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Stacking cannonballs – solving maths with code
Numberphile have recently done a video looking at the maths behind stacking cannonballs – so in this post I’ll look at the code needed to solve this problem.
Triangular based pyramid.
A triangular based pyramid would have:
1 ball on the top layer
1 + 3 balls on the second layer
1 + 3 + 6 balls on the third layer
1 + 3 + 6 + 10 balls on the fourth layer.
Therefore a triangular based pyramid is based on the sum of the first n triangular numbers.
The formula for the triangular numbers is:
and the formula for the sum of the first n triangular numbers is:
We can simplify this by using the identity for the sum of the first n square numbers and also the identity for the sum of the first n natural numbers:
Therefore:
and the question we want to find out is whether there is triangular based pyramid with a certain number of cannonballs which can be rearranged into a triangular number i.e.:
here n and m can be any natural number. For example if we choose n = 3 and m = 4 we see that we have the following:
Therefore we can have a triangular pyramid of height 3, which has 10 cannonballs. There 10 cannonballs can then be rearranged into a triangular number.
Square based pyramids and above.
For a square based pyramid we would have:
1 ball on the top layer
1 + 4 balls on the second layer
1 + 4 + 9 balls on the third layer
1 + 4 + 9 + 16 balls on the fourth layer.
This is the sum of the first n square numbers. So the formula for the square numbers is:
and the sum of the first n square numbers is:
For a pentagonal based pyramid we have:
1 ball on the top layer
1 + 5 balls on the second layer
1 + 5 + 12 balls on the third layer
1 + 5 + 12 + 22 balls on the fourth layer.
This is the sum of the first n pentagonal numbers. So the formula for the pentagonal numbers is:
and the formula for the first n pentagonal numbers is:
For a hexagonal based pyramid we have:
The formula for the first n hexagonal numbers:
and the formula for the sum of the first n hexagonal numbers:
For a k-agon based pyramid we have
and the formula for the sum of the first n k-agon numbers:
Therefore the general case is to ask if a k-agonal pyramid can be rearranged into a k-agon number i.e:
Computers to the rescue
We can then use some coding to brute force some solutions by running through large numbers of integers and seeing if any values give a solution. Here is the Python code. Type it (taking care with the spacing) into a Python editor and you can run it yourself.
You can then change the k range to check larger k-agons and also change the range for a and b. Running this we can find the following. (The first number is the value of k, the second the height of a k-agonal pyramid, the third number the k-agon number and the last number the number of cannonballs used).
Solutions:
3 , 3 , 4 , 10
3 , 8 , 15 , 120
3 , 20 , 55 , 1540
3 , 34 , 119 , 7140
4 , 24 , 70 , 4900
6 , 11 , 22 , 946
8 , 10 , 19 , 1045
8 , 18 , 45 , 5985
10 , 5 , 7 , 175
11 , 25 , 73 , 23725
14 , 6 , 9 , 441
14 , 46 , 181 , 195661
17 , 73 , 361 , 975061
20 , 106 , 631 , 3578401
23 , 145 , 1009 , 10680265
26 , 190 , 1513 , 27453385
29 , 241 , 2161 , 63016921
30 , 17 , 41 , 23001
32 , 298 , 2971 , 132361021
35 , 361 , 3961 , 258815701
38 , 430 , 5149 , 477132085
41 , 204 , 1683 , 55202400
41 , 505 , 6553 , 837244045
43 , 33 , 110 , 245905
44 , 586 , 8191 , 1408778281
50 , 34 , 115 , 314755
88 , 15 , 34 , 48280
145, 162, 1191, 101337426
276, 26, 77, 801801)
322, 28, 86, 1169686
823, 113, 694, 197427385
2378, 103, 604, 432684460
31265, 259, 2407, 90525801730
For example we can see a graphical representation of this. When k is 6, we have a hexagonal pyramid with height 11 or the 22nd hexagonal number – both of which give a solution of 946. These are all the solutions I can find – can you find any others? Leave a comment below if you do find any others and I’ll add them to the list! |
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### Course: 6th grade foundations (Eureka Math/EngageNY)>Unit 4
Lesson 2: Topic B: Foundations
# Dividing a decimal by a power of 10
Sal talks about why moving the decimal point to the left when dividing by a power of 10 makes sense. Created by Sal Khan and Monterey Institute for Technology and Education.
## Want to join the conversation?
• If I had 72.3 times 23.5 how would I do that
• I am not going to judge how easy multiplying decimals is easy or not, but I will show you an O.P strategy of solving these kind of problems.
When solving these questions you should ignore the decimal points and multiply it like a whole number.
Then the number of decimals spaces from the original spaces will go to the answer.
That sounded pretty hard, but it's not.
Lets solve your problem 72.3 × 23.5.
SO lets ignore the decimals and multiply 723 × 235.
You should get 169905.
Now in the original equation there were two numbers and each of them had one place to the left so we will make it two places to the left because of the two numbers.
You should have gotten 1699.0
• How many times do you need to divide 4379.9 by ten to get 4.3799?
• You need to divide 4379,9 by 1000 to get 4,3799. That would mean you will need to divide it by 10 three times (10^3 = 1000).
An easy trick to remember, is to count how many spaces you have to move the point to the left. Here, you have to move it to the left 3 times (from 4379.9 to 4.3799), so you have to divide it by 10 three times.
Likewise, to go from 4.3799 to 4379.9, you have to move the point 3 spaces, but to the right this time. This means that you have to multiply 4.3799 by 10 three times to get 4379.9.
• Does that mean if you do 0.001 divided by 100 I would just move the decimal right 2 times?
• Move the decimal place left two times. 0.001 divided by 100 = 0.00001. Place value goes up and down by powers of 10. Every time you multiply by 10, you go up a place value and every time you divide by 10, you go down a place value. When you divide 0.001 by 100, you’re basically dividing 0.001 by 10 twice, so you go down two place values.
• Please show a pattern that is easy to understand .. When multiplying a decimal by a power of ten
• Well, a pattern is:
0.001 * 10 = 0.01
0.001 * 100 = 0.1
0.001 * 1000 = 1
So, every time you multiply a number by ten you move the decimal place 1 place value to the right. And the same applies to numbers you usually don't see with decimal points, such as 1, 2, 3, etc.:
1*10 = ?(I know you probably already know this, but I'm providing an example)
Remember:
1.0 * 10 =
10.0
• why do you keep dividing the same number?
• Yea I don't know why he dose that
• why do we get a lower number
• Because when we divide, we change the whole number into groups of the whole number. For example, 4 divided by 2 means that only 2 groups of 2 will fit into 4, so 2 is the answer. Hope this helps! :D
• Do you always need to put a zero after the number after the decimal? For example, when your product for an answer is 3.20, do you have to keep that extra zero? Please reply A.S.A.P.
• No, 3.20 can just be written as 3.2. The zero on the end does not change the number. |
If you find any difficulty in understanding three-digit numbers, here you can have good knowledge of three-digit numbers like its definition, comparison of three-digit numbers. Learn How to arrange 3 digit numbers in ascending order and descending order. For a better understanding of this concept check the examples on comparison of three-digit numbers.
## What are Three-Digit Numbers?
Three-digit numbers have only three digits. In three-digit numbers, the numbers are placed at one’s, ten’s, and hundred’s place. In the right of the number the last digit is one’s place, then the second digit is ten’s place and to the left of it, there is a hundred’s place. The digits have their face value in a given number. Three-digit numbers are from 100 to 999.
For example, in 635 the place value of 6 is 600, 3 is 30, and 5 is 5. In other words, we can write this as six hundred thirty-five.
### How to Compare Three-Digit Numbers?
Know the procedure on how to compare 3 digit numbers by going through the below-listed steps. They are along the lines
(i) The numbers which have less than three digits are always smaller than the numbers having three digits as:
128 > 73 , 120 > 7 or 7 < 120 , 58 < 158
175 > 65 , 529 > 59 , 703 > 8 , etc.
(ii) If both the numbers have the same number (three) of digits, then the digits of the hundred-place and tens place are compared.
a) If the third digit from the right (Hundred-place digit) of a number is greater than the third digit from the right (Hundred-place digit) of the other number then the number having the greater is the greater one.
855>713, 984>981, 100>9,100>99.
b) If the numbers have the same third digit from the right, then the digits at ten’s place are compared and follow the rules of comparison of two-digit numbers.
967 > 929 , 586 > 567 , 462 > 449
c) If the digits at Hundred-place and ten’s place are equal, then follow the rules of comparison of single-digit numbers.
958 > 953 , 876 > 872 , 634 > 631
### Comparing 3 Digit Numbers Examples
Example1:
Compare three digit numbers 534, 345
Solution:
In 534 ,5 is in hundreds place.
3 is in the tens place.
4 is in one’s place.
In 345,3 is in the hundreds place.
4 is in the tens place.
5 is in one’s place.
since two numbers have three digits compare the hundreds place of two numbers.
5 is greater than 3.
so 534 greater than(>) 345.
Example 2.
Compare three-digit numbers 583, 526
Solution:
In 583,5 is in the hundreds place.
8 is in the tens place.
3 is in one’s place.
In 526, 5 is in the hundreds place.
2 is in the tens place.
6 is in one’s place.
since the hundreds place of both the numbers are same compare tens place.
8>2
So 583>526.
Numbers can be arranged in two ways.1) Ascending order 2) Descending Order.
### Ascending Order
Ascending order means the numbers are arranged from smallest to largest. the smallest number comes first and then the largest numbers.
#### How to arrange 3-Digit Numbers in Ascending Order?
1. Arrange the numbers 100,150,567,120,852,480 in ascending order.
The numbers in ascending order are 100,120,150,480,567,852.
2. Arrange the numbers 354,764,120,967,534,423 in ascending order.
The numbers in ascending order are 120, 354, 423, 534, 764, 967.
3. Arrange the numbers 220,560,420,678,168,934 in ascending order.
The numbers in ascending order are 168, 220,420,560,678,934.
### Descending Order
Descending order means the numbers are arranged from largest to smallest. the largest number comes first and then the smallest numbers.
#### How to arrange 3-Digit Numbers in Descending Order?
1. Arrange the numbers 345,567,987,213,621,789 in descending order.
The numbers in descending order are 987,789,621,567,345,213.
2. Arrange the numbers 150,533,189,256,876,323 in descending order.
The numbers in descending order are 876,533,323,256,189,150.
3. Arrange the numbers 100,623,345,750,923,420 in descending order.
The numbers in descending order are 923, 750, 623, 420, 345, 100.
### FAQ’s on Three Digits Numbers Comparison
1. What is the greatest three-digit number?
The greatest three-digit number is 999.
2. What is the smallest three-digit number?
The smallest three-digit number is 100.
3. Is a three-digit number greater than any single-digit number?
Yes, any Three-digit number is greater than any single-digit number.
4. Is a three-digit number greater than any two-digit number?
Yes, a three-digit number is greater than any two-digit number. |
# Evaluate the following integrals:
Question:
Evaluate the following integrals:
Solution:
Given $I=\int \frac{x-1}{\sqrt{x^{2}+1}} d x$
Integral is of form $\int \frac{p x+q}{\sqrt{2 x^{2}+b x+c}} d x$
Writing numerator as $\mathrm{px}+\mathrm{q}=\lambda\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right)\right\}+\mu$
$\Rightarrow p x+q=\lambda(2 a x+b)+\mu$
$\Rightarrow x-1=\lambda(2 x)+\mu$
Let $x-1=1 / 2(2 x)-1$ and split,
$\Rightarrow \int \frac{\mathrm{x}-1}{\sqrt{\mathrm{x}^{2}+1}} \mathrm{dx}=\int\left(\frac{2 \mathrm{x}}{2 \sqrt{\mathrm{x}^{2}+1}}-\frac{1}{\sqrt{\mathrm{x}^{2}+1}}\right) \mathrm{dx}$
$=\int \frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+1}} \mathrm{dx}-\int \frac{1}{\sqrt{\mathrm{x}^{2}+1}} \mathrm{dx}$
Consider $\int \frac{x}{\sqrt{x^{2}+1}} d x$
Let $u=x^{2}+1 \rightarrow d x=\frac{1}{2 x} d u$
$\Rightarrow \int \frac{\mathrm{x}}{\sqrt{\mathrm{x}^{2}+1}} \mathrm{dx}=\int \frac{1}{2 \sqrt{\mathrm{u}}} \mathrm{du}$
$=\frac{1}{2} \int \frac{1}{\sqrt{\mathrm{u}}} \mathrm{du}$
We know that $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$
$\Rightarrow \frac{1}{2} \int \frac{1}{\sqrt{u}} \mathrm{du}=\frac{1}{2}(2 \sqrt{\mathrm{u}})$
$=\sqrt{\mathrm{u}}=\sqrt{\mathrm{x}^{2}+1}$
Consider $\int \frac{1}{\sqrt{x^{2}+1}} d x$
We know that $\int \frac{1}{\sqrt{x^{2}+1}} d x+c=\sinh ^{-1} x+c$
$\Rightarrow \int \frac{1}{\sqrt{x^{2}+1}} d x=\sinh ^{-1}(x)$
Then,
$\Rightarrow \int \frac{x-1}{\sqrt{x^{2}+1}} d x=\int \frac{x}{\sqrt{x^{2}+1}} d x-\int \frac{1}{\sqrt{x^{2}+1}} d x$
$=\sqrt{x^{2}+1}-\sinh ^{-1}(x)+c$
$\therefore I=\int \frac{x-1}{\sqrt{x^{2}+1}} d x=\sqrt{x^{2}+1}-\sinh ^{-1}(x)+c$ |
# Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 17
Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 17 Textbook Exercise Important Questions and Answers.
## Maharashtra State Board Class 5 Maths Solutions Chapter 5 Fractions Problem Set 17
Question 1.
Write the proper number in the box.
Here 20 = 2 x 10
Here 15 = 3 x 5
Here 18 = 9 x 2
hehce,
Here 40 + 5 = 8,
hence,
Here 26 ÷ 2 = 13,
hence,
Here 6 ÷ 2 = 3,
hence,
Here 4 ÷ 4 = 1,
hence,
Here 25 ÷ 5 = 5,
hence,
Question 2.
Find an equivalent fraction with denominator 18, for each of the following fractions.
Solution:
Question 3.
Find an equivalent fraction with denominator 5, for each of the following fractions.
Solution:
Question 4.
From the fractions given below, pair off the equivalent fractions.
Solution:
Question 5.
Find two equivalent fractions for each of the following fractions.
Solution:
Like fractions and unlike fractions
Fractions such as $$\frac{1}{7}, \frac{4}{7}, \frac{6}{7}$$ whose denominators are equal, are called ‘like fractions’.
Fractions such as $$\frac{1}{3}, \frac{4}{8}, \frac{9}{11}$$ which have different denominators are called unlike fractions’.
Converting unlike fractions into like fractions
Example (1) Convert 5/6 and 7/9 into like fractions.
Here, we must find a common multiple for the numbers 6 and 9.
Multiples of 6 : 6, 12, 18, 24, 30, 36, ……..
Multiples of 9 : 9, 18, 27, 36, 45 ……..
Here, the number 18 is a multiple of both 6 and 9. So, let us make 18 the denominator of both fractions.
Thus, 15/18 and 1418 are like fractions, respectively equivalent to 5/6 and 7/9.
Here, 18 is a multiple of both 6 and 9. We could also choose numbers like 36 and 54 as the common denominators.
Example (2) Convert 4/8 and 5/16 into like fractions.
As 16 is twice 8, it is easy to make 16 the common denominator.
$$\frac{4}{8}=\frac{4 \times 2}{8 \times 2}=\frac{8}{16}$$ Thus, 8/16 and 5/16 are the required like fractions.
Example (3) Find a common denominator for 4/7 and 3/4.
The number 28 is a multiple of both 7 and 4. So, make 28 the common denominator. . Therefore, 16/28 and 21/28 are the required like fractions. |
# 1095 in words
1095 in words is written as One Thousand and Ninety Five. 1095 represents the count or value. The article on Counting Numbers can give you an idea about count or counting. The number 1095 is used in expressions that relate to money, distance, length and others. Let us consider an example for 1095. “A famous singer has sung One Thousand and Ninety Five songs”. Another example would be ”The distance between Bangalore and Nashik is almost One Thousand and Ninety Five kilometers”.
1095 in words One Thousand and Ninety Five One Thousand and Ninety Five in Numbers 1095
## How to Write 1095 in Words?
We can convert 1095 to words using a place value chart. This can be done as follows. The number 1095 has 4 digits, so let’s make a chart that shows the place value up to 4 digits.
Thousands Hundreds Tens Ones 1 0 9 5
Thus, we can write the expanded form as:
1 × Thousand + 0 × Hundred + 9 × Ten + 5 × One
= 1 × 1000 + 0 × 100 + 9 × 10 + 5 × 1
= 1095
= One Thousand and Ninety Five.
1095 is the natural number that is succeeded by 1094 and preceded by 1096.
1095 in words – One Thousand and Ninety Five.
Is 1095 an odd number? – Yes.
Is 1095 an even number? – No.
Is 1095 a perfect square number? – No.
Is 1095 a perfect cube number? – No.
Is 1095 a prime number? – No
Is 1095 a composite number? – Yes
## Solved Example
1. Write the number 1095 in expanded form
Solution: 1 × 1000 + 0 × 100 + 9 × 10 + 5 × 1
We can write 1095 = 1000 + 0 + 90 + 5
= 1 × 1000 + 0 × 100 + 9 × 10 + 5 × 1
## Frequently Asked Questions on 1095 in words
### How to write 1095 in words?
1095 in words is written as One Thousand and Ninety Five.
### State True or False. 1095 is divisible by 5?
True. 1095 is divisible by 5.
### Is 1095 a perfect square number?
No. 1095 is not a perfect square number. |
# Convert from standard form to slope intercept form
To convert from standard form to slope intercept form you just need to solve for y using the standard form of the equation.
Example #1:
Convert the equation 4x + 2y = 12 to slope-intercept form
4x + 2y = 12
Subtract 4x from each side of the equation.
4x - 4x + 2y = 12 - 4x
0 + 2y = 12 - 4x
2y = 12 - 4x
2y = 12 + -4x
2y = -4x + 12
Divide each side of the equation by 2 so the equation can be written in slope-intercept form.
2y / 2 = (-4x + 12) / 2
y = -4x / 2 + 12 / 2
y = -2x + 6
The slope-intercept form is y = mx + b and m is the slope.
Comparing y = mx + b with y = -2x + 6, we can see that the slope is m = -2
The slope of the line is -2.
Example #2:
Convert the equation of -5x + 3y = 8 to slope intercept form
-5x + 3y = 8
Add 5x to each side of the equation.
-5x + 5x + 3y = 8 + 5x
0 + 3y = 8 + 5x
3y = 5x + 8
Divide each side of the equation by 3 so the equation can be written in slope-intercept form.
3y / 3 = (5x + 8) / 3
y = 5x / 3 + 8 / 3
y = (5 / 3)x + 8/3
The slope of the line is 5 / 3.
## Convert from standard form to the slope-intercept form: the general form
Example #3:
Find the slope of Ax + By = C
Ax + By = C
Subtract Ax from each side of the equation.
Ax - Ax + By = C - Ax
0 + By = C - Ax
By = C - Ax
By = C + -Ax
By = -Ax + C
Divide each side of the equation by B so the equation can be written in slope-intercept form.
By / B = (-Ax + C) / B
y = -Ax / B + C / B
y = (-A / B)x + C / B
The slope of the line is -A / B.
100 Tough Algebra Word Problems.
If you can solve these problems with no help, you must be a genius!
Recommended |
# Right Triangle Trigonometry
## Presentation on theme: "Right Triangle Trigonometry"— Presentation transcript:
Right Triangle Trigonometry
Pre-Calculus Day 1 Right Triangle Trigonometry
Today’s Objective Review right triangle trigonometry from Geometry and expand it to all the trigonometric functions Begin learning some of the Trigonometric identities
What You Should Learn Evaluate trigonometric functions of acute angles. Use fundamental trigonometric identities. Use a calculator to evaluate trigonometric functions. Use trigonometric functions to model and solve real-life problems.
Plan 4.1 Right Triangle Trigonometry Homework
Definitions of the 6 trig functions Reciprocal functions Co functions Quotient Identities Homework
Right Triangle Trigonometry
Trigonometry is based upon ratios of the sides of right triangles. The ratio of sides in triangles with the same angles is consistent. The size of the triangle does not matter because the triangles are similar (same shape different size).
The six trigonometric functions of a right triangle, with an acute angle , are defined by ratios of two sides of the triangle. θ hyp opp The sides of the right triangle are: adj the side opposite the acute angle , the side adjacent to the acute angle , and the hypotenuse of the right triangle.
Trigonometric Functions
θ hyp The trigonometric functions are opp adj sine, cosine, tangent, cotangent, secant, and cosecant. sin = cos = tan = csc = sec = cot = opp hyp adj Note: sine and cosecant are reciprocals, cosine and secant are reciprocals, and tangent and cotangent are reciprocals. Trigonometric Functions
Reciprocal Functions Another way to look at it…
sin = 1/csc csc = 1/sin cos = 1/sec sec = 1/cos tan = 1/cot cot = 1/tan
Given 2 sides of a right triangle you should be able to find the value of all 6 trigonometric functions. Example: 5 12
Example: Six Trig Ratios
Calculate the trigonometric functions for . Calculate the trigonometric functions for . 4 3 5 The six trig ratios are sin = sin α = cos = cos α = tan = What is the relationship of α and θ? tan α = cot = cot α = sec = They are complementary (α = 90 – θ) sec α = csc = csc α = Example: Six Trig Ratios
Example: Using Trigonometric Identities
Trigonometric Identities are trigonometric equations that hold for all values of the variables. We will learn many Trigonometric Identities and use them to simplify and solve problems. Example: Using Trigonometric Identities
Quotient Identities hyp opp θ adj adj sin = cos = tan = opp adj
The same argument can be made for cot… since it is the reciprocal function of tan.
Quotient Identities
Problem-Solving Strategies
Scenario 1) You are given 2 sides of the triangle. Find the other side and the two non-right angles. OR 20 c 15 13 k 12 1A. Use the Pythagorean theorem to find the 3rd side. 1B. Use an inverse trig function to get an angle. Then use that angle to calculate the 3rd angle. Sum of the angles = 180º
Example: Given 1 Trig Function, Find Other Functions
Example: Given sec = 4, find the values of the other five trigonometric functions of . Draw a right triangle with an angle such that 4 = sec = = . adj hyp θ 4 1 Use the Pythagorean Theorem to solve for the third side of the triangle. sin = csc = = cos = sec = = 4 tan = = cot = Example: Given 1 Trig Function, Find Other Functions
Problem-Solving Strategies
26 a Scenario 2) You are given an angle and a side. Find the other angle and the two other sides. 1A. Use 2 different trig ratios from the given angle to get each of the other two sides. 1B. Use the sum of the angles to get the 3rd angle.
Problem-Solving Strategies
24 25 7 Scenario 3) You are given all 3 sides of the triangle. Find the two non-right angles. 1. Use 2 different trig ratios to get each of the angles.
Homework for tonight Page 227: complete the even numbered problems. Show your work. Cover your textbook
Using the calculator Function Keys Reciprocal Key Inverse Keys
Using Trigonometry to Solve a Right Triangle
A surveyor is standing 115 feet from the base of the Washington Monument. The surveyor measures the angle of elevation to the top of the monument as 78.3. How tall is the Washington Monument? Figure 4.33
Applications Involving Right Triangles
The angle you are given is the angle of elevation, which represents the angle from the horizontal upward to an object. For objects that lie below the horizontal, it is common to use the term angle of depression.
Solution where x = 115 and y is the height of the monument. So, the height of the Washington Monument is y = x tan 78.3 115( ) 555 feet.
Homework Section 4.1, pp. 238 # 40 – 54 evens |
## September 24, 2006
### Scribe Post: Day 11
Good day to you all, I'm Christian, your scribe. Sorry I wasn't able
to post this right away. There were very important matters I had to
attend to AND I forgot how to put posts up. Thanks MARK for
reminding me how. Thanks to LINDSAY and CHARLOTTE too for teaching me how to put exponents and subscripts here. I'll go on with our lesson shortly.
INTRODUCTION TO LOGARITHMS '06
Over the week we learned about functions, inverse functions. It is
easy most of the time to solve for variables because we only need
to use simple operations. For example, when solving for x in the
equation
(x + 5) / 2 = 6, we:
Step 1: Multiply both sides by 2
x + 5 = 12
Step 2: Subtract 5 from both sides
x = 7
In this case we only used multiplication and subtraction to solve for x. What happens if the variable we're looking for is an exponent? We can use our basic operations as well, but first we must understand what logarithms are, and when we should use them. Logarithms are exponents. When we are solving for an exponent in an equation, we are essentially solving the logarithm for that exponent. Look at the general equation for logarithms:
Logba = c
For this one, b represents the base, a represents the power, and c the exponent. Notice that there is an equal sign between the log expression and c. What does that mean? LOGS ARE EXPONENTS! If logarithms are exponents, to work with a log, we must understand this: In theory, since logs are exponents, the rules of logs and exponents must be the same right? Yes. The rules of logarithms describe what's happening to the exponents. Hence we must understand the rules of exponents.
RULES OF EXPONENTS
We learned these rules in grade 12 pre-calculus. This will just be a quick review. As a note again, and to make things clear here, I followed up the "cases" with the general equation in logarithms. As I mentioned, the rules of logarithms describe what happens to the exponents. We simply apply the exponent rules, and from there we can already manipulate our logs.
Case 1: When we multiply terms with the same base, we add the exponents.
(bx)(by) = bx+y
Let's apply this to some value:
This rule states that:
(34)(35) = 34+5
= 39
In logs, this means that:
log
x + logy = log(xy)
Case 2: When we divide terms with the same base, we subtract the exponents.
(bx) / (by) = bx-y
Again, applying to specific values, this rule states that:
(35) / (34) = 35-4
= 31
In logs, this means that:
log(x/y) = logx - logy
Case 3: When we have a base raised to an exponent and the whole term is raised to another exponent, we multiply the exponents.
(bx)y = b(x)(y)
= bxy
The rule states that:
(3^4)5 = 3(4)(5)
= 320
In logs, this means that:
logb(x)y = (logbx)(logby)
= ylogb(x)
Case 4: The Change of Base Law. This law states that if we have a log written in any base, we can switch the bases. We can easily solve logs of base "10" and base "e", because these are programmed in our calculator.
Take out your calculator. Turn it on. Clear the screen. If we press the log button,
log( will come out right? What we do next is we enter the power, close the bracket, enter, and we get our value. What do you notice in terms of the structure of a log? We did NOT enter the base! What happens then if we are given bases of 2, 900, or 3.2562435234? We apply the Change of Base Law.
This law states that:
logba = log(a) / log(b)
For example:
log34 = log(4) / log (3)
YAY! Then we had bus ridership, and that was our class. Sorry again for posting late. Please visit this.
The next scribe is Anh.
Feel free to add and comment to this. Be nice, though, I have a very fragile ego.
christian said...
I FORGOT TO CHOOSE WHO THE NEXT SCRIBE SHOULD BE!!! Again, apologies. I can't find the edit button so I don't know how to change my post now. I'd just work with the assumption that when people see this post, they'd look at the comments too. So I guess I'll just mention here who it is. The next scribe is ANH.
lindsay said...
good job on your post christian! i needed a review of those log laws. i liked how you used colour in a meaningful way. it wasn't over the top and easy to understand! it was very organized which made it easy to follow. yay!! 1000 points for you christian!
be happy.
linger said...
HIII CHRISTIAN!
nice post. yes it was up late, but it's nicely done =) thanks for explaining it well. your explanation was clear, easy to understand and to follow. there's nothing confusing about logs on your post... one thing i have to add; in case 3 where you put(b^x)y = b ^(x)(y) , shouldn't it be b^(x) * b^(y)= b^(x)(y) ? so you're just missing a b before the first y, but other than that good job, keep it up ;)
linger said...
haha ok i just made a mistake. forget about the part of my comment with the b and the x's and y's. i just realized that the y is an exponent, so you don't need another b. oh myyy... ok well that's all -_- "
jann said...
Super job on the post Christian! Your post was well explained. It would really help me in understanding logs. Good job! |
ABCD is a square. Two arcs are drawn with A and B as centers, and radius equal to the side of square. If arcs intersects at point E, find the angle ∠ACE.
30°
Step by Step Explanation:
1. We are given, ABCD is a square with a diagonal AC.
We know,
• All the angles of a square is equal to 90°.
• The diagonal of a square bisects its angles.
So, ∠A = 90°
and ∠BAC = 45° (AC is a diagonal)
2. Now, join A and B to E.
We know,
AE is the radius of the circle with center A.
BE is the radius of the circle with center B.
AB is the side of the square.
We are given that radius is equal to the side of the square.
So, ΔABE is an equilateral triangle
⇒ Each angle of ΔABE is 60°. (All the angles of an equilateral triangle are 60°.)
⇒BAE = 60°
3. We can see,
∠CAE = ∠BAE − ∠BAC = 60° − 45° = 15°
4. As, ∠B = 90°
Reflex ∠B = 360° − 90° = 270° (Angle at a point is 360°)
Now, ∠CEA = 135° (The angle subtended by a chord at the center of the circle is double the angle subtended by the same chord at the circumference.)
5. Consider ΔCEA:
∠CEA = 135°
∠CAE = 15°
∠CEA + ∠CAE + ∠ACE = 180° (Sum of angles of a triangle are 180°)
135° + 15° + ∠ACE = 180°
∠ACE = 180° − (135° + 15°)
∠ACE = 180° − (150°)
∠ACE = 30°
You can reuse this answer |
## April 30, 2020
### Powerful Sequences
Number sequences suggest all sorts of intriguing puzzles and patterns.
Consider, for example, the sequence of counting numbers: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, …
Take out every second number, leaving just the odd integers: 1, 3, 5, 7, 9, 11, 13, 15, …
Then, form the cumulative totals of the odd numbers, as follows:
1, (1 + 3) = 4, (4 + 5) = 9, (9 + 7) = 16, (16 + 9) = 25, (25 + 11) = 36, (36 + 13) = 49, (49 + 15) = 64, …
Out pops the sequence of consecutive squares: 1 = 12, 4 = 22, 9 = 32, 16 = 42, 25 = 52, 36 = 62, 49 = 72, 64 = 82, and so on.
This seemingly magical transformation of one sequence into another was studied by German mathematician Alfred Moessner in the early 1950s. He found a host of such relationships between different number sequences.
Again starting with the sequence of counting numbers, suppose you take away every third number (multiples of 3): 1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, …
Then, add up what's left to get cumulative totals:
1, (1 + 2) = 3, (3 + 4) = 7, (7 + 5) = 12, (12 + 7) = 19, (19 + 8) = 27, (27 + 10) = 37, (37 + 11) = 48, (48 + 13) = 61, (61 + 14) = 75, (75 + 16) = 91, …
You end up with the following sequence: 1, 3, 7, 12, 19, 27, 37, 48, 61, 75, 91, …
Remove every second number in the new list: 1, 7, 19, 37, 61, 91, …
Calculate the cumulative totals of the remaining numbers. What do you end up with?
1, (1 + 7) = 8, (8 + 19) = 27, (27 + 37) = 64, (64 + 91) = 125, (125 + 91) = 216, …
You get the sequence of cubes: 1 = 13, 8 = 23, 27 = 33, 64 = 43, 125 = 53, 216 = 63, and so on.
If you go through the same procedure again, this time striking out every fourth number at the start, the result should now come as no surprise. You end up with the sequence of fourth powers: 1 = 14, 16 = 24, 81 = 34, 256 = 44, …
In general, taking out the nth number and following the appropriate procedure gives a sequence of nth powers. What happens if you take out the so-called triangular numbers:
1, (1 + 2) = 3, (1 + 2 + 3) = 6, (1 + 2 + 3 + 4) = 10, (1 + 2 + 3 + 4 + 5) = 15, …, (1 + 2 + 3 + 4 + . . . + n)
and as before, calculate cumulative totals, then take out the first, third, sixth, tenth, fifteenth, and so on numbers from the new list, then continue on, as above?
Notice that the unique numbers down the left-hand side are the factorial numbers: 1, (1 x 2) = 2, (1 x 2 x 3) = 6, (1 x 2 x 3 x 4) = 24, (1 x 2 x 3 x 4 x 5) = 120, or in general, (1 x 2 x 3 x ... x n). Somehow, the recipe turns addition into multiplication. For more information about a related sequence, see Moessner’s factorial triangle at the On-Line Encyclopedia of Integer Sequences.
I first came across these surprising sequence transformations when Richard K. Guy described them at a meeting on recreational mathematics held in 1986 at the University of Calgary. This material—and much, much more—is included in a fascinating book by Guy and John H. Conway titled The Book of Numbers (Springer, 1996).
If you want to stretch your mind from the integers to the surreal, this is the book to read!
Originally posted November 18, 1996
## April 29, 2020
### Punctured Polyhedra
A tetrahedron has four triangular faces, four vertices, and six edges.
Consider what happens when a vertex of one tetrahedron pierces the face of a second tetrahedron to form a new, more complicated polyhedron. In the resulting geometric form, one triangular face has a triangular "hole" where the face was pierced. Mathematicians describe such a punctured face as being "multiply connected."
A while ago, mathematician John H. Conway wondered whether a polyhedron could have such a polygonal hole passing through each and every face and remain a polyhedron. A bit later, the term "holyhedron" was coined to describe this form, should it exist.
Here's the mathematical question that Conway posed: Is there a polyhedron in Euclidean three-dimensional space that has only finitely many plane faces, each of which is a closed connected subset of the appropriate plane whose relative interior in that plane is multiply connected?
Conway's specifications exclude polyhedra in which a hole's sides extend all the way to a pierced face's edges.
Examples of unacceptable faces.
They also explicitly exclude such structures as an infinite lattice of interpenetrating tetrahedra, where each face of each tetrahedron is pierced by the vertex of another.
For a long time, no one could come up with an example, even in principle, that met Conway's precise specifications for what he meant by a holyhedron. Conversely, no one could say why it was impossible to construct one.
It was, however, possible to come up with structures that met less stringent requirements. Artist and independent scholar George W. Hart created a number of such examples, which looked like holyhedra but failed to meet all of the mathematical criteria specified by Conway.
When Jade P. Vinson first heard about the holyhedron problem, he had just arrived at Princeton University as a graduate student. "The problem intrigued me, and the timing was perfect," Vinson said. "Since the problem required no special background, I could get started right away; and since I hadn't started on anything else yet, it received my undivided attention."
That effort paid off handsomely. Vinson's surprising solution appeared in the article "On Holyhedra," published in the January 2000 issue of the journal Discrete and Computational Geometry.
To solve the problem, Vinson opted to use polyhedra that have more vertices than faces. "The first key idea … was to find a simple, repetitive arrangement of polyhedra so that there is a large excess of unused vertices over unpierced faces," Vinson said. "The second key idea was to 'trade' several unused vertices in an inconvenient location for a single new vertex in a better location."
Given these two ideas, "it is possible to construct a holyhedron with a very large number of sides," he remarked.
Vinson's careful manipulations produced a monstrous holyhedron with 78,585,627 faces. "The current construction is hard to visualize," Vinson admitted. Simple cardboard models give just the roughest idea of how it all fits together.
Conway had offered a reward of \$10,000—divided by the number of faces—for finding a holyhedron, so Vinson's initial effort netted him a minuscule return. Conway at the time speculated that someone may yet find a holyhedron with far fewer faces—perhaps only 100 or so. In 2003, Don Hatch came up with a holyhedron of 492 faces.
It would be even nicer, Vinson said, if the discoverer could also construct a convincing cardboard model of such a three-dimensional structure.
Originally posted June 19, 2000
## April 28, 2020
### Growing Sprouts
The game of sprouts has a way of growing on you.
This two-person, pencil-and-paper game is simple enough that children can play it. Yet its intricacies provide much food for mathematical thought.
The players start with a number of dots scattered across a sheet of paper. A move consists of drawing either a line connecting two dots or a loop starting and ending at the same dot. With each move, the player then places an additional dot somewhere along the new line or loop.
The line (or loop) may be of any shape, but it must not cross itself, cross a previously drawn line, or pass through a previously made dot. Furthermore, no dot may have more than three lines emanating from it. Indeed, a new dot placed on a line starts off with two connections already made.
Players take turns drawing curves. The winner is the last person able to play.
At first glance, it may seem that a game could keep sprouting forever. However, because each turn makes two connections to dots and opens up only one new opportunity for a link, the number of moves has a definite limit. In fact, you can prove that a game starting with n dots must end after a maximum of 3n − 1 moves.
Initially, there are no lines, so the dots have a total of 3n possible links. Each move uses up two of those openings, at the beginning and end of the drawn curve, but also adds a new dot with one opening. Thus, each move decreases the number of openings by one. Because a move requires filling two openings, the game can't continue when only one opening remains. Hence, no game can last beyond 3n − 1 moves.
You can also show that every game must go at least 2n moves. Thus, a game starting with three dots must end on or before the eighth move and must last at least six moves.
A typical three-dot sprouts game.
For a small number of dots, one can diagram all the possible moves and determine which player is guaranteed a win. The second player can always win a game starting with two or six dots. The first player is guaranteed a win in games involving three, four, or five dots.
In 1991, David Applegate, Guy Jacobson, and Daniel Sleator used a lot of computer power to push the analysis of sprouts out to 11 dots. They found that the second player wins in games involving seven or eight dots, while the first player wins in games involving nine, 10, or 11 dots.
There's an interesting pattern here. The researchers conjectured that the first player has a winning strategy if the number of dots divided by six produces a remainder of three, four, or five. Hence, their prediction for a game involving 12 dots is a win for the second player (the remainder is zero after dividing 12 by six).
Games can sprout all sorts of unexpected growth patterns, making formulation of a winning strategy a tricky proposition. Toward the end of a game, however, you can often see how to draw closed curves that divide the plane into regions in such a way as to lead to a win.
The game has also attracted the attention of mathematicians, who have investigated the game in terms of graph theory and topology. It's possible to try sprouts, for example, on a doughnut-shaped surface or in higher dimensions.
Sprouts was invented in 1967 by mathematicians John H. Conway and Michael S. Paterson, when both Conway and Paterson were at the University of Cambridge. "The day after sprouts sprouted, it seemed that everyone was playing it," Conway once wrote. "At coffee or tea times, there were little groups of people peering over ridiculous to fantastic sprout positions."
Piers Anthony picked up on the sprouts craze in his science-fiction novel Macroscope (Avon, 1969). "Sprouts is an intellectual game that has had an underground popularity with scientists for a number of years," one character in the novel noted. "The rules are simple. All you do is connect the dots."
Anthony then proceeded to illustrate with a sample three-dot game that runs for six moves, with a win for the first player.
"It's not a game," protested another character in Macroscope. "There's no element of chance or skill."
Nonetheless, the possibilities are sufficiently vast that sprouts and its variants offer a great of deal of enjoyment for both player and mathematician.
Original version posted April 7, 1997
## April 27, 2020
### Computing in a Surreal Realm
Surreal numbers on the front page of a major daily newspaper?
It happened in 1996 when the Washington Post reported the winners of the Westinghouse Science Talent Search, which recognizes outstanding scientific and mathematical research by high school students. The headline and subhead read: "Complex Calculations Add Up to No. 1: Md. Math Whiz Makes Sense of the Surreal to Take Prestigious National Prize."
The student was Jacob A. Lurie, then a senior at Montgomery Blair High School in Silver Spring, Md. His project concerned "recursive surreal numbers."
Even among mathematicians, the study of surreal numbers is an obscure pastime. Only a handful have occupied themselves in recent years exploring the peculiarities of a number system that includes different kinds of infinities and vanishingly small quantities.
The notion of surreal numbers goes back several decades. Mathematician John H. Conway, then at Cambridge University, was trying to understand how to play Go, a challenging board game popular in China and Japan.
After studying the game carefully, Conway decided that Go could be interpreted as the sum of a large number of smaller, simpler games. Conway then applied the same idea to other games of strategy, including checkers and dominoes, and he came to the conclusion that certain types of games appear to behave like numbers with distinctive properties.
A variant of the game nim illustrates this relation between games and numbers. In standard nim, counters or other objects are divided into three piles, and each player in turn may remove any number of counters from any one pile. The player who removes the last counter is the winner.
The variant that Conway considered has each counter "owned" by one or the other of the two players. Moreover, a player may take a set of counters from a pile only if the lowest counter removed is one of his or her own.
Initial setup for a two-color variant of nim.
Suppose each player uses counters of a different color. It's possible to start with piles that are all one color, that alternate in color, or in some other arrangement. Each possible arrangement of colored counters, representing a game, has a certain numeric measure and a definite outcome. It turns out that each game can be associated with a particular number.
Conway's insight linking games and numbers led him to define a new family of numbers constructed out of mathematical sets related to sequences of binary choices. In other words, these numbers correspond to different patterns of yes or no decisions—like the piles of counters of two different colors in Conway's nim variant.
Remarkably, this new way of generating numbers takes in the entire system of real numbers, which comprises the integers (positive and negative whole numbers along with zero), the rational numbers (integral fractions), and irrational numbers (such as the square root of 2). But it also goes beyond the reals, providing a way to represent numbers "bigger" than infinity or "smaller" than the smallest fraction.
In 1972, Conway happened to explain his new number system to computer scientist Donald E. Knuth. Knuth found the notion fascinating. In the following year, he wrote a short novel introducing Conway's theory. Knuth gave his novel the title Surreal Numbers: How Two Ex-Students Turned on to Pure Mathematics and Found Happiness.
To capture the all-encompassing nature of Conway's numbers, Knuth dubbed them "surreal," using the French preposition "sur" (meaning "above") to modify "real."
Conway's surreal numbers incorporate the idea that there exist different sizes of infinity, a notion investigated more than a century ago by Georg Cantor (1845-1918). For example, the natural, or counting, numbers get larger and larger without limit. But there are infinitely many real numbers between any two natural numbers.
To make this distinction more precise, mathematicians describe the natural numbers as being a family with omega (ɷ) members. Real numbers, then, are an even bigger family. In fact, there are many infinities in addition to the two represented by the natural numbers and the real numbers.
In the surreal system, it's possible to talk about whether ɷ is odd or even. You can also add 1 to infinity, divide infinity in half, and take its square root or logarithm. Equally accessible and amenable to manipulation are the infinitesimals—the numbers generated by the reciprocals of these infinities (for example, 1/ɷ).
What can you do with surreal numbers? It's still hard to say because very little research has been done on them. Only a few mathematicians, notably Martin Kruskal (1925-2006) and Leon Harkleroad, have taken them seriously enough to put in the time and effort to explore the possibilities.
To Lurie, such a wide open field presented both a considerable challenge and a great opportunity. Inspired by a description of surreal numbers in Conway's book On Numbers and Games (CRC Press, 2000), Lurie focused on surreal numbers that can be defined by the repeated, step-by-step processes characteristic of computation.
A sophisticated extension of work done earlier by Harkleroad, Lurie's effort delved into the sorts of computations possible within the realm of the surreals. Examples from the theory of combinatorial games illustrated some of the results that emerged from his pioneering studies.
Newspaper accounts could only hint at the rich mathematical background underlying Lurie's remarkable piece of work. Lurie himself helped spread the word. Self-assured and articulate, he patiently explained his ideas to all comers at a public display of the 1996 Westinghouse Science Talent Search projects at the National Academy of Sciences in Washington, D.C.
Echoing Conway's game-based approach, Lurie described surreal numbers as those used to measure the advantage that one player has over another in certain types of simple games.
Here's how Lurie explained one of his games. Suppose there are some bottles of Coke and some of Pepsi on a table. One player loves Coke, and the other loves Pepsi. They take turns drinking their respective sodas until one player runs out of sodas to drink. That person dies of thirst and the other one lives.
If there are more Cokes on the table, the Coke lover lives and the Pepsi lover dies. If there are more Pepsis, the situation is reversed. If there are equal amounts, whoever drinks first will run out first and lose. In this game, the advantage of one player over the other is measured by the difference between the number of bottles of Coke and the number of bottles of Pepsi,
Lurie's example illustrated how it's possible to assign a number to a given game to measure advantage. Generalizing this notion to a broader range of games and analogous situations requires the introduction of surreal numbers. Lurie investigated to what extent computers could manipulate surreal numbers.
His prize-winning research paper ended with a list of questions concerning surreal numbers that no one had yet answered. As so often happens in mathematical research, every hard-won answer suggests many more thought-provoking questions.
Original version posted March 18,1996
## April 26, 2020
### John Horton Conway (1937-2020)
Mathematician John Horton Conway died on April 11, 2020, at the age of 82. Brilliant but quirky, he made significant contributions to a wide range of fields, from knot theory and finite groups to number theory and combinatorial games. Fascinated by the mathematics underlying games and puzzles, he was very active in the realm of recreational mathematics.
Over the years, I encountered Conway and heard him speak many times, both at Princeton University and at a variety of conferences, notably in recent years at the Gathering4Gardner events in Atlanta. I wrote about his "Game of Life" and the invention of cellular automata in my first book, The Mathematical Tourist: Snapshots of Modern Mathematics (W.H. Freeman, 1988).
The Game of Life
Imagine an immense checkerboard grid stretching as far as the eye can see. Most of the checkerboard's squares, or cells, are empty; a few are occupied by strange beings—creatures very sensitive to their immediate neighbors. Their individual fates teeter on numbers. Too many neighbors means death by overcrowding and too few death by loneliness. A cozy trio of neighbors leads to a birth and a pair of neighbors to comfortable stability.
At each time step, this cellular universe shuffles itself. Births and deaths change old patterns into new arrangements. The patterns evolve—sometimes into a static array that simple marks time, sometimes into a sequence of shapes repeated again and again, sometimes into a chain of arrangements that propagates throughout the checkerboard universe.
The mathematical game called "Life" generates a remarkably diverse array of thought-provoking patterns and scenarios. Invented in 1970 by the British mathematician John H. Conway, it vividly demonstrates how a set of simple rules can lead to a complex world displaying a rich assortment of interesting behavior.
Conway's aim was to create a cellular pastime based on the simplest possible set of rules that would still make the game unpredictable. Moreover, he wanted the rules to be complete enough so that once started, the game could play itself. Growth and change would occur in jumps, one step inexorably leading to the next.
The result would be a little universe founded on logic, in which everything would be predestined, but there would be no obvious way for a spectator or player to determine the fate of future generations except by letting the game play itself out.
To find appropriate rules, Conway and his students at Cambridge University investigated hundreds of possibilities. They did thousands of calculations, looking at innumerable special cases to expose hidden patterns and underlying structures. They tried triangular, square, and hexagonal lattices, scribbling across acres of paper. They used large numbers of poker chips, coins, shells, and stones in their search for a viable balance between life and death.
The game they came up with is played on a infinite grid of square cells. Each cell is surrounded by eight neighbors, four along its sides, four at its corners. It is initially marked as either occupied or vacant, creating some sort of arbitrary starting configuration.
Changes occur in jumps, with each cell responding according to the rules. Any cell having two occupied cells as neighbors stays in its original state. A cell that is alive stays alive, and one that is empty stays empty. Three living neighbors adjacent to an empty cell leads to tricellular mating. A birth takes place, filling the empty cell. In such a neighborhood, a cell already alive continues to live. However, an occupied cell surrounded by four or more living cells is emptied. Unhappily, death also occurs if none or only one of an isolated living cell's neighbors is alive.
A cell's eight nearest neighbors have a strong influence on its destiny.
These simple rules engender a surprisingly complex world that displays a wide assortment of interesting events and patterns—a microcosm that captures elements of life, birth, growth, evolution, and death.
The game was first introduced to the public in October 1970, in Martin Gardner's "Mathematical Games" column in Scientific American. It aroused tremendous interest, and the game became an additive passion for many people.
Because it was relatively easy to implement as a computer program, it also quickly became a favorite computer exercise. All kinds of people—students and professors, amateurs and professionals—spent years of computer time following the evolution of countless starting patterns.
"Life" aficionados gleefully pursued elusive arrangements and searched for unusual types of behavior. Many different forms evolved on the checkerboard and were painstakingly cataloged, sporting evocative names such as pipe, horse, snake, honeycomb cell, ship, loaf, frog, danger signal, glider, beacon, powder keg, spaceship, toad, pinwheel, and gun.
Some of these arrangements vegetated in a single contented state, and others pulsated, switching back and forth between one configuration and another.
One simple pattern evolves over time into a sequence that alternates between two different forms.
The possibilities were endless, and the game presented a variety of intriguing mathematical puzzles. For example, are there patterns that can have no predecessor? Several such "Garden of Eden" arrays were eventually discovered.
Example of a "Garden of Eden" array.
Other investigations revealed that while a given pattern leads to only one sequel pattern, it can have several possible predecessors.
Different starting states can lead to identical vegetating states.
Thus, a particular configuration can have a number of different pasts but only one future. That makes it difficult for a viewer, glued to a computer screen, to backtrack if a particularly interesting pattern appears fleetingly during the course of a run. There is no guaranteed way to travel backward in time to recreate a past "Life."
The computer also brought animation to the game. A rapidly computed sequence of generations could be viewed as pulsating shapes, creeping growths, lingering dusts, fragmenting forms, and chaotically dancing figures.
Other enthusiasts adapted Conway's game for surfaces other than the infinite plane. Players can follow the game on the surface of a cylinder, a torus, or even a Möbius strip. They can also pursue their creatures of "Life" on structures in three and higher dimensions.
The attraction of Conway's original game, and the chief reason for its popularity, is that although it is completely predictable on a cell-by-cell basis, the large-scale evolution of patterns defies intuition. Will a pattern grow without limit? Will it settle into a single stable object? Will it send off a shipload of colonists?
Conway managed to balance the system's competing tendencies for growth and death so precariously that "Life" is always full of surprises.
See also "Wild Beasts around the Corner," "Computing in a Surreal Realm," "Growing Sprouts," "Powerful Sequences," and "Punctured Polyhedra,"
## April 25, 2020
### Golden Blossoms, Pi Flowers
In the head of a sunflower, the tiny florets that turn into seeds typically arrange themselves in intersecting families of spirals, one set winding clockwise and the other set winding counterclockwise. Count the number of spirals of a certain type and you are likely to get 21, 34, 55, 89, or 144. Indeed, if 34 spirals curve in one direction, there will be either 21 or 55 spirals curving in the other direction.
These numbers all belong to a sequence named for the 13th-century Italian mathematician Fibonacci. Each consecutive number is the sum of the two numbers that precede it. Thus, 1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8, 5 + 8 = 13, and so on. The ratios of successive terms of the Fibonacci sequence get closer and closer to a specific irrational number, often called the golden ratio. It can be calculated as (1 + √5)/2, or 1.6180339887… . For instance, the ratio 55/34 is 1.617647…, and the next ratio, 89/55, is 1.6181818… .
Fibonacci numbers (and the golden ratio) come up surprisingly often in nature, from the number of petals in various flowers to the number of spirals among the scales of a pine cone.
Pine cones and pineapples, for instance, have rows of diamond-shaped markings, or scales, which spiral around both clockwise and counterclockwise. If you count the number of these spirals, you are likely to find 8, 13, or 21. How do these numbers and the golden ratio arise?
In the June 2002 Mathematics Magazine, Michael Naylor described a simple mathematical model of how a sunflower produces its florets (and seeds). The idea underlying the model is that a sunflower produces florets one by one at the flower's center, and these push the other florets outward.
"Each seed settles into a location that turns out to have a specific constant angle of rotation relative to the previous seed," Naylor remarked. "It is this rotating seed placement that creates the spiraling patterns in the seed pod."
To simulate these spiraling patterns, Naylor described the location of any seed, k, using polar coordinates: r = √[k] and θ = k α, where r is the radial distance, α is the angle from the zero-degree line, k is seed number (starting with 1 at the center) and θ is the angle between any two seeds (which is constant).
Suppose that the seed angle is 45º (or 1/8 of a complete rotation). Seed 1 would be located at a distance of √1 and an angle of 45º. Seed 2 would be located 45º from the first seed, or 2 x 45º = 90º from the zero-degree line at a distance of √2 from the origin. Seed 3 would be located at 3 x 45º at a distance of √3, and so on. Note that seed 9 would fall on the same line as the first seed, starting a new cycle.
Forty-five-degree spiral.
When you plot these locations for 100 seeds, you can readily detect a spiral near the center, but a radial pattern of eight spokes becomes the dominant one farther away from the center.
"Notice how close together the seeds become and how much space there is between rows of seeds," Naylor commented. "This is not a very even distribution of seeds."
You could try to get a better distribution by choosing a different seed angle, say 15º or 48º. However, if this angle is a rational fraction of one revolution, you would end up with distinct spokes, and the seed distribution would still be quite uneven.
What about a seed angle derived from the golden ratio, an irrational number? In this case, the angle would be about 0.618 revolutions or roughly 222.5º.
Golden-ratio spiral.
"Notice how well distributed the seeds appear; there is no clumping of seeds and very little wasted space," Naylor observed. "Even though the pattern grows quite large, the distances between neighboring seeds appear to stay nearly constant."
Why do Fibonacci numbers arise out of such a "golden" pattern?
If you number the seeds consecutively from the center, you find that the seeds closest to the zero-degree line are numbered 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, and so on—all Fibonacci numbers. Indeed, the numbered seeds converge on the zero line, alternating above and below it. That's just how ratios of pairs of consecutive Fibonacci numbers converge to the golden ratio, alternately less than and greater than the golden ratio.
"The larger the Fibonacci numbers involved, the closer their ratio to [the golden ratio] and therefore the closer the seeds lie to the zero degree line," Naylor remarked. "It is for this reason that seeds in each spiral arm in a golden flower differ by multiples of a Fibonacci number."
What happens with other irrational numbers? Would they work just as well?
Naylor generated seed arrays in which the seed angle was derived from the irrational number pi, the ratio of a circle's circumference to its diameter. In this case, the seed angle is about 0.14159 revolutions, or roughly 50.97º.
Pi spirals: 500 seeds (left) and 2000 seeds (right).
This time, the seed distribution is quite uneven. For relatively small numbers of seeds, seven spiral arms dominate the resulting pattern. Curiously, for patterns made up of thousands of seeds, an additional set of 113 spiral arms becomes apparent at longer distances from the center. Why these particular numbers arise is related to rational approximations of pi's value: 22/7 and 355/113.
Intriguingly, an angle of rotation related to the square root of 2 produces a remarkably even distribution of seeds. The resulting pattern also features a number of distinctive families of spirals—families that correspond to the numbers 1, 2, 3, 5, 7, 12, 17, 29, 41, 70, 99, and so on.
Root-2 spiral.
The numbers of this sequence are the numbers in the so-called Columns of Pythagoras. Start with 1 at the top of each of two columns of integers. Given a row with the numbers a and b, the first entry in the next row is a + b = c and the second entry in that row is a + c = d. The ratio of the numbers in each row converges to the irrational, numerical value of the square root of 2.
There's much more in Naylor's delightful article. It provides not only a thought-provoking introduction to mathematical modeling in the service of phyllotaxis but also a wonderful excuse for playing with patterns on a computer.
Originally posted September 2, 2002
References:
Naylor, M. 2002. Golden, √2, and π flowers: A spiral story. Mathematics Magazine 75(June):163-172.
## April 24, 2020
### Fermat's Natural Spirals
A typical daisy is a star-like flower. It features a fringe of white or colored petals and a central disk of tubular florets. Each floret is itself a tiny flower.
This daisy has 21 white petals and a yellow central disk of tubular florets. Photo by Kenneth Peterson
The tightly packed florets at a daisy's center have an intriguing arrangement. The florets get larger at greater distances from the center. And there are hints of clockwise and anticlockwise spirals in the pattern.
A close-up view of a daisy's disk of florets reveals intriguing spiral patterns. Photo by Kenneth Peterson
One way to model such a pattern is to start with a curve called Fermat's spiral. This curve is also known as a parabolic spiral. It's given by the polar equation
r = k a½
where r is the distance from the origin, k is a constant that determines how tightly wound the spiral is, and a is the polar angle.
This type of spiral has the property of enclosing equal areas with every turn.
Fermat's spiral.
By placing points (disks or polygons) centered at regular angular intervals along such a spiral, you can create a variety of intriguing patterns—depending on the angle you choose to use. Using the angle 222.49 degrees (a value related to the golden ratio, 1.618034. . . ), you get a pattern with an even packing of polygons (or disks). It closely resembles a daisy's florets.
By placing points at regular angular intervals along Fermat's spiral, you get a pattern resembling that of a daisy's central florets. Courtesy of Robert Krawczyk
By choosing other angles, you get intriguing variants. Each choice gives a different pattern of secondary spirals, some winding clockwise and others anticlockwise, which form an interlocking system. Robert Dixon explores some of these possibilities in his book Mathographics.
Using larger numbers of points and smaller angles produces patterns with a variety of secondary spirals and, often, with radial lines that become evident toward the edges.
Robert J. Krawczyk has taken the generation of these patterns another step further, creating striking images of eerie ripple patterns. He calls these circular designs "Fermat's spiral mandalas."
Krawczyk starts by combining several spirals to create one complex pattern.
Four different individual spirals (top) can be combined in various ways to create new, complex patterns (bottom). Courtesy of Robert Krawczyk
By placing points at fixed angular intervals along these curves, he gets very elaborate patterns that show a variety of features.
An example of a Fermat's spiral mandala. Courtesy of Robert Krawczyk.
To finish his images, Krawczyk enhances the texture and gives them a coppery glow. Examples of such mandalas can be seen at his website.
Courtesy of Robert Krawczyk.
A mandala's complex circular design, with its symmetrical and radial balance, is intended to draw the eye to its center, Krawczyk says. Fermat's spiral, in particular, "is a natural basis for this inward draw."
For those who prefer a more natural look, all it takes is a close view of a daisy's central disk.
Photo by Kenneth Peterson
Originally posted September 5, 2005
## April 23, 2020
### A Lawyer's Math Library
"Strangely enough, anyone wishing to write about Galois in Paris would do well to journey to Louisville, Kentucky."Leopold Infeld, Whom the Gods Love:The Story of Evariste Galois
French mathematician Evariste Galois (1811-1832), whose death in a duel at the age of 20 cut short a remarkably productive career, is just one of many mathematicians represented in a little-known collection of rare mathematical and astronomical books at the University of Louisville library.
A visitor can leaf through the wrinkled, yellowed pages, stiff with age, of the first printed edition of Euclid's Elements (Elementa Geometria) (1482), through Narratio Prima, in which Copernicus's pupil Georg Joachim Rheticus (1514-1576) announced the Copernican sun-centered concept of the solar system, and through a copy of Isaac Newton's Principia, with Newton's own handwritten corrections on the errata leaf.
The man who assembled this notable collection was an attorney and a mathematics enthusiast. Born in 1873 into a prominent Kentucky family, William Marshall Bullitt throughout his long life believed firmly in the value of mathematics.
