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# Worked out Problems on Ratio and Proportion
Worked out problems on ratio and proportion are explained here in detailed description using step-by-step procedure. Solved examples involving different questions related to comparison of ratios in ascending order or descending order, simplification of ratios and also word problems on ratio proportion.
Sample questions and answers are given below in the worked out problems on ratio and proportion to get the basic concepts of solving ratio proportion.
1. Arrange the following ratios in descending order.
2 : 3, 3 : 4, 5 : 6, 1 : 5
Solution:
Given ratios are 2/3, 3/4, 5/6, 1/5
The L.C.M. of 3, 4, 6, 5 is 2 × 2 × 3 × 5 = 60
Now, 2/3 = (2 × 20)/(3 × 20) = 40/60
3/4 = (3 × 15)/(4 × 15) = 45/60
5/6 = (5 × 10)/(6 × 10) = 50/60
1/5 = (1 × 12)/(5 × 12) = 12/60
Clearly, 50/60 > 45/60 > 40/60 > 12/60
Therefore, 5/6 > 3/4 > 2/3 > 1/5
So, 5 : 6 > 3 : 4 > 2 : 3 > 1 : 5
2. Two numbers are in the ratio 3 : 4. If the sum of numbers is 63, find the numbers.
Solution:
Sum of the terms of the ratio = 3 + 4 = 7
Sum of numbers = 63
Therefore, first number = 3/7 × 63 = 27
Second number = 4/7 × 63 = 36
Therefore, the two numbers are 27 and 36.
3. If x : y = 1 : 2, find the value of (2x + 3y) : (x + 4y)
Solution:
x : y = 1 : 2 means x/y = 1/2
Now, (2x + 3y) : (x + 4y) = (2x + 3y)/(x + 4y) [Divide numerator and denominator by y.]
= [(2x + 3y)/y]/[(x + 4y)/2] = [2(x/y) + 3]/[(x/y) + 4], put x/y = 1/2
We get = [2 (1/2) + 3)/(1/2 + 4) = (1 + 3)/[(1 + 8)/2] = 4/(9/2) = 4/1 × 2/9 = 8/9
Therefore the value of (2x + 3y) : (x + 4y) = 8 : 9
More solved problems on ratio and proportion are explained here with full description.
4. A bag contains $510 in the form of 50 p, 25 p and 20 p coins in the ratio 2 : 3 : 4. Find the number of coins of each type. Solution: Let the number of 50 p, 25 p and 20 p coins be 2x, 3x and 4x. Then 2x × 50/100 + 3x × 25/100 + 4x × 20/100 = 510 x/1 + 3x/4 + 4x/5 = 510 (20x + 15x + 16x)/20 = 510 ⇒ 51x/20 = 510 x = (510 × 20)/51 x = 200 2x = 2 × 200 = 400 3x = 3 × 200 = 600 4x = 4 × 200 = 800. Therefore, number of 50 p coins, 25 p coins and 20 p coins are 400, 600, 800 respectively. 5. If 2A = 3B = 4C, find A : B : C Solution: Let 2A = 3B = 4C = x So, A = x/2 B = x/3 C = x/4 The L.C.M of 2, 3 and 4 is 12 Therefore, A : B : C = x/2 × 12 : x/3 × 12 : x/4 = 12 = 6x : 4x : 3x = 6 : 4 : 3 Therefore, A : B : C = 6 : 4 : 3 6. What must be added to each term of the ratio 2 : 3, so that it may become equal to 4 : 5? Solution: Let the number to be added be x, then (2 + x) : (3 + x) = 4 : 5 ⇒ (2 + x)/(5 + x) = 4/5 5(2 + x) = 4(3 + x) 10 + 5x = 12 + 4x 5x - 4x = 12 - 10 x = 2 7. The length of the ribbon was originally 30 cm. It was reduced in the ratio 5 : 3. What is its length now? Solution: Length of ribbon originally = 30 cm Let the original length be 5x and reduced length be 3x. But 5x = 30 cm x = 30/5 cm = 6 cm Therefore, reduced length = 3 cm = 3 × 6 cm = 18 cm More worked out problems on ratio and proportion are explained here step-by-step. 8. Mother divided the money among Ron, Sam and Maria in the ratio 2 : 3 : 5. If Maria got$150, find the total amount and the money received by Ron and Sam.
Solution:
Let the money received by Ron, Sam and Maria be 2x, 3x, 5x respectively.
Given that Maria has got $150. Therefore, 5x = 150 or, x = 150/5 or, x = 30 So, Ron got = 2x =$ 2 × 30 = $60 Sam got = 3x = 3 × 60 =$90
Therefore, the total amount $(60 + 90 + 150) =$300
9. Divide $370 into three parts such that second part is 1/4 of the third part and the ratio between the first and the third part is 3 : 5. Find each part. Solution: Let the first and the third parts be 3x and 5x. Second part = 1/4 of third part. = (1/4) × 5x = 5x/4 Therefore, 3x + (5x/4) + 5x = 370 (12x + 5x + 20x)/4 = 370 37x/4 = 370 x = (370 × 4)/37 x = 10 × 4 x = 40 Therefore, first part = 3x = 3 × 40 =$120
Second part = 5x/4
= 5 × 40/4
= $50 Third part = 5x = 5 × 40 =$ 200
10. The first, second and third terms of the proportion are 42, 36, 35. Find the fourth term.
Solution:
Let the fourth term be x.
Thus 42, 36, 35, x are in proportion.
Product of extreme terms = 42 ×x
Product of mean terms = 36 X 35
Since, the numbers make up a proportion
Therefore, 42 × x = 36 × 35
or, x = (36 × 35)/42
or, x = 30
Therefore, the fourth term of the proportion is 30.
More worked out problems on ratio and proportion using step-by-step explanation.
11. Set up all possible proportions from the numbers 8, 12, 20, 30.
Solution:
We note that 8 × 30 = 240 and 12 × 20 = 240
Thus, 8 × 30 = 12 × 20 ………..(I)
Hence, 8 : 12 = 20 : 30 ……….. (i)
We also note that, 8 × 30 = 20 × 12
Hence, 8 : 20 = 12 : 30 ……….. (ii)
(I) can also be written as 12 × 20 = 8 × 30
Hence, 12 : 8 = 30 : 20 ……….. (iii)
Last (I) can also be written as
12 : 30 = 8 : 20 ……….. (iv)
Thus, the required proportions are 8 : 12 = 20 : 30
8 : 20 = 12 : 30 12 : 8 = 30 : 20 12 : 30 = 8 : 20
12. The ratio of number of boys and girls is 4 : 3. If there are 18 girls in a class, find the number of boys in the class and the total number of students in the class.
Solution:
Number of girls in the class = 18
Ratio of boys and girls = 4 : 3
According to the question,
Boys/Girls = 4/5
Boys/18 = 4/5
Boys = (4 × 18)/3 = 24
Therefore, total number of students = 24 + 18 = 42.
13. Find the third proportional of 16 and 20.
Solution:
Let the third proportional of 16 and 20 be x.
Then 16, 20, x are in proportion.
This means 16 : 20 = 20 : x
So, 16 × x = 20 × 20
x = (20 × 20)/16 = 25
Therefore, the third proportional of 16 and 20 is 25.
`
Ratio and Proportion
What is Ratio and Proportion?
Worked out Problems on Ratio and Proportion
Practice Test on Ratio and Proportion
Ratio and Proportion - Worksheets
Worksheet on Ratio and Proportion |
# How do you find the derivative of 1/(2x)?
##### 2 Answers
Jul 9, 2016
$- \frac{1}{2 {x}^{2}}$
#### Explanation:
Differentiate using the $\textcolor{b l u e}{\text{power rule}}$
$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{d}{\mathrm{dx}} \left(a {x}^{n}\right) = n a {x}^{n - 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
Rewrite the function as.
$\frac{1}{2 x} = \frac{1}{2} \times \frac{1}{x} = \frac{1}{2} \times {x}^{-} 1 = \frac{1}{2} {x}^{-} 1$
$\Rightarrow \frac{d}{\mathrm{dx}} \left(\frac{1}{2} {x}^{-} 1\right) = - 1 \times \frac{1}{2} {x}^{- 1 - 1} = - \frac{1}{2} {x}^{-} 2$
$\Rightarrow \frac{d}{\mathrm{dx}} \left(\frac{1}{2 x}\right) = - \frac{1}{2} {x}^{-} 2 = - \frac{1}{2 {x}^{2}}$
Jul 9, 2016
ALTERNATIVE APPROACH
#### Explanation:
By the quotient rule:
$\left(\frac{1}{2 x}\right) ' = \frac{\left(0 \times 2 x\right) - \left(1 \times 2\right)}{2 x} ^ 2$
$\left(\frac{1}{2 x}\right) ' = - \frac{2}{4 {x}^{2}}$
$\left(\frac{1}{2 x}\right) ' = - \frac{1}{2 {x}^{2}}$
Hopefully this helps! |
# Beginning Algebra Tutorial 23
Beginning Algebra
Tutorial 23:Slope
Learning Objectives
After completing this tutorial, you should be able to:
1. Find the slope given a graph or two points.
2. Know the relationship between slopes of parallel lines.
3. Know the relationship between slopes of perpendicular lines.
Introduction
This tutorial takes us a little deeper into linear equations. We will be looking at the slope of a line. We will also look at the relationship between the slopes of parallel lines as well as perpendicular lines. Let's see what you can do with slopes.
Tutorial
Slope
The slope of a line measures the steepness of the line.
Most of you are probably familiar with associating slope with "rise over run".
Rise means how many units you move up or down from point to point. On the graph that would be a change in the y values.
Run means how far left or right you move from point to point. On the graph, that would mean a change of x values.
Positive slope:
Note that when a line has a positive slope it goes up left to right.
Negative slope:
Note that when a line has a negative slope it goes down left to right.
Zero slope:
slope = 0
Note that when a line is horizontal the slope is 0.
Undefined slope:
slope = undefined
Note that when the line is vertical the slope is undefined.
Slope Formula Given Two Points
Given two points and
The subscripts just indicate that these are two different points. It doesn't matter which one you call point 1 and which one you call point 2 as long as you are consistent throughout that problem.
Note that we use the letter m to represent slope.
Example 1: Find the slope of the straight line that passes through (-5, 2) and (4, -7).
*Plug in x and y values into slope formula
*Simplify
Make sure that you are careful when one of your values is negative and you have to subtract it as we did in line 2. 4 - (-5) is not the same as 4 - 5.
The slope of the line is -1.
Example 2: Find the slope of the straight line that passes through (1, 1) and (5, 1).
*Plug in x and y values into slope formula
*Simplify
It is ok to have a 0 in the numerator. Remember that 0 divided by any non-zero number is 0.
The slope of the line is 0.
Example 3: Find the slope of the straight line that passes through (3, 4) and (3, 6).
*Plug in x and y values into slope formula
*Simplify
Since we did not have a change in the x values, the denominator of our slope became 0. This means that we have an undefined slope. If you were to graph the line, it would be a vertical line, as shown above.
The slope of the line is undefined.
Parallel Lines and Their Slopes
In other words, the slopes of parallel lines are equal.
Note that two lines are parallel if there slopes are equal and they have different y-intercepts.
Perpendicular Lines and Their Slopes
In other words, perpendicular slopes are negative reciprocals of each other.
Practice Problems
These are practice problems to help bring you to the next level. It will allow you to check and see if you have an understanding of these types of problems. Math works just like anything else, if you want to get good at it, then you need to practice it. Even the best athletes and musicians had help along the way and lots of practice, practice, practice, to get good at their sport or instrument. In fact there is no such thing as too much practice.
To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that problem. At the link you will find the answer as well as any steps that went into finding that answer.
Practice Problems 1a - 1d: Find the slope of each line if it exists.
1b.
1c.
1d.
Practice Problems 2a - 2b: Find the slope of the straight line that passes through the given points.
2a. (3, 5) and (-1, -8)
2b. (4, 2) and (4, -2)
Need Extra Help on these Topics?
The following is a webpage that can assist you in the topics that were covered on this page:
http://www.purplemath.com/modules/slope.htm
This webpage helps you with slope.
Go to Get Help Outside the Classroom found in Tutorial 1: How to Succeed in a Math Class for some more suggestions.
Last revised on July 31, 2011 by Kim Seward. |
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# TutorMe Blog
## What Is an Inequality and How to Show It on a Number Line?
Inactive
Jana Russick
April 07, 2021
In math, an inequality shows the relationship between two values in an algebraic expression that are not equal. Inequality signs can indicate that one variable of the two sides of the inequality is greater than, greater than or equal to, less than, or less than or equal to another value.
Whether a sign is greater than or less than depends on the direction of the inequality sign. If the open part of the sign is turned towards the left side, >, the value on the left side of the sign is greater than the right. If it's turned towards the right side, <, the value on the left is less than the value on the right.
Some inequality symbols will have a line underneath them: ≥ and ≤. This means that the two sides of an inequality expression could potentially be equal. However, not enough is known to prove this.
As you can see in the mathematical expression above, x, is greater than or equal to 7. Because x can be many values, saying it’s equal to 7 wouldn't be a true statement. That's why we must use .
## Inequalities on a Number Line
Whenever a linear inequality has a variable and a real number, you can express it on a number line. Here's how to use number lines to show x is greater than positive number 3 and less than or equal to negative number -1:
Any number line showing a linear inequality must have an open circle for < and > and a closed circle for ≤ and ≥.
### Using Interval Notation
When we know an inequality is between two numbers, you can write it in interval notation. Interval notation expresses the location range of an inequality by using brackets for ≥ and ≤ signs and parentheses for < and > signs.
Here's how you would show that y is less than or equal to -4 and 2 is greater than y:
As you can see, we use an open circle to show that y is less than 2 and a closed circle to show that y is equal to or greater than -4. In linear notation, this is written as:
Linear Notation: [-4, 2)
## Answering ‘What Is an Inequality?’
Understanding the concept of inequalities allows us to better understand linear equations and the number line. When we know that a variable, like x or y, is within a specific range of values, we can represent it by shading that range of numbers on the number line.
On this line, we use an open circle for greater than or less than values and a closed circle for equal to or less than and equal to or greater than values. If we know the inequality is between two numbers, we can use brackets and parentheses to show the possible range of the inequality’s values in linear notation.
### More Math Homework Help
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# 1445 Virgács for Your Boots Tonight
Tomorrow is Mikulás (Saint Nicholas Day) in Hungary. Children will awake to find candy, fruit, or nuts in their polished shoes or boots because every boy and every girl has been at least a little bit good all year long.
Because they have also been at least a little bit naughty, they will also find virgács in those same shoes or boots. Virgács are little twigs that have been spray-painted gold and tied together at the top with red ribbon.
Santa is so busy this time of year, that I thought I would give him a helping hand. I’ve made some virgács for YOUR boots or shoes!
Start at the top of the puzzle and work your way down cell by cell to solve this Level 3 puzzle. Oh, but I’ve been just a little bit naughty making this puzzle: you will need to look at later clues to figure out what factors to give to 40. Will clue 40 use a 5 or a 10? Look at clues 60 and 90, and you will have only one choice for that answer. Then you can forgive my tiny bit of naughtiness.
Now I’ll tell you a few facts about the puzzle number, 1445:
• 1445 is a composite number.
• Prime factorization: 1445 = 5 × 17 × 17, which can be written 1445 = 5 × 17²
• 1445 has at least one exponent greater than 1 in its prime factorization so √1445 can be simplified. Taking the factor pair from the factor pair table below with the largest square number factor, we get √1445 = (√289)(√5) = 17√5
• The exponents in the prime factorization are 1 and 2. Adding one to each exponent and multiplying we get (1 + 1)(2 + 1) = 2 × 3 = 6. Therefore 1445 has exactly 6 factors.
• The factors of 1445 are outlined with their factor pair partners in the graphic below.
1445 is the sum of two squares in THREE different ways:
31² + 22² = 1445
34² + 17² = 1445
38² + 1² = 1445
1445 is the hypotenuse of SEVEN Pythagorean triples:
76-1443-1445 calculated from 2(38)(1), 38² – 1², 38² + 1²
221-1428-1445 which is 17 times (13-84-85)
477-1364-1445 calculated from 31² – 22², 2(31)(22), 31² + 22²
612-1309-1445 which is 17 times (36-77-85)
680-1275-1445 which is (8-15-17) times 85
805-1200-1445 which is 5 times (161-240-289)
867-1156-1445 which is (3-4-5) times 289 and can
also be calculated from 34² – 17², 2(34)(17), 34² + 17² |
# Pascal's Triangle2
Pascal's Triangle
Field: Algebra
Image Created By: Math Forum
Website: Mathforum.org
Pascal's Triangle
Pascal's Triangle
# Basic Description
Pascal's triangle is a triangular arrangement of specific numbers which have interesting patterns. We start out with 1 and 1, 1, for the first two rows. To construct each entry in the next row, we add the two numbers immediately above it to the right and to the left. If there's only one number diagonally above the entry to the left and to the right, then we enter just that number. We can continue doing this endlessly. This pattern is just one of the many patterns within the triangle. The hockey stick pattern below is another example of one of the most interesting and fun patterns within the triangle.
#### Hockey Stick Pattern
Hockey stick pattern http://ptri1.tripod.com/
Look at the numbers highlighted in red in the image on the left. We can see that the sum of the ones in a line equals the entry not in a line. 1+6+21+56=84. This pattern holds for the other "hockey stick" selections.
# A More Mathematical Explanation
Each entry in a Pascal's triangle is identified by a row and a place. The rows are labeled starting f [...]
Each entry in a Pascal's triangle is identified by a row and a place. The rows are labeled starting from zero, so the first row would be row 0, the second one would be row 1, the third, row 2 and so forth. Places are given to each entry starting from the first number after 1, from the left to the right. 70 for example would be identified as row 8, place 4.
### Properties
• The triangle is bordered by 1's on the right and left edges.
• The next line of numbers in diagonal order after the edge numbers are natural numbers 1,2,3,4...
• The next set of numbers inwards after the natural numbers are
• After the triangular numbers we have tetrahedral numbers in order 1,4,10,20...
• The next d-diagonal contains the next higher dimensional "d-simplex" numbers.
• The first number after 1 in each row divides all the other numbers in that row if and only if it is prime.
### More patterns within Pascal's triangle
Pascal's triangle contains a number of smaller patterns within it. Some of these patterns include:
#### Magic 11's
If a row is made into a single number (omitting the spaces), the resulting number is equal to 11 raised to the $n^th$ power, where $n$ is the row number. For example:
Row # Actual Row Single Number Formula
0 1 1 $11^0$
1 1 1 11 $11^1$
2 1 2 1 121 $11^2$
3 1 3 3 1 1,331 $11^3$
4 1 4 6 4 1 14,641 $11^4$
The further we go down Pascal's triangle, the more complicated the procedure becomes.
#### Horizontal Sum
If you add the numbers in each row horizontally, then look at the pattern formed by the resultant numbers, you will notice that th numbers double each time, but all are in the power of 2.
#### Fibonacci Sequence
The Fibonacci sequence is a list of Fibonacci numbers. Each Fibonacci number is generated by adding the previous two consecutive terms in the sequence. The first two Fibonacci numbers are 0 and 1. This pattern can be located in Pascal's triangle. The sum of the entries in each row, as shown in the animation demonstrating patterns, gives a Fibonacci number.
The numbers that result from summing the rows from top to bottom form the .
#### Sierpenski Triangle
Sierpenski triangle generated by coloring odd and even numbers with Pascal's triangle with two different colors http://en.wikipedia.org/wiki/Pascal's_triangle
If we are to color the odd and even numbers in Pascal's triangle with two distinct colors, we would observe an interesting recursive pattern seen in the Sierpinski triangle. You could try this out in the animation and see what happens.
These are just a few of the patterns observed in Pascal's triangle. Other patterns include:
#### Animation Demonstrating Patterns
The Flash animation below allows you to explore some of the patterns present in Pascal's Triangle:
### Applications of Pascal's Triangle
Pascal's triangle can be used to determine the combinations of heads and tails we can have depending on the number of tosses. From the possible outcomes, we can calculate the probability of any combination.
For example, if we toss a coin twice, we will have the combination HH once, TH twice and TT once, thus the possible outcomes would be in the order 1 2 1.This is also the same as the second row of Pascal's triangle. In general, if we toss a coin $x$ times, the combination of possible outcomes would be the horizontal entries in the $x$th row of the Pascal's triangle.
Tosses Possible Outcomes Pascals Triangle
1 H T 1 1
2 HH
HT TH
TT
1 2 1
3 HHH
HHT HTH THH HTT THT TTH HTT THT TTH
TTT
1 3 3 1
##### What is the probability of getting exactly 2 heads with 3 coin tosses?
To answer this, we could use Pascal's triangle. The third row of Pascal's triangle has the entries 1 3 3 1=8 possible outcomes. Therefore we have 8 possible outcomes. 3 of the possible 8 outcomes give exactly 2 heads, therefore the probility of getting exactly 2 heads from 3 coin tosses is 3/8=37.5%.
#### In Algebra
Pascal's triangle can be used to determine binomial coefficients in binomial expansions. For example
$(x + y)^2 = x^2 + 2xy + y^2 =1{x^2}{y^0} + {2x^1}{y^1} +{1x^0}{y^2}$
In the above example, notice that the coefficients in the expansion are exactly the same as the entries in row two in Pascal's triangle.
This is summarized in the binomial theorem which states that in general,
$(x + y)^n$ = ${a_0}{x^n}$ $+$ ${a_1}{x^{n-1}}$ $+$ ${a_2}{x^{n-2}}{y^2}$$+$$+$ ${a_{n-1}}{xy^{n-1}}$ $+$ ${a_n}{y^n}$
where $a_i$ are the entries in row $n$ in Pascal's triangle.
## References
Hockey stick pattern http://ptri1.tripod.com/ Triangular numbers http://www.mathsisfun.com/numberpatterns.html#triangular Horizontal sum http://www.mathsisfun.com/pascals-triangle.html Fibonacci sequence http://mathforum.org/dr.math/faq/faq.pascal.triangle.html Sierpenski Triangle http://en.wikipedia.org/wiki/Pascal's_triangle |
# The Root of the Matter: Approximating Roots with the Greeks - Method, Motivation, and Verification
Author(s):
Matthew Haines and Jody Sorensen (Augsburg University)
### Method and Motivation
In his work On Mathematics Useful for the Understanding of Plato [4], Theon proposed a method which can be used to provide ever closer rational approximations of $$\sqrt{2}.$$ This method is sometimes referred to as Theon's ladder. The method starts with $$x_0=1,$$ $$y_0=1.$$ Our estimate of $$\sqrt{2},$$ which will be the ratio $$\displaystyle{\frac {y} {x}},$$ thus starts as 1. The numbers $$x$$ and $$y$$ are sometimes referred to as side and diagonal numbers – more on that to come. [5] To get a better estimate of $$\sqrt{2},$$ we let $$x_1=x_0+y_0$$ and $$y_1 = 2x_0+y_0.$$ Thus we get that $$x_1=2$$ and $$y_1=3,$$ so our new estimate is $$\displaystyle{\frac {3} {2}} = 1.5.$$ The recursive formula is $x_{n+1} = x_n+y_n \ \ \ \ \ \ \ \ y_{n+1} = 2x_n+y_n .$ Continuing in this way for a few more steps gives us the results in Table 1.
$\begin{array}{r|r|r|r} n & x_n & y_n & y_n/x_n \\\hline 0 & 1 & 1 & 1 \\\hline 1 & 2 & 3 & 1.5 \\\hline 2 & 5 & 7 & 1.4 \\\hline 3 & 12 & 17 & 1.4167 \\\hline 4 & 29 & 41 & 1.4138\\\hline 5 & 70 & 99 & 1.4143\\\hline 6 & 169 & 239 & 1.4142\end{array}$
Table 1. Side and diagonal numbers
So it appears that Theon's method approximates $$\sqrt{2} = 1.4142....$$ Note that the denominators (or sides), $$x_n,$$ are the Pell numbers, and their properties are well-studied by number theorists.
We cannot be sure how (or why) Theon came up with this procedure. One possible origin is numerical. [1] If $$\sqrt{2}$$ were rational, then $$\displaystyle\sqrt{2} = {\frac{y}{x}}$$ for some positive integers $$x$$ and $$y.$$ Squaring and simplifying gives us $$2x^2 =y^2.$$ So, does that equation have integer solutions? Let's look at the list of possibilities for $$y^2$$ and $$2x^2$$ in Table 2.
$\begin{array}{r|cccccccc} n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\\hline n^2 & 1 & 4 & \fbox{9} & 16 & 25 & 36 & \fbox{49} & 64 \\\hline 2n^2 & 2 & \fbox{8} & 18 & 32 & \fbox{50} & 72 & 98 & 128\end{array}$
Table 2. Values of $$y^2$$ and $$2x^2$$
From this tiny table we don't see any places where a square is equal to two times a square, but we do see some places where they are close. For example if $$x=5$$ and $$y=7,$$ then $$2x^2 = 50 = y^2+1.$$ This indicates that $$\displaystyle{\frac {7}{5}}$$ is a decent approximation of $$\sqrt{2}.$$ If we continue this table and list the possibilities as fractions, we get this list: ${\frac{1}{1}}, \,{\frac{3}{2}},\, {\frac{7}{5}},\, {\frac{17}{12}},\, \dots.$ These are all values of $$x$$ and $$y$$ where $$2x^2=y^2 \pm 1.$$ If you look for a pattern here, you see that the new denominator is the sum of the previous numerator and denominator, or $$x_{n+1} = x_n+y_n.$$ The new numerator is this new denominator plus the old denominator, or $$y_{n+1} = x_{n+1}+x_n = x_n+y_n+x_n = 2x_n+y_n.$$ Testing this out gives the next term as $$\displaystyle{\frac{41}{29}}$$ which satisfies $$2\cdot29^2 = 1682=41^2+1.$$ So Theon could have found this pattern by inspection.
There are also possible geometric motivations for this process, which explain why $$x$$ and $$y$$ are referred to as "side" and "diagonal" numbers. Suppose we start with an isosceles right triangle $$ABC$$ whose leg (side) is of length $$x$$ and hypotenuse (diagonal) of length $$y,$$ as in Figure 1. Note that this figure has the added point $$H$$ to create the square $$ABHC$$ to emphasize the names "side and diagonal numbers." Extend the two sides by $$y$$ to create an isosceles right triangle with sides of length $$x+y.$$
Figure 1. An isosceles right triangle with sides of length $$x+y$$ and diagonal $$\overline{DG}$$ of length $$2x+y$$
We can see that the larger triangle has a hypotenuse of length $$2x+y$$ by noticing that $\triangle BCA \cong \triangle DBE \cong \triangle GCF$ and that $$BCFE$$ is a rectangle. Thus $$DE=FG=x$$ and $$EF=y.$$
This triangle motivates the idea that if the side is extended from $$x$$ to $$x+y,$$ then the diagonal should be extended from $$y$$ to $$2x+y,$$ which fits Theon's method. We should note that no integers $$x$$ and $$y$$ exist that are sides and diagonals of the isosceles right triangle. This approach is simply a motivation and not a proof.
### Verification
Let's prove that Theon's method gives a method for finding better and better rational approximations of $$\sqrt{2}.$$ Suppose we start with $$x$$ and $$y$$ satisfying $$2x^2-y^2 = \pm 1$$ (note that $$x_0=y_0=1$$ satisfies this) and let $$x^*=x+y, y^*=2x+y.$$ Then
\begin{eqnarray*} 2(x^*)^2-(y^*)^2 &=& 2(x+y)^2-(2x+y)^2 \\ &=& -2x^2+y^2 = \mp 1.\end{eqnarray*}
This shows that as we iterate, $$2x^2-y^2$$ remains $$\pm 1.$$ Now if $$2x_n^2-y_n^2 = \pm 1,$$ then $$\displaystyle 2 = {\frac {y_n^2} {x_n^2}} \pm {\frac {1} {x_n^2}}.$$ If we start with positive values for $$x_0$$ and $$y_0,$$ then as we iterate, $$x_n$$ gets bigger by at least one at each step, so $$x_n$$ goes to $$\infty.$$ Therefore we can say that $$\displaystyle {\frac {1} {x_n^2}}$$ will go to $$0^{+}.$$ This means that $$\displaystyle\left({\frac{y_n} {x_n}} \right)^2$$ will tend to 2. Assuming that $$\displaystyle \lim_{n\rightarrow \infty} {\frac {y_n} {x_n}}$$ exists, then $$\displaystyle {\frac {y_n} {x_n}}$$ goes to $$\sqrt 2,$$ as desired.
So, in about 100 CE, Theon of Smyrna developed (for an unknown reason) an iterative method that can be used to approximate $$\sqrt 2$$ by rational numbers. With our modern mathematical tools, we will look at Theon's method through the lenses of Geometry and Linear Algebra. |
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# Find the value of $x$in the following proportion $5:15 = 4:x$
Last updated date: 18th Jun 2024
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Hint: A proportion is a mathematical comparison between two numbers. Often, these numbers can represent a comparison between things or people. For example, say you walked into a room full of people. You want to know how many boys there are in comparison to how many girls there are in the room. You would write that comparison in the form of a proportion.
A ratio is a way to compare two quantities by using division as in miles per hour.
A proportion on the other hand is an equation that says that two ratio are equivalent
If one number in a proportion is unknown you can find that number by solving the proportion
It is formulate as:
$a:b::c:d$
$\Rightarrow \dfrac{a}{b} = \dfrac{c}{d}$
$\Rightarrow a \times d = b \times c$
Therefore,
Given
$5:15 = 4:x......(1)$
We need to find the value of $x$in proportion.
We know that if two number are in proportion i.e $a:b::c:d$ or $a:b = c:d,$then we can write them as$\dfrac{a}{b} = \dfrac{c}{d}$
Hence $(1) \Rightarrow \dfrac{5}{{15}} = \dfrac{4}{x}$
Cross multiplying, we get
$5 \times x = 15 \times 4$
$\Rightarrow x = \dfrac{{15 \times 4}}{5}$
As 5 was in multiplication LHS it will be in division on RHS
$\Rightarrow x = 3 \times 4$
$\Rightarrow x = 12$
Hence the value of $x$ is 12.
Note: For four numbers a, b, c, d if $a:b = c:d$ then $b:a = d:c,$ it is known as invert and properly
For four numbers a ,b, c, d if $a:b = c:d$ then $a:c = b:d$if the second and third terms interchange their places, it is known as alternator property.
We can also use component do and dividend property to simply the proportion problems. |
# Factoring: Tricks to Make It Easy
08.05.2018
Episode #3 of the course Foundations of mathematics by John Robin
Today, as we continue to build our foundation of mathematics, we will turn to one of the most fundamental things we can do with numbers: factoring.
You might recall this term from school. What I’m going to show you today is how factoring is much easier to do when you think about a number and how to decompose it, since every number that is not prime is composite, meaning it can be broken down into a string of prime numbers multiplied together.
Exponent Notation
First off, let’s develop notation to make our life easier. As an example, let’s work with 16. In the list from Lesson 1, recall that it “decomposed” into 2x2x2x2. If we did this with 32, we would find it decomposed into 2x2x2x2x2. This is getting cumbersome. 128 is 2x2x2x2x2x2x2. We run into the same problem as numbers get larger and larger. The number 288 is 2x2x2x2x2x3x3. Have I convinced you yet that we need an easier way to do this?
It will be helpful now to rewrite our list of the natural numbers 2 through 19:
2, 3, 2×2, 5, 2×3, 7, 2x2x2, 3×3, 2×5, 11, 2x2x3, 13, 2×7, 3×5, 2x2x2x2, 17, 2x3x3, 19, ….
Now I’m going to write out the list of numbers from 2 through 19 again, using an easier notation:
2, 3, 22, 5, 2(3), 7, 23, 32, 2(5), 11, 22(3), 13, 2(7), 3(5), 24, 17, 2(32), 19, …
I did two things:
1. I introduced something called an exponent, which you might also remember from school. In the case of 16, which was 2x2x2x2, the prime number 2 was multiplied four times, so we write 24. The exponent is an instruction: It tells you, “Multiply this number by itself as many times as this number” (4, in our case).
2. I replaced the “x” with a bracket. When you see two numbers next to each other like this: a(b), you assume it means axb. 5(11) means 5 multiplied by 11. We can do this with several numbers multiplied together. For example, 2x3x5x11 would become 2(3)(5)(11). This is also called a product, i.e., a string of numbers multiplied together.
Putting this all together, 2x2x3x3x3x5x5x11x13x17x17 would become 22(33)(52)(11)(13)(172). This product multiplies to 111,582,900.
And looking at this in reverse: 111,582,900 is a composite number that can be broken down into a product of prime numbers: 22(33)(52)(11)(13)(172).
How to Factor Any Number
Take the number 6972. Always start by dividing the smallest prime, 2, as many times as possible:
6972 ÷ 2 = 2(3486)
2(3486) ÷ 2 = 22(1743)
We can’t divide a third time by 2. If you try, you’ll end up with a decimal. So, we move onto the next prime, 3:
22(1743) ÷ 3 = 22(3)(581)
We can’t go any further with 3—same problem, we’ll get a decimal. If we try 5, we get the same problem. So, we move onto the next prime, 7:
22(3)(581) ÷ 7 = 22(3)(7)(83)
We can’t go any further with 7. You’ll find the same thing with the next primes: 11, 13, 17, 19, 23, 29, 31, and so on.
Quick trick: When you reach the late stage and have only a smaller two- to three-digit number at the end, take the square root of that number. For example, √83 = ~9.1. All primes greater than √83 will not divide, so you can stop at the closest one (here, it’s 7).
So just like that, we’ve found that 6972 = 22(3)(7)(83), and we can form all the factors of this number by multiplying these primes together in any way we want, i.e., 4 (=22), 6 (=2×3), 12 (=22x3), 14 (=2×7), 21 (=3×7), 166 (=2×83), etc.
This will work with any number. You can divide by the primes in order of smallest to largest, then as your number on the right shrinks each time, take the square root to find out when you can stop.
Whenever you look at a number now, think about how you can break it down into a product of primes. As you’ll see tomorrow when we move on to fractions, approaching numbers like this will give you true power, especially when it comes to simplifying fractions.
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# In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown.
Question:
In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins/loses.
Solution:
When a die is thrown, then probability of getting a six $=\frac{1}{6}$ then, probability of not getting a six $=1-\frac{1}{6}=\frac{5}{6}$
If the man gets a six in the first throw, then
probability of getting a six $=\frac{1}{6}$
If he does not get a six in first throw, but gets a six in second throw, then
probability of getting a six in the second throw $=\frac{5}{6} \times \frac{1}{6}=\frac{5}{36}$
If he does not get a six in the first two throws, but gets in the third throw, then
probability of getting a six in the third throw $=\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}=\frac{25}{216}$
probability that he does not get a six in any of the three throws $=\frac{5}{6} \times \frac{5}{6} \times \frac{5}{6}=\frac{125}{216}$
In the first throw he gets a six, then he will receive Re 1.
If he gets a six in the second throw, then he will receive Re (1 - 1) = 0
If he gets a six in the third throw, then he will receive Rs(-1 - 1 + 1) = Rs (-1),
that means he will lose Re 1 in this case.
Expected value $=\frac{1}{6} \times 1+\left(\frac{5}{6} \times \frac{1}{6}\right) \times 0+\left(\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6}\right) \times(-1)=\frac{11}{216}$
So, he will loose Rs $\frac{11}{216}$. |
# Prove $\ \forall n \ge 2, \ 2^{n} + 3^{n} < 4^{n}$
Question:
Prove $\ \forall n \ge 2, \ 2^{n} + 3^{n} < 4^{n}$
My attempt:
Base case is trivial.
Suppose $\ n \ge 2$ and $\ 2^{n} + 3^{n} < 4^{n}$
Then,
$2^{n+1} + 3^{n+1} = 2.2^{n} + 3.3^{n} = 2.2^{n} + 2.3^{n} + 3^{n} = 2(2^{n} + 3^{n}) + 3^{n} <2(4^{n}) + 3^{n}$, by I.H.
I am stuck here. how do I show that this expression is $< 4^{n+1}$?
• Hints: Is $3^n<4^n$? Is $3(4^n)<4(4^n)$? – Michael Burr Sep 16 '17 at 20:25
• Another proof could use that the equivalent inequality $(2/4)^n+(3/4)^n<1$ after dividing both sides by $4^n$, where $4^n>0$, holds because it holds for $n=2$ and $(2/4)^n$ and $(3/4)^n$ are both strictly decreasing. – user236182 Sep 16 '17 at 20:32
You can say $$2\cdot 2^n + 3\cdot 3^n < 4\cdot 2^n + 4\cdot 3^n = 4(2^n+3^n) < 4\cdot 4^n = 4^{n+1}.$$
multiplying $$2^n+3^n<4^n$$ by $$4^n$$ we get $$2^n\cdot 2^{2n}+3^n\cdot 2^{2n}<4^{n+1}$$ and we have $$2^n\cdot 2<2^{3n}$$ since $$1<2^{2n-1}$$ for $n\geq 2$ and $$3^n\cdot 3<3^n\cdot 2^{2n}$$ since $$3<2^{2n}$$ therefore we have $$2^{n+1}+3^{n+1}<2^n\cdot 2^{2n}+3^n\cdot 2^{2n}<4^{n+1}$$
Hint: for $\,0 \lt a \lt 1\,$ the sequence $\,a^n\,$ is decreasing with $\,n\,$ since $\,a^{n+1} = a \cdot a^n \lt 1 \cdot a^n\,$, so:
$$\left(\frac{2}{4}\right)^n+\left(\frac{3}{4}\right)^n \le \left(\frac{2}{4}\right)^2+\left(\frac{3}{4}\right)^2 = \frac{13}{16} \lt 1 \quad\quad\text{for}\;\; \forall n \ge 2$$ |
back to algebra2 video lessons
Quadratic equation real roots parameter range
Question:
The quadratic equation 2x2 - 5x + m = 0 has real roots. What is the range of m?
Solution:
The quadratic equation ax2 + bx + c = 0 has real roots when b2 - 4ac >= 0
Compare the standard quadratic equation ax2 + bx + c = 0 with the given quadratic equation 2x2 - 5x + m = 0. We get, a = 2, b= -5 and c = m
b2 - 4ac = (-5)2 - 4 × 2 m = 25 - 8m >= 0
8m <= 25
m <= 25/8
Case 1: when m = 25/8, the quadratic equation is: 2x2 - 5x + 25/8 = 0. We use the formula to find roots of the quadratic equation.
x1,2 = [-b +- square root of (b2 - 4ac)]/2a
= [-(-5) +- square root of ((-5)2 - 4 × 2 × 25/8)]/2 × 2
= [5 +- square root of (25 - 25)]/4
= 5/4
Therefore, when m = 25/8, x1 = x2 = 5/4
Case 2: now we choose a value of m which is less than 25/8. Choose m = -25/8.
x1,2 = [-b +- square root of (b2 - 4ac)]/2a
= [-(-5) +- square root of ((-5)2 - 4 × 2 × -25/8)]/2 × 2
= [-(-5) +- square root of ((-5)2 + 4 × 2 × 25/8)]/2 × 2
= [5 +- square root of (25 + 25)]/4
= [5 +- 5 square root of (1 + 1]/4
= (5 +- 5 square root of 2)/4
= (5/4) (1 +- square root of 2)
So, x1 = (5/4) (1 + square root of 2) and x2 = (5/4) (1 - square root of 2).
Therefore, when b2 - 4ac = 0, the quadratic equation has two equal real roots. When b2 - 4ac > 0, the quadratic equation has two different real roots. Watch the video for more details. |
# What diagonals are congruent and bisect each other?
## What diagonals are congruent and bisect each other?
If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram. If a quadrilateral is a rectangle, then it is a parallelogram. If a parallelogram is a rectangle, then its diagonals are congruent. If one angle of a parallelogram is a right angle, then the parallelogram is a rectangle.
Do squares have congruent diagonals?
The diagonals of a square are the same length (congruent).
Do square diagonals bisect?
In other words, a square is a regular polygon with four sides. The diagonals of a square bisect one another and are perpendicular (illustrated in red in the figure above). In addition, they bisect each pair of opposite angles (illustrated in blue).
### What shapes congruent diagonals?
A B
in these quadrilaterals, the diagonals bisect each other paralellogram, rectangle, rhombus, square
in these quadrilaterals, the diagonals are congruent rectangle, square, isosceles trapezoid
in these quadrilaterals, each of the diagonals bisects a pair of opposite angles rhombus, square
How do you know if diagonals bisect each other?
The diagonals of a parallelogram bisect each other. In any parallelogram, the diagonals (lines linking opposite corners) bisect each other. That is, each diagonal cuts the other into two equal parts. In the figure above drag any vertex to reshape the parallelogram and convince your self this is so.
Is diagonal of square equal to sides?
All four angles of a square are equal (each being 360°/4 = 90°, a right angle). All four sides of a square are equal. The diagonals of a square are equal.
## Are diagonals of square equal?
All four sides of a square are equal. The diagonals of a square are equal.
What does it mean if diagonals are congruent?
The diagonals are congruent and bisect each other (divide each other equally). Opposite angles formed at the point where diagonals meet are congruent. A rectangle is a special type of parallelogram whose angles are right.
Are the diagonals of rectangle equal?
A rectangle is a quadrilateral in which all angles are right angles. A rectangle is a parallelogram, so its opposite sides are equal. The diagonals of a rectangle are equal and bisect each other.
### Why are the diagonals of a square bisect each other?
In a square, the diagonals bisect each other. This is a general property of any parallelogram. And as a square is a special parallelogram, which has all the parallelogram’s basic properties, this is true for a square as well.
How are the diagonals of quadrilaterals always congruent?
A B these quadrilaterals always have perpend rhombus, square if you divide a square into four right t 45 degrees the diagonals of a rhombus… are not always congruent, but they are a the diagonals of a rectangle… are not always perpendicular, but they a
How are the diagonals of a square and a parallelogram alike?
A parallelogram’s diagonals bisect each other. The diagonals of a square are perpendicular and congruent, bisect each other, and bisect opposite angles. A square is a parallelogram that is both a rhombus and a rectangle. and therefore shares all of their characteristics.
## How are the diagonals of a square and a rectangle related?
The diagonals of a square are perpendicular and congruent, bisect each other, and bisect opposite angles. A square is a parallelogram that is both a rhombus and a rectangle. and therefore shares all of their characteristics. A square, rhombus and rectangle share all of their characteristics. |
# Proportions
In this lesson, you will learn to write proportions and solve problems using proportions. You will also learn to use the properties of similar figures and proportions to solve for unknown lengths, perimeters, and areas.
First, we need to look at how units are used in our ratio problems. Take a look at the problem below.
Heather knew that she was 60 inches tall and that her younger brother, Ryan, was 5 feet tall. She claimed that she was 12 times as tall as Ryan because 5 times 12 is 60. Ryan disagreed with Heather and claimed that they were the same height. Can you see how important it is to label your units when they are different?
Remember that a ratio is the comparison of two quantities by division. A ratio can be written as a fraction, in words, or with colon notation. Please use the fraction form in this class so we can properly set up and solve problems.
• 22 miles to 1 gallon
• 22 miles : 1 gallon
A ratio should never be written as a mixed number and must have units labeled when these units are different.
The way that a ratio is written tells which value is written as the numerator and which is the denominator. For example, if you compare 60 miles to 1 hour, you will write these values as . The units in a ratio help you remember the order in which to write it, especially when you are comparing ratios. If you are comparing the same types of things, you do not have to use the units. For example, if you compare 12 feet to 8 feet, you can write . You can simplify the fractions to make comparisons easier, but remember not to use mixed numbers. This ratio will simplify to .
Try writing a few on your own to make sure you understand the concept.
## Writing Ratios Practice
1. What is the ratio of 120 miles to 3 hours?
2. Write a ratio comparing the time it takes to drive from Atlanta to Richmond (7 hrs) to the time it takes to fly (1 hour).
3. Write the ratio of 6 feet to 9 feet.
4. Write the ratio of 90 seconds to 15 seconds.
5. Write a ratio of 17 days to 51 days.
Sometimes ratios involve two different units representing time and two different units representing distance. For example, the ratio of 8 feet to 2 yards is the same as the ratio of 8 feet to 6 feet. In cases such as these, we need to be sure the units are the same when we write the ratios.
The ratio of 2 hours to 120 minutes would be written as . The ratio of 6 inches to 3 feet could be written as . The ratio of four eggs to two dozen eggs could be written as eggs.
In your vocabulary words, we see that a proportion is an equation that states that two ratios are equivalent. There is more than one strategy for finding equivalent fractions and solving proportions. First, you can use the Identity Property of Multiplication.
can be solved as
To check your work, simply cross multiply to confirm that these ratios are proportional.
7 * 96 = 672 and 12 * 56 = 672
Let’s look at the same problem and cross multiply to set up an equation to find x.
Cross multiply to get 7 times x and 12 times 56. Write this as an equation.
7x = 672 Now, simply solve this equation by dividing both sides by 7. x = 96.
There will be times when using the Identity Property of Multiplication will be the simpler solution, but when numbers are easily factored, cross multiplying to find the missing value, or cross multiplying to check for proportionality is the best way to go. Just remember to use your units of measure appropriately as you solve these types of ratio problems.
Using the same format, ratios of corresponding sides of similar figures can be used to check for proportionality and to solve for missing side lengths. Remember that similar figures have the same shape, where corresponding angles are congruent and corresponding sides are proportional. Take a look at the figures below.
It appears that these two figures are similar shapes, meaning their angles are equal and their corresponding sides are proportional. Corresponding sides are sides in different polygons that are in the same relative position. In the figure above, . These sides are in the same relative positions in two different polygons. Also, . We can use this information to write ratios to check for proportionality. If these corresponding sides are proportional, then it looks as though we have similar shapes. Simply substitute side lengths for the side labels make sure that you set up the proportion correctly.
BA = 4 and ED = 8, so write the ratio as .
AC = 10 and DF=20, so write the ratio as .
Now, write these ratios as a proportion and cross multiply to check for proportionality.
; 4 * 20 = 80 and 8 * 10 = 80, so the shapes are proportional.
NOTE: You can also simplify each ratio and get – this also shows that these are proportional side lengths creating similar shapes.
Using the same process, you can find missing side lengths from similar shapes.
First identify the corresponding sides and set up a proportion.
Now, cross multiply and set up an equation
54x = 1512
Divide both sides by 54 to find x.
x = 28. The missing side length is 28.
You can also simplify 54 over 36 prior to cross multiplying.
Now that you have seen how to write and solve proportions with ratios and corresponding side lengths, we need to look at proportion word problems and how to set those up. Check out the following example.
A five-pound pack of flour costs \$1.80 and contains 12 cups of flour. Rachel is making cookies where the recipe calls for 2 cups of flour. Determine the cost for the flour needed for the cookies. Set up a proportion to solve the problem.
Set up the problem as you read. \$1.80 is your first number and corresponds to 12 cups of flour. and the next number you see corresponds to the denominator in your first ratio, so 2 will be the denominator in your second ratio; .
. Now, cross multiply and set up an equation.
x = \$.30 (2 cups of flour will be valued at 30 cents).
## Solve Proportions Practice
Set up a proportion for each word problem and solve.
1. Mrs. Johnson drove 105 miles in two and a half hours. What was Mrs. Johnson’s speed in miles per hour?
2. A restaurant charges a single price for its buffet. The total bill for a table of 6 having the buffet was \$294. Each of the 8 people at a second table had the same buffet. What was the total bill at the second table?
3. On the Milestone test, Jacob answered the first 20 questions in 5 minutes. There are 76 questions on the test. If he continues to answer questions at the same rate, how long will it take him to complete the test from start to finish?
4. Will knows that a 45-ounce pitcher can hold enough lemonade for 6 people. At this rate, how many ounces of lemonade will he need to serve 26 people?
5. Two rectangular desks are similar. The larger one is 42 inches long and 18 inches wide. The smaller one is 35 inches long. What is the width of the smaller desk?
6. The official size of a basketball court in the NBA is 94 feet by 50 feet. The basketball court in the school gym is 47 feet long. How wide must it be to be similar to an NBA court?
(source) |
# Dear Family, your Operation Maths guide to Directed Numbers
## Dear Family, your Operation Maths guide to Directed Numbers
Category : Uncategorized
Dear Family, below is a brief guide to understanding the topic of directed numbers (fifth and sixth classes only), as well as some practical suggestions as to how you might support your children’s understanding at home. Also below, are a series of links to digital resources that will help both the children, and you, learn more about directed numbers.
#### Understanding Directed Numbers
Directed Numbers are numbers with both size and direction; one direction is positive, and the other is negative. For example, temperature is typically described in a number of degrees either above zero (positive values) or below zero (negative values). Positive and negative numbers are also referred to as integers.
In Operation Maths, the children are first introduced to directed numbers in fifth class, where the focus is on the children appreciating where directed numbers can be encountered in real life, for example:
• Temperature
• Bank statements/ money accounts: having money is shown as positive (+) and owing money or overdrafts are shown as negative (-)
• Elevations above and below sea level
• Floors below the ground floor in a large building are often labeled as -1, -2 etc
• Golf scores are written as above and below par
• Goal difference in soccer league tables
• Depths in a swimming pool
In school, the children are encouraged to use the words positive and negative, rather than plus and minus; for example for the value -6 we should say negative six rather than minus six. This is particularly important for when the children start adding positive and negative numbers (in Operation Maths 6): for example (–3) + (+9) should be read as ‘negative three add/plus positive nine’ rather than ‘minus 3 plus plus 9’. It is also important that the children recognise that positive numbers can be written either with, or without, the positive sign, therefore we can assume that any number, without a sign, is positive.
#### Practical Suggestions for all Children
• Draw your child’s attention to wherever they or you encounter directed numbers (see above for possible examples).
• If your child is having difficulty visualising, comparing, ordering etc directed numbers, encourage them to think of a real example. One of the most-relatable of these is that of temperatures and the thermometer. Look at a real thermometer or use an online virtual example such as this one.
• As explained above, encourage your child to use the words positive and negative, rather than plus and minus, when describing directed numbers.
#### Digital Resources for Fifth and Sixth Classes
Integers: Video lesson that introduces integers (positive and negative numbers) and where they occur in real life
Mashup Math – Elevation above or below sea-level: A video lesson that explores elevation as examples of positive and negative numbers.
Negative Numbers: A series of video lessons from White Rose Maths, including Introducing Negative Numbers (Year 4), Negative Numbers (Year 4), Negative Numbers (Year 5), Negative Numbers in Context (Year 6), Negative Numbers (Year 6), Add and Subtract Integers (Year 6).
Khan Academy Negative Numbers: A series of videos and practice questions exploring negative numbers. Afterwards, for more of a challenge, look at Integers and Whole Numbers. You can also register for a free Khan Academy account to record your progress and explore other topics/grades.
Maths Frame – Empty Number Line: Practice placing positive and negative integers in the correct position; choose -5 to 5 initially and then move to the next levels to challenge yourself.
What is the Temperature? Identify the temperature shown on the thermometers. A way to practice directed numbers in a real-life context. Includes values above and below zero, and has options for various ranges.
Caterpillar Ordering: Choose Ordering and then -10 to 10 to order integers.
Coconut Ordering: Hit the numbers in order of size. Select ‘numbers’ and then choose from numbers from -10 to 10 to order integers.
Number Lines in Disguise: A challenge and interactive game (scroll down) from NRICH; Can you work out the number marked by the dot? Includes positive and negative numbers.
That Quiz: Inequalities for comparing numbers, Arithmetic for addition calculations involving directed numbers and Number Line for identifying numbers by their position. In each of these activities there is the option to include negative numbers along with positive; just make sure the the “negative” option is selected on the left hand side.
