevent the network from overfitting.
In reality, dropout is a form of ensemble learning. Dropout trains an ensemble of networks where various neurons have been removed and then averages the results, just as before. Below is an image that may help visualize what dropout does to a network.
Dropout can be applied to input units and hidden units. The hyperparameter of dropout at a given layer is the probability with which a neuron is dropped. Furthermore, another major benefit of dropout is that the computational cost of using it is relatively low. Finding the correct probability will require parameter tuning because a probability too low and dropout will have no effect, while too high and the network will be unable to learn anything.
Overall, dropout makes more robust models and is a standard technique employed in deep neural networks.
As a final section, let's go over a problem in optimization that plagued the deep learning community for decades.
As we've seen, deeper neural networks can be a lot more powerful than their shallow counterparts. Deeper layers of neurons add more layers of abstraction for the network to work with. Deep neural networks are vital to visual recognition problems. Modern deep neural networks built for visual recognition are hundreds of layers deep.
However, you may think that you can take what you have learned so far build a very deep neural network and expect it to work. However, to your surprise you may see that adding more layers at a certain point does not seem to help and even reduces the accuracy. Why is this the case?
The answer is in unstable gradients. This problem plagued deep learning up until 2012, and its relatively recent solutions are responsible for much of the deep learning boom. The cause of the unstable gradient problem can be formulated as different layers in the neural network having vastly different learning rates. And this problem only gets worse with the more layers that are added. The vanishing gradient problem occurs when earlier layers learn slower than later layers. The exploding gradient problem is the opposite. Both of these issues deal with how the errors are backpropagated through the network.
Let's recall the equation for backpropagating the error terms through the network. $$\delta^{m} = f'(\textbf{z}^{m}) \left( \textbf{W}^{m + 1} \right)^{T} \delta^{m+1}$$ Now let's say the network has 5 layers. Let's compute the various error terms recursively through the network. $$\delta^5 = \frac{\partial J}{\partial \textbf{z}^5}$$ $$\delta^4 = f'(\textbf{z}^4)(\textbf{W}^{5})^T \frac{\partial J}{\partial \textbf{z}^5}$$ $$\delta^3 = f'(\textbf{z}^3)(\textbf{W}^{4})^T f'(\textbf{z}^4)(\textbf{W}^{5})^T \frac{\partial J}{\partial \textbf{z}^5}$$ $$\delta^2 = f'(\textbf{z}^2)(\textbf{W}^{3})^T f'(\textbf{z}^3)(\textbf{W}^{4})^T f'(\textbf{z}^4)(\textbf{W}^{5})^T \frac{\partial J}{\partial \textbf{z}^5}$$ $$\delta^1 = f'(\textbf{z}^1)(\textbf{W}^{2})^T f'(\textbf{z}^2)(\textbf{W}^{3})^T f'(\textbf{z}^3)(\textbf{W}^{4})^T f'(\textbf{z}^4)(\textbf{W}^{5})^T \frac{\partial J}{\partial \textbf{n}^5}$$ The term for $$\delta^1$$ is massive, and this is only for a five layer deep network. Imagine what it would be for a 100 layer deep network! The important take away is that all of the terms are being multiplied together in a giant chain.
For a while, the sigmoid function was believed to be a powerful activation function. The sigmoid function and its derivative are shown below.
Say we were using the sigmoid function for our five layer neural network. That would mean that $$f'$$ is the function shown in red. What is the maximum value of that function? It's around 0.25. What types of values are we starting with for the weights? Small random values. The key here is that the values start small. The cause of the vanishing gradient problem should now start becoming clear. Because of the chain rule, we are recursively multiplying by terms less far less than one, causing the error terms to shrink and shrink going backwards in the network.
With this many multiplication terms, it would be something of a magical balancing act to manage all the terms so that the overall expression does not explode or shrink significantly.
How do we fix this problem? The answer is actually pretty simple. Just use the ReLU activation function instead of the sigmoid activation function. The ReLU function and its derivative are shown below.
As you can see, its derivative is either 0 or 1, which alleviates the unstable gradient problem. This function is also much easier to compute.