Lurline Jochum, Bullitt's secretary from 1927 until his death in 1957, once recalled, "When a young man from law school would come into the office and want a job, the first thing [Bullitt] would say is: How much mathematics have you had? He felt that if you had a good mathematical background, then you had a good reasoning power."
While an undergraduate at Princeton University, Bullitt himself took mathematics courses in preparation for his subsequent legal career. Later, he studied at the University of Louisville law school and established a lucrative practice in Louisville, specializing in actuarial and constitutional law. His clients included several of the country's largest insurance companies. He even came up with a mathematical formula that helped him win several insurance cases, beginning with an important case for the New York Life Insurance Company.
Bullitt also served as Solicitor General of the United States for a brief period under President William Howard Taft (1857-1930).
At the same time, Bullitt kept up with developments in mathematics and astronomy by attending meetings of the American Mathematical Society and other groups and by corresponding with mathematicians and scientists, including Albert Einstein (1879-1955). His friends included astronomer Harlow Shapley (1885-1972) and mathematicians George D. Birkhoff (1884-1944), Eric Temple Bell (1883-1960), and Richard Courant (1888-1972).
Bullitt's goal of collecting "the most important original works of the most prominent mathematicians of all time" was established during a parlor game instigated by his friend, the prominent mathematician G.H. Hardy (1877-1947).
Like everything else he did, Bullitt went about his new project systematically. He asked Bell, Shapley, and others for lists of what they considered to be the most important books that he could collect. He wrote to mathematicians at various colleges all over the United States to get their comments on the lists. When he was ready, he notified rare-book dealers of his needs and even traveled personally to Germany and France to locate many of the works on his final list.
Starting his project in 1936, Bullitt didn't miss much in gathering first-edition works by the greatest mathematicians of all time. His final purchase for the collection, Niels H. Abel's 1824 Mémoire sur les Équations Algébriques, occurred in 1951. He paid \$500—a sum he termed "outrageous."
Bullitt kept most of his collection in his law office, locking away some of the more valuable books in the office vault. In addition, he maintained a good selection of mathematics books in a magnificent library at Oxmoor, his family home located just outside of Louisville.
Visitors to Oxmoor can remember browsing through the library's mathematics books and Bullitt's habit of sometimes testing his visitors by posing mathematical puzzles.
One special feature of the collection attracted a few scholars even when Bullitt was still alive. Bullitt managed to assemble the most complete collection of the works of Galois to be found outside of France. This included copies of hard-to-find, contemporary newspaper clippings, many unpublished items, and other documents.
When University of Toronto physicist Leopold Infeld (1898-1968) decided to write a biography of Galois, he visited Oxmoor and spent several days examining the collection. Infeld, a socialist, later described the visit—his first encounter with an American millionaire and the accompanying lifestyle—in his autobiography, Why I Left Canada: Reflections on Science and Politics.
"I still remember that in the bathroom the toilet paper was rose-colored and perfumed," Infeld wrote. "The window frames creaked so much in the wind that I was unable to sleep in the midst of all the abundance and luxury."
When Bullitt died, his widow donated the more valuable books to the University of Louisville, although schools such as Harvard would have liked to obtain the collection. Later, the remainder of the collection also went to the university library, and the current checklist contains about 370 items.
The collection is very rich in the authors that it covers, and it includes some extremely rare items. At the same time, most of the material is available elsewhere to mathematicians and interested historians in other forms or later editions.
Such a resource is useful, however, when historians want to check original editions of mathematical works. In later editions, particularly during the 19th century, changes made by editors often obscured an author's original intent.
The William Marshall Bullitt Collection of Rare Mathematics and Astronomy at the University of Louisville Ekstrom Library gives visitors a chance to trace the mathematical formulas and geometrical diagrams of ancient authors, to puzzle out cryptic Greek and Latin phrases, and to contemplate some of the greatest achievements in mathematics. It affords an opportunity to touch a heritage.
Original version posted May 20, 2002
Reference:
Davitt, R.M. 1989. William Marshall Bullitt and his amazing mathematical collection. Mathematical Intelligencer 11(No. 4):26-33.
## April 22, 2020
### Aaron Riker and the Occupation of Atlanta
This excerpt from the Civil War journal of Aaron Denton Riker (1830-1914) of the 66th Ohio Volunteer Infantry describes his regiment's role in the occupation of Atlanta in Georgia (account edited for spelling, punctuation, and consistency).
At that time in the fall of 1864, the regiment was part of the Army of the Cumberland, commanded by Maj. Gen. George Henry Thomas. The regiment (Lieut. Col. Eugene Powell) was in the first brigade (Col. Charles Candy) of the second division (Brig. Gen. John W. Geary) of the twentieth army corps (XX Corps), led by Gen. Henry W. Slocum,
Atlanta, Georgia, September-November 1864
September 2nd, 1864
The news of the capture of Atlanta reached us at 10 o’clock today. One brigade of the 3rd Division of 20th Corps were the first to enter the city… . We remained at Pace’s Ferry until Sunday morning, the 3rd of September, when we also took up the line of march for the city, at which place we arrived in the afternoon, camping in the west part of town.
I have been over a portion of the city today, which I find terribly torn up by our shells in the northern portion of town. From Main Street out, scarce a house is left whole. Many of them have been struck dozens of times and are literally torn to pieces.
Ruined buildings on Peachtree Street in Atlanta. Library of Congress
The fortifications around Atlanta are of the strongest that engineering skills could make them. The outer fortifications consist of a regular chain of forts of large size, linked together by a strong line of breastworks, all of which are guarded by abatis on the outer side.
Elaborate Confederate fortifications guarding Atlanta. Library of Congress
Inside of this main line, there are other forts and earthworks. At a distance of one hundred feet back from the main line, a regular line of pits are dug, where the machinery for heavy siege guns was erected, on which and extending a short distance above ground was mounted their heavy guns—64 pounders. These they were unable to get away. They burned the works on which they were mounted, spiked the guns, and left them. We find eight of these heavy guns.
On the surrender of the city, our forces found a large amount of stores of most every description. The boys have found large quantities of tobacco, which they have eagerly appropriated to their own use.
The city of Atlanta is built on very broken and […] soil. The streets are laid out very irregularly, so much so that some of our boys have remarked they could go to any portion of the city seemingly on the same street.
Panoramic view of the city of Atlanta, as photographed in October 1864 from the top of the Female Seminary, extending from the Atlanta Medical College on the southeast around by the south to a point on Peachtree Street a little north of west. Library of Congress
I have as yet failed to find the beautiful portion of Atlanta but for where we take private residences, some of which cannot be beat for beauty of architecture or situation. Most of these residences have the most beautiful lawns I have ever seen, with all kinds of the most beautiful shrubbery, some of which is now in full bloom. These lawns seem to have been cultivated with all the ingenuity of taste, reminding us of what we have read in legends of the fairies.
Union officers standing in front of the Atlanta house that had served as the headquarters of Confederate Gen. John B. Hood. Library of Congress
I notice some of the finest machine shops I have ever visited. The depot is very large, with three tracks running through. Three trains can be run in at one time. The engine house is built of bricks, covering near an acre of ground.
Railroad tracks and engine house in Atlanta. Library of Congress
There are also several large foundries, blacksmith and wagon shops; in fact, shops for the casting of all kinds of machinery, both for the army and for general use. Most of the machinery has been removed previous to the occupation of the city by our troops.
The wealth of Atlanta has been immense. Now, however, the city is a complete wreck, and years must elapse ere it will again revive and flourish as formerly. The machine shops are being fitted up for government use.
The society of Atlanta has been quite aristocratic. Here, like all other southern cities, the poorer class of both black and white have been the slaves of the rich. And we found, too, that all the white population of males have been or are now in the Southern army. Those who we now find living in the city have all been conscripted and have, in nine cases out of ten, been in arms against us.
Those who reside here now are all, or nearly so, of the wealthy class, who have remained behind to look after their property. As soon as General [William Tecumseh] Sherman announced the campaign over, he then issued an order to the effect that Atlanta was to be used strictly as a military post and that all citizens not in government employ would be required to leave the city between the 12th and 23rd of September.
Maj. Gen. William T. Sherman, leaning on the breach of a cannon, with members of his staff at a Union fort near Atlanta. Library of Congress
All who could prove their loyalty and choose to go north could do so. Those families having husbands and friends in the Rebel army must go south. A truce was entered into between Major General Sherman, commanding the Division of the Mississippi, and General [John Bell] Hood, commanding Rebel forces, for the purpose of the removal of all families and their effects going south.
Transportation is now being furnished to families going south. They are hauled in wagons to Rough and Ready, a point agreed upon being half way between the two armies, distant from here six miles. Our trains are there met by trains from the Rebel armies, and the goods transferred.
Refugee wagons near the Atlanta train depot. Library of Congress
No distinction is made between rich or poor. All must come under the same head. All share the same mode of conveyance and journey the same road, once more to try the fortunes of treason. Those whose loyalty is undoubted are furnished railroad transportation north of Nashville to try their fortunes in a land of plenty.
At the provost marshal's office, Atlanta citizens could obtain passes to go either north or south. Library of Congress
Some of the people are loud in their denunciation of this order, thinking great injustice is being done them in having to leave their once peaceful and quiet homes. They choose rather to seek protection under the flag of our Union (which they have tried to pull down), then to again launch out on the perilous seas of disunion. I believe the order just and proper, and it must and will be enforced.
There are [four] railroads centering in Atlanta: the Western & Atlantic Railroad [connecting north to Chattanooga, Tennessee], the Georgia Railroad [connecting east to Augusta, Georgia, and the Confederate Powderworks on the Savannah River], the Macon and Western Railroad [connecting south to Macon and Savannah, Georgia], and the Atlanta and West Point Railroad [connecting west to West Point, Georgia]. Then there are connections with these railroads leading over almost the entire south.
As a military point, it is second to none in the South. On the railroad leading east is where the Rebels destroyed their large rolling mill and eighty cars loaded with ammunition. This destruction of property took place half a mile from the city.
Ruins of railroad cars carrying ammunition and the Atlanta Rolling Mill, destroyed near Atlanta during the Confederate evacuation of the city. Library of Congress
. . .
October 1st
The truce entered into between Generals Sherman and Hood for the removal of families from Atlanta existed on the part of General Sherman, made and kept in good faith. But we have good reason to doubt [it] was most grossly violated on the part of Hood, the commander of the Rebel force.
It became known to Sherman about the time the truce ended that a large Rebel force had crossed the Chattahoochee River and was making for some point on our line of communication. We are well aware this force must have moved during the suspension of hostilities.
Thus again we see the utter failure of the Rebels to carry on a war as a Christian Nation. They have openly violated the most solemn pledge existing between armies during hostilities. The truce was accepted on their part in seeming good faith, but to their shame be it said has been most grossly violated.
When General Sherman became aware of this foul treachery, detachments from the Armies of the Tennessee and Cumberland were sent by rail to the rear to protect our communications. Major General [George Henry] Thomas went back also with the detachments.
We learn now that the main Rebel army have struck the railroad at Big Shanty, holding and destroying the same. With the exception of the 20th Corps, our army are under orders to move. Troops are being sent to the neighborhood of Marietta and Kennesaw Mountain.
October 2nd
Troops are being sent back today as fast as possible. Fortunately, three of four trains of cars happen to be on this side of the break in the railroad.
October 4th
But little is known of the extent of damage being done to the railroad; there [are] no bridges in the vicinity of Big Shanty. General Sherman and staff left Atlanta this morning for the scene of our present difficulties, the general remarking as he rode away that he was going back to grind Hood to powder. At four o’clock this afternoon, his headquarters were in Marietta. General Thomas with his force is north of the Rebel force and, consequently, we have no communication with him. Our cavalry are reported in rear of the enemy.
October 5th
General Sherman moved with his army this morning from around Marietta. A battle is expected hourly. No communication with Thomas yet.
October 6th
We have no news since morning. Then a report was prevalent that yesterday fighting was going on at Allatoona. There does not seem to be any large force of the enemy left behind. One man was wounded prowling around our pickets on the night of the 4th. Since then all has been quiet.
Heavy details are at work fortifying the city on the east and south sides. The 20th Corps are charged with the defense and holding of Atlanta at all hazards. We are well supplied with all kinds of rations except salt meat. There is none for issue. Forage for animals is also very scarce.
Union soldiers at work refortifying a former Confederate fort. Library of Congress
We hope to have communication open again in a short time. The enemy are making a desperate effort to retake Atlanta. Time will show whether they are or have been successful with all their treachery. We feel confident we are fully able to hold the city, and we are willing to trust the matter of our communications with General Sherman.
October 7th
Official dispatches from General Sherman have been received here today stating the enemy were driven from the railroad and were severely repulsed in two successive charges made by them on our forces. Thirteen hundred prisoners are reported captured by our forces. Five miles of railroad has been effectively destroyed. Material is on hand for the repair of the railroad and communication will soon be open again by rail.
We begin to feel the need of having communication opened again. Our animals are suffering for want of forage. The ration for the troops still holds good with the exception of salt meat. That part of the ration has entirely run out. We are reduced also to half a pound of fresh beef per day. The ration of bread is now 1½ pounds per day. Double rations of vegetables are also allowed for the present.
The work on the fortifications progresses finely. Large details are kept busy day and night. The Rebel prisoners confined in the barracks here are marched out each morning armed with pick and shovel and set to work on the fortifications.
New fortifications constructed by Union forces in Atlanta. Library of Congress
I presume a great hue and cry will be raised against setting these prisoners to work. But we think their treachery deserves far worse punishment than this. There is nothing they will not stop to do in the treatment of the prisoners the fate of war has thrown in their hands, and we think a little work will not hurt them.
October 9th
A Sabbath stillness pervades Atlanta. The day is cool and very pleasant. I have felt that this is the Sabbath Day. All places of business have been closed. Both soldier and citizen seem to have that regard for the Sabbath they should. We have news of a fight at Allatoona, in which our forces were victorious. One hundred and fifty of the enemy were buried by our forces. We captured 450 prisoners; our loss reported at 600.
October 10th
Last night we received a dispatch announcing the capture of Richmond; no particulars given as yet. We also have a dispatch by signal that Sherman has defeated Hood at Lost Mountain. All quiet here.
October 11th
The election for state and county officers in the states of Ohio, Indiana, and Pennsylvania was held today. Ohio soldiers were the only soldiers that voted in 66th Regiment. Stands 163 Union votes; no opposition. News of the capture of Richmond contradicted; no new official [report] from General Sherman.
October 14th
Railroad again repaired. Received a very large mail today. No supplies coming yet by rail. A foraging expedition returned today having been out four days. They brought in six hundred wagonloads of forage. Weather very fine. Sherman on the move. Rebels making desperate efforts to cut our communications. All quiet around Atlanta.
October 16th
Our communications again cut. Fighting reported; no definite news. Railroad said to be badly damaged. An attack on Atlanta expected. Troops prepared for any emergency.
November 1st
But little has transpired of which we have an official notice since the 16th. The enemy has been driven back from our line of communication. They are now in Alabama. General Sherman has advanced his army 30 miles beyond Rome, where his is now preparing for a new campaign.
The 20th Corps have orders to prepare for a fifty-days campaign and to be ready by the 4th to enter upon the duties which may be assigned us. Our communications with Nashville are now open and trains are now running through. The weather is cold and rainy.
November 8th
We are still in Atlanta though we are expecting orders hourly to march. We are well aware that it is General Sherman’s intention to evacuate Atlanta. All is bustle and excitement here. Everything movable is being sent to the rear as fast as trains can carry it away. Citizens are going away as fast as they can get transportation. All is confusion with them; they do not like the idea of again falling into Rebel hands and yet transportation is very hard to get now owing to the crowded state of the cars.
The polls are open today, and the soldiers are voting for President of the United States. I feel today that the fate of our country hangs on this day’s vote, how important that every man should vote right.
November 9th
This morning at daylight, cannonading was heard to the east of us. At sunup, the Rebel cavalry, as it proved to be, shifted around to the south of us and made an attack on the 1st Brigade, 2nd Division, 20th Corps. The cannonading and musketry was very brisk for half an hour, when the enemy fell back on the Sandtown Road in haste, leaving two killed, one mortally wounded, and one well prisoner in our hands. Our loss was one man killed.
The 66th Ohio Regiment was paid today, 7 months and a half’s pay. This brings our pay up to the 31st day of August. The election results stand 335 Lincoln votes and 131 McClellan, majority for Lincoln 204 votes. We have a dispatch this evening that Lincoln has carried every state except Kentucky and New Jersey.
The weather is warm and showery. The roads are getting muddy. I fear we shall have bad weather for campaigning as this is the season for rains in the South.
November 12th
This has been a very windy day. The roads are in very good condition; weather fair.
Last night was a very exciting night in the city. There seems to have been a concerted plan to burn the city without authority. The intention of the authorities has not been to burn the city and, even was that the intention, we were not ready for the conflagration.
Union efforts to blow up buildings of military significance before the evacuation of the city engulfed other buildings. Library of Congress
At about seven o’clock, a frame building near the Junction of the Macon and Chattanooga roads was set on fire by some unknown person. The flames from that communicated with a half roundhouse used as an engine house. This house was of brick, with a frame and tile roof. The fire burned very slowly, and the flames would have been arrested there had the incendiary’s torch not have been applied to a row of frame buildings on the opposite side of the street.
These were fired, and the flames spread rapidly to other buildings in the vicinity. The headquarters of the 66th Ohio [Regiment] were in a corner building fronting on Hunter and Forsythe Streets. On Forsythe Street running north are a row of one-story frame and brick buildings running to Alabama Street.
On the corner of Forsythe and Alabama Streets was a large two-story frame house. West of this house on Alabama Street stood another two-story frame house. Some rascal carried fire to this building. When the flames soon reached the corner building, guards were placed around other buildings nearby to prevent anyone from entering them.
There, fortunately for us, happened to be one vacant lot between the corner building and other buildings connecting with our headquarters, and these buildings were saved by the vigilance of Col. Powell, who acted on his own responsibility. No orders came to him during all this time, and he not knowing whether he was acting right or not, yet deeming his duty to be to arrest the progress of the flames, he acted according to his own judgement, which proved to be right as no authority was given to fire any building.
Fires in Atlanta. Library of Congress
There happened to be an engine in the city, which was brought out, and water was freely used on buildings that would have taken fire and done immense damage to the command here. The troops were ordered under arms and heavy patrols sent out over the city to arrest every man found on the streets without authority.
The fire of which I have spoken was not the only one in the city. Near General Slocum’s headquarters on Peachtree Street, a building was fired, and in the northeast part of the city several houses were fired and consumed. Then on Marietta Street in the northwest part of town, a dozen or so buildings were burned. In the south part of town, several buildings were fired but were found out in time to arrest the flames.
Who was the author of all this I have not learned. It is thought that some Rebel sympathizers were the chiefs, as men were seen in the vicinity of the fires dressed in citizen’s clothes and mounted on horseback. Our patrols could not get near enough to arrest them.
Had shells from the enemy’s batteries been thrown among us, they could not have created greater consternation. At midnight, the excitement had subsided and heavy patrols were out all night. This evening, the Michigan engineers are engaged in tearing down the Roundhouse. They have it about torn down at sundown. Details are also at work tearing up the railroads.
Ruins of the Atlanta Roundhouse (engine house). Library of Congress
November 13th
The work of tearing up and burning the railroad commenced early this morning. One building has been burned this morning near the Roundhouse. General Sherman has offered a reward of five hundred dollars for the evidence to lead to the conviction of those who were engaged in firing the city Saturday night.
Union soldiers tearing up railroad tracks. Library of Congress
The work of tearing up the railroad still goes on. This evening, most of the [public buildings] are torn down. Atlanta begins to wear a desolate and lonely appearance.
Atlanta train depot in ruins. Library of Congress
The last train of cars left for the North yesterday morning. We now are cut off from all communication with the North. What our fate must be, time will show. We are on the eve of an important move.
Thoughts of home, dear home, come up in my mind tonight. In imagination, I see happy faces seated around the fireside, chatting or singing the songs I love. I anticipate the happy time when I, too, can meet dear ones in my happy home.
See also "Aaron Riker at the Siege of Atlanta," "Aaron Riker at Cedar Mountain," "Aaron Riker in Dumfries," "Aaron Riker at Chancellorsville," and "Aaron Riker at Gettysburg."
Aaron Denton Riker (1830-1914) of Champaign County, Ohio, enlisted as a private in the 66th Ohio Volunteer Infantry on October 11, 1861. The regiment was mustered in for three years service on December 17, 1861, under the command of Colonel Charles Candy. In April 1862, while in Strasburg, Virginia, during the Shenandoah campaign, Riker was assigned to the commissary department, handling supplies for the troops. In October of that year, he found himself in charge of the regiment's commissary and subsequently attained the rank of sergeant while his regiment was stationed in Dumfries, Virginia.
Riker was mustered out of the regiment in 1865 as a first lieutenant.
Aaron D. Riker, Columbus, Ohio, July 27, 1865.
Riker kept a journal recounting his experiences during the Civil War. The journal is now housed at the George J. Mitchell Department of Special Collections and Archives, Bowdoin College Library, Brunswick, Maine. JournalTranscript. |
## THE DISTRIBUTIVE LAW
If we desire to main point a amount by one more number, one of two people we can multiply every term the the sum by the number prior to we include or we can very first add the terms and also then multiply. Because that example,
In either situation the an outcome is the same.
You are watching: Factor of x^2-x-6
This property, which we very first introduced in ar 1.8, is referred to as the distributive law. In symbols,
a(b + c) = abdominal muscle + ac or (b + c)a = ba + ca
By using the distributive law to algebraic expression containing parentheses, we can attain equivalent expressions without parentheses.
Our first example requires the product of a monomial and also binomial.
Example 1 create 2x(x - 3) without parentheses.
Solution
We think that 2x(x - 3) together 2x and then apply the distributive regulation to obtain
The above method works equally too with the product of a monomial and also trinomial.
Example 2 compose - y(y2 + 3y - 4) without parentheses.
Solution
Applying the distributive property yields
When simplifying expressions including parentheses, we very first remove the parentheses and then incorporate like terms.
Example 3 simplify a(3 - a) - 2(a + a2).
We start by removed parentheses come obtain
Now, combining prefer terms yields a - 3a2.
We deserve to use the distributive residential property to rewrite expression in i beg your pardon the coefficient of an expression in bracket is +1 or - 1.
Example 4 write each expression without parentheses.a. +(3a - 2b)b. -(2a - 3b)
Solution
Notice the in example 4b, the authorize of every term is adjusted when the expression is created without parentheses. This is the same an outcome that we would certainly have obtained if we offered the steps that we introduced in ar 2.5 to leveling expressions.
## FACTORING MONOMIALS indigenous POLYNOMIALS
From the symmetric building of equality, we know that if
a(b + c) = ab + ac, then abdominal muscle + ac = a(b + c)
Thus, if there is a monomial factor common to all terms in a polynomial, we can write the polynomial as the product the the common factor and another polynomial. Because that instance, since each term in x2 + 3x consists of x as a factor, we can write the expression as the product x(x + 3). Rewriting a polynomial in this way is called factoring, and also the number x is stated to it is in factored "from" or "out of" the polynomial x2 + 3x.
To aspect a monomial indigenous a polynomial:Write a collection of parentheses came before by the monomial common to each term in the polynomial.Divide the monomial variable into each term in the polynomial and also write the quotient in the parentheses.Generally, us can discover the usual monomial variable by inspection.
Example 1 a. 4x + 4y = 4(x + y) b. 3xy -6y - 3y(x - 2)
We can inspect that us factored appropriately by multiplying the factors and verifyingthat the product is the initial polynomial. Using example 1, we get
If the common monomial is hard to find, we deserve to write every term in prime factored type and keep in mind the typical factors.
Example 2 variable 4x3 - 6x2 + 2x.
equipment We deserve to write
We currently see the 2x is a common monomial variable to all 3 terms. Then we element 2x out of the polynomial, and also write 2x()
Now, we divide each ax in the polynomial by 2x
and compose the quotients inside the parentheses come get
2x(2x2 - 3x + 1)
We can examine our price in instance 2 by multiply the components to obtain
In this book, we will restrict the typical factors to monomials consist of of numerical coefficients that are integers and to integral powers of the variables. The an option of authorize for the monomial factor is a matter of convenience. Thus,
-3x2 - 6x
can it is in factored either together
-3x(x + 2) or together 3x(-x - 2)
The an initial form is usually an ext convenient.
Example 3Factor out the common monomial, consisting of -1.
a. - 3x2 - 3 xyb. -x3 - x2 + x equipment
Sometimes it is practically to create formulas in factored form.
Example 4 a. A = p + PRT = P(1 + RT) b. S = 4kR2 - 4kr2 = 4k(R2 - r2)
## 4.3BINOMIAL assets I
We have the right to use the distributive legislation to multiply two binomials. Although over there is small need to multiply binomials in arithmetic as shown in the example below, the distributive law also applies to expressions containing variables.
We will now apply the above procedure for an expression include variables.
Example 1
Write (x - 2)(x + 3) without parentheses.
Solution First, use the distributive building to obtain
Now, integrate like state to attain x2 + x - 6
With practice, friend will have the ability to mentally include the second and 3rd products. Theabove procedure is sometimes called the foil method. F, O, I, and L stand for: 1.The product the the first terms.2.The product of the external terms.3.The product that the inner terms.4.The product the the last terms.
The FOIL method can also be supplied to square binomials.
Example 2
Write (x + 3)2 without parentheses.Solution
First, rewrite (x + 3)2 together (x + 3)(x + 3). Next, apply the FOIL technique to get
Combining choose terms yieldsx2 + 6x + 9
When we have a monomial factor and also two binomial factors, it is simplest to first multiply the binomials.
Example 3
create 3x(x - 2)(x + 3) without parentheses.Solution First, multiply the binomials to obtain3x(x2 + 3x - 2x - 6) = 3x(x2 + x - 6)
Now, use the distributive legislation to get 3x(x2 + x - 6) = 3x3 + 3x2 - 18x
Common Errors
Notice in instance 2
Similarly,
In general,
## 4.4FACTORING TRINOMIALS i
In section 4.3, we saw exactly how to find the product of two binomials. Now we will reverse this process. That is, provided the product of 2 binomials, we will find the binomial factors. The procedure involved is another example the factoring. As before,we will certainly only think about factors in i beg your pardon the terms have integral numerical coefficients. Such components do not always exist, however we will study the situations where they do.
Notice that the first term in the trinomial, x2, is product (1); the last term in thetrinomial, 12, is product and also the center term in the trinomial, 7x, is the sum of commodities (2) and also (3).In general,
We use this equation (from best to left) come factor any trinomial of the kind x2 + Bx + C. We uncover two numbers whose product is C and also whose amount is B.
Example 1 variable x2 + 7x + 12.Solution we look for 2 integers whose product is 12 and also whose sum is 7. Think about the following pairs of components whose product is 12.
We see that the only pair of determinants whose product is 12 and also whose amount is 7 is 3 and also 4. Thus,
x2 + 7x + 12 = (x + 3)(x + 4)
Note that as soon as all terms of a trinomial are positive, we require only think about pairs of positive factors because we are searching for a pair of components whose product and also sum are positive. That is, the factored hatchet of
x2 + 7x + 12would it is in of the type
( + )( + )
When the very first and third terms the a trinomial room positive yet the middle term is negative, we require only think about pairs of negative factors due to the fact that we are looking for a pair of determinants whose product is positive but whose amount is negative. That is,the factored type of
x2 - 5x + 6
would be of the form
(-)(-)
Example 2 factor x2 - 5x + 6.
Solution due to the fact that the third term is positive and also the middle term is negative, we discover two negative integers who product is 6 and also whose sum is -5. We list the possibilities.
We check out that the only pair of factors whose product is 6 and also whose sum is -5 is -3 and also -2. Thus,
x2 - 5x + 6 = (x - 3)(x - 2)
When the first term the a trinomial is positive and also the 3rd term is negative,the indications in the factored form are opposite. That is, the factored type of
x2 - x - 12
would be of the type
(+)(-) or (-)(+)
Example 3
Factor x2 - x - 12.
Solution we must uncover two integers who product is -12 and also whose amount is -1. We list the possibilities.
We check out that the just pair of factors whose product is -12 and also whose amount is -1 is -4 and 3. Thus,
x2 - x - 12 = (x - 4)(x + 3)
It is less complicated to aspect a trinomial totally if any type of monimial factor typical to each term the the trinomial is factored first. For example, we can aspect
12x2 + 36x + 24
as
A monomial deserve to then it is in factored from this binomial factors. However, first factoring the typical factor 12 from the initial expression yields
12(x2 + 3x + 2)
Factoring again, we have actually
12(* + 2)(x + 1)
which is said to be in totally factored form. In such cases, the is not vital to variable the numerical aspect itself, the is, we carry out not create 12 together 2 * 2 * 3.
instance 4
element 3x2 + 12x + 12 completely.
SolutionFirst we factor out the 3 native the trinomial to gain
3(x2 + 4x + 4)
Now, we aspect the trinomial and also obtain
3(x + 2)(x + 2)
The approaches we have developed are additionally valid because that a trinomial such as x2 + 5xy + 6y2.
Example 5Factor x2 + 5xy + 6y2.
Solution We discover two positive factors whose product is 6y2 and also whose sum is 5y (the coefficient of x). The two determinants are 3y and also 2y. Thus,
x2 + 5xy + 6y2 = (x + 3y)(x + 2y)
once factoring, the is best to create the trinomial in descending powers of x. If the coefficient the the x2-term is negative, variable out a an unfavorable before proceeding.
Example 6
Factor 8 + 2x - x2.
Solution We an initial rewrite the trinomial in descending powers of x to get
-x2 + 2x + 8
Now, us can element out the -1 come obtain
-(x2 - 2x - 8)
Finally, we element the trinomial come yield
-(x- 4)(x + 2)
Sometimes, trinomials space not factorable.
Example 7
Factor x2 + 5x + 12.
Solution we look for two integers who product is 12 and whose amount is 5. Native the table in example 1 on page 149, we see that there is no pair of determinants whose product is 12 and whose sum is 5. In this case, the trinomial is not factorable.
Skill at factoring is typically the result of substantial practice. If possible, execute the factoring process mentally, writing your prize directly. Girlfriend can check the outcomes of a administrate by multiply the binomial factors and also verifying that the product is same to the offered trinomial.
## 4.5BINOMIAL commodities II
In this section, we usage the procedure developed in ar 4.3 to multiply binomial factors whose first-degree terms have numerical coefficients various other than 1 or - 1.
Example 1
Write as a polynomial.
a. (2x - 3)(x + 1)b. (3x - 2y)(3x + y)
Solution
We an initial apply the FOIL technique and then integrate like terms.
As before, if we have actually a squared binomial, we an initial rewrite it together a product, then apply the silver paper method.
Example 2
a. (3x + 2)2 = (3x + 2)(3x + 2) = 9x2 + 6x + 6x + 4 = 9x2 + 12x + 4
b. (2x - y)2 = (2x - y)(2x - y) = 4x2 - 2xy - 2xy + y2 - 4x2 - 4xy + y2
As you may have actually seen in section 4.3, the product of 2 bionimals may have actually no first-degree hatchet in the answer.
Example 3
a. (2x - 3)(2x + 3) = 4x2 + 6x - 6x - 9 = 4x2 -9
b. (3x - y)(3x + y) - 9x2 + 3xy - 3xy - y2= 9x2 - y2
When a monomial factor and also two binomial determinants are gift multiplied, the iseasiest to multiply the binomials first.
Example 4
Write 3x(2x - l)(x + 2) together a polynomial.
Solution We first multiply the binomials to get3x(2x2 + 4x - x - 2) = 3x(2x2 + 3x - 2)Now multiplying by the monomial yields3x(2x2) + 3x(3x) + 3x(-2) = 6x3 + 9x2 - 6x
## 4.6FACTORING TRINOMIALS II
In section 4.4 us factored trinomials of the kind x2 + Bx + C wherein the second-degree term had actually a coefficient the 1. Currently we want to extend our factoring techniquesto trinomials that the type Ax2 + Bx + C, where the second-degree term has actually acoefficient other than 1 or -1.
First, we think about a check to recognize if a trinomial is factorable. A trinomial ofthe form Ax2 + Bx + C is factorable if we can find two integers who product isA * C and also whose sum is B.
Example 1
Determine if 4x2 + 8x + 3 is factorable.
Solution We examine to view if there space two integers whose product is (4)(3) = 12 and also whosesum is 8 (the coefficient that x). Consider the adhering to possibilities.
Since the factors 6 and 2 have a amount of 8, the value of B in the trinomialAx2 + Bx + C, the trinomial is factorable.
Example 2
The trinomial 4x2 - 5x + 3 is no factorable, due to the fact that the above table mirrors thatthere is no pair of factors whose product is 12 and whose sum is -5. The check tosee if the trinomial is factorable have the right to usually be done mentally.
Once we have identified that a trinomial the the form Ax2 + Bx + C is fac-torable, we continue to uncover a pair of factors whose product is A, a pair that factorswhose product is C, and an plan that yields the suitable middle term. Weillustrate by examples.
Example 3
Factor 4x2 + 8x + 3.
Solution Above, we determined that this polynomial is factorable. We currently proceed.
1. We take into consideration all pairs of determinants whose product is 4. Since 4 is positive, just positive integers should be considered. The possibilities space 4, 1 and 2, 2.2. We think about all pairs of factors whose product is 3. Since the center term is positive, take into consideration positive pairs of components only. The possibilities room 3, 1. We create all possible arrangements the the factors as shown.
3. We pick the arrangement in which the sum of commodities (2) and also (3) returns a middle term the 8x.
Now, we think about the administer of a trinomial in i m sorry the constant term is negative.
Example 4
Factor 6x2 + x - 2.
Solution First, we test to view if 6x2 + x - 2 is factorable. Us look for two integers that havea product that 6(-2) = -12 and also a sum of 1 (the coefficient the x). The integers 4 and-3 have a product of -12 and a sum of 1, for this reason the trinomial is factorable. We nowproceed.
We think about all pairs of components whose product is 6. Due to the fact that 6 is positive, only positive integers must be considered. Climate possibilities space 6, 1 and 2, 3.We consider all bag of factors whose product is -2. The possibilities space 2, -1 and -2, 1. We compose all possible arrange ments that the factors as shown.We choose the setup in i m sorry the amount of products (2) and also (3) returns a middle term the x.
With practice, you will be able to mentally inspect the combinations and also will notneed to write out all the possibilities. Paying attention to the indicators in the trinomialis specifically helpful for mentally eliminating possible combinations.
It is easiest to factor a trinomial written in descending powers of the variable.
Example 5
Factor.
a. 3 + 4x2 + 8x b. X - 2 + 6x2
Solution Rewrite every trinomial in descending powers of x and also then follow the options ofExamples 3 and also 4.
a. 4x2 + 8x + 3 b. 6x2 + x - 2
As we stated in section 4.4, if a polynomial consists of a usual monomial factorin each of that is terms, we should element this monomial native the polynomial beforelooking for various other factors.
Example 6
Factor 242 - 44x - 40.
Solution We an initial factor 4 from each term to get
4(6x2 - 11x - 10)
We then factor the trinomial, come obtain
4(3x + 2)(2x - 5)
ALTERNATIVE an approach OF FACTORING TRINOMIALS
If the above "trial and error" technique of factoring does not yield fast results, analternative method, which we will certainly now show using the previously example4x2 + 8x + 3, might be helpful.
We know that the trinomial is factorable because we uncovered two number whoseproduct is 12 and also whose sum is 8. Those numbers are 2 and 6. We currently proceedand usage these number to rewrite 8x as 2x + 6x.
We now aspect the first two terms, 4*2 + 2x and also the last two terms, 6x + 3.A typical factor, 2x + 1, is in each term, so us can aspect again.This is the same an outcome that we acquired before.
## 4.7FACTORING THE difference OF two SQUARES
Some polynomials happen so typically that that is beneficial to acknowledge these specialforms, which in tum allows us to straight write their factored form. Watch that
In this section we room interested in city hall this connection from ideal to left, indigenous polynomial a2 - b2 come its factored type (a + b)(a - b).
The difference of 2 squares, a2 - b2, amounts to the product the the amount a + b and the difference a - b.
Example 1
a. X2 - 9 = x2 - 32 = (x + 3)(x - 3) b. X2 - 16 = x2 - 42 = (x + 4)(x - 4)
Since
(3x)(3x) = 9x2
we deserve to view a binomial such as 9x2 - 4 as (3x)2 - 22 and also use the above methodto factor.
Example 2
a.9x2 - 4 = (3x)2 - 22= (3x + 2)(3x - 2)b.4y2 - 25x2 = (2y)2 - (5x)2= (2y + 5x)(2y - 5x)
As before, we always factor the end a usual monomial first whenever possible.
Example 3
a.x3 - x5 = x3(l - x2) = x3(1 + x)(l - x)b.a2x2y - 16y = y(a2x2 - 16) = y<(ax)2 - 42>= y(ax - 4 )(ax + 4)
## 4.8EQUATIONS involving PARENTHESES
Often we need to solve equations in i m sorry the variable occurs within parentheses. Wecan solve these equations in the usual manner after ~ we have actually simplified lock byapplying the distributive regulation to remove the parentheses.
Example 1
Solve 4(5 - y) + 3(2y - 1) = 3.
Solution We first apply the distributive law to get
20 - 4y + 6y - 3 = 3
Now combining favor terms and also solving for y yields
2y + 17 = 3
2y = -14
y=-l
The same technique can be used to equations involving binomial products.
Example 2
Solve (x + 5)(x + 3) - x = x2 + 1.
Solution First, we apply the FOIL technique to remove parentheses and also obtain
x2 + 8x + 15 - x = x2 + 1
Now, combining choose terms and also solving because that x yields
x2 + 7x + 15 = x2 + 1
7x = -14
x = -2
## 4.9WORD problems INVOLVING NUMBERS
Parentheses are useful in representing commodities in i beg your pardon the variable is containedin one or much more terms in any factor.
Example 1
One essence is three much more than another. If x to represent the smaller integer, representin regards to x
a. The larger integer.b. Five times the smaller sized integer.c. 5 times the larger integer.
Solution a. X + 3b. 5x c. 5(x + 3)
Let united state say we recognize the amount of 2 numbers is 10. If we stand for one number byx, then the 2nd number should be 10 - x as said by the following table.
In general, if we understand the amount of two numbers is 5 and x to represent one number,the various other number need to be S - x.
Example 2
The sum of two integers is 13. If x to represent the smaller sized integer, stand for in termsof X
a. The bigger integer.b. 5 times the smaller sized integer.c. Five times the larger integer.
Solution a. 13 - x b. 5x c. 5(13 - x)
The next example concerns the id of continually integers that was consid-ered in ar 3.8.
Example 3
The difference of the squares of two consecutive weird integers is 24. If x representsthe smaller sized integer, represent in terms of x
a. The bigger integerb. The square the the smaller integer c. The square the the larger integer.
Solution
a. X + 2b. X2 c. (x + 2)2
Sometimes, the mathematical models (equations) because that word difficulties involveparentheses. We can use the strategy outlined on page 115 to attain the equation.Then, we proceed to deal with the equation by first writing equivalently the equationwithout parentheses.
Example 4
One integer is five much more than a second integer. 3 times the smaller integer plustwice the larger amounts to 45. Find the integers.
Solution
Steps 1-2 First, we create what we want to discover (the integers) as word phrases. Then, we represent the integers in terms of a variable.The smaller sized integer: x The bigger integer: x + 5
Step 3 A map out is not applicable.
Step 4 Now, we write an equation the represents the condition in the problemand get
3x + 2(x + 5) = 45
Step 5 applying the distributive legislation to eliminate parentheses yields
Step 6 The integers space 7 and also 7 + 5 or 12.
## 4.10 APPLICATIONS
In this section, we will examine number of applications the word problems that command toequations the involve parentheses. When again, we will follow the six measures out-lined on page 115 when we deal with the problems.
COIN PROBLEMS
The an easy idea of problems involving coins (or bills) is that the value of a numberof coins of the exact same denomination is same to the product of the worth of a singlecoin and the total number of coins.
A table favor the one shown in the next instance is beneficial in addressing coin problems.
Example 1
A collection of coins consisting of dimes and quarters has actually a value of \$5.80. Thereare 16 much more dimes 보다 quarters. How numerous dimes and also quarters room in the col-lection?
Solution
Steps 1-2 We an initial write what we desire to discover as word phrases. Then, werepresent each expression in terms of a variable.The number of quarters: x The number of dimes: x + 16
Step 3 Next, us make a table showing the number of coins and also their value.
Step 4 currently we have the right to write an equation.
Step 5 solving the equation yields
Step 6 There space 12 quarters and 12 + 16 or 28 dimes in the collection.
INTEREST PROBLEMS
The simple idea of addressing interest problems is the the amount of attention i earnedin one year at basic interest equates to the product of the price of attention r and theamount that money ns invested (i = r * p). Because that example, \$1000 invested for one yearat 9% returns i = (0.09)(1000) = \$90.
A table prefer the one displayed in the next example is helpful in addressing interestproblems.
Example 2
Two investments produce an yearly interest of \$320. \$1000 an ext is invested at11% than at 10%. Just how much is invested at each rate?
Solution
Steps 1-2 We very first write what we desire to uncover as word phrases. Then, werepresent each phrase in regards to a variable. Amount invest at 10%: x Amount invest at 11%: x + 100
Step 3 Next, us make a table mirroring the amount of money invested, therates of interest, and also the amounts of interest.
Step 4 Now, we have the right to write an equation relating the attention from each in-vestment and also the full interest received.
Step 5 To resolve for x, very first multiply every member by 100 to obtain
Step 6 \$1000 is invest at 10%; \$1000 + \$1000, or \$2000, is invest at11%.
MIXTURE PROBLEMS
The basic idea of resolving mixture troubles is the the amount (or value) that thesubstances being blended must same the lot (or value) of the last mixture.
A table like the ones presented in the following examples is helpful in solvingmixture problems.
Example 3
How lot candy worth 80c a kilogram (kg) have to a grocer blend through 60 kg ofcandy worth \$1 a kilogram to make a mixture precious 900 a kilogram?
Solution
Steps 1-2 We an initial write what we desire to find as a native phrase. Then, werepresent the phrase in regards to a variable.Kilograms of 80c candy: x
Step 3 Next, we make a table reflecting the types of candy, the amount of each,and the total values that each.
Step 4 We can now create an equation.
Step 5 solving the equation yields
Step 6 The grocer should use 60 kg of the 800 candy.
Another form of mixture problem is one that requires the mixture that the two liquids.
Example 4
How many quarts that a 20% equipment of acid have to be added to 10 quarts the a 30%solution of acid to achieve a 25% solution?
Solution
Steps 1-2 We very first write what we want to uncover as a word phrase. Then, werepresent the phrase in regards to a variable.
Number that quarts that 20% solution to it is in added: x
Step 3 Next, we make a table or illustration showing the percent of every solu-tion, the amount of each solution, and also the amount of pure acid in eachsolution.
Step 4 We can now create an equation relating the quantities of pure mountain beforeand after combine the solutions.
Step 5 To solve for x, an initial multiply every member through 100 to obtain
20x + 30(10) = 25(x + 10)20x + 300 = 25x + 250 50 = 5x 10 = x
Step 6 add 10 quarts that 20% equipment to develop the preferred solution.
## CHAPTER SUMMARY
Algebraic expression containing parentheses have the right to be composed without clip byapplying the distributive regulation in the forma(b + c) = abdominal muscle + ac
A polynomial that includes a monomial factor typical to all terms in thepolynomial deserve to be composed as the product of the usual factor and anotherpolynomial by using the distributive legislation in the formab + ac = a(b + c)
The distributive law deserve to be used to multiply binomials; the FOIL an approach suggeststhe four products involved.
Given a trinomial the the type x2 + Bx + C, if there space two numbers, a and also b,whose product is C and also whose amount is B, then x2 + Bx + C = (x + a)(x + b) otherwise, the trinomial is not factorable.
A trinomial of the type Ax2 + Bx + C is factorable if there space two numbers whoseproduct is A * C and also whose amount is B. |
# 2D shapes – circle
The circle is a basic two-dimensional shape in geometry that consists of all the points in a plane that are at a constant distance, called the radius, from a given point, called the center.
The circle is often symbolized by the letter “O” or “∘”.
Some properties of circles include:
• Circumference: The circumference is the distance around the circle. It is equal to 2π times the radius, where π (pi) is a mathematical constant approximately equal to 3.14.
• C = 2πr – where C is the circumference, π (pi) is a mathematical constant approximately equal to 3.14, and r is the radius of the circle.
• C = πd – if you know the diameter of the circle (the distance across the circle through its center), you can also use this formula, where d is the diameter of the circle.
• Diameter: The diameter is a line segment that passes through the center of the circle and has endpoints on the circle. The diameter is twice the radius.
• d = 2r – where d is the diameter of the circle, and r is the radius of the circle.
• Radius: The radius is a line segment that connects the center of a circle to any point on the circle. It is half the length of the diameter.
• r = C / 2π – where r is the radius, C is the circumference of the circle, and π (pi) is a mathematical constant approximately equal to 3.14.
• r = d / 2 – where d is the diameter of the circle.
• Area: The area of a circle is π times the square of the radius.
• Chord: A chord is a line segment that connects two points on the circle. The longest chord in a circle is the diameter.
• Tangent: A tangent is a line that intersects the circle at exactly one point. The tangent is perpendicular to the radius at the point of intersection.
• Segment: A segment in a circle refers to a part of the circle that is bounded by two points on the circle.
• Sector: A sector is a region of the circle that is bounded by two radii and an arc.
• Arc: An arc is a portion of the circumference of a circle. The length of an arc is proportional to the angle it subtends at the center of the circle. |
# Statistics Assignment Help With Rank Correlation
## Spearman’s Rank Correlation Coefficient
The spearman’s rank coefficient of correlation was developed by Charles Edward Spearman.
### What is Spearman correlation coefficient?
Let us suppose that a group of n individuals is arranged in order of merit or proficiency in possession of two characteristics A and B. these ranks in two characteristics will, in general, be different. For example, if we consider the relation between intelligence and beauty, it is not necessary that a beautiful individual is intelligent also. Let (xi , yi ); i=1,2,………,n be the ranks of the ith individual in two characteristics A and B respectively. Pearsonian coefficient of correlation between the ranks xi’s and yi ‘s is called the Spearman rank correlation coefficient between A and B for that group of individuals.
### The formula for correlation of rank coefficient is given as:
rk = 1 – [6 ∑D2 / N3 – N]
Where D = R1 – R2, between the paired items in the two rank series.
The value of rank correlation coefficient tells us about the degree of agreement between the 2 ranks.
#### See derivation for the rank coefficient of correlation by spearman below:
Assuming that no two individuals are bracketed equal in either classification, each of the variables X and Y takes the values 1,2,………..,n
Hence
Which is the spearman's formula for the rank correlation coefficient.
### Properties of Spearman’s Rank correlation Coefficient
-1 ≤ rk ≤ +1
Rank Correlation Coefficient Example
Calculate the Rank Correlation Coefficient in each of the following cases:
X R1 Y R2
10 1 5 1
20 2 6 2
30 3 7 3
To calculate the rank correlation coefficient, first we will determine the value of D = R1 – R2 in each of the entries:
X R1 Y R2 D D2
10 1 5 1 0 0
20 2 6 2 0 0
30 3 7 3 0 0
Then the Spearman’s rank correlation coefficient is calculated using the formula as:
rk = 1 – [6 ∑D2 / N3 – N]
= 1- 6(0)
= +1
Thus the value of rank correlation coefficient equal to +1 implies that there is complete agreement in the order of ranks and the ranks are in the same direction.
Let us calculate the rank correlation coefficient in another example:
X R1 Y R2
10 1 7 3
20 2 6 2
30 3 5 1
To calculate the rank correlation coefficient, first we will determine the value of D = R1 – R2 in each of the entries:
X R1 Y R2 D D2
10 1 7 3 -2 4
20 2 6 2 0 0
30 3 5 1 +2 4
Then the Spearman’s rank correlation coefficient is calculated using the formula as:
rk = 1 – [6 ∑D2 / N3 – N]
= 1 – (6*8)/9-3
= -1
Thus the value of rank correlation coefficient equal to -1 implies that there is complete agreement in the order of ranks and the ranks are in opposite direction |
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# Twelve more than twice a certain number is six fewer than the three times the number. What is the number ? (A) $6$ (B) $12$ (C) $16$ (D) $18$
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Hint: In the given question, we are required to write an algebraic equation based on the information given to us. First, we break down the information given to us to try and identify what the algebraic equation must look like. You should always know what you are dealing with. A sum means that you are adding something, so you are going to use the $+$ sign and so on. Then, after framing a mathematical equation, we try to solve it by using the method of transposition.
The statement “twelve more than twice a certain number” means first multiplying a number by $2$ and then adding $12$ to the product.
So, let us assume that the number is x.
Then, we get the algebraic expression $\left( {2x + 12} \right)$ .
Now, we are given that the expression is six fewer than the three times the number assumed.
Now, we know that three times the number is $3x$ .
So, framing the mathematical equation, we get,
$\Rightarrow \left( {2x + 12} \right) = \left( {3x - 6} \right)$
Now, we have to solve the above equation using the transpositions and simplification methods.
Shifting all the terms consisting x to the right side of the equation and all the constants to the left side, we get,
$\Rightarrow 12 + 6 = 3x - 2x$
Simplifying the expression further, we get,
$\Rightarrow x = 18$
So, the value of x is $18$ .
Hence, option (D) is correct.
So, the correct answer is “Option D”.
Note: An expression is a sentence with a minimum of two numbers and at least one math operation. This math operation can be addition, subtraction, multiplication, and division.
An equation will always use an equivalent $\left( = \right)$ operator between the two terms. Method of transposition involves doing the exact same thing on both sides of an equation with the aim of bringing like terms together and isolating the variable or the unknown term in order to simplify the equation and finding value of the required parameter. |
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Unit Rate Calculator
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Unit rates are common in our everyday experience in the real world. While a rate is an expression that compares those quantities in relation to one unit. For example cents per pound is a unit rate; the amount of cents per one pound. The unit rate will be the price per unit, such as dollars per pound, cents per ounce, or dollars per foot.
To find a unit rate, divide the numerator by the denominator to get a denominator of 1.
Thus for each of the selected resources the following arithmetic calculation is performed in unit rate estimating method.
Unit rate = Ratio comparing 2 quantities of different kinds of units.
A rate is a comparison of two quantities that have different units. A rate is written as a fraction. A unit rate is a rate in which the number in the denominator is 1.
You can see a calculator with two different default quantity with different units given below. On clicking "Calculate", the calculator will compare the two quantities and find the equal ratio.
## Steps for Unit Rate Calculator
Step 1 :
Read the problem and find the values given.
Step 2 :
Do the indicated division so that the denominator is a 1 unit measurement.
Step 3 :
Simplify the fraction.
## Problems on Unit Rate Calculator
1. ### Write the unit rate for $\frac{2.04}{3\ cans\ of\ peas}$
Step 1 :
The given fraction value is $\frac{2.04}{3\ cans\ of\ peas}$
Step 2 :
Do indicated division so that the denominator is 1 unit.
$\frac{2.04}{3\ cans\ of\ peas} = \frac{0.68}{1\ can\ of\ peas}$
Step 3 :
The unit rate is 68cent per cane.
The answer is 68cent per cane.
2. ### Write the unit rate for $\frac{802.5\ miles}{12.5\ hours}$
Step 1 :
The given fraction value is $\frac{802.5\ miles}{12.5\ hours}$
Step 2 :
Do indicated division so that the denominator is 1 unit.
$\frac{802.5\ miles}{12.5\ hours} = \frac{64.2\ miles}{12.5\ hours}$
Step 3 :
The unit rate is 64.2 miles per hour. |
# Math Snap
## $-6+\{14+2[60-9(1+3)]\}$
#### STEP 1
Assumptions1. We are given the expression $-6+\{14+[60-9(1+3)]\}$. . We need to simplify this expression following the order of operations, which is parentheses, brackets, braces, multiplication and division (from left to right), and addition and subtraction (from left to right). This is often abbreviated as PEMAS.
#### STEP 2
First, we simplify the expression inside the parentheses. $-6+\{14+2[60-9(1+)]\}$
#### STEP 3
Calculate the value inside the parentheses. $1+3 =$Replace the value inside the parentheses in the original expression. $-6+\{14+2[60-9()]\}$
#### STEP 4
Next, we simplify the expression inside the brackets by performing the multiplication. $60-9(4) =60-36 =24$Replace the value inside the brackets in the original expression. $-6+\{14+2[24]\}$
#### STEP 5
Now, we simplify the expression inside the braces by performing the multiplication. $14+2[24] =14+48 =62$Replace the value inside the braces in the original expression. $-+\{62\}$
##### SOLUTION
Finally, we simplify the entire expression by performing the addition. $-6+62 =56$The simplified value of the given expression is56. |
# For a circle with a diameter of 6 meters, what is the measurement of a central angle (in degrees) subtended by an arc with o length of 7 3 π meters?
140
Step-by-step explanation:
If it is an arc of 7pi/3, and the circumference is 6*pi or 6pi, then the arc takes up 7/18 of the circle, and since a circle has 360 degrees, the answr is 7/18 * 360 or 140.
..............................
Step-by-step explanation:
We know that,
where,
θ = central angle in radian.
Given,
diameter = 6 m, so radius = 3 m.
Putting the values,
Answer : The measurement of a central angle is,
Step-by-step explanation :
Formula used for angle subtended by an arc is:
where,
s = arc length =
Now put all the given values in the above formula, we get:
Thus, the measurement of a central angle is,
Step-by-step explanation:
The measurement of a central angle (in degrees) subtended by an arc is .
Step-by-step explanation:
as
meters
so
meters
and
meters
Using the formula
Simplify
Therefore, the measurement of a central angle (in degrees) subtended by an arc is .
The measurement of the angle subtended by an arc with a length of 5/2 pi meters is 149.542°
Step-by-step explanation:
Here, the diameter of the circle = 6 m
So, radius = D / 2 = 6 / 2 = 3 m
Also, the length of the arc = ( ) meters
Putting π = 3.14, we get
The length S of the arc =
or, S = 7.85 m
Let us assume the arc subtends angle Ф at the center of the circle.
⇒ S = r Ф
or, Ф =
Now, 1 Radian = 57.2958 Degrees
⇒ 2. 62 Radian = 2.61 x ( 57.2958 Degrees) = 149.542 °
or, Ф = 149.542°
Hence, the measurement of the angle subtended by an arc with a length of 5/2 pi meters is 149.542°
Arc length = (5/2) * PI meters = 7.853981634 meters
circle circumference = 2 * PI * radius = 2 *PI * 3
18.8495559215
arc length = (7.85381634 / 18.8495559215) * 360 =
0.4166578975 *360 =
150 degrees
Arc length is given by:
C=θ/360πd
thus plugging in the values we get:
5/2π=θ/360π*6
dividing through by 6π we get:
5/12=θ/360
multiplying both sides by 360
θ=150°
Arc length=5/2 pi
diameter=6m
Circumference=pi x d=pi x 6m= 6pi
Circumference/arc length=6pi/(5/2pie)=360degree/central angle
6pie x central angle=360degree x 5/2pie
Central angle=360degree x 5/2pie/6pie
Central angle=360degree x 5/12
Central angle=150degree
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# What is the vertex of y= -3x^2-5x-(3x-2)^2?
Dec 30, 2017
The vertex is $\left(\frac{7}{24} , - \frac{143}{48}\right)$.
#### Explanation:
First expand ${\left(3 x - 2\right)}^{2} = 9 {x}^{2} - 12 x + 4$.
Substituting that in we have:
$y = - 3 {x}^{2} - 5 x - \left(9 {x}^{2} - 12 x + 4\right)$
Distribute the negative:
$y = - 3 {x}^{2} - 5 x - 9 {x}^{2} + 12 x - 4$
Collect like terms:
$y = - 12 {x}^{2} + 7 x - 4$
The vertex is $\left(h , k\right)$ where $h = - \frac{b}{2 a}$ and $k$ is the value of $y$ when $h$ is substituted.
$h = - \frac{7}{2 \left(- 12\right)} = \frac{7}{24}$.
$k = - 12 {\left(\frac{7}{24}\right)}^{2} + 7 \left(\frac{7}{24}\right) - 4 = - \frac{143}{48}$ (I used a calculator...)
The vertex is $\left(\frac{7}{24} , - \frac{143}{48}\right)$.
Dec 30, 2017
$\left(\frac{7}{24} , - \frac{143}{48}\right)$
#### Explanation:
$\text{we require to express in standard form}$
$\Rightarrow y = - 3 {x}^{2} - 5 x - \left(9 {x}^{2} - 12 x + 4\right)$
$\textcolor{w h i t e}{\Rightarrow y} = - 3 {x}^{2} - 5 x - 9 {x}^{2} + 12 x - 4$
$\textcolor{w h i t e}{\Rightarrow y} = - 12 {x}^{2} + 7 x - 4 \leftarrow \textcolor{b l u e}{\text{in standard form}}$
$\text{given the equation of a parabola in standard form then}$
$\text{the x-coordinate of the vertex is}$
${x}_{\textcolor{red}{\text{vertex}}} = - \frac{b}{2 a}$
$\text{here } a = - 12 , b = 7 , c = - 4$
$\Rightarrow {x}_{\textcolor{red}{\text{vertex}}} = - \frac{7}{- 24} = \frac{7}{24}$
$\text{substitute this value into the equation for y}$
$y = - 12 {\left(\frac{7}{24}\right)}^{2} + 7 \left(\frac{7}{24}\right) - 4 = - \frac{143}{48}$
$\Rightarrow \textcolor{m a \ge n t a}{\text{vertex }} = \left(\frac{7}{24} , - \frac{143}{48}\right)$ |
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# Adding Numbers
The mathematical operation that combines two separate quantities into single sum is called addition. The mathematical symbol used for addition is "+".
The answer for adding two
numbers is called as sum.
For example, adding 3 + 3 is equal to 6.
Related Calculators Adding Binary Numbers Calculator Adding Complex Numbers Calculator Adding Hexadecimal Numbers Calculator Adding Mixed Numbers Calculator
## How to Add Numbers
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Below are the examples on adding numbers -
### Solved Examples
Question 1: Addition of two digit numbers 25 and 63 ?
Solution:
Step 1:
Let us write the given problem as 25 + 63
Write the given numbers one under the other like,
2 5
6 3
Step 2:
Add the one’s place we get,
5 + 3 = 8
Step 3:
Add the tens place with the carry of one’s place, we get,
2 + 6 = 8
Step 4:
Re-group the numbers, which is grouping the tens place and one’s place we get,
Here, 8 is in the tens place and 8 is in the one’s place.
= 88
Therefore, the solution for adding 25 and 63 is 88.
Question 2: Addition of two digits numbers 35 and 37 ?
Solution:
Step 1:
Let us write the given problem as 35 + 37.