Number Line App: This virtual tool can be used to explore the position of values on a number line,including positive and negative numbers; also to model addition and subtraction involving positive and negative numbers.
Directed Numbers Counters: These double sided counters can be used to model addition and subtraction involving positive and negative numbers (ensure that the “sign” option is ticked).
Integers (Directed Numbers): a selection of practice games from ixl.com. You can do a number of free quizzes each day without having a subscription. Start with the second class games and work up through the activities.
Integers: Practice games from Math Games. Choose your class level. |
# Introduction to Euclid’s Geometry RD Sharma Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.1
In this chapter, we provide RD Sharma Solutions for Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.1 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.1 pdf, free RD Sharma Solutions for Class 9 Solutions Chapter 7 Introduction to Euclid’s Geometry Ex 7.1 book pdf download. Now you will get step by step solution to each question.
## Introduction to Euclid’s Geometry RD Sharma Class 9 Solutions
### RD Sharma Solutions Class 9 Chapter 7 Introduction to Euclid’s Geometry Ex 7.1
Question 1.
Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case:
(i) -2x + 3y = 12
(ii) x-(frac { y }{ 2 }) -5 = 0
(iii) 2x + 3y = 9.35
(iv) 3x = -7y
(v) 2x + 3 = 0
(vi) y – 5 = 0
(vii) 4 = 3x
(viii) y = (frac { x }{ 2 })
Solution:
(i) -2x + 3y = 12
⇒ -2x + 3y – 12 = 0
Here a -2, b = 3, c = – 12
(ii) x – (frac { y }{ 2 }) -5 = 0
Here a = 1, b =(frac { 1 }{ 2 }) ,c = -5
(iii) 2x + 3y = 9.35
⇒ 2x + 3y – 9.35 = 0
Here a = 2, b = 3, c = – 9.35
(iv) 3x = -7y
⇒ 3x + 7y + 0 = 0
Here a = 3, b = 7,c = 0
(v) 2x + 3 = 0
⇒ 2x + 0y + 3 = 0
Here a = 2, b = 0, c = 3
(vi) y-5 = 0 ⇒ ox+y-5 = 0
Here a = 0, b = 1, c = -5
(vii) 4 = 3x
⇒ 3x – 4 = 0
⇒ 3x + 0y – 4 = 0
Here a = 3, b = 0, c = -4
(Viii) y= (frac { x }{ 2 })
⇒ (frac { x }{ 2 }) – y+ 0 = 0
⇒ x-2y + 0 = 0
Here a = 1, y = -2, c = 0
Question 2.
Write each of the following as an equation in two variables.
(i) 2x = -3
(ii) y = 3
(iii) 5x = (frac { 7 }{ 2 })
(iv) y =(frac { 3 }{ 2 })x
Solution:
(i) 2x = -3⇒ 2x + 3 = 0
⇒ 2x + 0y + 3 = 0
(ii) y= 3 ⇒ y-3=0
⇒ 0x+ y-3 = 0
(iii) 5x =(frac { 7 }{ 2 }) ⇒ 10x = 7
⇒ 10x + 0y – 7 = 0
(iv) y=(frac { 3 }{ 2 })x⇒2y = 3x
⇒ 3x – 2y + 0 = 0
Question 3.
The cost of ball pen is ₹5 less than half of the cost of fountain pen. Write this statement as a linear equation in two variables.
Solution:
Let cost of a fountain pen = ₹x
and cost of ball pen = ₹y
∴ According to the condition,
y = (frac { x }{ 2 }) -5⇒ 2y = x – 10
⇒ x – 2y – 10 = 0
### Introduction to Euclid’s Geometry RD Sharma Class 9 Solutions
2 (i)
Infinitely many
2 (ii)
one
3 (i)
one
3 (ii)
PQ,QR,PR
All Chapter RD Sharma Solutions For Class 9 Maths
*************************************************
I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on 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.
If these solutions have helped you, you can also sharencertsolutionsfor.com to your friends. |
# Complex conjugates and dividing complex numbers
## Complex conjugates
In order to divide a complex number by another, we need to understand the concept of a complex conjugate.
Definition: A complex conjugate of a complex number $$z = x+yi$$, where $$x$$ and $$y$$ are real, is defined as $$\overline{z} = x-yi.$$
In other words, the imaginary part changes sign.
Worked exercise: Evaluate $$\overline{-2-i}$$ and $$\overline{9}$$.
Solution: $$\overline{-2-i} = -2 + i$$ and $$\overline{9} = 9$$
Notice that the conjugate of a real number is itself.
## Dividing complex numbers
Complex conjugates are useful because when you multiply a complex number by its conjugate, you get a real number. This allows us to simplify a fraction of complex numbers by multiplying the top and bottom by the conjugate of the denominator, to “realise the denominator”.
Worked exercise: Find the conjugate of $$z = 4+i$$ and show that $$z \overline{z}$$ is a real number. Hence, evaluate $$\frac{3-4i}{4+i}.$$
Solution: The conjugate of $$4 + i$$ is $$4-i$$. We have $$z \overline{z} = (4+i)(4-i) = 4^2-i^2 = 17.$$ To evaluate the quotient, we have $$\frac{3-4i}{4+i} = \frac{3-4i}{4+i} \times \frac{4-i}{4-i} = \frac{(3-4i)(4-i)}{17} = \frac{8-19i}{17}.$$
## Problems
1. Evaluate the following:
1. $$\overline{5+6i}$$
2. $$\overline{i}$$
3. $$\overline{-4+6i}$$
4. $$\overline{8+8i}$$
5. $$\overline{ \overline{1-i} + \overline{4-3i} }$$
2. Simplify the following to $$x+yi$$ form:
1. $$\displaystyle \frac{1+i}{1-i}$$
2. $$\displaystyle\frac{-2-i}{-5-4i}$$
3. $$\displaystyle\frac{9+6i}{1+4i}$$
4. $$\displaystyle\frac{5-7i}{-10-i}$$
5. $$\displaystyle\frac{9+2i}{5+8i}$$ |
+0
Rhombus
0
263
2
ABCD is a rhombus. If angle A is 129 degrees and the perimeter of ABCD is 60, then find the length of each diagonal.
Feb 20, 2022
#1
+2666
0
Draw the rhombus, its diagonals and label the points.
You know the hypotenuse of one of the smaller triangles formed by the diagonals 15.
The diagonals split the triangle into right triangles with one angle of 64.5 degrees, and a hypotenuse of 15.
To find the longer side, we do $$15\times \sin(64.5) \approx13.53877927$$
Multiply this by 2, and one diagonal is: $$30 \times \sin(64.5) \approx \color{brown}\boxed{27.07755853}$$
The other length of the original right triangle is: $$15 \times \cos(64.5) \approx {6.457666452}$$
Multiply this by 2, and the other diagonal is: $${30 \times \cos(64.5)} \approx \color{brown}\boxed{12.9153329}$$
Feb 20, 2022
#2
+23239
+1
Another way to get the same answer that BuilderBoi has:
Since the perimeter of the rhombus is 60, each side is 15.
Let's start by drawing the rhombus:
each side is 15;
angle(C) = angle(A) = 129o;
angle(B) = angle(D) = 180o - 129o = 51o.
To find the longer diagonal, DB, use the Law of Cosines on triangle(ABD):
DB2 = DA+ AB2 - 2·DA·AB·cos(A)
DB2 = 152 + 152 - 2·15·15·cos(129)
DB2 = 733.194176
DB = 27.08
To find the shorter diagonal, AC, use the Law of Cosines on triangle(CDA):
AC2 = CD2 + DA2 - 2·CD·DA·cos(51)
Finish this to get the length of the shorter diagonal.
Feb 21, 2022
edited by geno3141 Feb 21, 2022 |
# How do you factor x² - 4x - 45?
Jul 6, 2015
Find two factors of $45$ whose difference is $4$. The numbers $9$ and $5$ work.
Hence ${x}^{2} - 4 x - 45 = \left(x - 9\right) \left(x + 5\right)$
#### Explanation:
$\left(x - a\right) \left(x + b\right) = {x}^{2} - \left(a - b\right) x - a b$
So given ${x}^{2} - 4 x - 45$, look for $a$ and $b$
such that $a b = 45$ and $a - b = 4$.
Jul 6, 2015
Alternatively, complete the square to find:
${x}^{2} - 4 x - 45 = {\left(x - 2\right)}^{2} - {7}^{2} =$
$\left(x - 2 - 7\right) \left(x - 2 + 7\right) = \left(x - 9\right) \left(x + 5\right)$
#### Explanation:
${x}^{2} - 4 x - 45$
$= {x}^{2} - 4 x + 4 - 4 - 45$
$= {\left(x - 2\right)}^{2} - 49$
$= {\left(x - 2\right)}^{2} - {7}^{2}$
This is a difference of squares, so we can use the identity:
${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$
Let $a = x - 2$ and $b = 7$
Then
${\left(x - 2\right)}^{2} - {7}^{2}$
$= \left(x - 2 - 7\right) \left(x - 2 + 7\right) = \left(x - 9\right) \left(x + 5\right)$ |
# Find the general solution of each of the following equations:
Question:
Find the general solution of each of the following equations:
(i) $\cos x=\frac{-1}{2}$
(ii) $\operatorname{cosec} x=-\sqrt{2}$
(iii) $\tan x=-1$
Solution:
To Find: General solution.
Formula used: $\cos \theta=\cos \alpha \Rightarrow \theta=2 \mathrm{n} \pi \pm \alpha, \mathrm{n} \in I$
By using above formula, we have
$\cos x=\frac{-1}{2}=-\cos \left(\frac{\pi}{3}\right)=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \left(\frac{2 \pi}{3}\right) \Rightarrow x=2 n \pi \pm \frac{2 \pi}{3}, n \in I$
So general solution is $x=2 n \pi \pm \frac{2 \pi}{3}$ where $n \in$ ।
(ii) Given: $\operatorname{cosec} x=-\sqrt{2}$
We know that $\operatorname{cosec} \theta \times \sin \theta=1$
So $\sin x=\frac{-1}{\sqrt{2}}$
Formula used: $\sin \theta=\sin \alpha \Rightarrow \theta=n \pi+(-1)^{n} \alpha, n \in$
By using above formula, we have
$\sin x=\frac{-1}{\sqrt{2}}=\sin \frac{5 \pi}{4} \Rightarrow x=n \pi+(-1)^{n} \cdot \frac{5 \pi}{4}$
So general solution is $x=n \pi+(-1)^{n} \cdot \frac{5 \pi}{4}$ where $n \in I$
(iii) Given: $\tan x=-1$
Formula used: $\tan \theta=\tan \alpha \Rightarrow \theta=n \pi+\alpha, n \in I$
By using above formula, we have
$\tan x=-1=\tan \frac{3 \pi}{4} \Rightarrow x=n \pi+\frac{3 \pi}{4}, n \in I$
So the general solution is $x=n \pi+\frac{3 \pi}{4}$ where $n \in$ । |
# Rate of change and slope practice and problem solving, also,...
If a line rises 4 units for every 1 unit that it runs, the slope is 4 divided by 1, or 4. Special Note: If we find the slope we can find the rate of change over that period. Time is always an x value. The terms gentle or steep describe a slope verbally, not mathematically.
At this site you will find descriptions advantages and disadvantages of drawing up a business plan fares of three competing cab companies and you are asked to suggest a new fare. Point 1 has coordinates x1, y1 and point 2 has coordinates x2, y2. Take a look at the graph below. So this is how you will most often see the slope formula written in algebra: Take a look at the following graph.
Slope Formula If you've never used this formula before, please visit our page on using the slope formula. By finding the slope of the line, we would be calculating the rate of change. Tangent Lines and Rates of Change In this chef de partie cover letter template we are going to take a look at two fairly important problems in the study of calculus.
First, both of these problems will lead us into the study of limits, which is the topic of this chapter after all. Also, do not worry about how I got the exact or approximate slopes. For example, compare a distance-time graph to a distance-time equation to determine which of two moving objects has greater speed.
Example 2: Its table of values?
In general, we will think of a line and a graph as being parallel at a point if they are both moving in the same direction at that point. Therefore, we should always take a look at what is happening on both sides of the point in question when doing this kind of process. The correct answer is the vertical change divided by the horizontal change between two points on a line.
Therefore, our two ordered pairs are 0, and 12, Check student understanding by reading their papers. We can also find the slope of a straight line if we know the coordinates of sncf ibm case study two points on that line.
Ask them to compare their results with their group members.
While observing the groups look for different ways groups answered questions 2, 3 and 4. This is all that we know about the tangent line. Calculating Slope from a Graph We can find the slope of a line on a graph by counting off the rise and the run between two points. National Bullying Prevention Month.
This gives us an "overview" of John's savings per month.
## Slope Formula
We can also find the slope of a straight line if we rate of change and slope practice and problem solving the coordinates of any two points on that line. Select the group or groups to present each solution. Walk around observing the groups work. The substitutions are as follows: A large number like this indicates a steep slope: Explanation from Regents Prep assessment preparation site [from the Internet archive, the Wayback Machine.
A small number like this represents a gentle slope. Today we will be extending our learning about the slope of a line and the rate of change.
The slope is the rate of change from one month to the next. That is the opposite of the slope formula.
## Rate of Change and Slope
The correct answer is the vertical change divided by the horizontal change between two points on a line. B Incorrect. A Correct. Section Point 1 has coordinates x1, y1 and point 2 has coordinates x2, y2.
## Rate of change and slope practice and problem solving a/b - Slope and Rate of Change
B Incorrect. This line goes up only 1 step while it travels 4 steps to the side. The ratio of rise over run describes the slope of all straight lines. Have the students determine the slope from A to B and the slope from B to C. Write sncf ibm case study as a ratio and compare it to the slope in the equation. Ask student how long it would take to walk 3 meters at this rate.
The terms gentle or steep describe a slope verbally, not mathematically. MathAlgebrarate of changeslopes. In most real u miami creative writing problems, your units will not be the same on the x and y axis.
Slope is the difference between the y-coordinates divided by the difference between the x-coordinates. GED Math Practice Test Slope Intercept Jeopardy rate of change and slope practice and problem solving This show is in Jeopardy format; with five categories each with five questions - good for a rate of change and slope practice and problem solving class review Straight Lines and Slopes - y-Intercept - Designed as a bellwork assignment, this show gives practice at reading slope and intercept from graphs [15 slides].
How did you use these to find the slope? Applet for Exploring Linear Functions http: Compare two different proportional relationships represented in different ways.
• So, looking at it now will get us to start thinking about it from the very beginning.
• National Dyslexia Awareness Month.
• Rate of change and slope practice and problem solving a/b
• The vertical change divided by the horizontal change between two points on a line.
Ask them to demonstrate a line with a positive slope, a negative slope, a line with a font size for term paper steep slope, a line with a less steep slope, a zero slope and no slope. This line goes up only 1 step while it travels 4 steps to the side. If a line rises 4 units for every 1 unit that it runs, the slope is 4 divided by 1, or 4.
## Find the slope of a line step-by-step
Ask students to recap what they learned rate of change and slope practice and problem solving slope from yesterday. The vertical change divided by the horizontal change between two points on a line. The slope is equal to So, in the first point above the graph and the line are moving in the same direction and so we will say they are parallel at that point.
A group may be selected based on a common misunderstanding that the teacher would like to clear up for the students.
C Incorrect. Before getting into this problem it would probably be best to define a tangent line. If a line rises 4 units for every 1 unit that it runs, the slope is what makes a good personal reflective essay divided by 1, or 4. So this is how you will most often see the slope formula written in algebra: The line passes through the points 1, 4 and -1, 8.
The y value tells us where the point is vertically. The rise is the font size for term paper distance between the two points, which is the difference between their y-coordinates.
## Example 2: Rate of Change
Look for understanding of the standards written in Write the learning objectives on the board. Round your answer to the nearest dollar. Cab fares - The charge for a cab ride is determined by the length of travel. Mathematicians commonly use the letter m to represent slope.
• Slope and Rate of Change
• Slope Calculator - Symbolab
• Solutions to Algebra 1 Common Core () :: Homework Help and Answers :: Slader
• Creative writing for beginners course modelo de curriculum vitae en wordpad para editar critical thinking slownik
National What makes a good personal reflective essay Awareness Month. The terms gentle or steep describe a slope verbally, not mathematically. Ask a student to restate the problem in their own words. Time is always an x value.
## Rate of Change and Slope
Do this a few times until the students get a feel for the speed of 1. Save Common Core Tags Close. John would like to find out how much money he saved per month for the year. The line passes through the points 1, 4 and -1, 8.
## DDoS protection.
Understand the connections between proportional relationships, lines, and linear crossmark research paper. Ask student how long it would take to walk 1 meter at this rate. We can now use the slope formula to find the slope of the line. In order rate of change and slope practice and problem solving find the tangent line we need either a second point or the slope of the tangent line.
The Hot Tub - This is a fun activity where students tell the story behind a graph and relate slope to rate of change.
This is most likely the initial year or the year rate of change and slope practice and problem solving house was built. Ask student to find the slope from A to C. Looking at these problems here will allow us to start to understand just what a limit is and what it can tell us rate of change and slope practice and problem solving a function.
Connecting Slope to Real Life Why do we need to find the slope of a line in real life? A small number like this represents a gentle slope. Argumentative essay on twitter how to shade the slope triangle by moving up or down and then to the right. |
# Latihan Tentang Ordinal 3
Soal-soal berikut ini adalah tentang aritmetika ordinal, yaitu bermain-main operasi jumlah, kali, perpangkatan dengan bilangan-bilangan ordinal.
Latihan 11. Periksa pernyataan berikut:
1. $\omega + 1 = \omega'$
2. $1 + \omega = \omega + 1$
3. $1 + 3 = 3 + 1$.
Latihan 12. Periksa pernyataan berikut:
1. $1 \times 3 = 3 \times 1$
2. $2 \times \omega = \omega + \omega$
3. $\omega \times 2 = \omega + \omega$
Latihan 13. Periksa pernyataan berikut:
1. $(\omega + 2) + \omega < (\omega + \omega) + 3$
2. $(\omega \times 2) + \omega < (\omega \times \omega) + 2$
3. $(\omega \times \omega) < \omega \times (2 \times \omega)$.
Latihan 14. Apakah berlaku $\omega \times (\omega + \omega) = (\omega \times \omega) + (\omega \times \omega)$.
Latihan 11.
1. $\omega + 1 = (\omega + 0)' = \omega'$. Benar.
2. $1 + \omega = \{ x: x \in 1+n \} = \omega \}$. Salah.
3. $1 + 3 = (1+2)' = (1+1)''=(1+0)'''=1'''=0''''$.
$3 + 1 = (3+0)' = 3' = 0''''$.
Benar.
Latihan 12.
1. $1 \times 3 = (1 \times 2) + 1 = (1 \times 2)' = ((1 \times 1) + 1)' = (1 \times 1)'' = ((1 \times 0) + 1)''$ $= (1 \times 0)''' = 0'''$.
$3 \times 1 = (3 \times 0) + 3 = 0 + 3 = (0 + 2)' = (0 + 1)'' = (0+0)''' = 0'''$.
Benar.
2. $2 \times \omega = \{ x: x \in 2 \times n \} = \omega$.
$\omega \ne \omega + \omega$.
Salah.
3. $\omega \times 2 = (\omega \times 1) + \omega = ((\omega \times 0) + \omega) + \omega = (0 + \omega) + \omega$ di mana $0 + \omega = \{ x : x \in 0+n \} = \omega$. Jadi, $\omega \times 2 = \omega + \omega$. Benar.
Latihan 13.
1. $(\omega + 2) + \omega = \omega + (2 + \omega) = \omega + \omega < (\omega + \omega) + 3$. Benar.
2. $(\omega \times 2) + \omega < (\omega \times 3) < (\omega \times \omega) < (\omega \times \omega)+3$. Benar.
3. $\displaystyle \omega \times (2 \times \omega) = \omega \times \bigcup_{\beta < \omega} 2 \times \beta = \omega \times \bigcup_{n=0}^\infty 2 \times n = \omega \times \omega$. Salah.
Latihan 14.
$\omega \times (\omega + \omega) = \omega \times (\omega \times 2) = (\omega \times \omega) \times 2 = (\omega \times \omega) + (\omega \times \omega)$. Perhatikan bukti ini menggunakan sifat asosiatif dari perkalian. |
# How do you complete the proportion 4/5=x/15?
Jan 13, 2017
Multiply each side of the equation by $15$ to find $x$. See full process below:
#### Explanation:
Multiply each side of the equation by $\textcolor{red}{15}$ to find $x$.
$\textcolor{red}{15} \times \frac{4}{5} = \textcolor{red}{15} \times \frac{x}{15}$
$\frac{60}{5} = \cancel{\textcolor{red}{15}} \times \frac{x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{15}}}}$
$12 = x$
Substituting $12$ for $x$ gives:
$\frac{4}{5} = \frac{12}{15}$ |
Experimental Probability – Explanation & Examples
Have you ever tossed a die multiple times hoping to get a 6 but get none? Since probability is the study of chance it makes sense that what we expect is not always what we get. This brings us to experimental probability and its definition.
Experimental probability is the probability determined based on the results from performing the particular experiment.
In this lesson we will go through:
• The meaning of experimental probability
• How to find experimental probability
What is experimental probability?
Theoretically, if you toss a die six times, you should expect to get one 6. However, if you’ve played any game involving dice you know that is not always the case. This is because the probability you get after performing an experiment may be different from what you expected. We can restate the definition of experimental probability as:
The ratio of the number of outcomes favorable to an event to the total number of trials of the experiment.
Experimental Probability can be expressed mathematically as:
$P(\text{E}) = \frac{\text{number of outcomes favorable to an event}}{\text{total number of trials of the experiment}}$
Let’s go back to the die tossing example. If after 12 throws you get one 6, then the experimental probability is $\frac{1}{12}$. You can compare that to the theoretical probability. The theoretical probability of getting a 6 is $\frac{1}{6}$. This means that in 12 throws we would have expected to get 6 twice.
Similarly, if in those 12 tosses you got a 1 five times, the experimental probability is $\frac{5}{12}$.
Does this mean that something is wrong with the theory of probability? Certainly not! As your number of trials of an experiment increases, you will notice that the experimental probability gets closer to the theoretical. For example, if you were to toss the die hundreds of times, the experimental probability will get closer or equal to the theoretical.
How do we find experimental probability?
Now that we understand what is meant by experimental probability, let’s go through how it is found.
To find the experimental probability of an event, divide the number of observed outcomes favorable to the event by the total number of trials of the experiment.
Let’s go through some examples.
Example 1: There are 20 students in a class. Each student simultaneously flipped one coin. 12 students got a Head. From this experiment, what was the experimental probability of getting a head?
Solution:
Number of coins showing Heads: 12
Total number of coins flipped: 20
$P(\text{Head}) = \frac{12}{20} = \frac{3}{5}$
Example 2: The tally chart below shows the number of times a number was shown on the face of a tossed die.
Number Frequency 1 4 2 6 3 7 4 8 5 2 6 3
a. What was the probability of a 3 in this experiment?
b. What was the probability of a prime number?
Solution:
First, sum the numbers in the frequency column to see that the experiment was performed 30 times. Then find the probabilities of the specified events.
a. Number of times 3 showed = 7
Number of tosses = 30
$P(\text{3}) = \frac{7}{30}$
b. Frequency of primes = 6 + 7 + 2 = 15
Number of trials = 30
$P(\text{prime}) = \frac{15}{30} = \frac{1}{2}$
Experimental probability can be used to predict the outcomes of experiments. This is shown in the following examples.
Example 3: The table shows the attendance schedule of an employee for the month of May.
a. What is the probability that the employee is absent?
b. How many times would we expect the employee to be present in June?
Sunday Monday Tuesday Wednesday Thursday Friday Saturday 1Present 2Present 3Present 4Present 5Present 6Present 7Present 8Present 9Present 10Present 11Present 12Present 13Present 14Present 15Present 16Present 17Absent 18Present 19Present 20Present 21Absent 22Absent 23Present 24Present 25Present 26Present 27Present 28Present 29Present 30Present 31Present
Solution:
a. The employee was absent three times and the number of days in this experiment was 31. Therefore:
$P(\text{Absent}) = \frac{3}{31}$
b. We expect the employee to be absent
$\frac{3}{31} × 30 = 2.9 ≈ 3$ days in June
Example 4: Tommy observed the color of cars owned by people in his small hometown. Of the 500 cars in town, 10 were custom colors, 100 were white, 50 were red, 120 were black, 100 were silver, 60 were blue, and 60 were grey.
a. What is the probability that a car is red?
b. If a new car is bought by someone in town, what color do you think it would be? Explain.
Solution:
a. Number of red cars = 50
Total number of cars = 500
$P(\text{red car}) = \frac{50}{500} = \frac{1}{10}$
b. Based on the information provided, it is most likely that the new car will be black. This is because it has the highest frequency and the highest experimental probability.
Now it is time for you to try these examples.
Practice Questions
The table below shows the colors of jeans in a clothing store and their respective frequencies. Use the table to answer the questions that follow.
Color of Jeans Frequency Blue 75 Black 60 Grey 45 Brown 25 White 20
1. What is the probability of selecting a brown jeans?
2. What is the probability of selecting a blue or a white jeans?
Question 3
On a given day, a fast food restaurant notices that it sold 110 beef burgers, 60 chicken sandwiches, and 30 turkey sandwiches. From this observation, what is the experimental probability that a customer buys a beef burger?
Question 4
Over a span of 20 seasons, a talent competition notices the following. Singers won 12 seasons, dancers won 2 seasons, comedians won 3 seasons, a poet won 1 season, and daring acts won the other 2 seasons.
a. What is the experimental probability of a comedian winning a season?
b. From the next 10 seasons, how many winners do you expect to be dancers?
Question 5
Try this at home! Flip a coin 10 times. Record the number of tails you get. What is your P(tail)?
Solutions
Question 1
Number of brown jeans = 25
Total Number of jeans = 125
$P(\text{brown}) = \frac{25}{125} = \frac{1}{5}$
Question 2
Number of jeans that are blue or white = 75 + 20 = 95
Total Number of jeans = 125
$P(\text{blue or white}) = \frac{95}{125} = \frac{19}{25}$
Question 3
Number of beef burgers = 110
Number of burgers (or sandwiches) sold = 200
$P(\text{beef burger}) = \frac{110}{200} = \frac{11}{20}$
Question 4
a. Number of comedian winners = 3
Number of seasons = 20
$P(\text{comedian}) = \frac{3}{20}$
b. First find the experimental probability that the winner is a dancer.
Number of winners that are dancers = 2
Number of seasons = 20
$P(\text{dancer}) = \frac{2}{20} = \frac{1}{10}$
Therefore we expect
$\frac{1}{10} × 10 = 1$ winner to be a dancer in the next 10 seasons.
Question 5
To find your P(tail) in 10 trials, complete the following with the number of tails you got.
$P(\text{tail}) = \frac{\text{number of tails}}{10}$ |
Percentage calculator v.1.
• Calculate the % of
Percentage calculator v.2.
• is what percent of
• %
Percentage calculator v.3.
• is % of what
Percentage change calculator
• Value 1: value 2:
• %
How to Calculate Percentages
For calculating percentage problems there are many formulas we can use. The most basic one would look like: X/Y = P x 100. The formulas listed below are all mathematical variations of this formula.
The three basic percentage problems would look like this. X and Y are number, P is the percentage:
a) Find P percent of X - Calculator 1
We will use the P% * X = Y formula here to find the percentage of a number. Let's calculate the 20% of 400:
• Let's convert the problem by using the percentage formula: P% * X = Y
• P is 20%, X is 400, so the equation is 20% * 400 = Y
• Convert 20% to a decimal by removing the percent sign and dividing by 100: 20/100 = 0.20
• Substitute 0.20 for 20% in the equation: 20% * 400= Y becomes 0.20 * 400 = Y
• Finally do the math: 0.20 * 400 = 80, Y = 80
• So 20% of 400 is 80
b) Find what percent of X is Y - Calculator 2
Here we will use the percentage formula: Y/X = P% Let's see what percent of 40 is 6?
• Convert the problem to an equation using the formula: Y/X = P%
• X is 40, Y is 6, so the equation is 6/40 = P%
• Do the math: 6/40 = 0.15
• The result will always be in decimal form, not percentage form. You need to multiply the result by 100 to get the percentage.
• Converting 0.15 to a percent: 0.15 * 100 = 15%
• So 15% of 40 is 6.
c) Find X if P percent of it is Y - Calculator 3
Let's use the percentage formula Y/P% = X to find out 10 is 20% of what?
• Convert the problem to an equation using the percentage formula: Y/P% = X
• Y is 10, P% is 20, so the equation is 10/20% = X
• Convert the percentage to a decimal by dividing by 100.
• Converting 20% to a decimal: 20/100 = 0.20
• Substitute 0.20 for 20% in the equation: 10/0.20 = X
• Do the math: 10/0.20 = X
• X = 50
• So 10 is 20% of 50
By purchasing products trough advertisements on the website, we might earn a small comission.
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# 2007 AIME I Problems/Problem 14
## Problem
A sequence is defined over non-negative integral indexes in the following way: $a_{0}=a_{1}=3$, $a_{n+1}a_{n-1}=a_{n}^{2}+2007$.
Find the greatest integer that does not exceed $\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}.$
## Solutions
### Solution 1
We are given that
$a_{n+1}a_{n-1}= a_{n}^{2}+2007$, $a_{n-1}^{2}+2007 = a_{n}a_{n-2}$.
Add these two equations to get
$a_{n-1}(a_{n-1}+a_{n+1}) = a_{n}(a_{n}+a_{n-2})$
$\frac{a_{n+1}+a_{n-1}}{a_{n}}= \frac{a_{n}+a_{n-2}}{a_{n-1}}$.
This is an invariant. Defining $b_{i}= \frac{a_{i}}{a_{i-1}}$ for each $i \ge 2$, the above equation means
$b_{n+1}+\frac{1}{b_{n}}= b_{n}+\frac{1}{b_{n-1}}$.
We can thus calculate that $b_{2007}+\frac{1}{b_{2006}}= b_{3}+\frac{1}{b_{2}}= 225$. Using the equation $a_{2007}a_{2005}=a_{2006}^{2}+2007$ and dividing both sides by $a_{2006}a_{2005}$, notice that $b_{2007}= \frac{a_{2007}}{a_{2006}}= \frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}> \frac{a_{2006}}{a_{2005}}= b_{2006}$. This means that
$b_{2007}+\frac{1}{b_{2007}}< b_{2007}+\frac{1}{b_{2006}}= 225$. It is only a tiny bit less because all the $b_i$ are greater than $1$, so we conclude that the floor of $\frac{a_{2007}^{2}+a_{2006}^{2}}{a_{2007}a_{2006}}= b_{2007}+\frac{1}{b_{2007}}$ is $224$.
### Solution 2
The equation $a_{n+1}a_{n-1}-a_n^2=2007$ looks like the determinant $$\left|\begin{array}{cc}a_{n+1}&a_n\\a_n&a_{n-1}\end{array}\right|=2007.$$ Therefore, the determinant of this matrix is invariant. Guessing that this sequence might be a linear recursion because of the matrix form given below, we define the sequence $b_n$ defined by $b_1=b_2=3$ and $b_{n+1}=\alpha b_n+\beta b_{n-1}$ for $n\ge 2$. We wish to find $\alpha$ and $\beta$ such that $a_n=b_n$ for all $n\ge 1$. To do this, we use the following matrix form of a linear recurrence relation
$$\left(\begin{array}{cc}b_{n+1}&b_n\\b_n&b_{n-1}\end{array}\right)=\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)\left(\begin{array}{cc}b_{n}&b_{n-1}\\b_{n-1}&b_{n-2}\end{array}\right).$$
When we take determinants, this equation becomes
$$\text{det}\left(\begin{array}{cc}b_{n+1}&b_n\\b_n&b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&b_{n-1}\\b_{n-1}&b_{n-2}\end{array}\right).$$
We want $$\text{det}\left(\begin{array}{cc}b_{n+1}&b_n\\b_n&b_{n-1}\end{array}\right)=2007$$ for all $n$. Therefore, we replace the two matrices by $2007$ to find that
$$2007=\text{det}\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)\cdot 2007$$ $$1=\text{det}\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)=-\beta.$$ Therefore, $\beta=-1$. Computing that $a_3=672$, and using the fact that $b_3=\alpha b_2-b_1$, we conclude that $\alpha=225$. Clearly, $a_1=b_1$, $a_2=b_2$, and $a_3=b_3$. We claim that $a_n=b_n$ for all $n\ge 1$. We proceed by induction. If $a_k=b_k$ for all $k\le n$, then clearly, $$b_nb_{n-2}-b_{n-1}^2=a_na_{n-2}-a_{n-1}^2=2007.$$ We also know by the definition of $b_{n+1}$ that
$$\text{det}\left(\begin{array}{cc}b_{n+1}&b_n\\b_n&b_{n-1}\end{array}\right)=\text{det}\left(\begin{array}{cc}\alpha&\beta\\1&0\end{array}\right)\text{det}\left(\begin{array}{cc}b_{n}&b_{n-1}\\b_{n-1}&b_{n-2}\end{array}\right).$$
We know that the RHS is $2007$ by previous work. Therefore, $b_{n+1}b_{n-1}-b_n^2=2007$. After substuting in the values we know, this becomes $b_{n+1}a_{n-1}-a_n^2=2007$. Thinking of this as a linear equation in the variable $b_{n+1}$, we already know that this has the solution $b_{n+1}=a_{n+1}$. Therefore, by induction, $a_n=b_n$ for all $n\ge 1$. We conclude that $a_n$ satisfies the linear recurrence $a_{n+1}=225a_n-a_{n-1}$.
It's easy to prove that $a_n$ is a strictly increasing sequence of integers for $n\ge 3$. Now
$$\frac{a_{2007}^2+a_{2006}^2}{a_{2007}a_{2006}}=\frac{a_{2007}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}=\frac{225a_{2006}-a_{2005}}{a_{2006}}+\frac{a_{2006}}{a_{2007}}.$$
$$=225+\frac{a_{2006}}{a_{2007}}-\frac{a_{2005}}{a_{2006}}=225+\frac{a_{2006}^2-a_{2005}a_{2007}}{a_{2005}a_{2006}}.$$
$$=225-\frac{2007}{a_{2005}a_{2006}}.$$
The sequence certainly grows fast enough such that $\frac{2007}{a_{2005}a_{2006}}<1$. Therefore, the largest integer less than or equal to this value is $224$.
### Solution 3 ( generalized )
This is a more elementary and rigorous solution to a slightly generalized version. The defining recursive sequence is generalized to
$$a_{n+1}a_{n-1} = a_n^2 + 9k, ---------(1)$$ where $k$ is a positive integer and $a_0 = a_1 = 3.$
Lemma 1 : For $n \geq 1$, $$a_{n+1} = ( k + 2)a_n - a_{n-1}. ---------(2)$$ We shall prove by induction. From (1), $a_2 = 3k + 3$. From the lemma, $a_2 = (k + 2) 3 - 3 = 3k + 3.$ Base case proven. Assume that the lemma is true for some $t \geq 1$. Then, eliminating the $a_{t-1}$ using (1) and (2) gives
$$(k+2)a_ta_{t+1} = a_t^2 + a_{t+1}^2 + 9k. ---------(3)$$
It follows from (2) that
$$(k+2)a_{t+1} - a_t =\frac{(k+2)a_{t+1}a_t - a_t^2}{a_t} =\frac{a_{t+1}^2 + 9k}{a_t} =a_{t+2},$$
where the last line followed from (1) for case $n = t+1$.
Lemma 2 : For $n \geq 0,$ $$a_{n+1} \geq a_{n}.$$ Base case is obvious. Assume that $a_{t+1} \geq a_{t}$ for some $t \geq 0$. Then it follows that $$a_{t+2} =\frac{a_{t+1}^2 + 9k}{a_t} = a_{t+1}(\frac{a_{t+1}}{a_t} ) + 9k \geq a_{t+1} + 9k > a_{t+1}.$$
This completes the induction.
Lemma 3 : For $n \geq 1,$ $$a_n a_{n+1} > 9k$$
Using (1) and Lemma 2, for $n \geq 1,$ $$a_{n+1}a_n \geq a_{n+1}a_{n-1} = a_n^2 + 9k > 9k$$
Finally, using (3), for $n \geq 1,$ $$\frac{a_n^2 + a_{n+1}^2}{a_n a_{n+1}} =\frac{(k+2)a_n a_{n+1} - 9k}{a_n a_{n+1}} = k+2 -\frac{9k}{a_n a_{n+1}}.$$ Using lemma 3, the largest integer less than or equal to this value would be $k + 1$.
### Solution 4 (pure algebra)
We will try to manipulate $\frac{a_0^2+a_1^2}{a_0a_1}$ to get $\frac{a_1^2+a_2^2}{a_1a_2}$.
$\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_0+\frac{a_1^2}{a_0}}{a_1} = \frac{a_0+\frac{a_1^2+2007}{a_0}-\frac{2007}{a_0}}{a_1}$ Using the recurrence relation, $= \frac{a_0+a_2-\frac{2007}{a_0}}{a_1} = \frac{a_0a_2+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2}$ Applying the relation to $a_0a_2$, $= \frac{a_1^2+2007+a_2^2-\frac{2007a_2}{a_0}}{a_1a_2} = \frac{a_1^2+a_2^2}{a_1a_2}+\frac{2007}{a_1a_2}-\frac{2007}{a_0a_1}$
We can keep on using this method to get that $\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{2005}a_{2006}}+\frac{2007}{a_{2005}a_{2006}}-\ldots-\frac{2007}{a_{0}a_{1}}$
This telescopes to $\frac{a_0^2+a_1^2}{a_0a_1} = \frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}}+\frac{2007}{a_{2006}a_{2007}}-\frac{2007}{a_{0}a_{1}}$
or $\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = \frac{a_0^2+a_1^2}{a_0a_1}+\frac{2007}{a_{0}a_{1}}-\frac{2007}{a_{2006}a_{2007}}$
Finding the first few values, we notice that they increase rapidly, so $\frac{2007}{a_{2006}a_{2007}} < 1$. Calculating the other values, $\frac{a_{2006}^2+a_{2007}^2}{a_{2006}a_{2007}} = 2+223-\frac{2007}{a_{2006}a_{2007}}$.
The greatest number that does not exceed this is $224$
### Solution 5 (using limits)
Let's start by computing the first couple terms of the given sequence so we know what we're working with. It's given that $a_0 = a_1 = 0$, and solving for $a_2$ and $a_3$ using the given relation we get $a_2 = 672 = 3(224)$ and $a_3 = 3(224^{2} + 223)$, respectively. It will be clear why I decided to factor these expressions as I did momentarily.
Next, let's see what the expression $\frac{a_{n}^{2}+a_{n + 1}^{2}}{a_{n}a_{n + 1}}$ looks like for small values of $n$. For $n = 1$, we get $\frac{1 + 224^2}{224}$, the floor of which is clearly $224$ because the $1$ in the numerator is insignificant. Repeating the procedure for $n + 1$ is somewhat messier, but we end up getting $\frac{224^4 + 224^2\cdot223\cdot2 + 224^2 + 223^2}{224^3 + 224\cdot223}$. It's not too hard to see that $224^4$ is much larger than the sum of the remaining terms in the numerator, and that $224^3$ is similarly a lot greater than the other term in the denominator. In fact, the second-largest term in the numerator is only barely larger than $224^3$, while the second-largest term in the denominator is smaller than $224^2$. Thus, the floor of this expression will come out to be $224$ as well.
Now we must consider whether this finding holds true for the rest of the sequence we're examining. First of all, notice that each time $n$ increases by $1$, the degrees of both the numerator and denominator increase by $2$, because we are squaring the $n+1th$ term in the numerator while the same term, appearing in the denominator, is being generated in part by squaring the term before it (seen in the relation $a_{n+1}^2 = (\frac{a_{n}^2 + 2007}{a_{n-1}})^2$). Thus, the degree of the numerator will always be one greater than the degree of the denominator; we have only to show that the other terms in the expression remain sufficiently small enough so as not to disturb the $\approx224:1$ ratio between the two.
For the non-greatest terms in the expression to offset this ratio for values of $n$ in the ballpark of $2006$, they would have to have massive coefficients, because or else they are dwarfed by the additional $224$ attached to the leading term. Thankfully, this is not the case. Since we are simply squaring previous terms repeatedly to get new terms, we end up getting a sequence that is approximately equal to $\frac{224^k+224^{k-1}+224^{k-2} . . . }{224^{k-1}+224^{k-2} . . . }$ for all $k\geq2$, whose $\lim_{k\to\infty}=$ $\boxed{224}$.
~anellipticcurveoverq |
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Ema Condori-Teves
May 3, 2017
Math III H
Summation Notation
Summation notation is also known as Sigma notation because of the capital greek letter sigma
that is used as an operator which sums the outputs of an expression. Summation provides a way
to precisely express any sum in a sequence of numbers when added up together. It is typically
expressed in this manner; where k represents the index variable or the index of summation, a
represents the starting value and b represents the ending value. f(k) or the function of k
describes each term in the sum for each value of k between a and b, f(k) will be some value
which gives one term in the sum. The basics of summation was discovered by Gauss. An
example of what he discovered is explained below.
If you were to add every number from 1-100 with each other you can find many patterns that will
help solve this summation with a few steps. One way this can be viewed is in pairs. The biggest
and smallest number can be added to equal a specific value. Then as long as the next smallest
and biggest number are paired until their is no more available pairs within this set; it will equal
the same value. So the idea of this is to multiply that value with the amount of pairs in the set of
numbers to find the sum of all these numbers.
The problem can also be viewed as finding the area of a rectangle. This might not sound like it's
related to finding the summation, but it is the same concept as above. A rectangle can be formed
when a visual of this set of numbers is made as shown below. When the width of this triangular
visual is split in half, and one half placed on top of the other, it pairs the numbers in the same
manner as the first explanation above. The height of this rectangle is the first value in the set
added with the last value in the set. The width is half of the total digits that are being
added.When you multiply the width by the height you find the area which is the same as the sum
of all these numbers.
A connection of summation to another topic that was discussed in class this year is the use of
summation or sigma notation to find area under the curve. The method to solve area under the
curve I will be describing is through splitting the width to equal sizes to form rectangles that will
give an area slightly larger than the actual answer. This is known as right end point rectangle
method. To find the different heights for each of these rectangles you will use summation
notation, to find the area under the curve you will multiple the width by each different height and
add each area together for the area under the curve. The more rectangles used, the less error. For
a more accurate result still using this method, you would do both left and right end point
rectangle methods and take the average from them. Another method that could be used is similar
to the area of the one rectangle used for summation. I will walk you through this application of
summation applied to area under the curve below.
The discovery of summation shows how a problem that seems to be time consuming, tedious,
and have the potential for many errors, was simply and accurately solved by Johann Carl
Friedrich Gauss during his elementary years as a student. Carl Gauss was a German
mathematician born 1777. As the story goes, despite no proof of being entirely accurate, Gauss
teacher gave his students a time consuming assignment to add all the numbers between 1 and
100, thinking he would be able to assign the student busy work and not have to worry about them
for a while. Well, Gauss turned in his slate first with only the answer written on the slate. And to
the teacher's surprise, this answer was correct. So how did Gauss find the method to accurately
find the answer so quickly with little work? He easily found a pattern that would work for all of
these summation type of problems. Each number had a pair, if the numbers were to be listed 1-
100 and then 100-1, the numbers would pair to equal 101 no matter what. So he took this number
and multiplied it by the number of pairs. He discovered this in elementary, and was able to find
out the equation for finding the sum in a sequence of numbers. It was even said that at the age of
3, Gauss had corrected his father's payroll calculation for his employees. He would get upset
when a mistake was made, and his father tested his hypothesis of whether his cries really came
from his mathematical errors, and made a mistake on purpose and sure enough Gauss got
upset.When Gauss was older, many mathematicians would present new discoveries they had |
# Coordinates
Coordinates are sets of numbers describing positions in space, as distances from some reference point. We need as many numbers as the space has dimensions to uniquely determine the location of each point.
## Intro
So you're in for a game of chess? Alright, you start. You spend quite a bit of time thinking about your first move. But then, eventually, you move your knight to...
How should you describe that position? I mean, it's kinda awkward to be saying: "I move my left knight two steps forward and one step to the right". Your left, or my left? And should I assume that forward is your forward?
Coordinates are used pretty much everywhere. For example, when your GPS informs you about your position, it's essentially taking a bunch of coordinates and translating them into "you're 100 metres from the McDonald's at the corner". Awesome, right?
## Concept
Coordinates allow us to express positions. They tell us which point in space we refer to.
And this is super handy. Rather than saying "move to the left - no, not that much! - just a tiny bit", say something like "move to position ". That'll do.
But a position mustn't refer to an actual position. For example, mathematicians might let the axis represent time and the axis represent average stock yield. Then your position, if you may call it as such, is the average stock yield after a given time.
## Math
Cartesian coordinate is the coordinate system we most often use it is defined with and . There are other coordinate systems that are equally valid, such as the polar coordinates and .
In the polar coordinate system we instead define every point as a distance from origo we also have that is defined as the angle that the point have relative the axis.
We can also define and with and
## Coordinate transformations
Coordinate systems are used to describe the location of points in space. In order to do so, they must define some measures of position in the system.
We always need as many measures, called coordinates, as there are dimensions in the space.
### Coordinates in 2D
The most classic example is the Cartesian coordinate system in 2 dimensions, formed by the and axes as we are used to. In this system, we denote a point as , describing its length of displacement from the origin in the directions of both axes.
But is this the only choice we have for a coordinate system in 2D? Certainly not.
Look at the system below with and axes, and compare with the Cartesian one. Both are able to describe points in the same space, and the use of and is simply a convention.
In fact, we can easily swap between the two, and convert the representation in one coordinate system to another through a coordinate transformation.
Consider the point in the Cartesian coordinate system. If we define the transformation from - to -coordinates as
we see that it becomes after the transformation. The point is still in the exact same place, it is just represented in another coordinate system.
The reason we want to be able to represent a point differently is because it is sometimes more convenient to use one over the other.
### Coordinates in 3D
If the choice of coordinate axes is ambiguous, how have we decided to extend the Cartesian coordinate system from 2 dimensions to 3?
Although not necessary, we generally want the axes to be perpendicular to each other. Note, however, that there are two opposite options for the -axis to be perpendicular to both the - and -axis. By convention, we therefore turn to the right-hand rule:
The right-hand rule
Using your right hand, let your index finger point in the direction of the -axis, and your middle finger in the direction of the -axis.
Then, the direction of the -axis in the Cartesian coordinate system of 3 dimensions is defined as the one in which your thumb points.
Just as for coordinate systems in 2D, we can define a different one and transform points between them as we like. In the following lecture notes, we will examine two useful alternative coordinate systems in 3D.
## Cylindrical coordinates
### Polar coordinates
As an alternative to Cartesian coordinates using distances along the - and -axis respectively, we can use polar coordinates to represent a point in two dimensions.
This form uses only one distance from the origin, as well as an angle , measured in the counter-clockwise direction from the positive -axis. Hence, a point in polar form is written as .
The coordinate transformation from Cartesian to polar coordinates looks as follows:
### Cylindrical coordinates
If we equip the polar coordinate system with a -axis, we get a coordinate system in 3 dimensions called cylindrical coordinates.
Cylindrical coordinates are polar coordinates with a -axis
As per usual, we use our right hand to define the directional relationships in the coordinate system:
The right hand rule for direction of rotation:
Using the right hand, let the thumb point in the direction of the -axis, and curl your fingers.
Then, if you twist the hand in the direction that the fingers point, this will define the direction of rotation for the angle .
To visualize what cylindrical coordinates can look like in practice, imagine a car driving up a circular ramp in a parking garage. The position of the car, taken from the center of the ramp as the origin, can be described using the radius of the ramp, the current angle to some defined horizontal direction (the regular -axis), and the height given by the -axis.
This example highlights one of the pros of cylindrical coordinates. Since the radius remains unchanged during the drive up the ramp, the only changing quantities in this particular case are the angle and the height.
A point expressed in cylindrical coordinates takes the form , and the coordinate transformation from the Cartesian coordinate system in 3 dimensions look as follows:
From these formulas, we can derive a set of equations to convert coordinates the other way around too:
## Spherical coordinates
During your summer vacation, you went to Sri Lanka with your friends. It was a blast. Sunbathing, partying, discovering the region. As you were out kayaking, you found a hidden lagoon. But it wasn't marked out on the map, so you had no way of remembering the location. Too bad.
In order to describe positions on spherical surfaces, like the earth's surface, you can use (surprise surprise!) spherical coordinates. It would've been quite cumbersome to work out your position in normal coordinates. Spherical coordinates are tailored for this kind of problem.
See the picture below.
Normally, we say and . If so, then there's only one set of coordinates which corresponds to a given position. But we could, as we shall see further on, let and be any real number.
Finally, here's a heads up. When we indicate spherical coordinates, the order of and is important. If you don't get the order right, you might be haunted by Einstein's ghost. Just so you know. |
# Common Core: 4th Grade Math : Compare Two Fractions with Different Numerators and Different Denominators: CCSS.Math.Content.4.NF.A.2
## Example Questions
### Example Question #2731 : Isee Lower Level (Grades 5 6) Quantitative Reasoning
Select the symbol to correctly fill in the blank below.
__________
Explanation:
To compare fractions, we need to first make common denominators.
Now that we have common denominators, we can compare numerators. The fraction with the bigger numerator has the greater value.
### Example Question #2732 : Isee Lower Level (Grades 5 6) Quantitative Reasoning
Select the symbol to correctly fill in the blank below.
__________
Explanation:
To compare fractions, we need to first make common denominators.
Now that we have common denominators, we can compare numerators. The fraction with the bigger numerator has the greater value.
### Example Question #21 : Number & Operations: Fractions
Select the symbol to correctly fill in the blank below.
__________
Explanation:
To compare fractions, we need to first make common denominators.
Now that we have common denominators, we can compare numerators. The fraction with the bigger numerator has the greater value.
### Example Question #22 : Number & Operations: Fractions
Select the symbol to correctly fill in the blank below.
__________
Explanation:
To compare fractions, we need to first make common denominators.
Now that we have common denominators, we can compare numerators. The fraction with the bigger numerator has the greater value.
### Example Question #1 : How To Compare Fractions
Select the symbol to correctly fill in the blank below.
__________
Explanation:
To compare fractions, we need to first make common denominators.
Now that we have common denominators, we can compare numerators. The fraction with the bigger numerator has the greater value.
### Example Question #21 : How To Order Fractions From Least To Greatest Or From Greatest To Least
Select the symbol to correctly fill in the blank below.
__________
Explanation:
To compare fractions, we need to first make common denominators.
Now that we have common denominators, we can compare numerators. The fraction with the bigger numerator has the greater value.
### Example Question #1 : How To Compare Fractions
Select the symbol to correctly fill in the blank below.
__________
Explanation:
To compare fractions, we need to first make common denominators.
Now that we have common denominators, we can compare numerators. The fraction with the bigger numerator has the greater value.
### Example Question #2741 : Isee Lower Level (Grades 5 6) Quantitative Reasoning
Select the symbol to correctly fill in the blank below.
__________
Explanation:
To compare fractions, we need to first make common denominators.
Now that we have common denominators, we can compare numerators. The fraction with the bigger numerator has the greater value.
### Example Question #2 : Compare Two Fractions With Different Numerators And Different Denominators: Ccss.Math.Content.4.Nf.A.2
Select the symbol to correctly fill in the blank below.
__________
Explanation:
To compare fractions, we need to first make common denominators.
Now that we have common denominators, we can compare numerators. The fraction with the bigger numerator has the greater value.
