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# Circles
## Circles
Circles are the shape of perfection in geometry - they're symmetrical and fascinating! Understanding circles means diving into their properties, including radius, diameter, circumference, and area. Let's go on a roundabout journey of exploring these properties!
## Properties of a Circle
To master the art of working with circles, let's get acquainted with some key properties and their definitions.
• Radius: The distance from the center of the circle to any point on the circle. It's half the diameter!
• Diameter: The longest distance from one end of the circle to the other, passing through the center. It's twice the radius!
• Circumference: The distance around the circle. It's like the "perimeter" of the circle.
• Area: The amount of space inside the circle.
## Circumference of a Circle
To calculate the circumference (C) of a circle, we use the diameter (d) or the radius (r) with the formula:
$$C = \pi d = 2\pi r$$
### Circumference Example:
Find the circumference of a circle with radius 5 units.
Use the formula:
$$C = 2\pi r = 2\pi \times 5 = 10\pi \text{ units}$$
## Area of a Circle
To calculate the area (A) of a circle, we use the radius (r) with the formula:
$$A = \pi r^2$$
### Area Example:
Find the area of a circle with radius 5 units.
Use the formula:
$$A = \pi r^2 = \pi \times 5^2 = 25\pi \text{ square units}$$
## Circles Tips
Here are some tips to help you work with circles more effectively:
• Remember the formulas for circumference and area, they'll be your best friends when dealing with circles!
• In a circle, every radius has the same length - that's the beauty of symmetry!
• If you only know the diameter of the circle, remember to divide it by 2 to get the radius.
• Remember that the value of pi (π) is approximately 3.14 or 22/7. However, when doing calculations, it's often best to keep it as pi and only substitute the decimal at the end.
## Circles Practice
Below is a table of practice problems and their solutions. Use these examples to practice your skills on circles.
Problem Solution
Find the circumference of a circle with radius 7 units. $$C = 2\pi r = 2\pi \times 7 = 14\pi \text{ units}$$
Find the area of a circle with radius 7 units. $$A = \pi r^2 = \pi \times 7^2 = 49\pi \text{ square units}$$
Continue to practice these problems to become more comfortable with circles. In no time, you'll be spinning circles around these concepts! |
# How do you write the slope-intercept equation of the line perpendicular to y = 5/2 x - 2, which passes though the point (0, 2)?
Sep 20, 2016
$y = - \frac{2}{5} x + 2$
#### Explanation:
Thinks to remember:
[1]$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{y = \textcolor{b l u e}{m} x + b}$ is the slope-intercept form
$\textcolor{w h i t e}{\text{XXXX}}$with a slope of $\textcolor{b l u e}{m}$ and a y-intercept of $\textcolor{red}{b}$
[2]$\textcolor{w h i t e}{\text{XXX}}$If a line has a slope of $\textcolor{b l u e}{m}$
$\textcolor{w h i t e}{\text{XXXX}}$all lines perpendicular to it have a slope of color(blue)(""(-1/m))
Based on this we know that:
given a line $y = \textcolor{b l u e}{\frac{5}{2}} x - 2$
any line perpendicular to this will have a slope of $\textcolor{b l u e}{- \frac{2}{5}}$
and
any such perpendicular line will have the form:
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{b l u e}{- \frac{2}{5}} x + \textcolor{red}{b}$
where $\textcolor{red}{b}$ is the y-intercept;
that is $\textcolor{red}{b}$ is the value of $y$ when $\textcolor{\mathrm{da} r k m a \ge n t a}{x = 0}$
and
given $\left(\textcolor{\mathrm{da} r k m a \ge n t a}{x} , \textcolor{red}{y}\right) = \left(\textcolor{\mathrm{da} r k m a \ge n t a}{0} , \textcolor{red}{2}\right)$ is a point on the required perpendicular line
$\textcolor{w h i t e}{\text{XXX}} \Rightarrow \textcolor{red}{b} = \textcolor{red}{2}$
So the equation of the required perpendicular line is:
color(white)("XXXXXXXX")color(green)(y=color(blue)(-2/5)x+color(red)(2) |
# Section 3 Graphing - copy
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Category: Entertainment
## Presentation Description
Nature of Science, Section 3: Graphing. PowerPoint for students to view.
## Presentation Transcript
### Section 3:
Section 3 Communicating with Graphs
### PowerPoint Presentation:
A Visual Display A graph is a visual display of information or data. This is a graph that shows a girl walking her dog. 3 Communicating with Graphs
### PowerPoint Presentation:
A Visual Display Communicating with Graphs The horizontal axis, or the x -axis, measures time. Time is the independent variable because as it changes, it affects the measure of another variable. 3
### PowerPoint Presentation:
Line Graphs A line graph can show any relationship where the dependent variable changes due to a change in the independent variable. Communicating with Graphs 3
### PowerPoint Presentation:
Line Graphs Line graphs often show how a relationship between variables changes over time. Communicating with Graphs 3
### PowerPoint Presentation:
Constructing Line Graphs The most important factor in making a line graph is always using the x -axis for the independent variable. Communicating with Graphs The y -axis always is used for the dependent variable. 3
### PowerPoint Presentation:
Bar Graphs A bar graph is useful for comparing information collected by counting. For example, suppose you counted the number of students in every classroom in your school on a particular day and organized your data in a table. Communicating with Graphs 3
### PowerPoint Presentation:
Bar Graphs Communicating with Graphs As on a line graph, the independent variable is plotted on the x- axis and the dependent variable is plotted on the y - axis. 3
### PowerPoint Presentation:
Circle Graphs A circle graph, or pie graph, is used to show how some fixed quantity is broken down into parts. Communicating with Graphs The circular pie represents the total. The slices represent the parts and usually are represented as percentages of the total. 3
### PowerPoint Presentation:
Circle Graphs Communicating with Graphs To create a circle graph, you start with the total of what you are analyzing. This graph starts with 72 buildings in the neighborhood. 3
### PowerPoint Presentation:
Circle Graphs Communicating with Graphs For each type of heating fuel, you divide the number of buildings using each type of fuel by the total (72). 3
### PowerPoint Presentation:
Circle Graphs Communicating with Graphs You then multiply that decimal by 360 to determine the angle that the decimal makes in the circle. Eighteen buildings use steam. Therefore, 18 72 x 360 = 90 on the circle graph. You then would measure 90 on the circle with your protractor to show 25 percent. 3
### Review Question #1:
Review Question #1 Which graph would be appropriate for showing the growth rate of a plant for six weeks? Bar graph Circle graph Line graph |
# Number Operations and Calculation Methods
### Consecutive Numbers
##### Stage: 2 and 3 Challenge Level:
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
### Dicey Operations
##### Stage: 2 and 3 Challenge Level:
Who said that adding, subtracting, multiplying and dividing couldn't be fun?
### Countdown
##### Stage: 2 and 3 Challenge Level:
Here is a chance to play a version of the classic Countdown Game.
### Got It
##### Stage: 2 and 3 Challenge Level:
A game for two people, or play online. Given a target number, say 23, and a range of numbers to choose from, say 1-4, players take it in turns to add to the running total to hit their target.
### The Remainders Game
##### Stage: 2 and 3 Challenge Level:
A game that tests your understanding of remainders.
### Two and Two
##### Stage: 2 and 3 Challenge Level:
How many solutions can you find to this sum? Each of the different letters stands for a different number.
### Cryptarithms
##### Stage: 3 Challenge Level:
Can you crack these cryptarithms?
### Going Round in Circles
##### Stage: 3 Challenge Level:
Mathematicians are always looking for efficient methods for solving problems. How efficient can you be?
### Missing Multipliers
##### Stage: 3 Challenge Level:
What is the smallest number of answers you need to reveal in order to work out the missing headers?
### Number Daisy
##### Stage: 3 Challenge Level:
Can you find six numbers to go in the Daisy from which you can make all the numbers from 1 to a number bigger than 25?
### Can You Make 100?
##### Stage: 3 Challenge Level:
How many ways can you find to put in operation signs (+ - x ÷) to make 100?
### What Numbers Can We Make?
##### Stage: 3 Challenge Level:
Imagine we have four bags containing a large number of 1s, 4s, 7s and 10s. What numbers can we make?
### Consecutive Seven
##### Stage: 3 Challenge Level:
Can you arrange these numbers into 7 subsets, each of three numbers, so that when the numbers in each are added together, they make seven consecutive numbers?
### Place Your Orders
##### Stage: 3 Challenge Level:
Can you rank these sets of quantities in order, from smallest to largest? Can you provide convincing evidence for your rankings?
### Keep it Simple
##### Stage: 3 Challenge Level:
Can all unit fractions be written as the sum of two unit fractions?
### Method in Multiplying Madness?live
##### Stage: 2 and 3 Challenge Level:
Watch our videos of multiplication methods that you may not have met before. Can you make sense of them?
### Legs Eleven
##### Stage: 3 Challenge Level:
Take any four digit number. Move the first digit to the 'back of the queue' and move the rest along. Now add your two numbers. What properties do your answers always have?
### Remainders
##### Stage: 3 Challenge Level:
I'm thinking of a number. When my number is divided by 5 the remainder is 4. When my number is divided by 3 the remainder is 2. Can you find my number?
### Where Can We Visit?
##### Stage: 3 Challenge Level:
Charlie and Abi put a counter on 42. They wondered if they could visit all the other numbers on their 1-100 board, moving the counter using just these two operations: x2 and -5. What do you think?
### Impossibilities
##### Stage: 3 Challenge Level:
Just because a problem is impossible doesn't mean it's difficult...
### Thousands and Millions
##### Stage: 3 Challenge Level:
Here's a chance to work with large numbers...
### Overlaps
##### Stage: 3 Challenge Level:
Can you find ways to put numbers in the overlaps so the rings have equal totals?
### Egyptian Fractions
##### Stage: 3 Challenge Level:
The Egyptians expressed all fractions as the sum of different unit fractions. Here is a chance to explore how they could have written different fractions.
### The Greedy Algorithm
##### Stage: 3 Challenge Level:
The Egyptians expressed all fractions as the sum of different unit fractions. The Greedy Algorithm might provide us with an efficient way of doing this.
### How Many Miles to Go?
##### Stage: 3 Challenge Level:
A car's milometer reads 4631 miles and the trip meter has 173.3 on it. How many more miles must the car travel before the two numbers contain the same digits in the same order?
### Largest Product
##### Stage: 3 and 4 Challenge Level:
Which set of numbers that add to 10 have the largest product?
### Cinema Problem
##### Stage: 3 and 4 Challenge Level:
A cinema has 100 seats. Show how it is possible to sell exactly 100 tickets and take exactly £100 if the prices are £10 for adults, 50p for pensioners and 10p for children.
### The Legacy
##### Stage: 4 Challenge Level:
Your school has been left a million pounds in the will of an ex- pupil. What model of investment and spending would you use in order to ensure the best return on the money?
### Dating Made Easier
##### Stage: 4 Challenge Level:
If a sum invested gains 10% each year how long before it has doubled its value?
### The Root of the Problem
##### Stage: 4 and 5 Challenge Level:
Find the sum of the series.
### Number Operations and Calculation Methods - Short Problems
##### Stage: 3 and 4
A collection of short Stage 3 and 4 problems on number operations and calculation methods.
### Split Clock Face
##### Stage: 3 Short Challenge Level:
Use 2 straight lines to split the clock face into 3 parts so that the sums of the numbers in each of the parts are equal.
### Sum of 1s
##### Stage: 3 Short Challenge Level:
Can you find the last 5 digits of this sum?
### Latin Numberslive
##### Stage: 3 Challenge Level:
Can you create a Latin Square from multiples of a six digit number?
### Multiplication Mistake
##### Stage: 3 Short Challenge Level:
Jane accidentally multiplied by 54 instead of 45, and her answer was 198 too big. What number did she multiply 54 by?
### Last Digit
##### Stage: 3 Short Challenge Level:
What is the last digit in this calculation involving powers? |
# Systems of Equations & Inequalities
## Presentation on theme: "Systems of Equations & Inequalities"— Presentation transcript:
Systems of Equations & Inequalities
Algebra I Algebra I ~ Chapter 7 ~ Systems of Equations & Inequalities Lesson 7-1 Solving Systems by Graphing Lesson 7-2 Solving Systems Using Substitution Lesson 7-3 Solving Systems Using Elimination Lesson 7-4 Applications of Linear Systems Lesson 7-5 Linear Inequalities Lesson 7-6 Systems of Linear Inequalities Chapter Review
Solving Systems by Graphing
Lesson 7-1 Cumulative Review Chap 1-6
Solving Systems by Graphing
Lesson 7-1 Cumulative Review Chap 1-10
Solving Systems by Graphing
Lesson 7-1 Notes System of linear equations – Two or more linear equations together… One way to solve a system of linear equations is by… Graphing. Solving a System of Equations Step 1: Graph both equations on the same plane. (Hint: Use the slope and the y-intercept or x- & y-intercepts to graph.) Step 2: Find the point of intersection Step 3: Check to see if the point of intersection makes both equations true. Solve by graphing. Check your solution. y = x + 5 y = -4x Your turn… y = -1/2 x + 2 y = -3x - 3 ~ Try another one ~ x + y = 4 x = -1
Solving Systems by Graphing
Lesson 7-1 Notes Systems with No Solution When two lines are parallel, there are no points of intersection; therefore, the system has NO SOLUTION! y = -2x + 1 y = -2x – 1 Systems with Infinitely Many Solutions y = 1/5x + 9 5y = x + 45 Since they are graphs of the same line… There are an infinite number of solutions.
Solving Systems by Graphing
Lesson 7-1 Solving Systems by Graphing Homework Homework – Practice 7-1 #1-28 odd
Solving Systems by Substitution
Lesson 7-2 Practice 7-1
Solving Systems Using Substitution
Lesson 7-2 Notes Using Substitution Step 1: Start with one equation. Step 2: Substitute for y using the other equation. Step 3: Solve the equation for x. Step 4: Substitute solution for x and solve for y Step 5: Your x & y values make the intersection point (x, y). Step 6: Check your solution. y = 2x 7x – y = 15 Your turn… y = 4x – 8 y = 2x + 10 ~ Another example~ c = 3d – 27 4d + 10c = 120
Solving Systems Using Substitution
Lesson 7-2 Notes Using Substitution & the Distributive Property 3y + 2x = 4 -6x + y = -7 Step 1: Solve the equation in which y has a coefficient of 1… +6x x y = 6x -7 Step 2: Use the other equation (substitute using the equation from Step 1.) 3(6x – 7) + 2x = 4 18x – x = 4 20x = 25 x = 1 1/4 Step 3: Solve for the other variable Substitute 1 ¼ or 1.25 for x y = 6(1.25) – 7 y = y = 0.5 Solution is (1.25, 0.5)
Solving Systems Using Substitution
Lesson 7-2 Notes Your turn… 6y + 8x = 28 3 = 2x – y Solution is (2.3, 1.6) or (2 3/10, 1 3/5) A rectangle is 4 times longer than it is wide. The perimeter of the rectangle is 30 cm. Find the dimensions of the rectangle. Let w = width Let l = length l = 4w 2l + 2w = 30 Solve for l… l = 4(3) l = 12 Use substitution to solve. 2(4w) + 2w = 30 8w + 2w = 30 10w = 30 w = 3
Solving Systems Using Substitution
Lesson 7-2 Solving Systems Using Substitution Homework Homework ~ Practice 7-2 even
Solving Systems Using Elimination
Lesson 7-3 Practice 7-2
Solving Systems Using Elimination
Lesson 7-3 Notes Adding Equations Step 1: Eliminate the variable which has a coefficient sum of 0 and solve. Step 2: Solve for the eliminated variable. Step 3: Check the solution. 5x – 6y = -32 3x + 6y = 48 8x + 0 = 16 x = 2 Solution is (2, 7) Check 3(2) + 6(7) = 48 = 48 48 = 48 Your turn… 6x – 3y = 3 & -6x + 5y = 3 5x – 6y = - 32 5(2) – 6y = - 32 10 – 6y = -32 -6y = -42 y = 7
Solving Systems Using Elimination
Lesson 7-3 Notes Multiplying One Equation Step 1: Eliminate one variable. -2x + 15y = -32 7x – 5y = 17 Step 2: Multiply one equation by a number that will eliminate a variable. 3(7x – 5y = 17) Step 3: Solve for the variable 19x = 19 x = 1 Step 4: Solve for the eliminated variable using either original equation. -2(1) + 15y = -32 Solution (1, -2) -2x + 15y = -32 21x - 15y = 51 19x + 0 = 19 y = -32 15y = -30 y = -2
Solving Systems Using Elimination
Lesson 7-3 Notes Your turn… 3x – 10y = -25 4x + 40y = 20 Solution (-5, 1) Multiply Both Equations Step 1: Eliminate one variable. 4x + 2y = 14 7x – 3y = -8 Step 2: Solve for the variable 26x = 26 x = 1 Try this one… 15x + 3y = 9 10x + 7y = -4 3(4x + 2y = 14) 2(7x – 3y = -8) 12x + 6y = 42 14x – 6y = -16 26x + 0 = 26 Step 3: Solve for the eliminated variable 4(1) + 2y = 14 2y = 10 y = Solution (1, 5)
Solving Systems Using Elimination
Lesson 7-3 Solving Systems Using Elimination Homework Homework – Practice 7-3 odd
Applications of Linear Systems
Lesson 7-4 Practice 7-3
Applications of Linear Systems
Lesson 7-4 Notes
Applications of Linear Systems
Lesson 7-4 Homework Homework – Practice #6-10
Linear Inequalities Lesson 7-5 Practice 7-4
Notes Linear Inequalities Lesson 7-5
Using inequalities to describe regions of a coordinate plane: x < 1 y > x + 1 y ≤ - 2x + 4 Steps for graphing inequalities… (1) First graph the boundary line. (2) Determine if the boundary line is a dashed or solid line. Shade above or below the boundary line… (< below or > above) Graph y ≥ 3x - 1 Rewriting to Graph an Inequality Graph 3x – 5y ≤ 10 Solve for y… (remember if you divide by a negative, the inequality sign changes direction) then apply the steps for graphing an inequality. Graph 6x + 8y ≥ 12
Linear Inequalities Lesson 7-5 Homework Homework ~ Practice 7-5 odd
Systems of Linear Inequalities
Lesson 7-6 Practice 7-5
Systems of Linear Inequalities
Lesson 7-6 Practice 7-5
Systems of Linear Inequalities
Lesson 7-6 Practice 7-5
Systems of Linear Inequalities
Lesson 7-6 Notes Solve by graphing… x ≥ 3 & y < -2 You can describe each quadrant using inequalities… Quadrant I? Quadrant II? Quadrant III? Quadrant IV? Graph a system of Inequalities… (1) Solve each equation for y… (2) Graph one inequality and shade. (3) Graph the second inequality and shade. The solutions of the system are where the shading overlaps. Choose a point in the overlapping region and check in each inequality.
Systems of Linear Inequalities
Lesson 7-6 Notes Graph to find the solution… y ≥ -x + 2 & 2x + 4y < 4 Writing a System of Inequalities from a Graph Determine the boundary line for the pink region… y = x – 2 The region shaded is above the dashed line… so y > x – 2 Determine the boundary line for the blue region… y = -1/3x + 3 The region shaded is below the solid line… so y ≤ -1/3x + 3 Your turn…
Systems of Linear Inequalities
Lesson 7-6 Systems of Linear Inequalities Practice 7-6 Homework 7-6 odd
Systems of Linear Inequalities
Lesson 7-6 Systems of Linear Inequalities Practice 7-6
Systems of Linear Inequalities
Lesson 7-6 Systems of Linear Inequalities Practice 7-5
Systems of Linear Inequalities
Lesson 7-6 Systems of Linear Inequalities Practice 7-6
~ Chapter 7 ~ Algebra I Algebra I Chapter Review
~ Chapter 7 ~ Algebra I Algebra I Chapter Review |
# 2.8: Algebraic Equations to Represent Words
Difficulty Level: At Grade Created by: CK-12
The sum of two consecutive even integers is 34. What are the integers?
### Guidance
To translate a problem from words into an equation, look for key words to indicate the operation used in the problem.
Once the equation is known, to solve the problem you use the same rules as when solving equations with one variable. Isolate the variable and then solve for it making sure that whatever you do to one side of the equals sign you do to the other side. Drawing a diagram is also helpful in solving some word problems.
#### Example A
Two consecutive integers have a sum of 173. What are those numbers?
Let \begin{align*}x =\end{align*} integer 1
Then \begin{align*}x + 1 =\end{align*} integer 2 (Because they are consecutive, they must be separated by only one number. For example: 1, 2, 3, 4,... all are consecutive.)
You can square the operation in the question.
Two consecutive integers have a \begin{align*}\boxed{\text{sum}}\end{align*} of 173. What are those numbers?
\begin{align*}x + (x + 1) &= 173\\ x + x + 1 &= 173 && (\text{Remove the brackets})\\ 2x + 1 &= 173 && (\text{Combine like terms})\\ 2x + 1 {\color{red}-1} &= 173 {\color{red}-1} && (\text{Subtract} \ 1 \ \text{from both sides to isolate the variable})\\ 2x &= 172 && (\text{Simplify})\\ \frac{2x}{{\color{red}2}} &= \frac{172}{{\color{red}2}} && (\text{Divide both sides by} \ 2 \ \text{to solve for the variable})\\ x &= 86 && (\text{Simplify})\end{align*}
Therefore the first integer is 86 and the second integer is \begin{align*}(86 + 1) = 87\end{align*}. Check: \begin{align*}86 + 87=173\end{align*}.
#### Example B
When a number is subtracted from 35, the result is 11. What is the number?
Let \begin{align*}x =\end{align*} the number
You can square the operation in the question.
When a number is \begin{align*}\boxed{\text{subtracted}}\end{align*} from 35, the result is 11.
\begin{align*}35 - x &= 11\\ 35 {\color{red}- 35} - x &= 11 {\color{red}- 35} && (\text{Subtract} \ 35 \ \text{from both sides to isolate the variable})\\ -x &= -24 && (\text{Simplify})\\ \frac{-x}{{\color{red}-1}} &= \frac{-24}{{\color{red}-1}} && (\text{Divide both sides by} \ -1 \ \text{to solve for the variable})\\ x &= 24 && (\text{Simplify})\end{align*}
Therefore the number is 24.
#### Example C
When one third of a number is subtracted from one half of a number, the result is 14. What is the number?
Let \begin{align*}x =\end{align*} number
You can square the operation in the question.
When one third of a number is \begin{align*}\boxed{\text{subtracted}}\end{align*} from one half of a number, the result is 14.
\begin{align*}\frac{1}{2}x-\frac{1}{3}x=14\end{align*}
You need to get a common denominator in this problem in order to solve it. For this problem, the denominators are 2, 3, and 1. The LCD is 6. Therefore multiply the first fraction by \begin{align*}\frac{3}{3}\end{align*}, the second fraction by \begin{align*}\frac{2}{2}\end{align*}, and the third number by \begin{align*}\frac{6}{6}\end{align*}.
\begin{align*}\left({\color{red}\frac{3}{3}}\right) \frac{1}{2}x-\left({\color{red}\frac{2}{2}}\right) \frac{1}{3}x &= \left({\color{red}\frac{6}{6}}\right)14\\ \frac{3}{6}x-\frac{2}{6}x &= \frac{84}{6} && (\text{Simplify})\end{align*}
Now that the denominator is the same, the equation can be simplified to be:
\begin{align*}3x-2x &= 84\\ x &= 84 && (\text{Combine like terms})\end{align*}
Therefore the number is 84.
#### Concept Problem Revisited
The sum of two consecutive even integers is 34. What are the integers?
Let \begin{align*}x =\end{align*} integer 1
Then \begin{align*}x + 2 =\end{align*} integer 2 (Because they are even, they must be separated by at least one number. For example: 2, 4, 6, 8,... all have one number in between them.)
You can know write an algebraic expression to solve for the two integers. Remember that you are talking about the sum. You can square the operation in the question.
The \begin{align*}\boxed{\text{sum}}\end{align*} of two consecutive even integers is 34.
\begin{align*}x + (x + 2) &= 34\\ x + x + 2 &= 34 && (\text{Remove the brackets})\\ 2x + 2 &= 34 && (\text{Combine like terms})\\ 2x + 2 {\color{red}-2} &= 34 {\color{red}-2} && (\text{Subtract} \ 2 \ \text{from both sides to isolate the variable})\\ 2x &= 32 && (\text{Simplify})\\ \frac{2x}{{\color{red}2}} &= \frac{32}{{\color{red}2}} && (\text{Divide both sides by} \ 2 \ \text{to solve for the variable})\\ x &= 16 && (\text{Simplify})\end{align*}
Therefore the first integer is 16 and the second integer is \begin{align*}(16 + 2) = 18\end{align*}. Also \begin{align*}16 + 18\end{align*} is indeed 34.
### Vocabulary
Algebraic Equation
An algebraic equation contains numbers, variables, operations, and an equals sign.
Consecutive
The term consecutive means in a row. Therefore an example of consecutive numbers is 1, 2, and 3. An example of consecutive even numbers would be 2, 4, and 6. An example of consecutive odd numbers would be 1, 3, and 5.
### Guided Practice
1. What is a number that when doubled would equal sixty?
2. The sum of two consecutive odd numbers is 176. What are these numbers?
3. The perimeter of a rectangular frame is 48 in. What are the lengths of each side?
1. What is a number that when doubled would equal sixty?
Let \begin{align*}x =\end{align*} number
You can square the operation in the question.
What is a number that when \begin{align*}\boxed{\text{doubled}}\end{align*} would equal sixty?
\begin{align*}2x &= 60\\ \frac{2x}{{\color{red}2}} &= \frac{60}{{\color{red}2}} && (\text{Divide by} \ 2 \ \text{to solve for the variable})\\ x &= 30 && (\text{Simplify})\end{align*}
Therefore the number is 30.
2. The sum of two consecutive odd numbers is 176. What are these numbers?
Let \begin{align*}x =\end{align*} first number
Let \begin{align*}x + 2 =\end{align*} second number
You can square the operation in the question.
The \begin{align*}\boxed{\text{sum}}\end{align*} of two consecutive odd numbers is 176.
\begin{align*}x + (x + 2) &= 176\\ x + x + 2 &= 176 && (\text{Remove brackets})\\ 2x + 2 &= 176 && (\text{Combine like terms})\\ 2x+2 {\color{red}-2} &= 176 {\color{red}-2} && (\text{Subtract} \ 2 \ \text{from both sides of the equal sign to isolate the variable})\\ 2x &= 174 && (\text{Simplify})\\ \frac{2x}{{\color{red}2}} &-\frac{174}{{\color{red}2}} && (\text{Divide by} \ 2 \ \text{to solve for the variable})\\ x &= 87\end{align*}
Therefore the first number is 87 and the second number is \begin{align*}(87 + 2) = 89\end{align*}.
3. The perimeter of a rectangular frame is 48 in. What are the lengths of each side?
You have to remember that a square has 4 sides of equal length in order to solve this problem.
Let \begin{align*}s =\end{align*} side length
\begin{align*}s + s + s + s &= 48 && (\text{Write initial equation with four sides adding to the perimeter})\\ 4s &= 48 && (\text{Simplify})\\ \frac{4s}{{\color{red}4}} &= \frac{48}{{\color{red}4}} && (\text{Divide by} \ 4 \ \text{to solve for the variable})\\ s &= 12\end{align*}
Therefore the side length is 12 inches.
### Practice
1. The sum of two consecutive numbers is 159. What are these numbers?
2. The sum of three consecutive numbers is 33. What are these numbers?
3. A new computer is on sale for 30% off. If the sale price is $500, what was the original price? 4. Jack and his three friends are sharing an apartment for the next year while at university (8 months). The rent cost$1200 per month. How much does Jack have to pay?
5. You are designing a triangular garden with an area of 168 square feet and a base length of 16 feet. What would be the height of the triangle of the garden shape?
6. If four times a number is added to six, the result is 50. What is that number?
7. This week, Emma earned ten more than half the number of dollars she earned last week babysitting. If this week, she earned 100 dollars, how much did she earn last week?
8. Three is twenty-one divided by the sum of a number plus five.
9. Five less than three times a number is forty-six.
10. Hannah had $237 in her bank account at the start of the summer. She worked for four weeks and now she has$1685 in the bank. How much did Hannah make each week in her summer job?
11. The formula to estimate the length of the Earth's day in the future is found to be twenty–four hours added to the number of million years divided by two hundred and fifty. In five hundred million years, how long will the Earth's day be?
12. Three times a number less six is one hundred twenty-six.
13. Sixty dollars was two-thirds the total money spent by Jack and Thomas at the store.
14. Ethan mowed lawns for five weekends over the summer. He worked ten hours each weekend and each lawn takes an average of two and one-half hours. How many lawns did Ethan mow?
15. The area of a rectangular pool is found to be two hundred eighty square feet. If the base length of the pool is 20 feet, what is the width of the pool?
16. A cell phone company charges a base rate of $10 per month plus 5¢ per minute for any long distance calls. Sandra gets her cell phone bill for$21.20. How many long distance minutes did she use?
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# Higher Order Linear Differential equations
Document Sample
``` Higher Order Linear Differential equations
Higher Order Linear Differential equations
Differentiation is important and interesting part of mathematics. We use
differentiation process for calculating the rate of particular function with respect
to particular variable like f(x) is a function and d/dx(f(x) is a rate of f(x) with
respect to x.
Now we discuss different type of differentiation equations – For understand the
different type of differentiation equation we assume a function y = f(x) and when
we calculate differentiation of f(x) with respect to x, it produces first order
differentiation equation and these type of first order differentiation contains
dy/dx and example of first order differentiation equation are – if y = sin x, then
dy/dx = cos x When we differentiate dy/dx with respect to x, it produces second
order differentiation equation and these type of second order equation contains
d2y/dx2 form like – if y = sin-1 x,
then d2y/dx2 = x/(1-x2)3/2 When we differentiate d2y/dx2 with respect to x, it
produces third order differentiation equation and these type of third order
differentiation equation contains d3y/dx3 form like – if y = log(sin x), then third
order differentiation equation is d3y/dx3 = 2 sin x * cosec3 x Similarly this
process continues to produce n order differentiation equation and these type of
n order differentiation equation contains dny/dxn form and these type of
equations are called as a higher order differential equations.
A linear equation which contains these higher order derivatives are called as a
higher order linear differential equations.For calculating the higher order
differential equations, we use following steps –
Step 1: First of all we assume given function as a variable like y = f(x).
Step 2: After assuming that variable, we differentiate that variable with respect
to any suitable variable like we differentiate y with respect to x, it produces
dy/dx.
Step 3: According to the equation we again differentiate variable with respect to
that particular variable like if we differentiate dy/dx with respect to x again, it
produces (d2y/dx2), which is a second order differentiation equation.
Step 4: According to situation, we continue this differentiation until we get our
required equation.
Step 5: After all differentiate process, we put the values of dy/dx, d2y/dx2,
…………dny/dxn in that equation, which we want to prove. Now we take some
examples of higher order differential equation –
Thank You
TutorCircle.com
```
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# Boolean Algebra
Dive into the intriguing world of Boolean Algebra, a cornerstone of modern computing and essential for those studying or working in Engineering. This pivotal subject within Engineering Mathematics explores fundamental concepts, underpinning operations, and essential rules of Boolean Algebra. Discover through real-life applications and practical case scenarios how this mathematical technique is applied every day. The insightful knowledge shared here will excellently enhance your understanding of Boolean Algebra's role and ongoing significant impact in Engineering.
#### Create learning materials about Boolean Algebra with our free learning app!
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## Understanding Boolean Algebra
Boolean algebra refers to a mathematical structure that captures abstract algebraic structure, central to digital logic and computer science. Named after George Boole, an English mathematician in 19th-century, Boolean algebra is fundamental to the design and working of modern digital computing systems.
### The Basic Concept of Boolean Algebra
Before diving into the basics, let's start with the backbone of Boolean algebra: the Boolean variables. A Boolean variable is something that can take only two values, either true or false, represented by 1 and 0 respectively.
• There are three primary operations in Boolean algebra: OR, AND, and NOT.
• The OR operation, often denoted by '+', is the Boolean equivalent to the arithmetic addition. However, in Boolean algebra, 1 + 1 is not 2, but 1.
• The AND operation, denoted by '.' or simply by writing variables together, is somewhat equivalent to the arithmetic multiplication. Here, 1.1 equals 1, and everything else equals 0.
• The NOT operation, defined as $$\bar{A}$$ or A', converts a 0 to a 1 and a 1 to a 0. Essentially, it is the negation operation.
For example, given two Boolean variables $$A$$ and $$B$$, their OR operation $$A+B$$ equals true if atleast one of them is true, AND operation $$A.B$$ equals true only if both $$A$$ and $$B$$ are true, and NOT operation $$\bar{A}$$ equals true only if $$A$$ is false.
#### What is the Boolean Algebra Meaning?
Boolean algebra involves operations on the set {0,1}. It's a branch of mathematics that deals with operations and concepts that are applicable to binary logic and the design and manipulation of computers. It's not just about simple calculations but also concerned with logical operations and relationships between variables.
The inputs or variables in Boolean algebra are true or false values The result of a Boolean operation is also a true or false value
It is critical for students to understand that in Boolean algebra, you are dealing with the truth values of logic expressions rather than numeric quantities.
#### Relation of Boolean Algebra with Engineering Mathematics
Boolean algebra plays a significant role in engineering mathematics, notably in electrical engineering and computer science. In computer engineering, for instance, it is used extensively to simplify logic gates and circuits.
If you've got a complex digital circuit, you can leverage Boolean algebra to simplify it, rendering it using fewer gates, which results in less power consumption and increased speed.
It's not just hardware where Boolean algebra shines. In software engineering, it's indispensable in control structures, like if statements and loops, where binary decisions have to be made.
Moreover, it's a crucial part of search algorithms, database querying, and even artificial intelligence. For these reasons, mastering Boolean algebra can open many doors in the tech industry.
## Operations in Boolean Algebra
In Boolean algebra, there are key operations that give this system its fundamental utility in fields such as engineering and computer science. These operations are AND, OR and NOT. When you understand these operations, you can begin to comprehend and manipulate logical statements, creating the foundation of complex digital systems.
### Explanation of Boolean Algebra Operations
Let's delve deep into the Boolean Algebra operations. First up is the **AND** operation, denoted by '.' or simply by writing variables together. This operation encapsulates the conjunction of two or more variables.
A conjunction is true only if all the variables involved in the AND operation are true. In other words, if A and B are Boolean variables, then the AND operation $$A.B$$ equals true or '1' only if both $$A$$ and $$B$$ are true or '1'. Otherwise, it equals false or '0'.
The **OR** operation is akin to the disjunction of Boolean variables.
A disjunction is true if at least one of the variables involved in the OR operation is true. It is denoted by '+'. Hence, If A and B are Boolean variables, then the OR operation $$A+B$$ equals true or '1' if either $$A$$ or $$B$$ or both are true or '1'. If both are false or '0', then $$A+B$$ equals false or '0'.
Next, we have the **NOT** operation, which is the simplest of all.
The NOT operation, denoted by $$\bar{A}$$ or A', basically reverses the value of a Boolean variable. If $$A$$ is true or '1', $$\bar{A}$$ is false or '0' and vice versa. This operation is the equivalent of negation.
#### Case Scenario of Boolean Algebra Operations in Practice
Let's illustrate these operations with an example from digital circuit design. Suppose you're designing a simple home security system that goes off if either the main gate is opened or a window is opened when the system is activated. Using Boolean variables, let's say 'M' represents the main gate, 'W' represents the window, and 'S' represents the state of the system (active or not). The action for the alarm to go off can be represented as follows:
Alarm = S. (M + W)
In this scenario, the alarm will only go off when the security system is active, and either the main gate or a window is opened. This is a basic example but effectively demonstrates how Boolean algebra operates in real-world situations.
#### Boolean Algebra Questions relevant to Operations
Now, let's look at some possible questions related to Boolean operations that you might encounter during your engineering studies. 1. Simplify the expression $$(A + B). (A + \bar{B}). (\bar{A} + B)$$ 2. For the full Boolean expression $$A. \bar{B} + \bar{A}. B + B. \bar{B}$$, compute the output for all possible inputs A and B. 3. Simplify $$(\bar{A} + \bar{B}). (A + B)$$ using Boolean algebra laws. Remember, practising these problems will help you acclimatise to Boolean algebra operations, and flex your problem-solving muscles, both key to mastering the role of Boolean algebra in the applications where they are used.
## Rules Governed by Boolean Algebra
In order to work effectively with Boolean algebra, it's vital to understand the set of foundational rules it follows. These rules aren't simply arbitrary, they form the driving principles that make Boolean algebra such a powerful tool in digital and computer logic.
### A Deep Insight into Boolean Algebra Rules
Boolean algebra operates on a set of rules that dictate the manipulation and simplification of Boolean expressions. While these might look a lot like ordinary algebraic rules at first glance, remember that Boolean algebra deals with binary values {0, 1}, making the rules behave differently. Here are some rules fundamental to Boolean algebra:
1. Identity Laws: These laws state that any Boolean value ORed with '0' or ANDed with '1' results in the original Boolean value itself. Mathematically, $$A + 0 = A$$ and $$A . 1 = A$$.
2. Null Laws: According to these rules, any Boolean value ORed with '1' gives '1', and any Boolean value ANDed with '0' gives '0'. Formally, $$A + 1 = 1$$ and $$A . 0 = 0$$.
3. Involution Law: This rule specifies that if the NOT operation is applied twice on any Boolean variable, it returns the original value. That means $$\overline{\overline{A}} = A$$.
4. Complement Laws: A Boolean variable ORed with its negation results in '1', and a Boolean variable ANDed with its negation results in '0'. Speaking mathematically, $$A + \overline{A} = 1$$ and $$A . \overline{A} = 0$$.
5. Commutative Laws: The order in which variables are ANDed or ORed does not matter. Formally, $$A + B = B + A$$ and $$A . B = B . A$$.
6. Associative Laws: When three variables are involved, the operation can be done using any order of pairing. That is $$A + (B + C) = (A + B) + C$$ and $$A . (B . C) = (A . B) . C$$.
7. Distributive Laws: One operation can be distributed over another, much like in ordinary algebra. Formally, $$A . (B + C) = (A . B) + (A . C)$$ and $$A + (B . C) = (A + B) . (A + C)$$.
8. Absorption Laws: These rules absorb redundant terms. That is $$A . (A + B) = A$$ and $$A + (A . B) = A$$.
#### Examples to Understand Boolean Algebra Rules Better
Let's use an example to understand these rules appropriately. Consider three Boolean variables $$A, B, C$$. According to the **Identity Laws**, we can calculate
A + 0 = A (OR operation with 0)
A . 1 = A (AND operation with 1)
As per the **Complement Laws**, we find
A + \overline{A} = 1 (OR operation with negation)
A . \overline{A} = 0 (AND operation with negation)
The **Associative Laws** can be seen in
A + (B + C) = (A + B) + C
A . (B . C) = (A . B) . C
The **Absorption Rules** demonstrate that
A . (A + B) = A
A + (A . B) = A
By understanding and applying these rules, manipulating and simplifying Boolean expressions becomes straightforward.
#### Boolean Algebra Questions About Rules
Let's examine some questions regarding Boolean algebra rules: 1. For given Boolean variables A, B, C, simplify the expression $$A.1 + \overline{A}$$ using the identity and complement laws. 2. Given $$A + AB = ?$$ simplify the expression using the absorption law. By working through these questions, you'll get practical experience in using these explanation of the key Boolean algebra rules, and how they translate into the principles that govern digital logic systems. Remember, practice is the key to mastering the rules governed by Boolean Algebra.
## Practical Examples of Boolean Algebra
Boolean Algebra is not just confined to textbooks or the learning spaces, it plays a significant role in our daily lives and professional sectors, particularly in realms such as computer programming, electronics, and digital circuits. Understanding its practical applications can pique your curiosity, and imbibe a deepened appreciation for this unique branch of algebra.
### Various Boolean Algebra Examples
Moving further, now let's explore various practical examples where Boolean Algebra finds major application.
#### Boolean Algebra Applications in Daily Life
Did you realise that every time you use a digital device or even carry out a simple Internet search, you're making use of Boolean Algebra? That's correct! Let's delve into specific instances here. 1. Search engines: Internet searches optimise the effectiveness of Boolean Operations. For instance, when you're trying to find information on 'Chocolate Cookies without Nuts', the search engine uses the NOT operation excluding web pages relating to 'Nuts'. Similarly, searching for 'Pizza OR Burger Restaurants' gives results that include either 'Pizza' or 'Burger'. 2. Digital Watches: The LED or LCD display of your watch utilises Boolean Algebra to show time, where each segment of the display represents a Boolean variable. 3. Alarm Systems: Boolean Algebra is also part of your home- or office-alarm system. A common setup might be:
Alarm = Door Sensor AND Motion Sensor
indicating that the alarm will only ring when BOTH the door and motion sensors are triggered. 4. Elevators: The lift system in buildings uses Boolean Algebra to decide floor movements. For example, to move up, the internal request must be higher AND no outer downwards request should be higher than the current floor. 5. Microwave Ovens: When you set your oven's auto-cook function (like Auto Defrost), it uses Boolean logic to decide when to switch off, considering factors such as the weight of the food item and the time entered.
#### Boolean Algebra Examples: Real-World Engineering Applications
Now let's see how Boolean Algebra concepts are applied in specific engineering applications: 1. Computer Programming: In computer science, Boolean Algebra is fundamental. It's used in coding conditional statements, loops, and arrays. For instance, an IF condition has Boolean logic. Here's a simple Python code snippet demonstrating this:
if x > 10 and y > 10:
print("Both numbers are greater than 10")
2. Logic Gates: Boolean operations symbolise the working of logic gates, which form the building blocks for all kinds of digital circuits, like microprocessors, counters, or calculators. 3. Digital Circuit Design: Boolean algebra is a pillar in the design and optimisation of digital circuits. For instance, with Karnaugh maps, a pictorial tool are used to simplify Boolean expressions without long, tedious algebraic manipulations. 4. Data Compressions: Boolean operations offer a technique for data compression. A simple form of data compression can be performed using the AND, OR, and XOR logic gates. 5. Networking: Boolean Algebra also finds applications in computer networking. IP addressing, subnetting and network gates are some areas in networking where Boolean algebra is used. These real-world examples undoubtedly illuminate Boolean Algebra as a bedrock in our digital world, revolutionising how systems operate and improving the efficiency of technological devices. It's fascinating to see how these abstract concepts significantly contribute to our daily life and the marvels of engineering.
## Role of Boolean Algebra in Engineering
The role of Boolean Algebra in engineering is paramount, specifically in the spheres of computer science and electrical engineering. It is a cornerstone for understanding, designing, and simplifying digital circuits, logical gates, microprocessors, and materialising binary operations. Moreover, it's instrumental in creating algorithms, setting conditions in programming, and managing the world of data structures. Grasping the rules, postulates and principles of Boolean Algebra equips engineering students and professionals alike with the skillset to excel in the digital and computation domain.
### Boolean Algebra Applications in Engineering
Digging deeper into the realms of engineering, Boolean Algebra's footprint can be easily traced in multiple disciplines. Computer Science: Boolean Algebra is synonymous with computer programming. Be it forming logical conditions in IF, WHILE, FOR loops or even setting conditions for switch statements, Boolean Algebra is the driving force. It also plays a crucial role in handling arrays and complex data structures. For example, the condition inside an if-else or switch-case statement involves Boolean Algebra. Here's a simple C++ code snippet demonstrating this:
if (x < 10 && y < 10)
{
cout<<"Both numbers are less than 10"<
Electronic Engineering: In electronics, Boolean logic forms the basis of digital design. Complex circuits, control systems, and digital system designs owe their operation to Boolean Algebra. The Implementation of logic gates such as AND OR, NOT, XOR, NAND are practical embodiments of Boolean Algebra.
Electrical Engineering: In electrical engineering, Boolean Algebra assists in the designs of switching circuits, serving as basic elements for automation, time sequence control, and the likes.
Telecommunication: Boolean Algebra is highly relevant in encoding, encryption, and data compression, crucial components of modern telecommunications. Boolean operations offer a technique for data compression, either lossy or lossless. Understanding Boolean Algebra hence becomes indispensable in the field of Communication.
Case Studies Representing Use of Boolean Algebra in Engineering
Exploring some specific instances can significantly highlight the crucial role of Boolean Algebra in engineering.
Case Study 1: Logic Gates - Logic gates are a fundamental example of Boolean Algebra application. They are digital components that perform operations on one or more logic inputs to produce a single logic output. All digital systems can be represented by logic gate networks.
Boolean operations represent each gate functioning. For example, OR gates, AND gates, NOT gates. Logic circuits perform tasks depending on the state of their inputs.
Case Study 2: Networks - Boolean Algebra is utilised in computer networking as well. IP addressing, subnet masking, and network gates are some areas where Boolean Algebra demonstrates its relevance.
Case Study 3: Microelectronics Microprocessors - Boolean Algebra assists in the design and operation of microprocessors. Microprocessors are essentially composed of various control units, each containing a host of logic gates.
Future Scope of Boolean Algebra in Engineering
Diving into the future, the importance of Boolean Algebra in advanced technology fields is becoming more pronounced. As the world gets progressively digitalised, Boolean Algebra becomes an increasingly central component of technological innovations.
Data Science and Machine Learning: Data Science and Machine Learning are two prominent areas where Boolean Algebra shows tremendous relevance. Features in machine learning algorithms often involve Boolean values.
Quantum Computing: Boolean Algebra also has applications in the rapidly developing field of quantum computing. Quantum logic gates, much like traditional digital logic gates, are constructed using Boolean functions.
Cyber Security: It is instrumental in cyber security where data encryption and safety protocols rely heavily on Boolean principles. From network security to data encryption, Boolean Algebra acts as the linchpin.
Artificial Intelligence: With the advent of AI and advanced robotics, Boolean logic is invariably a focal point. AI and robotic decision-making systems incorporate Boolean operations at their core.
Internet of Things (IoT): The IoT involves a multitude of digital devices communicating with each other. Implicit in this is a task of decision-making and event-handling, harnessing the power of Boolean Algebra.
Thus, Boolean Algebra will undoubtedly continue to be an indispensable asset in the future of engineering, serving as a principal instrument for illuminating the path leading to innovative realms of technology and automation.
Boolean Algebra - Key takeaways
Boolean algebra plays a significant role in engineering mathematics, particularly in computer and electrical engineering. It is used to simplify logic gates and circuits, control structures, and is crucial to search algorithms, database querying, and artificial intelligence.
In Boolean algebra, the key operations are AND, OR, and NOT. The AND operation is true if all variables involved are true, the OR operation is true if at least one variable is true, and the NOT operation reverses the value of a Boolean variable.
Boolean algebra operates according to certain rules, including identity laws, null laws, involution law, complement laws, commutative laws, associative laws, distributive laws, and absorption laws.
Boolean algebra is used in practical, everyday situations such as search engines, digital watches, alarm systems, elevators, and microwave ovens. It also has broader applications in areas like computer programming, digital circuit design, data compressions, and computer networking.
Understanding and applying Boolean algebra is vital in engineering, particularly in computer science, electronic engineering, electrical engineering, and telecommunications, where it is used in coding conditional statements, designing digital circuits, managing data structures, implementing logic gates, and in encoding and encryption processes.
#### Flashcards in Boolean Algebra 54
###### Learn with 54 Boolean Algebra flashcards in the free StudySmarter app
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What is Boolean algebra?
Boolean algebra is a sub-discipline of mathematics that deals with operations on logical values. It involves variables that can take two values: true or false. It's used extensively in computer science, digital electronics and the formulation of logical expressions.
How can I simplify Boolean algebra?
To simplify Boolean algebra, use the laws of Boolean algebra, including the identity law, null law, idempotent law, complement law, involution law, and the laws of commutativity, associativity, distributivity, absorption, and De Morgan's laws. It often involves grouping, factoring terms, and eliminating.
Is Boolean algebra difficult?
The difficulty of Boolean Algebra can vary. It depends largely on your understanding of logical operations. If you're comfortable with logical thinking and problem-solving, you'll likely find it straightforward. However, without this base, it might be challenging.
What is Boolean algebra used for?
Boolean algebra is used for designing and troubleshooting digital circuits. It helps in simplifying expressions and equations in digital electronics, which are important for constructing computing systems, electrical networks, and communication systems, among others.
Who invented Boolean algebra?
Boolean algebra was invented by an English mathematician named George Boole in the mid-19th century.
## Test your knowledge with multiple choice flashcards
What is an OR operation in Boolean Algebra and how is it denoted?
What is the Identity Law in Boolean Algebra?
What is Boolean Algebra and who is it named after?
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]> Counting Measure
## 7. Counting Measure
### Definitions and Basic Properties
Suppose that $S$ is a finite set. If $A S$ then the cardinality of $A$ is the number of elements in $A$, and is denoted $A$. The function is called counting measure. Counting measure plays a fundamental role in discrete probability structures, and particularly those that involve sampling from a finite set. The set $S$ is typically very large, hence efficient counting methods are essential. The first combinatorial problem is attributed to the Greek mathematician Xenocrates.
In many cases, a set of objects can be counted by establishing a one-to-one correspondence between the given set and some other set. Naturally, the two sets have the same number of elements, but for some reason, the second set may be easier to count.
#### The Addition Rule
The addition rule of combinatorics is simply the additivity axiom of counting measure. If $A 1 A 2 A n$ is a collection of disjoint subsets of $S$ then
$i 1 n A i i 1 n A i$
#### Simple Properties
The counting rules in this subsection are simple consequences of the addition rule.
Show that $A S A$.
Hint: $A$ and $A$ are disjoint and their union is $S$.
Show that $B A B B A$.
Hint: $A B$ and $B A$ are disjoint and their union is $B$.
Show that if $A B$ then $B A B A$.
Hint: Apply the difference rule and note that $A B A$
Show that if $A B$ then $A B$.
Thus, is an increasing function, relative to the subset partial order on $S$, and the ordinary order on .
#### Inequalities
This subsection gives two inequalities that are useful for obtaining bounds on the number of elements in a set.
Suppose that $A 1 A 2 A n$ is a a finite collection of subsets of $S$. Prove Boole's inequality (named after George Boole):
$i 1 n A i i 1 n A i$
1. Define $B 1 A 1$ and $B i A i A 1 A i 1$ for $i 2 3 n$.
2. Show that $B 1 B 2 B n$ are pairwise disjoint and have the same union as $A 1 A 2 A n$.
3. Use the addition rule
4. Use the increasing property.
Intuitively, Boole's inequality holds because parts of the union have been counted more than once in the expression on the right.
Suppose that $A 1 A 2 A n$ is a finite collection of subsets of $S$. Prove Bonferroni's inequality (named after Carlo Bonferroni):
$i 1 n A i S i 1 n S A i$
Hint: Apply Boole's inequality to $A 1 A 2 A n$ and use DeMorgan's law.
#### The Inclusion-Exclusion Formula
Show that $A B A B A B$
1. Note first that $A B A B A$ and the latter two sets are disjoint.
2. Now use the addition rule and the difference rule.
Show that $A B C A + B + C − A B − A C − B C + A B C$
Hint: Use the inclusion-exclusion rule for two sets. You will use this rule three times.
Exercise 7 and Exercise 8 can be generalized to a union of $n$ sets; the generalization is known as the inclusion-exclusion formula.
Suppose that $A i i I$ is a collection of subsets of $S$ where $I n$. Show that
$i I A i k 1 n 1 k 1 J I J k j J A j$
The general Bonferroni inequalities, named again for Carlo Bonferroni, state that if sum on the right is truncated after $k$ terms ($k n$), then the truncated sum is an upper bound for the cardinality of the union if $k$ is odd (so that the last term has a positive sign) and is a lower bound for the cardinality of the union if $k$ is even (so that the last terms has a negative sign).
#### The Multiplication Rule
The multiplication rule of combinatorics is based on the formulation of a procedure (or algorithm) that generates the objects to be counted. Specifically, suppose that the procedure consists of $k$ steps, performed sequentially, and that for each $j$, step $j$ can be performed in $n j$ ways regardless of the choices made on the previous steps. Then the number of ways to perform the entire algorithm (and hence the number of objects) is $n 1 n 2 n k$.
The key to a successful application of the multiplication rule to a counting problem is the clear formulation of an algorithm that generates the objects being counted, so that each object is generated once and only once. That is, we must neither over-count nor under-count.
The first two exercises below give equivalent formulations of the multiplication principle.
Suppose that $S$ is a set of sequences of length $k$, and that we denote a generic element of $S$ by $x 1 x 2 x k$. Suppose that for each $j$, $x j$ has $n j$ different values, regardless of the values of the previous coordinates. Show that
$S n 1 n 2 n k$
Suppose that $T$ is an ordered tree with depth $k$ and that each vertex at level $i 1$ has $n i$ children for $i 1 2 k$. Show that the number of vertices at level $k$ is $n 1 n 2 n k$.
#### Product Sets
Suppose that $S i$ is a set with $n i$ elements for $i 1 2 k$. Show that
$S 1 S 2 S k n 1 n 2 n k$
Show that if $S$ is a set with $n$ elements, then $S k$ has $n k$ elements.
Show that the number of ordered samples of size $k$ that can be chosen with replacement from a population of $n$ objects is $n k$.
#### Functions and Bit Strings
Show that the number of functions from a set $A$ of $n$ elements into a set $B$ of $m$ elements is $m n$.
Hint: To construct $f A B$, there are $m$ choices for $f x$ for each $x A$.
Recall that the set of functions from a set $A$ into a set $B$ (regardless of whether the sets are finite or infinite) is denoted $B A$. The result in the previous exercise is motivation for this notation.
Elements of $0 1 n$ are sometimes called bit strings of length $n$. Show that the number of such strings is $2 n$.
#### Subsets
Suppose that $S$ is a set with $n$ elements. Show that there are $2 n$ subsets of $S$.
Hint: To construct $A S$, decide whether $x A$ or $x A$ for each $x S$.
Suppose that $A 1 A 2 A k$ is a collection of $n$ subsets of a set $S$. Show that there are $2 2 k$ different (in general) sets that can be constructed from the $k$ given sets, using the operations of union, intersection, and complement. These sets form the algebra generated by the given sets.
1. Show that there are $2 k$ pairwise disjoint sets of the form $B 1 B 2 B k$ where $B i A i$ or $B i A i$ for each $i$.
2. Argue that every set that can be constructed from $A 1 A 2 A k$ is a union of some (perhaps all, perhaps none) of the sets in (a).
In the Venn diagram applet, observe the diagram of each of the 16 sets that can be constructed from $A$ and $B$.
Suppose that $S$ is a set with $n$ elements and that $A$ is a subset of $S$ with $k$ elements. Show that the number of subsets of $S$ that contain $A$ is $2 n k$.
### Computational Exercises
A license number consists of two letters (uppercase) followed by five digits. How many different license numbers are there?
Suppose that a Personal Identification Number (PIN) is a four-symbol code word in which each entry is either a letter (uppercase) or a digit. How many PINs are there?
In the board game Clue, Mr. Boddy has been murdered. There are 6 suspects, 6 possible weapons, and 9 possible rooms for the murder.
1. The game includes a card for each suspect, each weapon, and each room. How many cards are there?
2. The outcome of the game is a sequence consisting of a suspect, a weapon, and a room (for example, Colonel Mustard with the knife in the billiard room). How many outcomes are there?
3. Once the three cards that constitute the outcome have been randomly chosen, the remaining cards are dealt to the players. Suppose that you are dealt 5 cards. In trying to guess the outcome, what hand of cards would be best?
An experiment consists of rolling a fair die, drawing a card from a standard deck, and tossing a fair coin. How many outcomes are there?
A fair die is rolled 5 times and the sequence of scores recorded. How many outcomes are there?
Suppose that 10 persons are chosen and their birthdays recorded. How many possible outcomes are there?
In the card game Set, each card has 4 properties: number (one, two, or three), shape (diamond, oval, or squiggle), color (red, blue, or green), and shading (solid, open, or stripped). The deck has one card of each (number, shape, color, shading) configuration. A set in the game is defined as a set of three cards which, for each property, the cards are either all the same or all different.
1. How many cards are in a deck?
2. How many sets are there?
A fair coin is tossed 10 times and the sequence of scores recorded.
1. How many sequences are there?
2. How many sequences are there that contain exactly 3 heads?
A string of lights has 20 bulbs, each of which may be good or defective. How many configurations are there?
The die-coin experiment consists of rolling a die and then tossing a coin the number of times shown on the die. The sequence of coin results is recorded.
1. How many outcomes are there?
2. How many outcomes are there with exactly 2 heads?
Run the die-coin experiment until exactly 2 heads occurs.
Suppose that a sandwich at a restaurant consists of bread, meat, cheese, and various toppings. There are 4 different types of bread (select one), 3 different types of meat (select one), 5 different types of cheese (select one), and 10 different toppings (select any). How many sandwiches are there?
#### The Galton Board
The Galton Board, named after Francis Galton, is a triangular array of pegs. Galton, apparently too modest to name the device after himself, called it a quincunx from the Latin word for five twelfths (go figure). The rows are numbered, from the top down, by $0 1 2$. Row $k$ has $k 1$ pegs that are labeled, from left to right by $0 1 k$. Thus, a peg can be uniquely identified by an ordered pair $i j$ where $i$ is the row number and $j$ is the peg number in that row.
A ball is dropped onto the top peg $0 0$ of the Galton board. In general, when the ball hits peg $i j$, it either bounces to the left to peg $i 1 j$ or to the right to peg $i 1 j 1$ The sequence of pegs that the ball hits is a path in the Galton board.
Show that there is a one-to-one correspondence between each pair of the following three collections:
1. Bit strings of length $n$
2. Paths in the Galton board from $0 0$ to a peg in row $n$.
3. Subsets of a set with $n$ elements.
Thus, from the previous exercise, each of these collections has $2 n$ elements.
Open the Galton board applet.
1. Move the ball from $0 0$ to $10 6$ along a path of your choice. Note the corresponding bit string and subset.
2. Generate the bit string $011100101$. Note the corresponding subset and path.
3. Generate the subset $2 4 5 9 12$. Note the corresponding bit string and path.
4. Generate all paths from $0 0$ to $4 2$. How many paths are there? |
The Component Method, Part 1
The Basics - Good Vectors from Bad
The component method of summing vectors is universally feared by introductory physics students, but is actually simple as long as you don't get too worried about trigonometric details. The foundation of the component method actually relies on a basic principle:
Vectors are easy to sum if they fall into two categories:
• The vectors point along the same line. In this case, the sum of the two vectors is just the summation of the lengths if they point in the same direction, and the subtraction of the two lengths if they point in opposite directions.
• The vectors are perpendicular to each other. Here, simple mathematical relationships can be used to solve for the resultant vector.
To fully appreciate the importance of these two principles, complete the following web assignment.
What are component vectors?
Consider the equation vector C = A + B, which simply states that "vector C is the sum of vectors A and B." This statement is an equality, which is a very strong statement -- it means that C can be replaced with the term A + B whenever we see fit.
We will take advantage of this property to greatly simplify vector addition by replacing vectors that are hard to sum (we will call them bad vectors) with vectors that are easy to sum (good vectors).
Now there are any number of vectors that sum to vector C, but we are going to choose A and B very carefully. Consider the following example, where we are trying to add the two vectors C and G. If you checked out the previous web assignment above you would appreciate how difficult it is to add these two vectors because they do not point along the same direction, nor are they perpendicular. Therefore, we consider them bad vectors.
But why not replace both bad vectors with good vectors that are easy to sum?
In the figure below, we have defined two directions with dashed lines we call the x-direction (horizontal) and y-direction (vertical). If we can somehow find replacement vectors for C and G that line up along these directions, then we will be guaranteed that all vectors will either be pointing along the same direction or perpendicular to each other.
But how do we find these good vectors?
Applet by Zona Land
In the applet above, we have isolated vector C (in black). Notice that vectors A and B (the red and blue vectors) point along the x-direction or y-direction and that they both sum to C.
Now using the mouse, drag vector C around and watch what happens to A and B. Notice that no matter which direction C is pointing, vectors A and B always form a perfect rectangle.
In the figure below, we have replaced both vectors C and G with their component vectors. We can erase vectors C and G if we wish-- they are no longer needed. We now have four vectors instead of two, but each of these four vectors is guaranteed to be either parallel or perpendicular to the others.
We now have four vectors instead of two (see the figure below), but each of these four vectors is guaranteed to be either parallel or perpendicular to the others.
Using our basic rules listed above, we can guess that we have now simplified the problem a great deal. And we have.
A Little Bit of Trig (and I Mean Little)
Now that we know we have to replace C with A and B, how do we find the lengths of A and B? (We already know their directions, right?) To find these lengths is simple, and requires nothing more than the ability to punch the "COS" and "SIN" buttons on your calculator.
But what are COS and SIN?
Notice in the applet above that vectors A and B are always shorter than C. The COS and SIN buttons simply take the length and direction of C and use this information to find the lengths of A and B. In the figure below we examine our earlier example, defining a direction of C of 58 degrees with respect to the x-direction (horizontal).
But which do I choose? COS or SIN?
The rule here is simple: Do you see the angle symbol that we have drawn in the figure? Well, if that angle sign touches the component vector for which you are trying to find its length, use COS.
If the angle sign touches, it's COS!
If you use COS on one component vector, use SIN for the other. In summary, COS and SIN are "shrink factors" -- the amount of shrinkage depends on the angle.
So in our example above, the length of vector B is given by C X COS(58). The reason? The angle symbol touches vector B, so we use the COS button. Since we used COS for B, we use SIN for A. In summary,
B = C X COS(58),
A = C X SIN(58).
Why are the vector symbols not boldfaced in the equations above? What do the SIN and COS buttons really mean?
Suppose that the length of vector C was 60 (in whatever units). Then,
B = 60 X COS(58) = 60 X 0.53 = 32,
A = 17 X SIN(58) = 60 X 0.85 = 51.
But how do I punch these calculations out on my calculator? Why don't I get those numbers when I use my calculator!?!?
