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# SSAT Middle Level Math : Venn Diagrams
## Example Questions
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### Example Question #1 : How To Find The Common Part With A Venn Diagram
Using the Venn diagram shown above, what fraction represents the amount of students that enjoy both reading and playing video games?
Explanation:
The total number of survey respondents is equal to percent. Since the common part of the Venn diagram represents the respondents that enjoy both video games and reading, the solution is:
Thus, percent is equal to
### Example Question #1 : How To Find The Common Part With A Venn Diagram
The Venn diagram above, shows the results from a recent survey. The respondents were asked whether they enjoy playing video games, reading, or both. What percentage of the survey respondents enjoy both video games and reading?
Explanation:
The total number of survey respondents is equal to percent.
Since the common part of the Venn diagram represents the respondents that enjoy both video games and reading, the solution is:
### Example Question #1 : How To Find The Common Part With A Venn Diagram
The above Venn diagram depicts the number of students that: only play at the Rec. Center after school, only play at the park after school, and the students that play at both locations after school.
How many students play at both the park and Rec. Center after school?
Explanation:
To solve this problem, locate the common part of this Venn diagram. Since each of the two outside portions of the Venn diagram represent students that only play at either the park or Rec. Center.
The students in the common area of the Venn diagram, play both at the park and at the Rec. Center.
### Example Question #1 : How To Find The Common Part With A Venn Diagram
Using the Venn diagram above, what fraction of the respondents play at both the Rec. Center and the park. (Keep in mind that the two outside portions of the Venn diagram represent respondents that only play at one location or the other).
Explanation:
To solve this problem, locate the common part of this Venn diagram. Since each of the two outside portions of the Venn diagram represent students that only play at either the park or Rec. Center. The students in the common area of the Venn diagram belong to the group that plays at both the park and the Rec. Center.
This means that the total number of students (the denominator) must equal:
### Example Question #1 : How To Find The Common Part With A Venn Diagram
The Venn diagram shown above represents the number of students that: only took math during summer school, only took social studies during summer school, and the number of students that took both math and social studies during summer school.
How many students took both math and social studies?
Explanation:
The common portion of this Venn diagram is equal to , which depicts the number of students that took both math and social studies during summer school.
### Example Question #1 : Venn Diagrams
What fraction of the above Venn diagram represents the number of students that took both math and social studies during summer school?
Explanation:
The common portion of this Venn diagram is equal to , which depicts the number of students that took both math and social studies during summer school.
The denominator of the fraction is equal to the total number of students: \
Thus,
### Example Question #1 : How To Find The Common Part With A Venn Diagram
Using the Venn diagram above shows whether respondents only like Super Hero , only like Super Hero , or if they like both Super Heroes. What percentage of respondents like both Super Heroes?
Explanation:
The total number of survey respondents is equal to percent. Since the common part of the Venn diagram represents the respondents that like both Super Heroes.
Thus, the solution is:
### Example Question #1 : How To Find The Common Part With A Venn Diagram
The Venn diagram shown above displays whether respondents only use Social Media site , only use Social Media site , or use both of the Social Media sites.
What percent of respondents use both Social Media sites?
Explanation:
The total number of survey respondents is equal to percent. Since the common part of the Venn diagram represents the respondents that use both of the Social Media sites, the solution is:
### Example Question #1 : Venn Diagrams
What fraction represents the amount of respondents that use both Social Media site and Social Media site
Explanation:
The total number of survey respondents is equal to percent. Since the common part of the Venn diagram represents the respondents that use both social media sites, the solution is:
### Example Question #1 : How To Find The Common Part With A Venn Diagram
The above Venn diagram shows the amount of survey respondents that only like tacos, only like tamales, and those that like both tacos and tamales.
What fraction of respondents like both tacos and tamales?
Explanation:
The total number of survey respondents is equal to percent, which equals
The total number of respondents that do not like both tacos and tamales: .
The remaining amount equals the respondents that like both tacos and tamales.
Thus, the solution is:
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# Wai recorded the length of each string needed for a knitting project. What is the total length of the string needed?
Step-by-step explanation:
In the given question some information is missing that is attachment of file which can be described as follows:
## Related Questions
Mr. Hoya brought 5 watermelons from his grocery store. The watermelons weighed 8.25 pounds, 9.25 pounds, 8.875 pounds, 9.625 pounds and 10.75 pounds. At the party, 153 people each ate a 0.25 pound serving of watermelon. Was the amount of leftover watermelon less than, greater than, or equal to 8.5 pounds? Explain how to solve the problem.
The first thing we must do for this case is to find the total weight of the watermelons.
We have then:
We now look for the amount of watermelons that people ate.
We have then:
Therefore, the amount of watermelon remaining is:
We note that the amount of watermelon remaining is equal to 8.5 pounds.
the amount of leftover watermelon is: equal to 8.5 pounds
The amount of leftover watermelon is equal to 8.5 pounds
Explanation:
1- getting the total weight of the watermelon:
The total weight of the watermelon will be the summation of the weights of the 5 watermelons he bought.
This means that:
Total weight = 8.25 + 9.25 + 8.875 + 9.625 + 10.75
Total weight = 46.75 pounds
2- getting the weight of the eaten watermelon:
We know that at the party, there were 153 people and each ate 0.25 of a pound of watermelon.
This means that:
amount eaten = 153 * 0.25
amount eaten = 38.25 pounds
3- getting the weight of the leftover:
weight of leftover = total weight - amount eaten
We have:
Total weight = 46.75 pounds
amount eaten = 38.25 pounds
Therefore:
weight of leftover = 46.75 - 38.25
weight of leftover = 8.5 pounds
Hope this helps :)
Two numbers, if the first one increases by 1, and the second one decreases by 1, then their product increases by 2020. If the first number decreases by 1, and the second one increases by 1, what value does the product decrease?
The product decreases 2022.
Step-by-step explanation:
(x + 1)(y - 1) = xy + 2020
xy - x + y - 1 = xy + 2020
-x + y = 2021
(x - 1)(y + 1) = xy + x - y - 1
+ 2021 = -x + y
----------------------------------
(x - 1)(y + 1) + 2021 = xy - 1
(x - 1)(y + 1) = xy - 2022
The product decreases 2022.
Identify the initial value, a, and base, b, of the function f(x)=ab^x if its graph passes through the points (0, 4) and (1, 20)
a = 4, b = 5
Step-by-step explanation:
Given the exponential function
f(x) = a
Use the given points to find a and b
Using (0, 4 ) , then
4 = a ( = 1 ) , thus
a = 4 , so
f(x) = 4
Using (1, 20 ) , then
20 = 4b ( divide both sides by 4 )
b = 5
Need help ASAP if y’all could pls help me out that would be great
hi! here’s what i’d put for the sentences to fill in the blanks:
- 1 ten is equal to 10 ones. 1 ten is 10 times the value of 1 one.
- 1 hundred is equal to 10 tens. 1 hundred is 10 times the value of 1 ten.
- 1 thousand is equal to 10 hundreds. 1 thousand is 10 times the value of 1 hundred.
- 1 ten-thousand is equal to 10 thousands. 1 ten-thousand is 10 times the value of 1 thousand.
- 1 hundred-thousand is equal to 10 ten-thousands. 1 hundred-thousand is 10 times the value of 1 ten-thousand.
Place Value Chart:
blank space before 3 is 2
hope this helped!:)
Which of the following have a common denominator of 182 1/2 and 4/9
7/10 and 1 3/8
11/18 and 1 3/4
1 5/6 and 7/12
WILL MARK BRAINIEST AND WILL GET 5 STARS
1.) and
Step-by-step explanation:
To see which have the common denominator of 18, you must convert the mixed factions to improper fractions first, then try to find the LCM. Use
1.)
Find the LCM:
2,4,6,8,9,10,12,14,16,18,20
9,18
They both have the LCM of 18.
:Done
Solve for x. Note: Figure not drawn to scale.AA
B
8
X
24.6°
C
D |
# McGraw Hill My Math Kindergarten Chapter 6 Check My Progress Answer Key
All the solutions provided in McGraw Hill My Math Kindergarten Answer Key PDF Chapter 6 Check My Progress will give you a clear idea of the concepts.
## McGraw-Hill My Math Kindergarten Chapter 6 Check My Progress Answer Key
Check My Progress Page No. (395 – 396)
Vocabulary Check
Question 1.
take away
Answer:
Explanation:
There are five rabbits and three rabbits are hopping away which is 5-3 = 2. So the number of rabbits left is 2.
Question 2.
are left
Answer:
Explanation:
There are four butterflies and one butterfly flew away which is 4-1 = 3. So the number of butterflies left is 3.
Concept Check
Question 3.
Answer:
Explanation:
There are eight boats and five boats are crossed which is 8-5 = 3. So the number of boats left is 3.
Directions: Use counters to model each subtraction story. Trace the counters to show your work. 1. Five rabbits are in the grass. Three rabbits hop away. How many rabbits are left? 2. Four butterflies are on a branch. One butterfly flies away. How many butterflies are left? Write the number. 3. Eight sailboats are on the water. Five sailboats sail away. How many sailboats are left? Write the number.
Question 4.
Answer:
Explanation:
There are five books and three books are crossed away which is 5-3 = 2. So the number of book left is 2.
Question 5.
Answer:
Explanation:
There are four frogs and two frogs are hopping away which is 4-2 = 2. So the number of frogs left is 2.
Question 6.
Answer:
Explanation:
There are seven toys and four toys are taken away which is 7-4 = 3. So the number of rabbits left is 3.
Directions: Use counters to model each subtraction story. Trace the counters to show your work. 4. There are five books on the shelf. Three books are taken. Draw an X on three books. How many books are left? 5. There are four frogs on a log. Two frogs hop off. How many frogs are left? Write the number. 6. There are seven toys in the wagon. Four toys are taken. How many toys are left? Write the number.
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# Motion In Two Dimensions Problems And Solutions Pdf
File Name: motion in two dimensions problems and solutions .zip
Size: 1752Kb
Published: 26.03.2021
Motion does not happen in isolation. To explore this idea further, we first need to establish some terminology. To discuss relative motion in one or more dimensions, we first introduce the concept of reference frames.
## Motion in Two Dimensions ANSWERS TO QUESTIONS
But, we have not developed a specific equation that relates acceleration and displacement. In this section, we look at some convenient equations for kinematic relationships, starting from the definitions of displacement, velocity, and acceleration. We first investigate a single object in motion, called single-body motion. Then we investigate the motion of two objects, called two-body pursuit problems. First, let us make some simplifications in notation. Taking the initial time to be zero, as if time is measured with a stopwatch, is a great simplification. When initial time is taken to be zero, we use the subscript 0 to denote initial values of position and velocity.
## AP Physics 1 : Motion in Two Dimensions
How high does the object get before beginning its descent? The velocity must be broken down into x horizontal and y vertical components. We can use the y component to find how high the object gets. To find vertical velocity, v y , use. Next we find how long it takes to reach the top of its trajectory using. Finally, find how high the object goes with. How far away does the object land?
Note that if the initial velocity is in a direction that is along neither axis, one must first break up the initial velocity into its components. In the last few chapters we have considered the motion of a particle that moves along a straight line with constant acceleration. In such a case, the velocity and the acceleration are always directed along one and the same line, the line on which the particle moves. Here we continue to restrict ourselves to cases involving constant acceleration constant in both magnitude and direction but lift the restriction that the velocity and the acceleration be directed along one and the same line. If the velocity of the particle at time zero is not collinear with the acceleration, then the velocity will never be collinear with the acceleration and the particle will move along a curved path. The curved path will be confined to the plane that contains both the initial velocity vector and the acceleration vector, and in that plane, the trajectory will be a parabola. The trajectory is just the path of the particle.
The negative solutions are not relavent to this problem. Page MFMcGraw - PHY Chap_04H - 2D & 3D - Revised 1/3/.
## Motion in Two Dimensions ANSWERS TO QUESTIONS
Motion does not happen in isolation. To explore this idea further, we first need to establish some terminology. To discuss relative motion in one or more dimensions, we first introduce the concept of reference frames.
To browse Academia. Skip to main content. By using our site, you agree to our collection of information through the use of cookies.
Stanley Kowalski. Peter Dourmashkin, Prof. David Litster, Prof. David Pritchard, Prof. Bernd Surrow.
### Sample Problems and Solutions
Projectile Motion Answers In his work, he came up with the three basic ideas that we still use to describe the physics of motion up to a point. Projectile Motion. Exploration Sheet Answer Key. Projectile motion worksheet. Our projectile is equivalent to an object moving up and down, and also sliding in the horizontal direction while executing this vertical motion. You have remained in right site to start getting this info. Therefore even though acceleration due to the force of gravity is present vertically,.
Kinematic worksheet 1 answers. An object goes from one point in space to another. Created with Geogebra Print labbutsasen. It will play here. If we look at the area under the curve, we can break it into a rectangle and a triangle. Some of the worksheets displayed are Kinematics practice problems, Kinematic equations work, Unit 2 kinematics work 2 stacks of kinematic curves, Physics kinematics objectives students will be able to, Topic 3 kinematics displacement velocity acceleration, Physics kinematics work solutions, Ib physics kinematics work, 1 confronting. Notice that Equations 1 and 2 have a common variable, t.
How high does the object get before beginning its descent? The velocity must be broken down into x horizontal and y vertical components. We can use the y component to find how high the object gets. To find vertical velocity, v y , use. Next we find how long it takes to reach the top of its trajectory using. Finally, find how high the object goes with.
#### 4.5 Relative Motion in One and Two Dimensions
How high does the object get before beginning its descent? The velocity must be broken down into x horizontal and y vertical components. We can use the y component to find how high the object gets. To find vertical velocity, v y , use. Next we find how long it takes to reach the top of its trajectory using.
How high does the object get before beginning its descent? The velocity must be broken down into x horizontal and y vertical components. We can use the y component to find how high the object gets. To find vertical velocity, v y , use. Next we find how long it takes to reach the top of its trajectory using. Finally, find how high the object goes with. How far away does the object land?
How high does the object get before beginning its descent? The velocity must be broken down into x horizontal and y vertical components. We can use the y component to find how high the object gets. To find vertical velocity, v y , use.
The chapter has the following three sections: "Parameters" explains how Cg programs handle parameters. Participate in the Sanfoundry Certification contest to get free Certificate of Merit. She starts at position A, travels 3. Kariithi, M. Exercise 1: Read the following sentences.
How high does the object get before beginning its descent? The velocity must be broken down into x horizontal and y vertical components. We can use the y component to find how high the object gets.
Thanks for visiting our website. Our aim is to help students learn subjects like physics, maths and science for students in school , college and those preparing for competitive exams. All right reserved. All material given in this website is a property of physicscatalyst. Average velocity Consider a particle moving along a curved path in x-y plane shown below in the figure Suppose at any time,particle is at the point P and after some time 't' is at point Q where points P and Q represents the position of particle at two different points.
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Observe that motion in two dimensions consists of horizontal and vertical Apply the principle of independence of motion to solve projectile motion problems. Solution. (1) Draw the three displacement vectors. Figure (2) Place the. |
## Precalculus: Mathematics for Calculus, 7th Edition
Published by Brooks Cole
# Chapter 1 - Section 1.5 - Equations - 1.5 Exercises: 56
#### Answer
$x=\dfrac{\sqrt{7}-1}{5}$ $x=-\dfrac{1+\sqrt{7}}{5}$
#### Work Step by Step
$(5x+1)^{2}+3=10$ We need to solve this by factoring, so let's take the $10$ to the left side of the equation to substract: $(5x+1)^{2}+3-10=0$ Simplify: $(5x+1)^{2}-7=0$ Remember the difference of two squares factoring formula: $a^{2}-b^{2}=(a-b)(a+b)$ In this equation $a=(5x+1)^{2}$ and $b=7$, so we can factor it and we get: $[(5x+1)-\sqrt{7}][(5x+1)+\sqrt{7}]=0$ Then, we set each factor equal to $0$ and solve them as individual equations. Let the $[(5x+1)-\sqrt{7}]=0$ be equation number 1 and $[(5x+1)+\sqrt{7}]=0$ be equation number 2: Solving for $x$ in equation number 1: $(5x+1)-\sqrt{7}=0$ $5x=\sqrt{7}-1$ $x=\dfrac{\sqrt{7}-1}{5}$ Solving for $x$ in equation number 2: $(5x+1)+\sqrt{7}=0$ $5x=-1-\sqrt{7}$ $x=-\dfrac{1+\sqrt{7}}{5}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback. |
7 Title
Intermediate Algebra
Tutorial 36: Practice Test on Tutorials 32 - 35
Problem 1a:
Answer: Our restriction here is that the denominator of a fraction can never be equal to 0. So, to find our domain, we want to set the denominator "not equal" to 0 to restrict those values. Our domain is all real numbers except -9 and 9, because they both make the denominator equal to 0, which would not give us a real number answer for our function.
Problem 2a:
Problem 2b:
Problem 2c:
Answer: Step 1: Find the LCD if needed. The first denominator has the following factor: The second denominator, 3 - x, looks like the first denominator except the signs are switched. We can rewrite this as Putting all the different factors together and using the highest exponent, we get the following LCD: Step 2: Write equivalent fractions using the LCD if needed. Now the two fractions have a common denominator, so we do not have to rewrite the rational expressions. Step 3: Combine the rational expressions AND Step 4: Reduce to lowest terms.
Problem 2d:
Answer: Step 1: Find the LCD if needed. The first denominator has the following two factors: The second denominator has the following factor: The third denominator has the following factor: Putting all the different factors together and using the highest exponent, we get the following LCD: Step 2: Write equivalent fractions using the LCD if needed. Since the first rational expression already has the LCD, then we do not need to change this fraction. Rewriting the second expression with the LCD: Rewriting the third expression with the LCD: Step 3: Combine the rational expressions AND Step 4: Reduce to lowest terms.
Problem 3a:
Answer: Note that I used method I described in Tutorial 34 (Complex Fractions) to work this problem. It is perfectly find to use method II here. Step 1: If needed, rewrite the numerator and denominator so that they are each a single fraction AND Step 2: Divide the numerator by the denominator by multiplying the numerator by the reciprocal of the denominator AND Step 3: If needed, simplify the rational expression.
Problem 3b:
Answer: Note that I used method II described in Tutorial 34: Complex Fractions to work this problem. It is perfectly fine to use method I here. Rewriting it with positive exponents we get: Step 1: Multiply the numerator and denominator of the overall complex fractions by the LCD of the smaller fractions AND Step 2: If needed, simplify the rational expression. The two denominators of the numerator's fractions have the following factors: a and b The two denominators of the denominator's fractions have the following factors: and Putting all the different factors together and using the highest exponent we get the following LCD for all the small fractions: Multiplying numerator and denominator by the LCD and simplifying we get:
Problem 4a:
Problem 4b:
Answer: Step 1: Set up the long division AND Step 2: Divide 1st term of divisor by first term of dividend to get first term of the quotient AND Step 3: Take the term found in step 1 and multiply it times the divisor AND Step 4: Subtract this from the line above AND Step 5: Repeat until done AND Step 6: Write out the answer. Final answer:
Last revised on July 17, 2011 by Kim Seward. |
## Rational Approximation of Pi
A calculator gives us rational approximations of irrational numbers, so we can “see” that $2\sqrt{3}>\pi$. Explain how a circle inscribed in a regular hexagon allows us to “see” that $2\sqrt{3}>\pi$ without any decimal approximations.
Source: NCTM Mathematics Teacher 2008
SOLUTION
Consider the circle of center $A$ inscribed in a regular hexagon of side length $a$. Triangle $ABC$ is an equilateral triangle. Triangle $AHC$ a 30-60-90 degrees triangle with hypotenuse of length $a$. $AH$ is also the radius of the circle
$r=\sqrt{3}(a/2)$
$a=\dfrac{2r}{\sqrt{3}}$
$=\dfrac{2r\sqrt{3}}{3}$
Method 1: Area of triangle $AHC$ > area of sector $AHC$
$\dfrac{1}{2}\left (r\dfrac{a}{2}\right )>\pi r^2\dfrac{30^\circ}{360^\circ}$
Simplify
$\dfrac{ar}{4}>\dfrac{\pi r^2}{12}$
Divide both sides of inequality by $r\neq 0$ and multiply by $12$
$3a>\pi r$
Substitute the value of $a$ into the above inequality
$3\dfrac{2r\sqrt{3}}{3}>\pi r$
Simplify
$2\sqrt{3}>\pi$
Method 2: Area of hexagon > area of circle
$6\left (\dfrac{1}{2}ar\right )>\pi r^2$
$3ar>\pi r^2$
Substitute the value of $a$ into the above inequality
$3\left (\dfrac{2r\sqrt{3}}{3}\right )r>\pi r^2$
Simplify
$2\sqrt{3}>\pi$ |
DEFINITIONS - Trigonometric Functions and Their Inverses - Functions - REVIEW OF MAJOR TOPICS - SAT SUBJECT TEST MATH LEVEL 2
## CHAPTER 1Functions
### DEFINITIONS
The general definitions of the six trigonometric functions are obtained from an angle placed in standard position on a rectangular coordinate system. When an angle is placed so that its vertex is at the origin, its initial side is along the positive x-axis, and its terminal side is anywhere on the coordinate system, it is said to be in standard position. The angle is given a positive value if it is measured in a counterclockwise direction from the initial side to the terminal side, and a negative value if it is measured in a clockwise direction.