### Example Question #331 : Fractions
Select the symbol to correctly fill in the blank below.
__________ |
# GMAT Challenge Question: Prime Time
It’s time again for another GMAT challenge question, and this one focuses on one of the quantitative section’s favorite themes: prime factors.
Please submit your answers in the comments field, and check back later today for the solution and a more-thorough explanation of prime factors!
What is the greatest prime factor of 12!11! + 11!10!?
(A) 7
(B) 11
(C) 13
(D) 17
(E) 19
UPDATE: Solution!
While it’s quite common for students to simply look at the numbers 12!, 11!, and 10! and recognize that the highest naturally-occurring prime number is 11, it’s important to recognize that this is an addition problem – the numbers 12!11! and 11!10! are combined to create a new number that may well have a higher prime factor than its factorial components.
When adding large numbers like factorials and exponents, as we discussed in this space last week, it’s often quite helpful to factor out common terms. In this case, it’s particularly important, because our entire goal is to break out the large sum into prime factors so that we can determine which is biggest. Each term has a common 11!, so by factoring that out we can get from:
12!11! + 11!10!
to
11! (12! + 10!)
Now, 12! includes a 10! – it’s essentially 12 * 11 * 10!, so we have a common 10! within the parentheses that can also be factored out, going from:
11! (12*11*10! + 10!)
to
11!10! (12*11 + 1)
At this point, the largest prime factor must be either the 11 outside the parentheses or a factor of the number within it, so it’s necessary to check the number within. 12*11 + 1 = 132 + 1 = 133. 133 is the product of 7*19, so 19 is a prime factor of 12!11! + 11!10!, and therefore the largest prime factor. Accordingly, E is the correct answer.
Plan on taking the GMAT soon? See how Veritas Prep’s GMAT prep courses can help you reach your maximum potential on the test. And, as always, be sure to find us on Facebook and follow us on Twitter! |
#### Factorials
The product of the first $$k$$ positive integers (counting numbers) is called $$k$$ factorial and is denoted $$k!$$ (Weiss 2010).
$k! = k \times (k - 1) \times ...\times 2 \times 1$
We also define $$0! = 1$$.
Let us consider an easy example and calculate the factorial of 6. Plugging into the equation from above this gives us:
$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$
In R we write…
6*5*4*3*2*1
## [1] 720
…or we use the in-built function factorial()
factorial(6)
## [1] 720
#### Binomial Coefficients
If $$n$$ is a positive integer and $$k$$ is a non negative integer less than or equal to $$n$$, then the binomial coefficient $${n \choose k}$$ is defined as:
${n \choose k}= \frac{n!}{k!(n - k)!}$
The binomial coefficients $${n \choose k}$$ is often read aloud as “n choose k”, because there are $${n \choose k}$$ ways to choose $$k$$ elements, disregarding their order, from a set of $$n$$ elements. Or in other words, the binomial coefficients refers to the number of combinations of $$n$$ things taken $$k$$ at a time without repetition. Please note, that the order of selection does not matter.
Let us try an example to gain some intuition.
Consider a simple word, such as “dog”, containing three different letters; d,o,g. How many possibilities exist to draw exactly one letter out those 3 letters? For sure, there are 3 possibilities to draw exactly 1 letter: “d”, “o”, or “g”. Thus we can write all the combinations as $${3 \choose 1}$$. What about two letters? How many combinations exist to draw exactly two letters out of 3 letters? The combinations are “do”, “dg”, and “og” (Please note that for example “og” and “go” are counted only as one combinations, as order does not matter right now). Accordingly, the answer is 3 which can be written as $${3 \choose 2}$$. One final question: How many combinations exist to draw exactly three letters out of 3 letters? Let us ask R!
First, we implement a naive approach, by applying the formula from above.
n <- 3 # the word "dog" has 3 letters
k <- 3 # we draw exactly three letters
factorial(n)/(factorial(k)*(factorial(n-k)))
## [1] 1
However, we may as well apply the built-in function called choose() to calculate the number of combinations. We do that for $$k = 1,2,3$$.
n <- 3
choose(n, 1)
## [1] 3
choose(n, 2)
## [1] 3
choose(n, 3)
## [1] 1
Now, that we are familiar with the concept let us consider a more complex example: The Department of Earth Sciences of FU Berlin asks all graduates to choose their favorite 4 courses of the curriculum. How many different answers could the students give, if the curriculum offers 24 courses to choose from? Any guesses? Let us ask R.
choose(24, 4)
## [1] 10626
The students could give 10,626 different answers.
#### Bernoulli Trials
Repeated trials of an experiment are called Bernoulli trials if the following three conditions are satisfied:
1. The experiment (each trial) has two possible outcomes, denoted $$s$$, for success, and $$f$$, for failure.
2. The trials are independent.
3. The probability of a success, called the success probability and denoted $$p$$, remains the same from trial to trial. |
# Solve my math problem for me
In this blog post, we will take a look at how to Solve my math problem for me. We will also look at some example problems and how to approach them.
## Solving my math problem for me
It’s important to keep them in mind when trying to figure out how to Solve my math problem for me. The cosine solver iteratively solves for the cosine of a given angle. It uses a fixed value as the starting point, then iteratively increases the cosine value by each iteration until it reaches the target value. The cosine solver is an excellent tool to use when solving problems involving the cosine function. Let's take a look at an example. Say you want to find out how long it takes to drive from one location to another. You can first use a straightedge and compass to determine the distance between your starting point and destination. Then, you can plug this distance into a formula that calculates the cosine of the angle between your two points to get your driving time. This is an example of finding the exact value of something using calculus, a branch of mathematics that deals with change in quantities over time. In addition to being useful for solving problems about geometry, the cosine solver can also be used for finding accurate values of trigonometric functions such as sine and tangent . While there are many different ways to solve these problems using different formulas, one common solution method is called Simpson's rule . This method involves first calculating the ratio of opposite leg lengths and then using this ratio to calculate the hypotenuse length. By applying this step-by-step process, you can eventually reach an accurate answer for any trigonometric function
Solve with steps is one of the most popular types of puzzles. In this type, you must solve each step in sequence to reach the final solution. Solving with steps puzzles are great for people who want a quick yet challenging brain workout while also providing a sense of accomplishment. If you’re new to solving with steps, start off by simply counting out each step and then visualize yourself making your way through the puzzle. Once you have all the steps down, it will only take a few extra seconds to complete the puzzle. People unfamiliar with solving with steps often end up trying to count out each and every step when they should be focusing on just one or two steps at a time. This can quickly lead to frustration and confusion if you have a lot of information to process at once. Instead, focus on just one or two key steps that you need to remember and try to encode them in your memory as quickly as possible so that you can easily recall them later on.
The best geometric sequence solver is a computer program that solves geometric sequences, such as those found in long multiplication problems. The program works by taking a list of numbers and linking them together to produce a longer list. This process is repeated until the sequence is solved. The best geometric sequence solver can work in several ways. It can use either brute force or brute force with some help from a human. It can also use sorting or other computer algorithms to determine the next number in the sequence and find the gap between it and the other numbers. Once all the numbers have been determined, they are combined into one long list, which represents the solution to the problem. There are two main types of geometric sequence solvers. One type uses brute force and tries every possible combination until one of them works. The other type uses brute force with some help from a human and tries every combination that meets certain requirements, such as being in order or not having too many digits. Many people prefer using a geometric sequence solver because it can be faster than using other strategies, such as counting or figuring out how many digits there are in each number in the problem. This makes it great for students who don’t have time to think through their problems carefully or for people who have trouble with math in general. However, some people dislike these programs because they can take longer than typical math problems
The y intercept is the value at which the y-axis intersects the line from x = 0 to x = 1. This is the value where the graph will be at its maximum value. In order for a curve to be plotted, the y intercept must be defined. In other words, if we want to plot a curve, then we must have an equation that defines it. When we enter an equation into our calculator, our computer will do all of the work and automatically determine y intercept. There are many ways to solve for y intercept on graph calculators. We can manually enter 0 as our x value and then enter 1 as our y value. The y-intercept will show up on your calculator next to “y=0”. We can also enter “y=1” and see what happens in our graphing software. You can also figure out the y-intercept by simply drawing a line from x = 0 to x = 1, and then identifying where that line meets the axis of your graph. When calculating for a curve, we must know both values (x and y) that we are looking for when plotting a curve on a graph. We also need to know what exactly our equation defines (i.e., curvy line or straight line).
Incredibly helpful and very accurate, the fact that it shows the solution with steps is just amazing. The camera feature is very good as well though the numbers need to be clear. Very useful and helpful for people who doesn't know how to solve math problems like me
Margaret Roberts
Very polished, thorough, clean, and easy to use. Still struggles with some types of problems, but I'm sure we have much more in store with respect to its capabilities. This is so helpful my teacher always gives me correction so now I won't have any more correction
Ulva Moore
Symbolic system of equation solver Answers to aleks math problems Solve by square roots Polynomial division solver Ratios of special triangles solver Instant math help |
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# 9.6: Center of Mass (Part 1)
Skills to Develop
• Explain the meaning and usefulness of the concept of center of mass
• Calculate the center of mass of a given system
• Apply the center of mass concept in two and three dimensions
• Calculate the velocity and acceleration of the center of mass
We have been avoiding an important issue up to now: When we say that an object moves (more correctly, accelerates) in a way that obeys Newton’s second law, we have been ignoring the fact that all objects are actually made of many constituent particles. A car has an engine, steering wheel, seats, passengers; a football is leather and rubber surrounding air; a brick is made of atoms. There are many different types of particles, and they are generally not distributed uniformly in the object. How do we include these facts into our calculations?
Then too, an extended object might change shape as it moves, such as a water balloon or a cat falling (Figure 9.26). This implies that the constituent particles are applying internal forces on each other, in addition to the external force that is acting on the object as a whole. We want to be able to handle this, as well.
Figure $$\PageIndex{1}$$: As the cat falls, its body performs complicated motions so it can land on its feet, but one point in the system moves with the simple uniform acceleration of gravity.
The problem before us, then, is to determine what part of an extended object is obeying Newton’s second law when an external force is applied and to determine how the motion of the object as a whole is affected by both the internal and external forces.
Be warned: To treat this new situation correctly, we must be rigorous and completely general. We won’t make any assumptions about the nature of the object, or of its constituent particles, or either the internal or external forces. Thus, the arguments will be complex.
### Internal and External Forces
Suppose we have an extended object of mass M, made of N interacting particles. Let’s label their masses as mj, where j = 1, 2, 3, …, N. Note that
$$M = \sum_{j = 1}^{N} m_{j} \ldotp \tag{9.19}$$
If we apply some net external force $$\vec{F}_{ext}$$ on the object, every particle experiences some “share” or some fraction of that external force. Let:
$$\vec{f}_{j}^{ext}$$ = the fraction of the external force that the jth particle experiences
Notice that these fractions of the total force are not necessarily equal; indeed, they virtually never are. (They can be, but they usually aren’t.) In general, therefore,
$$\vec{f}_{1}^{ext} \neq \vec{f}_{2}^{ext} \neq \cdots \neq \vec{f}_{N}^{ext} \ldotp$$
Next, we assume that each of the particles making up our object can interact (apply forces on) every other particle of the object. We won’t try to guess what kind of forces they are; but since these forces are the result of particles of the object acting on other particles of the same object, we refer to them as internal forces $$\vec{f}_{j}^{int}$$; thus:
$$\vec{f}_{j}^{int}$$ = the net internal force that the jth particle experiences from all the other particles that make up the object.
Now, the net force, internal plus external, on the jth particle is the vector sum of these:
$$\vec{f}_{j} = \vec{f}_{j}^{int} + \vec{f}_{j}^{ext} \ldotp \tag{9.20}$$
where again, this is for all N particles; j = 1, 2, 3, … , N. As a result of this fractional force, the momentum of each particle gets changed:
$$\begin{split} \vec{f}_{j} & = \dfrac{d \vec{p}_{j}}{dt} \\ \vec{f}_{j}^{int} + \vec{f}_{j}^{ext} & = \dfrac{d \vec{p}_{j}}{dt} \ldotp \end{split} \tag{9.21}$$
The net force $$\vec{F}$$ on the object is the vector sum of these forces:
$$\begin{split} \vec{F}_{net} & = \sum_{j = 1}^{N} (\vec{f}_{j}^{int} + \vec{f}_{j}^{ext}) \\ & = \sum_{j = 1}^{N} \vec{f}_{j}^{int} + \sum_{j = 1}^{N} \vec{f}_{j}^{ext} \ldotp \end{split} \tag{9.22}$$
This net force changes the momentum of the object as a whole, and the net change of momentum of the object must be the vector sum of all the individual changes of momentum of all of the particles:
$$\vec{F}_{net} = \sum_{j = 1}^{N} \frac{d \vec{p}_{j}}{dt} \ldotp \tag{9.23}$$
Combining Equation 9.22 and Equation 9.23 gives
$$\sum_{j = 1}^{N} \vec{f}_{j}^{int} + \sum_{j = 1}^{N} \vec{f}_{j}^{ext} = \sum_{j = 1}^{N} \frac{d \vec{p}_{j}}{dt} \ldotp \tag{9.24}$$
Let’s now think about these summations. First consider the internal forces term; remember that each $$\vec{f}_{j}^{int}$$ is the force on the jth particle from the other particles in the object. But by Newton’s third law, for every one of these forces, there must be another force that has the same magnitude, but the opposite sign (points in the opposite direction). These forces do not cancel; however, that’s not what we’re doing in the summation. Rather, we’re simply mathematically adding up all the internal force vectors. That is, in general, the internal forces for any individual part of the object won’t cancel, but when all the internal forces are added up, the internal forces must cancel in pairs. It follows, therefore, that the sum of all the internal forces must be zero:
$$\sum_{j = 1}^{N} \vec{f}_{j}^{ext} = 0 \ldotp$$
(This argument is subtle, but crucial; take plenty of time to completely understand it.)
For the external forces, this summation is simply the total external force that was applied to the whole object:
$$\sum_{j = 1}^{N} \vec{f}_{j}^{ext} = \vec{F}_{ext} \ldotp$$
As a result,
$$\vec{F}_{ext} = \sum_{j = 1}^{N} \frac{d \vec{p}_{j}}{dt} \ldotp \tag{9.25}$$
This is an important result. Equation 9.25 tells us that the total change of momentum of the entire object (all N particles) is due only to the external forces; the internal forces do not change the momentum of the object as a whole. This is why you can’t lift yourself in the air by standing in a basket and pulling up on the handles: For the system of you + basket, your upward pulling force is an internal force.
### Force and Momentum
Remember that our actual goal is to determine the equation of motion for the entire object (the entire system of particles). To that end, let’s define:
$$\vec{p}_{CM}$$ = the total momentum of the system of N particles (the reason for the subscript will become clear shortly)
Then we have
$$\vec{p}_{CM} = \equiv \sum_{j = 1}^{N} \vec{p}_{j} \ldotp$$
and therefore Equation 9.25 can be written simply as
$$\vec{F} = \frac{d \vec{p}_{CM}}{dt} \ldotp \tag{9.26}$$
Since this change of momentum is caused by only the net external force, we have dropped the “ext” subscript. This is Newton’s second law, but now for the entire extended object. If this feels a bit anticlimactic, remember what is hiding inside it: $$\vec{p}_{CM}$$ is the vector sum of the momentum of (in principle) hundreds of thousands of billions of billions of particles (6.02 x 1023), all caused by one simple net external force—a force that you can calculate.
### Center of Mass
Our next task is to determine what part of the extended object, if any, is obeying Equation 9.26.
It’s tempting to take the next step; does the following equation mean anything?
$$\vec{F} = M \vec{a} \tag{9.27}$$
If it does mean something (acceleration of what, exactly?), then we could write
$$M \vec{a} = \frac{d \vec{p}_{CM}}{dt}$$
and thus
$$M \vec{a} = \sum_{j = 1}^{N} \frac{d \vec{p}_{j}}{dt} = \frac{d}{dt} \sum_{j = 1}^{N} \vec{p}_{j} \ldotp$$
which follows because the derivative of a sum is equal to the sum of the derivatives.
Now, $$\vec{p}_{j}$$ is the momentum of the jth particle. Defining the positions of the constituent particles (relative to some coordinate system) as $$\vec{r}_{j}$$ = (xj, yj, zj), we thus have
$$\vec{p}_{j} = m_{j} \vec{v}_{j} = m_{j} \frac{d \vec{r}_{j}}{dt} \ldotp$$
Substituting back, we obtain
$$\begin{split} M \vec{a} & = \frac{d}{dt} \sum_{j = 1}^{N} m_{j} \frac{d \vec{r}_{j}}{dt} \\ & = \frac{d^{2}}{dt^{2}} \sum_{j = 1}^{N} m_{j} \vec{r}_{j} \ldotp \end{split}$$
Dividing both sides by M (the total mass of the extended object) gives us
$$\vec{a} = \frac{d^{2}}{dt^{2}} \left(\dfrac{1}{M} \sum_{j = 1}^{N} m_{j} \vec{r}_{j}\right) \ldotp \tag{9.28}$$
Thus, the point in the object that traces out the trajectory dictated by the applied force in Equation 9.27 is inside the parentheses in Equation 9.28.
Looking at this calculation, notice that (inside the parentheses) we are calculating the product of each particle’s mass with its position, adding all N of these up, and dividing this sum by the total mass of particles we summed. This is reminiscent of an average; inspired by this, we’ll (loosely) interpret it to be the weighted average position of the mass of the extended object. It’s actually called the center of mass of the object. Notice that the position of the center of mass has units of meters; that suggests a definition:
$$\vec{r}_{CM} = \frac{1}{M} \sum_{j = 1}^{N} m_{j} \vec{r}_{j} \ldotp \tag{9.29}$$
So, the point that obeys Equation 9.26 (and therefore Equation 9.27 as well) is the center of mass of the object, which is located at the position vector $$\vec{r}_{CM}$$.
It may surprise you to learn that there does not have to be any actual mass at the center of mass of an object. For example, a hollow steel sphere with a vacuum inside it is spherically symmetrical (meaning its mass is uniformly distributed about the center of the sphere); all of the sphere’s mass is out on its surface, with no mass inside. But it can be shown that the center of mass of the sphere is at its geometric center, which seems reasonable. Thus, there is no mass at the position of the center of mass of the sphere. (Another example is a doughnut.) The procedure to find the center of mass is illustrated in Figure 9.27.
Figure $$\PageIndex{2}$$: Finding the center of mass of a system of three different particles. (a) Position vectors are created for each object. (b) The position vectors are multiplied by the mass of the corresponding object. (c) The scaled vectors from part (b) are added together. (d) The final vector is divided by the total mass. This vector points to the center of mass of the system. Note that no mass is actually present at the center of mass of this system.
Since $$\vec{r}_{j} = x_{j} \hat{i} + y_{j} \hat{j} + z_{j} \hat{k}$$, it follows that:
$$r_{CM,x} = \frac{1}{m} \sum_{j = 1}^{N} m_{j} x_{j} \tag{9.30}$$
$$r_{CM,y} = \frac{1}{m} \sum_{j = 1}^{N} m_{j} y_{j} \tag{9.31}$$
$$r_{CM,z} = \frac{1}{m} \sum_{j = 1}^{N} m_{j} z_{j} \tag{9.32}$$
and thus
$$\vec{r}_{CM} = r_{CM,x} \hat{i} + r_{CM,y} \hat{j} + r_{CM,z} \hat{k}$$
$$r_{CM} = |\vec{r}_{CM}| = (r_{CM,x}^{2} + r_{CM,y}^{2} + r_{CM,z}^{2})^{1/2} \ldotp$$
Therefore, you can calculate the components of the center of mass vector individually.
Finally, to complete the kinematics, the instantaneous velocity of the center of mass is calculated exactly as you might suspect:
$$\vec{v}_{CM} = \frac{d}{dt} \left(\dfrac{1}{M} \sum_{j = 1}^{N} m_{j} \vec{r}_{j}\right) = \frac{1}{M} \sum_{j = 1}^{N} m_{j} \vec{v}_{j} \tag{9.33}$$
and this, like the position, has x-, y-, and z-components.
To calculate the center of mass in actual situations, we recommend the following procedure:
Problem-Solving Strategy: Calculating the Center of Mass
The center of mass of an object is a position vector. Thus, to calculate it, do these steps:
1. Define your coordinate system. Typically, the origin is placed at the location of one of the particles. This is not required, however.
2. Determine the x, y, z-coordinates of each particle that makes up the object.
3. Determine the mass of each particle, and sum them to obtain the total mass of the object. Note that the mass of the object at the origin must be included in the total mass.
4. Calculate the x-, y-, and z-components of the center of mass vector, using Equation 9.30, Equation 9.31, and Equation 9.32.
5. If required, use the Pythagorean theorem to determine its magnitude.
Here are two examples that will give you a feel for what the center of mass is.
Example 9.16
##### Center of Mass of the Earth-Moon System
Using data from text appendix, determine how far the center of mass of the Earth-moon system is from the center of Earth. Compare this distance to the radius of Earth, and comment on the result. Ignore the other objects in the solar system.
##### Strategy
We get the masses and separation distance of the Earth and moon, impose a coordinate system, and use Equation 9.29 with just N = 2 objects. We use a subscript “e” to refer to Earth, and subscript “m” to refer to the moon.
##### Solution
Define the origin of the coordinate system as the center of Earth. Then, with just two objects, Equation 9.29 becomes
$$R = \frac{m_{c} r_{c} + m_{m} r_{m}}{m_{c} + m_{m}} \ldotp$$
From Appendix D,
$$m_{c} = 5.97 \times 10^{24}\; kg$$
$$m_{m} = 7.36 \times 10^{22}\; kg$$
$$r_{m} = 3.82 \times 10^{5}\; m \ldotp$$
We defined the center of Earth as the origin, so re = 0 m. Inserting these into the equation for R gives
$$\begin{split} R & = \frac{(5.97 \times 10^{24}\; kg)(0\; m) + (7.36 \times 10^{22}\; kg)(3.82 \times 10^{8}\; m)}{(5.98 \times 10^{24}\; kg) + (7.36 \times 10^{22}\; kg)} \\ & = 4.64 \times 10^{6}\; m \ldotp \end{split}$$
##### Significance
The radius of Earth is 6.37 x 106 m, so the center of mass of the Earth-moon system is (6.37 − 4.64) x 106 m = 1.73 x 106 m = 1730 km (roughly 1080 miles) below the surface of Earth. The location of the center of mass is shown (not to scale).
Suppose we included the sun in the system. Approximately where would the center of mass of the Earth-moon-sun system be located? (Feel free to actually calculate it.)
Example 9.17
##### Center of Mass of a Salt Crystal
Figure 9.28 shows a single crystal of sodium chloride—ordinary table salt. The sodium and chloride ions form a single unit, NaCl. When multiple NaCl units group together, they form a cubic lattice. The smallest possible cube (called the unit cell) consists of four sodium ions and four chloride ions, alternating. The length of one edge of this cube (i.e., the bond length) is 2.36 x 10−10 m. Find the location of the center of mass of the unit cell. Specify it either by its coordinates (rCM,x, rCM,y, rCM,z), or by rCM and two angles.
Figure $$\PageIndex{3}$$: A drawing of a sodium chloride (NaCl) crystal.
##### Strategy
We can look up all the ion masses. If we impose a coordinate system on the unit cell, this will give us the positions of the ions. We can then apply Equation 9.30, Equation 9.31, and Equation 9.32 (along with the Pythagorean theorem).
##### Solution
Define the origin to be at the location of the chloride ion at the bottom left of the unit cell. Figure 9.29 shows the coordinate system.
Figure $$\PageIndex{4}$$: A single unit cell of a NaCl crystal.
There are eight ions in this crystal, so N = 8:
$$\vec{r}_{CM} = \frac{1}{M} \sum_{j = 1}^{8} m_{j} \vec{r}_{j} \ldotp$$
The mass of each of the chloride ions is
$$35.453u \times \frac{1.660 \times 10^{-27}\; kg}{u} = 5.885 \times 10^{-26}\; kg$$
so we have
$$m_{1} = m_{3} = m_{6} = m_{8} = 5.885 \times 10^{-26}\; kg \ldotp$$
For the sodium ions,
$$m_{2} = m_{4} = m_{5} = m_{7} = 3.816 \times 10^{-26}\; kg \ldotp$$
The total mass of the unit cell is therefore
$$M = (4)(5.885 \times 10^{-26}\; kg) + (4)(3.816 \times 10^{-26}\; kg) = 3.880 \times 10^{-25}\; kg \ldotp$$
From the geometry, the locations are
$$\begin{split} \vec{r}_{1} & = 0 \\ \vec{r}_{2} & = (2.36 \times 10^{-10}\; m) \hat{i} \\ \vec{r}_{3} & = r_{3x} \hat{i} + r_{3y} \hat{j} = (2.36 \times 10^{-10}\; m) \hat{i} + (2.36 \times 10^{-10}\; m) \hat{j} \\ \vec{r}_{4} & = (2.36 \times 10^{-10}\; m) \hat{j} \\ \vec{r}_{5} & = (2.36 \times 10^{-10}\; m) \hat{k} \\ \vec{r}_{6} & = r_{6x} \hat{i} + r_{6z} \hat{k} = (2.36 \times 10^{-10}\; m) \hat{i} + (2.36 \times 10^{-10}\; m) \hat{k} \\ \vec{r}_{7} & = r_{7x} \hat{i} + r_{7y} \hat{j} + r_{7z} \hat{k} = (2.36 \times 10^{-10}\; m) \hat{i} + (2.36 \times 10^{-10}\; m) \hat{j} + (2.36 \times 10^{-10}\; m) \hat{k} \\ \vec{r}_{8} & = r_{8y} \hat{j} + r_{8z} \hat{k} = (2.36 \times 10^{-10}\; m) \hat{j} + (2.36 \times 10^{-10}\; m) \hat{k} \ldotp \end{split}$$
Substituting:
$$\begin{split} |\vec{r}_{CM,x}| & = \sqrt{r_{CM,x}^{2} + r_{CM,y}^{2} + r_{CM,z}^{2}} \\ & = \frac{1}{M} \sum_{j = 1}^{8} m_{j} (r_{x})_{j} \\ & = \frac{1}{M} (m_{1} r_{1x} + m_{2} r_{2x} + m_{3} r_{3x} + m_{4} r_{4x} + m_{5} r_{5x} + m_{6} r_{6x} + m_{7} r_{7x} + m_{8} r_{8x}) \\ & = \frac{1}{3.8804 \times 10^{-25}\; kg} \Big[ (5.885 \times 10^{-26}\; kg)(0\; m) + (3.816 \times 10^{-26}\; kg)(2.36 \times 10^{-10}\;m) \\ & + (5.885 \times 10^{-26}\; kg)(2.36 \times 10^{-10}\;m) + (3.816 \times 10^{-26}\; kg)(2.36 \times 10^{-10}\;m) + 0 + 0 \\ & + (3.816 \times 10^{-26}\; kg)(2.36 \times 10^{-10}\;m) + 0 \Big] \\ & = 1.18 \times 10^{-10}\; m \ldotp \end{split}$$
Similar calculations give rCM,y = rCM,z = 1.18 x 10−10 m (you could argue that this must be true, by symmetry, but it’s a good idea to check).
##### Significance
As it turns out, it was not really necessary to convert the mass from atomic mass units (u) to kilograms, since the units divide out when calculating rCM anyway.
To express rCM in terms of magnitude and direction, first apply the three-dimensional Pythagorean theorem to the vector components:
$$\begin{split} r_{CM} & = \sqrt{r_{CM,x}^{2} + r_{CM,y}^{2} + r_{CM,z}^{2}} \\ & = (1.18 \times 10^{-10}\; m) \sqrt{3} \\ & = 2.044 \times 10^{-10}\; m \ldotp \end{split}$$
Since this is a three-dimensional problem, it takes two angles to specify the direction of $$\vec{r}_{CM}$$. Let $$\phi$$ be the angle in the x,y-plane, measured from the +x-axis, counterclockwise as viewed from above; then:
$$\phi = \tan^{-1} \left(\dfrac{r_{CM,y}}{r_{CM,x}}\right) = 45^{o} \ldotp$$
Let $$\theta$$ be the angle in the y,z-plane, measured downward from the +z-axis; this is (not surprisingly):
$$\theta = \tan^{-1} \left(\dfrac{R_{z}}{R_{y}}\right) = 45^{o} \ldotp$$
Thus, the center of mass is at the geometric center of the unit cell. Again, you could argue this on the basis of symmetry |
# Ratio and Proportion Aptitude Test Paper 4
16) Rs 7800 are distributed among A, B, and C. The share of "A" is the ¾ of the share of B, and the share of B is the 2/3 of the share of C. Find the difference between the share of B and C.
1. 1200
2. 1300
3. 1500
4. 800
Explanation:
The share of A: B is 3: 4
The share of B: C is 2: 3
Note: Whenever such form is given, multiply a to b, then b to b, and then b to c.
i.e., A: B: C = 3*2: 4*2: 4*3
Or, A: B: C = 6: 8: 12
Or, A: B: C = 3: 4: 6
Sun of ratios = 13
Now, the share of B = [4/13] * 7800 = 2400
Share of C = [6/13]* 7800 = 3600
The difference between the share of B and C = 3600- 2400 = 1200
17) A bag contains Rs 410 in the form of Rs 5, Rs 2, and Rs 1 coins. The number of coins is in the ratio 4: 6: 9. So, find the number of 2 Rupees coins.
1. 40
2. 50
3. 60
4. 70
Explanation:
ATQ, if the ratio of coins = 4: 6: 9
That means if Rs 5 coins are 4, Rs 2 coins are 6, and then Rs 1 coins are 9.
According to the given ratio, the ratio of amounts = 5*4: 6*2: 9*1 = 20: 12: 9
The sum of the ratios of the amounts = 20+12+9 = Rs 41
But ATQ, it is Rs 410, which means multiply each ratio by 10
i.e., new ratio = 40: 60: 90
Now, 40*5: 60*2: 90*1 = 200: 120: 90
The total amount in the form of two rupees coins = 120
So, the two rupees coins = 120/2= 60
18) The ratio of copper and zinc in a 63 kg alloy is 4: 3. Some amount of copper is extracted from the alloy, and the ratio becomes 10: 9. How much copper is extracted?
1. 8kg
2. 6kg
3. 12kg
4. 10kg
Explanation:
Amount of copper in alloy = [copper ratio/ sum of ratios]* total quantity of alloy
Copper = {4/7}* 63 = 36 kg
Similarly, the amount of zinc in alloy = [3/7] * 63 = 27 kg
Let the extracted copper from alloy = x kg
Remaining copper in alloy = 36-x
The new ratio = 10: 9
i.e., 36-x: 27 = 10: 9
36-x: 3 = 10: 1
36-x = 10* 3
36-x = 30
x = 36-30 = 6 kg.
Hence, the extracted amount of copper is 6 kg.
19) The ratio of land and water on earth is 1: 2. In the northern hemisphere, the ratio is 2: 3. What is the ratio in the southern hemisphere?
1. 1:11
2. 2:11
3. 3:11
4. 4:11
Explanation:
The ratio of land: water = 1: 2
In the northern hemisphere the ration of land: water = 2: 3
Note: Earth is divided equally into two hemispheres called northern hemisphere and southern hemisphere.
i.e., the northern hemisphere is 50% of the total earth.
We can say southern hemisphere = total area- northern hemisphere.
To make the northern hemisphere 50% of the total area
Multiply the ratio of the earth by 10 and northern hemisphere by 3
Now, earth's ratio Land: water = 1*10: 2*10 = 10: 20............... (i)
Northern hemisphere's ratio = 2*3: 3*3 = 6: 9...................... (ii)
Subtract equation i by ii
Hence, southern hemisphere = 10- 6: 20-9 = 4: 11
20) Vessels A and B contain mixtures of milk and water in the ratios 4: 5and 5: 1respectively. In what ratio should quantities of the mixture be taken from A and B to form a mixture in which milk to water is in the ratio 5: 4?
1. 2: 5
2. 4: 3
3. 5: 2
4. 2: 3
Explanation:
Let the quantity of vessels A and B in the ratio x: y to form a mixture of water and milk in the ratio 4: 5.
Or, Milk: water = 5: 4.
From vessel A, milk: water = 4x/9: 5x/9
From vessel B, milk: water = 5y/6: 1y/6
Now, ATQ quantities of the mixture should be taken from A and B.
Ratio of milk: water = [4x/9 +5y/6]: [5x/9+ 1y/6] = 5/4
Take the LCM of 9 and 6 = 54
Or, [24x + 45y]/54: [30x+9y]/54 = 5/4
Or, [8x + 15 y]: [10x + 3y] = 5/4
Or, [8x + 15y] * 4 = [10x + 3y] * 5
Or, (60-15) y = (50-32) x
Or, 45y = 18x
Or, 5y = 2x
Or, x: y = 5: 2
Hence, the quantity of mixture should be taken from A and B in the ratio 5: 2.
Ratio and Proportion Aptitude Test Paper 1
Ratio and Proportion Aptitude Test Paper 2
Ratio and Proportion Aptitude Test Paper 3
Ratio and Proportion Concepts |
# Solve algebra online
Are you ready to learn how to Solve algebra online? Great! Let's get started! Our website can solve math word problems.
## Solving algebra online
When you try to Solve algebra online, there are often multiple ways to approach it. Let's say you're a cashier and need to figure out how much change to give someone from a \$20 bill. You would take the bill and subtract it from 20, which would give you the amount of change owed. So, if someone gave you a \$20 bill, you would give them back \$16 in change since 20-4 equals 16. You can use this same method to solve problems with larger numbers as well. For example, if someone gave you a \$50 bill, you would take the bill and subtract it from 50, which would give you the amount of change owed. So, if someone gave you a \$50 bill, you would give them back \$40 in change since 50-10 equals 40. As you can see, this method is simple yet effective when trying to figure out how much change to give someone. Give it a try next time you're stuck on a math problem!
Solving inequality equations requires a different approach than solving regular equations. Inequality equations involve two variables that are not equal, so they cannot be solved using the same methods as regular equations. Instead, solving inequality equations requires using inverse operations to isolate the variable, and then using test points to determine the solution set. Inverse operations are operations that undo each other, such as multiplication and division or addition and subtraction. To solve an inequality equation, you must use inverse operations on both sides of the equation until the variable is isolated on one side. Once the variable is isolated, you can use test points to determine the solution set. To do this, you substitute values for the other variable into the equation and see if the equation is true or false. If the equation is true, then the point is part of the solution set. If the equation is false, then the point is not part of the solution set. By testing multiple points, you can determine the full solution set for an inequality equation.
A complex number solver is a mathematical tool that can be used to solve equations that involve complex numbers. Complex numbers are numbers that have both a real and imaginary component, and they can be represented in the form a+bi, where a is the real component and b is the imaginary component. Many equations that involve variables raised to a power or roots cannot be solved using real numbers alone, but can be solved by adding or subtracting complex numbers. A complex number solver can be used to find the value of an unknown variable in such an equation. In addition, a complex number solver can also be used to graph complex numbers on a coordinate plane. This can be helpful in visualizing the solutions to equations or in understanding the behavior of complex numbers.
Algebra is a branch of mathematics that allows us to solve for unknowns. For example, solving for x in the equation 3x = 9 would give us x = 3. However, solving for x when there is a fraction can be more tricky. In order to solve for x with fractions, we need to use a method called clearing the fraction. This involves multiplying both sides of the equation by the denominator, so that all fractions are eliminated. For example, if we have the equation 2x/3 = 8/9, we would multiply both sides by 3 to get 6x = 24. From there, we can solve for x as usual to find that x = 4. Solving for x with fractions may require some extra steps, but it is still relatively straightforward once you know the process.
## We will support you with math difficulties
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Literal equations solver How do you solve a compound inequality Fraction calculator algebra help Step by step equation solver How to solve by substitution |
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# Aclass has 15 boys and 16 girls. what is the probability that a boy's name is drawn at random? 15 over 31 16 over 31 15 over 16 16 over 15
Step-by-step explanation:
In order to determine the probability, we have to know the classical probability.
Classical probability is a simple form of probability that has equal odds of something happening. The probability of a simple event happening is the number of times the event can happen, divided by the number of possible events.
Given:
A=amount of favorable outcomes
B=amount of possible outcomes
E=event
So that, we have to determine:
the amount of possible outcomes:
B=15 boys + 16 girls =31
the amount of favorable outcomes:
A=15 boys
E= boy's name is drawn at random event
15/31
Step-by-step explanation: because there are 31 kids in all and 15 of them are boys so 15/31
1. certain
2.15/31
3.40%
4.unlikely
A. 15 over 31
I took a the test and got it right, thank you!
15/31 because all together there would be 31 names to be drawn and there are 15 boys so therefore there would be 15/31 chance it would be a boy
15 over 31 is the answer.
15 boys and 16 girlstotal is (15 + 16) = 31
probability that a boys name is drawn : boys/total ppl = 15/31
15 over 31
Step-by-step explanation:
When dealing with problems like this, it is always going to be the number they are looking for over the total number.
15 over 31
Step-by-step explanation:
When dealing with problems like this, it is always going to be the number they are looking for over the total number.
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if you need just tell me and i will come to the rescue.step-by-step explanation:...Read More
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y = - 6step-by-step explanation: given the point (- 2, y) lies on the graph then it is a solution of it's equation.substitute x = - 2 into the equation for value of yy = 3(- 2) = 3...Read More |
Question
# In an AP, the sum of the first ten terms is -150 and the sum of its next ten terms is -550. Find AP.
Hint: Sum of n terms of AP ${S_n} = \dfrac{n}{2}[2a + (n - 1)d]$.
Given sum of first ten terms = $- 150$
We know that Sum of n terms of AP ${S_n} = \dfrac{n}{2}[2a + (n - 1)d]$
Therefore,
$\ \Rightarrow {S_{10}} = \dfrac{{10}}{2}[2a + (10 - 1)d] \\ \Rightarrow - 150 = 5[2a + 9d] \\ \Rightarrow - 150 = 10a + 45d \to (1) \\ \$
And also given that sum of next ten terms =$- 550$ (which also includes sum of first ten terms value which has to be removed)
Sum of next ten terms = $- 550$
$\Rightarrow $- 150$ - 550$ = $\dfrac{{20}}{2}[2a + (20 - 1)d]$
$\ \Rightarrow - 700 = 10(2a + 19d) \\ \Rightarrow - 70 = 2a + 19d \to (2) \\ \$
Multiplying equation $(2) \times 5$ then we get
$\Rightarrow - 350 = 10a + 95d \to (3)$
On solving $(1)\& (3)$ we get
$d = - 4$
Now by substituting $'d'$ value in equation $(1)$ we get
$\ \Rightarrow - 150 = 10a + 45d \\ \Rightarrow - 150 = 10a + 45( - 4) \\ \Rightarrow 10a = - 150 + 180 \\ \Rightarrow 10a = 30 \\ \Rightarrow a = 3 \\ \$
Hence we got the value $a = 3,d = - 4$
We know that for an AP series $'a'$ be the first term and $'d'$ is the difference between the terms.
We also know that AP series will be of the form $a,a + d,a + 2d,a + 3d.......$
On substituting the $'a'$ and $'d'$ values
We get the values of series as 3,-1,-5,-9
Then the AP series will be $3, - 1, - 5, - 9....$
Note: In the above problem second condition i.e. sum of next ten terms includes sum of first 10 terms plus the other ten terms (where sum of first ten terms need to be subtracted from sum given for second condition) .Ignoring such simple condition will affect the answer. |
# Which System is Represented by the Graph
Which System is Represented by the Graph.
### Learning Outcomes
• Graph systems of equations
• Graph a organization of two linear equations
• Graph a system of two linear inequalities
• Evaluate ordered pairs as solutions to systems
• Determine whether an ordered pair is a solution to a organization of linear equations
• Make up one’s mind whether an ordered pair is a solution to a system of linear inequalities
• Classify solutions to systems
• Place what blazon of solution a organisation volition have based on its graph
The way a river flows depends on many variables including how big the river is, how much water it contains, what sorts of things are floating in the river, whether or non it is raining, and and so forth. If you want to best describe its menstruation, you lot must take into business relationship these other variables. A organisation of linear equations can help with that.
A
system of linear equations
consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. Yous will find systems of equations in every awarding of mathematics. They are a useful tool for discovering and describing how behaviors or processes are interrelated. It is rare to discover, for instance, a pattern of traffic period that that is just afflicted past weather. Accidents, fourth dimension of twenty-four hour period, and major sporting events are only a few of the other variables that tin can bear upon the flow of traffic in a city. In this section, nosotros will explore some basic principles for graphing and describing the intersection of ii lines that brand up a arrangement of equations.
## Graph a system of linear equations
In this section, we will look at systems of linear equations and inequalities in two variables. Commencement, nosotros volition practice graphing two equations on the same gear up of axes, and and so nosotros will explore the different considerations y’all need to make when graphing two linear inequalities on the same fix of axes. The same techniques are used to graph a organization of linear equations equally you have used to graph single linear equations. Nosotros tin can utilize tables of values, slope and
y-intercept, or
x– and
y-intercepts to graph both lines on the aforementioned set of axes.
For example, consider the following system of linear equations in two variables.
$\begin{array}{r}2x+y=-eight\\ ten-y=-1\end{array}$
Permit’south graph these using gradient-intercept form on the aforementioned set of axes. Recall that gradient-intercept form looks like $y=mx+b$, and then we will want to solve both equations for $y$.
First, solve for y in $2x+y=-8$ $\brainstorm{array}{c}2x+y=-8\\ y=-2x – viii\cease{array}$
2nd, solve for y in $x-y=-ane$ $\begin{array}{r}x-y=-ane\,\,\,\,\,\\ y=x+ane\finish{array}$
The system is at present written as
$\brainstorm{array}{c}y=-2x – 8\\y=x+1\end{array}$
Now y’all can graph both equations using their slopes and intercepts on the aforementioned set of axes, as seen in the figure below. Note how the graphs share one betoken in common. This is their point of intersection, a point that lies on both of the lines. In the adjacent section we will verify that this point is a solution to the system.
Read: Which Candidate Will Most Likely Get the Job
In the following instance, y’all volition be given a organisation to graph that consists of ii parallel lines.
### Instance
Graph the system $\brainstorm{array}{c}y=2x+1\\y=2x-three\terminate{array}$ using the slopes and y-intercepts of the lines.
In the next example, you lot will be given a arrangement whose equations look different, only later graphing, turn out to be the aforementioned line.
### Example
Graph the organisation $\begin{assortment}{c}y=\frac{1}{2}ten+2\\2y-ten=iv\end{assortment}$ using the x – and y-intercepts.
Graphing a system of linear equations consists of choosing which graphing method you want to utilise and drawing the graphs of both equations on the aforementioned set up of axes. When you lot graph a system of linear inequalities on the same set of axes, in that location are a few more things you volition need to consider.
## Graph a system of 2 inequalities
Call up from the module on graphing that the graph of a unmarried linear inequality splits the
coordinate plane
into ii regions. On ane side lie all the solutions to the inequality. On the other side, there are no solutions. Consider the graph of the inequality $y<2x+five$.
The dashed line is $y=2x+5$. Every ordered pair in the shaded area below the line is a solution to $y<2x+5$, as all of the points below the line will make the inequality true. If you doubt that, effort substituting the
ten
and
y
coordinates of Points A and B into the inequality—yous’ll see that they work. So, the shaded area shows all of the solutions for this inequality.
The boundary line divides the coordinate aeroplane in half. In this example, it is shown as a dashed line every bit the points on the line don’t satisfy the inequality. If the inequality had been $y\leq2x+5$, then the boundary line would accept been solid.
Let’southward graph some other inequality: $y>−x$. You can check a couple of points to determine which side of the boundary line to shade. Checking points M and N yield true statements. So, we shade the area above the line. The line is dashed as points on the line are not true.
To create a system of inequalities, you need to graph ii or more inequalities together. Let’s employ $y<2x+5$ and $y>−x$ since we accept already graphed each of them.
The royal area shows where the solutions of the ii inequalities overlap. This surface area is the solution to the
system of inequalities. Any point within this imperial region will exist true for both $y>−x$ and $y<2x+five$.
In the next example, you lot are given a system of two inequalities whose purlieus lines are parallel to each other.
### Examples
Graph the system $\brainstorm{assortment}{c}y\ge2x+1\\y\lt2x-3\cease{assortment}$
In the adjacent section, we will come across that points can be solutions to systems of equations and inequalities. We will verify algebraically whether a point is a solution to a linear equation or inequality.
## Determine whether an ordered pair is a solution for a arrangement of linear equations
The lines in the graph above are defined as
$\brainstorm{array}{r}2x+y=-eight\\ x-y=-1\end{assortment}$.
They cross at what appears to be $\left(-3,-two\right)$.
Using algebra, we tin can verify that this shared betoken is actually $\left(-3,-two\right)$ and non $\left(-2.999,-ane.999\right)$. By substituting the
x– and
y-values of the ordered pair into the equation of each line, you tin exam whether the betoken is on both lines. If the exchange results in a true statement, then you have plant a solution to the system of equations!
Read: Which Physical Property Can Be Measured Color Density Odor Shape
Since the solution of the system must be a solution to
all
the equations in the arrangement, you volition demand to check the point in each equation. In the following example, we volition substitute -3 for
x
and -ii for
y
in each equation to test whether it is actually the solution.
### Example
Is $\left(-3,-ii\right)$ a solution of the organisation
$\begin{assortment}{r}2x+y=-8\\ x-y=-i\end{array}$
### Instance
Is (iii, nine) a solution of the system
$\begin{array}{r}y=3x\\2x–y=half-dozen\finish{array}$
Is $(−2,4)$ a solution for the system
$\begin{array}{r}y=2x\\3x+2y=1\finish{assortment}$
Before you do any calculations, look at the indicate given and the first equation in the system. Can you predict the answer to the question without doing whatsoever algebra?
Remember that in order to be a solution to the organisation of equations, the values of the point must be a solution for both equations. Once yous discover i equation for which the bespeak is imitation, yous have determined that it is not a solution for the system.
Nosotros can use the same method to determine whether a point is a solution to a system of linear inequalities.
## Make up one’s mind whether an ordered pair is a solution to a system of linear inequalities
On the graph above, you can see that the points B and N are solutions for the system because their coordinates volition make both inequalities true statements.
In contrast, points K and A both prevarication outside the solution region (purple). While betoken M is a solution for the inequality $y>−10$ and bespeak A is a solution for the inequality $y<2x+5$, neither signal is a solution for the
system. The following example shows how to examination a point to see whether it is a solution to a system of inequalities.
### Example
Is the point (2, ane) a solution of the system $x+y>1$ and $2x+y<8$?
Here is a graph of the organization in the example higher up. Discover that (2, 1) lies in the majestic area, which is the overlapping expanse for the two inequalities.
### Example
Is the betoken (2, i) a solution of the organisation $x+y>one$ and $3x+y<4$?
Here is a graph of this system. Notice that (2, 1) is not in the majestic area, which is the overlapping expanse; it is a solution for i inequality (the ruby region), but it is not a solution for the second inequality (the blue region).
As shown above, finding the solutions of a system of inequalities can be done past graphing each inequality and identifying the region they share. Below, you lot are given more than examples that show the entire process of defining the region of solutions on a graph for a arrangement of two linear inequalities. The full general steps are outlined below:
• Graph each inequality every bit a line and determine whether it will be solid or dashed
• Make up one’s mind which side of each boundary line represents solutions to the inequality past testing a betoken on each side
• Shade the region that represents solutions for both inequalities
In this department nosotros have seen that solutions to systems of linear equations and inequalities can be ordered pairs. In the next section, nosotros will work with systems that accept no solutions or infinitely many solutions.
## Utilise a graph to classify solutions to systems
Recall that a linear equation graphs every bit a line, which indicates that all of the points on the line are solutions to that linear equation. There are an infinite number of solutions. As we saw in the last department, if you have a system of linear equations that intersect at one point, this point is a solution to the system. What happens if the lines never cross, as in the instance of parallel lines? How would yous depict the solutions to that kind of system? In this section, we will explore the three possible outcomes for solutions to a system of linear equations.
### Three possible outcomes for solutions to systems of equations
Recall that the solution for a arrangement of equations is the value or values that are true for
all
equations in the organisation. There are three possible outcomes for solutions to systems of linear equations. The graphs of equations within a organization can tell y’all how many solutions exist for that system. Expect at the images below. Each shows ii lines that make up a organisation of equations.
I Solution No Solutions Infinite Solutions
If the graphs of the equations intersect, and so there is one solution that is true for both equations. If the graphs of the equations exercise not intersect (for example, if they are parallel), so there are no solutions that are true for both equations. If the graphs of the equations are the same, then at that place are an infinite number of solutions that are true for both equations.
• One Solution: When a arrangement of equations intersects at an ordered pair, the organization has one solution.
• Infinite Solutions:
Sometimes the two equations will graph as the aforementioned line, in which instance we have an infinite number of solutions.
• No Solution:
When the lines that brand up a system are parallel, there are no solutions because the two lines share no points in common.
### Instance
Using the graph of $\begin{array}{r}y=x\\x+2y=half dozen\end{array}$, shown beneath, determine how many solutions the system has.
Using the graph of $\begin{array}{r}y=3.5x+0.25\\14x–4y=-4.5\end{assortment}$, shown below, determine how many solutions the system has.
### Example
How many solutions does the system $\begin{array}{r}y=2x+1\\−4x+2y=2\end{array}$ accept?
In the side by side section, we volition learn some algebraic methods for finding solutions to systems of equations. Recall that linear equations in one variable can have i solution, no solution, or many solutions and we can verify this algebraically. Nosotros will use the same ideas to allocate solutions to systems in ii variables algebraically.
## Which System is Represented by the Graph
Source: https://courses.lumenlearning.com/beginalgebra/chapter/introduction-to-systems-of-linear-equations/
## Which Book Citations Are Formatted Correctly Check All That Apply
By Vladimir Gjorgiev/Shutterstock Concealer is an essential part of any makeup routine. It’s many women’s … |
## Domino Sorting
Here are some dominoes taken out of the full set:
Sort them into two groups - one group with an odd number of spots and one group with an even number of spots.
Do you have any dominoes left over? Why or why not?
Now put the dominoes into pairs. The number of spots on each pair of dominoes must make a total of $5$.
How many pairs can you make?
Which dominoes are left over?
Can you pair them up in any different ways so that each pair adds to $5$?
Which dominoes are left over now?
Are there any dominoes which are always left over?
Can you explain why?
### Why do this problem?
This problem will help learners to become more familiar with odd and even numbers, and number bonds to five. It will also challenge them to justify findings.
### Possible approach
If you have an interactive whiteboard, you may find our Dominoes Environment useful for this problem.
You might like to start by giving pairs of children a whole set of dominoes to explore and ask them some open-ended questions such as:
• How can you sort them?
• Can you make a pattern?
• Can you make a snake?
• What did you notice?
Learners can then find the subset of dominoes that they need for this task and tackle it in pairs. It will provoke a lot of meaningful discussion and will give pupils the experience of having to argue mathematically.
In a plenary, focus on the "can you explain why?"- the beginning of an understanding of proof. The sooner we start children justifying their conclusions, the better mathematicians they'll turn out to be!
### Key questions
Can you think of some pairs of numbers that add to five?
Have you added up or counted the spots on each domino?
Which domino could you pair with this one so that there are five spots altogether?
### Possible extension
You might want some children to find all the different ways of making pairs that add to $5$. This could be by picking two and then replacing them, or by finding all the different combinations which could be made at the same time (the problem as written focuses on the latter). Whichever way, part of their task should be to convince you that they have not missed any pairs out.
Domino Join Up is a similar problem which gives practice in number bonds to six, and could be used as an extension to this problem.