According to these calculations, vector A should be longer than vector B, and nearly as long as the original vector C. And vector B should be just slightly longer than half of vector C. Looking at the figure above, this appears to be the case. (You can use the above applet to verify these results as well, although it requires some precise movement of the mouse.)
So all the trigonometry you are going to need at this point consists solely of finding the angle symbol, seeing which component vector it touches, and using the COS button to find the length of that vector. Use the SIN button for the other and voila, you have successfully broken the original vector into its componenent vectors.
Naturally, you would then do the same for vector G.
Before moving on, test your ability to break a vector into its components. |
## Monday, March 30, 2020
### Newton's Second Law and Final Velocity
Suppose you are faced with a problem such as:
A motorcycle stunt rider needs to reach a final velocity of 799 m/s in order to make the jump needed for the scene in the movie. If she has an initial velocity of 10 m/s and the total mass of her and the motorcycle is 125 kg and if she can create a force of 100 N, how much time will be needed to achieve the needed final velocity?
On the surface, it looks (exciting, but also) like this could be hard to do.
IT'S NOT!
The velocity formula is very easy:
vf = vi + (a)(t)
In the problem above, you are looking for t and vf and vi are given. And then there's that mass and force...
You need an acceleration(a), but you have mass and force. Thankfully Newton did that thing:
F = ma
So, we'll be doing two steps.
Step 1:
Use
F=ma to find a
Step 2:
Then use THAT in the velocity formula.
vf = vi + (a)(t)
EXAMPLES
Example 1
Let's take a look at another example (even easier!) and work it out:
A force of 50 N acts on a object with a mass of 10 kg. If it has an initial velocity of 20 m/s, what is its final velocity after an elapsed time of 8 seconds?
The question tells us that we need to find final velocity (vf). But there is no acceleration given. So, do step 1 (above):
Find a where:
F = 50 N
m = 10 kg
F = ma
50 N = (10 kg)(a)
50 N/10 kg = a
5 m/s/s = a
Next, use THAT calculated a to find final velocity (step 2 above):
Find vf where
vi = 20 m/s
t = 8 s
a = 5 m/s/s
vf = vi + (a)(t)
vf = 20 m/s + (5 m/s/s)(8 s)
vf = 20 m/s + 40 m/s
vf = 60 m/s
Example 2
How about seeing it worked out?
SUMMARY:
You gotta do it in steps!
This process requires doing the work in steps. Depending on what is given, you use the two formulas below:
F = ma
vf = vi + (a)(t)
So, read the problem, write down what is being looked for, write down what is given, THEN... use the two formulas. Two steps! |
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### Section 4-5 : Miscellaneous Functions
The point of this section is to introduce you to some other functions that don’t really require the work to graph that the ones that we’ve looked at to this point in this chapter. For most of these all that we’ll need to do is evaluate the function as some $$x$$’s and then plot the points.
#### Constant Function
This is probably the easiest function that we’ll ever graph and yet it is one of the functions that tend to cause problems for students.
The most general form for the constant function is,
$f\left( x \right) = c$
where $$c$$ is some number.
Let’s take a look at $$f\left( x \right) = 4$$ so we can see what the graph of constant functions look like. Probably the biggest problem students have with these functions is that there are no $$x$$’s on the right side to plug into for evaluation. However, all that means is that there is no substitution to do. In other words, no matter what $$x$$ we plug into the function we will always get a value of 4 (or $$c$$ in the general case) out of the function.
So, every point has a $$y$$ coordinate of 4. This is exactly what defines a horizontal line. In fact, if we recall that $$f\left( x \right)$$ is nothing more than a fancy way of writing $$y$$ we can rewrite the function as,
$y = 4$
And this is exactly the equation of a horizontal line.
Here is the graph of this function.
#### Square Root
Next, we want to take a look at $$f\left( x \right) = \sqrt x$$. First, note that since we don’t want to get complex numbers out of a function evaluation we have to restrict the values of $$x$$ that we can plug in. We can only plug in value of $$x$$ in the range $$x \ge 0$$. This means that our graph will only exist in this range as well.
To get the graph we’ll just plug in some values of $$x$$ and then plot the points.
$$x$$ $$f(x)$$
0 0
1 1
4 2
9 3
The graph is then,
#### Absolute Value
We’ve dealt with this function several times already. It’s now time to graph it. First, let’s remind ourselves of the definition of the absolute value function.
$f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}x&{{\mbox{if }}x \ge 0}\\{ - x}&{{\mbox{if }}x < 0}\end{array}} \right.$
This is a piecewise function and we’ve seen how to graph these already. All that we need to do is get some points in both ranges and plot them.
Here are some function evaluations.
$$x$$ $$f(x)$$
0 0
1 1
-1 1
2 2
-2 2
Here is the graph of this function.
So, this is a “V” shaped graph.
#### Cubic Function
We’re not actually going to look at a general cubic polynomial here. We’ll do some of those in the next chapter. Here we are only going to look at $$f\left( x \right) = {x^3}$$. There really isn’t much to do here other than just plugging in some points and plotting.
$$x$$ $$f(x)$$
0 0
1 1
-1 -1
2 8
-2 -8
Here is the graph of this function.
We will need some of these in the next section so make sure that you can identify these when you see them and can sketch their graphs fairly quickly.
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# Reduce the following equations into normal form.
Question:
Reduce the following equations into normal form. Find their perpendicular distances from the origin and angle between perpendicular and the positive x-axis.
(i) $x-\sqrt{3} y+8=0$
(ii) $y-2=0$
(iii) $x-y=4$
Solution:
(i) The given equation is $x-\sqrt{3} y+8=0$.
It can be reduced as:
$x-\sqrt{3} y=-8$
$\Rightarrow-x+\sqrt{3} y=8$
On dividing both sides by $\sqrt{(-1)^{2}+(\sqrt{3})^{2}}=\sqrt{4}=2$, we obtain
$-\frac{x}{2}+\frac{\sqrt{3}}{2} y=\frac{8}{2}$
$\Rightarrow\left(-\frac{1}{2}\right) x+\left(\frac{\sqrt{3}}{2}\right) y=4$
$\Rightarrow x \cos 120^{\circ}+y \sin 120^{\circ}=4$ ..(1)
Equation (1) is in the normal form.
On comparing equation (1) with the normal form of equation of line
$x \cos \omega+y \sin \omega=p$, we obtain $\omega=120^{\circ}$ and $p=4$
Thus, the perpendicular distance of the line from the origin is 4, while the angle between the perpendicular and the positive x-axis is 120°.
(ii) The given equation is – 2 = 0.
It can be reduced as 0.x + 1.= 2
On dividing both sides by $\sqrt{0^{2}+1^{2}}=1$, we obtain $0 \cdot x+1 \cdot y=2$
$\Rightarrow x \cos 90^{\circ}+y \sin 90^{\circ}=2 \ldots$ (1)
Equation (1) is in the normal form.
On comparing equation (1) with the normal form of equation of line
$x \cos \omega+y \sin \omega=p$, we obtain $\omega=90^{\circ}$ and $p=2$.
Thus, the perpendicular distance of the line from the origin is 2, while the angle between the perpendicular and the positive x-axis is 90°.
(iii) The given equation is $x-y=4$.
It can be reduced as $1 \cdot x+(-1) y=4$
On dividing both sides by $\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}$, we obtain
$\frac{1}{\sqrt{2}} x+\left(-\frac{1}{\sqrt{2}}\right) y=\frac{4}{\sqrt{2}}$
$\Rightarrow x \cos \left(2 \pi-\frac{\pi}{4}\right)+y \sin \left(2 \pi-\frac{\pi}{4}\right)=2 \sqrt{2}$
$\Rightarrow x \cos 315^{\circ}+y \sin 315^{\circ}=2 \sqrt{2}$ $\ldots(1)$
Equation (1) is in the normal form.
On comparing equation (1) with the normal form of equation of line
$x \cos \omega+y \sin \omega=p$, we obtain $\omega=315^{\circ}$ and $\mathrm{p}=2 \sqrt{2}$.
Thus, the perpendicular distance of the line from the origin is $2 \sqrt{2}$, while the angle between the perpendicular and the positive $x$-axis is $315^{\circ}$. |
# Difference between revisions of "2015 AIME I Problems/Problem 3"
## Problem
There is a prime number $p$ such that $16p+1$ is the cube of a positive integer. Find $p$.
## Solution
Let the positive integer mentioned be $a$, so that $a^3 = 16p+1$. Note that $a$ must be odd, because $16p+1$ is odd.
Rearrange this expression and factor the left side (this factoring can be done using $(a^3-b^3) = (a-b)(a^2+a b+b^2)$, or synthetic divison once it is realized that $a = 1$ is a root):
$a^3-1 = 16p$
$(a-1)(a^2+a+1) = 16p$
Because $a$ is odd, $a-1$ is even and $a^2+a+1$ is odd. If $a^2+a+1$ is odd, $a-1$ must be some multiple of $16$. However, for $a-1$ to be any multiple of $16$ other than $16$ would mean $p$ is not a prime. Therefore, $a-1 = 16$ and $a = 17$.
Then our other factor, $a^2+a+1$, is the prime $p$:
$(a-1)(a^2+a+1) = 16p$
$(17-1)(17^2+17+1) =16p$
$p = 289+17+1 = \boxed{307}$. |
# Complex Number and Cartesian Form Essay
Submitted By burrb
Words: 2117
Pages: 9
1. COMPLEX NUMBERS Revision
K.J.WILLIAMS
School of Mechanical & Auto Engineering
Faculty of Science, Engineering & Computing
KINGSTON UNIVERSITY
Learning Objectives
• To revise your knowledge of Complex
Numbers in preparation for further work on the application of complex numbers to engineering applications – A.C. electrical/ electronic systems
2
Origins of Complex Numbers
• Complex numbers arise naturally from problems in mathematics involving the roots of negative numbers. • eg :– in quadratic equations such as:• ax2 + bx + c = 0, the general solution is:• x = -b +/- √(b2 – 4ac)
2a
3
Origins of Complex Numbers
• If b2 happens to be less than 4ac, then the contents of the square root will be negative – eg. √-9
• Normally, we cannot have a solution to this! NOTE :– 3 x 3 = 9 and -3 x -3 = 9
• but no number when multiplied by itself has the answer – 9!
• There is no such number!
4
Origins of Complex Numbers
• This problem was known to mathematics for a long time and was a stumbling block for a while.
• Eventually, a solution was found which involved a dodge that separated the minus part from the positive part.
• eg:- -9 = (-1) x 9 so that √-9 = √(-1) x 9
5
Origins of Complex Numbers
Now:- -9 = 9 x (-1), so that
√-9 = √(9 x (-1) ),
= √9 x √(-1),
= 3 x √(-1).
Mathematicians gave a special symbol to the √(-1) and they called it i, where:• i = √(-1)
• So the answer to √-9 = i3 or 3i
6
Real & Imaginary Numbers
• Numbers such as 3i are called imaginary number, because normally, we don’t encounter them and after all, it is just a dodge to get out of a problem.
• The numbers we normally encounter in our everyday life are called real numbers,
These are numbers such as:- integers, fractions and decimals eg:- 100, 50, 2/3,
0.5 and so on.
7
Complex Numbers
• So now, in mathematics, we have two types of numbers:• Real Numbers:– 25, 1034, ¾, 0.6 and so on. • Imaginary Numbers:- 25i, 1034i, 3/4i, 0.6i
• and so on.
• Then, mathematicians considered the possibility of combing the two to form
Complex numbers.
8
Complex Numbers
• So Complex numbers are a combination of
Real and Imaginary numbers.
• Real Numbers:- 50, -0.6, 1/8.
• Imaginary numbers:- -50i, 0.6i, 1/8i.
• Complex Numbers:-(3 + 4i),(-2 -5i),(4-0.6i)
• Just as in algebra, we can use x to represent a real number, we can also use letters to represent complex numbers:• eg:- Z = 3 +4i
9
i or j
• Mathematicians originally used the letter i to indicate imaginary numbers.
• Later on, however, electrical engineers found that complex numbers could be used to analyse
A.C. electrical circuits. The problem was that engineers were already using the symbol “i” to represent current – so they changed the symbol for √(-1)from i to j. Nowadays, mathematicians still use i for imaginary numbers while electrical engineers prefer to use j to avoid confusion.
10
Cartesian (X-Y or Rectangular)
Coordinates
• Complex numbers can be thought of as two dimensional numbers since the real part and imaginary part are separate. We can plot complex numbers graphically in a two dimensional complex plane using
Cartesian (x-y rectangular) coordinates with Real numbers plotted horizontally and
Imaginary numbers plotted vertically. This is also known as an Argand Diagram.
11
Cartesian (x-y rectangular)
Coordinates
• Argand Diagram or Complex Plane
Z = 3+4i
Imaginary numbers 4i
0
3
Real numbers
12
Polar Coordinates
• In addition to using Cartesian coordinates for complex numbers, it is often convenient to represent them on a Polar
Plot of magnitude (or modulus) vs angle
(or argument). Even though we can do this, the complex number still consists of two distinct parts which can be easily recovered. 13
Modulus & Argument
• Using Cartesian (x/y, or rectangular) coordinates, a complex number is represented as a point on the complex plane (or Argand diagram).
• Using Polar coordinates the same complex number is represented by a vector with a magnetude (modulus) and an
angle(argument) |
# What is the probability of getting a number 8 in a single throw of a dice?
Contents
Total Number of combinations Probability
7 6 16.67%
8 5 13.89%
9 4 11.11%
10 3 8.33%
## What is the probability of getting a number 8 on on a single throw of a die?
8 is not present in sample space. so its probability is zero.
## What is the probability of obtaining a total of 8 points in a single throw with two dice?
The required probability is 5/36 .
## What is the probability of getting 7 in a single throw of dice?
Answer: There is no 7 in dice at all. So, 0/6 is the Probability of getting a 7 in a single throw of a dice.
## What is the probability of getting an even number?
Now as we know that the probability (P) is the ratio of favorable number of outcomes to the total number of outcomes. So the required probability of getting an even number is \$dfrac{1}{2}\$. Hence option (B) is correct.
## What is the probability of getting a total of 8?
As the chart shows the closer the total is to 7 the greater is the probability of it being thrown.
Probabilities for the two dice.
IT IS INTERESTING: What figurative language is in the Lottery by Shirley Jackson?
Total Number of combinations Probability
7 6 16.67%
8 5 13.89%
9 4 11.11%
10 3 8.33%
## What is the probability of getting a total of 9 in a single throw of two dice?
The probability of getting 9 as the sum when 2 dice are thrown is 1/9.
## What is the probability of getting a 7 in a regular die Brainly?
If you do so you will find that the sum is 7 for 6 of the possible outcomes. Thus the sum is a 7 in 6 of the 36 outcomes and hence the probability of rolling a 7 is 6/36 = 1/6.
## What is P even number?
(a) At least one of the dice shows an even number? P(at least one is even) = 1 – P(both are odd). And the probability that the first die shows an odd number is 1/2, as is the probability that the second does. Since the dice fall independently, P(both are odd) = P(first is odd)*P(second is odd) = (1/2)*(1/2) = 1/4. |
# Convert Scale Diagrams to Real Measurements
In this worksheet, students will apply scale factors to find the real life values of specific scaled elements or calculate the scale which has been used to create a scale diagram or model.
Key stage: KS 4
Year: GCSE
GCSE Subjects: Maths
GCSE Boards: AQA, Eduqas, Pearson Edexcel, OCR,
Curriculum topic: Ratio, Proportion and Rates of Change, Mensuration
Curriculum subtopic: Ratio, Proportion and Rates of Change Units and Measurement
Difficulty level:
#### Worksheet Overview
A lot of the time, it's not really the best idea to have things full size.
A map would be useless if it was full size, and a toy car doesn't really want to be the same size as a normal car, does it?
When we have an issue like this in maths, we use a scale diagram or a scale model.
What Is a Scale Diagram / Model?
A scale diagram is just a diagram where everything has been reduced by the same factor
It could be half the size, a tenth of the size, or anything else, but every element must be reduced by exactly the same factor.
How Are Scales Written?
Scales are written as ratios, such as 1: 100 or 1:50,000.
What Does a Scale Mean?
Scales are read from left to right.
For example, the scale 1:100 would mean that every 1 unit of length on the scale is the same as 100 units in real life.
So an element that was 2 cm long on a scale diagram, would be 200 cm long in real life.
Let's look at this concept in action with some examples now.
e.g. A model car is 1.5 cm tall. If the scale used to create it is 1:100, how tall would the car be in real life?
All we need to do here is multiply by the scale factor:
1.5 × 100 = 150 cm
So the car would be 150 cm or 1.5 m tall in real life.
e.g. A model is made of a 2 m tall man. If the model is 4 cm tall, what scale has been used?
The first thing we should notice here is that the units used are different, so we need to make them the same before we start:
There are 100 cm in 1 m, so 2 m = 200 cm.
Now, we need to write these numbers as a ratio (remember that the model comes first):
4:200
Our final step is to simplify this ratio:
4:200 ÷ 4
1:50
In this activity, we will apply scale factors to find the real life scale values of specific scaled elements or calculate the scale which has been used to create a scale diagram or model.
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## CAT 2022 : Time and Work Practice Exercise-3
CAT Quantitative Aptitude section consists of problems based on Time and Work which have a considerate amount of weightage. The problems are based on amount of work done by people and the respective time taken by them. Cumulative work or total time taken is given and you are asked the independent time taken and vice versa. In this post, we share CAT Time and Work Problems and Solutions that you can use to super-charge your preparation and have a taste of the problems asked for this area.
Question 1: If 10 men or 20 boys can make 260 mats in 20 days, then 8 men and 4 boys will make how many mats?
(1) 260
(2) 240
(3) 280
(4) 520
### Answer and Explanation
10 men = 20 boys
∴ 1 man = 2 boys
∴ 8 men + 4 boys
= (16 + 4) boys = 20 boys
Hence, 8 men and 5 boys will make 260 mates in 20 days.
Question 2: 20 men or 24 women can com¬plete a piece of work in 20 days. If 30 men and 12 women under-take to complete the work how much time will they take?
(1) 10 days
(2) 12 days
(3) 15 days
(4) 16 days
### Answer and Explanation
20 men ≡ 24 women
⇒ 5 men ≡ 6 women
∴ 30 men + 12 women = 30 men+10 men=40 Men
∴ M1D1 = M2D2
⇒ 20 Men× 20 Days = 40 Men × D2days
⇒ D2 = (20 x 20)/40 = 10 days
Question 3: 8 days are taken by one man and one woman together to complete the job. A man alone can complete the work in 10 days. In how many days can one woman alone complete the work?
(1) 140/9
(2) 30
(3) 40
(4) 42
### Answer and Explanation
Work done by 1 woman in 1 day
= 1/8 – 1/10 = (5-4)/40 = 1/40
∴ One woman will complete the wok in 40 days.
Question 4: 4 men and 6 women can com¬plete a work in 8 days, while 3 men and 7 women can complete it in 10 days. In how many days will 10 women complete it?
(1) 50
(2) 45
(3) 40
(3) 35
### Answer and Explanation
Let 1 man’s 1 day’s work = x and
1 woman’s 1 day’s work = y
Then, 4a + 6b = 1/8
3a + 7b = 1/10
On solving, we get b = 1/400
∴ 10 women’s 1 day’s work
= 10/400 = 1/40
∴ 10 women will finish the work in 40 days.
Question 5: A man, a woman and a boy can complete a work in 20 days, 30 days and 60 days respectively. How many boys must assist 2 men and 8 women so as to com¬plete the work in 2 days?
(1) 8
(2) 12
(3) 4
(4) 6
### Answer and Explanation
Extra tips for CAT Time and Work Problems and Solutions:
• You can brush up the basic to solve Time and work problems by going through your 11th and 12th text books as well as school textbooks. This topic is relatively easy and needs basic knowledge of arithmetic and a basic understanding of the relationship between time and work.
• The most important formula you need to remember for solving every time and work formula is:
If X can do some piece of work in say ‘n’ number of days, then X’s 1 day’s work is equal to 1/n.
• You must always remember that Time and work are inversely proportional to each other.
• Solve as many time and work problems as possible. Some of the problems can be solved by simple logic and do not need any solving.
• Time and Work problems are less time consuming and definitely worth attempting in the exam.
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# Intro to probability questions
• Jan 27th 2010, 02:42 AM
rozekruez
Intro to probability questions
If the probability of the Leafs defeating Senators in a hockey game is 3/7, what is the probability that the Leafs will win two consecutive games against the Senators?
A box has a group of 24 blocks in it. Some are red, some are yellow, and some are a mixture of the two colors. The probability of drawing a red block is 1/3. The probability of drawing a red and yellow block is 1/12. Determine the number of blocks with yellow on them.
• Jan 27th 2010, 02:48 AM
e^(i*pi)
Quote:
Originally Posted by rozekruez
If the probability of the Leafs defeating Senators in a hockey game is 3/7, what is the probability that the Leafs will win two consecutive games against the Senators?
Assuming the two events are independent then the second match has the same chance as the first. To find AND probabilities you should multiply each chance:
Spoiler:
$\left(\frac{3}{7}\right)^2 = \frac{9}{49}$
Quote:
A box has a group of 24 blocks in it. Some are red, some are yellow, and some are a mixture of the two colors. The probability of drawing a red block is 1/3. The probability of drawing a red and yellow block is 1/12. Determine the number of blocks with yellow on them.
As the outcomes are mutually exclusive we can say that
$P(R) + P(Y) + P(R,Y) = 1$
In this case it is consider the chance of not getting a yellow block and take it from 1.
Spoiler:
$P(Y or R,Y) = 1 - P(R) = 1 - \frac{1}{3} = \frac{2}{3}$
Multiply the outcome by 24 to get the total number of blocks which are not red/which have yellow on them.
Spoiler:
$\frac{2}{3} \times 24 = 16$
• Jan 27th 2010, 02:55 AM
Quote:
Originally Posted by rozekruez
A box has a group of 24 blocks in it. Some are red, some are yellow, and some are a mixture of the two colors. The probability of drawing a red block is 1/3. The probability of drawing a red and yellow block is 1/12. Determine the number of blocks with yellow on them.
Alternatively
$P(Red)=\frac{1}{3}=\frac{R}{24}$
$P(Red\ and\ Yellow)=\frac{1}{12}=\frac{M}{24}$
The number of reds can now be calculated along with the number of those of mixed colours. Subtract the sum of these from 24 to find the yellow amount. |
## Composition of Functions
### LEARNING OBJECTIVES
By the end of this lesson, you will be able to:
• Combine functions using algebraic operations.
• Create a new function by composition of functions.
• Evaluate composite functions.
• Find the domain of a composite function.
• Decompose a composite function into its component functions.
Suppose we want to calculate how much it costs to heat a house on a particular day of the year. The cost to heat a house will depend on the average daily temperature, and in turn, the average daily temperature depends on the particular day of the year. Notice how we have just defined two relationships: The cost depends on the temperature, and the temperature depends on the day.
Figure 1
Using descriptive variables, we can notate these two functions. The function $C\left(T\right)$ gives the cost $C$ of heating a house for a given average daily temperature in $T$ degrees Celsius. The function $T\left(d\right)$ gives the average daily temperature on day $d$ of the year. For any given day, $\text{Cost}=C\left(T\left(d\right)\right)$ means that the cost depends on the temperature, which in turns depends on the day of the year. Thus, we can evaluate the cost function at the temperature $T\left(d\right)$. For example, we could evaluate $T\left(5\right)$ to determine the average daily temperature on the 5th day of the year. Then, we could evaluate the cost function at that temperature. We would write $C\left(T\left(5\right)\right)$.
By combining these two relationships into one function, we have performed function composition, which is the focus of this section.
## Combining Functions Using Algebraic Operations
Function composition is only one way to combine existing functions. Another way is to carry out the usual algebraic operations on functions, such as addition, subtraction, multiplication and division. We do this by performing the operations with the function outputs, defining the result as the output of our new function.
Suppose we need to add two columns of numbers that represent a husband and wife’s separate annual incomes over a period of years, with the result being their total household income. We want to do this for every year, adding only that year’s incomes and then collecting all the data in a new column. If $w\left(y\right)$ is the wife’s income and $h\left(y\right)$ is the husband’s income in year $y$, and we want $T$ to represent the total income, then we can define a new function.
$T\left(y\right)=h\left(y\right)+w\left(y\right)$
If this holds true for every year, then we can focus on the relation between the functions without reference to a year and write
$T=h+w$
Just as for this sum of two functions, we can define difference, product, and ratio functions for any pair of functions that have the same kinds of inputs (not necessarily numbers) and also the same kinds of outputs (which do have to be numbers so that the usual operations of algebra can apply to them, and which also must have the same units or no units when we add and subtract). In this way, we can think of adding, subtracting, multiplying, and dividing functions.
For two functions $f\left(x\right)$ and $g\left(x\right)$ with real number outputs, we define new functions $f+g,f-g,fg$, and $\frac{f}{g}$ by the relations
$\begin{cases}\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)\hfill \\ \left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)\hfill \\ \text{ }\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)\hfill \\ \text{ }\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}\hfill \end{cases}$
### Example 1: Performing Algebraic Operations on Functions
Find and simplify the functions $\left(g-f\right)\left(x\right)$ and $\left(\frac{g}{f}\right)\left(x\right)$, given $f\left(x\right)=x - 1$ and $g\left(x\right)={x}^{2}-1$. Are they the same function?
### Solution
Begin by writing the general form, and then substitute the given functions.
$\begin{cases}\left(g-f\right)\left(x\right)=g\left(x\right)-f\left(x\right) \\ \left(g-f\right)\left(x\right)={x}^{2}-1-\left(x - 1\right)\\ \text{ }={x}^{2}-x \\ \text{ }=x\left(x - 1\right) \\\end{cases}$ $\begin{cases}\text{ }\left(\frac{g}{f}\right)\left(x\right)=\frac{g\left(x\right)}{f\left(x\right)} \\ \text{ }\left(\frac{g}{f}\right)\left(x\right)=\frac{{x}^{2}-1}{x - 1}\\ \text{ }=\frac{\left(x+1\right)\left(x - 1\right)}{x - 1}\text{ where }x\ne 1 \\ \text{ }=x+1 \end{cases}$
No, the functions are not the same.
Note: For $\left(\frac{g}{f}\right)\left(x\right)$, the condition $x\ne 1$ is necessary because when $x=1$, the denominator is equal to 0, which makes the function undefined.
### Try It 1
Find and simplify the functions $\left(fg\right)\left(x\right)$ and $\left(f-g\right)\left(x\right)$.
$f\left(x\right)=x - 1\text{ and }g\left(x\right)={x}^{2}-1$
Are they the same function?
Solution
## Create a Function by Composition of Functions
Performing algebraic operations on functions combines them into a new function, but we can also create functions by composing functions. When we wanted to compute a heating cost from a day of the year, we created a new function that takes a day as input and yields a cost as output. The process of combining functions so that the output of one function becomes the input of another is known as a composition of functions. The resulting function is known as a composite function. We represent this combination by the following notation:
$\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$
We read the left-hand side as $f$ composed with $g$ at $x,''$ and the right-hand side as $f$ of $g$ of $x.''$ The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol $\circ$ is called the composition operator. We use this operator mainly when we wish to emphasize the relationship between the functions themselves without referring to any particular input value. Composition is a binary operation that takes two functions and forms a new function, much as addition or multiplication takes two numbers and gives a new number. However, it is important not to confuse function composition with multiplication because, as we learned above, in most cases $f\left(g\left(x\right)\right)\ne f\left(x\right)g\left(x\right)$.
It is also important to understand the order of operations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses first, and then working to the outside. In the equation above, the function $g$ takes the input $x$ first and yields an output $g\left(x\right)$. Then the function $f$ takes $g\left(x\right)$ as an input and yields an output $f\left(g\left(x\right)\right)$.
Figure 2
In general, $f\circ g$ and $g\circ f$ are different functions. In other words, in many cases $f\left(g\left(x\right)\right)\ne g\left(f\left(x\right)\right)$ for all $x$. We will also see that sometimes two functions can be composed only in one specific order.
For example, if $f\left(x\right)={x}^{2}$ and $g\left(x\right)=x+2$, then
$\begin{cases}\text{ }f\left(g\left(x\right)\right)=f\left(x+2\right)\hfill \\ \text{ }={\left(x+2\right)}^{2}\hfill \\ \text{ }={x}^{2}+4x+4\hfill \end{cases}$
but
$\begin{cases}\text{ }g\left(f\left(x\right)\right)=g\left({x}^{2}\right)\hfill \\ \text{ }={x}^{2}+2\hfill \end{cases}$
These expressions are not equal for all values of $x$, so the two functions are not equal. It is irrelevant that the expressions happen to be equal for the single input value $x=-\frac{1}{2}$.
Note that the range of the inside function (the first function to be evaluated) needs to be within the domain of the outside function. Less formally, the composition has to make sense in terms of inputs and outputs.
### A General Note: Composition of Functions
When the output of one function is used as the input of another, we call the entire operation a composition of functions. For any input $x$ and functions $f$ and $g$, this action defines a composite function, which we write as $f\circ g$ such that
$\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$
The domain of the composite function $f\circ g$ is all $x$ such that $x$ is in the domain of $g$ and $g\left(x\right)$ is in the domain of $f$.
It is important to realize that the product of functions $fg$ is not the same as the function composition $f\left(g\left(x\right)\right)$, because, in general, $f\left(x\right)g\left(x\right)\ne f\left(g\left(x\right)\right)$.
### Example 2: Determining whether Composition of Functions is Commutative
Using the functions provided, find $f\left(g\left(x\right)\right)$ and $g\left(f\left(x\right)\right)$. Determine whether the composition of the functions is commutative.
### Solution
$f\left(x\right)=2x+1g\left(x\right)=3-x$
Let’s begin by substituting $g\left(x\right)$ into $f\left(x\right)$.$\begin{cases}f\left(g\left(x\right)\right)=2\left(3-x\right)+1\hfill \\ \text{ }=6 - 2x+1\hfill \\ \text{ }=7 - 2x\hfill \end{cases}$
Now we can substitute $f\left(x\right)$ into $g\left(x\right)$.
$\begin{cases}g\left(f\left(x\right)\right)=3-\left(2x+1\right)\hfill \\ \text{ }=3 - 2x - 1\hfill \\ \text{ }=-2x+2\hfill \end{cases}$
We find that $g\left(f\left(x\right)\right)\ne f\left(g\left(x\right)\right)$, so the operation of function composition is not commutative.
### Example 3: Interpreting Composite Functions
The function $c\left(s\right)$ gives the number of calories burned completing $s$ sit-ups, and $s\left(t\right)$ gives the number of sit-ups a person can complete in $t$ minutes. Interpret $c\left(s\left(3\right)\right)$.
### Solution
The inside expression in the composition is $s\left(3\right)$. Because the input to the s-function is time, $t=3$ represents 3 minutes, and $s\left(3\right)$ is the number of sit-ups completed in 3 minutes.
Using $s\left(3\right)$ as the input to the function $c\left(s\right)$ gives us the number of calories burned during the number of sit-ups that can be completed in 3 minutes, or simply the number of calories burned in 3 minutes (by doing sit-ups).
### Example 4: Investigating the Order of Function Composition
Suppose $f\left(x\right)$ gives miles that can be driven in $x$ hours and $g\left(y\right)$ gives the gallons of gas used in driving $y$ miles. Which of these expressions is meaningful: $f\left(g\left(y\right)\right)$ or $g\left(f\left(x\right)\right)?$
### Solution
The function $y=f\left(x\right)$ is a function whose output is the number of miles driven corresponding to the number of hours driven.
$\text{number of miles }=f\left(\text{number of hours}\right)$
The function $g\left(y\right)$ is a function whose output is the number of gallons used corresponding to the number of miles driven. This means:
$\text{number of gallons }=g\left(\text{number of miles}\right)$
The expression $g\left(y\right)$ takes miles as the input and a number of gallons as the output. The function $f\left(x\right)$ requires a number of hours as the input. Trying to input a number of gallons does not make sense. The expression $f\left(g\left(y\right)\right)$ is meaningless.
The expression $f\left(x\right)$ takes hours as input and a number of miles driven as the output. The function $g\left(y\right)$ requires a number of miles as the input. Using $f\left(x\right)$ (miles driven) as an input value for $g\left(y\right)$, where gallons of gas depends on miles driven, does make sense. The expression $g\left(f\left(x\right)\right)$ makes sense, and will yield the number of gallons of gas used, $g$, driving a certain number of miles, $f\left(x\right)$, in $x$ hours.
### Q & A
Are there any situations where $f\left(g\left(y\right)\right)$ and $g\left(f\left(x\right)\right)$ would both be meaningful or useful expressions?
Yes. For many pure mathematical functions, both compositions make sense, even though they usually produce different new functions. In real-world problems, functions whose inputs and outputs have the same units also may give compositions that are meaningful in either order.
### Try It 2
The gravitational force on a planet a distance r from the sun is given by the function $G\left(r\right)$. The acceleration of a planet subjected to any force $F$ is given by the function $a\left(F\right)$. Form a meaningful composition of these two functions, and explain what it means.
Solution
## Evaluating Composite Functions
Once we compose a new function from two existing functions, we need to be able to evaluate it for any input in its domain. We will do this with specific numerical inputs for functions expressed as tables, graphs, and formulas and with variables as inputs to functions expressed as formulas. In each case, we evaluate the inner function using the starting input and then use the inner function’s output as the input for the outer function.
## Evaluating Composite Functions Using Tables
When working with functions given as tables, we read input and output values from the table entries and always work from the inside to the outside. We evaluate the inside function first and then use the output of the inside function as the input to the outside function.
### Example 5: Using a Table to Evaluate a Composite Function
Using the table below, evaluate $f\left(g\left(3\right)\right)$ and $g\left(f\left(3\right)\right)$.
$x$ $f\left(x\right)$ $g\left(x\right)$
1 6 3
2 8 5
3 3 2
4 1 7
### Solution
To evaluate $f\left(g\left(3\right)\right)$, we start from the inside with the input value 3. We then evaluate the inside expression $g\left(3\right)$ using the table that defines the function $g:$ $g\left(3\right)=2$. We can then use that result as the input to the function $f$, so $g\left(3\right)$ is replaced by 2 and we get $f\left(2\right)$. Then, using the table that defines the function $f$, we find that $f\left(2\right)=8$.
$\begin{cases}g\left(3\right)=2\hfill \\ f\left(g\left(3\right)\right)=f\left(2\right)=8\hfill \end{cases}$
To evaluate $g\left(f\left(3\right)\right)$, we first evaluate the inside expression $f\left(3\right)$ using the first table: $f\left(3\right)=3$. Then, using the table for $g\text{,\hspace{0.17em}}$ we can evaluate
$g\left(f\left(3\right)\right)=g\left(3\right)=2$
The table below shows the composite functions $f\circ g$ and $g\circ f$ as tables.
$x$ $g\left(x\right)$ $f\left(g\left(x\right)\right)$ $f\left(x\right)$ $g\left(f\left(x\right)\right)$ 3 2 8 3 2
### Try It 3
Using the table below, evaluate $f\left(g\left(1\right)\right)$ and $g\left(f\left(4\right)\right)$.
$x$ $f\left(x\right)$ $g\left(x\right)$
1 6 3
2 8 5
3 3 2
4 1 7
Solution
## Evaluating Composite Functions Using Graphs
When we are given individual functions as graphs, the procedure for evaluating composite functions is similar to the process we use for evaluating tables. We read the input and output values, but this time, from the $x\text{-}$ and $y\text{-}$ axes of the graphs.
### How To: Given a composite function and graphs of its individual functions, evaluate it using the information provided by the graphs.
1. Locate the given input to the inner function on the $x\text{-}$ axis of its graph.
2. Read off the output of the inner function from the $y\text{-}$ axis of its graph.
3. Locate the inner function output on the $x\text{-}$ axis of the graph of the outer function.
4. Read the output of the outer function from the $y\text{-}$ axis of its graph. This is the output of the composite function.
### Example 6: Using a Graph to Evaluate a Composite Function
Using the graphs in Figure 3, evaluate $f\left(g\left(1\right)\right)$.
Figure 3
### Solution
Figure 4
To evaluate $f\left(g\left(1\right)\right)$, we start with the inside evaluation.
We evaluate $g\left(1\right)$ using the graph of $g\left(x\right)$, finding the input of 1 on the $x\text{-}$ axis and finding the output value of the graph at that input. Here, $g\left(1\right)=3$. We use this value as the input to the function $f$.
$f\left(g\left(1\right)\right)=f\left(3\right)$
We can then evaluate the composite function by looking to the graph of $f\left(x\right)$, finding the input of 3 on the $x\text{-}$ axis and reading the output value of the graph at this input. Here, $f\left(3\right)=6$, so $f\left(g\left(1\right)\right)=6$.
### Analysis of the Solution
Figure 5 shows how we can mark the graphs with arrows to trace the path from the input value to the output value.
Figure 5
### Try It 4
Using Figure 6, evaluate $g\left(f\left(2\right)\right)$.
Figure 6
Solution
## Evaluating Composite Functions Using Formulas
When evaluating a composite function where we have either created or been given formulas, the rule of working from the inside out remains the same. The input value to the outer function will be the output of the inner function, which may be a numerical value, a variable name, or a more complicated expression.
While we can compose the functions for each individual input value, it is sometimes helpful to find a single formula that will calculate the result of a composition $f\left(g\left(x\right)\right)$. To do this, we will extend our idea of function evaluation. Recall that, when we evaluate a function like $f\left(t\right)={t}^{2}-t$, we substitute the value inside the parentheses into the formula wherever we see the input variable.
### How To: Given a formula for a composite function, evaluate the function.
1. Evaluate the inside function using the input value or variable provided.
2. Use the resulting output as the input to the outside function.
### Example 7: Evaluating a Composition of Functions Expressed as Formulas with a Numerical Input
Given $f\left(t\right)={t}^{2}-{t}$ and $h\left(x\right)=3x+2$, evaluate $f\left(h\left(1\right)\right)$.
### Solution
Because the inside expression is $h\left(1\right)$, we start by evaluating $h\left(x\right)$ at 1.
$\begin{cases}h\left(1\right)=3\left(1\right)+2\\ h\left(1\right)=5\end{cases}$
Then $f\left(h\left(1\right)\right)=f\left(5\right)$, so we evaluate $f\left(t\right)$ at an input of 5.
$\begin{cases}f\left(h\left(1\right)\right)=f\left(5\right)\\ f\left(h\left(1\right)\right)={5}^{2}-5\\ f\left(h\left(1\right)\right)=20\end{cases}$
### Analysis of the Solution
It makes no difference what the input variables $t$ and $x$ were called in this problem because we evaluated for specific numerical values.
### Try It 5
Given $f\left(t\right)={t}^{2}-t$ and $h\left(x\right)=3x+2$, evaluate
A) $h\left(f\left(2\right)\right)$
B) $h\left(f\left(-2\right)\right)$
Solution
## Finding the Domain of a Composite Function
As we discussed previously, the domain of a composite function such as $f\circ g$ is dependent on the domain of $g$ and the domain of $f$. It is important to know when we can apply a composite function and when we cannot, that is, to know the domain of a function such as $f\circ g$. Let us assume we know the domains of the functions $f$ and $g$ separately. If we write the composite function for an input $x$ as $f\left(g\left(x\right)\right)$, we can see right away that $x$ must be a member of the domain of $g$ in order for the expression to be meaningful, because otherwise we cannot complete the inner function evaluation. However, we also see that $g\left(x\right)$ must be a member of the domain of $f$, otherwise the second function evaluation in $f\left(g\left(x\right)\right)$ cannot be completed, and the expression is still undefined. Thus the domain of $f\circ g$ consists of only those inputs in the domain of $g$ that produce outputs from $g$ belonging to the domain of $f$. Note that the domain of $f$ composed with $g$ is the set of all $x$ such that $x$ is in the domain of $g$ and $g\left(x\right)$ is in the domain of $f$.
### A General Note: Domain of a Composite Function
The domain of a composite function $f\left(g\left(x\right)\right)$ is the set of those inputs $x$ in the domain of $g$ for which $g\left(x\right)$ is in the domain of $f$.
### How To: Given a function composition $f\left(g\left(x\right)\right)$, determine its domain.
1. Find the domain of g.
2. Find the domain of f.
3. Find those inputs, x, in the domain of g for which g(x) is in the domain of f. That is, exclude those inputs, x, from the domain of g for which g(x) is not in the domain of f. The resulting set is the domain of $f\circ g$.
### Example 8: Finding the Domain of a Composite Function
Find the domain of
$\left(f\circ g\right)\left(x\right)\text{ where}f\left(x\right)=\frac{5}{x - 1}\text{ and }g\left(x\right)=\frac{4}{3x - 2}$
### Solution
The domain of $g\left(x\right)$ consists of all real numbers except $x=\frac{2}{3}$, since that input value would cause us to divide by 0. Likewise, the domain of $f$ consists of all real numbers except 1. So we need to exclude from the domain of $g\left(x\right)$ that value of $x$ for which $g\left(x\right)=1$.
$\begin{cases}\frac{4}{3x - 2}=1\hfill \\ 4=3x - 2\hfill \\ 6=3x\hfill \\ x=2\hfill \end{cases}$
So the domain of $f\circ g$ is the set of all real numbers except $\frac{2}{3}$ and $2$. This means that
$x\ne \frac{2}{3}\text{or}x\ne 2$
We can write this in interval notation as
$\left(-\infty ,\frac{2}{3}\right)\cup \left(\frac{2}{3},2\right)\cup \left(2,\infty \right)$
### Example 9: Finding the Domain of a Composite Function Involving Radicals
Find the domain of
$\left(f\circ g\right)\left(x\right)\text{ where}f\left(x\right)=\sqrt{x+2}\text{ and }g\left(x\right)=\sqrt{3-x}$
### Solution
Because we cannot take the square root of a negative number, the domain of $g$ is $\left(-\infty ,3\right]$. Now we check the domain of the composite function
$\left(f\circ g\right)\left(x\right)=\sqrt{3-x+2}\text{ or}\left(f\circ g\right)\left(x\right)=\sqrt{5-x}$
The domain of this function is $\left(-\infty ,5\right]$. To find the domain of $f\circ g$, we ask ourselves if there are any further restrictions offered by the domain of the composite function. The answer is no, since $\left(-\infty ,3\right]$ is a proper subset of the domain of $f\circ g$. This means the domain of $f\circ g$ is the same as the domain of $g$, namely, $\left(-\infty ,3\right]$.
### Analysis of the Solution
This example shows that knowledge of the range of functions (specifically the inner function) can also be helpful in finding the domain of a composite function. It also shows that the domain of $f\circ g$ can contain values that are not in the domain of $f$, though they must be in the domain of $g$.
### Try It 6
Find the domain of
$\left(f\circ g\right)\left(x\right)\text{ where}f\left(x\right)=\frac{1}{x - 2}\text{ and }g\left(x\right)=\sqrt{x+4}$
## Decomposing a Composite Function into its Component Functions
In some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions. There may be more than one way to decompose a composite function, so we may choose the decomposition that appears to be most expedient.
### Example 10: Decomposing a Function
Write $f\left(x\right)=\sqrt{5-{x}^{2}}$ as the composition of two functions.
### Solution
We are looking for two functions, $g$ and $h$, so $f\left(x\right)=g\left(h\left(x\right)\right)$. To do this, we look for a function inside a function in the formula for $f\left(x\right)$. As one possibility, we might notice that the expression $5-{x}^{2}$ is the inside of the square root. We could then decompose the function as
$h\left(x\right)=5-{x}^{2}\text{ and }g\left(x\right)=\sqrt{x}$
We can check our answer by recomposing the functions.
$g\left(h\left(x\right)\right)=g\left(5-{x}^{2}\right)=\sqrt{5-{x}^{2}}$
### Try It 7
Write $f\left(x\right)=\frac{4}{3-\sqrt{4+{x}^{2}}}$ as the composition of two functions.
Solution
## Key Equation
Composite function $\left(f\circ g\right)\left(x\right)=f\left(g\left(x\right)\right)$
## Key Concepts
• We can perform algebraic operations on functions.
• When functions are combined, the output of the first (inner) function becomes the input of the second (outer) function.
• The function produced by combining two functions is a composite function.
• The order of function composition must be considered when interpreting the meaning of composite functions.
• A composite function can be evaluated by evaluating the inner function using the given input value and then evaluating the outer function taking as its input the output of the inner function.
• A composite function can be evaluated from a table.
• A composite function can be evaluated from a graph.
• A composite function can be evaluated from a formula.
• The domain of a composite function consists of those inputs in the domain of the inner function that correspond to outputs of the inner function that are in the domain of the outer function.
• Just as functions can be combined to form a composite function, composite functions can be decomposed into simpler functions.
• Functions can often be decomposed in more than one way.
## Glossary
composite function
the new function formed by function composition, when the output of one function is used as the input of another
1. How does one find the domain of the quotient of two functions, $\frac{f}{g}?$
2. What is the composition of two functions, $f\circ g?$
3. If the order is reversed when composing two functions, can the result ever be the same as the answer in the original order of the composition? If yes, give an example. If no, explain why not.
4. How do you find the domain for the composition of two functions, $f\circ g?$
5. Given $f\left(x\right)={x}^{2}+2x\text{ }$ and $g\left(x\right)=6-{x}^{2}$, find $f+g,f-g,fg,\text{ }$ and $\text{ }\frac{f}{g}$. Determine the domain for each function in interval notation.
6. Given $f\left(x\right)=-3{x}^{2}+x\text{ }$ and $\text{ }g\left(x\right)=5$, find $f+g,f-g,fg$, and $\text{ }\frac{f}{g}$. Determine the domain for each function in interval notation.
7. Given $f\left(x\right)=2{x}^{2}+4x\text{ }$ and $\text{ }g\left(x\right)=\frac{1}{2x}$, find $f+g,f-g,fg,\text{ }$ and $\text{ }\frac{f}{g}$. Determine the domain for each function in interval notation.
8. Given $f\left(x\right)=\frac{1}{x - 4}$ and $g\left(x\right)=\frac{1}{6-x}$, find $f+g,f-g,fg,\text{ }$ and $\text{ }\frac{f}{g}$. Determine the domain for each function in interval notation.
9. Given $f\left(x\right)=3{x}^{2}$ and $g\left(x\right)=\sqrt{x - 5}$, find $f+g,f-g,fg,\text{ }$ and $\text{ }\frac{f}{g}$. Determine the domain for each function in interval notation.
10. Given $f\left(x\right)=\sqrt{x}$ and $g\left(x\right)=|x - 3|$, find $\frac{g}{f}$. Determine the domain of the function in interval notation.
11. Given $f\left(x\right)=2{x}^{2}+1$ and $g\left(x\right)=3x - 5$, find the following:
$f\left(g\left(2\right)\right)$
$f\left(g\left(x\right)\right)$
$g\left(f\left(x\right)\right)$
$\left(g\circ g\right)\left(x\right)$
$\left(f\circ f\right)\left(-2\right)$
For the following exercises, use each pair of functions to find $f\left(g\left(x\right)\right)$ and $g\left(f\left(x\right)\right)$. Simplify your answers.
12. $f\left(x\right)={x}^{2}+1,g\left(x\right)=\sqrt{x+2}$
13. $f\left(x\right)=\sqrt{x}+2,g\left(x\right)={x}^{2}+3$
14. $f\left(x\right)=|x|,g\left(x\right)=5x+1$
15. $f\left(x\right)=\sqrt[3]{x},g\left(x\right)=\frac{x+1}{{x}^{3}}$
16. $f\left(x\right)=\frac{1}{x - 6},g\left(x\right)=\frac{7}{x}+6$
17. $f\left(x\right)=\frac{1}{x - 4},g\left(x\right)=\frac{2}{x}+4$
For the following exercises, use each set of functions to find $f\left(g\left(h\left(x\right)\right)\right)$. Simplify your answers.
18. $f\left(x\right)={x}^{4}+6$, $g\left(x\right)=x - 6$, and $h\left(x\right)=\sqrt{x}$
19. $f\left(x\right)={x}^{2}+1$, $g\left(x\right)=\frac{1}{x}$, and $h\left(x\right)=x+3$
20. Given $f\left(x\right)=\frac{1}{x}$ and $g\left(x\right)=x - 3$, find the following:
$\left(f\circ g\right)\left(x\right)$
the domain of $\left(f\circ g\right)\left(x\right)$ in interval notation
$\left(g\circ f\right)\left(x\right)$
the domain of $\left(g\circ f\right)\left(x\right)$
$\left(\frac{f}{g}\right)x$
21. Given $f\left(x\right)=\sqrt{2 - 4x}$ and $g\left(x\right)=-\frac{3}{x}$, find the following:
a. $\left(g\circ f\right)\left(x\right)$
b. the domain of $\left(g\circ f\right)\left(x\right)$ in interval notation
22. Given the functions $f\left(x\right)=\frac{1-x}{x}\text{and}g\left(x\right)=\frac{1}{1+{x}^{2}}$, find the following:
a. $\left(g\circ f\right)\left(x\right)$
b. $\left(g\circ f\right)\left(\text{2}\right)$
23. Given functions $p\left(x\right)=\frac{1}{\sqrt{x}}$ and $m\left(x\right)={x}^{2}-4$, state the domain of each of the following functions using interval notation:
$\frac{p\left(x\right)}{m\left(x\right)}$
$p\left(m\left(x\right)\right)$
$m\left(p\left(x\right)\right)$
24. Given functions $q\left(x\right)=\frac{1}{\sqrt{x}}$ and $h\left(x\right)={x}^{2}-9$, state the domain of each of the following functions using interval notation.
$\frac{q\left(x\right)}{h\left(x\right)}$
$q\left(h\left(x\right)\right)$
$h\left(q\left(x\right)\right)$
25. For $f\left(x\right)=\frac{1}{x}$ and $g\left(x\right)=\sqrt{x - 1}$, write the domain of $\left(f\circ g\right)\left(x\right)$ in interval notation.
For the following exercises, find functions $f\left(x\right)$ and $g\left(x\right)$ so the given function can be expressed as $h\left(x\right)=f\left(g\left(x\right)\right)$.
26. $h\left(x\right)={\left(x+2\right)}^{2}$
27. $h\left(x\right)={\left(x - 5\right)}^{3}$
28. $h\left(x\right)=\frac{3}{x - 5}$
29. $h\left(x\right)=\frac{4}{{\left(x+2\right)}^{2}}$
30. $h\left(x\right)=4+\sqrt[3]{x}$
31. $h\left(x\right)=\sqrt[3]{\frac{1}{2x - 3}}$
32. $h\left(x\right)=\frac{1}{{\left(3{x}^{2}-4\right)}^{-3}}$
33. $h\left(x\right)=\sqrt[4]{\frac{3x - 2}{x+5}}$
34. $h\left(x\right)={\left(\frac{8+{x}^{3}}{8-{x}^{3}}\right)}^{4}$
35. $h\left(x\right)=\sqrt{2x+6}$
36. $h\left(x\right)={\left(5x - 1\right)}^{3}$
37. $h\left(x\right)=\sqrt[3]{x - 1}$
38. $h\left(x\right)=\left|{x}^{2}+7\right|$
39. $h\left(x\right)=\frac{1}{{\left(x - 2\right)}^{3}}$
40. $h\left(x\right)={\left(\frac{1}{2x - 3}\right)}^{2}$
41. $h\left(x\right)=\sqrt{\frac{2x - 1}{3x+4}}$
For the following exercises, use the graphs of $f$ and $g$ to evaluate the expressions.
42. $f\left(g\left(3\right)\right)$
43. $f\left(g\left(1\right)\right)$
44. $g\left(f\left(1\right)\right)$
45. $g\left(f\left(0\right)\right)$
46. $f\left(f\left(5\right)\right)$
47. $f\left(f\left(4\right)\right)$
48. $g\left(g\left(2\right)\right)$
49. $g\left(g\left(0\right)\right)$
For the following exercises, use graphs of $f\left(x\right)$ $g\left(x\right)$, and $h\left(x\right)$, to evaluate the expressions.
50. $g\left(f\left(1\right)\right)$
51. $g\left(f\left(2\right)\right)$
52. $f\left(g\left(4\right)\right)$
53. $f\left(g\left(1\right)\right)$
54. $f\left(h\left(2\right)\right)$
55. $h\left(f\left(2\right)\right)$
56. $f\left(g\left(h\left(4\right)\right)\right)$
57. $f\left(g\left(f\left(-2\right)\right)\right)$
For the following exercises, use the function values for $f\text{ and }g$ to evaluate each expression.
$x$ $f\left(x\right)$ $g\left(x\right)$ 0 7 9 1 6 5 2 5 6 3 8 2 4 4 1 5 0 8 6 2 7 7 1 3 8 9 4 9 3 0
58. $f\left(g\left(8\right)\right)$
59. $f\left(g\left(5\right)\right)$
60. $g\left(f\left(5\right)\right)$
61. $g\left(f\left(3\right)\right)$
62. $f\left(f\left(4\right)\right)$
63. $f\left(f\left(1\right)\right)$
64. $g\left(g\left(2\right)\right)$
65. $g\left(g\left(6\right)\right)$
For the following exercises, use the function values for $f\text{ and }g$ to evaluate the expressions.
$x$ $f\left(x\right)$ $g\left(x\right)$ -3 11 -8 -2 9 -3 -1 7 0 0 5 1 1 3 0 2 1 -3 3 -1 -8
66. $\left(f\circ g\right)\left(1\right)$
67. $\left(f\circ g\right)\left(2\right)$
68. $\left(g\circ f\right)\left(2\right)$
69. $\left(g\circ f\right)\left(3\right)$
70. $\left(g\circ g\right)\left(1\right)$
71. $\left(f\circ f\right)\left(3\right)$
For the following exercises, use each pair of functions to find $f\left(g\left(0\right)\right)$ and $g\left(f\left(0\right)\right)$.
72. $f\left(x\right)=4x+8,g\left(x\right)=7-{x}^{2}$
73. $f\left(x\right)=5x+7,g\left(x\right)=4 - 2{x}^{2}$
74. $f\left(x\right)=\sqrt{x+4},g\left(x\right)=12-{x}^{3}$
75. $f\left(x\right)=\frac{1}{x+2},g\left(x\right)=4x+3$
For the following exercises, use the functions $f\left(x\right)=2{x}^{2}+1$ and $g\left(x\right)=3x+5$ to evaluate or find the composite function as indicated.
76. $f\left(g\left(2\right)\right)$
77. $f\left(g\left(x\right)\right)$
78. $g\left(f\left(-3\right)\right)$
79. $\left(g\circ g\right)\left(x\right)$
For the following exercises, use $f\left(x\right)={x}^{3}+1$ and $g\left(x\right)=\sqrt[3]{x - 1}$.
80. Find $\left(f\circ g\right)\left(x\right)$ and $\left(g\circ f\right)\left(x\right)$. Compare the two answers.
81. Find $\left(f\circ g\right)\left(2\right)$ and $\left(g\circ f\right)\left(2\right)$.
82. What is the domain of $\left(g\circ f\right)\left(x\right)?$
83. What is the domain of $\left(f\circ g\right)\left(x\right)?$
84. Let $f\left(x\right)=\frac{1}{x}$.
a. Find $\left(f\circ f\right)\left(x\right)$.
b. Is $\left(f\circ f\right)\left(x\right)$ for any function $f$ the same result as the answer to part (a) for any function? Explain.
For the following exercises, let $F\left(x\right)={\left(x+1\right)}^{5}$, $f\left(x\right)={x}^{5}$, and $g\left(x\right)=x+1$.
85. True or False: $\left(g\circ f\right)\left(x\right)=F\left(x\right)$.
86. True or False: $\left(f\circ g\right)\left(x\right)=F\left(x\right)$.
For the following exercises, find the composition when $f\left(x\right)={x}^{2}+2$ for all $x\ge 0$ and $g\left(x\right)=\sqrt{x - 2}$.
87. $\left(f\circ g\right)\left(6\right);\left(g\circ f\right)\left(6\right)$
88. $\left(g\circ f\right)\left(a\right);\left(f\circ g\right)\left(a\right)$
89. $\left(f\circ g\right)\left(11\right);\left(g\circ f\right)\left(11\right)$
90. The function $D\left(p\right)$ gives the number of items that will be demanded when the price is $p$. The production cost $C\left(x\right)$ is the cost of producing $x$ items. To determine the cost of production when the price is \$6, you would do which of the following?
a. Evaluate $D\left(C\left(6\right)\right)$.
b. Evaluate $C\left(D\left(6\right)\right)$.
c. Solve $D\left(C\left(x\right)\right)=6$.
d. Solve $C\left(D\left(p\right)\right)=6$.
91. The function $A\left(d\right)$ gives the pain level on a scale of 0 to 10 experienced by a patient with $d$ milligrams of a pain-reducing drug in her system. The milligrams of the drug in the patient’s system after $t$ minutes is modeled by $m\left(t\right)$. Which of the following would you do in order to determine when the patient will be at a pain level of 4?
a. Evaluate $A\left(m\left(4\right)\right)$.
b. Evaluate $m\left(A\left(4\right)\right)$.
c. Solve $A\left(m\left(t\right)\right)=4$.
d. Solve $m\left(A\left(d\right)\right)=4$.
92. A store offers customers a 30% discount on the price $x$ of selected items. Then, the store takes off an additional 15% at the cash register. Write a price function $P\left(x\right)$ that computes the final price of the item in terms of the original price $x$. (Hint: Use function composition to find your answer.)
93. A rain drop hitting a lake makes a circular ripple. If the radius, in inches, grows as a function of time in minutes according to $r\left(t\right)=25\sqrt{t+2}$, find the area of the ripple as a function of time. Find the area of the ripple at $t=2$.
94. A forest fire leaves behind an area of grass burned in an expanding circular pattern. If the radius of the circle of burning grass is increasing with time according to the formula $r\left(t\right)=2t+1$, express the area burned as a function of time, $t$ (minutes).
95. Use the function you found in the previous exercise to find the total area burned after 5 minutes.
96. The radius $r$, in inches, of a spherical balloon is related to the volume, $V$, by $r\left(V\right)=\sqrt[3]{\frac{3V}{4\pi }}$. Air is pumped into the balloon, so the volume after $t$ seconds is given by $V\left(t\right)=10+20t$.
a. Find the composite function $r\left(V\left(t\right)\right)$.
b. Find the exact time when the radius reaches 10 inches.
97. The number of bacteria in a refrigerated food product is given by $N\left(T\right)=23{T}^{2}-56T+1$, $3<T<33$, where $T$ is the temperature of the food. When the food is removed from the refrigerator, the temperature is given by $T\left(t\right)=5t+1.5$, where $t$ is the time in hours.
a. Find the composite function $N\left(T\left(t\right)\right)$.
b. Find the time (round to two decimal places) when the bacteria count reaches 6752. |
Find the equation of the line which satisfy the given conditions passing through ,$$(2,2\sqrt{3})$$) and inclined with the x-axis at an angle of 75°.
Asked by Pragya Singh | 1 year ago | 110
##### Solution :-
Given: point (2, $$2\sqrt{3}$$) and θ = 75°
Equation of line: (y – y1) = m (x – x1)
where, m = slope of line = tan θ and (x1, y1) are the points through which line passes
m = tan 75°
75° = 45° + 30°
Applying the formula:
$$\dfrac{tan 45°+tan 30°}{1-tan 45°.tan 30°}$$
Substitute the values,
$$1+\dfrac{1}{\sqrt{3}}.1-\dfrac{1}{\sqrt{3}}$$
tan 45° = $$\dfrac{\sqrt{3}+1}{\sqrt{3}-1}$$
Let us rationalizing we get,
$$\dfrac{3+1+2\sqrt{3}}{3-1}$$
$$2+\sqrt{3}$$
We know that the point (x, y) lies on the line with slope m through the fixed point (x1, y1), if and only if, its coordinates satisfy the equation y – y1 = m (x – x1)
Then, y –$$2\sqrt{3}$$ = ( $$2+\sqrt{3}$$) (x – 2)
y – $$2\sqrt{3}$$ = 2 x – 4 + $$\sqrt{3}x-$$$$2\sqrt{3}$$
y = 2 x – 4 +$$\sqrt{3}x$$
$$2+\sqrt{3}$$) x – y – 4 = 0
The equation of the line is ($$2+\sqrt{3}$$) x – y – 4 = 0.
Answered by Abhisek | 1 year ago
### Related Questions
#### Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.
Find the angle between the line joining the points (2, 0), (0, 3) and the line x + y = 1.
#### Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices
Prove that the points (2, -1), (0, 2), (2, 3) and (4, 0) are the coordinates of the vertices of a parallelogram and find the angle between its diagonals.
#### Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0.
Find the acute angle between the lines 2x – y + 3 = 0 and x + y + 2 = 0. |
# 7 Covariance and Correlation
Research is not always interested in the contrast of two variables, but oftentimes the relationship of two variables. For instance, what is the relationship between climate science and ideology? The following packages are required for this lab:
1. tidyverse
2. psych
3. car
4. vcd
## 7.1 Covariance
Covariance is the measure of change in one variable associated to change in another variable. That is, the measure of how two random variables vary together.
Calculating covariance of two variables of a known population is trivial, as the product of variation of two variables. However, for samples, covariance is calculated via the following formula:
$cov(x,y)=\frac{\sum_{i=1}^n (x_{i}-\bar{x})(y_{i}-\bar{y})}{{n-1}}$
Where x and y are two random variables, i is each observation (row), and n is the sample size.
### 7.1.1 Covariance by Hand
To find covariance by hand, let’s construct a hypothetical data set with variables x and y. We only give each variable three values so that the calculation will be shorter.
x <- c(25, 27, 29)
y <- c(5, 15, 9)
First we calculate the difference of each value and the mean for the variables:
xdev <- x - mean(x)
ydev <- y - mean(y)
Next we find the product of the above differences:
xdev_ydev <- xdev * ydev
xdev_ydev
## [1] 9.333333 0.000000 -1.333333
We complete the numerator by finding the sum of the products:
sum_xdev_ydev <- sum(xdev_ydev)
Lastly, we complete the covariance calculation by dividing the numerator by the denominator. The denominator is calculated as one less than the sample size:
cov_xy <- sum_xdev_ydev / (3 - 1)
cov_xy
## [1] 4
### 7.1.2 Covariance in R
By using the cov() function, R calculates the covariance. We demonstrate the cov() function by confirming the previous section’s calculations:
cov(x, y)
## [1] 4
### 7.1.3 Covariance in Class Data Set
To demonstrate further we will calculate covariance for various pairs of variables within the class data set. First, suppose we are interested in the relationship between certainty that humans cause climate change (glbcc_cert) and the perceived risk of climate change (glbcc_risk). That is, is there a relationship between respondents’ certainty that humans cause climate change and their perceived risk of climate change? Our hypothesis could be that individuals that are more certain that climate change is a consequence of humans are likely more concerned about the associated risk.
cov(ds$glbcc_cert, ds$glbcc_risk, use = "complete.obs")
## [1] 3.092168
Note: The cov() function requires use="complete.obs" to remove NA entries.