Write the given numbers one under the other like,
3 5
3 7
Step 2:
Add the one’s place we get,
5 + 7 = 12 = 2 (carry 1)
Step 3:
Add the tens place with the carry of one’s place, we get,
1 + 3 + 3 = 7
Step 4:
Regroup the numbers, which is grouping the tens place and one’s place we get,
Here, 7 is in the tens place and 2 is in the one’s place.
= 72
Therefore, the solution for adding 35 and 37 is 72.
Question 3: Addition of two digit numbers 27 and 36?
Solution:
Step 1:
Let us write the given problem as 27 + 36.
Write the given numbers one under the other like,
2 7
3 6
Step 2:
Add the one’s place we get,
6 + 7 = 13 = 3 (carry 1)
Step 3:
Add the tens place with the carry of one’s place, we get,
1 + 2 + 3 = 6
Step 4:
Regroup the numbers, which is grouping the tens place and one’s place we get,
Here, 6 is in the tens place and 3 is in the one’s place.
6 + 3 = 63
Therefore, the solution for adding 78 and 47 is 125.
Question 4: Addition of two digit numbers 77 and 66?
Solution:
Step 1:
Let us write the given problem as 77 + 66.
Write the given numbers one under the other like,
7 7
6 6
Step 2:
Add the one’s place we get,
7 + 6 = 13 = 3 (carry 1)
Add the carry 1 to the tens place.
Step 3:
Add the carry 1 of one’s place with the tens place we get,
1 + 7 + 6 = 14
Step 4:
Regroup the numbers, which is grouping the tens place and one’s place we get,
Here, 14 is in the tens place and 3 is in the one’s place.
14 + 3 = 1 + 4 + 3 = 143
Therefore, the solution for adding 77 and 66 is 143.
## Addition Practice
Back to Top
### Practice Problems
Question 1: Adding 33 + 27 ?
Question 2: Adding 55 + 66 ?
Question 3: Adding 24 + 58 ?
Question 4: Adding 36 + 24 ?
Question 5: Adding 21 + 36 ?
More topics in Adding Numbers Additive Identity 3 Digit Addition Adding Two Digit Numbers Adding 4-digit Numbers Addition Properties Adding Integer Rules Mental Addition and Subtraction Commutative Law
NCERT Solutions NCERT Solutions NCERT Solutions CLASS 6 NCERT Solutions CLASS 7 NCERT Solutions CLASS 8 NCERT Solutions CLASS 9 NCERT Solutions CLASS 10 NCERT Solutions CLASS 11 NCERT Solutions CLASS 12
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# Angle Bisectors in Triangles
%
Progress
Practice Angle Bisectors in Triangles
Progress
%
Angle Bisectors in Triangles
What if you were told that $\overrightarrow{GJ}$ is the angle bisector of $\angle FGH$ ? How would you find the length of $FJ$ given the length of $HJ$ ? After completing this Concept, you'll be able to use the Angle Bisector Theorem to solve problems like this one.
### Watch This
First watch this video.
Next watch this video.
Then watch this video.
Finally, watch this video.
### Guidance
An angle bisector cuts an angle exactly in half. One important property of angle bisectors is that if a point is on the bisector of an angle, then the point is equidistant from the sides of the angle. This is called the Angle Bisector Theorem.
In other words, if $\overrightarrow{BD}$ bisects $\angle ABC, \ \overrightarrow{BA} \perp \overline{FD}$ , and, $\overrightarrow{BC} \perp \overline{DG}$ then $FD = DG$ .
The converse of this theorem is also true.
Angle Bisector Theorem Converse: If a point is in the interior of an angle and equidistant from the sides, then it lies on the bisector of that angle.
When we construct angle bisectors for the angles of a triangle, they meet in one point. This point is called the incenter of the triangle.
#### Example A
Is $Y$ on the angle bisector of $\angle XWZ$ ?
If $Y$ is on the angle bisector, then $XY = YZ$ and both segments need to be perpendicular to the sides of the angle. From the markings we know $\overline{XY} \perp \overrightarrow{WX}$ and $\overline{ZY} \perp \overrightarrow{WZ}$ . Second, $XY = YZ = 6$ . So, yes, $Y$ is on the angle bisector of $\angle XWZ$ .
#### Example B
$\overrightarrow{MO}$ is the angle bisector of $\angle LMN$ . Find the measure of $x$ .
$LO = ON$ by the Angle Bisector Theorem.
$4x - 5 &= 23\\4x &= 28\\x &=7$
#### Example C
$\overrightarrow{AB}$ is the angle bisector of $\angle CAD$ . Solve for the missing variable.
$CB=BD$ by the Angle Bisector Theorem, so we can set up and solve an equation for $x$ .
$x+7 &=2(3x-4)\\x+7 &=6x-8\\15 &=5x\\x &=3$
-->
### Guided Practice
1. $\overrightarrow{AB}$ is the angle bisector of $\angle CAD$ . Solve for the missing variable.
2. Is there enough information to determine if $\overrightarrow{A B}$ is the angle bisector of $\angle CAD$ ? Why or why not?
3. A $108^\circ$ angle is bisected. What are the measures of the resulting angles?
1. $CB=BD$ by the Angle Bisector Theorem, so $x=6$ .
2. No because $B$ is not necessarily equidistant from $\overline{AC}$ and $\overline{AD}$ . We do not know if the angles in the diagram are right angles.
3. We know that to bisect means to cut in half, so each of the resulting angles will be half of $108$ . The measure of each resulting angle is $54^\circ$ .
### Explore More
For questions 1-4, $\overrightarrow{AB}$ is the angle bisector of $\angle CAD$ . Solve for the missing variable.
Is there enough information to determine if $\overrightarrow{A B}$ is the angle bisector of $\angle CAD$ ? Why or why not?
1. In what type of triangle will all angle bisectors pass through vertices of the triangle?
2. What is another name for the angle bisectors of the vertices of a square?
3. Draw in the angle bisectors of the vertices of a square. How many triangles do you have? What type of triangles are they?
4. Fill in the blanks in the Angle Bisector Theorem Converse.
Given : $\overline{AD} \cong \overline{DC}$ , such that $AD$ and $DC$ are the shortest distances to $\overrightarrow{BA}$ and $\overrightarrow{BC}$
Prove : $\overrightarrow{BD}$ bisects $\angle ABC$
Statement Reason
1. 1.
2. 2. The shortest distance from a point to a line is perpendicular.
3. $\angle DAB$ and $\angle DCB$ are right angles 3.
4. $\angle DAB \cong \angle DCB$ 4.
5. $\overline{BD} \cong \overline{BD}$ 5.
6. $\triangle ABD \cong \triangle CBD$ 6.
7. 7. CPCTC
8. $\overrightarrow{B D}$ bisects $\angle ABC$ 8.
### Vocabulary Language: English
angle bisector
angle bisector
An angle bisector is a ray that splits an angle into two congruent, smaller angles.
Angle Bisector Theorem
Angle Bisector Theorem
The angle bisector theorem states that if a point is on the bisector of an angle, then the point is equidistant from the sides of the angle.
Angle Bisector Theorem Converse
Angle Bisector Theorem Converse
The angle bisector theorem converse states that if a point is in the interior of an angle and equidistant from the sides, then it lies on the bisector of that angle.
incenter
incenter
The incenter is the point of intersection of the angle bisectors in a triangle.
### Explore More
Sign in to explore more, including practice questions and solutions for Angle Bisectors in Triangles.
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# 2015 AMC 8 Problems/Problem 25
One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can be fitted into the remaining space?
$\mathrm{(A) \ } 9\qquad \mathrm{(B) \ } 12\frac{1}{2}\qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ } 15\frac{1}{2}\qquad \mathrm{(E) \ } 17$
$[asy] draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); [/asy]$
## Contents
### Solution 1
We draw a diagram as shown. $[asy] draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); path arc = arc((2.5,4),1.5,0,90); pair P = intersectionpoint(arc,(0,5)--(5,5)); pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; draw(P--Pp--Ppp--Pppp--cycle); [/asy]$ Let us focus on the big triangles taking up the rest of the space. The triangles on top of the unit square between the inscribed square, are similiar to the 4 big triangles by $AA.$ Let the height of a big triangle be $x$ then $\tfrac{x}{x-1}=\tfrac{5-x}{1}$. $$x=-x^2+6x-5$$ $$x^2-5x+5=0$$ $$x=\dfrac{5\pm \sqrt{(-5)^2-(4)(1)(5)}}{2}$$ $$x=\dfrac{5\pm \sqrt{5}}{2}$$ Which means $x=\dfrac{5-\sqrt{5}}{2}$ This means the area of each triangle is $\dfrac{5-\sqrt{5}}{2}*(5-\dfrac{5-\sqrt{5}}{2})*\dfrac{1}{2}=\dfrac{5}{2}$ This the area of the square is $25-(4*\dfrac{5}{2})=\boxed{C,~15}$
### Solution 2
We draw a diagram as shown: $[asy] pair Q,R,S,T; Q=(1.381966,0); R=(5,1.381966); S=(3.618034,5); T=(0,3.618034); draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); draw(Q--R--S--T--cycle); draw((1,1)--(4,1)--(4,4)--(1,4)--cycle,dashed); [/asy]$
We wish to find the area of the larger triangle. The area of the larger square is composed of the smaller square and the four triangles. The triangles have base $3$ and height $1$, so the combined area of the four triangles is $4 \cdot \frac 32=6$. The area of the smaller square is $9$. We add these to see that the area of the large square is $9+6=\boxed{\mathrm{(C) \ } 15}$.
### Solution 3
$[asy] draw((0,0)--(0,5)--(5,5)--(5,0)--cycle); filldraw((0,4)--(1,4)--(1,5)--(0,5)--cycle, gray); filldraw((0,0)--(1,0)--(1,1)--(0,1)--cycle, gray); filldraw((4,0)--(4,1)--(5,1)--(5,0)--cycle, gray); filldraw((4,4)--(4,5)--(5,5)--(5,4)--cycle, gray); path arc = arc((2.5,4),1.5,0,90); pair P = intersectionpoint(arc,(0,5)--(5,5)); pair Pp=rotate(90,(2.5,2.5))*P, Ppp = rotate(90,(2.5,2.5))*Pp, Pppp=rotate(90,(2.5,2.5))*Ppp; draw(P--Pp--Ppp--Pppp--cycle); [/asy]$
Let's find the area of the triangles and the unit squares: on each side, there are 2 triangles. They both have 1 leg of length 1, and let's label the other legs x for one of thr triangles and y for the other. Note that x+y=3. The area of each of the triangles is $\frac{x}{2}$ and $\frac{y}{2}$, and there are 4 of each. So now we need to find $(4)\frac{x}{2} + (4)\frac{y}{2}$.
$(4)\frac{x}{2} + (4)\frac{y}{2}$ $\Rightarrow~~4\left(\frac{x}{2}+ \frac{y}{2}\right)$ $\Rightarrow~~4\left(\frac{x+y}{2}\right)$ Remember that x+y=3, so substituting this in we find that the area of all of the triangles is $4\left(\frac{3}{2}\right) = 6$. The area of the 4 unit squares is 4, so the area of the square we need is $25- (4+6) = 15 \Rightarrow \boxed{(C)}$
2015 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 24 Followed byLast Problem 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 AJHSME/AMC 8 Problems and Solutions |
## What is the example of lowest term?
A fraction is said to written using lowest terms if its numerator and denominator are relatively prime, that is, they have no common factors other than 1. (Some books use “simplest form” to mean the same thing.) So, 59 is written in lowest terms, since 5 and 9 have no common factors other than 1.
## What is 2/4 in the lowest terms?
Chart
Fraction Reduced Form Decimal Value
24 12 0.5
26 13 0.3333
28 14 0.25
210 15 0.2
## What is 2/8 in the lowest term?
Reduce 2/8 to lowest terms
1. Find the GCD (or HCF) of numerator and denominator. GCD of 2 and 8 is 2.
2. 2 ÷ 28 ÷ 2.
3. Reduced fraction: 14. Therefore, 2/8 simplified to lowest terms is 1/4.
## How do you simplify?
To simplify any algebraic expression, the following are the basic rules and steps:
1. Remove any grouping symbol such as brackets and parentheses by multiplying factors.
2. Use the exponent rule to remove grouping if the terms are containing exponents.
3. Combine the like terms by addition or subtraction.
4. Combine the constants.
## How do you find the LCD?
Find the least common denominator ( LCD ) of two fractions
1. Factor each denominator into its primes.
2. List the primes, matching primes in columns when possible.
3. Bring down the columns.
4. Multiply the factors. The product is the LCM of the denominators.
5. The LCM of the denominators is the LCD of the fractions.
## What is the lowest term of 5 2?
• As an improper fraction. (numerator larger than denominator): 5/2 = 5/2
• As a mixed number (mixed fraction) (a whole number and a proper fraction, of the same sign): 5/2 = 2 1/2
• As a decimal number: 5/2 = 2.5.
• As a percentage: 5/2 = 250%
## What is 3 9 in the lowest terms?
Reduce 3/9 to lowest terms
1. Find the GCD (or HCF) of numerator and denominator. GCD of 3 and 9 is 3.
2. 3 ÷ 39 ÷ 3.
3. Reduced fraction: 13. Therefore, 3/9 simplified to lowest terms is 1/3.
## What is the lowest term for 1 2?
So we can divide both the numerator and denominator by 3. Can we simplify 1/2 any further? Nope, there’s no number, other than 1, that we can use to divide both the numerator and denominator. 1/2 is the lowest term of 3/6.
## What is 4/5 in the lowest term?
45 is already in the simplest form. It can be written as 0.8 in decimal form (rounded to 6 decimal places). Steps to simplifying fractions
• Find the GCD (or HCF) of numerator and denominator. GCD of 4 and 5 is 1.
• 4 ÷ 15 ÷ 1.
• Reduced fraction: 45. Therefore, 4/5 simplified to lowest terms is 4/5.
## What is the lowest term for 6 10?
Steps to simplifying fractions Therefore, 6/10 simplified to lowest terms is 3/5.
You might be interested: FAQ: How To Get Good Grades In Math?
## What is 3/4 in the lowest terms?
Therefore, we divide the numerator and denominator by 30. 90/120 written in lowest terms is 3/4.
## What is 7/8 in the lowest terms?
78 is already in the simplest form. It can be written as 0.875 in decimal form (rounded to 6 decimal places). Steps to simplifying fractions
• Find the GCD (or HCF) of numerator and denominator. GCD of 7 and 8 is 1.
• 7 ÷ 18 ÷ 1.
• Reduced fraction: 78. Therefore, 7/8 simplified to lowest terms is 7/8.
## What is the lowest term for 6 8?
Steps to simplifying fractions Therefore, 6/8 simplified to lowest terms is 3/4.
## What is 2 6 in the lowest terms?
Reduce 2/6 to lowest terms
1. Find the GCD (or HCF) of numerator and denominator. GCD of 2 and 6 is 2.
2. 2 ÷ 26 ÷ 2.
3. Reduced fraction: 13. Therefore, 2/6 simplified to lowest terms is 1/3. |
# Solving Equations with Variables on Both Sides
## Presentation on theme: "Solving Equations with Variables on Both Sides"— Presentation transcript:
Solving Equations with Variables on Both Sides
Warm Up Lesson Presentation Lesson Quiz Holt Algebra 1 Holt McDougal Algebra 1
Warm Up Simplify. 1. 4x – 10x 2. –7(x – 3) 3. –6x 4. 15 – (x – 2)
Solve. 5. 3x + 2 = 8 6. –6x –7x + 21 2x + 3 17 – x 2 28
Objective Solve equations in one variable that contain variable terms on both sides.
Vocabulary identity
To solve an equation with variables on both sides, use inverse operations to "collect" variable terms on one side of the equation. Helpful Hint Equations are often easier to solve when the variable has a positive coefficient. Keep this in mind when deciding on which side to "collect" variable terms.
Example 1: Solving Equations with Variables on Both Sides
Solve 7n – 2 = 5n + 6. 7n – 2 = 5n + 6 To collect the variable terms on one side, subtract 5n from both sides. –5n –5n 2n – 2 = 2n = Since n is multiplied by 2, divide both sides by 2 to undo the multiplication. n = 4
Check It Out! Example 1a Solve 4b + 2 = 3b. 4b + 2 = 3b To collect the variable terms on one side, subtract 3b from both sides. –3b –3b b + 2 = 0 – 2 – 2 b = –2
Check It Out! Example 1b Solve y = 0.7y – 0.3. To collect the variable terms on one side, subtract 0.3y from both sides. y = 0.7y – 0.3 –0.3y –0.3y = 0.4y – 0.3 Since 0.3 is subtracted from 0.4y, add 0.3 to both sides to undo the subtraction. = 0.4y Since y is multiplied by 0.4, divide both sides by 0.4 to undo the multiplication. 2 = y
To solve more complicated equations, you may need to first simplify by using the Distributive Property or combining like terms.
Example 2: Simplifying Each Side Before Solving Equations
Solve 4 – 6a + 4a = –1 – 5(7 – 2a). 4 – 6a + 4a = –1 –5(7 – 2a) Distribute –5 to the expression in parentheses. 4 – 6a + 4a = –1 –5(7) –5(–2a) 4 – 6a + 4a = –1 – a 4 – 2a = – a Combine like terms. Since –36 is added to 10a, add 36 to both sides. 40 – 2a = a + 2a a To collect the variable terms on one side, add 2a to both sides. = a
Example 2 Continued Solve 4 – 6a + 4a = –1 – 5(7 – 2a). 40 = 12a Since a is multiplied by 12, divide both sides by 12.
Check It Out! Example 2a Solve Distribute to the expression in parentheses. 1 2 To collect the variable terms on one side, subtract b from both sides. 1 2 3 = b – 1 Since 1 is subtracted from b, add 1 to both sides. 4 = b
Check It Out! Example 2b Solve 3x + 15 – 9 = 2(x + 2). 3x + 15 – 9 = 2(x + 2) Distribute 2 to the expression in parentheses. 3x + 15 – 9 = 2(x) + 2(2) 3x + 15 – 9 = 2x + 4 3x + 6 = 2x + 4 Combine like terms. –2x –2x To collect the variable terms on one side, subtract 2x from both sides. x + 6 = – – 6 Since 6 is added to x, subtract 6 from both sides to undo the addition. x = –2
An identity is an equation that is true for all values of the variable
An identity is an equation that is true for all values of the variable. An equation that is an identity has infinitely many solutions. Some equations are always false. These equations have no solutions.
Identities and False Equations
WORDS Identity When solving an equation, if you get an equation that is always true, the original equation is an identity, and it has infinitely many solutions. NUMBERS 2 + 1 = 2 + 1 3 = 3 ALGEBRA 2 + x = 2 + x –x –x 2 = 2
Identities and False Equations
When solving an equation, if you get a false equation, the original equation has no solutions. WORDS x = x + 3 –x –x 0 = 3 1 = 1 + 2 1 = 3 ALGEBRA NUMBERS Identities and False Equations
Example 3A: Infinitely Many Solutions or No Solutions
Solve 10 – 5x + 1 = 7x + 11 – 12x. 10 – 5x + 1 = 7x + 11 – 12x 10 – 5x + 1 = 7x + 11 – 12x Identify like terms. 11 – 5x = 11 – 5x Combine like terms on the left and the right. + 5x x Add 5x to both sides. = 11 True statement. The equation 10 – 5x + 1 = 7x + 11 – 12x is an identity. All values of x will make the equation true. All real numbers are solutions.
Example 3B: Infinitely Many Solutions or No Solutions
Solve 12x – 3 + x = 5x – 4 + 8x. 12x – 3 + x = 5x – 4 + 8x 12x – 3 + x = 5x – 4 + 8x Identify like terms. 13x – 3 = 13x – 4 Combine like terms on the left and the right. –13x –13x Subtract 13x from both sides. –3 = –4 False statement. The equation 12x – 3 + x = 5x – 4 + 8x is a false equation. There is no value of x that will make the equation true. There are no solutions.
Check It Out! Example 3a Solve 4y + 7 – y = 10 + 3y.
Identify like terms. 3y + 7 = 3y + 10 Combine like terms on the left and the right. –3y –3y Subtract 3y from both sides. 7 = False statement. The equation 4y + 7 – y = y is a false equation. There is no value of y that will make the equation true. There are no solutions.
Check It Out! Example 3b Solve 2c + 7 + c = –14 + 3c + 21.
Identify like terms. 3c + 7 = 3c + 7 Combine like terms on the left and the right. –3c –3c Subtract 3c both sides. 7 = True statement. The equation 2c c = –14 + 3c + 21 is an identity. All values of c will make the equation true. All real numbers are solutions.
Example 4: Application Jon and Sara are planting tulip bulbs. Jon has planted 60 bulbs and is planting at a rate of 44 bulbs per hour. Sara has planted 96 bulbs and is planting at a rate of 32 bulbs per hour. In how many hours will Jon and Sara have planted the same number of bulbs? How many bulbs will that be? Person Bulbs Jon 60 bulbs plus 44 bulbs per hour Sara 96 bulbs plus 32 bulbs per hour
Example 4: Application Continued
Let b represent bulbs, and write expressions for the number of bulbs planted. 60 bulbs plus 44 bulbs each hour the same as 96 bulbs 32 bulbs each hour When is ? b = b b = b To collect the variable terms on one side, subtract 32b from both sides. – 32b – 32b b = 96
Example 4: Application Continued
b = 96 Since 60 is added to 12b, subtract 60 from both sides. – – 60 12b = 36 Since b is multiplied by 12, divide both sides by 12 to undo the multiplication. b = 3
Example 4: Application Continued
After 3 hours, Jon and Sara will have planted the same number of bulbs. To find how many bulbs they will have planted in 3 hours, evaluate either expression for b = 3: b = (3) = = 192 b = (3) = = 192 After 3 hours, Jon and Sara will each have planted 192 bulbs.
Let g represent Greg's age, and write expressions for his age.
Check It Out! Example 4 Four times Greg's age, decreased by 3 is equal to 3 times Greg's age increased by 7. How old is Greg? Let g represent Greg's age, and write expressions for his age. three times Greg's age four times Greg's age is equal to increased by decreased by 7 . 3 4g – = g
Check It Out! Example 4 Continued
4g – 3 = 3g + 7 To collect the variable terms on one side, subtract 3g from both sides. –3g –3g g – 3 = Since 3 is subtracted from g, add 3 to both sides. g = Greg is 10 years old.
Lesson Quiz 1. 7x + 2 = 5x + 8 2. 4(2x – 5) = 5x + 4
Solve each equation. 1. 7x + 2 = 5x (2x – 5) = 5x + 4 3. 6 – 7(a + 1) = –3(2 – a) 4. 4(3x + 1) – 7x = 6 + 5x – 2 5. 6. A painting company charges \$250 base plus \$16 per hour. Another painting company charges \$210 base plus \$18 per hour. How long is a job for which the two companies costs are the same? 3 8 all real numbers 1 20 hours
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# Why does 10/35 = 2/7?
My question is not meant to be dumb. I'm working on a much more complicated problem, but I forgot some of the basics.
Often we perform steps in math without really knowing why we can do a certain step. Why does $\frac{10}{35} = \frac{2}{7}$? Obviously it simply reduces to $\frac{2}{7}$, but exactly why ? We simply divide both numerator and denominator by $5$ but isn't this really multiplying by $5$? I'm wondering operationally how we do this.
Is it simple $\frac{10}{35} \cdot \frac{5}{5}$? Then we cross reduce? If we think of it $\frac{10}{35}$ divided by $5$ then I think of this as $\frac{10}{35} \cdot \frac{1}{5}$ (the reciprocal), BUT $\frac{10}{35} \cdot \frac{1}{5} = \frac{2}{35}$ (dividing the $5$ by itself and the $10$ by $5$ to get $2$.) Is there an actual proof or law that allows us to do this ?
• $\frac{10}{35}=\frac{5 \cdot 2}{5 \cdot 7}$ then using the property of the division you get $\frac{5 \cdot 2}{5 \cdot 7}=\frac{5 }{5 } \cdot \frac{2}{7}=\frac{2}{7}$. Is this what you were asking? – chak Jan 1 '18 at 12:36
• "We simply divide both numerator and denominator by $5$ but isnt this really multiplying by $5$?" No, it's multiplying the numerator and denominator by $\frac{1}{5},$ which is the same thing as multiplying the whole thing by $\frac{1/5}{1/5}=1$. – spaceisdarkgreen Jan 1 '18 at 12:44
• It's simply $$\frac{10}{35}=\dfrac{10\cdot \frac{1}{5}}{35\cdot \frac{1}{5}}=\frac{2}{7}$$ – Fakemistake Jan 1 '18 at 12:44
• Maybe helpful: math.stackexchange.com/questions/1127483/… – Hans Lundmark Jan 1 '18 at 14:24
I think a right thinking here it's the following. $$\frac{10}{35}=\frac{2\cdot5}{7\cdot5}=\frac{2}{7}.$$
For the intuition: It really does make sense to manipulate fractions like that and in fact we do it all the time in every day life without realizing it. Saying that $\frac{10}{35}=\frac{2}{7}$ is like saying that whether you are at a party with 10 pizzas and 35 people or at a party with 7 pizzas and 2 people ,you will eat the same amount of pizza. This is the thinking that one tries to capture in the following axiomatic approach.
Aximoatic approach: As a mathematician you need to know what are the objects you are working with, same goes with rations. What is a fraction of two numbers? Well, to define them you say this: Let's make formal symbol of two natural number $\frac{p}{q}. You define two such symbols to be equal if and only if the obey the simple rule you are describing in your post, namely$\frac{p_1}{q_1}=\frac{p_2}{q_2}$if and only if$p_{1}\cdot q_2 =p_2 \cdot q_1$So a fraction is the set of all satisfying the above condition. Example: If you want to be extremely formal, when you write$\frac{7}{2}$you mean all the numbers$\frac{p}{q}$where$7q=2p$. Finally let me say that if you understand this concept well you are on your way to understanding a much more abstract and deep mathemacal idea called "Localization of Ring" which generalizes what we did with the natural numbers in more weird "arithmetic systems" called rings. You can search in wikipedia (here) to see a more rigorous presentation than what I am writing here. Here is about rings(here) and localization (here) • Of course, thanks ! – Nick A. Jan 1 '18 at 14:08 $$\frac{a}{b} = \frac{\frac{a}{\gcd(a,b)}}{\frac{b}{\gcd(a,b)}}$$ when$b \ne 0$in general. In this case, since$\gcd(10,35) = 5$, we can write $$\frac{10}{35} = \frac{\frac{10}{5}}{\frac{35}{5}} = \frac{2}{7}$$ For an algebraic proof, you can start from the algebraic rule for multiplying any two fractions:$\dfrac{a}{b} \times \dfrac{c}{d} = \dfrac{a \times c}{b \times d}$so$\dfrac{2}{7} = \dfrac{2}{7} \times 1 = \dfrac{2}{7} \times \dfrac{5}{5} = \dfrac{2 \times 5}{7 \times 5} = \dfrac{10}{35}$For a more intuitive argument, you can say that$5 \times 7 = 35$and thirty five thirty-fifths is one, so five thirty-fifths must be the same as one seventh, and so ten thirty-fifths is the same as two sevenths. We are not dividing nore multiplying the fraction by 5. We are dividing both the top and the bottom of$10/35\$ by 5. To make sense out of this we need a definition for equality of fractions. We define $$a/b=c/d$$ if and only if $$ad=bc$$ Therefore $$10/35=2/7$$due to the fact that$$10\times7=35\times2$$ |
# Maths LAB 4.1 Assessment Preparation
August 30, 2017
Question
LAB 4.1 (3.0 HOURS)
Assessment Preparation Checklist:
Perform the following tasks to prepare for this assessment:
? Read Chapter 2, Sections 2.1–2.7, of your textbook, Algebra & Trigonometry. These sections will help you understand how to interpret information from a graph and make predictions by modeling data with functions.
? Review the lesson for this module, which explains how to graph functions and obtain information about a function from its graph.
Title: Transformation, Composite, and Inverse Functions
This lab includes problems based on graphing various functions and application problems involving the slope and equations of lines.
Recommended Procedures:
Solve the following problems, providing detailed steps wherever required.
1. (10 points) What test can we use to determine if a graph is a function or not? Explain in your own words.
2. (25 points) The following graph displays the rates of a computer service technician:
The following piecewise function is described as the following:(????)={\$50 ???????? 0??????6043?????30 ???????? ????>60
a. Find the cost if the computer service technician works for 45 minutes.
b. Find the cost if the computer service technician works for 105 minutes.
3. (5 points) Write the equation to find the slope of an equation.
4. (5 points) Write the point-slope form.
5. (5 points) Write the slope-intercept form.
6. (20 points) The total sales of a company in 2008 was \$45,000 dollars. In 2011, it grew to \$84,000 dollars. Use????as the number of years since 2000 and????as the total sales.
Hint: Use the points (8, \$45,000) and (11, \$84,000).
a. Find the rate of change (slope).
b. Write an equation in both point-slope form and slope-intercept form, modeling the company’s sales.
c. What are the projected sales for the year 2016? (16 years after 2000).
7. (20 points) Let(????)=????+4and????(????)=????2+8?????3.Find each of the following functions:
a.(????+????)(????)
b.(?????????)(????)
c.(????????)(????)
d.?????????
Hint: (????(????))
8. (10 points) Find the inverse of(????)=?9????+8.
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# ADVANCED GEOMETRY 3.1/2 What are Congruent Figures? / Three ways to prove Triangles Congruent. Learner Objective: I will identify the corresponding congruent.
## Presentation on theme: "ADVANCED GEOMETRY 3.1/2 What are Congruent Figures? / Three ways to prove Triangles Congruent. Learner Objective: I will identify the corresponding congruent."— Presentation transcript:
ADVANCED GEOMETRY 3.1/2 What are Congruent Figures? / Three ways to prove Triangles Congruent. Learner Objective: I will identify the corresponding congruent parts of congruent figures and use three triangle congruence postulates to prove that triangles are congruent.
Learner Objective: I will identify the corresponding congruent parts of congruent figures and use three triangle congruence postulates to prove triangles congruent. In the last chapter, we defined congruent segments and congruent angles. What must be true about congruent segments? What must be true about congruent angles? How many segments does it take to make a triangle? How many angles are in a triangle? So what must be true if two triangles are congruent? A B C D E F
Learner Objective: I will identify the corresponding congruent parts of congruent figures and use three triangle congruence postulates to prove triangles congruent. In general, two geometric shapes are congruent if one of them can be placed on top of the other and fit exactly, point for point, side for side, angle for angle. Thus, congruent figures have the same _________ and _________. A B C D E F
Learner Objective: I will identify the corresponding congruent parts of congruent figures and use three triangle congruence postulates to prove triangles congruent. DEFINITION: If two triangles are congruent, then all pairs of corresponding parts (angles and segments) are congruent. If two triangles have all pairs of corresponding parts (angles and segments) congruent, then the triangles are congruent. DEFINITION: If two polygons are congruent, then all pairs of corresponding parts (angles and segments) are congruent. If two polygons have all pairs of corresponding parts (angles and segments) congruent, then the polygons are congruent. Congruent Triangles all pairs of corresponding parts congruent. means "...if and only if..." (reversible statement). means "...implies..." (If...then...).
Learner Objective: I will identify the corresponding congruent parts of congruent figures and use three triangle congruence postulates to prove triangles congruent. A B C D E F
A B C D E F G H Given: Conclude:
Learner Objective: I will identify the corresponding congruent parts of congruent figures and use three triangle congruence postulates to prove triangles congruent.
Postulate: Any segment or angle is congruent to itself. (Reflexive Property) E F G H J K AB C D
Learner Objective: I will identify the corresponding congruent parts of congruent figures and use three triangle congruence postulates to prove triangles congruent.
SSS Postulate: If there exists a correspondence between the vertices of two triangles such that three sides of one triangle are congruent to the corresponding three sides of the other triangle, then the two triangles are congruent (SSS ≅ ) A B C D E F
Learner Objective: I will identify the corresponding congruent parts of congruent figures and use three triangle congruence postulates to prove triangles congruent. SAS Postulate: If there exists a correspondence between the vertices of two triangles such that two sides and the included angle of one triangle are congruent to the corresponding two sides and the included angle of the other triangle, then the two triangles are congruent (SAS ≅ ) A B C D E F
Learner Objective: I will identify the corresponding congruent parts of congruent figures and use three triangle congruence postulates to prove triangles congruent. ASA Postulate: If there exists a correspondence between the vertices of two triangles such that two angles and the included side of one triangle are congruent to the corresponding two angles and the included side of the other triangle, then the two triangles are congruent (ASA ≅ ) A B C D E F
Learner Objective: I will identify the corresponding congruent parts of congruent figures and use three triangle congruence postulates to prove triangles congruent.
STATEMENTS REASONS
Learner Objective: I will identify the corresponding congruent parts of congruent figures and use three triangle congruence postulates to prove triangles congruent. STATEMENTS REASONS
Learner Objective: I will identify the corresponding congruent parts of congruent figures and use three triangle congruence postulates to prove triangles congruent. STATEMENTS REASONS
Learner Objective: I will identify the corresponding congruent parts of congruent figures and use three triangle congruence postulates to prove triangles congruent. STATEMENTS REASONS
Learner Objective: I will identify the corresponding congruent parts of congruent figures and use three triangle congruence postulates to prove triangles congruent. STATEMENTS REASONS
Learner Objective: I will identify the corresponding congruent parts of congruent figures and use three triangle congruence postulates to prove triangles congruent. HW: pg. 114 # 1-3 pg. 120 # 1-4, 6-8
3-1 Learner Objective: I will identify the corresponding congruent parts of congruent figures and use three triangle congruence postulates to prove triangles congruent.
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## Precalculus (6th Edition) Blitzer
The standard form of the expression ${{\left( -2+\sqrt{-100} \right)}^{2}}$ is $-96-40i$.
Consider the expression, ${{\left( -2+\sqrt{-100} \right)}^{2}}$ Rewrite the expression ${{\left( -2+\sqrt{-100} \right)}^{2}}$ as ${{\left( -2+\sqrt{100}\sqrt{-1} \right)}^{2}}$ As $\sqrt{-1}=i$ Therefore, ${{\left( -2+\sqrt{100}\sqrt{-1} \right)}^{2}}={{\left( -2+10i \right)}^{2}}$ Apply the square of the difference on the above expression. \begin{align} & {{\left( -2+10i \right)}^{2}}={{2}^{2}}-2\left( 2 \right)\left( 10i \right)+{{\left( 10i \right)}^{2}} \\ & =4-40i+100{{i}^{2}} \end{align} As ${{i}^{2}}=-1$ Therefore, \begin{align} & 4-40i+100{{i}^{2}}=4-40i+100\left( -1 \right) \\ & =4-40i-100 \end{align} So, ${{\left( -2+10i \right)}^{2}}=4-40i-100$ To subtract two complex numbers, combine the real numbers together and the terms containing $i$. \begin{align} & 4-40i-100=\left( 4-100 \right)-40i \\ & =-96-40i \end{align} Hence, the standard form of the expression ${{\left( -2+\sqrt{-100} \right)}^{2}}$ is $-96-40i$. |
#### Please solve RD Sharma class 12 chapter Derivative As a Rate Measure exercise 12.2 question 21 maths textbook solution
Answer: $6 \mathrm{~cm}^{3} / \mathrm{sec}$
Hint: The surface area of the bubble will be, $S A=4 \pi r^{2}$ .
Given: The surface area of a spherical bubble is increasing at the rate of 2 cm2/sec
Solution: To find rate of increase of its volume, when the radius is 6 cm.
Let the radius of the given spherical bubble be $r$ cm at any instant time.
It is given that the surface area of a spherical bubble is increasing at the rate of $2 \mathrm{~cm}^{2} / \sec \frac{d s}{d t}=2$
\begin{aligned} &S=4 \pi r^{2} \\\\ &\frac{d S}{d t}=8 \pi r \times \frac{d r}{d t} \end{aligned}
\begin{aligned} &\frac{d r}{d t}=\frac{1}{8 \pi r} \times \frac{d S}{d t} \\\\ &\frac{d r}{d t}=\frac{1}{8 \pi r} \times 2 \quad\left(\frac{d S}{d t}=2(\text { given })\right) \end{aligned}
\begin{aligned} &\frac{d r}{d t}=\frac{1}{8 \pi \times 6} \times 2 \quad(r=6(\text { given })) \\\\ &\frac{d r}{d t}=\frac{1}{24 \pi} \mathrm{cm} / \mathrm{sec} \end{aligned}
Now, Volume of sphere $=V=\frac{4}{3} \pi r^{3}$
$\frac{d V}{d t}=4 \pi r^{2} \times \frac{d r}{d t}$
Let’s put value of $r =6$ and $\frac{d r}{d t}=\frac{1}{24 \pi} \mathrm{cm} / \mathrm{sec}$
$\frac{d V}{d t}=4 \pi \times 6^{2} \times \frac{1}{24 \pi}=6 \mathrm{~cm}^{3} / \mathrm{sec}$ |
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# The A to Z of mathematics a basic guide - Sidebotham T.H.
Sidebotham T.H. The A to Z of mathematics a basic guide - Wiley publishing , 2002. - 489 p.
ISBN 0-471-15045-2
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Example 1. Solve \x = 4.
Solution. Write
x = 3.5 0.7 x 5 = 3.5 or 3|
5 x = 5 x 4 Multiply both sides by 5
3jc = 20 3 20
-x = β Divide both sides by 3
x =6.1 20 -r- 3 = 6.7 (to 1 dp) or 6Β§
Example 2. Solve
x 1
β + 2 = - The two operations are add 2 and divide by 3
1
Solution. Write
1
2 β 2 = - β 2 Subtract 2 from both sides of the equation
x 1
3 x - = β 1- x 3 Multiply both sides by 3.
Example 3. Solve
3x + 17 = β6 The two operations are add 17 and multiply by 3
LINEAR EQUATION 277
Solution. Write
3x T 17 β 17 = β6 β 17 Subtracting 17 from both sides of the equation 3x = -23 3 β23
-x = βDividing both sides of equation by 3 a: = -7.7 β23 = 3 = -7.7 (to 1 dp) or - 7Β§
Example 4. Solve
3x β 5
7
Solution. The three operations on x, in order to form this equation, are multiply by 3, subtract 5, and divide by 7. In solving the equation this order is reversed, and of course the appropriate inverse operations are applied. Write
(3x 5) _ ^ Inserting brackets shows that all of 3a: β 5 is divided by 7
7(3x - 5)
=4x7 Multiplying both sides of the equation by 7
(3x - 5) = 28
3a: β 5 = 28 The brackets are no longer needed
3a:β5 + 5 = 28+ 5 Adding 5 to both sides of the equation
3a: = 33
3 33
-x = β Dividing both sides of the equation by 3
x = 11
Example 5. Solve 6x - 3 = 2(x + 5).
Solution. Write
6a: β 3 = 2a: + 10 Expanding the brackets
6x β 3 + 3 = 2a: + 10 + 3 Adding 3 to both sides of the equation
6x = 2a: + 13
6x β 2x = 2a: + 13 β 2a: Subtracting 2a: from both sides to get a:βs on the
left side of equation
278 LITER
4x = 13
4x _ 13 T ~ ~4
x = 3.25
6x β 2x = 4x and 2x β 2x = 0 Dividing both sides by 4
Cross multiplying is a useful way of eliminating fractions in equations, and applies to equations that have one fraction on each side of the equals sign. Cross multiplying is based upon the fact that if | = |, then ad = be.
Example. Solve
3.x x β 2 T β 4
Solution. Write
3x x 4 = 5(x β 2)
12x = 5x β 10 12x β 5x = 5x β 10 β 5x Ix = -10 Ix -10
y
x
7
-1.4
Inserting brackets and cross multiplying Expanding the brackets Subtracting 5x from both sides
Dividing both sides by 7 β 10^7 = -1.4 (toi dp) or -1
References: Balancing an Equation, Brackets, Degree, Equations, Inverse Operation, Operations.
LINEAR GRAPH
A linear graph is a straight-line graph.
LINEAR PROGRAMMING
Reference: Inequality.
LITER
The liter (abbreviation 1) is a unit of volume and is used for expressing the volume of liquids and capacity in general. For example, in countries where the metric system is used, milk is often sold in 1-liter quantities. There are 1000 milliliters (abbreviation ml) in 1 liter. The size of 1 liter corresponds to a cube of sides 10 cm. This means
locus 279
that 1 liter = 1000 cubic centimeters (cm3). Other useful equivalent relationships are as follows:
1 milliliter = 1 cubic centimeter
1000 liters = 1 cubic meter
A volume of 1000 liters of water has a mass of 1 tonne, provided the water is at 4Β° C (for the distinction between the unit tonne and the U.S. ton, see the entry Gram). Rainfall of 1 millimeter is equivalent to 1 liter of water per square meter.
References: CGS Units, Metric Units, SI Units.
LOCUS
The plural of locus is loci. A locus is a path made up of a set of points, and the position of each point in the path obeys a certain rale. This concept is demonstrated in the following example (see figure a). Amanda is training her horse, George, for a show. She stands in one place, at the point A, holding the end of a rope, and the other end of the rope is tied to George (G). As George trots around her, Amanda turns so that she is always facing him, but keeps the rope tight and the same length. Different positions of George make up the set of points, and the path of George is the locus of G. In this example the locus of G is a circle. The rale for this locus is that each point G is the same distance from the fixed point A. This locus is made up of an infinite number of positions of George.
Like the circle above, many lines and curves in mathematics can be defined in terms of a locus. Some of them are described here.
The line that is the bisector of an angle can be defined as the locus of points that are equidistant from the βarmsβ of the angle. In figure b, three of the points of the locus, each at an equal distance from either arm, are marked on the angle bisector.
Example. Jo has a dog, Spot, who is occasionally tied up in the garden with the end of his leash free to pass along a length of wire AB which is fixed to the ground at each end. The length of the leash is 2 meters and the length of the wire is 6 meters. Draw the region in which Spot is free to move, assuming the leash does not get tangled up with the wire.
280 LOWEST COMMON MULTIPLE
Solution. This is a locus problem where the set of points covers a region, rather than represents a line. First we draw the locus of Spot at his furthest positions from the wire AB (see figure c). The locus of the furthermost position of Spot is made up of two semicircles each of radius 2 meters, and two straight and parallel line segments each of length 6 meters. The region over which Spot is free to move is shaded.
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# Key Eleven
## An Advanced Mathematics Lesson Starter Of The Day
Choose 4 keys on your calculator key pad that are positioned in the four corners of a rectangle. Use these keys to type in a 4-digit number going round your rectangle either clockwise or anticlockwise. Eg. 3146
Check that your 4-digit number is a multiple of eleven.
Can you prove that all 4-digit numbers formed this way are multiples of 11?
Is this strange result also true for numbers formed from keys on the corners of a parallelogram?
Note 1: The rectangles referred to above have horizontal or vertical sides.
Note 2: A divisibility test for eleven is to consider the alternating digit sum of the number. If this is divisible by eleven then the original number is also divisible by eleven. For example, if a four digit number has digits a, b, c and d then the alternating digit sum is a - b + c - d.
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Let the rectangle have a width of $$w$$ key spacings and a height of $$h$$ key spacings. (The example in the diagram above has $$w = 2$$ and $$h = 1$$ )
Let the number be formed by going around the rectangle in a clockwise direction starting from the top left (unlike the example in the diagram above).
Let the first digit of the 4-digit number be $$x$$.
The alternating digit sum is $$x - (x+w) + (x+w+h) - (x+h)$$
$$= x - x - w + x + w + h - x - h$$
$$= 0$$ (which is a multiple of 11 so the 4-digit number must also be a multiple of 11)
Also any cyclic permutation of the 4-digit number will be divisible by 11 (accounting for the example in the diagram above)
This should also be applicable to parallelograms
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Christmas Present Ideas
It is often very difficult choosing Christmas presents for family and friends but so here are some seasonal, mathematics-related gifts chosen and recommended by Transum Mathematics.
## Equate board game
Here's a great board game that will give any family with school-aged kids hours of worthwhile fun. Christmas is a time for board games but this one will still be useful at any time of year. Games can be adapted to suit many levels of Mathematical ability.
For Maths tutors working with just one or small groups of pupils this game has proved to be an excellent activity for a tutorial. Deciding on the best moves can spark pertinent discussions about mathematical concepts.
Equate looks a bit like Scrabble--for aspiring mathematicians, that is. Designed by a real mathematician, it works like this: You put down tiles on a board and make points by correctly completing simple equations. Your nine tiles include both numbers and mathematical symbols; you can add on to previous plays both vertically and horizontally. more...
## How Not To Be Wrong
The maths we learn in school can seem like an abstract set of rules, laid down by the ancients and not to be questioned. In fact, Jordan Ellenberg shows us, maths touches on everything we do, and a little mathematical knowledge reveals the hidden structures that lie beneath the world's messy and chaotic surface. In How Not to be Wrong, Ellenberg explores the mathematician's method of analyzing life, from the everyday to the cosmic, showing us which numbers to defend, which ones to ignore, and when to change the equation entirely. Along the way, he explains calculus in a single page, describes Gödel's theorem using only one-syllable words, and reveals how early you actually need to get to the airport.
What more could the inquisitive adult want for Christmas? This book makes a cosy, interesting read in front of the fire on those cold winter evenings. more...
## Graphic Display Calculator
This handheld device and companion software are designed to generate opportunities for classroom exploration and to promote greater understanding of core concepts in the mathematics and science classroom. TI-Nspire technology has been developed through sound classroom research which shows that "linked multiple representation are crucial in development of conceptual understanding and it is feasible only through use of a technology such as TI-Nspire, which provides simultaneous, dynamically linked representations of graphs, equations, data, and verbal explanations, such that a change in one representation is immediately reflected in the others.
For the young people in your life it is a great investment. Bought as a Christmas present but useful for many years to come as the young person turns into an A-level candidate then works their way through university. more...
The analytics show that more and more people are accessing Transum Mathematics via an iPad as it is so portable and responsive. The iPad has so many other uses in addition to solving Transum's puzzles and challenges and it would make an excellent gift for anyone.
The redesigned Retina display is as stunning to look at as it is to touch. It all comes with iOS, the world's most advanced mobile operating system. iPad Pro. Everything you want modern computing to be. more...
## Aristotle's Number Puzzle
It’s a bit of a tradition to give puzzles as Christmas Gifts to nieces and nephews. This puzzle is ideal for the keen puzzle solver who would like a challenge that will continue over the festive period (at least!).
This number puzzle involves nineteen numbers arranged into a hexagon. The goal of the puzzle is to rearrange the numbers so each of the fifteen rows add up to 38. It comes in a wooden style with an antique, aged look.
Keep the Maths in Christmaths with this reasonably priced stocking filler. more...
## The Story Of Maths [DVD]
The films in this ambitious series offer clear, accessible explanations of important mathematical ideas but are also packed with engaging anecdotes, fascinating biographical details, and pivotal episodes in the lives of the great mathematicians. Engaging, enlightening and entertaining, the series gives viewers new and often surprising insights into the central importance of mathematics, establishing this discipline to be one of humanity s greatest cultural achievements. This DVD contains all four programmes from the BBC series.
Marcus du Sautoy's wonderful programmes make a perfect Christmas gift more...
## Christmas Maths
This book provides a wealth of fun activities with a Christmas theme. Each photocopiable worksheet is matched to the Numeracy Strategy and compatible with the Scottish 5-14 Guidelines. This series is designed for busy teachers in the late Autumn term who are desperate for materials that are relevant and interesting and that can be completed with minimun supervision.
All the activities are suitable for use by class teachers, supply teachers, SEN teachers and classroom assistants and cover topics such as 'How many partridges did the true love give all together?' and 'Filling a sleigh with presents by rolling a dice!'. Children will have lots of fun working through the Christmas Maths themes but also gain valuable skills along the way.
A great source of ideas and another reasonably priced stocking filler. more...
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Your access to the majority of the Transum resources continues to be free but you can help support the continued growth of the website by doing your Amazon shopping using the links on this page. Below is an Amazon search box and some items chosen and recommended by Transum Mathematics to get you started.
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# How do you solve x^3>2x^2+x using a sign chart?
Aug 25, 2017
Solution : $- 0.4142 < x < 0 \ast \mathmr{and} \ast$ x >2.4142 **In interval notation:** (-0.4142,0) uu (2.4142,oo)
#### Explanation:
${x}^{3} > 2 {x}^{2} + x \mathmr{and} {x}^{3} - 2 {x}^{2} - x > 0 \mathmr{and} x \left({x}^{2} - 2 x - 1\right) > 0$ or
Roots of $\left({x}^{2} - 2 x - 1\right)$ are $x = \frac{2 \pm \sqrt{\left({2}^{2} - 4 \cdot 1 \cdot - 1\right)}}{2}$
or $x = 1 \pm \sqrt{2} \mathmr{and} x = 2.4142 , x = - 0.4142$
$\therefore x \left({x}^{2} - 2 x - 1\right) > 0 \mathmr{and} x \left(x - 2.4142\right) \left(x + 0.4142\right) > 0$
Critical points are $x = 0 , x = 2.4142 , x = - 0.4142$
Sign chart:
When $x < - 0.4142$ sign of $x \left(x - 2.4142\right) \left(x + 0.4142\right)$ is
(-) * (-) * (-) = (-) ; <0
When $- 0.4142 < x < 0$ sign of $x \left(x - 2.4142\right) \left(x + 0.4142\right)$ is
(-) * (-) * (+) = (+) ; >0
When $0 < x < 2.4142$ sign of $x \left(x - 2.4142\right) \left(x + 0.4142\right)$ is
(+) * (-) * (+) = (-) ; <0
When $x > 2.4142$ sign of $x \left(x - 2.4142\right) \left(x + 0.4142\right)$ is
(+) * (+) * (+) = (+) ; >0
Solution: $- 0.4142 < x < 0 \mathmr{and}$ x >2.4142
In interval notation: $\left(- 0.4142 , 0\right) \cup \left(2.4142 , \infty\right)$
graph{x^3-2x^2-x [-10, 10, -5, 5]} [Ans] |
Paul's Online Math Notes
[Notes]
Calculus I - Notes
Limits Previous Chapter Next Chapter Applications of Derivatives Derivatives of Trig Functions Previous Section Next Section Derivatives of Inverse Trig Functions
## Derivatives of Exponential and Logarithm Functions
The next set of functions that we want to take a look at are exponential and logarithm functions. The most common exponential and logarithm functions in a calculus course are the natural exponential function, , and the natural logarithm function, . We will take a more general approach however and look at the general exponential and logarithm function.
Exponential Functions
We’ll start off by looking at the exponential function,
We want to differentiate this. The power rule that we looked at a couple of sections ago won’t work as that required the exponent to be a fixed number and the base to be a variable. That is exactly the opposite from what we’ve got with this function. So, we’re going to have to start with the definition of the derivative.
Now, the is not affected by the limit since it doesn’t have any h’s in it and so is a constant as far as the limit is concerned. We can therefore factor this out of the limit. This gives,
Now let’s notice that the limit we’ve got above is exactly the definition of the derivative of at , i.e. . Therefore, the derivative becomes,
So, we are kind of stuck. We need to know the derivative in order to get the derivative!
There is one value of a that we can deal with at this point. Back in the Exponential Functions section of the Review chapter we stated that What we didn’t do however is actually define where e comes from. There are in fact a variety of ways to define e. Here are three of them.
Some Definitions of e.
e is the unique positive number for which
The second one is the important one for us because that limit is exactly the limit that we’re working with above. So, this definition leads to the following fact,
Fact 1
For the natural exponential function, we have .
So, provided we are using the natural exponential function we get the following.
At this point we’re missing some knowledge that will allow us to easily get the derivative for a general function. Eventually we will be able to show that for a general exponential function we have,
Logarithm Functions
Let’s now briefly get the derivatives for logarithms. In this case we will need to start with the following fact about functions that are inverses of each other.
Fact 2
If f(x) and g(x) are inverses of each other then,
So, how is this fact useful to us? Well recall that the natural exponential function and the natural logarithm function are inverses of each other and we know what the derivative of the natural exponential function is!
So, if we have and then,
The last step just uses the fact that the two functions are inverses of each other.
Putting this all together gives,
Note that we need to require that since this is required for the logarithm and so must also be required for its derivative. It can also be shown that,
Using this all we need to avoid is .
In this case, unlike the exponential function case, we can actually find the derivative of the general logarithm function. All that we need is the derivative of the natural logarithm, which we just found, and the change of base formula. Using the change of base formula we can write a general logarithm as,
Differentiation is then fairly simple.
We took advantage of the fact that a was a constant and so is also a constant and can be factored out of the derivative. Putting all this together gives,
Here is a summary of the derivatives in this section.
Okay, now that we have the derivations of the formulas out of the way let’s compute a couple of derivatives.
Example 1 Differentiate each of the following functions. (a) (b) (c) Solution (a) This will be the only example that doesn’t involve the natural exponential and natural logarithm functions. (b) Not much to this one. Just remember to use the product rule on the second term. (c) We’ll need to use the quotient rule on this one.
There’s really not a lot to differentiating natural logarithms and natural exponential functions at this point as long as you remember the formulas. In later sections as we get more formulas under our belt they will become more complicated.
Next, we need to do our obligatory application/interpretation problem so we don’t forget about them.
Example 2 Suppose that the position of an object is given by Does the object ever stop moving? Solution First we will need the derivative. We need this to determine if the object ever stops moving since at that point (provided there is one) the velocity will be zero and recall that the derivative of the position function is the velocity of the object. The derivative is, So, we need to determine if the derivative is ever zero. To do this we will need to solve, Now, we know that exponential functions are never zero and so this will only be zero at . So, if we are going to allow negative values of t then the object will stop moving once at . If we aren’t going to allow negative values of t then the object will never stop moving.
Before moving on to the next section we need to go back over a couple of derivatives to make sure that we don’t confuse the two. The two derivatives are,
It is important to note that with the Power rule the exponent MUST be a constant and the base MUST be a variable while we need exactly the opposite for the derivative of an exponential function. For an exponential function the exponent MUST be a variable and the base MUST be a constant.
It is easy to get locked into one of these formulas and just use it for both of these. We also haven’t even talked about what to do if both the exponent and the base involve variables. We’ll see this situation in a later section.
Derivatives of Trig Functions Previous Section Next Section Derivatives of Inverse Trig Functions Limits Previous Chapter Next Chapter Applications of Derivatives
[Notes]
© 2003 - 2018 Paul Dawkins |
# LCM (3,822; 1,234) = ? Calculate the least common multiple, LCM, by two methods: 1) The prime factorization of the numbers and 2) The Euclidean algorithm
## The least common multiple, LCM: the latest 5 calculated values
The LCM of 3,822 and 1,234 = ? Jul 05 21:58 UTC (GMT) The LCM of 18 and 149 = ? Jul 05 21:58 UTC (GMT) The LCM of 643 and 2 = ? Jul 05 21:58 UTC (GMT) The LCM of 4,851 and 847 = ? Jul 05 21:58 UTC (GMT) The LCM of 51 and 132,650 = ? Jul 05 21:58 UTC (GMT) The least common multiple, LCM: the list of all the operations
## The least common multiple (lcm). What it is and how to calculate it.
• The number 60 is a common multiple of the numbers 6 and 15 because 60 is a multiple of 6 (60 = 6 × 10) and also a multiple of 15 (60 = 15 × 4).
• There are infinitely many common multiples of 6 and 15.
• If the number "v" is a multiple of the numbers "a" and "b", then all the multiples of "v" are also multiples of "a" and "b".
• The common multiples of 6 and 15 are the numbers 30, 60, 90, 120, and so on.
• Out of these, 30 is the smallest, 30 is the least common multiple (lcm) of 6 and 15.
• Note: The prime factorization of a number: finding the prime numbers that multiply together to give that number.
• If e = lcm (a, b), then the prime factorization of "e" must contain all the prime factors involved in the prime factorization of "a" and "b" taken by the highest power.
• Example:
• 40 = 23 × 5
• 36 = 22 × 32
• 126 = 2 × 32 × 7
• lcm (40, 36, 126) = 23 × 32 × 5 × 7 = 2,520
• Note: 23 = 2 × 2 × 2 = 8. We are saying that 2 was raised to the power of 3. Or, shorter, 2 to the power of 3. In this example 3 is the exponent and 2 is the base. The exponent indicates how many times the base is multiplied by itself. 23 is the power and 8 is the value of the power.
• Another example of calculating the least common multiple, lcm:
• 938 = 2 × 7 × 67
• 982 = 2 × 491
• 743 = is a prime number and cannot be broken down into other prime factors
• lcm (938, 982, 743) = 2 × 7 × 67 × 491 × 743 = 342,194,594
• If two or more numbers have no common factors (they are coprime), then their least common multiple is calculated by simply multiplying the numbers.
• Example:
• 6 = 2 × 3
• 35 = 5 × 7
• lcm (6, 35) = 2 × 3 × 5 × 7 = 6 × 35 = 210 |
# Maths - Bilinear Multiplication
We start by reviewing linearity and then extend this to bilineararity.
## Linear Multiplication.
To represent linear space we only need two operations:
• Vector addition: to add two vectors we take the start of the second vector and move it to the end of the first vector. The addition of these two vectors is the vector from the start of the first vector to the end of the second vector.
• Scalar multiplication changes the length of a vector without changing its direction. That is we 'scale' it by the multiplying factor. So scalar multiplication involves multiplying a scalar (single number) by a vector to give another number.
If the space is linear we have:
a*(v +u) = a*v + a*u
where:
• v & u are vectors
• a is a scalar
In other words the two operations are distributive.
## Bilinear Multiplication
To have bilinearity we need to include a multiplication where both operands are vectors.
For a bilinear multiplication, if we keep one of the vectors constant and then the other vector varies in a linear way, as defined above.
For example, in the diagram above we take the product as the area formed from the two vectors. So if the area formed from V and U is V•U then if we multiply V by a scalar value 'a' we get the area a*V•U, so,
(a*V)•U = a*(V•U)
and similarily if we multiply U by b:
V•(b*U) = b*(V•U)
so combining these gives:
(a*V)•(b*U) = ab*(V•U)
An example of this type of multiplication is the vector dot product.
However this type of multiplication takes two vectors as input and produces a scalar value as an output. Is it possible to have a bilinear multiplication which takes two vectors as input and produces a vector as an output?
The requirement is:
(a V + a' V')×(b U + b' U') = ab(V×U) + ab'(V×U') +a'b(V'×U) +a'b'(V'×U')
this applies to the vector cross product.