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# 1.3 Calculating the Present Value
## Learning Objectives
• Calculate present value for compound interest
## Formula & Symbol Hub
#### Symbols Used
• $f$ or $EFF$ = Effective interest rate
• $FV$ = Future value or maturity value
• $i$ = Periodic interest rate
• $j$ or $I/Y$ = Nominal interest rate per year
• $m$ or $C/Y$ = Number of compounds per year or compounding frequency
• $n$ or $N$ = Total number of compound periods for the term
• $PV$ = Present value of principal
#### Formulas Used
• ##### Formula 1.1 – Total Number of Compounds
$n=m \times \mbox{time in years}$
• ##### Formula 1.2 – Periodic Interest Rate
$i=\frac{j}{m}$
• ##### Formula 1.3 – Future Value
$FV=PV \times (1+i)^n$
• ##### Formula 1.4 – Present Value
$PV=FV \times (1+i)^{-n}$
## Introduction
The principal of the loan or investment is called the present value ($PV$). The present value is the amount of money borrowed for a loan or the amount of money invested for an investment at the start of the term. The present value is the amount at some earlier point in time than when the future value is known, and so excludes the future interest.
## $\boxed{1.4}$ The Present Value Formula
By solving for $PV$ in the future value formula from the previous section, the present valuefor compound interest is
$\Large{\color{red}{F}}{\color{red}{V}}={\color{blue}{P}}{\color{blue}{V}}\times(1+{\color{green}{i}})^{\color{purple}{n}}$
${\color{red}{PV}}\;\text{is the Present Value:}$ the present value is the starting amount upon which compound interest is calculated.
${\color{blue}{FV}}\;\text{is the Future Value:}$ the future value includes the principal plus all of the interest accumulated over the term.
${\color{green}{i}}\;\text{is the Periodic Interest Rate:}$ the periodic interest rate is the interest rate per compounding period: $i=\frac{j}{m}$ where $j$ is the nominal interest rate and $m$ is the compounding frequency.
${\color{purple}{n}}\;\text{is the Total Number of Compounding Periods Over the Term:}$ $n=m \times t$ where $m$ is the compounding frequency and $t$ is the length of the term in years.
### Example 1.3.1
Castillo’s Warehouse will need to purchase a new forklift for its warehouse operations three years from now, when its new warehouse facility becomes operational. If the price of the new forklift is $\38,000$ and Castillo’s can invest its money at $7.25\%$ compounded monthly, how much money should it put aside today to achieve its goal?
Solution
The timeline for the investment is shown below.
Step 1: Write what you get from the question.
$\begin{eqnarray*} FV & = & \38,000 \\ j & = & 7.25\% \\ m & = & 12 \\ t & = & 3 \mbox{ years} \end{eqnarray*}$
Step 2: Calculate the periodic interest rate.
$\begin{eqnarray*} i & = & \frac{j}{m} \\ & = & \frac{7.25\%}{12} \\ & = & 0.60416...\% \\ & = & 0.0060416...\end{eqnarray*}$
Step 3: Calculate the total number of compoundings.
$\begin{eqnarray*} n & = & m \times t \\ & = & 12 \times 3 \\ & = & 36 \end{eqnarray*}$
Step 4: Calculate the present value.
$\begin{eqnarray*} PV & = & FV \times (1+i)^{-n} \\ & = & 38,000 \times (1+0.0060416...)^{-36} \\ & = & \30,592.06 \end{eqnarray*}$
Step 5: Write as a statement.
If Castillo’s Warehouse places $\30,592.06$ into the investment, it will earn enough interest to grow to $\38,000$ three years from now to purchase the forklift.
## Using a Financial Calculator
As in the previous section, a financial calculator can be used to solve for the present value in compound interest problems. You use the financial calculator in the same way as described previously, but the only difference is that the unknown quantity is $PV$ instead of $FV$.You must still load the other six variables into the calculator and apply the cash flow sign convention carefully.
## Using the TI BAII Plus Calculator to Find the Present Value for Compound Interest
The time value of money buttons are located in the $TVM$ row (the third row from the top) of the calculator. The five buttons located on the third row of the calculator are five of the seven variables required for time value of money calculations. This row’s buttons are different in colour from the rest of the buttons on the keypad. The other two variables are in a secondary menu above the $I/Y$ key and are accessed by pressing 2nd $I/Y$.