The calculated covariance of certainty and risk perception is positive, indicating the two variables are positively related. As one variable changes, the other variable will change in the same direction with a magnitude of 3.09. Increased certainty that humans cause climate change increases the perceived risk of climate change.
Suppose we are also interested in the relationship of income and perceived risk of climate change:
cov(ds$income, ds$glbcc_risk, use = "complete.obs")
## [1] -10826.91
The calculated covariance of income and perceived risk of climate change is negative, indicating the two variables are negatively related. As one variable changes, the other variable will change in the opposite direction with a magnitude of -10826.91.
An important follow-up question is: which variable is more strongly associated to perceived risk, certainty or income? We cannot compare magnitudes to the difference in scales and units. Income is measured on a scale inclusive of higher numbers compared to certainty. To compare strengths of association we need to standardize covariance.
## 7.2 Correlation
Correlation standardizes covariance on a scale of negative one to one, whereby the magnitude from zero indicates strength of relationship. Similar to covariance, a positive and negative value reflects the respective relationship. The formula for correlation is the following:
$r_{xy}=\frac{cov(x,y)}{s_xs_y}$
Where x and y are two random variables.
### 7.2.1 Correlation by Hand
To find covariance by hand, let’s use the data set we constructed earlier to calculate covariance.
Recall we calculated covariance manually via the following steps:
x <- c(25, 27, 29)
y <- c(5, 15, 9)
xdev <- x - mean(x)
ydev <- y - mean(y)
xdev_ydev <- xdev * ydev
sum_xdev_ydev <- sum(xdev_ydev)
cov_xy <- (1 / (3 - 1)) * sum_xdev_ydev
cov_xy
## [1] 4
For calculating correlation we need the product of the standard deviations of variables x and y for the denominator:
stnd.dev <- sd(x)*sd(y)
Now we find the quotient of the covariance numerator and standard deviations denominator:
cov_xy/stnd.dev
## [1] 0.3973597
The relationship is positive; however, the correlation coefficient is $$\approx$$ 0.40.
Returning to the class data set, we can find the correlation coefficient for certainty humans cause climate change and perceived risk of climate change from the class data set:
numerator <- cov(ds$glbcc_cert, ds$glbcc_risk, use = "complete.obs")
denominator <- sd(ds$glbcc_cert, na.rm = T) * sd(ds$glbcc_risk, na.rm = T)
numerator / denominator
## [1] 0.3699554
The correlation coefficient is $$\approx$$ 0.37.
Now let’s find the correlation coefficient for ideology and perceived risk from climate change.
First we will create a subset from the class data set ds for the variables of interest, absent of missing observations. Note: This subset includes additional variables, f.gender and f.party.2 for use later in this lab.
ds.sub <- ds %>%
dplyr::select("glbcc_risk", "ideol", "f.gender", "f.party.2") %>%
na.omit()
The perceived risk and ideology variables are assigned to x and y variables within the ds.sub object, as to follow the formula.
ds.sub$x <- ds.sub$glbcc_risk
ds.sub$y <- ds.sub$ideol
The x and y variables are then used to find the covariance, similar to the steps demonstrated earlier:
xbar <- mean(ds.sub$x) ybar <- mean(ds.sub$y)
x.m.xbar <- ds.sub$x - xbar y.m.ybar <- ds.sub$y - ybar
n <- length(ds.sub$x) n ## [1] 2388 cov.xy <- (sum(x.m.xbar * y.m.ybar)) / (n - 1) cov.xy ## [1] -3.134752 Next, we find the correlation coefficient using the covariance as the numerator and the product of both variable standard deviations as the denominator: sd.x <- sqrt((sum(x.m.xbar^2)) / (n - 1)) sd.x ## [1] 3.058544 sd.y <- sqrt((sum(y.m.ybar^2)) / (n - 1)) sd.y ## [1] 1.74181 sd.xy <- sd.x*sd.y cov.xy/sd.xy ## [1] -0.5884202 The correlation coefficient for the variables is $$\approx$$ -0.59. The manual calculation is confirmed using the cor() function in R: cor(ds.sub$glbcc_risk, ds.sub$ideol) ## [1] -0.5884202 Note: To calculate the correlation coefficient manually in one line of code: sum((x-mean(x))*(y-mean(y))) /(sqrt(sum((x-mean(x))^2))*sqrt(sum((y-mean(y))^2))) ### 7.2.2 Correlation Tests The previous section demonstrated the cor() function to confirm the manual calculation of the correlation coefficient. Using this function we can find the correlation coefficient of the income and perceived risk variables: cor(ds$income, ds$glbcc_risk, use = "complete.obs") ## [1] -0.05904785 The correlation coefficient of -0.06 informs us of two things: the relationship is negative and very weak. Note: The correlation coefficient is drawn from observations within a sample, and therefore is a random value. That is, if we were to collect multiple samples we would calculate different correlation coefficients for each sample collected. This leads to a new hypothesis: is there a relationship between income and perceived risk? To test this we employ Pearson’s product-moment correlation available via the cor.test() function. Our testing hypotheses are: -$$H_0$$: the true correlation coefficient is zero -$$H_1$$: the true correlation coefficient is not zero cor.test(ds$income, ds$glbcc_risk, use="complete.obs") ## ## Pearson's product-moment correlation ## ## data: ds$income and ds$glbcc_risk ## t = -2.83, df = 2289, p-value = 0.004695 ## alternative hypothesis: true correlation is not equal to 0 ## 95 percent confidence interval: ## -0.09975880 -0.01813952 ## sample estimates: ## cor ## -0.05904785 The correlation test yields a p-value < $$\alpha$$ = 0.05, thereby the null hypothesis is rejected such that the true correlation coefficient is not zero. Note: Despite rejecting the null hypothesis, our random correlation value is still quite small at -0.06. This indicates a very weak, or non-substance, relationship between income and perceived risk. To demonstrate a potentially substantive relationship we look at ideology and perceived risk of climate change using the cor.test() function: cor.test(ds$ideol, ds$glbcc_risk) ## ## Pearson's product-moment correlation ## ## data: ds$ideol and ds$glbcc_risk ## t = -36.633, df = 2511, p-value < 0.00000000000000022 ## alternative hypothesis: true correlation is not equal to 0 ## 95 percent confidence interval: ## -0.6150780 -0.5640865 ## sample estimates: ## cor ## -0.5901706 The default correlation test method for the cor.test() function is the Pearson test. The Spearman test is required for ordinal data via the method="spearman" argument within the cor.test() function. For calculation correlation with ordinal data. ### 7.2.3 Correlation Across Groups Suppose you are interested in correlations across multiple variables. The cor() function examines the correlation between each variable pair for an entire data set. As such, if you are interested in only the correlation for select variables then use the select() function and select your variablesof interest, then pipe them into the cor() function. Include drop_na(). To demonstrate using variables from the class data set: ds %>% dplyr::select(glbcc_risk, ideol, income, age) %>% drop_na() %>% cor() ## glbcc_risk ideol income age ## glbcc_risk 1.00000000 -0.59999737 -0.06079785 -0.06799270 ## ideol -0.59999737 1.00000000 0.03901982 0.08608048 ## income -0.06079785 0.03901982 1.00000000 -0.11761722 ## age -0.06799270 0.08608048 -0.11761722 1.00000000 This resulting matrix provides the correlation coefficients for two variables at a time. The correlation coefficients are defined by the intercept of the row and columns corresponding to each variable. Additionally, recalling that each correlation coefficient itself is a random value, a test is required for inference. The corr.test() provided by the psych package is an alternative to the cor.test() function previously used. The corr.test() function supports examining variable pairs simultaneously within a given data set. Use the print() function with the short=FALSE argument to view the complete test with confidence intervals and p-values. ds %>% dplyr::select(glbcc_risk, ideol, income, age) %>% drop_na() %>% corr.test %>% print(short = FALSE) ## Call:corr.test(x = .) ## Correlation matrix ## glbcc_risk ideol income age ## glbcc_risk 1.00 -0.60 -0.06 -0.07 ## ideol -0.60 1.00 0.04 0.09 ## income -0.06 0.04 1.00 -0.12 ## age -0.07 0.09 -0.12 1.00 ## Sample Size ## [1] 2275 ## Probability values (Entries above the diagonal are adjusted for multiple tests.) ## glbcc_risk ideol income age ## glbcc_risk 0 0.00 0.01 0 ## ideol 0 0.00 0.06 0 ## income 0 0.06 0.00 0 ## age 0 0.00 0.00 0 ## ## Confidence intervals based upon normal theory. To get bootstrapped values, try cor.ci ## raw.lower raw.r raw.upper raw.p lower.adj upper.adj ## glbc_-ideol -0.63 -0.60 -0.57 0.00 -0.63 -0.56 ## glbc_-incom -0.10 -0.06 -0.02 0.00 -0.11 -0.01 ## glbc_-age -0.11 -0.07 -0.03 0.00 -0.12 -0.02 ## ideol-incom 0.00 0.04 0.08 0.06 0.00 0.08 ## ideol-age 0.05 0.09 0.13 0.00 0.03 0.14 ## incom-age -0.16 -0.12 -0.08 0.00 -0.17 -0.06 ## 7.3 Visualizing Correlation The simplest form to visualize correlation is a scatter plot with a trend line. We will review the relationship between ideology and perceived risk from climate change. Build the basic visualization by using ggplot() and the geom_point functions, with ideology on the x axis and perceived risk about climate change on the y axis. Use the ds.sub dataset, because we removed the missing values. ggplot(ds.sub, aes(x = ideol, y = glbcc_risk)) + geom_point(shape = 1) Notice how this doesn’t really make sense? This is because there are thousands of observations being placed on a discrete set of values. To get a better idea of the relationship, we can tell R to jitter the points. “Jittering” provides a tiny bit of white noise and variance to the values, so that we can see where there is high overlap of observations. This provides a better picture of the relationship. This time use the geom_jitter function instead of the geom_point function: ggplot(ds.sub, aes(x = ideol, y = glbcc_risk)) + geom_jitter(shape = 1) Notice the apparent negative correlation. Verify this by checking the correlation: ds %>% dplyr::select(ideol, glbcc_risk) %>% cor(use = "complete.obs") ## ideol glbcc_risk ## ideol 1.0000000 -0.5901706 ## glbcc_risk -0.5901706 1.0000000 A trend line to the scatter plot helps to interpret directionality of a given variable pair. The next lab will further introduce trend lines, but for now we introduce it via the geom_smooth() function by including the method=lm argument. We also need to include geom_point() for each point. ggplot(ds.sub, aes(x = ideol, y = glbcc_risk)) + geom_point(shape = 1) + geom_smooth(method = lm) + geom_jitter(shape = 1) The ideology points can be differentiated by other variables, such as gender, to examine potential difference among gender by defining the color argument as follows: ggplot(ds.sub, aes(x = ideol, y = glbcc_risk, color = f.gender)) + geom_point(shape = 1) + geom_smooth(method = lm) + geom_jitter(shape = 1) ### 7.3.1 Another Example: Political Party Let’s look at this relationship broken down by political party. Perhaps you wanted to see if the relationship looks different for Republicans and Democrats. We can first subset the data for Republicans and Democrats: ds.dem <- filter(ds.sub, f.party.2 == "Dem") ds.rep <- filter(ds.sub, f.party.2 == "Rep") Next we investigate the correlation between ideology and perceived risk of climate change for Republicans and Democrats separately: cor.test(ds.rep$ideol, ds.rep$glbcc_risk) ## ## Pearson's product-moment correlation ## ## data: ds.rep$ideol and ds.rep$glbcc_risk ## t = -14.352, df = 1171, p-value < 0.00000000000000022 ## alternative hypothesis: true correlation is not equal to 0 ## 95 percent confidence interval: ## -0.4343892 -0.3369909 ## sample estimates: ## cor ## -0.3867682 cor.test(ds.dem$ideol, ds.dem$glbcc_risk) ## ## Pearson's product-moment correlation ## ## data: ds.dem$ideol and ds.dem$glbcc_risk ## t = -13.977, df = 859, p-value < 0.00000000000000022 ## alternative hypothesis: true correlation is not equal to 0 ## 95 percent confidence interval: ## -0.4833657 -0.3744104 ## sample estimates: ## cor ## -0.4304548 The correlation coefficient is slightly more negative for Democrats. To create a visualization that compares the two parties and the relationship between climate change risk and ideology, a few simple lines of code can get the job done. First filter the data to include Democrats and Republicans, select the variables of interest, drop NAs, pipe it all into ggplot2, then use facet_wrap() to create two visualizations, one for each party: ds %>% filter(f.party.2 == "Dem" | f.party.2 == "Rep") %>% dplyr::select(ideol, glbcc_risk, f.party.2) %>% na.omit() %>% ggplot(., aes(ideol, glbcc_risk)) + geom_jitter(shape = 1) + geom_smooth(method = lm) + facet_wrap(~ f.party.2, scales = "fixed") ### 7.3.2 One More Visualization We cap this lab off with creating one last visualization. First create a new subset of our data exclusive of all missing observations. Include variables for climate change risk and age. sub.ds <- filter(ds) %>% dplyr::select("glbcc_risk", "age") %>% na.omit() First we look at our age variable. describe(sub.ds$age)
## vars n mean sd median trimmed mad min max range skew kurtosis
## X1 1 2536 60.37 14.2 62 61.02 13.34 18 99 81 -0.39 -0.24
## se
## X1 0.28
Now find the correlation:
cor.test(sub.ds$glbcc_risk, sub.ds$age)
##
## Pearson's product-moment correlation
##
## data: sub.ds$glbcc_risk and sub.ds$age
## t = -3.6446, df = 2534, p-value = 0.0002732
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
## -0.11082410 -0.03338258
## sample estimates:
## cor
## -0.07221218
There is a slight negative correlation. We can interpret this as indicating that younger people are slightly more concerned about climate change. Now we construct the visualization:
ggplot(sub.ds, aes(y = glbcc_risk, x = age)) +
geom_point(shape = 20, color = "#e20000") +
geom_jitter(shape = 20, color = "#e20000") +
geom_smooth(method = lm) +
xlab("Age") +
ylab("Climate Change Risk") +
ggtitle("Age and Climate Change Risk") +
scale_y_continuous(breaks = c(0:10),
labels = c("0","1","2", "3", "4", "5", "6", "7", "8", "9", "10")) +
theme_bw() |
# Word Problems on Addition and Subtraction of Whole Numbers | Simple Addition and Subtraction Word Problems
Word Problems on Addition and Subtraction provided here covers different kinds of questions. Practice using the Addition and Subtraction Problems provided here and test where you stand in your preparation. Now we will see how to solve the word problems on addition and subtraction of whole numbers. Learn to apply the problem-solving approach used and try to solve related problems.
Also, Check:
## Addition and Subtraction of Whole Numbers Word Problems
Example 1:
Dany jogs 1200m and runs 1500 m. Find the total distance.
Solution:
Given
Dany jogs = 1200m
Dany runs = 1500m
Total distance covered by dany is = dany jogs + dany runs
= 1200 + 1500
= 2700 m
Therefore distance covered by dany is 2700 m
Example 2:
A book has 650 pages. Ted has read 220 pages so far. How many pages of this book are left to read by ted?
Solution:
Given
Number of pages in the book = 650
Number of pages read by ted = 220
Number of pages left to read by ted = Number of pages in the book – number of pages read by ted
= 650 – 220
= 430
Therefore 430 pages are left to read by ted.
Example 3:
A factory makes 287 blue bags and 346 green bags in a month. How many bags does the factory produce in a month?
Solution:
Given
Number of blue bags = 287
Number of green bags = 346
Total number of bags = Number of blue bags + number of green bags
= 287 + 346
= 633
Therefore factory produce 633 bags in the month,
Example 4:
Mary is the teacher of class first. She has 12 girls and 15 boys in her class.
i)How many students are there in the class?
ii)She has 30 boxes of crayons. How many more boxes of crayons are there than the number of students?
Solution:
Given
Number of girls = 12
Number of boys = 15
i)Total number of students = Number of girls + Number of boys
= 12 + 15
= 27
Therefore, the total number of students in her class is 27
ii)Number of boxes mary have = 30
Number of boxes she needs = number of boxes mary have – total number of students
= 30 – 27
= 3
Therefore, there are 3 more boxes of crayons than the number of students.
Example 5:
Jones bought a dozen pencils and erasers costing Rs.125.25 and Rs.110.06 respectively. She paid rs.500 to the shopkeeper. How much should she get back from the shopkeeper?
Solution:
Given
Cost of pencils = 125.25
Cost of erasers = 110.06
Total cost = cost of pencils + cost of erasers
= 125.25 + 110.06
= 235.31rs
Amount paid by jones to shopkeeper = 500.00
Total amount of pencils and erasers = 235.31
Difference = 500.00 – 235.31
= 264.69
Jones should get back rs 264.69 from the shopkeeper.
Example 6:
In a library, there are 700 books. Jane took 70 books and peter took 40 books. After few days, peter returned 23 books to the library. How many books are there in the library now?
Solution:
Total number of books = 700
Number of books jane took = 70
Number of books peter took = 40
Number of books in the library = Total number of books – ( Number of books jane took + Number of books peter took)
= 700 – (70 + 40)
= 590 books
After few days,
Number of books peter returned = 23
Number of books in library = 590 + 23
= 613 books
Therefore, the number of books in the library now = 613 books
Example 7:
A basketball team is getting ready for the next season. The team decided to practice for 3 hours on Mondays and Wednesdays and for 4 hours on Thursdays and Saturdays. How many hours will they be practicing every week?
Solution:
Given
Number of hours team practice on Mondays and Wednesdays = 3 + 3 = 6 hours
Number of hours team practice on Thursdays and Saturdays = 4 + 4 = 8 hours
Total number of hours team practice every week = 6 + 8 = 14 hours
Therefore, the Number of hours team practice every week is 14 hours.
Example 8:
Rosy is the manager of a restaurant. 15 waiters are supposed to be at work. But 5 of them are sick. Rosy managed to call two more waiters to come in to the work. How many waiters are working today?
Solution:
Given
Number of waiters = 15
Number of sick waiters = 5
Number of additional waiters = 2
Number of waiters working today = (Number of waiters – Number of sick waiters) + Number of additional waiters
= (15 – 5) + 2
= 12 waiters
Therefore, the Number of waiters working today = 12 waiters.
Example 9:
Ram works at xyz company and his monthly salary is 30,000. His room rent is 7000 and other expenses are upto 2500. He will get an additional amount of 4500 from the bank as interest. How much money will be left with ram at the end of the month?
Solution:
Given,
Ram’s monthly salary = 30,000
Ram’s bank interest = 4500
His total monthly income = Monthly salary + bank interest
= 30,000 + 4500
= 34,500
His monthly room rent = 7000
His other expenses = 2500
Ram’s Expenditure = Room rent + other expenses
= 7000 + 2500
= 9500
Ram’s monthly savings = Income – Expenditure
= 34,500 – 9500
= 25,000
Money left with ram at the end of the month = 25,000
Example 10:
The total sale of a shop in the month of may was Rs.8956321. If the sale for the first two weeks were Rs.2569865 and Rs.2175800, find the sale of the remaining weeks?
Solution:
Given
Sale of the first week = 2569865
Sale of the second week = 2175800
Sale in first two weeks = Sale in first week + Sale in the second week
= 2569865 + 2175800
= 4745665
Total sale of the shop = 8956321
Sale in first two weeks = 4745665
Sale of remaining weeks = total sale of the shop – sale in first two weeks
= 8956321 – 4745665
= 4210656
Therefore, the sale of the remaining weeks’ Rs. 4210656 |
All the solutions provided in McGraw Hill Math Grade 1 Answer Key PDF Chapter 7 Test as per the latest syllabus guidelines.
Read the number sentences. Write the sums.
Question 1.
5 + 7 = ____________
7 + 5 = ____________
5 + 7 = 12
7 + 5 = 12
Explanation:
If we change the pattern of addends the sum doesn’t change,
The sum remains the same.
Question 2.
12 + 6 = ____________
6 + 12 = ____________
12 + 6 = 18
6 + 12 = 18
Explanation:
If we change the pattern of addends the sum doesn’t change,
The sum remains the same.
Question 3.
0 + 8 = ____________
0 + 8 = 8
Explanation:
The sum of 0 and 8 is 8
If we add the zero to a number the number remains the same.
Question 4.
0 + 4 = ____________
4
Explanation:
The sum of 0 and 4 is 4
If we add the zero to a number the number remains the same.
Question 5.
5 + 1 + 2 = ____________
5 + 1 = 6
6 + 2 = 8
Explanation:
The sum of 5 and 1 is 6
and 6 and 2 is 8
Question 6.
3 + 6 + 4 = ____________
6 + 4 = 10
10 + 6 = 16
Explanation:
First we have to form the ten, then we have to add the other number.
Read the problems. Write the missing number.
Question 7.
5 + = 9
9 – 5 = ____________
5 + ____________ = 9
5 + 4 = 9
9 – 5 = 4
5 + 4 = 9
Explanation:
To find the missing number in addition we have to subtract
Question 8.
5 + 1 = 18
18 – 5 = ____________
5 + ____________ = 18
5 + 13 = 18
18 – 5 = 13
5 + 13 = 18
Explanation:
To find the missing number in addition we have to subtract
Look at each addition sentence. Write another fact with the same sum.
Question 9.
9 + 2 = 11
____________ + ____________ = 11
9 + 2 = 11
2 + 9 = 11
Explanation:
If we change the pattern of addends the sum doesn’t change,
The sum remains the same.
Question 10.
3 + 4 = 7
____________ + ____________ = 7
3 + 4 = 7
4 + 3 = 7
Explanation:
If we change the pattern of addends the sum doesn’t change,
The sum remains the same.
Look at each set of numbers. Write = if they are equal. Write not = if they are not equal.
Question 11.
17 ____________ 17
17 = 17
Explanation:
If the given numbers are equal they are represented with the symbol ‘=’
Question 12.
35 ____________ 30 + 5
30 + 5 = 35
35 = 35
Explanation:
If the given numbers are equal they are represented with the symbol ‘=’
Question 13.
8 + 3 ____________ 12 + 5
11 ≠17
Explanation:
If the given numbers are equal they are represented with the symbol ‘=’
If they are not equal we represent with â‰
Question 14.
9 – 3 ____________ 4 + 2
9 – 3 = 6
4 + 2 = 6
6 = 6
Explanation:
If the given numbers are equal they are represented with the symbol ‘=’
Question 15.
8 + 5 = ?
10 + 3 = ____________
so 8 + 5 = _____________
8 + 5 = 13
10 + 3 = 13
Explanation:
There are 8 green dots in ten frame
we have to add 5 more to get 13
Question 16.
9 + 6 = ?
10 + 5 = _____________
so 9 + 6 = _____________
10 + 5 = 15
9 + 6 = 15
Explanation:
There are 9 red dots in ten frame.
so added 6 to get 15
Count on to add. Count back to subtract.
Question 17.
4 + 7 = ___________
Explanation:
The sum of 4 and 7 is 11
4 + 7 = 11
Question 18.
18 – 12 = ___________
Explanation:
The difference of 18 and 12 is 6
18 – 12 = 6
Look at the chart. Write an addition sentence. Then write a subtraction sentence.
Question 19.
__________ + ___________ = ___________
2 + 4 = 6
4 + 2 = 6
Explanation:
If we change the pattern of addends the sum doesn’t change,
The sum remains the same.
Question 20.
__________ – ___________ = ___________
6 – 2 = 4
6 – 4 = 2
Explanation:
The difference of 6 and 2 is 4, 6 and 4 is 2
Add or subtract to solve. Write the sum or difference. Use objects to help.
Question 21.
6 play. 2 sleep. How many are there?
6 + 2 = ___________
6 + 2 = 8
Explanation:
Number of dogs that play = 6
Number of dogs that sleep = 2
so, there are 8 dogs in all.
Question 22.
Ben has 13 He gives 6 to Sara. How many does Ben have now?
13 – 6 ___________ |
# For any point P on the unit circle, what is the average of the distance between P and all points on the unit circle?
My main issue with this question stems from the different answers I get depending on how I partition my intervals when attempting to write this question as an integral. The way I understand calculating averages like this is to use the fact that the average of N values is the sum of the values divided by the total number of values (n), meaning as N goes to infinity, this sequence of averages converges to our desired average. One way to do this is to try and write it as a Riemann sum.
One way to do this is if we have a the distance as function of our angle, which if we simply treat our point as being (1,0) is: $$\ f(θ) =\sqrt{(1-\cos(θ))^{2}+(\sin(θ))^{2}} =\sqrt{2}\sqrt{1-\cos(θ)}$$
Since the circle is symmetrical, all the distances are equal so, we can just partition the angles into N intervals between 0 and pi and can write the average as a Riemann Sum, which becomes an integral as n goes to infinity. $$\dfrac{\sum_{i=0}^n\ f(θ_{i})}{n} =\dfrac{\sum_{i=0}^n\ f(θ_{i})}{\dfrac{\pi}{\Deltaθ}}= \dfrac{\sum_{i=0}^n\ f(θ_{i})\Deltaθ}{\pi}= \frac{1}{\pi}\int_0^{\pi}f(θ)dθ=\dfrac{4}{\pi}$$
However it seems another way we can do this is partition along the x axis from -1 t0 1 and integrate this function, which is the distance of point on the unit cirlce to (1,0) as function of x: $$f(x)=\sqrt{(1-x)^{2}+(\sqrt{1-x^{2}})^{2}}=\sqrt{2}\sqrt{1-x}$$
and write the Riemann sum as $$\dfrac{\sum_{i=0}^n\ f(x_{i})}{n} =\dfrac{\sum_{i=0}^n\ f(x_{i})}{\dfrac{2}{\Delta{x}}}= \dfrac{\sum_{i=0}^n\ f(x_{i})\Delta{x}}{2}= \frac{1}{2}\int_{-1}^{1}f(x)dx=\dfrac{4}{3}$$
It seems like both ways are valid when it comes to getting the average, so I don't know why they result in different values.
• The arc covered by $x\in[-1,-0.99]$ is much longer than the arc covered by $x\in[0,0.01]$, hence in your second method you weigh these intervals differently. Sep 13 '20 at 8:19
• you take a point $P$ and draw a diameter from the point through center. Now your starting point is opposite end of the diameter and you traverse on the circle to point $P$. The distance varies from $2$ to $0$ and the angle between $0$ and $\pi / 2$. Average distance will be $4/ \pi$. Sep 13 '20 at 8:47
the average value of discrete variable $$x_i$$ with probability $$P(x_i)$$ is $$\overline{x} = \sum_{i_1}^n P(x_i)x_i$$ I set a circle with radius $$R$$ such that the center on the $$x$$-axis while its circumference passes through the origin, in polar coordinates $$r(\theta) = 2R\cos(\theta) \hspace{1 cm} \theta \in [-\frac{\pi}{2},-\frac{\pi}{2}]$$ $$\overline{r} = \sum_{i=1}^n P(r_i)r_i = \int \frac{dP(r)}{d\theta}r(\theta)\,d\theta$$ The probability of choosing the second point on the circle is distributed uniformly over the circumference (the first point is at the origin) $$dP(r) = \frac{ds}{\int ds} = \frac{ds}{2 \pi R}$$ where $$ds$$ is the arclenght $$ds = \sqrt{(rd\theta)^2+(r'd\theta)^2} = \sqrt{(r)^2+(r')^2}d\theta = 2Rd\theta$$ $$\overline{r} =\int \frac{dP(r)}{d\theta}r(\theta)\,d\theta = \int \frac{2R}{2 \pi R}2R\cos(\theta)\, d\theta = \frac{2R}{\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos(\theta)\, d\theta = \frac{4R}{\pi} < 2R$$ |
# What is the Median in Math? The Answer Might Leave You Surprised
Abhijit Naik May 12, 2019
While 'median' refers to the middle value of an ordered set of values in math, 'mode' refers to the most frequently occurring random variable in the set. Read on to know more about median.
In mathematics, median is the middle value of an ordered set of values. It is also defined as the numeric value which separates the higher half of the data from the lower half.
## How is Median Calculated?
Median is calculated by arranging the numerical data in ascending order, i.e., starting from the lowest, and choosing the middle value in the set. While that works fine when you have odd number of observations, you have to calculate the arithmetic mean for the two middle values to find the median if you have even number of observations.
### Example 1: Odd Number of Observations
You are supposed to calculate the median weight for a group of people. In order to do this, you weigh seven people who weigh 94, 108, 120, 145, 88, 98, and 115 lbs respectively. The first thing to do is, arrange the data in ascending order; that will be: 88, 94, 98, 108, 115, 120, and 145.
Like we said, median is the middle value when it comes to odd number of observations. As you took seven samples into consideration, the middle value will be the fourth value in your data, which in this case is 108. So, the median weight of the group of seven people you studied is 108 lbs.
### Example 2: Even Number of Observations
Now let's consider you are calculating the median weight of a group of eight individuals. Let's assume that the data you compiled after weighing eight people in the group was 94, 108, 120, 145, 88, 98, 115, and 130. (We will simply add another observation to the previous example.)
After arranging it in the ascending order, you get a list in the following order: 88, 94, 98, 108, 115, 120, 130, and 145. There being no middle value, you cannot pinpoint a particular number and call it the median. Instead, you have to choose two middle values and find the arithmetic mean.
In this case the two middle values are 108 and 115. In order to find their arithmetic mean, you will have to add these two values and divide their sum by 2. When you add 108 + 115, the sum will be 223, and when you divide this 223 by 2, the sum will be 111.5, which will be the median weight for the group.
## Is Median the Same as Average?
Average, as the name suggests, is the average value which is calculated by adding all the values in the set of data and dividing it by the total number of values. Median, on the other hand, is the middle value in the set of data arranged in ascending order.
While the median for the example we discussed earlier (88, 94, 98, 108, 115, 120, 145) will be 108, the average for the same will be 109.71, i.e., the sum of all the values (768) divided by the total number of values (7).
Yet another important concept related to median is 'median income'―a statistical number compiled by the U.S. Department of Housing and Urban Development, which divides households into two equal segments; one with income more than the median and the other with income less than it.
Interestingly, median income is a better indicator of the average American income, because, unlike average, it is not affected by extreme values. |
# What is the equation of the line that passes through (0,-1) and is perpendicular to the line that passes through the following points: (8,-3),(1,0) ?
Feb 10, 2016
$7 x - 3 y + 1 = 0$
#### Explanation:
Slope of the line joining two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is given by
$\frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$ or $\frac{{y}_{1} - {y}_{2}}{{x}_{1} - {x}_{2}}$
As the points are $\left(8 , - 3\right)$ and $\left(1 , 0\right)$, slope of line joining them will be given by $\frac{0 - \left(- 3\right)}{1 - 8}$ or $\frac{3}{- 7}$
i.e. $- \frac{3}{7}$.
Product of slope of two perpendicular lines is always $- 1$. Hence slope of line perpendicular to it will be $\frac{7}{3}$ and hence equation in slope form can be written as
$y = \frac{7}{3} x + c$
As this passes through point $\left(0 , - 1\right)$, putting these values in above equation, we get
$- 1 = \frac{7}{3} \cdot 0 + c$ or $c = 1$
Hence, desired equation will be
$y = \frac{7}{3} x + 1$, simplifying which gives the answer
$7 x - 3 y + 1 = 0$ |
# How do you find a polynomial function that has zeros 1+sqrt3, 1-sqrt3?
Aug 14, 2017
$f \left(x\right) = {x}^{2} - 2 x - 2$
#### Explanation:
Since these are the zeros, we can make the following equation:
$\left(x - \left(1 + \sqrt{3}\right)\right) \left(x - \left(1 - \sqrt{3}\right)\right) = 0$
Or
$\left(x - 1 - \sqrt{3}\right) \left(x - 1 + \sqrt{3}\right) = 0$
When we expand this, we get
${x}^{2} - x + x \sqrt{3} - x + 1 - \sqrt{3} - x \sqrt{3} + \sqrt{3} - 3 = 0$
Combining like terms:
color(blue)(ulbar(|stackrel(" ")(" "f(x) = x^2 - 2x - 2" ")|)
Aug 15, 2017
$f \left(x\right) = {x}^{2} - 2 x - 2$
#### Explanation:
The simplest polynomial with distinct zeros $\alpha$ and $\beta$ is:
$\left(x - \alpha\right) \left(x - \beta\right) = {x}^{2} - \left(\alpha + \beta\right) x + \alpha \beta$
With $\alpha = 1 + \sqrt{3}$ and $\beta = 1 - \sqrt{3}$, we find:
$\left\{\begin{matrix}\alpha + \beta = \left(1 + \sqrt{3}\right) + \left(1 - \sqrt{3}\right) = \textcolor{red}{2} \\ \alpha \beta = \left(1 + \sqrt{3}\right) \left(1 - \sqrt{3}\right) = {1}^{2} - {\left(\sqrt{3}\right)}^{2} = 1 - 3 = \textcolor{b l u e}{- 2}\end{matrix}\right.$
So a suitable polynomial function would be:
$f \left(x\right) = {x}^{2} - \textcolor{red}{2} x \textcolor{b l u e}{- 2}$
Any polynomial function in $x$ with these two zeros will be a multiple (scalar or polynomial) of this $f \left(x\right)$. |
# Algebraic Identities
The algebraic equations which are valid for all values of variables in them are called algebraic identities. They are also used for the factorization of polynomials. In this way, algebraic identities are used in the computation of algebraic expressions and solving different polynomials. You have already learned about a few of them in the junior grades. In this article, we will recall them and introduce you to some more standard algebraic identities, along with examples.
## Standard Algebraic Identities List
All the standard Algebraic Identities are derived from the Binomial Theorem, which is given as:
$$\mathbf{(a+b)^{n} =\; ^{n}C_{0}.a^{n}.b^{0} +^{n} C_{1} . a^{n-1} . b^{1} + …….. + ^{n}C_{n-1}.a^{1}.b^{n-1} + ^{n}C_{n}.a^{0}.b^{n}}$$
Some Standard Algebraic Identities list are given below:
Identity I: (a + b)2 = a2 + 2ab + b2
Identity II: (a – b)2 = a2 – 2ab + b2
Identity III: a2 – b2= (a + b)(a – b)
Identity IV: (x + a)(x + b) = x2 + (a + b) x + ab
Identity V: (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Identity VI: (a + b)3 = a3 + b3 + 3ab (a + b)
Identity VII: (a – b)3 = a3 – b3 – 3ab (a – b)
Identity VIII: a3 + b3 + c– 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)
Example 1: Find the product of (x + 1)(x + 1) using standard algebraic identities.
Solution: (x + 1)(x + 1) can be written as (x + 1)2. Thus, it is of the form Identity I where a = x and b = 1. So we have,
(x + 1)2 = (x)2 + 2(x)(1) + (1)2 = x2 + 2x + 1
Example 2: Factorise (x4 – 1) using standard algebraic identities.
Solution: (x4 – 1) is of the form Identity III where a = x2 and b = 1. So we have,
(x4 – 1) = ((x2)2– 12) = (x2 + 1)(x2 – 1)
The factor (x2 – 1) can be further factorised using the same Identity III where a = x and b = 1. So,
(x4 – 1) = (x2 + 1)((x)2 –(1)2) = (x2 + 1)(x + 1)(x – 1)
Eample 3: Factorise 16x2 + 4y2 + 9z2 – 16xy + 12yz – 24zx using standard algebraic identities.
Solution: 16x2 + 4y2 + 9z2– 16xy + 12yz – 24zx is of the form Identity V. So we have,
16x2 + 4y2 + 9z2 – 16xy + 12yz – 24zx = (4x)2 + (-2y)2 + (-3z)2 + 2(4x)(-2y) + 2(-2y)(-3z) + 2(-3z)(4x)= (4x – 2y – 3z)2 = (4x – 2y – 3z)(4x – 2y – 3z)
Example 4: Expand (3x – 4y)3 using standard algebraic identities.
Solution: (3x– 4y)3 is of the form Identity VII where a = 3x and b = 4y. So we have,
(3x – 4y)3 = (3x)3 – (4y)3– 3(3x)(4y)(3x – 4y) = 27x3 – 64y3 – 108x2y + 144xy2
Example 5: Factorize (x3 + 8y3 + 27z3 – 18xyz) using standard algebraic identities.
Solution: (x3 + 8y3 + 27z3 – 18xyz)is of the form Identity VIII where a = x, b = 2y and c = 3z. So we have,
(x3 + 8y3 + 27z3 – 18xyz) = (x)3 + (2y)3 + (3z)3 – 3(x)(2y)(3z)= (x + 2y + 3z)(x2 + 4y2 + 9z2 – 2xy – 6yz – 3zx)
## Frequently Asked Questions on Algebraic Identities
### What are the three algebraic identities in Maths?
The three algebraic identities in Maths are:
Identity 1: (a+b)^2 = a^2 + b^2 + 2ab
Identity 2: (a-b)^2 = a^2 + b^2 – 2ab
Identity 3: a^2 – b^2 = (a+b) (a-b)
### What is the difference between an algebraic expression and identities?
An algebraic expression is an expression which consists of variables and constants. In expressions, a variable can take any value. Thus, the expression value can change if the variable values are changed. But algebraic identity is equality which is true for all the values of the variables.
### How to verify algebraic identity?
The algebraic identities are verified using the substitution method. In this method, substitute the values for the variables and perform the arithmetic operation. Another method to verify the algebraic identity is the activity method. In this method, you would need a prerequisite knowledge of Geometry and some materials are needed to prove the identity. |
## Solution
Here we have the solution for the Muddy Head Puzzle:-
• Let S be the statement that the son has a muddy forehead and let D be the statement that the daughter has a muddy forehead.
• When the mother says that at least one of the two children has a muddy forehead, she is stating that the disjunction S ∨ D is true.
• Both children will answer “No” the first time the question is asked because each sees mud on the other child’s forehead.
• That is, the son knows that D is true but does not know whether S is true, and the daughter knows that S is true, but does not know whether D is true.
• After the son has answered “No” to the first question, the daughter can determine that D must be true. This follows because when the first question is asked, the son knows that S ∨ D is true, but cannot determine whether S is true.
• Using this information, the daughter can conclude that D must be true, for if D were false, the son could have reasoned that because S ∨ D is true, then S must be true, and he would have answered “Yes” to the first question.
• The son can reason similarly to determine that S must be true. It follows that both children answer “Yes” the second time the question is asked.
• Assume that both children are honest and that the children answer each question simultaneously.
### Solution
Here we have the solution for the Muddy Head Puzzle:-
• Let S be the statement that the son has a muddy forehead and let D be the statement that the daughter has a muddy forehead.
• When the mother says that at least one of the two children has a muddy forehead, she is stating that the disjunction S ∨ D is true.
• Both children will answer “No” the first time the question is asked because each sees mud on the other child’s forehead.
• That is, the son knows that D is true but does not know whether S is true, and the daughter knows that S is true, but does not know whether D is true.
• After the son has answered “No” to the first question, the daughter can determine that D must be true. This follows because when the first question is asked, the son knows that S ∨ D is true, but cannot determine whether S is true.
• Using this information, the daughter can conclude that D must be true, for if D were false, the son could have reasoned that because S ∨ D is true, then S must be true, and he would have answered “Yes” to the first question.
• The son can reason similarly to determine that S must be true. It follows that both children answer “Yes” the second time the question is asked.
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# A square is inscribed in the circle ${{x}^{2}}+{{y}^{2}}-2x+4y-3=0$ with its sides parallel to the coordinate axes. One vertex of square is:A. (3, 4)B. (3, -4)C. (8, -5)D. (-8, 5)
Last updated date: 16th Sep 2024
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Hint: At first, convert the equation of circle into the form ${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}$ where $\left( {{x}_{1}},{{y}_{1}} \right)$ is the center and r is radius. Then, use the property that, diagonal of a square equals the diameter of the circle to find the length of the square's side. Then, consider the coordinate of any of the vertices of the square as (a, b) and from that find other in terms of a and b. After that, use the property that the midpoint of the diagonal of the square is the center of the circle.
In the question, we are said that, square is drawn inside or inscribed in a circle with a given equation ${{x}^{2}}+{{y}^{2}}-2x+4y-3=0$ with a given condition that is the sides of square parallel to the axis. For the given condition, we have to find one vertex of the square.
The given equation of circle is,
${{x}^{2}}+{{y}^{2}}-2x+4y-3=0$
We will further write the equation as,
\begin{align} & {{x}^{2}}-2x+1+{{y}^{2}}+4y+4-3-5=0 \\ & \Rightarrow {{\left( x-1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}-8=0 \\ \end{align}
So, the equation is formed as,
${{\left( x-1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}=8={{\left( 2\sqrt{2} \right)}^{2}}$
If the equation of circle is in form of,
${{\left( x-{{x}_{1}} \right)}^{2}}+{{\left( y-{{y}_{1}} \right)}^{2}}={{r}^{2}}$
Then, its center is $\left( {{x}_{1}},{{y}_{1}} \right)$ and radius is r.
As the equation is ${{\left( x-1 \right)}^{2}}+{{\left( y+2 \right)}^{2}}={{\left( 2\sqrt{2} \right)}^{2}}$ so, its center is $\left( 1,-2 \right)$ and radius is $2\sqrt{2}$
Let ABCD be the square inside the circle, whose center is O. So, it can be drawn as,
As we know that, the radius of the circle is $2\sqrt{2}$ so, the diameter will be twice as radius $4\sqrt{2}$.
The diameter of the circle is a diagonal of a square. Using the relation, diagonal of square $\sqrt{2}\times \text{side of square}$ we can find the side of the square.
Here, side is AB, so, its length will be $\dfrac{\text{diagonal}}{\sqrt{2}}\Rightarrow \dfrac{4\sqrt{2}}{\sqrt{2}}\Rightarrow 4$
Now, let’s suppose, coordinates of B be (a, b).
Now, as we know that, each side length is 4 and sides are parallel to axes, we can write other coordinates in terms of a and b too.
So, the coordinates of A, C and D are $\left( a-4,b \right);\left( a,b-4 \right);\left( a-4,b-4 \right)$.
We know that, midpoint of the diagonal of a square is the center of the circle. So, we can say that the midpoint of the coordinator of B and D is the center of the circle.
We will find midpoint using formula, $x'=\dfrac{{{x}_{1}}+{{x}_{2}}}{2}\,\,and\,\,y'=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}$
If $\left( x',y' \right)$ is the midpoint between $\left( {{x}_{1}},{{y}_{1}} \right)\,\,and\,\,\left( {{x}_{2}},{{y}_{2}} \right)$ we can find midpoint using formula, $x'=\dfrac{{{x}_{1}}+{{x}_{2}}}{2}\,\,and\,\,y'=\dfrac{{{y}_{1}}+{{y}_{2}}}{2}$.
Here, points are B(a, b) and D(a-4, b-4), so, its midpoint will be,
$\left( \dfrac{a+a-4}{2},\dfrac{b+b-4}{2} \right)\Rightarrow \left( \dfrac{2a-4}{2},\dfrac{2b-4}{2} \right)$
Now, as we also know that its midpoint is the center of the circle which is (1, -2).
So we can say,
$\left( \dfrac{\left( 2a-4 \right)}{2},\dfrac{\left( 2b-4 \right)}{2} \right)=\left( 1,-2 \right)$
So, we can say that,
\begin{align} & \dfrac{2a-4}{2}=1\,\,and\,\,\dfrac{2b-4}{2}=-2 \\ & \Rightarrow 2a-4=2\,\,and\,\,2b-4=-4 \\ \end{align}
Thus, on simplifying we can say that,
\begin{align} & 2a=6\,\,\Rightarrow a=3 \\ & and\,\,2b=0\,\,\Rightarrow b=0 \\ \end{align}
So, the coordinates of B are (3, 0).
The coordinates of A is (-1, 0), C is (3, -4) and D is (-1, -4).
Among the above coordinates only (3, -4) matches the option.
So, the correct answer is “Option B”.
Note: Students while solving the problem must know the properties of the square when inscribed in the circle just like the diameter of a circle is equal to the diagonal of the square. Also, these properties help to solve the questions more easily. |
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# A particle moves in a circular path of radius r. In half the period of revolution its displacement and distance covered are.\begin{align} & \text{A}\text{. }2r,\pi r \\ & \text{B}\text{. }r,\pi r \\ & \text{C}.\text{ }2r,2\pi r \\ & \text{D}\text{. }r\sqrt{2},\pi r \\ \end{align}
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Hint: We know that basically in circular motion the total displacement for half complete motion is the shortest distance between initial and final points that is diameter. And the distance covered is actual path travelled that is half the circumference.
The displacement of a particle for complete one revolution will be its diameter.
Hence displacement will be 2r where,
R is the radius of the circle.
The distance travelled by the particle for complete half revolution will be half the circumference.
Hence distance travelled will be $\pi r$.
So, the correct answer is “Option A”.
Distance or path length: it is the actual length travelled by the body between initial position and final position.
Distance is the scalar quantity because it has only magnitude and no direction. Distance covered is always positive or zero. The S.I unit of distance is m(metre).
Displacement: The displacement of the object is the change in position of an object in a fixed direction. It is the shortest path measured in the direction from initial point to final point. Displacement is the vector quantity and has both magnitude and direction. Displacement may be positive, negative or zero.
Characteristics of displacement: displacements have units of length.
The displacement of an object can be negative or zero.
Displacement is not dependent on the choice of the origin O of the position coordinates.
The actual distance travelled by the object is greater than or equal to the magnitude of displacement.
Note:
The displacement in circular motion will be 0 because displacement is the shortest path between initial and final position and in this case both the points coincide resulting in 0 displacement. And for distance travelled that is defined as actual path travelled which is the circumference of the circle. |
# Simplifying Expressions – Explanation & Examples
Learning how to simplify an expression is the most important step in understanding and mastering algebra. Simplification of expressions is a handy mathematics skill because it allows us to change complex or awkward expressions into simpler and compact forms. But before that, we must know what an algebraic expression is.
An algebraic expression is a mathematical phrase where variables and constants are combined using the operational (+, -, × & ÷) symbols. For example, 10x + 63 and 5x – 3 are examples of algebraic expressions.
In this article, we shall learn a few tricks on how to simplify any algebraic expression.
## How to Simplify Expressions?
Simplification of an algebraic expression can be defined as the process of writing an expression in the most efficient and compact form without affecting the value of the original expression.
The process entails collecting like terms, which implies adding or subtracting terms in an expression.
Let’s remind ourselves of some of the important terms used when simplifying an expression:
• A variable is a letter whose value is unknown in an algebraic expression.
• The coefficient is a numerical value used together with a variable.
• A constant is a term that has a definite value.
• Like terms are variables with the same letter and power. Like terms can sometimes contain different coefficients. For example, 6x2and 5x2 are like terms because they have a variable with a similar exponent. Similarly, 7yx and 5xz are unlike terms because each term has different variables.
To simplify any algebraic expression, the following are the basic rules and steps:
• Remove any grouping symbol such as brackets and parentheses by multiplying factors.
• Use the exponent rule to remove grouping if the terms are containing exponents.
• Combine the like terms by addition or subtraction
• Combine the constants
Example 1
Simplify 3x2 + 5x2
Solution
Since both terms in the expression are have same exponents, we combine them;
3x2 + 5x2 = (3 + 5) x2 = 8x2
Example 2
Simplify the expression: 2 + 2x [2(3x+2) +2)]
Solution
First work out any terms within brackets by multiplying them out;
= 2 + 2x [6x + 4 +2] = 2 + 2x [6x + 6]
Now eliminate the parentheses by multiplying any number outside it;
2 + 2x [6x + 6] = 2 + 12x 2 + 12x
This expression can be simplified by dividing each term by 2 as;
12x 2/2 + 12x/2 + 2/2 = 6 x 2 + 6x + 1
Example 3
Simplify 3x + 2(x – 4)
Solution
In this case, it is impossible to combine terms when they are still in parentheses or any grouping sign. Therefore, eliminate the parenthesis by multiplying any factor outside the grouping by all terms inside it.
Hence, 3x + 2(x – 4) = 3x + 2x – 8
= 5x – 8
When a minus sign is in front of a grouping, it normally affects all the operators inside the parentheses. This means that a minus sign in front of a group will change the addition operation to subtraction and vice versa.
Example 4
Simplify 3x – (2 – x)
Solution
3x – (2 – x) = 3x + (–1) [2 + (–x)]
= 3x + (–1) (2) + (–1) (–x)
= 3x – 2 + x
= 4x – 2
However, if there is only a plus sign comes before the grouping, then the parentheses are simply erased.
For example, to simplify 3x + (2 – x), the brackets are eliminated as shown below:
3x + (2 – x) = 3x + 2 – x
Example 5
Simplify 5(3x-1) + x((2x)/ (2)) + 8 – 3x
Solution
15x – 5 + x(x) + 8 – 3x
15x – 5 + x2 + 8 – 3x.
Now combine the like terms by adding and subtracting the terms;
x2 + (15x – 3x) + (8 – 5)
x2 + 12x + 3
Example 6
Simplify x (4 – x) – x (3 – x)
Solution
x (4 – x) – x (3 – x)
4x – x2 – x (3 – x)
4x – x2 – (3x – x2)
4x – x2 – 3x + x2 = x
### Practice Questions
1. Simplify the expression, 2st + 3t – s + 5t + 4s.
2. Simplify the expression, 2a – 4b +3ab -5a +2b.
3. Simplify the expression, x (2x + 3y -4) – x 2 + 4xy – 12.
4. Simplify the expression, 4(2x+1) – 3x.
5. Simplify the expression, 4(p – 5) +3(p +1).
6. Simplify the expression, [2x 3y2]3.
7. Simplify the expression, 6(p +3q) – (7 +4q).
8. Simplify the expression, 4rs -2s – 3(rs +1) – 2s.
9. Simplify the expression, [ (3 – x) (x + 2) + (-x + 4) (7x + 2) – (x – y) (2x – y)] – 3x2 – 7x + 5. |
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## NCERT Solutions for Class 6 Maths Chapter 9 Data Handling Ex 9.1
Exercise 9.1
Ex 9.1 Class 6 Maths Question 1.
In a Mathematics test, the following marks were obtained by 40 students. Arrange these marks in a table using tally marks.
(a) Find how many students obtained marks equal to or more than 7.
(b) How many students obtained marks below 4?
Solution:
From the given data, we have the following table.
(a) Number of students who obtained marks equal to or more than 7 = 5 + 4 + 3 = 12
(b) Number of students who obtained marks below 4 = 2 + 3 + 3 = 8.
Ex 9.1 Class 6 Maths Question 2.
Following is the choice of sweets of 30 students of Class VI.
Arrange the names of sweets in a table using tally marks.
Which sweet is preferred by most of the students?
Solution:
(a) We have the following table:
(b) Ladoo is preferred by most of the students, i.e., 11 students.
Ex 9.1 Class 6 Maths Question 3.
Catherine threw a dice 40 times and noted the number appearing each time as shown below:
Make a table and enter the data using tally marks. Find the number that appeared.
(а) The minimum number of times
(b) The maximum number of times
(c) Find those numbers that appear an equal number of times.
Solution:
We have the following table:
From the above table, we get
(a) The number 4 appeared 4 times which is the minimum.
(b) The number 5 appeared 11 times which is the maximum.
(c) The number 1 and 6 appear for the same number of times, i.e., 7.
Ex 9.1 Class 6 Maths Question 4.
Following pictograph shows the number of tractors in five villages.
Observe the pictograph and answer the following questions.
(a) Which village has the minimum number of tractors?
(b) Which village has the maximum number of tractors?
(c) How many more tractors village C has as compared to village B?
(d) What is the total number of tractors in all. the five villages?
Solution:
From the given pictograph, we have
(а) Village D has the minimum number of tractors, i.e., 3.
(б) Village C has the maximum number of tractors, i.e., 8.
(c) Village C has 3 tractors more than that of the village B.
(d) Total number of tractors in all the villages is 28.
Ex 9.1 Class 6 Maths Question 5.
The number of girl students in each class of a co-educational middle school is depicted by the pictograph:
Observe this pictograph and answer the following questions:
(a) Which class has the minimum number of girl students?
(b) Is the number of girls in Class VI less than the number of girls in Class V?
(c) How many girls are there in Class VII?
Solution:
(a) Class VIII has the minimum number of girl students
i,e. 1(frac { 1 }{ 2 }) x 4 = 6.
(b) No, number of girls in Class VI = 4 x 4 = 16 and number of girls in Class V = 2(frac { 1 }{ 2 }) x 4 = 10
So, number of girl students in Class VI is not less than that of in Class V.
(c) Number of girls in Class VII = 3 x 4 = 12
Ex 9.1 Class 6 Maths Question 6.
The sale of electric bulbs on different days of a week is shown below:
Observe the pictograph and answer the following questions:
(a) How many bulbs were sold on Friday?
(b) On which day were the maximum number of bulbs sold?
(c) On which of the days same number of bulbs were sold?
(d) On which of the days minimum number of bulbs were sold?
(e) If one big carton can hold 9 bulbs. How many cartons were needed in the given week?
Solution:
(a) Number of bulbs sold on Friday = 7 x 2 = 14
(b) On Sunday, the number of bulbs sold = 9 x 2 = 18 which is maximum in number.
(c) On Wednesday and Saturday, the same number of bulbs were sold, i.e., 4 x 2 = 8
(d) The minimum number of bulbs were sold on Wednesday and Saturday, i.e., 4 x 2 = 8
(e) Total number of bulbs sold in a week = 43
Number of cartons needed 5
= (43 x 2) ÷ 9 = 86 ÷ 9 = 9(frac { 5 }{ 9 }) = 10 cartons.
Ex 9.1 Class 6 Maths Question 7.
In a village six fruit merchants sold the following number of fruit baskets in a particular season:
Observe this pictograph and answer the following questions:
(a) Which merchant sold the maximum number of baskets?
(b) How many fruit baskets were sold by Anwar?
(c) The merchants who have sold 600 or more number of baskets are planning to buy a godown for the next season. Can you name them?
Solution:
(a) Martin sold the maximum number of fruit baskets, i.e., 9(frac { 1 }{ 2 }) x 100 = 950
(c) Number of fruit baskets sold by Anwar is 7 x 100 = 700.
(c) Anwar, Martin and Ranjit Singh have sold 600 or more fruit baskets and planning to buy a godown.
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# 1.6 Inverse Functions
A fashion designer is traveling to Milan for a fashion show. He asks his assistant, Betty, what 75 degrees Fahrenheit is in Celsius, and after a quick search on Google, she finds the formula $$C=\frac{5}{9}(F-32)$$. Using this formula, she calculates $$\frac{5}{9}(75-32) \approx 24$$ degrees Celsius. The next day, the designer sends his assistant the week’s weather forecast for Milan, and asks her to convert the temperatures to Fahrenheit.
At first, Betty might consider using the formula she has already found to do the conversions. After all, she knows her algebra, and can easily solve the equation for $$F$$ after substituting a value for $$C$$. For example, to convert 26 degrees Celsius, she could write:
$26=\dfrac{5}{9}(F-32)$
$26 \cdot \frac{9}{5}= F-32$
$26 \cdot \dfrac{9}{5}+32 \approx 79$
After considering this option for a moment, she realizes that solving the equation for each of the temperatures would get awfully tedious, and realizes that since evaluation is easier than solving, it would be much more convenient to have a different formula, one which takes the Celsius temperature and outputs the Fahrenheit temperature. This is the idea of an inverse function, where the input becomes the output and the output becomes the input.
Definition: Inverse Functions
If $$f(a)=b$$, then a function g(x) is an inverse of $$f$$ if $$g(b)=a$$.
The inverse of $$f(x)$$ is typically notated, $$f^{-1}(x)$$ which is read “f inverse of x”, so equivalently, if $$f(a)=b$$ then $$f^{-1}(b)=a$$.
Important: The raised -1 used in the notation for inverse functions is simply a notation, and does not designate an exponent or power of -1.
Example 1
If for a particular function, $$f(2)=4$$, what do we know about the inverse?
Solution
The inverse function reverses which quantity is input and which quantity is output, so if $$f(2)=4$$, then $$f^{-1}(4)=2$$. Alternatively, if you want to re-name the inverse function $$g(x)$$, then $$g(4) = 2$$.
Try it Now: 1
Given that $$h^{-1}(6)=2$$, what do we know about the original function $$h(x)$$?
Notice that original function and the inverse function undo each other. If (f(a)=b\), then $$f^{-1}(b)=a$$, returning us to the original input. More simply put, if you compose these functions together you get the original input as your answer.
and
Since the outputs of the function f are the inputs to , the range of f is also the domain of . Likewise, since the inputs to f are the outputs of , the domain of f is the range of .
Basically, like how the input and output values switch, the domain & ranges switch as well. But be careful, because sometimes a function doesn’t even have an inverse function, or only has an inverse on a limited domain. For example, the inverse of is , since a square “undoes” a square root, but it is only the inverse of f(x) on the domain [0,∞), since that is the range of .
Example 2
In this table, we have to blah blah blah
The function has domain and range , what would we expect the domain and range of to be?
We would expect to swap the domain and range of f, so would have domain and range .
Example 3
A function f(t) is given as a table below, showing distance in miles that a car has traveled in t minutes. Find and interpret
The inverse function takes an output of f and returns an input for f. So in the expression, the 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function f, 90 minutes, so. Interpreting this, it means that to drive 70 miles, it took 90 minutes.
Alternatively, recall the definition of the inverse was that if then . By this definition, if you are given then you are looking for a value a so that. In this case, we are looking for a t so that, which is when t = 90.
Try it Now: 2
Using the table below
Find and interpret the following
a.
b.
Example 4
A function g(x) is given as a graph below. Find and
To evaluate, we find 3 on the horizontal axis and find the corresponding output value on the vertical axis. The point (3, 1) tells us that
To evaluate , recall that by definition means g(x) = 3. By looking for the output value 3 on the vertical axis we find the point (5, 3) on the graph, which means g(5) = 3, so by definition.
Try it Now: 3
Using the graph in Example 4 above
a. find
b. estimate
Example 5
Returning to our designer’s assistant, find a formula for the inverse function that gives Fahrenheit temperature given a Celsius temperature.
A quick Google search would find the inverse function, but alternatively, Betty might look back at how she solved for the Fahrenheit temperature for a specific Celsius value, and repeat the process in general
By solving in general, we have uncovered the inverse function. If
Then
In this case, we introduced a function h to represent the conversion since the input and output variables are descriptive, and writing could get confusing.
It is important to note that not all functions will have an inverse function. Since the inverse takes an output of f and returns an input of f, in order for to itself be a function, then each output of f (input to ) must correspond to exactly one input of f (output of ) in order for to be a function. You might recall that this is the definition of a one-to-one function.
### Properties of Inverses
In order for a function to have an inverse, it must be a one-to-one function. In some cases, it is desirable to have an inverse for a function even though the function is not one-to-one. In those cases, we can often limit the domain of the original function to an interval on which the function is one-to-one, then find an inverse only on that interval.
If you have not already done so, go back to the toolkit functions that were not one-to-one and limit or restrict the domain of the original function so that it is one-to-one. If you are not sure how to do this, proceed to Example 6.
Example 6
The quadratic function is not one-to-one. Find a domain on which this function is one-to-one, and find the inverse on that domain.
We can limit the domain to to restrict the graph to a portion that is one-to-one, and find an inverse on this limited domain.
You may have already guessed that since we undo a square with a square root, the inverse of on this domain is
You can also solve for the inverse function algebraically. If , we can introduce the variable to represent the output values, allowing us to write. To find the inverse we solve for the input variable
To solve for x we take the square root of each side. and get , so . We have restricted x to being non-negative, so we’ll use the positive square root, or . In cases like this where the variables are not descriptive, it is common to see the inverse function rewritten with the variable x: . Rewriting the inverse using the variable x is often required for graphing inverse functions using calculators or computers.
y = x
Note that the domain and range of the square root function do correspond with the range and domain of the quadratic function on the limited domain. In fact, if we graph h(x) on the restricted domain and on the same axes, we can notice symmetry: the graph of is the graph of h(x) reflected over the line y = x.
Example 7
Given the graph of f(x) shown, sketch a graph of .
This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of (0,∞) and range of (-∞,∞), so the inverse will have a domain of (-∞,∞) and range of (0,∞).
Reflecting this graph of the line y = x, the point (1, 0) reflects to (0, 1), and the point (4, 2) reflects to (2, 4). Sketching the inverse on the same axes as the original graph:
### Important Topics of this Section
• Definition of an inverse function
• Composition of inverse functions yield the original input value
• Not every function has an inverse function
• To have an inverse a function must be one-to-one
• Restricting the domain of functions that are not one-to-one.
1.
2.a. . In 60 minutes, 50 miles are traveled.
b. . To travel 60 miles, it will take 70 minutes.
3. a.
b. (this is an approximation – answers may vary slightly)
### Section 1.6 Exercises
Assume that the function f is a one-to-one function.
1. If , find 2. If , find
3. If , find 4. If , find
5. If , find 6. If , find
7. Using the graph of shown
a. Find
b. Solve
c. Find
d. Solve
8. Using the graph shown
a. Find
b. Solve
c. Find
d. Solve
9. Use the table below to find the indicated quantities.
x 0 1 2 3 4 5 6 7 8 9 f(x) 8 0 7 4 2 6 5 3 9 1
a. Find
b. Solve
c. Find
d. Solve
10. Use the table below to fill in the missing values.
t 0 1 2 3 4 5 6 7 8 h(t) 6 0 1 7 2 3 5 4 9
a. Find
b. Solve
c. Find
d. Solve
For each table below, create a table for
11.
x 3 6 9 13 14 f(x) 1 4 7 12 16
12.
x 3 5 7 13 15 f(x) 2 6 9 11 16
For each function below, find
13. 14.