The vector cross product only applies in 1,3 and 7 dimensions. However there are ways to generalise the vector cross product to other numbers of dimensions. |
# 17.2 Surface Area of Prisms and Cylinders
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1 Name Class Date 17. Surface Area of Prisms and Cylinders Essential Question: How can you find the surface area of a prism or cylinder? Explore G.11.C Apply the formulas for the total and lateral surface area of three-dimensional figures, including prisms,... cylinders,... to solve problems using appropriate units of measure. Also G.10.B Developing a Surface Area Formula Resource Locker Surface area is the total area of all the faces and curved surfaces of a three-dimensional figure. The lateral area of a prism is the sum of the areas of the lateral faces. Consider the right prism shown here and the net for the right prism. Complete the figure by labeling the dimensions of the net. h a c b B In the net, what type of figure is formed by the lateral faces of the prism? Write an expression for the length of the base of the rectangle. How is the base of the rectangle related to the perimeter of the base of the prism? The lateral area L of the prism is the area of the rectangle. Write a formula for L in terms of h, a, b, and c. Module Lesson
2 F Write the formula for L in terms of P, where P is the perimeter of the base of the prism. G Let B be the area of the base of the prism. Write a formula for the surface area S of the prism in terms of B and L. Then write the formula in terms of B, P, and h. Reflect 1. Explain why the net of the lateral surface of any right prism will always be a rectangle.. Suppose a rectangular prism has length l, width w, and height h, as shown. Explain how you can write a formula for the surface area of the prism in terms of l, w, and h. h l w Explain 1 Finding the Surface Area of a Prism Lateral Area and Surface Area of Right Prisms The lateral area of a right prism with height h and base perimeter P is L = Ph. The surface area of a right prism with lateral area L and base area B is S = L + B, or S = Ph + B. h B Module Lesson
3 Example 1 Each gift box is a right prism. Find the total amount of paper needed to wrap each box, not counting overlap. Step 1 Find the lateral area. Lateral area formula L = Ph P = (8) + (6) = 8 cm = 8(1) Multiply. = 336 c m Step Find the surface area. Surface area formula S = L + B Substitute the lateral area. = (6)(8) Simplify. = 43 c m 8 cm 1 cm 6 cm B Image Credits: C Squared Studios/Photodisc/Getty Images Step 1 Find the length c of the hypotenuse of the base. Pythagorean Theorem c = a + b Substitute. = + Simplify. = Take the square root of each side. c = Step Find the lateral area. Lateral area formula L = Ph Substitute. = ( ) Multiply. = in 10 in. 4 in. 0 in. Module Lesson
4 Step 3 Find the surface area. Surface area formula S = L + B Substitute. = + 1_ Simplify. = i n Reflect 3. A gift box is a rectangular prism with length 9.8 cm, width 10. cm, and height 9.7 cm. Explain how to estimate the amount of paper needed to wrap the box, not counting overlap. Your Turn Each gift box is a right prism. Find the total amount of paper needed to wrap each box, not counting overlap in. 6 in. 18 in. 5 in. 3.6 in. 8.5 in. Module Lesson
5 Explain Finding the Surface Area of a Cylinder Lateral Area and Surface Area of Right Cylinders The lateral area of a cylinder is the area of the curved surface that connects the two bases. The lateral area of a right cylinder with radius r and height h is L = πrh. The surface area of a right cylinder with lateral area L and base area B is S = L + B, or S = πrh + πr. r r πr h h Example Each aluminum can is a right cylinder. Find the amount of paper needed for the can s label and the total amount of aluminum needed to make the can. Round to the nearest tenth. Step 1 Find the lateral area. 3 cm Lateral area formula L = πrh Substitute. L = π (3) (9) Multiply. = 54π cm Step Find the surface area. Surface area formula S = L + π r Substitute the lateral area and radius. = 54π + r (3) Simplify. = 7π cm Step 3 Use a calculator and round to the nearest tenth. 9 cm The amount of paper needed for the label is the lateral area, 54π c m. The amount of aluminum needed for the can is the surface area, 7π 6. c m. Module Lesson
6 B Step 1 Find the lateral area. Lateral area formula L = πrh in 5 in Substitute; the radius is half the diameter. = π ( ) ( ) Multiply. = π in Step Find the surface area. Surface area formula S = L + π r Substitute the lateral area and radius. = π + r ( ) Simplify. = π in Step 3 Use a calculator and round to the nearest tenth. The amount of paper needed for the label is the lateral area, π i n. The amount of aluminum needed for the can is the surface area, π i n. Reflect 6. In these problems, why is it best to round only in the final step of the solution? Your Turn Each aluminum can is a right cylinder. Find the amount of paper needed for the can s label and the total amount of aluminum needed to make the can. Round to the nearest tenth mm 15 cm 6 cm 7 mm Module Lesson
7 Explain 3 Finding the Surface Area of a Composite Figure Example 3 Find the surface area of each composite figure. Round to the nearest tenth. Step 1 Find the surface area of the right rectangular prism. Surface area formula S = Ph + B 4 ft 0 ft Substitute. = 80 (0) + (4) (16) Simplify. = 368 f t Step A cylinder is removed from the prism. Find the lateral area of the cylinder and the area of its bases. 4 ft 16 ft Lateral area formula L = πrh Substitute. = π (4) (0) Simplify. = 160π ft Base area formula B = π r Substitute. = π (4) Simplify. = 16π ft Step 3 Find the surface area of the composite figure. The surface area is the sum of the areas of all surfaces on the exterior of the figure. S = (prism surface area) + (cylinder lateral area) - (cylinder base areas) = π - (16π) = π ft Module Lesson
8 B Step 1 Find the surface area of the right rectangular prism. 3 cm cm Surface area formula S = Ph + B Substitute. = ( ) + ( ) ( ) Simplify. = cm Step Find the surface area of the cylinder. Lateral area formula L = πrh Substitute. = π ( ) ( ) Simplify. = π cm Surface area formula S = L + π r Substitute. = π + π ( ) Simplify. = π cm 9 cm 5 cm 4 cm Step 3 Find the surface area of the composite figure. The surface area is the sum of the areas of all surfaces on the exterior of the figure. S = (prism surface area) + (cylinder surface area) - (area of one cylinder base) = + π - π ( ) = + π cm Reflect 9. Discussion A student said the answer in Part A must be incorrect since a part of the rectangular prism is removed, yet the surface area of the composite figure is greater than the surface area of the rectangular prism. Do you agree with the student? Explain. Module Lesson
9 Your Turn Find the surface area of each composite figure. Round to the nearest tenth in 3 in 7 mm 3 in 5 in 6 mm 9 in 7 in 3 mm Elaborate 1. Can the surface area of a cylinder ever be less than the lateral area of the cylinder? Explain. 13. Is it possible to find the surface area of a cylinder if you know the height and the circumference of the base? Explain. 14. Essential Question Check-In How is finding the surface area of a right prism similar to finding the surface area of a right cylinder? Module Lesson
10 Evaluate: Homework and Practice Find the lateral area and surface area of each prism cm 3 cm Online Homework Hints and Help Extra Practice 3 ft 7 ft 5 cm cm 5 ft cm 10 cm 5 cm 5 cm 1 m m Module Lesson
11 Find the lateral area and surface area of the cylinder. Leave your answer in terms of π ft 4 ft 11 in. 7 in. Find the total surface area of the composite figure. Round to the nearest tenth ft 14 ft 6 ft 14 ft 8 ft 14 ft 14 ft 8 ft 1 ft Module Lesson
12 Find the total surface area of the composite figure. Round to the nearest tenth cm ft ft 0.5 ft cm 6 cm 9 cm ft 10 cm 1 ft 11. The greater the lateral area of a florescent light bulb, the more light the bulb produces. One cylindrical light bulb is 16 inches long with a 1-inch radius. Another cylindrical bulb is 3 inches long with a 3 -inch radius. Which bulb will produce more light? 4 1. Find the lateral and surface area of a cube with edge length 9 inches. 13. Find the lateral and surface area of a cylinder with base area 64π m and a height 3 meters less than the radius. Module Lesson
13 14. Biology Plant cells are shaped approximately like a right rectangular prism. Each cell absorbs oxygen and nutrients through its surface. Which cell can be expected to absorb at a greater rate? (Hint: 1 μm = 1 micrometer = meter) 7 µm 35 µm 10 µm 15 µm 11 µm 15 µm ge07se_c10l04003a AB 15. Find the height of a right cylinder with surface area 160π f t and radius 5 ft. 16. Find the height of a right rectangular prism with surface area 86 m, length 10 m, and width 8 m. 17. Represent Real-World Problems If one gallon of paint covers 50 square feet, how many gallons of paint will be needed to cover the shed, not including the roof? If a gallon of paint costs \$5, about how much will it cost to paint the walls of the shed? 1 ft 1 ft 18 ft 18 ft Module Lesson
14 18. Match the Surface Area with the appropriate coin in the table. Coin Diameter (mm) Thickness (mm) Surface Area (m m ) Penny Nickel Dime Quarter A B C D Algebra The lateral area of a right rectangular prism is 144 c m. Its length is three times its width, and its height is twice its width. Find its surface area. 0. A cylinder has a radius of 8 cm and a height of 3 cm. Find the height of another cylinder that has a radius of 4 cm and the same surface area as the first cylinder. Module Lesson
15 H.O.T. Focus on Higher Order Thinking 1. Analyze Relationships Ingrid is building a shelter to protect her plants from freezing. She is planning to stretch plastic sheeting over the top and the ends of the frame. Assume that the triangles in the frame on the left are equilateral. Which of the frames shown will require more plastic? Explain how finding the surface area of these figures is different from finding the lateral surface area of a figure. 10 ft 10 ft 10 ft 10 ft 10 ft ge07sec10l04004aa 1st pass 4/3/5 cmurphy ge07sec10l04005a 1st pass 4/1/5 cmurphy. Communicate Mathematical Ideas Explain how to use the net of a three-dimensional figure to find its surface area. 3. Draw Conclusions Explain how the edge lengths of a rectangular prism can be changed so that the surface area is multiplied by 9. Module Lesson
16 Lesson Performance Task A manufacturer of number cubes has the bright idea of packaging them individually in cylindrical boxes. Each number cube measures inches on a side. 1. What is the surface area of each cube?. What is the surface area of the cylindrical box? Assume the cube fits snugly in the box and that the box includes a top. Use 3.14 for π. Module Lesson
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# 1.12 Round numbers
Lesson
## Are you ready?
We've seen how to break up, or partition, numbers , up to the thousands. Understanding how to do this will help us with our estimating.
### Exploration
Enter a three digit number into the applet and drag the place value cards around to see what amount each digit represents.
The place value cards can be used to write the expanded form of the numbers.
### Examples
#### Example 1
Fill in the boxes with the missing numbers.
698=⬚+⬚+8
Worked Solution
Create a strategy
Write the first number in a place value table.
Apply the idea
Here is the complete table:
We have 6 hundreds which equals 600.
We have 9 tens which equals 90.
We have 8 units which equals 8.
Idea summary
We can use a place value table to write a number in expanded form.
## Round numbers to the nearest 10
Sometimes we may need to solve problems by estimating. We can see if our number is closer to the ten before it, or after it. Let's see how we round our number to the nearest ten.
Loading video...
### Examples
#### Example 2
Round 54 to the nearest ten.
Worked Solution
Create a strategy
Plot 54 on a number line and look for the closest 10.
Apply the idea
Based on the number line, 54 is closest to 50.
The rounded value is 50.
Idea summary
To round a number to the nearest 10, plot the number on a number line to see which ten is closest to the number.
## Rounding numbers in more ways
We can also round to the nearest hundred or thousand. Let's see how.
Loading video...
### Examples
#### Example 3
Round 370 to the nearest hundred.
Worked Solution
Create a strategy
Plot 370 on a number line and look for the closest 100.
Apply the idea
Based on the number line, 370 is closest to 400.
The rounded value is 400.
Idea summary
• If you are rounding a number to the nearest 10, think about whether it is closer to the one above, or the one below. For example, 20 or 30.
• If a number is exactly in the middle, round to the higher number.
• This is the same for hundreds.
### Outcomes
#### MA2-4NA
applies place value to order, read and represent numbers of up to five digits |
CLASS-7INTRODUCTION OF RATIONAL NUMBERS
RATIONAL NUMBERS -
Rational numbers are numbers that can be expressed as the quotient or fraction of two integers, where the denominator is not zero. In other words, a rational number is any number that can be written in the form ๐/๐โ, where ๐ and ๐ are integers and ๐ โ 0.
Here are some key properties of rational numbers:-
1. Closure under Operations:- The sum, difference, product, and quotient (except division by zero) of any two rational numbers is also a rational number.
2. Repeating or Terminating Decimals:- Every rational number can be represented as either a terminating decimal (e.g., 3/4 = 0.75) or a repeating decimal (e.g., 1/3 = 0.3โพ ).
3. Equivalent Fractions:- Rational numbers can have an infinite number of equivalent fractions. For example, 1/2 = 2/4 = 3/6 =โฆ.................
4. Ordering:- Rational numbers can be ordered on the number line. For any two rational numbers ๐/๐โ and ๐/๐ โ, if ๐/๐ < ๐/๐ โ, then ๐๐ < ๐๐.
Rational numbers are numbers that can be expressed as the quotient or fraction of two integers, where the denominator is not zero. In other words, a rational number is any number that can be written in the form ๐๐qpโ, where ๐p and ๐q are integers and ๐q is not equal to zero.
Rational numbers include integers, fractions, and terminating or repeating decimals. Here are some examples of rational numbers:
1. Integers: โ3, 0, 1, 7
1 3 5
2. Fractions: ------, ------, โ ------
2 4 6
1 3 7
3. Terminating decimals: 0.25 = -----, 0.6 = -----, โ1.75 = โ -----
4 5 4
1 2 1
4. Repeating decimals: 0.3โพ= -----, 0.6โพ= ------, 0.142857โพ= -----
3 3 7
It's important to note that every integer is also a rational number, as it can be expressed as a fraction with a denominator of 1 (e.g., 5 = 5โ/1). Additionally, irrational numbers, such as โ2โ or ฯ, cannot be expressed as fractions of integers and are not considered rational numbers.
Rational numbers are closed under addition, subtraction, multiplication, and division, meaning that the result of any arithmetic operation on rational numbers is also a rational number. They form an important part of the number system and are used extensively in mathematics, science, and everyday life.
ยช IMPORTANT TO KNOW:-
1. Every integer (positive, negative or zero) is a rational number.
2. Every fraction is a rational number.
The rational number(-p/q) can be written as โ (p/q).
p/q is a positive rational number and -p/q is a negative number. |
How can I calculate 7 5 2
Insert numbers in a term
Calculate the value of the term for \$\$ x = 1 \$\$.
\$\$ 4 * (x + 2) \$\$
To do this, you put in for the value.
\$\$4*(\$\$ \$\$+\$\$ \$\$2)\$\$
Now you can calculate the term. Always turn the Priority rules at.
\$\$4*(\$\$ \$\$+\$\$ \$\$2)\$\$
\$\$=4*3\$\$
\$\$=12\$\$
Terms are meaningful combinations of numbers, variables and arithmetic symbols.
Examples:
\$\$(5+3)\$\$
\$\$ x + 3 \$\$
\$\$1/2\$\$
\$\$ - 2 * x \$\$
Priority rules:
1. Always brackets first
2. Point before line calculation
3. calculate from left to right
Calculate terms for multiple values
For different values you calculate the term one after the other.
Calculate the values of the term for \$\$ x = \$\$,, and.
\$\$ 3 * x-4 \$\$
\$\$ x = \$\$: \$\$ x = \$\$:
\$\$3*\$\$ \$\$-\$\$ \$\$4\$\$ \$\$3*\$\$ \$\$-\$\$ \$\$4\$\$
\$\$=2\$\$ \$\$=8\$\$
\$\$ x = \$\$: \$\$ x = \$\$:
\$\$3*\$\$ \$\$-\$\$ \$\$4\$\$ \$\$3*\$\$ \$\$-\$\$ \$\$4\$\$
\$\$=23\$\$ \$\$=14\$\$
At four You set values four times a number for \$\$ x \$\$ and do the math four times out.
Calculate terms for negative numbers
You can also use \$\$ 0 \$\$ or negative numbers for \$\$ x \$\$.
Calculate the values of the term for \$\$ x = \$\$,, and.
\$\$ (x + 2) * 3 \$\$
\$\$ x = \$\$: \$\$ x = \$\$:
\$\$(\$\$ \$\$+\$\$ \$\$2)*3\$\$ \$\$(\$\$ \$\$+\$\$ \$\$2)*3\$\$
\$\$=6\$\$ \$\$=3\$\$
Make sure you are really counting on the negative number when asked.
\$\$ x = \$\$: \$\$ x = \$\$:
\$\$(\$\$ \$\$+\$\$ \$\$2)*3\$\$ \$\$(\$\$ \$\$+\$\$ \$\$2)*3\$\$
\$\$=-6\$\$ \$\$=18\$\$
Negative numbers do not automatically mean that the result is negative.
The results can be very different if, for example, you calculate with \$\$ 4 \$\$ instead of \$\$ - 4 \$\$.
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Share by \$\$ 0 \$\$?
You can't divide by zero when calculating terms either. Dividing by zero is not possible.
Example 1: Calculate the value of the term for \$\$ x = 0 \$\$.
\$\$ (2: x) * 3 \$\$
\$\$(2:\$\$ \$\$)*3\$\$
not solvable
You cannot calculate the term for \$\$ x = 0 \$\$. You can calculate it for all other numbers.
Example 2: Substitute \$\$ 2 \$\$ and \$\$ - 2 \$\$ for \$\$ x \$\$ and calculate the term.
\$\$ 4: (2 + x) \$\$
for \$\$ x = -2 \$\$ for \$\$ x = 2 \$\$
\$\$4:(2+(-2))\$\$ \$\$4:(2+2)\$\$
\$\$=4:\$\$ \$\$=4:4\$\$
not solvable \$\$=1\$\$
You cannot calculate the term for \$\$ x = -2 \$\$. For all other values (example \$\$ x = 2 \$\$) it works anyway.
When you encounter dividing by zero, write not solvable to your account.
A variable can appear several times in a term
If a variable occurs more than once in a term, put \$\$ x \$\$ in each the same value a.
Calculate the value of the term for \$\$ x = \$\$ and \$\$ x = \$\$.
\$\$ 4 * x + x \$\$
For \$\$ x = 2 \$\$:
\$\$4*\$\$ \$\$+\$\$ \$\$=10\$\$
For \$\$ x = 3 \$\$:
\$\$4*\$\$ \$\$+\$\$ \$\$=15\$\$
Even if you are supposed to use several different values for x, you can only ever use the same number.
Wrong: \$\$ 4 * x + x \$\$ \$\$ rarr \$\$ 4 · 2 +3
Different variables can be in one term
Terms can have multiple variables.
Calculate the value of the term for and.
\$\$+\$\$ \$\$2*\$\$
For and :
\$\$+\$\$ \$\$2*\$\$
\$\$=3+4\$\$
\$\$=7\$\$
Don't swap the values and the variables. If you use the \$\$ y \$\$ value for the \$\$ x \$\$ value, this usually gives a different result.
Wrong: \$\$ 2 + 2 * 3 = 8 \$\$
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Free The First Quadrant Year 6 Position and Direction Learning Video Clip | Classroom Secrets
MathsYear 6Autumn Block 4 (Position and Direction)01 The First Quadrant › Free The First Quadrant Year 6 Position and Direction Learning Video Clip
# Free The First Quadrant Year 6 Position and Direction Learning Video Clip
## Step 1: Free The First Quadrant Year 6 Position and Direction Learning Video Clip
Pirate Pearl is plotting and writing coordinates for her journey to a treasure island.
More resources for Autumn Block 4 Step 1
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[/s2If]
Discussion points for teachers
1. Plot the given coordinates on her map. Join the coordinate points together to show the route she takes.
Discuss which axis we read first on the grid. Discuss joining points together accurately using a ruler.
Points plotted accurately on the grid (as shown) and joined accurately to show Pearl’s route.
2. Write the coordinates for each point.
Discuss how to write and present the written coordinates. Reminder of which axis we read first.
A = (1, 3); B = (2, 5); C = ( 4, 7); D = (6, 6); E = (7, 4)
3. Write and plot six more coordinates which take Pearl safely through the rocks.
Discuss a good strategy to use: marking the route first and then writing each coordinate. Discuss making sure they start and end in the correct places.
Various answers, for example: (2, 2); (3, 3); (3, 5); (4, 7); (5, 6); (7, 6)
4. Plot and join the points to find out what shape her route made.
Discuss plotting the points accurately. Discuss how you will know if you have made a mistake.
She sailed in a trapezium.
5. Plot and write the coordinates for a different shaped route around the island.
Discuss what other shapes she could have sailed in around the island. Does it have to be a quadrilateral? This question is open-ended for the children to explore.
Various answers, for example: rectangle ; (1, 2), (1, 7), (7, 2), (7, 7)
National Curriculum Objectives
Mathematics Year 6: (6P2) Draw and translate simple shapes on the coordinate plane, and reflect them in the axes
Mathematics Year 6: (6P3) Describe positions on the full coordinate grid (all four quadrants)
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Question
# There is a circle with radius 6 cm. A chord is drawn in it. Find the angle subtended by the minor segment of the chord at the centre of the circle if the area of segment is 22.1 cm2.120 60 90 180
Solution
## The correct option is A 120 (since OS is perpendicular to PQ) OQ = OP (radius of the circle) ∠OSP = ∠OSQ = 90∘ OS is the common side to both the triangles Thus, ΔPSO is congruent to ΔQSO by RHS congruence. This means that corresponding ∠POS and ∠QOS are equal. So, ∠POS = ∠QOS = θ ∠POQ =∠POS +∠QOS = 2θ Also PS = SQ (corresponding sides of 2 congruent triangles) Area of segment PSQR = Area of sector POQR - Area of ΔPOQ Area of sector = 2θ360∘×πr2 (Where r is radius) = 2θ360∘×π(6)2 = 0.63θ In ΔOSQ, ∠OSQ = 90∘ sinθ = oppositehypotenuse (SQ is opposite side for angle θ since the opposite side has no common vertex with angle) = SQOQ ⇒ sinθ = SQ6 Hence, SQ = 6sinθ Similarly, OS = 6cosθ PQ = PS + SQ = 2SQ (Since PS = SQ) ⇒ PQ = 12sinθ Area of triangle OPQ = 12×base×height = 12×PQ×OS = 12×12sinθ×6cosθ = 36(sinθ)(cosθ) Area of segment = Area of sector – Area of triangle = 0.63θ - 36(sinθ)(cosθ) 22.1 = 0.63θ - 36(sinθ)(cosθ) ...........(1) Here we can get the value of θ by substituting values of answer options. But take care! The answer is not θ. What is asked is ∠POQ = 2θ By substituting 2θ = 120∘, θ = 60∘ When θ = 60∘ the above equation (1) is satisfied. Hence, the angle subtended by segment at centre is 120∘. Note that sin (120∘) = sin (90∘+30∘), which is sin (30∘) = 12. This is because sin (90∘ + θ ) = sin(θ) sin (180∘) = 0, when we substitute 2θ as 180∘
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# A triangle has sides A, B, and C. If the angle between sides A and B is (7pi)/12, the angle between sides B and C is pi/3, and the length of B is 1, what is the area of the triangle?
Jan 23, 2018
Area of the triangle is $1.62$ sq.unit.
#### Explanation:
Angle between Sides $A \mathmr{and} B$ is $\angle c = \frac{7 \pi}{12} = \frac{7 \cdot 180}{12} = {105}^{0}$
Angle between Sides $B \mathmr{and} C$ is $\angle a = \frac{\pi}{3} = \frac{180}{3} = {60}^{0} \therefore$
Angle between Sides $C \mathmr{and} A$ is $\angle b = 180 - \left(105 + 60\right) = {15}^{0}$
The sine rule states if $A , B \mathmr{and} C$ are the lengths of the sides
and opposite angles are $a , b \mathmr{and} c$ in a triangle, then:
A/sina = B/sinb=C/sinc ; B=1 :. B/sinb=C/sinc or
$\frac{1}{\sin} 15 = \frac{C}{\sin} 105 \mathmr{and} C = 1 \cdot \left(\sin \frac{105}{\sin} 15\right) \approx 3.73 \left(2 \mathrm{dp}\right)$
Now we know sides $B = 1 , C = 3.73$ and their included angle
$\angle a = {60}^{0}$. Area of the triangle is ${A}_{t} = \frac{B \cdot C \cdot \sin a}{2}$
$\therefore {A}_{t} = \frac{1 \cdot 3.73 \cdot \sin 60}{2} \approx 1.62 \left(2 \mathrm{dp}\right)$ sq.unit.
Area of the triangle is $1.62$ sq.unit [Ans] |
# Algebra 1 Word Problems
## Online Tutoring Is The Easiest, Most Cost-Effective Way For Students To Get The Help They Need Whenever They Need It.
Various day-to-day mathematical problems or the real world problems can be easily solved by first converting the problem in the form of an equation and then solving it by applying the rules of algebra. Algebra 1 word problems tool helps in solving the word problem step by step.
This can be illustrated by a few examples as below.
Example 1: If the cost of 12 pens is \$50, find the cost of 5 such pens.
Solution:
Here we cannot find the cost of 5 pens unless we find the cost of one pen, which is also not known, from the question.
We start by assuming the cost of one pen to be ‘x’,
So the cost of 12 pens would be 12 (x) = 12x……………….. (1)
But according to the question cost of 12 pens = \$50….…. (2)
So from the equation (1) and (2) above, we get
12x = 50
Dividing both sides by 12
12x/12 = 50
x = 4.16, i.e. the cost of one pen………… (3)
Now, cost of 5 pens would be 5x. But x = 4.16 as per equation (3)
So, cost of 5 pens = 5 X 4.16 = \$20.8
Example 2: Find the cost of a bag which was sold for \$200 at a profit of 20%.
Solution:
Let us assume the cost price (C.P) to be =x
We know that,
Selling price (S.P) = C.P + Profit
Or S.P = C.P + P
Or, S.P = x + 20% of x (because profit is 20% of the C.P)
Or, S.P = x + 20x/100
Or, S.P. =120x/100…………………………………..(1)
But according to the question S.P = 200……….(2)
From equation (1) and (2) above
= 200
Multiplying both sides by 100
Or, 120x/100 * 100 = 200 x 100
Or,120x = 20000
Dividing both sides by 120
120x/120 = 20000/120
X= 166.67, or the cost price of the bag was \$166.66 |
## 35.2 Linear equations: A review
An example of a regression equation is
$\hat{y} = -4 + 2x.$ Here, $$x$$ refers to the explanatory variable, $$y$$ refers to the observed response variable, and $$\hat{y}$$ refers to the predicted values of the response variable.
In general, the equation of a straight line is written as
$\hat{y} = {b_0} + {b_1} x$ where $$b_0$$ and $$b_1$$ are just numbers. Again, $$\hat{y}$$ refers to the predicted (not observed) values of $$y$$.
$$\hat{y}$$ is pronounced as ‘why hat’; the ‘caret’ above the $$y$$ is called a ‘hat,’ and designates a predicted value (of $$y$$).
Example 35.1 (Regression equations) In the regression equation $$\hat{y} = 15 - 102x$$, we have $$b_0=15$$ and $$b_1 = -102$$.
Think 35.1 (Regression coefficients) Consider the regression equation $$\hat{y} = -0.0047x + 2.1$$. What are the values of $$b_0$$ and $$b_1$$? (Look carefully!)
In this case, $$b_0 = 2.1$$ and $$b_1 = -0.0047$$. Remember that $$b_1$$ is the number multiplied by the $$x$$, and that $$b_0$$ is the number by itself. Some software and some journal papers write regression equations with the value of $$b_0$$ first, and some with the value of $$b_1$$ first. It’s not the order that’s important.
The numbers $$b_0$$ and $$b_1$$ are called regression coefficients, where
• $$b_0$$ is a number called the intercept. It is the predicted value of $$y$$ when $$x=0$$.
• $$b_1$$ is a number called the slope. It is, on average, how much the value of $$y$$ changes when the value of $$x$$ increases by 1.
We will use software to find the values of $$b_0$$ and $$b_1$$. However, we can roughly guess the values of the intercept by first drawing what looks like a sensible straight line through the data, and determining what that line predicts for the value of $$y$$ when $$x=0$$.
A rough guess of the slope can be made using the formula
$\text{slope} = \frac{\text{Change in y}}{\text{Corresponding change in x}} = \frac{\text{rise}}{\text{run}}.$ That is, a guess of the slope is the change in the value of $$y$$ (the ‘rise’) divided by the corresponding change in the value of $$x$$ (the ‘run’).
To demonstrate, consider the scatterplot in Fig. 35.1. I have drawn a sensible line on the graph to capture the relationship (your line may look a bit different). When $$x = 0$$, the regression line predicts the value of $$y$$ is about to be 2, so $$b_0$$ is approximately 2.
To guess the slope, use the ‘rise over run’ idea. The animation below may help explain the rise-over-run idea. When the value of $$x$$ increases from 1 to 5 (a change of $$5-1=4$$), the corresponding values of $$y$$ change from 5 to 17 (a change of $$17 - 5 = 12$$). Then, use the formula:
\begin{align*} \frac{\text{rise}}{\text{run}} &= \frac{17 - 5}{5 - 1}\\ &= \frac{12}{4} = 3. \end{align*} The value of $$b_1$$ is about $$3$$. The regression line is approximately $$\hat{y} = 2 + (3\times x)$$, usually written as
$\hat{y} = 2+3x.$
The intercept has the same measurement units as the response variable. For example, with the red-deer data the intercept is measured in ‘grams,’ the measurement units of the molar weight.
The measurement unit for the slope is the ‘measurement units of the response variable,’ per ‘measurement units of the explanatory variable.’ For example, with the red-deer data the slope has the units of ‘grams per year.’
Example 35.2 (Estimating regression parameters) A study examined the number of cyclones $$y$$ in the Australian region each year from 1969 to 2005, and the relationship with a climatological index called the Ocean Nino Index (ONI, $$x$$); see (Fig. 35.2),
When the value of $$x$$ is zero, the predicted value of $$y$$ is about 12; $$b_0$$ is about 12. (You may get something slightly different.) Notice that the intercept is the predicted value of $$y$$ when $$x=0$$, which is not at the left of the graph.
To guess the value of $$b_1$$, use the ‘rise over run’ idea. When $$x$$ is about $$-2$$, the predicted value of $$y$$ is about 17. When $$x$$ is about $$2$$, the predicted value of $$y$$ is about 8. So when the value of $$x$$ changes by $$2 - (-2) = 4$$, the value of $$y$$ changes by $$8 - 17 = -9$$ (a decrease of about 9). Hence, the value of $$b_1$$ is approximately $$-9/4 = -2.25$$. (You may get something slightly different.) Notice that the relationship has a negative direction, so the slope must be negative.
Using these guesses of $$b_0 = 12$$ and $$b_1 = -2.25$$, the regression line is approximately $\hat{y} = 12 - 2.25x.$
In this section, we have seen how to understand a linear regression equation, and how an equation can be used to describe a fitted line. The above method gives a very crude guess of the values of the intercept $$b_0$$ and the slope $$b_1$$. In practice, many reasonable lines could be drawn through a scatterplot of data. However, one of those lines is the ‘best fitting line’ in some sense17. Software calculates this ‘line of best fit’ for us.
### References
Dunn PK, Smyth GK. Generalized linear models with examples in R. Springer; 2018.
1. For those who want to know: The ‘line of best fit’ is the line such that the sum of the squared vertical distances between the observations and the line is as small as possible.↩︎ |
# Into Math Grade 1 Module 13 Lesson 4 Answer Key Solve Two-Digit Addition and Subtraction Problems
We included HMH Into Math Grade 1 Answer Key PDF Module 13 Lesson 4 Solve Two-Digit Addition and Subtraction Problems to make students experts in learning maths.
## HMH Into Math Grade 1 Module 13 Lesson 4 Answer Key Solve Two-Digit Addition and Subtraction Problems
I Can choose strategies to solve two-digit addition and subtraction problems.
Step It Out
Question 1.
A farmer has 90 pumpkins. She sells 40 pumpkins. How many pumpkins are left?
A. Explain how to solve the problem.
I will counting on to subtract
B. Draw and write to solve.
THINK: I can add to subtract.
4 tens + _________ tens = 9 tens
So, 9 tens – 4 tens = ___________ tens.
90 – 40 = __________
I am at 4, I need 5 tens more to get 9 tens
4 tens + 5 tens = 9 tens
So, 9 tens – 4 tens = 5 tens
90 – 40 = 50.
C. There are _________ pumpkins left.
There are 50 pumpkins left.
Question 2.
Laney has 64 red tulips. She gets 8 more tulips. How many tulips does she have?
A. Use a strategy to solve. Draw to show your work.
THINK: I can make a ten.
I have the quick picture of the numbers 64 and 8
I can make ten
There are 4 and 8 that is 12 ones
12 ones is 1 ten 2 ones
So, 6 tens + 1 tens = 7 tens
64 + 8 = 7 tens 2 ones
B. Laney has _________ tulips.
Laney has 72 tulips.
Turn and Talk What other strategies could you use to solve this problem? Explain.
I can count on or count back to solve the problem.
Check Understanding
Solve. Draw to show your thinking.
Question 1.
There are 60 cows in a field. 50 cows go to the barn. How many cows are still in the field?
_________ cows
Explanation:
There are 60 cows in a field
50 cows go to the barn
60 – 50 =
I drew 6 tens and took away 5 tens
1 ten is left
60 – 50 = 10
10 cows are still in the field.
Solve.
Question 2.
Mrs. Kelsie has 32 red apples and 21 green apples. How many apples does she have?
_________ apples
53 apples
Explanation:
Mrs. Kelsie has 32 red apples and
21 green apples
32 + 21 =
32 = 3 tens 2 ones
21 = 2 tens 1 ones
32 + 21 = 5 tens 3 ones
So, Kelsie have 53 apples.
Question 3.
Reason Marco counts 80 pears. He counts 20 fewer peaches than pears. How many peaches does Marco count?
____________ peaches
60 peaches
Explanation:
Marco counts 80 pears
He counts 20 fewer peaches than pears
80 – 20 =
Start at 80 count back by 10’s till 20
80 before, 70, 60
I reached at 60
80 – 20 = 60
So, Marco counts 60 peaches.
Question 4.
Construct Arguments Mr. Jones bikes 20 miles to his office. Describe how to find the number of miles he bikes to his office and home again. Solve the problem.
Question 5.
Margo finds 57 orange leaves and 7 red leaves. How many leaves does she find?
__________ leaves
64 leaves
Explanation:
Margo finds 57 orange leaves and 7 red leaves
57 + 7 =
Make ten:
7 ones + 7 ones = 1 ten 4 ones
5 ten + 1 ten and 4 ones
= 6 tens 4 ones
57 + 7 = 64
So, Margo finds 51 leaves.
Solve. Draw or write to explain.
Question 6.
Kathy sees 32 orange carrots and 5 purple carrots in the garden. How many carrots does she see?
_________ carrots
Explanation:
Kathy sees 32 orange carrots and 5 purple carrots in the garden
3 tens 2 ones + 5 ones
= 3 tens + (5+2) ones
= 3 tens 7 ones
32 + 5 = 37
So, Kathy have 37 carrots in all.
Question 7.
Reason Sam counts 50 potatoes. Jalen counts 30 potatoes. How many potatoes do Sam and Jalen count together?
_________ potatoes
Explanation:
Sam counts 50 potatoes
Jalen counts 30 potatoes
50 + 30 =
Start at 50 count on by 10’s
Count more 3 times
50 after 60, 70 and 80
50 + 30 = 80
So, Sam and Jalen count 80 potatoes all together.
Question 8.
Construct Arguments Mr. Caleb has 30 brown eggs. He has 20 more brown eggs than white eggs. How many white eggs does Mr. Caleb have?
_________ white eggs |
# The Angle of Elevation of the Top of a Vertical Tower from a Point on Ground is 60° . from Another Point 10 M Vertically Above the First, Its Angle of Elevation is 30° .Find the Height of the Tower. - Mathematics
The angle of elevation of the top of a vertical tower from a point on the ground is 60° . From another point 10 m vertically above the first, its angle of elevation is 30° .Find the height of the tower.
#### Solution
Let PQ be the tower
We have,
AB =10m,MAP = 30° and ∠PBQ = 60°
Also, MQ = AB =10m
Let BQ = x and PQ = h
So, AM = BQ = x and PM = PQ - MQ = h -10
In ΔBPQ,
tan 60° = (PQ)/(BQ)
⇒ sqrt(3) = h/x
⇒ x = h/sqrt(3) ..............(1)
Now , in Δ AMP
tan 30° = (PM) /( AM)
⇒ 1/ sqrt(3) = (h-10) /x
⇒h sqrt(3)-10 sqrt(3)=x
⇒ h sqrt(3) - 10 sqrt(3) = h/ sqrt(3) [ Using (1)]
⇒ 3h - 30=h
⇒ 3h - h = 30
⇒ 2h = 30
⇒ h 30/2
∴ h = 15 m
So, the height of the tower is 15 m.
Concept: Heights and Distances
Is there an error in this question or solution?
Chapter 14: Height and Distance - Exercises
#### APPEARS IN
RS Aggarwal Secondary School Class 10 Maths
Chapter 14 Height and Distance
Exercises | Q 21
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## Class 11 math (India)
### Course: Class 11 math (India)>Unit 15
Lesson 4: Solutions to select NCERT problems
# Select problems from exercise 16.3
Solutions to some problems of NCERT exercise.
In this article we will look at solutions of a few selected problems from exercise 16.3 of NCERT.
Problem 1:
A fair coin is tossed four times, and a person wins Re 1 for each head and loses Rs 1, point, 50 for each tail that turns up.
From the sample space calculate how many different amounts of money you can have after four tosses and the probability of having each of these amounts.
Solution:
On each toss the person may either win or lose. The final amount the person gets depends on the number of times he wins or loses.
There are a total of 5 possible cases. Below start text, W, end text and start text, L, end text denote a win and loss respectively.
Case start text, I, end text: 4 wins.
Outcome is start text, W, W, W, W, end text. Here the person wins Rs 4 in total.
Case start text, I, I, end text: 3 wins and 1 loss.
Outcomes are start text, W, W, W, L, end text, start text, W, W, L, W, end text, start text, W, L, W, W, end text and start text, L, W, W, W, end text. Here the person wins Rs 1, point, 50 in total.
Case start text, I, I, I, end text: 2 wins and 2 losses.
Outcomes are start text, W, W, L, L, end text, start text, W, L, W, L, end text etc. Here the person loses Re 1 in total.
Try it out
The bigger question is, how many outcomes are possible for this case?
We can list out all the outcomes but a shortcut would be to use permutation theory.
Basically, here we are arranging 4 objects, 2 identical start text, W, s, end text and 2 identical start text, L, s, end text in a line. Number of ways to do that is start fraction, 4, !, divided by, 2, !, 2, !, end fraction, equals, 6.
Case start text, I, V, end text: 1 win and 3 losses.
Try it out
Using permutation theory, find out the number of outcomes in this case.
There are 4 outcomes: start text, L, L, L, W, end text, start text, L, L, W, L, end text, start text, L, W, L, L, end text and start text, W, L, L, L, end text. Here the person loses Rs 3, point, 50 in total.
Case start text, V, end text: 4 losses.
Outcome is start text, L, L, L, L, end text. Here the person loses Rs 6 in total.
Now how do we find the probability for each case?
start text, P, end text, equals, start fraction, start text, N, o, point, space, o, f, space, f, a, v, o, u, r, a, b, l, e, space, o, u, t, c, o, m, e, s, end text, divided by, start text, T, o, t, a, l, space, n, o, point, space, o, f, space, o, u, t, c, o, m, e, s, end text, end fraction
Total number of outcomes is the sum of number of outcomes in each case equals, 1, plus, 4, plus, 6, plus, 4, plus, 1, equals, 16.
Required probabilities are
CaseResultNo. of favourable outcomesProbability
start text, I, end textWin Rs 41start fraction, 1, divided by, 16, end fraction
start text, I, I, end textWin Rs 1, point, 504start fraction, 1, divided by, 4, end fraction
start text, I, I, I, end textLose Re 16start fraction, 3, divided by, 8, end fraction
start text, I, V, end textLose Rs 3, point, 504start fraction, 1, divided by, 4, end fraction
start text, V, end textLose Rs 61start fraction, 1, divided by, 16, end fraction
Problem 2:
In a lottery, a person choses six different natural numbers at random from 1 to 20, and if these six numbers match with the six numbers already fixed by the lottery committee, he wins the prize. What is the probability of winning the prize in the game.
[Hint: order of numbers is not important.]
Solution:
What is the number of possible outcomes in this case?
The person can pick any combination of 6 numbers from 20 numbers. Number of possible choices is start superscript, 20, end superscript, start text, C, end text, start subscript, 6, end subscript.
Out of these, only 1 is the winning combination. The person wins if he picks this exact combination.
Probability of winning equals, start fraction, 1, divided by, start superscript, 20, end superscript, start text, C, end text, start subscript, 6, end subscript, end fraction, equals, start fraction, 1, divided by, start fraction, 20, !, divided by, 6, !, 14, !, end fraction, end fraction, equals, start fraction, 1, divided by, 38760, end fraction.
Problem 3:
Events start text, E, end text and start text, F, end text are such that start text, P, left parenthesis, n, o, t, space, E, space, o, r, space, n, o, t, space, F, right parenthesis, space, =, space, 0, point, 25, end text. State whether start text, E, end text and start text, F, end text are mutually exclusive.
Solution:
Mutually exclusive events can never happen simultaneously. In other words, if events start text, E, end text and start text, F, end text are mutually exclusive, start text, P, left parenthesis, E, space, a, n, d, space, F, right parenthesis, end text, equals, start text, P, end text, left parenthesis, start text, E, end text, \cap, start text, F, end text, right parenthesis, equals, 0.
Given start text, P, left parenthesis, n, o, t, space, E, space, o, r, space, n, o, t, space, F, right parenthesis, space, =, space, 0, point, 25, end text.
How can we convert this into an equation about probability of start text, E, space, a, n, d, space, F, end text?
See that
\begin{aligned} \text {not E or not F} &= \text E' \cup \text F' \\\\ \text E' \cup \text F' &= (\text E \cap \text F)' \quad \small \text{using DeMorgan's law} \end{aligned}
Therefore
\begin{aligned} \text P(\text E' \cup \text F') &= \text P(\text E \cap \text F)' \\\\ &= 1 - \text P(\text E \cap \text F) \quad \small \because \text P (\text A') = 1 - \text P (\text A) \end{aligned}
Given start text, P, end text, left parenthesis, start text, E, end text, prime, \cup, start text, F, end text, prime, right parenthesis, equals, 0, point, 75. So
\begin{aligned} 0.75 &= 1 - \text P(\text E \cap \text F) \\\\ \text P(\text E \cap \text F) &= 0.25 \end{aligned}
Because start text, P, end text, left parenthesis, start text, E, end text, \cap, start text, F, end text, right parenthesis, does not equal, 0, start text, E, end text and start text, F, end text are not mutually exclusive events.
Problem 4:
The probability that a student will pass the final examination in both English and Hindi is 0, point, 5 and the probability of passing neither is 0, point, 1. If the probability of passing the English examination is 0, point, 75, what is the probability of passing the Hindi examination?
Solution:
Let us make a Venn diagram of probabilities using given data.
start text, P, left parenthesis, H, i, n, d, i, right parenthesis, end text, equals, 0, point, 5, plus, start text, question mark, end text
Because sum of all probabilities should equal 1, we have
\begin{aligned} 0.25 + 0.5 + \text ? + 0.1 &= 1 \\\\ \text ? &= 0.15 \end{aligned}
Therefore start text, P, left parenthesis, H, i, n, d, i, right parenthesis, end text, equals, 0, point, 5, plus, 0, point, 15, equals, 0, point, 65.
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Math
Using numbers, number operators, and math functions.
Numeric values: 0, 1, 2, 6.7, 10.083…
Just numbers by themselves. Sometimes these are called numeric literals.
Integers: whole numbers
``````let num = 0;
num = 0;
num = 1;
num = 2;``````
Floating point: numbers with a fractional part
Numbers can have their fractional part too. The decimal point is between the digits of the number. But, floating point numbers have the decimal point at any spot between digits, like: 3.14159 or 651.75.
``````let num = 0
num = 6.7
num = 10.083``````
Arithmetic binary operation (+, -, *, /)
The operations for basic arithmetic: add, subtract, multiply, and divide.
``````let more = 0+1;
let less = 0-1;
let twice = 1*2;
let divide = 8/4;``````
Remainder (%)
This is a extra operator for division. You can find out how much is left over if one number doesn’t divide into the other number evenly.
We know that 4 / 2 = 2, so 2 divides into 4 evenly. But, 5 / 2 = 2 with a remainder of 1. So, the remainder operation, 5 % 2 = 1, gives the number that’s left over from a division operation.
``let remainder = 7%4;``
Exponent (**)
The exponent operator will multiply the number on the left by itself for the amount of times of the number on its right. That is, 4 ** 2 = 4 * 4 and 2 ** 3 = 2 * 2 * 2. The area of a square that has sides with a length of `5` is equal to one side multiplied by another. For a square, all sides are equal, so:
``````let side = 5;
let area = side * side;``````
But using the exponent operator, this is the same as:
``````let side = 5;
let area = side ** 2;``````
The volume of a cube is three sides multiplied together. The two volumes are the same:
``````let side = 5;
let volume1 = side * side * side;
let volume2 = side ** 3;``````
Square root
The square root of a number is another number that when multiplied by itself it becomes the original number. You know that `2 * 2` equals `4` so the square root of `4` is `2`. It’s called a square root because the area of a square is the length of two equal sides multiplied together. The root is the length of a side.
``````let side = 5
let square = side * side
let root = Math.sqrt(side)``````
Absolute value
When you want to know how much a number is without its sign (+/-). The absolute value of -5 is 5 and the absolute value of 5 is also 5. The absolute value is sometimes called the magnitude.
``let item = Math.abs(-5);``
Minimum and maximum of two values
You can get the smaller or the bigger of two numbers with the min() and max() functions.
• The minimum of 2 and 9: Math.min(2, 9) equals 2.
• The maximum of 3 and 9: Math.max(3, 9) equals 9.
``````let minval = Math.min(0, 1);
let maxval = Math.max(8, 2);``````
Round
If a number has a fractional part, you can make the number change to be the closest, next integer value. This is called rounding. Rounding the number `6.78` will make it be `7` and rounding `9.3` will give you `9`. If a number has a fractional part greater than or equal to `0.5`, the number will round up to the next whole integer value with the higher value. Otherwise, it will round down to the next lowest integer value.
For negative numbers, they round toward the absolute value (the absolute value of `-8` is `8`) of the number. So, `-5.23` rounds to `-5` and `-2.68` rounds to `-3`.
``let rounded = Math.round(5.5)``
Ceiling
To make a number change to the next higher whole number (integer), get the number’s ceiling value. The ceiling value for `1.234` is `2` since that is the next higher whole number. For the negative number of `-3.63`, its ceiling is `-3` since that’s the next higher whole number.
``let nextup = Math.ceil(8.435)``
Floor
To make a number change to the next lower whole number (integer), get the number’s floor value. The floor value for `8.76` is `8` since that is the next lower whole number. For the negative number of `6.17`, its floor is `-7` since that’s the next lower whole number.
``let nextdown = Math.floor(4.97)``
Truncate
The fractional part of a number is removed by truncating it. If a number has the value `54.234` its truncated value is `54`. Truncation works the same way for a negative number. The truncated value of `-34.913` is `-34`.
``let nonfraction = Math.trunc(87.23455)``
Random value
Make up any number from a minimum value to a some maximum value. If you want a random number up to 100, say: Math.randomRange(0, 100).
``let myRandom = Math.randomRange(0, 5);``
Trigonometry
Functions for finding numbers of angles, sides of triangles, and positions on a circle. These functions are also used to make information for wave signals.
A compass shows where North is and the direction to it is 0 degrees (which is also 360 degrees). If you walk directly to the East, then your direction will be 90 degrees on the compass. This is also the measurement of the angle between the path of your direction and the path to North. If point yourself to North and spin around to the right (clockwise) in one spot, you will see the compass change from 0 degrees, go up to 359 degrees, and back to 0 degrees. So, a full rotation is 360 degrees, or full circle.
Another way to measure angles is to use numbers that are special to the size of circles. A circle has two important numbers related to it. They are called the radius and _pi_. The radius is the distance from the center of the circle to its outside edge. A long time ago, people learned that no matter how big a circle is, if you divide the length of its outside edge by the length of its radius, you always get the same number back. They decided to call this number _pi_. It’s 2 times _pi_ actually, because they used the whole length across the circle which is two lengths of the radius.
Now, we don’t have to worry about the radius anymore, we can just use some part of 2_pi_ to tell where any spot on the edge of a circle is. Any part or all of 2_pi_ is called radians. We can use radians to measure an angle of direction too, just like with the compass. Except now, we spin 2_pi_ radians around the circle instead of 360 degrees.
What number is _pi_?
As it turns out, _pi_ is an irrational number. That means it has a fractional part that uses more digits than we can display. So, it’s best to use a constant in code for the value of _pi_. In case you’re curious, the first 32 digits of _pi_ are: 3.1415926535897932384626433832795.
Luckily, you can use the built-in constant Math.PI.
In code, you use radians for angle measures. All of the math functions for trigonometry use radians.
What’s 60 degrees in radians? Well, one degree is (2 * Math.PI / 360) radians. Or, to make it simple, (Math.PI / 180) radians. So, 60 degrees is (60 * Math.PI / 180) radians.
Sine
Get the length of the vertical (up or down) side of a right triangle at some angle. But, the sine value is the length of the vertical side divided by the length of the longest side, its hypotenuse.
What’s the sine of 60 degrees? Math.sin(60 * Math.PI / 180) equals 0.5. The vertical side of a right triangle is one half the length of the longest side when the opposite angle is 60 degrees.
``let ySide = Math.sin(60 * Math.PI / 180)``
Cosine
Get the length of the horizontal (left or right) side of a right triangle at some angle. But, the cosine value is the length of the horizontal side divided by the length of the longest side, its hypotenuse.
What’s the cosine of 45 degrees? Math.cos(45 * Math.PI / 180) equals 0.707. The length of the horizontal side of a right triangle is about 70 percent of the length of the longest side when the angle between them is 45 degrees.
``let xSide = Math.cos(45 * Math.PI / 180)`` |
# Solve for m 3/(16m^2-81)=-5/(4m^2+9m)
316m2-81=-54m2+9m
Factor each term.
Rewrite 16m2 as (4m)2.
3(4m)2-81=-54m2+9m
Rewrite 81 as 92.
3(4m)2-92=-54m2+9m
Since both terms are perfect squares, factor using the difference of squares formula, a2-b2=(a+b)(a-b) where a=4m and b=9.
3(4m+9)(4m-9)=-54m2+9m
Factor m out of 4m2+9m.
Factor m out of 4m2.
3(4m+9)(4m-9)=-5m(4m)+9m
Factor m out of 9m.
3(4m+9)(4m-9)=-5m(4m)+m⋅9
Factor m out of m(4m)+m⋅9.
3(4m+9)(4m-9)=-5m(4m+9)
3(4m+9)(4m-9)=-5m(4m+9)
3(4m+9)(4m-9)=-5m(4m+9)
Find the LCD of the terms in the equation.
Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.
(4m+9)(4m-9),m(4m+9)
Since (4m+9)(4m-9),m(4m+9) contain both numbers and variables, there are four steps to find the LCM. Find LCM for the numeric, variable, and compound variable parts. Then, multiply them all together.
Steps to find the LCM for (4m+9)(4m-9),m(4m+9) are:
1. Find the LCM for the numeric part 1,1.
2. Find the LCM for the variable part m1.
3. Find the LCM for the compound variable part 4m+9,4m-9,4m+9.
4. Multiply each LCM together.
The LCM is the smallest positive number that all of the numbers divide into evenly.
1. List the prime factors of each number.
2. Multiply each factor the greatest number of times it occurs in either number.
The number 1 is not a prime number because it only has one positive factor, which is itself.
Not prime
The LCM of 1,1 is the result of multiplying all prime factors the greatest number of times they occur in either number.
1
The factor for m1 is m itself.
m1=m
m occurs 1 time.
The LCM of m1 is the result of multiplying all prime factors the greatest number of times they occur in either term.
m
The factor for 4m+9 is 4m+9 itself.
(4m+9)=4m+9
(4m+9) occurs 1 time.
The factor for 4m-9 is 4m-9 itself.
(4m-9)=4m-9
(4m-9) occurs 1 time.
The factor for 4m+9 is 4m+9 itself.
(4m+9)=4m+9
(4m+9) occurs 1 time.
The LCM of 4m+9,4m-9,4m+9 is the result of multiplying all factors the greatest number of times they occur in either term.
(4m+9)(4m-9)
The Least Common Multiple LCM of some numbers is the smallest number that the numbers are factors of.
m(4m+9)(4m-9)
m(4m+9)(4m-9)
Multiply each term by m(4m+9)(4m-9) and simplify.
Multiply each term in 3(4m+9)(4m-9)=-5m(4m+9) by m(4m+9)(4m-9) in order to remove all the denominators from the equation.
3(4m+9)(4m-9)⋅(m(4m+9)(4m-9))=-5m(4m+9)⋅(m(4m+9)(4m-9))
Cancel the common factor of (4m+9)(4m-9).
Factor (4m+9)(4m-9) out of m(4m+9)(4m-9).
3(4m+9)(4m-9)⋅((4m+9)(4m-9)(m))=-5m(4m+9)⋅(m(4m+9)(4m-9))
Cancel the common factor.
3(4m+9)(4m-9)⋅((4m+9)(4m-9)m)=-5m(4m+9)⋅(m(4m+9)(4m-9))
Rewrite the expression.
3⋅m=-5m(4m+9)⋅(m(4m+9)(4m-9))
3m=-5m(4m+9)⋅(m(4m+9)(4m-9))
Simplify -5m(4m+9)⋅(m(4m+9)(4m-9)).
Cancel the common factor of m(4m+9).
Move the leading negative in -5m(4m+9) into the numerator.
3m=-5m(4m+9)⋅(m(4m+9)(4m-9))
Cancel the common factor.
3m=-5m(4m+9)⋅(m(4m+9)(4m-9))
Rewrite the expression.
3m=-5⋅(4m-9)
3m=-5⋅(4m-9)
Apply the distributive property.
3m=-5(4m)-5⋅-9
Multiply.
Multiply 4 by -5.
3m=-20m-5⋅-9
Multiply -5 by -9.
3m=-20m+45
3m=-20m+45
3m=-20m+45
3m=-20m+45
Solve the equation.
Move all terms containing m to the left side of the equation.
Add 20m to both sides of the equation.
3m+20m=45
23m=45
23m=45
Divide each term by 23 and simplify.
Divide each term in 23m=45 by 23.
23m23=4523
Cancel the common factor of 23.
Cancel the common factor.
23m23=4523
Divide m by 1.
m=4523
m=4523
m=4523
m=4523
The result can be shown in multiple forms.
Exact Form:
m=4523
Decimal Form:
m=1.95652173…
Mixed Number Form:
m=12223
Solve for m 3/(16m^2-81)=-5/(4m^2+9m)
### Solving MATH problems
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# RD Sharma Class 12 Ex 26.1 Solutions Chapter 26 Scalar Triple Product
Here we provide RD Sharma Class 12 Ex 26.1 Solutions Chapter 26 Scalar Triple Product for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 26.1 Solutions Chapter 26 Scalar Triple Product book pdf download. Now you will get step-by-step solutions to each question.
## RD Sharma Class 12 Ex 26.1 Solutions Chapter 26 Scalar Triple Product
Solution:
=
= 1 + 1 + 1
= 3
Solution:
=
=
= 2 – 1 – 2
= -1
### Question 2(i). Find , when
Solution:
=
= 2(-1 – 0) + 3(-1 + 3)
= -2 + 6
= 4
### Question 2(ii). Find, when
Solution:
= 1(1 + 1) + 2(2 + 0) + 3(2 – 0)
= 2 + 4 + 6
= 12
### Question 3(i). Find the volume of the parallelepiped whose coterminous edges are represented by vector
Solution:
Volume of a parallelepiped whose adjacent edges are is equal to
=
= 2(4 – 1) – 3(2 + 3) + 4(-1 – 6)
= 6 – 15 – 28
= -9 – 28
= -37
So, Volume of parallelepiped is | -37 | = 37 cubic unit.
### Question 3(ii). Find the volume of the parallelepiped whose coterminous edges are represented by vector
Solution:
Volume of a parallelepiped whose adjacent edges are is equal to
=
= 2(-4 – 1) + 3(-2 + 3) + 4(-1 – 6)
= -10 + 3 – 28
= -10 – 25
= -35
So, Volume of parallelepiped = | -35 | = 35 cubic unit.
### Question 3(iii). Find the volume of the parallelepiped whose coterminous edges are represented by vector
Solution:
Let a = 11, b = 2, c = 13
Volume of a parallelepiped whose adjacent edges are is equal to
=
= 11(26 – 0) + 0 + 0
= 286
Volume of a parallelepiped = | 286| = 286 cubic units.
### Question 3(iv). Find the volume of the parallelepiped whose coterminous edges are represented by vector
Solution:
Let
Volume of a parallelepiped whose adjacent edges are is equal to
=
= 1(1 – 2) – 1(-1 – 1) + 1(2 + 1)
= -1 + 2 + 3
= 4
Volume of a parallelepiped = |4| = 4 cubic units.
### Question 4(i). Show of the following triads of vector is coplanar :
Solution:
As we know that three vectors are coplanar if their = 0.
=
= 1(10 – 42) – 2(15 – 35) – 1(18 – 10)
= -32 + 40 – 8
= 0
So, the given vectors are coplanar.
### Question 4(ii). Show of the following triads of vector is coplanar :
Solution:
As we know that three vectors are coplanar if their = 0.
=
= -4(12 + 3) + 6(-3 + 24) – 2(1 + 32)
= -60 + 126 – 66
= 0
So, the given vectors are coplanar.
### Question 4(iii). Show of the following triads of vector is coplanar :
Solution:
As we know that three vectors are coplanar if their = 0.
=
= 1(15 – 12) + 2(-10 + 4) + 3(6 – 3)
= 3 – 12 + 9
= 0
So, the given vectors are coplanar.
### Question 5(i). Find the value of λ so that the following vector is coplanar:
Solution:
As we know that three vectors are coplanar if their = 0.
=
= 1(λ -1) + 1(2λ + λ) + 1(-2 – λ)
= λ – 1 + 3λ – 2 -λ
3 = 3λ
1 = λ
So, the value of λ is 1
### Question 5(ii). Find the value of λ so that the following vector is coplanar:
Solution:
As we know that three vectors are coplanar if their = 0.
=
= 2(10 + 3 λ) + 1(5 + 3 λ) + 1(λ – 2 λ)
= 20 + 6 λ + 5 + 3 λ – λ
-25 = 8 λ
λ = – 25 / 8
So, the value of λ is -25/8
### Question 5(iii). Find the value of λ so that the following vector is coplanar:
Solution:
Given:
As we know that three vectors are coplanar if their = 0.