Altogether, there are seven variables required to complete time value of money calculations. Note that $P/Y$ and $C/Y$ are not main button keys in the $TVM$ row. The P/Y and C/Y variables are located in the secondary function accessed by pressing 2nd $I/Y$.
Variable Meaning N Total number of compounding periods. This is the same value as $n$ in the future value/present value formulas: $N=\mbox{time in years} \times \mbox{compounding frequency}$ I/Y Interest rate per year (i.e. the nominal interest rate). The interest rate is entered in percent form (without the % sign). For example, $5\%$ is entered as $5$. PV Present value or principal. PMT Periodic annuity payment. For compound interest only calculations, PMT=$0$. (Note: in later chapters you will learn about annuities where PMT will not be $0$.) FV Future value or maturity value. P/Y Payment frequency for annuity payment. For compound interest only calculations, P/Y is set to the same value as C/Y. (Note: in later chapters you will learn about annuities where P/Y will be set to the frequency of the payments.) C/Y Compounding frequency. This is the value of $m$.
To enter values into the calculator:
• For the main button keys in the $TVM$ row (i.e. $N$, $I/Y$, $PV$, $PMT$, $FV$), enter the number first and then press the corresponding button.
• For example, to enter $N=34$, enter $34$ on the calculator and then press $N$.
• For $P/Y$ and $C/Y$, press 2nd $I/Y$. At the $P/Y$ screen, enter the value for $P/Y$ and then press ENTER. Press the down arrow to access the $C/Y$ screen. At the $C/Y$ screen, enter the value for $C/Y$ and then press ENTER. Press 2nd QUIT (the CPT button) to exit the menu.
• For example, to enter $P/Y=4$ and $C/Y=4$, press 2nd $I/Y$. At the $P/Y$ screen, enter $4$ and press ENTER. Press the down arrow. At the $C/Y$ screen, enter $4$ and press ENTER. Press 2nd QUIT to exit.
After all of the known quantities are loaded into the calculator, press CPT and then $PV$ to solve for the present value.
Video: Compound Interest (Present and Future Values) by Joshua Emmanuel [6:56] (Transcript Available).
### Example 1.3.2
Castillo’s Warehouse will need to purchase a new forklift for its warehouse operations three years from now, when its new warehouse facility becomes operational. If the price of the new forklift is $\38,000$ and Castillo’s can invest its money at $7.25\%$ compounded monthly, how much money should it put aside today to achieve its goal?
Solution
The timeline for the investment is shown below.
N $12 \times 3=36$ PV ? FV $38,000$ PMT $0$ I/Y $7.25$ P/Y $12$ C/Y $12$
$\displaystyle{PV=\30,592.06}$
### Try It
1) A debt of $\37,000$ is owed $21$ months from today. If prevailing interest rates are $6.55\%$ compounded quarterly, what amount should the creditor be willing to accept today?
Solution
N $4 \times 1.75=7$ PV ? FV $-37,000$ PMT $0$ I/Y $6.55$ P/Y $4$ C/Y $4$
$\displaystyle{PV=\33,023.56}$
## Present Value Calculations with Variable Changes
Addressing variable changes in present value calculations follows the same techniques as future value calculations discussed in the previous section. You must break the timeline into separate time segments, each of which involves its own calculations. Solving for the unknown $PV$ at the left of the timeline means you must start at the right of the timeline. You must work from right to left, one time segment at a time using the formula for $PV$ each time. Note that the present value for one time segment becomes the future value for the next time segment to the left.
1. Read and understand the problem. Identify the future value. Draw a timeline broken into separate time segments at the point of any change. For each time segment, identify any principal changes, the nominal interest rate, the compounding frequency, and the length of the time segment in years.
2. Starting with the future value in the last time segment (starting on the right), solve for the present value.
3. Let the present value calculated in the previous step become the future value for the next segment to the left. If the principal changes, adjust the future value accordingly.