15. 16.
17. 18.
For each function, find a domain on which f is one-to-one and non-decreasing, then find the inverse of f restricted to that domain.
19. 20.
21. 22.
23. If and , find
a.
b.
c. What does this tell us about the relationship between and ?
24. If and , find
a.
b.
c. What does this tell us about the relationship between and ?
### Contributors
• David Lippman (Pierce College)
• Melonie Rasmussen (Pierce College) |
# Thread: Speed of the Boat in Still Water
1. ## Speed of the Boat in Still Water
The speed of a stream is 3 mph. A boat travels 5 miles upstream in the same time it takes to travel 11 miles downstream. What is the speed of the boat in still water?
2. Speed up stream = x -3 (-3 because the stream movement will be negative in relation to the boat) and the boat will travel 5 miles
$\displaystyle x-3=5$
$\displaystyle x=5+3$
$\displaystyle x=8$
You can check using the other data.
Speed down stream = x+3 (because the stream and boat movement will add) and the boat will travel 11 miles
$\displaystyle x+3=11$
$\displaystyle x=11-3$
$\displaystyle x=8$
Now you see, the boat -on still water- travels at 8 mph.
3. ## perfectly done............
Originally Posted by Alienis Back
Speed up stream = x -3 (-3 because the stream movement will be negative in relation to the boat) and the boat will travel 5 miles
$\displaystyle x-3=5$
$\displaystyle x=5+3$
$\displaystyle x=8$
You can check using the other data.
Speed down stream = x+3 (because the stream and boat movement will add) and the boat will travel 11 miles
$\displaystyle x+3=11$
$\displaystyle x=11-3$
$\displaystyle x=8$
Now you see, the boat -on still water- travels at 8 mph.
4. Hello, magentarita!
Another approach . . .
We'll use: .$\displaystyle \text{Distance} \:=\:\text{Speed} \times \text{Time} \quad\Rightarrow\quad T \:=\:\frac{D}{S}$
The speed of a stream is 3 mph.
A boat travels 5 miles upstream in the same time it takes to travel 11 miles downstream.
What is the speed of the boat in still water?
Let $\displaystyle x$ = boat's speed in still water.
Going upstream, the current works against the boat. .The boat's speed is $\displaystyle x - 3$ mph.
. . It went $\displaystyle 5$ miles at $\displaystyle x-3$ mph. .This took: .$\displaystyle \frac{5}{x-3}$ hours.
Going downstream, the current works with the boat. .The boat's speed is $\displaystyle x+3$ mph.
. . It went $\displaystyle 11$ miles at $\displaystyle x+3$ mph. .This took: .$\displaystyle \frac{11}{x+3}$ hours.
These two times are equal: .$\displaystyle \frac{5}{x-3} \:=\:\frac{11}{x+3}\quad\hdots\quad There!$
5. ## ok............
Originally Posted by Soroban
Hello, magentarita!
Another approach . . .
We'll use: .$\displaystyle \text{Distance} \:=\:\text{Speed} \times \text{Time} \quad\Rightarrow\quad T \:=\:\frac{D}{S}$
Let $\displaystyle x$ = boat's speed in still water.
Going upstream, the current works against the boat. .The boat's speed is $\displaystyle x - 3$ mph.
. . It went $\displaystyle 5$ miles at $\displaystyle x-3$ mph. .This took: .$\displaystyle \frac{5}{x-3}$ hours.
Going downstream, the current works with the boat. .The boat's speed is $\displaystyle x+3$ mph.
. . It went $\displaystyle 11$ miles at $\displaystyle x+3$ mph. .This took: .$\displaystyle \frac{11}{x+3}$ hours.
These two times are equal: .$\displaystyle \frac{5}{x-3} \:=\:\frac{11}{x+3}\quad\hdots\quad There!$
What can I say? Your replies indicate that you are a teacher or were a teacher at one time.
,
,
,
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# Forty-three people went to a canteen, which sells soup and tea. If $18$ persons took soup only, $8$ took tea only and $5$ took nothing. Use a Venn-diagram to find how many took tea.
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Hint: First of all, we shall note some important terms from the given question and they are persons, soup and tea.
Here, it is given that a total of forty-three people went to a canteen. Out of forty-three, $18$ persons took soup only (i.e. they took soup not tea). And $8$ took tea only (i.e. they took tea not soup).Also, $5$ persons didn’t take anything (i.e. they didn’t buy soup and tea).
Now, our question is to calculate how many members took tea and also we are asked to represent it in a Venn-diagram. When we use circles to show the relationships among a group of things, this type of illustration is generally known as a Venn-diagram.
Formula used:
$n(A \cap B) = n(U) - n(A \cap \bar B) - n(\bar A \cap B) - n(\overline {A \cup B} )$
$n\left( B \right) = n\left( {A \cap B} \right) + n\left( {\bar A \cap B} \right)$
Let us name the set of all $43$ persons as$U$ .
Let $A$ be the set of all people who took only soup not tea.
Let $B$ be the set of all people who took only tea, not soup.
We shall denote the above assumptions mathematically using given information.
The total number of persons,
$n\left( U \right) = 43$
The number of persons, who took only soup,
$n\left( {A \cap \bar B} \right) = 18$
The number of persons, who took only tea,
$\;\;n\left( {\bar A \cap B} \right) = 8$
The number of persons, who took nothing,
$n\left( {\overline {A \cup B} } \right) = 5$
Now, we need to substitute these values in the first formula.
$n(A \cap B) = n(U) - n(A \cap \bar B) - n(\bar A \cap B) - n(\overline {A \cup B} )$
$= 43 - 18 - 8 - 5$
$= 12$
Using second formula, we get
$n\left( B \right) = n\left( {A \cap B} \right) + n\left( {\bar A \cap B} \right)$
$= 12 + 8$
$= 20$
Therefore, $20$ people took tea.
Then, we need to represent it in a Venn-diagram.
This is the required Venn-diagram.
Note: When we use circles to show the relationships among a group of things, this type of illustration is generally known as a Venn-diagram. Our question is to calculate how many members took tea and also we are asked to represent it in a Venn-diagram. Here, $20$ people took tea. |
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# What is a parallelogram with congruent and perpendicular diagonals?
## What is a parallelogram with congruent and perpendicular diagonals?
We know a square is a parallelogram with congruent sides and congruent angles, and has diagonals that are congruent and perpendicular.
Which parallelograms have perpendicular bisectors?
Theorem 7: A rhombus is a parallelogram. Theorem 8: A quadrilateral is a rhombus if and only if the diagonals are perpendicular bisectors of each other.
What is a parallelogram in which all sides are congruent?
One special kind of polygons is called a parallelogram. It is a quadrilateral where both pairs of opposite sides are parallel. If we have a parallelogram where all sides are congruent then we have what is called a rhombus.
### What is a parallelogram with perpendicular diagonals?
If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. • If one diagonal of a parallelogram bisects a pair of opposite angles, then the parallelogram is a rhombus.
What is a parallelogram with four congruent sides?
A rhombus is a parallelogram with all four sides congruent to each other. diamond-like shape. A square is a parallelogram with four congruent sides and four right angles. In other words, a square is a rectangle and a rhombus.
Do all parallelograms have 4 equal sides?
A parallelogram has two parallel pairs of opposite sides. A rectangle has two pairs of opposite sides parallel, and four right angles. It is also a parallelogram, since it has two pairs of parallel sides. A square has two pairs of parallel sides, four right angles, and all four sides are equal.
## Are all Rhombuses parallelograms?
Not every parallelogram is a rhombus, though any parallelogram with perpendicular diagonals (the second property) is a rhombus. In general, any quadrilateral with perpendicular diagonals, one of which is a line of symmetry, is a kite.
Is bisector of parallelogram perpendicular?
All of the properties of a parallelogram apply (the ones that matter here are parallel sides, opposite angles are congruent, and consecutive angles are supplementary). The diagonals bisect the angles. The diagonals are perpendicular bisectors of each other.
Do all parallelograms have opposite sides that are perpendicular?
Parallelogram are quadrilaterals with opposite sides parallel, but adjacent sides not perpendicular. If the sides have equal lengths, then it is called a rhombus.
### Is a parallelogram with all four sides congruent?
Do all parallelograms have four congruent sides?
It is a special case of a parallelogram that has four congruent sides and four right angles. A square is also a rectangle because it has two sets of parallel sides and four right angles. A square is also a parallelogram because its opposite sides are parallel. All four sides of a rhombus are congruent.
Does a parallelogram have congruent sides?
A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel . Opposite sides are congruent; Adjacent angles are supplementary; The diagonals bisect each other.
## When do the sides of a parallelogram become a square?
The diagonals are perpendicular to and bisect each other. A square is a special type of parallelogram whose all angles and sides are equal. Also, a parallelogram becomes a square when the diagonals are equal and right bisectors of each other. Click to see full answer
How is a rhombus a special type of parallelogram?
It is defined as a quadrilateral all of whose sides are congruent. It is a special type of parallelogram, and its properties (aside from those properties of parallelograms) include: Its diagonals divide the figure into 4 congruent triangles. Its diagonals are perpendicular bisectors of eachother.
Which is characteristic guarantees that a parallelogram is a rectangle?
Answer:- The characteristic guarantees that a parallelogram is a rectangle is “the diagonals are congruent”. The opposite sides are parallel. The opposite sides and angles are equal. The diagonals are perpendicular bisectors.
### Which is the sum of the interior angles of a parallelogram?
The sum of interior angles of a parallelogram is equal to 360°. The consecutive angles of a parallelogram should be supplementary (180°). The 7 important theorems on properties of a parallelogram are given below: A diagonal of a parallelogram divides the parallelogram into two congruent triangles. |
# Question: What Happens When A Line Is Dilated?
## How do you prove dilation?
Lesson Summary Dilations can be used to prove figures are similar by finding the scale factor between the two images and ensuring the sides are proportional.
To find the scale factor, use the lengths of corresponding sides, set up a ratio and divide..
## How do you find the horizontal dilation?
Horizontal dilation. For example, replacing x by x/0.5=x/(1/2)=2x has the effect of contracting toward the y-axis by a factor of 2. If A is negative, we dilate by a factor of |A| and then flip about the y-axis. Thus, replacing x by −x has the effect of taking the mirror image of the graph with respect to the y-axis.
## What happens when you dilate an angle?
Explanation: Dilation (scaling) is the main basis for similarity in geometry. It transforms any line into the one parallel to it. Therefore, an image of an angle scaled to another location is another angle with correspondingly parallel sides.
## What is horizontal dilation?
“horizontal dilation” A horizontal stretching is the stretching of the graph away from the y-axis. A horizontal compression (or shrinking) is the squeezing of the graph toward the y-axis.
## How do I find the slope of the line?
The slope will be the same for a straight line no matter which two points you pick as you know. All you need to do is to calculate the difference in the y coordinates of the 2 points and divide that by the difference of the x coordinates of the points(rise over run). That will give you the slope.
## Is it true dilations always increase the length of line segments?
While they scale distances between points, dilations do not change angles. … All lengths of line segments in the plane are scaled by the same factor when we apply a dilation.
## What stays the same after a dilation?
After a dilation, the pre-image and image have the same shape but not the same size. Sides: In a dilation, the sides of the pre-image and the corresponding sides of the image are proportional. Properties preserved under a dilation from the pre-image to the image.
## Does angle measure change under a dilation?
While they scale distances between points, dilations do not change angles. … All lengths of line segments in the plane are scaled by the same factor when we apply a dilation.
## What is true about the dilation?
Dilation is defined as a transformation of an image in which the size on this image changes but its shape does not change. 2. In dilation, to obtain the new dimensions of the image you must multiply the dimensions of the original image by a number called “Scale factor”.
## Does dilation preserve slope?
11. Dilating LINES y = mx + b (SLOPE ALWAYS PRESERVED… STAYS THE SAME!)
## What stays the same in a dilation?
In dilation, the image and the original are similar, in that they are the same shape but not necessarily the same size. They are not congruent because that requires them to be the same shape and the same size, which they are not (unless the scale factor happens to be 1.0).
## Is a horizontal stretch a dilation?
The image of a dilation is the same shape as the original figure, but is not necessarily the same size. Both the vertical length and horizontal length of a dilated figure are increased (or decreased) by the same factor. … If a figure is enlarged (or reduced) in only one direction, the change is referred to as a stretch.
## How do you dilate a line by scale factor?
One point on the line that you are dilating, another point on the target for that point. Find the scale factor by dividing the distance to the target by the distance to the point on the line that you are dilating.
## How do you dilate a figure by 1 3?
Perform a Dilation of 3 on point A (2, 1) which you can see in the graph below. Multiply the coordinates of the original point (2, 1), called the image, by 3. Image’s coordinates = (2 * 3, 1 * 3) to get the coordinates of the image (6, 3).
## What is the difference between horizontal and vertical dilation?
The difference occurs because vertical dilations occur when we scale the output of a function, whereas horizontal dilations occur when we scale the input of a function.
## What means dilation?
Dilation: The process of enlargement, stretching, or expansion. The word “dilatation” means the same thing. Both come from the Latin “dilatare” meaning “to enlarge or expand.” |
# Common Core: 1st Grade Math : Find the Missing Number in Addition and Subtraction Equations: CCSS.MATH.CONTENT.1.OA.D.8
## Example Questions
### Example Question #1 : Find The Missing Number In Addition And Subtraction Equations: Ccss.Math.Content.1.Oa.D.8
Fill in the blank.
__________
Explanation:
To find the missing piece of an addition problem, you can take the biggest number minus the smallest number. , which means .
### Example Question #2 : Find The Missing Number In Addition And Subtraction Equations: Ccss.Math.Content.1.Oa.D.8
Fill in the blank.
__________
Explanation:
To find the missing piece of an addition problem, you can take the biggest number minus the smallest number. , which means .
### Example Question #3 : Find The Missing Number In Addition And Subtraction Equations: Ccss.Math.Content.1.Oa.D.8
Fill in the blank.
__________
Explanation:
To find the missing piece of an addition problem, you can take the biggest number minus the smallest number. , which means .
### Example Question #4 : Find The Missing Number In Addition And Subtraction Equations: Ccss.Math.Content.1.Oa.D.8
Fill in the blank.
__________
Explanation:
To find the missing piece of an addition problem, you can take the biggest number minus the smallest number. , which means .
### Example Question #1 : Find The Missing Number In Addition And Subtraction Equations: Ccss.Math.Content.1.Oa.D.8
Fill in the blank.
__________
Explanation:
To find the missing piece of an addition problem, you can take the biggest number minus the smallest number. , which means .
### Example Question #6 : Find The Missing Number In Addition And Subtraction Equations: Ccss.Math.Content.1.Oa.D.8
Fill in the blank.
__________
Explanation:
To find the missing piece of an addition problem, you can take the biggest number minus the smallest number. , which means .
### Example Question #7 : Find The Missing Number In Addition And Subtraction Equations: Ccss.Math.Content.1.Oa.D.8
Fill in the blank.
__________
Explanation:
To find the missing piece of an addition problem, you can take the biggest number minus the smallest number. , which means .
### Example Question #8 : Find The Missing Number In Addition And Subtraction Equations: Ccss.Math.Content.1.Oa.D.8
Fill in the blank.
__________
Explanation:
To find the missing piece of an addition problem, you can take the biggest number minus the smallest number. , which means .
### Example Question #9 : Find The Missing Number In Addition And Subtraction Equations: Ccss.Math.Content.1.Oa.D.8
Fill in the blank.
__________
Explanation:
To find the missing piece of an addition problem, you can take the biggest number minus the smallest number. , which means .
### Example Question #10 : Find The Missing Number In Addition And Subtraction Equations: Ccss.Math.Content.1.Oa.D.8
Fill in the blank.
__________ |
# You will learn to find the perimeter and area for different shapes and the circumference of circles. s s Perimeter = 4s or s+s+s+s Area = s 2 l w P =
## Presentation on theme: "You will learn to find the perimeter and area for different shapes and the circumference of circles. s s Perimeter = 4s or s+s+s+s Area = s 2 l w P ="— Presentation transcript:
You will learn to find the perimeter and area for different shapes and the circumference of circles. s s Perimeter = 4s or s+s+s+s Area = s 2 l w P = 2l+2w or 2(l+w) or l+w+l+w A = lw b h s P = s 1 + s 2 + s 3 A = ½ bh r C = 2 r = d A = r 2
What is a perimeter? Perimeter is the distance all the way around an object. The perimeter of a circle is called the circumference.
How To Find Perimeter Find the length of each side of the object. Add all the lengths of the sides together. This total is the perimeter.
Lets’ find the perimeter. 8 + 4 + 8 + 4 = 24. This rectangle has a perimeter of 24.
How about this perimeter ? 6 + 5 + 5 + 6 + 12 = 34. This pentagon has a perimeter of 34.
Now lets’ try a circumference: c=2πr or c=πd r = radius (half the distance across a circle) d = diameter (the distance across a circle) Note that 2r = d
Find the circumference of a circle with a diameter of 12 cm. c = πd c = π 12 c = 3.1412 c = 37.68 cm 12 cm Now lets’ try a circumference:
Find the circumference of a circle with a radius of 3 meters. 3 m Lets’ try aother circumference: c = 2πr c = 2 π 3 c = 2 3.14 3 c = 18.84 m
What if we know the circumference ? If the circumference of a circle is approximately 50.3 cm, find the radius. (Use 3.14 for π) C = 2πr 50.3 = 6.28r 50.3 / 6.28 = r 8 cm r
Time for area. Area of a rectangle A = bh A = 4(2) A = 8 cm² 4 cm 2 cm
Not done yet!! Area of a triangle: A = ½ bh A = ½ (6)(7) A = 21 mm² 7 mm 6 mm
What if you are missing aside ? If the base is 16, and the area is 40, what is the height? 16 in H = ? 40 = ½ (16)h 40 = 8h 40/8 = h 5 in = h A = ½ bh
Here comes the area of a trapezoid A = ½ (b 1 +b 2 )h A = ½ (6 + 10)(5) A = ½(16)(5) A = 40 ft² b 2 = 10 ft b 1 = 6 ft 5 ft
And yet, here’s another area of a trapezoid A = ½ (8 + 12)(3) A = ½ (20)(3) A = 30 km² 8 km 12 km 3 km
If the area of the trapezoid is 24, and the height is 4 and base 1 is 8, what is the other base? B 2 = ? km 8 km 4 km Let’s find a missing side 24 = ½ (4)(8 + x) 24 = 2(8 + x) 24 = 16 + 2x 24 – 16 = 2x 8 = 2x 4 km = x A = ½ (h)(b 1 + b 2 )
Just one more, the area of a circle A = πr² A = 3.14(10)² A = 3.14(100) A = 314 m² 10 m
OK, one last area of a circle. Diameter = 8 in Radius = 4 in A = 3.14(4)² A = 3.14(16) A = 50.24 in² 8 in
Assignment
1.9 Perimeter, Circumference, and Area Geometry Find the perimeter and area of each rectangle. Label each measurement. 1. 6 in 3 in 2. 1 yd 12 yd 3. 1.65 cm 4. l = 4.5, w = 1.5, P = ? Find the missing measure in each formula if P = 2l + 2w and A = lw. 5. l = 2.2, w = 1.1, A = ? 6. l = 12, A = 30, w = ? 7. A = 3½, w = ½, l = ? 8. P = 13, w = 2.5, l = ? Find the circumference and area of each circle. Label each measurement. 9. 10. 15 cm 3 m
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# Lesson #23 Analysis of Variance. In Analysis of Variance (ANOVA), we have: H 0 : 1 = 2 = 3 = … = k H 1 : at least one i does not equal the others.
## Presentation on theme: "Lesson #23 Analysis of Variance. In Analysis of Variance (ANOVA), we have: H 0 : 1 = 2 = 3 = … = k H 1 : at least one i does not equal the others."— Presentation transcript:
Lesson #23 Analysis of Variance
In Analysis of Variance (ANOVA), we have: H 0 : 1 = 2 = 3 = … = k H 1 : at least one i does not equal the others Assumptions - - the k populations are independent - the k populations are Normally distributed - the k populations all have equal variances
Model: Y is the called the “dependent variable”, or “response”, or “outcome”. Y ij = + ( i – ) + (Y ij – i ) The variable (“factor”) whose levels define the k “treatments” is the “independent variable”. ii ij Y ij = + i + ij i = 1, 2, …, k j = 1, 2, …, n i
Let = n 1 + n 2 + … + n k = n. = n From each sample, we calculate – - the sample mean, - the sample variance, Also calculate the overall sample mean,
Sums of Squares SS TREATMENT = SS BETWEEN = SS AMONG = SS MODEL SS ERROR = SS RESIDUAL = SS WITHIN SS TOTAL = SS TREATMENT + SS ERROR
df TREATMENT = (k – 1) df ERROR = (n. – k) Degrees of freedom - Mean squares are sums of squares divided by df:
E ( MS ERROR ) = 2 Reject H 0 if F 0 = > F (k-1,n. -k),1-
The F-distribution has two parameters, called the numerator df (df 1 ) and denominator df (df 2 ). The distribution is positively skewed, and defined only for positive values. In this case, df 1 = (k - 1), df 2 = (n. – k) F (5,20),.95 = 2.71
ANOVA Table Source df SS MS F Treatment Error Total k-1 n. –k n. –1 SS TRT SS ERROR SS TOTAL MS TRT MS ERROR F0F0
3.500 4.125 3.875 3.500 4.250 4.000 3.625 4.250 4.625 4.500 4.875 5.000 3.500 4.125 3.750 4.750 2.750 3.750 3.500 3.375 3.625 White Pink Red nini 8 10 5 3.875 4.3 3.4 0.1116 0.2889 0.1516 n. = 23
nini 8 10 5 3.875 4.3 3.4 0.1116 0.2889 0.1516 n. = 23 + (10)(4.3 - 3.9565) 2 = (8)(3.875 - 3.9565) 2 + (5)(3.4 - 3.9565) 2 = 2.782 SS TRT SS ERROR = (7)(0.1116) + (9)(0.2889) + (4)(0.1516) = 3.988
Source df SS MS F Treatment (color) Error Total 2 20 22 2.782 3.988 6.770 1.391 0.1994 6.98 Reject H 0 if F 0 > F (2,20),.90 = 2.59 Reject H 0, conclude the average diameter is not the same for all three colors of roses.
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# How are you supposed to solve this?
0
347
1
If we express $$3x^2 - 6x - 2$$ in the form $$a(x - h)^2 + k$$, then what is $$a + h + k$$?
Jan 2, 2018
#1
+2340
+2
In order to solve this problem, I would use a similar strategy as I used here https://web2.0calc.com/questions/halp_34
There is a slight difference from this problem and the one I hyperlinked, though. The coefficient of the quadratic term is not one. In order to make it one, let's factor it out without changing the value of the expression.
$$3x^2-6x-2\Rightarrow3(x^2-2x)-2$$
Now, the problem is more or less identical to the one I hyperlinked you to. Let's find the value that we can use to manipulate the expression. This magical value is found exactly the same way as completing the square.
$$x^2-2x+\left(\frac{-2}{2}\right)^2\\ x^2-2x+1\\ (x-1)^2$$
This process shows that we need 1 to be inserted inside the parentheses.
$$3(x^2-2x+1)-2-3(1)$$
Since we added one inside the parentheses, we must compensate for that by subtracting the same amount somewhere else in the expression. We need to multiply by three because of the distributive property. Simplify from here and convert into the desired form.
$$3(x^2-2x+1)-2-3(1)\\ 3(x-1)^2-2-3\\ 3(x-1)^2-5$$
Notice the parallelism.
$$\textcolor{green}{3}(x-\textcolor{blue}{1})^2+(\textcolor{red}{-5})\\ \textcolor{green}{a}(x-\textcolor{blue}{h})^2+\textcolor{red}{k}$$
Just like before, a=3, h=1, and k=-5. Therefore, the sum of these values is $$a + h + k=3+1-5=-1$$
.
Jan 2, 2018 |
# Differentiation by Substitution
### Differentiation by Substitution:
• Differentiation by substitution is a technique used to simplify the process of finding derivatives by replacing complex expressions with simpler variables.
• It's based on the chain rule and often involves substituting one variable or expression with another to make the derivative calculation more manageable.
### Steps for Differentiation by Substitution:
1. Identify a complex part of the function that can be substituted with a simpler variable.
2. Choose an appropriate substitution:
• Select a new variable or expression to replace the complex part of the function.
• Ensure the substitution makes the differentiation process easier.
3. Compute the derivative using the substituted variable:
• Differentiate the function with respect to the new variable.
• Substitute back the original expression to express the derivative in terms of the original variable.
### Example:
Given the function $y=\mathrm{sin}\left(3{x}^{2}+5\right)$:
1. Identify the complex part: $3{x}^{2}+5$
2. Choose a substitution: Let $u=3{x}^{2}+5$.
3. Compute the derivative using substitution:
• Express $y$ as $y=\mathrm{sin}\left(u\right)$.
• Find $\frac{dy}{du}$. $\frac{dy}{du}=\mathrm{cos}\left(u\right)$
• Now, find $\frac{du}{dx}$ : $\frac{du}{dx}=6x$
• Apply the chain rule: $\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$. $\frac{dy}{dx}=\mathrm{cos}\left(u\right)\cdot 6x$
• Substitute back $u=3{x}^{2}+5$ and $\mathrm{cos}\left(u\right)=\mathrm{cos}\left(3{x}^{2}+5\right)$ : $\frac{dy}{dx}=6x\cdot \mathrm{cos}\left(3{x}^{2}+5\right)$
### Tips for Substitution in Differentiation:
• Choose effective substitutions:
• Opt for substitutions that simplify the expression and make differentiation easier.
• Chain Rule Application:
• Remember to apply the chain rule when differentiating the substituted expression.
• Substitution Reversal:
• Substitute back the original expression to express the final derivative in terms of the original variable.
### Advantages of Substitution in Differentiation:
• Simplification of Complex Functions:
• It helps handle intricate functions by replacing parts with simpler expressions.
• Ease in Calculation:
• Substitution often makes differentiation more straightforward, especially for functions involving nested or composite structures.
### Special Cases:
• Nested Functions:
• Particularly useful for functions within functions, making nested derivatives more manageable.
• Inverse Functions:
• Substitution can aid in finding derivatives of inverse functions by choosing appropriate substitutions. |
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# Do Now: - PowerPoint PPT Presentation
Aim: What good is the Unit Circle and how does it help us to understand the Trigonometric Functions?. Do Now:. A circle has a radius of 3 cm. Find the length of an arc cut off by a central angle of 270 0. Q II. Quadrant I. terminal side. 90 < < 180. 0 < < 90. terminal side.
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Presentation Transcript
Aim: What good is the Unit Circle and how does it help us to understand the Trigonometric Functions?
Do Now:
A circle has a radius of 3 cm. Find the length of an arc cut off by a central angle of 2700.
Q II to understand the Trigonometric Functions?
terminal side
90 < < 180
0 < < 90
terminal side
initial side
t.s.
Q III
Q IV
t.s
180 < < 270
270 < < 360
• An angle on the coordinate plane is in standard position when its vertex is at the origin and its initial side coincides with the nonnegative ray of the x-axis.
Angles in Standard Position
y
x
• An angle formed by a counterclockwise rotation
• has a positive measure.
• Angles whose terminal side lies on one of the axes is
• a quadrantal angle. i.e. 900, 1800, 2700, 3600, 4500 etc.
Q II to understand the Trigonometric Functions?
90 < < 180
0 < < 90
initial side
-
t.s.
Q III
Q IV
180 < < 270
270 < < 360
Co-terminal and Negative Angles
y
3000 =
x
600
• An angle formed by a clockwise rotation has a
• negative measure
• Angles in standard position having the same
• terminal side are co-terminal angles.
Q II to understand the Trigonometric Functions?
90 < < 180
0 < < 90
Q III
Q IV
180 < < 270
270 < < 360
• Angles whose terminal side rotates more than one
• revolution form angles with measures greater
• than 3600.
Angles Greater than 3600
y
4850
1250
x
• To find angles co-terminal with an another angle
1250 and 4850 are co-terminal
Model Problems to understand the Trigonometric Functions?
• Find the measure of an angle between 00 and 3600 co-terminal with
• 3850 b) 5750 c) -4050
• In which quadrant or on which axis, does the terminal side of each angle lie?
• a) 1500b) 5400 c) -600
215o
315o
25o
x-axis
QIV
QII
hypotenuse to understand the Trigonometric Functions?
side opp.
cos
Unit Circle
y
1
center at (0,0)
cos , sin
(x,y)
x
-1
1
-1
Aim: What good is the Unit Circle and how does it help us to understand the Trigonometric Functions?
Do Now:
Find the measure of an angle between 00 and 3600 co-terminal with an angle whose measure is -1250.
3 to understand the Trigonometric Functions?
Hypotenuse = 2 shorter leg
Longer leg = shorter leg
Value of Sine & Cosine: Quadrant I
y
1
center at (0,0)
cos 600, sin 600
(x,y)
600
x
-1
1
What is the value of coordinates (x,y)?
300-600-900 triangle
Sine and Cosine values for angles in
( to understand the Trigonometric Functions?x,y)
1200
3
Hypotenuse = 2 shorter leg
Longer leg = shorter leg
Sine values for angles in Quadrant II are positive.
Value of Sine & Cosine: Quadrant II
y
1
cos 1200, sin 1200
What is the value
of coordinates (x,y)?
1
60º is the
reference
angle
(180º-120º)
600
x
-1
1
directed distance
A reference angle for any angle in standard
position is an acute angle formed by the terminal
side of the given angle and the x-axis.
What is the cosine/sine of a 1200 angle?
300-600-900 triangle
Cosine values for angles in Quadrant II are negative.
side opp. to understand the Trigonometric Functions?
Value of Sine & Cosine: Quadrant III
y
1
What is the value
of coordinates (x,y)?
What is the cosine/sine
of a 2400 angle?
2400
directed distance
60º is the
reference
angle
(240º-180º)
x
-1
600
1
directed dist.
1
(x,y)
cos 2400, sin 2400
Sine and Cosine values for angles in
side opp. to understand the Trigonometric Functions?
(x,y)
Sine values for angles in Quadrant IV are negative.
Value of Sine & Cosine: Quadrant IV
y
1
What is the value
of coordinates (x,y)?
What is the cosine/sine
of a 3000 angle?
60º is the
reference
angle
(360º-300º)
3000
x
-1
600
1
directed dist.
1
cos 3000, sin 3000
Cosine values for angles in Quadrant IV are positive.
Unit Circle – 12 Equal Arcs to understand the Trigonometric Functions?
Periodic to understand the Trigonometric Functions?
Unit Circle – 8 Equal Arcs
Negative Angles Identities
y to understand the Trigonometric Functions?
x
Value of Sine & Cosine in Coordinate Plane
cos is +
sin is +
cos is –
sin is +
cos is +
sin is –
cos is –
sin is –
for any angle in standard
position is an acute angle formed by the terminal
side of the given angle and the x-axis.
The reference angle:
Model Problems to understand the Trigonometric Functions?
• Fill in the table
• Quad. Ref. sin cos
• 2360
• 870
• -1600
• -36
• 13320
• -3960
y to understand the Trigonometric Functions?
1
x
-1
1
-1
Regents Prep
On the unit circle shown in the diagram below, sketch an angle, in standard position, whose degree measure is 240 and find the exact value of sin 240o.
Aim: What good is the Unit Circle and how does it help us to understand the Trigonometric Functions?
Do Now:
Use the unit circle to find:
a. sin 1800 () b. cos 1800
to understand the Trigonometric Functions?
(-1,0)
Model Problems
Use the unit circle to find:
a. sin 1800 () b. cos 1800
(x, y) = (-1, 0)
sin 1800 = y
= 0
cos 1800 = x
= -1
sin to understand the Trigonometric Functions?
sin
cos
cos = 1
-1
( 1, tan)
( , )?
Tan
center at (0,0)
cos , sin
y
1
(x,y)
tan
1
x
-1
1
Trigonometric Values to understand the Trigonometric Functions?
+
+
Quadrant I to understand the Trigonometric Functions?
Q II
90 < < 180
0 < < 90
Q III
Q IV
180 < < 270
270 < < 360
Trigonometric Values - A C T S
y
S Sine is +
A
All are +
x
T
Tangent is +
C
Cosine is +
1 to understand the Trigonometric Functions?
y
r
y
x
-1
x
1
-1
Reciprocal Functions
Negative Angles Identities
csc = 1/y
sec = 1/x
cot = x/y
denominators 0
Need to Knows
When r = 1
sin = y
cos = x
tan = y/x
y to understand the Trigonometric Functions?
(x,y)
1
1
sin = y
2
2
450
x
-1
1
cos = x
In a 450-450-900 triangle, the length of the hypotenuse is times the length of a leg.
-1
(x)
length of hypo. =
Model Problems
Using the unit circle, find
cos 450 (/4)
sin 450
tan 450
450-450-900 triangle
A 450-450-900 triangle is an isosceles right triangle.
therefore x = y
cos = sin
y to understand the Trigonometric Functions?
1
1
sin = y
45º
x
-1
1
cos = x
=
=
-1
Model Problems
Using the unit
circle, find
cos 45º(/4)
sin 45º
tan 45º
(x,y)
cos 45º = x
tan 45 = 1
sin 45º = y
to understand the Trigonometric Functions?
0
30º
/6
45º
/4
60º
/3
90º
/2
sin
0
1
cos
1
0
tan
0
1
UND.
Trigonometric Values for Special Angles
Why is tan 90º undefined?
What is the slope of a line perpendicular to the x-axis?
= slope
What is the cos 510º (17 to understand the Trigonometric Functions?/6)?
• cos 30º =
• cos 510º=
Model Problems
What is the tan 135º (3/4)?
• 135º is in the 2nd quadrant
• 45º is reference angle (180 – 135 = 45)
• tan 45º = 1
• tangent is negative in 2nd quadrant
• tan 135º= -1
• 510º is in the 2nd quadrant
• (510 – 360 = 150)
• 30º is reference angle (180 – 150 = 30)
• cosine is negative in 2nd quadrant
≈ -.866…
Model Problems to understand the Trigonometric Functions?
Model Problems to understand the Trigonometric Functions?
Given:
sin 68o = 0.9272
cos 68o = 0.3746
Find cot 112o
• -0.3746 B) -2.4751
• C) -0.404 D) 1.0785
reference angle for 112o is 68o; 112o is in QII; tan and cot are negative in QII
WHAT ELSE DO WE KNOW?
Model Problems to understand the Trigonometric Functions?
Express sin 285º as the function of an angle
whose measure is less than 45º.
What do we know?
the sine of a IV quadrant
angle is negative
-sin 75º
reference angle for 285º is
(360 – 285) = 75º
> 45º
sine and cosine are co-functions
complement of 75º is 15º
< 45º
sin 285º =
= -cos 15º
-sin 75º
Trig Functions Using Radian Measures to understand the Trigonometric Functions?
Algebraically:
Find:
sin (π/3)
remember:
π/3
60º
sin 60º =
≈ .866…
Using the calculator:
Use the mode key:
change setting from degrees to radians
then hit:
sin
2nd
π
÷
ENTER
3
Display: .8660254083
y to understand the Trigonometric Functions?
1
1
-1
unit circle
1
x
-1
Un-unit circle
is any angle in standard position with (x, y) any point on the terminal side of and
r 1
4 to understand the Trigonometric Functions?
r
= 5
3
Model Problem
(-3, 4) is a point on the terminal side of . Find the sine, cosine, and tangent of .
Q II
-1 to understand the Trigonometric Functions?
r
= 2
Model Problem
is a point on the terminal side of . Find , the sine, cosine, and tangent of .
Q III
Model Problem to understand the Trigonometric Functions?
Tan = -5/4 and cos > 0, find sin and sec
When tangent is negative and cosine is positive angle is found in Q IV.
Model Problem to understand the Trigonometric Functions?
The terminal side of is in quadrant I and lies on the line y = 6x. Find tan ; find .
y = mx + b - slope intercept form of equation
m = slope of line
y = 6x
m = 6 = tan
Q I
Model Problem to understand the Trigonometric Functions?
The terminal side of is in quadrant IV and lies on the line 2x + 5y = 0. Find cos .
y = mx + bslope intercept
form of equation
tan = m = -2/5
y to understand the Trigonometric Functions?
1
1
45º
x
-1
1
-1
Templates |
# Myra can fill 18 glasses with 2 containers of iced tea. How many glasses can she fill with 3 containers of tea?
Dec 15, 2016
27
#### Explanation:
18 glasses to 2 containers: $\frac{18 \setminus \textrm{g l a s s e s}}{2 \setminus \textrm{c o n t a \in e r s}}$
Simplify by dividing top and bottom by 2:
$\frac{\setminus \stackrel{9}{\setminus \cancel{18}} \textrm{g l a s s e s}}{\setminus \stackrel{1}{\setminus \cancel{2}} \setminus \textrm{c o n t a \in e r s}} \setminus \Rightarrow \frac{9 \textrm{g l a s s e s}}{1 \setminus \textrm{c o n t a \in e r}}$
Now use this ratio in dimensional analysis to find how much glasses 3 containers would fill.
$3 \setminus \cancel{\setminus \textrm{c o n t a \in e r s}} \setminus \times \frac{9 \setminus \textrm{g l a s s e s}}{1 \setminus \cancel{\setminus \textrm{c o n t a \in e r}}} = \frac{3 \setminus \times 9 \setminus \textrm{g l a s s e s}}{1} = \frac{27 \setminus \textrm{g l a s s e s}}{1} = 27 \setminus \textrm{g l a s s e s}$ |
## The Average Value of a Function
Finding the average of a finite number of values is easy -- just add them up and divide by the number of values. For example, we know that the average of $n$ function values $f(x_1), \ f(x_2), \, \ldots, \, f(x_n)$ is given by
$$\frac{f(x_1) + f(x_2) + \cdots + f(x_n)}{n}$$
However, how can one find the average of all of the values a function assumes over an interval? Interestingly, this so-called average value of a function over $[a,b]$ can be expressed in terms of a limit of a Riemann sum, and consequently a definite integral!
Suppose $f$ is a function defined on $[a,b]$ and $a = x_0 \lt x_1 \lt x_2 \lt \cdots \lt x_n = b$ is a regular partition of $[a,b]$.
Let us denote the width of the $i^{th}$ sub-interval by $\displaystyle{\Delta x = \frac{b-a}{n}}$, which means $x_i = a + i\Delta x$.
$$\begin{array}{rcl} \displaystyle{\frac{f(x_1) + f(x_2) + \cdots + f(x_n)}{n}} &=& \displaystyle{\sum_{i=1}^n \frac{f(x_i)}{n}}\\ &=& \displaystyle{\sum_{i=1}^n f(x_i) \frac{\Delta x}{b-a} \quad \quad {\left(\small \textrm{since } \, \frac{1}{n} = \frac{\Delta x}{b-a}\right)}}\\ &=& \displaystyle{\frac{1}{b-a} \sum_{i=1}^n f(x_i) \Delta x} \end{array}$$
Note that the sum in the last expression is a Riemann sum. To average all of the function values assumed over $[a,b]$, we simply allow $n$ to increase without bound, recalling
$$\lim_{n \rightarrow \infty} \sum_{i=1}^n f(x_i) \Delta x = \int_a^b f(x)\,dx$$
Thus, if $f_{avg}$ denotes the aforementioned average value of $f$ over $[a,b]$, we have
$$f_{avg} = \frac{1}{b-a} \int_a^b f(x)\,dx$$ |
Modulus functions luxvis.com 11/12/2012
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1 All reasonable efforts have been made to make sure the notes are accurate. The author cannot be held responsible for any damages arising from the use of these notes in any fashion. Modulus functions Modulus functions and their graphs. luxvis.com 11/1/01
2 Modulus Functions The modulus function or otherwise known as the absolute value of a real number x is defined by the following x if x 0 x x if x0 It may also be defined as x x Properties of the Modulus Function Property The absolute value of x is written as x. It is defined by the following: Example ( a) if a 0 a ( a) if a <0 x can be thought of as the distance that x is from zero. For example, the distance that 5 is from zero is 5, whereas the distance that -3 is from zero is 3. So we can then say 5 5whereas 3 3 If a and b are both non negative or both non positive then equality a b a b If a 0 then x a is equivalent to a x a x 4 Then we get the following expression a x a If a 0 then x k a is equivalent to k a x k a a 4 Then we get the following expression a x a ab a b a b a b Sometimes x is referred to as magnitude of x, or the modulus of x, which can be thought to roughly mean the size of x Page 1 of 15
3 Graphs of modulus There are essentially a few ways to sketch modulus functions, namely we can use our graphic calculators (using the graph) or we can go from the definition method. Let s us examine the definition method Example 1: Sketch the graph of y x 3 Steps First always start with the definition Remember the definition a Method ( a) if a 0 ( a) if a <0 Now put in the required variables for the question at hand we get the following It pays to use the brackets so that we do not get confused Now we solve each expression separately. In this case we have ax 3 So using the definition we have the following ( x 3) if x 3 0 x 3 ( x 3) if x 3 0 y x 3 if x 3 0 Now to solve this above equation we will have to remember how to deal with inequalities. Remember the inequality changes if we divide or multiply by a negative number! So we solve the inequality x 30 x 3 So this equation looks like this y x 3 x 3 Now we solve the other inequality Now we look at the other part of the expression x 30 x 3 So this graph would be y x 3 for x 3 Page of 15
4 Now we sketch the graph Be careful when sketching the graphs Notice how the graph is positive for all values of x Now we could have just used our graphics calculators and we would had obtained the above graph quickly, however it is important to be able to do the maths. Let us look at a few more examples on using modulus functions. Example : y x Start with the definition always We use the definition of this modulus function a ( a) if a 0 ( a) if a <0 to see how to sketch Now replace the numbers with what we actually have Here in the place of ax ( x ) if x 0 So we have the following x ( x ) if x <0 Now separating the two expressions into two yx if x 0 which basically means x The other expression becomes y x x, which applies for the following x 0 x Page 3 of 15
5 Now we sketch the graph once again using both expressions Notice how the next graph looks a little different; in the sense the modulus signs only cover the x values Example 3: Sketch y x 4 Definition as always Remember the definition a a a if if a 0 a <0 Now use the expressions for our question x 4 if x 0 x x 4 if -x <0 Now do the first expression Now the graph of yx 4 is for x 0 Now the other expression While the graph of y x 4 if for x 0 Page 4 of 15
6 Notice how the graph has been moved downwards Using the graphics calculator to sketch the modulus functions Many different ways let us look at two ways Step-1:Start the calculator Step-:Press Main Step-3:Press keyboard-d Page 5 of 15
7 Step-4:Press the absolute button and input the equation directly into the calculator Step-5:Press Enter the button and you will get the following. And drag the equation to the next line Step-6:Now press the graph button on the top of the screen and you will get the following Step-7:drag the equation to the screen on the bottom Step-8:Now click on the screen below and resize and you get Step-9:That is your graph You could had also done it by directly input the function from the screen using the short word abs( x)- 4 and moving to the graph quickly Skill Builder Try the following questions to hone your skills Sketch the graph of each of the following modulus function. Make sure you include the domain and range of the function, a) Sketch the graph of y x 5 b) Sketch the graph of y x c) Sketch the graph of y x 1 d) Sketch the graph of y x1 5 e) Sketch the graph of y x 3 5 f) Sketch the graph of y x1 5 g) Sketch the graph of y4 x h) Sketch the graph of y x 1 Page 6 of 15
8 More difficult questions regarding modulus functions How do we sketch the following graph? Let us use the definition of modulus y x x a a a if if a 0 -a <0 x x x x if x x 0 x x if x x <0 Now this is where it gets difficult x x 0 We can factorise the above quadratic equation, x x x x 1 Now lets us sketch the normal graph of y x x Page 7 of 15
9 From the above graph we can see that the x-intercepts are x= - and x=1 Now let us consider the two expressions to see when they are true x x 0 x x1 0 To get the above expression to be positive both brackets must be positive or both brackets must be negative, therefore x or x 1 The second expression x x 1 0 x x x x if 0 To get the above expression 0 then x must be between - and 1, we write this like (-, 1) Then we can sketch the above graph taking into account the domain of the two expressions If you like at the two graph side by side notice what has actually happen. Page 8 of 15
10 The normal graph Notice the how the bottom is reflected in the x-axis Method : Simply sketch the graph. y x x The absolute value of a positive number is equal to that number, and the absolute value of a negative number is equal to the negative number and is therefore positive. So we simply sketch the normal graph and reflect in the x-axis the part of the graph that has a negative y value. Example: Sketch the graph of y xx 1 x 3 Let s have done it the fast way And if you reflect the Page 9 of 15
11 Skill builder a) Sketch the graph of y x x 3 10 b) Sketch the graph of y x x 8 c) Sketch the graph of y x 1 x 1 x 3 d) Sketch the graph of y x 1 x 3x 4 e) Sketch the graph of y x 1 x 3 f) Sketch the graph of y x x SOLVING MODULUS EQUATIONS We normally use the absolute value in Physics when we are looking at the magnitude of something which basically means the value without worrying about it direction like in the case of velocity. Simply put the absolute value means how far the number is from the zero on a number line. 6 its absolute value is 6 6 Its absolute value is 6 also, as it is 6 units away from zero on the number line. So in practice the 'absolute value' means to remove any negative sign and give its positive value. Example Problems Answer ( tricky question, notice that the negative sign is outside the modulus sign) Page 10 of 15
12 Properties of the modulus So when a number is positive or zero we leave it alone but if it is a negative number we change it to a positive We can show that by using the following: x ( x) if x 0 ( x) if x0 Which is the formal definition of what actually happens. So -17 will give us the following -(-17) and we get 17 Another property which is extremely useful is the following: z a z a Solve the following equation x 3 5 : x 3 5 x 3 5 x 3 5 x 3 5 x 8 x Notice how we split the problems into two parts and then proceed with the individual solution Let us return back to the definition of absolute value x is the distance of x from the zero. So the modulus of 3 3 and so is 3 3, so modulus is how far a number is from zero. In summary then we have the following definitions: Page 11 of 15
13 Absolute values means So if we are asked to find the absolute value of an number we just give a positive value To show the absolute value Sometimes calculators we show the absolute values be using the expression abs(-1), which means find the absolute value of -1 How far a number is from zero In Physics we tend to use the expression the magnitude of a number, it size For example the absolute value of -30 is 30 Absolute value of -5 is 5 We tend to ignore the negative and just state the number To show we are talking about the absolute value of a function we use the following x, we call them bars Sometimes absolute value is also written as "abs()", so abs(-1) = 1 is the same as -1 = 1 Case 1- Questions involving < inequality or the inequality The solution is always an interval and the pattern holds true. The inequality to solve What it means x a a x a x a a x a Examples involving case 1 type of problems Solve the following inequality x 5 This means that the solution is giving by the above definition 5 x 5 and that is your answer However if you wanted the complete working out you will need to do the following: We split it to put parts from the definition of the modulus which is namely the following ( x) if x 0 x ( x) if x0 x5, provided x0 And the next part of the equation becomes x 5 x 5 provided x 0 Now we have the two parts and we get the following inequality that provides us with the solution that is: 5 x 5 Now we could write this as an interval as 5,5 Another example to illustrate the basic idea is solve for x the following Let us use the quick way of solving this inequality It is of the form of case 1, so we get the following expression Page 1 of 15
14 x x x x x I could had separated the expressions and proceed with two equations but I decided in using the fast method to obtain the answer Case - the inequality is > or x a or x a x a Remember that x ( x) if x 0 ( x) if x0 So that means xa xa x a x a or x a So the solution is two inequalities not one. Do not try to combine them into one inequality as this would be a mistake. Example Solution x and x Solve x Solve x 3 5 Now we can write this as,, It would be a mistake to combine this and write is as x? Why? You cannot have x and x Hold true at the same time. First step solve this equation x 3 5 x x 1 Then next step solve this equation x 3 5 x 8 x 4 And the answer is x 1 x 4 Page 13 of 15
15 What about if we are asked to go backwards? Say for example- find the absolute value inequality that corresponds to the inequality of x 4? How would you proceed? Let me show you a basic series of steps that you can use to get to the correct answer? Step Sketch the number line and you will see the following Method Look at the end points The end points are - and 4 and they are separated by 6 units apart Now divide this by Dividing by we have 3 What is happening I want to adjust the inequality so it is -3 and 3 instead of - and 4 For this to happen I will need to subtract 1 from both sides of the inequality x 4 Does that look familiar 1 x x 1 3 This is where you can look at case 1 and you can rewrite this as x 1 3, which is the answer Skills builder Show the following on a number line Simplify each of the following inequalities and draw a number diagram Find the solution sets of the following inequalities 3 x 4 1 x 5 (3 x) ( x ) 3x 9 5x 7 5x11 x x x 3 1 3x 1 5 Page 14 of 15
16 MODULUS FUNCTIONS 1 What is modulus? 3 4 What is the meaning of 3? How do you sketch graphs of modulus What do you need to be careful when working with modulus? It can be thought of as distance from zero without worrying about positive or negative It is called the modulus of 3 or the magnitude of 3 or the size of three which is 3 of course Essentially you will need to split the graph up using the definition and then sketch each separately Watch out for the inequality signs carefully. If you multiply or divide by a negative number Also include the brackets! ( a) if a 0 a ( a) if a <0 y x 1 use the definition and then rewrite it as ( x1) x1 0 x 1 ( x1) x1 0 Now split the problem into two, first do the top part y ( x 1) x 1 0 and then do the second bottom part y ( x 1) x 1 0 Be careful to use a number line to help you with the domains and ranges also! Page 15 of 15
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## Day 54 - Lesson 4.5
##### Learning Targets
• Use the general multiplication rule to find probabilities.
• Use a tree diagram to model a chance process involving a sequence of outcomes.
• Calculate conditional probabilities using tree diagrams.
##### Activity:
Start by playing a few rounds of the game in front of the whole class. Be sure to show the students all five cards before the first draw, and all four cards before the second draw. Ask students if getting an Ace on the first and second draw are independent events. They should recognize that drawing cards without replacement gives us dependent events (the probability of the second card being an Ace depends on whether or not the first card is an Ace). We also found it helpful to do one probability calculation in front of the class, such as the probability of two aces.
Students the play the game 10 times and record the humber of wins. On the front white board, they recorded the number of wins from Aces, the number of wins from Kings, and the total number of wins. Once all pairs had recorded the data, we found a probability of winning for the whole class. Students then worked through the rest of the Activity in pairs.
For question #4, ask students why we are adding the probabilities. The answer is that we can win the game by getting two Aces OR getting two Kings and those two events are mutually exclusive (Lesson 4.2). For question #5, we are really just trying to find a conditional probability (Lesson 4.3).
Compare the answer for #4 (the calculated probability) with the experimental probability from #2. This reminds students that probability is really just a long-run relative frequency (Lesson 4.1). Also compare the answer for #5 to the data collected. What percent of the wins occurred by getting Kings?
##### Teaching Tip:
Help students to recognize that all the probabilities in the tree diagram for the 2nd card are actually conditional probabilities. Also point out the fact that the sum of all the probabilities at the end of a tree diagram must add up to 1. We require our students to calculate all of the end probabilities for any tree diagram that they create.
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# How to Count Numbers in Millions
How to Count Numbers in Millions : Welcome to our tutorial on how to count numbers in millions! In this tutorial, we will provide a step-by-step guide on how to read and write numbers in the millions. Millions can be a large number to wrap your head around, but with a little practice and the right techniques, you’ll be able to count in millions with ease. Whether you’re a student, a business professional, or just someone looking to improve your math skills, this tutorial is for you. So let’s get started!
## How to Count Numbers in Millions ?
To count numbers in the millions, you can use the following steps:
1. Determine the number of millions: The first step in counting numbers in the millions is to determine the number of millions. This can be done by dividing the number by one million (1,000,000). For example, if you want to count the number 3,456,789, you would divide it by one million to get 3.456789 millions.
2. Determine the number of thousands: Next, you can determine the number of thousands by dividing the number of millions by one thousand (1,000). For example, using the same number as above (3,456,789), you would divide 3.456789 by 1,000 to get 3456.789 thousands.
3. Determine the number of hundreds: To determine the number of hundreds, you can divide the number of thousands by one hundred (100). For example, using the same number as above (3,456,789), you would divide 3456.789 by 100 to get 34567.89 hundreds.
4. Determine the number of tens: To determine the number of tens, you can divide the number of hundreds by ten (10). For example, using the same number as above (3,456,789), you would divide 34567.89 by 10 to get 345678.9 tens.
5. Determine the number of ones: Finally, you can determine the number of ones by dividing the number of tens by one (1). For example, using the same number as above (3,456,789), you would divide 345678.9 by 1 to get 3456789 ones.
Using these steps, you can count any number in the millions. It is important to be able to count and manipulate large numbers in order to solve more complex mathematical problems.
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# Vector Calculator
Fill the table and press the make graph button to use vector calculator
Magnitude
Angle (°)
x
y
Dot Product
ab
0
bc
0
cd
0
=
0
0
0
0
Give Us Feedback
The vector calculator performs several calculations on up to 10 vectors. The list of its functions is as follows:
1. On entering magnitude and angle, it gives x and y components of the vector.
2. When you enter a second vector, it performs vector addition on the two vectors at the bottom.
3. On the right side, it also gives the dot product between two vectors.
## How to use the Vector Calculator?
To find the angle and magnitude of a vector using this calculator, follow these steps.
To find the vector components: Enter the magnitude and the angle.
To find the vector sum and dot product: Enter the magnitude and angle of second vectors.
## What is a vector?
“An object that has a magnitude and a direction.”
A vector has two points. First is the starting point which is called the tail and the ending point is called the head.
## How to find the components of a vector?
The components of a vector are calculated using magnitude and angle. The formulas used are;
X-Component = |F|cos????
Y-Component = |F|sin????
Example:
Find the components of force whose magnitude is 20N and is acting at an angle of 45 degrees.
Solution:
Step 1: Identify the values.
Magnitude = 20 N
Angle = 35 degrees
Step 2: Put in the formulas.
X-Component = |F|cos????
X-Component = |20|cos35
X-Component = 16.3830
Y-Component = |F|sin????
Y-Component = |20|sin45
Y-Component = 14.14
The same units are added together. You have to add the magnitude of the i unit vector of the first vector into the i unit vector of the second vector and so on.
Example:
6i 7j 1k and 3i 2j 0k
Solution:
6i 7j 1k
+ 3i 2j 0k
____________
9i 9j 1k
## How to multiply two vectors using a dot product?
The dot product is written in mathematical form as
A.B = |A||B|cos????
Example:
Multiply the vectors A and B.
A = 9i 3 2k and B = 1i 1j 1k
The angle between the vectors is 20 degrees.
Solution:
Find the magnitudes of A and B.
|A| = √ ((9)2+(3)2+(2)2)
|A| = (81+9+4)½
|A| = 9.69
|B| = √ ((1)2+(1)2+(1)2)
|B| = (1+1+1)½
|B| = 1.73
Now use it in the formula.
A.B = |9.69||1.73|cos 20
A.B = 15.752
Note: Enter the angle between two vectors as the angle of the first vector. The calculator will automatically assume the other vector along the x-axis. |
You will need
• - handle;
• paper
Instruction
1
Consider the scope of the definition of some elementary functions. If the function has the form y = a/b a region determination are all values except zero. The number a is any number. For example, to find the region of definition of the function y = 3/2x-1, it is necessary to find those values of x for which the denominator of the given fraction is not zero. To do this, find the values of x for which the denominator is equal to zero. For this Paranaita the denominator to zero and find the value, solving the resulting equation for x : 2x – 1 = 0; 2x = 1; x = ½; x = 0,5. It follows that the definition of the function will be any number other than 0.5 in.
2
To find the region of definition of the function radical expression with an even exponent, consider the fact that this expression must be greater than or equal to zero. Example: Find the area of definition of the function y = √3x-9. Referring to the above condition, the expression takes the form of the inequality: 3x – 9 ≥ 0. Solve the following: 3x ≥ 9; x ≥ 3. So, the scope of definition of this function will be all values of x greater than or equal to 3, i.e., x ≥ 3.
3
Finding the area of definition of the function radical expression with an odd exponent, you need to remember the rule that x can be any number if the radical expression is not a fraction. For example, to find the region of definition of the function y = 3√2x-5 , it is sufficient to indicate that x is any real number.
4
While the field definition logarithmic functions, remember that the expression standing under the sign of the logarithm must be a positive value. For example, find the area of definition of the function y = log2 (4x – 1). Given the above condition, find the area of definition of the function in the following way: 4x – 1 > 0; hence 4 > 1; x > 0,25. Thus, the scope of the definition of the function y = log2 (4x – 1) are all values of x > 0,25. |
# A bat and a ball cost \$1.10. The bat costs \$1 more than the ball. How much does the ball cost?
This is a problem that illustrates how solving equations using algebra can get a solution that may defy a false mathematical intuition.
Since the bat costs \$1 more than the ball, and the two of them together cost \$1.10, we would like to know the cost of the ball. We can set up an algebraic problem to solve this. Usually we let a variable represent what we are looking for.
Let x be the cost of the ball.
Then setup an equation that combines all the information given:
`x+(x+1)=1.10` since the ball plus the bat cost \$1.10. Now simplify the left side.
`x+x+1=1.1` multiply by 10 to get rid of decimals
`10x+10x+10=11` collect like terms
`20x=11-10` simplify the right side
`20x=1` divide
`x=1/20` express as money
`x=0.05`
The ball costs 5 cents.
Approved by eNotes Editorial Team
Let x be the cost of the ball.
Since the bat costs \$1 more than the ball, in math form it is:
cost of the bat = x + 1
So the total cost of the bat and the ball is:
cost of the bat + cost of ball = 1 .10
`x + 1 + x = 1.10`
Combining like terms yields:
`2x+ 1 = 1.10`
Then, solve for x. To do so, subtract both sides by 1.
`2x+1-1=1.10-1`
`2x=0.10`
And divide both sides by 2.
`(2x)/2=0.10/2`
`x= 0.05`
Hence, the ball costs \$0.05 .
Approved by eNotes Editorial Team |
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# Solutions of Quadratic Equations by Factorization
Course
#### notes
In general, a real number α is called a root of the quadratic equation ax^2 + bx + c = 0, a ≠ 0 if aα^2 + bα + c = 0. We also say that x = α is a solution of the quadratic equation, or that α satisfies the quadratic equation. Note that the zeroes of the quadratic polynomial ax^2 + bx + c=0 and the roots of the quadratic equation ax^2 + bx + c = 0 are the same.
By substituting arbitrary values for the variable and deciding the roots of quadratic equation is a time consuming process. Let us learn to use factorisation method to find the roots of the given quadratic equation.
x^2 - 4 x - 5 = (x - 5) (x + 1)
(x - 5) and (x + 1) are two linear factors of quadratic polynomial x^2 - 4 x - 5.
So the quadratic equation obtained from x^2 - 4 x - 5 can be written as (x - 5) (x + 1) = 0
x - 5 = 0 or x + 1 = 0
∴x = 5 or x = -1
∴5 and the -1 are the roots of the given quadratic equation.
While solving the equation first we obtained the linear factors. So we call this method as ’factorization method’ of solving quadratic equation
Ex.(1) m^2 - 14 m + 13 = 0
∴m^2 - 13 m - 1m + 13 = 0
∴m (m - 13) -1 (m - 13) = 0
∴(m - 13) (m - 1) = 0
∴m - 13 = 0 or m - 1 = 0
∴m = 13 or m = 1
∴13 and 1 are the roots of the given quadratic equation.
### Shaalaa.com
Quadratic Equation part 5 (Solution by Factorization) [00:11:01]
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Question Video: Finding the Value of an Unknown in a Quadratic Equation by Using the Relation between Its Coefficient and Its Roots Mathematics
If the roots of the equation 5π₯Β² β 2ππ₯ + 5 = 0 are equal, what are the possible values of π?
03:16
Video Transcript
If the roots of the equation five π₯ squared minus two ππ₯ plus five equals zero are equal, what are the possible values of π?
So, what we have here is a quadratic. And with a quadratic, what we can do is use the discriminant to help us decide whether our roots are gonna be equal, whether theyβre gonna have real roots, or theyβre gonna have no real roots. So, the discriminant is π squared minus four ππ. But what does this mean? Whatβs π, whatβs π, and whatβs π? Well, ππ and π are parts of our quadratic when we have it in the form that we have. So, weβve got ππ₯ squared plus ππ₯ plus π. So, π is the coefficient of π₯ squared, π is the coefficient of π₯, and π is our constant term, or our numerical value.
Well, with our quadratic, weβve got π equals five. And thatβs because thatβs our coefficient of π₯ squared. π is equal to negative two π. And be very careful here. Make sure that we include the negative, so the sign. And then, our π is gonna be equal to five. But how are these and our discriminant useful? Well, the discriminant is useful because it can tell us how many roots our quadratic will have. So, for instance, if we have π squared minus four ππ is less than zero, then there are no real roots. If π squared minus four ππ is greater than zero, then there are real roots. And if π squared minus four ππ equals zero, then the roots are going to be equal.
And Iβve added some sketches to show what the graphs would look like. So, we can see if it has no roots, that it wouldnβt cross the π₯-axis at all. If there are real roots, then itβll cross our π₯-axis in two distinct places. And if the roots were equal, it would just touch the π₯-axis. Okay, great, so, we now know what the discriminant is. And we know how it could be used. Letβs use it to solve our problem.
Well, in our question, weβre told that the roots are equal. So therefore, itβs this third scenario weβre interested in when π squared minus four ππ is equal to zero. So therefore, if we substitute in our values π, π, and π, we can have negative two π all squared, thatβs because thatβs our π, minus four multiplied by π, which is five, multiplied by five, which is π. And then, this is all equal to zero because, as we said, we have equal roots.