Let P(x,y) be any point on the terminal side of the angle, and let r represent the distance between O and P. The six trigonometric functions are defined to be:
TIP sin θ and cos θ are always between –1 and 1.
From these definitions it follows that:
The distance OP is always positive, and the x and y coordinates of P are positive or negative depending on which quadrant the terminal side of lies in. The signs of the trigonometric functions are indicated in the following table.
TIP All trig functions are positive in quadrant I. Sine and only sine is positive in quadrant II. Tangent and only tangent is positive in quadrant III. Cosine and only cosine is positive in quadrant IV. Just remember:All StudentsTake Calculus.
Each angle whose terminal side lies in quadrant II, III, or IV has associated with it an angle called its reference angle , which is formed by the x-axis and the terminal side.
Any trig function of = ± the same function of . The sign is determined by the quadrant in which the terminal side lies.
EXAMPLES
1. Express sin 320° in terms of .
Since the sine is negative in quadrant IV, sin 320° = –sin 40°.
2. Express cot 200° in terms of .
Since the cotangent is positive in quadrant III, cot 200° = cot 20°.
3. Express cos 130° in terms of .
Since the cosine is negative in quadrant II, cos 130° = –cos 50°.
Sine and cosine, tangent and cotangent, and secant and cosecant are cofunction pairs. Cofunctions of complementary angles are equal. If and are complementary, then trig () = cotrig () and trig () = cotrig () .
4. If both the angles are acute and sin (3x + 20°) = cos (2x – 40°), find x.
Since these cofunctions are equal, the angles must be complementary.
Therefore,
EXERCISES
1. Express cos 320° as a function of an angle between 0° and 90°.
(A) cos 40°
(B) sin 40°
(C) cos 50°
(D) sin 50°
(E) none of the above
2. If point P(–5,12) lies on the terminal side of in standard position, sin =
(A)
(B)
(C)
(D)
(E)
3. If and sin , then tan =
(A)
(B)
(C)
(D)
(E) none of the above
4. If x is an angle in quadrant III and tan (x – 30°) = cot x, find x.
(A) 240°
(B) 225°
(C) 210°
(D) 60°
(E) none of the above
5. If 90° < < 180° and 270° < < 360°, then which of the following cannot be true?
(A) sin = sin
(B) tan = sin
(C) tan = tan
(D) sin = cos
(E) sec = csc
6. Expressed as a function of an acute angle, cos 310° + cos 190° =
(A) –cos 40°
(B) cos 70°
(C) –cos 50°
(D) sin 20°
(E) –cos 70°
|
# Proof of the quadratic formula
The following is a proof of the quadratic formula. It will show you how the quadratic formula, which is widely used has been developed.
The evidence is performed using the standard form of a quadratic equation and the standard form of problems by completing the square
ax2 + bx + c = 0
Divide both sides of the equation by a kind you can fill the place
Subtract c/a leave both sides
Complete the square:
The coefficient of the second term is b / a
This coefficient divided by 2 and square the result to get (a b/2) 2
Add 2 (b/a) 2 on both sides:
Since the left side of the right side of the equation above is a perfect square, you can factor the left-hand side by using the coefficient of the first term (x) and the base of the last term(b/2a)
Add these two and raise at the second.
Then, square on the right side to get (b2) /(4a2)
Get the same denominator on the right side:
Now, take the square root of each side:
Simplify the left:
Rewrite the right side:
Subtract (b) / 2 (a) of both sides:
Adding the numerator, keeping the same denominator, we get the quadratic formula:
The + – between the b and the sign of the square root mean more or negative. In other words, most of the time, you will get two answers using the quadratic formula. |
# Asymptotes of a function
Printable View
• Sep 19th 2009, 11:31 PM
VitaX
Asymptotes of a function
y = (2x)/(x+1) When solving for the horizontal asymptote and substituting infinity in for x, infinity just crosses out and i'm left with 2/1 making the horizontal asymptote y = 2. right?
• Sep 20th 2009, 12:18 AM
mr fantastic
Quote:
Originally Posted by VitaX
y = (2x)/(x+1) When solving for the horizontal asymptote and substituting infinity in for x, infinity just crosses out and i'm left with 2/1 making the horizontal asymptote y = 2. right?
You cannot substitute infinity. It's not a number.
Approach 1:
$y = \frac{2x}{x + 1} = \frac{2}{1 + \frac{1}{x}}$.
$\lim_{x \rightarrow \pm \infty} \frac{2}{1 + \frac{1}{x}} = 2$.
Therefore $\lim_{x \rightarrow \pm \infty} y = 2$ and so $y = 2$ is the horizontal asymptote.
Approach 2:
$y = \frac{2x}{x + 1} = 2 \left( \frac{x}{x + 1} \right) = 2 \left( 1 - \frac{1}{x + 1}\right) = 2 - \frac{2}{x + 1}$
and by inspection the horizontal asymptote is $y = 2$.
Other approaches are possible.
• Sep 20th 2009, 12:42 AM
VitaX
Quote:
Originally Posted by mr fantastic
You cannot substitute infinity. It's not a number.
Approach 1:
$y = \frac{2x}{x + 1} = \frac{2}{1 + \frac{1}{x}}$.
$\lim_{x \rightarrow \pm \infty} \frac{2}{1 + \frac{1}{x}} = 2$.
Therefore $\lim_{x \rightarrow \pm \infty} y = 2$ and so $y = 2$ is the horizontal asymptote.
Approach 2:
$y = \frac{2x}{x + 1} = 2 \left( \frac{x}{x + 1} \right) = 2 \left( 1 - \frac{1}{x + 1}\right) = 2 - \frac{2}{x + 1}$
and by inspection the horizontal asymptote is $y = 2$.
Other approaches are possible.
You say you cannot substitute infinity because it is not a number. But my teacher has said on several occassions it is a number but not a measurable one. But when infinity is in the denominator of any equation example : 6/infinity that is 0. So perhaps I just understood you wrong.
• Sep 20th 2009, 12:52 AM
mr fantastic
Quote:
Originally Posted by VitaX
You say you cannot substitute infinity because it is not a number. But my teacher has said on several occassions it is a number but not a measurable one. But when infinity is in the denominator of any equation example : 6/infinity that is 0. So perhaps I just understood you wrong.
No, you have understood me perfectly. Infinity is not a number and cannot be substituted into an expression. To do so is sloppy and mathematically wrong.
When dealing with infinity, you have to take a limit. 6/infinity = 0 is a complete nonsense, $\lim_{x \rightarrow + \infty} \frac{6}{x} = 0$ is the correct way of treating it.
• Sep 20th 2009, 12:54 AM
VitaX
Quote:
Originally Posted by mr fantastic
No, you have understood me perfectly. Infinity is not a number and cannot be substituted into an expression. To do so is sloppy and mathematically wrong.
When dealing with infinity, you have to take a limit. 6/infinity = 0 is a complete nonsense, $\lim_{x \rightarrow + \infty} \frac{6}{x} = 0$ is the correct way of treating it.
Yes i understand I just didn't say take the limit of 1/x as x approaches infinity. I understand the terminology, just not familiar with latex. |
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# Common Multiples
What do the numbers 20 and 50 have in common? Well, both these numbers are divisible by 2, 5 and 10. We say this because no remainder will exist when we divide them by these three said numbers. This means that 20 and 50 are multiples of 2, 5 and 10 since
$\mathbf{2}\times 10=\mathbf{5}\times 4=\mathbf{10}\times 2=20$
$\text{and}$
$\mathbf{2}\times 25=\mathbf{5}\times 10=\mathbf{10}\times 5=50$
Looking at our derivation above, we can further infer that the numbers 2, 5 and 10 share two multiples, namely 20 and 50. These shared numbers are called common multiples. This article will demonstrate a method we can use to identify common multiples for a given set of numbers.
## Recap: Multiples
To ease ourselves into this topic, let us go through a quick overview of our previous topic on multiples.
A multiple of a non-zero integer $$A$$ is a non-zero integer $$C$$ that can be obtained by multiplying it with another number, say $$B$$.
In other words, $$C$$ is a multiple of $$A$$ if $$C$$ is in the multiplication table of $$A$$.
The multiple of a number, say $$a$$, is given by the general formula,
$\text{Multiple of}\ a=a\times z$
where $$z\in\mathbb{Z}$$. In other words,
if $$A\times B=C$$ then $$A$$ and $$B$$ are divisors (or factors) of $$C$$,
or $$C$$ is a multiple of $$A$$ (and also $$B$$).
To find a particular set of multiples for a given number, we can use the multiplication table.
As with our example above, the numbers 20 and 50 are multiples of 2, 5 and 10. The following table shows other multiples of 2, 5 and 10.
Number First 6 non-zero multiples 2 2, 4, 6, 8, 10, 12 5 5, 10, 15, 20, 25, 30 20 20, 40, 60, 80, 100, 120
A more in-depth explanation of multiples can be found in the topic called Multiples.
## Definition of a Common Multiple and Method
Let us now define a common multiple.
A common multiple is a multiple that is shared between two (or more) numbers.
Identifying a common multiple(s) for a given set of numbers is fairly straightforward. Given a set of numbers, you would simply execute two steps:
Step 1: List the multiples of each number given in the set;
Step 2: Pick out any identical multiples shared from the lists written in Step 1.
Recall that there are an infinite number of multiples for any integer. With this property in mind, a restriction may be introduced in Step 1. In most cases, the question will define an interval for which the common multiples are satisfied for a given set of numbers.
For example, you may get questions that use the phrase "find the first two common multiples of 2 and 3" or "list the common multiples of 2 and 3 between 1 and 10". However, an interval restriction is not necessary here. But, it is safe to say that no individual can list all the common multiples for a given set of numbers by hand. That would take yards of ink and paper!
Important note: Although zero is indeed a common multiple for any set of numbers, you would typically list down the non-zero common multiples only (we shall look into this in the next section).
Here is an example for finding common multiples for a given set of numbers.
List all the (non-zero) common multiples of 9, 12 and 15 between 1 and 100.
Solution
The interval restriction here is that we need to list the multiples of 9, 12 and 15 between 1 and 100. We shall begin by listing these multiples using the table below.
Number Multiples between 1 and 100. 9 9, 18, 27, 36, 45, 54, 63, 72, 81, 90, 99 12 12, 24, 36, 48, 60, 72, 84, 96 15 15, 30, 45, 60, 75, 90
Looking at the table above, there are no visible common multiples of 9, 12 and 15 for this interval. However, you can deduce the following ideas:
1. The common multiples of 9 and 12 are 36 and 72 for this interval;
2. The common multiples of 9 and 15 are 45 and 90 for this interval;
3. The common multiple of 12 and 15 is 60 for this interval.
Here is another worked example.
List the first 2 non-zero common multiples of 5 and 17.
Solution
The interval restriction here is that we need to list the first 2 non-zero multiples of 5 and 17.
Sometimes, listing multiples can be rather cumbersome, especially when the numbers are very far apart from each other. As with our case here, the difference between 5 and 17 is quite large, so listing the multiples of 5 may take a while until we can find one that is also a multiple of 17.
For situations like this, it is advised to list down the multiples of the larger number and test whether these multiples are also multiples of the smaller number. We do this by verifying that it is divisible by each other (this will be further explained in the next section).
For this example, let us write down the first few non-zero multiples of 17.
Multiples of 17: 17, 34, 51, 68, 85, 102, 119, 136, 153, 170, 187,...
From this list, we can observe that 85 and 170 are indeed divisible by 5 since $$5\times 17=85$$ and $$5\times 34=170$$. Thus, the first 2 non-zero common multiples of 5 and 17 are 85 and 170.
Let us look at one more example before moving on to the next section.
List all the common multiples of 11 and 13 between 130 and 300.
Solution
The interval restriction here is that we need to list the multiples of 11 and 13 between 130 and 300. As before, we will start by listing these multiples using the table below.
Number Multiples between 130 and 300 11 132, 143, 154, 165, 176, 187, 198, 209, 220, 231, 242, 253, 264, 275, 286, 297 13 130, 143, 156, 169, 182, 195, 208, 221, 234, 247, 260, 273, 286, 299
From the table above, observe that there are two common multiples of 11 and 13 between 130 and 300, namely 143 and 286.
The concept of a common multiple is primarily used to find the lowest common multiple (or LCM) between a given set of numbers. This is the smallest common multiple shared between two (or more) numbers. You can find a thorough discussion on this topic in the article: Lowest Common Multiple.
Try it yourself: Answer the following questions.
1. What are the first two non-zero common multiples of 16 and 27?
2. What are the common multiples of 9 and 12 between 22 and 140?
Solutions
Question 1: 432, 864
Question 2: 36, 72, 108
## Properties of Common Multiples
Before we move on to more examples involving common multiples, let us establish some important properties of common multiples.
Property Example A set of numbers can have more than one common multiple. 6 is a common multiple of 3 and 6. However, this is not the only common multiple of 3 and 6. The numbers 12, 18 and 24 are also some other common multiples of 3 and 6. A set of numbers can have an infinite number of common multiples. Common multiples of 5 and 8 include 40, 80, 120, 160, ... The values will keep increasing and the list will go on forever. The common multiple of a set of numbers is always greater than or equal to each of the numbers themselves (excluding 0). Common multiples of 2 and 4 between 1 and 10 are 4 and 8. Notice that the multiple 4 is greater than the given number 2 but equal to the given number 4. The multiple 8 however is greater than both the given numbers 2 and 4. The given set of numbers divides the common multiple without leaving a remainder. These numbers are called the factors. A common multiple of 8 and 17 is 136. Dividing 136 by each of these given numbers will not produce a remainder since $$8\times 17=136$$ and $$17\times 8=136$$. Every non-zero integer is a multiple of 0 since any non-zero integer multiplied by 0 equal 0. In most cases, we will only consider non-zero common multiples. Since $$7\times 0=0$$ and $$9\times 0=0$$ then 0 is a common multiple of 7 and 9.
## Examples of Common Multiples
We shall end this topic by looking at a few more worked examples concerning common multiples.
List all the common multiples of 6, 8 and 10 between 1 and 100.
Solution
To begin, we shall list the multiples of each given number between 1 and 100. This is shown in the table below.
Number Multiples between 1 and 100 6 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96 8 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96 10 10, 20, 30, 40, 50, 60, 70, 80, 90, 100
From the table above, we find that there are no common multiples of 6, 8 and 10 between 1 and 100. However, we can indeed conclude the following relationships:
1. The common multiples of 6 and 8 are 24, 48, 72 and 96 for this interval
2. The common multiples of 6 and 10 are 30, 60 and 90 for this interval
3. The common multiples of 8 and 10 are 4 and 80 for this interval
Let us now move on to another example.
List the first 4 non-zero common multiples of 2, 7 and 14.
Solution
First, note that 2 is relatively located far from 7 and 14 on the number line. Thus, it is sensible to list the multiples of 7 and 14 and compare their common multiples. From here, you will check whether these common multiples are also divisible by 2.
Multiples of 7: 7, 14, 21, 28, 35, 42, 49, 56, 63, 70,...
Multiples of 14: 14, 28, 42, 56, 70,...
Notice that the first 4 non-zero common multiples of 7 and 14 are 14, 28, 42 and 56. All four of these numbers are even which means that they are also divisible by 2. Thus, the first 4 non-zero common multiples of 2, 7 and 14 are 14, 28, 42 and 56.
We shall look at one last example involving common multiples.
List the first 3 non-zero common multiples of 3 and 19.
Solution
The difference between 3 and 19 is rather significant. So, as with our previous example, we will only list the multiples of 19 and verify whether they are also divisible by 3.
Multiples of 19: 19, 38, 57, 76, 95, 114, 133, 152, 171, 190,...
From this list, we find that the numbers 57, 114 and 171 are also divisible by 3 since $$3\times 19=57$$, $$3\times 138=114$$ and $$3\times 57=171$$. Therefore, the first 3 non-zero common multiples of 3 and 19 are 57, 114 and 171.
### Real-world Examples Involving Common Multiples
Here is an interesting question: can we apply common multiples in real-life situations? As a matter of fact, we can! In this section, we shall demonstrate two examples of real-world scenarios that encapsulate all that we have learnt from this discussion.
Polly and Hannah decide to take turns visiting their friend, Ben, at the hospital. Polly suggests visiting Ben every 3 days while Hannah visits him every 5 days. If both of them visited Ben today, how long will it be until the next time they see him the same day again?
Solution
Here, we simply need to find the first non-zero common multiple of days 3 and 5. We shall take today as the first multiple of days 3 and 5 which is day 0. Remember, every non-zero integer is a multiple of 0 since any non-zero integer multiples by 0 equal 0 (property 5 of common multiples).
Let us now write down the common multiples of 3 and 5:
Multiples of 3: 2, 6, 9, 12, 15, 18,...
Multiples of 5: 5, 10, 15, 20,...
From both these lists, we see that 15 is the first non-zero common multiple of 3 and 5. Hence, the next time both Polly and Hannah will visit Ben together is on day 15.
Rory and Tana are jogging around a circular running track. Rory takes 12 minutes to complete a lap while Tana takes 16 minutes. If both of them leave the starting point at the same time, write down the next two times both of them will pass the starting point together again.
Solution
Using a similar approach as the previous example, we need to locate the first two non-zero common multiples of minutes 12 and 16. Taking the first time they leave the starting point says minute 0, we can now list the multiples of 12 and 16.
Multiples of 12: 12, 24, 36, 48, 60, 72, 84, 96, 108,...
Multiples of 16: 16, 32, 48, 64, 80, 96, 112,...
Looking at the lists above, notice that 48 and 96 are the first two non-zero common multiples of 12 and 16. Thus, Rory and Tana will pass the starting together again at minutes 48 and 96.
## Common Multiples - Key takeaways
• A common multiple is a multiple that is shared between two numbers.
• To find the common multiples of a given set of numbers:
1. List the multiples of each number given in the set;
2. Pick out any identical multiples shared from the lists written in Step 1.
• Important properties of common multiples:
• A set of numbers can have more than one common multiple
• A set of numbers can have an infinite number of common multiples
• The common multiple of a set of numbers is always greater than or equal to each of the numbers themselves
• The given set of numbers divides the common multiple without leaving a remainder.
• Every non-zero integer is a multiple of 0 since any non-zero integer multiples by 0 equal 0.
A common multiple is a multiple that is shared between two (or more) numbers.
The characteristics/properties of common multiples are:
• a set of numbers can have more than one common multiple;
• a set of numbers can have an infinite number of common multiples;
• the common multiple of a set of numbers is always greater than or equal to each of the numbers themselves (excluding 0);
• the given set of numbers divides the common multiple without leaving a remainder;
• every non-zero integer is a multiple of 0 since any non-zero integer multiplied by 0 equal 0.
A common multiple is not to be solved but to be determined. And you can determine a common multiple of two (or more) numbers by executing two steps:
• Step 1: List the multiples of each number given in the set;
• Step 2: Pick out any identical multiples shared from the lists written in Step 1.
The rules of common multiples are the properties of common multiples.
Examples of common multiples of two (or more) number are:
• the first two non-zero common multiples of 16 and 27 are 432 and 864;
• the common multiples of 2, 7 and 14 between 1 and 60 are 14, 28, 42 and 56.
## Final Common Multiples Quiz
Question
What is a common multiple?
A common multiple is a multiple that is shared between two numbers
Show question
Question
What are the two steps in finding common multiples?
1. List the multiples of each number given in the set;
2. Pick out any identical multiples shared from the lists written in Step 1.
Show question
Question
A set of numbers can have more than one common multiple. Is this true or false?
True
Show question
Question
A set of numbers can have an infinite number of common multiples. Is this true or false?