=
= 1(2λ – 2) – 2(6 – 1) – 3(6 – λ)
= 2λ – 2 -12 + 2 -18 + 3λ
= 5λ – 30
30 = 5λ
λ = 6
So, the value of the λ is 6
### Question 5(iv). Find the value of λ so that the following vector is coplanar:
Solution:
Given:
So, to prove that these points are coplanar, we have to prove that = 0
=
= 1(0 + 5) – 3(0 – 5λ) + 0
= 5 + 15λ
-5 = 15λ
λ = – 1 / 3
### Question 6. Show that the four points having position vectors are not coplanar.
Solution:
Let us considered
OA =
OB =
OC =
OD =
AB = OB – OA =
AC = OC – OA =
CD = OD – OC =
AD = OD – OA =
So, to prove that these points are coplanar, we have to prove that
= 16(-160 – 24) + 25(-160 + 8) – 4(-144 + 64) ≠ 0
Hence, proved that the points are not coplanar.
### Question 7. Show that the points A (-1, 4, -3), B(3, 2, -5), C(-3, 8, -5), and D(-3, 2, 1) are coplanar
Solution:
Given:
A = (-1, 4, -3)
B = (3, 2, -5)
C = (-3, 8, -5)
D = (-3, 2, 1)
=
=
=
So, to prove that these points are coplanar, we have to prove that
Thus,
= 4[16 – 4] + 2[-8 -4] – 2[4 + 8]
= 48 – 24 – 24 = 0
Hence, proved.
### Question 8. Show that four points whose position vectors are
Solution:
Let us considered
OA =
OB =
OC =
OD =
Thus,
AB = OB – OA =
AC = OC – OA =
AD = OD – OA =
If the vectors AB, AC and AD are coplanar then the four points are coplanar
On simplifying, we get
= 10(70 + 12) + 12(-30 – 24) – 4(-6 + 28)
= 820 – 648 – 88
= 84 ≠ 0
So, the points are not coplanar.
### Question 9. Find the value of λ for which the four points with position vectors are coplanar
Solution:
Let us considered:
Position vector of A =
Position vector of B =
Position vector of C =
Position vector of D =
If the given vectors are coplanar, then the four points are coplanar
=
=
=
On simplifying, we get
4(50 – 25) – 6(15 + 20) + (λ + 1)(15 + 40) = 0
100 – 210 + 55 + 55λ = 0
55λ = 55
λ = 1
So, when the value of λ = 1, the given points are coplanar.
### Question 10. Prove that
Solution:
Given:
One solving the given equation we get
=
= 6 [ a b c ] – 6 [ a b c ]
= 0
Hence proved
### Question 11. are the position vectors of points A, B and C respectively, prove that is a vector perpendicular to the plane of triangle ABC.
Solution:
In the given triangle ABC,
If = AB
= BC
= AC
Then,
is perpendicular to the plane of the given triangle ABC
is perpendicular to the plane of the given triangle ABC
is perpendicular to the plane of the given triangle ABC
Hence, proved that
is a vector perpendicular to the plane of the given triangle ABC.
### Question 12(i). Let . Then, if c1 = 1 and c2 = 2, find c3 which makes coplanar.
Solution:
Given:
are coplanar only if = 0
0 – 1(C3) + 1(2) = 0
C3 = 2
So, when the value C3 = 2, then these points are coplanar.
### Question 12(ii). Let and . Then, if c2 = -1 and c3 =1, show that no value of c1 can make coplanar
Solution:
Given:
are coplanar only if = 0
So,
0 – 1 + 1 (C1) = 0
C1 = 1
Hence, prove that no value of Ccan make these points coplanar
### Question 13. Find λ for which the points A (3, 2, 1), B (4, λ, 5), C (4, 2, -2), and D (6, 5, -1) are coplanar
Solution:
Let us considered:
Position vector of OA =
Position vector of OB =
Position vector of OC =
Position vector of OD =
If the vectors AB, AC, and AD are coplanar, then the four points are coplanar
AB =
AC =
On simplifying, we get
1(9) – (λ – 2)(-2 + 9) + 4(3 – 0) = 0
9 – 7 λ + 14 + 12 = 0
7 λ = 35
λ = 5
Hence, the value of λ is 5. So the coplanar points are, A(3, 2, 1), B(4, 5, 5), C(4, 2, -2), and D(6, 5, -1)
I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.
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# General Values of Inverse Trigonometric Functions
We will learn how to find the general values of inverse trigonometric functions in different types of problems.
1. Find the general values of sin$$^{-1}$$ (- √3/2)
Solution:
Let, sin$$^{-1}$$ (- √3/2) = θ
Therefore, sin θ = - √3/2
⇒ sin θ = - sin (π/3)
⇒ sin θ = (- π/3)
Therefore, the general value of sin$$^{-1}$$ (- √3/2) = θ = nπ - (- 1)$$^{n}$$ π/3, where, n = 0 or any integer.
2. Find the general values of cot$$^{-1}$$ (- 1)
Solution:
Let, cot$$^{-1}$$ (- 1) = θ
Therefore, cot θ = - 1
⇒ cot θ = cot (- π/4)
Therefore, the general value of cot$$^{-1}$$ (- 1) = θ = nπ - π/4, where, n = 0 or any integer.
3. Find the general values of cos$$^{-1}$$ (1/2)
Solution:
Let, cos$$^{-1}$$ 1/2 = θ
Therefore, cos θ = 1/2
⇒ cos θ = cos (π/3)
Therefore, the general value of cos$$^{-1}$$ (1/2) = θ = 2nπ ± π/3, where, n = 0 or any integer.
4. Find the general values of sec$$^{-1}$$ (- 2)
Solution:
Let, sec$$^{-1}$$ (- 2) = θ
Therefore, sec θ = - 2
⇒ sec θ = - sec (π/3)
⇒ sec θ = sec (π - π/3)
⇒ sec θ = sec (2π/3)
Therefore, the general value of sec$$^{-1}$$ (- 2) = θ = 2nπ ± 2π/3, where, n = 0 or any integer.
5. Find the general values of csc$$^{-1}$$ (√2)
Solution:
Let, csc$$^{-1}$$ (√2) = θ
Therefore, csc θ = √2 .
⇒csc θ = csc (π/4)
Therefore, the general value of csc$$^{-1}$$ (√2 ) = θ = nπ + (- 1)$$^{n}$$ π/4, where, n = 0 or any integer.
6. Find the general values of tan$$^{-1}$$ (√3)
Solution:
Let, tan$$^{-1}$$ (√3) = θ
Therefore, tan θ = √3
⇒ tan θ = tan (π/3)
Therefore, the general value of tan$$^{-1}$$ (√3) = θ = nπ + π/3 where, n = 0 or any integer.
`
Inverse Trigonometric Functions |
# What Is the Result of 13 Multiplied by 3?
The answer to the question of what is the result of 13 multiplied by 3 is 39. It is easy to calculate this answer as multiplication is simply a short cut way of adding a number to itself multiple times. In this case, 13 multiplied by 3 is the same as adding 13 three times.
Multiplication is one of the four basic operations of arithmetic, along with addition, subtraction, and division. It is an important part of mathematics and it is used in many different applications, from calculating the cost of items to measuring the area of a room.
The result of 13 multiplied by 3 can be calculated using a variety of methods. The most basic way is to use the standard multiplication algorithm. This involves multiplying each digit of one number by each digit of the other and summing the products. In this case, the algorithm would be 1×3 + 3×3 = 9 + 9 = 18. This is the same result as if one were to add 13 three times.
Another way to calculate the result of 13 multiplied by 3 is to use the distributive property. This property states that when a number is multiplied by a sum, the product is equal to the sum of the products of each addend multiplied by the number. In this case, the distributive property would be 13x(3+0) = 13×3 + 13×0 = 39. This is the same result as the standard multiplication algorithm.
Finally, one can also use a calculator or other tool to quickly calculate the result of 13 multiplied by 3. This is often the quickest and most convenient way to answer the question and can be done with a simple calculation. Most calculators will have a multiplication function and can be used to quickly calculate the result without any manual calculations.
##
In conclusion, the answer to the question of what is the result of 13 multiplied by 3 is 39. This can be calculated using a variety of methods, from the standard multiplication algorithm to the distributive property or using a calculator. No matter which method is used, the result will be the same. |
# Average Absolute Deviation Formula
• Last Updated : 13 May, 2022
Average absolute deviation is defined as the mean of absolute values of the deviations from the data set values. It is also known as the mean absolute deviation. It is one of the measures of central tendency among mean. median, mode and standard deviations. It provides more accuracy and is easy to use in real-life situations. Its formula equals the sum of differences between the mean and data values divided by the number of values in data set. Its value is always greater than zero as it takes the modulus of differences.
Formula
M = Σ (xi – x̄)/n
where,
M is the average absolute deviation,
x̄ is the mean of data set,
Σ (xi – x̄) is the summation of deviations from mean,
n is the number of values in data set.
### Sample Problems
Problem 1. Calculate the average absolute deviation of the data set, 2, 6, 7, 4, 1.
Solution:
The data set is 2, 6, 7, 4, 1.
Here, n = 5.
Mean of the data, x̄ = (2 + 6 + 7 + 4 + 1)/5
= 20/5
= 4
Using the formula we get,
M = Σ (xi – x̄)/n
= [|4 – 2| + |4 – 6| + |4 – 7| + |4 – 4| + |4 – 1|]/5
= (2 + 2 + 3 + 0 + 3)/5
= 10/5
= 2
Problem 2. Calculate the average absolute deviation of the data set, 5, 8, 3, 0, 1, 7.
Solution:
The data set is 5, 8, 3, 0, 1, 7.
Here, n = 6.
Mean of the data, x̄ = (5 + 8 + 3 + 0 + 1 + 7)/6
= 24/6
= 4
Using the formula we get,
M = Σ (xi – x̄)/n
= [|4 – 5| + |4 – 8| + |4 – 3| + |4 – 0| + |4 – 1| + |4 – 7|]/6
= (1 + 4 + 1 + 4 + 3 + 3)/6
= 16/6
= 2.67
Problem 3. Calculate the average absolute deviation of the data set, 6, 9, 4, 2, 1, 8.
Solution:
The data set is 6, 9, 4, 2, 1, 8.
Here, n = 6.
Mean of the data, x̄ = (6 + 9 + 4 + 2 + 1 + 8)/6
= 30/6
= 5
Using the formula we get,
M = Σ (xi – x̄)/n
= [|5 – 6| + |5 – 9| + |5 – 4| + |5 – 2| + |5 – 1| + |5 – 8|]/6
= (1 + 4 + 1 + 3 + 4 + 3)/6
= 16/6
= 2.67
Problem 4. Calculate the sum of deviations from the mean if the average absolute deviation is 15 and the number of observations is 6.
Solution:
We have,
M = 15
n = 6
Using the formula we get,
M = S/n
=> S = Mn
= 15 (6)
= 90
Problem 5. Calculate the sum of deviations from the mean if the average absolute deviation is 12 and the number of observations is 9.
Solution:
We have,
M = 12
n = 9
Using the formula we get,
M = S/n
=> S = Mn
= 12 (9)
= 108
Problem 6. Calculate the number of observations if the sum of deviations from the mean is 336 and the average absolute deviation is 12.
Solution:
We have,
S = 336
M = 12
Using the formula we get,
M = S/n
=> n = S/M
= 336/12
= 28
Problem 7. Calculate the number of observations if the sum of deviations from the mean is 518 and the average absolute deviation is 14.
Solution:
We have,
S = 518
M = 14
Using the formula we get,
M = S/n
=> n = S/M
= 518/14
= 37
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# How do you calculate load voltage?
## How do you calculate load voltage?
Calculating an Electrical Load in a Simple Circuit Let Power = Voltage * Current (P=VI). Let Current = Voltage/Resistance (I=V/R). Apply Kirchoff’s Second Law, that the sum of the voltages around a circuit is zero. Conclude that the load voltage around the simple circuit must be 9 volts.
### When loads are connected in series voltage is?
In a series circuit, all voltage sources and loads are connected, end to end, to form a single path. If any component of a series circuit becomes open or burnt out, current will be unable to flow. This circuit shows 2 lamps in series connected to a battery.
What is the load in a series circuit?
Series Circuit The current will flow from a power source, such as a battery, into one or more electrical loads, such as a light bulb, and then back to the power source. In a series circuit, the same amount of amperage from the power source flows through each load.
Is voltage the same in series?
The sum of the voltages across components in series is equal to the voltage of the supply. The voltages across each of the components in series is in the same proportion as their resistances . This means that if two identical components are connected in series, the supply voltage divides equally across them.
## What is connected in series?
Components connected in series are connected along a single “electrical path”, and each component has the same current through it, equal to the current through the network. The voltage across the network is equal to the sum of the voltages across each component.
### What is the formula of load?
Multiply the mass of the object by the gravitational acceleration of the earth (9.8 m/sec2), and the height in meters. This equation is the object at rest’s potential energy. Potential energy is measured in joules; this is the load force.
What is the formula for calculating voltage drop?
Voltage drop of the circuit conductors can be determined by multiplying the current of the circuit by the total resistance of the circuit conductors: VD = I x R.
Why are loads not connected in series?
It is not safe to operate e-loads in series if the device under test can supply more voltage than the rating of any one e-load, as the whole voltage will eventually appear across one e-load. So, the only safe way to operate in series is to have every e-load be rated for the full voltage of the device under test.
## What are the 3 rules for a series circuit?
From this definition, three rules of series circuits follow: all components share the same current; resistances add to equal a larger, total resistance; and voltage drops add to equal a larger, total voltage. All of these rules find root in the definition of a series circuit.
### What are the 3 rules of electricity?
We’ve organized these principles into three basic rules: Rule 1 – Electricity will always want to flow from a higher voltage to a lower voltage. Rule 2 – Electricity always has work that needs to be done. Rule 3 – Electricity always needs a path to travel.
Does voltage stay the same in parallel?
In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents flowing through each component.
How is voltage applied to a series circuit?
Voltage applied to a series circuit is equal to the sum of the individual voltage drops. The voltage drop across a resistor in a series circuit is directly proportional to the size of the resistor. If the circuit is broken at any point, no current will flow. “1. The same current flows through each part of a series circuit.”
## How to calculate the resistance of a series circuit?
1 Components in a series circuit share the same current: I Total = I 1 = I 2 = . . . I n 2 The total resistance in a series circuit is equal to the sum of the individual resistances: RTotal = R 1 + R 2 + . . . R n 3 Total voltage in a series circuit is equal to the sum of the individual voltage drops E Total = E 1 + E 2 + . . . En
### Can a series of voltage sources be combined?
But we have to consider the polarity as shown below. In the case of AC voltage sources in series, the voltage sources can be added or combined together to form a single source provided that the angular frequency (ω) of the connected sources are identical.
How are series circuits related to Ohm’s law?
“1. The same current flows through each part of a series circuit.” In a series circuit, the amperage at any point in the circuit is the same. This will help in calculating circuit values using Ohm’s Law. You will notice from the diagram that 1 amp continually flows through the circuit. We will get to the calculations in a moment. |
# A bag contains 3 red marbles, 3 green ones, 1 lavender one, 3 yellows, and 4 orange marbles. HINT [See Example 7.] How many sets of five marbles include either the lavender one or exactly one yellow one but not both colors?
There are 840 sets of five marbles that include either the lavender one or exactly one yellow one but not both colors.
Step-by-step explanation:
We need to treat the case where we get the lavender marble, then the case where we get exactly one yellow marble, because according to the question, these cases are mutually exclusive.
Given
Red Marbles = 3
Green = 3
Lavender = 1
Yellow = 3
Oranges = 4
TOTAL = 14
Now
For when we get the lavender marble alone, we have
C(1, 1) = 1!/(0! 1!) = 1/1 = 1 way to do this.
The remaining four marbles are neither lavender nor yellow.
There are 14 - 4 = 10 such marbles, so there are
C(10, 4) = 10!/(6! 4!) = 3628800/17280 = 210 ways to select the last four marbles.
In total, there are
C(1, 1) × C(10, 4) = 1 × 210 = 210 ways to select five marbles where one of them is the lavender color and none of them is yellow color.
Again, we need to get exactly one yellow marble.
There are C(3, 1) = 3!/(2! 1!) = 6/2 = 3 ways to do this.
The other four marbles are neither yellow nor lavender, there are
14 - 4 = 10 such marbles, so we have
C(10, 4) = 210 ways to select the last four marbles.
In total, we have
C(3, 1) × C(10, 4) = 3 × 210 = 630 ways to get exactly one yellow marble.
Finally, using the principle of addition to find the total number of ways to choose a set of five marbles that include either the lavender one or exactly one yellow
one, but not both. We have 210 + 630 = 840 ways to do
this. |
# Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force F. The magnitude of the tension in the string between blocks B and C is T=3.00N. Assume that each block has mass m=0.400kg. What is the magnitude F of the force? What is the tension tab in the string between block A and block B?
This article aims to find the tension in the string between two blocks $A$ and $B$. This article uses the concept of how to find the tension in the string. Tension in physics is the force developed in a rope, string, or cable when an applied force stretches it. Tension acts along the length of the rope in the opposite direction to the force acting on it. Tension can sometimes be referred to as stress, strain, or tension.
The formula for the tension in a string is given as:
$T = ma$
Given data
$T = 3.00\: N$
$m = 0.400 \: kg$
The magnitude $F$ of the force is given by:
$T = m a$
$3.00 = ( 0.400 ) a$
$a = \dfrac { 3 }{ 0.400 }$
$a = 7.5 \dfrac {m}{s^{2}}$
This is the total acceleration; acceleration for the individual block is:
$a = \dfrac {7.5}{2} = 3.75 \dfrac {m}{s^{2}}$
Force $F$ can be found by using:
$a = \dfrac {F}{3m}$
$F = 3am$
$F = 3 (3.75)(0.400 )$
$F = 4.5\:N$
For the tension between block $A$ and $B$:
$T = ma$
$T = (0.400\:kg) (3.75 \dfrac {m}{s^{2}})$
$T = 1.5 \: N$
The tension for each block is $1.5 \: N$.
## Numerical Result
The tension for each block is $1.5 \: N$.
## Example
Three identical blocks connected by ideal strings are pulled along a frictionless horizontal surface by a horizontal force $F$.
Magnitude of the tension in string between blocks $B$ and $C$ is $T=5.00\:N$. Assume that each block has a mass of $m=0.500 \:kg$.
-What is the magnitude of the $F$ force?
-What is the tension in the string between block $A$ and block $B$?
Solution
Given data
$T = 5.00\: N$
$m = 0.500 \: kg$
The magnitude $F$ of the force is given by:
$T = m a$
$5.00 = ( 0.500 ) a$
$a = \dfrac { 5 }{ 0.500 }$
$a = 10 \dfrac { m }{s ^ { 2 }}$
This is the total acceleration; acceleration for the individual block is:
$a = \dfrac { 10 }{ 5 } = 2 \dfrac { m }{ s ^ { 2 }}$
Force $F$ can be found by using:
$a = \dfrac { F }{ 3 m }$
$F = 3 a m$
$F = 3 ( 2 )( 0.500 )$
$F = 3 \:N$
For the tension between block $A$ and $B$:
$T = ma$
$T = ( 0.500\:kg ) ( 2 \dfrac {m}{s ^ { 2 }} )$
$T = 1.0 \: N$
The tension for each block is $1.0 \:N$. |
# 5300 in Words
5300 in words is written as Five thousand three hundred. We can write the word form of the number 5300 using a place value chart effortlessly. However, we use the word form of 5300 in expressing a certain amount equivalent to 5300. Suppose a truck has 5300 packets of candies; we say it as “The truck contains five thousand three hundred packets of candies”. Let’s learn the process of converting the number 5300 into words.
5300 in words Five thousand three hundred Five thousand three hundred in Numbers 5300
## 5300 in English Words
We generally write numbers in words using the English alphabet. Thus, we can express 5300 in English words as “Five thousand three hundred”.
## How to Write 5300 in Words?
In this section, you will understand how to write 5300 in words using a four-column place value chart since it has four digits.
Thousands Hundreds Tens Ones 5 3 0 0
Here, ones = 0, tens = 0, hundreds = 3, and thousands = 5.
Let us expand these digits as per their place values.
5 × Thousand + 3 × Hundred + 0 × Ten + 0 × One
= 5 × 1000 + 3 × 100 + 0 × 10 + 0 × 1
= 5000 + 300
= Five thousand + Three hundred
= Five thousand three hundred
Hence, 5300 in words = Five thousand three hundred.
As we know, 5300 is a natural number that precedes 5301 and succeeds 5299.
5300 in words – Five thousand three hundred
Is 5300 an odd number? – No
Is 5300 an even number? – Yes
Is 5300 a prime number? – No
Is 5300 a composite number? – Yes
Is 5300 a perfect square number? – No
Is 5300 a perfect cube number? – No
## Frequently Asked Questions on 5300 in Words
Q1
### How do you write 5300 in words?
Generally, we write the number 5300 in words as Five thousand three hundred.
Q2
### How to write 5300 rupees in words on a cheque?
On a cheque, we generally write 5300 rupees in words as Five thousand three hundred rupees only.
Q3
### Write the value of 2800 + 2500 in words.
2800 + 2500 = 5300 Therefore, the value of 2800 + 2500, i.e., 5300 in words is five thousand three hundred. |
# Proof matrix $A=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^n , \text{ when } n \in \mathbb{N}$ [duplicate]
## Problem
Find generalitazion for matrix A exponents, when $$n\in\{1,2,3,\dots\},n \in \mathbb{N}$$
$$A^n=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^n , \text{ when } n \in \mathbb{N}$$
Proof generalization by induction.
## Attempt to solve
By computing a set of $$A$$ exponent's $$n\in \{\ 1,2,3,4 \}$$. It is possible to form generalization that is applicable for set $$n\in \{1,2,3,4 \}$$
$$A^1=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix},A^2=\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix},A^3=\begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix},A^4=\begin{bmatrix} 1 & 4 \\ 0 & 1 \end{bmatrix} \dots A^n =\begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}$$
### Induction proof
Induction hypothesis
Assume expression is valid when $$n=k$$
$$A^k = \begin{bmatrix} 1 & k \\ 0 & 1\end{bmatrix}$$
Base case
When $$n=1$$ $$A^1=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$ which is valid by definition.
Induction step
When $$n=k+1$$
$$A^{k+1} = \begin{bmatrix} 1 & k+1 \\ 0 & 1\end{bmatrix}$$
$$A^{k+1}=A^kA^1$$
By utilizing induction hypothesis we have
$$\implies A^{k+1}=\begin{bmatrix} 1 & k \\ 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$ By utilizing matrix multiplication we have $$\implies A^{k+1} \begin{bmatrix} 1\cdot 1 + k\cdot 0 & 1 \cdot 1 + 1 \cdot k \\ 0 \cdot 1 + 1 \cdot 0 & 0 \cdot 1 + 1 \cdot 1 \end{bmatrix}$$ $$\implies A^{k+1}=\begin{bmatrix} 1 & k+1 \\ 0 & 1 \end{bmatrix}$$
$$\tag*{\square}$$
## EDIT
The point of posting this was to have comment on if my solution seems correct or not. If you can notice something that doesn't look right, let me know !
## marked as duplicate by Mark McClure, ArsenBerk, Chris Custer, user10354138, Lee David Chung LinNov 13 '18 at 2:00
• It is correct, just that it is confusing the way you "define " $A$ in the first equation as you define $A$ to be what you later call $A^n$. – Surb Nov 12 '18 at 19:36
Your proof looks good to me. There is an alternative proof: Note that $$A=I+N$$, where $$N=\pmatrix{0&1\\ 0&0}$$ squares to zero. Therefore, by binomial expansion, $$A^n=(I+N)^n=I+nN+\binom{n}{2}N^2+\cdots+\binom{n}{n}N^n=I+nN=\pmatrix{1&n\\ 0&1}.$$ |
# SAT II Math II : Systems of Equations
## Example Questions
### Example Question #1 : Systems Of Equations
Where do the graphs of these two equations intersect?
Equation 1:
Equation 2:
Explanation:
One can find the points of intersection of these two functions by setting them equal to one another, essentially substituting in one equation for the -side of the other equation. This will tell us when the (output) will be the same in each equation for a given (input).
By simplifying this equation and setting it equal to zero, we can find the two -values that produce the same values in the system of the two equations.
Subtract from both sides of the equation, and add to both sides.
Factoring this last equation makes it easier to find the -values that will result in zero on the left side of the equation. Set the two parenthetical phrases equal to zero to find two separate -values that satisfy the equation. These values will be the values of the points of intersection between the two equations.
We know our factors multiply to , and the six times one factor plus the other is equal to
and , so and are our factors.
Substituting these two values into either of the two original equations results in the -values of the points of intersection.
are the points of intersection.
### Example Question #1 : Systems Of Equations
Find the solution:
Explanation:
To solve this system of equations, we must first eliminate one of the variables. We will begin by eliminating the variables by finding the least common multiple of the variable's coefficients. The least common multiple of 3 and 2 is 6, so we will multiply each equation in the system by the corresponding number, like
.
By using the distributive property, we will end up with
Now, add down each column so that you have
Then you solve for and determine that .
But you're not done yet! To find , you have to plug your answer for back into one of the original equations:
Solve, and you will find that .
### Example Question #2 : Systems Of Equations
Evaluate
The system has no solution.
Explanation:
Rewrite the two equations by setting and and substituting:
The result is a two-by-two linear system in terms of and :
This can be solved, among other ways, using Gaussian elimination on an augmented matrix:
Perform row operations until the left two columns show identity matrix . One possible sequence:
and .
In the former equation, substitute back for :
.
Taking the reciprocal of both sides, we get
.
### Example Question #4 : Systems Of Equations
Evaluate .
The system has no solution.
Explanation:
Rewrite the two equations by setting and and substituting:
The result is a two-by-two linear system in terms of and :
This can be solved, among other ways, using Gaussian elimination on an augmented matrix:
Perform row operations until the left two columns show identity matrix . One possible sequence:
and .
Substituting back in the second equation and solving for by cubing both sides:
### Example Question #3 : Systems Of Equations
Evaluate .
The system has no solution.
Explanation:
Rewrite the two equations by setting and and substituting:
In terms of and , this is a two-by-two linear system:
Rewrite this as an augmented matrix as follows:
Perform the following row operations to make the left two columns the identity matrix :
and .
Replacing with , then squaring both sides:
### Example Question #1 : Systems Of Equations
Solve for :
Explanation:
To solve for , we will need to eliminate the variable. To do so, we can choose either the substitution or the elimination method.
Let's choose the substitution method. Rewrite the second equation so that is isolated.
Subtract on both sides.
Divide by five on both sides.
Substitute this back to the first equation.
Distribute the terms.
Multiply both sides by 5 to eliminate the fractions.
Add 8 on both sides and combine like-terms on the left.
Divide by 18 on both sides.
Solve for : |
# Cuisenaire Rods
## Curriculum Goal
#### Kindergarten: Demonstrating Literacy and Mathematics Behaviour
• Demonstrate an understanding of numbers, using concrete materials to explore and investigate counting, quantity, and number relationships (#15).
## Context
• Students and teacher will be sitting on the floor with a working mat in front of them.
## Materials
• Working Mat for each student
• Cuisenaire rods
## Lesson
• Lay out the Cuisenaire Rods and explain that each rod can be assigned a number.
• Hold up one of the rods and ask the kids to grab the same colour of rod you have.
• Then say, If I gave this rod the number 2, what number do you think this rod would be? (hold up immediate next rod, 3). Wait to hear their answers and reasoning or strategies behind their response.
• Show the rod immediately before the 2nd rod and ask children what they think that rod may represent.
• Once again, wait to hear children’s answers and reasoning behind their response.
• Ask students what they think will happen if a #1 rod is put on top of another #1 rod.
• Allow children to try it. Students will discover that if you put the #1 rod on top of another #1 rod, it is the same height as the #2 rod.
• After establishing the #1, #2, #3 rods and the numbers they represent, skip the #4 rod and instead show the #5 rod.
• Ask students what number they think it will represent and how they figured it out.
• Fill in the rest of the numbers and ask students what they notice about all those rods lined up.
## Look Fors
• Are they able to express their reasoning?
• Are children able to grasp the pattern and guess at the next and previous term?
## Extension
Make different combinations of the same number
• Demonstrate that if you put #5 and #5 together (one on top of another) it makes 10.
• Then ask students if they can figure out how to make 10 by putting 2 rods together.
• Students can work in pairs to make as many different combinations as possible.
• Bring students back to the whole group and ask them to share one of the combinations they made.
• Students will be able to see the combinations that make 10 because they will all be the same height.
• Students may further extend the activity by creating their own combinations.
Roll a dice and make as many combinations of that number as you can
• Students roll two dice and add the numbers together.
• Using the Cuisenaire rods, students show different ways to make that number.
## Related Lessons
Foster an understanding of number relationships using manipulatives.
Encourage students to use manipulatives to compose and decompose numbers. |
Factors that 121 are the list of integers that us can split evenly right into 121. There space 3 components of 121 of i beg your pardon 121 itself is the biggest factor and also its positive components are 1, 11 and also 121. The Prime components of 121 space 1, 11, 121 and also its components in Pairs space (1, 121) and also (11, 11).
You are watching: What two numbers multiply to 121
Factors that 121: 1, 11 and 121Negative determinants of 121: -1, -11 and -121Prime components of 121: 11Prime administrate of 121: 11 × 11 = 112Sum of determinants of 121: 133
1 What are factors of 121? 2 How come Calculate components of 121? 3 Factors the 121 by element Factorization 4 Factors that 121 in Pairs 5 FAQs on factors of 121
## What are components of 121?
The aspect of a number is that number the divides it fully i.e. Leaving no remainder. In stimulate to uncover the factors of the number 121, we will need to perform division on 121 and find the numbers which division 121 completely, leave no remainders.
## How to calculation the factors of 121?
There room two common methods to find the components of 121.
Let us try to discover using the division method:
We begin with the number 1, then check for numbers 2, 3, 4, 5, 6, 7, etc up come 61( approximate fifty percent of 121) respectively.The number 1 and also the number chin would constantly be a factor of the provided number.Refer to the following table come check department 121 by its factors:
Divide Result 121÷ 1 Gives remainder 0So, 1 is a factor. 121 ÷ 11 Gives remainder 0So, 11 is a factor. 121 ÷ 121 Gives remainder 0So, 121 is a factor.
Hence, the components of 121 are 1, 11 , and 121. Explore factors using illustrations and interactive examples:
Tips and Tricks
Use the divisibility rule of various prime determinants to find the prime factorization of 121.The element factorization that 121 is 11 × 11 or 112, then add 1 come the exponent. I.e. 2 + 1 = 3. We have 3 distinct factors of 121.
## Factors of 121 by prime Factorization
Prime administrate is expressing the number as the product the its element factors.
Step 2: After finding the smallest prime element of the number 121 i.e., 11, divide 121 by 11 to achieve the quotient 11.Step 3: The derived quotient is a prime number itself. The department procedure have the right to thus be terminated in ~ this step.
So, the Prime factorization the 121 is: 121 = 11 × 11 = 112
## Factors that 121 in Pairs
Positive pair determinants of 121 : (1,11), (11,11), (1,121)Negative pair factors of 121 : (-1, -11), (-11, -11), (-1, -121)
Important Notes:
Only entirety numbers and also integers have the right to be convert to factors.Only composite numbers can have much more than two factors.The smallest factor of a number is 1 and the biggest aspect of a number would certainly be the number itself.
## Factors that 121 solved Examples
Example 1: Find the amount of all the components of 121.
Solution:
We understand the components of 121 are 1, 11, and also 121.Thus, the sum of all the determinants is 1 + 11 + 121 = 133.
Example 2: Find the common factors that 121 and also 242.
Solution:
The components of 121 space 1, 11, and 121. The determinants of 242 are 1, 2, 11, 121, 242.Hence, the typical factors that 121 and 242 room 1, 11, and 121.
Example 3: discover the product of all the prime factors of 121.
Solution:
Since the prime factors of 121 are 11. Therefore, the product the prime factors = 11 = 11.
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## FAQs on components of 121
### What are the factors of 121?
The factors of 121 space 1, 11, 121 and also its an unfavorable factors are -1, -11, -121.
### What is the sum of the determinants of 121?
Since all factors of 121 room 1, 11, 121 therefore, the sum of its factors is 1 + 11 + 121 = 133.
### What is the Greatest typical Factor of 121 and also 53?
The factors of 121 space 1, 11, 121 and also the factors of 53 are 1, 53. 121 and 53 have only one usual factor i beg your pardon is 1. This implies that 121 and 53 are co-prime.Hence, the Greatest usual Factor (GCF) of 121 and 53 is 1.
### What are the Pair components of 121?
The pair factors of 121 are (1, 121), (11, 11).
See more: What Kind Of Products Are Purchased With Discretionary Income Definition
### How numerous Factors that 66 are likewise common to the factors of 121?
Since the factors of 121 are 1, 11, 121, and the components of 66 space 1, 2, 3, 6, 11, 22, 33, 66.Hence, <1, 11> room the usual factors that 121 and also 66. |
## Skills Review for Properties of Power Series
### Learning Outcomes
• Use partial fraction decomposition for linear factors
• Simplify expressions using the power property of pxponents
In the Properties of Power Series section, we will look at how to combine, differentiate, and integrate power series. Here we will review how to use partial fraction decomposition for linear factors and how to use the power property of exponents.
## Use Partial Fraction Decomposition for Linear Factors
(also in Module 5, Skills Review for Infinite Series)
Partial fraction decomposition is used to break up one fraction into two.
\begin{align}\underset{\text{ }\\ \text{Simplified sum}}{\frac{x+7}{{x}^{2}-x - 6}}=\underset{\text{ }\\ \text{Partial fraction decomposition}}{\frac{2}{x - 3}+\frac{-1}{x+2}}\\ \text{ }\end{align}
We will investigate rational expressions with linear factors in the denominator where the degree of the numerator is less than the degree of the denominator. Regardless of the type of expression we are decomposing, the first and most important thing to do is factor the denominator.
### How To: Given a rational expression with distinct linear factors in the denominator, decompose it.
1. Use a variable for the original numerators, usually $A,B,$ or $C$, depending on the number of factors, placing each variable over a single factor. For the purpose of this definition, we use ${A}_{n}$ for each numerator
$\frac{P\left(x\right)}{Q\left(x\right)}=\frac{{A}_{1}}{\left({a}_{1}x+{b}_{1}\right)}+\frac{{A}_{2}}{\left({a}_{2}x+{b}_{2}\right)}+\cdots \text{+}\frac{{A}_{n}}{\left({a}_{n}x+{b}_{n}\right)}$
2. Multiply both sides of the equation by the common denominator to eliminate fractions.
3. Expand the right side of the equation and collect like terms.
4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.
### Example: Decomposing a Rational Function with Distinct Linear Factors
Decompose the given rational expression with distinct linear factors.
$\dfrac{3x}{\left(x+2\right)\left(x - 1\right)}$
### Try It
Find the partial fraction decomposition of the following expression.
$\dfrac{x}{\left(x - 3\right)\left(x - 2\right)}$
### Try It
Some fractions we may come across are special cases that we can decompose into partial fractions with repeated linear factors. We must remember that we account for repeated factors by writing each factor in increasing powers.
### How To: Given a rational expression with repeated linear factors, decompose it.
1. Use a variable like $A,B$, or $C$ for the numerators and account for increasing powers of the denominators.
$\dfrac{P\left(x\right)}{Q\left(x\right)}=\dfrac{{A}_{1}}{\left(ax+b\right)}+\dfrac{{A}_{2}}{{\left(ax+b\right)}^{2}}+ \text{. }\text{. }\text{. + }\dfrac{{A}_{n}}{{\left(ax+b\right)}^{n}}$
2. Multiply both sides of the equation by the common denominator to eliminate fractions.
3. Expand the right side of the equation and collect like terms.
4. Set coefficients of like terms from the left side of the equation equal to those on the right side to create a system of equations to solve for the numerators.
### Example: Decomposing with Repeated Linear Factors
Decompose the given rational expression with repeated linear factors.
$\dfrac{-{x}^{2}+2x+4}{{x}^{3}-4{x}^{2}+4x}$
### Try It
Find the partial fraction decomposition of the expression with repeated linear factors.
$\dfrac{6x - 11}{{\left(x - 1\right)}^{2}}$
## Simplify Expressions Using the Power Property for Exponents
### A General Note: The Power Rule of Exponents
For any real number $a$ and positive integers $m$ and $n$, the power rule of exponents states that
${\left({a}^{m}\right)}^{n}={a}^{m\cdot n}$
### Example: USING the Power Rule of Exponents
Write each of the following products with a single base. Do not simplify further.
1. ${\left({x}^{2}\right)}^{7}$
2. ${\left({\left(2t\right)}^{5}\right)}^{3}$
3. ${\left({\left(-3\right)}^{5}\right)}^{11}$ |
# Integer addition: Problem type 2
If the signs of the integers are same, we add the numbers and keep the sign
• If the signs are positive, we add the numbers and keep the positive sign.
• If the signs are negative, we add the numbers and keep the negative sign.
If the signs are different, we subtract the numbers and keep the sign of the number with larger absolute value.
23 + (−36)
### Solution
Step 1:
The signs of the numbers are different. So, we subtract the absolute values of the integers.
|−36| – |23| = 36 – 23 = 13
Step 2:
The sign of the number with larger absolute value (−36) is −.
We keep this sign with the difference obtained in above step
So, 23 + (−36) = − 13
−45 + (−58)
### Solution
Step 1:
The signs of the numbers are same. So, we add the absolute values of the integers.
|−45| +| − 58| = 45 + 58 = 103
Step 2:
The common sign of both numbers is −.
We keep this sign with the sum obtained in above step
So, −45 + (−58) = − 103 |
# Fractional Equivalents: How to Find Them
Fractional Equivalents: How to Find Them
Understanding and working with fractions can be a tricky concept for many students. Fractional equivalents are a helpful tool to help students identify and work with fractions. This blog post will discuss what fractional equivalents are, how to find them, and provide examples to help illustrate the concepts. We will also include a FAQ section to answer some of the most commonly asked questions about fractional equivalents. Finally, we will provide a summary and conclusion of the post.
Introduction to Fractional Equivalents
Fractional equivalents are fractions that are equal in value, but may have different numerators and denominators. For example, 1/2 is equivalent to 2/4, 4/8, and 8/16. All of these fractions are equal in value, but have different numerators and denominators. Fractional equivalents can be used to simplify fractions, find common denominators, and add, subtract, multiply, and divide fractions.
How to Find Fractional Equivalents
Finding fractional equivalents is a simple process. To find fractional equivalents, all you need to do is multiply or divide the numerator and denominator of the fraction by the same number. This will result in a fraction that is equal in value to the original fraction. For example, if you want to find the fractional equivalent of 3/4, you can multiply both the numerator and denominator by 2 to get 6/8. This is the fractional equivalent of 3/4.
Examples of Fractional Equivalents
To help illustrate the concept of fractional equivalents, let’s look at some examples.
Example 1:
Find the fractional equivalent of 2/3.
Solution:
The fractional equivalent of 2/3 is 4/6. This is because we can multiply both the numerator and denominator of the fraction by 2 to get 4/6.
Example 2:
Find the fractional equivalent of 5/6.
Solution:
The fractional equivalent of 5/6 is 10/12. This is because we can multiply both the numerator and denominator of the fraction by 2 to get 10/12.
Example 3:
Find the fractional equivalent of 7/8.
Solution:
The fractional equivalent of 7/8 is 14/16. This is because we can multiply both the numerator and denominator of the fraction by 2 to get 14/16.
FAQ Section
Q: What is the difference between a fraction and a fractional equivalent?
A: A fraction is a number that is written as a ratio of two numbers. A fractional equivalent is a fraction that is equal in value to the original fraction, but may have a different numerator and denominator.
Q: How do you find fractional equivalents?
A: To find fractional equivalents, all you need to do is multiply or divide the numerator and denominator of the fraction by the same number. This will result in a fraction that is equal in value to the original fraction.
Q: What are some examples of fractional equivalents?
A: Some examples of fractional equivalents are 1/2 = 2/4, 4/8, and 8/16; 3/4 = 6/8; 5/6 = 10/12; and 7/8 = 14/16.
Summary and Conclusion
In this blog post, we discussed what fractional equivalents are, how to find them, and provided examples to help illustrate the concepts. We also included a FAQ section to answer some of the most commonly asked questions about fractional equivalents.
Fractional equivalents are a helpful tool to help students identify and work with fractions. Understanding and working with fractions can be a tricky concept for many students, and fractional equivalents can make the process easier. By understanding how to find fractional equivalents, students can simplify fractions, find common denominators, and add, subtract, multiply, and divide fractions.
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# Question Video: Determining the Tools Required to Construct an Angle Bisector without a Protractor Mathematics
True or False: Using only a straightedge, you can construct an angle bisector.
01:38
### Video Transcript
True or False: Using only a straightedge, you can construct an angle bisector.
To determine whether this statement is true, weβll need to review the steps required to construct an angle bisector. Letβs sketch ray π΄π΅ and ray π΄πΆ to form an arbitrary angle π΅π΄πΆ. We recall that our very first step requires tracing a circle centered at π΄ that intersects the sides of the angle at two distinct points.
Now, we canβt construct a circle using just a straightedge. It will require a compass. We place the point of our compass on the vertex π΄. Now, we donβt have to trace the whole circle, just an arc that intersects ray π΄π΅ at point π· and ray π΄πΆ at point πΈ. The next step also requires a compass in order to trace two circles of the same radius centered at π· and πΈ. These circles will intersect at a point that we name point πΉ.
It is only in the final step that we use our straightedge to sketch the line π΄πΉ that bisects angle π΅π΄πΆ. As we have demonstrated, we can use a straightedge to construct an angle bisector, but not only a straightedge. Without a compass, we cannot complete the construction. Thus, the given statement is false.
We note that we could alternatively bisect an angle using only a ruler by constructing an isosceles triangle. We would have to measure the sides then connect the midpoint of the base to the opposite vertex. |
# How do you solve 3x + 2y + 2z =3, x + 2y - z =5 and 2x - 4y + z=0 using matrices?
##### 1 Answer
Sep 29, 2016
The answer is $x = 2 , y = \frac{1}{2} , z = - 2$. Please see the explanation for the row operations.
#### Explanation:
Put the row for x + 2y - z = 5 at the top, because the coefficient of x is 1:
|1 2 -1 |5|
Put the row for 3x + 2y + 2z = 3 next:
|1 2 -1 |5|
|3 2 2|3|
Put the row for 2x -4y + z = last:
|1 2 -1 |5|
|3 2 2|3|
|2 -4 1|0|
Multiply row 1 by -3 and add to row 2:
|1 2 -1 |5|
|0 -4 5|-12|
|2 -4 1|0|
Multiply row 1 by -2 and add to row 3:
|1 2 -1 |5|
|0 -4 5|-12|
|0 -8 3|-10|
Multiply row 2 by -2 and add to row 3:
|1 2 -1 |5|
|0 -4 5|-12|
|0 0 -7|14|
Divide row 3 by -7
|1 2 -1 |5|
|0 -4 5|-12|
|0 0 1|-2|
Multiply row 3 by -5 and add to row 2:
|1 2 -1 |5|
|0 -4 0|-2|
|0 0 1|-2|
Divide row 2 by -4:
|1 2 -1 |5|
|0 1 0|1/2|
|0 0 1|-2|
Add row 3 to row 1:
|1 2 0|3|
|0 1 0|1/2|
|0 0 1|-2|
Multiply row 2 by -2 and add to row 1:
|1 0 0|2|
|0 1 0|1/2|
|0 0 1|-2| |
# What numbers give you 60?
6 and 10 both are factors of 60. For example, 6 × 10 = 60. The pair factors of 60 would be the two numbers which, when multiplied together, result in 60.
## How do you do quick multiplication?
Instead of going through the usual multiplication process, in this case all you have to do is double the number you’re multiplying for each power of 2 in the other number. For example, the problem 12 x 8 is the same thing as 12 x 2^3 or 12 x 2 x 2 x 2.
## What are the factors of 16?
The factors of 16 are 1, 2, 4, 8, 16. You can also think about factors in terms of division: The factors of a number include all numbers that divide evenly into that number with no remainder. Consider the number 10. Since 10 is evenly divisible by 2 and 5, you can conclude that both 2 and 5 are factors of 10.
## What are the factors of 81 and 16?
The gcf of 81 and 16 can be obtained like this: The factors of 81 are 81, 27, 9, 3, 1. The factors of 16 are 16, 8, 4, 2, 1. The common factors of 81 and 16 are 1, intersecting the two sets above.
## Can 81 be divided?
When we list them out like this it’s easy to see that the numbers which 81 is divisible by are 1, 3, 9, 27, and 81.
## What two numbers make 16 and 60?
The GCF of 16 and 60 is 4. To calculate the GCF of 16 and 60, we need to factor each number (factors of 16 = 1, 2, 4, 8, 16; factors of 60 = 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60) and choose the greatest factor that exactly divides both 16 and 60, i.e., 4.
## Who invented math?
Archimedes is known as the Father of Mathematics. Mathematics is one of the ancient sciences developed in time immemorial. A major topic of discussion regarding this particular field of science is about who is the father of mathematics.
## Is math easy or hard?
Mathematics is considered as one of the toughest subjects by many students. However, many also consider as the most scoring subject. In fact, according to some students, one can easily score 100/100 (or full marks) in maths easily as compared to other subjects.
## Why is math so hard?
Math seems difficult because it takes time and energy. Many people don’t experience sufficient time to “get” math lessons, and they fall behind as the teacher moves on. Many move on to study more complex concepts with a shaky foundation. We often end up with a weak structure that is doomed to collapse at some point.
## What multiplied equals 42?
You can think of factors as the numbers that you multiply to get another number. In this case 2, 3, 6, and 7 are all numbers you can multiply to get 42. So, they are all factors of 42.
## What are the multiples of 6?
We can arrange the multiples of 6 in increasing order, 0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96,… so that they form a simple pattern increasing by 6 at each step. Because 6 is an even number, all its multiples are even.
## What are the prime numbers of 16?
The Prime Factors of 16 are 1, 2, 4, 8, 16 and its Factors in Pairs are (1, 16), (2, 8), and (4, 4).
## What are 16 and 81 divisible by?
For 16 and 81 those factors look like this: Factors for 16: 1, 2, 4, 8, and 16. Factors for 81: 1, 3, 9, 27, and 81.
## What numbers are 16 divisible by?
When we list them out like this it’s easy to see that the numbers which 16 is divisible by are 1, 2, 4, 8, and 16.
## Is 243 divisible by any number?
The factors of 243 and the multiples of 3 are 3, 9, 27, 81, 243. Hence, 3, 9, 27, 81, and 243 are the factors of 243 which are divisible by 3.
## What are 60 factors?
The number 60 has more than two factors, as it is a composite number. The factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.
## What is the GCF and LCM of 16 and 60?
Therefore, the LCM of 16 and 60 is 240. Example 2: Verify the relationship between GCF and LCM of 16 and 60. Hence, verified. Example 3: The GCD and LCM of two numbers are 4 and 240 respectively.
## What can equal 98?
Pair Factors of 98 If 1 × 98 = 98, then (1, 98) is a pair factor of 98. Therefore, the positive pair factors of 98 are (1, 98), (2, 49), and (7, 14). If -1 × -98 = 98, then (-1, -98) is a pair factor of 98. |
# Trigonometry Expression
Introduction
Trigonometry deals with the angle and side of triangles. Unknown angles or lengths are evaluated by using trigonometrical ratios such as sine, cosine, and tangent. Trigonometry used in the fields of navigation, astronomy and surveying, and usually involved working out an inaccessible distance.Trigonometry used to find the distance of the Earth from the Moon. Trigonometric ratios of acute angles as the ratio of the sides of a right angled triangle.
## Trigonometric functions on trigonometry expression:
Some trigonometry functions are given below:
SIN = OPPOSITE / HYPOTENUSE
COSEC = HYPOTENSE / OPPOSITE
The right angle sided of the triangle consists of any value of all sides and the lengths will be the same for all right triangles either the triangle may be Large or small; the ratio present only depend on the angles and not on the length which is actual the functions which in pair that is sine and cosine, tangent and cotangent, secant and co-secant called co functions.
## Examples for Trigonometry Expression:
Example 1:
Simplify the trigonometric expression [sin 6x - cos 6x] / [sin 4x - cos 4x].
Solution:
• Factor the denominator
[sin 6x - cos 6x] / [sin 4x - cos 4x]
= [sin 2x - cos 2x] [sin 2x + cos 2x] [sin 2x - cos 2x] / [sin 2x - cos 2x] [sin 2x - cos 2x] and simplify
= [sin 2x + cos 2x] = 1
Example 2:
Simplify the trigonometric expression [sec(x) sin 4x] / [[1 + sec(x)] Sin2x].
Solution:
• Substitute sec (x) is in the numerator by 1 / cos (x) and simplify.
[sec(x) sin 4x] / [[1 + sec(x)] Sin2x]
= sin 4x / [ [cos x (1 + sec (x) ] Sin2x]
= sin 2x / [ cos x + 1 ]
These are the trigonometry expression. |
# Surface Area of a Prism Formula
A prism is a polyhedron with 2 polygonal bases parallel to each other. The two polygonal bases are joined by lateral faces. The number of lateral faces are equal to the number of sides in the base. The lateral faces in a prism are perpendicular to the polygonal bases. Surface area of a prism is the sum of the area of the 2 polygonal bases and the area of the lateral faces. The surface area of a prism is measured in terms of square units.
The Surface Area of a Prism Formula is given as,
Surface Area Of A Rectangular Prism is A = 2 (wl + lh + hw)
Surface Area Of A Triangular Prism is A = bh + L (s1 + s2 + s3)
   Â
Where,
a – apothem length of the prism.
b – base length of the prism.
l – base width of the rectangular prism.
h – height of the prism.Â
Â
## Surface Area of a Prism Problems
Some solved problems on the surface area of a prism are given below :
### Solved Examples
Question 1: What will be the surface area of a triangular prism if the apothem length, base length and height are 7 cm, 10 cm and 18 cm respectively ?
Solution:
Â
Given,
a = 7 cm
b = 10 cm
h = 18 cm
Surface area of a triangular prism
= ab + 3bh
= (7 cm
$$\begin{array}{l}\times\end{array}$$
10 cm) + (3
$$\begin{array}{l}\times\end{array}$$
10 cm
$$\begin{array}{l}\times\end{array}$$
18 cm)
= 70 cm2 + 540 cm2
= 610 cm2
Â
Question 2: Calculate the surface area of a pentagonal prism with apothem length, base length and height as 12 cm, 15 cm, and 21 cm respectively ?
Solution:
Â
Given,
a = 12 cm
b = 15 cm
h = 21 cm
Surface area of a pentagonal prism
= 5ab + 5bh
= (5
$$\begin{array}{l}\times\end{array}$$
12 cm
$$\begin{array}{l}\times\end{array}$$
15 cm) + (5
$$\begin{array}{l}\times\end{array}$$
15 cm
$$\begin{array}{l}\times\end{array}$$
21 cm)
= 900 cm2 + 1575 cm2
= 2475 cm2
Â
More topics in Surface Area of a Prism Formula Surface Area of a Triangular Prism Formula Surface Area of a Rectangular Prism Formula |
# How to Divide by a Two‐Digit Number
Dividing by a two-digit number is a lot like single-digit division, but it does take a little longer and some practice. Since most of us haven't memorized our 47 times tables, this can take a little guesswork, but there's a handy trick you can learn to make it faster. It also gets easier with practice, so don't get frustrated if it seems slow at first.
Part 1
Part 1 of 2:
### Dividing by a Two-Digit Number
1. 1
Look at the first digit of the larger number. Write the problem as a long-division problem. Just like a simpler division problem, you can start by looking at the smaller number, and asking "Does it fit into the first digit of the larger number?"[1]
• Let's say you're solving 3472 ÷ 15. Ask "Does 15 fit into 3?" Since 15 is definitely larger than 3, the answer is "no," and we move on to the next step.
2. 2
Look at the first two digits. Since you can't fit a two-digit number into a one-digit number, we'll look at the first two digits of the dividend instead, just like we would in a regular division problem. If you still have an impossible division problem, you'll need to look at the first three digits instead, but we don't need to in this example:[2]
• Does 15 fit into 34? Yes, it does, so we can start calculating the answer. (The first number doesn't have to fit perfectly, it just needs to be smaller than the second number.)
3. 3
Use a little guesswork. Find out exactly how many times the first number fits into the other. You might know the answer already, but if you don't, try making a good guess and checking your answer with multiplication.[3]
• We need to solve 34 ÷ 15, or "how many times does 15 go into 34"? You're looking for a number you can multiply with 15 to get a number less than 34, but pretty close to it:
• Does 1 work? 15 x 1 = 15, which is less than 34, but keep guessing.
• Does 2 work? 15 x 2 = 30. This is still less than 34, so 2 is a better answer than 1.
• Does 3 work? 15 x 3 = 45, which is greater than 34. Too high! The answer must be 2.
4. 4
Write the answer above the last digit you used. If you set this up like a long division problem, this should feel familiar.
• Since you were calculating 34 ÷ 15, write the answer, 2, on the answer line above the "4."
5. 5
Multiply your answer by the smaller number. This is the same as a normal long division problem, except we'll be using a two-digit number.[4]
• Your answer was 2 and the smaller number in the problem is 15, so we calculate 2 x 15 = 30. Write "30" underneath the "34."
6. 6
Subtract the two numbers. The last thing you wrote went underneath the original larger number (or part of it). Treat this as a subtraction problem and write the answer on a new line underneath.[5]
• Solve 34 - 30 and write the answer underneath them on a new line. The answer is 4. This 4 is still "left over" after we fit 15 into 34 two times, so we'll need to use it in the next step.
7. 7
Bring down the next digit. Just like a regular division problem, we're going to keep calculating the next digit of the answer until we've finished.[6]
• Leave the 4 where it is and bring down the "7" from "3472" to make 47.
8. 8
Solve the next division problem. To get the next digit, just repeat the same steps you did above for the new problem. You can use guesswork again to find the answer:[7]
• We need to solve 47 ÷ 15:
• 47 is bigger than our last number, so the answer will be higher. Let's try four: 15 x 4 = 60. Nope, too high!
• We'll try three instead: 15 x 3 = 45. Smaller than 47 but close to it. Perfect.
• (If we ended up with a problem like 13 ÷ 15, with the first number smaller, we would need to bring down a third digit before we could solve it.)
9. 9
Continue using long division. Repeat the long division steps we used before to multiply our answer by the smaller number, write the result underneath the larger number, and subtract to find the next remainder.[8]
• Remember, we just calculated 47 ÷ 15 = 3, and now we want to find what's left over:
• 3 x 15 = 45, so write "45" underneath the 47.
• Solve 47 - 45 = 2. Write "2" underneath the 45.
10. 10
Find the last digit. As before, we bring down the next digit from the original problem so we can solve the next division problem. Repeat the steps above until you find every digit in the answer.[9]
• We've got 2 ÷ 15 as our next problem, which doesn't make much sense.
• Bring down a digit to make 22 ÷ 15 instead.
• 15 goes into 22 one time, so we write "1" at the end of the answer line.
• Our answer is now 231.
11. 11
Find the remainder. One last subtraction problem to find the final remainder, then we'll be done. In fact, if the answer to the subtraction problem is 0, you don't even need to write a remainder at all.[10]
• 1 x 15 = 15, so write 15 underneath the 22.
• Calculate 22 - 15 = 7.
• We have no more digits to bring down, so instead of more division we just write "remainder 7" or "R7" at the end of our answer.
• The final answer: 3472 ÷ 15 = 231 remainder 7
Part 2
Part 2 of 2:
### Making Good Guesses
1. 1
Round to the nearest ten. It's not always easy to see how many times a two-digit number goes into a larger one. One useful trick is to round to the nearest multiple of 10 to make guessing easier. This comes in handy for smaller division problems, or for parts of a long division problem.
• For example, let's say we're solving 143 ÷ 27, but we don't have a good guess at how many times 27 goes into 143. Let's pretend we're solving 143 ÷ 30 instead.
2. 2
Count by the smaller number on your fingers. In our example, we can count by 30s instead of counting by 27s. Counting by 30 is pretty easy once you get the hang of it: 30, 60, 90, 120, 150.
• If you find this difficult, just count by threes and add a 0 to the end.
• Count until you get higher than the larger number in the problem (143), then stop.
3. 3
Find the two most likely answers. We didn't hit 143 exactly, but we got two numbers close to it: 120 and 150. Let's see how many fingers we counted on to get them:
• 30 (one finger), 60 (two fingers), 90 (three fingers), 120 (four fingers). So 30 x four = 120.
• 150 (five fingers), so 30 x five = 150.
• 4 and 5 are the two most likely answers to our problem.
4. 4
Test those two numbers with the real problem. Now that we have two good guesses, let's try them out on the original problem, which was 143 ÷ 27:
• 27 x 4 = 108
• 27 x 5 = 135
5. 5
Make sure you can't get any closer. Since both our numbers ended up below 143, let's try getting even closer by trying one more multiplication problem:
• 27 x 6 = 162. This is higher than 143, so it can't be the right answer.
• 27 x 5 came closest without going over, so 143 ÷ 27 = 5 (plus a remainder of 8, since 143 - 135 = 8.)
## Community Q&A
Search
• Question
How to convert a mixed number fraction?
Donagan
To convert a mixed number to an improper fraction, multiply the denominator by the whole number, and add the original numerator. That gives you the new numerator. The denominator remains unchanged, and the whole number no longer exists.
• Question
How do I solve a problem that doesn't divide equally?
Donagan
Whatever is left over in the last step is shown as a "remainder" attached to the whole-number quotient.
• Question
I am 89 and don't know any math. Is this a problem?
Donagan
Probably not. If you've survived this long without resorting to math, there's no reason you couldn't continue to do so. Someone considerably younger than you, however, would be well advised to learn the fundamentals of math, because they can make life easier in many ways.
200 characters left
## Tips
• If you don't want to multiply by hand during the long division, try breaking up the problem into digits and solving each part in your head. For example, 14 x 16 = (14 x 10) + (14 x 6). Write down 14 x 10 = 140 so you don't forget. Then think: 14 x 6 = (10 x 6) + (4 x 6). Well, 10 x 6 = 60 and 4 x 6 = 24. Add 140 + 60 + 24 = 224 and you have the answer.
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## Warnings
• If, at any point, your subtraction results in a negative number, your guess was too high. Erase that entire step and try a smaller guess.
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• If, at any point, your subtraction results in a number larger than your divisor, your guess wasn't high enough. Erase that entire step and try a larger guess.