4. Calculate the present value of the next time segment.
5. Repeat the previous steps until you obtain the final present value from the leftmost time segment.
### HOW TO
To use your calculator efficiently in working through multiple time segments, follow a procedure similar to that for future value.
1. Load the calculator with all known compound interest variables for the last time segment (on the right).
2. Compute the present value at the beginning of the segment.
3. With the answer still on your display, adjust the principal if needed, change the cash flow sign by pressing the $\pm$ key, and then store the unrounded number back into the future value button by pressing $FV$. Change the $N$, $I/Y$, and $C/Y$ as required for the next segment.
4. Return to step $2$ for each time segment until you have completed all time segments.
### Example 1.3.3
Sebastien needs to have $\9,200$ saved up three years from now. The investment he is considering pays $7\%$ compounded semi-annually, $8\%$ compounded quarterly, and $9\%$ compounded monthly in successive years. To achieve his goal, how much money does he need to place into the investment today?
Solution
The timeline shows today through to the future value three years from now
Step 1: Calculate the present value at the start of the last segment on the right.
N $12 \times 1=12$ PV ? FV $9,200$ PMT $0$ I/Y $9$ P/Y $12$ C/Y $12$
$\displaystyle{PV_1=\8,410.991...}$
Step 2: Calculate the present value at the start of the second segment on the right. The present value from the first step becomes the future value for the second step: $PV_1=\8,410.991...=FV_2$.
N $4 \times 1=4$ PV ? FV $8,410.991...$ PMT $0$ I/Y $8$ P/Y $4$ C/Y $4$
$\displaystyle{PV_2=\7,770.4555...}$
Step 3: Calculate the present value at the end of the third segment on the right. The present value from the second step becomes the future value for the third step: $PV_2=\7,770.455...=FV_3$.
N $2 \times 1=2$ PV ? FV $7,770.455..$ PMT $0$ I/Y $7$ P/Y $2$ C/Y $2$
$\displaystyle{PV_3=\7,253.80}$
Step 4: Write as a statement.
Sebastien needs to place $\7,253.80$ into the investment today to have $\9,200$ three years from now.
### Try It
2) For the first $4.5$ years, a loan was charged interest at $4.5\%$ compounded semi-annually. For the next $4$ years, the rate was $3.25\%$ compounded annually. If the maturity value was $\45,839.05$ at the end of the $8.5$ years, what was the principal of the loan?
Solution
N $4$ $9$ PV $\color{blue}{40,334.378...}$ $\color{blue}{33,014.56}$ FV $-45,839.05$ $-40,334.378...$ PMT $0$ $0$ I/Y $3.25$ $4.5$ P/Y $1$ $2$ C/Y $1$ $2$
$\displaystyle{PV=\33,014.56}$
### Section 1.3 Exercises
1. A loan is repaid with $\14,000$. If the loan was taken out $14$ years ago at $9\%$ compounded semi-annually, how much money was borrowed? How much interest was paid on the loan?
Solution
$PV=\4,081.99$, $I=\9,9180.01$
2. In $9$ years and $3$ months, you want to have $\97,000$ in your savings account. How much money must you invest today if the savings account earns $6\%$ compounded monthly?
Solution
$\55,762.07$
3. Eight and a half years ago, Tom took out a loan. The interest rate on the loan was $4.5\%$ compounded semi-annually for the first four and half years and $3.25\%$ compounded annually for the last four years. Tom repaid the loan today with a payment of $\45,839.05$. How much money did Tom borrow? How much interest did Tom pay?
Solution
$PV=\33,014.56$, $I=\12,824.49$
4. George wants to invest some money today. In $6.5$ years, George wants to have $\7,223.83$ in his investment. The investment earns $8.05\%$ compounded semi-annually for the first $2$ years and $6$ months, then $7.95\%$ compounded quarterly for $1$ year and $3$ months, and then $7.8\%$ compounded monthly for $2$ years and $9$ months. How much money does George need to invest?
Solution
$\4,340$
5. Dovetail Industries needs to save $\1,000,000$ for new production machinery that it expects will be needed six years from today. If money can earn $8.35\%$ compounded monthly, how much money should Dovetail invest today?