### Possible support
Children would really benefit from having sets of dominoes to manipulate as this allows them to change their mind easily, so giving them more confidence to begin the task, and also prevents them from using any domino twice. |
# Analyzing Algorithm Control Structure
To analyze a programming code or algorithm, we must notice that each instruction affects the overall performance of the algorithm and therefore, each instruction must be analyzed separately to analyze overall performance. However, there are some algorithm control structures which are present in each programming code and have a specific asymptotic analysis.
Some Algorithm Control Structures are:
1. Sequencing
2. If-then-else
3. for loop
4. While loop
### 1. Sequencing:
Suppose our algorithm consists of two parts A and B. A takes time tA and B takes time tB for computation. The total computation "tA + tB" is according to the sequence rule. According to maximum rule, this computation time is (max (tA,tB)).
Example:
```Suppose tA =O (n) and tB = θ (n2).
Then, the total computation time can be calculated as
Computation Time = tA + tB
= (max (tA,tB)
= (max (O (n), θ (n2)) = θ (n2)
```
### 2. If-then-else:
The total time computation is according to the condition rule-"if-then-else." According to the maximum rule, this computation time is max (tA,tB).
Example:
```Suppose tA = O (n2) and tB = θ (n2)
Calculate the total computation time for the following:
Total Computation = (max (tA,tB))
= max (O (n2), θ (n2) = θ (n2)
```
### 3. For loop:
The general format of for loop is:
## Complexity of for loop:
The outer loop executes N times. Every time the outer loop executes, the inner loop executes M times. As a result, the statements in the inner loop execute a total of N * M times. Thus, the total complexity for the two loops is O (N2)
Consider the following loop:
If the computation time ti for ( PI) various as a function of "i", then the total computation time for the loop is given not by a multiplication but by a sum i.e.
Takes
If the algorithms consist of nested "for" loops, then the total computation time is
```For i ← 1 to n
{
For j ← 1 to n
{
P (ij)
}
}
```
Example:
Consider the following "for" loop, Calculate the total computation time for the following:
Solution:
The total Computation time is:
### 4. While loop:
The Simple technique for analyzing the loop is to determine the function of variable involved whose value decreases each time around. Secondly, for terminating the loop, it is necessary that value must be a positive integer. By keeping track of how many times the value of function decreases, one can obtain the number of repetition of the loop. The other approach for analyzing "while" loop is to treat them as recursive algorithms.
## Algorithm:
Example:
The running time of algorithm array Max of computing the maximum element in an array of n integer is O (n).
Solution:
The number of primitive operation t (n) executed by this algorithm is at least.
The best case T(n) =5n occurs when A [0] is the maximum element. The worst case T(n) = 7n-2 occurs when element are sorted in increasing order.
We may, therefore, apply the big-Oh definition with c=7 and n0=1 and conclude the running time of this is O (n). |
When equations have lots of decimals, like the one seen below, you may be able to solve it as it is written, but it will probably be easier to clear the decimals first. To clear an equation of decimals, multiply each term on both sides by the power of ten that will make all the decimals whole numbers.
Keeping this in view, what is 5/12 as a decimal?
Fraction to Decimal Conversion Tables
fraction = decimal
1/12 = 0.0835/12 = 0.4167/12 = 0.583
1/16 = 0.06253/16 = 0.18755/16 = 0.3125
11/16 = 0.687513/16 = 0.8125
1/32 = 0.031253/32 = 0.093755/32 = 0.15625
Likewise, how do we divide decimals? To divide decimal numbers: Multiply the divisor by as many 10's as necessary until we get a whole number. Remember to multiply the dividend by the same number of 10's.
Also, how do you clear decimals in an equation?
To clear an equation of decimals, multiply each term on both sides by the power of ten that will make all the decimals whole numbers. In our example above, if we multiply . 25 by 100, we will get 25, a whole number. Since each decimal only goes to the hundredths place, 100 will work for all three terms.
How do we multiply decimals?
Multiply the numbers just as if they were whole numbers.
1. Line up the numbers on the right – do not align the decimal points.
2. Starting on the right, multiply each digit in the top number by each digit in the bottom number, just as with whole numbers.
## How do you solve one step equations with fractions?
To solve one step equations involving fractions, we add, subtract, multiply, or divide a fraction from both sides of the equation to isolate the variable. That is, if a and b are fractions, and x is a variable, then we use the rules in the following table to solve for x in each of the one step equations.
## What are the four rules of decimals?
You should become efficient in using the four basic operations involving decimals—addition, subtraction, multiplication, and division.
## How do you solve a point addition?
1. Write down the numbers, one under the other, with the decimal points lined up.
2. Put in zeros so the numbers have the same length (see below for why that is OK)
## What is a decimal factor?
Multiplying decimals is the same as multiplying whole numbers except for the placement of the decimal point in the answer. When you multiply decimals, the decimal point is placed in the product so that the number of decimal places in the product is the sum of the decimal places in the factors.
## How do we change a decimal to a fraction?
Convert Decimals to Fractions
1. Step 1: Write down the decimal divided by 1, like this: decimal 1.
2. Step 2: Multiply both top and bottom by 10 for every number after the decimal point. (For example, if there are two numbers after the decimal point, then use 100, if there are three then use 1000, etc.)
3. Step 3: Simplify (or reduce) the fraction.
## Where is the tens place in a decimal?
Place Value for Decimals
For example, the number 0.1234 has a “1” in the tenths place, a “2” in the hundredths place, a “3” in the thousandths place, and a “4” in the ten thousandths place. |
Education
How Teachers Should Make Quadratic Equations Interesting For Students?
What could be better than finding a teacher for a complicated subject such as math? Most students do not like this tricky subject since they do not find it interesting. On the other hand, students who get creative teachers even love complicated subjects such as science, math, economics, etc.
When it comes to math, many complicated topics make students get frustrated. A quadratic equation is one of the topics in which students do not show much interest as they get confused while studying it. But teachers can simplify this topic if they follow a creative approach towards their teaching methods.
Here, we are going to emphasize a bit on how intelligent and creative teachers make it easy for the students. Let’s check out more about it in a detailed manner –
Significance Of Quadratic Equation Functions –
Well, we all know how quadratic function makes most of us frustrated because of being part of the school curriculum. These functions are a bit more advanced. They are known for imparting an important move. A function in which variable rates of change are called qualitatively. If output values are shown as X and Y, it could be easy to find the value. It could be quite tricky to understand if the translation was vertical or horizontal when it comes to translating linear functions. Though students could wonder why they should learn about y-intercept following y=mx+c representation especially if the ax+by=c which means they both pretty much hold equal importance.
Quadratic function plays an important role to remove these issues since it helps to figure out the direction of translations. It introduces intercepts on the x-axis. If you get good at it, it would also be helping you to do future related work going with functions and graphs.
Moreover, quadratics are also regarded as the only functions where students would be ideally accessible in algebraic as well as arithmetic manipulation to introduce the relationship between input/out values. It will also help in the context of understanding the different algebraic representations.
When you study how to find roots of quadratic equations you also get answers to different questions including the increase and decrease, upturns and downturns, rates of change, specific values, and maxima location, and so on. Understanding quadratics, students will also get access to the imaginary number to find answers to different mathematical issues.
Show Their Purpose In The Classroom –
To make this topic interesting for the students, they need to understand how to teach it in an ideal way. Teachers should arrange some question-related sessions where students should be free to ask anything coming to their minds related to this topic. Moreover, this topic is also quite important to understand for students who are supposed to go for high mathematics study or for them who would be using graphs in a wide array of subjects and purposes.
Creative teachers say that the quadratic factored form and interpretation and sketching of graphs should be taught together so that students could have more clarity about it. Moreover, it would be better if vertical translations and scaling can be used in the context of distinguishing between quadratics having the same roots indeed.
An ideal teacher much goes with curiosity, conjecture, and exploration following graphical software so that students could learn it in an ideal way. A teacher should encourage them to go with intellectual power so that they could easily create and identify quadrats which look different from their appearance.
Going with the factored form is also considered ideal since it imparts needed access to the affected area model of the quadratic express. This method will also help them to emphasize overall characteristics including linear equations in two variables instead of individual points. This way will truly make for them easy to understand functions in the form of objects in their own right instead of just a way of joining the dots.
Conclusion –
Every student deserves the best teacher so that complicated subjects do not remain hard-to-understand for them. |
# Class 7 Maths Chapter 4 Exercise 4.4 Pdf Notes NCERT Solutions
Class 7 Maths Chapter 4 Simple Equations Exercise 4.4 pdf notes:-
Exercise 4.4 Class 7 maths Chapter 4 Pdf Notes:-
## Ncert Solution for Class 7 Maths Chapter 4 Simple Equations Exercise 4.4 Tips:-
APPLICATIONS OF SIMPLE EQUATIONS TO PRACTICAL
SITUATIONS
We have already seen examples in which we have taken statements in everyday language
and converted them into simple equations. We also have learnt how to solve simple equations.
Thus we are ready to solve puzzles/problems from practical situations. The method is first
to form equations corresponding to such situations and then to solve those equations
to give the solution to the puzzles/problems.
TRY THESE
(i) When you multiply a number by 6 and subtract 5 from the product, you get 7. Can you tell what the number is?
(ii) What is that number one third of which added to 5 gives 8?
TRY THESE There are two types of boxes containing mangoes. Each box of the larger type contains
4 more mangoes than the number of mangoes contained in 8 boxes of the smaller type.
Each larger box contains 100 mangoes. Find the number of mangoes contained in the
smaller box?
WHAT HAVE WE DISCUSSED?
1.An equation is a condition on a variable such that two expressions in the variable should have equal value.
2.The value of the variable for which the equation is satisfied is called the solution of the equation.
3.An equation remains the same if the LHS and the RHS are interchanged.
4.In case of the balanced equation, if we
(i) add the same number to both the sides, or (ii) subtract the same number from
both the sides, or (iii) multiply both sides by the same number, or (iv) divide both
sides by the same number, the balance remains undisturbed, i.e., the value of the
LHS remains equal to the value of the RHS
5.The above property gives a systematic method of solving an equation. We carry out
a series of identical mathematical operations on the two sides of the equation in such
a way that on one of the sides we get just the variable. The last step is the solution of
the equation.
6.Transposing means moving to the other side. Transposition of a number has the same
effect as adding same number to (or subtracting the same number from) both sides of
the equation. When you transpose a number from one side of the equation to the
other side, you change its sign. For example, transposing +3 from the LHS to the
RHS in equation x + 3 = 8 gives x = 8 ā 3 (= 5). We can carry out the transposition
of an expression in the same way as the transposition of a number.
7.We have learnt how to construct simple algebraic expressions corresponding to
practical situations.
We also learnt how, using the technique of doing the same mathematical operation
(for example adding the same number) on both sides, we could build an equation
starting from its solution. Further, we also learnt that we could relate a given equation
to some appropriate practical situation and build a practical word problem/puzzle
from the equation. |
# Thread: Find area of a 'curved triangle'
1. ## Find area of a 'curved triangle'
Background
Originally, there was a pen of radius 10cm for her animals to safely roam around in (Figure 1).
She found that the animals often destroyed the vegetation on the floor before it could regrow. She decided to split the pen into 3 equal lengths of AB, BC and CD (Figure 2).
She found she could join them together to create an equilateral triangle that had a side length of 10cm within the 'curved triangle'. (Figure 3).
Questions
1) What is the area of the whole 'curved triangle'?
2) She found she could enclose it in a square, with the sides of the curved triangle just touching the sides of the square. She found that any square she drew around this curved triangle was the same size. Assuming this is true, what is the side length of the square?
2. Originally Posted by BG5965
Background
Originally, there was a pen of radius 10cm for her animals to safely roam around in (Figure 1).
She found that the animals often destroyed the vegetation on the floor before it could regrow. She decided to split the pen into 3 equal lengths of AB, BC and CD (Figure 2).
She found she could join them together to create an equilateral triangle that had a side length of 10cm within the 'curved triangle'. (Figure 3).
Questions
1) What is the area of the whole 'curved triangle'?
2) She found she could enclose it in a square, with the sides of the curved triangle just touching the sides of the square. She found that any square she drew around this curved triangle was the same size. Assuming this is true, what is the side length of the square?
to #1: The radius of the arcs didn't change. Therefore you'll get an equilateral triangle with the side-length of 10.
$\displaystyle a_{\Delta}=\dfrac14 \cdot r^2 \cdot \sqrt{3}$
To this equilateral triangle are attached 3 congruent circle segments. The equilateral triangle with one attached segment (painted grey) forms one sixth of a circle with radius 10. Therefore the area of the complete shape is:
$\displaystyle a = \dfrac14 \cdot r^2 \cdot \sqrt{3} + 3\cdot \left(\dfrac16 \cdot \pi r^2 - \dfrac14 \cdot r^2 \cdot \sqrt{3} \right) = \dfrac12 r^2(\pi - \sqrt{3})$
to #2:
Two adjacent sides of the square are tangents to adjacent arcs(with radius 10) while the opposite sides of the square pass through vertices of the curved triangle. The distance between a vertex of the curved triangle to the corresponding tangent side of the square must be a radius too. Therefore the square has a side-length of r.
3. ## square root
hi, why is there a square root of 3? Is it because of the three segments of the circle?$\displaystyle a_{\Delta}=\dfrac14 \cdot r^2 \cdot \sqrt{3}$
4. Originally Posted by smile4life
hi, why is there a square root of 3? Is it because of the three segments of the circle?$\displaystyle a_{\Delta}=\dfrac14 \cdot r^2 \cdot \sqrt{3}$
If you want tocalculate the area of an equilateral triangle you need the base and the height of the triangle.
Take the left triangle AFC. Use Pythagorean theorem:
$\displaystyle a^2 = h^2+\left(\dfrac a2 \right)^2~\implies~ h^2 = a^2-\dfrac14 a^2 = \dfrac34 a^2$
Therefore $\displaystyle h = \dfrac12 a \sqrt{3}$
The area of the triangle is then:
$\displaystyle A = \dfrac12 \cdot \underbrace{a}_{base} \cdot \underbrace{\dfrac12 \cdot a \cdot \sqrt{3}}_{height} = \dfrac{a^2}4 \sqrt{3}$
,
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# area of a squared triangle of a sinosidal curve
Click on a term to search for related topics. |
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# Dividing By 3 and 4 (Three and Four) Lesson Plan
Lesson Plans > Mathematics > Division
## Dividing By 3 and 4 (Three and Four) Lesson Plan
Heather Breaux
Division Unit
Lesson 6: Dividing by 3 and 4
Date:
Approximate Time: 60 minutes
Berks County Standards:
1. 2.1.3 Numbers, Number Systems and Number Relationships- C. Represent equivalent forms of the same number through the use of concrete objects, drawings, word names and symbols.
2. 2.1.3 Numbers, Number Systems and Number Relationships- L. Demonstrate knowledge of basic facts in four basic operations.
3. 2.2.3 Computation and Estimation- D. Demonstrate concept of division as repeated subtraction and as sharing.
NCTM Standards:
1. Number and Operations
7. Reasoning and Proof
8. Communication
Behavioral Objectives: Upon completion of this lesson, students will be able to...
1. Divide by 3 and 4
2. Know how to use multiplication to check their division facts
Materials/Technology Needs: 18 Enlarged cookies and 3 plates; Paddle Problem and What If problem in textbooks; dry erase boards and markers; tables 12 and 13 from p. 260; colored chalk; “Reteach 13.2” RW68; “Practice 13.2” PW68; “Memory” cards and baggies for groups of 3’s.
Anticipatory Set: Congratulate the students on a spectacular job of learning how to divide by 2 and 5 yesterday!! Begin the lesson by putting 18 enlarged cookies in a group and 3 paper plates on the board. Explain to the class that we need to separate the cookies into equal groups amongst the 3 plates so that Miss Sweigart, Miss Perritt, and I can each have a plate of cookies. We have to make sure that the plates have the same amount of cookies or we will have some angry teachers!! Ask what division sentence is being used and have a student come up and write it on the board. Ask, “How should we begin to solve this problem? Who can tell me how to do this by connecting what we learned yesterday to this problem?” (find a multiplication sentence because the two processes are inverse operations). Write 3 x ?= 18 (6). “So, how many cookies need to go on each plate so we can have happy teachers?” (6).
Procedure:
1. Introduce today’s topic as learning how to divide by 3 and 4, explaining that we are going to use the same procedures as yesterday, except just with new divisors. Turn the classes’ attention to the “Paddle Problem” on p. 260 of their textbooks. Have a volunteer read the problem and ask the class what division sentence is being used. Write the sentence on the board and have a student draw arrows and label each number in the sentence as the dividend, divisor, and quotient. Ask what the meaning of each number is in relation to the problem. Say, “If I was going to use a multiplication fact to help me solve this problem, what would I use?” ( 3 x ?= 24). Call on a student to fill in the question mark with 8. Say, “Because 3 x 8= 24, that tells us that 24 ÷ 3= 8 because of our knowledge about… (call on students for answers like fact families and inverse relationships).
Paddle Problem: The Traveler Scouts want to rent canoes. There are 24 people in the group. A canoe can hold 3 people. How many canoes should the group rent?
2. Have the students take out their dry erase boards and markers and solve the “What If” problem from p. 260 in the same fashion. Ask the students once again for the meaning of each number and have them write their answers on their boards. Have the students write the multiplication fact that helped them solve the division sentence and the answer to the question on their boards.
What If Problem: What if the group wants to rent rowboats instead? If each rowboat holds 4 people, how many rowboats should they rent?
3. Copy tables #12 and 13 from page. 261 onto the board and call on students to identify the number to fill in to make the number sentence true. Have the students explain their answer by writing the multiplication fact they used to check their work.
4. Pass out “Reteach 13.2” on p. 260. Read the top of the page as a class out loud and answer questions 1-6 together. For problems 7-23, have the students work in pairs, where they will individually solve on dry erase boards and compare each others work before coming upon an answer to write on their paper.
5. Split students into groups of threes, where each group will be given a baggie containing index cards with division sentences with missing quotients and index cards with multiplication facts on them. Each group will flip the cards upside-down and play the “Memory” game, where the object is to collect the most matches. Other students must double check all matches that are made, even if they are not their own, to be sure that it is indeed a match.
6. Hand out “Practice 13.2” PW68 on p. 260 for homework.
Accommodations Jessica will be seated at the front of the room so that she can see all problems on the board. I will provide her with enlarged worksheets to work with at her desk and for homework, as well as give her a larger version of the “Memory” game.
Closure: Write 18 ÷ 3; 12 ÷ 4; 28 ÷4; and 27÷3 on the left side of the board and 4 x 3; 3 x 9; 3 x 6; and 4 x 7 on the right side of the board. Call on students to draw a line using colored chalk to match the division sentences to the proper multiplication facts.
Key Questions:
1. “What process can we use to check our division facts?”
2. “Why can we use this process?”
3. “What two divisors did we work with a great deal today?”
Evaluation: I will be using “Reteach 13.2” problems 1-6, the table problems #12 and 13 from p. 261, and the students abilities to make matches during the “Memory” game as my formative assessment. The students’ homework assignment will be used as their summative assessment for the day.
Assignment: “Practice 13.2” PW68 on p. 260
Resources Used:
1. (2004). Harcourt Math Teacher Edition: Third Grade. Vol. 2.) Harcourt, Inc.
Frank D'Angelo, who submitted these articles, writes: [This] is an excellent unit submitted by a student in my Elementary Mathematics Methods Course.
Lesson by Frank D'Angelo
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Lesson 3: Rotations
# Rotations review
Review the basics of rotations, and then perform some rotations.
### What is a rotation?
A rotation is a type of transformation that takes each point in a figure and rotates it a certain number of degrees around a given point.
For example, this animation shows a rotation of pentagon $IDEAL$ about the point $\left(0,-1\right)$. You can see the angle of rotation at the bottom, which increases the further we rotate the figure from its original position.
The result of a rotation is a new figure, called the image. The image is congruent to the original figure.
## Performing rotations
Although a figure can be rotated any number of degrees, the rotation will usually be a common angle such as ${45}^{\circ }$or ${180}^{\circ }$.
If the number of degrees are positive, the figure will rotate counter-clockwise.
If the number of degrees are negative, the figure will rotate clockwise.
The figure can rotate around any given point.
Example:
Rotate $\mathrm{△}OAR$ ${60}^{\circ }$ about point $\left(-2,-3\right)$.
The center of rotation is $\left(-2,-3\right)$.
Rotation by ${60}^{\circ }$ moves each point about $\left(-2,-3\right)$ in a counter-clockwise direction. The rotation maps $\mathrm{△}OAR$ onto the triangle below.
## Practice
Problem 1
$\mathrm{△}NOW$ is rotated ${90}^{\circ }$ about the origin.
Draw the image of this rotation.
Want to try more problems like this? Check out this exercise.
## Want to join the conversation?
• Why is positive counter-clockwise? Is that just a rule or...
• It’s because you’re rotating in the order of the coordinate plane’s quadrants, which goes from right to left to down to right (*counterclockwise*)
• So how do you suppose the exact rotation on here? Or do you just guess how much to turn it ....
• I don't understand how to rotate about the origin, please explain.
• @garrettcummings22, I realize the frustration of these geometric principles, but these same principles are the foundations of graphic design, several types of engineering, carpentry, masonry, many forms of art. The reality is no one in grades 7-12 will ever know if they will use any of the math they are required to study. No one knows what their future holds. Reality also tells us that every math principle taught is a math concept actually used somewhere in real life. I have used several concepts, especially writing, solving, and graphing linear equations, Pythagorean Theorem, ratios and percents, and many other aspects of statistics throughout my many years of life and many occupations in life.
Good luck in all you do. I hope you find some purpose in your life for the math. I genuinely mean that.
• I don’t see how it’s possible to rotate any polygon free-hand,without the rotation tool.
• You could rotate each point of the polygon on its own, and then connect each point correctly. Remember that each point will be the same distance from the centre of rotation.
• I have to find the coordinates of a point on a shape after a rotation, but I wasn't given a graph or any image.
(Example):'triangle'NPQ has vertices N(-6,-4), P(-3,4), and Q(1,1). If the triangle is rotated 90 degrees about the origin, what are the coordinates of P'?
Is there a rule or formula to solve this or do I have to actually draw a graph, plot the points, rotate, ect.? Thanks.
• Great question! There are actually several helpful shortcuts for finding rotations. For rotating 90 degrees counterclockwise about the origin, a point (x, y) becomes (-y, x). So for example, your N would become (4, -6). |
# Arithmetic vs. Geometric Rate of Return. Calculations and Explanation
#### Table of Contents
When you’re calculating the rate of return for stock or for an index like the S&P 500 you have to be very careful to understand what type of return it is you’re calculating. Are you calculating an arithmetic return or a geometric compounded annual rate of return? That matters and I’m gonna explain why in this article.
## Arithmetic Rate of Return
So let’s take the arithmetic rate of return first, so let’s go ahead and calculate that if we would just take 25% return for year 1, we have the rate of return 25% we add that and then we subtract out 40 for year 2, as we have a loss of 40% and then we add 30 for year 3. These are our rates of return for some stock. Let’s say and then we divide that by 3 which is the number of years, and that’s going to give us a rate of return of 5% if we have used the arithmetic method.
## Geometric or Compounded Rate of Return
Now if we were to use the geometric or compounded rate of return we might get a very different rate of return. So let me go ahead and I’ll just calculate that, I have already written an article on Geometric or Compounded Rate of Return and if you don’t quite follow it I suggest you read the article that I wrote on that. So we’ll take the (1.25 ✕ 0.6 1.3) and the reason this, 0.6 might throw you off is basically when we lose 40% of our investment that’s the same way as saying we’re keeping 60% of it or 0.6. So we’re gonna take all that but then we’re going to raise that to power, we’re gonna raise that to a power which is 1 divided by the number of periods and there are three years here, so we’re gonna raise it to the 1/3 power. Then from that whole thing, we are gonna subtract 1. Now that is going to give us -.008. I’m just gonna round that number, it is basically -0.01 which is the same as saying -1%.
## What is the Difference Between Arithmetic and Geometric Return?
Now if you’ve noticed immediately in one situation when we do the arithmetic rate of return we actually have a positive 5% but when we use the geometric or compounded rate of return it’s actually negative. It’s not just a different rate of return it’s actually a negative number because we’ve lost money and you might be wondering how could this be. Well, basically the arithmetic rate of return does not consider volatility. It doesn’t take into consideration the fact, it’s just averaging these three rates of return but you have to remember for example after year 2 when you’ve had that negative 40% that’s affecting the balance. When you’re using the arithmetic return it’s not really considering any of the volatility and how the balance that earning, whatever the rate is it’s not considering how that balance changes over time. It’s just taking basically the average rate of return so that’s really what’s driving the difference.
## When should the Arithmetic Rate of Return be Used?
So you might wonder then why would we ever use the arithmetic rate of return? Well, we want to use the arithmetic rate of return when we’re looking at future performance. When we’re trying to say “What do we think is going to be the expected rate of return for this particular stock or this particular portfolio of assets?” When you’re looking you know forward-looking information you’re trying to compute an expected return, then it’s actually not going to be that problematic to use the arithmetic rate of return.
Now when you’re looking backward, when you’re looking at historical information so let’s say you’re talking about a mutual fund and you want to look at that mutual fund’s performance over time then you want to be using the compounded annual rate of return. You might even see something in your mutual fund prospectus it says something like your compounded annual growth rate (CAGR) etc and that’s this geometric rate of return. You can call it geometric or call it compounded but that’s the idea is, this is understanding that because of volatility the balance that is earning a rate of return is going to change over time, it changes from year 1 to year 2 from year 2 to year 3. The balance is changing and so you’re having a bigger or smaller balance the next year that’s earning that rate of return. So if you really want to evaluate a mutual fund or some stock or you know that’s SP 500 and you want to look at historical information then you really need to be looking at this geometric compounded rate of return rather than the arithmetical.
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# How do you differentiate f(x)=e^cos(-2lnx) using the chain rule?
Aug 25, 2016
$f ' \left(x\right) = \frac{2 \sin \left(- 2 \ln \left(x\right)\right) \times {e}^{\cos \left(- 2 \ln \left(x\right)\right)}}{x}$
#### Explanation:
Let's first consider a few derivative techniques and procedures:
If $F \left(x\right) = f \left(g \left(n x\right)\right)$, then the derivative of the function may be written as: $F ' \left(x\right) = f ' \left(g \left(x\right)\right) \times g ' \left(x\right) \times n$
*The derivative of ${e}^{k x}$ if $k {e}^{k x}$ *
The derivative of $\cos \left(n x\right)$ is $- n \sin \left(x\right)$
The derivative of $\ln \left(a x\right)$ is $\frac{1}{a x} \times a$
We can structure your question in a similar manner:
If we consider each "section" of the function separately, we can state that:
f'(x)=(d{e^(cos(-2ln(x)))}}/(d{cos(-2ln(x))}
xx (d{cos(-2ln(x)))}/(d{(-2ln(x))}
xx (d{-2ln(x))}/(d{x}
Applying our rules from above, we can derive such that:
$f ' \left(x\right) = \cos \left(- 2 \ln \left(x\right)\right) \times {e}^{\cos \left(- 2 \ln \left(x\right)\right)} \times - \sin \left(- 2 \ln \left(x\right)\right) \times - 2 \times \frac{1}{x}$
Simplifying this we get:
$f ' \left(x\right) = - 2 \times - \sin \left(- 2 \ln \left(x\right)\right) \times {e}^{\cos \left(- 2 \ln \left(x\right)\right)} \times \frac{1}{x}$
Simplifying once more we get:
$f ' \left(x\right) = \frac{2 \sin \left(- 2 \ln \left(x\right)\right) \times {e}^{\cos \left(- 2 \ln \left(x\right)\right)}}{x}$ |
Are You Teaching the Two Types of Division?
It’s time to introduce your class to the subject of division. You teach the vocabulary of division and then you begin problem-solving. Yet, when you problem solve, do you teach just one type of division problems or do you teach both? Most often teachers will introduce students to one kind and never even mention the other.
It’s quite common among teachers to introduce students to fair sharing. This is technically called Partitive division. However, the common core standards require us to teach both partitive and quotative division. Quotative division is creating sets–and often overlooked.
Teaching Partitive Division (or Fair Sharing)
While fair sharing division doesn’t need much introduction, I’ll still provide information about it so this post is complete. First, fair sharing is often “taught” to children before it’s formally taught in the classroom during math. For instance, my twins are five and they definitely can tell you about fair sharing when it comes to cookies, candy, and other things they want.
Fair sharing is essentially just repeated subtraction. Students simply take the dividend and evenly distribute it among the divisor. This is typically done in a concrete matter first, whether with base-ten blocks, goldfish crackers, or the like. Then, students should be moved into the semi-concrete, and finally into the abstract. You can read about how to teach math so students get it here. (See image below.)
In class, you would have students sort things into groups such as this activity below, where we sort bees into jars. In partitive division, the question would look like this:
You have 12 bees that need to be sorted evenly into 3 jars. How many bees will you place into each jar?
In this problem, the dividend is known (the 12 – total number of bees) and the number of groups is known (the divisor as the jars), but the number in each group is unknown until the problem is solved. (And the problem below is not completed. There are more bees off in the corner, but the image is cut off.)
Teaching Quotative Division (Measurement)
Quotative division is the least taught–especially at the concrete level. Just as in multiplication, this is the “sets” version. Just as in the partitive type, the student is told the total number of items, and then they need to divide the sets into groups. However, the number of the groups is not known. They only know how many are in each group.
When I teach my students to do this, I pull out graphing paper and color pencils. I do not worry about arrays. I don’t feel that is our focus at this time. First, I have students either cut out the dividend or outline it in heavy black. Then I have students color in each set, or the divisor, in a different color until they have completed the graphing paper or until they cannot complete anymore sets. We then discuss if there is a remainder or not. (See the chart above.)
As in the example above, in quotative division the problem would look like this:
You have 12 bees. You are going to put them in jars, with 3 in each jar. How many jars will you need?
Again, in this problem, the dividend is known (the total number of bees – 12), the number in each group is known (the sets – divisor of 3), but the number of groups is unknown.
And using the method of graph paper and color pencils that I use, it would look like this below:
We then took all of these together and created a quotative division craftivity to hang up and display. (See image below.)
I decided that since quotative division is the least concrete taught type, I would provide you with a free lesson that has all of the materials in this post. It includes the step-by-step lesson on how to teach it (in the math workshop format, but you can modify it to your liking), along with the independent activity that includes the craftivity. It also includes the anchor chart above. It is a small sample from my fourth-grade division math workshop (or you can find it on TpT here).
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# Collinear points
Alignments to Content Standards: A-REI.D.10
Consider three points in the plane, $P=(-4, 0), Q=(-1, 12)$ and $R=(4, 32)$.
1. Find the equation of the line through $P$ and $Q$.
2. Use your equation in (a) to show that $R$ is on the same line as $P$ and $Q$.
3. Show that $P, Q$ and $R$ are on the graph of the equation $y = x^3 + x^2 - 12 x$.
4. Is it possible for $P, Q$ and $R$ to all lie on a parabola of the form $y=ax^2+bx+c$?
## IM Commentary
This task leads students through a series of problems which illustrate a crucial interplay between algebra (e.g., being solutions to equations) and geometry (e.g., being points on a curve). Students first confirm that $P, Q$ and $R$ are collinear, and then explore other types of curves on which all three points could reside. In particular, they conclude that there does exist such a cubic curve, but no such parabola exists.
When factored, the cubic equation is $y = x(x-3)(x+4)$. It is easy to graph by hand since it has roots at $x=0, 3$ and $-4$ and also goes through $P, Q$ and $R$.
Part (d) is an exploration with multiple possible approaches. They might arrive at a negative answer geometrically, by noting that a secant to a parabola crosses the parabola at most twice. Alternatively, students might try an algebraic approach, discovering that all three points can satisfy the equation $y = a x^2 + b x + c$, only if $a=0$, recovering the original line from part (a).
Instructors might profitably use this task to instigate a discussion about the relationship between algebraic and geometric approaches, perhaps having students share methods of each type.
## Solution
1. Students may try to use slope-intercept form or slope-point form for the equation of a line. Since two points are given, they will find the slope first as rise-over-run or $$m=(y_2-y_1)/(x_2-x_1) = (12-0)/(-1-(-4)) = 12/3 =4.$$
If they use slope-intercept form, they will then plug one point into the equation, $y=4x+b$ and find $b=16$, so $y=4x+16$.
If they use slope-point form, they will plug a point into $y-y_1=4(x-x_1)$ to get either of the equivalent forms $y=4(x+4)$ or $y-12=4(x+1)$.
2. To see that $R$ is on the same line, students decide if the point $(x,y)=(4,32)$ satisfies their equation from the previous part. For example, given $y=4x+16$, students can verify that $32=4(4)+16$. We conclude that $R$ is on the same line as $P$ and $Q$.
Other solution approaches abound: Students might, for example, also check that the slope between $R$ and either $P$ or $Q$ is 4, so they must lie on the same line.
3. Students will most easily check this by confirming that the equation $y=x^3+x^3-12x$ is true after substituting in the $x$-coordinates and $y$-coordinates of each of $P$, $Q$ and $R$.
4. We can argue geometrically as follows: A straight line crosses a parabola at most twice, so since $P$, $Q$, and $R$ are colinear, it cannot be that all three of them are on any parabola. (Note that this answer requires the convention that lines do not count as parabolas. Without this convention, the "degenerate parabola" consisting of the line $y=4x+16$ itself does contain all three points.)
Algebraically, we can proceed by solving the system of equations we get by evaluating the expression $y=ax^2+bx+c$ at the $x$- and $y$-coordinates of $P$, $Q$, and $R$. This gives \begin{eqnarray*} 0&=&a (-4)^2+b(-4)+c\\ 12&=&a(-1)^2+b(-1)+c\\ 32&=&a(4)^3+b(4)+c \end{eqnarray*} Solving the resulting system gives $a=0, b=4$ and $c=16$.
It may be worth remarking that, in this case, the geometric approach proves a more general statement, as it applies to all parabolas, and not just those of the form $y=ax^2+bx+c$. |
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8.5 Arc Length ( Rectification ) The method of finding the length of the arc of a curve is called the rectification. For arc length, the function and its derivative must both be continuous on the closed interval. If y = f (x) and f’ (x) are continuous on [ a, b ], then the arc length ( L ) of f (x) on [ a,b ] is given by Similarly for, x = f (y) and f’ (y) are continuous on [ a,b ], then the arc length (L) of f (y) on [ a,b ] is given by Example 29 Find the length of the arc of f (x) = x3/2 on [ 0, 5 ]. Solution : both are continuous on [ 0, 5 ]. then length of arc of Example 30 Find the length of the arc of the parabola y2 = 12x cut off by the Latus rectum. Solution : y2 = 12 x ® Parabola. Comparing with y2 = 4 ax, we see that 4 a = 12 \ a = 3 \ AS = SL = 6 \ co-ordinates of L are (3,6) Example 31 Find the length of the curve y2 = (2 x - 1 )3, cut off by the line x = 4. Solution : y2 = (2 x - 1 )3 is a curve (i) symmetrical about the x-axis (ii) not passing through (0,0) (iii) it cuts the x-axis, where y = 0 \ x = ˝ \ It's vertex is at ( ˝ , 0) (iv) It does not cut the y-axis as, taking x = 0 we get y = ± i (v) No asymptotes. (vi) (2x - 1)3 positive \ y2 ³ 0 \ x ³ ˝ \ The curve lies in only 1st and 4th quadrants Now Index 8.1 Introduction8.2 Area 8.3 Volumes 8.4 Mean Value 8.5 Arc Length(Rectification) Chapter 1 |
# Understanding the Formula for A^3 + B^3
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Mathematics is a subject that can be intimidating for many people, but breaking down complex formulas into simpler terms can make it more manageable to grasp. One such formula that often confuses students is the formula for (A^3 + B^3), where (A) and (B) are any real numbers. In this blog post, we will delve into this formula, explore its significance, and break it down step by step to help you understand it better.
### The Formula for (A^3 + B^3)
The formula for (A^3 + B^3) is a fundamental algebraic expression that arises frequently in mathematics and has wide-ranging applications. Understanding this formula can not only help you solve mathematical problems more efficiently but also give you a deeper insight into the properties of numbers and operations.
### Breaking Down the Formula
Let’s break down the formula for (A^3 + B^3) into simpler terms:
1. Cubing the Numbers: The first step is to cube the individual numbers, (A) and (B). This means multiplying each number by itself three times. For example, if (A = 2) and (B = 3), then (A^3 = 2^3 = 2 \times 2 \times 2 = 8) and (B^3 = 3^3 = 3 \times 3 \times 3 = 27).
2. Adding the Cubes: Once you have found (A^3) and (B^3), the next step is to add these two values together. So, for the example above, (A^3 + B^3 = 8 + 27 = 35).
### Significance of the Formula
The formula for (A^3 + B^3) is significant because it is one of the basic formulas in algebra that helps in simplifying expressions and solving equations. It is a special case of the more general formula for the sum of cubes, which is (A^3 + B^3 = (A + B)(A^2 – AB + B^2)).
### Why is Understanding This Formula Important?
Understanding the formula for (A^3 + B^3) is essential for various reasons:
• Solving Equations: This formula is often used to simplify and solve equations in algebra and calculus.
• Identifying Patterns: By understanding this formula, you can start recognizing patterns in numbers and operations.
• Building a Foundation: Mastering this formula is crucial for building a strong foundation in mathematics and tackling more advanced topics in the future.
### Example Applications
Now, let’s look at a few examples of how the formula for (A^3 + B^3) is applied in different scenarios:
1. Volume of a Cubic Box: The volume of a cubic box can be expressed as the sum of the cubes of its dimensions. If the length, width, and height of a box are (A), (B), and (C) respectively, then the volume can be calculated as (A^3 + B^3 + C^3).
2. Sum of Consecutive Cubes: The formula for (A^3 + B^3) can be used to find the sum of consecutive cubes. For example, (1^3 + 2^3 = 1 + 8 = 9) and (2^3 + 3^3 = 8 + 27 = 35).
1. What is the formula for (A^3 + B^3)?
The formula for (A^3 + B^3) is (A^3 + B^3 = (A + B)(A^2 – AB + B^2)) where (A) and (B) are any real numbers.
2. How can I apply the formula for (A^3 + B^3) in real-life scenarios?
You can apply this formula in various situations such as calculating volumes, finding sums of consecutive cubes, and simplifying complex expressions in mathematics.
3. Why is it important to understand the formula for (A^3 + B^3)?
Understanding this formula is crucial for solving equations, identifying patterns in numbers, and building a strong foundation in mathematics.
4. Can the formula for (A^3 + B^3) be extended to more than two terms?
Yes, the formula can be extended to more than two terms. For example, (A^3 + B^3 + C^3 = (A + B + C)(A^2 + B^2 + C^2 – AB – AC – BC)).
5. Are there any shortcuts or tricks to simplify the calculation of (A^3 + B^3)?
One common trick is to recognize that (A^3 + B^3 = (A + B)(A^2 – AB + B^2)) and use this formula directly to simplify the expression.
In conclusion, understanding the formula for (A^3 + B^3) is not only beneficial for solving mathematical problems but also for developing a deeper appreciation for the elegance and patterns in mathematics. By breaking down the formula into simpler steps and exploring its applications, you can strengthen your mathematical skills and approach problem-solving with confidence.
Would you want to convert more people? Would your desired role include international travel or do the majority of it from within the office environment with some fieldwork mixed in every now and again?" |
# Class XI Chapter 5 Complex Numbers and Quadratic Equations Maths. Exercise 5.1. Page 1 of 34
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## Transcription
1 Question 1: Exercise 5.1 Express the given complex number in the form a + ib: Question 2: Express the given complex number in the form a + ib: i 9 + i 19 Question 3: Express the given complex number in the form a + ib: i 39 Page 1 of 34
2 Question 4: Express the given complex number in the form a + ib: 3(7 + i7) + i(7 + i7) Question 5: Express the given complex number in the form a + ib: (1 i) ( 1 + i6) Question 6: Express the given complex number in the form a + ib: Question 7: Express the given complex number in the form a + ib: Page 2 of 34
3 Question 8: Express the given complex number in the form a + ib: (1 i) 4 Question 9: Express the given complex number in the form a + ib: Page 3 of 34
4 Question 10: Express the given complex number in the form a + ib: Question 11: Find the multiplicative inverse of the complex number 4 3i Page 4 of 34
5 Let z = 4 3i Then, = 4 + 3i and Therefore, the multiplicative inverse of 4 3i is given by Question 12: Find the multiplicative inverse of the complex number Let z = Therefore, the multiplicative inverse of is given by Question 13: Find the multiplicative inverse of the complex number i Let z = i Therefore, the multiplicative inverse of i is given by Question 14: Page 5 of 34
6 Express the following expression in the form of a + ib. Page 6 of 34
7 Exercise 5.2 Question 1: Find the modulus and the argument of the complex number On squaring and adding, we obtain Since both the values of sin θ and cos θ are negative and sinθ and cosθ are negative in III quadrant, Thus, the modulus and argument of the complex number are 2 and respectively. Question 2: Page 7 of 34
8 Find the modulus and the argument of the complex number On squaring and adding, we obtain Thus, the modulus and argument of the complex number are 2 and respectively. Question 3: Convert the given complex number in polar form: 1 i 1 i Let r cos θ = 1 and r sin θ = 1 On squaring and adding, we obtain Page 8 of 34
9 the required polar form. This is Question 4: Convert the given complex number in polar form: 1 + i 1 + i Let r cos θ = 1 and r sin θ = 1 On squaring and adding, we obtain It can be written, Page 9 of 34
10 This is the required polar form. Question 5: Convert the given complex number in polar form: 1 i 1 i Let r cos θ = 1 and r sin θ = 1 On squaring and adding, we obtain required polar form. This is the Question 6: Convert the given complex number in polar form: 3 3 Let r cos θ = 3 and r sin θ = 0 On squaring and adding, we obtain Page 10 of 34
11 This is the required polar form. Question 7: Convert the given complex number in polar form: Let r cos θ = and r sin θ = 1 On squaring and adding, we obtain Page 11 of 34
12 This is the required polar form. Question 8: Convert the given complex number in polar form: i i Let r cosθ = 0 and r sin θ = 1 On squaring and adding, we obtain This is the required polar form. Page 12 of 34
13 Exercise 5.3 Question 1: Solve the equation x = 0 The given quadratic equation is x = 0 On comparing the given equation with ax 2 + bx + c = 0, we obtain a = 1, b = 0, and c = 3 Therefore, the discriminant of the given equation is D = b 2 4ac = = 12 Therefore, the required solutions are Question 2: Solve the equation 2x 2 + x + 1 = 0 The given quadratic equation is 2x 2 + x + 1 = 0 On comparing the given equation with ax 2 + bx + c = 0, we obtain a = 2, b = 1, and c = 1 Therefore, the discriminant of the given equation is D = b 2 4ac = = 1 8 = 7 Therefore, the required solutions are Page 13 of 34
14 Question 3: Solve the equation x 2 + 3x + 9 = 0 The given quadratic equation is x 2 + 3x + 9 = 0 On comparing the given equation with ax 2 + bx + c = 0, we obtain a = 1, b = 3, and c = 9 Therefore, the discriminant of the given equation is D = b 2 4ac = = 9 36 = 27 Therefore, the required solutions are Question 4: Solve the equation x 2 + x 2 = 0 The given quadratic equation is x 2 + x 2 = 0 On comparing the given equation with ax 2 + bx + c = 0, we obtain a = 1, b = 1, and c = 2 Therefore, the discriminant of the given equation is D = b 2 4ac = ( 1) ( 2) = 1 8 = 7 Therefore, the required solutions are Question 5: Solve the equation x 2 + 3x + 5 = 0 The given quadratic equation is x 2 + 3x + 5 = 0 On comparing the given equation with ax 2 + bx + c = 0, we obtain Page 14 of 34
15 a = 1, b = 3, and c = 5 Therefore, the discriminant of the given equation is D = b 2 4ac = =9 20 = 11 Therefore, the required solutions are Question 6: Solve the equation x 2 x + 2 = 0 The given quadratic equation is x 2 x + 2 = 0 On comparing the given equation with ax 2 + bx + c = 0, we obtain a = 1, b = 1, and c = 2 Therefore, the discriminant of the given equation is D = b 2 4ac = ( 1) = 1 8 = 7 Therefore, the required solutions are Question 7: Solve the equation The given quadratic equation is On comparing the given equation with ax 2 + bx + c = 0, we obtain a =, b = 1, and c = Therefore, the discriminant of the given equation is D = b 2 4ac = 1 2 = 1 8 = 7 Therefore, the required solutions are Page 15 of 34
16 Question 8: Solve the equation The given quadratic equation is On comparing the given equation with ax 2 + bx + c = 0, we obtain a =, b =, and c = Therefore, the discriminant of the given equation is D = b 2 4ac = Therefore, the required solutions are Question 9: Solve the equation The given quadratic equation is This equation can also be written as On comparing this equation with ax 2 + bx + c = 0, we obtain a =, b =, and c = 1 Therefore, the required solutions are Page 16 of 34
17 Question 10: Solve the equation The given quadratic equation is This equation can also be written as On comparing this equation with ax 2 + bx + c = 0, we obtain a =, b = 1, and c = Therefore, the required solutions are Page 17 of 34
18 NCERT Miscellaneous Solutions Question 1: Evaluate: Page 18 of 34
19 Question 2: For any two complex numbers z 1 and z 2, prove that Re (z 1 z 2 ) = Re z 1 Re z 2 Im z 1 Im z 2 Page 19 of 34
20 Question 3: Reduce to the standard form. Page 20 of 34
21 Question 4: If x iy = prove that. Page 21 of 34
22 Question 5: Convert the following in the polar form: (i), (ii) (i) Here, Let r cos θ = 1 and r sin θ = 1 Page 22 of 34
23 On squaring and adding, we obtain r 2 (cos 2 θ + sin 2 θ) = r 2 (cos 2 θ + sin 2 θ) = 2 r 2 = 2 [cos 2 θ + sin 2 θ = 1] z = r cos θ + i r sin θ This is the required polar form. (ii) Here, Let r cos θ = 1 and r sin θ = 1 On squaring and adding, we obtain r 2 (cos 2 θ + sin 2 θ) = r 2 (cos 2 θ + sin 2 θ) = 2 r 2 = 2 [cos 2 θ + sin 2 θ = 1] Page 23 of 34
24 z = r cos θ + i r sin θ This is the required polar form. Question 6: Solve the equation The given quadratic equation is This equation can also be written as On comparing this equation with ax 2 + bx + c = 0, we obtain a = 9, b = 12, and c = 20 Therefore, the discriminant of the given equation is D = b 2 4ac = ( 12) = = 576 Therefore, the required solutions are Question 7: Solve the equation Page 24 of 34
25 The given quadratic equation is This equation can also be written as On comparing this equation with ax 2 + bx + c = 0, we obtain a = 2, b = 4, and c = 3 Therefore, the discriminant of the given equation is D = b 2 4ac = ( 4) = = 8 Therefore, the required solutions are Question 8: Solve the equation 27x 2 10x + 1 = 0 The given quadratic equation is 27x 2 10x + 1 = 0 On comparing the given equation with ax 2 + bx + c = 0, we obtain a = 27, b = 10, and c = 1 Therefore, the discriminant of the given equation is D = b 2 4ac = ( 10) = = 8 Therefore, the required solutions are Question 9: Solve the equation 21x 2 28x + 10 = 0 The given quadratic equation is 21x 2 28x + 10 = 0 On comparing the given equation with ax 2 + bx + c = 0, we obtain Page 25 of 34
26 a = 21, b = 28, and c = 10 Therefore, the discriminant of the given equation is D = b 2 4ac = ( 28) = = 56 Therefore, the required solutions are Question 10: If find. Question 11: Page 26 of 34
27 If a + ib =, prove that a 2 + b 2 = On comparing real and imaginary parts, we obtain Hence, proved. Question 12: Let. Find Page 27 of 34
28 (i), (ii) (i) On multiplying numerator and denominator by (2 i), we obtain On comparing real parts, we obtain (ii) On comparing imaginary parts, we obtain Question 13: Find the modulus and argument of the complex number. Let, then Page 28 of 34
29 On squaring and adding, we obtain Therefore, the modulus and argument of the given complex number are respectively. Question 14: Find the real numbers x and y if (x iy) (3 + 5i) is the conjugate of 6 24i. Let It is given that, Page 29 of 34
30 Equating real and imaginary parts, we obtain Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain Putting the value of x in equation (i), we obtain Thus, the values of x and y are 3 and 3 respectively. Question 15: Find the modulus of. Question 16: Page 30 of 34
31 If (x + iy) 3 = u + iv, then show that. On equating real and imaginary parts, we obtain Hence, proved. Question 17: If α and β are different complex numbers with = 1, then find. Let α = a + ib and β = x + iy It is given that, Page 31 of 34
32 Question 18: Find the number of non-zero integral solutions of the equation. Page 32 of 34
33 Thus, 0 is the only integral solution of the given equation. Therefore, the number of nonzero integral solutions of the given equation is 0. Question 19: If (a + ib) (c + id) (e + if) (g + ih) = A + ib, then show that (a 2 + b 2 ) (c 2 + d 2 ) (e 2 + f 2 ) (g 2 + h 2 ) = A 2 + B 2. On squaring both sides, we obtain (a 2 + b 2 ) (c 2 + d 2 ) (e 2 + f 2 ) (g 2 + h 2 ) = A 2 + B 2 Hence, proved. Question 20: If, then find the least positive integral value of m. Page 33 of 34
34 Therefore, the least positive integer is 1. Thus, the least positive integral value of m is 4 (= 4 1). Page 34 of 34
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If the given matrix $A = \left[ {\begin{array}{*{20}{c}} 1&3&1 \\ 2&1&{ - 1} \\ 3&0&1 \end{array}} \right]$, then rank(A) is equal to$\left( a \right)4 \\ \left( b \right)1 \\ \left( c \right)2 \\ \left( d \right)3 \\$
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Hint: In this question, the rank of the matrix is equal to the number of non-zero rows in the matrix after reducing it to the echelon form. In echelon form we only apply row operation.
Given, $A = \left[ {\begin{array}{*{20}{c}} 1&3&1 \\ 2&1&{ - 1} \\ 3&0&1 \end{array}} \right]$
Now, we have to convert the above matrix into echelon form. Echelon forms the same upper triangular matrix. In echelon form we only apply row operation.
$A = \left[ {\begin{array}{*{20}{c}} 1&3&1 \\ 2&1&{ - 1} \\ 3&0&1 \end{array}} \right]$
Apply row operation, ${R_2} \to {R_2} - 2{R_1}$
$A = \left[ {\begin{array}{*{20}{c}} 1&3&1 \\ 0&{ - 5}&{ - 3} \\ 3&0&1 \end{array}} \right]$
Now apply row operation, ${R_3} \to {R_3} - 3{R_1}$
$A = \left[ {\begin{array}{*{20}{c}} 1&3&1 \\ 0&{ - 5}&{ - 3} \\ 0&{ - 9}&{ - 2} \end{array}} \right]$
Again, apply row operation, ${R_3} \to {R_3} - \dfrac{{9{R_2}}}{5}$
$A = \left[ {\begin{array}{*{20}{c}} 1&3&1 \\ 0&{ - 5}&{ - 3} \\ 0&0&{\dfrac{{17}}{5}} \end{array}} \right]$
We can see the above matrix is an upper triangular matrix. Now it is converted into echelon form so the rank of the matrix is equal to the number of non-zero rows.
$A = \left[ {\begin{array}{*{20}{c}} 1&3&1 \\ 0&{ - 5}&{ - 3} \\ 0&0&{\dfrac{{17}}{5}} \end{array}} \right]$
In this matrix there are no non zero rows so the rank of this matrix is 3.