So, what weβre gonna get is four π squared minus 100 equals zero. So, then, if we add 100 to each side of the equation, weβre gonna get four π squared is equal to 100. And then, what we need to do is divide both sides of the equation by four cause we want to find out what π is going to be. So, we get π squared is equal to 25. So, then, if we take the square root of both sides of the equation, weβre gonna get π is equal to positive or negative five.
And thatβs because if you square root π squared, you get π. And you square root 25, weβre gonna get positive or negative five. Thatβs because five multiplied by five gives us 25, and negative five multiplied by negative five gives us 25. So therefore, given the roots of the equation five π₯ squared minus two ππ₯ plus five equals zero are equal, the possible values of π are five and negative five. |
## Deriving simple divisibility rules
I learned something new today!
Many people probably know of these simple divisibility rules:
• An integer is divisible by 10 if it ends in a zero.
• An integer is divisible by five if it ends in zero or five.
• An integer is divisible by two if its final digit is.
• An integer is divisible by three or nine if the sum of its digits is.
• An integer is divisible by six if it is divisible by both two and three.
Slightly less well-known are these rules:
• An integer is divisible by four if the two-digit number formed from its last two digits is.
• An integer is divisible by eight if the three-digit number formed from its last three digits is.
That last one is simpler than it sounds—you don’t have to memorize all the three-digit numbers that are divisible by eight (that would be rather silly!), since it’s pretty easy to try dividing a three-digit number by two in your head three times. If you can divide it by two three times, it’s divisible by eight; if you hit an odd number before the third division, it’s not. For example, 29873458636 -> 636 -> 318 -> 159, oops, not divisible by eight.
You will notice that there is a nice simple divisibility rule listed above for every number up to ten… except seven. The only divisibility rule I had ever seen for seven was rather complicated and hard to perform in one’s head, making it rather useless. At least, I think it was, because I don’t remember it.
But just today, I read an excellent post over at Foxmaths! explaining—complete with derivation—a nice divisibility rule for seven. The rule is this: take the number, and subtract twice the final digit from the number formed by the remainder of the digits. The result will be divisible by seven if and only if the original number was. Of course, this process can be repeated until you’re left with something simple. For example, imagine starting with 3641: subtract twice 1 from 364, giving 364 – 2 = 362. Now subtract twice two from 36, giving 32, which is not divisible by seven. Therefore, 3641 isn’t, either. See, nice and simple!
The really neat thing is that this isn’t peculiar to seven—Fox goes on to show how the same technique can be used, in principle, to derive a similar divisibility test for any prime number! Go read it.
Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.
This entry was posted in computation, links, number theory. Bookmark the permalink.
### 14 Responses to Deriving simple divisibility rules
1. Fox says:
I was even able to show, though I don’t think I’ll post on this, that any divisibility rule you generate is going to be relatively simple with respect to the divisor. For example, the rule for 41 is you multiply the last digit by 4 and subtract it from the others. Neat stuff.
2. Fox says:
Oh, the link on your blogroll is kind of out of date …
3. Brent says:
Fox: neat stuff indeed. I’m inspired to play around with it a bit myself. I have a few other thoughts, maybe I’ll post more on it later. Re: the link, thanks for the heads-up, I’ve fixed it now!
4. Tim McKenzie says:
Yes, I wonder if the divisible-by-7 rule is still more trouble than it’s worth, though. It’s still interesting, either way.
A commenter on Fox’s site points out that the divisibility rule for 11 is relatively easy: alternately add and subtract the digits, and see if the result is divisible by 11. For example 37241398 -> 3 – 7 + 2 – 4 + 1 – 3 + 9 – 8 = -7 or something, so it’s not divisible by 11.
In base 6, we can get quite far with quite simple divisibility rules; a number is:
* divisible by 2 iff its last digit is,
* divisible by 3 iff its last digit is,
* divisible by 4 iff the number formed by its last two digits is,
* divisible by 5 iff the sum of its digits is,
* divisible by 6 iff its last digit is,
* divisible by 7 iff the alternating sum of its digits is (like 11 in base 10),
* divisible by 8 iff the number formed by its last three digits is,
* divisible by 9 iff the number formed by its last two digits is, and
* divisible by ten iff it’s divisible by both 2 and 5.
In base twelve, we can get even further, with only one appeal to Fox’s generalized divisibility trick, because we can use this trick for both 5 and 7 at the same time: a number 12*A + B is divisible by 5 or 7 iff A + 3*B is. Then, in general, a number is:
* divisible by 2, 3, 4, 6, or twelve iff its last digit is,
* divisible by 5 or 7 iff (see above),
* divisible by 8, 9, or sixteen iff the number formed by its last two digits is,
* divisible by ten iff it’s divisible by both 2 and 5,
* divisible by eleven iff the sum of its digits is,
* divisible by thirteen iff the alternating sum of its digits is,
* divisible by fourteen iff it’s divisible by both 2 and 7, and
* divisible by fifteen iff it’s divisible by both 3 and 5.
5. Brent says:
Tim: heh, well, I guess that depends what it’s worth, doesn’t it? =) I can do the divisible by 7 rule iteratively in my head much faster than I could do a trial division by 7 in my head, which is not true for the alternating-sums-of-groups-of-three rule which is the other rule I’ve seen for 7. But anyway, thanks for the analysis of base 6 and base 12! Memorizing rules for base 6 and base 12, though, is definitely more trouble than it’s worth. 😉
6. I happened to learn the seven method sometime in the last year, having seen a harder one in fourth grade (unless it was the same and I just wasn’t ready for it). It’s useful for testing whether a number is prime, which is something I do now and then for diversion.
7. Jonathan says:
Brent,
My 7th grade math teacher taught us the divisibility by 7 rule you list here, not the sums of groups of 3. I guess I had a stronger foundation in junior high level math than you . . . how remarkable you managed to overtake me eventually 😉
Having not used it once since then I’d forgotten it anyway. As an homage to 7, though, the best single-digit number, I do set my watch seven minutes fast: I like to have a fast watch to help me get places on time, and I’ve always found 7 to be the hardest number to subtract.
8. silverpie says:
Pondering base 16, we get the following:
2, last digit
3, sum of digits
4, last digit
5, sum of digits
6, 2 and 3
7, use Fox’s trick, 16A+B is congruent to A+4B mod 7 (and mod 9)
8, last digit
9, same as 7
10, 2 and 5
11, Fox’s trick again, but this time 16A+B is congruent to A-2B mod 11
12, 3 and 4
13, yet another Fox trick, with 16A+B congruent to A-4B mod 13
14, 2 and 7
15, last digit
In fact, I see a generalization; for any multiple-of-four base 4X, A+XB provides a test for 2X-1 and 2X+1.
9. Brent says:
silverpie: nice analysis! I should point out one technicality, however, which is that it is incorrect to say, e.g. “16A+B is congruent to A+4B mod 7”. In fact, it is only the case that 16A+B is divisble by 7 if and only if A+4B is. This is not the same as being congruent. For example, if we take A=2 and B=4, we have 16*2 + 4 = 36, which is 1 mod 7, and 2 + 4*4 = 18, which is 4 mod 7. If you read Fox’s original post carefully you will see the distinction and why it arises.
10. JAM says:
There is also a nice rule for divisibility by 11. If one adds and sutracts alternate digits then if the result is divisible by 11 then the original number is. This follows form the fact that (our bas) 10=-1 modulo 11. The adding digits test for divisibility by 9 (and 3) follows from the fact that 10=1 modulo 9 (3). In general in base B we always have a nice test for divisibility by B-1 (adding digits) and B+1 (adding and subtracting alternate digits). This makes base 12 a “better” choice of number system than 10 as we have a test for divisibility by 13, which is more useful that a test for 9 or 3 (both of which give nice results in base 12 anyway).
11. kiwi says:
Here’s another test for 7 … starting from the leftmost digit and proceeding to the right, triple what you have and add the next digit …
1342 -> 1 *3+3=6 *3+4=22 *3+2=68 and 68 -> 6 *3+8=26 which is not divisisible by 7 so neither is the original number.
You can simplify by dropping multiples of 7 at will …
1342 -> 1 *3+2=6 *3+4=22=1(mod 7) *3+2=5 so dividing 1342 by 7 leaves remainder 5. This is better than the above test because it preserves the remainder.
12. Jennifer says:
I recently found another test for 11.
Take the number in the ones place and subtract it from the truncated number. If the difference is divisible by 11, so is the original number
Consider the number 847.
84-7 = 77
Yes! Divisible by 11
1342
134-2 = 132 (Some will see that it is divisible here, otherwise…)
13-2 = 11
Yes! Divisible by 11
757
75-7 = 68
No. Not Divisible by 11.
Can be a tad bit cumbersome for longer numbers, but it works well for 3 to 5 digit numbers.
13. Brent says:
Jennifer: good work! In fact, this test for 11 is also an instance of the general method that Fox originally wrote about. Like the rule for 7, it doesn’t preserve the remainder mod 11, but it does preserve divisibility/non-divisibility by 11.
14. Ron says:
divisibility by 11 is also easy.
Starting with the number abcdefg, where a, b, c, etc. are all, independently, single digits;
Let X = the sum of every other digit; i.e., X = a+ c + e ; and
let Y = the sum of the other series of every other digit, i.e., Y = b + d + f;
If the absolute value of X minus Y is divisible by 11, so is abcdefg!
It works for any number, without limit to the number of digits in abcdefg….! |
# VHS A1 1.1 Variables
33 %
67 %
Information about VHS A1 1.1 Variables
Education
Published on July 31, 2008
Author: cashill
Source: authorstream.com
Slide 1: Lessons from the Math Zone: Variables Click to Start Lesson Slide 2: Lessons from the Math Zone Variables © Copyright 2006, Don Link Permission Granted for Educational Use Only Slide 3: Variables Vary x = 5 7 6 8 9 10 10 15 20 25 30 40 50 100 200 300 1,000 2,000 3,000 4,000 5,000 1.0 1.1 1.2 1.3 1.4 1.5 2.5 3.5 4.5 5.5 5.5 0.01 0.02 3.33 –1 –2 –3 –4 –5 –1.5 –2.6 –3.7 –4.8 –5.9 –6.0 Slide 4: Algebraic Expressions An Algebraic Expression is an expression that contains at least one Variable. x + 14 3x + 1 9 – x Slide 5: 9 – Using Variables x = + 14 3 + 1 =22 =25 x x =1 x =2 x One way to use variables is to substitute their numerical value into an algebraic expression. (8) (8) (8) (8) 8 Slide 6: 9 – One way to use variables is to substitute their numerical value into an algebraic expression Using Variables x = 8 + 14 3 + 1 =22 =25 x x =1 x =2 x Way to Go! Slide 7: Let’s Practice x – 3 = 1 x = 4 2x – 3 = 5 9x + 3 = 39 10 – 2x = 2 x + x = 8 x ÷ 2 = 2 2(x + 5) = 18 (x +10)×(x – 2) = 28 3x + 4x – 2 = 26 5x = 20 Slide 8: Let’s Practice x – 3 = 1 x = 4 2x – 3 = 5 9x + 3 = 39 10 – 2x = 2 x + x = 8 x ÷ 2 = 2 2(x + 5) = 18 (x +10)×(x – 2) = 28 3x + 4x – 2 = 26 5x = 20 High Five! Slide 9: Let’s Practice with Two Variables a+b–2 = 6 a = 3 2b – 3a = 1 5a ÷ b = 3 ab = 15 a(b + 5) = 30 (a +10)×(b – 2) = 39 3ab + 4a – 2 = 55 5a–b = 10 b = 5 Slide 10: Let’s Practice with Two Variables a+b–2 = 6 a = 3 2b – 3a = 1 5a ÷ b = 3 ab = 15 a(b + 5) = 30 (a +10)×(b – 2) = 39 3ab + 4a – 2 = 55 5a–b = 10 b = 5 Good Job! Slide 11: Another way to use a variable is as an unknown value in an algebraic equation. Unknown Variables 5 + x = 8 An Equation Claims that the Left Side = the Right Side Unknown Left Side Right Side Slide 12: Unknown Variables 5 + x = 8 An Equation If: Left Side Right Side, = then the “equation is .” true If: Left Side Right Side, ≠ then the “equation is .” false To find the unknown value that makes the equation true, we “solve the equation for x.” Slide 13: 5 + = 8 If x = 7 , the equation is 7 Unknown Variables True? False? True? False? x 12 8 3 If x = 3 , the equation is = 8 Slide 14: Solving an Equation Solving an equation means finding a value for the unknown that makes the equation true. 5 + x = 8 What value for x will make this equation true? 3 is a solution to the equation. Slide 15: Let’s Practice Using Different Variables x – 3 = 10 4w – 3 = 5 2δ ÷ 3 = 6 20 – □ = 2 5a = 20 x = 13 a = 4 w = 2 δ = 9 □ = 18 Slide 16: Let’s Practice Using Different Variables x – 3 = 10 4w – 3 = 5 2δ ÷ 3 = 6 20 – □ = 2 5a = 20 x = 13 a = 4 w = 2 δ = 9 □ = 18 Cookin’! Slide 17: x + 17 = 25 8 x = 8 x – 3 = 4 Variables: Summary and Review 1 Substitute the numerical value of a variable into an algebraic expression 2 “Solve the equation for x.” x = 4 1 Slide 18: For Printing
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# Measure and Classify Angles
Sep 12, 2022
## Key Concepts
• Name angles
• Measure and classify angles
• Find angle measures
• Identify congruent angles
• Double an angle measure
### Introduction
In this chapter, we will learn to define, classify, draw, name, and measure angles, use the protractor and angle addition postulates and double an angle measure.
### Angle
An angle consists of two different rays (sides) that share the same endpoint (vertex).
Angle ABC, ∠ABC, or ∠B
Sides:
Sides are the rays.
Vertex:
It is the point where the two rays meet.
Example:
In the above example, the angle with sides AB and AC- can be named or . Point A is the vertex of the angle.
### Name angles
We can follow the rules below while naming angles:
• Use three capital letters – Vertex in the middle
• Can use one capital letter if it is the vertex and it is obvious which angle you are referring to
• Can use the number located inside the angle
Example 1:
Name this angle in three different ways:
The above angle can be named as,
1. ∠DRY
2. ∠YRD
3. ∠R
Example 2:
Name this angle in four different ways:
The above angle can be named as,
1. ∠WET
2. ∠TEW
3. ∠E
4. ∠1
Example 3:
Name the three angles in the diagram.
The three angles are,
WXY, or ∠YXW
YXZ, or ∠ZXY
WXZ, or ∠ZXW
Why should you NOT label any of these angles “Angle X“?
You should not name any of these angles ∠X because all three angles have X as their vertex.
### POSTULATE 3 Protractor Postulate
The m∠AOB is equal to the absolute value of the difference between the real numbers for OA and OB m∠AOB = |55° – 180°|
m∠AOB = |–125°|
m∠AOB = 125°
## Measure and classify angles
The angle is measured using a protractor in degrees. It is the smallest amount of rotation about the vertex from one side to the other.
Example:
• m∠APB = 60°
• m∠APC = 100°
• m∠BPC = 40°
### Types of angles
• Acute angle: between 0° and 90°
• Right angle: exactly 90°°
• Obtuse Angle: between 90° ° and 180°°
• Straight angle: exactly 180°°
Smaller angles can be added together to form larger angles if they share a common vertex.
If B is in the interior of ∠AOC, then the m∠AOC is equal to the sum of m∠AOB and m∠BOC.
mAOC = mAOB + mBOC
### Find angle measures
Example:
Given that m∠LKN = 145° find m∠LKM and m∠MKN.
STEP 1: Write and solve an equation to find the value of x.
STEP 2: Evaluate the given expressions when x = 23.
m∠LKM = (2x + 10)°= (2.23 + 10)° = 56°
m∠MKN = (4x – 3)°= (4.23 – 3)° = 89°
So, m∠LKM = 56°
and m∠MKN = 89°.
### Congruent Angles
Congruent angles have the same angle measure.
Example:
Can be marked using the same number of hash marks:
#### Identify congruent angles
In the given picture, identify the angles that are congruent. If m∠DEG = 157° then find m∠GKL?
Solution:
There are two pairs of congruent angles:
∠DEF ≅ ∠JKL and ∠DEG ≅ ∠GKL.
Because ∠DEG ≅ ∠GKL, m∠DEG = m∠GKL. So, m∠GKL = 157°.
### Double an angle measure
In the below diagram, YW bisects ∠XYZ, and m∠XYW = 18°. Find m∠XYZ.
Solution:
By the Angle Addition Postulate, m∠XYZ = m∠XYW + m∠WYZ. Because YW bisects ∠XYZ, you know that ∠XYW ≅ ∠WYZ.
So, m∠XYW = m∠WYZ, and you can write
m∠XYZ = m∠ XYW + m∠WYZ = 18° + 18° = 36°.
## Exercise
1. Write three names for the angle shown. Then name the vertex and sides of the angle.
1. Name three different angles in the diagram given below.
1. Classify the angle with the given measure as acute, obtuse, right, or straight.
a) m∠W = 180 b) m∠X = 30
1. Classify the angle with the given measure as acute, obtuse, right, or straight.
a) m∠Y = 90 b) m∠Z = 95
1. Use the diagram to find the angle measure. Then classify the angle.
1. ∠BOC =
2. ∠AOB =
3. ∠DOB =
4. ∠DOE =
1. Use the diagram to find the angle measure. Then classify the angle.
1. ∠AOC =
2. ∠BOE =
3. ∠EOC =
4. ∠COD =
1. Fill in the blanks with the help of the below given picture.
1. Angles with the same measure are congruent angles. This means that if _____________________, then __________________. You can also say that if _______________________, then ______________________.
1. Find the indicated angle measure
1. Given m∠WXZ = 80 , find m∠YXZ,
1. In the below diagram, bisects ∠XWY, and m∠XWZ = 52 . Find m∠YWZ.
### What have we learned
• To define, classify, draw, name, and measure angles.
• To use the protractor and angle addition postulates.
• To double an angle measure
### Concept Map
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […] |
LCM of 4 and 9 is the smallest number among all typical multiples of 4 and 9. The first couple of multiples that 4 and also 9 room (4, 8, 12, 16, 20, 24, 28, . . . ) and also (9, 18, 27, 36, . . . ) respectively. There space 3 commonly used techniques to uncover LCM the 4 and 9 - by element factorization, through listing multiples, and by division method.
You are watching: What is the least common multiple of 4 and 9
1 LCM that 4 and 9 2 List that Methods 3 Solved Examples 4 FAQs
Answer: LCM the 4 and also 9 is 36.
Explanation:
The LCM of two non-zero integers, x(4) and also y(9), is the smallest positive integer m(36) that is divisible through both x(4) and also y(9) without any remainder.
The methods to discover the LCM of 4 and 9 are described below.
By Listing MultiplesBy element Factorization MethodBy division Method
### LCM of 4 and 9 by Listing Multiples
To calculation the LCM of 4 and also 9 by listing out the typical multiples, we deserve to follow the given listed below steps:
Step 1: list a few multiples of 4 (4, 8, 12, 16, 20, 24, 28, . . . ) and 9 (9, 18, 27, 36, . . . . )Step 2: The typical multiples from the multiples of 4 and also 9 are 36, 72, . . .Step 3: The smallest usual multiple that 4 and 9 is 36.
∴ The least typical multiple that 4 and also 9 = 36.
### LCM that 4 and 9 by prime Factorization
Prime administrate of 4 and 9 is (2 × 2) = 22 and also (3 × 3) = 32 respectively. LCM the 4 and 9 deserve to be derived by multiplying prime determinants raised to their respective highest possible power, i.e. 22 × 32 = 36.Hence, the LCM of 4 and 9 by prime factorization is 36.
### LCM that 4 and 9 by division Method
To calculation the LCM the 4 and also 9 by the department method, we will certainly divide the numbers(4, 9) by their prime determinants (preferably common). The product of these divisors offers the LCM of 4 and also 9.
Step 3: proceed the actions until only 1s space left in the last row.
The LCM that 4 and 9 is the product of all prime numbers on the left, i.e. LCM(4, 9) by department method = 2 × 2 × 3 × 3 = 36.
☛ likewise Check:
## FAQs top top LCM of 4 and also 9
### What is the LCM of 4 and 9?
The LCM that 4 and also 9 is 36. To uncover the least common multiple (LCM) of 4 and also 9, we require to discover the multiples of 4 and also 9 (multiples that 4 = 4, 8, 12, 16 . . . . 36; multiples that 9 = 9, 18, 27, 36) and choose the the smallest multiple that is specifically divisible through 4 and 9, i.e., 36.
### How to uncover the LCM of 4 and also 9 by prime Factorization?
To uncover the LCM of 4 and also 9 using prime factorization, we will uncover the prime factors, (4 = 2 × 2) and (9 = 3 × 3). LCM of 4 and 9 is the product that prime determinants raised to your respective highest possible exponent among the numbers 4 and also 9.⇒ LCM that 4, 9 = 22 × 32 = 36.
### What is the Relation in between GCF and LCM of 4, 9?
The complying with equation can be supplied to to express the relation in between GCF and also LCM of 4 and also 9, i.e. GCF × LCM = 4 × 9.
### If the LCM that 9 and also 4 is 36, find its GCF.
LCM(9, 4) × GCF(9, 4) = 9 × 4Since the LCM of 9 and also 4 = 36⇒ 36 × GCF(9, 4) = 36Therefore, the greatest common factor (GCF) = 36/36 = 1.
See more: How Do You Say I Love You In Philippines, I Love You In Tagalog Language
### What is the least Perfect Square Divisible through 4 and 9?
The the very least number divisible by 4 and also 9 = LCM(4, 9)LCM that 4 and 9 = 2 × 2 × 3 × 3 ⇒ the very least perfect square divisible by each 4 and 9 = 36 Therefore, 36 is the required number. |
Now that we have learnt how to how geometric sequences and series, we can apply them to real life scenarios.
## Growth
Geometric growth is found in many real life scenarios such as population growth and the growth of an investment.
Geometric growth occurs when the common ratio is greater than 1, that is $r>1$.
The common ratio can be found by adding the percentage increase and 1, that is $r=1+%$increase.
Say that the percentage increase was 3%. The common ratio can be found by adding the percentage increase (of 3%) and 1.
This would be the original amount plus an extra 3%.
The reasoning for this is as follows:
Original amount + 3% of the original amount
= original amount (1+3%)
=original amount (1+0.03)
=1.03 x original amount.
### Example 1
The rabbit population in a Victorian town was estimated to be 320,000 in 2012. Scientists believe that this will increase by two percent each year.
a) What will the rabbit population be in 2015? Round your answer to the nearest decimal place.
Start by finding the first term.
$a=320,000$
Find the common ratio.
$1+2%=1+0.02$
$r=1.02$
Find the value of $n$.
2012 is the first term, so $n=1$
2013 is the second term, so $n=2$
2015 is the fourth term, so $n=4$
Use the rule.
$t_n=ar^{n-1}$
$t_4=320,000\times 1.02^{4-1}$
$t_4=320,000\times 1.0612$
$t_4=339,586.56$
There will be 339,587 rabbits in 2015.
b) In which year will the rabbit population reach 400,000?
Use the equation where $t_n=400,000$
$t_n=ar^{n-1}$
$400,000=320,000\times (1.02)^{n-1}$
$1.25=(1.02)^{n-1}$
$log_{10}(1.02)^{n-1}=log_{10}(1.25)$
$n-1=\dfrac {log_{10}(1.25)}{log_{10}(1.02)}$
$n-1=11.26838$
$n=12.27$
The rabbit population will reach 400,000 during the 12th year, 2023.
## Decay
Geometric decay is found in real life instances such as depreciation and population decreases.
Geometric decay occurs when the common ratio is less than 1, that is $r<1$.
The common ratio can be found by subtracting the percentage increase from 1, that is $r=1-%$increase.
Say that the percentage decrease was 3%. The common ratio can be found by subtracting the percentage decrease (of 3%) from 1.
This would be the original amount minus 3% of the original amount.
The reasoning for this is as follows:
Original amount – 3% of the original amount
= original amount (1- 3%)
=original amount (1- 0.03)
=0.97 x original amount.
### Example 2
A photocopier was purchased for $13,000 in 2014. The photocopier decreases in value by 20% of the previous year’s value. a) What is an expression for the value of the photocopier, $V_n$, after $n$ years? We know that this is a geometric sequence as there is a 20% decrease on the previous year’s value. Find $a$ and $r$. $a=13,000$ $r=1-20%$ $r=1-0.2$ $r=0.8$ Use the equation to find the expression. Take care to use $V_n$ instead of $t_n$ as this is what the question asked. $V_n=ar^{n-1}$ $V_n=13,000\times 0.8^{n-1}$ Make sure that you write your answer as a full sentence. The expression $V_n=13,000\times 0.8^{n-1}$ gives the value of the photocopier. b) What is the value of the photocopier after three years? Use the expression found in part a and substitute the value of $n$. $V_n=13,000\times 0.8^{n-1}$ $V_3=13,000\times 0.8^{3-1}$ $V_3=13,000\times 0.64$ $V_3=8,320$ Write the answer as a full sentence, make sure to include the dollar sign as we are referring to money. The value of the photocopier after three years is$8,320. |
Q: Solve for x: (x + 3)(x – 4) < 0
A:
Step 1: Find the zeroes
Since the quadratic is already factored, this isn’t too tricky. If it wasn’t factored, you’d have to factor first! (always make sure there is a 0 on one side of the equation / inequality before proceeding).
OK, so what are the zeroes?
x + 3 = 0 or x – 4 = 0
The zeroes are x = -3 or x = 4
So, draw a number line and plot the zeroes on the number line:
Now, you have to test each “interval” that is separated by the “zeroes”. There are three intervals to test.
Interval 1: The numbers to the left of -3 –> in interval notation this is (-infinity, -3)
Interval 2: The numbers between -3 and 4 –> in interval notation this is (-3, 4)
Interval 3: The numbers to the right of 4 –> in interval notation this is (4, infinity)
Step 2: Test each interval
Pick any number on interval 1 and test it into the original inequality. I’ll pick -5:
(x + 3)(x – 4) < 0
(-5 + 3)(-5 – 4) < 0
(-2)(-9) < 0
18 < 0 <— this is false, so numbers on this interval [interval 1] are not part of the solution.
Pick any number on interval 2 and test it into the original inequality. I’ll pick 0:
(x + 3)(x – 4) < 0
(0 + 3)(0 – 4) < 0
(3)(-4) < 0
-12 < 0 <— this is true, so numbers on this interval [interval 2] are a part of the solution.
Pick any number on interval 3 and test it into the original inequality. I’ll pick 5:
(x + 3)(x – 4) < 0
(5 + 3)(5 – 4) < 0
(8)(1) < 0
8 < 0 <— this is false, so numbers on this interval [interval 3] are not part of the solution.
SO: The only interval that “worked” was interval 2. Therefore, the solution is all number between -3 and 4.
In interval notation, we write that like: (-3, 4)
In inequality notation, we write that like: -3 < x < 4 |
Home >
### Math behind the Tiles
Why do 7 numbers create 28 tiles? After all, with six numbers, we get 36 combinations with two dice. Let's start with the simplest situation. If we have one number, we get one tile. One number, 1 tile 0,0 Two numbers, 3 tiles 0,0 0.1 1,1 Three numbers, 6 tiles 0,0 0,1 0,2 1,1 1,2 2,2 Four numbers, 10 tiles 0,0 0,1 0,2 0,3 1,1 1,2 1,3 2,2 2,3 3,3 Five numbers, 15 tiles 0,0 0,1 0,2 0,3 0,4 1,1 1,2 1,3 1,4 2,2 2,3 2,4 3,3 3,4 4,4 Six numbers, 21 tiles 0,0 0,1 0,2 0,3 0,4 0,5 1,1 1,2 1,3 1,4 1.5 2,2 2,3 2,4 2.5 3,3 3,4 3,5 4,4 4,5 5,5 Seven numbers, 28 tiles 0,0 0,1 0,2 0,3 0,4 0,5 0,6 1,1 1,2 1,3 1,4 1.5 1,6 2,2 2,3 2,4 2.5 2,6 3,3 3,4 3,5 3,6 4,4 4,5 4,6 5,5 5,6 6,6 BONUS See the ""Math in Backgammon" in the bonus part of this book. RETURN Some people might assume that we need 42 tiles to capture the combinations because there are seven numbers or 7 x 7. After all, they learned when analyzing moves in backgammon that there are 36 combinations for the six-sided dice. The difference becomes clear when you count the duplicated combinations. Look at the combinations that are mirror reflections of each other: 1,6 is the same as 6,1. When we add the duplications, we find that the number goes up to 49.In the table showing seven numbers, we see there are seven tiles with doubles and 21 tiles with pairs of different numbers. We could have 49 tiles, but 42 of them would have duplicates. This step-by-step analysis, looking at what is in front of us, is part of the benefits tat come to a class of students who are GUIDED in the use of dominoes. WIthout questions, students might not know how to analyze a situation.one way to teach this is to create a grid |
# Create Unique and Ordered Pythagorean Triplets: A How-To Guide
Posted on
Are you tired of boring math problems? Do you want to challenge yourself and create something unique? Look no further than the world of Pythagorean triplets! These sets of three numbers are more than just a formula; they can be beautiful, ordered, and customized to your preferences. With this how-to guide, you’ll learn the tips and tricks to create your own unique Pythagorean triplets that will leave your friends and classmates in awe.
Whether you’re a beginner or a math enthusiast, this guide is perfect for anyone looking to spice up their math skills. With step-by-step instructions and examples, you’ll be able to master the art of Pythagorean triplets in no time. Imagine impressing your teacher with a new, creative approach to a classic problem, or using your newfound expertise to solve even the toughest of equations.
But this guide isn’t just about solving math problems – it’s about exploring your creative side. The possibilities for Pythagorean triplets are endless, with each set being unique and personal to you. You can choose the sizes and patterns to suit your style or challenge yourself by designing complex triplets that require a multitude of calculations. The only limit is your imagination!
So, what are you waiting for? Dive into the world of Pythagorean triplets and let your creativity soar. Who knows, you might just discover a passion for math that you never knew you had. With this how-to guide, anything is possible. Keep reading to unlock the secrets of creating unique and ordered Pythagorean triplets!
“Generating Unique, Ordered Pythagorean Triplets” ~ bbaz
## Introduction
Pythagorean triplets are an interesting phenomenon in mathematics that have puzzled mathematicians for centuries. Pythagorean triplets are a set of three integers that satisfy the Pythagorean theorem. This theorem states that the sum of the squares of the lengths of the two shorter sides of a triangle is equal to the square of the length of the longest side of the triangle. In this article, we will discuss how to create unique and ordered Pythagorean triplets using a simple method.
## The Basics of Pythagorean Triplets
To understand the process of creating unique and ordered Pythagorean triplets, it is important to first understand the basics of Pythagorean triplets. There are three types of Pythagorean triplets that exist – primitive, non-primitive, and special. Primitive Pythagorean triplets are those in which the three integers do not have any common factors. Non-primitive Pythagorean triplets are those in which the three integers share common factors. Special Pythagorean triplets are those in which one of the integers is a multiple of another integer.
## Create Unique Pythagorean Triplets
Creating unique Pythagorean triplets involves using a simple formula that involves finding square roots and integers. First, select two random integers, x and y, such that x > y. Next, using the formula a = x² – y² and b = 2xy, solve for c, which is the hypotenuse of the triangle. This will give you a unique Pythagorean triplet that cannot be multiplied by a scalar to obtain another Pythagorean triplet.
## Example of Creating Unique Pythagorean Triplets
For example, if we choose x = 5 and y = 3, we can use the formula a = x² – y² = 22, b = 2xy = 30, and c = √(a² + b²) = √(484 + 900) = 34. This gives us the unique Pythagorean triplet (22, 30, 34).
## Create Ordered Pythagorean Triplets
The process of creating ordered Pythagorean triplets involves taking the Pythagorean triplets that we created earlier and multiplying them by a scalar value to obtain other Pythagorean triplets. To create ordered Pythagorean triplets, we need to make sure that the integers are in a specific order in each triplet. One way to do this is to order the triplets based on the odd number of the triplet. This will ensure that the triplets are ordered in a consistent manner.
## Example of Creating Ordered Pythagorean Triplets
Using the same example as earlier (x = 5 and y = 3), we first create the unique Pythagorean triplet (22, 30, 34). Next, we multiply this triplet by a scalar value of 2 to obtain the triplet (44, 60, 68). We then order the triplets based on the odd number, which gives us the ordered Pythagorean triplets (22, 30, 34), (30, 40, 50), (44, 60, 68), (56, 42, 70), and so on.
## Comparison between Unique and Ordered Pythagorean Triplets
Unique Pythagorean Triplets Ordered Pythagorean Triplets
Do not have a specific order Are ordered based on the odd number of the triplet
Can only be multiplied by a scalar value to obtain other Pythagorean triplets Can be multiplied by a scalar value to obtain other Pythagorean triplets in a consistent manner
Can be easily created using a simple formula Requires extra steps to ensure the triplets are ordered correctly
## Conclusion
Creating unique and ordered Pythagorean triplets is a fascinating mathematical concept that can be easily understood with a little bit of practice. These triplets have practical applications in many fields, including engineering, physics, and computer science. Understanding how to create these triplets can help you in your studies or in your work, and it can also provide a sense of satisfaction and curiosity as you explore this interesting mathematical phenomenon.
Thank you for taking the time to read our guide on creating unique and ordered Pythagorean triplets! We hope that you found it informative and helpful in your explorations of mathematics. As demonstrated in the article, Pythagorean triplets can be a fascinating area of study, with countless applications in fields ranging from architecture to physics.
By understanding the concepts covered in this guide, you will be better equipped to create your own Pythagorean triplets and explore the endless possibilities that they offer. Whether you are a student, a researcher or simply an avid lover of math, this guide is a valuable resource that you can return to time and time again.
Once again, we would like to extend our gratitude for your readership. We value your participation in our community and welcome any feedback or questions you may have. Above all, we wish you continued success in your mathematical pursuits and hope that this guide has sparked your curiosity to explore this fascinating branch of mathematics further.
People also ask about Create Unique and Ordered Pythagorean Triplets: A How-To Guide:
1. What are Pythagorean triplets?
2. Pythagorean triplets are sets of three positive integers that satisfy the Pythagorean theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the lengths of the other two sides.
3. What is the formula for Pythagorean triplets?
4. The formula for Pythagorean triplets is a^2 + b^2 = c^2, where a, b, and c are positive integers.
5. What is the difference between unique and ordered Pythagorean triplets?
6. Unique Pythagorean triplets are sets of three positive integers that satisfy the Pythagorean theorem, where no two triplets have the same combination of numbers. Ordered Pythagorean triplets are sets of three positive integers that satisfy the Pythagorean theorem, where the order of the numbers in the triplet matters.
7. How can I create unique and ordered Pythagorean triplets?
8. To create unique and ordered Pythagorean triplets, you can use the Euclid’s formula method. The formula is as follows:
• Choose any two positive integers m and n, where m > n.
• Calculate a = m^2 – n^2, b = 2mn, and c = m^2 + n^2.
• The result is a unique and ordered Pythagorean triplet (a, b, c). |
# ISEE Middle Level Math : Probability
## Example Questions
### Example Question #51 : Probability
Find the probability of rolling a multiple of 3 on a dice.
Explanation:
To find the probability of an event, we will use the following formula:
Now, given the event of rolling a multiple of 3 on a dice, we can determine the following.
because there are 2 multiples of 3 on a dice:
• 3
• 6
Now, we can also determine the following:
because there are 6 different numbers we could potentially roll:
• 1
• 2
• 3
• 4
• 5
• 6
Knowing this, we can substitute into the formula. We get
Therefore, the probability of rolling a multiple of 3 on a dice is
### Example Question #52 : Probability
Find the probability of picking an even number between 1 and 10 (1 and 10 are included).
Explanation:
To find the probability of an event, we will use the following formula:
Now, given the event of picking an even number between 1 and 10, we can determine the following.
because there are 5 even numbers between 1 and 10:
• 2
• 4
• 6
• 8
• 10
Now, we can also determine the following:
because there are 10 different numbers we could potentially pick:
• 1
• 2
• 3
• 4
• 5
• 6
• 7
• 8
• 9
• 10
Knowing this, we can substitute into the formula. We get
Therefore, the probability of picking an even number between 1 and 10 is .
### Example Question #51 : Probability
Find the probability of rolling a factor of 5 on a dice.
Explanation:
To find the probability of an event, we will use the following formula:
Now, given the event of rolling a factor of 5, we can calculate the following.
because there are 2 factors of 5 on a dice:
• 1
• 5
We can also calculate the following.
because there are 6 different possible numbers we can roll on a dice:
• 1
• 2
• 3
• 4
• 5
• 6
Knowing this, we can substitute into the formula. We get
Therefore, the probability of rolling a factor of 5 on a dice is .
### Example Question #561 : Data Analysis And Probability
Find the probability of rolling a multiple of 2 on a dice.
Explanation:
To find the probability of an event, we will use the following formula:
Now, given the event of rolling a multiple of 2, we can calculate the following.
because there are 3 multiples of 2 on a dice:
• 2
• 4
• 6
We can also calculate the following.
because there are 6 different possible numbers we can roll on a dice:
• 1
• 2
• 3
• 4
• 5
• 6
Knowing this, we can substitute into the formula. We get
Therefore, the probability of rolling a multiple of 2 on a dice is .
### Example Question #52 : Outcomes
Find the probability of drawing an Ace from a deck of cards.
Explanation:
To find the probability of an event, we will use the following formula:
Now, given the event of drawing an Ace from a deck of cards, we can calculate the following.
because there are 4 Aces in a deck of cards that we could draw:
• Ace of Hearts
• Ace of Diamonds
• Ace of Clubs
Now, we can calculate he following.
because there are 52 total cards within a deck that we could potentially draw.
Knowing this, we can substitute into the formula. We get
Therefore, the probability of drawing an Ace from a deck of cards is .
### Example Question #871 : Isee Middle Level (Grades 7 8) Mathematics Achievement
Find the probability of rolling a factor of 2 on a dice.
Explanation:
To find the probability of an event, we will use the following formula:
Now, given the event of rolling a factor of 2 on a dice, we can calculate the following.
because there are 2 factors of 2 on a dice:
• 1
• 2
Now, we can calculate he following.
because there are 6 possible numbers we could potentially roll on a dice:
• 1
• 2
• 3
• 4
• 5
Knowing this, we can substitute into the formula. We get
Therefore, the probability of rolling a factor of 2 on a dice is .
### Example Question #872 : Isee Middle Level (Grades 7 8) Mathematics Achievement
A bag contains the following:
• 3 blue pens
• 2 red pens
• 5 black pens
• 2 pencils
Find the probability of grabbing a pencil out of the bag.
Explanation:
To find the probability of an event, we will use the following formula:
Now, given the event of drawing a pencil out of the bag, we can determine the following:
because there are 2 pencils in the bag.
Now, we can determine the following:
because there are 12 total objects in the bag we could potentially draw:
• 3 blue pens
• 2 red pens
• 5 black pens
• 2 pencils
Now, knowing this, we can substitute into the formula. We get
Therefore, the probability of drawing a pencil out of the bag is .
### Example Question #51 : How To Find The Probability Of An Outcome
A classroom contains 8 boys and 12 girls.
Find the probability the teacher calls on a boy.
Explanation:
To find the probability of an event, we will use the following formula:
Now, given the event of the teacher calling on a boy, we can determine the following:
because there are 8 boys in the classroom.
Now, we can determine the following:
because there are 20 total students in the class the teacher could potentially call on:
• 8 boys
• 12 girls
Now, knowing this, we can substitute into the formula. We get
Therefore, the probability of the teacher calling on a boy is .
### Example Question #562 : Data Analysis And Probability
A bag contains the following:
• 5 red pens
• 3 blue pens
• 4 black pens
Find the probability of grabbing a blue pen from the bag.
Explanation:
To find the probability of an event, we will use the following formula:
Now, given the event of grabbing a blue pen from the bag, we can calculate the following:
because there are 3 blue pens in the bag.
Now, we can calculate the following:
because there are 12 pens in the bag we could potentially grab:
• 5 red pens
• 3 blue pens
• 4 black pens
Knowing this, we can substitute into the formula. We get
Therefore, the probability of grabbing a blue pen from the bag is .
### Example Question #52 : How To Find The Probability Of An Outcome
Find the probability of rolling a factor of 6 on a dice.
Explanation:
To find the probability of an event, we will use the following formula:
Now, given the event of rolling a factor of 6 on a dice, we can calculate the following:
because there are 4 factors of 6 on a dice:
• 1
• 2
• 3
• 6
Now, we can calculate the following:
because there are 6 possible outcomes we could potentially get when rolling a dice:
• 1
• 2
• 3
• 4
• 5
• 6
Knowing this, we can substitute into the formula. We get
Therefore, the probability of rolling a factor of 6 on a dice is . |
# The area of a rhombus is 480 cm2, and one of its diagonals measures 48 cm.
Question:
The area of a rhombus is 480 cm2, and one of its diagonals measures 48 cm. Find
(i) the length of the other diagonal,
(ii) the length of each of its sides, and
(iii) its perimeter.
Solution:
It is given that,
Area of rhombus = 480 cm2.
One of the diagonal = 48 cm.
(i) Area of the rhombus $=\frac{1}{2} \times d_{1} \times d_{2}$
$\Rightarrow 480=\frac{1}{2} \times 48 \times d_{2}$
$\Rightarrow 480=24 \times d_{2}$
$\Rightarrow d_{2}=\frac{480}{24}$
$\Rightarrow d_{2}=20 \mathrm{~cm}$
$\Rightarrow 480=\frac{1}{2} \times 48 \times d_{2}$
$\Rightarrow 480=24 \times d_{2}$
$\Rightarrow d_{2}=\frac{480}{24}$
$\Rightarrow d_{2}=20 \mathrm{~cm}$
Hence, the length of the other diagonal is 20 cm.
(ii) We know that the diagonals of the rhombus bisect each other at right angles.
In right angled ∆ABO,
$A B^{2}=A O^{2}+O B^{2}$ (Pythagoras Theorem)
$\Rightarrow A B^{2}=24^{2}+10^{2}$
$\Rightarrow A B^{2}=576+100$
$\Rightarrow A B^{2}=676$
$\Rightarrow A B=26 \mathrm{~cm}$
Hence, the length of each of the sides of the rhombus is 26 cm.
(iii) Perimeter of the rhombus = 4 × side
= 4 × 26
= 104 cm
Hence, the perimeter of the rhombus is 104 cm. |
# 9th-RATIONAL NUMBERS
Published on May 31, 2013
Author: allwynasir
Source: authorstream.com
PowerPoint Presentation: 5 7 2 1 Rational Numbers S.JASMINE SUGIRTHA 9 th CLASS S.JASMINE SUGIRTHA 1 MATHS PowerPoint Presentation: Rational and Irrational Numbers Rational Numbers A rational number is any number that can be expressed as the ratio of two integers. Examples All terminating and repeating decimals can be expressed in this way so they are irrational numbers. a b 4 5 2 2 3 = 8 3 6 = 6 1 2.7 = 27 10 0.625 = 5 8 34.56 = 3456 100 -3 = 3 1 - 0.3 = 1 3 0.27 = 3 11 0.142857 = 1 7 0.7 = 7 10 S.JASMINE SUGIRTHA 2 MATHS Rational: Rational and Irrational Numbers Rational Numbers A rational number is any number that can be expressed as the ratio of two integers. All terminating and repeating decimals can be expressed in this way so they are irrational numbers. a b Show that the terminating decimals below are rational. 0.6 3.8 56.1 3.45 2.157 6 10 38 10 561 10 345 100 2157 1000 Rational S.JASMINE SUGIRTHA 3 MATHS PowerPoint Presentation: Rational and Irrational Numbers Rational Numbers A rational number is any number that can be expressed as the ratio of two integers. All terminating and repeating decimals can be expressed in this way so they are irrational numbers. a b To show that a repeating decimal is rational. Example 1 To show that 0.333… is rational. Let x = 0.333… 10 x = 3.33… 9 x = 3 x = 3/9 x = 1/3 Example 2 To show that 0.4545… is rational. Let x = 0.4545… 100 x = 45.45… 99 x = 45 x = 45/99 x = 5/11 S.JASMINE SUGIRTHA 4 MATHS PowerPoint Presentation: Rational and Irrational Numbers Rational Numbers A rational number is any number that can be expressed as the ratio of two integers. All terminating and repeating decimals can be expressed in this way so they are irrational numbers. a b Question 1 Show that 0.222… is rational. Let x = 0.222… 10 x = 2.22… 9 x = 2 x = 2/9 Question 2 Show that 0.6363… is rational. Let x = 0.6363… 100 x = 63.63… 99 x = 63 x = 63/99 x = 7/11 S.JASMINE SUGIRTHA 5 MATHS PowerPoint Presentation: Rational and Irrational Numbers Rational Numbers A rational number is any number that can be expressed as the ratio of two integers. All terminating and repeating decimals can be expressed in this way so they are irrational numbers. a b 999 x = 273 x = 273/999 9999 x = 1234 x = 1234/9999 Question 3 Show that 0.273is rational. Let x = 0.273 1000 x = 273.273 x = 91/333 Question 4 Show that 0.1234 is rational. Let x = 0.1234 10000 x = 1234.1234 S.JASMINE SUGIRTHA 6 MATHS PowerPoint Presentation: Rational and Irrational Numbers Rational Numbers A rational number is any number that can be expressed as the ratio of two integers. All terminating and repeating decimals can be expressed in this way so they are irrational numbers. a b By looking at the previous examples can you spot a quick method of determining the rational number for any given repeating decimal. 0.1234 1234 9999 0.273 273 999 0.45 45 99 0.3 3 9 S.JASMINE SUGIRTHA 7 MATHS PowerPoint Presentation: Rational and Irrational Numbers Rational Numbers A rational number is any number that can be expressed as the ratio of two integers. All terminating and repeating decimals can be expressed in this way so they are irrational numbers. a b 0.1234 1234 9999 0.273 273 999 0.45 45 99 0.3 3 9 Write the repeating part of the decimal as the numerator and write the denominator as a sequence of 9’s with the same number of digits as the numerator then simplify where necessary. S.JASMINE SUGIRTHA 8 MATHS PowerPoint Presentation: Rational and Irrational Numbers Rational Numbers A rational number is any number that can be expressed as the ratio of two integers. All terminating and repeating decimals can be expressed in this way so they are irrational numbers. a b 1543 9999 628 999 32 99 7 9 0.1543 0.628 0.32 0.7 Write down the rational form for each of the repeating decimals below. S.JASMINE SUGIRTHA 9 MATHS Irrational: Irrational a b Rational and Irrational Numbers Irrational Numbers An irrational number is any number that cannot be expressed as the ratio of two integers. 1 1 Pythagoras The history of irrational numbers begins with a discovery by the Pythagorean School in ancient Greece. A member of the school discovered that the diagonal of a unit square could not be expressed as the ratio of any two whole numbers. The motto of the school was “All is Number” (by which they meant whole numbers). Pythagoras believed in the absoluteness of whole numbers and could not accept the discovery. The member of the group that made it was Hippasus and he was sentenced to death by drowning. S.JASMINE SUGIRTHA 10 MATHS PowerPoint Presentation: 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 Rational Numbers Irrational Numbers S.JASMINE SUGIRTHA 11 MATHS PowerPoint Presentation: a b Rational and Irrational Numbers Irrational Numbers An irrational number is any number that cannot be expressed as the ratio of two integers. 1 1 Pythagoras Intuition alone may convince you that all points on the “Real Number” line can be constructed from just the infinite set of rational numbers, after all between any two rational numbers we can always find another. It took mathematicians hundreds of years to show that the majority of Real Numbers are in fact irrational. The rationals and irrationals are needed together in order to complete the continuum that is the set of “Real Numbers”. S.JASMINE SUGIRTHA 12 MATHS PowerPoint Presentation: a b Rational and Irrational Numbers Irrational Numbers An irrational number is any number that cannot be expressed as the ratio of two integers. 1 1 Pythagoras Surds are Irrational Numbers We can simplify numbers such as into rational numbers. However, other numbers involving roots such as those shown cannot be reduced to a rational form. Any number of the form which cannot be written as a rational number is called a surd. All irrational numbers are non-terminating , non-repeating decimals. Their decimal expansion form shows no pattern whatsoever. Other irrational numbers include and e , ( Euler’s number ) S.JASMINE SUGIRTHA 13 MATHS PowerPoint Presentation: Rational and Irrational Numbers Multiplication and division of surds. For example: and also for example and S.JASMINE SUGIRTHA 14 MATHS PowerPoint Presentation: Rational and Irrational Numbers Example questions Show that is rational rational Show that is rational rational a b S.JASMINE SUGIRTHA 15 MATHS PowerPoint Presentation: Rational and Irrational Numbers Questions a e State whether each of the following are rational or irrational. b c d f g h irrational rational rational irrational rational rational rational irrational S.JASMINE SUGIRTHA 16 MATHS PowerPoint Presentation: Rational and Irrational Numbers Combining Rationals and Irrationals Addition and subtraction of an integer to an irrational number gives another irrational number, as does multiplication and division. Examples of irrationals S.JASMINE SUGIRTHA 17 MATHS PowerPoint Presentation: Rational and Irrational Numbers Combining Rationals and Irrationals Multiplication and division of an irrational number by another irrational can often lead to a rational number. Examples of Rationals 21 26 8 1 -13 S.JASMINE SUGIRTHA 18 MATHS PowerPoint Presentation: Rational and Irrational Numbers Combining Rationals and Irrationals Determine whether the following are rational or irrational. (a) 0.73 (b) (c) 0.666…. (d) 3.142 (e) (f) (g) (h) (i) (j) (j) (k) (l) irrational rational rational rational irrational irrational irrational rational rational irrational irrational rational rational S.JASMINE SUGIRTHA 19 MATHS END: END THANK U S.JASMINE SUGIRTHA 20 MATHS
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The power rule underlies the Taylor series as it relates a power series with a function's derivatives You can see why this works if you study the example shown. This rule means that you multiply the exponents together and keep the base unchanged. Power Rule. The Power Rule for Exponents. Since differentiation is a linear operation on the space of differentiable functions, polynomials can also be differentiated using this rule. It's Binary! If you're seeing this message, it means we're having trouble loading external resources on our website. Examples. In this tutorial, you'll see how to simplify a monomial raise to a power. Solution: Each factor within the parentheses should be raised to the 2 nd power: (7a 4 b 6) 2 = 7 2 (a 4) 2 (b 6) 2. The Power Rule, one of the most commonly used rules in Calculus, says: The derivative of x n is nx (n-1) Example: What is the derivative of x 2? The "power rule" tells us that to raise a power to a power, just multiply the exponents. To apply the rule… Students learn the power rule, which states that when simplifying a power taken to another power, multiply the exponents. Tutorial 1: Power Rule for Differentiation In the following tutorial we illustrate how the power rule can be used to find the derivative function (gradient function) of a function that can be written $$f(x)=ax^n$$, when $$n$$ is a positive integer. TOP : Product with same base . Examples of the power rule in effect are shown below: x 6 = 6x 5 x 8 = 8x 7 x 3 = 3x 2 x 8 = 8x 7. Raising a Positive Power to a Positive Power. The Power of the Rule of Three in Speech Writing Public Speaking , Speech Writing A well delivered speech, a speech which leaves an impact, always has these small nuances that make it … You may also need the power of a power rule too. You could use the power of a product rule. The Power Rule for exponents states that when we raise a power to a power, we can multiply the exponents. To simplify (6x^6)^2, square the coefficient and multiply the exponent times 2, to get 36x^12. General Sol Example 2 Grade 11 and 12 - … Here you see that 5 2 raised to the 3rd power is equal to 5 6. For any positive number x and integers a and b: $\left(x^{a}\right)^{b}=x^{a\cdot{b}}$.. Take a moment to contrast how this is different from the product rule for exponents found on the previous page. When raising an exponential expression to a new power, multiply the exponents. The power of product rule states that: Laws of Exponents - people.sunyulster.edu. The Power of a Quotient Rule states that the power of a quotient is equal to the quotient obtained when the numerator and denominator are each raised to the indicated power separately, before the division is performed. Power Rule. Power Rule of Exponents (a m) n = a mn. To simplify (6x^6)^2, square the coefficient and multiply the exponent times 2, to get 36x^12 How to use the power rule for derivatives. In calculus, the power rule is used to differentiate functions of the form () =, whenever is a real number. Power examples to rule power. The Tempest includes elements of both tragedy and comedy. The story is set on a remote island, where Prospero, the rightful Duke of Milan, schemes to restore his daughter Miranda to her proper place using manipulation and illusion. - The Power of a Power Rule states (b m) n is equal to b mn. \end{gather*} Taking a number to the power of $\frac{1}{2}$ undoes taking a number to the power … For any real number n, the product of the exponent times x with the exponent reduced by 1 is the derivative of a power of x, which is known as the power rule. In algebraic form, this rule look like this: . By doing so, we have derived the power rule for logarithms which says that the log of a power is equal to the exponent times the log of the base.Keep in mind that, although the input to a logarithm may not be written as a power, we may be able to change it to a power. For example, (x^2)^3 = x^6. A power of a power means you are taking an expression that is already raised to an exponent and raising it to yet another exponent! The way I remember the power rule is take the exponent of a function and move it to the front (to multiply the rest by, including any coefficients), and then take the exponent down a level. Power raised to a power you multiply. It is useful when finding the derivative of a function that is raised to the nth power. 14 interactive practice Problems worked out step by step For example, d/dx x 3 = 3x (3 – 1) = 3x 2 . Thus if we have n 1 x n 2 = n 1 n 2 x n 2-1. The Power Rule is surprisingly simple to work with: Place the exponent in front of “x” and then subtract 1 from the exponent. The a represents the number and n and m represent the powers. Quotient to a power. If there is a coefficient in front of the base, then this coefficient gets multiplied by the value of the exponent. How to use the power rule for derivatives. So "101" is replaced by 1 a, 0 b and 1 c to get us {a,c} Like this: The general power rule is a special case of the chain rule. This video explains the power rule for exponents and is followed by a few examples. Ontario Tech University is the brand name used to refer to the University of Ontario Institute of Technology. Usually the first shortcut rule you study for finding derivatives is the power rule. For a number n, the power rule states: Let’s start with some really easy examples to see it in action. To find a power of a product, find the power of each factor and then multiply. Power of a Power Exponent Rules Help Fun Game Tips: - The fourth power of the third power of two can be written as (2 3) 4. When one power is raised to another, we multiply exponents: This is true for all kinds of exponents, positive and negative (and as we will see later, fractional). Long solution: Short solution: Have a look at some more worked examples: ( ) = = = The reason is that it is a simple rule to remember and it applies to all different kinds of functions. POWER OF A QUOTIENT. Example: Simplify: (7a 4 b 6) 2. Negative powers. If you can write it with an exponents, you probably can apply the power rule. The power rule for integrals allows us to find the indefinite (and later the definite) integrals of a variety of functions like polynomials, functions involving roots, and even some rational functions. The power of power rule \eqref{power_power} allows us to define fractional exponents. Suppose f (x)= x n is a power function, then the power rule is f ′ (x)=nx n-1.This is a shortcut rule to obtain the derivative of a power function. Power of Product Rule. 905.721.8668. Taking a monomial to a power isn't so hard, especially if you watch this tutorial about the power of a monomial rule! It was written around 1610 and it's generally considered Shakespeare's final play as well as the last of his romance plays. The quotient rule tells us that we can divide two powers with the same base by subtracting the exponents. i.e. The general power rule states that this derivative is n times the function raised to the (n-1)th power times the derivative of the function. Notice that we used the product rule for logarithms to simplify the example above. Here is an example of this rule in action. Derivation: Consider the power function f (x) = x n. Zero Rule How Do You Take the Power of a Monomial? Examples of the Power of Three in Headlines. : #(a/b)^n=a^n/b^n# For example: #(3/2)^2=3^2/2^2=9/4# You can test this rule by using numbers that are easy to manipulate: Raising One Power to Another. The law of exponents for a power of an indicated quotient may be developed from the following example : Therefore, The law is stated as follows: The power of a quotient is equal to the quotient obtained when the dividend and divisor are each raised to the indicated power separately, before the division is performed.. Use the power rule to differentiate functions of the form xⁿ where n is a positive integer. B − a times = xa − b. And here is the most amazing thing. So the power rule "works'' in this case, but it's really best to just remember that the derivative of any constant function is zero. To create the Power Set, write down the sequence of binary numbers (using n digits), and then let "1" mean "put the matching member into this subset". Power Rule (Powers to Powers): (a m) n = a mn, this says that to raise a power to a power you need to multiply the exponents.There are several other rules that go along with the power rule, such as the product-to-powers rule and the quotient-to-powers rule. Examples of this include: 7x 8 = 56x 7 10x 6 = 60x 5 Take a moment to contrast how this is different from the product rule …. For example, rule \eqref{power_power} tells us that \begin{gather*} 9^{1/2}=(3^2)^{1/2} = 3^{2 \cdot 1/2} = 3^1 = 3. You can simplify (2 3) 4 = (2 3)(2 3)(2 3)(2 3) to the single power 2 12. Quotient with same base. What is the Power Rule for Exponents? (3 2) 5 = 3 10 The Power Rule - Multiplying Exponents Learn the power rule, which states that when simplifying a power taken to another power, multiply the exponents. Use the power rule to differentiate functions of the form xⁿ where n is a positive integer. Practice problems. Three adjectives, verbs or nouns can add extra power to your headlines: A Brief Guide to Fixing Your Old, Neglected, and Broken Content; How to Stay Healthy, Happy and Combative in Impossible Political Times; 37 Tips for … For x 2 we use the Power Rule with n=2: The derivative of x 2 = 2 x (2-1) = 2x 1 = 2x: Answer: the derivative of x 2 is 2x 2000 Simcoe Street North Oshawa, Ontario L1G 0C5 Canada. Example: quotient to a Power: When raising a fraction to a power, distribute the power to each factor in the numerator and denominator of the fraction. Example: In the example, we have added 2 and 3 together to give us 5. Quotient Rule. So, the solution is 3 to the power of 5. Exercises 3.1 Find the derivatives of the given functions. You will see in a minute why the number of members is a power of 2. Another power, multiply the exponents in the example, ( x^2 ^3! If you watch this tutorial about the power of each factor and then multiply it written. 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# Chris Fox's Engineering Section
## Trigonometry
"The square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the other two sides"
Pythagoras' Rule
Trigonometry is concerned with calculating the sides and angles of triangles. However, as will be seen, it can also be applied to any situation involving circular motion.
Trigonometry is a long-established branch of mathematics in my country; Angles came to England in the 5th century.
The degree (°), is commonly used as a unit of plane angle; there are 360 in a circle. An angle of 90° is termed a right angle; an angle measuring between 0 and 90° is an acute angle, and one measuring between 90 and 180° is an obtuse angle. The SI unit of plane angle is the radian (rad; c). 1c is the angle formed at the centre of a circle by two radii that cut off an arc equal in length to the radius; there are therefore 2πc in a circle. Because the circumference of a circle is given by C=2πr and the area by A=πr², if radians are used the arc length l=rθ and the sector area A=0.5r²θ.
Consider a right-angled triangle (One in which two sides join at a right angle). One of the angles other than the right angle is denoted by θ. The side connecting this angle to the right angle is termed the adjacent side; the side opposite the angle θ is termed the opposite side; and the side opposite the right angle is termed the hypotenuse.
The three basic functions of trigonometry are the sine, cosine and tangent. sin θ = opposite/hypotenuse, cos θ = adjacent/hypotenuse and tan θ = opposite/adjacent. Substitution will show that tan θ = sin θ/cos θ.
Three other functions are the cosecant (cosec), secant (sec) and cotangent (cot). These are the reciprocals of the sine, cosine and tangent functions respectively.
Now consider a circle of radius r, whose centre lies at the origin of a graph. The angle θ denotes the anticlockwise turn from the x-axis to a particular radius. The co-ordinates of the other end of the radius are (r cos θ, r sin θ). This gives rise to the polar co-ordinate system, based on r and θ. Because x=r cos θ and y=r sin θ, r²=x²+y² (Pythagoras' Rule) and θ=tan-1 (y/x). Tan-1 denotes the inverse tangent, so that θ=tan -1 (tan θ).
If the radius is the hypotenuse of a right-angled triangle of which θ is an angle, using Pythagoras' Rule and cancelling shows that sin²θ+cos²θ=1 (a convenient way of writing (sin θ)² ... ). Dividing throughout by cos²θ gives tan²θ+1=sec²θ, and dividing by sin²θ gives 1+cot²θ=cosec²θ.
Consider a triangle with an angle A in the bottom left corner and a right angle in the bottom right corner, rotated anticlockwise through an angle B. Because the angles of a triangle total 180°, and because the angles on a straight line total 180°, the opposite side forms an angle B with the vertical. The B angles can each be considered part of another triangle, whose hypotenuse is also one side of the first. If the hypotenuse of the first triangle has length 1, the lengths of the opposite and adjacent sides are, respectively, sin A and cos A. Considering the co-ordinates of the top corner of the first triangle, and given that opposite sides of a rectangle have equal length:
sin (A+B) = sin A cos B + cos A sin B
cos (A+B) = cos A cos B - sin A sin B |
# Question
$H_0:\ \mbox{X has PDF}\ f_{X|\theta}(x|\theta_0)= \begin{cases} 2x & \mbox{for }0 \le x \le 1 \\ 0 & \mbox{else} \end{cases}$
$H_1:\ \mbox{X has PDF}\ f_{X|\theta}(x|\theta_1)= \begin{cases} 1 & \mbox{for }0 \le x \le 1 \\ 0 & \mbox{else} \end{cases}$
Find the probability of type I errors and type II errors using the ML rule.