True
Show question
Question
Common multiples are always less than or equal to each of the numbers in a given set. Is this true or false?
False
Show question
Question
A remainder is always present when you divide a given set of numbers by its common multiples. Is this true or false?
False
Show question
Question
Every non-zero integer is a multiple of 0. Is this true or false?
True
Show question
Question
What are the first two non-zero common multiples of 17 and 22?
374, 748
Show question
Question
What are the first three non-zero common multiples of 5 and 16?
80, 160, 240
Show question
Question
What are the first two non-zero common multiples of 12 and 47?
564, 1128
Show question
Question
What are the first three non-zero common multiples of 81 and 108?
324, 648, 972
Show question
Question
What are the first four non-zero common multiples of 6, 10 and 14?
210, 420, 630, 840
Show question
Question
What are the common multiples of 5 and 11 between 30 and 130?
55, 110
Show question
Question
What are the common multiples of 23 and 25 between 200 and 500?
225, 450
Show question
Question
What are the common multiples of 32 and 48 between 200 and 400?
192, 288, 384
Show question
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# Multiplication Flash Cards 6-9
Learning multiplication following counting, addition, and subtraction is good. Youngsters find out arithmetic through a normal progression. This growth of studying arithmetic is usually the subsequent: counting, addition, subtraction, multiplication, and ultimately department. This document results in the question why find out arithmetic in this particular sequence? More importantly, why find out multiplication after counting, addition, and subtraction but before department?
## The next facts answer these questions:
1. Youngsters learn counting initial by associating visible objects using their fingertips. A tangible case in point: How many apples exist from the basket? Far more abstract instance is how older are you?
2. From counting numbers, the next rational stage is addition combined with subtraction. Addition and subtraction tables can be quite valuable training tools for children as they are graphic instruments making the changeover from counting easier.
3. Which should be figured out after that, multiplication or section? Multiplication is shorthand for addition. At this stage, kids possess a business knowledge of addition. Therefore, multiplication will be the next reasonable method of arithmetic to learn.
## Overview fundamentals of multiplication. Also, assess the fundamentals how to use a multiplication table.
Let us assessment a multiplication example. Employing a Multiplication Table, flourish 4 times about three and obtain an answer a dozen: 4 x 3 = 12. The intersection of row three and line 4 of a Multiplication Table is a dozen; twelve is definitely the solution. For the kids beginning to understand multiplication, this can be effortless. They may use addition to eliminate the issue thus affirming that multiplication is shorthand for addition. Case in point: 4 by 3 = 4 4 4 = 12. It is an exceptional summary of the Multiplication Table. An added benefit, the Multiplication Table is visible and demonstrates to understanding addition.
## Where can we get started understanding multiplication using the Multiplication Table?
1. Initially, get informed about the table.
2. Start out with multiplying by a single. Begin at row number 1. Relocate to line primary. The intersection of row a single and line one is the perfect solution: a single.
3. Repeat these techniques for multiplying by one particular. Multiply row 1 by columns 1 by way of twelve. The replies are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 correspondingly.
4. Perform repeatedly these methods for multiplying by two. Increase row two by posts one particular by means of five. The replies are 2, 4, 6, 8, and 10 correspondingly.
5. Allow us to hop in advance. Repeat these methods for multiplying by 5. Flourish row several by posts one particular through 12. The responses are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 correspondingly.
6. Now we will improve the amount of problems. Recurring these actions for multiplying by 3. Increase row 3 by posts a single through twelve. The responses are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 correspondingly.
7. In case you are comfortable with multiplication so far, try out a test. Remedy these multiplication troubles in your mind then assess your responses to the Multiplication Table: grow 6 and 2, grow nine and 3, increase one particular and 11, increase 4 and several, and increase six and 2. The issue replies are 12, 27, 11, 16, and 14 respectively.
If you obtained several out of 5 various problems appropriate, build your very own multiplication exams. Determine the solutions in your thoughts, and check them utilizing the Multiplication Table. |
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)
$(-2,1)$
Isolate $t$ in eq.$1$ $-5s+t=11\qquad$ ... add $5s$ $t=11+5s\qquad$(*) Replace $t$ with $11+5s$ in the other equation: $4s+12(11+5s)=4\qquad$ ... and solve for $s$. Simplify $4s+132+60s=4$ $64s+132=4\qquad$ ... subtract 132 $64s=-128\qquad$ ... divide with $64$ $s=-2$ Now, back-substitute into (*) $t=11+5(-2)=11-10=1$ Form an ordered pair $(s,t)$ as the solution to the system $(-2,1)$ |
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# 45°-45°-90° Triangle Explained with Examples
Since we know the special right triangles and right Triangles, it is time to review them independently. Let’s learn more about the 45 ° -45 ° -90 ° triangle.
### Quick discussion on 45 ° -45 ° -90 ° Triangle
It is a special right triangle that has one 90-degree angle and two 45-degree angles. The side lengths of this Triangle remain in the ratio of;
n: n: n √ 2 = 1:1: √ 2: Side 1: Side 2: Hypotenuse
Half of a square is the 45 ° -45 ° -90 ° right Triangle. This is because the square has each angle equivalent to 90 °, and when it is reduced diagonally, one angle stays the same as 90 °, and the other 2 90 ° angles are bisected (cut into half) as well as become 45 ° each.
The diagonal of a square end up being the two sides of a square, and the hypotenuse of a right triangle is the two sides of a right triangle.
The 45 ° -45 ° -90 ° right Triangle is occasionally described as an isosceles right triangle because it has two equivalent side lengths and two equivalent angles.
### We can determine the 45 ° -45 ° -90 ° right triangle hypotenuse as adheres to:
Let’s assume in the isosceles right Triangle, side one and side two be x.
Apply the Pythagorean Theory a2 + b2 = c2, where an and b are sides one and two, and c is the hypotenuse.
x2 + x2 = 2×2
Locate the square root of each term in the formula
√ x2 + √ x2 = √( 2×2).
x + x = x √ 2.
As a result, the 45 °; 45 °; 90 ° triangle hypotenuse is x √ 2.
### How to Resolve a 45 ° -45 ° -90 ° Triangle?
To aid show what the Triangle with 45 45 90 as its angles appears like, along with clarifying the values you’ll have to deal with moving forward, we’ll use the example below. It shows a conventional 45-45 90 triangle that can help you understand the proportions that transpire when this Triangle is utilized.
Provided the 45 ° -45 ° -90 ° triangle length of one side, you can quickly compute the various absent side lengths without turning to the Pythagorean Theorem or trigonometric approaches functions.
Computations of a 45 ° -45 ° -90 ° right triangle come under 2 opportunities:
### Case 1
To compute the hypotenuse length, multiply the given length by 2 when offered the length of one side.
### Example.
When provided the hypotenuse length of a 45 ° -45 ° -90 ° triangle, you can determine the side lengths by just dividing the hypotenuse by 2.
Note: The 45 ° -45 ° -90 ° triangles can be resolved using the 1:1: 2 ratio approach.
### Example
The 45 °, 45 °, and 90 ° triangle hypotenuses are 6 2 mm. Determine the length of its base and also elevation.
### Solution
The 45 °, 45 °, and 90 ° triangle ratio is n: n: n 2. So, we have;
n 2 = 6 2 mm
Square both sides of the formula.
(n 2) 2 = (6 2) 2 mm
2n2 = 36 * 2
2n2 = 72
n2 = 36
Locate the square root.
n = 6 mm
Thus, the height and base of the right Triangle are 6 mm each.
### Proving the particular triangle theorem
There are two ways we can use to verify the 45-45-90 triangle theory. With 45-45-90 triangles, we are given the angles and the proportions of the length of the sides. This tells us that we will have to solve simple dimensions of the 45-45-90 Triangle to support the 45-45-90 triangle theory.
Recognizing these items of info, we can check our job by working backwards to show that the size of the sides remains in conformity with the ratios. Keep in mind that in the isosceles Triangle, half the work is carried out merely by locating the length of the opposite sides of the 45-45-90 Triangle. These values will be equivalent!
### Two Cents
Special triangles are a way to obtain precise values for trigonometric equations. Many trig questions you’ve done up until now have needed your round answers. When numbers are rounded, it suggests that your solution isn’t exact, which mathematicians would prefer to avoid. Special triangles take those lengthy numbers that call for rounding and come up with accurate ratio answers.
There are only a few angles that offer clean and excellent trigonometric worths. But also for the ones that do, you will need to remember their angles’ values in examinations and examinations. These are the ones you’ll most commonly use in mathematics troubles too. For a listing of all the various special triangles, you will experience mathematics.
One of these triangles is the 45-45 90 triangle. It is an isosceles triangle with two equivalent sides. Since you’ll likewise find that this Triangle is right-angled, we know that the third side that is not equal to the others is the hypotenuse. You again take place to know a good formula to determine the size of the hypotenuse (the Pythagorean Theorem), and we’ll reveal how You will utilize it. If you want to look at even more examples of the 45 45 90 triangles, look at this online reference for a special right triangle. |
Medical
# How to Do a Diamond Problem in Math?
## What Is Diamond Math?
Due to the distinctive method that they are made, diamond math problems are also known as “diamond math.” The majority of diamond issues are represented by an actual four-sided diamond that has been divided into four smaller diamonds by a huge X in the centre.
In the diamond at the bottom, one number is inscribed, and in the diamond at the top, another number. Since the student must fill up these two spaces, the two diamonds on the left and right are left blank.
Remember that not all diamond issues are presented in this precise manner; occasionally, you’ll encounter them with simply a giant X to denote the four portions and no diamond form in the vicinity. Both approaches are acceptable, although the drawn diamond is the more typical one.
## How to do Diamond Problems
The simplest method for solving diamond problem is to use an online diamond problem solver. Finding the number of choices for the vacant cells is done by factoring the highest number. There are a great number of ways to combine whole numbers to get a total; if negative integers are permitted, the number is truly infinite. This makes starting with the lowest number considerably more difficult.
Make a list of all the possible numbers that, when multiplied together, provide the required result (such as 3 and 4 if the product is 12.) Using your list, try adding the two numbers to check whether they match the intended amount (for example, 3 + 4 if the desired sum is 7).
Write the two numbers you’ve discovered as a match in the two blank cells. Since the numbers in the diamond problem are merely in a collection and not truly in a mathematical problem, it makes no difference what order they are stated in. Even if they were, addition and multiplication are the only operations that let you arrange the numbers in any sequence and yet obtain the same result.
### Example # 01:
Fix the diamond issue:
Solution: Currently, we have
Multiple pairings of nine are possible:
3*3 \s9*1 \s-3*-3 \s-9*-1
Finding each pair’s total separately:
3+3 = 6
9+1 = 10 -3+ (-3) = -3-3 = -6 -9+ (-1) = -9-1 = -10
The total is only provided by the first pair, therefore we have:
1/factor = 3
2 + 3 = a factor
## Diamond Math Problem Rules:
The student must enter numbers into the two blank slots. The value in the bottom cell must equal the total of the two numbers. The same numbers are added in the top-left cell after multiplying the aggregate data.
It may be necessary to use both positive and negative numbers, depending on the pupils’ level of proficiency (which would result in negative numbers in the top or bottom cells, a big hint to the students.) However, it is advised that you begin by sticking with all positive numbers if kids are still establishing this skill.
## How do I use a calculator to solve a diamond problem?
Simply enter the first two digits and press the calculate button to quickly determine the other two.
## What standards does this Diamond Problem Calculator support?
If the user provides Two Factors, One Factor, and the Product or Sum, Product, and Sum, Diamond Problem Solver will handle three alternative circumstances. The last two are simple to locate Employees leave a company.
## Where can I get the Step-by-Step Diamond Problem-Solving Procedure?
On this page, you may get the Step-by-Step Guide for Solving Diamond Problems.
If the elements are 12 and 5, what is the product and sum?
Product: (12 * 5) = 60
Sum = 12+5 = 17
## A diamond—is it a square?
Without a doubt, the definition states that if the diamond and square closely resemble each other, we may conclude that the square is also a diamond.
## What does diamond issue in the programming language c#?
When two classes split out from a single parent class, a phenomena known as the “diamond issue” is said to have occurred. The combinations of the first two classes are then produced to get the fourth class.
## Conclusion:
The solution to the diamond issue is a common method used in mathematical analysis. This method is often used by experts and students to understand the problems that arise during calculations. Try using our top diamond issue solution for totally no cost if you want quick and immediate results. |
### Welcome to our community
#### agentmulder
##### Active member
Moderator edit: This commentary topic pertains to the following tutorial:
I agree with everything MarkFL has posted so i'll just give a quick 'scheme' leaving out the details.
Another approach is to factor by grouping, consider,
$6x^2 - 17x - 45$
$(6)(-45) = -270$
Now find 2 integers that multiply to -270 and sum to -17,
-27 with 10 work
rewrite $-17x$ as $-27x + 10x$
$6x^2 - 27x + 10x - 45$
use grouping, paying mind to the signs,
$(6x^2 - 27x) + (10x -45)$
factor,
$3x(2x - 9) + 5(2x - 9)$
$(2x - 9)(3x + 5)$
You can multiply this out using FOIL to confirm the factoring is correct.
Last edited by a moderator:
#### mathmaniac
##### Well-known member
$6x^2 - 17x - 45$
$(6)(-45) = -270$
Now find 2 integers that multiply to -270 and sum to -17,
-27 with 10 work
How to do it:
We have to find p and q such that pq=6 x (-45)= 2 x 3 x 3 x 3 x 5
and p+q=-17
We observe that 17 is a prime.So p and q cannot have common factors (otherwise p+q will be have a common factor).
So either p or q will have all the 3 "3"s.Let it be p.
We have a 2 and 5 remaining which can go into either p or q.Let us first try both of them in q.
So we have p=27 and q=10
One and only one of p and q is negative and also the greater of p and q has to have the sign "-" (because the sum is negative).
So we have p + q = -27 + 10 = -17 and it works.If it hadn't worked then we had to exchange the 2 and 5 a couple of times until we get it right.
If p + q had a common factor,then we had to allow it to be the GCD of p and q and continue this.
#### agentmulder
##### Active member
How to do it:
We have to find p and q such that pq=6 x (-45)= 2 x 3 x 3 x 3 x 5
and p+q=-17
We observe that 17 is a prime.So p and q cannot have common factors (otherwise p+q will be have a common factor).
So either p or q will have all the 3 "3"s.Let it be p.
We have a 2 and 5 remaining which can go into either p or q.Let us first try both of them in q.
So we have p=27 and q=10
One and only one of p and q is negative and also the greater of p and q has to have the sign "-" (because the sum is negative).
So we have p + q = -27 + 10 = -17 and it works.If it hadn't worked then we had to exchange the 2 and 5 a couple of times until we get it right.
If p + q had a common factor,then we had to allow it to be the GCD of p and q and continue this.
These are nice observations, i especially like the part of all 3's must belong to p or q. As the product AC gets large your observations can in fact reduce the work involved trying to find p and q.
#### Petrus
##### Well-known member
Let's say we want to factor $$\displaystyle p(x)=-4x^2+24x-32$$ we can use this formula $$\displaystyle p(x)=k*(x-a)(x-b)$$
we want to calculate crittical point (without derivate). In Sweden we learned a formula caled "pq-formula"(too calculate critical point in second polynom) as you can see there cant be any number or negative number infront of $$\displaystyle x^2$$ so we start to break out -4 then we got $$\displaystyle -4(x^2-6x+8)$$
now we calculate the critical point $$\displaystyle x^2-6x+8=0$$ and we get $$\displaystyle x_1=4$$ and $$\displaystyle x_2=2$$ now we use the formula $$\displaystyle p(x)=k*(x-a)(x-b)$$ where k is our constant and a,b our crit point so we got $$\displaystyle p(x)=-4*(x-4)(x-2)$$ |
Lesson Video: Range of a Data Set | Nagwa Lesson Video: Range of a Data Set | Nagwa
# Lesson Video: Range of a Data Set Mathematics • Third Year of Preparatory School
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In this video, we will learn how to find the range of a data set.
12:56
### Video Transcript
In this video, we will learn how to find the range of a data set. We will begin by looking at a definition of the range and then answer a variety of questions.
The range of a data set is the difference between the largest and smallest values. We can therefore calculate the range by subtracting the smallest data value from the largest data value. The importance of calculating the range is it tells us how spread-out the data is. We will now look at some examples to practice finding and using the range.
The number of goals scored by 12 soccer players in a season are 13, 11, 12, five, five, nine, six, 11, eight, five, six, and 19. State whether the following statement is true or false. The range of the data is 14 goals.
We can calculate the range of any set of data by subtracting the smallest value from the largest value. Whilst we could find these values from the list by inspection, it is often useful to rewrite the data set in numerical order first. The smallest value in the data set is five. And there are three of these. There are two sixes in the data set. The next lowest value is eight. Continuing our list from smallest to largest, we have 9, 11, 11, 12, 13, and 19. The smallest value is equal to five and the largest value is 19. We can therefore calculate the range by subtracting five from 19. This is equal to 14. The statement in the question said that the range of the data is 14 goals, which is true.
Our next question involves working out the range from a set of data represented on a line plot.
The graph shows the weights, in kilograms, of emperor penguins at a zoo. Is the range of the weights 24 kilograms?
The range of any data set can be calculated by subtracting the smallest value from the largest value. As the weights on the line plot are already in order, we can see that the smallest value is 23 kilograms. The largest value is 49 kilograms. We can therefore calculate the range by subtracting 23 from 49. 40 minus 20 is equal to 20, and nine minus three is equal to six. Therefore, 49 minus 23 equals 26. The range of values of the emperor penguins is 26 kilograms. This means that the answer to the question โIs the range of the weights 24 kilograms?โ is no.
Let the greatest element in a set be 445 and the range of the set be 254. What is the smallest element of this set?
We know that, in order to calculate the range of any set, we subtract the smallest element from the largest element. In this question, we are given the largest or greatest element and the range of the set and need to calculate the smallest element. We can do this by using the formula or by using the number line. Substituting in our values gives us 254 is equal to 445 minus ๐ฅ, where ๐ฅ is the smallest element of the set. Adding ๐ฅ to both sides of this equation gives us ๐ฅ plus 254 is equal to 445. We can then subtract 254 from both sides of this equation to work out the value of ๐ฅ. ๐ฅ is equal to 191 as 445 minus 254 is 191. We can therefore conclude that the smallest element of the set is 191.
As already mentioned, we could also have calculated this using a number line. We know that the greatest value is 445. The range is the difference between the greatest or largest value and the smallest value. In this case, we are told it is 254. The difference between the smallest and greatest value is 254, which means we can subtract this from 445 to calculate the smallest value. Once again, this gives us an answer of 191.
We will now look at a couple of more complicated problems to finish this video.
The following figure demonstrates the number of glasses of water a group of people consume per day. Describe how the range would change if an additional data value of one was added to the data set.
The graph tells us that eight people consume zero glasses of water per day. Five people consume one glass; two people consume two glasses; six people, three glasses; one person, four glasses; and, finally, seven people consume five glasses of water per day.
We know that, in order to calculate the range of a data set, we subtract the smallest value from the largest value. In this question, it will be the number of glasses. It doesnโt matter how many different people consume that number of glasses. The smallest value is therefore zero. As thereโre some people, in this case, eight, that consume no glasses of water. The largest value is equal to five as this is the greatest number of glasses of water that anybody consumes. We can therefore calculate the range of the original data by subtracting zero from five. This is equal to five.
We are then told that there is an additional data value of one that is added to the data set. This means that there are now six people that consumed one glass of water per day. This does not impact the smallest or largest value though. So the new range is still equal to five minus zero. Adding the extra data value does not alter the range. We can therefore conclude that the range would remain unchanged at five.
Michael has the following data: Six, eight, ๐, eight, eight, and nine. If the range is three, which number could ๐ be? Is it (A) three, (B) four, (C) five, (D) six, or (E) 13?
We recall that we can calculate the range by subtracting the smallest value from the largest value. In this question, we will consider what the largest value and smallest values are when ๐ takes each of the five options. One of these options will have a range of three which will be the correct answer. When ๐ is equal to three, our list of values in ascending order are three, six, eight, eight, eight, and nine. As the largest value is nine and the smallest value is three, the range will be equal to nine minus three. As this is equal to six, option (A) is not correct.
When ๐ is equal to four, the smallest value is four and the largest value is nine. This time the range would be equal to nine minus four, which is equal to five. Once again, this is not correct. When ๐ is equal to five the smallest number is five. The largest number is still nine. The range in this case is equal to four, which once again is not correct. When ๐ is equal to six, we have two sixes. We could write these in either order. Our list is now six, six, eight, eight, eight, and nine. As the smallest number in this set of date of is six and the largest is nine, the range is equal to nine minus six. This is equal to three, which suggests that option (D) is correct.