Hence, Rank(A)=3
So, the correct option is (d).
Note: Whenever we face such types of problems we use some important points. First we convert matrix into echelon form by using some row operations then observe how many non- zero rows in echelon form matrix and we know the number of non-zero rows in echelon form is equal to rank of matrix. |
Counting Principle MathBitsNotebook.com Terms of Use Contact Person: Donna Roberts
When dealing with the occurrence of more than one event (a compound event), it is important to be able to quickly determine how many possible outcomes exist.
What will be the size of the sample space?
For example, if ice cream sundaes come in 5 flavors with 4 possible toppings, how many different sundaes can be made with one flavor of ice cream and one topping?
In an attempt to find the "count" of how many sundaes are possible, a tree diagram could be drawn or a list of all possible sample space entries could be created. Unfortunately, these processes can be very time consuming when dealing with a large number of options. There is a short cut method! To count the combinations of ice cream and toppings, simply multiply the number of options for each event: 5 • 4 = 20 possible sundaes. This simple multiplication process is known as the Counting Principle.
The Fundamental Counting Principle: If there are a ways for one event to occur, and b ways for a second event to occur, then there are a • b ways for both to occur.
NOTE: The Counting Principle only works when all of the choices are independent of each other.
One choice does not depend upon another choice.
If one choice affects another choice, then this simple multiplication process will not yield the correct answer.
Examples: 1. Activities: roll a die and flip a coin There are 6 ways to roll a die and 2 ways to flip a coin. There are 6 • 2 = 12 ways to roll a die and flip a coin. (The sample space will contain a count of 12 possible outcomes.) 2. Activities: draw two cards from a standard deck of 52 cards without replacing the cards There are 52 ways to draw the first card. There are 51 ways to draw the second card. There are 52 • 51 = 2,652 ways to draw the two cards. (The sample space will contain a count of 2,652 possible outcomes.) The Counting Principle also works for more than two activities. 3. Activities: a coin is tossed five times There are 2 ways to flip each coin. There are 2 • 2 • 2 • 2 •2 = 32 arrangements of heads and tails. (The sample space will contain a count of 32 possible outcomes.) 4. Activities: a die is rolled four times There are 6 ways to roll each die. There are 6 • 6 • 6 • 6 = 1,296 possible outcomes. (The sample space will contain a count of 1,296 possible outcomes.) So, when does the Counting Principle NOT work? 5. Activities: You are buying a new skateboard. There are 2 styles: wood or polymer. There are 4 colors: red, black, blue, orange. There are 3 designs: line art, tattoo, splash There are 2 • 4 • 3 = 24 possible choices of skateboards. Yes, find and dandy!! BUT... the salesman says he is sorry but they are all sold out of all orange polymer skateboards. Now, how many choices do you have? Your choices are no longer independent of each other. The color options now depend upon which style board is chosen, wood or polymer. Well, you still have all of the wood skateboard options which will be 1 • 4 • 3 = 12 choices. Under the polymer skateboards, you now have 1• 3 • 3 = 9 choices. This gives a total of 21 choices.
The Counting Principle is easy! Simply MULTIPLY the number of ways each event can occur to find the sample space count. (events are independent)
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## Systems of Linear Inequalities
### Learning Outcomes
• Identify solutions to systems of linear inequalities
## Graph a System of Two Inequalities
Remember from the module on graphing that the graph of a single linear inequality splits the coordinate plane into two regions. On one side lie all the solutions to the inequality. On the other side, there are no solutions. Consider the graph of the inequality $y<2x+5$.
The dashed line is $y=2x+5$. Every ordered pair in the shaded area below the line is a solution to $y<2x+5$, as all of the points below the line will make the inequality true. If you doubt that, try substituting the x and y coordinates of Points A and B into the inequality; you will see that they work. So, the shaded area shows all of the solutions for this inequality.
The boundary line divides the coordinate plane in half. In this case, it is shown as a dashed line as the points on the line do not satisfy the inequality. If the inequality had been $y\leq2x+5$, then the boundary line would have been solid.
Now graph another inequality: $y>−x$. You can check a couple of points to determine which side of the boundary line to shade. Checking points M and N yield true statements. So, we shade the area above the line. The line is dashed as points on the line are not true.
To create a system of inequalities, you need to graph two or more inequalities together. Let us use $y<2x+5$ and $y>−x$ since we have already graphed each of them.
The purple area shows where the solutions of the two inequalities overlap. This area is the solution to the system of inequalities. Any point within this purple region will be true for both $y>−x$ and $y<2x+5$.
In the following video examples, we show how to graph a system of linear inequalities and define the solution region.
In the next section, we will see that points can be solutions to systems of equations and inequalities. We will verify algebraically whether a point is a solution to a linear equation or inequality.
## Determine Whether an Ordered Pair is a Solution to a System of Linear Inequalities
On the graph above, you can see that the points B and N are solutions for the system because their coordinates will make both inequalities true statements.
In contrast, points M and A both lie outside the solution region (purple). While point M is a solution for the inequality $y>−x$ and point A is a solution for the inequality $y<2x+5$, neither point is a solution for the system. The following example shows how to test a point to see whether it is a solution to a system of inequalities.
### Example
Is the point $(2, 1)$ a solution of the system $x+y>1$ and $2x+y<8$?
Here is a graph of the system in the example above. Notice that $(2, 1)$ lies in the purple area which is the overlapping area for the two inequalities.
### Example
Is the point $(2, 1)$ a solution of the system $x+y>1$ and $3x+y<4$?
Here is a graph of the system above. Notice that $(2, 1)$ is not in the purple area which is the overlapping area; it is a solution for one inequality (the red region), but it is not a solution for the second inequality (the blue region).
In the following video, we show another example of determining whether a point is in the solution of a system of linear inequalities.
As shown above, finding the solutions of a system of inequalities can be done by graphing each inequality and identifying the region they share. Below, you are given more examples that show the entire process of defining the region of solutions on a graph for a system of two linear inequalities. The general steps are outlined below:
• Graph each inequality as a line and determine whether it will be solid or dashed.
• Determine which side of each boundary line represents solutions to the inequality by testing a point on each side.
• Shade the region that represents solutions for both inequalities.
## Systems with No Solutions
In the next example, we will show the solution to a system of two inequalities whose boundary lines are parallel to each other. When the graphs of a system of two linear equations are parallel to each other, we found that there was no solution to the system. We will get a similar result for the following system of linear inequalities.
### Examples
Graph the system $\begin{array}{c}y\ge2x+1\\y\lt2x-3\end{array}$
## Summary
• Solutions to systems of linear inequalities are entire regions of points.
• You can verify whether a point is a solution to a system of linear inequalities in the same way you verify whether a point is a solution to a system of equations.
• Systems of inequalities can have no solutions when boundary lines are parallel. |
Nikoismusic.com Helpful tips How do you know if two vectors are linearly dependent?
# How do you know if two vectors are linearly dependent?
## How do you know if two vectors are linearly dependent?
We have now found a test for determining whether a given set of vectors is linearly independent: A set of n vectors of length n is linearly independent if the matrix with these vectors as columns has a non-zero determinant. The set is of course dependent if the determinant is zero.
### How do you solve linearly dependent vectors?
Let A = { v 1, v 2, …, v r } be a collection of vectors from Rn . If r > 2 and at least one of the vectors in A can be written as a linear combination of the others, then A is said to be linearly dependent.
#### Which vectors are linearly dependent?
Two vectors are linearly dependent if and only if they are collinear, i.e., one is a scalar multiple of the other. Any set containing the zero vector is linearly dependent. If a subset of { v 1 , v 2 ,…, v k } is linearly dependent, then { v 1 , v 2 ,…, v k } is linearly dependent as well.
Can 3 vectors in R4 be linearly independent?
Solution: No, they cannot span all of R4. Any spanning set of R4 must contain at least 4 linearly independent vectors. Our set contains only 4 vectors, which are not linearly independent. The dimension of R3 is 3, so any set of 4 or more vectors must be linearly dependent.
Can 2 vectors in R3 be linearly independent?
Two vectors are linearly dependent if and only if they are parallel. Hence v1 and v2 are linearly independent. Vectors v1,v2,v3 are linearly independent if and only if the matrix A = (v1,v2,v3) is invertible. Four vectors in R3 are always linearly dependent.
## What is the difference between linearly dependent and linearly independent vectors?
A set of two vectors is linearly dependent if at least one vector is a multiple of the other. A set of two vectors is linearly independent if and only if neither of the vectors is a multiple of the other.
### Are linearly dependent vectors parallel?
A set of two vectors is linearly dependent if one is parallel to the other, If any two of the vectors are parallel, then one is a scalar multiple of the other. A scalar multiple is a linear combination, so the vectors are linearly dependent.
#### Can a single vector be linearly independent?
▶ If v = 0 then the only scalar c such that cv = 0 is c = 0. Hence, 1vl is linearly independent. A set consisting of a single vector v is linearly dependent if and only if v = 0. Therefore, any set consisting of a single nonzero vector is linearly independent.
Can 2 vectors in R4 be linearly independent?
Solution: No, they cannot span all of R4. Any spanning set of R4 must contain at least 4 linearly independent vectors. Our set contains only 4 vectors, which are not linearly independent. −3 5 , v3 = −1 0 5 , v4 = −2 3 0 , v5 = 5 −2 −3 .
Can 3 vectors span all of R4?
Solution: A set of three vectors can not span R4. To see this, let A be the 4 × 3 matrix whose columns are the three vectors. This matrix has at most three pivot columns. This means that the last row of the echelon form U of A contains only zeros.
## Can 3 vectors in R3 be linearly dependent?
Two vectors in R3 are linearly dependent if they lie in the same line. Three vectors in R3 are linearly dependent if they lie in the same plane.
### How do you find the linear combination of a vector?
Linear Combination of Vectors. A linear combination of two or more vectors is the vector obtained by adding two or more vectors (with different directions) which are multiplied by scalar values. Write the vector = (1, 2, 3) as a linear combination of the vectors: = (1, 0, 1), = (1, 1, 0) and = (0, 1, 1).
#### Are linearly dependent vectors always coplanar?
If the scalar triple product of any three vectors is zero, then they are considered as coplanar. If any three vectors are linearly dependant, they are coplanar. Vectors are considered coplanar if amongst them no more than two vectors are linearly independent vectors.
Are the columns linearly dependant or independent?
The columns of matrix A are linearly independent if and only if the equation Ax 0 has only the trivial solution. Special Cases Sometimes we can determine linear independence of a set with minimal effort. 1. ASetofOneVector Consider the set containing one nonzero vector: v1
What does linearly independent mean?
linearly independent(Adjective) (Of a set of vectors or ring elements) whose nontrivial linear combinations are nonzero. |
# Common Core: High School - Geometry : Congruence
## Example Questions
### Example Question #7 : Definition Of Congruency For Triangles: Ccss.Math.Content.Hsg Co.B.7
Which theorem can be used to prove triangle congruency between triangle A and B?
AAS
AAA
SAS
ASA
SSS
SSS
Explanation:
For this particular problem there are two geometric theorems that could potentially be used to prove that the triangles are similar.
The geometric theorems that could be used:
1. Hypotenuse and One Leg of a Right Triangle (HL)
2. Side, Side, Side (SSS)
To use SSS or HL, the Pythagorean theorem will need to be used to calculate the missing side.
For these triangles the HL is the most evident to use but is not an option in the answer selections, therefore, SSS is the correct answer.
### Example Question #8 : Definition Of Congruency For Triangles: Ccss.Math.Content.Hsg Co.B.7
Identify the missing term in the statement.
The __________ geometric theorem deals with proving congruency among right triangles; specifically when the length of one leg and the length of the hypotenuse are known.
Side, Angle, Side
Angle, Angle, Side (AAS)
Hypotenuse and One Leg (HL)
Angle, Angle, Angle (AAA)
Side, Side, Side (SSS)
Hypotenuse and One Leg (HL)
Explanation:
The statement, "The __________ geometric theorem deals with proving congruency among right triangles; specifically when the length of one leg and the length of the hypotenuse are known." is describing the geometric theorem known as the Hypotenuse and One Leg theorem. When abbreviated this is seen as, HL.
Therefore, the missing term is, Hypotenuse and One Leg (HL)
### Example Question #11 : Definition Of Congruency For Triangles: Ccss.Math.Content.Hsg Co.B.7
Identify the missing term in the statement.
The __________ geometric theorem can be used to identify whether triangles that each have three known side lengths are congruent.
Side, Side, Side (SSS)
Side, Angle, Side (SAS)
Angle, Side, Angle (ASA)
Angle, Angle, Side (AAS)
Hypotenuse and One Leg (HL)
Side, Side, Side (SSS)
Explanation:
The statement, "The __________ geometric theorem can be used to identify whether triangles that each have three known side lengths are congruent." is describing the geometric theorem known as the Side, Side, Side theorem. When abbreviated this is seen as, SSS.
Therefore, the missing term is, Side, Side, Side (SSS).
### Example Question #12 : Definition Of Congruency For Triangles: Ccss.Math.Content.Hsg Co.B.7
Identify the missing term in the statement.
When two triangles have two known angles and a known side length that is in between the angles, the geometric theorem that can be used to prove congruency is known as __________
Side, Side, Side (SSS)
Angle, Angle, Angle (AAA)
Angle, Angle, Side (AAS)
Angle, Side, Angle (ASA)
Side, Angle, Side (SAS)
Angle, Side, Angle (ASA)
Explanation:
The statement, "When two triangles have two known angles and a known side length that is in between the angles, the geometric theorem that can be used to prove congruency is known as __________. " is describing the geometric theorem known as the Angle, Side, Angle theorem. When abbreviated this is seen as, ASA.
Therefore, the missing term is, Angle, Side, Angle (ASA).
### Example Question #1 : Prove Line And Angle Theorems: Ccss.Math.Content.Hsg Co.C.9
What is the supplement of the complement of ?
Explanation:
In order to solve this problem, we need to break down each word.
We need to first find the complement of .
The complement is
Since we are given an angle of we can substitute it for , and solve for .
Now since we need to find the supplement of the answer, we just got.
The supplement is
Now we simply substitute the answer we just got for .
### Example Question #1 : Prove Line And Angle Theorems: Ccss.Math.Content.Hsg Co.C.9
What is the supplement of the complement of ?
Explanation:
In order to solve this problem, we need to break down each word.
We need to first find the complement of
The complement is
Since we are given an angle of we can substitute it for , and solve for .
Now since we need to find the supplement of the answer, we just got.
The supplement is
Now we simply substitute the answer we just got for .
### Example Question #2 : Prove Line And Angle Theorems: Ccss.Math.Content.Hsg Co.C.9
What is the supplement of the complement of ?
Explanation:
In order to solve this problem, we need to break down each word.
We need to first find the complement of .
The complement is
Since we are given an angle of we can substitute it for , and solve for .
Now since we need to find the supplement of the answer, we just got.
The supplement is
Now we simply substitute the answer we just got for .
### Example Question #81 : High School: Geometry
What is the supplement of the complement of ?
Explanation:
In order to solve this problem, we need to break down each word.
We need to first find the complement of .
The complement is
Since we are given an angle of we can substitute it for , and solve for.
Now since we need to find the supplement of the answer, we just got.
The supplement is
Now we simply substitute the answer we just got for .
### Example Question #4 : Prove Line And Angle Theorems: Ccss.Math.Content.Hsg Co.C.9
What is the supplement of the complement of ?
Explanation:
In order to solve this problem, we need to break down each word.
We need to first find the complement of
The complement is
Since we are given an angle of we can substitute it for , and solve for .
Now since we need to find the supplement of the answer, we just got.
The supplement is
Now we simply substitute the answer we just got for .
### Example Question #82 : High School: Geometry
What is the supplement of the complement of ?
Explanation:
In order to solve this problem, we need to break down each word.
We need to first find the complement of
The complement is
Since we are given an angle of we can substitute it for , and solve for .
Now since we need to find the supplement of the answer, we just got.
The supplement is
Now we simply substitute the answer we just got for |
TRIANGLE WORKSHEET SOLUTION 3
3. Determine if the following set of points are collinear or not.
(i) (4, 3), (1, 2) and (-2, 1)
Solution :
Let A(4, 3), B(1, 2) and C(-2, 1) be the vertices of the triangle.
If the three points are collinear then area of triangle will be zero
x1 = 4 x2 = 1 x3 = -2
y1 = 3 y2 = 2 y3 = 1
= (1/2)[(8 + 1 – 6) – (3 - 4 + 4)]
= (1/2)[3 – 3]
= 0
Hence the given points are collinear.
(ii) (-2, -2), (-6, -2) and (-2, 2)
Solution :
Let A (-2,-2) B (-6,-2) and C (-2,2) are the vertices of the triangle
If the three points are collinear then area of triangle will be zero
x1 = -2 x2 = -6 x3 = -2
y1 = -2 y2 = -2 y3 = 2
= (1/2)[(4 - 12 + 4) – (12 + 4 - 4)]
= (1/2)[-4 – 12]
= (1/2)(-16)
= -8 ≠ 0
Hence the given points are not collinear.
(iii) (-3/2, 3) (6, -2) and (-3, 4)
Solution :
Let A (-3/2,3) B (6,-2) and C (-3,4) are the vertices of the triangle
If the three points are collinear then area of triangle will be zero
x1 = -3/2 x2 = 6 x3 = -3
y1 = 3 y2 = -2 y3 = 4
= (1/2)[(3 + 24 - 9) – (18 + 6 - 6)]
= (1/2)[(27 - 9) - (18)]
= (1/2)(18 - 18)
= 0
Hence the given points are collinear.
4. 4) In each of the following, find the value of k for which the given points are collinear.
(i) (k, -1) ( 2, 1) and (4, 5)
Solution :
If the given points are collinear then the area of triangle is zero
(1/2) [(k + 10 – 4) – (-2 + 4 + 5k)] = 0
[(k + 6) – (2+ 5k)] = 0 x 2
(k + 6 – 2 - 5k) = 0
-4 k + 4 = 0
- 4k = -4
K = (-4)/(-4)
K = 1
Hence, the value of k is 1
(ii) (2, -5) ( 3, -4) and (9, k)
Solution :
If the given points are collinear then the area of triangle is zero
(1/2) [(-8 + 3k – 45) – (-15 - 36 + 2k)] = 0
[(3k - 53) – (-51+ 2k)] = 0 x 2
(3k - 53 + 51- 2k) = 0
k - 2 = 0
k = 2
Hence the value of k is 2
(iii) (k, k) (2, 3) and (4, -1)
Solution :
If the given points are collinear then the area of triangle is zero
(1/2) [(3k - 2 + 4k) – (2k + 12 - k)] = 0
[(7k - 2) – (k+12)] = 0 x 2
(7k - 2 – k - 12) = 0
6 k - 14 = 0
6k = 14
K = 14/6
K = 7/3
Hence the value of k is 7/3.
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# Plus Two Maths Chapter Wise Questions and Answers Chapter 7 Integrals
weierstrass substitution for integrations, intro
weierstrass substitution for integrations, intro
Plus Two Maths Chapter Wise Questions and Answers Chapter 7 Integrals
Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 7 Integrals
### Plus Two Maths Integrals Three Mark Questions and Answers
Question 1.
Integrate the following. (3 Score each)
1. ∫sin x sin 2x sin 3 xdx
2. ∫sec2x cos22x dx
1. We have sinxsin2xsin3x
= 1/2 (2sinxsin3x) sin2x
= 1/2 (cos2x – cos4x) sin2x
= 1/4 (2sin2xcos2x – 2cos4xsi n2x)
= 1/4 [sin4x – (sin6x – sin2x)]
= 1/4(sin4x + sin2x – sin6x)
∫sin x sin 2x sin 3 xdx
= $$\frac{1}{4}$$ ∫(sin 4x + sin 2x – sin 6x)dx
= –$$\frac{1}{16}$$ cos4x – $$\frac{1}{8}$$ cos2x + $$\frac{1}{24}$$ cos6x + c.
2. sec2x cos22x = $$\frac{\left(2 \cos ^{2} x-1\right)^{2}}{\cos ^{2} x}$$
= $$\left(\frac{2 \cos ^{2} x}{\cos x}-\frac{1}{\cos x}\right)^{2}$$ = (2cosx – secx)2
= 4cos2x + sec2x – 4
= 2(1 + cos2x) + sec2x – 4
= 2cos2x + sec2x – 2
∫sec2 x cos2 2x dx = ∫(2 cos 2x + sec2 x – 2)dx
= sin 2x + tan x – 2x + c.
Question 5.
(i) If f (x) is an odd function, then $$\int_{-a}^{a} f(x)$$ = ?
(a) 0
(b) 1
(c) 2$$\int_{0}^{a} f(x)$$ dx
(d) 2a
Evaluate
(ii) $$\int_{-\pi / 2}^{\pi / 2} \sin ^{99} x \cdot \cos ^{100} x d x$$
(iii) $$\int_{-1}^{1} e^{|x|} d x$$
(i) (a) 0.
(ii) Here, f(x) = sin99x.cos100x .then,
f(-x) = sin99(- x).cos100(- x) = – sin99 x. cos100 x = -f(x)
∴ odd function ⇒ $$\int_{-\pi / 2}^{\pi / 2} \sin ^{99} x \cdot \cos ^{100} x d x=0$$.
(iii) Here, f(x) = e|x|, f(-x) = e|-x| = e|x| = f(x)
∴ even function.
we have |x| = x, 0 ≤ x ≤ 1
Question 6.
1. Show that cos2 x is an even function. (1)
2. Evaluate $$\int_{-\pi / 4}^{\pi / 4} \cos ^{2} x d x$$ (2)
1. Let f(x) = cos2x ⇒ f(-x) = cos2 (-x) = cos2 x = f(x) even.
2.
Question 8.
Find the following integrals.
Question 9.
Find the following integrals.
1. $$\int \frac{1}{3+\cos x} d x$$
2. $$\int \frac{2 x}{x^{2}+3 x+2} d x$$
1. $$\int \frac{1}{3+\cos x} d x$$
Put t = tanx/2 ⇒ dt = 1/2 sec2 x/2 dx
2. $$\int \frac{2 x}{x^{2}+3 x+2} d x$$ = $$\int \frac{2 x}{(x+2)(x+1)} d x$$
2x = A(x + 1) + B (x + 2)
when x = -1, -2 = B ; B = -2
when x = -2, -4 = -A ; A = 4
= 4log(x + 2) – 2log (x + 1) + C.
### Plus Two Maths Integrals Four Mark Questions and Answers
Question 4.
1. Choose the correct answer from the bracket.
∫ex dx = — (e2x + c, e-x + c, e2x + c) (1)
2. Evaluate: ∫ ex sinxdx
1. ex + c
2. I = ∫ex sinxdx = sinx.ex – ∫cos x.exdx
= sin x.ex – (cos x.ex – ∫(- sin x).ex dx)
= sinx.ex – cosxex – ∫sinx.exdx
= sin x.ex – cos xex – I
2I = sin x.ex – cos xex
I = $$\frac{1}{2}$$ex(sinx – cosx) + c.
Question 5.
(i) f(x)∫g(x) dx – ∫(f'(x)∫g(x) dx)dx (1)
(a) ∫f'(x)g{x)dx
(b) ∫f(x)g'(x)dx
(c) ∫$$\frac{f(x)}{g(x)}$$dx
(d) ∫f(x)g(x)dx
(ii) Integrate sin-1$$\sqrt{\frac{x}{a+x}}$$dx w.r.to x. (3)
(i) (d) ∫f(x)g(x)dx
(ii) ∫sin-1$$\sqrt{\frac{x}{a+x}}$$dx,
Put x = a tan2θ, θ = tan-1$$\sqrt{\frac{x}{a}}$$
⇒ dx = 2a tanθ sec2θ dθ
I = ∫sin-1$$\left(\frac{\tan \theta}{\sec \theta}\right)$$ 2a tanθ sec2θ dθ
= ∫sin-1(sinθ)2a tanθ sec2θ dθ
= 2a∫θ tanθ sec2θ dθ
Put tanθ = t, θ = tan-1 t ⇒ sec2θ dθ = dt
= 2a ∫ tan-1 t (t) dθ
2. ∫sec x(sec x + tan x)dx = ∫(sec2 x + sec x. tan x)dx
= tanx + secx + c.
3. ∫e3xdx = $$\frac{e^{3 x}}{3}$$ + c.
4. ∫(sin x + cos x)dx = sin x – cosx + c.
Question 9.
Consider the integral I = $$\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x$$?
1. What substitution can be given for simplifying the above integral? (1)
2. Express I in terms of the above substitution. (1)
3. Evaluate I. (2)
1. Substitute sin-1 x = t.
2. We have, sin-1 x = t ⇒ x = sint
Differentiating w.r.t. x; we get,
$$\frac{1}{\sqrt{1-x^{2}}}$$dx = dt
∴ I = ∫t sin t dt.
3. I = ∫t sin t dt = t.(-cost) -∫(-cost)dt = -t cost + sint + c
= -sin-1 x. cos (sin-1 x) + sin(sin-1 x) + c
x – sin-1 x.cos(sin-1 x) + c.
Question 10.
Evaluate $$\int_{0}^{\pi / 4} \log (\tan x) d x$$.
Question 11.
Find the following integrals.
1. $$\int \frac{\sec ^{2} x}{\cos e c^{2} x} d x$$ (2)
2. $$\int \frac{1}{x^{2}-6 x+13} d x$$ (2)
1. $$\int \frac{\sec ^{2} x}{\cos e c^{2} x} d x$$ = $$\int \frac{\sin ^{2} x}{\cos ^{2} x} d x$$ = ∫tan2 xdx
= ∫(sec2x – 1)dx = tanx – x + c.
2. $$\int \frac{1}{x^{2}-6 x+13} d x$$
Question 13.
(i) ∫sin2x dx = ? (1)
(a) 2 cos x + c
(b) -2 sin x + c
(c) $$\frac{\cos 2 x}{2}$$ + c
(d) $$-\frac{\cos 2 x}{2}$$ + c
(ii) Evaluate ∫ex sin 2x dx (3)
(i) (d) $$-\frac{\cos 2 x}{2}$$ + c.
(ii) Consider I = ∫ex sin 2x dx
= ∫sin 2x. exdx = sinx.ex – 2∫cos 2x. exdx
= sin 2x.ex – 2 (cos 2x.ex + 2∫sin 2x. exdx)
= sin 2x. ex – 2 cos 2x ex – 4 ∫sin 2x. exdx
= sin 2x. ex – 2 cos 2x ex – 4I
5 I = sin 2x. ex – 2 cos 2x ex
I = $$\frac{e^{x}}{5}$$ (sin 2x – 2 cos 2x).
### Plus Two Maths Integrals Six Mark Questions and Answers
Question 1.
(i) Fill in the blanks. (3)
(a) ∫ tan xdx = —
(b) ∫ cos xdx = —
(c) ∫$$\frac{1}{x}$$dx = —
(ii) Evaluate ∫sin3 xcos2 xdx (3)
(i) (a) log|secx| + c
(b) sinx + c
(c) log|x| + c.
(ii) ∫sin3 xcos2 xdx = ∫sin2 xcos2 x sin xdx
= ∫(1 – cos2 x)cos2 x sin xdx
Put cos x = t ⇒ – sin xdx = dt
∴ ∫(1 – cos2 x)cos2 xsin xdx = -∫(1 – t2 )t2dt
= ∫(t4 – t2)dt = $$\frac{t^{5}}{5}-\frac{t^{3}}{3}$$ + c
= $$\frac{\cos ^{5} x}{5}-\frac{\cos ^{3} x}{3}$$ + c.
Question 5.
1. Evaluate the as $$\int_{0}^{2}$$x2dx the limit of a sum. (3)
2. Hence evaluate $$\int_{-2}^{2}$$x2dx (1)
3. If $$\int_{0}^{2}$$ f(x)dx = 5 and $$\int_{-2}^{2}$$ f(x)dx = 0, then $$\int_{-2}^{0}$$ f(x)dx = …….. (2)
1. Here the function is f(x) = x2, a = 0, b = 2 and h = $$\frac{b-a}{n}=\frac{2}{n}$$
$$\int_{0}^{2}$$x2dx =
Question 11.
1. Find ∫$$\frac{1}{x^{2}+a^{2}}$$dx (1)
2. Show that 3x + 1 = $$\frac{3}{4}$$(4x – 2) + $$\frac{5}{2}$$ (2)
3. Evaluate $$\int \frac{3 x+1}{2 x^{2}-2 x+3} d x$$ (3)
1. ∫$$\frac{1}{x^{2}+a^{2}}$$dx = 1/a tan-1 x/a + c.
2. 3x + 1 = A $$\frac{d}{d x}$$(2×2 – 2x + 3) + B
= A(4x – 2) + B
3 = 4A; A = 3/4
1 = -2A + B
1 = -3/2 + B, B = 1 + 3/2 = 5/2
∴ 3x + 1 = 3/4(4x – 2) + 5/2
3.
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# Question: What Is The Value Of Root 24?
## How do you simplify square root of 24?
We should try to reduce √24 to the root of a number with a perfect square multiplied by some other whole number.Let’s consider the factors of 24:1,4,6,8,12,24.√24=√4⋅6=√4⋅√6 as √ab=√a⋅√b.√4⋅√6=2√6..
## What is the value of root?
A square root of a number b is the solution of the equation x2=b . It is a number that when multiplied by itself gives you b . Every positive number b has two square roots , denoted √b and −√b .
## What is equal to the square root of 200?
The factor of 200 that we can take the square root of is 100. We can write 200 as (100)(2) and then use the product rule of radicals to separate the two numbers.
## What is the value of Root 5?
2.2360Therefore, the value of root 5 is, √5 = 2.2360… You can find the value of the square root of all the non-perfect square number with the help of the long division method. This is the old method which gives the exact value of the root of numbers.
## Is 23 a perfect square?
A number is a perfect square (or a square number) if its square root is an integer; that is to say, it is the product of an integer with itself. Here, the square root of 23 is about 4.796. … Anyway, 23 is a prime number, and a prime number cannot be a perfect square.
## Is the square root of 24 a rational number?
Since 24 is a natural number, but the square root of 24 isn’t a natural number (it’s an irrational number that never terminates or repeats [4.898 979 486 …]), 24 is NOT a perfect square.
## How do you find the value of Root 23?
√23=[4;¯¯¯¯¯¯¯¯¯¯¯¯¯¯1,3,1,8]=4+11+13+11+18+11+13+11+…
## What is a square root of 24?
4.89897948557Answer and Explanation: The square root of 24 is 4.89897948557. Rounded, that would give you a square root of 4.899. As you can see, the square root contains a decimal.
## How do you simplify square roots without a calculator?
Use a process of averaging.Then, divide your number by one of those square root numbers. Take the answer, and find the average of it and the number you divided by (average is just the sum of those two numbers divided by two). … Sound complicated? … Check your work by multiplying your answer (in this case 3.1623) by itself.
## What is the formula for square root?
When a value is multiplied by itself to give the original number then that number is a square root. It is represented by a radical symbol \sqrt{}. “Square Root” is often used to refer to the principal square root….Solution:FORMULAS Related LinksCg Of TrapeziumVolume Of A Triangular Pyramid4 more rows
## What is the root square of 25?
625Table of Squares and Square RootsNUMBERSQUARESQUARE ROOT245764.899256255.000266765.099277295.19696 more rows
## Is 24 a perfect cube?
A perfect cube is a number that is the cube of an integer. Some examples of perfect cubes are 1, 8, 27, 64, 125, 216, 343, ..
## What is the value of root 7?
2.64To find the square root of 7 we can use the long division method which is simple and easy. Therefore, the value of the square root of 7 is 2.64. |
# Circle Theorems and Angles
GCSELevel 4-5Level 6-7Cambridge iGCSE
## Circle Theorems
Circle theorems are properties of circles that allow us to consider and work out angles within the geometry of a circle.
You need to be able to identify, utilise, and describe each theorem.
## Rule 1 – Angles in a Semicircle
The angle extended from the diameter is always a right angle.
This may be written as ‘a diameter subtends a right-angle at the circumference’.
Level 4-5GCSECambridge iGCSE
## Rule 2 – Tangent/Radius Angle
The angle where a tangent meets a radius is always a right angle.
This may be written as ‘the tangent and radius that meet are perpendicular’.
Level 4-5GCSECambridge iGCSE
## Rule 3 – Segment Angles
Angles drawn from the same chord are equal when touching the circumference.
The lines may or may not pass through the centre.
Level 6-7GCSECambridge iGCSE
## Rule 4 – Centre Angle
The angle at the centre is twice the size of the angle at the circumference.
Level 6-7GCSECambridge iGCSE
## Rule 5 – Cyclic Quadrilateral
A cyclic quadrilateral is a four-sided shape within a circle, with each corner touching the circumference.
In a cyclic quadrilateral, opposite angles add to $180\degree$.
$\textcolor{purple}{w}+\textcolor{red}{x}=180\degree$
$\textcolor{blue}{y}+\textcolor{limegreen}{z}=180\degree$
Level 6-7GCSECambridge iGCSE
## Rule 6 – Alternate Segment Theory
The angle between the tangent to the circle and the side of the triangle is equal to the opposite interior angle. This is the same for the angle on the other side of the tangent:
$\textcolor{limegreen}{x}=\textcolor{limegreen}{x}$
$\textcolor{red}{y}=\textcolor{red}{y}$
Level 6-7GCSECambridge iGCSE
## Rule 7 – Tangents from the Same Point
Tangents from the same point to the circumference are equal in length:
$AB = BC$
Level 6-7GCSECambridge iGCSE
## Rule 8 – Perpendicular Bisector of a Chord
The perpendicular bisector to a chord will always pass through the centre of the circle.
This can be any length chord anywhere in the circle.
Level 6-7GCSECambridge iGCSE
## Rule 9 – Equal Chords are Equidistant from the Centre
Chords of equal length of a circle are equidistant from the centre.
Taking a look at the diagram to the left.
$AB=CD$ and are both chords of the same circle.
This means they are both equal distance from the centre
Therefore $EO=OF$
Level 6-7GCSECambridge iGCSE
## Example 1: Using Circle Theorems
Use one of the circle theorems to calculate the size of angles $x$ and $y$.
[3 marks]
We can use rule 5 – this is a cyclic quadrilateral, so the opposite angles add to $180\degree$.
So $x$ and $118$ add to $180$,
$180 - 118 = x\\$ $x = 62\degree\\$
And $y$ and $124$ add to $180$,
$180 - 124 = y\\$ $y = 56\degree\\$
Level 6-7GCSECambridge iGCSE
## Example 2: Using Multiple Theorems
The line $AC$ passes through the centre of the circle.
Use circle theorems to work out the angles $x\degree$ and $y\degree$
[3 marks]
Firstly, we can use rule 5, this is a cyclic quadrilateral, so opposite angles add to $180\degree$
We can use this to work out $y$,
$180-105=75\degree\\$ $y=75$
Now, we can use rule 1 – as $AC$ passes through the centre, this is a diameter of the circle, so the angle that extends from the diameter is a right angle. Therefore, $B$ and $D$ are right angles, so,
$x=90$
Level 6-7GCSECambridge iGCSE
## Circle Theorems and Angles Example Questions
We know from rule 7 that tangents from the same point are the same length, so $BE$ is also $6\text{ cm}$
Gold Standard Education
Using rule 3, we know angles from the same chord are equal when touching the circumference. So:
$x=16\degree\\$ $y=22\degree\\$
We know that perpendicular bisectors to chords pass through the centre (rule 8), so the point where $EI$ and $BG$ cross must be the centre of the circle.
This means we can use rule 4, as we have an angle at the centre and an angle at the circumference. Using this rule, we know $x$ will be half the size of $48$, and hence, is $24\degree$. |
# How do you simplify (x^2+3x+2)/(x^2-1)?
May 15, 2018
(x²+3x+2)/(x²-1)=(x+2)/(x-1)
#### Explanation:
(x²+3x+2)/(x²-1)
$= \frac{\cancel{\left(x + 1\right)} \left(x + 2\right)}{\left(x - 1\right) \cancel{\left(x + 1\right)}}$
$= \frac{x + 2}{x - 1}$
May 15, 2018
First we should split the middle term
By the video... you should have understood that we need to make $3 x$ in two number that one is divisible by 2 and the other is divisible itself
$\frac{{x}^{2} + x + 2 x + 2}{{x}^{2} - 1}$
You should always remember that
$1 = {1}^{2}$
$\frac{x \left(x + 1\right) + 2 \left(x + 1\right)}{{x}^{2} - {1}^{2}}$
${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$
$\frac{\left(x + 2\right) \cancel{\left(x + 1\right)}}{\left(x - 1\right) \cancel{\left(x + 1\right)}}$
$\frac{x + 2}{x - 1}$ |
Equivalent fractions room the fractions having the exact same value. Same fraction can be stood for in plenty of ways. Let united state take the adhering to example.
You are watching: Write two equivalent fractions for 2 3
The shaded part in picture (ii) is stood for by fraction $$\frac24$$. In photo (iii) the same component is represented by fraction $$\frac48$$. SO, the portion represented by this shaded portions are equal. Together fractions are referred to as equivalent fractions.
We to speak that $$\frac12$$ = $$\frac24$$ = $$\frac48$$
Hence, because that a given fraction there can be plenty of equivalent fractions.
Making equivalent Fractions:
We have seen in the above example that $$\frac12$$, $$\frac24$$ and $$\frac48$$ are equivalent fractions.
Therefore, $$\frac12$$ deserve to be created as $$\frac12$$ = $$\frac1 × 22 × 2$$ = $$\frac1 × 32 × 3$$ = $$\frac1 × 42 × 4$$ and also so on.
Hence, one equivalent portion of any kind of given fraction can be obtained by multiply its numerator and denominator by the same number.
Same way, once the numerator and denominator of a portion are divided by the exact same number, we gain its equivalent fractions.
$$\frac12$$ = $$\frac1 ÷ 12 ÷ 1$$ = $$\frac24$$ = $$\frac2 ÷ 24 ÷ 2$$ = $$\frac36$$ = $$\frac3 ÷ 36 ÷ 3$$
We have,
2/4 = (1 × 2)/(2 × 2)3/6 = (1 × 3)/(2 × 3)4/8 = (1 × 4)/(2 × 4)We observe the 2/4, 3/6 and 4/8 are obtained by multiplying the numerator and also denominator of 1/2 by 2, 3 and also 4 respectively.Thus, one equivalent portion of a given fraction can be acquired by multiplying its numerator and denominator through the exact same number (other 보다 zero).2/4 = (2÷ 2)/(4 ÷ 2) = 1/23/6 = (3÷ 3)/(6 ÷ 3) = 1/24/8 = (4 ÷ 4)/(8 ÷ 4) = 1/2We observe the if we division the numerators and denominators the 2/4, 3/6 and also 4/8 each by their typical factor 2, we get an equivalent portion 1/2.Thus, an equivalent portion of a given portion can be obtained by splitting its numerator and denominator by their typical factor (other 보다 1), if ant.
Note:
(i) multiply its numerator (top) and also denominator (bottom) by the exact same number (other than 0).(ii) splitting its numerator (top) and also denominator (bottom) by their typical factor (other 보다 1).For Example:1. Write 3 equivalent fraction of 3/5.Equivalent fractions of 3/5 are:(3 × 2)/(5× 2) = 6/10,(3 × 3)/(5 × 3) = 9/15,(3 × 4)/(5 × 4) = 12/20Therefore, tantamount fractions that 3/5 space 6/10, 9/15 and 12/20.
2. Write following three equivalent portion of $$\frac23$$.
We main point the numerator and the denominator by 2.
We get, $$\frac2 × 23 × 2$$ = $$\frac46$$
Next, us multiply the numerator and also the denominator through 3. We get
$$\frac2 × 33 × 3$$ = $$\frac69$$.
Next, we multiply the numerator and also the denominator by 4. We get
$$\frac2 × 43 × 4$$ = $$\frac812$$.
Therefore, equivalent fractions of $$\frac23$$ room $$\frac46$$, $$\frac69$$ and also $$\frac812$$.
3. Write 3 equivalent portion of 1/4.Equivalent fractions of 1/4 are:(1× 2)/(4× 2) = 2/8,(1 × 3)/(4 × 3) = 3/12,(1× 4)/(4× 4) = 4/16Therefore, tantamount fractions of 1/4 are 2/8, 3/12 and 4/16.4. Write 3 equivalent fraction of 2/15.Equivalent fountain of 2/15 are:(2× 2)/(15 × 2) = 4/30,(2 × 3)/(15 × 3) = 6/45,(2× 4)/(15 × 4) = 8/60Therefore, indistinguishable fractions the 2/15 space 4/30, 6/45 and 8/60.5. Write 3 equivalent portion of 3/10.Equivalent fountain of 3/10 are:(3× 2)/(10× 2) = 6/20,(3 × 3)/(10 × 3) = 9/30,(3× 4)/(10× 4) = 12/40Therefore, equivalent fractions of 3/10 space 6/20, 9/30 and 12/40.
See more: D Is Hawaii Close To The Equator, Distance Between Hawaii And Equator
● Fraction
Representations of fractions on a Number Line
Fraction together Division
Types that Fractions
Conversion of blended Fractions into Improper Fractions
Conversion of improper Fractions into Mixed Fractions
Equivalent Fractions
Fractions in lowest Terms
Like and also Unlike Fractions
Comparing like Fractions
Comparing unlike Fractions
Addition and also Subtraction of favor Fractions
Addition and also Subtraction of unlike Fractions
Inserting a portion between Two offered Fractions |
# How To Calculate Percentage [for Beginners]
As you navigate through life, you will frequently come across situations where you need to calculate a percentage. Whether it’s figuring out how much of a discount you’re getting on a sale or determining the increase in your salary, knowing how to calculate percentages is an essential skill. In this article, we will provide a comprehensive guide on how to calculate percentages, including different methods and practical examples.
A percentage is a measure of the relationship between two values, typically expressed as a fraction of 100. It is commonly used in various fields, including finance, business, education, and science. Knowing how to calculate percentages is not only helpful in everyday life but also crucial in many professional settings.
In this article, we will explain the basics of percentage calculation and provide examples of various scenarios where percentage calculation is necessary. B
y the end of this article, you will be able to calculate percentages confidently and accurately.
## What is a Percentage?
A percentage is a fraction of 100. It is used to compare one value to another and express the relationship between them. For example, if you scored 80 out of 100 on a test, your percentage score would be 80%. Similarly, if a product costs \$50 and is on sale for \$40, the percentage discount would be 20%.
## Common Uses of Percentage
A percentage is used in a wide range of applications, including:
• Determining discounts and sales prices in retail
• Calculating interest rates and loan payments in finance
• Measuring growth and changes in business and economics
• Analyzing data in scientific research
## Basic Percentage Calculation
• ### Percentage Formula
The basic formula for calculating a percentage is:
percentage = (part / whole) x 100
Examples of Basic Percentage Calculation
Example 1: If you scored 75 out of 100 on a test, what is your percentage score?
percentage = (75 / 100) x 100 = 75%
Example 2: If a pizza has 8 slices and you eat 4 slices, what percentage of the pizza have you eaten?
percentage = (4 / 8) x 100 = 50%
• ### Percentage Change Calculation
Percentage Change Formula
The percentage change formula is used to determine the percentage increase or decrease between two values. The formula is:
percentage change = ((new value – old value) / old value) x 100
Examples of Percentage Change Calculation
Example 1: If the price of a product increased from \$50 to \$60, what is the percentage increase?
percentage increase = ((60 – 50) / 50) x 100 = 20%
Example 2: If a company’s revenue decreased from \$100,000 to \$80,000, what is the percentage decrease?
percentage decrease = ((80,000 – 100,000) / 100,000) x 100 = -20%
Note that in the second example, the result is negative because the value has decreased. ## 5. Percentage Increase/Decrease Calculation ### Percentage Increase/Decrease Formula The formula for calculating percentage increase or decrease is similar to the percentage change formula. The formula is:
percentage increase/decrease = ((new value – old value) / old value) x 100
The only difference is that if the result is negative, it represents a decrease, and if it is positive, it represents an increase. ### Examples of Percentage Increase/Decrease Calculation Example 1: If a company’s revenue increased from \$100,000 to \$120,000, what is the percentage increase?
percentage increase = ((120,000 – 100,000) / 100,000) x 100 = 20%
Example 2: If the number of visitors to a website decreased from 1,000 to 800, what is the percentage decrease?
percentage decrease = ((800 – 1,000) / 1,000) x 100 = -20%
## 6. Percentage Points Calculation ### Percentage Points Formula Percentage points refer to the difference between two percentages. The formula for calculating percentage points is:
percentage points = new percentage – old percentage
### Examples of Percentage Points Calculation Example 1: If a political candidate’s approval rating increased from 40% to 50%, what is the percentage point increase?
percentage points = 50% – 40% = 10%
Example 2: If the pass rate for a test increased from 60% to 80%, what is the percentage point increase?
percentage points = 80% – 60% = 20%
## Tips for Calculating Percentages
• Always convert fractions to decimals before calculating percentages
• Double-check your calculations to avoid errors
• Use a calculator or spreadsheet for complex calculations –
• Be mindful of the context and units of measurement when calculating percentages
• Round your final answer to the nearest whole number or decimal place, depending on the precision required ##
## Conclusion
Knowing how to calculate percentages is a useful skill that can be applied in many different situations. Whether you’re calculating grades, discounts, or changes in values, understanding the basic formulas and tips for calculating percentages can help you make accurate calculations.
By following the guidelines and examples provided in this article, you can become proficient in percentage calculation and apply this skill in your personal and professional life.
ALSO SEE: How to Get Bet9ja Old Mobile
## FAQs
• ### What is the difference between a percentage and a percentage point?
A percentage is a fraction of 100, while a percentage point refers to the difference between two percentages.
• ### Can percentages be negative?
Yes, percentages can be negative if the value has decreased.
• ### How do I calculate a percentage of a number?
Multiply the number by the percentage as a decimal (e.g., 50% = 0.5) to get the percentage of the number.
• ### How do I calculate a percentage increase or decrease?
Use the formula: percentage increase/decrease = ((new value – old value) / old value) x 100
• ### What is a common use of percentage in finance?
Percentage is commonly used to calculate interest rates and loan payments. |
# Productive Talk Moves
## 5 Talk Moves That Every Teacher Should Know and Use in the Classroom
1. Revoicing: Restate student questions to engage students and reinforce appropriate language.
2. Rephrasing: Have students restate their peers ideas in their own words to increase student involvement and encourage student interaction, thereby, enhancing the classroom discussion.
3. Reasoning: Get other students' perspective on their peers' ideas or questions.
4. Elaborating: Challenge students to expand upon their classmates ideas.
5. Waiting: Give students time to ponder or think about their ideas or answers before providing solutions (Van DeWalle, Karp, & Bay-Williams, 2016).
## Example of a 7th Grade Math Problem
Problem: In a bag of small balls 1/4 are green, 1/8 are blue, 1/12 are yellow and the remaining 26 white. How many balls are blue?
Solution
First, find the fraction of the green, blue and yellow balls by finding a common denominator and adding them together.
1/4 + 1/8 + 1/12 = 6 / 24 + 3 / 24 + 2 / 24 = 11 / 24
Next subtract the 11 / 24 balls from the total number of balls
24 / 24 - 11 / 24 = 13 / 24
Thus, the fraction 13 / 24 corresponds to the remaining 26 balls. Using x as the total number of balls then
(13 / 24) of x = 26 balls or (13 / 24) × x = 26
x = 26 × (24 / 13) = 48 , total number of balls
The fraction of blue balls is 1 / 8 of x, so that corresponds to 1/ 8 of 48 =6
Answer: 6 balls are blue
(Analyzemath.com, n.d.)
## Using the 5 Talk Moves to Solve the Problem
• Talk Move 1: A common error that students make when solving a problem like this is finding a common denominator, so using Revoicing teachers can clarify students questions while reinforcing appropriate mathematical language. An example of a teacher prompt using revoicing to solve the above problem could be asking a student "How did he/she decide to use 24 as a common denominator?"
• Talk Move 2: When using Rephrasing a teacher may want to ask another student to restate their classmate's reasoning for the use of 24 as a common denominator in their own words to enhance the classroom discussion.
• Talk Move 3: An example of a teacher prompt using Reasoning to engage students could be to ask the class "If anyone else used a different common denominator to solve this problem or if the rest of the of the class agreed with the learner's reasoning to use 24 as the common denominator?" This allows the teacher to get other students' thoughts on using this technique to solve the problem.
• Talk Move 4: A teacher prompt using Elaborating to solve the problem may include asking students "If they have another example of how the problem can be solved?" This can encourage student participation, and challenge learners to think of other ways that the problem can be solved while making connections to the similarity and differences between the two approaches.
• Talk Move 5: The process of Waiting is an essential element in letting students think things through on their on, so a teacher prompt using this technique may include telling students "To take time and really think about another way that this problem can be approached, or asking students to work together to come up with another strategy to solve this problem." This helps students understand that there can be more than one way to solve a math question, while also engaging students in higher order thinking about the problem solving process. |
# NCERT Solutions for Class 9 Maths Chapter 13 Surface Areas and Volumes
Q. How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?
Ans.
$$\text{Given, radius of bowl (r)} =\frac{10.5}{2}\\ = 5.25 cm\\ \text{Volume of hemisphere (V)} =\frac{2}{3}\pi r^2\\ =\frac{2}{3}×\frac{22}{7}×{(5.25)^3}\\ =\frac{2}{3}×\frac{22}{7}×{(5.25)×(5.25)×(5.25)}\\ = 303.1875 cm^3\\ \text{Hence, the hemispherical bowl can hold of milk 0.303 litre.}$$
Q. How many square metres of canvas is required for a conical tent whose height is 3.5 m and the radius of the base is 12 m?
Ans.
$$\text{Given, height of conical tent, h = 3.5 m}\\ \text{Radius of the base of conical tent, r = 12 m}\\ \text{Slant height, l} =\sqrt[]{h^2+r^2}\\ =\sqrt[]{{(3.5)}^2+{(12)}^2}\\ =\sqrt[]{{12.25}+{144}}\\ =\sqrt[]{{156.25}}\\ = 12.5 m\\ \text{Canvas required = Curved surface area of cone}\\(conical tent)\\ = \pi rl=\frac{22}{7}12×12.5\\ = 471.42 m^2\\ \text{Hence, the canvas required to make a conical tent is 471.42 m2.}$$
Q. The radius of a spherical balloon increases from 7 cm to 14 cm as air is being pumped into it. Find the ratio of surface area of the balloon in two cases.
Ans.
$$\text{ When, radius} (r_1) = 7 cm\\ \text{Surface area} = 4\pi r_1^2 =4×\frac{22}{7}×7×7= 616 cm^2\\ \text{When, radius} (r_2) = 14 cm\\ New\space\text{Surface area} = 4\pi r_2^2 =4×\frac{22}{7}×14×14= 2464cm^2\\ \text{Required ratio} =\frac{616}{2466}=\frac{1}{4}\space or\space1 : 4$$
Q. A hemispherical tank is made up of an iron sheet of1 cm thickness. If the inner radius is 1 m, then find the volume of the iron used to make the tank.