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Updated: January 31, 2023
Views: 494,293
Article SummaryX
To divide by a two-digit number, try to determine whether that number would fit into the first 2 digits of the dividend. If it does, write down how many times it will go into those numbers. Follow the same rules of division that you would with a smaller number. Calculate the remainder, if any, and add the remainder as a prefix to the next number in the dividend. Continue solving the problem to either a whole number with a remainder or a decimal. For tips on estimating the answer, read on! |
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# Counting Techniques
## Number of ways to choose (and potentially arrange) k objects from a group of n objects.
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Practice Counting Techniques
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Counting with Permutations and Combinations
Sometimes it makes sense to count the number of ways for an event to occur by looking at each possible outcome. However, when there are a large number of outcomes this method quickly becomes inefficient. If someone asked you how many possible regular license plates there are for the state of California, it would not be feasible to count each and every one. Instead, you would need to use the fact that on the typical California license plate there are four numbers and three letters. Using this information, about how many license plates could there be?
#### Watch This
http://www.youtube.com/watch?v=qJ7AYDmHVRE James Sousa: The Counting Principle
#### Guidance
Consider choice $A$ with three options $(A_1,A_2,A_3)$ and choice $B$ with two options $(B_1,B_2)$ . If you had to choose an option from $A$ and then an option from $B$ , the overall total number of options would be $3 \cdot 2=6$ . The options are $A_1B_1,A_1B_2,A_2B_1,A_2B_2,A_3B_1,A_3B_2$ .
You can see where the six comes from by making a decision chart and using the Fundamental Counting Principle. First, determine how many decisions you are making. Here, there are only two decisions to make (1: choose an option from $A$ ; 2: choose an option from $B$ ), so you will have two “slots” in your decision chart. Next, think about how many possibilities there are for the first choice (in this case there are 3) and how many possibilities there are for the second choice (in this case there are 2). The Fundamental Counting Principle says that you can multiply those numbers together to get the total number of outcomes.
Another type of counting question is when you have a given number of objects, you want to choose some (or all) of them, and you want to know how many ways there are to do this. For example, a teacher has a classroom of 30 students, she wants 5 of them to do a presentation, and she wants to know how many ways this could happen. These types of questions have to do with combinations and permutations. The difference between combinations and permutations has to do with whether or not the order that you are choosing the objects matters.
• A teacher choosing a group to make a presentation would be a combination problem, because order does not matter.
• A teacher choosing 1 st , 2 nd , and 3 rd place winners in a science fair would be a permutation problem, because the order matters (a student getting 1 st place vs. 2 nd place are different outcomes).
Recall that the factorial symbol, !, means to multiply every whole number up to and including that whole number together. For example, $5!=5 \cdot 4 \cdot 3 \cdot 2 \cdot 1$ . The factorial symbol is used in the formulas for permutations and combinations.
Combination Formula: The number of ways to choose $k$ objects from a group of $n$ objects is --
$_nC_k=\dbinom{n}{k}=\frac{n!}{k! \left(n-k\right)!}$
Permutation Formula: The number of ways to choose and arrange $k$ objects from a group of $n$ objects is --
$_nP_k=k!\dbinom{n}{k}=k! \cdot \frac{n!}{k!\left(n-k\right)!}=\frac{n!}{\left(n-k\right)!}$
Notice that in both permutation and combination problems you are not allowed to repeat your choices. Any time you are allowed to repeat and order does not matter, you can use a decision chart. (Problems with repetition where order does not matter are more complex and are not discussed in this text.)
Whenever you are doing a counting problem, the first thing you should decide is if the problem is a decision chart problem, a permutation problem, or a combination problem. You will find that permutation problems can also be solved with decision charts. The opposite is not true. There are many decision chart problems (ones where you are allowed to repeat choices) that could not be solved with the permutation formula.
Note: Here you have only begun to explore counting problems. For more information about combinations, permutations, and other types of counting problems, consult a Probability text.
Example A
You are going on a road trip with 4 friends in a car that fits 5 people. How many different ways can everyone sit if you have to drive the whole way?
Solution: A decision chart is a great way of thinking about this problem. You have to sit in the driver’s seat. There are four options for the first passenger seat. Once that person is seated there are three options for the next passenger seat. This goes on until there is one person left with one seat.
$1 \cdot 4 \cdot 3 \cdot 2 \cdot 1=24$
Example B
How many different ways can the gold, silver and bronze medals be awarded in an Olympic event with 12 athletes competing?
Solution: Since the order does matter with the three medals, this is a permutation problem. You will start with 12 athletes and then choose and arrange 3 different winners.
$_{12}P_3=\frac{12!}{\left(12-3\right)!}=\frac{12!}{9!}=\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot \ldots}{9 \cdot \ldots}=12 \cdot 11 \cdot 10=1320$
Note that you could also use a decision chart to decide how many possibilities are there for gold (12) how many possibilities are there for silver (11 since one already has gold) and how many possibilities are there for bronze (10). You can use a decision chart for any permutation problem.
$12 \cdot 11 \cdot 10=1320$
Example C
You are deciding which awards you are going to display in your room. You have 8 awards, but you only have room to display 4 awards. Right now you are not worrying about how to arrange the awards so the order does not matter. In how many ways could you choose the 4 awards to display?
Solution: Since order does not matter, this is a combination problem. You start with 8 awards and then choose 4.
$_8C_4=\dbinom{8}{4}=\frac{8!}{4!\left(8-4\right)!}=\frac{8 \cdot 7 \cdot 6 \cdot 5}{4 \cdot 3 \cdot 2 \cdot 1}=7 \cdot 2 \cdot 5=70$
Note that if you try to use a decision chart with this question, you will need to do an extra step of reasoning. There are 8 options I could choose first, then 7 left, then 6 and lastly 5.
$8 \cdot 7 \cdot 6 \cdot 5=1680$
This number is so big because it takes into account order, which you don’t care about. It is the same result you would get if you used the permutation formula instead of the combination formula. To get the right answer, you need to divide this number by the number of ways 4 objects can be arranged, which is $4!=24$ . This has to do with the connection between the combination formula and the permutation formula.
Concept Problem Revisited
A license plate that has 3 letters and 4 numbers can be represented by a decision chart with seven spaces. You can use a decision chart because order definitely does matter with license plates. The first spot is a number, the next three spots are letters and the last three spots are numbers. Note that when choosing a license plate, repetition is allowed.
$10 \cdot 26 \cdot 26 \cdot 26 \cdot 10 \cdot 10 \cdot 10=26^3 \cdot 10^4=175,760,000$
This number is only approximate because in reality there are certain letter and number combinations that are not allowed, some license plates have extra symbols, and some commercial and government license plates have more numbers, fewer letters or blank spaces.
#### Vocabulary
A combination is the number of ways of choosing $k$ objects from a total of $n$ objects (order does not matter).
A permutation is the number of ways of choosing and arranging $k$ objects from a total of $n$ objects (order does matter)
A decision chart is a sequence of numbers that multiply together where each number represents the number of possible options for that slot.
#### Guided Practice
1. There are 20 hockey players on a pro NHL team, two of which are goalies. In how many different ways can 5 skaters and 1 goalie be on the ice at the same time?
2. In how many different ways could you score a 70% on a 10 question test where each question is weighted equally and is either right or wrong?
3. How many different 4 digit ATM passwords are there? Assume you can repeat digits.
1. The question asks for how many on the ice, implying that order does not matter. This is combination problem with two combinations. You need to choose 1 goalie out of a possible of 2 and choose 5 skaters out of a possible 18.
$\dbinom{2}{1}\dbinom{18}{5}=2 \cdot \frac{18!}{5! \cdot 13!}=17136$
2. The order of the questions you got right does not matter, so this is a combination problem.
$\dbinom{10}{7}=\frac{10!}{7!3!}=120$
3. Order does matter. There are 10 digits and repetition is allowed. You can use a decision chart for each of the four options.
$10 \cdot 10 \cdot 10 \cdot 10=10,000$
#### Practice
Simplify each of the following expressions so that they do not have a factorial symbol.
1. $\frac{7!}{3!}$
2. $\frac{110!}{105!5!}$
3. $\frac{52!}{49!}$
4. In how many ways can you choose 3 objects from a set of 9 objects?
5. In how many ways can you choose and arrange 4 objects from a set of 15 objects?
First, state whether each problem is a permutation/decision chart problem or a combination problem. Then, solve.
6. Suppose you need to choose a new combination for your combination lock. You have to choose 3 numbers, each different and between 0 and 40. How many combinations are there?
7. You just won a contest where you can choose 2 friends to go with you to a concert. You have five friends who are available and want to go. In how many ways can you choose the friends?
8. You want to construct a 3 digit number from the digits 4, 6, 8, 9. How many possible numbers are there?
9. There are 12 workshops at a conference and Sam has to choose 3 to attend. In how many ways can he choose the 3 to attend?
10. 9 girls and 5 boys are finalists in a contest. In how many ways can 1 st , 2 nd , and 3 rd place winners be chosen?
11. For the special at a restaurant you can choose 3 different items from the 10 item menu. How many different combinations of meals could you get?
12. You visit 12 colleges and want to apply to 4 of them. In how many ways could you choose the four to apply to?
13. For the 12 colleges you visited, you want to rank your top five. In how many ways could you rank your top 5?
14. Explain why the following problem is not strictly a permutation or combination problem : The local ice cream shop has 12 flavors. You decide to buy 2 scoops in a dish. In how many ways could you do this if you are allowed to get two of the same scoop?
15. Your graphing calculator has the combination and permutation formulas built in. Push the MATH button and scroll to the right to the PRB list. You should see ${_nP}_r$ and $_nC_r$ as options. In order to use these: 1) On your home screen type the value for $n$ ; 2) Select ${_nP}_r$ or $_nC_r$ ; 3) Type the value for $k$ ( $r$ on the calculator). Use your calculator to verify that ${_{10}C}_5=252$ .
### Vocabulary Language: English
!
!
The factorial of a whole number $n$ is the product of the positive integers from 1 to $n$. The symbol "!" denotes factorial. $n! = 1 \cdot 2 \cdot 3 \cdot 4...\cdot (n-1) \cdot n$.
combination
combination
Combinations are distinct arrangements of a specified number of objects without regard to order of selection from a specified set.
decision chart
decision chart
A decision chart is a tree-shaped sequence of numbers used to determine the probability of occurrence of a given event.
Event
Event
An event is a set of one or more possible results of a probability experiment.
factorial
factorial
The factorial of a whole number $n$ is the product of the positive integers from 1 to $n$. The symbol "!" denotes factorial. $n! = 1 \cdot 2 \cdot 3 \cdot 4...\cdot (n-1) \cdot n$.
Fundamental Counting Principle
Fundamental Counting Principle
The Fundamental Counting Principle states that if an event can be chosen in p different ways and another independent event can be chosen in q different ways, the number of different arrangements of the events is p x q.
Independent Events
Independent Events
Two events are independent if the occurrence of one event does not impact the probability of the other event.
Permutation
Permutation
A permutation is an arrangement of objects where order is important.
Sample Space
Sample Space
In a probability experiment, the sample space is the set of all the possible outcomes of the experiment. |
Matrix transpose AT = 15 33 52 −21 A = 135−2 532 1 Example Transpose operation can be viewed as flipping entries about the diagonal. While certain “natural” properties of multiplication do not hold, many more do. MATRIX MULTIPLICATION. The determinant of a 4×4 matrix can be calculated by finding the determinants of a group of submatrices. $$\begin{pmatrix} e & f \\ g & h \end{pmatrix} \cdot \begin{pmatrix} a & b \\ c & d \end{pmatrix} = \begin{pmatrix} ae + cf & be + df \\ ag + ch & bg + dh \end{pmatrix}$$ Distributive law: A (B + C) = AB + AC (A + B) C = AC + BC 5. The following are other important properties of matrix multiplication. Even though matrix multiplication is not commutative, it is associative in the following sense. The last property is a consequence of Property 3 and the fact that matrix multiplication is associative; A square matrix is called diagonal if all its elements outside the main diagonal are equal to zero. Equality of matrices proof of properties of trace of a matrix. Proof of Properties: 1. Given the matrix D we select any row or column. 19 (2) We can have A 2 = 0 even though A ≠ 0. But first, we need a theorem that provides an alternate means of multiplying two matrices. Definition The transpose of an m x n matrix A is the n x m matrix AT obtained by interchanging rows and columns of A, Definition A square matrix A is symmetric if AT = A. The proof of this lemma is pretty obvious: The ith row of AT is clearly the ith column of A, but viewed as a row, etc. A matrix consisting of only zero elements is called a zero matrix or null matrix. (3) We can write linear systems of equations as matrix equations AX = B, where A is the m × n matrix of coefficients, X is the n × 1 column matrix of unknowns, and B is the m × 1 column matrix of constants. Notice that these properties hold only when the size of matrices are such that the products are defined. Multiplicative Identity: For every square matrix A, there exists an identity matrix of the same order such that IA = AI =A. Example 1: Verify the associative property of matrix multiplication … i.e., (AT) ij = A ji ∀ i,j. For the A above, we have A 2 = 0 1 0 0 0 1 0 0 = 0 0 0 0. Multiplicative identity: For a square matrix A AI = IA = A where I is the identity matrix of the same order as A. Let’s look at them in detail We used these matrices The first element of row one is occupied by the number 1 … A diagonal matrix is called the identity matrix if the elements on its main diagonal are all equal to $$1.$$ (All other elements are zero). If $$A$$ is an $$m\times p$$ matrix, $$B$$ is a $$p \times q$$ matrix, and $$C$$ is a $$q \times n$$ matrix, then $A(BC) = (AB)C.$ This important property makes simplification of many matrix expressions possible. Subsection MMEE Matrix Multiplication, Entry-by-Entry. For sums we have. 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While certain “ natural ” properties of multiplication do not hold, many more do following are important.
## properties of matrix multiplication proof
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# Matrix Scalar Multiplication: Product of a Scalar and a Matrix
Matrix multiplication usually falls into one of two types or classifications. The first one is called Scalar Multiplication, also known as the “Easy Type“; where you simply multiply a number into each and every entry of a given matrix.
The second one is called Matrix Multiplication which is discussed in a separate lesson.
In this lesson, we will focus on the “Easy Type” because the approach is extremely simple or straightforward.
## “Formula” of Scalar Multiplication (Easy Type)
Here’s the simple procedure as shown by the formula above.
Take the number outside the matrix (known as the scalar) and multiply it by each and every entry or element of the matrix.
### Examples of Scalar Multiplication
Directions: Given the following matrices, perform the indicated operation. Apply scalar multiplication as part of the overall simplification process.
Example 1: Perform the indicated operation for $2A$.
I will take the scalar 2 (similar to the coefficient of a term) and distribute it by multiplying it to each entry of matrix $A$. In case you forgot, you may review the general formula above.
Since matrix $A$ is
then $2A$ is solved by…
That’s all there is to it. Done!
Example 2: Perform the indicated operation for $-3B$.
I will do the same thing similar to Example 1. No big deal! Multiply the negative scalar, $−3$, into each element of matrix $B$.
Since matrix $B$ is
then matrix $-3B$ is solved by…
Did you arrive at the same final answer? If not, please recheck your work to make sure that it matches the correct answer.
Example 3: Perform the indicated operation for $-2D + 5F$.
To solve this problem, I need to apply scalar multiplication twice and then add their results to get the final answer.
• First, find the value of matrix $-2D$
I know that matrix $D$ is
Therefore, $-2D$ is obtained as follows using scalar multiplication.
• Second, find the value of $5F$
Matrix $F$ is given as
That means $5F$ is solved using scalar multiplication.
• Now, I can solve for $-2D+5F$ by adding the values of matrices $-2D$ and $5F$, as shown above. Check out the solution to review how to add and subtract matrices.
That’s it!
Example 4: What is the difference of $4A$ and $3C$?
At this point, you should have mastered already the skill of scalar multiplication. The very first step is to find the values of $4A$ and $3C$, respectively. Then we subtract the newly formed matrices, that is, $4A-3C$.
Finding the value of $4A$
Finding the value of $3C$
Finally, we subtract $4A$ by $3C$.
You may also be interested in these related math lessons or tutorials: |
Why Radian Measure Makes Life Easier In Mathematics And Physics
The two most commonly used measures for angles are degrees and radians. There are 360 degrees in a full circle (a right angle is 90 degrees), and $2\pi$ radians in a full circle (there are $\pi/2$ radians in a right angle), so there are about 57 degrees in a radian.
Students typically learn about degrees before they learn about radians, which brings up the question: Why learn about radians if degrees are good enough for measuring angles? There are two reasons, but both are grounded in the fact that the radian is a unitless measure, which I’ll explain in the following paragraph.
The radian is defined to be (see the diagram) the ratio of the length of the arc of a circle (indicated by $s$ in the diagram) to the length of the radius of the circle (indicated by $r$ in the diagram), where each length is measured in the same unit. Therefore, when you divide $s$ by $r$ to get the radian measure of the angle, the units for the two lengths cancel, and you end up with a measure that has no units:
$\theta = \dfrac{s}{r}$
When $s$ is the entire circumference of the circle, the corresponding angle is that of the entire circle. Since the circumference of a circle is $s = 2\pi r$, the angle of a full circle is $\dfrac{2\pi r}{r} = 2\pi$.
Unless there is a good reason, one is free to use any measure whatsoever for angles, such as degrees, radians, or one of the less common ones. There are two good reasons to use radians:
1. If you are working with the derivative of a trigonometric function, then it is preferable to use radian measure for angles, because then derivative formulas (and limit formulas) are easier. For example, using radians, the derivative formulas for sine and cosine are:
$\dfrac{{\rm d}}{{\rm d}x} \sin x = \cos x$
and
$\dfrac{{\rm d}}{{\rm d}x} \cos x = -\sin x$
However, if degrees are used, the derivative formulas for sine and cosine are:
$\dfrac{{\rm d}}{{\rm d}x} \sin x = \left ( \dfrac{\pi}{180} \right ) \cos x$
and
$\dfrac{{\rm d}}{{\rm d}x} \cos x = - \left ( \dfrac{\pi}{180} \right ) \sin x$
The latter pair of formulas are a pain because of the additional factors, so why wouldn’t we use the simpler first pair of formulas? It’s the laziness principle that is used so often in mathematics … we always do the easiest thing. Therefore, we use radians whenever we are dealing with derivatives of trigonometric functions. (The derivatives of the other trigonometric functions are similarly simplified if radians are used … work them out, if you wish!)
2. In describing rotational motion, one rearranges the definition of radian measure to relate the linear displacement to the angular displacement:
$s = r \theta$
Differentiating both sides of this relation with respect to time (and assuming a constant radius), one obtains a relation between the linear velocity $v$ and the angular velocity $\omega$ for motion in a circle:
$v = r \omega$
If $\theta$ is measured in radians, then $\omega$ is measured in radians per second, and then the speed comes out in a natural unit; for example, if $r$ is measured in metres, then $v$ comes out in m/s. However, if $\theta$ were measured in degrees, then the units for $v$ would be m·degrees/s, which is an unnatural unit, and hard to understand.
Differentiating the relation from the previous display with respect to time once more, we get a relation between the linear acceleration and angular acceleration for an object moving in a circle:
$a = r \alpha$
A similar comment about units is also relevant here: Measuring $\alpha$ in radians/s2 means that if $r$ is measured in metres, then $a$ comes out in natural and easily understood units, m/s2. But if $\alpha$ is measured in degrees/s2, or in revolutions/s2, then the units for $a$ come out an awkward mess.
Finally, there is something arbitrary about degree measure (why 360? why not 100, or 1000, or some other round number?) that makes it aesthetically not pleasing. On the other hand, there is something aesthetically pleasing about the definition of radian measure, as it is a simple ratio. This makes it a natural measure, wouldn’t you say?
Which leads me to think about the use of the word “natural” in mathematics … it would be nice to discuss this another time.
p.s.: By the way, did you notice that the factors in the formulas for the derivatives of sine and cosine in degree measure are the same as the factor that is used to convert degrees to radians? This makes sense if you think about the conversion from degrees to radians as a scale change (i.e., a change of variable) and then use the chain rule for differentiation.
Update, 20 September 2012: Ted Burke notes in a comment to this post that Euler’s formula
$e^{i\theta} = \cos \theta + i\sin \theta$
is valid only when the angle $\theta$ is measured in radians. If the angle is measured in degrees, Euler’s formula would have to be written in the considerably uglier form
$e^{i\pi \theta / 180} = \cos \theta + i\sin \theta$
This is another situation in which radian measure is more convenient than degree measure.
Update, 28 February 2016: Parker Harris contributed a very interesting comment, which leads me to doubt that “unitless” is a good term for the radian. Perhaps “ignorable unit” is a better term, as discussed in my response to Parker’s comment.
Thanks very much to Parker for his thought-provoking comment!
I have taught mathematics and physics since the mid 1980s. I have also been a textbook writer/editor since then. Currently I am working independently on a number of writing and education projects while teaching physics at my local university. I love math and physics, and love teaching and writing about them. My blog also discusses education, science, environment, etc. https://qedinsight.wordpress.com Further resources, and online tutoring, can be found at my other site http://www.qedinfinity.com
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26 Responses to Why Radian Measure Makes Life Easier In Mathematics And Physics
1. Richard Hayes says:
This a very nice post and came up when I posed the question to my colleagues (the type of basic question which you sometimes have to be old(er) to pose) ‘why radians?. I started thinking about the answer when I got lazy and did a google instead. I am teaching a class on ac circuits and phasors and realized that I wasnt sure (had forgotten?) why I had to use the radian setting on the calculator to plot the sine wave, i.e. why the sine formula is x(t) = sin( 2 PI f t)
• Thanks, Richard! The word “phasor” brings back some fond memories. All the best in your teaching!
2. batchloaf says:
Thanks, a very good explanation! One other thing that I was reminded of while reading your post is that radian angles make Euler’s formula particularly elegant.
• Thanks, Ted! And it’s a good point about Euler’s formula, which I’ll add to the post. All the best!
3. Jennifer says:
I agree that students would benefit from learning radian measure earlier. Think of how beautiful the interior angle sums for polynomials are in radians! Triangle: pi, Quadrilateral: 2pi, Pentagon: 3pi… My precal kids got so flustered that I ended up making radian scale protractors for them! You can check them out on my website, proradian.net.
4. Jennifer says:
Reblogged this on jensilver.
5. tomcircle says:
Reblogged this on Singapore Maths Tuition and commented:
Why radian is preferred in Math? Very good explanation here…
6. Skye says:
Very nice explanation! So radians are used because it’s more beautiful!
7. kyledelaney1988 says:
I’m still struggling to understand why the derivative of sine only works for radians. I had always thought that radians and degrees were both arbitrary units of measurement, and just now I’m discovering that I’ve been wrong all along! I’m guessing that when you differentiate sine, the step that only works for radians is when you replace sin(dx) with just dx, because as dx approaches 0 then sin(dx) equals dx because sin(0) equals 0. But isn’t the same true for degrees? As dx approaches 0 degrees then sin(dx degrees) still approaches 0. But I’ve come to the understanding that sin(dx degrees) approaches 0 almost 60 times slower, so if sin(dx radians) can be replaced with dx then sin(dx degrees) would have to be replaced with pi/180 times dx degrees.
But the question remains of why it works perfectly for radians. How do we know that we can replace sin(dx) with just dx without any kind of conversion applied like we need for degrees? It’s not good enough to just say that we can see that sin(dx) approaches dx as sin gets very small. Mathematically we can see that sin(.00001) is pretty darn close to .00001 when we’re using radians. But let’s say we had a unit of measurement “sixths” where there are 6 of them in a full circle, pretty close to radians. It would also look like sin(dx sixths) approaches dx when it gets very small, but we know we’d have to replace sin(dx sixths) with pi/6 dx sixths when differentiating. So how do we know that radians work out so magically, and why do they?
• These are good questions.
One answer to why radians work out so nicely is that you just do the calculation (of the derivative of the sine function) in arbitrary angle units, and then see which one comes out the simplest. As you suggest, you could use “sixths” or any other angle units. When you try this, you find that the derivative formula comes out simplest for angles measured in radians. This is motivation to use radian measure in calculus, but if you wished you could use any other angle measure you wish, provided that you accepted the specific derivative formula for the sign function that comes with that angle measure.
You can find the derivation of the derivative of the sine function in many places; search for a derivation involving a limit calculation. If you go through the details carefully, you’ll note that the limit of sine theta divided by theta is 1 only when theta is measured in radians. For other angle measures, this limit is not 1, and so the derivative formula must be multiplied by this factor.
OK, that’s one answer. Another answer is geometrical. Think of the slope of a line, which is defined as rise divided by run. If you change the units for the run, you will get a different value for the slope.
A derivative can be interpreted as the slope of the tangent line to the graph of the function in question. Using different angle measures amounts to changing the units of the run, which changes the slope. The units for the slope are (rise units)/(run units). Changing the angle measure changes the units of the run, but not the rise.
Does this make sense?
• kyledelaney1988 says:
Yes, I understand all that about slopes, but that’s not really what my question was about. I understand why it would be different depending on which units you use. I just don’t understand how to mathematically prove that radians are those perfect units. I guess I need to look further into proving limits.
• OK, I get your question now. There is a geometric argument that you might find compelling:
Draw a circle centred at the origin. Now draw a ray from the origin angled off into the first quadrant somewhere, intersecting the circle. The purpose of doing this is to convince ourselves that the limit of sin (dx) divided by dx as dx approaches 0 is 1.
Drop a vertical line segment from the point of intersection of the ray and the circle down to the x-axis. This produces a right triangle, formed from the vertical line segment (of length Y), the line segment along the x-axis (of length X) and the segment along the radius of the circle (of length R). Label the angle as dx. Then in the right triangle, sin (dx) = Y/R.
Note that the arc of the circle subtended by the angle is not very different in length from Y; furthermore, the approximation gets better and better as the angle gets smaller. Thus, in the limit as dx approaches 0, Y can be replaced by the arc of the circle subtended by the angle, which we can call S. Thus, as dx approaches 0, sin (dx) gets closer and closer to S/R.
But, if we measure the angle in radians, then S/R = dx, an exact relation that is the definition of radian measure. Thus, in the limit as dx approaches 0, sin (dx) approaches dx, and so sin (dx) divided by dx approaches 1.
If we used some other angle measure, then we would have to modify S/R = dx by some factor, and the same factor would appear in the derivative formula.
• kyledelaney1988 says:
Okay, I get it now, thank you. And I also found a detailed proof here: http://oregonstate.edu/instruct/mth251/cq/Stage4/Lesson/sinProof.html
8. Hello Mr. Bushell,
If you can point to an error at a specific location in the post, then I’m happy to hear it. Otherwise take your verbal abuse elsewhere.
9. Parker says:
If you have a moment, I’d love some help understanding why radians have a better claim at being unitless than degrees. Here’s what I’m thinking now: To calculate and angle in radians, you divide s/r, which is, say, feet/feet, which cancel out. Suppose I want to calculate he same angle in Parkers, a unit of angle I’ll define as a full rotation. I’d divide s (feet) by 2*pi* (r (also feet)), getting again a ratio, so Parkers are unitless as well? Just as radians are defined as the ratio of arclength to radius, Parkers are defines as ratio of arclength to circumference. To calculate degrees I’d divide s by 2pi and then divide that by 360. Again the units I use are going to be feet/feet, aren’t they? I can’t quite make sense of exactly why but this seems related to the relationship between force, distance, and energy, and power. Energy can have units length*force, for example ft*lb. Power, or energy per second, can have units ft*lb/s. However a more common unit of power is the horsepower, equal to 550 ft*lb/s. The 550 is arbitrary and inconvenient, but I don’t think it changes anything fundamental. What do you think?
• I think you’ve made a very important point, which leads me to conclude that “unitless” doesn’t capture what is special about radian measure. That is, I don’t think “unitless” is the right word, and I’ll have to cast for a better word.
Here’s the difference between the radian and the Parker (or degree, or any other angle measure): In the formula $v = r\omega$, if you measure $\omega$ in radians per second, and if $r$ is measured in metres, then multiplying the two values gives you a speed that is in metres per second. On the other hand, if you measure $\omega$ in Parkers per second, and if $r$ is measured in metres, then multiplying the two values does not give you the correct speed in metres per second; you’ll have to do some kind of conversion to convert from metres.Parkers/second to m/s.
So the key difference is that when you multiply (or divide) some combination of units by a radian, you can just ignore the radian in the final unit, as if it were not even there. You can’t do this with degrees, Parkers, or any other angle unit. This property is what I tried to capture with the term “unitless,” but it’s not clear that this is the best word. But it’s hard for me to summarize this property of radians with a single word.
Maybe “ignorable unit” would be an acceptable term, because that captures the fact that in a product or quotient of units, if the radian is included, it can be safely ignored without introducing error, whereas this is not the case for the degree, the Parker, etc.
Thank-you for a very good comment!
• kyledelaney1988 says:
I’m glad to hear you say all that. I’ve encountered many mathematicians who maintain that angles measured in radians are indeed unitless and thus that angles themselves are dimensionless, i.e. not a dimension that is distinct from ratios or spatial dimensions on a graph.
10. Songwhe Huh says:
The natural constant ( e ) is the key to understand the mystery of radians.
e^x function is very special because it is invariant under differentiation with x.
For the e^ix, with geometrical explanation on complex plane, it is the point on a circle with radius 1 and the arc distance from the reference point (1+i0) is x. ( it can be proved by invariant property)
So, now we can define angle as the arc length of unit circle from (1,0) without length unit.
Then we can interpret the Euler formula as the definition of sine and cosine with radian unit.
With that definition and simple complex algebra, you can intuitively see differentiation of sin(x) is cos(x).
11. Hardik Ghoshal says:
Thus makes it easier to study complex number arguments.
12. Ted says:
Radians are great for hand calculations, but when it comes to computing there are some advantages to using cycles or degrees:
1. Trigonometric libraries are slowed down when calculating large numbers, where cycles could make it trivial to perform the calculation: simply drop the whole number, and do the calculation on the whole number part.
2. 360 divides nicely by factors of 10, 9, 8. This means the degree measure for any of those angles can be stored exactly as a finite binary number.
As for Euler’s formula, cos t + j sin t = e^jt is kind of nice, but when you start doing Fourier transforms, you see e^(j*2*pi…) terms, and the reason the 2*pi is there is because you’re using radians instead of cycles. Use cycles, and the term goes away.The Fourier transform (along with its variants like cosine transform) is widely used in engineering and science for everything from jpeg image compression to radio frequency modulation. A silly thing that slows down trig functions in FFT calculation is the common use of radians for cosine and sine in standard libraries where it’s completely unnecessary.
13. Mark Girard says:
Comment:
When illustrating to students the connection between the unit circle and its map into a sine or cosine function, one could say that the distance of the wave line y=sinx is the radian value of x.
— So, the distance from the start of the sine function to the end of its first iteration at 2pi (360) is 2pi. It would seem so, but do not have any formal proof for it.
14. lockedmoon says:
Do you know of a logical decomposition of radian vs degree vs arc distance?
There is no SI unit of “degree” or even “angle”.
Length along the arc is clearly a unit of distance. So is length of the radius. So one divided by the other is a simple number. This defines the angle.
But you can have an angle without a distance. “Angle” could be a unit of rotation. Any ideas why we don’t have an SI unit of rotation?
• Rodney says:
You can also define angle by the area of the sector by saying by using r^2 as a unit of measure. To find the angle you can just divide the area swept out by the angle you are measuring by r^2. You can use radius/2 as a unit which is half radians. It is totally arbitrary. I can divide the circle into 4 equal pieces and call each piece some made up name pegrees or something and use it as a measure of angle and 90 degrees would now be equal to 1 pegrees. Totally arbitrary. The whole point is that radians just give you cleaner looking mathematical results that is where the term naturalness comes from in regards to radians…And also the whole unit arc length is just to justify the naturalness of radians.. That is it. Dont get hung up on that.
15. Rodney says:
From what I understand every arbitrary unit you use to measure angle requires the set of real numbers. It is just that you get different domains depending on the arbitrary unit you use to measure angle. And the important point is that they are all just as valid as the units of radians. However, radians just give you nicer looking mathematical results but it doesn’t mean the results you get from other arbitrary units are not valid. You will always have that pi/X factor where X depends on the arbitrary unit you use. To Songwhe Huh the choice of units dictates the form of the derivative of the sine and cosine not e. From your argument you assumed radians already for the euler formula and pretty much proved what you already assumed. |
# 30-60-90 triangle
A 30°-60-90 triangle is a special right triangle in which the length of the hypotenuse is twice the length of the shorter leg. The length of the longer leg is the square root of 3 times the length of the shorter leg.
Hypotenuse = 2 × shorter leg
Longer leg = √3 × shorter leg
Looking carefully at the figure above, you may have observed the following ratios:
Suppose we start from the smallest angle to the largest angle and from the shortest side to the longest side
The angles of a 30-60-90 triangle are in the ratio 1:2:3
The sides of a 30-60-90 triangle are in the ratio 1:√3:2
## 30-60-90 triangle proof
Start with an equilateral triangle ABC where each interior angle is equal to 60 degrees.
Since segment BD is the perpendicular bisector of segment AC, we now have two 30-60-90 triangles.
Let segment AD equal to s and use the Pythagorean theorem to find the length of segment BD.
Notice that AC = s + s = 2s. Since AB = AC, AB = 2s.
s2 + BD2 = (2s)2
s2 - s2 + BD2 = (2s)2 - s2
BD2 = (2s)2 - s2
BD2 = 4s2 - s2
BD2 = 3s2
BD = √(3s2)
BD = s√3
## Examples of 30-60-90 triangles
Example #1: 1, √3, 2
Short leg: 1
long leg: √3
Hypotenuse: 2
Example #2: 2, 2√3, 4
Short leg: 2
long leg: 2√3
Hypotenuse: 4
Example #3: 5, 5√3, 10
Short leg: 5
long leg: 5√3
Hypotenuse: 10
## Using the length of one side to solve a 30-60-90 triangle
Example #4:
The longer leg of a 30-60-90 triangle is 5. Find the lengths of the other sides.
Longer leg = √3 × shorter leg
5 = √3 × shorter leg
Shorter leg = 5/√3
Multiply both numerator and denominator of 5/√3 by √3
Shorter leg = 5(√3) / √3(√3)
Shorter leg = 5(√3) / √9
Shorter leg = 5(√3) / 3
Hypotenuse = 2 × shorter leg
Hypotenuse = 2 × [5(√3) / 3]
Hypotenuse = 10√3 / 3
## Using the 30-60-90 triangle to solve real-world problems
Example #5:
The warning sign below is an equilateral triangle. Suppose the height of the sign is 1 meter. Find the length s of each side of the sign.
Since the triangle is an equilateral triangle, the altitude divides the triangle into two 30-60-90 triangles as shown in the figure below.
Notice that the length of the shorter leg is equal to (1/2)s or 0.5s since the altitude also bisects the base.
1 = √3 × shorter leg
1 = √3 × 0.5s
Divide both sides of the equation by 0.5
1/0.5 = (√3)s
2 = (√3)s
Divide both sides of the equation by (√3)
2/(√3) = s
s = 1.15 m
The length of each side of the sign is about 1.2 m
Example #6:
An escalator lifts people to the second floor, 40 feet above the first floor. The escalator rises at a 30 degrees angle. How far does the person travel from the bottom to the top of the escalator?
The figure above describes example #6 with a 30-60-90 triangle. Notice that the shorter side is 40 feet and it is opposite to the angle that measures 30 degrees.
The distance from the bottom of the escalator to the top is the length of the hypotenuse of the 30-60-90 triangle.
This distance is twice the length of the shorter side. Therefore, the distance you travel is 80 feet.
## Area of a 30-60-90 triangle
To find the area of a 30-60-90 triangle, just make sure you get the height of the triangle correctly.
Keep in mind that either the longer side or the shorter side could be the height of the triangle. Just use the formula below:
Longer leg = √3 × shorter leg
If the shorter leg of the triangle is given, then the height = √3 × shorter leg and the shorter leg is the base.
If the longer leg of the triangle is given, then the height = longer leg / √3 and the base is the longer leg.
Example #7:
Find the area of the 30-60-90 triangle below
Since the longer leg is the base, height = 8/√3 = 4.618
Area = 8 × 4.618
Area = 36.944
The area of the triangle is 36.944 cm2
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# Question e2eed
May 27, 2017
Marie has decreased her calorie intake by 38.1%.
#### Explanation:
First, we need the amount that Marie's calorie intake has decreased
$2100 - 1300 = 800 \text{ cal}$
Marie is eating $800$ less calories than before.
Now we need to ask the question,:
"What fraction of the initial calorie intake is $800$?"
$\frac{800}{2100} = 0.3810$
$800$ calories is $0.3810$ of the initial intake, so Marie is eating $0.3810$ less than she used to.
To convert a fraction to a percentage, multiply by 100%
0.3810*100%=38.1"%"
Marie is eating 38.1% less calories than she used to.
May 27, 2017
38% decrease
#### Explanation:
To calculate the percentage change:
Step 1. Work out the difference between the two numbers you are comparing. New number minus old number.
$1300 - 2100 = - 800$
Step 2. Divide the difference by the old, original number. If your answer is a negative number then this is a percentage decrease. If it's positive, then it is a percent increase.
$\frac{- 800}{2100} \approx - 0.38$
Step 3. Multiply by 100%. Notice negative sign doesn't just magically disappear. Rather, it gets turned into the word "decrease".
-0.38xx100%=38%color(white)(aaa)#decrease. |
# Study Guide for Problem #12
Problem #(12), By Anil K. Dipu
The number of major earthquakes in a year is approximately normally distributed with a mean of 20.8 and a standard deviation of 4.5
a) Find the probability that in a given year there will be less than 21 earthquakes.
b) Find the probability that in a given year there will be between 18 and 23 earthquakes.
Step 1. Part (a) List all the given facts for the first statement.
Given:
The statement says that the situation is approximately normally distributed for the number of earthquakes in a year.
This simply means that we can use the “Standard Normal Distribution Curve” for µ=0 and make it equal to the mean that it gives us to use from the statement which is 20.8 as the new µ (“Miu”-mean). Thus the curve is balanced on both sides. (see example below)
The mean is now 20.8, (µ=20.8).
The standard deviation sigma, is equal to 4.5 (σ=4.5) which we will use to find the Probability with the mean (µ=20.8)
Step 2. Analyze problem (a). And draw a standard Normal distribution curve. And label the µ and X.
In both a and b, we do not have an exact sample of population “n”, or success “p” so we do not have to use the subtraction or addition of 0.5.
Therefore we can use the normal method to find the probability with the Z score from the Standard distribution Table which only gives the probability as an area to the left of Z and X.
“Z” represents a result for a part of “1” under the normal standard distribution curve.
The “X” is a piece of data that relates to an area or fraction under the curve. It belongs to entire data series that can be near or away from the mean (µ). Though X is not area, it is a continuous a number.
For problem a, the probability must be less than the X of 2. Therefore X can be understood as less than 21
(X= less than 21). The probability is not X and it is not 21. It is an area under the curve which is less than 21.
To find the area that of X= less than 21 we draw a standard distribution curve and label the Miu which is 20.8 as the middle or center of the data. We mark where x is located.
As you will see, 21 is a bit further on the right side of (µ=20.8), but we need an area that is less than 21.
Since we are using a table which shows areas to the left of Z and X we have to shade or mark the area that is less than 21. The area will be on the left.
This also means that we Do Not have to subtract the final value of area from 1 since the area that is already on the left is less than the X of 21.
Step 3. Find Z and Area for probability.
We now use the regular method to find Z in order to find the value of Probability as an area which means a X=probability is less than 21.
Z= (X- µ)/ σ = (21-20.8)/ 4.5= 0.2/4.5 = 0.0444444444444444 =0.4
(round to the nearest two decimal places after the “point or period” for z since the table uses a few decimals)
(A positive, “+” number will mean that is an area of Z which is greater than the Miu for 20.8).
Step 4. Find value of Probability with the Z score
The table that we are going to use is Table 4. We can see that 0.04 of z is an area of .5160 (=.516) which is understood as a Probability less than 21 which also on the left under the curve for normal and standard distribution.
Step 5. Part (b) List all the given facts for the first statement.
Given:
Standard Deviation is σ =4.5 and mean is µ=20.8
X is 18 and 23. The probability is supposed to be between these two data points.
Step 6. Draw the standard distribution table with the Miu and two X’s labeled.
From the standard normal distribution curve we can see that the area is between the two data points for x of 18 and 23. Therefore we need to try find the value of that area.
(Note 1: That part b did not say less than or more than so still do not need to worry about subtracting our final answer from 1, Note 2: Also since the table will always show an area that is to the left for X and Z, we are only looking at areas that concern the middle portion between the two X’s of 18 and 23. With corresponding Z’s)
Step 7. Find the two z scores, with their corresponding areas in order to find the actual difference for area of Probability.
For the first Z we use the first x of 18 and subtract Miu of 20.8, then by standard deviation 4.5.
We will now proceed as follows (18 – 20.8)/4.5 = -.062, with area of 0.2672 for A1
(For a negative value of Z, use Table 1 for area of Z less than Miu)
The second Z is the same but we use 23 for the second X.
The process will be written as (23 – 20.8)/4.5 = .49, with area of 0.6879 for A2
Step 8. Find the probability between the two X’s of 18 and 23.
The method for probability between two values of X will be stated as follows: A2-A1
(The larger Area subtract the smaller Area)
Then we can use the table which corresponds to the new Z to find the Probability of area under the curve.
The new area is calculated by 0.6879 – 0.2676 = 0.4203 |
Interpretation of Linear Functions in Context
Interpretation of linear functions in context is a key concept tested on the SAT Math section. Linear functions are used to model relationships between two variables, such as time and distance, or temperature and humidity. Understanding how to interpret linear functions in real-world contexts is essential for success on the SAT Math section. To help students prepare for this type of question, we have created an SAT practice test that focuses specifically on interpretation of linear functions in context.
As always, should you have any trouble with the questions on this test, don't hesitate to reach out for personalized SAT tutoring to help you ace this section.
Good luck!
Question 1:
A store offers a discount of \$20 per item when a customer purchases more than one item. If the price of one item is \$100, which of the following linear equations represents the total cost, C, for purchasing n items, where n > 1?
-------------
1. C = 100n - 20
2. C = 100n - 20(n - 1)
3. C = 80n
4. C = 80n - 20
Question 2:
A car rental company charges \$50 per day for renting a car and an additional one-time insurance fee of \$30. Which of the following linear equations represents the total cost, T, for renting a car for d days?
-------------
1. T = 50d + 30
2. T = 30d + 50
3. T = 50d
4. T = 80d
Question 3:
The cost of a taxi ride is \$3 for the first mile and \$2 for each additional mile. Which of the following linear equations represents the total cost, C, for a taxi ride of x miles?
-------------
1. C = 3x
2. C = 3 + 2(x - 1)
3. C = 2x + 1
4. C = 5x
Question 4:
A company produces T-shirts and sells them for \$25 each. It costs the company \$5 per T-shirt for materials and an additional \$3,000 for equipment. Which of the following linear equations represents the total profit, P, after selling x T-shirts?
-------------
1. P = 25x - 5x - 3,000
2. P = 25x - 5(x + 3,000)
3. P = 20x - 3,000
4. P = 20x + 3,000
Question 5:
A student earns \$10 per hour at a part-time job. If she earns \$100 per week from this job, how many hours did she work per week?
-------------
1. 5 hours
2. 10 hours
3. 20 hours
4. 30 hours
Question 6:
A phone company charges a flat rate of \$25 per month for basic service and \$0.10 per text message. If a customer's total bill was \$40 for one month, how many text messages were sent?
-------------
1. 150
2. 160
3. 200
4. 250
Question 7:
A company's revenue is represented by the linear function R(x) = 50x, where x is the number of items sold. The company's costs are represented by the linear function C(x) = 30x + 500. What is the break-even point, where the company's revenue equals its costs?
-------------
1. 25 items
2. 30 items
3. 50 items
4. 75 items
Question 8:
A parking garage charges \$10 for the first hour and \$5 for each additional hour. Which of the following linear equations represents the total cost, C, for parking at the garage for h hours, where h > 1?
-------------
1. C = 10h
2. C = 10 + 5(h - 1)
3. C = 5h + 5
4. C = 5h + 10
Question 9:
A factory produces 500 units of a product per day at a cost of \$2 per unit. It sells the product for \$5 per unit. Which of the following linear equations represents the daily profit, P, after selling x units?
-------------
1. P = 5x - 2x
2. P = 3x - 500
3. P = 3x
4. P = 2x + 500
Question 10:
A train travels at a constant speed of 60 miles per hour. Which of the following linear equations represents the distance, D, the train has travelled after t hours?
-------------
1. D = 60t
2. D = 60 + t
3. D = 60/t
4. D = t/60 |
# Random Variables. Chapter 2. Random Variables 1
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## Transcription
1 Random Variables Chapter 2 Random Variables 1
2 Roulette and Random Variables A Roulette wheel has 38 pockets. 18 of them are red and 18 are black; these are numbered from 1 to 36. The two remaining pockets are green and are numbered 0 and 00. The wheel is spun and people bet on where a small ball placed in the wheel will land. We can conduct an experiment in which we bet on a number once in Roulette with possible outcomes we win or we lose. Random Variables 2
3 Roulette and Random Variables Most people would probably be less interested in whether they win or lose than in how much they win or lose. We might also like to know how much we would expect to win if the experiment is repeated 5 or 6 times and how much our winnings might vary over multiple repetitions of the experiment. Random Variables 3
4 Roulette and Random Variables The questions 1. How much can we win or lose on a particular play? 2. How much would we expect to win if we played 5 times? 3. How much will the amount we win vary? and others can be addressed mathematically if we convert the experimental outcomes win and lose into numbers. Random Variables 4
5 Roulette and Random Variables A random variable assigns a numerical value to each outcome in an experiment. For instance, we could define a random variable X= the number on a pocket that the ball lands on in the roulette wheel. (We d let 0 and 00 both be counted as 0). This isn t very interesting, but we could do it. A better ideas would be to define a random variable X = the amount we win (or lose) in one \$1 bet on a number in Roulette. If we lose, X=-1 (remember we bet \$1) and if we win, X=35. Random Variables 5
6 Roulette and Random Variables Sometimes the experimental outcomes map naturally to a set of numbers. If we conduct an experiment to measure the height of a person in this class, the possible values for the random variable X = height of individual will all be between about 60 and 80 inches. At other times, the mapping is less natural, but still useful. For instance, we may conduct an experiment to determine how many members of our class watch Grey s Anatomy. The possible outcomes for each person are watches and does not watch. In this case, we can assign a 1 to the yeses and a 0 to the nos. Random Variables 6
7 Roulette and Random Variables Define a random variable for each of the following experimental outcomes what are the values the random variable can take on? Number of children Area of yard Favorite grocery store Temperature Sex (M, F) There are two types of random variables that are of particular interest to us. What is a natural way to classify random variables? Random Variables 7
8 Roulette and Random Variables The random variable associated with the roulette experiment is called discrete random variable because the number of outcomes it can assume can be counted. Some random variables correspond to outcomes that cannot be counted (e.g. things that are measured) these are called continuous random variables. Is the random variable that represents earnings in roulette discrete or continuous? What is the probability that X=-1 (the probability that we lose)? What is the probability that X=35 (the probability that we win)? Random Variables 8
9 The Distribution of A Discrete R.V. A random variable is usually denoted by an uppercase letter from near the end of the alphabet (e.g. X, Y, Z), while the specific values the random variable takes on are denoted by the corresponding lowercase letters. The values of a discrete random variable along with their associated probabilities constitute a probability mass function (pmf). These functions are usually represented in tables or graphs as shown below. Random Variables 9
10 The Distribution of A Discrete R.V. In a probability mass function, we denote the probability associated with the ith outcome x i as p i, that is P(X = x i )= p i. These values must satisfy 0 p i 1 and Σ i p i = 1. Verify that the probability mass function for the amount we win in roulette is a legitimate pmf, i.e. it satisfies the above conditions. A probability mass function is sometimes referred to as the distribution of a discrete random variable. Random Variables 10
11 The Distribution of A Discrete R.V. Example: Suppose we randomly select a USU undergraduate student. Let X= year in college, then P(X=1)=0.17, P(X=2)=0.21, P(X=3)=0.22. What is P(X=4)? P(X=5)?. Construct the pmf for X=year in college in tabular and graph forms. Suppose we randomly select a USU student and want to know the probability he/she is a freshman or a sophomore. If X=year in college, we can express the probability as P(X 2). A function of the form F(x)= P(X x), is called the cumulative distribution function (cdf) of the random variable X. Random Variables 11
12 The Distribution of A Discrete R.V. The cdf for X=year in college. x F(x) - <x<1 0 1 x< x< x<4.6 4 x< 1 Random Variables 12
13 The Distribution of A Discrete R.V. The pmfand cdffor X=year in college. Random Variables 13
14 The Distribution of A Discrete R.V. ~ Find the CDF (in tabular and graphical forms) for X = the amount we win (or lose) in one \$1 bet on a number in Roulette. ~ A game is played in which a fair die is tossed. If the roll is even, the player earns a dollar amount corresponding to the die roll, otherwise, he pays that amount. Calculate the probability mass function and the cumulative distribution function of the player s net earnings. Random Variables 14
15 Expectation and Variance of a Discrete R.V. The expected value (also expectation or mean), μ, of a random variable can be thought of as a weighted average and is defined as follows: For a discrete random variable, with pmf P(X=x i )=p i, the expected value E(X)=Σ i p i x i. The expected value for X = Net winnings in one \$1 bet on a number in Roulette is -1(37/38)+35(1/38) = -1/ This means that if we bet \$1 on a number in roulette many times, we can expect to lose about \$0.05 per play. Random Variables 15
16 Expectation and Variance of a Discrete R.V. Obviously, we won t win -\$0.05 on every play, rather we will win - \$1 or \$35, the variance tells us about the spread of the values of the random variable. The variance (denoted Var(X) or σ 2 ) of a random variable X measures the variability in the values taken on by the random variable. It is defined to be Var(X)=E((X-E(X)) 2 ). This can also be expressed as Var(X)=E(X 2 )-(E(X)) 2. The standard deviation of a random variable X is the positive square root of the variance and is therefore denoted by σ. Example 1: Find the variance of X = Net winnings in one \$1 bet on a number in Roulette Example 2: Find the expected value and standard deviation of X=year in college. Random Variables 16
17 The Distribution of A Continuous R.V. What problems might we encounter in attempting to identify a probability mass function for a continuous random variable? A continuous random variable can be described by a probability density function f(x). f(x) 0 and S f(x)dx=1. The graphs below are of probability density functions of continuous random variables. Random Variables 17
18 The Distribution of A Continuous R.V. We can use the probability density function to find the probability that X lies in an interval, i.e. P(a<x<b). The probability density function of a continuous random variable is also called its distribution. The area under the curve over a given interval gives the probability that the random variable takes on a value in that region. Note that P(X=x)=f(x)=0 for all values of x. Random Variables 18
19 The Distribution of A Continuous R.V. For pdf f(x), we can compute the probability over b an interval as P ( a x b) = f ( x) dx This has two possible interpretations: 1. The probability that an individual randomly selected from the underlying population will have a value of X that falls between a and b. 2. The proportion of individuals in the underlying population that have a value of X between a and b. a Random Variables 19
20 The Distribution of A Continuous R.V. Example (Insect 1): The lifetime in months, X, of a particular insect has the probability distribution shown at the right and given by f 1 x 2 ( x) = e, x 2 Verify that this is a legitimate probability density function. 0 Random Variables 20
21 The Distribution of A Continuous R.V. 1. On the plot, diagram the area that represents the probability that a given insect lives no longer than 6 months. 2. Diagram the area that represents the proportion who live between 2 and 8 months. 3. Compute the probabilities that you diagrammed Random Variables 21
22 The Distribution of A Continuous R.V. Example (Insect 2): Suppose a second type of insect is know to live no longer than 4 months. The probability density function associated with the lifetime Y of this insect is given below: A y 2 f ( y) = e,0 y 4 2 ~ What must A be in order for this to be a valid pdf? The definition of the cumulative distribution function for a continuous random variable is the same as for a discrete random variable: F(x)= P(X x). Random Variables 22
23 The Distribution of A Continuous R.V. Percentiles, quantiles, and quartiles provide information about the spread of a random variable. For and random variable with cdf F(x), the pth quantile is the value of x for which F(x)=p. This is also referred to as the px100th percentile. The lower and upper quartiles refer to the.25, and.75 quantiles or the 25 th and 75 th percentiles. For a random variable X with CDF F(X), the median is the value of X for which F(x)=0.5. The interquartile range (IQR) is the distance between the lower and upper quartiles. That is if x 1 is the lower quartile and x 2 is the upper quartile, IQR=x 2 -x 1. Random Variables 23
24 The Distribution of A Continuous R.V. 1. Find and plot the cumulative distribution function for lifetime of insect Use the cdf to compute the probabilities that you diagrammed on the plot of the pdf. 3. What is the median lifetime? 4. What is the 75th percentile of lifetime? Random Variables 24
25 Expectation and Variance Suppose we d like to know the average lifetime for an insect of the first type (insect 1). As with discrete random variable, the expected value, μ, of a continuous random variable can be thought of as a weighted average. For a continuous random variable with probability density function f(x), the expected value E(X)= S xf(x)dx. The variance of a continous random variable X is defined in the same way as for a discrete random variable, that is, Var(X)=E((X-E(X)) 2 ) or Var(X)=E(X 2 )-(E(X)) 2. ~ Find the expected value and variance of X as defined for insect 1. Random Variables 25
26 Expectation and Variance The top figure at the right shows the distributions of two random variables with equal variances but different means (expected values). The bottom figure shows the distributions of two random variables with equal means, but different variances. Random Variables 26
27 Combinations and Functions of R.V.s Suppose we want to play roulette for bigger stakes, specifically we want to bet \$100 on a number instead of \$1. We can define a new random variable Y= the amount we win in one \$100 bet on a number in roulette or Y=100X. Then we can use what we know about X to find a new expectation and variance. Suppose, instead, we want to keep the stakes the same, but we want to place three \$1 bets instead of just one. Again we can define a new random variable Z= the amount we win in three \$1 bets on a number in roulette or Z=X+X+X. Random Variables 27
28 Combinations and Functions of R.V.s Let X be a random variable such that E(X)=μ and Var(X)=σ 2. Suppose we re interested in a random variable Y defined as Y = a + bx, where a and b are constants. In other words, Y is a linear function of X. Then E(Y)=E(a+bX)=E(a)+bE(X)=a+bμ, and Var(Y)=Var(a+bX)=Var(a)+b 2 Var(X)=0+b 2 σ 2 =b 2 σ 2. Random Variables 28
29 Combinations and Functions of R.V.s Example: The random variables X, Y, and Z are independent with E(X)=1, Var(X)=3, E(Y)=2, Var(Y)=6, E(Z)=3, Var(Z)=4. Calculate the expected values and variances of the following random variables: 1. 2Z 2. Y X + Y - 5Z - 7 Example: The length X of rainbow trout in a particular fishing hole, has expected value 35 cm with a variance of 10 cm. To translate to inches, let Y=.39(X). What are the expected value and variance of rainbow trout length in inches? Random Variables 29
30 Combinations and Functions of R.V.s Let X 1 and X 2 be independent random variables such that E(X 1 )=μ 1 and Var(X 1 )=σ 12, and E(X 2 )=μ 2 and Var(X 2 )=σ 22. Let Y be a linear combination of X 1 and X 2, Y=c 1 X 1 +c 2 X 2, where c 1 and c 2 are constants. Then E(Y)=E(c 1 X 1 + c 2 X 2 )=c 1 E(X 1 )+c 2 E(X 2 )=c 1 μ 1 +c 2 μ 2, and Var(Y)=Var(c 1 X 1 + c 2 X 2 )= c 12 Var(X 1 )+c 22 Var(X 2 )=c 12 σ 12 +c 22 σ 22. The expectation of Y given above holds regardless of the dependence of X 1 and X 2. However, the variance formula holds only if X 1 and X 2 are independent. Random Variables 30
31 Combinations and Functions of R.V.s Example 1: The random variables X, Y, and Z are independent with E(X)=1, Var(X)=3, E(Y)=2, Var(Y)=6, E(Z)=3, Var(Z)=4. Calculate the expected values and variances of the following random variables: 1. 3X-2Y 2. 5X+2Y+Z-12 Example 2: Recall the insect 1 lifetime example. Suppose that we randomly sample 10 insects of this type and record the length of time they live. What is the expected combined lifetime of these 10 insects in months? in weeks (assume 4 weeks/month)? in years? What is the standard deviation of their combined lifetime in months? in weeks? in years? Random Variables 31
32 Combinations and Functions of R.V.s If X 1, X n is a sequence of independent random variables each with expectation μ and variance σ 2. The average of these random variables is X = X n X n What are the expected value and variance of? X Random Variables 32
33 Combinations and Functions of R.V.s Example: Consider again the insect1 example. Suppose that we randomly select 20 insects and compute the average lifetime. ~ What is the expected value of the average? ~ What is the variance of the average? ~ By what factor do we need to increase our sample size to divide the standard deviation of the average in half? Random Variables 33
34 Combinations and Functions of R.V.s There are not general rules to relate expectations and variances of a random variable X to a nonlinear function Y of that random e.g. Y=X 2, Y=e x. In this case, we can construct the cumulative distribution function of Y from that of X. Consider the random variables X and Y such that f x (x)=x for 0 x 1 and Y=X 1/2. F x (x)=(1/2)x 2 F y (Y)=P(Y y)=p(x 1/2 y)=p(x y 2 )= F x (y 2 )=(1/2)(y 2 ) 2 =(1/2)(y 4 ) Therefore f y (y)=2y 3. Random Variables 34
35 Combinations and Functions of R.V.s Example: Suppose that the random variable X has probability density f x (x)=1 for 2 x 4. Find the probability density function of the random variable Y in the following cases: 1. Y=1/X 2. Y=3 x 3. Y=X 2 Random Variables 35
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# 1.What is the slope of the line that passes through the pair of points (3,8) and (9,5)?
Options
-2
2
1/2
-1/2
2.Write an equation in point-slope form for the line
Through the given point with the given slope. (8,3);m=5
Options
Y+3=5(x-8)
Y+3=5x-8
Y-3=5(x-8)
Y-3=5(x+5)
I think question 1 would be a and question 2 would be c can u help please:)
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## Of course! It's my pleasure to help.
1. To find the slope of the line passing through the points (3,8) and (9,5), we can use the formula:
slope = (y2 - y1) / (x2 - x1)
So, substituting the coordinates we have:
slope = (5 - 8) / (9 - 3)
Simplifying:
slope = -3 / 6
Which simplifies further to:
slope = -1/2
Therefore, the slope of the line is -1/2. So option d, -1/2, is the correct answer.
2. To write the equation in point-slope form for the line passing through the point (8,3) with the slope of 5, we can use the formula:
y - y1 = m(x - x1)
Substituting the values, we get:
y - 3 = 5(x - 8)
Simplifying further:
y - 3 = 5x - 40
Finally, rearranging the equation to the standard form, we get:
y = 5x - 37
Therefore, the correct equation in point-slope form is Y+3=5(x-8), so option a is the correct answer.