We will check option (E) just to make sure. This time, ๐ is equal to 13. This means that the smallest number is six and the largest number is 13. The range is the difference between these values. 13 minus six is equal to seven. So this answer is also incorrect. This means that the correct answer is option (D). If the range is three, the number from the list that ๐ could be is six.
There are a few other numbers that were not one of the options that ๐ could be. As long as six remains the smallest number and nine remains the largest number, the range will always be three. This means that ๐ could be any one of the four integers, six, seven, eight, or nine. In this question, the only one of those that was listed as an option was six, which is why this is the only correct answer.
We will now summarize the key points from this video. The range of a data set is the difference between the largest and smallest values. We can therefore calculate the range of any data set by subtracting the smallest value from the largest value. The range of a set of data tells us how spread-out the data is. This means that adding any extra values to a data set doesnโt always change the range.
For example, letโs consider the data set four, seven, 10, 10, and 13. The smallest value here is four, and the largest value is 13. This means that the range is equal to nine as 13 minus four equals nine. Adding in any extra values between four and 13 inclusive will not affect the range. For example, if we added the number eight, the smallest number is still four and the largest number is 13. This means that the range is still nine.
It is also important to note that it doesnโt matter how many of each value we have. If we added any extra fours or extra 13s to this list, the range would still be nine. As well as using a list of data values, we can also calculate the range from a frequency table or graph.
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How to solve this logarithm equation? [closed]
How can I solve it?
$$\frac{2}{\log_{8}(x-1)} - \frac{2}{\log_{8}(x )} =1$$
I don't have idea how to solve it and I will be happy for help about this exercise.
closed as off-topic by Namaste, Stefan4024, José Carlos Santos, Zhanxiong, Parcly TaxelDec 6 '17 at 3:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Stefan4024, José Carlos Santos, Zhanxiong, Parcly Taxel
If this question can be reworded to fit the rules in the help center, please edit the question.
• Multiply the fractions crosswise, and rewrite $1$ as a log – R3E1W4 Dec 5 '17 at 16:06
• @R3E1W4: I fail to see how this helps... – Yves Daoust Dec 5 '17 at 16:21
• Where do this equation come from? – Raffaele Dec 5 '17 at 16:37
Combine the fractions: $$\frac{2\log_8 x - 2\log_8 (x-1)}{\log_8 x \log_8 (x-1)} =1 \\ \log_8 x - \log_8 (x-1) = \frac12 \log_8 x \log_8 (x-1)$$ Then use properties of the logarithm: $$\log_8\left( \frac{x}{x-1} \right) = \log_8 \sqrt{ (x-1)^{\log_8 x }}$$ Now raise $8$ to both sides of this equation and square both sides: $$\frac{x^2}{(x-1)^2} = (x-1)^{\log_8 x}$$ Handle the denominator on the left by adding to the exponent on the right: $$x^2 = (x-1)^{2+\log_8 x}$$
This has a unique real positive solution at roughly $x = 3.7093175$, but there is no solution in closed form using only elementary functions.
One can shift the equation by using: \begin{align} \log_{b}(x) &= \frac{\log_{d}(x)}{\log_{d}(b)} \\ x &\to t + \frac{1}{2} \end{align} to obtain $$\frac{1}{\ln\left(t - \frac{1}{2}\right)} - \frac{1}{\ln\left(t + \frac{1}{2}\right)} = \frac{1}{6 \, \ln(2)}.$$ The solution for $t$ is $t \approx 3.20931751$ and yields $x \approx 3.70931751$. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are reading an older version of this FlexBook® textbook: CK-12 Texas Instruments Trigonometry Student Edition Go to the latest version.
# 4.3: What’s the Difference?
Difficulty Level: At Grade Created by: CK-12
This activity is intended to supplement Trigonometry, Chapter 3, Lesson 5.
ID: 12556
Time Required: 15–20 minutes
## Activity Overview
Students explore the angle difference formula for cosine. Students will apply the formula and compare their results to interactive unit circle diagrams that assist the student in visualizing the problems involved. The derivations of the angle difference and sum formulas for cosine are optional extensions included with this activity.
Topic: Cosine Difference Identity
• Angle Sum and Difference Identity Derivation (optional extensions)
• Unit Circle
• Sine and Cosine values
• Verification of Equivalence by Graphing
Teacher Preparation and Notes
• If the extensions are used during class, the activity will take approximately 30–45 minutes to complete.
• It will be necessary to load the UNITCIRC Cabri jr. files to the graphing calculators before beginning this activity.
• The first and second problems engage students in an exploration of the difference formula for cosine. Problem 1 is devoted to unit circle review and developing an understanding of the angle difference diagram included in the activity.
• Problem 2 engages students in the application of the angle difference formula for cosine. Students find the cosine for angles such as $15^\circ$ from well-known angles on the unit circle, such as $45^\circ$ and $60^\circ$.
• The extensions of this activity have students derive the angle sum and difference formulas for cosine.
Associated Materials
## Problem 1 – Exploring the Angle Difference Formula for Cosine
One of the great things about using the unit circle is that the $y-$coordinate is always the sine of the angle and the $x-$coordinate is always the cosine of the angle.
The Cabri Jr. file titled UNITCIRC is useful in exploring this topic and then for exploring the angle difference formula for cosine.
Students will press ALPHA to grab points $A$ and $B$ and move them around the circle. Remind them that their results are limited to the resolution of the sketch. In other words, just because the $x-$ and $y-$coordinates are only displayed to the tenth, they really go on for a long while.
In this part of the activity, the students answer a variety of questions related to the angle difference diagram.
## Problem 2 – Applying the Angle Difference Formula
Students find cosine values for angle measures such as $15^\circ$ and $105^\circ$, which take advantage of angles with well known values (for many students) of sine and cosine.
Again, remind the students about the Cabri Jr. application only measuring the nearest tenth to account for any discrepancies between their calculated results and their graphical results. Also, be sure to have the students set their graphing calculators to Degree mode.
## Extension #1– Deriving the Angle Difference Formula for Cosine
Students use the Law of Cosines to derive the angle difference formula for cosine. A unit circle representation is provided to help students visualize the problem and to provide the necessary background to set up the derivation.
Guidance regarding how to begin the derivation will be helpful to students. Show students how to set up their work for the first derivation and students should be able to follow that example for the remaining derivations in this activity.
## Extension #2 – Angle Sum Formula for Cosine
It is important to take some time here to discuss with students why $\cos(-y) = \cos (y)$ and $\sin (-y) = -\sin (y)$ and explain these two situations involved in this formula derivation.
## Solutions
1. cosine
2. sine
3. 0.98
4. -0.17
5. 0.34
6. 0.94
7. 0.98
8. 0.17
9. answers may vary—relationship is not easy to quickly obtain from the interactive graph page
10. 0.97
11. 0.26
12. -0.26
13. $(AB)^2 &= AO^2 + BO^2 - 2 \cdot AO \cdot BO \cdot \cos(AOB)\\&= 1 + 1 - 2\cos(\alpha - \beta)\\&= 2 - 2\cos(\alpha - \beta)$
14. $(AB)^2 &= (\cos(\alpha) - \cos(\beta))^2 + (\sin(\alpha) - \sin(\beta))^2\\&= \cos^2(\alpha) - 2\cos(\alpha)\cos(\beta) + \cos^2(\beta) + \sin^2(\alpha) - 2\sin(\alpha)\sin(\beta) + \sin^2(\beta)\\&= 1 - 2\cos(\alpha)\cos(\beta) + 1 - 2\sin(\alpha)\sin(\beta)\\&= 2 - 2\cos(\alpha)\cos(\beta) - 2\sin(\alpha)\sin(\beta)$
15. $\cos(\alpha - \beta) = \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta)$
16. $\cos(\alpha + \beta) = \cos(\alpha)\cos(\beta) - \sin(\alpha)\sin(\beta)$
Feb 23, 2012
Nov 04, 2014 |
# X Coordinate
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A 2-dimension coordinate plane consists of two axis perpendicular to each other known as the ‘X-axis’ and the ‘Y-axis’. The horizontal axis is known as the X-axis and the vertical axis is known as the Y-axis and they intersect at the origin. A point in the coordinate plane is represented as (x, y) where ‘x’ is called as the x-coordinate and ‘y’ is called as the y-coordinate. The ‘x-coordinate’ represents the distance on the X-axis of the given point and similarly the ‘y-coordinate’ represents the distance on the Y-axis of the given point from the origin.
Example 1: Given the equation of a line y = 2x – 4. Find the ‘x-coordinate’ of the point where the line crosses the X-axis.
Given: y= 2x – 4
In order to find the point where the line crosses the X-axis, we can plug-in y = 0 because the value of the y-coordinate on the X-axis is 0.
This implies: y = 2x – 4 ==>0= 2x – 4.
This gives: 2x = 4 ==>x= 4/2 = 2.
Therefore the point where the line crosses the x-axis is (2, 0)
The x-coordinate of the point is 2.
Example 2: Given the equation of a line y = x – 5. Find the ‘x-coordinate’ of the point where the line crosses the X-axis.
Given: y= x - 5
In order to find the point where the line crosses the X-axis, we can plug-in y = 0 because the value of the y-coordinate on the X-axis is 0.
This implies: y = x – 5 ==> 0= x - 5
This gives: x = 5.
Therefore the point where the line crosses the x-axis is (5, 0)
The x-coordinate of the point is 5. |
# Limits
September 26th, 2009 by ben
Question from the tryout:
Find:
$\dfrac{\displaystyle\lim_{x \to 0} \dfrac{\sin \dfrac{x}{5}}{\sin 2x} \cdot \lim_{x \to 3^{-}} \lfloor x \rfloor}{\displaystyle\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3} + \lim_{x \to\infty} \dfrac{3 - 4x^2}{2x^2 + 7}}$
This problem just looks scary, doesn’t it? Well, let’s split up the four limits.
First limit, the top left.
$\displaystyle\lim_{x \to 0} \dfrac{\sin \dfrac{x}{5}}{\sin 2x}$
l’Hôpital it! Bwahaha.
$\displaystyle\lim_{x \to 0} \dfrac{\dfrac{1}{5}\cos \dfrac{x}{5}}{2 \cos 2x} = \dfrac{\dfrac{1}{5}}{2} = \boxed{\dfrac{1}{10}}$
Now for the top right:
$\displaystyle\lim_{x \to 3^{-}} \lfloor x \rfloor$
The limit of the floor function approaching from the negative side is the lower number, so this equals $\boxed{2}$.
Bottom left:
$\displaystyle\lim_{x \to 3} \dfrac{x^2 - 9}{x - 3}$
Indeterminate form (zero over zero), so once again we can use l’Hôpital’s rule on it.
$\displaystyle\lim_{x \to 3} \dfrac{2x}{1} = \boxed{6}$
Last limit is the bottom right one.
$\displaystyle\lim_{x \to\infty} \dfrac{3 - 4x^2}{2x^2 + 7}$
This one is $\frac{\infty}{\infty}$, so once again we must use l’Hôpital’s rule. This time we apply it twice in succession (because I’m not sure whether or not you can just cancel out the $x$s or not since it’s technically dividing by infinity; ask your local calculus teacher).
$\displaystyle\lim_{x \to \infty} \dfrac{-8x}{4x} = - \dfrac{8}{4} = \boxed{-2}$
Now, we put these solved limits back into the original problem:
$\dfrac{\dfrac{1}{10} \cdot 2}{6 + (-2)} = \dfrac{\dfrac{1}{5}}{4} = \boxed{\dfrac{1}{20}}$
Andy? We both fail. XD; |
Question Video: Finding the Moment of a Couple Equivalent to a System of Two Forces Acting on a Regular Hexagon | Nagwa Question Video: Finding the Moment of a Couple Equivalent to a System of Two Forces Acting on a Regular Hexagon | Nagwa
# Question Video: Finding the Moment of a Couple Equivalent to a System of Two Forces Acting on a Regular Hexagon Mathematics • Third Year of Secondary School
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π΄π΅πΆπ·π»π is a regular hexagon with sides of length 5 cm. Two forces of the same magnitude 13 N are acting along πΆπ΅ and ππ» respectively. Determine the magnitude of the moment of the couple.
03:19
### Video Transcript
π΄π΅πΆπ·π»π is a regular hexagon with sides of length five centimeters. Two forces of the same magnitude 13 newtons are acting along πΆπ΅ and ππ», respectively. Determine the magnitude of the moment of the couple.
We will begin by sketching the regular hexagon as shown. We are told that it has side length five centimeters. There are two forces of magnitude 13 newtons acting along πΆπ΅ and ππ», respectively. These two forces form a couple, as they are a pair of parallel forces with equal magnitude and opposite direction which do not lie on the same line of action. We are asked to calculate the magnitude of the moment of this couple.
This is equal to πΉπ sin π, where πΉ is the magnitude of the forces β in this question, 13 newtons β π is the distance between their points of action β in this case, the line segment ππΆ β and π is the angle that the forces make with this line segment. If we let the center of the hexagon be point π, we can use our properties of hexagons to help determine the length of ππΆ and the angle π. The line segment π΄π΅ is parallel and equal in length to the line segment ππΆ. This means that ππΆ is equal to five centimeters. Likewise, the line segment π»π· is the same length and parallel to the line segment ππ. This is also equal to five centimeters, meaning that the line segment ππΆ is equal to 10 centimeters.
Triangle πππ» is therefore equilateral. And we know that the angles in an equilateral triangle equal 60 degrees. The measure of angle π is therefore equal to 60 degrees. And since the two forces form a couple, this is also true of the angle π΅πΆπ. We now have values of πΉ, π, and π. The magnitude of the moment of the couple is equal to 13 multiplied by 10 multiplied by the sin of 60 degrees. Since the sin of 60 degrees is root three over two, this simplifies to 130 multiplied by root three over two, which in turn is equal to 65 root three.
The magnitude of the moment of the couple is therefore equal to 65 root three newton-centimeters.
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Ch. 5 Solutions
# Ch. 5 Solutions - CHAPTER 5 PROBLEM SOLUTIONS Problem 4. A...
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CHAPTER 5 PROBLEM SOLUTIONS Problem 4. A car leaves the road traveling at 110 km/h and hits a tree, coming to a complete stop in 0.14 s. What average force does a seatbelt exert on a 60-kg passenger during this collision? Solution Assume that the seatbelt holds the passenger firmly to the seat, so that the passenger also stops in 0.14 s without incurring any secondary impact. Then the passenger’s average acceleration is a av = (0 ! v 0 ) = t , and the average net force on the passenger, while coming to rest, is F av = ma av = ! m v 0 = t = ! (60 kg)(110 = 3.6)(m/s) = (0.14 s) = ! 13.1 kN , or about 1.5 tons. (Here, we used the one-dimensional form of Newton’s second law. The minus sign indicates that the force is opposite to the direction of the initial velocity. It is reasonable to assume that this is a component of the force exerted by the seatbelt.) Problem 7. Object A accelerates at 8.1 m/s 2 when a 3.3-N force is applied. Object B accelerates at 2.7 m/s 2 when the same force is applied. (a) How do the masses of the two objects compare? (b) If A and B were stuck together and accelerated by the 3.3-N force, what would be the acceleration of the composite object? Solution In this idealized one-dimensional situation, the applied force of F = 3.3 N is the only force acting. (a) When applied to either object, Newton’s second law gives F = m A a A and F = m B a B , so m B = m A = a A = a B = (8.1 m/s 2 ) = (2.7 m/s 2 ) = 3. (For constant net force, mass is inversely proportional to acceleration.) (b) When F is applied to the combined object, F = ( m A + m B ) a . Since F = m A a A , and m B = 3 m A , one finds a = F = ( m A + m B ) = m A a A = 4 m A = 1 4 (8.1 m/s 2 ) = 2.03 m/s 2 . (Note: It was not necessary to calculate the masses, which are m A = (3.3 N) = 2 ) = 0.407 kg , and m B = (3.3 N) = (2.7 m/s 2 ) = 1.22 kg.) Problem 9. By how much does the force required to stop a car increase if the initial speed is doubled and the stopping distance remains the same? Solution The average net force on a car of given mass is proportional to the average acceleration, F av » a av . To stop a car in a given distance, ( x ! x 0 ), a av = (0 ! v 0 2 ) = 2( x ! x 0 ), so F av » v 0 2 . Doubling v 0 quadruples the magnitude of F av , a fact that is important to remember when driving at high speeds. Problem 11. The maximum braking force of a 1400-kg car is about 8.0 kN. Estimate the stopping distance when the car is traveling (a) 40 km/h; (b) 60 km/h; (c) 80 km/h; (d) 55 mi/h. Solution The maximum braking acceleration is a = F = m = ! 8 . 0 " 10 3 N = 1400 kg = ! 5 . 71 m / s 2 . (We expressed the braking force as negative because it is opposite to the direction of motion.) (a) On a straight horizontal road, a car traveling at a velocity of v 0 x = 40 km/h can stop in a minimum distance found from Equation 2-11 and the maximum deceleration just calculated: x ! x 0 = ! v 0 x 2 = 2 a = ! (40 m = 3.6 s) 2 = 2( ! 5.71 m/s 2 ) = 10.8 m. For the other initial velocities, the stopping distance is (b) 24.3 m, (c) 43.2 m, and (d) 52.9 m.
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## Class 8 Math (Assamese)
### Course: Class 8 Math (Assamese)>Unit 2
Lesson 1: Solving equations with variable on one side
# Equation with variables on both sides: fractions
To solve the equation (3/4)x + 2 = (3/8)x - 4, we first eliminate fractions by multiplying both sides by the least common multiple of the denominators. Then, we add or subtract terms from both sides of the equation to group the x-terms on one side and the constants on the other. Finally, we solve and check as normal. Created by Sal Khan and Monterey Institute for Technology and Education.
## Want to join the conversation?
• so it is x=-16?
• Yes, but as I saw the comment from a year ago the same thought popped up in my head. Where are you know. Your comment was from 9 years ago, and your description saws you were 14 so you would have to be 23, or so. Time flies, wow
• I am struggling on how to transfer varibles to one side of the equation. Any tips?
• Hi, Rebecca;
Transferring Variables
Transferring variables might look like a complex subject to tackle at first glance, but it actually proves itself to be much simpler--you just need to understand it.
In an equation, the left hand side (LHS--the left expression) and the right hand side (RHS--the right expression) are equal. Now, a very significant tip to take note is that since both sides are equal, both must be treated equally. But how do we do that?
Here is an example:
Billy has two baskets of equally filled apples. The left basket has 2 packs of 4 apples and 2 pears while the right basket has 3 packs of 2 apples and 2 pears. On the way home, Billy decided to eat 3 fruits from each basket. How many fruits are left in each basket?
``2(4+2)=3(2+2)``
I've done the equation for the original number of fruits in each basket, but after Billy took 3 from each basket, I am left to modify my equation. But how do we do that?
``2(4+2)-3=3(2+2)-3``
We subtract 3 from both sides! This would mean that both baskets, being originally equal, would still be equal when Billy goes home to eat the rest.
``2(4+2) -3 =3(2+2) -32(6)-3=3(4)-312-3=12-39=9∴ There are 9 fruits left in each basket``
Now what does this have to do with transferring variables? Transferring variables are basically like what we did above, except they're not numbers yet.
This time, let's say we don't actually know how many pears there are in each pack, considering the fact that the number of pears are equal in all of the packs in both baskets.
``2(4+x)=3(2+x)``
First, let us simplify the equation.
``2(4+x)=3(2+x)8+2x=6+3x``
Here we are--transferring variables! This time, think about what we did to the equation when Billy decided to eat 3 fruits from both baskets; we subtract the variable from both sides!
``8+2x=6+3x8+2x -3x =6+3x -3x8+2x-3x=6``
Notice that when we transfered `3x` from the RHS to the LHS, it turned negative. When we transfer variables to the other side, its sign becomes opposite! That's how easy it is!
Now, let's solve the equation!
``8+2x-3x=62x-3x=6-8-x=-2-1(-x)=-1(-2)x=2∴ There are 2 pears in each pack.``
Ta-da! The same equation!
(Sorry if I was very lengthened about such a simple subject. I like to explain thoroughly)
• can you multiply the denominator on both sides
• If you multiply both sides by an integer, it's always multiplying the numerator and therefore making the number larger. If you did otherwise, by multiplying the denominator, then the number would be smaller, which is rather a division. I hope my explanation makes sense and is helpful.