Ans.
$$\text{Given, inner radius of hemispherical tank (r) = 1m}\\ = 100 cm\\ \text{Thickness of the iron sheet} = 1 cm\\ \text{Outer radius of hemispherical tank (r′) = 100 + 1}\\ = 101 cm\\ \text{Volume of the hemispherical tank} =\frac{2}{3}\pi(r′^3-r^3)\\ =\frac{2}{3}×\frac{27}{7}[{(101)^3-(100)^3}]\\ =\frac{44}{21}×[1030301×1000000]\\ =\frac{44}{21}×[30301]\\ = 63487.81 cm^3\\ = 0.063 m^3\space [ 1 m^3 = (10)^6 cm^3]\\ \text{Hence, the volume of the iron used to make the tank is 0.063 m3}$$ |
# 2017 USAJMO Problems/Problem 6
## Problem
Let $P_1, \ldots, P_{2n}$ be $2n$ distinct points on the unit circle $x^2 + y^2 = 1$ other than $(1,0)$. Each point is colored either red or blue, with exactly $n$ of them red and exactly $n$ of them blue. Let $R_1, \ldots, R_n$ be any ordering of the red points. Let $B_1$ be the nearest blue point to $R_1$ traveling counterclockwise around the circle starting from $R_1$. Then let $B_2$ be the nearest of the remaining blue points to $R_2$ traveling counterclockwise around the circle from $R_2$, and so on, until we have labeled all the blue points $B_1, \ldots, B_n$. Show that the number of counterclockwise arcs of the form $R_i \rightarrow B_i$ that contain the point $(1,0)$ is independent of the way we chose the ordering $R_1, \ldots, R_n$ of the red points.
## Solution
I define a sequence to be, starting at $(1,0)$ and tracing the circle counterclockwise, and writing the color of the points in that order - either R or B. For example, possible sequences include $RB$, $RBBR$, $BBRRRB$, $BRBRRBBR$, etc. Note that choosing an $R_1$ is equivalent to choosing an $R$ in a sequence, and $B_1$ is defined as the $B$ closest to $R_1$ when moving rightwards. If no $B$s exist to the right of $R_1$, start from the far left. For example, if I have the above example $RBBR$, and I define the 2nd $R$ to be $R_1$, then the first $B$ will be $B_1$. Because no $R$ or $B$ can be named twice, I can simply remove $R_1$ and $B_1$ from my sequence when I choose them. I define this to be a move. Hence, a possible move sequence of $BBRRRB$ is: $BBR_1RRB_1\implies B_2BRR_2\implies B_3R_3$
Note that, if, in a move, $B_n$ appears to the left of $R_n$, then $\stackrel{\frown}{R_nB_n}$ intersects $(1,0)$
Now, I define a commencing $B$ to be a $B$ which appears to the left of all $R$s, and a terminating $R$ to be a $R$ which appears to the right of all $B$s. Let the amount of commencing $B$s be $j$, and the amount of terminating $R$s be $k$, I claim that the number of arcs which cross $(1,0)$ is constant, and it is equal to $\text{max}(j,k)$. I will show this with induction.
Base case is when $n=1$. In this case, there are only two possible sequences - $RB$ and $BR$. In the first case, $\stackrel{\frown}{R_1B_1}$ does not cross $(1, 0)$, but both $j$ and $k$ are $0$, so $\text{max}(j,k)=0$. In the second example, $j=1$, $k=1$, so $\text{max}(j,k)=1$. $\stackrel{\frown}{R_1B_1}$ crosses $(1,0)$ since $B_1$ appears to the left of $R_1$, so there is one arc which intersects. Hence, the base case is proved.
For the inductive step, suppose that for a positive number $n$, the number of arcs which cross $(1,0)$ is constant, and given by $\text{max}(j, k)$ for any configuration. Now, I will show it for $n+1$.
Suppose I first choose $R_1$ such that $B_1$ is to the right of $R_1$ in the sequence. This implies that $\stackrel{\frown}{R_1B_1}$ does not cross $(1,0)$. But, neither $R_1$ nor $B_1$ is a commencing $B$ or terminating $R$. These numbers remain constant, and now after this move we have a sequence of length $2n$. Hence, by assumption, the total amount of arcs is $0+\text{max}(j,k)=\text{max}(j,k)$.
• Here is a counter-case. $BRR_{1}B_{1}RR$ : j = 1, k = 2 => $BRRR$ : j = 1, k = 3. These numbers may not remain constant.
Thus, this solution probably doesn't work.
Now suppose that $R_1$ appears to the right of $B_1$, but $B_1$ is not a commencing $B$. This implies that there are no commencing $B$s in the series, because there are no $B$s to the left of $B_1$, so $j=0$. Note that this arc does intersect $(1,0)$, and $R_1$ must be a terminating $R$. $R_1$ must be a terminating $R$ because there are no $B$s to the right of $R_1$, or else that $B$ would be $B_1$. The $2n$ length sequence that remains has $0$ commencing $B$s and $k-1$ terminating $R$s. Hence, by assumption, the total amount of arcs is $1+\text{max}(0,k-1)=1+k-1=k=\text{max}(j,k)$.
Finally, suppose that $R_1$ appears to the right of $B_1$, and $B_1$ is a commencing $B$. We know that this arc will cross $(1,0)$. Analogous to the previous case, $R_1$ is a terminating $R$, so the $2n$ length sequence which remains has $j-1$ commencing $B$s and $k-1$ terminating $R$s. Hence, by assumption, the total amount of arcs is $1+\text{max}(j-1,k-1)=1+\text{max}(j,k)-1=\text{max}(j,k)$.
There are no more possible cases, hence the induction is complete, and the number of arcs which intersect $(1,0)$ is indeed a constant which is given by $\text{max}(j,k)$.
-william122
2017 USAJMO (Problems • Resources) Preceded byProblem 5 Last Problem 1 • 2 • 3 • 4 • 5 • 6 All USAJMO Problems and Solutions |
Home | | Maths 7th Std | Number System
Number System
Learning Objectives ● To recall the decimal notation and to understand the place value in decimals. ● To learn the concepts of decimals as fractions with denominators of tens and its powers. ● To represent decimal numbers on number line.
Chapter 1
NUMBER SYSTEM
Learning Objectives
● To recall the decimal notation and to understand the place value in decimals.
● To learn the concepts of decimals as fractions with denominators of tens and its powers.
● To represent decimal numbers on number line.
Recap
Decimal Numbers
Kala and Kavin went to a stationery shop to buy pencils. Their conversation is given below.
Kala : What is the price of a pencil?
Shopkeeper : The price of a pencil is four rupees and fifty paise.
Kala : Ok sir. Give me a pencil.
Kavin : We usually express the bill amount in rupees and paise as decimals. So the price of a pencil can be expressed as ₹4.50. Here 4 is the integral part and 50 is the decimal part. The dot represents the decimal point.
(After a week of time in the class room)
Teacher : We have studied about fractions and decimal numbers in earlier classes. Let us recall decimals now. Kala and kavin, can you measure the length of your pencils?
Kavin : Both the pencils seem to be of same length. Shall we measure and check?
Kala : Ok Kavin. The length of my pencil is 4 cm 3 mm (Fig. 1.1).
Kavin : Length of my pencil is 4 cm 5 mm (Fig. 1.2).
Kala : Can we express these lengths in terms of centimetres?
Kavin : Each centimetre is divided into ten equal parts known as millimetres. Do you remember that we have studied about tenths. I can say the length of my pencil as 4 and 5 tenths of a cm.
Kala : Since 1 mm = 1/10 cm or one tenth of a cm, it can be represented as 4.5 cm.
Kavin : Then the length of your pencil is 4.3 cm. Is it right?
Teacher : Both of you are right. Now we will further study about decimal numbers.
Try these
1. Observe the following and write the fraction of the shaded portion and mention in decimal form also.
(i) 4/8 = 0.5
(ii) 3/10 = 0.3
(iii) 5/10 = 0.5
2. Represent the following fractions in decimal form by converting denominator into ten or powers of 10.
3. Give any two life situations where we use decimal numbers.
The length of shirt cloth is 2.50 m.
The cost of one packet of chocolate is ₹ 20.60
Introduction
Consider the following situation. Ravi has planned to celebrate pongal festival in his native place Kanthapuram. He has purchased dress materials and groceries for the celebration. The details are furnished below.
Bill-1
Bill-2
What do you observe in the bills shown above? The prices are usually represented in decimals. But the quantities of length are represented in terms of metre and centimetre and that of weight are represented in terms of kilograms and grams. To express the quantities in terms of higher units, we use the concept of decimals.
MATHEMATICS ALIVE- Number system in Real Life
Tags : Term 2 Chapter 1 | 7th Maths , 7th Maths : Term 2 Unit 1 : Number System
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
7th Maths : Term 2 Unit 1 : Number System : Number System | Term 2 Chapter 1 | 7th Maths |
# Absolute Value Equations and Inequalities
#### Whats is Absolute value?
Absolute value (or Modulus) represented by |x| of a real number x is the non-negative value of x without regard to its sign.
For example:
Absolute value of -3 => |-3| = 3
Absolute value of 7 => |7| = 7
#### Absolute value as distance?
We can also think of absolute value of a number as its distance from zero. For example, consider this number line,
Here, the integer 3 is at a distance of 3 units from number 0. Similarly number -3 is also at distance of 3 units from 0. Since distance is always positive, we can say that,
The Distance of 3 from 0 => |3| = 3 & The Distance of -3 from 0 => |-3| = 3
Now, if |x| = 3, then x is at a distance of 3 units from 0. From the number line, we get that the absolute of x = 3 or x = -3
In general, if |x| = a => x = a or x = -a. (Since x is at a distance of ‘a’ units from 0).
#### If |x| = a is the distance of x from 0, then what is |x-a|?
Distance of x from a number ‘a’ on the number line can be represented by |x-a|.
1. Solve the equation |x-2| = 3.
|x-2| = 3 implies x is at a distance of 3 units from 2. Representing this on the number line,
The numbers on the number line which are at a distance of 3 units from 2 are 5 and -1. Therefore, x = 5 or x = -1 are the values of the equation |x-2| = 3.
2. Solve the inequality |x-2| < 3
|x-2| = 3 implies x is at a distance less than 3 units from 2. Representing this on the number line,
From the number line, we see that all the points between -1 and 5 are at a distance less than 3 units from 2. Hence, the values taken by x is given by -1 < x < 5.
#### If you know what is |x| = a, can you explain what is meant by |x-a| + |x-b| ?
From our understanding of absolute value so far, we know that
|x – a| represents the distance of x from a. Similarly |x – b| represents the distance of x from b. Implying that |x-a| + |x-b| is the sum of the distances of x from both a and b! :). This is evident from the number line,
Solve for x, |x+2| + |x-3| = 7
=> |x-(-2)| + |x-3| = 7 i.e we have to calculate the sum of the distances of x from -2 and 3.
Representing this on the number line,
From the image, we see that 4 is at distance of 1 unit from 3 and 6 units from -2. Hence, the sum of the distances of 4 from 3 and -2 is 7.
Similarly the sum of the distances of -3 from 4 and 3 is 7.
Therefore, x = 4 or x = -3 are the values of the equation |x+2| + |x-3| = 7.
#### When is the distance of x from ‘a’ and ‘b’ minimum?
We know that the sum of the distances of x from a and b is represented in terms of absolute value by |x – a| + |x – b|.
Now, |x – a| + |x – b| is minimum when x lies between a and b. The minimum value is given by |b-a|.
What is the minimum value of |x+2| + |x-3|?
=> Minimum value = |3-(-2)| = 5, which occurs for -2 ≤ x ≤ 3
## Solving Absolute Value Equations
### Math Tricks Workout
Please do try our android app – Math Tricks Workout. The app is developed to improve mental arithmetic using a series of left to right fast math workouts.
Scan the QR code below or click on it for more details. |
# -x-4y=-10 in ordered pai
-x-4y=-10 in ordered pai
## This Post Has 7 Comments
Step-by-step explanation:
As the given equation is
$-x-4y=-10$
We have to find the ordered pair which satisfies the equation.
So,
$-x-4y=-10$........[A]
Plug x = 0 in Equation [A] to find the y-intercept
$-(0)-4y=-10$
$-4y=-10$
$\mathrm{Divide\:both\:sides\:by\:}-4$
$\frac{-4y}{-4}=\frac{-10}{-4}$
$y=\frac{5}{2}$
So,
when x = 0, then y = 5/2
In other words, (0, 5/2) is the ordered pair of the equation $-x-4y=-10$.
Putting the value of x = 0, and y = 5/2 in Equation [1] for validation of the ordered pair (0, 5/2)
$-x-4y=-10$
$-0-4(\frac{5}{2} )=-10$
$-(\frac{20}{2} )=-10$
$-10=-10$ ∵ L.H.S = R.H.S
Hence, (0, 5/2) is one of the ordered pairs which is a solution to the equation $-x-4y=-10$.
Similarly, x-intercept can be found by putting y = 0 in $-x-4y=-10$.
So,
$-x-4y=-10$
$-x-4(0)=-10$
$-x=-10$
$x=10$
So, (10, 0) is another ordered pair (x-intercept) which satisfies the equation $-x-4y=-10$.
The graph is also attached for the equation $-x-4y=-10$ where you can easily figure out x and y intercepts.
Keywords: ordered pair, x-intercept, y-intercept, graph, solution
#learnwithBrainly
$What ordered pair is a solution to the equation -x-4y=-10$
2. kris22elizondop9v1bb says:
(0,3/2) (1,9/4) ( 2,2)
3. blairjaneaoyrfvp says:
D
Step-by-step explanation:
we let that,
(3,2)=(x,y)
given the equation that,
x-4y=10
so,we substitute the values of (x,y)=(3,2)it become
3-4(2)=-5
and test another coordinates(-3,3)
-3-4(3)=-15
botha answers doesn't give 10 as the answer so the answer is D(neither)
4. dbegay36 says:
x=−4y+10
Step-by-step explanation:
Let's solve for x.
−x−4y=−10
Step 1: Add 4y to both sides.
−x−4y+4y=−10+4y
−x=4y−10
Step 2: Divide both sides by -1.
5. computer15 says:
(6,0)(6,1)(6,2)
idk if i’m right sorry
6. kingteron5870 says:
x=−4y+10
Step-by-step explanation:
7. briizy says:
1/4
Step-by-step explanation:
Use the slope-intercept form
y
=
m
x
+
b
to find the slope
m
.
m
=
1
4 |
Partitioning Rectangles
1 / 10
# Partitioning Rectangles - PowerPoint PPT Presentation
Partitioning Rectangles. Lesson 6.14:. More or Less. Let’s count by tens. How many dimes are shown? What is the value of 6 dimes? 60 cents. More or Less. What is 5 cents more? 65 cents. Give the number sentence. 60 cents + 5 cents = 65 cents. More or Less. What is 10 cents less?
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### Partitioning Rectangles
Lesson 6.14:
More or Less
• Let’s count by tens.
• How many dimes are shown?
• What is the value of 6 dimes?
• 60 cents
More or Less
• What is 5 cents more?
• 65 cents.
• Give the number sentence.
• 60 cents + 5 cents = 65 cents.
More or Less
• What is 10 cents less?
• Give the number sentence.
• 65 cents – 10 cents = 55 cents
More or Less
• Let’s count by tens.
• What is the value of 7 dimes?
• What is the value if I remove 3 dimes?
• Give the number sentence.
• What is the value?
• Give the number sentence.
Concept Development
• Today we’re going to use the BLM 6.14 for our lesson! We’ll use the sentence frames to record our answers and to speak in complete sentences.
• Pass out the template, BLM 6.14, and scissors.
• Cut Rectangle A into rows and complete Problem 1. Share your responses and thinking with your partner.
• Cut Rectangle B into columns and complete Problem 2. Share again.
• Put both rectangles back together again.
Concept Development
• Move the columns of Rectangle B so they are sitting directly on top of the rows of Rectangle A with no gaps or overlaps.
• They both show the same amount, and they’re both the same size and shape.
• We can see the same rectangle different ways: it’s 2 rows of 4, 4 columns of 2 or, 8 squares.
• You’ve recognized that we can decompose the same rectangle into rows, columns or unit squares.
Concept Development
• Take both your rows of 4 and cut them to show 4 twos instead of 2 fours.
• Imagine we were going to put 2 rows on top to make the exact same rectangle. Talk to your partner.
• Explain what those rectangles would be.
• It would be 2 rows of 8. They would be 2 of the long skinny rectangles.
• We can decompose this rectangle into 2 rows of 8 or 8 columns of 2.
Concept Development
• Cut out all your squares from Rectangles A and B.
• How many squares do you have now?
• 16 |
# 2 Times Table
In 2 times table we will learn how to read and write multiplication table of 2.
We read two times table as:
One time two is 2
Two times two is 4
Three times two is 6
Four times two is 8
Five times two is 10
Six times two is 12
Seven times two is 14
Eight times two is 16
Nine times two is 18
Ten times two is 20
Eleven times two is 22
Twelve times two is 24
We write 2 times table as:
1 × 2 = 2
2 × 2 = 4
3 × 2 = 6
4 × 2 = 8
5 × 2 = 10
6 × 2 = 12
7 × 2 = 14
8 × 2 = 16
9 × 2 = 18
10 × 2 = 20
11 × 2 = 22
12 × 2 = 24
Let us follow the how to read and write multiplication of 2 times table chart:
2 Times Table
This is the easiest way to follow 2 times table in the chart.
Now, let us count the dots on the dice and write the multiplications.
1. What is 2 × 2 × 2 × 2 × 2 × 2 × 2?
Answer: 2 × 7 = 14
2. Skip count in 2's and color the boxes.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
2. 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
7 × 2 = _____
3.
4. There are 5 rows of 2 pineapples. How many pineapples are there in all?
4. 5 × 2 = 10
5. Fill in the blanks using 2 times table:
(i) 1 more than 13 = 7 × ___
(ii) 3 times ___ = 6
(iii) 5 times ___ = 10
(iv) 1 less than 23 = 2 × ___
(v) 9 times 2 = ___
(vi) 5 more than 13 = ___ × ___
5. (i) 1 more than 13 = 7 × 2
(ii) 3 times 2 = 6
(iii) 5 times 2 = 10
(iv) 1 less than 23 = 2 × 11
(v) 9 times 2 = 18
(vi) 5 more than 13 = 9 × 2
6. There are 15 lollipops in a packet. How many lollipops are there in 2 such packets?
Solution:
No. of lollipops in 1 packet = 15
No. of packets = 2
Therefore, total no. of lollipops in 2 packets = 15 × 2
Answer: There will be 30 lollipops in 2 packets.
7. Find the product of 13 and 2.
13 and 2
13
× 2
26
The product of 13 and 2 is 26.
8. John eats 2 chocolates every day. How many chocolates does he eat in a week?
Solution:
No. of chocolates he eats everyday = 2
No. of days in a week = 7
Therefore, total no. of chocolates he eats in a week = 2 × 7 = 14
Answer: John eats 14 chocolates in a week.
9. Using the multiplication time’s table of 2, calculate 2 times 11.
Solution:
According to the 2 times tables, 2 times 11 = 2 × 11 = 22
Therefore, 2 times 11 = 22
10.
Ron ate 2 apples in the breakfast every day. How many apples will he eat in 13 days?
No. of apples he ate every day = 2
No. of days = 13
Therefore, total no. of apples he eat in 13 days = 2 × 13 = 26
Answer: Ron eats 26 apples in 13 days.
## You might like these
• ### 5 Times Table | Read and Write Multiplication Table of 5 | Times Table
In 5 times table we will learn how to read and write multiplication table of 5. We read five times table as: One time five is 5 Two times five is 10 Three times five is 15 Four times five is 20
• ### 4 Times Table | Read and Write Multiplication Table of 4 | Four Table
In 4 times table we will learn how to read and write multiplication table of 4. We read four times table as: One time four is 4 Two times four is 8 Three times four is 12 Four times four is 16
• ### 3 Times Table | Read and Write Multiplication Table of 3 | Times Table
In 3 times table we will learn how to read and write multiplication table of 3. We read three times table as: One time three is 3 Two times three is 6 Three times three is 9 Four times three is 12
• ### 10 Times Table | Read and Write Multiplication Table of 10|Times Table
In 10 times table we will learn how to read and write multiplication table of 10. We read ten times table as: One time ten is 10 Two times ten is 20 Three times ten is 30 Four times ten is 40
• ### 0 Times Table | Multiplication of 0 Times Table Chart | Table of 0
In 0 times table we will learn how to read and write multiplication table of 0. One time zero is 0 Two times zero is 0 Three times zero is 0 Four times zero is 0 Five times zero is 0
• ### 1 Times Table | Read One Times Table | Write Multiplication of 1
In 1 times table we will learn how to read and write multiplication table of 1. We read one times table as: One time one is 1 Two times one is 2 Three times one is 3 Four times one is 4
• ### 16 Times Table | Read and Write Multiplication Table of 16|Times Table
In 16 times table we will learn how to read and write multiplication table of 16. We read sixteen times table as: One time sixteen is 16 Two times sixteen are 32 Three times sixteen are 48
• ### 17 Times Table |Read and Write Multiplication Table of 17 |Times Table
In 17 times table we will learn how to read and write multiplication table of 17. We read seventeen times table as: One time seventeen is 17 Two times seventeen are 34 Three times seventeen
• ### 14 Times Table | Read and Write Multiplication Table of 14|Times Table
In 14 times table we will learn how to read and write multiplication table of 14. We read fourteen times table as:One time fourteen is 14 Two times fourteen are 28 Three times fourteen are 42
• ### 13 Times Table | Read and Write Multiplication Table of 13|Times Table
In 13 times table we will learn how to read and write multiplication table of 13. We read thirteen times table as: One time thirteen is 13 Two times thirteen are 26 Three times thirteen are 39
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## Answered question
2021-08-20
Solve the recursion:
${A}_{k}=5{A}_{k-1}-6{A}_{k-2}$
### Answer & Explanation
Clara Reese
Skilled2021-08-21Added 120 answers
Step 1
Concept Used:
To determine which general solution to use for a recurrence relation of degree 2, determinants can be helpful
from the quadratic equation $a{x}^{2}+bx+c=0$ the discriminant is ${b}^{2}-4ac$
Case 1 if ${b}^{2}-4ac>0$
we get two distinct roots ${r}_{1}$ and ${r}_{2}$ then general solution is ${a}_{n}={\alpha }_{1}{\left({r}_{1}\right)}^{n}+{\alpha }_{2}{\left({r}_{2}\right)}^{n}$
Case 2 if ${b}^{2}-4ac=0$
we get one root with multiplicity 2 ${r}_{0}$ then general solution is ${a}_{n}={\alpha }_{1}{\left({r}_{0}\right)}^{n}+{\alpha }_{2}\left(n{r}_{0}\right\}{\right)}^{n}$
Step 2
Given:
Given recurrence relation
Solution:
Given ${A}_{k}=5{A}_{k-1}-6{A}_{k-2}$
Therefore characteristic equation is ${r}^{2}-5r+6=0$
Determinant: ${b}^{2}-4ac={\left(-5\right)}^{2}-4\left(1\right)\left(6\right)=25-24=1>0$
Since our determinant is greater than 0 we know that we get two distinct real roots
We can factor ${r}^{2}-5r+6=0$ into $\left(r-3\right)\left(r-2\right)=0$
So our roots are ${r}_{1}=3$ and ${r}_{2}=2$
Hence general solution is ${A}_{k}={\alpha }_{1}{\left(3\right)}^{k}+{\alpha }_{2}{\left(2\right)}^{k}$
Now we find ${\alpha }_{1}$ and ${\alpha }_{2}$ by using given initial conditions
Now for ${A}_{1}=1$
${A}_{1}={\alpha }_{1}{\left(3\right)}^{1}+{\alpha }_{2}{\left(2\right)}^{1}$
1) $⇒3{\alpha }_{1}+2{\alpha }_{2}=1$
Now for ${A}_{2}=-1$
${A}_{2}={\alpha }_{1}{\left(3\right)}^{2}+{\alpha }_{2}{\left(2\right)}^{2}$
2) $⇒9{\alpha }_{1}+4{\alpha }_{2}=-1$
By solving equation 1 and 2 we get
$\mathrm{¬}\left\{9{\alpha }_{1}\right\}+6{\alpha }_{2}=3$
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# Segment Relationships in Circles
Instructor: Yuanxin (Amy) Yang Alcocer
Amy has a master's degree in secondary education and has taught math at a public charter high school.
After reading this lesson, you'll understand how useful segments and circles are. You'll learn that when you have segments in a circle, you get some very interesting relationships that help you figure out various values.
## Segments in Circles
In this lesson, you'll learn about the relationships that form when you combine segments and circles together. A segment is a line that has a beginning and an end. A circle is a flat round shape. There are two types of segments you can have that cross your circle. A secant segment is a segment that intersects the edge of your circle twice. A tangent segment is a segment that touches the edge of your circle once. A tangent segment never passes through a circle.
There are actually three different relationships your segments and circles can have. These three different relationships are actually theorems as they are proven relationships that work all the time. Let's take a look at these three.
## Intersecting Segments
The first scenario is when you have two secant segments that intersect each other inside the circle.
When you have intersecting segments such as these, the relationship is that the product of the segment pieces of one segment is equal to the product of the segment pieces of the other.
• (segment piece a) * (segment piece b) = (segment piece c) * (segment piece d)
• a * b = c * d
You can use this relationship to help you solve problems that ask you find the length of a missing segment piece. For example, if you are given values for a, b, and d, but not c, you'll be asked to use this relationship to help you figure out the length of segment piece c.
To answer this problem, you plug in your values for a, b, and d and then use algebra to solve for your variable c. Looking at the above picture, you see that your a value is 4, your b value is 6, and your d value is 8. You plug these into your relationship and then you can solve for your c value.
• a * b = c * d
• 4 * 6 = c * 8
• 24 = 8c
• c = 3
And you are done!
## Two Secants
The next scenario is when you have two secant segments that intersect outside the circle.
The a and the c are the segment pieces that are outside the circle. The b and d are the whole segment including the a and c segment pieces. So, b includes a, and d includes c. The relationship here is that the product of a whole secant segment with its external part is equal to the product of the other whole secant segment with its external part.
• a * b = c * d
If you compare this to the first relationship, they are very similar. You use this relationship just like you would use the first.
For example, say you are given this information:
You can use this second relationship to help you find the missing value d. Your a is 5 and your b is 8. Your c is 4. So, plugging these in and solving for d, you get this:
• a * b = c * d
• 5 * 8 = 4 * d
• 40 = 4d
• d = 10
Your d is 10.
## One Secant and One Tangent
The third relationship is when you have one secant segment and one tangent segment.
The relationship here is that the product of the whole secant segment with its external part is equal to the square of the tangent segment.
• b * c = a2
Just like with the other relationships, this is best used to find missing values.
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# If E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD, show that ar (EFGH) = 1/2 ar (ABCD)
Given: E,F,G and H are respectively the mid-points of the sides of a parallelogram ABCD
Construction: Let us join HF.
To prove: ar(EFGH) = $$\frac{1}{2}$$ ar(ABCD)
Proof: In parallelogram ABCD,
( Opposite sides of a parallelogram are equal and parallel)
Multiplying both sides by $$\frac{1}{2}$$ we get,
$$\Rightarrow$$ $$\frac{1}{2}$$ x AD = $$\frac{1}{2}$$ x BC
$$\Rightarrow$$ AH = BF and AH || BF
$$\therefore$$ ABFH is a parallelogram.
Since, $$\triangle{HEF}$$ and parallelogram ABFH are on the same base HF and between the same parallel lines AB and HF,
$$\therefore$$ Area ($$\triangle{HEF}$$) = $$\frac{1}{2}$$ Area (ABFH) ...(i)
Similarly, it can be proved that,
Area ($$\triangle{HGF}$$) = $$\Rightarrow$$ Area (HDCF) ...(ii)
On adding equations (i) and (ii), we get,
$$\Rightarrow$$ Area ($$\triangle{HEF}$$) + Area ($$\triangle{HGF}$$) = $$\frac{1}{2}$$ Area (ABFH) + $$\frac{1}{2}$$ Area (HDCF)
$$\Rightarrow$$ Area [($$\triangle{HEF}$$) + ($$\triangle{HGF}$$)] = [Area (ABFH) + Area (HDCF)]
$$\therefore$$ Area (EFGH) = $$\frac{1}{2}$$ Area (ABCD)
Hence, proved. |
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# Shear Force and Bending Moments Part 1
Example Problem
A beam is loaded and supported as shown in Fig. 1. For this beam a. Draw complete shear force and bending moment diagrams. b. Determine the equations for the shear force and the bending moment as functions of x. Fig. 1
Solution
Overall Equilibrium
We start by drawing a free-body diagram (Fig. 2) of the beam and determining the support reactions. Summing moments about the left end of the beam
MA = 7RC - 2 [ 4 10 ]
gives - 4(16) - 9(19) = 0
RC = 45 kN
## Drawing the Shear Force Diagram
Sometimes we are not so much interested in the equations for the shear force and bending moment as we are in knowing the maximum and minimum values or the values at some particular point. In these cases, we want a quick and efficient method of generating the shear force and bending moment diagrams (graphs) so we can easily find the maximum and minimum values. That is the subject of this first part of the problem.
Concentrated Force
The 30-kN concentrated force (support reaction) at the left end of the beam causes the shear force graph to jump up (in the direction of the force) by 30 kN (the magnitude of the force) from 0 kN to 30 kN.
Fig. 3
The downward distributed load causes the shear force graph to slope downward (in the direction of the load). Since the distributed load is constant, the slope of the shear force graph is constant (dV/dx = w = constant). The total change in the shear force graph between points A and B is 40 kN (equal to the area under the distributed load between points A and B) from +30 kN to -10 kN. Fig. 4 We also need to know where the shear force becomes zero. We know that the full 4 m of the distributed load causes a change in the shear force of 40 kN. So how much of the distributed load will it take to cause a change of 30 kN (from +30 kN to 0 kN)? Since the distributed load is uniform, the area (change in shear force) is just 10 b = 30, which gives b = 3 m. That is, the shear force graph becomes zero at x = 3 m (3 m from the beginning of the uniform distributed load).
Concentrated Force
Fig. 5
The 16-kN concentrated force at B causes the shear force graph to jump down (in the direction of the force) by 16 kN (the magnitude of the force) from -10 kN to -26 kN.
Since there are no loads between points B and C, the shear force graph is constant (the slope dV/dx = w = 0) at -26 kN.
Fig. 6
Concentrated Force
The 45-kN concentrated force (support reaction) at C causes the shear force graph to jump up (in the direction of the force) by 45 kN (the magnitude of the force) from -26 kN to +19 kN.
Fig. 7
Since there are no loads between points C and D, the shear force graph is constant (the slope dV/dx = w = 0) at +19 kN.
Fig. 8
Concentrated Force
The 19-kN concentrated force at D causes the shear force graph to jump down (in the direction of the force) by 19 kN (the magnitude of the force) from +19 kN to 0 kN.
Fig. 9
## Drawing the Bending Moment Diagram
Since there are no concentrated moments acting on this beam, the bending moment diagram (graph) will be continuous (no jumps) and it will start and end at zero.
## Decreasing Shear Force
The bending moment graph starts out at zero and with a large positive slope (since the shear force starts out with a large positive value and dM/dx = V ). As the shear force decreases, so does the slope of the bending moment graph. At x = 3 m the shear force becomes zero and the bending moment is at a local maximum (dM/dx = V = 0 ) For values of x greater than 3 m (3 < x < 4 m) the shear force is negative and the bending moment decreases (dM/dx = V < 0). The shear force graph is linear (1st order function of x ), so the bending moment graph is a parabola (2nd order function of x ).
Fig. 10
The change in the bending moment between x = 0 m and x = 3 m is equal to the area under the shear graph between those two points. The area of the triangle is
M = (1/2)(30 3) = 45 kNm So the value of the bending moment at x = 3 m is M = 0 + 45 = 45 kNm. The change in the bending moment between x = 3 and x = 4 m is also equal to the
area under the shear graph M = (1/2)(-10 1) = -5 kNm So the value of the bending moment at x = 4 m is M = 45 - 5 = 40 kNm.
## Constant Shear Force
Although the bending moment graph is continuous at x = 4 m, the jump in the shear force at x = 4 m causes the slope of the bending moment to change suddenly from dM/dx = V = -10 kNm/m to dM/dx = -26 kNm/m. Since the shear force graph is constant between x = 4 m and x = 7 m, the bending moment graph has a constant slope between x = 4 m and x = 7 m (dM/dx = V = -26 kNm/m). That is, the bending moment graph is a straight line. The change in the bending moment between x = 4 m and x = 7 m is equal to the Fig. 11 area under the shear graph between those two points. The area of the rectangle is just M = (-26 3) = -78 kNm. So the value of the bending moment at x = 7 m is M = 40 - 78 = -38 kNm.
## Constant Shear Force
Again the bending moment graph is continuous at x = 7 m. The jump in the shear force at x = 7 m causes the slope of the bending moment to change suddenly from dM/dx = V = -26 kNm/m to dM/dx = +19 kNm/m. Since the shear force graph is constant between x = 7 m and x = 9 m, the bending moment graph has a constant slope between x = 7 m and x = 9 m (dM/dx = V = +19 kNm/m). That is, the bending moment graph is a straight line. The change in the bending moment between x = 7 m and x = 9 m is equal to the Fig. 12 area under the shear graph between those two points. The area of the rectangle is just M = (+19 2) = +38 kNm. So the value of the bending moment at x = 7 m is M = -38 + 38 = 0 kNm.
## Determining the Shear Force and Bending Moment Equations
Sometimes we are not so much interested in the graphs of the shear force and bending moment as we are in knowing the equations. In particular, we need to integrate the equation for the bending moment to determine the shape of beam and how much the beam will bend as a result of the loads. That is the subject of the second part of this problem.
## Shear Force and Bending Moments Part 2
Solution (Cont.)
Determining the Shear Force and Bending Moment Equations
The easiest way to get the equations for the shear force and bending moment as functions of the position x is to use equilibrium.
0 m < x < 4 m
Figure 13 shows a free-body diagram of the left end of the beam to an arbitrary position, 0 m < x < 4 m. The righthand portion of the beam that has been discarded exerts a shear force and a bending moment on the left-hand portion of the beam as shown. Summing forces in the vertical direction F = 30 - (10x ) - V = 0
Fig. 13 3a 3b
gives
V = 30 - (10x ) kN
(0 m < x < 4 m)
Summing moments about a point on the "cut end" of the beam Mcut = M + (10x )(x/2) - 30x = 0 gives M = 30x - (5x 2) kNm (0 m < x < 4 m)
4a 4b
4 m < x < 7 m
Figure 14 shows a free-body diagram of the left end of the beam to an arbitrary position, 4 m < x < 7 m. The right-hand portion of the beam that has been discarded exerts a shear force and a bending moment on the left-hand portion of the beam as shown. Summing forces in the vertical direction F = 30 - (10 4) - 16 - V = 0
Fig. 14 5a 5b
gives
V = -26 kN
(4 m < x < 7 m)
Summing moments about a point on the "cut end" of the beam Mcut = M + (10 4)(x - 2) + 16(x - 4) - 30x = 0 gives M = 144 - 26x kNm (4 m < x < 7 m)
6a 6b
7 m < x < 9 m
Figure 15 shows a free-body diagram of the left end of the beam to an arbitrary position, 7 m < x < 9 m. The right-hand portion of the beam that has been discarded exerts a shear force and a bending moment on the left-hand portion of the beam as shown. Summing forces in the vertical direction F = 30 - (10 4) - 16 - 45 - V = 0
Fig. 15 7a 7b
gives
V = 19 kN
(7 m < x < 9 m)
Summing moments about a point on the "cut end" of the beam Mcut = M + (10 4)(x - 2) + 16(x - 4) - 30x - 45(x - 7) = 0 gives M = 171 - 19x kNm (7 m < x < 9 m)
8a 8b
It is easily verified that these equations have the appropriate character to match the shear force and bending moment diagrams developed in the first part of this problem. It is also easily verified that these equations match the previous graphs at the points x = 0 m, x = 3 m, x = 4 m, x = 7 m, and x = 9 m. Finally, note that these equations satisfy the load-shear force-bending moment relationships dV/dx = w dM/dx = V |
# How do you use synthetic division to divide x^4-3x^3+3x^2-2x+3 by x^2-3x+2?
Jul 24, 2015
color(red)((x^4-3x^3+3x^2-2x+3)/(x^2-3x+2) = x^2+1+(x+1)/(x^2-3x+2)
#### Explanation:
Step 1. Write only the coefficients of $x$ in the dividend inside an upside-down division symbol.
Step 2. Negate the coefficients in the divisor and put every coefficient but the first one diagonally at the left.
Step 3. Drop the first coefficient of the dividend below the division symbol.
Step 4. Multiply the drop-down by the divisor, and put the result diagonally in the next columns.
Step 5. Add down the column.
Step 6. Repeat Steps 4 and 5 until you would go past the entries at the top with the next diagonal.
The quotient is ${x}^{2} + 1 + \frac{x + 1}{{x}^{2} - 3 x + 2}$.
Check:
$\left({x}^{2} - 3 x + 2\right) \left({x}^{2} + 1 + \frac{x + 1}{{x}^{2} - 3 x + 2}\right) = \left({x}^{2} - 3 x + 2\right) \left({x}^{2} + 1\right) + x + 1 = {x}^{4} - 3 {x}^{3} + 2 {x}^{2} + {x}^{2} - 3 x + 2 + x + 1 = {x}^{4} - 3 {x}^{3} + 3 {x}^{2} - 2 x + 3$ |
# 3.3Two Basic Rules of Probability
Introductory Business Statistics3.3 Two Basic Rules of Probability
When calculating probability, there are two rules to consider when determining if two events are independent or dependent and if they are mutually exclusive or not.
## The Multiplication Rule
If A and B are two events defined on a sample space, then: $P(A ∩B) = P(B)P(A | B)P(A∩B)=P(B)P(A|B)$. We can think of the intersection symbol as substituting for the word "and".
This rule may also be written as: $P(A| B) = P(A∩B) P(B) P(A|B)= P(A∩B) P(B)$
This equation is read as the probability of A given B equals the probability of A and B divided by the probability of B.
If A and B are independent, then $P ( A | B ) = P ( A ) P(A|B)=P(A)$. Then $P ( A ∩ B ) = P ( A | B ) P ( B ) P(A∩B)=P(A|B)P(B)$ becomes $P ( A ∩ B ) = P ( A ) ( B ) P(A∩B)=P(A)(B)$ because the $P ( A | B ) = P ( A ) P(A|B)=P(A)$ if A and B are independent.
One easy way to remember the multiplication rule is that the word "and" means that the event has to satisfy two conditions. For example the name drawn from the class roster is to be both a female and a sophomore. It is harder to satisfy two conditions than only one and of course when we multiply fractions the result is always smaller. This reflects the increasing difficulty of satisfying two conditions.
If A and B are defined on a sample space, then: $P ( A ∪ B ) = P ( A ) + P ( B ) - P ( A ∩ B ) P(A∪B)=P(A)+P(B)-P(A∩B)$. We can think of the union symbol substituting for the word "or". The reason we subtract the intersection of A and B is to keep from double counting elements that are in both A and B.
If A and B are mutually exclusive, then $P ( A ∩ B ) = 0 P(A∩B)=0$. Then $P ( A ∪ B ) = P ( A ) + P ( B ) - P ( A ∩ B ) P(A∪B)=P(A)+P(B)-P(A∩B)$ becomes $P ( A ∪ B ) = P ( A ) + P ( B ) P(A∪B)=P(A)+P(B)$.
## Example 3.14
Klaus is trying to choose where to go on vacation. His two choices are: A = New Zealand and B = Alaska
• Klaus can only afford one vacation. The probability that he chooses A is P(A) = 0.6 and the probability that he chooses B is P(B) = 0.35.
• $P ( A ∩ B ) = 0 P(A∩B)=0$ because Klaus can only afford to take one vacation
• Therefore, the probability that he chooses either New Zealand or Alaska is $P ( A ∪ B ) = P ( A ) + P ( B ) = 0.6 + 0.35 = 0.95P(A∪B)=P(A)+P(B)=0.6+0.35=0.95$. Note that the probability that he does not choose to go anywhere on vacation must be 0.05.
## Example 3.15
Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. A = the event Carlos is successful on his first attempt. P(A) = 0.65. B = the event Carlos is successful on his second attempt. P(B) = 0.65. Carlos tends to shoot in streaks. The probability that he makes the second goal $||$ that he made the first goal is 0.90.
### Problem
1. What is the probability that he makes both goals?
2. What is the probability that Carlos makes either the first goal or the second goal?
3. Are A and B independent?
4. Are A and B mutually exclusive?
## Try It 3.15
Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. C = the event that Helen makes the first shot. P(C) = 0.75. D = the event Helen makes the second shot. P(D) = 0.75. The probability that Helen makes the second free throw given that she made the first is 0.85. What is the probability that Helen makes both free throws?
## Example 3.16
A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly.
### Problem
1. What is the probability that the member is a novice swimmer?
2. What is the probability that the member practices four times a week?
3. What is the probability that the member is an advanced swimmer and practices four times a week?
4. What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why not?
5. Are being a novice swimmer and practicing four times a week independent events? Why or why not?
## Try It 3.16
A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is taking a gap year?
## Example 3.17
Felicity attends Modesto JC in Modesto, CA. The probability that Felicity enrolls in a math class is 0.2 and the probability that she enrolls in a speech class is 0.65. The probability that she enrolls in a math class $||$ that she enrolls in speech class is 0.25.
Let: M = math class, S = speech class, M$||$S = math given speech
### Problem
1. What is the probability that Felicity enrolls in math and speech?
Find P(M $∩∩$ S) = P(M$||$S)P(S).
2. What is the probability that Felicity enrolls in math or speech classes?
Find P(M $∪∪$ S) = P(M) + P(S) - P(M $∩∩$ S).
3. Are M and S independent? Is P(M$||$S) = P(M)?
4. Are M and S mutually exclusive? Is P(M $∩∩$ S) = 0?
## Try It 3.17
A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P(B) = 0.40, P(D) = 0.30 and P(D$||$B) = 0.5.
1. Find P(B $∩∩$ D).
2. Find P(B $∪∪$ D).
## Example 3.18
Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let B = woman develops breast cancer and let N = tests negative. Suppose one woman is selected at random.
### Problem
1. What is the probability that the woman develops breast cancer? What is the probability that woman tests negative?
2. Given that the woman has breast cancer, what is the probability that she tests negative?
3. What is the probability that the woman has breast cancer AND tests negative?
4. What is the probability that the woman has breast cancer or tests negative?
5. Are having breast cancer and testing negative independent events?
6. Are having breast cancer and testing negative mutually exclusive?
## Try It 3.18
A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is going to college and plays sports?
## Example 3.19
### Problem
Refer to the information in Example 3.18. P = tests positive.
1. Given that a woman develops breast cancer, what is the probability that she tests positive. Find P(P$||$B) = 1 - P(N$||$B).
2. What is the probability that a woman develops breast cancer and tests positive. Find P(B $∩∩$ P) = P(P$||$B)P(B).
3. What is the probability that a woman does not develop breast cancer. Find P(B′) = 1 - P(B).
4. What is the probability that a woman tests positive for breast cancer. Find P(P) = 1 - P(N).
## Try It 3.19
A student goes to the library. Let events B = the student checks out a book and D = the student checks out a DVD. Suppose that P(B) = 0.40, P(D) = 0.30 and P(D$||$B) = 0.5.
1. Find P(B′).
2. Find P(D $∩∩$ B).
3. Find P(B$||$D).
4. Find P(D $∩∩$ B′).
5. Find P(D$||$B′).
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A symmetric matrix is always a square matrix. If A be any square matrix and if A = AT, then matrix A is called symmetric matrix.
Let A = [ aij ] be a square matrix and if A = AT, it means if we interchange rows and columns of A, then the matrix remains unchanged i.e. aij = aji, for all i and j. So, we say that, in a symmetric matrix, all elements are symmetric with respect to main diagonal.
A = $\begin{bmatrix} 2 &-1 &0 \\ -1 &-2 &3 \\ 0 &3 &4 \end{bmatrix}$ AT = $\begin{bmatrix} 2 &-1 &0 \\ -1 &-2 &3 \\ 0 &3 &4 \end{bmatrix}$
Here A =AT. So, A is a symmetric matrix. All diagonal matrices are symmetric because in diagonal matrix all non-diagonal elements are zero.
## Skew Symmetric Matrix
A square matrix say B = [ bij ] is said to skew-symmetric, if and only if B = -BT
i.e. [ bij ] = - [ bji ].
Let A = $\begin{bmatrix} 0&a &b \\ -a &0 &-c \\ -b &c &0 \end{bmatrix}$
Now AT = $\begin{bmatrix} 0&-a &-b \\ a &0 &c \\ b &-c &0 \end{bmatrix}$
so A = AT
For symmetric and skew-symmetric matrices, the necessary condition is that the matrices should be square matrices. And for skew-symmetric matrix, all the diagonal elements are zero.
## Inverse of Symmetric Matrix
To find the inverse of symmetric matrix, let A = $\begin{bmatrix} 1 &-3 &5 \\ -3 &2 &-1 \\ 5 &-1 &4 \end{bmatrix}$
We know that A-1 =$\frac{adjA}{\left | A \right |}$
First we have to find out I A I = ?
For this we use first row expansion method:
I A I = 1 $\begin{vmatrix} 2 &-1 \\ -1 &4 \end{vmatrix}$ - (-3) $\begin{vmatrix} -3 &-1 \\ 5 &4 \end{vmatrix}$ + 5 $\begin{vmatrix} -3&2 \\ 5&-1 \end{vmatrix}$
= 1(8 - 1) + 3(-12 + 5) + 5(3 - 10)
= 7 -21 - 35
= - 49 $\neq$ 0
Here I A I $\neq$ 0, hence inverse of matrix is exists.
For adj A, first calculate cofactor matrix of A say B and after that taking transpose of B,
which is said to be adj A.
Cofactor matrix of A = B = $\begin{bmatrix} 7 &-7 &-7 \\ 7 &-21 &-14 \\ -7 &-14 &-7 \end{bmatrix}$
Now adj A = BT = $\begin{bmatrix} 7 &7 &-7 \\ -7 &-21 &-14 \\ -7 &-14 &-7 \end{bmatrix}$
For A -1 = $\frac{1}{\left | A \right |}$ x adj A
= $\frac{1} {-49}$ x $\begin{bmatrix} 7 &7 &-7 \\ -7 &-21 &-14 \\ -7 &-14 &-7 \end{bmatrix}$
or A -1 = $\begin{bmatrix} 1/7 &1/7 &-1/7 \\ -1/7 &-3/7 &-2/7\\ -1/7 &-2/7 &-1/7 \end{bmatrix}$
## Determinant of Symmetric Matrix
Let C = $\begin{bmatrix} 1 &0 &2 \\ 0 &2 &-4 \\ 2 &-4 &3 \end{bmatrix}$be a square matrix. For det C, we choose first row for expansion.
I C I = 1 $\begin{vmatrix} 2 &-4 \\ -4&3 \end{vmatrix}$ - 0 $\begin{vmatrix} 0&-4 \\ 4 &3 \end{vmatrix}$ + 2 $\begin{vmatrix} 0 &2 \\ 2 &-4 \end{vmatrix}$
or I C I = 1( 6 - 16 ) + 0 + 2( 0 - 4)
or I C I = -10 - 8
or I C I = - 18
It is clear that finding the determinant of a symmetric matrix is the same as finding determinant of any square matrix.
## Properties of Symmetric Matrices
Listed below are some of the Symmetric Matrix Properties
Property 1
: If A and B are two symmetric matrix and (A + B)T = AT + BT = A + B
then A + B is a symmetric matrix.
Proof: Let A = $\begin{bmatrix} 3&2 &-1 \\ 2 &2 &5\\ -1 &5 &4 \end{bmatrix}$ and B = $\begin{bmatrix} 7 &0 &4 \\ 0&2 &-6 \\ 4 &-6 &1 \end{bmatrix}$ are two symmetric matrices of order 3x3.
Then AT = $\begin{bmatrix} 3&2 &-1 \\ 2 &2 &5\\ -1 &5 &4 \end{bmatrix}$ and BT = $\begin{bmatrix} 7 &0 &4 \\ 0&2 &-6 \\ 4 &-6 &1 \end{bmatrix}$
AT + BT = $\begin{bmatrix} 3&2 &-1 \\ 2 &2 &5\\ -1 &5 &4 \end{bmatrix}$ + $\begin{bmatrix} 7 &0 &4 \\ 0&2 &-6 \\ 4 &-6 &1 \end{bmatrix}$
= $\begin{bmatrix} 10 &2 &3\\ 2&4&-1 \\ 3 &-1 &5 \end{bmatrix}$ ...................................................(1)
( A + B ) = $\begin{bmatrix} 3&2 &-1 \\ 2 &2 &5\\ -1 &5 &4 \end{bmatrix}$ + $\begin{bmatrix} 7 &0 &4 \\ 0&2 &-6 \\ 4 &-6 &1 \end{bmatrix}$
= $\begin{bmatrix} 10 &2 &3\\ 2&4 &-1 \\ 3 &-1 &5 \end{bmatrix}$ ..................................................(2)
( A + B )T = $\begin{bmatrix} 10 &2 &3\\ 2&4 &-1 \\ 3 &-1 &5 \end{bmatrix}$ ................................................(3)
From (1), (2) and (3), we have ( A + B )T = AT + BT = A + B
Hence A + B is a symmetric matrix.
Property 2: If C and D are symmetric matrices, then (CD)T = DTCT = DC $\neq$ CD
Usually CD is not symmetric.
Proof: Let C = $\begin{bmatrix} 1 &2 &3\\ 2&-1 &4 \\ 3 &4&0 \end{bmatrix}$ and D = $\begin{bmatrix} -1 &3 &2\\ 3&0 &5 \\ 2 &5 &1 \end{bmatrix}$ are symmetric matrices of order 3x3.
Then CD = $\begin{bmatrix} 1 &2 &3\\ 2&-1 &4 \\ 3 &4&0 \end{bmatrix}$ x $\begin{bmatrix} -1 &3 &2\\ 3&0 &5 \\ 2 &5 &1 \end{bmatrix}$
= $\begin{bmatrix} 11 &18&15\\ 3&26 &3 \\ 9&9 &26 \end{bmatrix}$ ................................................(1)
( CD )T = $\begin{bmatrix} 11 &3 &9\\ 18&26 &9\\ 15&3 &26 \end{bmatrix}$ ..................................................(2)
Here CT = $\begin{bmatrix} 1 &2 &3\\ 2&-1 &4 \\ 3 &4&0 \end{bmatrix}$ , DT = $\begin{bmatrix} -1 &3 &2\\ 3&0 &5 \\ 2 &5 &1 \end{bmatrix}$
DTCT = $\begin{bmatrix} -1 &3 &2\\ 3&0 &5 \\ 2 &5 &1 \end{bmatrix}$ x $\begin{bmatrix} 1 &2 &3\\ 2&-1 &4 \\ 3 &4&0 \end{bmatrix}$
= $\begin{bmatrix} 11 &3 &9\\ 18&26 &9\\ 15&3 &26 \end{bmatrix}$ ................................................(3)
CD = $\begin{bmatrix} 1 &2 &3\\ 2&-1 &4 \\ 3 &4&0 \end{bmatrix}$ x $\begin{bmatrix} -1 &3 &2\\ 3&0 &5 \\ 2 &5 &1 \end{bmatrix}$
= $\begin{bmatrix} 11 &3 &9\\ 18&26 &9\\ 15&3 &26 \end{bmatrix}$ ...............................................(4)
From (1), it is clear that matrix CD is not symmetric and (CD)T = DTCT = DC $\neq$ CD
Property 3: If A is any symmetric matrix then B = ATA is symmetric and
( ATA)T = AT (AT)T = ATA
Proof: For this let A2x2 = $\begin{bmatrix} 1 &-2 \\ -2 &1 \end{bmatrix}$ be a matrix, then AT = $\begin{bmatrix} 1 &-2 \\ -2 &1 \end{bmatrix}$
Compute B = ATA = $\begin{bmatrix} 1 &-2 \\ -2 &1 \end{bmatrix}$ x $\begin{bmatrix} 1 &-2 \\ -2 &1 \end{bmatrix}$
= $\begin{bmatrix} 5 &-4\\ -4 &5 \end{bmatrix}$ ..........................................................(a)
So matrix B is a symmetric matrix of order 2x2.