To find the ML Rule we say pick $H_1\!$ if $f_{X|\theta}(x|\theta_1)>f_{X|\theta}(x|\theta_0)\!$
Or in otherwords pick $H_1\!$ if $1>2x\!$ Thus,
$\mbox{ML Rule: } \begin{cases} \mbox{say }H_1 &\mbox{if }x<1/2 \\ \mbox{say }H_0 &\mbox{else} \end{cases}$
Type I Error: A Type I error is the $Pr[\mbox{say } H_1|H_0]\!$ this is equivalent to saying $Pr[x<1/2|\theta=\theta_0]\!$ we calculate this using integration
$Pr[x<1/2|\theta=\theta_0] = \int_{0}^{1/2}2x\, dx=1/4\!$
Type II Error:A Type II error is the $Pr[\mbox{say } H_0|H_1]\!$ this is equivalent to saying $Pr[x\ge1/2|\theta=\theta_1]$ we calculate this using integration
$Pr[x\ge1/2|\theta=\theta_1] = \int_{1/2}^{1}1\, dx=1/2\!$
## Alumni Liaison
Correspondence Chess Grandmaster and Purdue Alumni
Prof. Dan Fleetwood |
## Introductory Algebra for College Students (7th Edition)
The angle we are looking for is $53$ degrees.
We know that complementary angles add up to $90$ degrees. Let us set one angle equal to $x$. Since the other angle is 16 degrees more than its complement, we will set this angle as $x + 16$. The equation we can use is: $$x + (x + 16) = 90$$ Distribute what is in parentheses to simplify the problem: $$x + x + 16 = 90$$ Simplify by adding the $x$ terms together: $$2x + 16 = 90$$ Subtract $16$ from both sides of the equation to isolate constants on the other side of the equation: $$2x = 74$$ Solve for $x$ by dividing both sides of the equation by $2$: $$x = 37$$ If one angle is $37$ degrees, then we can figure out the other angle is $16$ more than this angle: 2nd angle = $37 + 16$ The 2nd angle is $53$ degrees. |
# Algebra II : Transformations
## Example Questions
← Previous 1
### Example Question #1 : Transformations
How is the graph of different from the graph of ?
is narrower than and is shifted up 3 units
is wider than and is shifted down 3 units
is wider than and is shifted to the right 3 units
is narrower than and is shifted down 3 units
is narrower than and is shifted to the left 3 units
is narrower than and is shifted down 3 units
Explanation:
Almost all transformed functions can be written like this:
where is the parent function. In this case, our parent function is , so we can write this way:
Luckily, for this problem, we only have to worry about and .
represents the vertical stretch factor of the graph.
• If is less than 1, the graph has been vertically compressed by a factor of . It's almost as if someone squished the graph while their hands were on the top and bottom. This would make a parabola, for example, look wider.
• If is greater than 1, the graph has been vertically stretched by a factor of . It's almost as if someone pulled on the graph while their hands were grasping the top and bottom. This would make a parabola, for example, look narrower.
represents the vertical translation of the graph.
• If is positive, the graph has been shifted up units.
• If is negative, the graph has been shifted down units.
For this problem, is 4 and is -3, causing vertical stretch by a factor of 4 and a vertical translation down 3 units.
### Example Question #2 : Transformations
Which of the following transformations represents a parabola shifted to the right by 4 and halved in width?
Explanation:
Begin with the standard equation for a parabola: .
Horizontal shifts are represented by additions (leftward shifts) or subtractions (rightward shifts) within the parenthesis of the term. To shift 4 units to the right, subtract 4 within the parenthesis.
The width of the parabola is determined by the magnitude of the coefficient in front of . To make a parabola narrower, use a whole number coefficient. Halving the width indicates a coefficient of two.
### Example Question #3 : Transformations
Which of the following represents a standard parabola shifted up by 2 units?
Explanation:
Begin with the standard equation for a parabola: .
Vertical shifts to this standard equation are represented by additions (upward shifts) or subtractions (downward shifts) added to the end of the equation. To shift upward 2 units, add 2.
### Example Question #4 : Transformations
Which of the following transformation flips a parabola vertically, doubles its width, and shifts it up by 3?
Explanation:
Begin with the standard equation for a parabola: .
Inverting, or flipping, a parabola refers to the sign in front of the coefficient of the . If the coefficient is negative, then the parabola opens downward.
The width of the parabola is determined by the magnitude of the coefficient in front of . To make a parabola wider, use a fractional coefficient. Doubling the width indicates a coefficient of one-half.
Vertical shifts are represented by additions (upward shifts) or subtractions (downward shifts) added to the end of the equation. To shift upward 3 units, add 3.
### Example Question #5 : Transformations
Which of the following shifts a parabola six units to the right and five downward?
Explanation:
Begin with the standard equation for a parabola: .
Vertical shifts to this standard equation are represented by additions (upward shifts) or subtractions (downward shifts) added to the end of the equation. To shift downward 5 units, subtract 5.
Horizontal shifts are represented by additions (leftward shifts) or subtractions (rightward shifts) within the parenthesis of the term. To shift 6 units to the right, subtract 6 within the parenthesis.
### Example Question #6 : Transformations
Which of the following transformations represents a parabola that has been flipped vertically, shifted to the right 12, and shifted downward 4?
Explanation:
Begin with the standard equation for a parabola: .
Horizontal shifts are represented by additions (leftward shifts) or subtractions (rightward shifts) within the parenthesis of the term. To shift 12 units to the right, subtract 12 within the parenthesis.
Inverting, or flipping, a parabola refers to the sign in front of the coefficient of the . If the coefficient is negative, then the parabola opens downward.
Vertical shifts are represented by additions (upward shifts) or subtractions (downward shifts) added to the end of the equation. To shift downward 4 units, subtract 4.
### Example Question #7 : Transformations
Which of the following transformations represents a parabola that has been shifted 4 units to the left, 5 units down, and quadrupled in width?
Explanation:
Begin with the standard equation for a parabola: .
Horizontal shifts are represented by additions (leftward shifts) or subtractions (rightward shifts) within the parenthesis of the term. To shift 4 units to the left, add 4 within the parenthesis.
The width of the parabola is determined by the magnitude of the coefficient in front of . To make a parabola wider, use a fractional coefficient. Doubling the width indicates a coefficient of one-fourth.
Vertical shifts are represented by additions (upward shifts) or subtractions (downward shifts) added to the end of the equation. To shift downward 5 units, subtract 5.
### Example Question #8 : Transformations
If the function is shifted left 2 units, and up 3 units, what is the new equation?
Explanation:
Shifting left 2 units will change the y-intercept from to .
The new equation after shifting left 2 units is:
Shifting up 3 units will add 3 to the y-intercept of the new equation.
### Example Question #9 : Transformations
If , what is ?
It is the parabola shifted to the right by 1.
It is the parabola reflected across the x-axis.
It is the same as .
It is the parabola reflected across the y-axis.
It is the function.
It is the parabola reflected across the x-axis.
Explanation:
It helps to evaluate the expression algebraically.
. Any time you multiply a function by a -1, you reflect it over the x axis. It helps to graph for verification.
This is the graph of
and this is the graph of
### Example Question #10 : Transformations
If , what is ?
It is the graph rotated about the origin.
It is the graph reflected across the y-axis.
It is the graph reflected across the x-axis.
It is the graph.
It is the graph shifted 1 to the right.
It is the graph reflected across the y-axis.
Explanation:
Algebraically, .
This is a reflection across the y axis.
This is the graph of
And this is the graph of
← Previous 1 |
Tangent Circles
by:
Brandt Hacker
In this assignment we are asked to examine the following scenario: Given two circles and a point on one of the circles. Construct a circle tangent to the two circles with one point of tangency being the designated point.
In order to tackle this question we must first understand what tangent circles are. Two circles are said to be tangent to one another if they intersect at a single point. If you would like to see a further explanation of this concept as well as several illustrations of tangent circles, click http://en.wikipedia.org/wiki/Tangent_circles.
The problem asks us to construct a circle tangent to two other circles. We will attempt to do this by first drawing our two circles. Then, we will draw a line through center of circle A that passes through the point C which we had drawn on the circumference of circle A. After constructing the line through points A and C, then construct a circle with the same radius as circle B, centered at C.
1. 2.
The two original circles are pictured in green. We will use the newly created red circle to try and accomplish the original goal of creating a circle that is tangent to both circle A and circle B. After completing the constructions seen above, we will then mark the point of intersection between circle C and line AC outside the circle, letŐs call this point D. After creating point D, we will then want to make a perpendicular bisector of the segment running from B to C, the intersection o this perpendicular bisector and line AC is the center of our tangent circle (pictured in black).
3. 4.
Below is an image of the tangent circle with the rest of the construction hidden:
We will now use this construction to examine what happens to the center of the tangent circle as we rotate point C around circle A. For GeometerŐs Sketchpad file that allows you do this on your own, click HERE(A7tangentinside).
For the first scenario we will look at what happens as we rotate point C around circle A when circle B is inside circle A. We will trace point E, the center of the tangent circle, as we rotate point C around circle A. When traced we see that point E creates an ellipse with points B and A, the centers of the two original circles, as the foci of the ellipse.
Secondly, we will look at what happens as we rotate point C around circle A when circle B intersects circle A. We will trace point E, the center of the tangent circle, as we rotate point C around circle A. When traced we see that point E, once again, creates an ellipse with points B and A, the centers of the two original circles, as the foci of the ellipse.
Lastly, we will look at what happens as we rotate point C around circle A when circle B is located outside circle A. We will trace point E, the center of the tangent circle, as we rotate point C around circle A. When traced we see that this time, point E creates a hyperbola with points A and B as the foci of the hyperbola. |
Tea Story
# Problem Statement
Origin The Mathematical Association of Thailand Contest 2007, High School Level, Problem 35 (write down the answer only). Note that this is a loose translation.
Eighty students with IDs 1, 2, ..., 80 take turns in the order of increasing ID to write numbers on a blackboard. During the first round, each student writes down her own ID. During the subsequent rounds, each student computes the arithmetic mean of the numbers on the board, and then replaces the number she wrote in the previous round with the computed result. As they perform this process, all numbers on the board converge to the same constant. Find that constant.
# Solution
After the contest, I went to China with my family during the New Year's holiday. The Chinese tour we went with imposed mandatory visits to local factories (sometimes called "scams") as usual. One of them is a tea factory.
The staff prepared the following tools for making tea:
First, the staff made a jug of tea by pouring hot water through the tea filter. To double its concentration, she poured the tea through the filter once more.
Suppose each jug holds 2 units of tea. The numbers outside parentheses are concentrations, while the numbers inside are quantities.
Next, the staff made less concentrated tea in the first jug, and then poured away half of it (= 1 unit); this way, the concentration was preserved.
Then the staff mixed the tea by alternatingly pouring 1 unit of tea from one jug into the other.
Observe that after each pour, the new concentration is the average of the two old concentrations.
Now suppose she continued mixing until the two jugs have roughly the same concentration $x$. This is the classic mixture problem: the total mass (= volume × concentration) must remain constant. So we get $$1\times 1 + 2\times 2 = (1+2)\times x$$ which makes $x = 5/3$.
Now imagine 80 jugs. Each jug $i$ ($i = 1,\dots,80$) starts with $i$ units of tea with concentration $i$. Then at each step, we find the jug $J$ with 1 unit of tea, and then pour 1 unit of tea from each of the other jugs into $J$. Now $J$ will have 80 units of tea, while the other jugs which had $2,3,\dots,80$ units of tea will now have $1,2,\dots,79$ units. Moreover, the concentration in $J$ will become the arithmetic mean of the previous concentrations in all 80 jugs. Essentially, the concentrations exactly match the numbers the students write on the board.
So if all concentrations converge to $x$, we get $$1\times 1 + \dots + 80\times 80 = (1 + \dots + 80)\times x$$ giving $x = 161/3$. (Luckily, we don't have to prove that the numbers converge here.)
# Epilogue
I wrote this explanation in my (now defunct) blog in 2008 and also in my college essay. The illustrations above were salvaged from my old CD-R, but the original text was lost.
Exported: 2017-06-10T03:10:49.476000 |
# How do you simplify (8sqrt2)/(2sqrt8)?
Mar 19, 2018
$\implies \pm 2$
#### Explanation:
We can't factor out a perfect square out of the numerator, so it'll remain the same. Let's see what we can do with the denominator.
We can factor $\sqrt{8}$ into $\sqrt{2} \cdot \sqrt{4}$. Thus we have:
$\frac{8 \cancel{\sqrt{2}}}{2 \cancel{\sqrt{2}} \cdot \sqrt{4}}$
$\sqrt{2}$ obviously cancels with itself, and we get:
$\frac{8}{2 \sqrt{4}}$
Which can be simplified to
$\frac{8}{2 \cdot \pm 2}$
$\implies \frac{8}{\pm 4}$
$\implies \pm 2$ |
How to Integrate by Parts when One Function Is a Polynomial
Author Info
Updated: January 2, 2017
Integration by parts is a technique used to integrate a product of two functions. The formula is
${\displaystyle \int u\mathrm {d} v=uv-\int v\mathrm {d} u}$
where we choose a ${\displaystyle u}$ and ${\displaystyle \mathrm {d} v}$ and calculate ${\displaystyle \mathrm {d} u}$ and ${\displaystyle v.}$ When one function is a polynomial, then we choose ${\displaystyle u}$ to be the polynomial, so that we get an integral that is simpler to evaluate.
We could do this multiple times, but we will also show a tabular method that does this more efficiently.
Method 1 of 2: Repeated Integration by Parts
1. 1
Evaluate the integral below. The integrand is a product of two functions: ${\displaystyle x^{2}}$ and ${\displaystyle \sin x.}$
• ${\displaystyle \int x^{2}\sin x\mathrm {d} x}$
2. 2
Choose and . Differentiate ${\displaystyle u}$ to find ${\displaystyle \mathrm {d} u=2x\mathrm {d} x}$ and integrate ${\displaystyle \mathrm {d} v}$ to get ${\displaystyle v=-\cos x.}$ We can drop the constant of integration from finding ${\displaystyle v}$ because it will eventually drop out in the end. Then we just plug all of these into the formula.
• ${\displaystyle \int x^{2}\sin x\mathrm {d} x=-x^{2}\cos x+\int 2x\cos x\mathrm {d} x}$
• Notice that the second integral is positive because of the ${\displaystyle -\cos x.}$
3. 3
Choose and . Notice that we need to use integration by parts on the second integral. So we will have to choose a new ${\displaystyle u}$ and ${\displaystyle \mathrm {d} v.}$ We then find that ${\displaystyle \mathrm {d} u=2\mathrm {d} x}$ and ${\displaystyle v=\sin x.}$ Plug these into the formula. Make sure to keep track that these new variables apply only to the second integral.
• ${\displaystyle \int x^{2}\sin x\mathrm {d} x=-x^{2}\cos x+2x\sin x-\int 2\sin x\mathrm {d} x}$
• We now see that the integral on the right is something that we can directly evaluate.
4. 4
Evaluate the integral on the right. The antiderivative of ${\displaystyle \sin x}$ is ${\displaystyle -\cos x.}$ Make sure to add the constant of integration ${\displaystyle C.}$
• ${\displaystyle \int x^{2}\sin x\mathrm {d} x=-x^{2}\cos x+2x\sin x+2\cos x+C}$
Method 2 of 2: Tabular Integration
Example 1
1. 1
Evaluate the integral below. This is a similar integral as before, with the only modification being the ${\displaystyle x^{3}}$ instead of ${\displaystyle x^{2}.}$ If we wanted to integrate by parts manually, we would have to do it three times, and it only increases with the degree of the polynomial. Obviously, there is a more efficient method to integrating something like this.
• ${\displaystyle \int x^{3}\sin x\mathrm {d} x}$
2. 2
Write out a table with two columns. The first column denotes ${\displaystyle u}$ and its derivatives. The second column denotes ${\displaystyle \mathrm {d} v}$ and its antiderivatives. Write them out until you reach 0.
• ${\displaystyle {\begin{array}{c|c}x^{3}&\sin x\\3x^{2}&-\cos x\\6x&-\sin x\\6&\cos x\\0&\sin x\end{array}}}$
3. 3
Multiply the two columns while alternating signs. The way to do it is by multiplying the 1st entry (in the ${\displaystyle u}$ column) with the 2nd entry (in the ${\displaystyle \mathrm {d} v}$ column), the 2nd entry with the 3rd entry, etc. with alternating signs. So the first pairing is positive, the second pairing is negative, the third pairing is positive, etc.
• ${\displaystyle (+)(x^{3})(-\cos x)-(3x^{2})(-\sin x)+(6x)(\cos x)-(6)(\sin x)+C}$
4. 4
Simplify. Make sure to keep track of your negative signs, and don't forget to add ${\displaystyle C}$ if you are evaluating an indefinite integral.
• ${\displaystyle \int x^{3}\sin x\mathrm {d} x=-x^{3}\cos x+3x^{2}\sin x+6x\cos x-6\sin x+C}$
Example 2
1. 1
Evaluate the integral below. We show another example where the tabular method is a good method to avoid directly integrating by parts.
• ${\displaystyle \int x^{2}e^{-2x}\mathrm {d} x}$
2. 2
Write out the table. ${\displaystyle u=x^{2}}$ and ${\displaystyle \mathrm {d} v=e^{-2x}\mathrm {d} x.}$ So take derivatives of ${\displaystyle u}$ until you reach 0 and take integrals of ${\displaystyle e^{-2x}}$ until you reach the same row as 0.
• ${\displaystyle {\begin{array}{c|c}x^{2}&e^{-2x}\\2x&-{\frac {1}{2}}e^{-2x}\\2&{\frac {1}{4}}e^{-2x}\\0&-{\frac {1}{8}}e^{-2x}\end{array}}}$
3. 3
Pair the entries while alternating signs. Just like before, multiply entry 1 in column 1 with entry 2 in column 2, entry 2 with entry 3, etc. all while alternating signs.
• ${\displaystyle (+)(x^{2})\left(-{\frac {1}{2}}e^{-x^{2}}\right)-(2x)\left({\frac {1}{4}}e^{-2x}\right)+(2)\left(-{\frac {1}{8}}e^{-2x}\right)+C}$
4. 4
Simplify. We can pull out an ${\displaystyle e^{-2x}}$ and a ${\displaystyle -{\frac {1}{4}}.}$
• ${\displaystyle \int x^{2}e^{-2x}\mathrm {d} x=-{\frac {1}{4}}e^{-2x}(2x^{2}+2x+1)+C}$
Community Q&A
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7.3 Unit circle
Page 1 / 11
In this section you will:
• Find function values for the sine and cosine of and
• Identify the domain and range of sine and cosine functions.
• Find reference angles.
• Use reference angles to evaluate trigonometric functions.
Looking for a thrill? Then consider a ride on the Singapore Flyer, the world’s tallest Ferris wheel. Located in Singapore, the Ferris wheel soars to a height of 541 feet—a little more than a tenth of a mile! Described as an observation wheel, riders enjoy spectacular views as they travel from the ground to the peak and down again in a repeating pattern. In this section, we will examine this type of revolving motion around a circle. To do so, we need to define the type of circle first, and then place that circle on a coordinate system. Then we can discuss circular motion in terms of the coordinate pairs.
Finding trigonometric functions using the unit circle
We have already defined the trigonometric functions in terms of right triangles. In this section, we will redefine them in terms of the unit circle. Recall this a unit circle is a circle centered at the origin with radius 1, as shown in [link] . The angle (in radians) that $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ intercepts forms an arc of length $\text{\hspace{0.17em}}s.\text{\hspace{0.17em}}$ Using the formula $\text{\hspace{0.17em}}s=rt,$ and knowing that $\text{\hspace{0.17em}}r=1,$ we see that for a unit circle, $\text{\hspace{0.17em}}s=t.$
The x- and y- axes divide the coordinate plane into four quarters called quadrants. We label these quadrants to mimic the direction a positive angle would sweep. The four quadrants are labeled I, II, III, and IV.
For any angle $\text{\hspace{0.17em}}t,$ we can label the intersection of the terminal side and the unit circle as by its coordinates, $\text{\hspace{0.17em}}\left(x,y\right).\text{\hspace{0.17em}}$ The coordinates $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ will be the outputs of the trigonometric functions $\text{\hspace{0.17em}}f\left(t\right)=\mathrm{cos}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}f\left(t\right)=\mathrm{sin}\text{\hspace{0.17em}}t,$ respectively. This means and
Unit circle
A unit circle has a center at $\text{\hspace{0.17em}}\left(0,0\right)\text{\hspace{0.17em}}$ and radius $\text{\hspace{0.17em}}1.\text{\hspace{0.17em}}$ In a unit circle, the length of the intercepted arc is equal to the radian measure of the central angle $\text{\hspace{0.17em}}t.$
Let $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ be the endpoint on the unit circle of an arc of arc length $\text{\hspace{0.17em}}s.\text{\hspace{0.17em}}$ The $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ coordinates of this point can be described as functions of the angle.
Defining sine and cosine functions from the unit circle
The sine function relates a real number $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ to the y -coordinate of the point where the corresponding angle intercepts the unit circle. More precisely, the sine of an angle $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ equals the y -value of the endpoint on the unit circle of an arc of length $\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ In [link] , the sine is equal to $\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$ Like all functions, the sine function has an input and an output. Its input is the measure of the angle; its output is the y -coordinate of the corresponding point on the unit circle.
The cosine function of an angle $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ equals the x -value of the endpoint on the unit circle of an arc of length $\text{\hspace{0.17em}}t.\text{\hspace{0.17em}}$ In [link] , the cosine is equal to $\text{\hspace{0.17em}}x.$
Because it is understood that sine and cosine are functions, we do not always need to write them with parentheses: $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is the same as $\text{\hspace{0.17em}}\mathrm{sin}\left(t\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}t\text{\hspace{0.17em}}$ is the same as $\text{\hspace{0.17em}}\mathrm{cos}\left(t\right).\text{\hspace{0.17em}}$ Likewise, $\text{\hspace{0.17em}}{\mathrm{cos}}^{2}t\text{\hspace{0.17em}}$ is a commonly used shorthand notation for $\text{\hspace{0.17em}}{\left(\mathrm{cos}\left(t\right)\right)}^{2}.\text{\hspace{0.17em}}$ Be aware that many calculators and computers do not recognize the shorthand notation. When in doubt, use the extra parentheses when entering calculations into a calculator or computer.
A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5) and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
The sequence is {1,-1,1-1.....} has
how can we solve this problem
Sin(A+B) = sinBcosA+cosBsinA
Prove it
Eseka
Eseka
hi
Joel
June needs 45 gallons of punch. 2 different coolers. Bigger cooler is 5 times as large as smaller cooler. How many gallons in each cooler?
7.5 and 37.5
Nando
find the sum of 28th term of the AP 3+10+17+---------
I think you should say "28 terms" instead of "28th term"
Vedant
the 28th term is 175
Nando
192
Kenneth
if sequence sn is a such that sn>0 for all n and lim sn=0than prove that lim (s1 s2............ sn) ke hole power n =n
write down the polynomial function with root 1/3,2,-3 with solution
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions |
# 16.3 Adding and subtracting vectors (Page 2/2)
Page 2 / 2
## Resultant vectors
1. Harold walks to school by walking 600 m Northeast and then 500 m N $40{}^{\circ }$ W. Determine his resultant displacement by using accurate scale drawings.
2. A dove flies from her nest, looking for food for her chick. She flies at a velocity of $2\mathrm{m}·\mathrm{s}{}^{-1}$ on a bearing of $135{}^{\circ }$ and then at a velocity of $1,2\phantom{\rule{2pt}{0ex}}\mathrm{m}·\mathrm{s}{}^{-1}$ on a bearing of $230{}^{\circ }$ . Calculate her resultant velocity by using accurate scale drawings.
3. A squash ball is dropped to the floor with an initial velocity of $2,5\phantom{\rule{2pt}{0ex}}\mathrm{m}·\mathrm{s}{}^{-1}$ . It rebounds (comes back up) with a velocity of $0,5\phantom{\rule{2pt}{0ex}}\mathrm{m}·\mathrm{s}{}^{-1}$ .
1. What is the change in velocity of the squash ball?
2. What is the resultant velocity of the squash ball?
Remember that the technique of addition and subtraction just discussed can only be applied to vectors acting along a straight line. When vectors are not in a straight line, i.e. at an angle to each other, the following method can be used:
## A more general algebraic technique
Simple geometric and trigonometric techniques can be used to find resultant vectors.
A man walks 40 m East, then 30 m North. Calculate the man's resultant displacement.
1. As before, the rough sketch looks as follows:
2. Note that the triangle formed by his separate displacement vectors and his resultant displacement vector is a right-angle triangle. We can thus use the Theorem of Pythagoras to determine the length of the resultant. Let ${x}_{R}$ represent the length of the resultant vector. Then:
$\begin{array}{ccc}\hfill {x}_{R}^{2}& =& {\left(40\phantom{\rule{2pt}{0ex}}\mathrm{m}\right)}^{2}+{\left(30\phantom{\rule{2pt}{0ex}}\mathrm{m}\right)}^{2}\hfill \\ \hfill {x}_{R}^{2}& =& 2\phantom{\rule{4pt}{0ex}}500\phantom{\rule{2pt}{0ex}}{\mathrm{m}}^{2}\hfill \\ \hfill {x}_{R}& =& 50\phantom{\rule{2pt}{0ex}}\mathrm{m}\hfill \end{array}$
3. Now we have the length of the resultant displacement vector but not yet its direction. To determine its direction we calculate the angle $\alpha$ between the resultant displacement vector and East, by using simple trigonometry:
$\begin{array}{ccc}\hfill tan\alpha & =& \frac{\mathrm{opposite side}}{\mathrm{adjacent side}}\hfill \\ \hfill tan\alpha & =& \frac{30}{40}\hfill \\ \hfill \alpha & =& {tan}^{-1}\left(0,75\right)\hfill \\ \hfill \alpha & =& 36,{9}^{\circ }\hfill \end{array}$
4. The resultant displacement is then 50 m at $36,9{}^{\circ }$ North of East.
This is exactly the same answer we arrived at after drawing a scale diagram!
In the previous example we were able to use simple trigonometry to calculate the resultant displacement. This was possible since thedirections of motion were perpendicular (north and east). Algebraic techniques, however, are not limited to cases where the vectors to be combined are along the same straight line or at right angles to oneanother. The following example illustrates this.
A man walks from point A to point B which is 12 km away on a bearing of $45{}^{\circ }$ . From point B the man walks a further 8 km east to point C. Calculate the resultant displacement.
1. $B\stackrel{^}{A}F={45}^{\circ }$ since the man walks initially on a bearing of $45{}^{\circ }$ . Then, $A\stackrel{^}{B}G=B\stackrel{^}{A}F={45}^{\circ }$ (parallel lines, alternate angles). Both of these angles are included in the rough sketch.
2. The resultant is the vector AC. Since we know both the lengths of AB and BC and the included angle $A\stackrel{^}{B}C$ , we can use the cosine rule:
$\begin{array}{ccc}\hfill A{C}^{2}& =& A{B}^{2}+B{C}^{2}-2·AB·BCcos\left(A\stackrel{^}{B}C\right)\hfill \\ & =& {\left(12\right)}^{2}+{\left(8\right)}^{2}-2·\left(12\right)\left(8\right)cos\left({135}^{\circ }\right)\hfill \\ & =& 343,8\hfill \\ \hfill AC& =& 18,5\phantom{\rule{2pt}{0ex}}\mathrm{km}\hfill \end{array}$
3. Next we use the sine rule to determine the angle $\theta$ :
$\begin{array}{ccc}\hfill \frac{sin\theta }{8}& =& \frac{sin{135}^{\circ }}{18,5}\hfill \\ \hfill sin\theta & =& \frac{8×sin{135}^{\circ }}{18,5}\hfill \\ \hfill \theta & =& {sin}^{-1}\left(0,3058\right)\hfill \\ \hfill \theta & =& 17,{8}^{\circ }\hfill \end{array}$
To find $F\stackrel{^}{A}C$ , we add $45{}^{\circ }$ . Thus, $F\stackrel{^}{A}C=62,{8}^{\circ }$ .
4. The resultant displacement is therefore 18,5 km on a bearing of $062,8{}^{\circ }$ .
## More resultant vectors
1. A frog is trying to cross a river. It swims at $3\phantom{\rule{2pt}{0ex}}\mathrm{m}·s{}^{-1}$ in a northerly direction towards the opposite bank. The water is flowing in a westerly direction at $5\phantom{\rule{2pt}{0ex}}\mathrm{m}·s{}^{-1}$ . Find the frog's resultant velocity by using appropriate calculations. Include a rough sketch of the situation in your answer.
2. Sandra walks to the shop by walking 500 m Northwest and then 400 m N $30{}^{\circ }$ E. Determine her resultant displacement by doing appropriate calculations.
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yeah
Joseph
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no can't
Lohitha
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
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da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
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what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
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Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
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Alexandre
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is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
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Damian
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What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
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Rafiq
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Mahi
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Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
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write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
how did you get the value of 2000N.What calculations are needed to arrive at it
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Richard is sitting on his chair and reading a newspaper three (3) meters away from the door
The fundamental frequency of a sonometer wire streached by a load of relative density 's'are n¹ and n² when the load is in air and completly immersed in water respectively then the lation n²/na is
Properties of longitudinal waves |
# 8.02 Special right triangles and exact value trigonometric ratios
Lesson
### $45^\circ-45^\circ-90^\circ$45°−45°−90° Triangles
While the Pythagorean Theorem can apply to any kind of right triangle, there are particular types of right triangles whose side lengths and angles have helpful properties.
The first kind of triangle that we want to look at is the $45^\circ-45^\circ-90^\circ$45°45°90° triangle.
The first thing to notice about this kind of triangle is that it has two congruent angles and is, therefore, an isosceles triangle. This, in turn, means that the two sides of the triangle that are not the hypotenuse are equal in length. That is, $a=b$a=b in the diagram above.
Knowing this fact means that we can also use the Pythagorean Theorem to find out what the length of the hypotenuse should be. For example, let's look at the case where $a=b=1$a=b=1.
We can then find the value of $c$c in the following way.
$1^2+1^2$12+12 $=$= $c^2$c2 Using the Pythagorean Theorem $1+1$1+1 $=$= $c^2$c2 Simplifying the powers $c^2$c2 $=$= $2$2 Evaluating the addition $c$c $=$= $\sqrt{2}$√2 Taking the positive square root of both sides
So, here we have a $45^\circ-45^\circ-90^\circ$45°45°90° with all side lengths filled in:
Note that any triangle with angle measures of $45^\circ$45°, $45^\circ$45° and $90^\circ$90° is necessarily similar to this triangle. Using what we know about similar triangles, this means that the sides of any $45^\circ-45^\circ-90^\circ$45°45°90° triangle are in the ratio $1:1:\sqrt{2}$1:1:2.
Summary
In any $45^\circ-45^\circ-90^\circ$45°45°90° triangle, the hypotenuse is $\sqrt{2}$2 times as long as either of the other sides of the triangle.
That is, the ratio of side lengths in a $45^\circ-45^\circ-90^\circ$45°45°90° triangle is always $1:1:\sqrt{2}$1:1:2.
#### Practice questions
##### Question 1
Consider the triangle below.
1. Find the exact value of $a$a.
2. Find the exact value of $c$c.
##### Question 2
Consider the triangle below.
1. Find the exact value of $a$a.
2. Find the value of $b$b.
##### Question 3
Consider the triangle below.
1. Find the exact value of $b$b.
2. Find the exact value of $c$c.
### $30^\circ-60^\circ-90^\circ$30°−60°−90° Triangles
Just like the $45^\circ$45°-$45^\circ$45°-$90^\circ$90° triangle, the $30^\circ$30°-$60^\circ$60°-$90^\circ$90° triangle is another special type of right triangle. We can construct a $30^\circ$30°-$60^\circ$60°-$90^\circ$90° triangle by starting with an equilateral triangle and cutting it into two halves.
Here is an equilateral triangle:
To form a $30^\circ$30°-$60^\circ$60°-$90^\circ$90° triangle, we draw in an altitude from any vertex. In fact, we get two congruent $30^\circ$30°-$60^\circ$60°-$90^\circ$90° triangles by doing so:
In order to look at the relationships between the side lengths of such a triangle, let's suppose that the initial equilateral triangle $\triangle ABC$ABC had side lengths of $2$2 units each. Since the altitude $\overline{AD}$AD bisects the side $\overline{BC}$BC, this means that the length of the short side $\overline{BD}$BD will be $1$1 unit. Here is this information on the $30^\circ$30°-$60^\circ$60°-$90^\circ$90° triangle $\triangle ABD$ABD:
There is only one unknown side length left on this triangle, $AD$AD, which has been labeled $x$x on the diagram above. Since $\triangle ABD$ABD is a right triangle, we can use the Pythagorean Theorem to find this length:
$1^2+x^2$12+x2 $=$= $2^2$22 Using the Pythagorean Theorem $1+x^2$1+x2 $=$= $4$4 Simplifying $x^2$x2 $=$= $3$3 Subtracting $1$1 from both sides $x$x $=$= $\sqrt{3}$√3 Taking the positive square root of both sides
Here is a $30^\circ$30°-$60^\circ$60°-$90^\circ$90° with all side lengths filled in:
Note that any triangle with angle measures of $30^\circ$30°, $60^\circ$60° and $90^\circ$90° is necessarily similar to this triangle. Using what we know about similar triangles, this means that the sides of any $30^\circ$30°-$60^\circ$60°-$90^\circ$90° triangle are in the ratio $1:\sqrt{3}:2$1:3:2.
Summary
In any $30^\circ$30°-$60^\circ$60°-$90^\circ$90° triangle, the longer leg is $\sqrt{3}$3 times as long as the shorter leg, and the hypotenuse is $2$2 times as long as the shorter leg.
That is, the ratio of side lengths in a $30^\circ$30°-$60^\circ$60°-$90^\circ$90° triangle is always $1:\sqrt{3}:2$1:3:2.
#### Practice questions
##### Question 4
Consider the right triangle below.
1. Use the Pythagorean Theorem to find the missing side length $b$b.
2. Using the fact that the sum of the interior angle measures of a triangle is $180^\circ$180°, find the value of $\theta$θ.
##### Question 5
Consider the triangle below.
1. Determine the value of $c$c.
2. Determine the value of $b$b.
##### Question 6
Consider the triangle below.
1. Determine the value of $a$a.
2. Determine the value of $b$b.
### Trigonometric ratios for special triangles
You'll have noticed by now that when you find angles using trigonometric ratios, you often get long decimal answers. If, for example, you put $\cos30^\circ$cos30° into the calculator, you will see an answer of $0.86602$0.86602... which we'd have to round. However, when you take cos, sin or tan of some angles, you can express the answer as an exact number, rather than a decimal. It just may include irrational numbers. We often use these exact ratios in relation to $30^\circ$30°, $45^\circ$45° and $60^\circ$60°.
Let's use our special triangles to find the exact value trig ratios for $30^\circ$30°$45^\circ$45° and $60^\circ$60°.
Special triangle with $45^\circ$45°
Special triangle with $30^\circ$30° and $60^\circ$60°
Using what we know about trigonometric ratios, we can find the exact values for sine, cosine and tangent.
For example, using the second triangle:
$\sin\theta$sinθ $=$= $\frac{\text{opposite }}{\text{hypotenuse }}$opposite hypotenuse $\sin30^\circ$sin30° $=$= $\frac{1}{2}$12
This table is one way to display the information in the exact value triangles. You can choose which method you prefer to help you remember these exact ratios.
sin cos tan
$30^\circ$30° $\frac{1}{2}$12 $\frac{\sqrt{3}}{2}$32 $\frac{1}{\sqrt{3}}$13
$45^\circ$45° $\frac{1}{\sqrt{2}}$12 $\frac{1}{\sqrt{2}}$12 $1$1
$60^\circ$60° $\frac{\sqrt{3}}{2}$32 $\frac{1}{2}$12 $\sqrt{3}$3
You may be asked to rationalize the denominators as we generally don't like to have radicals in the denominator. In that case, the table becomes:
sin cos tan
$30^\circ$30° $\frac{1}{2}$12 $\frac{\sqrt{3}}{2}$32 $\frac{\sqrt{3}}{3}$33
$45^\circ$45° $\frac{\sqrt{2}}{2}$22 $\frac{\sqrt{2}}{2}$22 $1$1
$60^\circ$60° $\frac{\sqrt{3}}{2}$32 $\frac{1}{2}$12 $\sqrt{3}$3
#### Practice questions
##### QUESTION 7
Given that $\sin\theta=\frac{1}{2}$sinθ=12, we want to find the value of $\cos\theta$cosθ.
1. First, find the value of $\theta$θ.
2. Hence, find the exact value of $\cos30^\circ$cos30°.
##### Question 8
You are given that $\tan\theta=\frac{1}{\sqrt{3}}$tanθ=13.
1. First, find the value of $\theta$θ.
2. Hence, find exact the value of $\sin\theta$sinθ.
##### Question 9
Given that $\cos\theta=\frac{1}{\sqrt{2}}$cosθ=12, we want to find the value of $\tan\theta$tanθ.
1. First find the value of $\theta$θ.
2. Hence, find the exact value of $\tan45^\circ$tan45°. |
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### Linear Equations In Two Variables Class 9th Mathematics Part I MHB Solution
##### Linear Equations In Two Variables Class 9th Mathematics Part I MHB Solution
###### Practice Set 5.1
Question 1.
By using variables x and y form any five linear equations in two variables.
a) x + y = 5
b) x + 2y = 6
c) 2x + y = 4
d) 3x + 4y = 8
e) 5x + 9y = 1
Question 2.
Write five solutions of the equation x + y = 7.
(a)
Let x = 1,
As, x + y = 7
⇒ 1 + y = 7
⇒ y = 6
Hence, solution is x = 1 and y = 6.
(b)
Let x = 2,
As, x + y = 7
⇒ 2 + y = 7
⇒ y = 5
Hence, solution is x = 2 and y = 5.
(c)
Let x = 3,
As, x + y = 7
⇒ 3 + y = 7
⇒ y = 4
Hence, solution is x = 3 and y = 4.
(d)
Let x = 4,
As, x + y = 7
⇒ 4 + y = 7
⇒ y = 3
Hence, solution is x = 4 and y = 3.
(e)
Let x = 5,
As, x + y = 7
⇒ 5 + y = 7
⇒ y = 2
Hence, solution is x = 5 and y = 2.
Question 3.
Solve the following sets of simultaneous equations.
i. x + y = 4; 2x - 5y = 1
ii. 2x + y =5; 3x – y =5
iii. 3x – 5y =16; x - 3y = 8
iv. 2y – x =0; 10x + 15y =105
v. 2x +3y + 4 = 0; x – 5y = 11
vi. 2x – 7y =7; 3x + y =22
(i)
x + y = 4 eq.[1]
2x - 5y = 1 eq.[2]
We can write eq.[1] as,
x = 4 - y eq.[3]
Substituting eq.[3] in eq.[2],
⇒ 2(4 - y) - 5y = 1
⇒ 8 - 2y - 5y = 1
⇒ -7y = -7
⇒ y = 1
Substituting 'y' in eq.[3]
⇒ x = 4 - 1
⇒ x = 3
Hence, solution is x = 3 and y = 1.
(ii)
2x + y = 5 eq.[1]
3x - y = 5 eq.[2]
We can write eq.[1] as,
y = 5 - 2x eq.[3]
Substituting eq.[3] in eq.[2],
⇒ 3x - (5 - 2x) = 5
⇒ 3x - 5 + 2x = 5
⇒ 5x = 10
⇒ x = 2
Substituting 'x' in eq.[3]
⇒ y = 5 - 2(2)
⇒ y = 1
Hence, solution is x = 2 and y = 1.
(iii)
3x - 5y = 16 eq.[1]
x - 3y = 8 eq.[2]
We can write eq.[2] as,
x = 8 + 3y eq.[3]
Substituting eq.[3] in eq.[1],
⇒ 3(8 + 3y) - 5y = 16
⇒ 24 + 9y - 5y = 16
⇒ 4y = -8
⇒ y = -2
Substituting 'y' in eq.[3]
⇒ x = 8 + 3(-2)
⇒ x = 8 - 6 = 2
Hence, solution is x = 2 and y = -2
(iv)
2y - x = 0 eq.[1]
10x + 15y = 105 eq.[2]
We can write eq.[1] as,
x = 2y eq.[3]
Substituting eq.[3] in eq.[2],
⇒ 10(2y) + 15y = 105
⇒ 20y + 15y = 105
⇒ 35y = 105
⇒ y = 3
Substituting 'y' in eq.[3]
⇒ x = 2(3)
⇒ x = 6
Hence, solution is x = 6 and y = 3.
(v)
2x + 3y + 4 = 0 eq.[1]
x - 5y = 11 eq.[2]
We can write eq.[2] as,
x = 11 + 5y eq.[3]
Substituting eq.[3] in eq.[1],
⇒ 2(11 + 5y) + 3y + 4 = 0
⇒ 22 + 10y + 3y + 4 = 0
⇒ 13y + 26 = 0
⇒ 13y = -26
⇒ y = -2
Substituting 'y' in eq.[3]
⇒ x = 11 + 5(-2)
⇒ x = 11 - 10 = 1
Hence, solution is x = 1 and y = -2.
(vi)
2x - 7y = 7 eq.[1]
3x + y = 22 eq.[2]
We can write eq.[2] as,
y = 22 - 3x eq.[3]
Substituting eq.[3] in eq.[1],
⇒ 2x - 7(22- 3x) = 7
⇒ 2x - 154 + 21x = 7
⇒ 23x = 161
⇒ x = 7
Substituting 'x' in eq.[3]
⇒ y = 22 - 3(7)
⇒ y = 22 - 21 = 1
Hence, solution is x = 7 and y = 1.
###### Practice Set 5.2
Question 1.
In an envelope there are some 5 rupee notes and some 10 rupee notes. Total amount of these notes together is 350 rupees. Number of 5 rupee notes are less by 10 than number of 10 rupee notes. Then find the number of 5 rupee and 10 rupee notes.
Let the number of 5 rupees notes = x
Let the number of 10 rupees notes = y
Given, Total amount is 350 Rupees
⇒ 5x + 10y = 350 eq.[1]
Also,
Number of 5 rupees notes are less by 10 than number of 10 rupees note,
y = x - 10
⇒ x = y + 10 eq.[2]
Putting [2] in [1]
⇒ 5(y + 10) + 10y = 350
⇒ 5y + 50 + 10y = 350
⇒ 15y = 300
⇒ y = 20
Then, x = y + 10
⇒ x = 20 + 10
⇒ x = 30.
Answer: 30 notes of Rs 5 and 20 notes of Rs. 10.
Question 2.
The denominator of a fraction is 1 more than twice its numerator. If 1 is added to numerator and denominator respectively, the ratio of numerator to denominator is
1 : 2. Find the fraction.
Let the numerator be 'x' and denominator be 'y'
Given,
The denominator of a fraction is 1 more than twice its numerator
⇒ y =2 x + 1
⇒ y - 2 x = 1 ....... (1)
If 1 is added to numerator and denomination, the ratio of the numerator to denominator becomes 1:2.
⇒ 2(x + 1) = y + 1
⇒ 2x + 2 = y + 1
⇒ y - 2x = 1 ....... (2)
As (1) and (2) are the same, there can be infinitely many solutions for x and y.
One such solution is:
x = 4 and y = 9
Now,
y - 2 x = 9 - 2(4)
= 9 - 8
= 1
Question 3.
The sum of ages of Priyanka and Deepika is 34 years. Priyanka is elder to Deepika by 6 years. Then find their today's ages.
Let the ages of Priyanka and Deepika be 'x' and 'y' respectively.
Given,
Sum of ages is 34
⇒ x + y = 34
⇒ y = 34 - x eq.[1]
Also, Priyanka is elder to Deepika by 6 years
⇒ x = y + 6
Using eq.[1] we have
⇒ x = 34 - x + 6
⇒ 2x = 40
⇒ x = 20
Putting this value in eq.[1]
⇒ y = 34 - 20 = 14 years.
Hence, Age of Priyanka = x = 20 Years
Age of Deepika = y = 14 years.
Question 4.
The total number of lions and peacocks in a certain zoo is 50. The total number of their legs is 140. Then find the number of lions and peacocks in the zoo.
Let the number of lions be 'x' and peacocks be 'y'
Given, Total no of lions and peacocks is 50
⇒ x + y = 50
⇒ x = 50 - y eq.[1]
Also, Total no of their legs is 140, as lion has four legs and peacocks has 2 legs
⇒ 4x + 2y = 140
⇒ 4(50 - y) + 2y = 140
⇒ 200 - 4y + 2y = 140
⇒ 2y = 60
⇒ y = 30
Using this in eq.[1]
⇒ x = 50 - 30 = 20
Therefore,
No of lions, x = 20
No of peacocks, y = 30
Question 5.
Sanjay gets fixed monthly income. Every year there is a certain increment in his salary. After 4 years, his monthly salary was Rs. 4500 and after 10 years his monthly salary became 5400 rupees, then find his original salary and yearly increment.
Let the original salary be 'x' and yearly increment be 'y'
After 4 years, his salary was Rs. 4500
⇒ x + 4y = 4500
⇒ x = 4500 - 4y eq.[1]
After 10 years, his salary becomes 5400
⇒ x + 10y = 5400
⇒ 4500 - 4y + 10y = 5400
⇒ 6y = 900
⇒ y = 150
Putting this in eq.[1],
⇒ x = 4500 - 4(150)
⇒ x = 4500 - 600 = 3900
Hence, his original salary was Rs. 3900 and increment per year was 150 Rs.
Question 6.
The price of 3 chairs and 2 tables is 4500 rupees and price of 5 chairs and 3 tables is 7000 rupees, then find the price of 2 chairs and 2 tables.
Let the price of one chair be 'x' and one table be 'y'.
Given,
Price of 3 chairs and 2 tables = 4500 Rs
⇒ 3x + 2y = 4500
Multiplying by 3 both side,
⇒ 9x + 6y = 13500
⇒ 6y = 13500 - 9x eq.[1]
Price of 5 chairs and 3 tables = 7000 Rs
⇒ 5x + 3y = 7000
Multiplying by eq.[2] both side,
⇒ 10x + 6y = 14000
⇒ 10x + 13500 - 9x = 14000 eq.[From 1]
⇒ x = 500
Putting this in eq.[1]
⇒ 6y = 13500 - 9(500)
⇒ 6y = 13500- 4500
⇒ 6y = 9000
⇒ y = 1500
Also, Price of 2 chairs and 2 tables = 2x + 2y
= 2(500) + 2(1500)
= 1000 + 3000 = 4000 Rs.
Question 7.
The sum of the digits in a two-digits number is 9. The number obtained by interchanging the digits exceeds the original number by 27. Find the two-digit number.
Let the unit digit be 'x'
Let the digit at ten's place be 'y'
The original number will be 10y + x
Given,
Sum of digits = 9
⇒ x + y = 9
⇒ x = 9 - y eq.[1]
Also,
If the digits are interchanged,
Reversed number will be = 10x + y
As, reversed number exceeds the original number by 27,
⇒ (10x + y) - (10y + x) = 27
⇒ 10x + y - 10y - x = 27
⇒ 9x - 9y = 27
⇒ x - y = 3
⇒ 9 - y - y = 3 eq.[using 1]
⇒ -2y = -6
⇒ y = 3
Using this in eq.[1]
⇒ x = 9 - 3 = 6
Hence the original number is 10y + x = 10(3) + 6 = 30 + 6 = 36.
Question 8.
In ΔABC, the measure of angle A is equal to the sum of the measures of ∠B and ∠C. Also the ratio of measures of ∠B and ∠C is 4 : 5. Then find the measures of angles of the triangle.
Given that, In ΔABC
∠A = ∠B + ∠C eq.[1]
Let ∠B = x and ∠C = y
Then,
∠A = x + y
In ΔABC, By angle sum property of triangle
∠A + ∠B + ∠C = 180°
⇒ x + y + x + y = 180
⇒ 2x + 2y = 180
⇒ x + y = 90
⇒ x = 90 - y eq.[2]
Also, Given that
⇒ 5x = 4y
From eq.[2]
⇒ 5(90 - y) = 4y
⇒ 450 - 5y = 4y
⇒ 9y = 450
⇒ y = 50°
Putting this in eq.[2]
⇒ x = 90 - 50 = 40°
Therefore, we have
∠A = x + y = 40° + 50° = 90°
∠B = x = 40°
∠C = y = 50°
Question 9.
Divide a rope of length 560 cm into 2 parts such that twice the length of the smaller part is equal to 1/3 of the larger part. Then find the length of the larger part.
Let the length of smaller part be 'x' cm and larger part be 'y' cm.
Length of rope = 560 cm
⇒ x + y = 560
⇒ y = 560 - x eq.[1]
Also,
Twice the length of smaller part is equal to of the larger part
⇒ 6x = y
⇒ 6x = 560 - x
⇒ 7x = 560
⇒ x = 80
Using this in eq.[1]
⇒ y = 560 - 80 = 480
Therefore,
Length of smaller part = 'x' cm = 80 cm
Length of larger part = 'y' cm = 480 cm
Question 10.
In a competitive examination, there were 60 questions. The correct answer would carry 2 marks, and for incorrect answer 1 mark would be subtracted. Yashwant had attempted all the questions and he got total 90 marks. Then how many questions he got wrong?
Let the no of questions he got wrong be 'x'
And the no of questions he got right be 'y'
As, he attempted all the questions,
⇒ x + y = 60
⇒ y = 60 - x eq.[1]
Also, he carries 2 for each corrects question and (-1) for each wrong question, also he got 90 marks
⇒ y(2) + x(-1) = 90
⇒ 2y - x = 90
⇒ 2(60 - x) - x = 90 eq.[Using 1]
⇒ 120 - 2x - x = 90
⇒ -3x = -30
⇒ x = 10
⇒ he got 10 wrong questions.
###### Problem Set 5
Question 1.
Choose the correct alternative answers for the following questions.
If 3x + 5y =9 and 5x + 3y =7 then What is the value of x + y?
A. 2
B. 16
C. 9
D. 7
3x + 5y = 9 eq.[1]
5x + 3y = 7 eq.[2]
Adding eq.[1] and eq.[2] we get
3x + 5y + 5x + 3y = 9 + 7
⇒ 8x + 8y = 16
Dividing both side by 8, we get
⇒ x + y = 2
Question 2.
'When 5 is subtracted from length and breadth of the rectangle, the perimeter becomes 26.' What is the mathematical form of the statement?
A. x – y = 8
B. x + y = 8
C. x + y =23
D. 2x + y = 21
Let the length be 'x' and breadth be 'y' units.
Perimeter of triangle = 2(x + y) units
If 5 is subtracted from length and breadth
Perimeter = 26 units eq.[Given]
⇒ 2( x - 5 + y - 5) = 26
⇒ 2(x + y - 10) = 26
⇒ x + y - 10 = 13
⇒ x + y = 23
Question 3.
Ajay is younger than Vijay by 5 years. Sum of their ages is 25 years. What is Ajay's age?
A. 20
B. 15
C. 10
D. 5
Let Ajay's age be 'x' years and Vijay's age be 'y' years.
Given, Ajay is younger than Vijay by 5 years
⇒ x = y - 5 eq.[1]
Also, Sum of their ages is 25 years,
⇒ x + y = 25
From eq.[1]
⇒ y - 5 + y = 25
⇒ 2y = 30
⇒ y = 15
Putting this in eq.[1]
⇒ x = 15 - 5 = 10
Age of Ajay = x = 10 Years
Age of Vijay = y = 15 Years
Question 4.
Solve the following simultaneous equations.
i. 2x + y = 5; 3x - y = 5
ii. x - 2y = -1; 2x - y = 7
iii. x + y = 11; 2x - 3y = 7
iv. 2x + y = -2; 3x - y = 7
v. 2x - y = 5; 3x + 2y = 11
vi. x - 2y = -2; x + 2y = 10
(i)
2x + y = 5
⇒ y = 5 - 2x eq.[1]
3x - y = 5
Using eq.[1] we have
⇒ 3x - (5 - 2x) = 5
⇒ 3x - 5 + 2x = 5
⇒ 5x = 10
⇒ x = 2
Using 'x' in eq.[1]
⇒ y = 5 - 2(2)
⇒ y = 5 - 4 = 1 cm
(ii)
x - 2y = -1
⇒ x = 2y - 1 eq.[1]
2x - y = 7
Using eq.[1], we have
⇒ 2(2y - 1) - y = 7
⇒ 4y - 2 - y = 7
⇒ 3y = 9
⇒ y = 3
Using this value in eq.[1]
⇒ x = 2(3) - 1
⇒ x = 5
(iii)
x + y = 11
⇒ y = 11 - x eq.[1]
2x - 3y = 7
Using eq.[1], we have
⇒ 2x - 3(11 - x) = 7
⇒ 2x - 33 + 3x = 7
⇒ 5x = 40
⇒ x = 8
Using this in eq.[1]
⇒ y = 11 - 8
⇒ y = 3
(iv)
2x + y = -2
⇒ y = -2x - 2 eq.[1]
3x - y = 7
Using eq.[1]
3x - (-2x - 2) = 7
⇒ 3x + 2x + 2 =7
⇒ 5x = 5
⇒ x = 1
Using this in eq.[1]
⇒ y = -2(1) - 2
⇒ y = -2 - 2 = -4
(v)
2x - y = 5
⇒ y = 2x - 5 eq.[1]
3x + 2y = 11
Using eq.[1]
⇒ 3x + 2(2x - 5) = 11
⇒ 3x + 4x - 10 = 11
⇒ 7x = 21
⇒ x = 3
Using this in eq.[1]
⇒ y = 2(3) - 5
⇒ y = 1
(vi)
x - 2y = -2
x = 2y - 2 eq.[1]
x + 2y = 10
using eq.[1], we have
⇒ 2y - 2 + 2y = 10
⇒ 4y = 12
⇒ y = 3
Using this in eq.[1]
⇒ x = 2(3) - 2
⇒ x = 4
Question 5.
By equating coefficients of variables, solve the following equations.
i. 3x - 4y = 7; 5x + 2y = 3
ii. 5x + 7y = 17; 3x - 2y = 4
iii. x - 2y = -10; 3x - 5y = -12
iv. 4x + y = 34; x + 4y = 16
(i)
3x - 4y = 7 eq.[1]
5x + 2y = 3 eq.[2]
Multiplying eq.[2] by 2 both side, we get
10x + 4y = 6 eq.[3]
Adding eq.[1] and eq.[3], we get
3x - 4y + 10x + 4y = 7 + 6
⇒ 13x = 13
⇒ x = 1
Putting this in eq.[1], we get
3(1) - 4y = 7
⇒ -4y = 7 - 3
⇒ -4y = 4
⇒ y = -1
(ii)
5x + 7y = 17 eq.[1]
3x - 2y = 4 eq.[2]
Multiplying eq.[1] by 3 both side and Multiplying eq.[2] by 5 both side we get,
15x + 21y = 51 eq.[3]
15x - 10y = 20 eq.[4]
Subtracting eq.[4] from eq.[3], we get
15x + 21y - 15x + 10y = 51 - 20
⇒ 31y = 31
⇒ y = 1
Putting this in eq.[1], we get
5x + 7(1) = 17
⇒ 5x = 10
⇒ x = 2
(iii)
x - 2y = -10 eq.[1]
3x - 5y = -12 eq.[2]
Multiplying eq.[1] by 3
3x - 6y = -30 eq.[3]
Subtracting eq.[2] from eq.[3], we get
3x - 6y - 3x + 5y = -30 + 12
⇒ -y = -18
⇒ y = 18
Putting this in eq.[1], we get
x - 2(18) = -10
⇒ x - 36 = -10
⇒ x = 26
(iv)
4x + y = 34 eq.[1]
x + 4y = 16 eq.[2]
Multiplying eq.[2] by 4 both side, we get
4x + 16y = 64 eq.[3]
Subtracting eq.[3] from eq.[1], we get
4x + 16y - 4x - y = 64 - 34
⇒ 15y = 30
⇒ y = 2
Putting this in eq.[2], we get
x + 4(2) = 16
⇒ x + 8 = 16
⇒ x = 8
Question 6.
Solve the following simultaneous equations.
i.
ii.
iii.
(i)
⇒ 4x + 3y = 48 eq.[1]
⇒ 2x - 3y = 4 eq.[2]
Adding eq.[1] and eq.[2], we get
⇒ 4x + 3y + 2x - 3y = 48 + 4
⇒ 6x = 52
Using this in eq.[1], we have
⇒ 104 + 9y = 144
⇒ 9y = 40
(ii)
⇒ x + 15y = 39
⇒ x = 39 - 15y eq.[1]
⇒ 4x + y = 38
Using eq.[1], we have
⇒ 4(39 - 15y) + y = 38
⇒ 156 - 60y + y = 38
⇒ 59y = 118
⇒ y = 2
Putting this value in eq.[2]
⇒ x = 39 - 15(2)
⇒ x = 39 - 30
⇒ x = 9
(iii)
⇒ 2y + 3x = 13xy eq.[1]
⇒ 5y - 4x = -2xy eq.[2]
Multiplying eq.[1] by 4 both side, and Multiplying eq.[2] by 3 both side, we get
8y + 12x = 52xy eq.[3]
15y - 12x = -6xy eq.[4]
⇒ 8y + 12x + 15y - 12x = 52xy - 6xy
⇒ 23y = 46xy
⇒ 1 = 2x
Putting this in eq.[1]
Question 7.
A two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to this number, the sum is equal to the number obtained by interchanging the digits. Find the number.
Let the unit digit be 'x'
Let the digit at ten's place be 'y'
The original number will be 10y + x
Given, number is 3 more than 4 times the sum of its digits
⇒ 10y + x = 4(x + y) + 3
⇒ 10y + x = 4x + 4y + 3
⇒ 6y - 3x = 3
⇒ 2y - x = 1
⇒ x = 2y - 1 eq.[1]
Also,
If the digits are interchanged,
Reversed number will be = 10x + y
As, reversed number exceeds the original number by 18,
⇒ (10x + y) - (10y + x) = 18
⇒ 10x + y - 10y - x = 18
⇒ 9x - 9y = 18
⇒ x - y = 2
⇒ 2y - 1 - y = 2 eq.[using 1]
⇒ y = 3
Using this in eq.[1]
⇒ x = 2(3) - 1 = 5
Hence the original number is 10y + x = 10(3) + 5 = 30 + 5 = 35.
Question 8.
The total cost of 6 books and 7 pens is 79 rupees and the total cost of 7 books and 5 pens is 77 ruppees. Find the cost of 1 book and 2 pens.
Let the cost of one book be 'x' rupees and cost of one pen be 'y' rupees.
Cost of 6 books and 7 pens = 79 Rs
⇒ 6x + 7y = 79 eq.[1]
Cost of 7 books and 5 pens = 77 Rs
⇒ 7x + 5y = 77 eq.[2]
Multiplying eq.[1] by 5 both side, and Multiplying eq.[2] by 7 both side, we get
⇒ 30x + 35y = 395 eq.[3]
⇒ 49x + 35y = 539 eq.[4]
Subtracting eq.[3] from eq.[4], we get
⇒ 49x + 35y - 30x - 35y = 539 - 395
⇒ 19x = 144
Using this in eq.[1]
⇒ 864 + 19×7y = 79×19
⇒ 19×7y = 79×19 – 864
⇒
⇒ y = 5
& 6x + 7y = 79
⇒ 6x + 35 = 79
⇒ 6x = 44
⇒ x = 7
Hence, the cost of 1 pen & 2 books = Rs 1(y) + 2x
= 5 + 14 = Rs 19.
Question 9.
The ratio of incomes of two persons is 9 : 7. The ratio of their expenses is 4 : 3. Every person saves rupees 200, find the income of each.
As the ratio of incomes is 9 : 7,
Let income of first person = 9x
Income of second person = 7x
Also, ratio of incomes is 4 : 3,
Let expenses of first person = 4y
Expenses of second person = 3y
Each person saves 200 Rs,
⇒ 9x - 4y = 200 eq.[1]
⇒ 7x - 3y = 200 eq.[2]
Multiplying eq.[1] by 3 both side and Multiplying eq.[2] by 4 both side, we get
⇒ 27x - 12y = 600 eq.[3]
⇒ 28x - 12y = 800 eq.[4]
Subtracting eq.[3] from eq.[4], we get
⇒ 28x - 12y - (27x - 12y) = 800 - 600
⇒ 28x - 12y - 27x + 12y = 200
⇒ x = 200
Income of first person = 9x = 9(200) = 1800 Rs
Income of second person = 7x = 7(200) = 1400 Rs
Question 10.
If the length of a rectangle is reduced by 5 units and its breadth is increased by 3 units, then the area of the rectangle is reduced by 8 square units. If length is reduced by 3 units and breadth is increased by 2 units, then the area of rectangle will increase by 67 square units. Then find the length and breadth of the rectangle.
Let the length be 'x' and breadth be 'y'
Area of rectangle = length × breadth
Area of rectangle = xy
First case:
Length = x - 5
As, area is reduced by 8 sq. units
⇒ xy - (x - 5)(y + 3) = 8
⇒ xy - (xy + 3x - 5y - 15) = 8
⇒ xy - xy - 3x + 5y + 15 = 8
⇒ 3x - 5y = 7 eq.[1]
Second case:
Length = x - 3
As, the area is increased by 67 units
⇒ (x - 3)(y + 2) - xy = 67
⇒ xy + 2x - 3y - 6 - xy = 67
⇒ 2x - 3y = 73 eq.[2]
Multiplying eq.[1] by 2 both side, and Multiplying eq.[2] by 3 both side, we get
⇒ 6x - 10y = 14 eq.[3]
⇒ 6x - 9y = 219 eq.[4]
Subtracting eq.[3] from eq.[4]
⇒ 6x - 9y - 6x + 10y = 219 - 14
⇒ y = 205
Using this in eq.[1]
⇒ 3x - 5(205) = 7
⇒ 3x - 1025 = 7
⇒ 3x = 1032
⇒ x = 344
Hence, length = x = 344 units
Breadth = y = 219 units.
Question 11.
The distance between two places A and B on road is 70 kilometers. A car starts from A and the other from B. If they travel in the same direction, they will meet after 7 hours. If they travel towards each other they will meet after 1 hour, then find their speeds.
Let the speed of car at place A is x km/h and that of car at place B is y km/h
If they travel in same direction, they will meet after 7 hours, i.e. the difference of distance covered by them in 7 hours will be equal to distance b/w A and B.
As, distance = speed × time, and distance from A to B is 70 km
⇒ 7x - 7y = 70
⇒ x - y = 10
⇒ x = y + 10 eq.[1]
If they, travel in opposite direction, they will meet after 1 hour i.e. sum of distance travelled by both cars will be equal to the distance b/w A and B.