• in the test i got the problem...
16 - 2t = 3/2t + 9
and i converted the fraction to the decimal 1.5
so...
16 - 2t = 1.5t + 9
-16 -16
-2t = 1.5t - 7
-1.5t -1.5t
0.5t = -7
i devided both sides by 0.5 and got -14, so i punched the answer in and they said the correct answer was actually 2. what did i do wrong?
• hello, so what you did wrong was simply a subtracting mistake. you can totally just convert your fraction into a decimal and it will still work. So lets start from the beginning,
16 - 2t = 3/2t +9
so you convert the fraction into the decimal
16 - 2t = 1.5t + 9
then you subtracted 16 from both sides which is right,
16 - 2t = 1.5t +9
-16 -16
-2t = 1.5t -7
you were right up to this step. now we subtract 1.5t from both sides
-2t = 1.5t -7
-1.5t -1.5
you get...
-3.5t = -7
which equals 2!
so you only messed up in the step where you add a -2t to a -1.5t.
if you do not understand why we add these together look at Khan's video "adding negative numbers example"
hope this helps
• But arent what you do to one side you must do to the other??
• I still didn't understand where the 8 came from, can someone please explain it again differently? How do you get to that conclusion?
• The 8 is the lowest common multiple of the 2 denominators (4 and 8). Use the same process you would use to select the smalles common denominator for those 2 fractions.
Multiples of 4: 4, 8, 12, 16, etc.
Multiples of 8: 8, 16, 24, etc.
The first multiple in common is 8.
Hope this helps.
• Another way to solve this "quickly" is this: 3/4x+2=3/8x-4, what number you need to multiply 3/4x for so denominator becomes 8 as well?
You multiply by 2 and get 6/8x+2=3/8x-4
1st step 6/8x-3/8+2=3/8x-3/8x-4
2nd step 3/8x+2-2=-4-2
3/8x/3/8=-6/3/8 (0.375)
x=-16
• The third step is much easier to multiply by the reciprocal rather than dividing, so
3/8 x • 8/3 = 6 • 8/3, since 6/3 =2, you get 16 faster.
If you are going to do it quickly, try doing everything as simply as possible, but this is great for those not afraid of fractions.
• this helped me a lot. thank you!
• my question is how do you solve this problem with one fraction? |
# How to Find Surface Area of Cones: A Comprehensive Guide
## I. Introduction
If you’re a student of mathematics, you may have encountered the problem of finding the surface area of cones. Calculating the surface area of a cone can seem daunting, but with a little bit of practice and the right approach, it can be a breeze. In this article, we provide a comprehensive guide to help you understand and solve this problem.
## II. Step-by-Step Guide
The surface area of a cone is the total area that the curved surface of the cone occupies. This can be calculated using the formula:
SA = πr² + πr√(r² + h²)
Where SA is the surface area, r is the radius of the base, and h is the height of the cone.
We will now walk you through an example problem step-by-step:
Example problem: Find the surface area of a cone with radius 4 cm and height 6 cm.
Solution:
1. Start by calculating the slant height of the cone, which can be done using the Pythagorean theorem:
l = √(r² + h²) = √(4² + 6²) = √(16 + 36) = √52 = 7.21 cm
2. Next, plug in the values for the radius, height, and slant height to the formula:
SA = πr² + πr√(r² + h²) = π4² + π4(7.21) = 50.27 + 90.12 = 140.39 cm²
It’s important to remember that the surface area includes both the curved surface and the base of the cone. So if you’re asked to find the total surface area, you need to add the area of the base as well:
Total Surface Area = Surface Area of Curved Surface + Area of Base
Total Surface Area = πr² + πr√(r² + h²) + πr²
Now that you know how to calculate the surface area of a cone, it’s time to practice on your own. Here are a few additional example problems:
Example problem 1: Find the surface area of a cone with radius 3 m and height 8 m.
Example problem 2: Find the surface area of a cone with radius 5 cm and height 12 cm.
Example problem 3: Find the surface area of a cone with radius 6 ft and height 10 ft.
## III. Visual Approach
Sometimes, it can be difficult to visualize a mathematical concept in your head. That’s why it’s helpful to include diagrams, images, and videos to assist with understanding. Let’s break down the formula visually.
The formula for surface area of a cone involves 2 parts: the curved part and the base. The curved part can be represented as a sector of a circle:
The area of this sector can be calculated as:
Area of sector = (θ/360) x πr²
The value of θ can be found using trigonometry:
θ = 2 x tan⁻¹(l/r)
Where l is the slant height of the cone and r is the radius of the base
The base of the cone can be represented as a circle:
The area of the base can be calculated using the formula:
Area of base = πr²
By adding the areas of the curved part and the base, we can find the total surface area of the cone:
Total Surface Area = Area of sector + Area of base
Total Surface Area = (θ/360) x πr² + πr²
It’s also helpful to illustrate how the surface area of a cone changes as the height and radius of the cone change. Here are a few diagrams to help with visualization:
As the radius of the cone gets bigger, the surface area increases.
As the height of the cone gets bigger, the surface area increases.
## IV. Real-Life Applications
The surface area of cones has a number of real-life applications. Some of the most common scenarios where surface area of cones is used include:
• Calculating the amount of paint or material needed to cover a conical object
• Determining the volume of a container with a conical shape
• Designing buildings and structures with curved surfaces, such as domes
• Calculating the surface area of volcanoes
In various industries, surface area of cones is used to solve problems and make decisions. For example:
• In manufacturing, the surface area of cones is used to calculate the amount of paint needed to paint a conical object
• In architecture, the surface area of cones is used to design buildings with pointed roofs or curved walls
• In engineering, the surface area of cones is used to design airplane wings with curved surfaces
## V. Interactive Activities
Learning doesn’t have to be boring. There are plenty of interactive resources available online that can help you practice and master the surface area of cones. Here are a few links to get you started:
These resources can be useful in helping you improve your understanding and retention of the material. Use them to practice solving problems and reinforce your knowledge of surface area of cones.
## VI. Common Mistakes and Tips
When finding surface area of cones, there are a few common mistakes that students often make. Here are some tips to help you avoid these mistakes:
• Don’t forget to include the base of the cone in your calculation
• Make sure you’re using the correct units of measurement
• Double-check your calculations for accuracy
• Practice, practice, practice!
## VII. Conclusion
Calculating the surface area of cones may seem challenging, but with the right approach, it’s a manageable task. In this article, we provided a comprehensive guide to help you understand and solve this problem. By following our step-by-step instructions, using visual representations, exploring real-life applications, and practicing with interactive activities, you can become proficient at finding surface area of cones.
Remember to avoid common mistakes, double-check your work, and practice regularly. |
# How do you graph f(x) = 3^(x+1)?
Aug 1, 2015
You have an Exponential Function.
#### Explanation:
This is an Exponential Function which domain will be all the real $x$ while the range will be all the $y > 0$.
Basically the graph of your function will occupy only the first and second quadrant.
When $x$ becomes very large positively your function also becomes large, it tends to INFINITY; for example, if you have $x = 100$ then $y = {3}^{100 + 1} = 5 \times {10}^{47}$!!!!.
On the other hand when $x$ becomes very large negatively your function becomes very small, it tends to ZERO; for example, if you have $x = - 100$ then $y = {3}^{- 100 + 1} = 6 \times {10}^{-} 48$!!!!.
In plotting your function we can focus our attention around the origin choosing values of $x$ not too big or small to allow us to actually "see", on the graph, the points we have.
Let us try with:
$x = - 3$ then $y = {3}^{- 3 + 1} = {3}^{- 2} = 0.11$;
$x = - 2$ then $y = {3}^{- 2 + 1} = {3}^{- 1} = 0.33$;
$x = - 1$ then $y = {3}^{- 1 + 1} = {3}^{0} = 1$;
$x = 0$ then $y = {3}^{0 + 1} = {3}^{1} = 3$;
$x = 1$ then $y = {3}^{1 + 1} = {3}^{2} = 9$;
$x = 2$ then $y = {3}^{2 + 1} = {3}^{3} = 27$;
Plotting these points you get: |
# Quadratic Equations Set – 3
1. I. 3x2 + 20x + 32 = 0,
II. 3y2 – 4y – 4 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Option B
Solution:
3x2 + 20x + 32 = 0
3x2 + 12x + 8x + 32 = 0
So x = -4, -8/3
3y2 – 4y – 4 = 0
3y2 – 6y + 2y – 4 = 0
So y = -2/3, 2
Put all values on number line and analyze the relationship
-4… -8/3… -2/3… 2
2. I. 4x2 – 12x + 5 = 0,
II. 6y2 – 13y + 6 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Option E
Solution:
4x2 – 12x + 5 = 0
4x2 – 2x – 10x + 5 = 0
So x = ½, 5/2
6y2 – 13y + 6 = 0
6y2 – 4y – 9y + 6 = 0
So y = 2/3, 3/2
Put all values on number line and analyze the relationship
1/2… 2/3… 3/2…. 5/2
3. I. 32 – 14x + 16 = 0,
II. 4y2 – 5y – 6 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Option C
Solution:
32 – 14x + 16 = 0
32 – 6x – 8x + 16 = 0
So x = 8/3, 2
4y2 – 5y – 6 = 0
4y2 – 8y + 3y – 6 = 0
So y = -3/4, 2
Put all values on number line and analyze the relationship
-3/4 …. 2 ….8/3
4. I. 5x2 – 8x – 4 = 0,
II. 5y2 – 23y – 10 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Option E
Solution:
5x2 – 8x – 4 = 0
5x2 – 10x + 2x – 4 = 0
So x = -2/5, 2
5y2 – 23y – 10 = 0
5y2 – 25y + 2y – 10 = 0
So y = -2/5, 5
Put all values on number line and analyze the relationship
-2/5 …. 2….. 5
5. I. 3x2 + 13x + 14 = 0,
II. 4y2 + 9y + 2 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Option D
Solution:
3x2 + 13x + 14 = 0
3x2 + 6x + 7x + 14 = 0
So x = -7/3, -2
4y2 + 9y + 2 = 0
4y2 + 8y + y + 2 = 0
So y = -2, -1/4
Put all values on number line and analyze the relationship
-7/3 …. -2…. -1/4
6. I. 3x2 + 8x + 5 = 0,
II. 5y2 – 7y – 6 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Option B
Solution:
3x2 + 8x + 5 = 0
3x2 + 3x + 5x + 5 = 0
So x = -5/3, -1
5y2 – 7y – 6 = 0
5y2 – 7y – 6 = 0
So y = -3/5, 2
Put all values on number line and analyze the relationship
-5/3…. -1…. -3/5…. 2
7. I. 3x2 ¬¬+ 16x + 20 = 0,
II. 3y2 + 14y + 16 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relationship cannot be determined
Option E
Solution:
3x2 ¬¬+ 16x + 20 = 0
3x2 ¬¬+ 6x + 10x + 20 = 0
So x = -10/3, -2
3y2 + 14y + 16 = 0
3y2 + 6y + 8y + 16 = 0
So y = -8/3, -2
Put all values on number line and analyze the relationship
-10/3…. -8/3…. -2
8. I. 4x2 – 9x + 2 = 0,
II. 3y2 – 16y + 21 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Option B
Solution:
4x2 – 9x + 2 = 0
4x2 – 8x – x + 2 = 0
So x = 1/4, 2
3y2 – 16y + 21 = 0
3y2 – 9y – 7y + 21 = 0
So y = 7/3, 3
Put all values on number line and analyze the relationship
1/4…. 2…. 7/3… 3
9. I. 3x2 + 5x + 2 = 0,
II. 3y2 + 11y + 10 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relation cannot be established
Option A
Solution:
3x2 + 5x + 2 = 0
3x2 + 3x + 2x + 2 = 0
So x = -1, -2/3
3y2 + 11y + 10 = 0
3y2 + 6y + 5y + 10 = 0
So y = -2, -5/3
Put all values on number line and analyze the relationship
-2….. -5/3…. -1….. -2/3
10. I. 4x2 – 9x + 2 = 0,
II. 2y2 – 19y + 35 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relationship cannot be determined |
# Representations of a Line
Related Topics:
Lesson Plans and Worksheets for Grade 8
Lesson Plans and Worksheets for all Grades
More Math Lessons for Grade 8
Common Core For Grade 8
Examples, solutions, worksheets, videos, and lessons to help Grade 8 students learn how to graph a line specified by a linear function.
### New York State Common Core Math Grade 8, Module 6, Lesson 3
Download Worksheets for Grade 8, Module 6, Lesson 3
Lesson 3 Student Outcomes
• Students graph a line specified by a linear function.
• Students graph a line specified by an initial value and rate of change of a function and construct the linear function by interpreting the graph.
• Students graph a line specified by two points of a linear relationship and provide the linear function.
Lesson 3 Summary
When the rate of change, b, and an initial value, a, are given in the context of a problem, the linear function that models the situation is given by the equation y=a + bx.
The rate of change and initial value can also be used to graph the linear function that models the situation.
When two or more ordered pairs are given in the context of a problem that involves a linear relationship, the graph of the linear function is the line that passes through those points. The linear function can be represented by the equation of that line.
Lesson 3 Classwork
Example 1: Rate of Change and Initial Value given in the Context of the Problem
A truck rental company charges a \$150 rental fee, in addition to a charge of \$0.50 per mile driven. In this problem, you will graph the linear function relating the total cost of the rental in dollars, C, to the number of miles driven, m, on the axes below.
a. If the truck is driven zero miles, what will be the cost to the customer? How will this be shown on the graph?
b. What is the rate of change that relates cost to number of miles driven? Explain what it means within the context of the problem.
c. On the axes given, graph the line that relates C to m.
d. Write the linear function that models the relationship between number of miles driven and total rental cost?
Exercises 1 - 5
Jenna bought a used car for \$18,000. She has been told that the value of the car is likely to decrease by \$2,500 for each year that she owns the car. Let the value of the car in dollars be V and the number of years Jenna has owned the car be t.
1. What is the value of the car when t = 0? Show this point on the graph.
2. What is the rate of change that relates V to t? (Hint: Is it positive or negative? How can you tell?)
3. Find the value of the car when
a. t = 1?
b. t = 2?
c. t = 7?
4. Plot the points for the values you found in Exercise 3, and draw the line (using a straight-edge) that passes through those points.
5. Write the linear function that models the relationship between the number of years Jenna has owned the car and the value of the car.
Exercises 6 - 10
An online bookseller has a new book in print. The company estimates that if the book is priced at \$15.00 then 800 copies of the book will be sold per day, and if the book is priced \$20 at then 550 copies of the book will be sold per day.
6. Identify the ordered pairs given in the problem. Then plot both on the graph.
7. Assume that the relationship between the number of books sold and the price is linear. (In other words, assume that the graph is a straight line.) Using a straight-edge, draw the line that passes through the two points.
8. What is the rate of change relating number of copies sold to price?
9. Based on the graph, if the company prices the book at \$18, about how many copies of the book can they expect to sell per day?
10. Based on the graph, approximately what price should the company charge in order to sell 700 copies of the book per day?
Lesson 3 Exit Ticket
1. A car starts a journey with 8 gallons of fuel. The car will consume 0.04 gallons for every mile driven. Let A represent the amount of gas in the tank (in gallons) and m represent the number of miles driven.
a. How much gas is in the tank if 0 miles have been driven? How would this be represented on the axes above?
b. What is the rate of change that relates the amount of gas in the tank to the number of miles driven? Explain what it means within the context of the problem.
c. On the axes above, graph the line that relates A to m.
d. Write the linear function that models the relationship between the number of miles driven and the amount of gas in the tank.
2. Andrew works in a restaurant. The graph below shows the relationship between the amount Andrew earns and the number of hours he works.
a. If Andrew works for7 hours, approximately how much does he earn?
b. Estimate how long Andrew has to work in order to earn \$6a?
c. What is the rate of change of the function given by the graph? Interpret the value within the context of the problem.
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Question Video: Identifying a Graph with a Given Number of Zeros and Maxima from a Set of Graphs | Nagwa Question Video: Identifying a Graph with a Given Number of Zeros and Maxima from a Set of Graphs | Nagwa
# Question Video: Identifying a Graph with a Given Number of Zeros and Maxima from a Set of Graphs
Which graph has three real zeros and two local maxima?
02:25
### Video Transcript
Which graph has three real zeros and two local maxima?
Let’s begin by recalling the definitions of the zeros of a graph and local maxima. Firstly, the zeros of a function 𝑓 of 𝑥 are the values of 𝑥 for which the function 𝑓 of 𝑥 is equal to zero. On a graph of that function, the zeros are its 𝑥-intercepts, the 𝑥-values at which the graph either crosses or touches the 𝑥-axis. Considering graph (a) first of all, we can see that it intercepts the 𝑥-axis in four places. And so the function shown in graph (a) has four real zeros.
Looking at graph (b), we can see that it intercepts the 𝑥-axis twice and it touches the 𝑥-axis once. And so graph (b) has three real zeros. Graph (c) intercepts the 𝑥-axis in three places, and so graph (c) also has three real zeros. Notice that the question mentions three real zeros. It’s possible that there may be other values of 𝑥 for which any of these functions is equal to zero. But if these are nonreal values, then we aren’t concerned. We’re looking for three real zeros, and so we’ve narrowed our options down to (b) and (c).
Next, let’s recall what we mean by local maxima. A local maximum is a point on the graph of a function at which two things are true. Firstly, the first derivative or slope of the function, 𝑓 prime of 𝑥, is equal to zero. And the slope changes from positive to negative as the 𝑥-values increase. This means that the shape of the graph of the function around this value will look like this. This is called a local maximum because the function at that particular point is at its largest value in the region immediately surrounding that point.
Looking at graph (b) first of all then, we can identify two local maxima at these points here. In graph (c), however, we only identify a single local maximum. For completeness, let’s just observe that there are two local maxima in graph (a). Graph (b) is therefore the only graph which has the right number of real zeros and the right number of local maxima. And so our answer is (b).
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Calculating the mean - MathBootCamps
## Calculating the mean
The mean is a way of measuring the center of a data set. That is, it is a way of trying to describe the typical data value. For symmetric data sets, it does a good job of this. But, for skewed data sets or data sets with outliers it can be a bit misleading. Before we see how it is calculated, let’s talk about notation.
## Notation
The mean is represented in two different ways, depending on whether or not it represents the mean of a sample or the mean of a population. Both are calculated the same, and the difference between the two is only important in some settings.
Population Mean Sample Mean
$\mu$
(This is the Greek letter Mu. Read this as “mew”.)
$\bar{x}$
For this guide, we will assume that we are working with sample data, so we will use $\bar{x}$.
## Calculation Using the Formula
When you learned how to find the average, you were likely taught the arithmetic average. This is where you add up all the values and then divide by however many values were in the data set. The mean is calculated in the exact same way.
### Example
Find the mean of the data set below.
10 12 11 8 14 12
Using the idea above:
$\bar{x} = \dfrac{10 + 12 + 11 + 8 + 14 + 12}{6} \approx 11.2$
So, the mean for this data set is approximately 11.2. Looking at the original data, this does a good job of describing the center of this data set.
## Calculation Using a TI83 or TI84 Graphing Calculator
The mean can easily be found using the function 1varstats on Ti83/84 graphing calculators. Here, we will go through the steps one by one. If you want to see a video of how to do this, scroll down to the bottom (or click here)!
### Step 1: Enter your data in L1
To enter data in your calculator, press [STAT] and then go to 1: Edit by pressing [ENTER] or [1].
Now to enter the data, type each number and press enter. Note that if you already have data in your list, highlight the very top where it says L1 and press [CLEAR] followed by [ENTER].
### Step 2: Calculate 1-var-stats
Once your data is in the list, press [STAT] again and then go to the CALC menu. From here, choose 1: 1varstats.
Now press enter twice and you will get a list of summary statistics. The first value that comes up is the mean!
Note: if you have a newer calculator, the menu looks a little different now. Instead of pressing enter twice, you will press enter and then have a 1-var stats menu come up. Just click CALCULATE at this menu and you will have the same information come up as you see above.
As you can see we get the exact same value as we did above. This is very nice for working with larger data sets. |
What are complex numbers?Thanx.
Feb 9, 2016
Complex numbers are numbers of the form $a + b i$ where $a$ and $b$ are real numbers and $i$ is defined as $i = \sqrt{- 1}$.
(The above is a basic definition of complex numbers. Read on for a little more about them.)