Solution
$\606,976.63$
6. A debt of $\37,000$ is owed $21$ months from today. If prevailing interest rates are $6.55\%$ compounded quarterly, what amount should the creditor be willing to accept today?
Solution
$\33,023.56$
7. Rene wants to invest a lump sum of money today to make a $\35,000$ down payment on a new home in five years. If he can place his money in an investment that will earn $4.53\%$ compounded quarterly in the first two years followed by $4.76\%$ compounded monthly for the remaining years, how much money does he need to invest today?
Solution
$\27,736.24$
8. In August 2004, Google Inc. made its initial stock offering. The value of the shares grew to $\531.99$ by July 2011. What was the original value of a share in August 2004 if the stock has grown at a rate of $26.8104\%$ compounded monthly?
Solution
$\85$
9. If a three-year and seven-month investment earned $\8,879.17$ of interest at $3.95\%$ compounded monthly, what amount was originally placed into the investment?
Solution
$\58,499.97$
10. A lottery ticket advertises a $\1$ million prize. However, the fine print indicates that the winning amount will be paid out on the following schedule: $\250,000$ today, $\250,000$ one year from now, and $\100,000$ per year thereafter. If money can earn $9\%$ compounded annually, what is the value of the prize today?
Solution
$\836,206.54$
• Option A: $\520,000$ today plus $\500,000$ one year from now.
• Option B: $\200,000$ today, $\250,000$ six months from now, and $\600,000 15$ months from now.
• Option C: $\70,000$ today, $\200,000$ in six months, then four quarterly payments of $\200,000$ starting six months later.Your company is selling some real estate and has received three potential offers:
11. Prevailing interest rates are expected to be $6.75\%$ compounded semi-annually in the next year, followed by $6.85\%$ compounded quarterly afterwards. Rank the three offers from best to worst based on their values today.
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## Week 3: Assignment: 3.3
NCC is a leading organization in the fruit industry and it has failed to find the bottleneck in its Receiving Plant 1. It is faced with the challenge of long queues at RP1 leading to reduced operational efficiency. This paper aims to determine the bottleneck of its operations, calculate overtime, and the time wasted while trucks are waiting to unload the berries. It will further provide recommendations on the best ways to improve the operations at the organization.
1. Process Flow Diagram:
• Determine the bottleneck of the operation(s)
Number of dumpers = 5
Average time used = 7.5 minutes
Average = 75 barrels
Therefore, the total dumping capacity,
= 75* 5 *(60) / 7.5
=3000 barrels/hour
Individual bins’ capacity 1-16 = 250
So, total Individual bins’ capacity 1-16 = 6*250
= 1,500 barrels
Individual bins’ capacity 17-24 = 250
So, Individual bins’ capacity 17-24 = 8*250
= 2,000 barrels
Individual bins’ capacity 25, 26 and 27 = 400
So, Individual bins’ capacity 25, 26 and 27 = 3*400
= 1,200 barrels
Total dryers = 3
One dryer capacity/hour = 200 barrels
Total capacity = 600 barrels/hour
Available Jumbo separators = 3
Capacity for one separator/hour = 400 barrels
Total capacity/hour = 1,200 barrels
Wet berries (de-chaffing capacity/hour) = 3,000 barrels
Dry berries (de-chaffing capacity/hour) = 1,500 barrels
Total de-chaffing capacity/hour = 4,500 barrels
Therefore, drying capacity is the bottleneck operation because it has the lowest capacity per hour (600 barrels).
• Overtime Required
Operations commence at 11.00 AM while dumping bins start at 7.00 AM. The time difference for the two operations is 4 hours.
Dry Berries
In one hour, there is 450 barrels
So in 4 hours, there will be,
4 hours* 450 barrels = 1,800 barrels
Dumping bins total capacity = 4,000 barrels.
This means that within the 4 hours, there is no hitch in the processing of dry berries.