I hope that clears things up! Let me know if there's anything else I can assist you with.
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## 1. To find the slope of the line that passes through the points (3,8) and (9,5), you can use the formula:
slope (m) = (y2 - y1) / (x2 - x1)
Substituting the coordinates, we have:
m = (5 - 8) / (9 - 3)
m = -3 / 6
m = -1/2
So, the slope of the line is -1/2. Therefore, option d, -1/2, is the correct answer.
2. To write an equation in point-slope form, we can use the formula:
y - y1 = m(x - x1)
where (x1, y1) is the given point, and m is the given slope.
Substituting the values from the given point (8,3) and slope m=5, we have:
y - 3 = 5(x - 8)
This equation represents the line passing through the point (8,3) with slope 5. Simplifying the equation, we get:
y - 3 = 5x - 40
Option a, Y + 3 = 5(x - 8), is the correct equation in point-slope form.
3539617
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0
## Sure! Let's go through each question step by step:
1. To find the slope of the line passing through the points (3,8) and (9,5), you can use the formula for calculating the slope:
slope = (y2 - y1) / (x2 - x1)
Let's assign the coordinates: (x1, y1) = (3, 8) and (x2, y2) = (9, 5)
Now, substitute the values into the formula:
slope = (5 - 8) / (9 - 3)
= -3 / 6
= -1/2
Therefore, the slope of the line passing through the given points is -1/2.
Looking at the options, you chose option a (slope = -2), but the correct answer is option d (-1/2).
2. To write an equation in point-slope form for the line passing through the point (8,3) with a slope of 5, you can use the following formula:
y - y1 = m(x - x1)
Let (x1, y1) = (8, 3) and m = 5.
Now, substitute the values into the formula:
y - 3 = 5(x - 8)
Simplifying further:
y - 3 = 5x - 40
Therefore, the equation in point-slope form for the given line is y - 3 = 5x - 40.
Among the options you provided, you chose option c, which is correct.
1. -1/2
2. Y - 3 = 5x - 40
3768174
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## 1.
slope = (y2 - y1)/(x2 - x1)
so , please recalculate the slope, it is not a
I think you have it "upside down"
2. correct |
# 0.7 1.8 acceleration (Page 4/8)
Page 4 / 8
The graphs of position, velocity, and acceleration vs. time for the trains in [link] and [link] are displayed in [link] . (We have taken the velocity to remain constant from 20 to 40 s, after which the train decelerates.)
## Calculating average velocity: the subway train
What is the average velocity of the train shown below, if it takes 5.00 min to make its trip?
Strategy
Average velocity is displacement divided by time. It will be negative here, since the train moves to the left and has a negative displacement.
Solution
1. Identify the knowns. ${x\prime }_{f}=3\text{.75 km}$ , ${x\prime }_{0}=\text{5.25 km}$ , $\Delta t=\text{5.00 min}$ .
2. Determine displacement, $\Delta x\prime$ . We found $\Delta x\prime$ to be $-\text{1.5 km}$
3. Solve for average velocity.
$\stackrel{-}{v}=\frac{\Delta x\prime }{\Delta t}=\frac{-\text{1.50 km}}{\text{5.00 min}}$
4. Convert units.
$\stackrel{-}{v}=\frac{\Delta x\prime }{\Delta t}=\left(\frac{-1\text{.}\text{50 km}}{5\text{.}\text{00 min}}\right)\left(\frac{\text{60 min}}{1 h}\right)=-\text{18}\text{.0 km/h}$
Discussion
The negative velocity indicates motion to the left.
## Sign and direction
Perhaps the most important thing to note about these examples is the signs of the answers. In our chosen coordinate system, plus means the quantity is to the right and minus means it is to the left. This is easy to imagine for displacement and velocity. But it is a little less obvious for acceleration. Most people interpret negative acceleration as the slowing of an object. In fact, a negative acceleration will increase a negative velocity. For example, the train moving to the left could speed up by an acceleration to the left. In this case, both $v$ and $a$ are negative. The plus and minus signs give the directions of the accelerations. If acceleration has the same sign as the velocity, the object is speeding up. If acceleration has the opposite sign as the velocity, the object is slowing down.
An airplane lands on a runway traveling east. Describe its acceleration.
If we take east to be positive, then the airplane has negative acceleration, as it is accelerating toward the west. It is also decelerating: its acceleration is opposite in direction to its velocity.
## Section summary
• Acceleration is the rate at which velocity changes. In symbols, average acceleration $\stackrel{-}{a}$ is
$\stackrel{-}{a}=\frac{\Delta v}{\Delta t}=\frac{{v}_{f}-{v}_{0}}{{t}_{f}-{t}_{0}}\text{.}$
• The SI unit for acceleration is ${\text{m/s}}^{2}$ .
• Acceleration is a vector, and thus has a both a magnitude and direction.
• Acceleration can be caused by either a change in the magnitude or the direction of the velocity.
• Instantaneous acceleration $a$ is the acceleration at a specific instant in time.
• Deceleration is an acceleration with a direction opposite to that of the velocity.
## Conceptual questions
Is it possible for speed to be constant while acceleration is not zero? Give an example of such a situation.
Is it possible for velocity to be constant while acceleration is not zero? Explain.
Give an example in which velocity is zero yet acceleration is not.
If a subway train is moving to the left (has a negative velocity) and then comes to a stop, what is the direction of its acceleration? Is the acceleration positive or negative?
## Problems&Exercises
A cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00 s. What is its acceleration?
$4\text{.}\text{29}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}$
A commuter backs her car out of her garage with an acceleration of $1\text{.}{\text{40 m/s}}^{2}$ . (a) How long does it take her to reach a speed of 2.00 m/s? (b) If she then brakes to a stop in 0.800 s, what is her acceleration?
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
how did you get the value of 2000N.What calculations are needed to arrive at it
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# 1.1 Real numbers: algebra essentials  (Page 4/35)
  Page 4 / 35
## Differentiating the sets of numbers
Classify each number as being a natural number ( N ), whole number ( W ), integer ( I ), rational number ( Q ), and/or irrational number ( Q′ ).
1. $\sqrt{36}$
2. $\frac{8}{3}$
3. $\sqrt{73}$
4. $-6$
5. $3.2121121112\dots$
N W I Q Q′
a. $\text{\hspace{0.17em}}\sqrt{36}=6$ X X X X
b. $\text{\hspace{0.17em}}\frac{8}{3}=2.\overline{6}$ X
c. $\text{\hspace{0.17em}}\sqrt{73}$ X
d. –6 X X
e. 3.2121121112... X
Classify each number as being a natural number ( N ), whole number ( W ), integer ( I ), rational number ( Q ), and/or irrational number ( Q′ ).
1. $-\frac{35}{7}$
2. $0$
3. $\sqrt{169}$
4. $\sqrt{24}$
5. $4.763763763\dots$
N W I Q Q'
a. $\text{\hspace{0.17em}}-\frac{35}{7}$ X X
b. 0 X X X
c. $\text{\hspace{0.17em}}\sqrt{169}$ X X X X
d. $\text{\hspace{0.17em}}\sqrt{24}$ X
e. 4.763763763... X
## Performing calculations using the order of operations
When we multiply a number by itself, we square it or raise it to a power of 2. For example, $\text{\hspace{0.17em}}{4}^{2}=4\cdot 4=16.\text{\hspace{0.17em}}$ We can raise any number to any power. In general, the exponential notation $\text{\hspace{0.17em}}{a}^{n}\text{\hspace{0.17em}}$ means that the number or variable $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is used as a factor $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ times.
In this notation, $\text{\hspace{0.17em}}{a}^{n}\text{\hspace{0.17em}}$ is read as the n th power of $\text{\hspace{0.17em}}a,\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ is called the base and $\text{\hspace{0.17em}}n\text{\hspace{0.17em}}$ is called the exponent . A term in exponential notation may be part of a mathematical expression, which is a combination of numbers and operations. For example, $\text{\hspace{0.17em}}24+6\cdot \frac{2}{3}-{4}^{2}\text{\hspace{0.17em}}$ is a mathematical expression.
To evaluate a mathematical expression, we perform the various operations. However, we do not perform them in any random order. We use the order of operations . This is a sequence of rules for evaluating such expressions.
Recall that in mathematics we use parentheses ( ), brackets [ ], and braces { } to group numbers and expressions so that anything appearing within the symbols is treated as a unit. Additionally, fraction bars, radicals, and absolute value bars are treated as grouping symbols. When evaluating a mathematical expression, begin by simplifying expressions within grouping symbols.
The next step is to address any exponents or radicals. Afterward, perform multiplication and division from left to right and finally addition and subtraction from left to right.
Let’s take a look at the expression provided.
$24+6\cdot \frac{2}{3}-{4}^{2}$
There are no grouping symbols, so we move on to exponents or radicals. The number 4 is raised to a power of 2, so simplify $\text{\hspace{0.17em}}{4}^{2}\text{\hspace{0.17em}}$ as 16.
$\begin{array}{l}\hfill \\ \begin{array}{l}24+6\cdot \frac{2}{3}-{4}^{2}\hfill \\ 24+6\cdot \frac{2}{3}-16\hfill \end{array}\hfill \end{array}$
Next, perform multiplication or division, left to right.
$\begin{array}{l}\hfill \\ \begin{array}{l}24+6\cdot \frac{2}{3}-16\hfill \\ 24+4-16\hfill \end{array}\hfill \end{array}$
Lastly, perform addition or subtraction, left to right.
Therefore, $\text{\hspace{0.17em}}24+6\cdot \frac{2}{3}-{4}^{2}=12.$
For some complicated expressions, several passes through the order of operations will be needed. For instance, there may be a radical expression inside parentheses that must be simplified before the parentheses are evaluated. Following the order of operations ensures that anyone simplifying the same mathematical expression will get the same result.
## Order of operations
Operations in mathematical expressions must be evaluated in a systematic order, which can be simplified using the acronym PEMDAS :
P (arentheses)
E (xponents)
M (ultiplication) and D (ivision)
A (ddition) and S (ubtraction)
Given a mathematical expression, simplify it using the order of operations.
1. Simplify any expressions within grouping symbols.
2. Simplify any expressions containing exponents or radicals.
3. Perform any multiplication and division in order, from left to right.
4. Perform any addition and subtraction in order, from left to right.
write down the polynomial function with root 1/3,2,-3 with solution
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
the polar co-ordinate of the point (-1, -1)
prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x
tanh`(x-iy) =A+iB, find A and B
B=Ai-itan(hx-hiy)
Rukmini
what is the addition of 101011 with 101010
If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
Jack
extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3 |
# NCERT Exemplar Problems Class 8 Mathematics Chapter 3 Square-Square Root and Cube-Cube Root
Uncategorized
## NCERT Exemplar Problems Class 8 Mathematics Chapter 3 Square-Square Root and Cube-Cube Root
Question1 196 is the square of
(a) 11 (b) 12
(c) 14 (d) 16
Solution.
(c) Square of 11 = 11 x 11 = 121
Square of 12 = 12 x 12 = 144
Square of 14 = 14 x 14 = 196
Clearly, 196 is the square of 14
Question 2 Which of the following is a square of an even number?
(a) 144 (b) 169
(c) 441 (d) 625
Solution.
Thus, 144 is a square of an even number.
Alternate Method
We know that, square of an even number is always an even number. Hence, 169, 441 and 625 are not even numbers. So, only 144 is an even number, which is the square of 12.
Question 3 A number ending in 9 will have the unit’s place of its square as
(a) 3 (b) 9
(c) 1 (d) 6
Solution.
Question 4 Which of the following will have 4 at the unit’s place?
(a) 142 (b) 622 (c) 272 (d)352
Solution.
Question 5 How many natural numbers lie between 52 and 62?
(a) 9 (b) 10 (c)11 (d) 12
Solution. (b) The natural numbers lying between 52 and 62, i.e. between 25 and 36 are 26, 27, 28, 29, 30, 31,32, 33, 34 and 35.
Hence, 10 natural numbers lie between 52 and 62.
Question 6 Which of the following cannot be a perfect square?
(a) 841 (b) 529 (c) 198 (d) All of these
Solution.(c) We know that, a number ending with digits 2, 3, 7 or 8 can never be a perfect square. So, 198 cannot be written in the form of a perfect square.
Question 7 The one’s digit of the cube of 23 is
(a) 6 (b) 7 (c) 3 (d) 9
Solution. (b) We know that, the cubes of the numbers ending with digits 3 and 7, have 7 and 3 at one’s digit, respectively.
So, the one’s digit of the cube of 23 is 7.
Question 8 A square board has an area of 144 sq units. How long is each side of the board?
(a) 11 units (b) 12 units (c) 13 units (d) 14 units
Solution.
Question 9
Solution.
Question 10 If one member of a Pythagorean triplet is 2m, then the other two members are
Solution.
Question 11 The sum of successive odd numbers 1, 3, 5, 7, 9, 11, 13 and 15 is
(a) 61 (b) 64 (c) 49 (d) 36
Solution. (b) We know that, the sum of first n odd natural numbers is n2.
Given odd numbers are 1,3, 5, 7, 9,11,13 and 15.
So, number of odd numbers, n = 8
The sum of given odd numbers =n2 = (8)2 = 64
Question 12 The sum of first n odd natural numbers is
(a) 2n +1 (b) n2 (c) n21 (d) n2 +1
Solution.
Question 13 Which of the following numbers is a perfect cube?
(a) 243 (b) 216 (c) 392 (d) 8640
Solution.(b) For option (a) We have, 243
Resolving 243 into prime factors, we have
243= 3 x 3 x 3 x 3 x 3
Grouping the factors in triplets of equal factors, we get
243 = (3 x 3 x 3) x 3 x 3
Clearly, in grouping, the factors in triplets of equal factors, we are left with two factors 3 x 3.
Therefore, 243 is not a perfect cube.
For option (b) We have, 216 Resolving 216 into prime factqrs, we have
216 = 2 x 2 x 2 x 3 x 3 x 3
Grouping the factors in triplets of equal factors, we get 216 = (2 x 2 x 2) x (3 x 3 x 3)
Clearly, in grouping, the factors of triplets of equal factors, no factor is left over.
So, 216 is a perfect cube.
For option (c) We have, 392
Resolving 392 into prime factors, we get
392 = 2 x 2 x 2 x 7 x 7
Grouping the factors in triplets of equal factors, we get
392 = (2 x 2 x 2) x 7 x 7
Clearly, in grouping, the factors in triplets of equal factors, we are left with two factors 7 x 7.
Therefore, 392 is not a perfect cube.
For option (d) We have, 8640
Resolving 8640 into prime factors, we get
8640=2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 5
Grouping the factors in triplets of equal factors, we get
8640 = (2 x 2 x 2) x (2 x 2 x 2) x (3 x 3 x 3) x 5
Clearly, in grouping, the factors in triplets of equal factors, we are left with one factor 5. Therefore, 8640 in not a perfect cube.
After solving, it is clear that option (b) is correct.
Question 14 The hypotenuse of a right angled triangle with its legs of lengths 3x x 4x is
(a) 5X (b )7x (c) 16x (d) 25x
Solution.
Question 15 The next two numbers in the number pattern 1, 4, 9,16, 25,… are
(a) 35, 48 (b) 36, 49 (c) 36, 48 (d) 35, 49
Solution. (b) We have, 1,4, 9,16, 25, ….
The number pattern can be written as (1)2, (2)2, (3)2, (4)2, (5)2
Hence, the next two numbers are (6)2 and (7)2, i.e. 36 and 49.
Question 16 Which among 432 , 672 , 522 , 592 would end with digit 1?
(a) 432 (b)672 (c)522 (d)592
Solution.
Question 17 A perfect square can never have the following digit in its one’s place.
(a) 1 (b) 8 (c) 0 (d) 6
Solution.
(b) We know that, a number ending with digits 2, 3, 7 or 8 can never be a perfect square. Clearly, a perfect square can never have the digit 8 in its one’s place.
Question 18 Which of the following numbers is not a perfect cube?
(a) 216 (b) 567 (c) 125 (d) 343
Solution.
(b) 216=6 x 6 x 6, 567 = 3 x 3 x 3 x 3 x 7
125 = 5 x 5 x 5, 343 = 7 x 7 x 7
Clearly, 567 is not a perfect cube, because in grouping, the factors in triplets of equal factors, we are left with two factors 3 x 7.
Question 19
Solution.
Question 20
Solution.
Question 21 A perfect square number having n digits, where n is even, will have square root with
Solution. (b) A perfect square number having n digits, where n is even, will have square root with n/2 digit.
Question 22
solution.
Question 23
Solution.
Question 24
Solution.
Fill in the Blanks
In questions 25 to 48, fill in the blanks to make the statements true.
Question 25 There are________perfect squares between 1 and 100.
Solution.8
There are 8 perfect squares between 1 and 100, i.e. 4, 9,16, 25, 36, 49, 64 and 81.
Question 26 There are________ perfect cubes between 1 and 1000.
Solution.8
There are 8 perfect cubes between 1 and 1000, i.e. 8, 27,64,125, 216, 343 and 729.
Question 27 The unit’s digit in the square of 1294 is________
Solution. 6
We know that, the unit’s digit of the square of a number having digit .at unit’s place as 4 or 6 is 6.
Hence, the units digit in the square of 1294 is 6 as 4 x 4 = 16.
Question 28 The square of 500 will have zeroes.
Solution. four
The square of 500 = (500)2
= 500 x 500 = 250000
Hence, the square of 500 will have four zeroes.
Question 29 There are natural numbers between n2 and (n + l)2
Solution.
Question 30 The square root of 24025 will have________digits.
Solution.
Question 31 The square of 5.5 is________
Solution. 30.25
Square of 5.5= (5.5)2 = 55 x 5.5= 30.25
Question 32 The square root of 5.3 x 5.3 is________
Solution.
Question 33 The cube of 100 will havfe________zeroes.
Solution. 6
Cubeof100 = 1003
= 100x100x100 = 1000000
Question 34 1m2 =________ cm2.
Solution.
Question 351m3 =________ cm3.
Solution.
Question 36 One’s digit in the cube of 38 is________
Solution.
Question 37 The square of 0.7 is________
Solution.0.49
Square of 0.7 = (0.7)2 = 07 x 07 = 0.49
Question 38 The sum of first six odd natural numbers is________
Solution.
Question 39 The digit at the one’s place of 572 is________
Solution.
Question 40 The sides of a right angled triangle whose hypotenuse is 17cm, are________and________
Solution.
As, hypotenuse of right angled triangle is 17 cm.
Question 41
Solution.
Question 42 (1.2)3=________
Solution.
Question 43 The cube of an odd number is always an________number.
Solution.odd
We know that, the cubes of all odd natural numbers are odd.
Question 44 The cube root of a number x is denoted by________
Solution.
Question 45 The least number by which 125 be multiplied to make it a perfect square, is________
Solution.
Question 46 The least number by which 72 be multiplied to make it a perfect cube, is________
Solution. 3
Resolving 72 into prime factors, we get
72=2 x 2 x 2 x 3 x 3
Grouping the factors in triplets of equal factors, we get
72 = (2 x 2 x 2) x 3 x 3
We find that 2 occurs as a prime factor of 72 thrice, but 3 occurs as a prime factor only twice. Thus, if we multiply 72 by 3, 3 will also occurs as a prime factor thrice and the product will be 2 x 2 x 2 x 3 x 3 x 3, which is a perfect cube.
Hence, the least number, which should be multiplied with 72 to get perfect cube, is 3.
Question 47 The least number by which 72 be divided to make it a perfect cube, is________
Solution. 9
Resolving 72 into prime factors, we get
72=2 x 2 x 2 x 3 x 3
Grouping the factors in triplets of equal factors, we get
72 = (2 x 2 x 2) x 3 x 3
Clearly, if we divide 72 by 3 x 3, the quotient would be 2 x 2 x 2, which is a perfect cube. Hence, the least number by which 72 be divided to make it, a perfect cube, is 9.
Question 48 Cube of a number ending in 7 will end in the digit________
Solution 3
We know that, the cubes of the numbers ending in digits 3 or 7 ends in digits 7 or 3, respectively.
i.e 7 x 7 x 7 = 343
Hence, the cube of a number ending in 7 will end in the digit 3.
True/False
In questions 49 to 86, state whether the statements are True or False.
Question 49 The square of 86 will have 6 at the unit’s place.
Solution True
We know that, the unit’s digit of the square of a number having digit at unit’s place as 4 or 6 is 6.
Question 50 The sum of two perfect squares is a perfect square.
Solution False
e.g. 16 and 25 are the perfect squares, but 16 + 25 = 41 is not a perfect square.
Question 51 The product of twtfperfect squares is a perfect square.
Solution True
e.g. If 4 and 25 are the perfect square, then 4 x 25 = 100 is also a perfect square.
Clearly, the product of two perfect squares is a perfect square.
Question 52 There is no square number between 50 and 60.
Solution True
Numbers between 50 and 60 are 51,52, 53, 54, 55, 56, 57, 58 and 59.
We observed that there is no square number between 50 and 60.
Question 53 The square root of 1521 is 31.
Solution Falsie %
As, the square of 31 = (31)2 = 31 x 31 = 961
Question 54 Each prime factor appears 3 times in its cube.
Solution True
If a3 is the cube and m is one of the prime factors of a. Then, m appears three times in a3.
Question 55 The square of 2.8 is 78.4.
Solution False
The square of 2.8 = (2.8)2 = 2.8×2.8 = 7.84
Question 56 The cube of 0.4 is 0.064.
Solution True
Cube of 0.4 = (0.4)2 = 0.4 x 0.4 x 0.4 = 0.064
Question 57 The square root of 0.9 is 0.3.
Solution False
As, the square of 0.3 = (0.3)2 = 0.3 x 0.3 =0.09
Question 58 The square of every natural number is always greater than the number itself.
Solution False
1 is a natural number and square of 1 is not greater than 1.
Question 59 The cube root of 8000 is 200.
Solution
Question60 There are five perfect cubes between 1 and 100.
Solution False
There are eight perfect cubes between 1 and 100, i.e. 8,27,64,125,216,343,512 and 729.
Question 61 There are 200 natural numbers between 1002 and 1012.
Solution.
Question 62 The sum of first n odd natural numbers is n2.
Solution.
Question 63 1000 is a perfect square.
Solution False
1000 = 2 x 2 x 2 x 5 x 5 x 5 = 2 2 x 52x 2 x 5 Clearly, it is not a perfect square, because it has two unpaired factors 2 and 5.
Question 64 A perfect square can have 8 as its unit’s digit.
Solution False
A perfect square can never have 8 as its unit’s digit.
Question65
Solution
Question 66 All numbers of a Pythagorean triplet are odd.
Solution False
3, 4 and 5 are the numbers of Pythagorean triplet as 52 = 4 2 + 3 2 where, 4 is not an odd number.
Question 67 For an integer a, a3 is always greater than a2.
Solution
Qustion68
Solution
Question 69 Let x and y be natural numbers. If x divides y, then x3 divides y3.
Solution.
Question 70 If a2 ends in 5, then a3 ends in 25:
Solution.
Question 71 If a2 ends in 9, thena3 ends in 7.
Solution
Question72
Solution.
Question 73 Square root of a number x is denoted by 4x.
Solution True
Square root of a number x is denoted by 4x.
Question 74 A number having 7 at its one’s place will have 3 at the unit’s place of its square.
Solution False
Square of 7 = 7 x 7 = 49
Square of 17 = 17 x 17 = 289
Square of 27 = 27 x 27 = 729
and so on.
Question 75 A number having 7 at its one’s place will have 3 at the one’s place of its cube.
Solution True
Cube of 7 = 7 x 7 x 7 =343
Cube of 17 = 17 x 17 x 17=4913
Cube of 27 = 27 x 27 x 27 = 19683
and so on.
Question 76 The cube of a one-digit number cannot be a two-digit number.
Solution
Question 77 Cube of an even number is odd.
Solution. False
We know that, the cube of an even number is always an even number,
e.g. 2 is an even number. Then, 23 = 2 x 2 x 2 = 8
Clearly, 8 is also an even number.
Question 78 Cube of an odd number is even.
Solution. False
We know that, the cube of an odd number is always an odd number,
e.g. 3 is an odd number. Then, 33 = 3 x 3 x 3 = 27
Clearly, 27 is not an even number.
Question 79 Cube of an even number is even.
Solution. True
We know that, the cube of an even number is always an even number,
e.g. 4 is an even number. Then, 43 = 4 x 4 x 4 = 64
Clearly, 64 is also an even number.
Question 80 Cube of an odd number is odd.
Solution. True
We know that, the cube of an odd number is always an odd number,
e.g. 9 is an odd number.
Then,93 =9 x 9 x 9 = 729
Clearly, 729 is also an odd number.
Question 81 999 is a perfect cube.
Solution.
Question 82 363 x 81 is a perfect cube.
Solution.
Question 83 Cube roots of 8 are + 2 and – 2 .
Sol. False
Cube root of 8 is 2 only and cube root of – 8 is – 2.
Question 84
Solution.
Question 85 There is no cube root of a negative integer.
Solution.
Question 86 Square of a number is positive, so the cube of positive.
Solution.
Question 87 Write the first five square numbers.
Solution. First five square numbers are 12,22, 32, 42 and 52, i.e. 1, 4, 9,16and 25.
Question 88 Write cubes of first three multiples of 3.
Solution. Since, the first three multiples of 3 are 3, 6 and 9.
Hence, the cubes of first three multiples of 3 are (3)3, (6)3 and (9)3, i.e. 27, 216 and 729.
Question 89 Show that 500 is not a perfect square.
Solution.
Question 90 Express 81 as the sum of first nine consecutive odd numbers.
Solution. 81= (9)2 =1+3+ 5+ 7 + 9+ 11 + 13+ 15+ 17 = Sum of first nine consecutive odd numbers
Question 91 Using prime factorisation, find which of the following are perfect squares.
(a) 484 (b) 11250
(c) 841 (d) 729 .
Solution.
So, 484 is a perfect square.
(b) Prime factors of 11250 = 2 x (3 x 3) x (5 x 5) x (5 x 5)
As grouping, 2 has no pair.
So, 11250 is not a perfect square,
(c) Prime factors of 841 = (29 x 29)
As grouping, there is no unpaired factor left over. So, 841 is a perfect square.
(d) Prime factors of 729 = (3 x 3) x (3 x 3) x (3 x 3)
As grouping, there is no unpaired factor left over.
So, 729 is a perfect square.
Question 92 Using prime factorisatioji, find which of the following are perfect cubes,
(a) 128 (b) 343 (c) 729 (d) 1331
Solution.(a) We have, 128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
Since, 2 remains after grouping in triplets.
So, 128 is not a perfect cube.
(b) We have, 343 = 7 x 7 x 7
Since, the prime factors appear in triplets.
So, 343 is a perfect cube.
(c) We have, 729 = 3 x 3 x 3 x 3 x 3 x3
Since, the prime factors appear in triplets.
So, 729 is a perfect cube.
(d) We have, 1331 =11x11x11
Since, the prime factors appear in triplets.
So, 1331 is a perfect cube.
Question 93 Using distributive law, find the squares of (a) 101 (b) 72
Solution. (a) We have, 1012 = 101 x 101
= 101(100+ 1)= 10100+ 101 = 10201
(b) We have, 722 = 72 x 72 = 72 x (70 + 2)
= 5040+ 144= 5184
Question 94 Can a right angled triangle with sides 6cm, 10cm and 8cm be formed? Give reason.
Solution.
Question 95 Write the Pythagorean triplet whose one of the numbers is 4.
Solution.
Question 96 Using prime factorisation, find the square roots of (a) 11025 (b) 4761
Solution.
Question 97 Using prime factorisation, find the cube roots of
(a) 512
(b) 2197
Solution.
Question 98 Is 176 a perfect square? If not;- find the smallest number by which it should be multiplied to get a perfect square.
Solution.
Question 99 Is 9720 a perfect cube? If not, find the smallest number by which it should be divided to get a perfect cube.
Solution.Prime factors of 9720 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x5
The prime factors 3 and 5 do not appear in group of triplets.
So, 9720 is not a perfect cube.
If we divide the number by 3 x 3 x 5, then the prime factorisation of the quotient will not contain 3 x 3 x 5 = 45.
Question 100 Write two Pythagorean triplets, each having one of the numbers as 5.
Solution.
Question 101 By what smallest number should 216 be divided, so that the quotient
is’ a perfect square? Also, find the square root of the quotient.
Solution.
Question 102 By what smallest number should 3600 be multiplied, so that the quotient is a perfect cube. Also, find the cube root of the quotient.
Solution. Prime factors of 3600 = 2x2x2x2x3x3x5x5
Grouping the factors into triplets of equal factors, we get
Question 103 Find the square root of the following by long division method.
(a) 1369 (b) 5625
Solution.
Question 104 Find the square root of the following by long division method. : (a) 27.04 (b) 1.44
Solution.
Question 105 What is the least number, that should be subtracted from 1385 to get a perfect square? Also, find the square root of the perfect square.
Solution.
Question 106 What is the least number that should be added to 6200 to make it a perfect square?
Solution.
Question 107 Find the least number of four digits that is a perfect square.
Solution.
Question 108 Find the greatest^umber of three digits that is a perfect square.
Solution.
Question 109 Find the least square, number, which is exactly divisible by 3, 4, 5, 6 and 8.
Solution.
Question 110 Find the length of the side of a square, if the length of its diagonal is 10 cm.
Solution.
Question 111 A decimal number is multiplied by itself. If the product is 51.84, then find the number.
Solution.
Question 112 Find the decimal fraction, which when multiplied by itself, gives 84.64.
Solution.
Question 113 A farmer wants to plough his square field of side 150m. How much area will he have to plough?
Solution.
Question 114 What will be the number of unit squares on each side of a square graph paper, if the total number of unit squares is 256?
Solution.
Question 115 If one side of a cube is 15m in Length, then find its volume.
Solution.
Question 116 The dimensions of a rectangular field are 80m and 18m. Find the length of its diagonal.
Solution.
Question 117 Find the area of a square field, if its perimeter is 96 m.
Solution.
Question 118 Find the length of each side of a cube, if its volume is 512 cm3.
Solution.
Question 119 Three numbers are in the ratio 1:2:3 and the sum of their cubes is 4500. Find the numbers.
Solution.
Question 120 How many square metres of carpet will be required for a square room of side 6.5m to be carpeted?
Solution.
Question 121 Find the side of a square, whose area is equal to the area of a rectangle with sides 6.4m and 2.5m.
Solution.
Question 122 Difference of two perfect cubes is” 189. If the cube root of the smaller of the two numbers is 3, then find the cube root of the larger number.
Solution.
Question 123 Find the number of plants in each row, if 1024 plants are arranged, so
that number of plants in a row is the same as the number of rows.
Solution.
Question 124 A hall has a capacity of 2704 seats. If the number of rows is equal to the number of seats in each row, then find the number of seats in each row.
Solution.
Question 125 A General wishes to draw up his 7500 soldiers in the form of a square. After arranging, he found out that some of them are left out. How many soldiers were left out?
Solution.
Question 126 8649 students were sitting in a lecture room in such a manner that there were as many students in the row as there were rows in the lecture room. How many students were there in each row of the lecture room?
Solution.
Question 127 Rahul walks 12m North from his house and turns West to walk 35m to reach his friend’s house. While returning, he walks diagonally from his friend’s house to reach back to his house. What distance did he walk, while returning?
Solution.
Question 128 A 5.5m long ladder is leaned against a wall. The ladder reaches the wall to a height of 4.4m. Find^the distance between the wall and the foot of the ladder.
Solution.
Question 129 A king wanted to reward his advisor, a wiseman of the kingdom. So, he asked the wiseman to name his own reward. The wiseman thanked the king, but said that he would ask only for some gold coins each day for a month. The coins were to be counted out in a pattern of one coin for the first day, 3 coins for the second day, 5 coins for the third day and so on for 30 days. Without making calculations, find how many coins will the advisor get in that month?
Solution.
Question 130 Find three numbers in the ratio 2 : 3 : 5, the sum of whose squares is 608.
Solution.
Question 131 Find the smallest square number divisible by each of the numbers 8, 9 and 10.
Solution.
Question 132
Solution.
Question 133 Find the square root of 324 by the method of repeated subtraction.
Solution.
Question 134 Three numbers are in the ratio 2 : 3 : 4. The sum of their cubes is 0.334125. Find the numbers.
Solution.
Question 135
Solution.
Question 136
Solution.
Question137
Solution.
Question 138
A perfect square number has four digits, none of which is zero. The- digits from left to right have valfles, that are even, even, odd, even. Find the number.
Solution.
Suppose abed is a perfect square.
where, a = even number
b = even number
c = odd number
d = even number
Hence, 8836 is one of the number which satisfies the given condition.
Question 139 Put three different numbers in the circles, so that when you add the numbers at the end of each line, you always get a perfect square.
Solution.
Question 140 The perimeters of two squares are 40m and 96m, respectively. Find the perimeter of another square equal in area to the sum of the first two squares.
Solution.
Question 141 A three-digit perfect square is such that, if it is viewed upside down, the number seen is also a perfect square. What is the number?
[Hint The digits 1, 0 and 8 stay the same when viewed upside down, whereas 9 becomes 6 and 6 becomes 9]
Solution. Three-digit perfect squares are 196 and 961, which looks same when viewed upside down.
Question 142 13 and 31 is a strange pair of numbers, such that their squares 169 and 961 are also mirror of each other. Can yqu find two other such pairs?
Solution.
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# How do you find the measure of a central angle?
## How do you find the measure of a central angle?
For finding the central angle in radians, we have to divide the arc length by the length of the radius of the circle.
## How do you find the central angle example?
In the figure above, ∠PZQ,∠QZR , and ∠RZP are central angles. Sum of Central Angles: The sum of the measures of the central angles of a circle with no points in common is 360° . For example, in the figure above, m∠PZR+m∠RZQ+m∠QZP=360° .
Does a central angle Measure 90?
By the inscribed angle theorem, the measure of an inscribed angle is half the measure of the intercepted arc. The measure of the central angle ∠POR of the intercepted arc ⌢PR is 90°.
### Is a diameter a central angle?
Diameter: Any straight line segment that passes through the center of a circle and whose endpoints lay on the circle. Central angle: the angle at the center of a circle between two radii. Inscribed angle: the angle of a vertex on a circle between its chords.
### Is the central angle the same as the arc?
Central angles are subtended by an arc between those two points, and the arc length is the central angle of a circle of radius one (measured in radians). The central angle is also known as the arc’s angular distance.
How do you solve central angle problems?
You can also use the radius of the circle and the arc length to find the central angle. Call the measure of the central angle θ. Then: θ = s ÷ r, where s is the arc length and r is the radius.
#### How do you find the central angle of a class 8?
So the area of 20% = $\dfrac{{2 \times 22}}{{7 \times 5}}$ = 1.26°. So, our answer is $\dfrac{{2\pi }}{5}$ in radian and 1.26° in degree. Note: In a pie chart the complete 360° represents 100% of the data. The internal angle of a circle is 360° or $2\pi$.
#### Can a central angle be 180 degrees?
A convex central angle is a central angle that measures less than 180°. Reflex angles measure more than 180° and less than 360°. Check out this second image to notice the reflex angle that is created on the other side of the convex central angle. A reflex central angle is indicated and measures more than 180 degrees.
Can a diameter be a central angle?
When a diameter goes through the center of a circle, then the central angle subtended by the semicircle arc is simply 180°, no doubt about that. And, if the theory is right, any inscribed angle whose chords’ endpoints are at the corresponding endpoints of this diameter should be twice less — it should be 90°.
## Is angle equal to arc?
The measure of an arc refers to the arc length divided by the radius of the circle. The arc measure equals the corresponding central angle measure, in radians.
## Why is x2 y2 r2?
Thus, using the theorem of Pythagoras, x2 + y2 = r2 , and this is the equation of a circle of radius r whose centre is the origin O(0, 0). The equation of a circle of radius r and centre the origin is x2 + y2 = r2 .
What is the formula of central angle in pie chart?
So, our answer is 2π5 in radian and 1.26° in degree. Note: In a pie chart the complete 360° represents 100% of the data. The internal angle of a circle is 360° or 2π. It is to be noted that not only the angle is divided this way, but the area of the region is also divided.
### How do you calculate central angle?
From the arc length, the central angle can be calculated. Indeed, one formula that can help in determining the central angle states that the arc length (s) is equal to the radius times the central angle, or s = r × θ, where the angle, theta, must be measured in radians.
### What is the measure of a central angle?
All central angles would add up to 360° (a full circle), so the measure of the central angle is 360 divided by the number of sides. The measure of the central angle thus depends only on the number of sides.
How to find central angle formula?
A central angle is an angle with a vertex at the centre of a circle, whose arms extend to the circumference. You can imagine the central angle being at the tip of a pizza slice in a large circular pizza. You can find the central angle of a circle using the formula: θ = L / r. where θ is the central angle in radians, L is the arc length and r is the radius.
#### What is the measure of the central angle formula?
Indeed, one formula that can help in determining the central angle states that the arc length (s) is equal to the radius times the central angle, or s = r × θ, where the angle, theta, must be measured in radians. So to solve for the central angle, theta, one need only divide the arc length by the radius, or s ÷ r = θ. |
# Net Force: A Guide for HSC Physics Students
This topic is part of the HSC Physics course under the section Forces.
### HSC Physics Syllabus
• explore the concept of net force and equilibrium in one-dimensional and simple two-dimensional contexts using: (ACSPH050)
– vector addition by resolution into components
### What is Net Force?
In the HSC Physics course, the concepts of net force and equilibrium are essential for understanding the behaviour of objects in motion or at rest.
### Net Force in HSC Physics
Net force is the vector sum of all individual forces acting on an object. It determines the overall effect of the forces and dictates whether the object will accelerate, decelerate, or remain at a constant velocity. The net force can be calculated using vector addition in both one-dimensional and two-dimensional contexts.
One-dimensional Net Force
In a one-dimensional context, forces act along a straight line. The net force can be calculated by adding the magnitudes of forces acting in the same direction and subtracting those acting in the opposite direction. This concept is essential for understanding linear motion in the HSC Physics course.
For example, the net force in diagram above is given by
$$F_{net}=150 - 100$$
$$F_{net}=50 \hspace{1mm} N$$
Therefore, the net force is 50 N to the right.
Two-dimensional Net Force
In a two-dimensional context, forces act in a plane, and their components need to be considered separately along the x and y axes. The net force can be calculated by vector addition, combining the x-components of the forces and the y-components separately to determine the net force in each direction.
Let's consider an example where a mass is acted by multiple forces as shown below.
Before we can determine the net force in x and y axes, the 150 N force vector must be resolved into its x and y components.
$$F_{x-component}=150\cos{40º}$$
$$F_{y-component}=150\sin{40º}$$
The net force in the x-direction in the diagram above is given by
$$F_x = 150\cos{40º} - 100$$
$$F_x = 14.9 \hspace{1mm} N$$
Net force in the x-direction is 14.9 N to the right.
The net force in the y-direction is given by
$$F_y = 150\sin{40º} - 80$$
$$F_y = 16.4 \hspace{1mm} N$$
Net force in the y-direction is 16.4 N upwards.
The net force in the x and y direction can be added to find the resultant net force. The magnitude of this resultant net force is given by Pythagorean theorem.
$$F_{resultant}^2=14.9^2 + 16.4^2$$
$$F_{resultant} = \sqrt{14.9^2 + 16.4^2}$$
$$F_{resultant} = 21.1 \hspace{1mm} N$$
The direction of the resultant net force can be determined using trigonometry.
$$\tan{\theta}=\frac{16.4}{14.9}$$
$$\theta = 48º$$
Therefore, the resultant net force is 21.1 N at 48º above the horizontal or x-axis. |
Home » Online Formulas » Law of Cosines Formula
# Law of Cosines Formula
If the two sides and angles of the triangle are given, then the unknown side and angles can be calculated using the cosine law. Law of cosine is another formula used to find out the unknown side of the triangle.
The Law of Cosine Formula is,
$\large a^2=b^2+c^2-2(bc)Cos\;A$ $\large b^2=a^2+c^2-2(ac)Cos\;B$ $\large c^2=a^2+b^2-2(ab)Cos\;C$The cosine law can be derived out of Pythagoras Theorem.
The Pythagorean theorem can be derived from the cosine law. In the case of a right triangle the angle, θ = 90°. So, the value of cos θ becomes 0 and thus the law of cosines reduces to $c^2=a^2+b^2$
## Law of Cosines Problem
Some solved problem on the law of cosines are given below:
### Solved Examples
Question 1: Given the sides of the triangle b = 7 cm; c = 8 cm and the angle A = 45o. Calculate the unknown sides and angles ?
Solution:
Given,
b = 7 cm
c = 8 cm
A = 45o
The law of cosines formula is: a2 = b2 + c2 – 2bc cos A
a2 = (7 cm)2 + (8 cm)2 – 2(7 cm)(8 cm) cos 45
a2 = 49 cm2 + 64 cm2 – (112 cmx 0.707)
a2 = 49 cm2 + 64 cm2 – 79.18 cm2
a2 = 33.82
a = 5.82 cm
b2= a2 + c2 – 2ac cos B72 = (5.82)2 + 82 – 2(5.82)(8) cos B
49 = 33.8724 + 64 – 93.12 cos B
93.12 cos B = 48.8724
Cos B = 48.8724/93.12
B = 58.3o
c2 = a2+ b2 – 2ab cos C
82 = (5.82)2 + 72 -2(5.82)(7) cos C
64 = 33.8724 + 49 – 81.48 cos C
81.48 cos C = 18.8724
Cos C = 18.8724/81.48
C = 76.6o |
# NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.1
These NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.1 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-1-ex-1-1/
## NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions 1.1
Class 12 Maths Chapter 1 Exercise 1.1 Question 1.
Determine whether each of the following relations are reflexive, symmetric and transitive.
i. Relation R in the set A= {1, 2, 3,… 13, 14} defined as R = {(x, y): 3x – y = 0}
ii. Relation R in the set N of natural numbers defined as R = {(x, y): y = x + 5 and x < 4}
iii. Relation R in the set A= (1, 2, 3, 4, 5, 6} as R = {(x, y): y is divisible by x}
iv. Relation R in the set Z of all integers defined as R = {(x, y): x -y is an integer}
v. Relation R in the set A of human beings in a town at a particular time given by
a. R = {(x, y) : x and y work at the same place}
b. R= {(x, y) : x and y live in the same locality}
c. R = {(x, y) : x is exactly 7 cm taller than y}
d. R = {(x, y) : x is wife of y}
e. R = {(x, y) : x is father of y}
Solution:
i. Reflexive R = {(1, 3),(2, 6), (3, 9), (4, 12)}
(1.1) ∈ R
∴ R is not reflexive.
Symmetric
(1, 3) ∈ R but (3, 1) ∉ R
∴ R is not symmetric.
Transitive
(1, 3) ∈ R and (3,9) ∈ R but (1, 9) ∉ R
∴ R is not transitive.
Hence R is neither reflexive, nor symmetric nor transitive.
ii. R = {(1, 6),(2, 7), (3, 8)}
(1, 1) ∉ R
∴ R is not reflexive.
(1, 6) ∈ R but (6, 1) ∉ R
∴ R is not symmetric. In R there does not exist ordered pairs of the form (x, y) and (y, z)
∴ R is transitive.
iii. R = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2.2) , (2,4), (2,6), (3,3), (3,6), (4,4), (5,5), (6,6)}
Reflexive
(x, x) ∈ R for all x ∈ A, since x is divisible by x.
∴ R is reflexive
Symmetric
(1,2) ∈ R but (2, 1) ∈ R
∴R is not symmetric.
Transitive
If y is divisible by x and z is divisible by y, then z is divisible by x. i.e., (x, y) ∈ R
(y, z) ∈R ⇒ (x, z) ∈R for all x, y, z ∈ A
∴R is transitive
iv. Reflexive
x – x = 0, which is an integer
i.e., (x, x)∈ R for every x ∈ Z
∴ R is reflexive
Symmetric
(x, y) ∈ R
⇒ x – y is an integer
⇒ y – x is an integer
⇒ (y, x) ∈ R
∴ R is symmetric
Transitive
(x, y) ∈ R, (y, z) ∈ R ⇒ x – y is an integer and y – z is an integer.
⇒ (x – y) + (y – z) is an integer
⇒ x – z is an integer
⇒ (x, z) ∈ R for all x, y, z ∈ Z
∴ R is transitive.
v. a. Reflexive
(x, x) ∈ R for every x ∈ A, since x and x work at the same place.
∴R is reflexive.
Symmetric
Let (x, y) ∈ R
⇒ x and y work at the same place
⇒ y and x work at the same place
⇒ (y, x) ∈ R for all x, y ∈ A
∴ R is symmetric
Transitive
Let (x, y), (y, z) ∈ R
⇒ x and y work at the same place;
y and z work at the same place.
⇒ x and z work at the same place
⇒ (x, z) ∈ R
∴ R is transitive
Hence R is reflexive, symmetric and transitive.
b. Reflexive, symmetric and transitive
(Similar to v.a)
c. Reflexive
Since x cannot 7 cm taller than x, (x, x) ∉ R.
∴ R is not reflexive
Symmetric
(x, y) ∈ R ⇒ x is exactly 7 cm taller than y
⇒ y cannot be exactly 7 cm taller than x.
⇒ (y, x) ∉ R
∴ R is not symmetric
Transitive
Let (x, y),(y, z) ∈ R ⇒ x is exactly 7 cm
taller than y and y is exactly 7 cm taller than z.
⇒ x is exactly 14 cm taller than z.
⇒ (x, z) ∉ R ∴ R is not transitive.
d. Reflexive
(x, x) ∉ R as x cannot be the wife of x.
∴ R is not reflexive
Symmetric
Let (x, y) ∈ R ⇒ x is the wife of y
⇒ y is not the wife of x ⇒ (y, x) ∉ R
∴ R is not symmetric
Transitive
Let (x, y) ∈ R ⇒ x is the wife of y ⇒ y
cannot be the wife of any z
⇒ (y, z) ∉ R
That is, there does not exist any (x, y) and (y, z) in R.
∴ R is transitive
e. Reflexive
(x, x) ∉ R as x cannot be the father of x.
∴ R is not reflexive
Symmetric
(x, y) ∈ R ⇒ x is the father of y.
⇒ y is not the father of x
⇒ (y, x) ∉ R
∴ R is not symmetric
Transitive
Let (x, y), (y, z) ∈ R ⇒ x is the father of y
and y is the father of z
⇒ x cannot be the father of z
⇒ (x, z) ∉ R
∴ R is not transitive
Chapter 1 Math Class 12 Question 2.
Show that the relation R in the set R of real numbers, defined as R= {(a, b); a<b²} is neither reflexive nor symmetric nor transitive.
Solution:
Reflexive
Let a = $$\frac { 1 }{ 2 }$$. Then a² = $$\frac { 1 }{ 4 }$$
⇒ $$\frac { 1 }{ 2 }$$ is not less than or equal to ($$\frac { 1 }{ 2 }$$)²
⇒ ($$\frac { 1 }{ 2 }$$, $$\frac { 1 }{ 2 }$$) ∉ R
∴ R is not reflexive.
Symmetric
Since 1 ≤ 2², (1, 2) ∈ R
But 2 is not less than or equal to 1²
⇒ (2, 1) ∈ R
i.e., (1, 2) ∈ R and (2, 1) ∉ R
∴ R is not symmetric
Transitive
1 ≤ (- 2)² and – 2 ≤ (0)²
But 1 is not less than or equal to 0²
i.e., (1, – 2), (- 2, 0) ∈ R but (1, 0) ∉ R
∴ R is not transitive
Class 12 Maths Ex 1.1 Question 3.
Check whether the relation R defined in the set {1, 2, 3,4, 5, 6} as R = {(a, b) : b = a + 1} is reflexive, symmetric or transitive.
Solution:
R={(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}
(1, 1) ∉ R
∴ R is not reflexive
(1, 2) ∈ R and (2, 1) ∉ R
∴ R is not symmetric
(3, 4) ∈ R , (4, 5) ∉ R but (3, 5) ∉ R
∴ R is not transitive
Class 12 Ex 1.1 Question 4.
Show that the relation R in R defined as R = {(a, b); a < b), is reflexive and transitive but not symmetric.
Solution:
Reflexive
(a, a) ∈ R as a < a ∴ R is reflexive
Symmetric
(1, 2) ∈ R ⇒ 1 ≤ 2 ⇒ 2 ≤ 1
⇒ (2, 1) ∉ R
∴ R is not symmetric
Transitive
(a, b), (b, c) ∈ R a ≤ b , b ≤ c
⇒ a ≤ c which is true
∴ R is transitive
Maths Class 12 Chapter 1 Exercise 1.1 Question 5.
Check whether the relation R in R defined by R = {(a,b); a ≤ b³} is reflexive, symmetric or transitive.
Solution:
Reflexive
Let a = $$\frac { 1 }{ 2 }$$. Then $$\frac { 1 }{ 2 }$$ is not less than ($$\frac { 1 }{ 2 }$$)³.
i.e., ($$\frac { 1 }{ 2 }$$, $$\frac { 1 }{ 2 }$$) ∉ R
∴ R is not reflexive.
Symmetric
Since 1 ≤ 2³, (1, 2) ∈ R
But 2 is not less than or equal to L3.
⇒ (2, 1) ∉ R
i.e., (1, 2) ∈ R and (2, 1) ∉ R
∴ R is not symmetric
Transitive
9 ≤ 3³ and 3 ≤ 2³, i.e., (9, 3), (3, 2) ∈ R
But 9 is not less than or equal to 2³
i.e., (9, 2) ∉ R
i.e., (9, 3),(3, 2) ∈ R but (9, 2) ∉ R
∴ R is not transitive.
Question 6.
Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.
Solution:
(1, 1) ∉ R ∴ R is not reflexive
Symmetric
(1, 2), (2, 1) ∈ R ⇒ (2, 1), (1, 2) ∈ R
∴ R is symmetric
Transitive
(1, 2), (2, 1) ∈ R but (1, 1) ∉ R
∴ R is not transitive
Question 7.
Show that the relation R in the set A of all the books in a library of a college, given by ,R = {(x, y) : x and y have same number of pages} is an equivalence relation.
Solution:
Reflexive
(x, x) ∈ R, since book x and x have same number of pages.
∴ R is reflexive
Symmetric
If (x, y) ∈ R, then books x and y have same number of pages i.e., books y and x have same number of pages. Hence (y, x) ∈ R.
∴ R is symmetric
Transitive
Let (x, y, (y, z) ∈ R ⇒ books x, y and books y, z have same number of pages.
⇒ books x and z have same number of pages.
⇒ (x, z) ∈ R
∴ R is symmetric
R is trarisitive R is reflexive, symmetric and transitive. Hence R is an equivalence relation.
Question 8.
Show that the relation R in the set A= {1, 2, 3, 4, 5} given by R = {(a, b) : |a – b| is even}, is an equivalence relation. Show that all the elements of {1, 3, 5} are related to each other and all the elements of {2, 4} are related to each other. But no element of {1, 3, 5} is related to any element of {2, 4}.
Solution:
R= {(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (3, 1), (3, 3), (3, 5), (4, 2), (4, 4), (5, 1), (5, 3),(5, 5)}
Reflexive
Since |a – a| = 0 which is even (a, a) ∈ R
∴ R is reflexive
Symmetric
Let (d, b) ∈ R ⇒ |a – b| is even
⇒ |b – a| is even
⇒ (b, a) ∈ R for all a, b ∈ A
∴ R is symmetric
Transitive
Let (a, b), (b, c) ∈ R
⇒ |a – b| is even and |b – c| is even
⇒ |a – c| is even
∴ R is transitive
Hence R is an equivalence relation.
The elements of {1, 3, 5} are related to each other, since |1 – 1|, |1 – 3|, |1 – 5| etc. are even numbers.
The elements of {2, 4} are related to each other, since |2 – 2|, |2 – 4|, |4 – 2|, |4 – 4| are even numbers.
No elements of {1, 3, 5} is related to any element of {2, 4}, since the absolute value of the difference of any element from {1, 3, 5} with any element from {2, 4} is not even.
Question 9.
Show that each of the relation R in the set A= {x ∈ Z : 0 ≤ x ≤ 12}, given by
i. R = {(a, b): |a – b| is a multiple of 4}
ii. R = {(a, b): a = b} is an equivalence relation. Find the set of all elements related to 1 in each case.
Solution:
i. Reflexive
|a – a| = 0 is a multiple of 4, a ∈ A
⇒ (a, a) ∈ R
∴ R is reflexive
Symmetric
Let (a, b) ∈R ⇒ |a – b| is a multiple of 4
⇒ |b – a| is a multiple of 4
⇒ (b, a) ∈R
∴ R is symmetric
Transitive
Let (a, b), (b, c) ∈ R
⇒ |a – b| and |b – c| are multiple of 4
⇒ a- b and b – c are multiple of 4
Now a – c = a – b + b – c
= Multiple of 4 + multiple of 4
= Multiple of 4
∴ |a – c| is a multiple of 4 ⇒ (a, c) ∈R
i.e., (a, b), (b, c) ∈ R ⇒ (a, c) ∈ R
∴R is transitive
Hence R is an equivalence relation.
The set of elements related to 1 is {1, 5,9}
ii. Reflexive
Let a ∈ A. Then a = a
⇒ (a, a) ∈R for all a ∈ A
∴ R is reflexive
Symmetric
Let (a, b) ∈ R ⇒ a – b
⇒ b = a
⇒ (b, a) ∈ R
∴ R is symmetric
Transitive
Let {a, b), (b, c) ∈ R ⇒ a = b and b = c
⇒ a = c
⇒ (a, c) ∈ R
∴ R is transitive.
Hence R is an equivalence relation.
The set of elements related to 1 is {1}
Question 10.
Give an example of a relation, which is
i. Symmetric but neither reflexive nor transitive.
ii. Transitive but neither reflexive nor symmetric.
iii. Reflexive and symmetric but not transitive’
iv. Reflexive and transitive but not symmetric.
v. Symmetric and transitive but not reflexive.
Solution:
Let A= {1, 2, 3}
i. Let R= {(1, 2), (2, 1)}
Reflexive
R is not reflexive, since (1, 1) ∉ R
Symmetric
R is symmetric, since (1, 2), (2, 1) ∈ R
⇒ (2, 1),(1, 2) ∈ R
Transitive
R is not transitive, since (1, 2), (2, 1) ∈R but (1, 1) ∉ R
∴ R is symmetric, but neither reflexive nor transitive.
ii. Let R= {(1,2), (1,3), (3, 2)}
Reflexive
R is not reflexive, since (1, 1) ∉ R
Symmetric
R is not symmetric, since(1, 2) ∈ R, but (2, 1) ∉ R
Transitive
R is transitive, since (1, 3), (3, 2) ∈ R and
(1, 2) ∈ R
∴ R is transitive but neither reflexive nor symmetric.
iii. Let R = {(1, 1), (2, 2), (3, 3), (1, 2),
(2,1) , (2,3), (3,2)}
Reflexive
R is reflexive, since (1,1), (2,2), (3,3) ∈ R
Symmetric
R is symmetric, since (1, 2) ∈ R
⇒ (2, 1) ∈ R(2, 3) ∈ R ⇒ (3, 2) ∈R
Transitive
R is not transitive, since (1, 2), (2, 3) ∈ R but(1, 3) ∉ R
∴ R is reflexive and symmetric but not transitive.
iv. Let R= {(1, 1), (2, 2), (3, 3), (1, 2)}
Reflexive
R is reflexive, since (1, 1), (2, 2), (3, 3) ∈ R
Symmetric
R is not symmetric, since (1, 2) ∈ R but (2, 1) ∉ R
Transitive
R is transitive, since (1, 1), (1, 2) ∈ R and , (1, 2) ∈ R
v. Let R= {(1, 2), (2, 1), (1, 1), (2, 2)}
Reflexive
R is not reflexive, since (3, 3) ∉ R
Symmetric
R is symmetric, since (1, 2) ∈ R ⇒ (2, 1) ∈ R
Transitive
R is transitive, since (1, 2), (2, 1) ∈ R ⇒ (1, 1) ∈ R
and (2, 1 )(1, 2) ∈ R ⇒ (2, 2) ∈ R
Question 11.
Show that the relation R in the set A of points in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the point Q from the origin}, is an equivalence relation. Further, show that the set of all points related to a point P ≠ (0, 0) is the circle passing through P with origin as centre.
Solution:
Let O be the origin
Then R = {(P, Q) : OP = OQ}
Reflexive
Let P be a point in the plane Then OP = OP
⇒ (P, P) ∈ R for all P
∴ R is reflexive
Symmetric
Let (P, Q) ∈ R
⇒ OP = OQ
⇒ OQ = OP
⇒ (Q, P) ∈ R
∴ R is symmetric.
Transitive
Let (P, Q), (Q, S) ∈ R
⇒ OP = OQ and OQ = OS
⇒ OP = OS
⇒ (P, S) ∈ R
∴ R is transitive
Hence R is an equivalence relation.
Let P ≠ (0,0) be a point in the plane. Consider the circle with centre at origin and radius OP. Then the set of points on this circle are related to P since the distance from the origin to any point on the circle is OP.
Hence the set of points related to P is the circle passing through P with origin as centre.
Question 12.
Show that the relation R defined in the set A of all triangles as R = {(T1, T2) : T1 is similar to T2}, is an equivalence relation. Consider three right angled triangles T1 with sides 3,4, 5, T2 with sides 5, 12, 13 and T3 with sides 6, 8,10. Which triangles among T1, T1 and T3 are related?
Solution:
R = {(T1, T2): T1 is similar to T2
Reflexive
Let T ∈ A
Since T is similar to T, (T, T) ∈ R
∴ R is reflexive
Symmetric
Let (T1,T2) ∈ R
⇒ T1 is similar to T2
⇒ T2 is similar to T1
⇒ (T2,T1) ∈ R
∴ R is symmetric
Transitive
Let (T1, T2), (T2, T3) ∈ R
⇒ T1 is similar to T2 and T2 is similar to T3
⇒ T1 is similar to T3
⇒ (T1, T3) ∈ R
∴ R is transitive
Hence R is an equivalence relation.
Two triangles are similar if their sides are proportional. The sides 3, 4, 5 of triangle T1 is proportional to the sides 6, 8, 10 of triangle T3
($$\frac { 3 }{ 6 }$$ = $$\frac { 4 }{ 8 }$$ = $$\frac { 5 }{ 10 }$$)
i. e., the sides of and T1 are proportional. Hence T1 is related to T3.
Question 13.
Show that the relation R defined in the set A of all polygons as R = {(P1, P2): P1 and P2 have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3, 4 and 5?
Solution:.
R = {(P1, P2): P1 and P2 have same number of sides}
Reflexive
Let P ∈ A
Since P and P have same number of sides,
(P, P) ∈ R for all P.