Much like how we denote the set of real numbers as $\mathbb{R}$, we denote the set of complex numbers as $\mathbb{C}$. Note that all real numbers are also complex numbers, as any real number $x$ may be written as $x + 0 i$.
Given a complex number $z = a + b i$, we say that $a$ is the real part of the complex number (denoted $\text{Re} \left(z\right)$) and $b$ is the imaginary part of the complex number (denoted $\text{Im} \left(z\right)$).
Performing operations with complex numbers is similar to performing operations on binomials. Given two complex numbers ${z}_{1} = {a}_{1} + {b}_{1} i$ and ${z}_{2} = {a}_{2} + {b}_{2} i$
${z}_{1} + {z}_{2} = {a}_{1} + {b}_{1} i + {a}_{2} + {b}_{2} i = \left({a}_{1} + {a}_{2}\right) + \left({b}_{1} + {b}_{2}\right) i$
${z}_{1} - {z}_{2} = {a}_{1} + {b}_{1} i - \left({a}_{2} + {b}_{2} i\right) = \left({a}_{1} - {a}_{2}\right) + \left({b}_{1} - {b}_{2}\right) i$
${z}_{1} \times {z}_{2} = \left({a}_{1} + {b}_{1} i\right) \left({a}_{2} + {b}_{2} i\right)$
$= {a}_{1} {a}_{2} + {a}_{1} {b}_{2} i + {a}_{2} {b}_{1} i + {b}_{1} {b}_{2} {i}^{2}$
$= {a}_{1} {a}_{2} + {a}_{1} {b}_{2} i + {a}_{2} {b}_{1} i - {b}_{1} {b}_{2}$ (remember $i = \sqrt{- 1}$)
$= \left({a}_{1} {a}_{2} - {b}_{1} {b}_{2}\right) + \left({a}_{1} {b}_{2} + {a}_{2} {b}_{1}\right) i$
${z}_{1} \div {z}_{2} = \frac{{a}_{1} + {b}_{1} i}{{a}_{2} + {b}_{2} i}$
$= \frac{\left({a}_{1} + {b}_{1} i\right) \left({a}_{2} - {b}_{2} i\right)}{\left({a}_{2} + {b}_{2} i\right) \left({a}_{2} - {b}_{2} i\right)}$
$= \frac{\left({a}_{1} {a}_{2} + {b}_{1} {b}_{2}\right) + \left({a}_{2} {b}_{1} - {a}_{1} {b}_{2}\right) i}{{a}_{2}^{2} + {b}_{2}^{2}}$
$= \frac{{a}_{1} {a}_{2} + {b}_{1} {b}_{2}}{{a}_{2}^{2} + {b}_{2}^{2}} + \frac{{a}_{2} {b}_{1} - {a}_{1} {b}_{2}}{{a}_{2}^{2} + {b}_{2}^{2}} i$
For division, we used the fact that $\left(a + b i\right) \left(a - b i\right) = {a}^{2} + {b}^{2}$. Given a complex number $z = a + b i$ we call $a - b i$ the complex conjugate of $z$ and denote it $\overline{z}$ It is a useful property (as seen above) that $z \overline{z}$ is always a real number.
The complex numbers have many useful applications and attributes, but one which is often encountered early is their use in factoring polynomials. If we limit ourselves to only real numbers, a polynomial such as ${x}^{2} + 1$ cannot be factored further, however if we allow for complex numbers, then we have ${x}^{2} + 1 = \left(x + i\right) \left(x - i\right)$.
In fact, if we allow for complex numbers, then any single-variable polynomial of degree $n$ may be written as the product of $n$ linear factors (possibly with some being the same). This result is known as the fundamental theorem of algebra, and, as the name indicates, is very important to algebra and has broad application. |
Jump to a New ChapterAnatomy of SAT DS&PEssential ConceptsEssential StrategiesTest-Taking StrategiesThe 9 Most Common MistakesConclusionPractice Set 1: Multiple ChoicePractice Set 2: Grid-InsPosttest
Types of DS&P Items Tackling Graph(ics) Tackling Data Puzzlers
Tackling No Problem with Probability Tackling What the #!*@?
Tackling What the #!*@?
Permutation and combination items are the test-makers’ last-ditch effort to bring down your score. The test-takers’ usual reaction to these items is “What the #!*@?” Believe it or not, this visceral reaction can actually help you identify a permutation or combination item. When you find yourself frustrated beyond belief, there is a very good chance you’re looking at one of these items.
Once you’ve collected yourself, follow the three-step method below for solving What the #!*@? items. Not only will it lead you to the correct answer but to a new found feeling of empowerment from shredding the SAT math section.
Step 1: Does order matter? If YES, go to step 2 ONLY. If NO, go to steps 2 and 3.
Step 2: Multiply the factorial of the number of things being arranged to the number of spaces designated for arrangement.
Step 3: Divide by the factorial of the number of spaces in the group.
What the #!*@? in Slow Motion
Now let’s look at the steps more closely, starting with a fairly common What the #!*@? item.
17. Susanne has eleven different medals from her two years of competitive swimming. Unfortunately, the mounting frame she wishes to place them on only has room for two. How many different combinations of medals can Susanne place on her frame? (A) 21 (B) 33 (C) 55 (D) 66 (E) 110
Step 1: Does order matter? If YES, go to step 2 ONLY. If NO, go to steps 2 and 3.
Susanne is placing two medals onto a mounted frame. Does it matter in which order the medals are placed? No, the word combinations in the item tells us that order does not matter. Recall from our study of the essential concepts that combinations deal with groups of things, while permutations deal with arrangements. In permutations, order matters.
Now let’s move on and complete steps 2 and 3.
Step 2: Multiply the factorial of the number of things being arranged to the number of spaces designated for arrangement.
This step is actually easier than it sounds. How many medals does Susanne have to arrange? 11. How many spaces on her frame does she have for arrangement? 2. So all we do is multiply the first two values in the factorial of 11 (11!):
Susanne has 11 medals to choose from for the first space and 10 to choose from for the second space.
Step 3: Divide by the factorial of the number of spaces in the group.
Because there are two positions on the frame, we go ahead and divide by the factorial of 2, (2!). So now our entire equation looks like this: . That’s it, that’s all! The correct answer is C, 55.
Guided Practice
Try this one on your own.
17. Kevin had a very busy Saturday afternoon. He just bought a new spice rack that holds four spice jars. When he got home, he remembered that he has six different types of spices. How many different arrangements of spices, from left to right, are possible for Kevin’s new spice rack? (A) 15 (B) 60 (C) 120 (D) 180 (E) 360
Step 1: Does order matter? If YES, go to step 2 ONLY. If NO, go to steps 2 and 3.
Is the item asking about groups or arrangements? If it is a group, then order does not matter. If it is not a group, then it must be an arrangement, and order does matter.
Step 2: Multiply the factorial of the number of things being arranged to the number of spaces designated for arrangement.
Remember your factorials. Find the number of spaces in the arrangement or group.
Step 3: Divide by the factorial of the number of spaces in the group.
How many people or things are in the group? That is the number you’ll use to create a factorial.
Guided Practice: Explanation
Step 1: Does order matter? If YES, go to step 2 ONLY. If NO, go to steps 2 and 3.
The item is asking for arrangements, from left to right. This means that order does matter. Now we go to step 2 only.
Step 2: Multiply the factorial of the number of things being arranged to the number of spaces designated for arrangement.
How many things are being arranged? Six spices. How many spaces on the spice rack are there for them? Four. Now multiply the factorial of the number of spices (6!) down to the number of spaces available. It should look like this:
That’s the number of arrangements, answer E. I hope Kevin doesn’t ask us to help.
Step 3: Divide by the factorial of the number of spaces in the group.
This step isn’t necessary for this item, so move on.
Independent Practice
The explanation for this item is on the following page.
18. The student council has five vacancies since the seniors graduated. Nine talented and aspiring young freshmen have applied to fill the vacancies. How many different groups of freshmen are possible to fill the vacancies? (A) 60 (B) 120 (C) 126 (D) 7,560 (E) 15,120
Independent Practice: Explanation
Step 1: Does order matter? If YES, go to step 2 ONLY. If NO, go to steps 2 and 3.
The key word in this item is groups, which tells us that order does not matter. On to steps 2 and 3.
Step 2: Multiply the factorial of the number of things being arranged to the number of spaces designated for arrangement.
How many people or things are being arranged? Nine freshmen. How many spaces are there on the student council for them? Five. The factorial of 9, (9!), multiplied down five spaces looks like this:
Step 3: Divide by the factorial of the number of spaces in the group.
How many freshmen are going to be in the group? Five. So we are going to divide by the factorial of 5, (5!). Here’s our entire equation:
That’s it! There is the possibility of 126 groups of 5 freshmen to fill the vacancies of the student council. Our answer is choice C, 126.
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# The 4th Dimension
The journey into the 4th Dimension 4/19/12 Nature of Math The world we live in today is a world of 3-dimensions filled with objects that are zero, one and two dimensions. We all walk around in our 3-dimensional world thinking there could be no other dimensions. But would you believe me if I were to tell you that there is a 4th dimension that lies past our daily experience? The truth is that there is a 4th dimension and it’s not that far away, the crazy thing is that there could be an infinite number of other dimensions out there as well that we will never see and that our minds cannot even begin to fathom.
In order for you to fully understand the possibility of a 4th and possibly other dimensions I first need to explain the three prior and how they work. Let’s start with 0 –dimensions. When we classify an object’s dimensions we classify it according to the number of degrees of freedom it has. Therefore a 0-dimensional object would have zero degrees of freedom and would be represented as a point. With 0-dimensions you do not need any information to locate a point within that dimension. This is true because any 0-dimensional object has no length width or height.
Now think about taking that point and simply sweeping it to the left like you are drawing a line with the point. By sweeping the point in a line you have just taken a zero dimensional point and created a 1-dimensional line. All of 1-dimensional space is a line. Within a line there is only one degree of freedom, or one direction in which the line is capable of moving. It may seem like a line should be classified as a 2-dimensional object because it can move left and right but really it is based on how many different directions the line travels, which is one left and right. Now think about a line as your street.
Your specific house would only be one point on that street and in order to find it you would only need to tell one number in order for it to be found. Now if we take the 1 dimensional object and try and make it into a 2-dimensional object all we have to do is repeat the same process as before, take the line and find a new direction it can move. In the case of the second dimension we are going to take the line and move it vertically (perpendicular to the original line) in a sweeping motion, thus creating a plane. Along with creating a plane you have also just created the 2-dimension.
Inside of this world of 2-dimensions you now have the freedom to go left and right and up and down. 2-dimensional objects are all around us, squares, triangles, circles. A 2 dimensional world would be one where everything is flat, people would not be able to see depth or width we could only see what is in front of us in our flat world. Crazy huh? You may have noticed the trend by now on how we move into new dimensions by simply sweeping the current dimension in a new direction. So when we take our 2-dimensional plane and sweep it up and down it will form a cube. This creates the move from 2nd to 3rd dimension.
The 3rd dimension is one in which I am assuming you are fairly familiar with considering we live in a 3-dimensional world. Anything in our world that is tangible would be something 3-dimensional, so for example your cat, your favorite pants, a cube. Three dimensional object now have the ability not only to move left and right or up and down on a plane but can also incorporate depth and width into the picture You may now be asking “well if we live in the 3rd dimension where is the 4th dimension and what is it? ” Great Question! We would make the 4th dimension the same way as we have made all of the others.
Simply take the third dimension (for our sake lets say a cube) and slide it into a new direction perpendicular to all three previous directions. This may seem tough because we do not know any other directions aside from the three we are confined to in our world. But supposing we drag our 3-d cube in this new direction, then the 3-d cube now becomes 4-dimensional. We know this is true because in order to locate a point on this new figure we would need four different directions. This shows that there very well could be a 4th dimension out there somewhere.
Some people believe that time could be the 4th dimension, but it is still not determined. Also it could mean that there are hundreds of millions of other possibilities for dimensions as well. And while we can’t see the fourth dimension or wrap our minds around it we can now depict it to others. I chose this topic for my presentation because dimensions were my favorite topic of the class this year. They intrigued me, and made me question a lot about the world I perceive and live in. Also dimensions made me want to further pursue math and see how it relates to me in other ways that I didn’t realize.
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# Determine whether the given value is a solution of the equation.
Determine whether the given value is a solution of the equation.
$$\displaystyle{\frac{{{x}-{a}}}{{{x}-{b}}}}={\frac{{{a}}}{{{b}}}}{\left({b}\ne{0}\right)}$$
(a)x=0
(b)x=b
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Corben Pittman
Step 1
The given equation is $$\displaystyle{\frac{{{x}-{a}}}{{{x}-{b}}}}={\frac{{{a}}}{{{b}}}}$$
Step 2
Substitute x=0 in given equation,
$$\displaystyle{L}.{H}.{S}={\frac{{{x}-{a}}}{{{x}-{b}}}}$$
$$\displaystyle={\frac{{{0}-{a}}}{{{0}-{b}}}}$$
$$\displaystyle={\frac{{-{a}}}{{-{b}}}}$$
$$\displaystyle={\frac{{{a}}}{{{b}}}}$$
=R.H.S
So, the given value x=0 is a solution of the equation $$\displaystyle{\frac{{{x}-{a}}}{{{x}-{b}}}}={\frac{{{a}}}{{{b}}}}$$.
Step 3
Substitute x=b in given equation,
$$\displaystyle{L}.{H}.{S}={\frac{{{x}-{a}}}{{{b}-{b}}}}$$
$$\displaystyle={\frac{{{x}-{a}}}{{{0}}}}$$
we cannot divide by zero, so it is undefined.
So, the given value x=b is not a solution of the equation $$\displaystyle{\frac{{{x}-{a}}}{{{x}-{b}}}}={\frac{{{a}}}{{{b}}}}$$.
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# (i). A lens produces a magnification of $- 0.5$ . Is this a converging or diverging lens? If the focal length of the lens is $6cm$, draw a ray diagram showing the image formation in this case.(ii). A girl was playing with a thin beam of light from a laser torch by directing it from different directions on a convex lens vertically. She was surprised to see that in a particular direction, the beam of light continued to move along the same direction after passing through the lens. State the reason for her observation. Draw a ray diagram to support your answer.
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Hint: You can start by writing down the nature of the lens by looking at the sign of the magnification (negative means a convex/converging lens). Then use the equation $m = \dfrac{v}{{\left( { - u} \right)}}$ to get a relation between $v$ and $u$ . Then put these values in the equation $\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$ . Use the value of $v$ and $u$ obtained to make a diagram. Then in the second part, you have to mention where the light must strike the lens in to not get refracted (at the optical center) and then draw a corresponding diagram.
Complete step-by-step solution:
(i). We are given that the lens produces a magnification of $- 0.5$. We know that a concave lens does not have a negative magnification and only convex lenses have negative magnification. A convex lens is converging in nature.
Hence, the given lens is converging.
We are given in the problem
$f = + 6cm$
$m = - 0.5$
We know that,
$m = \dfrac{v}{{\left( { - u} \right)}} = - 0.5$
$v = + 0.5u$
We know by the lens formula
$\dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$
$\Rightarrow \dfrac{1}{6} = \dfrac{1}{{0.5u}} - \dfrac{1}{{\left( { - u} \right)}}$
$\Rightarrow \dfrac{1}{6} = \dfrac{1}{{0.5u}} + \dfrac{1}{{\left( { u} \right)}}$
Now on taking LCM,
$\Rightarrow \dfrac{1}{6} = \dfrac{1+0.5}{{0.5u}}$
$\Rightarrow \dfrac{1}{6} = \dfrac{3}{{u}}$
$\Rightarrow u = 18cm$
$\therefore v = 0.5u = 9cm$
The diagram of the given situation is given below
(ii). In the problem, we are told that the girl points a laser towards a convex lens that is held vertically. If she points in a particular direction the beam of light does not get refracted. This situation can only occur if the laser is directly pointed at the optical center. Any light ray that strikes at the optical center passes through without any refraction.
The diagram of the given situation is given below
Note: In the first part of the solution above, we mentioned that you can tell whether the lens is convex or concave (hence converging or diverging) by looking at the sign of the magnification. Negative magnification means a convex lens. You could have also predicted nature by looking at the sign of the focal length as the only convex lens has a positive focal length, but we did not do so because here the sign of the focal length was not provided. |
# Area of Trapezium
## Area of Trapezium
A trapezium is a quadrilateral having a pair of its opposite sides parallel. Area of trapezium is measured in terms of its two parallel sides and the height between them. Area of trapezium is given by the half of the product of the sum of parallel sides and height i.e. Area of trapezium = ½ (l1 + l2) × h, where l1 and l2 are two parallel sides of the trapezium and h is the height between them.
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### Derivation of the Formula for Area of Trapezium
Let, ABCD in the given figure be a trapezium having AD//BC. AC is a diagonal. AEBC drawn. Diagonal AC divides the trapezium ABCD into two triangles ΔABC and ΔACD. Since perpendicular distance between two parallel lines is same everywhere, the height of both the triangles is same as the height of two parallel lines.
Let, length of two parallel lines be l1 and l2 i.e. BC = l1 AD = l2 and height between two parallel sides = AE = h
Now,
Area of trapezium ABCD = area of triangle ABC + area of triangle ACD
= ½ × BC × AE + ½ × AD × AE
= ½ (BC + AD) × AE
= ½ (l1 + l2) × h
Area of trapezium = ½ (l1 + l2) × h, where l1 and l2 are two parallel sides and h is the height between them.
### Workout Examples
Example 1: Calculate the area of given trapezium.
Solution: Here,
l1 = 9cm
l2 = 5cm
h = 4cm
Area of trapezium = ½ (l1 + l2) × h
= ½ (9cm + 5cm) × 4cm
= ½ × 14cm × 4cm
= 28cm2
Example 2: In the given figure, ABCD is a trapezium in which AD//BC, ABC = 90°, AB = 15cm and DC = 17cm. If the area of trapezium ABCD is 204cm2, calculate the length of AD.
Solution: DEBC drawn,
From the figure,
AB = DE = 15cm, DC = 17cm
In right angle ΔDEC,
Now,
Area of trapezium ABCD = 204 cm2
i.e. ½ (AD + BC) × AB = 204
or, ½ (AD + BE + EC) × 15 = 204 |
# Division of Complex Numbers
## Theorem
Let $z_1 := a_1 + i b_1$ and $z_2 := a_2 + i b_2$ be complex numbers such that $z_2 \ne 0$.
The operation of division is performed on $z_1$ by $z_2$ as follows:
$\dfrac {z_1} {z_2} = \dfrac {a_1 a_2 + b_1 b_2} {a_2^2 + b_2^2} + i \dfrac {a_2 b_1 - a_1 b_2} {a_2^2 + b_2^2}$
## Proof 1
$\ds \frac {z_1} {z_2}$ $=$ $\ds z_1 \paren {z_2}^{-1}$ Definition of Complex Division $\ds$ $=$ $\ds \paren {a_1 + i b_1} \dfrac {a_2 - i b_2} {a_2^2 + b_2^2}$ Inverse for Complex Multiplication $\ds$ $=$ $\ds \frac {\paren {a_1 a_2 + b_1 b_2} + i \paren {a_2 b_1 - a_1 b_2} } {a_2^2 + b_2^2}$ Definition of Complex Multiplication $\ds$ $=$ $\ds \frac {a_1 a_2 + b_1 b_2} {a_2^2 + b_2^2} + i \frac {a_2 b_1 - a_1 b_2} {a_2^2 + b_2^2}$
$\blacksquare$
## Proof 2
$\ds \frac {z_1} {z_2}$ $=$ $\ds \frac {z_1 \overline {z_2} } {\cmod {z_2}^2}$ Complex Division as Product with Conjugate over Square of Modulus $\ds$ $=$ $\ds \dfrac {\paren {a_1 + i b_1} \paren {a_2 - i b_2} } {\cmod {z_2}^2}$ Definition of Complex Conjugate $\ds$ $=$ $\ds \dfrac {\paren {a_1 + i b_1} \paren {a_2 - i b_2} } {\paren {\sqrt { {a_2}^2 + {b_2}^2} }^2}$ Definition of Complex Modulus $\ds$ $=$ $\ds \frac {\paren {a_1 a_2 + b_1 b_2} + i \paren {a_2 b_1 - a_1 b_2} } { {a_2}^2 + {b_2}^2}$ Definition of Complex Multiplication and simplification $\ds$ $=$ $\ds \frac {a_1 a_2 + b_1 b_2} { {a_2}^2 + {b_2}^2} + i \frac {a_2 b_1 - a_1 b_2} { {a_2}^2 + {b_2}^2}$
$\blacksquare$
## Also presented as
The operation of complex division on $z_1$ by $z_2$ can also be presented as:
$\dfrac {z_1} {z_2} = \dfrac {a_1 a_2 + b_1 b_2 + i \paren {a_2 b_1 - a_1 b_2} } { {a_2}^2 + {b_2}^2}$ |
# What is 51/144 as a decimal?
## Solution and how to convert 51 / 144 into a decimal
51 / 144 = 0.354
Convert 51/144 to 0.354 decimal form by understanding when to use each form of the number. Both represent numbers between integers, in some cases defining portions of whole numbers In certain scenarios, fractions would make more sense. Ex: baking, meal prep, time discussion, etc. While decimals bring clarity to others including test grades, sale prices, and contract numbers. Once we've decided the best way to represent the number, we can dive into how to convert 51/144 into 0.354
## 51/144 is 51 divided by 144
Teaching students how to convert fractions uses long division. The great thing about fractions is that the equation is already set for us! The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. Now we divide 51 (the numerator) into 144 (the denominator) to discover how many whole parts we have. Here's how you set your equation:
### Numerator: 51
• Numerators sit at the top of the fraction, representing the parts of the whole. Any value greater than fifty will be more difficult to covert to a decimal. 51 is an odd number so it might be harder to convert without a calculator. Large numerators make converting fractions more complex. Let's look at the fraction's denominator 144.