Now ( ATA)T = $\begin{bmatrix} 5 &-4\\ -4 &5 \end{bmatrix}$ ...........................................................(b)
Again AT (AT)T = $\begin{bmatrix} 1 &-2 \\ -2 &1 \end{bmatrix}$ x $\begin{bmatrix} 1 &-2 \\ -2 &1 \end{bmatrix}$
= $\begin{bmatrix} 5 &-4\\ -4 &5 \end{bmatrix}$ .....................................................(c)
Using (a), (b) and (c), we get ( ATA)T = AT (AT)T = ATA |
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### Course: Digital SAT Math>Unit 12
Lesson 3: Operations with polynomials: advanced
# Operations with polynomials | Lesson
A guide to operations with polynomials on the digital SAT
## What are polynomial expressions?
A polynomial expression has one or more terms with a coefficient, a variable base, and an exponent.
$3{x}^{4}$ is a
. We'll also frequently see
and
.
• $3{x}^{4}+2x$ is a binomial. The exponent of the term $2x$ is $1$ ($x={x}^{1}$).
• $3{x}^{4}+2x+7$ is a trinomial. $7$ is a constant term. We can also think of $7$ as an exponential term with an exponent of $0$. Since ${x}^{0}=1$, $7$ is equivalent to $7{x}^{0}$.
In this lesson, we'll learn to add, subtract, and multiply polynomials.
You can learn anything. Let's do this!
## How do I add and subtract polynomials?
### What should I be careful of when adding and subtracting polynomials?
While we can add and subtract any polynomials, we can only combine like terms, which must have:
• The same variable base
• The same exponent
For example, we can combine the terms $2{x}^{3}$ and $4{x}^{3}$ because they have the same variable base, $x$, and the same exponent, $3$. However, we cannot combine the terms $2{x}^{2}$ and $2{x}^{3}$ because they have different exponents, $2$ and $3$.
When we combine like terms, only the coefficients change. Both the base and the exponent remain the same. For example, when adding $2{x}^{3}$ and $4{x}^{3}$, the ${x}^{3}$ part of the terms remain the same, and we add only $2$ and $4$ when combining the terms:
$\begin{array}{rl}2{x}^{3}+4{x}^{3}& =\left(2+4\right){x}^{3}\\ \\ & =6{x}^{3}\end{array}$
When subtracting polynomials, make sure to distribute the negative sign as needed. For example, when subtracting the polynomial $-2{x}^{2}-7$, the negative sign from the subtraction is distributed to both $-2{x}^{2}$ and $-7$, which means:
$\begin{array}{rl}& 5{x}^{2}-\left(-2{x}^{2}-7\right)\\ \\ & =5{x}^{2}+\left(-1\right)\left(-2{x}^{2}\right)+\left(-1\right)\left(-7\right)\\ \\ & =5{x}^{2}+2{x}^{2}+7\\ \\ & =\left(5+2\right){x}^{2}+7\\ \\ & =7{x}^{2}+7\end{array}$
Subtracting $-2{x}^{2}-7$ is equivalent to adding $2{x}^{2}+7$!
To add or subtract two polynomials:
1. Group like terms.
2. For each group of like terms, add or subtract the coefficients while keeping both the base and the exponent the same.
3. Write the combined terms in order of decreasing power.
### Try it!
TRY: Match the equivalent expressions
Because $9{x}^{2}$ and $3{x}^{2}$ have
and
, the two terms
into a single term.
Because $4{y}^{4}$ and $4y$ have
, the two terms
into a single term.
TRY: Match the equivalent expressions
Match each polynomial expression below with an equivalent expression.
## How do I multiply polynomials?
### Multiplying binomials
Multiplying binomialsSee video transcript
### What should I be careful of when multiplying polynomials?
When multiplying two polynomials, we must make sure to distribute each term of one polynomial to all the terms of the other polynomial. For example:
$\begin{array}{rl}& \left(ax+b\right)\left(cx+d\right)\\ \\ =& \phantom{\rule{0.167em}{0ex}}\left(ax\right)\left(cx+d\right)+\left(b\right)\left(cx+d\right)\\ \\ =& \phantom{\rule{0.167em}{0ex}}\left(ax\right)\left(cx\right)+\left(ax\right)\left(d\right)+\left(b\right)\left(cx\right)+\left(b\right)\left(d\right)\end{array}$
The total number of products we need to calculate is equal to the product of the number of terms in each polynomial. Multiplying two binomials requires $2\cdot 2=4$ products, as shown above. Multiplying a monomial and a trinomial requires $1\cdot 3=3$ products; multiplying a binomial and a trinomial requires $2\cdot 3=6$ products.
When multiplying two binomials, we can also use the mnemonic FOIL to account for all four multiplications. For $\left(ax+b\right)\left(cx+d\right)$:
1. Multiply the First terms ($ax\cdot cx$)
2. Multiply the Outer terms ($ax\cdot d$)
3. Multiply the Inner terms ($b\cdot cx$)
4. Multiply the Last terms ($b\cdot d$)
When multiplying terms of polynomial expressions with the same base:
1. Multiply the coefficients, or multiply the coefficient and the constant.
2. Keep the base the same.
$\begin{array}{rl}a{x}^{m}\cdot b{x}^{n}& =ab\cdot {x}^{m+n}\\ \\ a\cdot b{x}^{n}& =ab\cdot {x}^{n}\end{array}$
To multiply two polynomials:
1. Distribute the terms.
2. Multiply the distributed terms according to the exponent rules above.
3. Group like terms.
4. For each group of like terms, add or subtract the coefficients while keeping both the base and the exponent the same.
5. Write the combined terms in order of decreasing power.
#### Let's look at some examples!
What is the product of $2x-1$ and $x-5$ ?
What is the product of $3x$ and ${x}^{2}-4x+9$ ?
### Try it!
TRY: multiply two terms
When multiplying $3x$ and $2{x}^{2}$, we
the coefficients of the terms and
the exponents of $x$.
$\left(3x\right)\left(2{x}^{2}\right)=\phantom{\rule{0.167em}{0ex}}$
TRY: Multiply two binomials using FOIL
Use the table below to FOIL $\left(8x-3\right)\left({x}^{2}+1\right)$.
TermExpressionProduct
First$8x\cdot {x}^{2}$
Outer
$8x$
Inner$-3\cdot {x}^{2}$
Last
$-3$
Which of the following is the sum of ${x}^{2}+5$ and $2{x}^{2}+4x$ ?
Practice: subtract two polynomials to find a coefficient
$\left(9{x}^{2}+5x-1\right)-\left(6{x}^{2}-4x\right)=a{x}^{2}+bx+c$
The equation above is true for all $x$, where $a$, $b$ and $c$ are constants. What is the value of $b$ ?
Practice: multiply two binomials
Which of the following is equivalent to $\left(x+3\right)\left(2x-5\right)$ ?
Practice: multiply two binomials with symbolic coefficients
$\left(ax+3\right)\left(bx+2\right)=9{x}^{2}+21x+6$
In the equation above, $a$ and $b$ are constants. What is the value of $ab$ ?
## Things to remember
The mnemonic FOIL for multiplying two binomials:
1. Multiply the First terms
2. Multiply the Outer terms
3. Multiply the Inner terms
4. Multiply the Last terms
$\begin{array}{rl}a{x}^{m}\cdot b{x}^{n}& =ab\cdot {x}^{m+n}\\ \\ a\cdot b{x}^{n}& =ab\cdot {x}^{n}\end{array}$
## Want to join the conversation?
• polynomials are a breath of fresh air after radicals and rationals
• You're dang right!
• I believe the previous lesson was way harder, might wanna reconsider the structure a bit, so that each lesson would help you understand the next one better. great work anyway
• 7 days remaining to my SAT
• Same! Wish me luck 😭
• I have my SAT exam in 9 hours. Good luck to me.
• I hope it went well.
• This was so easy I was singing Hamilton while doing this. May all math questions be likewise!
• After the last few lessons, it's great to not be confused for once. |
Home · Algebra · Basic properties of logarithms
Basic properties of logarithms
The logarithm of a given number b is the exponent to which another fixed number, the base a, must be raised, to produce that number b $(a>0, \ a\neq 1, \ b>0)$.
$$\displaystyle \log_ab=c, \ a^c=b$$
Logarithmic identities
Logarithm power rules are very useful whenever expressions involving logarithmic functions need to be simplified:
$$\displaystyle \alpha \log _ax=\log _a x^{\alpha}$$ $$\frac{1}{\alpha}\log _ax=\log _{a^{\alpha}}x$$ $$\displaystyle a^{\log_ax}=x$$
The base of logarithm can be changed in other. It's very helpful, when we need to calculate logarithm with calculator which has only standard 10 and e bases. Change of base formula:
$$\displaystyle \log_ax=\frac{\log_bx}{\log_ba}$$
Sum and difference of two logarithms with the same base properties (or product to sum and quotient to difference) formulas:
$$\log_ax+\log_ay=\log_axy$$ $$\log_ax-\log_ay=\log_a\frac{x}{y}$$
Solved examples
Solve this expression $\log_2{8}+3\log_3{9}$
$$\log_2{8}+3\log_3{9}=3+3\cdot 2=9$$
Simplify this expression $2\log_2{3}+\log_2{9}-2\log_2{27}$
$$2\log_2{3}+\log_2{9}-2\log_2{27}=2\log_2{3}+\log_2{3^2}-2\log_2{3^3}=2\log_2{3}+2\log_2{3}-2\cdot 3 \log_2{3}=-2\log_2{3}$$
Solve this expression $27^{\log_3{5}}+25^{\log_5{2}}$
$$27^{\log_3{5}}+25^{\log_5{2}}=3^{3\log_3{5}}+5^{2\log_5{2}}=3^{\log_3{5^3}}+5^{\log_5{2^2}}=3^{\log_3{125}}+5^{\log_5{4}}=125+4=129$$
Convert $\log_2{5}$ to an expression with logarithms having a base of 10.
$$\log_2{5}=\frac{\lg{5}}{\lg{2}}$$
Solve this expression $\log_2{3}+\log_2{6}-\log_2{9}$
$$\log_2{3}+\log_2{6}-\log_2{9}=\log_2{\frac{3\cdot 6}{9}}=\log_2{2}=1$$ |
# 4.1 Linear functions
Page 1 / 27
In this section you will:
• Represent a linear function.
• Determine whether a linear function is increasing, decreasing, or constant.
• Interpret slope as a rate of change.
• Write and interpret an equation for a linear function.
• Graph linear functions.
• Determine whether lines are parallel or perpendicular.
• Write the equation of a line parallel or perpendicular to a given line.
Just as with the growth of a bamboo plant, there are many situations that involve constant change over time. Consider, for example, the first commercial maglev train in the world, the Shanghai MagLev Train ( [link] ). It carries passengers comfortably for a 30-kilometer trip from the airport to the subway station in only eight minutes http://www.chinahighlights.com/shanghai/transportation/maglev-train.htm .
Suppose a maglev train travels a long distance, and maintains a constant speed of 83 meters per second for a period of time once it is 250 meters from the station. How can we analyze the train’s distance from the station as a function of time? In this section, we will investigate a kind of function that is useful for this purpose, and use it to investigate real-world situations such as the train’s distance from the station at a given point in time.
## Representing linear functions
The function describing the train’s motion is a linear function , which is defined as a function with a constant rate of change. This is a polynomial of degree 1. There are several ways to represent a linear function, including word form, function notation, tabular form, and graphical form. We will describe the train’s motion as a function using each method.
## Representing a linear function in word form
Let’s begin by describing the linear function in words. For the train problem we just considered, the following word sentence may be used to describe the function relationship.
• The train’s distance from the station is a function of the time during which the train moves at a constant speed plus its original distance from the station when it began moving at constant speed.
The speed is the rate of change. Recall that a rate of change is a measure of how quickly the dependent variable changes with respect to the independent variable. The rate of change for this example is constant, which means that it is the same for each input value. As the time (input) increases by 1 second, the corresponding distance (output) increases by 83 meters. The train began moving at this constant speed at a distance of 250 meters from the station.
## Representing a linear function in function notation
Another approach to representing linear functions is by using function notation. One example of function notation is an equation written in the slope-intercept form of a line, where $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ is the input value, $\text{\hspace{0.17em}}m\text{\hspace{0.17em}}$ is the rate of change, and $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ is the initial value of the dependent variable.
In the example of the train, we might use the notation $\text{\hspace{0.17em}}D\left(t\right)\text{\hspace{0.17em}}$ where the total distance $\text{\hspace{0.17em}}D\text{\hspace{0.17em}}$ is a function of the time $\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ The rate, $\text{\hspace{0.17em}}m,\text{\hspace{0.17em}}$ is 83 meters per second. The initial value of the dependent variable $\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$ is the original distance from the station, 250 meters. We can write a generalized equation to represent the motion of the train.
sin theta=3/4.prove that sec square theta barabar 1 + tan square theta by cosec square theta minus cos square theta
I want to know trigonometry but I can't understand it anyone who can help
Yh
Idowu
which part of trig?
Nyemba
functions
Siyabonga
trigonometry
Ganapathi
differentiation doubhts
Ganapathi
hi
Ganapathi
hello
Brittany
Prove that 4sin50-3tan 50=1
f(x)= 1 x f(x)=1x is shifted down 4 units and to the right 3 units.
f (x) = −3x + 5 and g (x) = x − 5 /−3
Sebit
what are real numbers
I want to know partial fraction Decomposition.
classes of function in mathematics
divide y2_8y2+5y2/y2
wish i knew calculus to understand what's going on 🙂
@dashawn ... in simple terms, a derivative is the tangent line of the function. which gives the rate of change at that instant. to calculate. given f(x)==ax^n. then f'(x)=n*ax^n-1 . hope that help.
Christopher
thanks bro
Dashawn
maybe when i start calculus in a few months i won't be that lost 😎
Dashawn
what's the derivative of 4x^6
24x^5
James
10x
Axmed
24X^5
Taieb
secA+tanA=2√5,sinA=?
tan2a+tan2a=√3
Rahulkumar
classes of function
Yazidu
if sinx°=sin@, then @ is - ?
the value of tan15°•tan20°•tan70°•tan75° -
NAVJIT
0.037 than find sin and tan?
cos24/25 then find sin and tan |
# Math Snap
## If $f$ is a continuous, decreasing function on $[1, \infty)$ and $\lim _{x \rightarrow \infty} f(x)=0$, then $\int_{1}^{\infty} f(x) d x$ is convergent. Select one: True False
#### STEP 1
Assumptions 1. $f$ is a continuous function on the interval $[1, \infty)$. 2. $f$ is a decreasing function on the interval $[1, \infty)$. 3. $\lim_{x \to \infty} f(x) = 0$. 4. We need to determine if the improper integral $\int_{1}^{\infty} f(x) \, dx$ is convergent.
#### STEP 2
Recall the definition of an improper integral. The integral $\int_{1}^{\infty} f(x) \, dx$ is defined as: $\int_{1}^{\infty} f(x) \, dx = \lim_{b \to \infty} \int_{1}^{b} f(x) \, dx$
#### STEP 3
Since $f$ is continuous and decreasing on $[1, \infty)$, and $\lim_{x \to \infty} f(x) = 0$, we need to analyze the behavior of the integral $\int_{1}^{b} f(x) \, dx$ as $b \to \infty$.
#### STEP 4
Consider the integral $\int_{1}^{b} f(x) \, dx$ for a finite $b$. Since $f$ is continuous on $[1, b]$, the integral $\int_{1}^{b} f(x) \, dx$ is well-defined and finite for any finite $b$.
#### STEP 5
To determine the convergence of $\int_{1}^{\infty} f(x) \, dx$, we need to show that $\lim_{b \to \infty} \int_{1}^{b} f(x) \, dx$ exists and is finite.
#### STEP 6
Since $f$ is decreasing and $\lim_{x \to \infty} f(x) = 0$, for any $\epsilon > 0$, there exists a $M > 1$ such that for all $x > M$, $0 \leq f(x) < \epsilon$.
#### STEP 7
Split the integral $\int_{1}^{b} f(x) \, dx$ into two parts: $\int_{1}^{M} f(x) \, dx$ and $\int_{M}^{b} f(x) \, dx$. $\int_{1}^{b} f(x) \, dx = \int_{1}^{M} f(x) \, dx + \int_{M}^{b} f(x) \, dx$
#### STEP 8
The first part, $\int_{1}^{M} f(x) \, dx$, is a finite integral because $f$ is continuous on $[1, M]$.
#### STEP 9
For the second part, $\int_{M}^{b} f(x) \, dx$, since $f(x)$ is decreasing and $f(x) < \epsilon$ for $x > M$, we have: $\int_{M}^{b} f(x) \, dx < \int_{M}^{b} \epsilon \, dx = \epsilon (b - M)$
#### STEP 10
As $b \to \infty$, $\epsilon (b - M) \to \infty$ if $\epsilon$ is not zero. However, since $f(x)$ approaches 0, the integral $\int_{M}^{b} f(x) \, dx$ does not grow unboundedly.
#### STEP 11
To rigorously show convergence, we use the comparison test. Since $f(x)$ is positive, decreasing, and $\lim_{x \to \infty} f(x) = 0$, we can compare $f(x)$ to a known convergent integral.
#### STEP 12
Consider the integral $\int_{1}^{\infty} \frac{1}{x^p} \, dx$ for $p > 1$. This integral is known to converge.
#### STEP 13
Since $f(x)$ is decreasing and $\lim_{x \to \infty} f(x) = 0$, there exists some $x_0 \geq 1$ such that for all $x \geq x_0$, $f(x) \leq \frac{1}{x^p}$ for some $p > 1$.
#### STEP 14
Thus, for $x \geq x_0$, $\int_{x_0}^{\infty} f(x) \, dx \leq \int_{x_0}^{\infty} \frac{1}{x^p} \, dx$
#### STEP 15
Since $\int_{x_0}^{\infty} \frac{1}{x^p} \, dx$ converges for $p > 1$, by the comparison test, $\int_{x_0}^{\infty} f(x) \, dx$ also converges.
##### SOLUTION
Therefore, the integral $\int_{1}^{\infty} f(x) \, dx$ converges because the finite part $\int_{1}^{x_0} f(x) \, dx$ is finite and the remaining part $\int_{x_0}^{\infty} f(x) \, dx$ converges. The statement is True. |
# How do you sketch the general shape of f(x)=-x^3+x^2+1 using end behavior?
Nov 10, 2016
The end behavior of the function is:
As $\rightarrow + \infty$, $f \left(x\right) \rightarrow - \infty$
As $\rightarrow - \infty$, $f \left(x\right) \rightarrow + \infty$
#### Explanation:
If the degree is even, the ends of the graph point in the same direction. If the leading coefficient of an even degree is positive, the ends point up. If the leading coefficient is negative, the ends point down.
As $x \rightarrow + \infty$, $f \left(x\right) \rightarrow + \infty$
As $x \rightarrow - \infty$, $f \left(x\right) \rightarrow + \infty$
As $x \rightarrow + \infty$, $f \left(x\right) \rightarrow - \infty$
As $x \rightarrow - \infty$, $f \left(x\right) \rightarrow - \infty$
If the degree is odd, the ends of the graph point in opposite directions. If the leading coefficient of an odd degree is positive, the ends of the graph point down on the left and up on the right. If the leading coefficient is negative, the ends of the graph point up on the left and down on the right.
As $x \rightarrow + \infty$, $f \left(x\right) \rightarrow + \infty$
As $x \rightarrow - \infty$, $f \left(x\right) \rightarrow - \infty$
As $x \rightarrow + \infty$, $f \left(x\right) \rightarrow - \infty$
As $\rightarrow - \infty$, $f \left(x\right) \rightarrow + \infty$
In the case of $f \left(x\right) = \textcolor{b l u e}{- 1} {x}^{\textcolor{red}{3}} + {x}^{2} + 1$, the degree is $\textcolor{red}{3}$, an odd number, and the leading coefficient is $\textcolor{b l u e}{- 1}$.
So, the end behavior of the function is:
As $\rightarrow + \infty$, $f \left(x\right) \rightarrow - \infty$
As $\rightarrow - \infty$, $f \left(x\right) \rightarrow + \infty$
The graph of the function looks like:
graph{-x^3+x^2+1 [-10, 10, -5, 5]} |
# How do you solve Log(x-9) = 3-Log(100x) ?
Jul 3, 2018
$\textcolor{b l u e}{x = 10}$
#### Explanation:
$\log \left(x - 9\right) = 3 - \log \left(100 x\right)$
By the laws of logarithms:
$\log \left(a b\right) = \log \left(a\right) + \log \left(b\right) \textcolor{w h i t e}{888} \left[1\right]$
$\log \left(100 x\right) = \log \left(100\right) + \log \left(x\right)$
Assuming these are base 10 logarithms:
$\log \left(100\right) + \log \left(x\right) = 2 + \log \left(x\right)$
We now have:
$\log \left(x - 9\right) = 3 - 2 - \log \left(x\right)$
$\log \left(x - 9\right) = 1 - \log \left(x\right)$
Using $\left[1\right]$
$\log \left(x - 9\right) + \log \left(x\right) = 1$
$\log \left(x \left(x - 9\right)\right) = 1$
$\log \left({x}^{2} - 9 x\right) = 1$
${10}^{\log \left({x}^{2} - 9 x\right)} = {10}^{1}$
${x}^{2} - 9 x = 10$
${x}^{2} - 9 x - 10 = 0$
Factor:
$\left(x + 1\right) \left(x - 10\right) = 0 \implies x = - 1 \mathmr{and} x = 10$
Checking solutions.
$x = - 1$
$\log \left(\left(- 1\right) - 9\right) = 3 - \log \left(100 \left(- 1\right)\right)$
$\log \left(- 10\right) = 3 - \log \left(- 100\right)$
Logarithms are only defined for real numbers if for:
$\log \left(x\right)$
$x > 0$
Therefore $- 1$ is not a solution.
For $x = 10$
$\log \left(10 - 9\right) = 3 - \log \left(100 \left(10\right)\right)$
$\log \left(1\right) = 3 - \log \left(1000\right)$
$0 = 3 - 3$
$0 = 0$
So $x = 10$ is the only solution. |
Informative line
Different Tests On Series And P Series
Learn rules of convergence & divergence test series, integral test for convergence. Practice convergence and divergence of 'P' series.
Theorem on Convergence of a Series, Test of Divergence
• If the series $$\sum\limits_{n=1} ^\infty\; a_n$$ converges then $$\lim\limits _{n\to \infty}\; a_n = 0$$
• This means that the terms of a converging series must be very close to 0 for very large values of n.
• This will not mean that if $$\lim\limits _{n\to \infty }\; a_n = 0$$ then series converges. If $$\lim\limits _{n\to \infty }\; a_n = 0$$ then the series may or may not converge.
• If $$\lim\limits _{n\to \infty }\; a_n$$ does not exist or if $$\lim\limits _{n\to \infty }\; a_n \neq 0$$ then series $$\sum\limits_{n=1}^\infty \; a_n$$ is divergent.
e.g. Consider the series
$$\sum \limits _{n=1}^\infty\dfrac{n-2}{4n + 5} ,$$ this series is divergent as
$$\lim\limits _{n\to \infty} \; a_n = \lim\limits _{n\to \infty }\; \dfrac{n-2}{4n +5} = \lim\limits _{n\to \infty}\; \dfrac{1-\dfrac{2}{n}}{4+\dfrac{5}{n }}\;\;=\dfrac{1}{4}\; \neq\; 0$$
So if $$\lim\limits _{n\to \infty}\; a_n \neq 0$$ we say that the series $$\sum\limits ^\infty_{n=1}\; a_n$$ is divergent where as if $$\lim\limits _{n\to \infty }\; a_n = 0$$ nothing can be said about convergence or divergence of the series.
Which of the following series is divergent ?
A $$\sum\limits_{n=1}^\infty \;\; \left(\dfrac{1}{2}\right)^n$$
B $$\sum\limits_{n=1}^\infty \;\; 2 × \dfrac{3^n}{4^{n-1}}$$
C $$\sum\limits_{n=1}^\infty \;\; ln \left(\dfrac{3n^2+1}{4n^2+3}\right)$$
D $$\sum\limits_{n=1}^\infty \;\; \left(\dfrac{5}{7}\right)^n$$
×
If $$\lim\limits _{n\to\infty}\; a_n \neq 0$$ then series $$\sum\limits _1^\infty\;a_n$$ will be divergent and geometric series is convergent if |r| < 1
Consider the option
$$(a)\;\;\;\;\;\sum\limits _{n=1}^\infty\; \left(\dfrac{1}{2}\right)^n\; = \dfrac{1}{2} +\dfrac{1}{2^2} +\dfrac{1}{2^3}+ .....$$
This is a geometric series with $$a= \dfrac{1}{2}, \; r= \dfrac{1}{2}$$
$$\therefore$$ it is convergent
$$(b)\;\;\;\;\sum \limits _{n=1}^\infty\; 2 × \dfrac{3^n}{4 ^ {n-1}} \;\; = \sum \limits ^\infty_{n=1}\; 8 × \dfrac{3^n}{4^n}\; = 8 \; \sum \limits _{n=1}^\infty\; \left(\dfrac{3}{4}\right)^n$$
This again is a geometric series with $$a = \dfrac{3}{4}$$
$$\therefore$$ It is convergent
$$(c)\;\;\;\;\; \sum\limits^\infty_{n=1}\; ln \left(\dfrac{3n^2 +1}{4n^2+3}\right)$$
$$\lim\limits _{n\to \infty}\; a_n = \lim\limits _{n\to \infty}\; ln \left(\dfrac{3n^2 +1}{4n^2 +3}\right) = \lim\limits _{n\to \infty}\; ln \dfrac{3+\dfrac{1}{n^2}}{4+\dfrac{3}{n^2}}$$
$$= ln \dfrac{3}{4} \neq 0$$
$$\therefore$$ Series is divergent
$$(d)\;\;\;\;\; \sum\limits_{n=1}^\infty\; \left(\dfrac{5}{7}\right)^n = \dfrac{5}{7}+ \left(\dfrac{5}{7}\right)^2+ \left(\dfrac{5}{7}\right)^3+ .....$$
This is a geometric series with $$a= \dfrac{5}{7}, \;\;r=\dfrac{5}{7}$$
$$\therefore$$ It is convergent
Which of the following series is divergent ?
A
$$\sum\limits_{n=1}^\infty \;\; \left(\dfrac{1}{2}\right)^n$$
.
B
$$\sum\limits_{n=1}^\infty \;\; 2 × \dfrac{3^n}{4^{n-1}}$$
C
$$\sum\limits_{n=1}^\infty \;\; ln \left(\dfrac{3n^2+1}{4n^2+3}\right)$$
D
$$\sum\limits_{n=1}^\infty \;\; \left(\dfrac{5}{7}\right)^n$$
Option C is Correct
Rules of Convergence
If $$\sum\ a_n$$ and $$\sum \; b_n$$ are convergent series then the following series are also convergent
$$(1)\;\;\;\; \sum \; ca_n$$ where c is a constant
$$(2)\;\;\;\; \sum \;(a_n +b_n)$$
$$(3)\;\;\;\; \sum \;(a_n-b_n)$$ and we have
$$\sum \limits ^\infty_{n=1}\;ca_n = c\sum\limits^\infty_{n=1}\; a_n$$
$$\sum \limits ^\infty_{n=1} (a_n +b_n) = \sum \limits ^\infty_{n=1}\; a_n + \sum \limits ^\infty_{n=1} b_n$$
$$\sum \limits ^\infty_{n=1} (a_n -b_n) = \sum \limits ^\infty_{n=1}\; a_n - \sum \limits ^\infty_{n=1} b_n$$
The above properties of convergent series follow directly for corresponding limits laws, for sequences.
Find the sum of the series $$\sum\limits^\infty_{n=2} \left(\dfrac{5}{n^2-1}+ \dfrac{1}{2^n}\right)$$
A $$\dfrac{5}{2}$$
B $$\dfrac{17}{4}$$
C $$\dfrac{13}{4}$$
D $$\dfrac{17}{6}$$
×
$$\sum\limits_{n=1}^\infty\; (a_n +b_n) = \sum\limits_{n=1}^\infty\; a_n + \sum\limits_{n=1}^\infty b_n$$
If $$\sum\;a_n$$ and $$\sum\;b_n$$ are both convergent
Consider the given series
$$\sum\limits_{n = 2}^\infty \left(\dfrac{5}{n^2 -1} +\dfrac{1}{2^n}\right)= \sum\limits_{n = 2}^\infty \dfrac {5}{n^2 -1 } +\sum\limits_{n = 2}^\infty \dfrac{1}{2^n}$$
Consider the first series
$$\sum\limits^\infty_{n = 2} \; \dfrac{5}{n^2-1}\;\;= 5 \sum\limits^\infty_{n=2} \; \dfrac{1}{n^2 -1}$$
$$a_n = \dfrac{1}{n^2-1} = \dfrac{1}{2} \left[\dfrac{1}{n-1}-\dfrac{1}{n+1}\right]$$
$$\therefore \;\; \sum\limits _{n=2}^\infty\; \dfrac{1}{n^2-1}$$
$$=\dfrac{1}{2} \left[\left(\dfrac{1}{1}-\dfrac{1}{3}\right) +\left(\dfrac{1}{2}-\dfrac{1}{4}\right) +\left(\dfrac{1}{3}-\dfrac{1}{5}\right)+...+...\left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)+...\right]$$
$$\dfrac{1}{2} \underbrace {\left[1 + \dfrac{1}{2} - \dfrac{1}{n} - \dfrac{1}{n+1}\right]}_\text{All other terms cancel out}\; _\text{where}\;\;n\to \infty$$
$$= \dfrac{1}{2} \left(\dfrac{3}{2}\right)= \dfrac{3}{4}$$
$$\therefore\; \sum \limits ^\infty_{n=2} \;\;\dfrac{5}{n^2-1} = 5 × \dfrac{3}{4} = \dfrac{15}{4}$$
Consider the second series
$$\sum\limits _{n = 2}^\infty\; \dfrac{1}{2^n} = \dfrac{1}{2^2} + \dfrac{1}{2^3} + \dfrac{1}{2^4} + .....$$
This is a geometric series with $$a = \dfrac{1}{4}\; , r =\dfrac{1}{2}$$
$$\therefore$$ $$S = \dfrac{a}{1-r} = \dfrac{\dfrac{1}{4}}{1-\dfrac{1}{2}}\; = \dfrac{1}{2}$$
$$\therefore\; \sum\limits ^\infty_{n=2} \;\; \dfrac{5}{n^2 - 1} + \dfrac{1}{2^n}\; = \dfrac{15}{4} \; \; + \;\;\dfrac{1}{2}\;\;=\;\;\dfrac{17}{4}$$
Find the sum of the series $$\sum\limits^\infty_{n=2} \left(\dfrac{5}{n^2-1}+ \dfrac{1}{2^n}\right)$$
A
$$\dfrac{5}{2}$$
.
B
$$\dfrac{17}{4}$$
C
$$\dfrac{13}{4}$$
D
$$\dfrac{17}{6}$$
Option B is Correct
Estimating the Sum of Series
• After confirming that a series $$\sum\; a_n$$ is convergent by integral test, we are interested in finding an approximation to the sum s of of the series $$\sum\; a_n$$.
• Any partial sum Sn is an approximation to S, but how good is this approximation , to find this we find the remainder.
$$R_n =S-S_n =a_{n+2}+ a_{n+3}+......$$
Rn = error made when Sn is used as an approximate to the sum S.
• Consider the following figure
$$R_n = a_{n+1} + a_{n+2} + a_{n+3} + ...... \leq\; \int \limits_n^\infty f (x)\; dx$$
(Note that integral is shaded area + some extra area )
Also $$R_n = a_{n+1} + a_{n+2}+ a_{n+3}.....\geq \; \int\limits_{n+1}^\infty\; f (x)\;dx$$
$$\therefore\;\;\; \int\limits_{n+1}^\infty\; f(x)\; dx \; \leq \; R_n \leq \int\limits_n^\infty\; f (x) \; dx$$
• Suppose $$f(k) = a_k$$ where f is a continuous, positive decreasing function for $$x\geq n$$ and $$\sum\;a_n$$ is convergent if $$R_n = S-S_n$$ then
$$\int\limits_{n+1}^\infty\; f(x) dx \; \leq\; R_n \leq \; \int \limits_n^\infty f (x) \; dx$$
For the series $$\sum\limits _{n=1}^\infty\; \dfrac{1}{n^2}$$ find a value of n such that Sn ( the partial sum to n terms ) is such that Rn = S–Sn is less than 0.001 ?
A n > 150
B n > 420
C n > 575
D n > 1000
×
If $$R_n = S -S_n$$ then
$$\int\limits_{n+1}^\infty\; f(x) dx \leq R_n \leq \; \int\limits_n^\infty \; f (x) \; dx$$
In this case $$f (x) = \dfrac{1}{x^2}$$
$$\therefore\;\; R_n \leq\; \int\limits^\infty_n \; \dfrac{1}{x^2}\; dx = \dfrac{-1}{x}\Big]^\infty_n\; =- \left[0- \dfrac{1}{n}\right] = \dfrac{1}{n}$$
Now we desire
$$R_n < 0.001 \;\;$$
$$\Rightarrow\;\; \dfrac{1}{n} < 0.001$$
$$\Rightarrow\;\; n > \dfrac{1}{0.001}$$
$$\Rightarrow\;\; n > 1000$$
For the series $$\sum\limits _{n=1}^\infty\; \dfrac{1}{n^2}$$ find a value of n such that Sn ( the partial sum to n terms ) is such that Rn = S–Sn is less than 0.001 ?
A
n > 150
.
B
n > 420
C
n > 575
D
n > 1000
Option D is Correct
Comparison Tests
• In comparison test we compare a given series with a series that is known to be convergent or divergent.
• Suppose $$\sum\;a_n$$ and $$\sum\; b_n$$ are series with positive terms.
• If $$\sum\; b_n$$ is convergent and $$a_n\leq b_n$$ for all n, then $$\sum\;a_n$$ is also convergent.
• If $$\sum\; b_n$$ is divergent and $$a_n\geq b_n$$ for all n, then $$\sum\;a_n$$ is also divergent.
• The above statement is called comparison test
e.g. $$\sum\limits_{n=1}^\infty\; \dfrac{1}{3^n +1}$$ is convergent because $$\sum\limits _{n=1}^\infty \; \dfrac{1}{3^n}\;\;=\dfrac{1}{2}$$ is convergent and $$\dfrac{1}{3^n+1}\; <\; \dfrac{1}{3^n } \forall\; n.$$
Using comparison test, what can be said about the series $$\sum\limits_{n=1}^\infty \; \left(\dfrac{8^n}{2+9^n}\right)$$ .
A The series is convergent
B The series is divergent
C $$S= \sum \limits_{n=1}^\infty\; \dfrac{8^n}{2+9^n}\; > 8$$
D The series is a geometric series
×
Suppose $$\sum\;a_n$$ and $$\sum\; b_n$$ are series with positive terms.
If $$\sum\; b_n$$ is convergent and $$a_n\leq b_n$$ for all n, then $$\sum\;a_n$$ is also convergent.
If $$\sum\; b_n$$ is divergent and $$a_n\geq b_n$$ for all n, then $$\sum\;a_n$$ is also divergent.
In this case $$a_n = \dfrac{8^n}{2 +9 ^n}$$, consider for all n.
$$b_n = \dfrac{8^n}{9^n} = \left(\dfrac{8}{9}\right)^n$$
clearly bn > an for all n.
$$\sum \limits_{n =1}^\infty \left(\dfrac{8}{9}\right)^n = \dfrac{8}{9} +\left(\dfrac{8}{9}\right)^2+.......$$
This is a geometric series with $$S= \dfrac{8/9}{1-8/9} = 8$$
$$\left(r = 8/9, \;\;a=8/9,\;\;S= a/1-r\right)$$
$$\therefore\;\; \sum \; b_n$$ is convergent
By comparison test $$\sum\; a_n$$ is also convergent and $$S= \sum \limits _{n=1}^\infty\; a_n< 8$$
Using comparison test, what can be said about the series $$\sum\limits_{n=1}^\infty \; \left(\dfrac{8^n}{2+9^n}\right)$$ .
A
The series is convergent
.
B
The series is divergent
C
$$S= \sum \limits_{n=1}^\infty\; \dfrac{8^n}{2+9^n}\; > 8$$
D
The series is a geometric series
Option A is Correct
The Integral Test for Convergence of Series
• In general the partial sum of most of series is very difficult to find so, if we do not have an expression of Sn we can not test for convergence of series.
• Integral test is a test that enables us to determine whether series is convergent or divergent without explicitly founding the sum.
• Consider the series
$$S= \dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+......$$
There is no formula for Sn . Look at Sn for very various values of n.
n $$S_n = \sum \limits_{i = 1}^n\; 1/i^2$$ 5 1.4636 10 1.5498 100 1.6350 500 1.6429 1000 1.6439 5000 1.6447
If looks as if the series is convergent and converges to a value near 1.64.
Now consider the curve $$y =\dfrac{1}{x^2}$$
The series $$\dfrac{1}{1^2}+\dfrac{1}{2^2} +......=$$ sum of all shaded areas.
Now consider $$I= \int\limits_1^\infty \; \dfrac{1}{x^2}\; dx = \dfrac{-1}{x}\Big]^\infty_1 = 1$$
This is area under the curve $$y = \dfrac{1}{x^2}$$ from 1 to $$\infty$$
$$\therefore$$ Series value $$<1+ \int\limits_1^\infty\;\dfrac{1}{x^2}\; dx$$
$$<1+1<2$$
$$\therefore$$ Series is convergent
This leads to the integral test
• Suppose f is a continuous, positive, decreasing function on $$[1,\infty]$$ and let $$a_n = f (x)$$. Then the series $$\sum\limits_{n=1}^\infty\; a_n$$ is convergent if and only if $$\int\limits_1^\infty\; f (x) \; dx$$ is convergent.
If $$\int\limits_1^\infty\; f (x) \; dx$$ is convergent then $$\sum\limits _{n= 1} ^ \infty\; a_n$$ is convergent.
If $$\int\limits_1^\infty\; f (x) \; dx$$ is divergent then $$\sum\limits _{n= 1} ^ \infty\; a_n$$ is divergent.
Using Integral test, find which of the following series is convergent.
A $$\sum\limits_{n=1}^\infty\; \dfrac{n^2}{n^3 +1}$$
B $$\sum\limits_{n=1}^\infty\; \dfrac{1}{\sqrt{n+4}}$$
C $$\sum\limits_{n=1}^\infty\; n^2e^{-n^3}$$
D $$\sum\limits_{n=1}^\infty\; \dfrac{n}{n^2+1}$$
×
If $$\int\limits_1^\infty\; f (x) \; dx$$ is convergent then $$\sum\limits _{n= 1} ^ \infty\; a_n$$ is convergent.
If $$\int\limits_1^\infty\; f (x) \; dx$$ is divergent then $$\sum\limits _{n= 1} ^ \infty\; a_n$$ is divergent.
Consider the options
$$(a)\;\;\;\sum\limits_{n=1}^\infty\;\dfrac{n^2}{n^3+1}\; \rightarrow\; a_n = \dfrac{n^2}{n^3+1}\; \Rightarrow\; f (x) = \dfrac{x^2}{x^3+1}$$
$$I = \int\limits_1^\infty\; \dfrac{x^2}{x^3 +1}\; dx \; \rightarrow\; put \;x^ 3 +1=t$$
$$\Rightarrow \; 3x^2 \; dx\;=\;dt$$
when $$x=\infty\rightarrow\; t = \infty\;\;\\\;x=1 \rightarrow\; t=2$$
$$\therefore\; I = \dfrac{1}{3} \int \limits_2^\infty \; \dfrac{1}{t} \; dt= \dfrac{1}{3} \; lnt \Big ]^\infty_2\; =\infty$$
$$\therefore \; I$$ is divergent
$$\therefore$$ Series is also divergent
$$(b)\;\;\;\sum\limits_{n=1}^\infty \dfrac{1}{\sqrt{n+4}}\; \rightarrow a_n= \dfrac{1}{\sqrt{n+4}}$$
$$\Rightarrow\; f (x) = \dfrac{1}{\sqrt {x+4}}$$
$$I = \int\limits_1^\infty \; \dfrac{1}{\sqrt{x+4}} \; dx = \dfrac{(x+4)^{1/2}}{1/2}\Bigg ]_1^\infty = \infty$$
$$\therefore\; I$$ is divergent
$$\Rightarrow$$ Series is divergent
$$(c)\;\;\;\;\sum\limits_{n=1}^\infty \; n^2 e ^{-n^3}\;\Rightarrow \; a_n = n^2 e ^{-n^3}$$
$$\Rightarrow f (x) = x^2 e ^{-x^3}$$
$$I = \int\limits_1^\infty x^2 e ^{-x^3}\; dx \; \rightarrow \; put\; x^3 =t$$
$$\Rightarrow 3x^2 \; dx = dt$$
Where $$x = \infty, \; t=\infty\;\;\; x=1,\; t= 1$$
$$\therefore\;\; I = \dfrac{1}{3}\; \int \limits _1^\infty\; e^{-t} \; dt = \dfrac{1}{3} \; (-e^{-t})\Big]^\infty_1$$
$$= \dfrac{-1}{3} [0-e^{-1}] = \dfrac{1}{3e}$$
$$\therefore\; I$$ is convergent
$$\Rightarrow$$ Series is convergent
$$(d)\;\;\;\; \sum\limits_{n=1} ^\infty \;\; \dfrac{n}{n^2 + 1}\;\; \rightarrow\; a_n = \dfrac{n}{n^2 +1}= f(n)$$
$$\Rightarrow\; f(x) = \dfrac{x}{x^2 +1}$$
$$I = \int\limits_1^\infty\; \dfrac{x}{x^2 +1}\; dx\; \rightarrow \; put\; x^2 +1 = t$$
$$\Rightarrow\; 2x \; dx = dt$$
Where $$x = \infty, \;\; t = \infty\;,\;\;\;x=1, \; t=2$$
$$\therefore\;\;I = \dfrac{1}{2}\; \int \limits _2^\infty\; \dfrac{dt}{t} = \dfrac{1}{2}\; lnt \; \Big ]^\infty_2\;=\infty$$
$$\therefore\;\; I$$ is divergent
$$\Rightarrow$$ Series is divergent
$$\therefore$$ Correct option is 'C'
Using Integral test, find which of the following series is convergent.
A
$$\sum\limits_{n=1}^\infty\; \dfrac{n^2}{n^3 +1}$$
.
B
$$\sum\limits_{n=1}^\infty\; \dfrac{1}{\sqrt{n+4}}$$
C
$$\sum\limits_{n=1}^\infty\; n^2e^{-n^3}$$
D
$$\sum\limits_{n=1}^\infty\; \dfrac{n}{n^2+1}$$
Option C is Correct
Convergence and Divergence of 'P' Series
• The series $$\sum\limits_{n=1}^\infty\; \dfrac{1}{n^p}$$ is called the p series.
• If p < 0 then $$\lim\limits_{n\to\infty}\; \dfrac{1}{n^p} = \infty$$, so the p series diverges by test of divergence.
• If p = 0 then $$\lim\limits_{n\to \infty}\; \dfrac{1}{n^p} 1 \neq 0$$, so the p series diverge by test of divergence.
• If p > 0 then $$f(x) = \dfrac{1}{x^p}$$ is continuous, positive and decreasing on $$[1, \infty]$$
$$\int\limits_1^\infty\; \dfrac{1}{x^p} = \dfrac{x^{-p+1}}{-p+1}\Bigg]_1^\infty\; =\infty\;\; if \; 0 <p<1$$
$$\therefore$$ By integral test p series converges for p > 1 and diverges if $$p\leq 1$$
The p series $$\sum\limits_{n=1}^\infty\; \dfrac{1}{n^p}$$ is convergent if p > 1 and divergent if $$p\;\leq\;1$$.
Euler found the sum of p series with p = 4 $$\sum\limits_{n= 1}^\infty\; \dfrac{1}{n^4}\; =\dfrac{\pi^4}{90}$$ Assuming this result, find the value of $$\sum\limits _{n=1}^\infty \left(\dfrac{3}{n}\right)^4$$
A $$\dfrac{8 \pi ^4}{5}$$
B $$\dfrac{9\pi ^4}{10}$$
C $$\dfrac{10\pi ^4}{9}$$
D $$\dfrac{7\pi ^4}{8}$$
×
The p series $$\sum\limits_{n=1}^\infty\; \dfrac{1}{n^p}$$ is convergent if p > 1 and divergent if $$p\;\leq\;1$$.
Given that $$\sum\limits_{n=1}^\infty\; \dfrac{1}{n^4} = \dfrac{\pi^4}{90}$$
$$\therefore\;\;\sum\limits_{n=1}^\infty\; \left(\dfrac{3}{n}\right)^4 = 81 \; \sum\limits_{n=1}^\infty\; \dfrac{1}{n^4}\;=81 × \dfrac{\pi^4}{90}\;=\; \dfrac{9 \pi^4}{10}$$
Euler found the sum of p series with p = 4 $$\sum\limits_{n= 1}^\infty\; \dfrac{1}{n^4}\; =\dfrac{\pi^4}{90}$$ Assuming this result, find the value of $$\sum\limits _{n=1}^\infty \left(\dfrac{3}{n}\right)^4$$
A
$$\dfrac{8 \pi ^4}{5}$$
.
B
$$\dfrac{9\pi ^4}{10}$$
C
$$\dfrac{10\pi ^4}{9}$$
D
$$\dfrac{7\pi ^4}{8}$$
Option B is Correct |
>
Published on January 26th, 2013 In category Education | Maths
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# How to solve problems on ratio, proportion and variation
Ratio of the two terms is denoted by a:b and is measured by a/b. The a in the numerator is called as antecedent and the denominator b is called as consequent.
The ratio basically means how many multiples or parts is a of b.
Proportion: The numbers are in proportion when two ratios of the two terms are equal to ratio of other two terms, if a:b=c:d then a,b,c and d are in proportions
Variations: The 2 quantities are said to be in direct or inverse proportion (variation) when a change in one quantity directly effects the other one. There can be direct effect that is known as direct proportion ( Like increase in “a” causes increase in “b”) and Inverse proportion when ( increase in “a” causes decrease in b).
Have understood what is meant by Ratio, Proportion and Variations: we will consider some important points and rules related to these subjects
Ratios: Some important points are as follows:
• Both the antecedent and consequent should be in the same unit of measure
• If the units are given in different measures then we need to combine units by compounding. e.g. ( m:n) and (a:b) then compounded ratio is (mxa)/(nxb)
• Ratio would remain unchanged when a constant number is multiplied or divided by antecedent or consequent
Some imp. Rules for Ratios
Rules of Invertendo, Atlterendo, Componendo, Dividendo and Compendo and Dividendo in Ratios
Important Rules for Proportions:
1. If a:b=c:d then ad = bc
2. If in case we have a:b=b:c then this is a case of continued proportions and we can have b2 = ac, thus we can say b is mean of “a” and “c” and can be called as mean proportional
The concept of mean proportional and continued proportions can be useful finding one of the missing ratios in a set of 4 ratios for e.g.
Important rules on Variations are as follows:
If you liked my contribution, please donate, a little that will be generated will all be donated for noble cause. [paypal-donation] |
Arithmetic Sequences and Series
Home > Lessons > Arithmetic Sequences and Series Updated June 30th, 2019
Introduction
Sequences of numbers that follow a pattern of adding a fixed number from one term to the next are called arithmetic sequences. The following sequences are arithmetic sequences:
Sequence A: 5 , 8 , 11 , 14 , 17 , ...
Sequence B: 26 , 31 , 36 , 41 , 46 , ...
Sequence C: 20 , 18 , 16 , 14 , 12 , ...
For sequence A, if we add 3 to the first number we will get the second number. This works for any pair of consecutive numbers. The second number plus 3 is the third number: 8 + 3 = 11, and so on.
For sequence B, if we add 5 to the first number we will get the second number. This also works for any pair of consecutive numbers. The third number plus 5 is the fourth number: 36 + 5 = 41, which will work throughout the entire sequence.
Sequence C is a little different because we need to add -2 to the first number to get the second number. This too works for any pair of consecutive numbers. The fourth number plus -2 is the fifth number: 14 + (-2) = 12.
Because these sequences behave according to this simple rule of addiing a constant number to one term to get to another, they are called arithmetic sequences. So that we can examine these sequences to greater depth, we must know that the fixed numbers that bind each sequence together are called the common differences. Sometimes mathematicians use the letter d when referring to these types of sequences.
Mathematicians also refer to generic sequences using the letter a along with subscripts that correspond to the term numbers as follows:
Generic Sequence: a1, a2, a3, a4, ...
This means that if we refer to the fifth term of a certain sequence, we will label it a5. a17 is the 17th term. This notation is necessary for calculating nth terms, or an, of sequences.
The d-value can be calculated by subtracting any two consecutive terms in an arithmetic sequence.
d = an - an - 1
...where n is any positive integer greater than 1.
In order for us to know how to obtain terms that are far down these lists of numbers, we need to develop a formula that can be used to calculate these terms. If we were to try and find the 20th term, or worse to 2000th term, it would take a long time if we were to simply add a number -- one at a time -- to find our terms.
If a 5-year-old was asked what the 301st number is in the set of counting numbers, we would have to wait for the answer while the 5-year-old counted it out using unnecessary detail. We already know the number is 301 because the set is extremely simple; so, predicting terms is easy. Upon examining arithmetic sequences in greater detail, we will find a formula for each sequence to find terms.
1. Let's examine sequence A so that we can find a formula to express its nth term.
If we match each term with it's corresponding term number, we get:
n 1 2 3 4 5 . . . Term 5 8 11 14 17 . . .
The fixed number, called the common difference (d), is 3; so, the formula will be an = dn + c or an = 3n + c, where c is some number that must be found.
For sequence A above, the rule an = 3n + c would give the values...
3×1 + c = 3 + c
3×2 + c = 6 + c
3×3 + c = 9 + c
3×4 + c = 12 + c
3×5 + c = 15 + c
If we compare these values with the ones in the actual sequence, it should be clear that the value of c is 2. Therefore the formula for the nth term is...
an = 3n + 2.
Now if we were asked to find the 37th term in this sequence, we would calculate for a37 or 3(37) + 2 which is equal to 111 + 2 = 113. So, a37 = 113, or the 37th term is 113. Likewise, the 435th term would be a435 = 3(435) + 2 = 1307.
2. Let's take a look at sequence B.
n 1 2 3 4 5 . . . Term 26 31 36 41 46 . . .
The fixed number, d, is 5. So the formula will be an = dn + c or an = 5n + c .
For the sequence above, the rule an = 5n + c would give the values...
5×1 + c = 5 + c
5×2 + c = 10 + c
5×3 + c = 15 + c
5×4 + c = 20 + c
5×5 + c = 25 + c
If we compare these values with the numbers in the actual sequence, it should be clear that the value of c is 21. Therefore, the formula for the nth term is...
an = 5n + 21.
If we wanted to calculate the 14th term, we would calculate for
a14 = 5(14) + 21 = 70 + 21 = 91. If we needed the 40th term, we would calculate a40 = 5(40) + 21 = 200 + 21 = 221. The general formula is very handy.
3. Now let's do the third and final example....
n 1 2 3 4 5 . . . Term 20 18 16 14 12 . . .