∴ R is reflexive
Symmetric
Let (P1, P2) ∈ R
⇒ Number of sides of P1 = Number of sides of P2
⇒ Number of sides of P1 = Number of sides of P1
⇒ (P2, P1) ∈ R
∴ R is symmetric
Transitive
Let (P1, P2), (P2, P3) ∈ R
⇒ P1, P2 have same number of sides and P2, P3 have same number of sides.
⇒ P1, P3 have same number of sides
⇒ (P1, P3) ∈ R
∴ R is transitive
Hence R is an equivalence relation.
The right triangle with sides 3, 4, 5 is a polygon having 3 sides.
∴ T he set of elements of A related to T is the set of triangles in the plane.
Question 14.
Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1, L2) : L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.
Solution:.
Reflexive
(L1, L1) ∈R since the line L1 is parallel to itself.
∴ R is reflexive
Symmetric
(L1, L2) ∈ R ⇒ L1 is parallel to L2
⇒ L2 is parallel to L1 ⇒ (L2, L1) ∈ R
∴ R is symmetric.
Transitive
(L1, L2) ∈R, (L2, L3) ∈ R
L1, L2 are parallel and L2, L3 are parallel
⇒ L1 and L3 are parallel. ⇒ (L1, L3) ∈ R
∴ R is transitive.
Hence R is an equivalence relation.
The set of all lines related to the line y = 2x + 4 is the line y = 2x + c where C ∈ R.
Question 15.
Let R be the relation in the set (1, 2, 3, 4} given by R = {(1, 2), (2, 2), (1, 1), (4, 4), (1, 3) , (3, 3), (3, 2)}. Choose the correct answer.
a. R is reflexive and symmetric but not transitive.
b. R is reflexive and transitive but not symmetric.
c. R is symmetric and transitive but not reflexive.
d. R is an equivalence relation.
Solution:
b. R is reflexive and transitive but not symmetric.
(1, 2) ∈ R but (2, 1) ∉ R
∴ R is not symmetric
Question 16.
Let R be the relation in the set N given by R = {{a, b) : a = b – 2, b > 6}. Choose the correct answer.
a. (2, 4) ∈ R
b. (3, 8) ∈ R
c. (6, 8) ∈ R
d. (8, 7) ∈ R
Solution:
Ans. c
If b = 8, a = 6 – 2 = 8 – 2 = 6
∴ (6, 8) ∈ R is correct.
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Have you ever wondered why rounding numbers is important? From calculating the cost of groceries to estimating distances, rounding plays a crucial role in our daily lives. In this blog post, we’ll dive into the art of rounding to the nearest tenth, equipping high school students with the skills they need to tackle real-life calculations with ease. So, let’s dig in and master the art of rounding!
## Understanding Decimal Places
Before we delve into rounding, it’s essential to understand the concept of decimal places. In a decimal number, each digit has a specific place value, ranging from the ones place to the right of the decimal point. For example, in the number 3.25, the digit 3 is in the ones place, 2 is in the tenths place, and 5 is in the hundredths place.
## The Rule of Five for Rounding
To round numbers effectively, we follow the rule of five. When rounding to the nearest tenth, we analyze the digit to the right of the tenths place. If it is five or greater, we round up; if it is four or less, we round down. For example, if we have the number 3.45, we would round it down to 3.4. However, if we have the number 3.55, we would round it up to 3.6.
## Step-by-Step Guide to Rounding
Let’s break down the process of rounding numbers to the nearest tenth into simple steps:
1. Identify the target digit, which is the digit in the tenths place.
2. Look at the digit to the right of the target digit.
3. Apply the rule of five: If the digit is five or greater, round the target digit up. If it is four or less, keep the target digit as is.
4. Adjust the numbers to the right of the target digit to zero.
Let’s practice this step-by-step guide with some examples and exercises.
## Practical Examples and Exercises
1. Example: Round 7.83 to the nearest tenth.
– The target digit is 8.
– The digit to the right is 3, which is less than 5.
– Round the target digit down.
– The rounded number is 7.8.
2. Exercise: Round 9.46 to the nearest tenth.
– The target digit is 4.
– The digit to the right is 6, which is greater than 5.
– Round the target digit up.
– The rounded number is 9.5.
By practicing these examples and exercises, you’ll become more comfortable with rounding to the nearest tenth and gain confidence in your calculations.
## Common Mistakes and How to Avoid Them
As with any skill, there are common mistakes to be aware of while rounding. Here are a few tips to help you avoid them:
1. Don’t forget to consider the digit to the right of the target digit. This digit determines whether you round up or down.
2. Be cautious when rounding numbers that end in 5. Remember, if the digit to the right is 5, you round up.
3. Double-check your final answer to ensure it matches the given place value.
By keeping these tips in mind, you’ll be able to round numbers accurately and avoid common errors.
## Conclusion
Congratulations! You’ve mastered the art of rounding to the nearest tenth. Understanding decimal places, applying the rule of five, and following a step-by-step guide will empower you to tackle real-life calculations with ease. Whether you’re calculating distances, estimating costs, or solving mathematical problems, rounding to the nearest tenth will be a valuable skill in your toolkit.
Remember, practice makes perfect. Continuously challenge yourself with more examples and exercises to solidify your understanding. With time and dedication, rounding numbers will become second nature, and you’ll navigate the world of calculations with confidence.
So go forth, high school students, and conquer the realm of rounding to the nearest tenth! Embrace the power of precision and accuracy in your everyday calculations. Happy rounding! |
# Balancing Redox Equations Using Oxidation Numbers Worksheet
A Rational Phone numbers Worksheet can help your youngster become more familiar with the concepts powering this ratio of integers. Within this worksheet, students will be able to remedy 12 different troubles related to rational expressions. They may learn to increase two or more figures, group them in couples, and figure out their products and services. They will likely also training simplifying logical expressions. When they have learned these principles, this worksheet will be a important resource for advancing their reports. Balancing Redox Equations Using Oxidation Numbers Worksheet.
## Logical Phone numbers can be a ratio of integers
There are two forms of figures: rational and irrational. Logical numbers are defined as entire figures, while irrational phone numbers tend not to perform repeatedly, and also have an limitless amount of numbers. Irrational phone numbers are no-zero, low-terminating decimals, and rectangular origins that are not excellent squares. These types of numbers are not used often in everyday life, but they are often used in math applications.
To establish a realistic amount, you need to realize such a realistic quantity is. An integer is a total number, along with a rational number can be a ratio of two integers. The proportion of two integers will be the amount on top split through the variety on the bottom. If two integers are two and five, this would be an integer, for example. There are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They could be created into a fraction
A realistic variety carries a numerator and denominator which are not absolutely no. This means that they can be conveyed being a small percentage. Along with their integer numerators and denominators, realistic figures can furthermore have a bad value. The bad benefit should be located left of and its definite value is its extended distance from absolutely no. To make simpler this case in point, we will say that .0333333 is a portion that could be written like a 1/3.
As well as bad integers, a reasonable quantity can be manufactured right into a fraction. As an example, /18,572 can be a rational number, when -1/ will not be. Any small percentage composed of integers is rational, given that the denominator will not contain a and will be written as being an integer. Also, a decimal that ends in a position can be another realistic variety.
## They make sense
Despite their label, realistic numbers don’t make very much feeling. In mathematics, these are individual entities having a special size in the number line. Consequently when we add up something, we are able to buy the size and style by its proportion to the authentic number. This keeps true regardless if there are actually limitless realistic amounts in between two particular amounts. If they are ordered, in other words, numbers should make sense only. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
In real life, if we want to know the length of a string of pearls, we can use a rational number. To obtain the time period of a pearl, as an example, we might matter its breadth. One particular pearl weighs in at 10 kgs, which is actually a reasonable amount. Additionally, a pound’s weight is equal to ten kgs. Thus, we should certainly divide a pound by 15, with out worry about the size of an individual pearl.
## They could be indicated as being a decimal
You’ve most likely seen a problem that involves a repeated fraction if you’ve ever tried to convert a number to its decimal form. A decimal amount could be created being a numerous of two integers, so four times 5 is the same as seven. A similar problem involves the frequent small percentage 2/1, and both sides should be divided up by 99 to obtain the right solution. But how would you make the conversion? Below are a few examples.
A reasonable variety may also be written in great shape, including fractions along with a decimal. A good way to signify a rational quantity inside a decimal is usually to divide it into its fractional equivalent. There are actually 3 ways to separate a rational amount, and all these ways yields its decimal counterpart. One of these brilliant methods is always to break down it into its fractional equal, and that’s what’s known as the terminating decimal. |
In this video you are going to learn how
to write numbers in scientific notation and in standard form. Alright let’s do
this! hello-hello Melissa Maribel here and I help
students like you understand what you just learned in class so you stress less
and you graduate faster. Hit that subscribe button and we’ll pass
chemistry together. On to scientific notation. Scientific notation is just an
easier way to write really small or really large numbers. You will have two
different forms, standard form is just the expanded form of that number or the
number itself and scientific notation is just the condensed form of that same
number. The proper set up for scientific notation, you will always have one
non-zero number followed by a decimal then after that decimal you will have
one or more numbers. Doesn’t matter if it’s 0 or not and that would have been
times 10 to the some sort of exponent. So you’ll see for example we have 2.30×10 to the 4th. That is an example of your proper scientific notation. If
you wanted to then write a really really small number into scientific notation,
remember that all small numbers will have negative exponents when you put
them in scientific notation. So all numbers less than 1 have negative
exponents. Let’s try one. Write 0.0007211 in scientific notation. This is your clue
this is a very small number or it’s less than 1. So it will have a negative
exponent. What we’re going do, take that decimal place and we’re going to
move it all the way to that 7 because that is your first non-zero number. So
moving this over 1, 2, 3, & 4 putting our decimal place there, we know
that we will have a negative 4 as our exponent because that’s how many places
we move to get to that first nonzero number.
So our proper scientific notation is then 7.211×10 to the negative
4th. When we’re writing numbers that are already in scientific notation and going
back to standard form that exponent tells you whether or not
to move to the right or to the left. So if it’s negative you’re actually going
to move to the left and if it’s positive you’re gonna move to the right. 8.72×10 to -3rd we’re gonna write that in standard form. You’ll see that we have a -3 as our exponent so a -3 recall that tells us it’s going to be a
number that is less than one so we’re going to move the decimal place to the
left three. So I’m going to move this over one two and three and my decimal
place is here. I’m going to place a zero wherever I had any missing spaces and
our final answer is 0.00872. When you’re converting
really large numbers into scientific notation large numbers have positive
exponents. All numbers larger than one have positive exponents. For example
let’s write this ridiculously large number into scientific notation. So
clearly this is larger than 1 so we know it will have a positive exponent.
Your decimal place is always after the 0 or always after that last value
since there wasn’t initially a decimal place and we’re going to move this all
the way over to this 9 because that’s our first non-zero number. So I’m going
to just count one, two, three, four, five, six and seven. So our setup will be 9.0 and then following with all of the numbers that we had before times
10 to the positive 7. Seven because we move that decimal place over seven
times. Let’s write 1.489×10 to the 5th in standard form. Remember standard form now, is just the
expanded form of this number. That 5 tells us it’s a very large number. So
from that decimal place we’re going to actually move over to the right five
times. One, two, three, four, five, placing that decimal place here which
really it’s gone. This would have been zero and zero. The reason I say that that
decimal place is gone is because it’s now a really large number and there’s no
need to put a decimal place since there is no numbers following. So this would be
our correct number in standard form. Sometimes you have numbers that look
like they’re in scientific notation but they actually have two numbers in front
of that decimal place and we know that that’s not correct. So how do you change
them back? If you’re trying to move that decimal place to the left, you will add
how many decimal places you’re moving to that exponent. Now if you’re going the
other way if you’re going to the right, you’re actually going to subtract. I know
it’s always the opposite of what you’re thinking or what you’re used to, but
let’s try some. Though this number looks like it’s in scientific notation, it’s
not because remember we have to have one non-zero number in front of that decimal
place but instead we have two. So we have to move this decimal place over one,
whenever you move over one to the left you will add 1 to that exponent. So
instead we have 1.2409×10^-5 because we added a 1 to that exponent. Now we’re trying to move this decimal
place over one. When we move that over once to the right we will subtract this
exponent by one making it a negative four so when we move over to the left
you add and when we move to the right we subtract. Now it’s just you me and these
practice problems. For practice problems with
step-by-step answers, check the description box. Also let me know if you
want to see more “How to pass Chemistry” videos by liking and subscribing. And
see you next week.
## 18 thoughts on “Scientific Notation and Standard Form Explained with Practice Problems | How to Pass Chemistry”
1. Blue Cloud says:
Hi Melissa. To answer your question i'm studying chemistry because it's a very interesting subject. All these chemicals get mixed together and BOOM you get a change in colour…explosion…a gas etc. I'm thinking about taking a (BS) degree in chemistry. The thing is i'm not really sure about it because as you already know i have to be very good at maths and calculations. Maths has always been a problem for me. I have to look over and over again to understand something. I know it will cause a huge problem for me specially when it comes to practical chemistry. I'm not even sure if chemistry is the right major for me. I'm so confused and my mind is all over the place. University is such a big step for me and i don't know what to do. Sorry for the very long comment, but i have no one to talk to about this since i'm homeschooled.
2. Cameron Igar says:
Hi Melissa, I am studying Chemistry because I have to take it in High School LOL. I am currently a Junior. Chemistry is not my favorite subject at all-mostly because it's science and math combine- but your videos are the first ones I go to when I feel stuck. They are EXTREMELY helpful! Can you do a video on electron configurations sometime in the future? Would be very much appreciated:)
3. Oliver Jones says:
Thank you so much! I missed some school and have tons of work to catch up on and your channel (especially this video) Is so much help!
4. arcelia goldsberry says:
Can you do equations with scientific notation with the equation having subtracting an addition or multiplication with addition
5. msapril yang says:
MELISSA! I'm taking chemistry this semester and I'm not so good with math. I hope I pass this course using your helpful videos. If I don't pass then I believe Nursing ain't for me. ๐
6. Florence Quintana says:
Thank you so much! I relied on calculators so much but you make it so easy to learn without using a Calculator Thanks! Do you have a channel on how to convert metric conversions for volume, length, mass and height?
7. Destiny F. says:
You're awesome, God has really blessed you with an amazing gift!!!
8. kenia ojeda says:
thank you! i was struggling to understand this and now it makes so much sense:)
9. Jacob Rabon says:
kenz and destiny yall are correct…love u melissa
10. Study Time says:
this. i need!
11. Midnighto _ says:
We are using this video to study chemistry ??
12. Breanna Ricketts says:
This helped a lot!
13. Kumar Siddhartha says:
nice explanation keep making more video like this and thank you for your effort.
14. Gaurav Saharan says:
You looking so good
15. Dryer Machine says: |
# Solving Proportions With Mixed Numbers
Save
Next Video:
Solving Proportions....5
In order to make solving proportions with mixed numbers easier, simply turn the mixed number into an improper fraction. Solve proportions with mixed numbers by using cross multiplication with help from a math teacher in this free video on proportions in math.
Part of the Video Series: Fractions & Proportions
Promoted By Zergnet
## Video Transcript
So, how do you solve proportions with mixed numbers? Doesn't that make the problem that much harder? Hi, I'm Jimmy Chang. I've been teaching college math for nine years, and I'm here to address that. When you put mixed numbers into a proportion problem it does add another layer to the problem, but really, not by much. What you have to keep in mind, though, is if you turn the mixed number into an improper fraction it makes the problem actually that much easier, so here's an example. Suppose you have 1and a half over 2 equal to x over 3. Now, obviously you have a mixed number in the form of 1 and a half. You definitely don't want to solve it with it at a mixed number form. You definitely want to turn this into an improper fraction, so as long as you remember how to do that you'll be good to go. So remember, take the denominator and multiply it with the whole number, so 2 times 1 is 2, and then you add the numerator, so 1 and a half is really 3 halves. So, you have three halves over 2 equals 2x over 3. Now, where you go from there is you actually solve it the same way like you would any proportion problem, and that is by cross multiplication. Take the 2, multiply it with the x and you will have 2x, and then take the 3 and multiply it by the 3 halves. Now, 3 times 3 halves; that will give you 9 over 2 equals to 2x. Now, your objective then is to solve for x, and because the 2 is multiplied with the x just divide both sides by 2. Now, you have x on one side, but the thing is to figure out 9 halves divided by 2. Since you're dividing by a whole number and you have a fraction on top give this a fraction as well, so you have 9 over 2 divided by 2 over 1. So, as a result you'll have 9 halves, and remember from arithmetic rules if you're dividing by 2 over 1 you have to multiply by the reciprocal, one half, so you have x is equal to whatever 9 halves times 1 half is, and you will get 9 over 4 is equal to x, and then, that is your final answer. So, mixed numbers and proportion problems; they do make the problem a little bit longer, but not much more difficult than that. As long as you convert it to a improper fraction mixed numbers are just like any other problem. So, I'm Jimmy Chang, and that is how you solve proportion problems using mixed numbers.
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# How much should I scale $dx$ and $dy$ individually to get a vector of required magnitude
I have a $dx$ and a $dy$ and I need to create a vector of magnitude $35.5$ in that $(dx, dy)$ direction. How much should I scale $dx$ and $dy$?
Scale by $35.5 / \sqrt{dx^2 + dy^2}$.
Example: if $dx = 1$ and $dy = 2$, then the denominator is $\sqrt{5} \approx 2.236$, so you'd multiply each of $dx$ and $dy$ by $35.5/2.236 \approx 15.88$ to get approximately $$[15.88, 31.75]$$ as your new vector.
The length of the original vector will be $\sqrt{dx^2+dy^2}$, which I call $L$. Suppose you want the vector to be of length $M$.
Then you need to scale the components by $\dfrac{M}{L}$.
Example. Suppose you have $dx=3$ and $dy=-4$ and you want the vector to have length $M=10$.
The length of the vector currently is $L=\sqrt{3^2+(-4)^2}=5$. Thus, we need to scale them components by $\dfrac{10}{5}=2$.
The new vector would be $dx'=3\times2=6$ and $dy'=-4\times2=-8$.
So you get basically...: $\vec{v} = dx\vec{i} + dy\vec{j}$ , and $||\vec{v}|| = \sqrt{(dx)^2+(dy)^2} = 35.5$. Now let $dx = kdx', dy = kdy'$, then $k = 35.5$, and that $\vec{v'} = dx'\vec{i} + dy'\vec{j}$ is the unit vector in the direction of $\vec{v}$, and $||\vec{v'}|| = 1$ |
# Balls in an Urn
• Oct 30th 2009, 11:26 AM
Aquafina
Balls in an Urn
Twenty balls are placed in an urn. Five are red, five green, five yellow and five blue. Three balls are drawn from the urn at random without replacement. Write down expressions for
the probabilities of the following events. (You need not calculate their numerical values.)
a) Exactly one of the balls drawn is red.
b) The three balls drawn have different colours.
c) The number of blue balls drawn is strictly greater than the number of yellow balls
drawn.
• Oct 30th 2009, 02:24 PM
Soroban
Hello, Aquafina!
Quote:
Twenty balls are placed in an urn: 5 red, 5 green, 5 yellow, 5 blue.
Three balls are drawn from the urn at random without replacement.
There are: . ${20\choose3} \,=\,1140$ possible outcomes.
Quote:
Write expressions for the probabilities of the following events.
(You need not calculate their numerical values.)
a) Exactly one of the balls drawn is red.
There are: 5 Reds and 15 Others.
We want 1 Red and 2 Others. .There are: . ${5\choose1}{15\choose2} \:=\:525$ ways.
. . Therefore: . $P(\text{exactly 1 Red}) \:=\:\frac{525}{1140} \:=\:\frac{35}{76}$
Quote:
b) The three balls drawn have different colours.
Select 3 of the 4 colors: . ${4\choose3} \:=\:4$ ways.
To pick one of each of the colors, there are: . ${5\choose1}{5\choose1}{5\choose1} \:=\:125$ ways.
Hence, there are: . $4\cdot125 \:=\:500$ ways to get 3 colors.
Therefore: . $P(\text{3 colors}) \:=\:\frac{500}{1140} \:=\:\frac{25}{57}$
Quote:
c) The number of blue balls drawn is strictly greater than the number of yellow balls drawn.
There are two cases to consider:
. . [1] Two Blue and one Yellow: . ${5\choose2}{5\choose1} \:=\:50$ ways
. . [2] Three Blue: . ${5\choose3} \:=\:10$ ways
Hence, there are: . $50 + 10 \:=\:60$ ways to have more Blue than Yellow.
. . Therefore: . $P(\text{Blue} > \text{Yellow}) \:=\:\frac{60}{1140} \;=\;\frac{1}{19}$
• Oct 30th 2009, 11:19 PM
Aquafina
Quote:
Originally Posted by Soroban
Hello, Aquafina!
There are: . ${20\choose3} \,=\,1140$ possible outcomes.
There are: 5 Reds and 15 Others.
We want 1 Red and 2 Others. .There are: . ${5\choose1}{15\choose2} \:=\:525$ ways.
. . Therefore: . $P(\text{exactly 1 Red}) \:=\:\frac{525}{1140} \:=\:\frac{35}{76}$
Select 3 of the 4 colors: . ${4\choose3} \:=\:4$ ways.
To pick one of each of the colors, there are: . ${5\choose1}{5\choose1}{5\choose1} \:=\:125$ ways.
Hence, there are: . $4\cdot125 \:=\:500$ ways to get 3 colors.
Therefore: . $P(\text{3 colors}) \:=\:\frac{500}{1140} \:=\:\frac{25}{57}$
There are two cases to consider:
. . [1] Two Blue and one Yellow: . ${5\choose2}{5\choose1} \:=\:50$ ways
. . [2] Three Blue: . ${5\choose3} \:=\:10$ ways
Hence, there are: . $50 + 10 \:=\:60$ ways to have more Blue than Yellow.
. . Therefore: . $P(\text{Blue} > \text{Yellow}) \:=\:\frac{60}{1140} \;=\;\frac{1}{19}$
Hi I got the same answers for i) and ii) but not iii). First of all, can you explain your method in ii?
I basically did:
P(diff colours) = 5/20 * 5/19 * 5/18 * 3! * 4C3
I get why you multiplied by 4, thats my 4C3, but how do you get 125 ways?
for iii)
I did P(1 blue, no yellow) as well
Also, P(2 blue, no yellow)
• Oct 31st 2009, 01:07 AM
ANDS!
Quote:
I get why you multiplied by 4, thats my 4C3, but how do you get 125 ways?
You have 5 reds. Each of the 5 reds can be paired with 5 greens lets say. So you have 25 Red-Green combos. Each of the 25 Red-Green combos can be paired with one of the 5 yellows. So you have 125 Red-Green-Blue combos. But lets get even more general than that. You have 125 Color1-Color2-Color3 combos. Now if you only had three colors to choose from, this problem would be done. But you don't have only three colors, you have 4 - so you need to know how many unique combinations of colors you have. So, you have:
Red-Green-Blue, Red-Green-Yellow, Red-Blue-Yellow, and Green-Yellow-Blue. Each of those three color combinations has 125 different arragements, because as we established above, there are 125 ways of arranging Color1-Color2-Color3. Therefore there are 500 total ways of arranging 3 out of 4 colors (where there are 5 unique objects of each color).
• Oct 31st 2009, 03:43 AM
Aquafina
Quote:
Originally Posted by ANDS!
You have 5 reds. Each of the 5 reds can be paired with 5 greens lets say. So you have 25 Red-Green combos. Each of the 25 Red-Green combos can be paired with one of the 5 yellows. So you have 125 Red-Green-Blue combos. But lets get even more general than that. You have 125 Color1-Color2-Color3 combos. Now if you only had three colors to choose from, this problem would be done. But you don't have only three colors, you have 4 - so you need to know how many unique combinations of colors you have. So, you have:
Red-Green-Blue, Red-Green-Yellow, Red-Blue-Yellow, and Green-Yellow-Blue. Each of those three color combinations has 125 different arragements, because as we established above, there are 125 ways of arranging Color1-Color2-Color3. Therefore there are 500 total ways of arranging 3 out of 4 colors (where there are 5 unique objects of each color).
Thanks.
for iii)
I did P(1 blue, no yellow) as well
Also, P(2 blue, no yellow)
Why aren't these included in the answer given in the post above?
• Oct 31st 2009, 05:22 AM
ANDS!
Soroban answered that for you - just labeled it part C. Well I mean he/she is using the labels you are - lol. But it's there. Just add your probabilities and you will get the same answer they did.
• Oct 31st 2009, 05:44 AM
Aquafina
Soroban only considered these 2 cases:
Quote:
Originally Posted by Soroban
There are two cases to consider:
. . [1] Two Blue and one Yellow: . ${5\choose2}{5\choose1} \:=\:50$ ways
. . [2] Three Blue: . ${5\choose3} \:=\:10$ ways
Hence, there are: . $50 + 10 \:=\:60$ ways to have more Blue than Yellow.
. . Therefore: . $P(\text{Blue} > \text{Yellow}) \:=\:\frac{60}{1140} \;=\;\frac{1}{19}$
Not the ones I mentioned as well. When I had worked the answer out, I included these 2 cases, along with the cases I mentioned to get the answer.
I want to know why those are not needed.
• Oct 31st 2009, 08:25 AM
ANDS!
I think I understand your question. You are wondering why he doesn't consider you drawing 1 or 2 blue balls and the rest red or green? You are right. You would need to take those draws into consideration (assuming that haven't stated that its a given red and green balls WERENT drawn).
• Oct 31st 2009, 10:23 AM
Aquafina
Quote:
Originally Posted by ANDS!
I think I understand your question. You are wondering why he doesn't consider you drawing 1 or 2 blue balls and the rest red or green? You are right. You would need to take those draws into consideration (assuming that haven't stated that its a given red and green balls WERENT drawn). |
# Divisibility Rules for 6 - PowerPoint PPT Presentation
1 / 7
Divisibility Rules for 6. The divisibility rule for 6 is as follows: If the number can be divided by BOTH 2 and 3 Ex. Determine whether the number 876 453 is divisible by 6 Solution: the last digit is not even, so the number is not divisible by 2. 8+7+6+4+5+3=33 3+3=6 6/2=3
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Divisibility Rules for 6
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## Divisibility Rules for 6
• The divisibility rule for 6 is as follows:
• If the number can be divided by BOTH 2 and 3
• Ex. Determine whether the number 876 453 is divisible by 6
• Solution:
• the last digit is not even, so the number is not divisible by 2.
• 8+7+6+4+5+3=33
• 3+3=6
• 6/2=3
• Since 3 is a factor of 33, the number is divisible by 3
• Therefore, 6 is not a factor of 876 453 because 2 is not a factor
### Example #1
• Determine whether the number 839 729 384 192 is divisible by 6.
• Solution:
• 8+3+9+7+2+9+3+8+4+1+9+2= 65
• 6+5=11
• Therefore, 6 is not a factor of 839 729 384 192 because 3 is not a factor
### Example#2
• Determine whether the number 45 672 978 is divisible by 6
• Solution:
• 4+5+6+7+2+9+7+8=48
• 4+8= 12
• 12/3=4
• Therefore, 6 is a factor of 45 672 978 because 2 and 3 are factors.
• Explain whether the number 186 is divisible by 6.
• 1+8+6= 15
• 1+5=6
• 6/3=2
• Yes it is.
• Explain whether the number 283 945 is divisible by 6.
• 2+8+3+9+4+5= 31
• 3+1=4
• 4/3
• No
### Practice
• Page 13 #1
• STOP and CHECK
• Continue onto #3, 5, 6, 7, 8 |
Motion in two dimensions
9-15-99
Sections 3.5 - 3.7
Extending things from 1 dimension
In 1 dimension, we wrote down some general equations relating velocity to displacement, and relating acceleration to the change in velocity. We also wrote down the four equations that apply in the special case where the acceleration is constant. We're going to do the same thing in 2 dimensions, and the equations will look similar; this shouldn't be surprising because, as we will see, a two (or three) dimensional problem can always be broken down into two (or three) 1-dimensional problems.
When we're dealing with more than 1 dimension (and we'll focus on 2D, but we could use these same equations for 3D), the position is represented by the vector r. The velocity will still be represented by v and the acceleration by a. In general, the average velocity will be given by:
The instantaneous velocity is given by a similar formula, with the condition that a very small time interval is used to measure the displacement.
A similar formula gives the average acceleration:
Again, the instantaneous acceleration is found by measuring the change in velocity over a small time interval.
The constant acceleration equations
When the acceleration is constant, we can write out four equations relating the displacement, initial velocity, velocity, acceleration, and time for each dimension. Like the 1D equations, these apply under the following conditions:
1. the acceleration is constant
2. the motion is measured from t = 0
3. the equations are vector equations, but the variables are not normally written in bold letters. The fact that they are vectors comes in, however, with positive and negative signs.
If we focus on two dimensions, we get four equations for the x direction and four more for the y direction. The four x equations involve only the x-components, while the four y equations involve only the y-components.
One thing to notice is that the time, t, is the only thing that doesn't involve an x or a y. This is because everything else is a vector (or a component of a vector, if you'd rather look at it that way), but time is a scalar. Time is the one thing that can be used directly in both the x and y equations; everything else (displacement, velocity, and acceleration) has to be split into components.
This is important!
Something that probably can't be emphasized enough is that even though an object may travel in a two-dimensional path (often following a parabola, in the standard case of an object moving under the influence of gravity alone), the motion can always be reduced to two independent one-dimensional motions. The x motion takes place as if the y motion isn't happening, and the y motion takes place independent of whatever is happening in the x direction.
One good example of this is the case of two objects (e.g., baseballs) which are released at the same time. One is dropped so it falls straight down; the other is thrown horizontally. As long as they start at the same height, both objects will hit the ground at the same time, no matter how fast the second one is thrown.
What we're ignoring
We will generally neglect the effect of air resistance in most of the problems we do. In some cases that's just fine. In other cases it's not so fine. A feather and a brick dropped at the same time from the same height will not reach the ground at the same time, for example. This has nothing to do with the weight of the feather compared to the brick. It's simply air resistance; if we took away all the air and dropped the feather and brick, they would hit the ground at exactly the same time.
So, remember that we're often analyzing ideal cases, especially this early in the semester. In reality, things might be a little different because of factors we're neglecting at the moment.
An example
Probably the simplest way to see how to apply these constant acceleration equations is to work through a sample problem. Let's say you're on top of a cliff, which drops vertically 150 m to a flat valley below. You throw a ball off the cliff, launching it at 8.40 m/s at an angle of 20° above the horizontal.
(a) How long does it take to reach the ground?
(b) How far is it from the base of the cliff to the point of impact?
It's a good idea to be as systematic as possible when it comes to analyzing the situation. Here's an example of how to organize the information you're given. First, draw a diagram.
Then set up a table to keep track of everything you know. It's important to pick an origin, and to choose a coordinate system showing positive directions. In this case, the origin was chosen to be the base of the cliff, with +x being right and +y being up. You don't have to choose the origin or the positive directions this way. Pick others if you'd like, and stick with them (an origin at the top of the cliff, and/or positive y-direction down would be two possible changes).
Now that everything's neatly organized, think about what can be used to calculate what. You know plenty of y-information, so we can use that to find the time it takes to reach the ground. One way to do this (definitely not the only way) is to do it in two steps, first calculating the final velocity using the equation:
This gives vy2 = 2.8732 + 2 (-9.8) (-150) = 2948.3 m2 / s2 . Taking the square root gives: vy = +/- 54.30 m/s.
Remember that the square root can be positive or negative. In this case it's negative, because the y-component of the velocity will be directed down when the ball hits the ground.
Now we can use another equation to solve for time:
So, -54.30 = 2.873 - 9.8 t, which gives t = 5.834 seconds. Rounding off, the ball was in the air for 5.83 s.
We can use this time for part (b), to get the distance traveled in the x-direction during the course of its flight. The best equation to use is:
So, from the base of the cliff to the point of impact is 46.0 m. |
Ohio Resource Center
Lessons
Comparing Connecting Cubes 4: Balancing
Discipline
Mathematics
Grades
Kindergarten, 1, 2
Share This Resource
Professional Commentary
This lesson features two activities using connecting cubes on a pan balance to model subtraction. Students may also enjoy the pan balance applet linked to the site. The resource includes a list of children's counting stories, recording sheets, discussion questions, suggestions for assessment, and questions for teacher reflection. (sw)
Common Core State Standards for Mathematics
Kindergarten
Operations and Algebraic Thinking
Understand addition as putting together and adding to, and understand subtraction as taking apart and taking from.
K.OA.A.1
Represent addition and subtraction with objects, fingers, mental images, drawings, sounds (e.g., claps), acting out situations, verbal explanations, expressions, or equations.
K.OA.A.2
Solve addition and subtraction word problems, and add and subtract within 10, e.g., by using objects or drawings to represent the problem.
Grade 1
Operations and Algebraic Thinking
Represent and solve problems involving addition and subtraction.
1.OA.A.1
Use addition and subtraction within 20 to solve word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using objects, drawings, and equations with a symbol for the unknown number to represent the problem.
Add and subtract within 20.
1.OA.C.6
Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use strategies such as counting on; making ten (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a ten (e.g., 13 – 4 = 13 – 3 – 1 = 10 – 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 – 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
Work with addition and subtraction equations.
1.OA.D.7
Understand the meaning of the equal sign, and determine if equations involving addition and subtraction are true or false. For example, which of the following equations are true and which are false? 6 = 6, 7 = 8 – 1, 5 + 2 = 2 + 5, 4 + 1 = 5 + 2.
Grade 2
Operations and Algebraic Thinking
Represent and solve problems involving addition and subtraction.
2.OA.A.1
Use addition and subtraction within 100 to solve one- and two-step word problems involving situations of adding to, taking from, putting together, taking apart, and comparing, with unknowns in all positions, e.g., by using drawings and equations with a symbol for the unknown number to represent the problem.
Add and subtract within 20.
2.OA.B.2
Fluently add and subtract within 20 using mental strategies. By end of Grade 2, know from memory all sums of two one-digit numbers.
Ohio Mathematics Academic Content Standards (2001)
Number, Number Sense and Operations Standard
Benchmarks (K–2)
H.
Model, represent and explain subtraction as comparison, take-away and part-to-whole.
Grade Level Indicators (Grade Kindergarten)
10.
Model and represent addition as combining sets and counting on, and subtraction as take-away and comparison. For example:
a. Combine and separate small sets of objects in contextual situations; e.g., add or subtract one, two, or another small amount.
b. Count on (forward) and count back(backward) on a number line between 0 and 10.
Grade Level Indicators (Grade 1)
11.
Model, represent and explain subtraction as take-away and comparison. For example: a. Model and explain subtraction using physical materials in contextual situations. b. Draw pictures to model subtraction. c. Write number sentences to represent subtraction. d. Explain that subtraction of whole numbers yields an answer smaller than the original number.
12.
Use conventional symbols to represent the operations of addition and subtraction.
Grade Level Indicators (Grade 2)
6.
Model, represent and explain subtraction as comparison, take-away and part-to-whole; e.g., solve missing addend problems by counting up or subtracting, such as "I had six baseball cards, my sister gave me more, and I now have ten. How many did she give me?" can be represented as 6 + ? = 10 or 10 - 6 = ?.
Principles and Standards for School Mathematics
Number and Operations Standard
Understand meanings of operations and how they relate to one another
Expectations (Pre-K–2)
understand various meanings of addition and subtraction of whole numbers and the relationship between the two operations; |
Question Video: Expanding a Binomial of 𝑛th Degree | Nagwa Question Video: Expanding a Binomial of 𝑛th Degree | Nagwa
# Question Video: Expanding a Binomial of 𝑛th Degree Mathematics • Third Year of Secondary School
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Expand (7 + 2𝑥)³.
04:57
### Video Transcript
Expand seven plus two 𝑥 all cubed.
In order to expand this expression, we’re going to be using the binomial theorem. So here’s the binomial theorem and we’re actually gonna have a look at how we use it. Actually, it looks quite complicated, but what it actually means is that our 𝑎 term will be descending in power and our 𝑏 term is going to be ascending in powers. I’m gonna look at how that works now.
But before I do, I wanna quickly draw your attention to this part which is a binomial coefficient. And what this tells us is what each term in our expansion is actually gonna be multiplied by, and it’s gonna be worked out in this way. But this shows us that our binomial coefficient is equal to 𝑛 factorial over 𝑛 minus 𝑘 factorial multiplied by 𝑘 factorial.
And a factorial is the product of all positive integers less than or equal to 𝑛. So I’m gonna show you a quick example of how we actually find a binomial coefficient. Okay, so here’s an example just using that formula above are for the binomial coefficient four, two. So it’s got four factorial, which is four multiplied by three multiplied by two multiplied by one, over 𝑛 minus 𝑘 factorial. So in this case that’s going to be four minus two factorial, so that’s two factorial, so two multiplied by one, and then multiplied by 𝑘 factorial.
So in this case, it’s two factorial so that’s two multiplied by one, which gives us a binomial coefficient of six. It’s also though worth mentioning this other notation, 𝑛 𝐶 𝑘, or on some calculators, 𝑛 𝐶 𝑟. And this is just another way of representing the binomial coefficient.
So if you’re actually using a calculator, what you type in is four then your 𝑛 𝐶 𝑟 button and then two, and then what that will give you is your binomial coefficient of six. Great! So now we actually know what the binomial theorem is and the parts within it. We can now go on and expand our expression.
So we’ve got seven plus two 𝑥 all cubed is equal to, well, here’s our first term. Well we’ve got our binomial coefficient — three, zero — where we’ve got three because that is the actual power of our brackets, so that’s our 𝑛, and then we’ve got zero because that’s our term number.
What’s also worth bearing in mind is that any binomial coefficient with a zero at the bottom is always equal to one, and we have that multiplied by seven cubed. And then the next part — which we actually probably wouldn’t usually write down, but I’ve just put it in to show you what actually happens — is we’ve got two 𝑥, our two 𝑥 term, to the power of zero. Well anything to the power of zero is just one so it says that it’d just be like multiplying it by one.
And that’s because our powers begin at zero and then they work their way up. And then we can see our next term where, as you can see, the two 𝑥 has had its powers raised. It’s now just two 𝑥 or two the power of one and our seven term has actually decreased in power since instead of being seven cubed, it’s now seven squared.
And the same with our next term, our power of seven has decreased yet again so it’s now just seven, seven to the power of one, and our power of two 𝑥 has increased so it’s now two 𝑥 squared, which then brings us to our final term — which, again, I wouldn’t usually write this, but I’m just trying to show you what’s happening. We’d have seven to the power of zero, which again will be one, and then we’d have two 𝑥 cubed.
And then by either using a calculator or the formula we have here for binomial coefficient, we can actually simplify and work out each term. And then we get our first two terms, which is 343 because that’s seven cubed, plus 294𝑥. And then I draw your attention to the third term, which is plus 84𝑥 squared. Well this is actually where the most common mistakes are actually made.
Okay, so first of all, we get three, because that’s what our binomial coefficient worked out to be, multiplied by seven and then multiplied by four 𝑥 squared, not two 𝑥 squared; this is where people make the most common mistake because they forget to square the coefficient of 𝑥 as well as the 𝑥 itself.
So it’s multiplied by four 𝑥 squared, which gives us 84𝑥 squared. That then leaves us with our final term, which is eight 𝑥 cubed. So then we can say our final answer is, when we expand seven plus two 𝑥 all cubed, it is equal to eight 𝑥 cubed plus 84𝑥 squared plus 294𝑥 plus 343. And I’ve actually written it that way because convention says that we usually work in descending powers of 𝑥.
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# What is 6 divided by 7 as a fraction?
## What is 6 divided by 7 as a fraction?
6 divided by 7 equals 67 or 0.86, rounded to the nearest hundredth. When you divide, you have two numbers the dividend and the…
## What is the fraction of 42 is 6?
Steps to simplifying fractions Therefore, 42/6 simplified to lowest terms is 7/1.
How do you write 42 as a fraction?
The denominator is the bottom number of the fraction. Now we have a fraction 42/100, which can be reduced to 21/50. This means that 42% written as a fraction is 21/50.
### What is 42 7 as a fraction?
Therefore, 42/7 simplified to lowest terms is 6/1.
### What is 7 divided by 9 long?
0.7777
7 divided by 9 is 0.7777… (repeating).
What is 6 out of 7 as a percentage?
What is this? What is this? Now we can see that our fraction is 85.714285714286/100, which means that 6/7 as a percentage is 85.7143%.
#### What is the simplest form of 6 42?
The simplest form of 642 is 17.
#### How do you solve 42 divided by 6?
42 divided by 6 is 7.
What is 42 100 as a decimal?
42 / 100 = 0.42.
## What is 42 as a reduced fraction?
2150
Since 42% is basically 0.42, we can turn that into a fraction. We can do that by converting it into 42100 , andsimplifying the fraction. We can divide by 2 on the top and bottom, and we get 2150 as our answer.
## How do you write 42 divided by 7?
Using a calculator, if you typed in 42 divided by 7, you’d get 6. You could also express 42/7 as a mixed fraction: 6 0/7.
What is the lowest term of 63 9?
Reduce 63/9 to lowest terms
• Find the GCD (or HCF) of numerator and denominator. GCD of 63 and 9 is 9.
• 63 ÷ 99 ÷ 9.
• Reduced fraction: 71. Therefore, 63/9 simplified to lowest terms is 7/1.
### How are fractions divided in a fraction calculator?
The process for dividing fractions is similar to that for multiplying fractions. In order to divide fractions, the fraction in the numerator is multiplied by the reciprocal of the fraction in the denominator. The reciprocal of a number a is simply .
### Do you need a common denominator for multiplying fractions?
Multiplying fractions is fairly straightforward. Unlike adding and subtracting, it is not necessary to compute a common denominator in order to multiply fractions. Simply, the numerators and denominators of each fraction are multiplied, and the result forms a new numerator and denominator.
Which is the correct way to do fraction subtraction?
To complete an addition (or subtraction) problem, multiply the numerators and denominators of each fraction in the problem by whatever value will make the denominators 12, then add the numerators. Fraction subtraction is essentially the same as fraction addition. A common denominator is required for the operation to occur.
#### How to calculate fractions in a Big Number Calculator?
Big Number Fraction Calculator 1 Addition: Unlike adding and subtracting integers such as 2 and 8, fractions require a common denominator to undergo these operations. 2 Subtraction: Fraction subtraction is essentially the same as fraction addition. 3 Multiplication: Multiplying fractions is fairly straightforward. |
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1 answerLast reply by: Khanh NguyenFri Oct 2, 2015 9:53 PMPost by Khanh Nguyen on August 1, 2015 mm,mn uohij
### The Coordinate Plane
• The coordinate plane contains the x- and y-axes
• The x-axis and y-axis break the coordinate plane into four sections, called quadrants
• 1st quadrant: upper right
• 2nd quadrant: upper left
• 3rd quadrant: bottom left
• 4th quadrant: bottom right
### The Coordinate Plane
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• The Coordinate System 0:05
• The Coordinate Plane
• Quadrants, Origin, and Ordered Pair
• The Coordinate Plane 7:02
• Write the Coordinates for Points A, B, and C
• Extra Example 1: Graph Each Point on the Coordinate Plane 9:03
• Extra Example 2: Write the Coordinate and Quadrant for Each Point 11:05
• Extra Example 3: Name Two Points From Each of the Four Quadrants 13:13
• Extra Example 4: Graph Each Point on the Same Coordinate Plane 17:47
### Transcription: The Coordinate Plane
Welcome back to Educator.com.0000
For the next lesson, we are going to go over the coordinate plane.0002
This right here is called the coordinate plane; it consists of two lines like this.0008
This right here, the horizontal line, is called the x-axis.0016
The vertical line is the y-axis; together they make up the coordinate plane.0024
Think of the coordinate plane as like a map.0034
It is a map; we are going to place points along this map.0037
We have to label each point based on the x-axis and the y-axis.0042
That is the coordinate plane.0048
These axes, the x-axis and the y-axis, break the coordinate plane into four sections.0052
One, two, three, four, those four sections are called quadrants.0064
The first quadrant is this space right here; this is called quadrant one.0074
Going over to this side, this is quadrant two; then down, this is quadrant three.0086
The last one is quadrant four; those are the four sections of the coordinate plane.0097
Each axis is numbered; it is labelled.0108
Here, this point right here is going to start off at 0.0112
It is going to go 1, 2, 3, 4, 5, and so on; 1, 2, 3, 4, 5.0116
Again this is 0; this way, then it is going to go negative.0126
See how it goes positive this way.0133
If you go the other way, it is going to go negative; -1, -2, -3, -4, -5, and so on.0134
Same thing for the y-axis.0143
If you go upwards, it is going to go positive.0146
It is going to go 1, 2, 3, 4, 5.0151
The arrows represent continuation; it is going to keep going on forever.0158
It doesn't just stop at 5; it is going to keep going.0165
Down this way, it is going to be negative; -1, -2, -3, -4, -5.0168
Next right here, a number on the x-axis paired with the number on the y-axis is going to give us an ordered pair.0183
You are going to look for the number on the x and the number on the y.0198
Together it is going to map out your location on the coordinate plane.0202
That point is called an ordered pair; it is going to be written like this.0209
In parentheses, you are going to have two numbers.0216
The first number is going to be your x number; that is called the x-coordinate.0219
The second number is going to be your y coordinate.0231
It is always going to be x and then y.0239
X always has to come first; (x, y) for short.0242
(x, y), that is going to be your ordered pair.0247
That is going to map out your location on the coordinate plane.0252
This point where the x-axis and y-axis meet, that is called the origin.0259
That is right there.0267
If you look at the x-number, the x-number, it is 0.0269
The x-number is 0; the y-number is also 0.0275
The origin is (0, 0); that is the ordered pair for the origin.0280
If I put a point right here on the coordinate plane, that ordered pair, what is the x-number?0287
If you look, this point is paired with an x-number and a y-number.0301
This point is going to have a x-number of 3 and a y-number of 1.0309
The ordered pair for this point is (3, 1).0321
Be careful that you don't mix those up, switch them up.0324
It is not (1, 3); that is not the same thing.0328
(1, 3), if I say (1, 3), because I said 1 first, that means I am talking about 1 on the x-axis.0331
(1, 3) would be 1 on the x; 3 on the y.0341
Where do they meet?--right there, that would be (1, 3).0348
(1, 3) and (3, 1) are not the same thing; be careful with that.0352
Always make sure the first number is your x-coordinate on the x-axis, the horizontal line.0356
The second number is your y-axis, the number on your vertical line.0363
That is the coordinate plane.0369
Again we have the x-axis, the y-axis; each of the four sections are called quadrants.0370
It starts off with this right here, quadrant one, quadrant two, quadrant, three, and quadrant four.0379
Remember when you are writing out the numbers, this number is always 0.0385
It is going to go positive to the right and negative to the left.0392
On the y-axis, positive when you go up, negative when you go down.0396
Everything together, the coordinate plane which is like the map, the ordered pairs, the quadrants,0402
all of this that have to do with these points, the plane, all that is called the coordinate system.0408
It is like a system of all these things together.0418
Again a coordinate plane, each of these are just marked for you.0425
We have three points, A, B, and C.0431
Here is A; here is B; here is C; we want to write the coordinates.0435
In other words, we want to know the ordered pairs.0439
Find the ordered pairs of A, B, and C.0442
We know that this is the x and that is the y.0446
Let's start with A, this point right here.0451
What is the x-number that makes up this point?--1.0455
What is the y-number?--1.0459
For A, my x-number, the number on my x-axis, is 1.0461
The number on the y-axis is 1.0468
The ordered pair or the coordinates for A is (1, 1).0473
I am just going to write... because this is the name of the point.0480
A is what that point is labelled.0483
You can just write it in front of that ordered pair.0486
Next is B; again I am looking for the x-number first.0491
The x-number, the number on this x-axis that makes up this B is -1.0497
The y-number, -2.0505
They are negative numbers; but it is fine.0513
You are just listing them out like that, same way as when you have positive numbers, (-1, -2).0515
The next one is C; the x-number for C is 2.0525
The y-number for C is -1.0533
Those are the coordinate for A, B, and C.0540
Let's do our examples now; graph each point on the coordinate plane.0543
The first one, A is (4, 2).0549
Remember this is the x-number; this is the y-number.0554
I am going to label this as x; this as y.0559
4, you are going to look for 4 on the x-axis.0564
4; this is -4; here is 4.0567
Then look for 2 on the y-axis; here is 2.0572
Where do they meet?--right there.0575
This is A; I can label that point as A.0581
Next, B is (-3, 0); the x is -3; the y is 0.0586
-3 on the x is right here; 0 on the y.0596
Where is 0?--0 is right there.0602
(-3, 0), that means I am not moving any up or any down because there is no y.0606
If the y was 1, I would have to move up 1.0611
If y was -1, I would have to move down 1.0614
But y is 0 so I stay here; that is my B.0617
Next, x, y; 2 is x; y is 1.0629
Where do they meet?--right there; that is C.0636
Next is (-4, -2).0643
-4 on the x is right there; -2 on the y is right there.0647
Right there, they intersect at that point right there; that point is called D.0654
That is it for this example.0663
The next one, write the coordinates and quadrant for each point.0667
It is like the same thing that we just did a couple examples ago.0672
First let's start with A; remember this is my x, this is my y.0676
Look for the x that makes up this point right here.0682
A is -1; that is my x; y is a -3.0688
We also have to state what quadrant it is in.0698
Remember that this right here, this section, any points that fall in this section right here is quadrant one.0702
This section is quadrant two; quadrant three; and quadrant four.0709
If a point is on let's say right here, that doesn't fall under any quadrants.0720
That wouldn't be considered quadrant one or quadrant four.0727
In that case, you are just going to say that it is on the x-axis.0729
Or in this case, if a point lands right here.0734
Then it is not in quadrant three or four; it is actually on the y-axis.0739
This quadrant, point A is in quadrant three; let's just say quadrant three.0745
B, point B is... x is 1, y is 2; (1, 2); that is in quadrant one.0753
C is right there; 3 as my x; -1 as my y.0768
That is in quadrant four.0776
D, x is -3; y is 3; that is quadrant two.0780
Example three, name two points from each of the four quadrants.0795
It is a little bit hard to do when you don't have the coordinate plane.0801
Let's draw out a coordinate plane.0805
We don't have to draw it too big because we are not going to be plotting any points.0812
Here is the x; here is the y.0817
The trick here is to figure out whether my x-coordinates are going to be positive or negative0830
and my y-coordinates are going to be positive or negative.0840
Quadrant one, if you look here, all the points that have this as my x and this as my y,0843
any points that pair up with this part of my x, that part of my y, is going to be in quadrant one.0852
I know that here as I get to these numbers, it is going to be positive.0861
This is +1, +2, +3, so on.0866
My x-coordinate is going to be positive; it is a positive number; positive.0870
My y-coordinate is also going to be positive because these are positive numbers here; positive, positive.0878
Any ordered pair that has a positive number paired with a positive y-number is going to be in quadrant one.0888
For quadrant one, I have to name two points.0898
Any ordered pair with a positive x, positive y, is in quadrant one.0904
(1, 2), that is in quadrant one; what else?0911
(3, 4), that is also in quadrant one.0916
Look at quadrant two; quadrant two, all points with this part of my x.0920
Those are my negative numbers.0931
A negative x with this part of my y is going to be in quadrant two.0934
Then when I list my points in quadrant two,0944
it is going to be a negative x number with a positive y number.0948
Does that make sense?--it is negative.0955
All these numbers are negative here.0957
This is -1, -2, -3; it has to be a negative x-coordinate.0961
Then paired with a positive y coordinate is going to be in quadrant two.0968
(-1, +2) is right there; it is in quadrant two.0976
Again negative x, positive y; (-3, 4).0984
Quadrant three, again any numbers or any ordered pairs paired with a ?x with this part of the y-axis.0995
That is negative also; negative and a negative is going to be in quadrant three.1007
This is going to be a negative x, negative y.1014
(-1, -2) here is -1; here is -2.1020
Another one, (-3, -4).1026
For quadrant four, again positive x; how about the y?1033
It is going to be any ordered pairs paired with this side and this side.1041
Positive x, negative y; positive x, 1; negative y, -2.1045
Positive x, 3; negative y, -4.1056
Those two points are going to fall under quadrant four.1061
The fourth example, we are just going to graph more points on this coordinate plane.1069
I didn't number them; go ahead and number them right now.1075
1, 2, 3, 4, let's go up to 5; -1, -2, -3, -4, -5; 1, 2, 3, 4, 5.1078
The first point, A; I know this is going to be my x-number; this my y-number.1098
0 on my x-number is right here; on this right here, 0.1106
3 on my y-number is right here.1113
That means my y is going to be 3 and my x is going to be 0.1116
Meaning I am not going to have -1.1121
I am not going to have 1 as my x; I am going to have 0.1124
It is going to be that point right there; this is A.1126
The next one is (-2, -1); -2 on my x; -1 on my y.1132
They meet right there; here is B.1140
C is going to be -5 and then 0 as my y.1147
Remember if my y is 1, I go up 1.1154
If my y is -1, I go down 1; right there.1156
But it is 0; my y is 0; that means I don't move up or down.1161
I stay put; that is my C.1164
My last point D is going to be 4 and -6.1168
-6 on my y-axis is right here; they meet right there.1177
That point is labelled D.1187
That is it for this lesson; thank you for watching Educator.com.1195 |
# Problems on Dice MCQ Quiz - Objective Question with Answer for Problems on Dice - Download Free PDF
Most examinations such as GATE, SBI PO, RRB JE, and SSC CGL feature the Logical Reasoning section. Various tricky questions may be asked on topics like Missing Numbers, Arrangement-Patterns, syllogism etc. Familiarise yourself with the Problems on Dice MCQs and get tips to solve them. These Problems on Dice question answers will help you in acing any exam you applied for. Take the Problems on Dice quiz and get explanations for solutions of all the Problems on Dice objective questions.
# Which of the following cubes can be created by folding the given figure?
Option 2 :
## Problems on Dice MCQ Question 1 Detailed Solution
Opposite surface of the open dice can be found by the below methods:-
• Alternate surface are opposite to each other.
• No two opposite surface are touched by side or by corners
So By using the above rule we can find the opposite surface of open dice.
Below are the opposite surfaces:-
As we know that, no two opposite surfaces are adjacent to each other.
So in option 1, @ and & are opposite surfaces but shown adjacent. Hence wrong.
In option 3, © and # are opposite surfaces but shown adjacent. Hence wrong.
3. #
4. @
Option 4 : @
## Problems on Dice MCQ Question 12 Detailed Solution
‘<’ is present in 3rd dice.
The common symbol in 2nd and 3rd dice is ‘%’.
So, keeping ‘%’ constant and then rotating in clockwise direction, we get the symbols opposite to each other:
% @ \$ # < &
So, ‘@’ is opposite to ‘<’.
Hence, ‘@’ is the correct answer.
# The different of the same dice are shown. Select the number that will be on the face opposite to the face showing the number '1'
1. 3
2. 6
3. 5
4. 4
Option 2 : 6
## Problems on Dice MCQ Question 13 Detailed Solution
By comparing fig. (1) and fig. (3)
‘2’ is opposite to ‘4’.
‘3’ is opposite to ‘5’.
And,
‘1’ is opposite to ‘6’.
Hence, “6” is the correct answer.
# The expanded form of a dice is given below. If it is changed to a dice after folding, then which dice can be formed?
Option 2 :
## Problems on Dice MCQ Question 14 Detailed Solution
By observing the figures it is clear that the die in figure (2) can be formed if the expanded form of the dice is folded.
# Three different positions of the same dice are shown. Select the number that will be on the face opposite to the one having 2?
1. 5
2. 6
3. 4
4. 1
Option 3 : 4
## Problems on Dice MCQ Question 15 Detailed Solution
From the above question figure it can be concluded that:
from cube 1 and 2 : 1, 3, 6, and 5 are adjacent to 4. Therefore,2 is opposite to 4.
from cube 2 and 3: 3 is opposite 6.
Then 1 will be opposite to 5. The open cube thus formed will be as follows:
Hence, the number that will be on the face opposite to the one having 2 is 4.
as in the dice, 2 is not mentioned so we can choose 4 as its opposite side.
# A biased six-faced dice when thrown, is thrice as likely to show and odd number than an even number. If it is thrown twice, then the probability that the sum of two numbers thrown is odd will be:
1. 3/8
2. 1/4
3. 1/8
4. 1/2
Option 1 : 3/8
## Problems on Dice MCQ Question 16 Detailed Solution
Concept:
If A and B are two independent events, then the probability of A and B will be
P(A∩B) = P(A) × P(B)
Calculation:
Given The dice is thrice as likely to show an odd number than an even number.
Let the odd number be event A and the even number be event B.
P(A) = 3 P(B)
let P(A) = p
P(B) = 3p
Now p + 3p = 1
p = 1/4
When thrown twice, the sum of two numbers thrown is odd when one number is even and the other is odd.
Possible chances are:
1) Even in 1st throw and Odd in 2nd throw = $$\frac {1}{4} \times \frac {3}{4} = \frac {3}{16}$$
2) Odd in 1st throw and Even in 2nd throw = $$\frac {3}{4} \times \frac {1}{4} = \frac {3}{16}$$
∴ The total probability will be:
$$=\frac {3}{16} + \frac {3}{16} = \frac {3}{8}$$
# Two positions of a cube are shown below. What will come opposite to face containing ‘B’?
1. Q
2. T
3. S
4. Q or T
Option 1 : Q
## Problems on Dice MCQ Question 17 Detailed Solution
In the case of two dices, we’ll write the face of dices in the anticlock wise direction starting from N,
Dice Same face Opposite face 1 N S B 2 N T Q
Hence, we can say that face ‘Q’ will be opposite to face ‘B’.
# Two different positions of the same dice are shown. Select the number that will be on the face opposite to the one having '2', if the number on the face opposite to '3' is odd.
1. 6
2. 1
3. 3
4. 5
Option 1 : 6
## Problems on Dice MCQ Question 18 Detailed Solution
The number on the face opposite to ‘3’ is odd.
‘1’ is opposite to ‘3’
‘4’ is opposite to ‘5’
And,
‘2’ is opposite to ‘6’
Hence, “6” is the correct answer.
# Based on the figures of the dice below, which number will be opposite to 6?
1. 3
2. 4
3. 5
4. 6
Option 1 : 3
## Problems on Dice MCQ Question 19 Detailed Solution
As you can see, the common number visible in both the dice is 4. Rotate the dice in position II by 180 degrees keeping all the 4 sides same. So we find 3 will be opposite to 6.
Hence, 3 is the right answer.
# Which of the cube in the given options can’t be made from the below sheet of paper?
1. A
2. B
3. C
4. D
Option 3 : C
## Problems on Dice MCQ Question 20 Detailed Solution
When we will fold a given sheet of paper, then faces (1, 5), (2, 4) and (3, 6) will lie opposite to each other. But, in figure C, face 2 and 4 are lying adjacent to each other.
So, figure C can’t be made from the given sheet of paper.
Therefore, answer is figure ‘C’. |
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