Wet Berries
In one hour, there is 1050 barrels
Dryer capacity/hour = 600 barrels
So in 4 hours, there will be,
4 hours* 1050 barrels = 4,200 barrels
Dumping bins total capacity = 3,200 barrels
From the given data, the drying capacity is less than the required capacity and thus there is waiting in order for trucks to unload. There is hourly added 450 barrels after 11.00 AM. This means that for a day, the barrels,
= 4,200 + (8 *450)
= 7,800 barrels (wet berries)
Total barrels in truck at end of day = 4,600 barrels
Total barrels in wet dumping bins at end of day = 3,200 barrels
Total time to unload = 4,600/600,
Where 600 is the processing rate.
= 7.67 hours
This means that the earliest time for all the trucks to unload is 2.40 AM the following day.
Loading process further takes another 5 and a half hours to drive the process to 8.00 AM the next day. Therefore, the total overtime during peak days is approximately 13 hours.
Overtime = 13 hours
• Truck Delay and Time Lost
Assuming that a truck will unload within 7.5 minutes and 18 trucks unload per hour, there will a back-up challenge of approximately 1 hour 15 minutes. Consequently, considering that truck 1 arrives at 7.00 Am and plant operations begin at 11.00AM, truck one waits for at least 4 hours to unload.
• Recommendations
• This case strongly recommends a re-scheduling of the arrival time for the trucks. According to calculations;
Total deliveries= 610,185 barrels
Standard deliveries/truck = 20-400 barrels
Average = 75 barrels
Therefore, the total truck load per season will be,
= 610,185/75
= 8,136
Time used to dump berries = 5-10 minutes
So, average time used = 7.5 minutes
So in a day,
= (8 hours * 60 minutes) /7.5 minutes
=64
=64 trucks per dumper in a day
Total dumpers = 5
Total trucks/day = 64 *5
=320
So, in one hour = 320/8
No. of trucks that can be scheduled per hour = 40 trucks/hr.
1. The organization should also better re-schedule its labor to reduce the total overtime cash paid.
At Receiving Plant 1, there is an approximately 12,000 overtime labor hours.
The average normal time pay rate = \$6.50 per hour
Therefore,
Total overtime pay = \$39,000
1. Introduction of 2 additional dryers at the Receiving plant No. 1 (RP1). This will help lower the output time as well as in increasing the volume.
Currently, there are 3 dryers with a capacity of 600bbls per hour.
Therefore, 5 dryers will be 1000bbls/hour
3 dryers utilize 175% (bottleneck of the process)
• dryers = 105%, (not the bottleneck of the process anymore)
• The dechaffing process should commence at 7.00 AM and not 11.00 AM, especially in peak periods. This is important in reducing overhead cost through the reduction of the unloading time.
• Acquire light meter systems
From calculations,
Net benefit = premium saving cost – cost of the system
168750- 20000
Net benefit= \$148,750
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# Lars Eighner's Homepage
## LarsWiki
### Comments for The Derivative of Arc Cotangent
Cotangent
##### Contents
A previous result is: {$${d \over {d\theta}} \cot\theta = - \csc^2\theta$$}
Attention is called to the diagram which illustrates the Pythagorean identity:
{$$csc^2\theta = 1 + cot^2\theta$$}
{$y = \cot(x)$} and {$y=\operatorname{arccot}(x)$}
The demonstration proceeds by implicit differentiation:
{\begin{align} \cot(\operatorname{arccot}\theta) &= \theta \tag{definition of inverse} \cr \text{Let } y &= \operatorname{arccot}\theta \cr \cot(y) &= \theta \cr {d \over {d\theta}} \left( \cot(y) \right) &= {d \over {d\theta}} \theta \cr {d \over {d\theta}}cot(y){d \over {d\theta}}y &= 1 \tag{Chain rule} \cr -\csc^2{d \over {d\theta}}y &= 1 \tag{previous result} \cr {d \over {d\theta}}y &= - {1 \over {csc^2(y)}} \cr {d \over {d\theta}}y & = - {1 \over {1 + cot^2(y)}} \tag{Pythagorean identity} \cr {d \over {d\theta}}\operatorname{arccot}\theta &= - {1 \over {1 + cot^2(\operatorname{arccot}\theta)}} \tag{value y} \cr \therefore \quad {d \over {d\theta}}\operatorname{arccot}\theta &= - {1 \over {1 + \theta^2}}\end{align}}
{$y = \cot(x)$}, {$y=\operatorname{arccot}(x)$}, {$y = \cot(x)$} and {$y=\operatorname{arccot}(x)$}.