⇒ x + y = 70
Using eq.[1], we have
⇒ y + 10 + y = 70
⇒ 2y = 60
⇒ y = 30
Using this in eq.[1], we have
x = 30 + 10 = 40
Hence,
Speed of car at A = x = 40 km/h
Speed of car at B = y = 30 km/h
Question 12.
The sum of a two-digit number and the number obtained by interchanging its digits is 99. Find the number.
Let the unit digit be 'x' and digit at ten's place be 'y'
Original Number = 10y + x
Number obtained by interchanging digits = 10x + y
Given,
10y + x + 10x + y = 99
⇒ 11x + 11y = 99
⇒ x + y = 9
If x = 1, y = 8 and number is 18
If x = 2, y = 7 and number is 27
If x = 3, y = 6 and number is 36
If x = 4, y = 5 and number is 45
If x = 5, y = 4 and number is 54
If x = 6, y = 3 and number is 63
If x = 7, y = 2 and number is 72
If x = 8, y = 1 and number is 81
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# Vedic Maths Shortcut Technique for Finding Square of a Number
Friends, today we shall discuss a simple vedic math method for multiplications. This method is called "Ekadhikena Purvena" that means by one more than the previous one. In vedic maths there are several methods for multiplication of two numbers. One of the methods explained below is for finding out the square of a number (multiplying a number with the same number) having 5 in the units place.
Example 1 : 35 x 35 = ?
The method explained in the schools gets the answer as shown below.
• The present method following the sutra "Ekadhikena Purvena" involves the procedure of filling up the blanks.
• For the problem given above, let us have four blanks (the number of blanks depend on the base of the number).
• initially, multiply the digits in the units places.
5 x 5 = 25
• put the number 25 in the two blanks on RHS (Right Hand Side)
• Next select one of the two given numbers. That is, select a 35. Pic up the digit that is preceding (purva of) 5, that is 3.
3 + 1 = 4
• Multiply this 4 with the digit 3 appearing in the second number 35.
3 x 4 = 12
• Put this 12 in the two blanks that are on the LHS of the result.
Present status :
Lets do one more problem
45 x 45 = ?
1. Initially : 5 x 5 = 25
2. Ekadhika : 4 + 1 = 5
3. Next : 4 x 5 = 20
4. Result : 45 x 45 = 20 | 25 = 2025
Lets try another example
75 x 75 = ?
Result : 5625
Now you may ask. How to apply this method for three digit numbers ?
Same process, have a look
115 x 115 = ?
1. Initially : 5 x 5 = 25
2. Ekadhika : 11 + 1 = 12
3. Next : 11 x 12 = 132
4. Result : 115 x 115 = 132 | 25 = 13225
That's all for now friends. Tomorrow we shall discuss more vedic math shortcut techniques for multiplication. Happy Reading :)
Rupali Shete (M.Sc., M.Phil., Ph.D) is a gold medalist from Osmania University (Hyderabad), one of the working members of Rananujan Mathematics Academy and Institute of Scientific Research on Vedas (ISERVE), Currently working in University of Pune.
1. for finding square of a number ending with 5 u can use this method for e.g. 35 add 3 with 1 then multiply with that same number i.e. 3*4 =12 multiply with 100 and add 25 (3*4)*100+25 1225
shared by Krishnananth Saju
2. Thanks friends. We need more tricks like these..
3. nice mam
bt how to find the squares that are not ending with 5???
4. what about other 2 digits n 3 digits number square
5. square of 123. divide the number with two parts 12 & 3.
square of 12/2*12*3/square of 3.
144/72/9
144+7 29=15129 |
# Triangle Proofs - Hypotenuse Leg (Part 4)
What does the Hypotenuse - Leg Theorem say?
If the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and one leg of another right triangle, then the two right triangles are congruent.
So we only have to show that in two right triangles that the hypotenuse is congruent to the other hypotenuse and show that one leg is congruent to another corresponding leg in order to show that the two triangles are congruent.
Here is an example:
In this example, let's assume we are given that ABEF and CBDF.
Notice that both triangles are right triangles. The hypotenuse of triangle ABC is CB and the hypotenuse of triangle DEF is DF. So we can see that the hypotenuses, CB and DF, are congruent from the given statements.
In addition, we have been told that AB is congruent with EF.
Therefore, we have two congruent triangles, with congruent hypotenuses and congruent legs, showing that these two triangles are congruent. So we can say that ΔABC ≅ ΔEFD.
The Hypotenuse - Leg theorem can be used to prove more than just congruent triangles by including the CPCTC move. Recall that CPCTC represents "corresponding parts of congruent triangles are congruent."
Here is another example:
Given: <W is congruent to <Y and XZ is an altitude.
Prove: WZ is congruent to YZ.
Let's take a look at how to use Hypotenuse - Leg to complete this informal proof.
The altitude XZ is perpendicular to WY, which means that <XZW and <XZY are right angles. Therefore, both triangle XZW and triangle XZY are both right triangles. In addition, because <W is congruent to <Y, we know that triangle WXY is isosceles and therefore WX is congruent to YX. This is helpful because WX and YX are also the hypotenuses of triangles WXZ and YXZ. We also have XZ which is a leg in both triangle WXZ and triangle YXZ, and XZ is of course congruent to XZ.
Now we have a right angle in both triangles, congruent hypotenuses and congruent legs. Therefore, triangle WXZ is congruent to triangle YXZ. And because these triangles are congruent, WZ is congruent to YZ because corresponding parts of congruent triangles are congruent.
Summing it All Up
So the main point here is that if you have two right triangles and the hypotenuses are congruent and one pair of corresponding legs are congruent, then the two triangles are congruent. This can then be used to prove other things in a triangle diagram.
Related Links: Math Geometry Triangles Congruent Triangles
To link to this Triangle Proofs - Hypotenuse Leg (Part 4) page, copy the following code to your site: |
We shall now discuss problems that can be solved with quadratic equations.
To solve such problems, we first formulate the quadratic equation from the information we have, and then we
solve it.
Consider these examples.
Example 1:
The product of two consecutive numbers is 72. Find the numbers.
Solution :
Let the smaller number = x
Let the larger number = x+1
Their product = x(x+1)
We are given that
x(x + 1) = 72
⇒ x2 + x = 72
⇒ x2 + x - 72 = 0
Factorizing
x2 + 9x - 8x - 72 = 0
x(x + 9) - 8(x + 9) = 0
(x - 8) (x + 9) = 0
⇒ x - 8 = 0 or x + 9 = 0
⇒ x= 8 or x = - 9
⇒ x + 1 = 8 + 1 or x + 1 = -9 + 1
= 9 = -8
Therefore, the two consecutive numbers are 8, 9 or -9, -8
Example 2:
A man walks a distance of 48 kilometers in a certain amount of time. If he increased his speed by 2 km/hr, he
would have reached his destination four hours earlier. Find his usual speed.
Solution :
Let the man's usual speed = x km/hr
Time taken to walk 48 km = (48/x) hrs
Time = Distance/Speed
Increased speed = (x + 2) km/hr
Time taken to walk 48 km =48/x+2
At a speed of (x + 2) km/hr, the man reaches his destination 4 hours earlier.
⇒ 96 = 4(x2 + 2x)
96 = 4x2+ 8x
Transposing
4x2 + 8x - 96 = 0
⇒ 4(x2 + 2x - 96) = 0
⇒ x2 + 2x - 24 = 0/4
⇒ x 2+ 2x - 24 = 0
Factorizing
x2 + 6x - 4x - 24 = 0
x(x + 6) - 4 (x+6) = 0
⇒ (x - 4) (x + 6) = 0
⇒ x - 4 = 0 or x + 6 = 0
⇒ x = 4 or x = -6.
Since the man cannot walk at a negative
pace, we reject x = -6
Speed = x =4km/hr
#### Try this question
Ann can row her boat at a speed of 5 km/hr in still water. If it takes her one hour more to row her boat 5.25 km
upstream than to return downstream, find the speed of the stream.
Speed of boat = 5 km/hr
Let the speed of the stream current = x km/hr.
Speed of the boat upstream (against the current)
= (5 - x) km/hr
Speed of the boat downstream (with the current)
= (5 + x) km/hr.
Let t1 = time taken to travel 5.25 km upstream
then t1 = 5.25/5-x since time = Distance/Speed
Let t2 = time taken to travel 5.25 km downstream then
t2 = 5.25/5+x
Now, t1> t2
Since it takes 1 hour more to travel upstream than downstream
t1 = t2 + 1
5.25 [2x] = 1(25-x2)
10.50 x = 25 - x2
Transposing
x2 + 10.5x - 25 = 0
⇒ 2x2 + 21x - 50 = 0 * 2
⇒ 2 x2 + 21x - 50 = 0
⇒ 2x2 + 25x - 4x - 50 = 0
⇒ x(2x + 25) - 2(2x + 25) = 0
⇒ (x-2) (2x + 25) = 0
⇒ x-2 = 0 or 2x + 25 = 0
⇒ x = 2 or 2x = - 25
or x=-25/2
= -12.5 km/hr.
Since the speed of the stream cannot be negative we reject
x = -12.5km/hr.
Speed of the stream = 2 km/hr. |
# 0.3 Radical concepts -- radical equations
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This module contains methods on solving radical equations.
When solving equations that involve radicals , begin by asking yourself: is there an $x$ under the square root? The answer to this question will determine the way you approach the problem.
If there is not an $x$ under the square root—if only numbers are under the radicals—you can solve much the same way you would solve with no radicals at all.
## Radical equation with no variables under square roots
$\sqrt{2}x+5=7–\sqrt{3}x$ Sample problem: no variables under radicals $\sqrt{2}+\sqrt{3}x=7-5$ Get everything with an $x$ on one side, everything else on the other $x\left(\sqrt{2}+\sqrt{3}\right)=2$ Factor out the $x$ $x=\frac{2}{\sqrt{2}+\sqrt{3}}$ Divide, to solve for $x$
The key thing to note about such problems is that you do not have to square both sides of the equation . $\sqrt{2}$ may look ugly, but it is just a number—you could find it on your calculator if you wanted to—it functions in the equation just the way that the number 10, or $\frac{1}{3}$ , or π would.
If there is an $x$ under the square root, the problem is completely different. You will have to square both sides to get rid of the radical. However, there are two important notes about this kind of problem.
1. Always get the radical alone, on one side of the equation , before squaring.
2. Squaring both sides can introduce false answers —so it is important to check your answers after solving!
Both of these principles are demonstrated in the following example.
## Radical equation with variables under square roots
$\sqrt{x+2}+\mathrm{3x}=\mathrm{5x}+1$ Sample problem with variables under radicals $\sqrt{x+2}=\mathrm{2x}+1$ Isolate the radical before squaring! $x+2={\left(\mathrm{2x}+1\right)}^{2}$ Now, square both sides $x+2={\mathrm{4x}}^{2}+\mathrm{4x}+1$ Multiply out. Hey, it looks like a quadratic equation now! $x+2={\mathrm{4x}}^{2}+\mathrm{4x}+1$ As always with quadratics, get everything on one side. $\left(\mathrm{4x}-1\right)\left(x+1\right)=0$ Factoring: the easiest way to solve quadratic equations. $x=\frac{1}{4}$ or $x=-1$ Two solutions . Do they work? Check in the original equation !
Check $x=\frac{1}{4}$ Check $x=–1$ $\sqrt{\frac{1}{4}+2}+3\left(\frac{1}{4}\right)\stackrel{?}{=}5\left(\frac{1}{4}\right)+1$ $\sqrt{-1+2}+3\left(-1\right)\stackrel{?}{=}5\left(-1\right)+1$ $\sqrt{\frac{1}{4}+\frac{8}{4}}+\frac{3}{4}\stackrel{?}{=}\frac{5}{4}+1$ $\sqrt{1}-3\stackrel{?}{=}-5+1$ $\sqrt{\frac{9}{4}}+\frac{3}{4}\stackrel{?}{=}\frac{5}{4}+\frac{4}{4}$ $1-3\stackrel{?}{=}-5+1$ $\frac{3}{2}+\frac{3}{4}\stackrel{?}{=}\frac{5}{4}+\frac{4}{4}$ $-2=-4$ Not equal! $\frac{9}{4}=\frac{9}{4}$
So the algebra yielded two solutions: $\frac{1}{4}$ and –1. Checking, however, we discover that only the first solution is valid. This problem demonstrates how important it is to check solutions whenever squaring both sides of an equation.
If variables under the radical occur more than once, you will have to go through this procedure multiple times. Each time, you isolate a radical and then square both sides.
## Radical equation with variables under square roots multiple times
$\sqrt{x+7}-x=1$ Sample problem with variables under radicals multiple times $\sqrt{x+7}=\sqrt{x}+1$ Isolate one radical. (I usually prefer to start with the bigger one.) $x+7=x+2\sqrt{x}+1$ Square both sides. The two -radical equation is now a one -radical equation. $6=2\sqrt{x}$ $3=x$ Isolate the remaining radical, then square both sides again.. $9=x$ In this case, we end up with only one solution. But we still need to check it.
$\sqrt{9+7}-\sqrt{9}\stackrel{?}{=}1$ $\sqrt{16}-\sqrt{9}\stackrel{?}{=}1$ $4-3=1$
Remember, the key to this problem was recognizing that variables under the radical occurred in the original problem two times . That cued us that we would have to go through the process—isolate a radical, then square both sides—twice, before we could solve for $x$ . And whenever you square both sides of the equation, it’s vital to check your answer(s)!
Why is it that—when squaring both sides of an equation—perfectly good algebra can lead to invalid solutions? The answer is in the redundancy of squaring. Consider the following equation:
$–5=5$ False. But square both sides, and we get...
25 = 25 True. So squaring both sides of a false equation can produce a true equation.
To see how this affects our equations, try plugging $x=\mathrm{-1}$ into the various steps of the first example.
## Why did we get a false answer of x=–1 in example 1?
$\sqrt{x+2}+\mathrm{3x}=\mathrm{5x}+1$ Does $x=\mathrm{-1}$ work here? No, it does not. $\sqrt{x+2}=\mathrm{2x}+1$ How about here? No, $x=\mathrm{-1}$ produces the false equation 1=–1. $x+2={\left(\mathrm{2x}+1\right)}^{2}$ Suddenly, $x=\mathrm{-1}$ works. (Try it!)
When we squared both sides, we “lost” the difference between 1 and –1, and they “became equal.” From here on, when we solved, we ended up with $x=\mathrm{-1}$ as a valid solution.
Test your memory: When you square both sides of an equation, you can introduce false answers. We have encountered one other situation where good algebra can lead to a bad answer . When was it?
Answer: It was during the study of absolute value equations, such as $|\mathrm{2x}+3|=\mathrm{-11x}+42$ . In those equations, we also found the hard-and-fast rule that you must check your answers as the last step.
What do these two types of problem have in common? The function $|x|$ actually has a lot in common with ${x}^{2}$ . Both of them have the peculiar property that they always turn $\mathrm{-a}$ and $a$ into the same response. (For instance, if you plug –3 and 3 into the function, you get the same thing back.) This property is known as being an even function . Dealing with such “redundant” functions leads, in both cases, to the possibility of false answers.
The similarity between these two functions can also be seen in the graphs: although certainly not identical, they bear a striking resemblance to each other. In particular, both graphs are symmetric about the y-axis, which is the fingerprint of an “even function”.
#### Questions & Answers
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
is it 3×y ?
J, combine like terms 7x-4y
im not good at math so would this help me
how did I we'll learn this
f(x)= 2|x+5| find f(-6)
f(n)= 2n + 1
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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# Sin 150: Understanding The Meaning And Importance
## Introduction
Sin 150 is a term that refers to a specific angle measurement in trigonometry. In this article, we’ll explore the meaning and importance of sin 150 in mathematics and real-world applications.
## What is Sin 150?
Sin 150 is the sine of the angle 150 degrees. Sine is one of the six trigonometric functions and is defined as the ratio of the opposite side to the hypotenuse in a right triangle. In other words, sin 150 is the ratio of the length of the side opposite the angle to the length of the hypotenuse.
## Calculating Sin 150
To calculate sin 150, we first need to convert the angle to radians. We know that 180 degrees is equal to pi radians, so 150 degrees is equal to 5pi/6 radians. Once we have the angle in radians, we can use the sine function on a calculator or by hand to find the value of sin 150, which is approximately -0.866.
## Real-World Applications
While sin 150 may seem like a purely theoretical concept, it has real-world applications in fields such as engineering, physics, and astronomy. For example, engineers may use trigonometry to calculate the angles and distances needed for building structures or designing machines. Astronomers may use trigonometry to calculate the distances between celestial objects.
## Trigonometry Tips
If you’re struggling with trigonometry, don’t worry – it’s a complex subject that takes time and practice to master. Here are some tips to help you get started: – Practice, practice, practice! The more problems you solve, the more comfortable you’ll become with trigonometric concepts. – Memorize the basic trigonometric identities, such as sin^2 x + cos^2 x = 1 and tan x = sin x / cos x. – Use visual aids, such as diagrams or graphs, to help you understand the relationships between angles and sides in a right triangle. – Don’t be afraid to ask for help from a teacher or tutor if you’re struggling.
## Conclusion
In conclusion, sin 150 is an important concept in trigonometry that has real-world applications in various fields. By understanding the meaning and importance of sin 150, you can improve your understanding of trigonometry and its applications. Remember to practice and seek help when needed, and you’ll be on your way to mastering trigonometry in no time. |
# 2009 AIME I Problems/Problem 12
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
In right $\triangle ABC$ with hypotenuse $\overline{AB}$, $AC = 12$, $BC = 35$, and $\overline{CD}$ is the altitude to $\overline{AB}$. Let $\omega$ be the circle having $\overline{CD}$ as a diameter. Let $I$ be a point outside $\triangle ABC$ such that $\overline{AI}$ and $\overline{BI}$ are both tangent to circle $\omega$. The ratio of the perimeter of $\triangle ABI$ to the length $AB$ can be expressed in the form $\frac {m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
## Solution 1
Let $O$ be center of the circle and $P$,$Q$ be the two points of tangent such that $P$ is on $BI$ and $Q$ is on $AI$. We know that $AD:CD = CD:BD = 12:35$.
Since the ratios between corresponding lengths of two similar diagrams are equal, we can let $AD = 144, CD = 420$ and $BD = 1225$. Hence $AQ = 144, BP = 1225, AB = 1369$ and the radius $r = OD = 210$.
Since we have $\tan OAB = \frac {35}{24}$ and $\tan OBA = \frac{6}{35}$ , we have $\sin {(OAB + OBA)} = \frac {1369}{\sqrt {(1801*1261)}},$$\cos {(OAB + OBA)} = \frac {630}{\sqrt {(1801*1261)}}$.
Hence $\sin I = \sin {(2OAB + 2OBA)} = \frac {2*1369*630}{1801*1261}$. let $IP = IQ = x$ , then we have Area$(IBC)$ = $(2x + 1225*2 + 144*2)*\frac {210}{2}$ = $(x + 144)(x + 1225)* \sin {\frac {I}{2}}$. Then we get $x + 1369 = \frac {3*1369*(x + 144)(x + 1225)}{1801*1261}$.
Now the equation looks very complex but we can take a guess here. Assume that $x$ is a rational number (If it's not then the answer to the problem would be irrational which can't be in the form of $\frac {m}{n}$) that can be expressed as $\frac {a}{b}$ such that $(a,b) = 1$. Look at both sides; we can know that $a$ has to be a multiple of $1369$ and not of $3$ and it's reasonable to think that $b$ is divisible by $3$ so that we can cancel out the $3$ on the right side of the equation.
Let's see if $x = \frac {1369}{3}$ fits. Since $\frac {1369}{3} + 1369 = \frac {4*1369}{3}$, and $\frac {3*1369*(x + 144)(x + 1225)}{1801*1261} = \frac {3*1369* \frac {1801}{3} * \frac {1261*4}{3}} {1801*1261} = \frac {4*1369}{3}$. Amazingly it fits!
Since we know that $3*1369*144*1225 - 1369*1801*1261 < 0$, the other solution of this equation is negative which can be ignored. Hence $x = 1369/3$.
Hence the perimeter is $1225*2 + 144*2 + \frac {1369}{3} *2 = 1369* \frac {8}{3}$, and $BC$ is $1369$. Hence $\frac {m}{n} = \frac {8}{3}$, $m + n = 11$.
## Solution 2
As in Solution $1$, let $P$ and $Q$ be the intersections of $\omega$ with $BI$ and $AI$ respectively.
First, by pythagorean theorem, $AB = \sqrt{12^2+35^2} = 37$. Now the area of $ABC$ is $1/2*12*35 = 1/2*37*CD$, so $CD=\frac{420}{37}$ and the inradius of $\triangle ABI$ is $r=\frac{210}{37}$.
Now from $\triangle CDB \sim \triangle ACB$ we find that $\frac{BC}{BD} = \frac{AB}{BC}$ so $BD = BC^2/AB = 35^2/37$ and similarly, $AD = 12^2/37$.
Note $IP=IQ=x$, $BP=BD$, and $AQ=AD$. So we have $AI = 144/37+x$, $BI = 1225/37+x$. Now we can compute the area of $\triangle ABI$ in two ways: by heron's formula and by inradius times semiperimeter, which yields
$rs=210/37(37+x) = \sqrt{(37+x)(37-144/37)(37-1225/37)x}$
$210/37(37+x) = 12*35/37 \sqrt{x(37+x)}$
$37+x = 2 \sqrt{x(x+37)}$
$x^2+74x+1369 = 4x^2 + 148x$
$3x^2 + 74x - 1369 = 0$
The quadratic formula now yields $x=37/3$. Plugging this back in, the perimeter of $ABI$ is $2s=2(37+x)=2(37+37/3) = 37(8/3)$ so the ratio of the perimeter to $AB$ is $8/3$ and our answer is $8+3=\boxed{011}$
## Solution 3
As in Solution $2$, let $P$ and $Q$ be the intersections of $\omega$ with $BI$ and $AI$ respectively.
Recall that the distance from a point outside a circle to that circle is the same along both tangent lines to the circle drawn from the point.
Recall also that the length of the altitude to the hypotenuse of a right-angle triangle is the geometric mean of the two segments into which it cuts the hypotenuse.
Let $x = \overline{AD} = \overline{AQ}$. Let $y = \overline{BD} = \overline{BP}$. Let $z = \overline{PI} = \overline{QI}$. The semi-perimeter of $ABI$ is $x + y + z$. Since the lengths of the sides of $ABI$ are $x + y$, $y + z$ and $x + z$, the square of its area by Heron's formula is $(x+y+z)xyz$.
The radius $r$ of $\omega$ is $\overline{CD}/2$. Therefore $r^2 = xy/4$. As $\omega$ is the in-circle of $ABI$, the area of $ABI$ is also $r(x+y+z)$, and so the square area is $r^2(x+y+z)^2$.
Therefore $$(x+y+z)xyz = r^2(x+y+z)^2 = \frac{xy(x+y+z)^2}{4}$$ Dividing both sides by $xy(x+y+z)/4$ we get: $$4z = (x+y+z),$$ and so $z = (x+y)/3$. The semi-perimeter of $ABI$ is therefore $\frac{4}{3}(x+y)$ and the whole perimeter is $\frac{8}{3}(x+y)$. Now $x + y = \overline{AB}$, so the ratio of the perimeter of $ABI$ to the hypotenuse $\overline{AB}$ is $8/3$ and our answer is $8+3=\boxed{011}$
## Solution 4
We shall yet again let $P$ and $Q$ be the intersections of $AI$ and $BI$ to $\omega$, respectively. We want to find the perimeter of $ABI$, which is $AD+BD+BQ+QI+IP+PA$. We can easily find $AD$ and $BD$ using the fact that $ABC$, $ACD$, and $BCD$ are all similar triangles. We get $AD=\frac{144}{37}$ and $\frac{1225}{37}$. Since $AP$ and $AD$ are tangents to $\omega$, $AP=AD=\frac{144}{37}$, and similarly $BQ=BD=\frac{1225}{37}$. We now wish to find $IP$ and $IQ$.
Let the center of the given circle be $O$. We know that $\angle AOP=\angle AOD$, $\angle BOQ=\angle BOD$, and $\angle IOQ=\angle IOP$. Since all six angles sum to $360^{\circ}$, $\angle AOP+\angle BOQ+\angle IOP=180^{\circ}$. If we knew the radius of circle $\omega$ now, then we could find $\tan{\angle AOP}$ and $\tan{\angle BOQ}$, and then we can use the sum (or difference) of tangents formula to find $\tan{\angle IOP}$, which reveals $IP$. This means we should find the radius of $\omega$. We can easily see that the height of triangle $ABC$ from $C$ has length $\frac{12*35}{37}$, so the radius of $\omega$ is $\frac{210}{37}$. Now we shall proceed with the above plan.
$\tan{\angle AOP}=\frac{144}{210}$. $\tan{\angle BOQ}=\frac{1225}{210}$.
$\tan{\angle IOP}=\tan{(180^{\circ}-\angle AOP-\angle BOQ)}=-\tan{(\angle AOP+\angle BOQ)}$
$=-\frac{\frac{144}{210}+\frac{1225}{210}}{1-\frac{144}{210}*\frac{1225}{210}}=-\frac{1369}{210-\frac{144*1225}{210}}=\frac{1369}{\frac{144*1225}{210}-210}=\frac{37*37}{35*18}$.
Therefore $OP=\frac{210}{37},IP=\frac{37}{3}$, and the perimeter of $AIB$ is $2*\frac{37}{3}+2*\frac{144}{37}+2*\frac{1225}{37}=37*\frac{8}{3}$. Since $AB=37$, the desired ratio is $\frac{8}{3}$, and $8+3=\boxed{011}$.
## Solution 5
This solution is not a real solution and is solving the problem with a ruler and compass.
Draw $AC = 4.8, BC = 14, AB = 14.8$. Then, drawing the tangents and intersecting them, we get that $IA$ is around $6.55$ and $IB$ is around $18.1$. We then find the ratio to be around $\frac{39.45}{14.8}$. Using long division, we find that this ratio is approximately 2.666, which you should recognize as $\frac{8}{3}$. Since this seems reasonable, we find that the answer is $\boxed{11}$ ~ilp |
# 1995 AIME Problems/Problem 11
## Problem
A right rectangular prism $P_{}$ (i.e., a rectangular parallelpiped) has sides of integral length $a, b, c,$ with $a\le b\le c.$ A plane parallel to one of the faces of $P_{}$ cuts $P_{}$ into two prisms, one of which is similar to $P_{},$ and both of which have nonzero volume. Given that $b=1995,$ for how many ordered triples $(a, b, c)$ does such a plane exist?
## Solution
Let $P'$ be the prism similar to $P$, and let the sides of $P'$ be of length $x,y,z$, such that $x \le y \le z$. Then
$$\frac{x}{a} = \frac{y}{b} = \frac zc < 1.$$
Note that if the ratio of similarity was equal to $1$, we would have a prism with zero volume. As one face of $P'$ is a face of $P$, it follows that $P$ and $P'$ share at least two side lengths in common. Since $x < a, y < b, z < c$, it follows that the only possibility is $y=a,z=b=1995$. Then,
$$\frac{x}{a} = \frac{a}{1995} = \frac{1995}{c} \Longrightarrow ac = 1995^2 = 3^25^27^219^2.$$
The number of factors of $3^25^27^219^2$ is $(2+1)(2+1)(2+1)(2+1) = 81$. Only in $\left\lfloor \frac {81}2 \right\rfloor = 40$ of these cases is $a < c$ (for $a=c$, we end with a prism of zero volume). We can easily verify that these will yield nondegenerate prisms, so the answer is $\boxed{040}$. |
# Fraction calculator
This fraction calculator performs basic and advanced fraction operations, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. The calculator helps in finding value from multiple fractions operations. Solve problems with two, three, or more fractions and numbers in one expression.
## The result:
### 13/4 * 3 = 21/4 = 5 1/4 = 5.25
Spelled result in words is twenty-one quarters (or five and one quarter).
### How do we solve fractions step by step?
1. Conversion a mixed number 1 3/4 to a improper fraction: 1 3/4 = 1 3/4 = 1 · 4 + 3/4 = 4 + 3/4 = 7/4
To find a new numerator:
a) Multiply the whole number 1 by the denominator 4. Whole number 1 equally 1 * 4/4 = 4/4
b) Add the answer from the previous step 4 to the numerator 3. New numerator is 4 + 3 = 7
c) Write a previous answer (new numerator 7) over the denominator 4.
One and three quarters is seven quarters.
2. Multiple: 7/4 * 3 = 7 · 3/4 · 1 = 21/4
Multiply both numerators and denominators. Result fraction keep to lowest possible denominator GCD(21, 4) = 1. In the following intermediate step, it cannot further simplify the fraction result by canceling.
In other words - seven quarters multiplied by three is twenty-one quarters.
#### Rules for expressions with fractions:
Fractions - use a forward slash to divide the numerator by the denominator, i.e., for five-hundredths, enter 5/100. If you use mixed numbers, leave a space between the whole and fraction parts.
Mixed numerals (mixed numbers or fractions) keep one space between the integer and
fraction and use a forward slash to input fractions i.e., 1 2/3 . An example of a negative mixed fraction: -5 1/2.
Because slash is both sign for fraction line and division, use a colon (:) as the operator of division fractions i.e., 1/2 : 1/3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
### Math Symbols
SymbolSymbol nameSymbol MeaningExample
+plus signaddition 1/2 + 1/3
-minus signsubtraction 1 1/2 - 2/3
*asteriskmultiplication 2/3 * 3/4
×times signmultiplication 2/3 × 5/6
:division signdivision 1/2 : 3
/division slashdivision 1/3 / 5
:coloncomplex fraction 1/2 : 1/3
^caretexponentiation / power 1/4^3
()parenthesescalculate expression inside first-3/5 - (-1/4) |
# Slope Intercept Form To Standard Form
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
Slope Intercept Form To Standard Form – There are many forms used to represent a linear equation, among the ones most frequently found is the slope intercept form. The formula of the slope-intercept find a line equation assuming that you have the straight line’s slope , and the yintercept, which is the coordinate of the point’s y-axis where the y-axis crosses the line. Read more about this particular linear equation form below.
## What Is The Slope Intercept Form?
There are three basic forms of linear equations: the traditional one, the slope-intercept one, and the point-slope. While they all provide the same results when utilized, you can extract the information line that is produced quicker by using the slope intercept form. Like the name implies, this form makes use of an inclined line where the “steepness” of the line is a reflection of its worth.
The formula can be used to find the slope of straight lines, the y-intercept, also known as x-intercept where you can apply different formulas available. The line equation of this particular formula is y = mx + b. The slope of the straight line is symbolized through “m”, while its y-intercept is represented by “b”. Every point on the straight line is represented as an (x, y). Note that in the y = mx + b equation formula the “x” and the “y” have to remain as variables.
## An Example of Applied Slope Intercept Form in Problems
In the real world in the real world, the slope-intercept form is frequently used to depict how an object or problem evolves over the course of time. The value that is provided by the vertical axis represents how the equation addresses the extent of changes over the amount of time indicated via the horizontal axis (typically in the form of time).
One simple way to illustrate using this formula is to find out how much population growth occurs in a specific area in the course of time. If the population in the area grows each year by a specific fixed amount, the point value of the horizontal axis will rise one point at a moment for every passing year, and the amount of vertically oriented axis is increased to represent the growing population by the fixed amount.
You can also note the beginning value of a problem. The beginning value is at the y’s value within the y’intercept. The Y-intercept is the place at which x equals zero. If we take the example of a problem above, the starting value would be at the point when the population reading starts or when the time tracking begins along with the associated changes.
Thus, the y-intercept represents the point when the population is beginning to be recorded in the research. Let’s assume that the researcher begins with the calculation or the measurement in the year 1995. This year will be the “base” year, and the x = 0 point would be in 1995. Thus, you could say that the population of 1995 represents the “y”-intercept.
Linear equation problems that use straight-line formulas are nearly always solved this way. The starting value is depicted by the y-intercept and the change rate is represented through the slope. The principal issue with the slope intercept form typically lies in the horizontal interpretation of the variable in particular when the variable is attributed to an exact year (or any kind of unit). The most important thing to do is to make sure you are aware of the meaning of the variables. |
# Which addition can be performed without regrouping?
## Which addition can be performed without regrouping?
How to Add without Regrouping. Place the addends one on top of the other so that the place values fall in the same columns. Add each column together separately, starting with the 1s place. The sums go below each column, underneath the line.
### What is double digit addition?
Adding double-digit numbers is just like adding single-digit values, you just go through the process of addition more than once. When you add two-digit numbers you add the columns together, not the entire numbers. For example, you wouldn’t want to try and add 78 and 57 as complete numbers in your head.
What does it mean without regrouping?
Addition without regrouping is when the digits add up to a number that is 9 or less. The answer can simply be written below each place value column. There is no carrying of tens or hundreds. When talking about adding numbers, regrouping means the same as carrying.
What is multiplication without regrouping?
Lesson Summary Solving multi-digit multiplication problems can be done with or without regrouping, which is placing your numbers in another group. Without regrouping, you use basic facts and other tricks to find the product, or the answer to the multiplication problem.
## How does regrouping explain addition?
Addition with regrouping is a technique used in Maths when adding together two or more numbers of any size. It is used with the column method of addition, where sums are arranged vertically, and numbers are added one column at a time. You may also hear regrouping referred to as “carrying over”.
### How do you teach regrouping?
You stack the numbers one on top of the other, line them up by place value, and add each column. If one column makes a sum greater than or equal to 10, you regroup the number and place the extra ten in the column to the left. (This process is sometimes called carrying.)
What does addition without regrouping mean?
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# Fallacy of the Isosceles Triangle
Throughout the last of couple of weeks, we've been having some fun with geometrical fallacies. We've seen what makes them work and how to spot the wrong arguments that constitute the fallacious proofs. Today we're going to end the series with a classic geometrical fallacy; one that is possibly the most popular when it comes to geometric fallacies: the "proof" that all triangles are isosceles.
This fallacy has been attributed to Charles Dodgson who is better known as Lewis Carroll. Most people know Carroll as the author of the children's book, Alice's Adventure In Wonderland. But what some people don't know is Carroll was also a noted mathematician and logician who loved some recreational mathematics.
This post is going to be pretty much like the earlier ones. I first present you with the "proof". Then I ask you think about what went wrong and why. Finally I walk you through the steps and reveal the invalid argument. So let's get on with it!
Imgur
Let $ABC$ be a triangle. Draw the angle bisector of $\angle A$ and the perpendicular bisector of $BC$. Notice that if they are the same line, $\triangle ABC$ is isosceles. If they are not the same line, they're going to intersect at a point. Let's call that $D$.
Let $E$ be the midpoint of $BC$. $F$ and $G$ are the feet of perpendiculars from $D$ to $AB$ and $AC$ respectively. Join $B, D$ and $C, D$.
Notice that $\triangle BDE$ and $\triangle CDE$ are congruent [SAS], because $BE=CE$, $DE=DE$ and $\angle DEB=\angle DEC$ since they are equal to a right angle.
So, now we can write $BD=CD$ $\cdots (1)$.
Triangles $ADF$ and $ADG$ are also congruent [AAS] as $\angle DFA=\angle DGA=90^\circ$, $\angle FAD=\angle GAD$ [remember that $AD$ was the angle bisector of $\angle BAC$] and $AD=AD$.
From that we can write $DF=DG$ $\cdots (2)$
And $AF=AG$ $\cdots (3)$.
Now we move on to triangles $DBF$ and $DCG$.
From $(1)$ and $(2)$ and from the fact that $\angle DFB$ and $\angle DGC$ are right angles, we can conclude that triangles $DBF$ and $DCG$ are also congruent.
And that means $FB=GC$ $\cdots (4)$.
Now add $(3)$ and $(4)$ together and see what happens.
$AF+FB=AG+GC$
$\Rightarrow AB=AC$
That means $\triangle ABC$ is isosceles! How did that happen? You can use the similar arguments to prove $BC=AC$ and prove that all triangles are actually equilateral!
Now comes the slightly over-asked question: what went wrong? Go back to the steps and try to figure it out. Don't continue reading if you don't want to know the solution without giving it a shot.
I take it that you've thought about the fallacy for a while. Have you figured it out? If repeatedly talking about fallacies taught us one thing, it's this: drawing a sufficiently accurate picture always helps. And that advice isn't restricted only to fallacies like this. You can use this even while solving normal geometry problems. An accurate figure helps you make good educated guesses [for example, "the lines sure look like they're parallel, let's prove it! And that angle looks like a right angle. Can I prove it?...]. So, let's take a look at a more accurate picture:
Imgur
This actually makes sense. The point $D$ is actually outside the triangle. We never proved where $D$ was. We only assumed [incorrectly] that it was inside. This also tells us that everything up to the last step was absolutely correct. The lengths were equal. The triangles were congruent as well. Even $AF+FB=AG+GC$ was correct. But the transition to the final step was not. We made an incorrect assumption that $AG+GC=AC$. If $AB>AC$, $AC=AG-GC$, not $AG+GC$. This is what made us arrive at a wrong conclusion.
So, there you have it! This is the last post of the geometric fallacies series. I hope you had as much fun reading the posts as I had writing them. And thank you for staying with me till the end. If you missed one or two posts, you can check out my feed here or try the #mathematicalfallacies tag.
Until next time!
Note by Mursalin Habib
6 years, 4 months ago
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That was it! The last post on the geometrical fallacies series! I hope you enjoy this one as well. Feel free to post any feedback in the comments section.
And thank you, thank you, thank you for staying with me.
- 6 years, 4 months ago
Your initial diagram can be made to look more convincing by drawing a triangle closer to isosceles. However, the accurate diagram would be harder to draw.
Other than that, great posts! I loved them.
This makes me want to post more, although I am pretty busy... :P
- 6 years, 4 months ago
Thanks, Daniel, for your kind words. It really does mean a lot!
And about making the diagram more convincing: if my triangle were any closer to an isosceles triangle, point $D$ would be really close to $BC$. And people don't draw diagrams that accurately as well. Ask people to draw a diagram based on the instructions above and if they don't use compasses or rulers, they are going get a picture that is quite similar to mine, specially if you add the line "let the angle bisector of $\angle A$ and the perpendicular bisector of $BC$ intersect at $D$" [this psychologically influences them to assume that $D$ is inside the triangle].
It's interesting that people are discussing about why the picture's wrong when the more important thing is why the argument's wrong. You don't have to draw pictures 100% accurately make geometric arguments.
Once again, thanks for your feedback! I appreciate it a lot! And I'm looking forward to your posts. Keep them coming :)
- 6 years, 4 months ago
How do you think of such things , can you suggest me from where do I have to start?
- 5 years, 4 months ago
I believe he was sharing a well-known incorrect proof. I don't think he was the one who invented it.
- 5 years, 4 months ago
Yes, I think I mentioned that this fallacy's been attributed to Lewis Carroll.
- 5 years, 4 months ago
Sorry I commented here , I was reading all the notes of the set geometric fallacies one by one , i was astonished to see this fallacy - Are you inside or outside
Thanks
- 5 years, 4 months ago
Cool!
- 6 years, 4 months ago
I read your PDF on it in Bengali a month ago and figured out the mistake there instantly when I looked at the picture. The difference between that one and this one is that in that one, you bisected $\angle A$ correctly, but did not point out the midpoint of $BC$ accurately. So anyone could catch the mistake instantly. In fact, before reading the whole, I quickly did a GeoGebra check and found out that the point $D$ is actually outside the triangle.
Comparing to that one, I like the picture in this one as you pointed out the midpoint of $BC$ accurately, but did not bisect $\angle A$ accurately. It's far more difficult to catch the mistake from bisected angle than from the midpoint.
And...... Great posts! I had fun!
- 6 years, 4 months ago
Where on earth did you read that!?
By the way, the point of this [and the other posts on fallacies] was not to find the mistakes in the pictures. The point was to find which arguments were invalid and learn that you should never ever assume anything without proof. You line of thought should be like this: "Yeah, the picture looks fishy, but then again you can't draw anything 100% accurately. So, what is wrong with the arguments presented?" Do you draw diagrams with compasses and rulers whenever you solve a geometry problem or do you just draw something that looks okay.
So, why am I saying this? I'm saying this to emphasize that the "proof" isn't wrong because the picture is wrong. The proof is wrong because we made an implicit assumption. Yes, the inaccurate picture helped us make that assumption but that is not what's wrong. I hope you understand this.
By the way, the picture from the pdf: that was from Wikipedia. So, that's Wikipedia's fault, not mine:)
- 6 years, 4 months ago
"I present you wit the 'proof.'" Typo paragraph 3 :)
- 6 years, 4 months ago
It seems like I can't break the habit of making stupid typos. Thanks for pointing it out!
- 6 years, 4 months ago
This is very interesting. But you don't give any proof or reason to support the fact that D is outside the triangle. The proof is actually an extension of the 'angle bisector theorem' which basically proves that the bisector will never cross the perpendicular bisector of the opposite side (unless it's an isosceles, in which case the angle bisector is also the perpendicular bisector).
- 2 years, 6 months ago
asked me in kvpy interview
- 2 years, 3 months ago |
The weights (in kg.) of 15 students of a class are:38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47$(i)$. Find the mode and median of this data.$(ii)$. Is there more than one mode?
Given:
The weights (in kg.) of 15 students of a class are:
38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47
To do:
We have to find:
(i) The mode and median of this data.
(ii) If there is more than one mode.
Solution:
Total number of students $=15$
Weights of $15$ students$=$38, 42, 35, 37, 45, 50, 32, 43,43, 40, 36, 38, 43, 38, 47
Arranging in ascending order, we get,
32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50
(i) Mode is the observation that occurs the highest number of times in a data.
Thus, 38 and 43 occur the highest number of times
$\therefore$ Mode$=38$ and $43$
We know that the median is the middle observation of a given data.
Also, out of the 15 observations given, the 8th observation is the mid data.
Hence, median$=40$ [8th observation]
(ii) Yes, there are two modes $38$ and $43$.
Updated on: 10-Oct-2022
60 Views |
# 1.7: Tangent Planes and Normal Lines
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## Tangent Planes
Let $$z = f(x,y)$$ be a function of two variables. We can define a new function $$F(x,y,z)$$ of three variables by subtracting $$z$$. This has the condition
$F(x,y,z) = 0.\nonumber$
Now consider any curve defined parametrically by
$x = x(t), \;\;\; y = y(t), \;\;\; z = z(t).\nonumber$
We can write,
$F(x(t), y(t), z(t)) = 0.\nonumber$
Differentiating both sides with respect to $$t$$, and using the chain rule gives
$F_x(x, y, z) x' + F_y(x, y, z) y' + F_z(x, y, z) z' = 0\nonumber$
Notice that this is the dot product of the gradient function and the vector $$\langle x',y',z'\rangle$$,
$\nabla F \cdot \langle x', y', z'\rangle = 0.\nonumber$
In particular the gradient vector is orthogonal to the tangent line of any curve on the surface. This leads to:
Definition: Tangent Plane
Let $$F(x,y,z)$$ define a surface that is differentiable at a point $$(x_0,y_0,z_0)$$, then the tangent plane to $$F ( x, y, z )$$ at $$( x_0, y_0, z_0)$$ is the plane with normal vector
$\nabla \, F(x_0,y_0,z_0) \nonumber$
that passes through the point $$(x_0,y_0,z_0)$$. In particular, the equation of the tangent plane is
$\nabla \, F(x_0,y_0,z_0) \cdot \langle x - x_0 , y - y_0 , z - z_0 \rangle = 0. \nonumber$
Example $$\PageIndex{1}$$
Find the equation of the tangent plane to
$z = 3x^2 - xy \nonumber$
at the point $$(1,2,1)$$.
Solution
We let
$F(x,y,z) = 3x^2 - xy - z\nonumber$
then
$\nabla F = \langle 6x - y, -x, -1\rangle . \nonumber$
At the point $$(1,2,1)$$, the normal vector is
$\nabla F(1,2,1) = \langle 4, -1, -1\rangle . \nonumber$
Now use the point normal formula for a plan
$\langle 4, -1, -1\rangle \cdot \langle x - 1, y - 2, z - 1\rangle = 0\nonumber$
or
$4(x - 1) - (y - 2) - (z - 1) = 0.\nonumber$
Finally we get
$4x - y - z = 1.\nonumber$
## Normal Lines
Given a vector and a point, there is a unique line parallel to that vector that passes through the point. In the context of surfaces, we have the gradient vector of the surface at a given point. This leads to the following definition.
Definition: Normal Line
Let $$F(x,y,z)$$ define a surface that is differentiable at a point $$(x_0,y_0,z_0)$$, then the normal line to $$F(x,y,z)$$ at $$(x_0,y_0,z_0)$$ is the line with normal vector
$\nabla \, F(x_0,y_0,z_0) .\nonumber$
that passes through the point $$(x_0,y_0,z_0)$$. In Particular the equation of the normal line is
$x(t) = x_0 + F_x(x_0,y_0,z_0) t, \nonumber$
$y(t) = y_0 + F_y(x_0,y_0,z_0) t, \nonumber$
$z(t) = z_0 + F_z(x_0,y_0,z_0) t. \nonumber$
Example $$\PageIndex{2}$$
Find the parametric equations for the normal line to
$x^2yz - y + z - 7 = 0 \nonumber$
at the point $$(1,2,3)$$.
Solution
We compute the gradient:
$\nabla F = \langle 2xyz, x^2z - 1, x^2y + 1\rangle = \langle 12, 2, 3\rangle .\nonumber$
Now use the formula to find
$x(t) = 1 + 12t, \;\;\; y(t) = 2 + 2t, \;\;\; z(t) = 3 + 3t.\nonumber$
The diagram below displays the surface and the normal line.
## Angle of Inclination
Given a plane with normal vector n the angle of inclination, $$q$$ is defined by
$\cos q = \dfrac{|\textbf{n} \cdot k|}{ ||\textbf{n} ||}. \nonumber$
More generally, if $$F(x,y,z) = 0$$ is a surface, then the angle of inclination at the point $$(x_0,y_0,z_0)$$ is defined by the angle of inclination of the tangent plane at the point with
$\cos\,q = \dfrac{ | \nabla F(x_0, y_0, z_0) \cdot \textbf{k}| }{|| \nabla F(x_0, y_0, z_0)||}. \nonumber$
Example $$\PageIndex{3}$$
Find the angle of inclination of
$\dfrac{x^2}{4} + \dfrac{y^2}{4} + \dfrac{z^2}{8} = 1\nonumber$
at the point $$(1,1,2)$$.
Solution
First compute
$\nabla F = \langle \dfrac{x}{2}, \dfrac{y}{2}, \dfrac{z}{4}\rangle .\nonumber$
Now plug in to get
$\nabla F(1,1,2) = \langle \dfrac{1}{2}, \dfrac{1}{2}, \dfrac{1}{2} \rangle .\nonumber$
We have
$|\langle \dfrac{1}{2} , \dfrac{1}{2} , \dfrac{1}{2} \rangle \cdot \hat{\textbf{k}} | = \dfrac{1}{2} .\nonumber$
Also,
$||\langle \dfrac{1}{2} , \dfrac{1}{2} , \dfrac{1}{2} \rangle || = \dfrac{\sqrt{3}}{2} .\nonumber$
Hence
$\cos q = \dfrac{\frac{1}{2}}{( \frac{\sqrt{3}}{2} )} = \dfrac{1}{\sqrt{3}} . \nonumber$
So the angle of inclination is
$q = \cos^{-1}(\dfrac{1}{\sqrt{3}}) = 0.955 \text{ radians} .\nonumber$
## The Tangent Line to a Curve
Example $$\PageIndex{4}$$
Find the tangent line to the curve of intersection of the sphere
$x^2 + y^2 + z^2 = 30\nonumber$
and the paraboloid
$z = x^2 + y^2\nonumber$
at the point $$(1,2,5)$$.
Solution
We find the gradient of the two surfaces at the point
$\nabla(x^2 + y^2 + z^2) = \langle 2x, 2y, 2z\rangle = \langle 2, 4,10\rangle \nonumber$
and
$\nabla (x^2 + y^2 - z) = \langle 2x, 2y, -1\rangle = \langle 2, 4, -1\rangle .\nonumber$
These two vectors will both be perpendicular to the tangent line to the curve at the point, hence their cross product will be parallel to this tangent line. We compute
$\begin{vmatrix} \hat{\textbf{i}} & \hat{\textbf{j}} & \hat{\textbf{k}} \\ 2 & 4 & 10 \\ 2 & 4 & -1 \end{vmatrix} = -44 \hat{\textbf{i}} + 22 \hat{\textbf{j}}. \nonumber$
Hence the equation of the tangent line is
$x(t) = 1 - 44t y(t) = 2 + 22t z(t) = 5.\nonumber$
## Contributors and Attributions
This page titled 1.7: Tangent Planes and Normal Lines is shared under a not declared license and was authored, remixed, and/or curated by Larry Green. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 8.3: Zero, Negative, and Fractional Exponents
Difficulty Level: At Grade Created by: CK-12
In the previous lessons, we have dealt with powers that are positive whole numbers. In this lesson, you will learn how to solve expressions when the exponent is zero, negative, or a fractional number.
Exponents of Zero: For all real numbers \begin{align*}\chi, \chi \neq 0, \chi^0=1\end{align*}.
Example: \begin{align*}\frac{\chi^4}{\chi^4} = \chi^{4-4} = \chi^0 = 1\end{align*}. This example is simplified using the Quotient of Powers Property.
## Simplifying Expressions with Negative Exponents
The next objective is negative exponents. When we use the quotient rule and we subtract a greater number from a smaller number, the answer will become negative. The variable and the power will be moved to the denominator of a fraction. You will learn how to write this in an expression.
Example: \begin{align*}\frac{x^4}{x^6} =x^{4-6}=x^{-2}=\frac{1}{x^2}\end{align*}. Another way to look at this is \begin{align*}\frac{\chi \cdot \chi \cdot \chi \cdot \chi}{\chi \cdot \chi \cdot \chi \cdot \chi \cdot \chi \cdot \chi}\end{align*}. The four \begin{align*}\chi\end{align*}s on top will cancel out with four \begin{align*}\chi\end{align*}s on bottom. This will leave two \begin{align*}\chi\end{align*}s remaining on the bottom, which makes your answer look like \begin{align*}\frac{1}{\chi^2}\end{align*}.
Negative Power Rule for Exponents: \begin{align*}\frac{1}{\chi^n} = \chi^{-n}\end{align*} where \begin{align*}\chi \neq 0\end{align*}
Example: \begin{align*}\chi^{-6} \gamma^{-2}= \frac{1}{\chi^6} \cdot \frac{1}{\gamma^2} = \frac{1}{\chi^6 \gamma^2}\end{align*}. The negative power rule for exponents is applied to both variables separately in this example.
Multimedia Link: For more help with these types of exponents, watch this http://www.phschool.com/atschool/academy123/english/academy123_content/wl-book-demo/ph-241s.html - PH School video or visit the http://www.mathsisfun.com/algebra/negative-exponents.html - mathisfun website.
Example 1: Write the following expressions without fractions.
(a) \begin{align*}\frac{2}{x^2}\end{align*}
(b) \begin{align*}\frac{x^2}{y^3}\end{align*}
Solution:
(a) \begin{align*}\frac{2}{x^2}=2x^{-2}\end{align*}
(b) \begin{align*}\frac{x^2}{y^3}=x^2y^{-3}\end{align*}
Notice in Example 1(a), the number 2 is in the numerator. This number is multiplied with \begin{align*}\chi^{-2}\end{align*}. It could also look like this, \begin{align*}2 \cdot \frac{1}{\chi^2}\end{align*} to be better understood.
## Simplifying Expressions with Fractional Exponents
The next objective is to be able to use fractions as exponents in an expression.
Roots as Fractional Exponents: \begin{align*}\sqrt[m]{a^n}=a^{\frac{n}{m}}\end{align*}
Example: \begin{align*}\sqrt{a}=a^{\frac{1}{2}}, \sqrt[3]{a}=a^{\frac{1}{3}}, \sqrt[5]{a^2}=(a^2)^{\frac{1}{5}}=a^{\frac{2}{5}}=a^{\frac{2}{5}}\end{align*}
Example 2: Simplify the following expressions.
(a) \begin{align*}\sqrt[3]{\chi}\end{align*}
(b) \begin{align*}\sqrt[4]{\chi^3}\end{align*}
Solution:
(a) \begin{align*}\chi^{\frac{1}{3}}\end{align*}
(b) \begin{align*}\chi^{\frac{3}{4}}\end{align*}
It is important when evaluating expressions that you remember the Order of Operations. Evaluate what is inside the parentheses, then evaluate the exponents, then perform multiplication/division from left to right, then perform addition/subtraction from left to right.
Example 3: Evaluate the following expression.
(a) \begin{align*}3 \cdot 5^2 - 10 \cdot 5+1\end{align*}
Solution: \begin{align*}3 \cdot 5^2-10 \cdot 5+1=3 \cdot 25-10 \cdot 5+1=75-50+1=26\end{align*}
## Practice Set
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both.
Simplify the following expressions. Be sure the final answer includes only positive exponents.
1. \begin{align*}x^{-1} \cdot y^2\end{align*}
2. \begin{align*}x^{-4}\end{align*}
3. \begin{align*}\frac{x^{-3}}{x^{-7}}\end{align*}
4. \begin{align*}\frac{1}{x}\end{align*}
5. \begin{align*}\frac{2}{x^2}\end{align*}
6. \begin{align*}\frac{x^2}{y^3}\end{align*}
7. \begin{align*}\frac{3}{xy}\end{align*}
8. \begin{align*}3x^{-3}\end{align*}
9. \begin{align*}a^2b^{-3}c^{-1}\end{align*}
10. \begin{align*}4x^{-1}y^3\end{align*}
11. \begin{align*}\frac{2x^{-2}}{y^{-3}}\end{align*}
12. \begin{align*}a^{\frac{1}{2}} \cdot a^{\frac{1}{3}}\end{align*}
13. \begin{align*}\left(a^{\frac{1}{3}}\right)^2\end{align*}
14. \begin{align*}\frac{a^{\frac{5}{2}}}{a^{\frac{1}{2}}}\end{align*}
15. \begin{align*}\left(\frac{x^2}{y^3}\right)^{\frac{1}{3}}\end{align*}
16. \begin{align*}\frac{x^{-3}y^{-5}}{z^{-7}}\end{align*}
17. \begin{align*}(x^{\frac{1}{2}} y^{-\frac{2}{3}})(x^2 y^{\frac{1}{3}})\end{align*}
18. \begin{align*}\left(\frac{a}{b}\right)^{-2}\end{align*}
19. \begin{align*}(3a^{-2}b^2c^3)^3\end{align*}
20. \begin{align*}x^{-3} \cdot x^3\end{align*}
Simplify the following expressions without any fractions in the answer.
1. \begin{align*}\frac{a^{-3}(a^5)}{a^{-6}}\end{align*}
2. \begin{align*}\frac{5x^6y^2}{x^8y}\end{align*}
3. \begin{align*}\frac{(4ab^6)^3}{(ab)^5}\end{align*}
4. \begin{align*}\left(\frac{3x}{y^{\frac{1}{3}}}\right)^3\end{align*}
5. \begin{align*}\frac{4a^2b^3}{2a^5b}\end{align*}
6. \begin{align*}\left(\frac{x}{3y^2}\right)^3 \cdot \frac{x^2y}{4}\end{align*}
7. \begin{align*}\left(\frac{ab^{-2}}{b^3}\right)^2\end{align*}
8. \begin{align*}\frac{x^{-3}y^2}{x^2y^{-2}}\end{align*}
9. \begin{align*}\frac{3x^2y^{\frac{3}{2}}}{xy^{\frac{1}{2}}}\end{align*}
10. \begin{align*}\frac{(3x^3)(4x^4)}{(2y)^2}\end{align*}
11. \begin{align*}\frac{a^{-2}b^{-3}}{c^{-1}}\end{align*}
12. \begin{align*}\frac{x^{\frac{1}{2}}y^{\frac{5}{2}}}{x^{\frac{3}{2}}y^{\frac{3}{2}}}\end{align*}
Evaluate the following expressions to a single number.
1. \begin{align*}3^{-2}\end{align*}
2. \begin{align*}(6.2)^0\end{align*}
3. \begin{align*}8^{-4} \cdot 8^6\end{align*}
4. \begin{align*}(16^{\frac{1}{2}})^3\end{align*}
5. \begin{align*}5^0\end{align*}
6. \begin{align*}7^2\end{align*}
7. \begin{align*}\left(\frac{2}{3}\right)^3\end{align*}
8. \begin{align*}3^{-3}\end{align*}
9. \begin{align*}16^{\frac{1}{2}}\end{align*}
10. \begin{align*}8^{\frac{-1}{3}}\end{align*}
In 43 – 45, evaluate the expression for \begin{align*}x=2, y=-1, z=3\end{align*}.
1. \begin{align*}2x^2-3y^3+4z\end{align*}
2. \begin{align*}(x^2-y^2)^2\end{align*}
3. \begin{align*}\left(\frac{3x^2y^5}{4z}\right)^{-2}\end{align*}
4. Evaluate \begin{align*}x^24x^3y^44y^2\end{align*} if \begin{align*}x=2\end{align*} and \begin{align*}y=-1\end{align*}.
5. Evaluate \begin{align*}a^4(b^2)^3+2ab\end{align*} if \begin{align*}a=-2\end{align*} and \begin{align*}b=1\end{align*}.
6. Evaluate \begin{align*}5x^2-2y^3+3z\end{align*} if \begin{align*}x=3, \ y=2,\end{align*} and \begin{align*}z=4\end{align*}.
7. Evaluate \begin{align*}\left(\frac{a^2}{b^3}\right)^{-2}\end{align*} if \begin{align*}a=5\end{align*} and \begin{align*}b=3\end{align*}.
8. Evaluate \begin{align*}3 \cdot 5^5 - 10 \cdot 5+1\end{align*}.
9. Evaluate \begin{align*}\frac{2 \cdot 4^2-3 \cdot 5^2}{3^2}\end{align*}.
10. Evaluate \begin{align*}\left(\frac{3^3}{2^2}\right)^{-2} \cdot \frac{3}{4}\end{align*}.
Mixed Review
1. A quiz has ten questions: 7 true/false and 3 multiple choice. The multiple choice questions each have four options. How many ways can the test be answered?
2. Simplify \begin{align*}3a^4 b^4 \cdot a^{-3} b^{-4}\end{align*}.
3. Simplify \begin{align*}(x^4 y^2 \cdot xy^0)^5\end{align*}.
4. Simplify \begin{align*}\frac{v^2}{-vu^{-2} \cdot u^{-1} v^4}\end{align*}.
5. Solve for \begin{align*}n: -6(4n+3)=n+32\end{align*}.
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$\newcommand{\dollar}{\} \DeclareMathOperator{\erf}{erf} \DeclareMathOperator{\arctanh}{arctanh} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&}$
## Section2.3The Product and Quotient Rules
###### Motivating Questions
• How does the algebraic structure of a function guide us in computing its derivative using shortcut rules?
• How do we compute the derivative of a product of two basic functions in terms of the derivatives of the basic functions?
• How do we compute the derivative of a quotient of two basic functions in terms of the derivatives of the basic functions?
• How do the product and quotient rules combine with the sum and constant multiple rules to expand the library of functions we can differentiate quickly?
So far, we can differentiate power functions $x^n\text{,}$ exponential functions $a^x\text{,}$ and the two fundamental trigonometric functions $\sin(x)$ and $\cos(x)\text{.}$ With the sum rule and constant multiple rules, we can also compute the derivative of combined functions.
###### Example2.34
Differentiate
\begin{equation*} f(x) = 7x^{11} - 4 \cdot 9^x + \pi \sin(x) - \sqrt{3}\cos(x)\text{.} \end{equation*}
Hint
Use a combination of all the differentiation rules we've learned in this chapter.
$f'(x) = 77x^{10} - 4 \cdot 9^x \ln(9) + \pi \cos(x) + \sqrt{3} \sin(x)\text{.}$
Solution
Because $f$ is a sum of basic functions, we can now quickly say that $f'(x)=77x^{10}-4\cdot9^x\ln(9)+\pi\cos(x)+\sqrt{3}\sin(x)\text{.}$
What about a product or quotient of two basic functions, such as $p(z) = z^3 \cos(z) \text{,}$ or $q(t) = \frac{\sin(t)}{2^t} \text{?}$
While the derivative of a sum is the sum of the derivatives, it turns out that the rules for computing derivatives of products and quotients are more complicated.
###### Example2.35
Let $f$ and $g$ be the functions defined by $f(t) = 2t^2$ and $g(t) = t^3 + 4t\text{.}$
1. Determine $f'(t)$ and $g'(t)\text{.}$
2. Let $p(t) = 2t^2 (t^3 + 4t)$ and observe that $p(t) = f(t) \cdot g(t)\text{.}$ Rewrite the formula for $p$ by distributing the $2t^2$ term. Then compute $p'(t)$ using the sum and constant multiple rules.
3. True or false: $p'(t) = f'(t) \cdot g'(t)\text{.}$
4. Let $q(t) = \frac{t^3 + 4t}{2t^2}$ and observe that $q(t) = \frac{g(t)}{f(t)}\text{.}$ Rewrite the formula for $q$ by dividing each term in the numerator by the denominator and simplify to write $q$ as a sum of constant multiples of powers of $t\text{.}$ Then compute $q'(t)$ using the sum and constant multiple rules.
5. True or false: $q'(t) = \frac{g'(t)}{f'(t)}\text{.}$
Hint
1. Use a combination of the power rule, the constant multiple rule, and the sum rule.
2. $p(t)$ is a polynomial.