### Denominator: 144
• Unlike the numerator, denominators represent the total sum of parts, located at the bottom of the fraction. Larger values over fifty like 144 makes conversion to decimals tougher. And it is nice having an even denominator like 144. It simplifies some equations for us. Have no fear, large two-digit denominators are all bark no bite. So grab a pen and pencil. Let's convert 51/144 by hand.
## Converting 51/144 to 0.354
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 144 \enclose{longdiv}{ 51 }$$
To solve, we will use left-to-right long division. This method allows us to solve for pieces of the equation rather than trying to do it all at once.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 144 \enclose{longdiv}{ 51.0 }$$
We've hit our first challenge. 51 cannot be divided into 144! Place a decimal point in your answer and add a zero. This doesn't add any issues to our denominator but now we can divide 144 into 510.
### Step 3: Solve for how many whole groups you can divide 144 into 510
$$\require{enclose} 00.3 \\ 144 \enclose{longdiv}{ 51.0 }$$
We can now pull 432 whole groups from the equation. Multiple this number by our furthest left number, 144, (remember, left-to-right long division) to get our first number to our conversion.
### Step 4: Subtract the remainder
$$\require{enclose} 00.3 \\ 144 \enclose{longdiv}{ 51.0 } \\ \underline{ 432 \phantom{00} } \\ 78 \phantom{0}$$
If your remainder is zero, that's it! If you still have a remainder, continue to the next step.
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals is a necessity. They each bring clarity to numbers and values of every day life. And the same is true for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But they all represent how numbers show us value in the real world. Without them, we’re stuck rounding and guessing. Here are real life examples:
### When you should convert 51/144 into a decimal
Contracts - Almost all contracts leverage decimal format. If a worker is logging hours, they will log 1.35 hours, not 1 and 51/144 hours. Percentage format is also used in contracts as well.
### When to convert 0.354 to 51/144 as a fraction
Pizza Math - Let's say you're at a birthday party and would like some pizza. You aren't going to ask for 1/4 of the pie. You're going to ask for 2 slices which usually means 2 of 8 or 2/8s (simplified to 1/4).
### Practice Decimal Conversion with your Classroom
• If 51/144 = 0.354 what would it be as a percentage?
• What is 1 + 51/144 in decimal form?
• What is 1 - 51/144 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.354 + 1/2? |
# BASIC MATHEMATICS for Class 11 MATHS
## Introduction
Hello bachcho, I hope you all are fine. Aaj hum JEE ki teyaari shuru krne vale hai. Aaj hum maths ka first chapter jiska naam hai BASIC MATHEMATICS padhenge. Iss chapter mein hum maths ke basic concepts jo humne abhi tak padhe hai, unn concepts prr discussion karenge. Hum Iss chapter me Number system aur indices ke baare me discussion karne vale hai. Iske sath sath hum, Set theory, Wavy curve Method, Modulus Function ,Exponential and logarithmic functions details mein padhenge aur samjhenge practically questions kis tarha se solve kiye jate hai.. Bachcho ye chapter padhne ke baad aapke sabhi concepts clear ho jayenge aur aap jee ke questions ko easily solve krr lengye. Chalo phir shuru krte hai first chapter ka first topic Number System.
## Number System
1. Natural numbers :
The counting numbers 1,2,3,4,.... are called Natural Numbers. The set of natural numbers is denoted by N. Thus N={1,2,3,4,....}.
2. Whole numbers :
Natural numbers including zero are called whole numbers. The set of whole numbers is denoted by W. Thus W={0,1,2,.........}.
3. Integers : The numbers ...3,2,1,0,1,2,3....are called integers and the set is denoted by I or Z. Thus I(orZ)={..3,2,1,0,1,2,3...}
Note
(a)Positive integers I+={1,2,3....}=N
(b)Negative integers I={.....,3,2,1}.
(c) Non-negative integers (whole numbers) ={0,1,2,......}.
(d)Non-positive integers ={......,3,2,1,0}.
4. Even integers :
Integers which are divisible by 2 are called even integers.
Example :0,±2,±4,.......
5. Odd integers : Integers which are not divisible by ‘2’ are called odd integers.
Example : ±1,±3,±5,±7......
6. Prime numbers :
Natural numbers which are divisible by 1 and itself only are called prime numbers.
Example :2,3,5,7,11,13,17,19,23,29,31,........
7. Composite number :
Let a be a natural number, a is said to be composite if, it has atleast three distinct factors.
Example : 4,6,8,9,10,12,14,15..........
Note :
(a)1 is neither a prime number nor a composite number.
(b)Numbers which are not prime are composite numbers (except 1).
(c)4 is the smallest composite number.
(d)2 is the only even prime number
8. Co-prime numbers :Two natural numbers (not necessarily prime) are called coprime, if there H.C.F (Highest common factor) is one.
Example :(1,2),(1,3),(3,4),(3,10),(3,8),(5,6),(7,8)(15,16) etc. These numbers are also called as relatively prime numbers.
Note :(a)Two prime number(s) are always co-prime but converse need not be true.
(b)Consecutive natural numbers are always co-prime numbers.
9. Twin prime numbers :If the difference between two prime numbers is two, then the numbers are called twin prime numbers.
Example:{3,5},{5,7},{11,13},{17,19},{29,31}
Note :Number between twin prime numbers is divisible by 6 (except (3,5)).
10. Rational numbers :All the numbers that can be represented in the form pq, where p and q are integers and q0, are called rational numbers and their set is denoted by Q. Thus Q={pq:p,qIandq0}. It may be noted that every integer is a rational number since it can be written as p1. It may be noted that all recurring decimals are rational numbers.
Note :Maximum number of different decimal digits in pq is equal to q, i.e. 119 will have maximum of 9 different decimal digits.
11. Irrational numbers :
The numbers which cannot be expressed in pq form where p,qI and q0 i.e. the numbers which are not rational are called irrational numbers and their set is denoted by Qc . (i.e. complementary set of Q) e.g. 2,1+3 etc. Irrational numbers can not be expressed as recurring decimals.
Note : e2.71 is called Napier's constant and π3.14 are irrational numbers.
12. Real numbers :
Numbers which can be expressed on number line are called real numbers. The complete set of rational and irrational numbers is the set of real numbers and is denoted by R. Thus R=QQc .
All real numbers follow the order property i.e. if there are two distinct real numbers a and b then either a<b or a>b.
Note :(a)Integers are rational numbers, but converse need not be true.
(b)Negative of an irrational number is an irrational number.
(c)Sum of a rational number and an irrational number is always an irrational number e.g. 2+3
(d)The product of a non-zero rational number & an irrational number will always be an irrational number.
(e)If aQ and bQ, then ab= rational number, only if a=0.
(f)Sum, difference, product and quotient of two irrational numbers need not be a irrational number or we can say, result may be a rational number also.
## Example 1
If (22002192.31+2n) is the perfect square of a natural number , then find the sum of digits of 'n' .
We have given
22002192×31+2n.
So we can take 2192 common.
2192[25631+2k] [taking n192=k]
Since 2192 is a perfect square.
To bacchon we have to make (225+2k) as a perfect square.
So, we will use hit and trial method.
Let k=1(225+2)=227 which is not a perfect square.
Now, let k=2(225+4)=229 which is still not a perfect square.
Now, let k=6(225+64)=289 which is a perfect square of 17.
So here we got the value of k.
K=6
Lakin n192=k=6
n=6+192=198
To bacchon according to the question sum of digits of n=1+9+8=18.
## Example 2
Find all 3-digit numbers which are the sums of the cubes of their digits.
Let us suppose the number be xyz.
According to question
xyz=x3+y3+z3.
Sum of cubes of x,yandz should be xyz.
We know that, smallest 3-digit number is 100=1×100+0×10+0×1
Similarly greatest three digit number will be 999.
To hum yaha kya dekhte hai ki, our range lies between 100999.
So, the numbers that satisfy this equation is
(i) 15313+53+33=1+125+27=153
(ii) 37033+73+03=27+343+0=370
Similary two more numbers are 371and407.
## Example 3
Find the number of natural numbers less than 107 which are exactly divisible by 7.
Let the natural numbers less than 107 be
1,2,3,4,5,.......................106.
To bacchon hume ye pata karna hai ki how many numbers are there which are exactly divisible by 7. Matlab us number ko 7 se divide karne par remainder 0 aana chaiye.
To uske liye humare pas ek formula hai i.e.
Let us suppose our first value be a=1 and last value be an=106.
Let d=7 and n is the required number.
an=a+(n1)d
106=1+(n1)×7
1061=(n1)×7
1057=n1
n=15+1=16
So there are 16 such numbers which are divisible be 7.
## Example 4
An eight digit number is a multiple of 73 and 137. If the second digit from left is 7, what is the 6th digit from the left of the number ?
We have given 15n2+8n+6n
15n2n+8nn+6n
15n+8+6n
We can see that 15n+8 will always be a natural number for any value of natural number.
Now, for 6n to be natural number, 6 should be completely divisible by n.
possible values of n are 1,2,4and6.
total 4 such numbers are possible.
## Example 5
The sum of all values of integers n for which n29n1 is also an integer is
Given n29n1
To bacchon hum ise aise likh sakhte hai
n218n1
n21n18n1
(n1)(n+1)n18n1
=n+18n1
To bacchon aap yha dekh sakte hai ki kise bhi integer value of n ke liye (n+1) will always be an integer.
Now, for 8n1 to be an integer 8 should be divisible be |n1| and jaisa ki hum jante hai divisor of 8 are 1,2,3,4and8
Now if (n1)>0
n=2,3,5,9
Similarly for (n1)<0
n=0,1,3,and7.
Sum of all possible values of n=2+3+5+9+0+(1)+(3)+(7)=8
## Indices
Indices ka matlab hai ek number apne aap se kitni baar multiply ho sakta hai. Jaise 'a' koi bi ek non zero real or imaginary number hai to the power 'm' where m is a positive integer, a is called the base and m is called the index, m ko hum power or exponent bhi khte hai. Toh chlo ab hum indicies ke kuch Laws ke baare mein padhte hai jo ki humme algebraic expression ko solve krne mein help karenge.
## Definition of Indices
If 'a' is any non zero real or imaginary number and 'm' is the positive integer, then am=a.a.a..................a (m times). Here a is called the base and m is called the index, power or exponent.
## Laws of Indices :
1. a0=1,(a0)
Kabhi bhi kisi bhi constant ki power agar zero hogi to uska result one ke equal aaye ga. Ye kuch matter nahi krta ki base ki kya value kya hai. For Example : - 90=1
2. am=1am,(a0)
Jab bhi kisi index ki power agar negative hai iska matlab vo iska inverse hai i.e. = 1am
3. am.an=am+n where m and n are rational numbers, (a0)
Jab bhi koi do indices multiplication mein ho aur unka base same ho that means unki power add ho jayengi. For Example : - 97=197
4. aman=amn where m and n are rational numbers, (a0)
Jab bhi do indices ka base same ho aur vo division mein ho toh unki powers subtract ho jayengi. For Example : - 9492=942=92
5. (am)n=amn
Jab bhi kisi base ki power mein raise to power ho toh yeh dono powers result mein multiply ho jayengi. For Example : - (92)3=92.3=96
6. apq=qap
agar kisi base ki power fraction me di hai to, isse hum radical form me bi likh sakte hai.
For Example : - 943=((94)3)=394
## Example 6
a(a+b2ba)1+b(a+b2ab)1
=a(2baa+b)+b(2aba+b) [because a1=1a]
=a(2ba)+b(2ab)a+b
=2ab(a+b)a+b
=2ab
## Example 7
Solve
322332+23+1232
We have given 322332+23+1232
Now we can rationalize both the terms.
322332+2332233223+12323+23+2
=18+121261812+23(3+2)32
=301266+6+261
=526+6+26
=11.
## Example 8
If both a and b are rational numbers, then find a and b.3+737=ab7
We have given 3+737=ab7
Now we can rationalize the L.H.S i.e.
= 3+737×3+73+7
=9+6+6797
=16+672
=8+37
Ab ise R.H.S ke sath equate and cmpare krenge. i.e.
8+37=ab7
On comparing a=8 and b=3.
## Example 9
If 143k=41k+k+1=ab then a and b are respectively.
We have given
143k=41k+k+1=ab
Multiplying numerator and denominator by k+1k i.e.
143k=4k+1k(k+k+1)×(k+1k)
=143k=4k+1kk+1k
=143k=4k+1k1
=143k=4[k+1k]
Now expanding the summation
(54)+(65)+(76)........................+(144143)
Now hum yha dekh sakte hai ki bahut sare terms aapas me cancle ho jayenge so finally we have
4+144
=2+12
=10.
So according to the question 10=ab
On comparing L.H.SandR.H.S.
a=10andb=0.
## Example 10
(461+462+463+464) is divisible by
(a) 3
(b) 11
(c) 13
(d) 17
We have given (461+462+463+464)
Taking 461 common
=461[1+41+42+43]
=461[1+4+16+64]
=85×461
Now ab hum yha 85 ko factor karke likh sakte hai i.e.
=17×5×461
Yha hum dekh sakte hai ki 17 is a factor which means (461+462+463+464) is divisible by 17.
So the correct option is d.
## Example 11
If x+y+z=12andx2+y2+z2=96and1x+1y+1z=36, then the value x3+y3+z3 divisible by prime number is________.
Given (x+y+z)2=144
x2+y2+z2+2(xy+yz+zx)=144
96+2(xy+yz+xz)=144 [since x2+y2+z2=96]
xy+yz+zx=144962=24
Given
1x+1y+1z=36
Taking L.C.M
yz+zx+xyxyz=36
36xyz=24
xyz=2436=23
Now x3+y3+z33xyz=(x+y+z)(x2+y2+z2(xy+yz+zx))
x3+y3+z33×(23)=12×[9624]
x3+y3+z32=12×72
=x3+y3+z3=864
Now hum ye dekhte hai ki 864 is a even number which can only be divisible by a prime number which is 2.
## Definition of Set
A set is a collection of well defined objects which are distinct from each other . Set are generally denoted by capital letters A,B,C,.................. etc. and the elements of the set by a,b,c........... etc.
If a is an element of a set A, then we write aA and say a belongs to A. If a does not belong to A then we write aA.
e.g. The collection of first five prime natural numbers is a set containing the elements 2,3,5,7,11.
N = Set of all natural numbers = {1,2,3,4,....}
W = Set of all whole numbers = {0,1,2,3,....}
Z or I set of all integers = {....3,2,1,0,1,2,3,....}.
## Methods to write a Set : -
1. Tabulation Method or Roster : In this method a set is described by listing elements, separated by commas and enclose then by curly brackets.
e.g. The set of vowels of English Alphabet may be described as {a,e,i,o,u}
Example : A set of days of week {Sun,Mon,Tues,Wed,Thru}
2. Set builder Method : In this case we write down a property or rule p Which gives us all the element of the set.
Example : If A={1,2,3,4} then in set builder method,
A={x:xNandx4} where N is the Set of natural numbers.
## Types of Sets
• Null Set or Empty Set : A set having no element in it is called an Empty Set or a null Set or void Set. It is denoted by ϕ or {}.
Example : A={x:xNand5<x<6}=ϕ
• Singleton Set : A set consisting of a single element is called a singleton Set.
Example : Then set {0}, is a singleton Set.
• Finite Set : - A set which has only finite number of elements is called a finite set.
Example : A = {a, b, c}.
• Order of a finite set : - The number of elements in a finite set is called the order of the set A and is denoted by O(A) or n(A). It is also called cardinal number of the set.
Example : A={a,b,c,d}n(A)=4
• Infinite Set : - A set which has an infinite number of elements is called an infinite set.
Example : A = {1, 2, 3, 4, ....} is an infinite set.
• Equal Set : - Two sets A and B are said to be equal if every element of A is a member of B, and every element of B is a member of A. If sets A and B are equal. We write A=B and A and B are not equal then AB.
Example : A={1,2,6,7} and B={6,1,2,7} A=B.
• Equivalent Sets : - Two finite sets A and B are equivalent if their number of elements are same i.e. n(A)=n(B).
Example : A={1,3,5,7},B={a,b,c,d}
n(A)=4 and n(B)=4 n(A)=n(B)
Note : Equal set always equivalent but equivalent sets may not be equal.
• Subset : Let A and B be two sets if every element of A is an element B, then A is called a subset of B if A is a subset of B. we write AB.
Example : A = {1, 2, 3, 4} and B = {1, 2, 3, 4, 5, 6, 7} in A sub B
The symbol stands for implies.
• Proper subset : If A is a subset of B and AB then A is a proper subset of B and we write AB
Note-1 : Every set is a subset of itself i.e. AA for all A
Note-2 : Empty set ϕ is a subset of every set
Note-3 : Clearly NWZQRC
Note-4 : The total number of subsets of a finite set containing n elements is 2n
• Universal set : A set consisting of all possible elements which occur in the discussion is called a Universal Set and is denoted by U
Note : All sets are contained in the universal set.
Example : If A={1,2,3},B={2,4,5,6},C={1,3,5,7}thenU={1,2,3,4,5,6,7}canbetakenastheUniversalset.<b><li>PowerSet:<b>LetAbeanyset.ThesetofallsofAiscaldpowersetofAandisde¬edbyP(A).<br><b>Examp:<b>LetA = {1, 2}thenP(A) = {phi, {1}, {2}, {1, 2}}<li>Twofitesetshavemandnementsrespectively.Thetalνmberofementsthepowersetofthefirstsetis56morethanthetalνmberofementsthepowersetofthesecondsetfdthevalueofmandnrespectively.Answer:2^m-2^n=56=2^3.7=2^3(2^3-1)2^m-2^n=2^6-2^3m=6, n=3
## Some Operations On Sets
• Union of two sets(U):
AB={x:xAorxB}
e.g. A={1,2,3},B={2,3,4} then AB={1,2,3,4}
• Intersection of two sets :
AB={x:xAandxB}
e.g. A={1,2,3,},B={2,3,4} then AB={2,3}
• Difference of two sets :
AB={x:xAandxB}
e.g. A={1,2,3},B={2,3,4};AB={1}
• Complement of a sets:
A'={x;xAandxU}=UA
e.g. U={1,2,....,10},<br>A={1,2,3,4,5}
then A={6,7,8,9,10}
• De-Morgan Laws :
(AB)'=A'B'
(AB)'=A'B'
• A(BC)=(AB)(AC)
A(BC)=(AB)(AC)
• Distributive Laws :
A(BC)=(AB)(AC)
A(BC)=(AB)(AC)
• Commutative Laws :
AB=BA
AB=BA
• Associative Laws:
(AB)C=A(BC)
(AB)C=A(BC)
• Aϕ=ϕ;AU=A
Aϕ=A;AU=U
• Disjoint Sets :
If AB=ϕ, then A,B are disjoint.
e.g. if A={1,2,3},B={7,8,9} then AB=ϕ
• Note : AA'=ϕ therefore A,A' are disjoint
• Symmetric Difference of Sets :
AB=(AB)(BA)
=(AB)(AB)
## Example 12
Let R be the real line. Consider the following subsets of the plane R×R.
S={(x,y):y=x+1 and 0<x<2},
T={(x,y):xy is an integer }.