Sources:
Recommended:
Category: Math Calculus Trigonometry
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### December 24, 2018
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# What is the mistake in this solution?
Problem: Solve the initial value problem $$y’=3y; y(0)=a>0$$.
My solution: Exploiting variable separable method, $${dy\over y}=3\, dx$$ $$\implies\ln |y|=3x+C$$ where $$C$$ is the constant of integration. Now using $$y(0)=a$$ and $$a>0$$, $$\ln a=C$$. $$\therefore\ln |y|=3x+\ln a$$ $$\implies |y|=e^{3x+\ln a}$$ $$\implies y=\pm ae^{3x}$$
But clearly $$y=-ae^{3x}$$ isn’t a solution for the given problem.
So what went wrong?
There is no error, as $$-ae^{3x}$$ is not a solution, violating the initial condition, and $$+ae^{3x}$$ is a solution, you have found the unique solution.
It is always possible that some non-equivalent step increases the number of intermediate solution candidates, so that you have to compare back with the original problem to select the true solutions.
You could also have done the preliminary analysis of the ODE in that $$y=0$$ is a constant, stationary solution of the ODE, so that by uniqueness no other solution can change its sign (as a sign change would imply the crossing of two solutions). Then because of $$y(0)=a>0$$, the solution of the IVP is positive, so that you can resolve the absolute value in $$\ln|y|$$ immediately to $$\ln y$$, removing the ambiguity.
Or staying with your solution method, observe that by dividing by $$y$$ the result is only valid as long as $$y\ne 0$$ to avoid division by zero. By the initial condition this means the obtained solution is only valid where $$y>0$$, again giving $$|y|=y$$. In the finished solution you can then find that the solution function has no zero crossings, thus is valid everywhere.
• Can you please explain why $y=0$ being a solution for the ODE implies that no solution can change sign. – Atom Jul 20 at 13:58
• Because the right side is smooth, all solutions are (locally) unique. If some solution is zero somewhere, it is already the zero solution everywhere. – LutzL Jul 20 at 14:17
• Not getting your second statement. And how does it lead to what I’m asking? Please help. – Atom Jul 20 at 14:34
• Assume you have some arbitrary solution where you only know that it has a zero somewhere. Then by uniqueness you automatically know that it is already the zero solution. Changing sign requires by IVT that there is a zero somewhere, which is thus impossible. – LutzL Jul 20 at 14:39
• No, what I'm saying is that a scalar ODE of the form $y'=f(y)=yg(y)$ has the zero function as solution and thus any IVP with an non-zero initial value can have no zeros. More generally, in an autonomous scalar ODE $y'=f(y)$ any solution is bounded by the stationary solutions at the roots of $f(y)=0$, if an initial value is between two such roots, then the full solution stays between these roots. – LutzL Jul 20 at 14:53
The implications are only flowing forward, not backwards. You know, from your argument, $$((y' = 3y) \land (y(0) = a > 0)) \implies (y = ae^{3x}) \lor (y = -ae^{3x}).$$ You don't know the converse: $$(y = ae^{3x}) \lor (y = -ae^{3x}) \implies ((y' = 3y) \land (y(0) = a > 0)).$$ In fact, the converse is false, as you noted. If $$y = -ae^{3x}$$, then the right side is not satisfied.
This is a common thing in many equation-solving methods: an infinite set of potential solutions is reduced to a small finite set of possible solutions, but none of the potential solutions are actually proven by the method. There may even be no solutions! It is important to verify the potential solutions you obtain from such a solving method, otherwise you might settle on erroneous possible solutions.
• I remember once trying to prove the existence of a solution of an ODE. A friend (and a more senior mathematics student) plugged it into Matlab to obtain an approximate solution, and argued that this proved the existence of a solution. This was an example of a similar fallacy. Most numerical procedures for approximating solutions to DEs tend to blindly assume that solutions exist. Just because something that looked like a suitable curve popped out of the algorithm, doesn't mean that there was a solution that was being approximated. – Theo Bendit Jul 20 at 13:44
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