The common difference is -2. So the formula will be -2n + c, where c is a number that must be found.
For sequence C, the rule -2n + c would give the values...
-2×1 + c = -2 + c
-2×2 + c = -4 + c
-2×3 + c = -6 + c
-2×4 + c = -8 + c
-2×5 + c = -10 + c
If we compare these values with the numbers in the actual sequence, it should be clear that the value of c is 22. Therefore, the formula for the nth term is...
an = -2n + 22.
If for some reason we needed the 42nd term, we would calculate for a42 = -2(42) + 22 = -84 + 22 = -62. Similarly, a90 = -2(90) + 22 = -180 + 22 = -158.
ideo: Finding the nth Term of an Arithmetic Sequence
uizmaster: Finding Formula for General Term
uizmaster: Finding the nth Term
It may be necessary to calculate the number of terms in a certain arithmetic sequence. To do so, we would need to know two things.
We would need to know a few terms so that we could calculate the common difference and ultimately the formula for the general term. We would also need to know the last number in the sequence.
Once we know the formula for the general term in a sequence and the last term, the procedure is relatively uncomplicated. Set them equal to each other. Since the formula uses the variable n to calculate terms, we can also use it to determine the term number for any given term.
1. If we again look at sequence A above, let's use the formula that was found to calculate term values, an = 3n + 2. If we knew that 47 was a number in the sequence -- 5, 8, 11, 14, 17, ..., 47 -- we would set the number 47 equal to the formula an = 3n + 2, we would get 47 = 3n + 2. Solving this equation yields n = 15. This means that there are 15 terms in the sequence and that the 15th term, a15, is equal to 47.
2. Let's look at a portion of sequence C. If the sequence went from 20 to -26, we would have: 20, 18, 16, 14, 12, ...,-26. We would use the formula for the general term, an = -2n + 22, and set it equal to the last term, -26. We would get -26 = -2n + 22 and algebra would allow us to arrive at n = 24. This means that there are 24 terms in the sequence and that a24 = -26.
Given our generic arithmetic sequence...
a1, a2, a3, a4, ...
...we can add the terms, called a series, as follows.
a1 + a2 + a3 + a4 + ... + an.
Given the formula for the general term an = dn + c, there exists a formula that can add such a finite list of these numbers. It requires three pieces of information. The formula is...
Sn = ½n(a1 + an)
...where Sn is the sum of the first n numbers, a1 is the first number in the sequence and an is the nth number in the sequence.
If you would like to see a derivation of this arithmetic series sum formula, watch this video.
Usually problems present themselves in either of two ways. Either the first number and the last number of the sequence are known or the first number in the sequence and the number of terms are known.
1. Starting with the easier of the two cases, let's take a portion of sequence A. If we were dealing with 5, 8, 11, 14, 17, ... , 128, then we would know that a1 = 5 and an = 128. If we knew the number of terms in this sequence, we would be able to use the formula. Finding n becomes our next task. Since we know the formula for the general term, an = 3n + 2, we can use it to find the number of terms in this sequence. We set the last term equal to the formula and solve for n. We get 128 = 3n + 2, which means that n = 42 and a42 = 128. Now we can plug the information into the sum formula and get S42 = ½(42)(5 + 128) = (21)(133) = 2793, which must be the sum of the first 42 terms in the sequence.
2. For the more difficult situation, let's take a finite portion of sequence B. If we had 26, 31, 36, 41, 46, ... and knew that there were 50 terms in the sequence, then we have a1 = 26 and n = 50. We would have to develop a formula for the nth term so we could calculate a50, the last term in the sequence. Since we already calculated the formula above, we can use it to calculate a50. It is an = 5n + 21 is the formula so a50 = 5(50) + 21 = 250 + 21 = 271. Now we can plug the numbers into the formula and gain a solution. S50 = ½(50)(26 + 271) = 25(297) = 7425. This means that the sum of the first 50 terms is 7425.
ideo: Arithmetic Sequence: Finding the Sum
uizmaster: Finding the Sum of a Series: Given a1 and an
uizmaster: Finding the Sum of a Series: Given a1 and n
Instructional Videos Watch these instructional videos. Interactive Quizmasters After reading the lesson, try our quizmaster. MATHguide has developed numerous testing and checking programs to solidify skills demonstrated in this lesson. The following quizmasters are available: uizmaster: Finding Formula for General Term uizmaster: Finding the nth Term uizmaster: Finding the Sum of a Series: Given a1 and an uizmaster: Finding the Sum of a Series: Given a1 and n uizmaster: Sum of First N Multiples of M uizmaster: Sum of First N Multiples of M #2
Related Lessons and Quizmasters After reading the lesson, try a related lesson or quizmaster. |
# Tamilnadu Board Class 9 Maths Solutions Chapter 2 Real Numbers Ex 2.6
## Tamilnadu State Board Class 9 Maths Solutions Chapter 2 Real Numbers Ex 2.6
Question 1.
Simplify the following using addition and subtraction properties of surds:
(i) 5√3 + 18√3 – 2√3
(ii) 4$$\sqrt [ 3 ]{ 5 }$$ + 2$$\sqrt [ 2 ]{ 5 }$$ – 3$$\sqrt [ 3 ]{ 5 }$$
(iii) 3$$\sqrt { 75 }$$ + 5$$\sqrt { 48 }$$ – $$\sqrt { 243 }$$
(iv) 5$$\sqrt [ 3 ]{ 40 }$$ + 2$$\sqrt [ 3 ]{ 625 }$$ – 3$$\sqrt [ 3 ]{ 320 }$$
Solution:
(i) 5√3 + 18√3 – 2√3 = (5 + 18 – 2)√3 = 21√3
(ii) 4$$\sqrt [ 3 ]{ 5 }$$ + 2$$\sqrt [ 2 ]{ 5 }$$ – 3$$\sqrt [ 3 ]{ 5 }$$ = (4 + 2 – 3)$$\sqrt [ 3 ]{ 5 }$$ = 3$$\sqrt [ 3 ]{ 5 }$$
(iii) 3$$\sqrt { 75 }$$ + 5$$\sqrt { 48 }$$ – $$\sqrt { 243 }$$
(iv) 5$$\sqrt [ 3 ]{ 40 }$$ + 2$$\sqrt [ 3 ]{ 625 }$$ – 3$$\sqrt [ 3 ]{ 320 }$$
Question 2.
Simplify the following using multiplication and division properties of surds :
(i) √3 x √5 x √2
(ii) √35 ÷ √7
(iii) $$\sqrt[3]{27} \times \sqrt[3]{8} \times \sqrt[3]{125}$$
(iv) (7√a – 5√b)(7√a + 5√b)
(v) $$[\sqrt{\frac{225}{729}}-\sqrt{\frac{25}{144}}] \div \sqrt{\frac{16}{81}}$$
Solution:
Question 3.
If √2 =1.414, √3= 1.732, √5= 2.236, √10 = 3.162, then find the values of the following correct to 3 places of decimals.
(i) √40 – √20
(ii) $$\sqrt { 300 }$$ + $$\sqrt { 90 }$$ – √8
Solution:
Question 4.
Arrange surds in descending order :
(i) $$\sqrt[3]{5}, \sqrt[9]{4}, \sqrt[6]{3}$$
(ii) $$\sqrt[2]{\sqrt[3]{5}}, \sqrt[3]{\sqrt[4]{7}}, \sqrt{\sqrt{3}}$$
Solution:
(i) $$\sqrt[3]{5}, \sqrt[9]{4}, \sqrt[6]{3}$$
5$$\frac { 1 }{ 3 }$$
∴ The order of the surds $$\sqrt[3]{5}, \sqrt[9]{4}, \sqrt[6]{3}$$ are 3, 9, 6.
4$$\frac { 1 }{ 9 }$$
3$$\frac { 1 }{ 6 }$$ l.c.m of 3, 9, 6 is 18
∴ The descending order of $$\sqrt[3]{5}, \sqrt[9]{4}, \sqrt[6]{3}$$ is $$(15625)^{\frac{1}{18}}>(27) \frac{1}{18}>16^{\frac{1}{18}} \text { i.e. } \sqrt[3]{5}>\sqrt[6]{3}>\sqrt[9]{4}$$
(ii) $$\sqrt[2]{\sqrt[3]{5}}, \sqrt[3]{\sqrt[4]{7}}, \sqrt{\sqrt{3}}$$
The order of the surds $$\sqrt[2]{\sqrt[3]{5}}, \sqrt[3]{\sqrt[4]{7}}, \sqrt{\sqrt{3}}$$ are 6, 12, 4
l.c.m of 6, 12, 4 is 12
Question 5.
Can you get a pure surd when you find
(i) the sum of two surds
(ii) the difference of two surds
(iii) the product of two surds
(iv) the quotient of two surds
Justify each answer with an example.
Solution:
Question 6.
Can you get a rational number when you compute
(i) the sum of two surds
(ii) the difference of two surds
(iii) the product of two surds
(iv) the quotient of two surds
Justify each answer with an example.
Solution:
(i) Yes (5 – √3) + (5 + √3) = 10, a rational number
(ii) Yes $$(5+\sqrt[3]{7})-(-6+\sqrt[3]{7})=11$$, a rational number
(iii) Yes (5 + √3 ) (5 – √3 ) = 25 – 3 = 22, a rational number
(iv) Yes $$\frac{5 \sqrt{3}}{\sqrt{3}}=5$$ ,a rational number |
Chapter 7 Probability. Example of a random circumstance. Random Circumstance. What does probability mean?? Goals in this chapter
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1 Homework (due Wed, Oct 27) Chapter 7: #17, 27, 28 Announcements: Midterm exams keys on web. (For a few hours the answer to MC#1 was incorrect on Version A.) No grade disputes now. Will have a chance to do that in writing at end of quarter. Grades are curved at end of quarter, not now. Material gets harder from here on! Practice problems with answers will be posted on website. Chapter 7 Probability Today: 7.1 to 7.3, small part of 7.4 Fi Fri: Finish i h74 7.4, Skip Section 7.6 Mon: Section 7.7 and supplemental material on intuition and probability Random Circumstance A random circumstance is one in which the outcome is unpredictable. Could be unpredictable because: It isn't determined yet or We have incomplete knowledge Example of a random circumstance Sex of an unborn child is unpredictable, so it is a random circumstance. We can talk about the probability that a child will be a boy. Why is it unpredictable? Before conception: It isn't determined yet After conception: We have incomplete knowledge Goals in this chapter Understand what is meant by probability Assign probabilities to possible outcomes of random circumstances. Learn how to use probability wisely What does probability mean?? What does it mean to say: The probability of rain tomorrow is.3. The probability that a coin toss will land heads up is ½. The probability that humans will survive to the year 3000 is.8. Is the word probability interpreted the same way in all of these? 1
2 Two basic interpretations of probability (Summary box, p. 237) Interpretation 1: Relative frequency Used for repeatable circumstances The probability of an outcome is the proportion of time that outcome does or will happen in the long run. Interpretation 2: Personal probability (subjective) Most useful for one-time events The probability of an outcome is the degree to which an individual believes it will happen. Two methods for determining relative frequency probability 1. Make an assumption about the physical world or 2. Observe the relative frequency of an outcome over many repetitions. Repetitions can be: a. Over time,, such as how often a flight is late b. Over individuals, by measuring a representative sample from a larger population and observing the relative frequency of an outcome or category of interest, such as the probability that a randomly selected person has a certain gene. How relative frequency probabilities are determined, Method 1: Make an assumption about the physical world. Examples: Flip a coin, probability it lands heads = ½. We assume the coin is balanced in such a way that t it is equally likely to land on either side. Draw a card from a shuffled, regular deck of cards, probability of getting a heart = ¼ We assume all cards are equally likely to be drawn How relative frequency probabilities are determined, Method 2a (over time): Observe the relative frequency of an outcome over many repetitions (long run relative frequency) Probability that a flight will be on time: According to United Airline s website, the probability that Flight 436 from SNA to Chicago will be on time (within 14 minutes of the stated time) is Based on observing this particular flight over many, many days; it was on time on 90% of those days. The relative frequency on time =.9 How relative frequency probabilities are determined, Method 2b (over individuals): Measure a representative sample and observe the relative frequency of possible outcomes or categories for the sample Probability that an adult female in the US believes in life after death is about.789. [It s.72 for males] Based on a national survey that asked 517 women if they believed in life after death 408 said yes Relative frequency is 408/517 =.789 Note about methods 2a and 2b: Usually these are just estimates of the true probability, based on n repetitions or n people in the sample. So, they have an associated margin of error with them. Example: Probability that an adult female in the US believes in life after death =.789, based on n = 517 women. 1 Margin of error is 517 =
3 Personal Probability Especially useful for one-time only events The personal probability of an outcome is the degree to which an individual believes it will happen. Examples: What is the probability that you will get a B in this class? We can t base the answer on relative frequency! LA Times, 10/8/09, scientists have determined that the probability of the asteroid Apophis hitting the earth in 2036 is 1 in 250,000. In 2004, they thought the probability was.027 that it would hit earth in Expert opinion has been updated. Notes about personal probability Sometimes the individual is an expert, and combines subjective information with data and models, such as in assessing the probability bilit of a magnitude 7 or higher h earthquake in our area in the next 10 years. Sometime there is overlap in these methods, such as determining probability of rain tomorrow uses similar pasts. Clicker questions not for credit! The probability that the winning Daily 3 lottery number tomorrow evening will be 777 is 1/1000. Which interpretation is best? A. Rel. freq. based on physical assumption B. Rel. freq. based on long run over time C. Rel. freq. based on representative sample D. Personal probability Clicker questions not for credit! The probability that the SF Giants will win the World Series this year is.40. Which interpretation is best? A. Rel. freq. based on physical assumption B. Rel. freq. based on long run over time C. Rel. freq. based on representative sample D. Personal probability Clicker questions not for credit! The probability that the plaintiff in a medical malpractice suit will win is.29. (Based on an article in USA Weekend) Which interpretation is best? A. Rel. freq. based on physical assumption B. Rel. freq. based on long run over time C. Rel. freq. based on representative sample D. Personal probability Section 7.3: Probability definitions and relationships We will use 2 examples to illustrate: 1. Daily 3 lottery winning number. Outcome = 3 digit number, from 000 to Choice of 3 parking lots, you always try lot 1, then lot 2, then lot 3. Lot 1 works 30% of the time, you aren t late Lot 2 works 50% of the time, you are late Lot 3 always works, so you park there 20% of the time, and when you do, you are very late! 3
4 Definitions of Sample space and Simple event The sample space S for a random circumstance is the collection of unique, non-overlapping outcomes. A simple event is one outcome in the sample space. Ex 1: S = {000, 001, 002,, 999} Simple event: 659 There are 1000 simple events. Ex 2: S = {Lot 1, Lot 2, Lot 3} Simple event: Lot 2 There are 3 simple events. Definition and notation for an event Definition: An event is any subset of the sample space. (One or more simple events) Notation: A, B, C, etc. Ex 1: A = winning number begins with 00 A = {000, 001, 002, 003,, 009} B = all same digits = {000, 111,, 999} Ex 2: A = late for class = {Lot 2, Lot 3} Probability of Events: Notation and Rules (for all interpretations and methods) Notation: P(A) = probability of the event A Rules: Probabilities are always assigned to simple events such that these 2 rules must hold: 1. 0 P(A) 1 for each simple event A 2. The sum of probabilities of all simple events in the sample space is 1. The probability of any event is the sum of probabilities for the simple events that are part of it. Special Case: Assigning Probabilities to Equally Likely Simple Events P(A) = probability of the event A Remember, Conditions for Valid Probabilities: Each probability is between 0 and 1. The sum of the probabilities over all possible simple events is 1. Equally Likely Simple Events If there are k simple events in the sample space and they are all equally likely, then the probability of the occurrence of each one is 1/k. Example: California Daily 3 Lottery Random Circumstance: A three-digit winning lottery number is selected. Sample Space: {000, 001, 002, 003,..., 997, 998, 999}. There are 1000 simple events. Probabilities for Simple Event: Probability that any specific three-digit number is a winner is 1/1000. Physical assumption: all three-digit numbers are equally likely. Event A = last digit is a 9 = {009, 019,..., 999}. P(A) = 100/1000 = 1/10. Event B = three digits are all the same = {000, 111, 222, 333, 444, 555, 666, 777, 888, 999}. Since event B contains 10 events, P(B) = 10/1000 = 1/100. Example 2: Simple events are not equally likely Simple Event Probability Park in Lot 1.30 Park in Lot 2.50 Park in Lot 3.20 Event A = late for class = {Lot 2, Lot 3} P(A) = =.70 Note that these sum to 1 4
5 Probability in daily language: People often express probabilities as percents, proportions, probabilities: United flight 436 from SNA to Chicago is late 10 percent of fthe time. The proportion of time United flight 436 is late is.1. The probability that United flight 436 from SNA to Chicago will be late is.1. These are all equivalent. RELATIONSHIPS BETWEEN EVENTS Defined for events in the same random circumstance only: Complement of an event Mutually exclusive events = disjoint events Defined for events in the same or different random circumstances: Independent events Conditional events Definition and Rule 1 (apply to events in the same random circumstance): Definition: One event is the complement of another event if: They have no simple events in common, AND They cover all simple events Notation: The complement of A is A C RULE 1: P(A C ) = 1 P(A) Ex 2: Random circumstance = parking on one day A = late for class, A C = on time P(A) =.70, so P(A C ) = 1.70 =.30 Complementary Events, Continued Rule 1: P(A) + P(A C ) = 1 Example: Daily 3 Lottery A = player buying single ticket wins A C = player does not win P(A) = 1/1000 so P(A C ) = 999/1000 Example: On-time flights A = flight you are taking will be on time A C = flight will be late Suppose P(A) =.80, then P(A C ) = 1.80 =.20. Mutually Exclusive Events Two events are mutually exclusive, or equivalently disjoint, if they do not contain any of the same simple events (outcomes). (Applies in same random circumstance.) Example: Daily 3 Lottery A = all three digits are the same (000, 111, etc.) B = the number starts with 13 (130, 131, etc.) The events A and B are mutually exclusive (disjoint), but they are not complementary. (No overlap, but don t cover all possibilities.) Independent and Dependent Events Two events are independent of each other if knowing that one will occur (or has occurred) does not change the probability that the other occurs. Two events are dependent if knowing that one will occur (or has occurred) changes the probability that the other occurs. The definitions can apply either to events within the same random circumstance or to events from two separate random circumstances. 5
6 EXAMPLE OF INDEPENDENT EVENTS Events in the same random circumstance: Daily 3 lottery on the same draw A = first digit is 0 B = last digit is 9 Knowing first digit is 0, P(B) is still 1/10. Events in different random circumstances: Daily 3 lottery on different draws A = today s winning number is 191 B = tomorrow s winning number is 875 Knowing today s # was 191, P(B) is still 1/1000 Mutually exclusive or independent? If two events are mutually exclusive (disjoint), they cannot be independent: If disjoint, then knowing A occurs means P(B) = 0 In independent, knowing A occurs gives no knowledge of P(B) Example of mutually exclusive (disjoint): A = today s winning number is 191, B = today s winning number is 875 Example of independent: A = today s winning number is 191 B = tomorrow s winning number is 875 Conditional Probabilities The conditional probability of the event B, given that the event A will occur or has occurred, is the long-run relative frequency with which event B occurs when circumstances are such that A also occurs; written as P(B A). P(B) = unconditional probability event B occurs. P(B A) = probability of B given A = conditional probability event B occurs given that we know A has occurred or will occur. EXAMPLE OF CONDITIONAL PROBABILITY Random circumstance: Observe one randomly selected child A = child slept in darkness as infant [Use total column.] P(A) = 172/479 =.36 B = child did not develop myopia [Use total row] P(B) = 342/479 =.71 P(B A) = P(no myopia slept in dark) [Use darkness row] = 155/172 =.90 P(B) NOTES ABOUT CONDITIONAL PROBABILITY 1. P(B A) generally does not equal P(B). 2. P(B A) = P(B) only when A and B are independent events 3. In Chapter 6, we were actually testing if two types of events were independent. 4. Conditional probabilities are similar to row and column proportions (percents) in contingency tables. (Myopia example on previous page.) 6
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# Math
This month is all about practicing counting money and telling time! Quiz me at home with the extra change in your pocket,
or before we leave for school on the analog clock!
(Remember: The little hand is the boss and says "SAY MY NAME FIRST!" Then we count by 5's all the way around the clock with the big hand!
We are working hard on our measurement unit in math! We have learned that there are all sorts of ways to measure things:
-inches
-centimeters
-pounds
-kilograms
-cups
-pints
-quarts
-degrees F
We have also learned a super fun way to remember the number of cups, pints, and quarts we use!
This month we will be talking about measurement. What types of things can we use to measure objects? In class, we came up with the following things:
*measuring cup (liquids)
*ruler (solid)
*yard stick (solid)
*measuring tape (solid) When working at home, experiment with different measurement tools.
Help make a cake at home and measure out the ingredients! Remember to bring Miss Snowden a piece to try! :)
Key Vocabulary words(we will continue to work with shapes and fractions as the year draws to an end):
Polygon: a closed figure shape with straight edges
Congruent: two figures that are the same shape and same size
Line of symmetry: a line the cuts a shape in half so that both sides are exactly equal
Face: the flat, 2D part of a 3D shape
Edge: the place where two faces meet
Vertices: corners
We will also be working on identifying 2D and 3D shapes, quiz your student at home about their knowledge of each of the shapes below!
More on top?
No need to stop.
More on the floor?
Go next door,
Get 10 more.
To add (OR SUBTRACT) 2 digit numbers here are our steps:
1- Draw a dotted line down the middle.
2-Start at the ones place and add or subtract from top to bottom.Ask yourself, do I need to borrow? Do I need to carry over a ten?
3-Move to the tens place and add or subtract from top to bottom.
PLEASE CONTINUE TO WORK WITH YOUR CHILD ON THEIR BASIC FACTS: Knowing and memorizing these facts is vital to their success the rest of the year!
Counting up and back is a great strategy to use when figuring out our addition and subtraction facts. To count up, we just have to remember to grab the bigger number and count up!
For example: 8 + 3 = ?
We grab the 8, since it is the bigger number, then count up 3 times: 9, 10, 11
So: 8 + 3 = 11
One way that we can help ourselves memorize our facts faster is through nightly flash cards. We often use these flash cards in our center time with a partner. We are getting really good at it! |
# Real Numbers: Euclid’s Division Lemma
EUCLID’S DIVISION LEMMA :
Euclid was the first Greek Mathematician who gave a new way of thinking the study of geometry. He also made important contributions to the number theory. Euclid’s Lemma is one of them. It is a proven statement which is used to prove other statements.
Let ‘a’ and ‘b’ be any two positive integers. Then, there exist unique integers q and r such that
;
Now, we say ‘a’ as dividend, ‘b’ as divisor, ‘q’ as quotient and ‘r’ as remainder.
Dividend = (divisor $\displaystyle \times$ quotient) + remainder
Example: Let 578 be divided by 16.
36 is as quotient and 2 as remainder.
In this case :
Dividend = 578
Divisor = 16
Quotient = 36
And remainder = 2
Dividend = (quotient $\displaystyle \times$ divisor) + remainder
Hence, 578 = (36 $\displaystyle \times$ 16) + 2
To get the HCF of two positive integers, let ‘c’ and ‘d’, with c > d, follow as :
(i) Applying Euclid’s Lemma,
$\displaystyle c=dq+r;\,0\le r
(ii) If r = 0, d is the HCF of c and d.
If $\displaystyle r\ne 0,$ then applying the division lemma to d and r.
(iii) We can continue the process till the remainder is zero. The divisor at this stage will be the required HCF.
This algorithm works because HCF (c,d) = HCF (d,r)
Where the symbol HCF (c,d) denotes the HCF of c and d etc.
Example: Use Euclid’s algorithm to find HCF of 425 and 40
Solution: Let a = 425 and b = 40
By Euclid’s division lemma; we have
$\displaystyle 425=40\times 10+25$ …. (i)
By using the above theorem, we observe that the common divisors of a = 425 and b = 40 are also the common divisors of b = 40 and r1 = 25 and vice-versa.
Applying Euclid’s division lemma on divisor b = 40 and remainder r1 = 25,
We get $\displaystyle 40=25\times 1+15$ …. (ii)
$\displaystyle b=q_{2}^{r},+{{r}_{2}}$; where q2 = 1 and r2 = 15
Again using the above theorem, we find that, the common divisors of r1 = 25 and r2 = 15 are the common divisors of b = 40 and r1 = 25 and vice-versa. But the common divisors of b = 40 and r1 = 25 are the common divisors of a = 425 and b = 40 and vice-versa.
Applying Euclid’s division lemma on r1 = 25 and r2 = 15, we get
$\displaystyle 25=15\times 1+10$ …. (iii)
$\displaystyle \Rightarrow$ r1 = r2q3 + r3, where q3 = 1 and r3 = 10
Again by using the above theorem, we find that common divisors of r2 = 15 and r3 = 10 are the common divisors of a = 425 and b = 40 and vice-versa.
Now, Using Euclid’s division lemma on r2 = 15 and r3 = 10, we get
$\displaystyle 15=10\times 1+5$ …. (iv)
$\displaystyle \Rightarrow$ $\displaystyle r{{ & }_{2}}={{r}_{3}}\times {{q}_{4}}+{{r}_{4}}$; where q4 = 1 and r4 = 5
Again by using the above theorem, we find that, the common divisors of r2 = 15 and r3 = 10 are the common divisors of a = 425 and b = 40 and vice-versa.
Using Euclid’s division lemma on r3 = 10 and r4 = 5, we get
$\displaystyle 15=5\times 3+0$ …. (v)
Hence, r4 = 5 is a divisor of r3 = 10 and r4 = 5. Also, it is the greatest common divisor (or HCF) of r3 and r4. So, r4 = 5 is the greatest common divisor (or HCF) of a = 425 and b = 40. We also observe that r4 = 5 is the last non-zero remainder in the above process of repeated application of Euclid’s division lemma on the divisor and the remainder in the next step.
The set of equation (i) to (v) is called Euclid’s division algorithm for 425 and 40. The last divisor, or the last but non-zero remainder 4 is the HCF (or GCD) of 425 and 40.
The above process of finding HCF can also be carried out by successive divisions as follows:
(i) Euclid’s division lemma and algorithm are so closely interlinked that it is often called division algorithm.
(ii) Euclid’s Division Algorithm is stated for only positive integers except zero. i.e., $\displaystyle b\ne 0$.
Example 3:
A sweet seller has 840 kaju barfis and 260 badam barfis. He wants to stack them in such a way that each stack has the same number, and they take up the least area of the tray. What is the number of barfis that can be placed in each stack for this purpose?
Solution. The area of the tray that is used up will be the least. For this, we find HCF (840, 260). Then this number will give the maximum number of barfis in each stack and the number of stacks will then be the least.
Now, applying Euclid’s algorithm to find their HCF, we have
840 = 260 $\displaystyle \times$ 3 + 60
260 = 60 $\displaystyle \times$ 4 + 20
60 = 20 $\displaystyle \times$ 3 + 0
Hence, HCF of 840 and 260 is 20.
So, the sweet seller can make stacks of 10 for both kinds of barfi |
Chapter 13 Class 9 Surface Areas and Volumes
Class 9
Important Questions for Exam - Class 9
Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
### Transcript
Question 8 A solid cube of side 12 cm is cut into eight cubes of equal volume. What will be the side of the new cube? Also, find the ratio between their surface areas. Here cube of side 12 cm is divided into 8 cubes of side a cm. Given that Their volumes are equal Volume of big cube of side 12 cm = Volume of 8 cubes of side a cm (Side of big cube)3 = 8 × (Side of small cube) 3 (12)3 = 8 × a 3 (12 × 12 × 12) = 8 × a 3 1/8×12×12×12 = a 3 a 3 = 1/8×12×12×12 a3 = 1/(2 × 2 × 2)×12×12×12 a3 = 6 ×6×6 cm3 a3 = 63 cm3 a = 6 cm ∴ Side of small cube = 6 cm Ratio of their surface areas = (𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑐𝑢𝑏𝑒 𝑜𝑓 𝑠𝑖𝑑𝑒 12 𝑐𝑚)/(𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑐𝑢𝑏𝑒 𝑜𝑓 𝑠𝑖𝑑𝑒 6 𝑐𝑚) = 6(𝑠𝑖𝑑𝑒 𝑜𝑓 𝑏𝑖𝑔 𝑐𝑢𝑏𝑒)2/6(𝑠𝑖𝑑𝑒 𝑜𝑓 𝑠𝑚𝑎𝑙𝑙 𝑐𝑢𝑏𝑒)2 = (6 × 12 × 12)/(6 × 6 × 6) = 4/1 So the ratio of surface area is 4 : 1. |
# In triangle ABC, ∠C=90°. Let BC= a, CA= b, AB= c and let 'p' be the length of the perpendicular from 'C' on AB, prove that: 1. cp = ab 2. 1/p^2=1/a^2+1/b^2 - Geometry
Sum
In triangle ABC, ∠C=90°. Let BC= a, CA= b, AB= c and let 'p' be the length of the perpendicular from 'C' on AB, prove that:
1. cp = ab
2. 1/p^2=1/a^2+1/b^2
#### Solution
1. Area of a triangle = (1/2) x Base x Height
A(ΔABC) = (1/2) x AB x CD
A(ΔABC) = (1/2) x cp .......(i)
Area of right angle triangle ABC = A(ΔABC) = (1/2) x AC x BC
A(ΔABC) = (1/2) x ba ........(ii)
From (i) and (ii)
cp=ba⇒ cp⇒ab ..........(iii)
2. We have,
cp=ab ..........From(iii)
p = ab/c
Square both sides of the equation.
We get, p^2=(a^2b^2)/c^2
1/p^2=c^2/(a^2b^2)" ..................(iv).....[By invertendo]"
In right angled triangle ABC,
AB2 = AC2 + BC2 ................[By Pythagoras’ theorem]
c2 = b2 + a2 ............(v)
c^2/(a^2b^2) = b^2/(a^2b^2) + a^2/(a^2b^2)..........[Dividing throughout by a^2b^2]
c^2/(ab)^2 = 1/a^2 + 1/b^2 .........(iii)
c^2/(cp)^2 = 1/a^2 + 1/b^2 ...........[From (ii) and (iii)]
c^2/(c^2p^2) = 1/a^2 + 1/b^2
1/p^2 = 1/a^2 + 1/b^2
Concept: Right-angled Triangles and Pythagoras Property
Is there an error in this question or solution? |
Notes On Basic Concepts of a Circle - CBSE Class 9 Maths
There are many objects around us that are circular in shape. A circle is defined as the collection of all the points on a plane that are at equal distance from a given fixed point on the plane. This fixed point is called the centre of the circle and the fixed distance is called the radius. A circle divides the plane on which it lies into three parts. They are (i) inside the circle, which is also called the interior of the circle; (ii) the circle and (iii) outside the circle, which is also called the exterior of the circle. The circle and its interior together form the circular region. Circumference The circumference of a circle is the actual length around the circle. It is the length of the edge of the circle traced around the centre of the circle. Radius A line segment joining the centre of a circle with any point on its circumference is called the radius of the circle. Chord A line that joins two points on the circumference of a circle is called a chord. Diameter A chord that passes through the centre of a circle is called the diameter of the circle. A diameter divides a circle into two equal segments, each is called a semicircle. Diameter is the longest chord of a circle. The diameter of a circle is twice the radius. Arc The part of the circumference of a circle between two given points is called an arc.The shorter arc between two given points on the circumference of a circle is called as the minor arc whereas the longer arc is called as the major arc. Segment A chord of a circle divides the circle into two regions called the segments of the circle. The region bounded by the chord and the minor arc intercepted by the chord is called the minor segment. The region bounded by the chord and the major arc intercepted by the chord is called the major segment. Sector The region between two radii of a circle and any of the arcs between them is called a sector. Sector corresponding to the minor arc is called the minor sector. Sector corresponding to the major arc is the major segment.
#### Summary
There are many objects around us that are circular in shape. A circle is defined as the collection of all the points on a plane that are at equal distance from a given fixed point on the plane. This fixed point is called the centre of the circle and the fixed distance is called the radius. A circle divides the plane on which it lies into three parts. They are (i) inside the circle, which is also called the interior of the circle; (ii) the circle and (iii) outside the circle, which is also called the exterior of the circle. The circle and its interior together form the circular region. Circumference The circumference of a circle is the actual length around the circle. It is the length of the edge of the circle traced around the centre of the circle. Radius A line segment joining the centre of a circle with any point on its circumference is called the radius of the circle. Chord A line that joins two points on the circumference of a circle is called a chord. Diameter A chord that passes through the centre of a circle is called the diameter of the circle. A diameter divides a circle into two equal segments, each is called a semicircle. Diameter is the longest chord of a circle. The diameter of a circle is twice the radius. Arc The part of the circumference of a circle between two given points is called an arc.The shorter arc between two given points on the circumference of a circle is called as the minor arc whereas the longer arc is called as the major arc. Segment A chord of a circle divides the circle into two regions called the segments of the circle. The region bounded by the chord and the minor arc intercepted by the chord is called the minor segment. The region bounded by the chord and the major arc intercepted by the chord is called the major segment. Sector The region between two radii of a circle and any of the arcs between them is called a sector. Sector corresponding to the minor arc is called the minor sector. Sector corresponding to the major arc is the major segment.
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# How do you find the domain and range of f(x)= 1/x + 5/(x-3)?
Sep 23, 2015
Domain:
All real numbers $x$ except $0$ and $3$.
In other words, $x \ne 0$ and $x \ne 3$
Range:
All real numbers.
In other words, $- \infty < f \left(x\right) < + \infty$
#### Explanation:
It is traditionally assumed that functions like this, unless specifically mentioned otherwise, are defined for real numbers as argument, having values also among real numbers.
Domain of a real function is a set of values where this function is defined.
The function $f \left(x\right) = \frac{1}{x} + \frac{1}{x - 3}$ is defined for all real values of argument $x$ except those where the denominator of one of its two terms equals to $0$.
This happens only for $x = 0$, in which case $\frac{1}{x}$ becomes undefined, and for $x = 3$, in which case $\frac{1}{x - 3}$ becomes undefined.
Therefore, the domain of this function is:
all real values except $0$ and $3$.
It can be written as
$x \ne 0$ AND $x \ne 3$
Alternatively, it can be written as
$- \infty < x < 0$ OR $0 < x < 3$ OR $3 < x < + \infty$.
One more way:
$\left(- \infty , 0\right) \cup \left(0 , 3\right) \cup \left(3 , + \infty\right)$
Range of a real function is a set of values that this function can take while its argument takes all the values from the domain.
To determine the range, let's try to resolve an equation
$f \left(x\right) = a$
for any value $a$. Every value for which this equation has a solution (a value of $x$ from the domain) belongs to a range.
So, let's try to find all $a$, for which there is a solution of the equation
$\frac{1}{x} + \frac{1}{x - 3} = a$
We assume that $x \ne 0$ and $x \ne 3$ since we have already excluded these values of argument, they cannot be solutions, even if we come to these values for some $a$.
Multiplying the equation by $x \left(x - 3\right)$, we obtain
$x - 3 + x = a x \left(x - 3\right)$
or
$a {x}^{2} + \left(- 3 a - 2\right) x + 3 = 0$
The quadratic equation above has a solution if its discriminant is not negative.
The discriminant of this equation is
${\left(- 3 a - 2\right)}^{2} - 4 \cdot a \cdot 3 = 9 {a}^{2} + 12 a + 4 - 12 a = 9 {a}^{2} + 4$
As we see, the discriminant $9 {a}^{2} + 4$ is always positive for any real $a$. Therefore, any real value can be a value of our function. In other words, the range of our function is all real values.
An interesting exercise would be to graph this function. I suggest to add two graphs, $y = \frac{1}{x}$ and $y = \frac{1}{x - 3}$. The result would look like
graph{1/x + 1/(x-3) [-10, 10, -5, 5]} |
### Converting from One Metric Unit to Another
Skills you need to do this include:
1) memorize the metric prefixes names and symbols
2) determine which of two prefixes represents a larger amount
3) determine the exponential "distance" between two prefixes
4) significant figure rules
5) scientific notation
Here are two typical metric conversion problems:
1) Convert 2.50 μg to picograms.
2) Convert 0.080 cm to km.
The explanation below will focus on the first problem. Try to set up the second on as you read through the first example. The answers are provided below.
There is second type of metic conversion, one that involves converting both the numerator and the denominator. You can go to another tutorial which discusses the second type.
Also:
Special note on a particular conversion that is sometimes not mentioned in class.
It has to do with the fact that 1 mL equals 1 cm3 and 1 L equals 1 dm3.
Just below I discuss how to construct a conversion factor. There is an important point about the numerator and the denominator of the conversion factor. Here it is:
The numerator and the denominator of a conversion factor BOTH describe the same amount.
In essence, a conversion factor is equal to one. This is because the numerator and the denominator both describe exactly the same amount. Here's an example:
5280 feet / 1 mile
Both 5280 feet and 1 mile describe exactly the same distance.
Here's a metric example:
1 kg / 1000 g
1000 g and 1 kg both have exactly the same amount of mass.
So, when you are done making a given conversion, you haven't changed the amount, you've just changed the way it is written. For example, 1 kg is differently written from 1000 g, but they both describe the same amount of stuff.
The key skill in solving these problems is to construct a conversion factor. This conversion factor will make the old unit go away (micrograms and km in the top two examples) and create the new unit (pm and cm) in its place. Along with this change, there will be a change in the value of the number.
Let's focus on the first example: Convert 2.50 μg to picograms
STEP ONE: Write the value (and its unit) from the problem, then in order write: 1) a multiplication sign, 2) a fraction bar, 3) an equals sign, and 4) the unit in the answer. Put a gap between 3 and 4. All that looks like this:
The fraction bar will have the conversion factor. There will be a number and a unit in the numerator and the denominator.
STEP TWO: Write the unit from the problem in the denominator of the conversion factor, like this:
STEP THREE: Write the unit expected in the answer in the numerator of the conversion factor.
STEP FOUR: Examine the two prefixes in the conversion factor. In front of the LARGER one, put a one.
There is a reason for this. I'll get to it in a second.
STEP FIVE: Determine the absolute distance between the two prefixes in the conversion unit. Write it as a positive exponent in front of the other prefix.
Now, multiply and put into proper scientific notation format. Don't forget to write the new unit. Sometimes, the exponential number is in the denominator. You must move it to the numerator and when you do so, remember to change the sign. Also, DO NOT move the unit with it. That unit has been cancelled and is no longer there.
Here are all five steps for the second example, put into one image:
Note that the old unit cancels, since it appears in the numerator and denominator of two parts of a multiplication problem.
Why a one in front of the larger unit? I believe it is easier to visualize how many small parts make up one bigger part, like 1000 m making up one km. Going the other way, visualizing what part a larger unit is of one smaller unit, is possible, but requires more sophistication. For example, how many meters are in one nanometer? The answer is 0.000000001 or 10¯9. You may be able to handle a conversion involving what part a larger unit is of a smaller unit and that is just fine.
The conversion factor I have been discussing above is sometimes called a "unitary rate." Unitary in this case simply means that the conversion factor equals one. Look at the conversion factor in the example, the 106 pg / 1 μm. The numerator and the denominator both describe the same amount, as in 106 pg equals 1 μm. The word unitary is used to identify the fact that both values describe the same amount.
What I have done is describe a system where the unit with the one (1 μm in my example) can be in either the numerator or denominator of the conversion unit. You may have a teacher that forces the one to be only in the denominator. What that means is that you will have to decide if you are going from a large unit to a small unit (making the numerator exponent positive) or going from a small one to a large one (making the numerator unit negative).
1) If you do the conversion correctly, the numerical part and the unit will go in opposite directions. If the unit goes from smaller (mm) to larger (km), then the numerical part goes from larger to smaller. There will never be a correct case where number and unit both go larger or both go smaller.
2) A common mistake is to put the one in front of the SMALLER unit. This results in a wrong answer. Put the one in front of the LARGER unit.
3) Sometimes you see this:
106 pg / μm
In that case, there is an assumed one in front of the unit with no number. In other words, 106 pg / 1 μm means the same thing as 106 pg / μm. In closing: many times, the 1 will be assumed when it is in the denominator. Very seldom is the 1 assumed when in the numerator. Almost always, the 1 is used explicitly when it is in the numerator.
The answers to the two examples above are 2.50 x 106 pg and 8.0 x 10¯7 km.
Five Conversion Examples
1) 0.75 kg to milligrams
2) 1500 millimeters to km
3) 2390 g to kg
4) 0.52 km to meters
5) 65 kg to g
Did you read that special note about a conversion sometimes not mentioned in class? Here's the link.
Example #6: Convert 1.80 mL to cm3.
1.80 mL times (1 cm3 / 1 mL) = 1.80 cm3
Example #7: Convert 1.80 L to dm3
1.80 L times (1 dm3 / 1 L) = 1.80 dm3
Be careful with the next two problems.
Example #8: Convert 1.80 L to cm3
1.80 L times (1000 cm3 / 1 L) = 1.80 x 103 cm3
Example #9: Convert 1.80 mL to dm3.
1.80 mL times (1 dm3 / 1000 mL) = 0.00180 dm3
One dm3 equals 1000 mL, just like 1 L equals 1000 cm3. You have to be careful when looking at the interplay between L, mL, cm3, and dm3.
Example #10: Convert 4.35 10¯9 m3 to μL
Solution:
The plan is to change cubic meters to cubic centimeters because 1 cm3 equals 1 mL. Then, change mL to μL. The key point is that 1 m = 100 cm.
1) Imagine 4.35 10¯9 m3 like this:
4.35 10¯9 m x 1 m x 1 m
2) Now, replace each m with 100 cm:
(4.35 10¯9 x 100 cm) x 100 cm x 100 cm
3) Multiply it out to get cm3
0.00435 cm3
4) This equals 0.00435 mL because 1 cm3 = 1 mL.
5) Now, change mL to μL:
0.00435 mL times (___ μL / ___ mL)
6) 1 mL is bigger than 1 μL:
0.00435 mL times (___ μL / 1 mL)
7) How many μL in one mL?
The milli prefix means 10¯3 and the μ prefix means 10¯6. The absolute exponential distance between the two units is 103:
0.00435 mL times (103 μL / 1 mL) = 4.35 μL |
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Every Question Helps You Learn
# Fractions (Easy)
This 11-plus Maths quiz introduces the ideas of an 'improper fraction' and a 'mixed fraction'. In this quiz we are going to do a few simple calculations with fractions. An improper fraction is a fraction in which the numerator is greater than or equal to the denominator, e.g. 53 and 99. A mixed fraction (also called a mixed number) is a number that has a whole number part and a fractional part, which is not written as a decimal, e.g. 235. Here are a couple of tips.
TIP 1: Don't forget: the numerator is the number at the top and the denominator is the number at the bottom of a fraction.
TIP 2: To convert a mixed fraction to an improper fraction, follow these steps:
1. Multiply the whole number part by the denominator.
2. Add this result to the numerator.
3. Write the fraction with step 2 in the numerator and keep the original denominator.
1.
What is 335 as an improper fraction?
455
93
185
183
To convert a mixed fraction to an improper fraction, follow these steps: 1. Multiply the whole number part by the denominator. 2. Add this result to the numerator. 3. Write the fraction with step 2 in the numerator and keep the original denominator. STEP 1: 3 × 5 = 15. STEP 2: 15 + 3 = 18. STEP 3: 18?5
2.
What is the improper fraction 165 as a mixed fraction?
3165
315
1615
1065
To convert an improper fraction to a mixed fraction, follow these steps: 1. Divide the numerator by the denominator. 2. Note the whole number remainder. 3. Write the number from step 1 as the whole number in front of the fractional part AND write the fractional part with the remainder in the numerator and keep the original denominator. STEP 1: 16 ÷ 5 = 3. STEP 2: Remainder 1. STEP 3: 31?5
3.
If you are asked to find 18 of something, what is the best method of doing it?
Subtract 8
Multiply by 8
Divide by 8
The word 'of' means multiply; therefore, one-eighth OF something means multiply by 1?8 BUT multiplying by an eighth is the same as dividing by eight because 1?8 × 32 = 32?8. The same applies in similar situations
4.
What is 12 of 14?
34
18
116
132
1?2 × 1?4 = 1?8. You multiply the numbers in the denominator together, and you multiply the numbers in the numerator together to form a single fraction
5.
Mary says that you can have one fifth of her sweets. If she has 125 sweets, how many sweets will you get?
75
25
45
65
The word 'of' means multiply; therefore one-fifth OF something means multiply by 1?5 BUT multiplying by a fifth is the same as dividing by five: 125 ÷ 5 = 25
6.
What is 639 as an improper fraction?
193
189
543
573
To convert a mixed fraction to an improper fraction, follow these steps: 1. Multiply the whole number part by the denominator. 2. Add this result to the numerator. 3. Write the fraction with step 2 in the numerator and keep the original denominator. STEP 1: 6 × 9 = 54. STEP 2: 54 + 3 = 57. STEP 3: 57?9 BUT you must now REDUCE the fraction to its simplest form: divide the numerator and the denominator by 3 to get 19?3
7.
What is 5418 as a mixed fraction?
3318
3118
113
3
54?18 = 3: REDUCE the fraction to its simplest form: divide the numerator and the denominator by 18. Be on the LOOKOUT for this sort of thing. Note: we could have written 31?1 but this is pointless
8.
Harry lost three-quarters of half the money his aunt had given him. If half of the money was £64, how much did Harry lose?
£96
£32
£24
£48
3?4 × £64 = (3 × £64) ÷ 4 = 192 ÷ 4 = £48
9.
First, Anne had 416 of the cake. Then, James had 916 and finally, Henry had 616 of the cake. Is there anything wrong here?
Anne should have been given more cake because she is a girl
No. Bon appétit!
Yes. Henry didn't get the stated amount of cake
Yes. there won't be any cake left for Henry
4?16 + 9?16+ 6?16 = 19?16 which is more than 1 cake! In fact, Henry only got 3?16 of the cake, which was less than Anne got
10.
What is one seventh of 49?
42
56
7
343
1?7 × 49 = 49 ÷ 7 = 7
Author: Frank Evans |
# RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3
## RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3
These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3
Other Exercises
Question 1.
In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.
Solution:
Given : In ∆ABC and ∆DEF,
∠B = ∠E = 90°
∠C = ∠F
AB = DE
To prove : ∆ABC = ∆DEF
Proof : In ∆ABC and ∆DEF,
∠B = ∠E (Each = 90°)
∠C = ∠F (Given)
AB = DE (Given)
∆ABC = ∆DEF (AAS axiom)
Question 2.
If the bisector of the exterior vertical angle of a triangle be parallel to the base. Show that the triangle is isosceles.
Solution:
Given : In ∆ABC, AE is the bisector of vertical exterior ∠A and AE $$\parallel$$ BC
To prove : ∆ABC is an isosceles
Proof: ∵ AE $$\parallel$$ BC
∴ ∠1 = ∠B (Corresponding angles)
∠2 = ∠C (Alternate angle)
But ∠1 = ∠2 (∵ AE is the bisector of ∠CAD)
∴ ∠B = ∠C
∴ AB = AC (Sides opposite to equal angles)
∴ ∆ABC is an isosceles triangle
Question 3.
In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.
Solution:
Given : In ∆ABC, AB = AC
∠A = 2(∠B + ∠C)
To calculate: Base angles,
Let ∠B = ∠C = x
Then ∠A = 2(∠B + ∠C)
= 2(x + x) = 2 x 2x = 4x
∵ Sum of angles of a triangle = 180°
∴ 4x + x + x – 180° ⇒ 6x = 180°
⇒ x= $$\frac { { 180 }^{ \circ } }{ 6 }$$ = 30° o
∴ ∠B = ∠C = 30 and ∠A = 4 x 30° = 120
Question 4.
Prove that each angle of an equilateral triangle is 60°. [NCERT]
Solution:
Given : ∆ABC is an equilateral triangle
Proof: In ∆ABC,
AB = AC (Sides of an equilateral triangle)
∴ ∠C = ∠B …(i)
(Angles opposite to equal angles)
Similarly, AB = BC
∴ ∠C = ∠A …(ii)
From (i) and (ii),
∠A = ∠B = ∠C
But ∠A + ∠B + ∠C = 180° (Sum of angles of a triangle)
∴ ∠A = ∠B = ∠C = $$\frac { { 180 }^{ \circ } }{ 3 }$$= 60°
Question 5.
Angles A, B, C of a triangle ABC are equal to each other. Prove that ∆ABC is equilateral.
Solution:
Given : In ∆ABC, ∠A = ∠B = ∠C
To prove : ∆ABC is an equilateral
Proof: In ∆ABC,
∴ ∠B = ∠C (Given)
∴ AC = AB …(i) (Sides opposite to equal angles)
Similarly, ∠C = ∠A
∴ BC =AB …(ii)
From (i) and (ii)
AB = BC = CA
Hence ∆ABC is an equilateral triangle
Question 6.
ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Solution:
In ∆ABC, ∠A = 90°
AB =AC (Given)
∴ ∠C = ∠B (Angles opposite to equal sides)
But ∠B + ∠C = 90° (∵ ∠B = 90°)
∴ ∠B = ∠C = $$\frac { { 90 }^{ \circ } }{ 2 }$$ = 45°
Hence ∠B = ∠C = 45°
Question 7.
PQR is a triangle in which PQ = PR and S is any point on the side PQ. Through S, a line is drawn parallel to QR and intersecting PR at T. Prove that PS = PT.
Solution:
Given : In ∆PQR, PQ = PR
S is a point on PQ and PT || QR
To prove : PS = PT
Proof : ∵ST || QR
∴ ∠S = ∠Q and ∠T = ∠R (Corresponding angles)
But ∠Q = ∠R (∵ PQ = PR)
∴ PS = PT (Sides opposite to equal angles)
Question 8.
In a ∆ABC, it is given that AB = AC and the bisectors of ∠B and ∠C intersect at O. If M is a point on BO produced, prove that ∠MOC = ∠ABC.
Solution:
Given : In ∆ABC, AB = AC the bisectors of ∠B and ∠C intersect at O. M is any point on BO produced.
To prove : ∠MOC = ∠ABC
Proof: In ∆ABC, AB = BC
∴ ∠C = ∠B
∵ OB and OC are the bisectors of ∠B and ∠C
∴ ∠1 =∠2 = $$\frac { 1 }{ 2 }$$∠B
Now in ∠OBC,
Ext. ∠MOC = Interior opposite angles ∠1 + ∠2
= ∠1 + ∠1 = 2∠1 = ∠B
Hence ∠MOC = ∠ABC
Question 9.
P is a point on the bisector of an angle ∠ABC. If the line through P parallel to AB meets BC at Q, prove that triangle BPQ is isosceles.
Solution:
Given : In ∆ABC, P is a point on the bisector of ∠B and from P, RPQ || AB is draw which meets BC in Q
To prove : ∆BPQ is an isosceles
Proof : ∵ BD is the bisectors of CB
∴ ∠1 = ∠2
∵ RPQ || AB
∴ ∠1 = ∠3 (Alternate angles)
But ∠1 == ∠2 (Proved)
∴ ∠2 = ∠3
∴ PQ = BQ (sides opposite to equal angles)
∴ ∆BPQ is an isosceles
Question 10.
ABC is a triangle in which ∠B = 2∠C, D is a point on BC such that AD bisects ∠BAC = 72°.
Solution:
Given: In ∆ABC,
∠B = 2∠C, AD is the bisector of ∠BAC AB = CD
To prove : ∠BAC = 72°
Construction : Draw bisector of ∠B which meets AD at O
Hope given RD Sharma Class 9 Solutions Chapter 12 Heron’s Formula Ex 12.3 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. |
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