3. Multiply your answers from (a) together; does the result match what you found in (b)?
4. Start by splitting the fraction into two pieces, then simplify.
5. Divide your answers from (a) in the correct order; does this match what you found in (d)?
1. $f'(t)=4t$ and $g'(t)=3t^2+4\text{.}$
2. Since $p(t)=2t^5+8t^3\text{,}$ then $p'(t)=10t^4+24t^2\text{.}$
3. False, $p'(t)\neq f'(t)\cdot g'(t)\text{.}$
4. Since $q(t)=\frac12t+2t^{-1}\text{,}$ then $q'(t)=\frac12-2t^{-2}\text{.}$
5. False, $q'(t)\neq\frac{g'(t)}{f'(t)}\text{.}$
Solution
1. Using the power rule and the constant multiple rule, we see that
\begin{equation*} f'(t)=\frac{d}{dt}[2t^2]=2\frac{d}{dt}[t^2]=2(2t)=4t\text{.} \end{equation*}
Using the same rules together with the sum rule, we see that
\begin{equation*} g'(t)=\frac{d}{dt}[t^3+4t]=\frac{d}{dt}[t^3]+4\frac{d}{dt}[t]=(3t^2)+4(1)=3t^2+4\text{.} \end{equation*}
2. Observe that $p(t)=2t^2(t^3+4t)=2t^5+8t^3\text{.}$ Since $\frac{d}{dt}[t^5]=5t^4$ and $\frac{d}{dt}[t^3]=3t^2\text{,}$ we can use the sum and constant multiple rules to say that the derivative of $p$ with respect to $t$ is
\begin{equation*} p'(t)=\frac{d}{dt}[2t^5+8t^3]=2(5t^4)+8(3t^2)=10t^4+24t^2\text{.} \end{equation*}
3. From our answers in (a), we know
\begin{equation*} f'(t)\cdot g'(t)=(4t)\cdot(3t^2+4)=12t^3+16t\text{.} \end{equation*}
This does not match what we found in (b) as the formula for $p'(t)=10t^4+24t^2\text{.}$ We conclude that even though $p(t)=f(t)\cdot g(t)\text{,}$ it is not the case that $p'(t)=f'(t)\cdot g'(t)\text{.}$ In other words, the derivative of a product of two functions need not be the product of the derivatives.
4. To put $q(t)$ into a form we can differentiate, notice that
\begin{equation*} q(t)=\frac{t^3+4t}{2t^2}=\frac{t^3}{2t^2}+\frac{4t}{2t^2}=\frac12t+2t^{-1}\text{.} \end{equation*}
Now since $\frac{d}{dt}[t]=1$ and $\frac{d}{dt}[t^{-1}]=-t^{-2}\text{,}$ we use the sum and constant multiple rules to find
\begin{equation*} q'(t)=\frac12(1)+2(-t^{-2})=\frac12-2t^{-2}\text{.} \end{equation*}
5. From our answers in (a), we know
\begin{equation*} \frac{g'(t)}{f'(t)}=\frac{3t^2+4}{4t}=\frac34t+t^{-1}\text{.} \end{equation*}
This does not match what we found in (d) as the formula for $q'(t)=\frac12-2t^{-2}\text{.}$ We conclude that even though $q(t)=\frac{g(t)}{f(t)}\text{,}$ it is not the case that $q'(t)=\frac{g'(t)}{f'(t)}\text{.}$ In other words, the derivative of a quotient of two functions need not be the quotient of the derivatives.
### SubsectionThe Product Rule
As part (b) of Example2.35 shows, it is not true in general that the derivative of a product of two functions is the product of the derivatives of those functions. To see why this is the case, we consider a situation involving functions with physical context.
###### Example2.36
Say that an investor is regularly purchasing stock in a particular company. Let $N(t)$ represent the number of shares owned on day $t\text{,}$ where $t = 0$ represents the first day on which shares were purchased. Let $S(t)$ give the value of one share of the stock on day $t\text{;}$ note that the units on $S(t)$ are dollars per share. To compute the total value of the stock on day $t\text{,}$ we take the product
\begin{equation*} V(t) = N(t) \, \text{shares} \cdot S(t) \, \text{dollars per share}\text{,} \end{equation*}
Observe that over time, both the number of shares and the value of a given share will vary. The derivative $N'(t)$ measures the rate at which the number of shares is changing, while $S'(t)$ measures the rate at which the value per share is changing. How do these respective rates of change affect the rate of change of the total value function?
To help us understand the relationship among changes in $N\text{,}$ $S\text{,}$ and $V\text{,}$ let's consider some specific data.
• Suppose that on day 100, the investor owns 520 shares of stock and the stock's current value is $27.50 per share. This tells us that $N(100) = 520$ and $S(100) = 27.50\text{.}$ • On day 100, the investor purchases an additional 12 shares (so the number of shares held is rising at a rate of 12 shares per day). • On that same day the price of the stock is rising at a rate of 0.75 dollars per share per day. In calculus notation, the latter two facts tell us that $N'(100) = 12$ (shares per day) and $S'(100) = 0.75$ (dollars per share per day). At what rate is the value of the investor's total holdings changing on day 100? Observe that the increase in total value comes from two sources: the growing number of shares and the rising value of each share. • If only the number of shares is increasing (and the value of each share is constant), the rate at which the total value would rise is the product of the rate at which the number of shares is changing and the current value of the shares. That is, the rate at which the total value would change when the share value is constant is given by \begin{equation*} N'(100) \cdot S(100) = 12 \, \frac{\text{shares}}{\text{day}} \cdot 27.50 \, \frac{\text{dollars}}{\text{share}} = 330 \, \frac{\text{dollars} }{\text{day} }\text{.} \end{equation*} Note particularly how the units make sense and show the rate at which the total value $V$ is changing, measured in dollars per day. • If instead the number of shares is constant, but the value of each share is rising, the rate at which the total value would rise is the product of the number of shares and the rate of change of share value. In this case, the total value is rising at a rate of \begin{equation*} N(100) \cdot S'(100) = 520 \, \text{shares} \cdot 0.75 \, \frac{\text{dollars per share} }{\text{day} } = 390 \, \frac{\text{dollars} }{\text{day} }\text{.} \end{equation*} • Of course, when both the number of shares and the value of each share are changing, we have to include both of these sources. In that case the rate at which the total value is rising is \begin{equation*} V'(100) = N'(100) \cdot S(100) + N(100) \cdot S'(100) = 330 + 390 = 720 \, \frac{\text{dollars} }{\text{day} }\text{.} \end{equation*} We expect the total value of the investor's holdings to rise by about$720 on the 100th day.5While this example highlights why the product rule is true, there are some subtle issues to recognize. For one, if the stock's value really does rise exactly $0.75 on day 100, and the number of shares really rises by 12 on day 100, then we would expect that $V(101) = N(101) \cdot S(101) = 532 \cdot 28.25 = 15,029\text{.}$ If, as noted above, we expect the total value to rise by$720, then with $V(100) = N(100) \cdot S(100) = 520 \cdot 27.50 = 14,300\text{,}$ then it seems we should find that $V(101) = V(100) + 720 = 15,020\text{.}$ Why do the two results differ by 9? One way to understand why this difference occurs is to recognize that $N'(100) = 12$ represents an instantaneous rate of change, while our (informal) discussion has also thought of this number as the total change in the number of shares over the course of a single day. The formal proof of the product rule reconciles this issue by taking the limit as the change in the input tends to zero.
Next, we expand our perspective from the specific example above to the more general and abstract setting of a product $P$ of two differentiable functions, $f$ and $g\text{.}$ If $P(x) = f(x) \cdot g(x)\text{,}$ our work above suggests that $P'(x) = f'(x) g(x) + f(x) g'(x)\text{.}$ Indeed, a formal proof using the limit definition of the derivative can be given to show that the following rule, called the product rule, holds in general.
###### Product Rule
If $f$ and $g$ are differentiable functions, then their product $P(x) = f(x) \cdot g(x)$ is also a differentiable function, and
\begin{equation*} P'(x) = f'(x) g(x) + f(x) g'(x)\text{.} \end{equation*}
In light of Example2.36 involving shares of stock, the product rule also makes sense intuitively: the rate of change of $P$ should take into account both how fast $f$ and $g$ are changing, as well as how large $f$ and $g$ are at the point of interest. In words, the product rule says: if $P$ is the product of two functions $f$ (the first function) and $g$ (the second), then the derivative of $P$ is the derivative of the first times the second, plus the first times the derivative of the second. It is often a helpful mental exercise to say this phrasing aloud when executing the product rule.
###### Example2.37
If $P(z) = z^3 \cdot \cos(z)\text{,}$ we can use the product rule to differentiate $P\text{.}$ The first function is $z^3$ and the second function is $\cos(z)\text{.}$ By the product rule, $P'$ will be given by the derivative of the first, $3z^2\text{,}$ times the second, $\cos(z)\text{,}$ plus the first, $z^3\text{,}$ times the derivative of the second, $-\sin(z)\text{.}$ That is,
\begin{equation*} P'(z) = 3z^2\cos(z)+z^3(-\sin(z)) = 3z^2 \cos(z)-z^3\sin(z)\text{.} \end{equation*}
###### Example2.38
Use the product rule to answer each of the questions below. Throughout, be sure to carefully label any derivative you find by name. It is not necessary to algebraically simplify any of the derivatives you compute.
1. Let $m(w)=3w^{17} 4^w\text{.}$ Find $m'(w)\text{.}$
2. Let $h(t) = (\sin(t) + \cos(t))t^4\text{.}$ Find $h'(t)\text{.}$
3. Determine the slope of the tangent line to the curve $y = f(x)$ at the point where $a = 1$ if $f$ is given by the rule $f(x) = e^x \sin(x)\text{.}$
4. Find an equation for the tangent line $L$ to the graph of $y = g(x)$ at the point where $a = -1$ if $g$ is given by the rule $g(x) = (x^2 + x) 2^x\text{.}$
Hint
1. Let the first function be $3w^{17}\text{.}$
2. Let the first function be $(\sin(t) + \cos(t))\text{.}$
3. Remember that the slope of the tangent line to $y = f(x)$ at $(a,f(a))$ is given by $f'(a)\text{.}$
4. Recall that the tangent line to $y=g(x)$ at $(a,g(a))$ is given by the equation $y-g(a)=g'(a)(x-a)\text{.}$ To denote this line by a function $L\text{,}$ we can write $L(x)=g'(a)(x-a)+g(a)\text{.}$
1. $m'(w) = 51w^{16} \cdot 4^w + 3w^{17}\cdot 4^w \ln(4)\text{.}$
2. $h'(t) = (\cos(t) - \sin(t)) \cdot t^4 + (\sin(t) + \cos(t))\cdot 4t^3\text{.}$
3. $f'(1) = e(\sin(1) + \cos(1)) \approx 3.756\text{.}$
4. $L(x) = -\frac{1}{2}(x+1)\text{.}$
Solution
1. By the product rule,
\begin{equation*} m'(w) = 51w^{16}\cdot 4^2+3w^{17}\cdot 4^2\ln(4)\text{.} \end{equation*}
2. By the product rule,
\begin{equation*} h'(t) = (\cos(t) - \sin(t)) \cdot t^4 + (\sin(t) + \cos(t))\cdot 4t^3\text{.} \end{equation*}
3. To determine the slope of the tangent line at $a = 1\text{,}$ we want to find $f'(1)\text{.}$ Since $f(x) = e^x \sin(x)\text{,}$ the product rule tells us that $f'(x) = e^x \cdot \sin(x) + e^x \cdot \cos(x)\text{.}$ Thus, $f'(1) = e^1 \cdot \sin(1) + e^1\cdot \cos(1) = e(\sin(1) + \cos(1)) \approx 3.756\text{.}$
4. First, observe that $g(-1) = ((-1)^2 - 1) \cdot 2^{-1} = 0\text{,}$ so we are looking for the tangent line to $g$ at the point $(-1,0)\text{.}$ Furthermore, by the product rule, $g'(x) = (2x + 1) \cdot 2^x + (x^2+x)\cdot 2^x\ln(2)\text{,}$ and hence
\begin{align*} g'(-1) \amp= (2(-1)+ 1) \cdot 2^{-1} + ((-1)^2 - 1)\cdot 2^{-1}\ln(2) \\ \amp= (-1)\frac{1}{2} +0 = -\frac{1}{2}\text{.} \end{align*}
Therefore, $L(x)-0=-\frac{1}{2}(x-(-1))\text{,}$ or
\begin{equation*} L(x) = -\frac{1}{2}(x+1)\text{.} \end{equation*}
### SubsectionThe Quotient Rule
Because quotients and products are closely linked, we can use the product rule to understand how to take the derivative of a quotient. Let $Q(x)$ be defined by $Q(x) = \frac{f(x)}{g(x)}\text{,}$ where $f$ and $g$ are both differentiable functions. It turns out that $Q$ is differentiable everywhere that $g(x) \ne 0\text{.}$ We would like a formula for $Q'$ in terms of $f\text{,}$ $g\text{,}$ $f'\text{,}$ and $g'\text{.}$ Multiplying both sides of the equation $Q = \frac{f}{g}$ by $g\text{,}$ we observe that
\begin{equation*} f(x) = Q(x) \cdot g(x)\text{.} \end{equation*}
Now we can use the product rule to differentiate $f\text{:}$
\begin{equation*} f'(x) = Q'(x) g(x) + Q(x)g'(x)\text{.} \end{equation*}
We want to know a formula for $Q'\text{,}$ so we solve this equation for $Q'(x)\text{.}$ Then
\begin{equation*} Q'(x) g(x) = f'(x) - Q(x) g'(x)\text{,} \end{equation*}
and dividing both sides by $g(x)\text{,}$ we have
\begin{equation*} Q'(x) = \frac{f'(x) - Q(x) g'(x)}{g(x)}\text{.} \end{equation*}
Finally, we recall that $Q(x) = \frac{f(x)}{g(x)}\text{.}$ Substituting this expression in the preceding equation, we have
\begin{align*} Q'(x) =\mathstrut \amp \frac{f'(x) - \frac{f(x)}{g(x)} g'(x)}{g(x)}\\ =\mathstrut \amp \frac{f'(x) - \frac{f(x)}{g(x)} g'(x)}{g(x)} \cdot \frac{g(x)}{g(x)}\\ =\mathstrut \amp \frac{f'(x)g(x) - f(x) g'(x)}{[g(x)]^2}\text{.} \end{align*}
This calculation gives us the quotient rule.
###### Quotient Rule
If $f$ and $g$ are differentiable functions, then their quotient $Q(x) = \frac{f(x)}{g(x)}$ is also a differentiable function for all $x$ where $g(x) \ne 0\text{,}$ and
\begin{equation*} Q'(x) = \frac{f'(x)g(x) - f(x) g'(x)}{[g(x)]^2}\text{.} \end{equation*}
As with the product rule, it can be helpful to think of the quotient rule verbally. If a function $Q$ is the quotient of a top function $f$ and a bottom function $g\text{,}$ then $Q'$ is given by the derivative of the top times the bottom, minus the top times the derivative of the bottom, all over the bottom squared.6If you have studied differentiation before, you may have seen this rule as $\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}\text{,}$ with the mnemonic "low-d-high minus high-d-low all over low squared." If this version is more comfortable for you, you are welcome to use it instead.
###### Example2.39
If $Q(t) = \frac{\sin(t)}{2^t}\text{,}$ we call $\sin(t)$ the top function and $2^t$ the bottom function. By the quotient rule, $Q'$ is given by the derivative of the top, $\cos(t)\text{,}$ times the bottom, $2^t\text{,}$ minus the top, $\sin(t)\text{,}$ times the derivative of the bottom, $2^t \ln(2)\text{,}$ all over the bottom squared, $(2^t)^2\text{.}$ That is,
\begin{equation*} Q'(t) = \frac{\cos(t)2^t - \sin(t) 2^t \ln(2)}{(2^t)^2}\text{.} \end{equation*}
In this particular example, it is possible to simplify $Q'(t)$ by removing a factor of $2^t$ from both the numerator and denominator, so that
\begin{equation*} Q'(t) = \frac{\cos(t) - \sin(t) \ln(2)}{2^t}\text{.} \end{equation*}
In general, we must be careful in doing any such simplification, as we don't want to follow a correct execution of the quotient rule with an algebraic error.
###### Example2.40
Use the quotient rule to answer each of the questions below. Throughout, be sure to carefully label any derivative you find by name. That is, if you're given a formula for $f(x)\text{,}$ clearly label the formula you find for $f'(x)\text{.}$ It is not necessary to algebraically simplify any of the derivatives you compute.
1. Let $r(z)=\frac{3^z}{z^4 + 1}\text{.}$ Find $r'(z)\text{.}$
2. Let $v(t) = \frac{\sin(t)}{\cos(t) + t^2}\text{.}$ Find $v'(t)\text{.}$
3. Determine the slope of the tangent line to the curve $y=R(x)$ at the point where $x=0\text{,}$ if $\displaystyle R(x) = \frac{x^2 - 2x - 8}{x^2 - 9}\text{.}$
4. When a camera flashes, the intensity $I$ of light seen by the eye $t$ milliseconds after the initial flash point is given by the function
\begin{equation*} I(t) = \frac{100t}{e^t}\text{,} \end{equation*}
where $I$ is measured in candles. Compute $I'(0.5)\text{,}$ $I'(2)\text{,}$ and $I'(5)\text{;}$ include appropriate units on each value; and discuss the meaning of each.
Hint
1. When applying the quotient rule, use parentheses around the bottom function, $z^4 + 1\text{,}$ to ensure that when you compute the derivative of the top times the bottom that the rule is applied correctly.
2. When applying the quotient rule, use parentheses around the bottom function, $\cos(t) + t^2\text{,}$ and its derivative to ensure that the rule is applied correctly.
3. Remember one of the key interpretations of the derivative.
4. Let the top function be $100t$ and simply use the constant multiple rule to find its derivative.
1. $r'(z)=\frac{3^z \ln(3)(z^4+1) - 3^z(4z^3)}{(z^4 + 1)^2}\text{.}$
2. $v'(t) = \frac{\cos(t)(\cos(t) + t^2) - \sin(t)(\sin(t) + 2t)}{(\cos(t) + t^2)^2}\text{.}$
3. $R'(0) = \frac{2}{9}\text{.}$
4. $I'(0.5) = \frac{50}{e^{0.5}} \approx 30.327\text{,}$ $I'(2) = \frac{-100}{e^{2}} \approx -13.534\text{,}$ and $I'(5) = \frac{-400}{e^4} \approx -2.695\text{,}$ each in candles per millisecond.
Solution
1. By the quotient rule,
\begin{equation*} r'(z)=\frac{3^z \ln(3) (z^4+1) - 3^z(4z^3)}{(z^4 + 1)^2}\text{.} \end{equation*}
2. By the quotient rule,
\begin{equation*} v'(t) = \frac{\cos(t)(\cos(t) + t^2) - \sin(t)(\sin(t) + 2t)}{(\cos(t) + t^2)^2}\text{.} \end{equation*}
3. We first compute $R'(x)\text{.}$ By the quotient rule,
\begin{equation*} R'(x) = \frac{(2x - 2)(x^2 - 9) - (x^2 - 2x - 8)(2x)}{(x^2 - 9)^2}\text{.} \end{equation*}
From this, it follows that $R'(0) = \frac{(-2)(-9)-(-8)(0)}{(-9)^2} = \frac{2}{9}\text{,}$ which is the slope of the tangent line to the curve at the point where $x = 0\text{.}$
4. By the quotient rule and algebraic simplification,
\begin{equation*} I'(t) = \frac{100e^t - 100te^t}{(e^t)^2} = \frac{100-100t}{e^t}\text{.} \end{equation*}
Thus, $I'(0.5) = \frac{50}{e^{0.5}} \approx 30.327\text{,}$ $I'(2) = \frac{-100}{e^{2}} \approx -13.534\text{,}$ and $I'(5) = \frac{-400}{e^4} \approx -2.695\text{,}$ each measured in candles per millisecond. These results show that at $t = 0.5\text{,}$ the intensity of the flash is increasing rapidly, while at $t = 2$ and $t = 5\text{,}$ the intensity is decreasing, with the intensity decreasing more rapidly when $t = 2\text{.}$
### SubsectionThe Product and Quotient Rule Using Tables and Graphs
In addition to being used to finding the derivatives of functions given by equations, the product and quotient rule can be useful for finding the derivatives of functions given by tables and graphs. The following examples illustrate this process.
###### Example2.41
Suppose the functions $f$ and $g$ are described by Table2.42 below, and the function $h$ is given by the formula $h(x)=f(x)g(x)\text{.}$ Then we can calculate $h'(2)=f'(2)g(2)+f(2)g'(2)=4 \times 5 + 2 \times 6=32\text{.}$
###### Example2.43
Suppose the function $f$ is described by Table2.44 below, the function $g$ is described by the graph in Figure2.45, and the function $h$ is given by the formula $h(x)=\frac{f(x)}{g(x)}\text{.}$ Then for certain values of $x\text{,}$ we can calculate
\begin{equation*} h'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}\text{.} \end{equation*}
Suppose we want to find $h'(0)\text{.}$ Table2.44 tells us that $f(0)=4$ and $f'(0)=5\text{.}$ From visual inspection of Figure2.45, we also notice that $g(0)=2$ and $g'(0)=1\text{.}$ Therefore
\begin{equation*} h'(0)=\frac{f'(0)g(0)-f(0)g'(0)}{(g(0))^2}=\frac{5\times 2-4 \times 1}{(2)^2}=\frac{6}{4}=\frac{3}{2}. \end{equation*}
$x$ -2 -1 0 1 2 $f(x)$ 3 2 4 3 2 $f'(x)$ 1 2 5 2 4
### SubsectionCombining Rules
In order to apply the derivative shortcut rules correctly we must recognize the fundamental structure of a function.
###### Example2.46
Determine the derivative of the function
\begin{equation*} f(x) = x\sin(x) + \frac{x^2}{\cos(x) + 2}\text{.} \end{equation*}
How do we decide which rules to apply? Our first task is to recognize the structure of the function. This function $f$ is a sum of two slightly less complicated functions, so we can apply the sum rule7When taking a derivative that involves the use of multiple derivative rules, it is often helpful to use the notation $\frac{d}{dx} \left[ ~~\right]$ to wait to apply subsequent rules. This is demonstrated in both Example2.46 and Example2.47, and has come up previously in solutions to various examples in this chapter. to get
\begin{align*} f'(x) =\mathstrut \amp \frac{d}{dx} \left[ x\sin(x) + \frac{x^2}{\cos(x) + 2} \right]\\ =\mathstrut \amp \frac{d}{dx} \left[ x\sin(x) \right] + \frac{d}{dx}\left[ \frac{x^2}{\cos(x) + 2} \right]\text{.} \end{align*}
Now, the left-hand term above is a product, so the product rule is needed there, while the right-hand term is a quotient, so the quotient rule is required. Applying these rules respectively, we find that
\begin{align*} f'(x) =\mathstrut \amp \left( \sin(x)+x \cos(x) \right) + \frac{ 2x(\cos(x) + 2) - x^2(-\sin(x))}{(\cos(x) + 2)^2}\\ =\mathstrut \amp \sin(x) + x \cos(x) + \frac{2x\cos(x) + 4x + x^2\sin(x)}{(\cos(x) + 2)^2}\text{.} \end{align*}
###### Example2.47
Differentiate
\begin{equation*} s(y) = \frac{y \cdot 7^y}{y^2 + 1}\text{.} \end{equation*}
The function $s$ is a quotient of two simpler functions, so the quotient rule will be needed. To begin, we set up the quotient rule and use the notation $\frac{d}{dy}$ to indicate the derivatives of the numerator and denominator. Thus,
\begin{equation*} s'(y) = \frac{\frac{d}{dy}\left[ y \cdot 7^y \right]\cdot(y^2 + 1) - y \cdot 7^y \cdot \frac{d}{dy}\left[y^2 + 1 \right]}{(y^2 + 1)^2}\text{.} \end{equation*}
Now, there remain two derivatives to calculate. The first one, $\frac{d}{dy}\left[ y \cdot 7^y \right]$ calls for use of the product rule, while the second, $\frac{d}{dy}\left[y^2 + 1 \right]$ needs only the sum rule. Applying these rules, we now have
\begin{equation*} s'(y) = \frac{[7^y+y \cdot 7^y \ln(7)](y^2 + 1) - y \cdot 7^y [2y]}{(y^2 + 1)^2}\text{.} \end{equation*}
While some simplification is possible, we are content to leave $s'(y)$ in its current form.
Success in applying derivative rules begins with recognizing the structure of the function, followed by the careful and diligent application of the relevant derivative rules. The best way to become proficient at this process is through continuous practice!
###### Example2.48
Use relevant derivative rules to answer each of the questions below. Throughout, be sure to use proper notation and carefully label any derivative you find by name.
1. Let $f(r) = (5r^3 + \sin(r))(4^r - 2\cos(r))\text{.}$ Find $f'(r)\text{.}$
2. Let $\displaystyle p(t) = \frac{\cos(t)}{t^6 \cdot 6^t}\text{.}$ Find $p'(t)\text{.}$
3. Let $g(z) = 3z^7 e^z - 2z^2 \sin(z) + \frac{z}{z^2 + 1}\text{.}$ Find $g'(z)\text{.}$
4. A moving particle has its position in feet at time $t$ in seconds given by the function $s(t) = \frac{3\cos(t) - \sin(t)}{e^t}\text{.}$ Find the particle's instantaneous velocity at the moment $t = 1\text{.}$
5. Suppose that $f(x)$ and $g(x)$ are differentiable functions and it is known that $f(3) = -2\text{,}$ $f'(3) = 7\text{,}$ $g(3) = 4\text{,}$ and $g'(3) = -1\text{.}$ If $p(x) = f(x) \cdot g(x)$ and $\displaystyle q(x) = \frac{f(x)}{g(x)}\text{,}$ calculate $p'(3)$ and $q'(3)\text{.}$
Hint
1. Observe that $f$ is fundamentally a product. Which is the first function? The second?
2. Note that $p$ has the overall structure of a quotient.
3. Think about how $g$ is a sum of three functions. What is the structure of each of the three functions in the sum?
4. How is the velocity of a moving object related to its position?
5. Since we know $p(x) = f(x) \cdot g(x)\text{,}$ it follows $p'(x) = f'(x)g(x)+f(x) g'(x)$ for any value of $x\text{.}$
1. $f'(r) = [15r^2 + \cos(r)](4^r - 2\cos(r))+(5r^3 + \sin(r))[4^r \ln(4) + 2\sin(r)]\text{.}$
2. $p'(t) = \frac{[-\sin(t)]t^6 \cdot 6^t - \cos(t) [6t^5\cdot 6^t + t^6 \cdot 6^t \ln(6)] }{(t^6 \cdot 6^t)^2}\text{.}$
3. $g'(z) = 3 [7z^6e^z+z^7 e^z] - 2[2z\sin(z)+z^2 \cos(z)] + \frac{(z^2+1) - z(2z)}{(z^2 + 1)^2}\text{.}$
4. $s'(1) = \frac{-2\sin(1)-4\cos(1)}{e} \approx -1.414$ feet per second.
5. $p'(3) = 30$ and $q'(3) = \frac{13}{8}\text{.}$
Solution
1. Using the product rule, followed by the sum and constant multiple rule, observe that
\begin{align*} f'(r) =\mathstrut \amp \frac{d}{dr}[5r^3 + \sin(r)](4^r - 2\cos(r))+(5r^3 + \sin(r))\frac{d}{dr}[4^r - 2\cos(r)] \\ =\mathstrut \amp [15r^2 + \cos(r)](4^r - 2\cos(r))+(5r^3 + \sin(r))[4^r \ln(4) + 2\sin(r)]\text{.} \end{align*}
2. We use the quotient rule on $p\text{,}$ followed by the product rule to differentiate the denominator, finding that
\begin{align*} p'(t) =\mathstrut \amp \frac{\frac{d}{dt}[\cos(t)]t^6 \cdot 6^t - \cos(t) \frac{d}{dt}[t^6 \cdot 6^t]}{(t^6 \cdot 6^t)^2}\\ =\mathstrut \amp \frac{[-\sin(t)]t^6 \cdot 6^t - \cos(t) [6t^5 \cdot 6^t + t^6\cdot 6^t \ln(6)]}{(t^6 \cdot 6^t)^2}\text{.} \end{align*}
3. Using the sum and constant multiple rules, it follows first that
\begin{equation*} g'(z) = 3 \frac{d}{dz}[z^7 e^z] - 2\frac{d}{dz}[z^2 \sin(z)] + \frac{d}{dz}\left[ \frac{z}{z^2 + 1} \right]\text{.} \end{equation*}
Applying the product rule in the first two terms and the quotient rule in the third, we find that
\begin{equation*} g'(z) = 3 [7z^6e^z+z^7 e^z] - 2[2z\sin(z)+z^2 \cos(z)] + \frac{(z^2+1) - z(2z)}{(z^2 + 1)^2}\text{.} \end{equation*}
4. The particle's instantaneous velocity at the moment $t = 1$ is given by $s'(1)\text{.}$ We use the quotient rule to find $s'(t)\text{,}$ and then simplify by removing a common factor of $e^t$ to get
\begin{align*} s'(t) =\mathstrut \amp \frac{(-3\sin(t) - \cos(t))e^t-(3\cos(t) - \sin(t))e^t}{(e^t)^2}\\ =\mathstrut \amp \frac{-2\sin(t)-4\cos(t)}{e^t}\text{.} \end{align*}
Thus, $s'(1) = \frac{-2\sin(1)-4\cos(1)}{e^1} \approx -1.414\text{,}$ which is the particle's instantaneous velocity in feet per second at the moment $t = 1\text{.}$
5. Since $p(x) = f(x) \cdot g(x)\text{,}$ the product rule tells us
\begin{equation*} p'(x) = f'(x)g(x)+f(x)g'(x)\text{,} \end{equation*}
and since $\displaystyle q(x) = \frac{f(x)}{g(x)}\text{,}$ by the quotient rule we know
\begin{equation*} q'(x) = \frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}\text{.} \end{equation*}
Using the given information ($f(3) = -2\text{,}$ $f'(3) = 7\text{,}$ $g(3) = 4\text{,}$ and $g'(3) = -1$) we now see that
\begin{equation*} p'(3) = f'(3)g(3)+f(3)g'(3) = (7)(4)+(-2)(-1) = 30 \end{equation*}
and
\begin{equation*} q'(3) = \frac{f'(3)g(3)-f(3)g'(3)}{g(3)^2} = \frac{(7)(4) - (-2)(-1)}{4^2} = \frac{13}{8}\text{.} \end{equation*}
As the algebraic complexity of the functions we are able to differentiate continues to increase, it is important to remember that the meaning of the derivative remains the same. Regardless of the structure of the function $f\text{,}$ the value of $f'(a)$ tells us the instantaneous rate of change of $f$ with respect to $x$ at the moment $x = a\text{,}$ as well as the slope of the tangent line to $y = f(x)$ at the point $(a,f(a))\text{.}$
### SubsectionSummary
• If a function is a sum, product, or quotient of simpler functions, then we can use the sum, product, or quotient rules to differentiate it in terms of the simpler functions and their derivatives.
• The product rule tells us that if $P$ is a product of differentiable functions $f$ and $g$ according to the rule $P(x) = f(x) g(x)\text{,}$ then
\begin{equation*} P'(x) = f'(x)g(x)+f(x)g'(x)\text{.} \end{equation*}
• The quotient rule tells us that if $Q$ is a quotient of differentiable functions $f$ and $g$ according to the rule $Q(x) = \frac{f(x)}{g(x)}\text{,}$ then whenever $g(x)\neq0\text{,}$
\begin{equation*} Q'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}\text{.} \end{equation*}
• Along with the constant multiple and sum rules, the product and quotient rules enable us to compute the derivative of any function that consists of sums, constant multiples, products, and quotients of basic functions. For instance, if $F$ has the form
\begin{equation*} F(x) = \frac{2a(x) - 5b(x)}{c(x) \cdot d(x)}\text{,} \end{equation*}
then $F$ is a quotient, in which the numerator is a sum of constant multiples (of the functions $a$ and $b$), and the denominator is a product (of the functions $c$ and $d$). Thus the derivative of $F$ can be found by first applying the quotient rule, and then using the sum and constant multiple rules to differentiate the numerator, and the product rule to differentiate the denominator.
### SubsectionExercises
Let $f$ and $g$ be differentiable functions for which the following information is known: $f(2) = 5\text{,}$ $g(2) = -3\text{,}$ $f'(2) = -1/2\text{,}$ $g'(2) = 2\text{.}$
1. Let $h$ be the new function defined by the rule $h(x) = g(x) \cdot f(x)\text{.}$ Determine $h(2)$ and $h'(2)\text{.}$
2. Find an equation for the tangent line to $y = h(x)$ at the point $(2,h(2))$ (where $h$ is the function defined in (a)).
3. Let $r$ be the function defined by the rule $r(x) = \frac{g(x)}{f(x)}\text{.}$ Is $r$ increasing, decreasing, or neither at $a = 2\text{?}$ Why?
4. Estimate the value of $r(2.06)$ (where $r$ is the function defined in (c)) by using the tangent line to the graph of $y=r(x)$ at the point $(2,r(2))\text{.}$
Consider the functions $r(t) = t^t$ and $s(t) = \arccos(t)\text{,}$ for which you are given the facts that $r'(t) = t^t(\ln(t) + 1)$ and $s'(t) = -\frac{1}{\sqrt{1-t^2}}\text{.}$ Do not be concerned with where these derivative formulas come from. We restrict our interest in both functions to the domain $0 \lt t \lt 1\text{.}$
1. Let $w(t) = t^t \arccos(t)\text{.}$ Determine $w'(t)\text{.}$
2. Find an equation for the tangent line to $y = w(t)$ at the point $(\frac{1}{2}, w(\frac{1}{2}))\text{.}$
3. Let $v(t) = \frac{t^t}{\arccos(t)}\text{.}$ Is $v$ increasing or decreasing at the instant $t = \frac{1}{2}\text{?}$ Why?
Let functions $p$ and $q$ be the piecewise linear functions given by their respective graphs in Figure2.49. Use the graphs to answer the following questions.
1. Let $r(x) = p(x) \cdot q(x)\text{.}$ Determine $r'(-2)$ and $r'(0)\text{.}$
2. Are there values of $x$ for which $r'(x)$ does not exist? If so, which values, and why?
3. Find an equation for the tangent line to $y = r(x)$ at the point $(2,r(2))\text{.}$
4. Let $z(x) = \frac{q(x)}{p(x)}\text{.}$ Determine $z'(0)$ and $z'(2)\text{.}$
5. Are there values of $x$ for which $z'(x)$ does not exist? If so, which values, and why?
A farmer with large land holdings has historically grown a wide variety of crops. With the price of ethanol fuel rising, he decides that it would be prudent to devote more and more of his acreage to producing corn. As he grows more and more corn, he learns efficiencies that increase his yield per acre. In the present year, he used 7000 acres of his land to grow corn, and that land had an average yield of 170 bushels per acre. At the current time, he plans to increase his number of acres devoted to growing corn at a rate of 600 acres/year, and he expects that right now his average yield is increasing at a rate of 8 bushels per acre per year. Use this information to answer the following questions.
1. Say that the present year is $t = 0\text{,}$ that $A(t)$ denotes the number of acres the farmer devotes to growing corn in year $t\text{,}$ $Y(t)$ represents the average yield in year $t$ (measured in bushels per acre), and $C(t)$ is the total number of bushels of corn the farmer produces. What is the formula for $C(t)$ in terms of $A(t)$ and $Y(t)\text{?}$ Why?
2. What is the value of $C(0)\text{?}$ What does it measure?
3. Write an expression for $C'(t)$ in terms of $A(t)\text{,}$ $A'(t)\text{,}$ $Y(t)\text{,}$ and $Y'(t)\text{.}$ Explain your thinking.
4. What is the value of $C'(0)\text{?}$ What does it measure?
5. Based on the given information and your work above, estimate the value of $C(1)\text{.}$
Let $f(v)$ be the gas consumption (in liters/km) of a car going at velocity $v$ (in km/hour). In other words, $f(v)$ tells you how many liters of gas the car uses to go one kilometer if it is traveling at $v$ kilometers per hour. In addition, suppose that $f(80)=0.05$ and $f'(80) = 0.0004\text{.}$
1. Let $g(v)$ be the distance the same car goes on one liter of gas at velocity $v\text{.}$ What is the relationship between $f(v)$ and $g(v)\text{?}$ Hence find $g(80)$ and $g'(80)\text{.}$
2. Let $h(v)$ be the gas consumption in liters per hour of a car going at velocity $v\text{.}$ In other words, $h(v)$ tells you how many liters of gas the car uses in one hour if it is going at velocity $v\text{.}$ What is the algebraic relationship between $h(v)$ and $f(v)\text{?}$ Hence find $h(80)$ and $h'(80)\text{.}$
3. How would you explain the practical meaning of these function and derivative values to a driver who knows no calculus? Include units on each of the function and derivative values you discuss in your response. |
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# Angle-Angle-Side Triangles
## Law of sines: proportion based on ratios of sides and sines of the opposite angles.
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Angle-Angle-Side Triangles
You and a friend decide to go fly kites on a breezy Saturday afternoon. While sitting down to make your kites, you are working on make the best shape possible to catch the breeze. While your friend decides to go with a diamond shaped kite, you try out making a triangle shaped one. While trying to glue the kite together, you make the first and second piece lock together with a $70^\circ$ angle. The angle between the first and third pieces is $40^\circ$ . Finally, you also have measured the length of the second piece and found that it is 22 inches long.
Is there a way to find out, using math, what the length of the third side will be?
Keep reading, and you'll be able to answer this question at the end of this Concept.
### Guidance
The Law of Sines states: $\frac{\sin A}{a} = \frac{\sin B}{b}$ . This is a ratio between the sine of an angle in a triangle and the length of the side opposite that angle to the sine of a different angle in that triangle and the length of the side opposing that second angle.
The Law of Sines allows us to find many quantities of interest in triangles by comparing sides and interior angles as a ratio. One case where we can to use the Law of Sines is when we know two of the angles in a triangle and a non-included side (AAS).
#### Example A
Using $\triangle{GMN}, \angle{G} = 42^\circ, \angle{N} = 73^\circ$ and $g = 12$ . Find $n$ .
Since we know two angles and one non-included side $(g)$ , we can find the other non-included side $(n)$ .
$\frac{\sin 73^\circ}{n} & = \frac{\sin 42^\circ}{12} \\n \sin 42^\circ & = 12 \sin 73^\circ \\n & = \frac{12 \sin 73^\circ}{\sin 42^\circ} \\n & \approx 17.15$
#### Example B
Continuing on from Example A, find $\angle{M}$ and $m$ .
Solution: $\angle{M}$ is simply $180^\circ - 42^\circ - 73^\circ = 65^\circ$ . To find side $m$ , you can now use either the Law of Sines or Law of Cosines. Considering that the Law of Sines is a bit simpler and new, let’s use it. It does not matter which side and opposite angle you use in the ratio with $\angle{M}$ and $m$ .
Option 1: $\angle{G}$ and $g$
$\frac{\sin 65^\circ}{m} & = \frac{\sin 42^\circ}{12} \\m \sin 42^\circ & = 12 \sin 65^\circ \\m & = \frac{12 \sin 65^\circ}{\sin 42^\circ} \\m & \approx 16.25$
Option 2: $\angle{N}$ and $n$
$\frac{\sin 65^\circ}{m} & = \frac{\sin 73^\circ}{17.15} \\m \sin 73^\circ & = 17.15 \sin 65^\circ \\m & = \frac{17.15 \sin 65^\circ}{\sin 73^\circ} \\m & \approx 16.25$
#### Example C
A business group wants to build a golf course on a plot of land that was once a farm. The deed to the land is old and information about the land is incomplete. If $AB$ is 5382 feet, $BC$ is 3862 feet, $\angle{AEB}$ is $101^\circ,\angle{BDC}$ is $74^\circ,\angle{EAB}$ is $41^\circ$ and $\angle{DCB}$ is $32^\circ$ , what are the lengths of the sides of each triangular piece of land? What is the total area of the land?
Solution: Before we can figure out the area of the land, we need to figure out the length of each side. In $\triangle ABE$ , we know two angles and a non-included side. This is the AAS case. First, we will find the third angle in $\triangle ABE$ by using the Triangle Sum Theorem. Then, we can use the Law of Sines to find both $AE$ and $EB$ .
$\angle{ABE} & = 180 - (41 + 101) = 38^\circ \\\frac{\sin 101}{5382} & = \frac{\sin 38}{AE} && \frac{\sin 101}{5382} = \frac{\sin 41}{EB} \\AE (\sin 101) & = 5382 (\sin 38) && EB (\sin 101) = 5382 (\sin 41) \\AE & = \frac{5382(\sin 38)}{\sin 101} && EB = \frac{5382(\sin 41)}{\sin 101} \\AE & = 3375.5\ feet && EB \approx 3597.0\ feet$
Next, we need to find the missing side lengths in $\triangle DCB$ . In this triangle, we again know two angles and a non-included side (AAS), which means we can use the Law of Sines. First, let’s find $\angle{DBC} = 180 - (74 + 32) = 74^\circ$ . Since both $\angle{BDC}$ and $\angle{DBC}$ measure $74^\circ$ , $\triangle DCB$ is an isosceles triangle. This means that since $BC$ is 3862 feet, $DC$ is also 3862 feet. All we have left to find now is $DB$ .
$\frac{\sin 74}{3862} & = \frac{\sin 32}{DB}\\DB (\sin 74) & = 3862 (\sin 32)\\DB & = \frac{3862(\sin 32)}{\sin 74}\\DB & \approx 2129.0\ feet$
Finally, we need to calculate the area of each triangle and then add the two areas together to get the total area. From the last section, we learned two area formulas, $K = \frac{1}{2} \ bc \sin A$ and Heron’s Formula. In this case, since we have enough information to use either formula, we will use $K = \frac{1}{2} \ bc \sin A$ since it is less computationally intense.
First, we will find the area of $\triangle ABE$ .
$\triangle ABE$ :
$K & = \frac{1}{2}(3375.5)(5382)\sin 41 \\K & = 5,959,292.8\ ft^2$
$\triangle DBC$ :
$K & = \frac{1}{2}(3862)(3862)\sin 32 \\K & = 3, 951,884.6\ ft^2$
The total area is $5,959,292.8 + 3,951,884.6 = 9,911,177.4\ ft^2$ .
### Vocabulary
Angle Angle Side Triangle: An angle angle side triangle is a triangle where two of the angles and the non-included side are known quantities.
### Guided Practice
1. Find side "d" in the triangle below with the following information: $e = 214.9, D = 39.7^\circ, E = 41.3^\circ$
2. Find side "o" in the triangle below with the following information: $M = 31^\circ, O = 9^\circ, m = 15$
3. Find side "q" in the triangle below with the following information: $Q = 127^\circ, R = 21.8^\circ, r = 3.62$
Solutions:
1. $\frac{\sin 41.3^\circ}{214.9} = \frac{\sin 39.7^\circ}{d}, d = 208.0$
2. $\frac{\sin 9^\circ}{o} = \frac{\sin 31^\circ}{15}, o = 4.6$
3. $\frac{\sin 127^\circ}{q} = \frac{\sin 21.8^\circ}{3.62}, q = 7.8$
### Concept Problem Solution
Since you know two angles and one non-included side of the kite, you can find the other non-included side using the Law of Sines. Set up a ratio using the angles and side you know and the side you don't know.
$\frac{\sin 70^\circ}{x} & = \frac{\sin 40^\circ}{22} \\x & = \frac{22 \sin 70^\circ}{\sin 40^\circ} \\x & \approx 32.146$
The length of the dowel rod on the unknown side will be approximately 32 inches.
### Practice
In $\triangle ABC$ , $m\angle A=50^\circ$ , $m\angle B=34^\circ$ , and a=6.
1. Find the length of b.
2. Find the length of c.
In $\triangle KMS$ , $m\angle K=42^\circ$ , $m\angle M=26^\circ$ , and k=14.
1. Find the length of m.
2. Find the length of s.
In $\triangle DEF$ , $m\angle D=52^\circ$ , $m\angle E=78^\circ$ , and d=23.
1. Find the length of e.
2. Find the length of f.
In $\triangle PQR$ , $m\angle P=2^\circ$ , $m\angle Q=79^\circ$ , and p=20.
1. Find the length of q.
2. Find the length of r.
In $\triangle DOG$ , $m\angle D=50^\circ$ , $m\angle G=59^\circ$ , and o=12.
1. Find the length of d.
2. Find the length of g.
In $\triangle CAT$ , $m\angle C=82^\circ$ , $m\angle T=4^\circ$ , and a=8.
1. Find the length of c.
2. Find the length of t.
In $\triangle YOS$ , $m\angle Y=65^\circ$ , $m\angle O=72^\circ$ , and s=15.
1. Find the length of o.
2. Find the length of y.
In $\triangle HCO$ , $m\angle H=87^\circ$ , $m\angle C=14^\circ$ , and o=19.
1. Find the length of h.
2. Find the length of c.
### Vocabulary Language: English
Angle Angle Side Triangle
Angle Angle Side Triangle
An 'angle angle side triangle' is a triangle where two of the angles and the non-included side are known quantities. |
# High School Math Tip 13: Using the Discriminant of a Quadratic
The discriminant of a quadratic is very useful when finding the solution directly is not required and/or is too algebraic intensive.
To illustrate this, consider the following example:
Let the equation of a line be
for constants m and b.
Find the condition(s) on m and b such that the line is a tangent to a circle centred at the origin with radius 5.
To solve this, we know that the circle must thave equation
Solving this simultaneously with the equation of the line gives:
The solution to this equation gives us the x-coordinates of the point(s) of intersection between the line and the circle. In order for this line to be a tangent, there must be only one real solution. This means that the discriminant of this quadratic is zero:
Notice that this equation does not always give real solutions for m and b.
To ensure that we get real solutions for m and b, we observe that
Hence:
Essentially this means that in order for the line to be a tangent, it cannot be inside the circle, which makes sense.
Hence, as long as this condition on b is satisfied, solving for m using
will give us the equation of the tangent.
The discriminant is also useful in finding the maximum range for a projectile.
Let the launch velocity be V, the launch angle be θ and acceleration due to gravity be g.
Assuming that air resistance is negligible and that the projectile is launched from ground level, then the Cartesian equation of the projectile is:
The range is defined to be the horizontal displacement when the projectile hits the ground (y=0):
Since we are not interested in the initial horizontal displacement of zero, we can divide everything by x:
which usually has two real solutions for θ. But if the range is maximised, this θ must have only one real solution. Hence we require the discriminant to be zero:
since we are only interested in a positive horizontal displacement.
In order to find the optimal launch angle so that this maximum range is achieved, we can substitute this back into our quadratic equation for θ:
This technique can also be extended to cases where we are not launching from ground level.
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[–][🍰] 2 points3 points (5 children)
(n-2)!/n!= 1/(n(n-1))
I think?
Edit: changed my mind
[–] 1 point2 points (4 children)
It's going to be more complicated than that. If someone earlier in the chain picks your name, then it changes the probability. For example, consider 3 people. Person 1 must pick person 2 for the game to stay alive, so that is a 1/2 chance. Person 2 can't pick their own name anymore, and must pick person 1 for the game to stay alive, again another 1/2 chance. So the probability is 1/4 which is not equal to 1 / 6.
The probability for 4 people I believe is 5/36. I don't quite know what the general formula is though....
[–][🍰] 1 point2 points (3 children)
Since the last person has to pick their own...could we just figure out how many different ways (n-1) could arrange themselves in this secret santa fashion so no one picks themselves...and then once we have that, divide by the same equation but do a +1 in the denominator for every term? Does that make sense?
[–] 1 point2 points (2 children)
That makes sense, and that was my first thought. The challenge is that some permutations are more likely than others. For example, let's take 3 people. If we allow them to pick their own names, they can arrange themselves into 3! ways. We can clearly see that the chance of the last person getting their own name is 1/n. Now let's strip out those permutations where someone picks their own name, we are left with (213), (231), and (312). Given there is only one permutation with person 3 getting their name, we might think the probability remains 1/3. But (312) is now twice as likely as the other two permutations, so the true probability is 1/4.
If there were some way of throwing out the bad permutations, and updating the probability of the other permutations then that would work nicely.
[–] 0 points1 point (1 child)
I disagree. Why is 312 twice as likely?
[–] 0 points1 point (0 children)
To figure this out draw the probability tree:
Player 1 has a 50/50 chance between drawing 2 or 3.
{draws 3}(1/2): Since Player 2 can't draw his/her own name, and 3 is already taken, player 2 must draw 1, and player 3 is left with 2. Prob((312)) = 1/2.
{draws 2}(1/2): Player 2 has a 50/50 chance of drawing 1 or 3. If he/she draws 1, then we get (213) (with probability 1/4) and if he/she draws 3 we get (231) (with probability 1/4).
The logic is that if player 1 draws 3, then player 2 is 'bound' to pick player 1 since player 2's name is still in the pot, whereas if player 1 draws 2 then player 2 has two options.
If you still don't see it, simulate a few draws and you'll find that (312) pops up twice as often.
[–] 2 points3 points (0 children)
Well, if you start with the senario where everyone just picks a name and you allow other people to pick their own name, then the odds that the last person would pick their own name would be 1/n
But if you make it more realistic where people prior to the last person pick again if they get their own name it becomes a little more complicated.
I don't know the formula so I'll try to work it out. If you start with 3 A,B and C A starts and can pick B or C B is next and can pick A or the remaining from above but there is a 1/2 chan C is forced to pick the final which would be 1/4 chance their own.
So for each pick the person lets say the 5th, they have either 5 options to pick from or 4 if their name is still in the mix.
What is the chance they are still in the mix? it is the number of people before then and the odds that they were picked in each event.
So for the first person the odds of picking any number is 1/(N-1) The next is: 1/ ((N-1) x (1/(N-1)) + 1/((N) x (N-1)) ...
And After looking at my working I have made myself really confused.
Someone save me!
[–] 1 point2 points (0 children)
Wouldn't you know it, someone much smarter than me details a recursive solution:
http://www.spontaneoussymmetry.com/blog/archives/232
I wonder if a closed formula solution exists?
[–] 0 points1 point (4 children)
Why not code it?
``````selsan <- function(who,persons) {
if (length(persons)==1) return(persons)
sel <- sample(persons[persons!=who],1)
return(c(sel,selsan(who+1,persons[persons!=sel])))
}
#selsan(1,1:5)
finselsan <- function(n){
selsan(1,1:n)[n]
}
nrep=1e4
sa <- sapply(1:nrep,function(x) finselsan(8))
table(sa)/nrep
sa
1 2 3 4 5 6 7 8
0.1114 0.1138 0.1216 0.1333 0.1356 0.1594 0.1292 0.0957
``````
Either I made an error, or you have less than 1/n chance. At the same time, the last person has best chance to get the third to last person
[–] 0 points1 point (3 children)
I don't understand R, so I don't know what you are doing - but this is how I implemented it in python. We know the answer for N=3 has to be 1/4 and N=4 has to be 5/36:
https://gist.github.com/4010541
N = 3: 0.249998
N = 4: 0.139264
N = 5: 0.131944
N = 6: 0.112344
N = 7: 0.100284
N = 8: 0.090528
N = 9: 0.081776
[–] 0 points1 point (2 children)
I don't know python, but I guess you simulated for N=3 to 9 the chances the last would have to get him/herself. In contrast I calculated which person the last one would draw for N=8.
So your 0.090528 is my 0.0957 while the others don't correspond.
Edit: is your's a simulation or that recursive formula?
[–] 0 points1 point (1 child)
Got it, that's interesting that there is so much probability on person 6. Any intuition why you get the pattern you did? I wouldn't have guessed that it would be shifted towards person #6 .
Mine is just a simulation as well, don't have the energy to actually code up the recursion.
[–] 0 points1 point (0 children)
I guess that by the time the 6th person gets to draw there is still a good chance he/she can draw him/herself. In these occasions a present for 6 is pushed onto person 7 and 8.
[–] 0 points1 point (0 children)
Playing around with this formula-wise:
It seems like we could generate the boundaries of the probability with a little work. So, in the first case, assume that Person 1 can choose any entry except her own. The odds of selecting Person N (the last person) is 1/(n-1).
For Person 2, there are two possibilities from the previous state. Either Person 1 chose Person 2's name, or they didn't. In the case that they didn't, Person 2's odds of selecting Person N are 1/(n-2). In the case that Person 1 did select Person 2's name, Person 2's odds of selecting Person N are 1/(n-1). So first, we can extend the case in which everyone's name is picked by the person before them.
By the time we get to Person N-1 (the second to last person) there are only two options remaining. They can either pick Person 1, or Person N. Therefore, their odds of picking Person N are 1/(n-(i-1)) where i is number of the current person. So in the first boundary case, the odds of someone else selecting Person N is equal to:
( 1/(n-1) ) ((i=2 -> i=n) 1/(n-(i-1)) )
*(For those not mathematically inclined, the upper case pi symbol indicates that the operation afterward should be repeated for every i and then the results of each operation should be multiplied by one another. In this case, beginning at i = 2)
This first case is the “least restrictive” case as it leaves the largest possible set to choose from for each subsequent Person. We can establish the “most restrictive” case by using the opposite principles. Let’s go back to our divergence at Person 2. In this case, assume the first event, that is, that Person 1 did not choose Person 2’s name. In this case, as was mentioned above, the odds of Person 2 selecting Person N are 1/(n-2). In order to get the most mileage out of our restrictive case, let’s assume that Person 1 selects Person N-1. Person 2 selects Person 1, Person 3 selects Person 2, and so on.
Once again we proceed to the end of our list. In this case, the exception from the front end of the previous case has emerged at the back end of the current case. That is, in this case, the odds of Person N-1 selecting Person N are still 1/(n-(i-1)). However, the odds of Person N-2 selecting Person N are 1/(n-i). Therefore, our formula for the restrictive case is equal to:
((i=1 -> i=n-2) 1/(n-i) ) ( 1/(n-(n-2)) )
Thoughts?
[–] -4 points-3 points (2 children)
0
Let's start with the base case of 2 people. If the first person (A) draws their own name, they must redraw until they get person 2 (B)'s name. There's no way for B to ever get a chance to draw unless someone's already drawn their name.
[–] 1 point2 points (1 child)
What if the group is an uneven number?
In this case it is possible to create a closed cycle, with the remainder picking last.
For the case n=3, A draws B, B draws A and C can only draw C.
[–] -1 points0 points (0 children)
What you mean I can't draw general conclusions from a single case? Of course you're correct, but I like my logic. It doesn't even need to be an odd number. Take the case of 4 people (A, B, C, D). A draws B, B draws C, C draws A, D's left with D. |
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# 11+ Verbal Reasoning
## Type 17 – Maths equations (Complete the sum.)
​
​
In this question type you are asked to complete the mathematical sum by inserting the correct answer in the brackets. The answer will go on one side of the equal sign in order to complete the sum. We work from left to right recording our answers as we go!!
​
Look at this first example and then we can work our way through the question until we get the answer: -
Complete this sum by writing the correct number in the brackets.
2 + 2 = 3 + (?)
​
We know that the calculations on both sides of the equal sign (=) must have the same total and “balance” each other.
If we start with the left-hand side of the equal sign we can calculate (work out) that
2 + 2 = 4
​
The right-hand side of the equal sign must also equal 4.
So we can ask ourselves the following question.
3 plus what equals 4?
Clearly in this simple example the answer is 1 and this is our answer.
Look at this second example and then we can work our way through the question until we get the answer: -
Complete this sum by writing the correct number in the brackets.
200 ÷ 2 + 10 = 55 X (?)
​
We know that the calculations on both sides of the equal sign (=) must have the same total and “balance” each other.
If we start with the left-hand side of the equal sign we can calculate (work out) that
200 ÷ 2 = 100 and then when we add 10 we get 110
​
The right-hand side of the equal sign must also equal 110.
So we can ask ourselves the following question.
55 multiplied by what equals 110?
If we know our multiplication facts we will work out that the answer is 2 and this is our answer.
Look at this third example and then we can work our way through the question until we get the answer: -
Complete this sum by writing the correct number in the brackets.
(97 - 7) ÷ (?) = 15 X 2
​
We know that the calculations on both sides of the equal sign (=) must have the same total and “balance” each other.
The brackets with the question mark (?) is on the left-hand side of the equal sign this time so we cannot complete the sum until we know what the left-hand side is equal to. Therefore we need to start on the right-hand side of the sum.
If we start with the right-hand side of the equal sign we can calculate (work out) that
15 X 2 = 30
​
The other side of the equal sign must also equal 30.
So now we must calculate the sum on the left-hand side of the equal sign, doing the part of the question that is in the brackets first (97 - 7).
​
The answer to this is 90.
Now we can ask ourselves the following question.
What do I divide 90 by in order to get 30 and “balance” the sum?
If we know our division facts we will work out that the answer is 3 and this is our answer.
​
THINGS TO LOOK OUT FOR!!!!
​
Make sure that you do your calculations very carefully in order to avoid simple errors.
Remember that the totals on both sides of the equal sign must be exactly the same
Avoid guessing the answer – though if you are completely stuck a guess is better than a blank answer space.
There is not really any “trick” or “technique” that can be taught for this question type as it relies very strongly on the child having a good understanding of mathematical concepts and being able to use them accurately.
​
NB When you are completing a Verbal Reasoning test in Multiple-Choice format the correct answer will always be one of the choices that you are given on your answer sheet. If your answer does not match one of the answers that you are given then your answer is wrong.
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# Online trigonometry
Online trigonometry is a software program that supports students solve math problems. We can help me with math work.
## The Best Online trigonometry
Math can be a challenging subject for many learners. But there is support available in the form of Online trigonometry. There are a number of ways to solve equations involving synthetic division, but one of the most popular is to use a synthetic division solver. This tool can be found online or in many math textbooks, and it can be a great help in solving complex equations. Synthetic division solvers work by breaking down an equation into smaller pieces, which makes it easier to solve. In addition, they often include step-by-step instructions that can make the process of solving an equation much simpler. If you're struggling with an equation that involves synthetic division, a synthetic division solver can be a valuable resource.
In mathematics, a function is a rule that assigns a unique output to every input. A function can be represented using a graph on a coordinate plane. The input values are plotted on the x-axis, and the output values are plotted on the y-axis. A function is said to be a composite function if it can be written as the composition of two or more other functions. In other words, the output of the composite function is equal to the input of one of the other functions, which is then evaluated to produce the final output. For example, if f(x) = x2 and g(x) = 2x + 1, then the composite function h(x) = f(g(x)) can be graphed as follows: h(x) = (2x + 1)2. As you can see, solving a composite function requires you to first solve for the innermost function, and then work your way outwards. This process can be summarized using the following steps: 1) Identify the innermost function; 2) Substitute the input value into this function; 3) Evaluate the function to find the output; 4) substitute this output value into the next outermost function; 5) repeat steps 2-4 until all functions have been evaluated. By following these steps, you can solve any composite function.
An equation is a mathematical statement that two things are equal. For example, the equation 2+2=4 states that two plus two equals four. In order to solve for x, one must first identify what x represents in the equation. In the equation 2x+4=8, x represents the unknown quantity. In order to solve for x, one must use algebraic methods to determine what value x must be in order to make the equation true. There are many different methods that can be used to solve for x, but the most common method is to use algebraic equations. Once the value of x has been determined, it can be plugged into the original equation to check if the equation is still true. For example, in the equation 2x+4=8, if x=2 then 2(2)+4=8 which is true. Therefore, plugging in the value of x allows one to check if their solution is correct. While solving for x may seem like a daunting task at first, with a little practice it can be easily mastered. With a little perseverance and patience anyone can learn how to solve for x.
Fractions can be a tricky concept, especially when you're dealing with fractions over fractions. But luckily, there's a relatively easy way to solve these types of problems. The key is to first convert the mixed fraction into an improper fraction. To do this, simply multiply the whole number by the denominator and add it to the numerator. For example, if you have a mixed fraction of 3 1/2, you would convert it to 7/2. Once you've done this, you can simply solve the problem as two regular fractions. So, if you're trying to solve 3 1/2 divided by 2/5, you would first convert it to 7/2 divided by 2/5. Then, you would simply divide the numerators (7 and 2) and the denominators (5 and 2) to get the answer: 7/10. With a little practice, solving fractions over fractions will become easier and more intuitive.
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1 / 6Step-by-step explanation:There are 6 ways we have the right to roll doubles out of a possible 36 rolls (6 x 6), for a probcapacity of 6/36, or 1/6, on any type of roll of 2 fair dice."},"id":56597523,"content":"
### Tbelow are 6 means we can roll doubles out of a feasible 36 rolls (6 x 6), for a probcapacity of 6/36, or 1/6, on any kind of roll of 2 fair dice.Step-by-action explanation:If you roll two six-sided dice, what are the odds of rolling doubles?To calculate your opportunity of rolling doubles, add up all the feasible methods to roll doubles (1,1; 2,2; 3,3; 4,4; 5,5; 6,6). There are 6 methods we have the right to roll doubles out of a possible 36 rolls (6 x 6), for a probcapacity of 6/36, or 1/6, on any type of roll of 2 fair dice. So you have actually a 16.7% probability of rolling doubles via 2 fair six-sided dice.You deserve to likewise think of a sample area of all 36 possible outcomes, with the doubles highlighted:">" data-test="answer-box-list">Answer:1 / 6Step-by-step explanation:Tbelow are 6 ways we deserve to roll doubles out of a feasible 36 rolls (6 x 6), for a probcapacity of 6/36, or 1/6, on any kind of roll of two fair dice.You are watching: Probability of rolling doubles with 2 diceAdvertisementAdvertisementhimanshujc7himanshujc7Tbelow are 6 means we deserve to roll doubles out of a feasible 36 rolls (6 x 6), for a probability of 6/36, or 1/6, on any roll of 2 fair dice.
Step-by-step explanation:
If you roll 2 six-sided dice, what are the odds of rolling doubles?
To calculate your chance of rolling doubles, add up all the feasible means to roll doubles (1,1; 2,2; 3,3; 4,4; 5,5; 6,6). Tright here are 6 ways we can roll doubles out of a possible 36 rolls (6 x 6), for a probability of 6/36, or 1/6, on any type of roll of 2 fair dice. So you have a 16.7% probcapacity of rolling doubles via 2 fair six-sided dice.
You can also think of a sample area of all 36 feasible outcomes, via the doubles highlighted: |
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