Which one of the following is true ?
a. Neither S not T is an equivalence relation on R.
b. Both S and T are equivalence relations on R.
c. S is an equivalence relation on R but T is not.
d. T is an equivalence relation on R but S is not.
y=x+1
y<3
Let S is symmetric
So (x,y)R<br>(y,x)R
but (y,x) is not satisfying 0<x<2
So S is not an equivalence relation.(for verification take y=2 and x=1)
T(x,y)=xy is integer
(x,x) is an integer always
If (x,y)R, then yx is also an integer.
(y,x)IR
If (x,y)IR&(y,z)IR, then (xy) is an integer, (yz) is an integer, sum of xy & yz=xz is also an integer.
Hence T is equivalence relation. Option (d) is correct.
## Example 13
Let R be the real line. Consider the following subsets of the plane R×R.
S={(x,y):y=x+1 and 0<x<2},
T={(x,y):xy is an integer }.
Which one of the following is true ?
a. Neither S not T is an equivalence relation on R.
b. Both S and T are equivalence relations on R.
c. S is an equivalence relation on R but T is not.
d. T is an equivalence relation on R but S is not.
y=x+1
y<3
Let S is symmetric
So (x,y)R<br>(y,x)R
but (y,x) is not satisfying 0<x<2
So S is not an equivalence relation.(for verification take y=2 and x=1)
T(x,y)=xy is integer
(x,x) is an integer always
If (x,y)R, then yx is also an integer.
(y,x)IR
If (x,y)IR&(y,z)IR, then (xy) is an integer, (yz) is an integer, sum of xy & yz=xz is also an integer.
Hence T is equivalence relation. Option (d) is correct.
## Example 14
If A, B and C are three sets such that AB=AC and AB=AC, then
a.A=B b.A=C
c.B=C d.AB=ϕ
Let xC
Suppose xAxAC
xAB (AC=AB)
Thus xB
Again suppose xAxCA
xBAxB
Hence CB ...(1)
Similarly we can show that BC ...(2)
Combining (1) and (2) we get B=C.
Option (c) is correct.
## Example 15
Let R be the set of real numbers.
Statement 1:A={(x,y)R×R:yx is an integer} is an equivalence relation on R.
Statement 2:B={(x,y)R×R:x=αy for some rational number alpha} is an equivalence relation on R.
a. Statement 1 is false, statement 2 is true
b. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1
c. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1
d. Statement 1 is true, statement 2 is false
yx is an integer
xx=0 is also an integer
A is reflexive
xy is also an integer. Hence, it is symmetric.
For (x,y)R×R and (y,z)R×R. yx and zy are both integers, then sum of them is also an integer. Hence, transitive.
If a set is symmetric, transitive and reflexive then it also has equivalence relation.
Clearly, A is an equivalence relation
Now B,
xy=α is rational but yx need not be rational i.e. 01 is rational but 10 is not rational.
Hence, B doesn't show equivalence relationship.
Correct option is (d) statement 1 is true, statement 2 is false.
## Example 16
Let S be a non-empty subset of R. Consider the following statement :
P: There is a rational number xS such that x>0.
Which of the following statements is the negation of the statement P?
a. xS and x0x is not rational.
b. There is a rational number xS such that x0.
c. There is no rational number xS such that x0
d. Every rational number xS satisfies x0.
P: there is a rational number xS such that x>0
P: Every rational number xS satisfies x0
Hence, option (d) is correct.
## Example 17
A function f from the set of natural numbers to integers defined by
f(n)=n12, when n is odd
=n2, when n is even
is
a. one-one but not onto
b. onto but not one-one
c. one-one and onto both
d. neither one-one nor onto
One-One test for f:
Let x1 and x2 be any two elements in the domain (N).
Case I : When both x1 and x2 are even.
Let f(x1)=f(x2)
x12=x22
x1=x2
x1=x2
Case II : When both x1 and x2 are odd.
Let f(x1)=f(x2)
x112=x212
x1=x2
From, all cases, we can say that f is one-one
Onto test of f :
Codomain of f=Z={...,3,2,1,0,1,2,3,...}
Range of f={...,212,22,112,02,112,22,312,...}
Range of f={...,2,1,1,0,0,1,1,...}
Co-domain of f= Range of f
Or we can say that since, each negative integer is an image of even natural number and positive integer is an image of odd natural number.
f is onto.
Option (c) is correct.
## Venn diagram
A Venn diagram is a graphical representation of the elements in a set or a group. It is a diagram that shows all the possible logical relationships between sets or groups. In Venn diagrams, the elements of the sets are written in their respective circles.
In below figure U={1,2,3,...,10} is the universal set of which A={2,4,6,8,10} and B={4,6} are subsets, and also BA.
## Example 18
There are 200 individuals with a skin disorder, 120 had been exposed to the chemical C1? 50 to chemical C2, and 30 to both the chemicals C1and C2. Find the number of individuals exposed to(i) Chemical C1but not chemical C2
By using venn diagram
(i) 90
## Wavy Curve Method
The wavy curve method is used for solving inequalities in the form f(x)=(xa1)n1(xa2)n2.(xak)n=k(xb1)m1(xb2)m2..(xbp)mp>0(<0,=<0,or0) where n1,n2,....,nk,m1,m2,....,mp are natural numbers and the numbers a1,a2,..,ak;b1,b2,.....,bp are any real numbers such that a i b j, where i=1,2,3,,k and j=1,2,3,,p.The wavy curve method is also called the method of intervals. Rules
1. Firstly we have to simplify and factorize inequality
2. Make sure, the coefficient of x is/are positive
3. Then, find all the critical points
4. Put + sign for right most places and alternatively change sign as move towards left on the number line
## Example 19
If (y25y+3)(x2+x+1)<2x for all xR, then find the interval in which y lies.
We have, (y25y+3)(x2+x+2)<2x,xR
y25y+3<2xx2+x+1 (because x2+x+1>0xR)
L.H.S. must be less then least value of R.H.S. Now let's find the range of R.H.S
Let 2xx2+x+1=p
or px2+(p2)x+p=0
Since x is real, we have
(p2)24p20
or 2p23
The minimum value of 2xx2+x+1 is 2. So,
y25y+3<2
y25y+3<0
y(552,5+52)
## Example 20
Find the values of k for which x2+kx+1x2+x+1<2,xR
We have
|x|<aa<x<+a
The given inequality implies
2<x2+kx+1x2+x+1<2 ...(1)
Now, x2+x+1=(x+12)2+(34) is +ve for all values of x.
Multiplying (1) by x2+x+1, we get
2(x2+x+1)<x2+kx+1<2(x2+x+1)
This yields two inequalities, viz.,
3x2+(2+k)x+3)>0
and x2+(2k)x+1<0
For above inequalities to be true for all values of x, their discriminants must be negative. Hence,
(2+k)236<0 and (2k)24<0
(k+8)(k4)<0andk(k4)<0
8<k<4and0<k<4
Therefore, 0<k<4.
## Example 21
Solve the following inequality
x43x3+2x2x2x30>0
We have, x43x3+2x2x2x30>0
(x2)(x23x+2)x2x30>0
On factorizing both numerator and denominator we get,
x2(x1)(x2)(x+5)(x6)>0
Our inequality becomes +ve when x(6,), -ve when x(2,6), +ve when x(1,2), -ve when x(5,1) and +ve when (,5)
So, we can conclude that the required result is x(,5)(1,2)(6,).
## Example 22
Complete the solution of inequality (x+2)(x+3)(x2)(x3)1 is
We have,
(x+2)(x+3)(x2)(x3)10
(x+2)(x+3)(x2)(x3)(x2)(x3)0
10x(x2)(x3)0
From the ab |
The square source of 225 is expressed together √225 in the radical kind and together (225)½ or (225)0.5 in the exponent form. The square source of 225 is 15. It is the hopeful solution the the equation x2 = 225. The number 225 is a perfect square.
You are watching: How to find square root of 225
Square root of 225: 15Square source of 225 in exponential form: (225)½ or (225)0.5Square source of 225 in radical form: √225
1 What Is the Square root of 225? 2 Is Square source of 225 reasonable or Irrational? 3 How to find the Square root of 225? 4 FAQs on Square root of 225
## What Is the Square root of 225?
For real numbers a and b, if a2= b, we deserve to express a =√bThis way that a is the 2nd root the b.15 × 15 = 225 and also -15 ×-15 = 225.Therefore, √25 = 15
## Is the Square root of 225 reasonable or Irrational?
225 have the right to be expressed together the proportion of 2 integers.As 225 have the right to be expressed as 225 = 225/1.Thus 225 is rational.
## How to uncover the Square root of 225?
The square source of 225 deserve to be discovered by assorted methods.
### Square source of 225 by prime Factorization
Get the element factorization done because that the number 225 by the laddermethod or the factor tree technique to acquire the factors.Aftergetting the prime factors, arrange them in such a way that they canbe expressed together a product the number x number.√225 = √(5 × 5 × 3 × 3)Squaring on both the sides, us get, √225 =√(52 × 32)This gives, 5 × 3 = 15
### Square root of 225by Long department Method
Here room the steps to uncover the square source of 225
Step 1: compose the pair that digits beginning from one's place. Here 25 is the pair.Step 2: On recognize a divisor "n"such that n× nresults in the product ≤2. We find1 × 1 = 1, follow the procedure oflong division and obtainthe remainder. Here it is 1.Step 3: Now,bring down the next pair of numbers. Below it's 25. Multiply the quotient 1 by 2 and write that in the new divisor's place. Below it's2.Step 4: uncover a divisor "n"such the n × nresults in the product ≤125. Get the next quotient location as 5. Currently we get our brand-new divisor together 25, as 5 × 125 = 225. Divide and get the remainder. Right here we gain 0. Thus finding the square root of 225 by long division is completed.Therefore, 15 is the square source of 225.
Explore Square roots utilizing illustrations and interactive examples
Important Notes:
Finding the square root is an inverse procedure of squaring the number.Irrational numbers can not be expressed together a proportion of two integers. Example: π and also √sqrt 2. This makes 225 is rational together it can be expressed together 225/1.225 is a perfect square.
Tips and also Tricks
Subtract native 225 by successive odd herbal numbers, till you achieve 0. The variety of times friend subtract provides you the square source of 225.Check by the recognized multiplication reality that 225 = 15 × 15.
## Square root of 225 fixed Examples
Example 1:Mark desires to fencehis square backyard. He to know his backyard is 225 square feet. How numerous feet of fencing will note need?
Solution:
To fence he demands to know each side of his backyard.All the sides of the fence are equal.Area = next × next = 225 sq feet.As, 15 × 15 = 225Each side will certainly be = 15 feet
He needs to fence 4 × 15 = 60 sq feet.
Example 2:James wants to purchase a brand-new rug for his living room. In thestore, he find a square rug that has actually an area that 25 sq feet.a. Just how long is each side that the rug?b. How many of those rugs are required to cover an area of 225 square feet?
Solution
a. Area is 25 sq feet = side × sidesq feet.The length of each side of the rug will certainly be 5 feet.
b. To cover an area of 225 square feet, he requirements 225÷ 5= 45 rugs.
Example 3: If the area of a circle is 225π in2. Discover the radius that the circle.
Solution:
Let 'r' be the radius of the circle.⇒ Area that the circle = πr2 = 225π in2⇒ r = ±√225 inSince radius can't be negative,⇒ r = √225The square root of 225 is 15.⇒ r = 15 in
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## FAQs top top the Square source of 225
### What is the value of the Square root of 225?
The square source of 225 is 15.
### Why is the Square root of 225 a reasonable Number?
Upon prime factorizing 225 i.e. 32 × 52, we find that every the prime components are in also power. This implies that the square source of 225 is a positive integer. Therefore, the square source of 225 is rational.
### Evaluate 5 to add 3 square root 225
The given expression is 5 + 3 √225. We understand that the square source of 225 is 15. Therefore, 5 + 3 √225 = 5 + 3 × 15 = 5 + 45 = 50
### If the Square source of 225 is 15. Find the value of the Square root of 2.25.
Let us stand for √2.25 in p/q kind i.e. √(225/100) = 15/10 = 1.5. Hence, the worth of √2.25 = 1.5
### Is the number 225 a Perfect Square?
The prime factorization the 225 = 32 × 52. Here, all the numbers are in the power of 2. This implies that the square root of 225 is a optimistic integer. Therefore, 225 is a perfect square.
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### What is the Square of the Square source of 225?
The square the the square root of 225 is the number 225 itself i.e. (√225)2 = (225)2/2 = 225. |
# mean and range
Views:
Category: Education
## Presentation Description
mean and range
By: 8056123833 (100 month(s) ago)
sir/madam, pls send this ppt to my id for my teaching. my id : kaksamuel_82@yahoo.com
## Presentation Transcript
### Mean, Median, Mode & Range :
Mean, Median, Mode & Range By Paul Hodson
### Definition :
Definition Mean Means Average
### Definition :
Definition Mean – the average of a group of numbers. 2, 5, 2, 1, 5 Mean = 3
### Mean is found by evening out the numbers :
Mean is found by evening out the numbers 2, 5, 2, 1, 5
### Mean is found by evening out the numbers :
Mean is found by evening out the numbers 2, 5, 2, 1, 5
### Mean is found by evening out the numbers :
Mean is found by evening out the numbers 2, 5, 2, 1, 5 mean = 3
### How to Find the Mean of a Group of Numbers :
How to Find the Mean of a Group of Numbers Step 1 – Add all the numbers. 8, 10, 12, 18, 22, 26 8+10+12+18+22+26 = 96
### How to Find the Mean of a Group of Numbers :
How to Find the Mean of a Group of Numbers Step 2 – Divide the sum by the number of addends. 8, 10, 12, 18, 22, 26 8+10+12+18+22+26 = 96 How many are there?
### How to Find the Mean of a Group of Numbers :
How to Find the Mean of a Group of Numbers Step 2 – Divide the sum by the number of addends. 6) 96 sum 1 6 3 6 6 6 3
### How to Find the Mean of a Group of Numbers :
How to Find the Mean of a Group of Numbers The mean or average of these numbers is 16. 8, 10, 12, 18, 22, 26
### What is the mean of these numbers? :
What is the mean of these numbers? 7, 10, 16 11
### What is the mean of these numbers? :
What is the mean of these numbers? 2, 9, 14, 27 13
### What is the mean of these numbers? :
What is the mean of these numbers? 1, 2, 7, 11, 19 8
### What is the mean of these numbers? :
What is the mean of these numbers? 26, 33, 41, 52 38
### Definition :
Definition Median is in the Middle
### Definition :
Definition Median – the middle number in a set of ordered numbers. 1, 3, 7, 10, 13 Median = 7
### How to Find the Median in a Group of Numbers :
How to Find the Median in a Group of Numbers Step 1 – Arrange the numbers in order from least to greatest. 21, 18, 24, 19, 27 18, 19, 21, 24, 27
### How to Find the Median in a Group of Numbers :
How to Find the Median in a Group of Numbers Step 2 – Find the middle number. 21, 18, 24, 19, 27 18, 19, 21, 24, 27
### How to Find the Median in a Group of Numbers :
How to Find the Median in a Group of Numbers Step 2 – Find the middle number. 18, 19, 21, 24, 27 This is your median number.
### How to Find the Median in a Group of Numbers :
How to Find the Median in a Group of Numbers Step 3 – If there are two middle numbers, find the mean of these two numbers. 18, 19, 21, 25, 27, 28
### How to Find the Median in a Group of Numbers :
How to Find the Median in a Group of Numbers Step 3 – If there are two middle numbers, find the mean of these two numbers. 21+ 25 = 46 2) 46 23 median
### What is the median of these numbers? :
What is the median of these numbers? 16, 10, 7 10 7, 10, 16
### What is the median of these numbers? :
What is the median of these numbers? 29, 8, 4, 11, 19 11 4, 8, 11, 19, 29
### What is the median of these numbers? :
What is the median of these numbers? 31, 7, 2, 12, 14, 19 13 2, 7, 12, 14, 19, 31 12 + 14 = 26 2) 26
### What is the median of these numbers? :
What is the median of these numbers? 53, 5, 81, 67, 25, 78 60 53 + 67 = 120 2) 120 5, 25, 53, 67, 78, 81
### Definition :
Definition Mode is the most Popular
### Definition :
Definition A la mode – the most popular or that which is in fashion. Baseball caps are a la mode today.
### Definition :
Definition Mode – the number that appears most frequently in a set of numbers. 1, 1, 3, 7, 10, 13 Mode = 1
### How to Find the Mode in a Group of Numbers :
How to Find the Mode in a Group of Numbers Step 1 – Arrange the numbers in order from least to greatest. 21, 18, 24, 19, 18 18, 18, 19, 21, 24
### How to Find the Mode in a Group of Numbers :
How to Find the Mode in a Group of Numbers Step 2 – Find the number that is repeated the most. 21, 18, 24, 19, 18 18, 18, 19, 21, 24
### Which number is the mode? :
Which number is the mode? 29, 8, 4, 8, 19 8 4, 8, 8, 19, 29
### Which number is the mode? :
Which number is the mode? 1, 2, 2, 9, 9, 4, 9, 10 9 1, 2, 2, 4, 9, 9, 9, 10
### Which number is the mode? :
Which number is the mode? 22, 21, 27, 31, 21, 32 21 21, 21, 22, 27, 31, 32
### Definition :
Definition Range is the distance Between
### Definition :
Definition Range – the difference between the greatest and the least value in a set of numbers. 1, 1, 3, 7, 10, 13 Range = 12
### How to Find the Range in a Group of Numbers :
How to Find the Range in a Group of Numbers Step 1 – Arrange the numbers in order from least to greatest. 21, 18, 24, 19, 27 18, 19, 21, 24, 27
### How to Find the Range in a Group of Numbers :
How to Find the Range in a Group of Numbers Step 2 – Find the lowest and highest numbers. 21, 18, 24, 19, 27 18, 19, 21, 24, 27
### How to Find the Range in a Group of Numbers :
How to Find the Range in a Group of Numbers Step 3 – Find the difference between these 2 numbers. 18, 19, 21, 24, 27 27 – 18 = 9 The range is 9
### What is the range? :
What is the range? 29, 8, 4, 8, 19 29 – 4 = 25 4, 8, 8, 19, 29
### What is the range? :
What is the range? 22, 21, 27, 31, 21, 32 32 – 21 = 11 21, 21, 22, 27, 31, 32
### What is the range? :
What is the range? 31, 8, 3, 11, 19 31 – 3 = 28 3, 8, 11, 19, 31
### What is the range? :
What is the range? 23, 7, 9, 41, 19 41 – 7 = 34 7, 9, 23, 19, 41 |
# A small farm has ducks and goats. There are 11 animals in all and 34 legs in all (wings don't count!). How many goats does the farm have? How many ducks?
5 ducks, 6 goats
#### Explanation:
We have some Goats and some Ducks. There are eleven animals in all:
$G + D = 11$
Goats have 4 legs and Ducks 2. There are 34 legs altogether:
$4 G + 2 D = 34$
I'm going solve the first equation for $G$ in terms of $D$, then substitute in:
$G = 11 - D$
$4 \left(11 - D\right) + 2 D = 34$
and now solve for $D$:
$44 - 4 D + 2 D = 34$
$2 D = 10$
color(blue)(ul(bar(abs(color(black)(D=5))))
which will mean that color(blue)(ul(bar(abs(color(black)(G=6)))):
$G + D = 11$
$G + 5 = 11 \implies G = 6$
And now let's check our leg count:
$4 G + 2 D = 34$
$4 \left(6\right) + 2 \left(5\right) = 34$
$24 + 10 = 34$
34=34color(white)(000)color(green)sqrt |
# How do you use the binomial series to expand (x+1)^20?
Nov 9, 2016
${\left(x + 1\right)}^{20} = {\sum}_{k = 0}^{20} \left(\begin{matrix}20 \\ k\end{matrix}\right) {x}^{20 - k}$
#### Explanation:
${\left(a + b\right)}^{n} = {\sum}_{k = 0}^{n} \left(\begin{matrix}n \\ k\end{matrix}\right) {a}^{n - k} {b}^{k}$
where ((n),(k)) = (n!)/((n-k)! k!)
Putting $a = x$, $b = 1$ and $n = 20$ we get:
${\left(x + 1\right)}^{20} = {\sum}_{k = 0}^{20} \left(\begin{matrix}20 \\ k\end{matrix}\right) {x}^{20 - k}$
$\textcolor{w h i t e}{{\left(x + 1\right)}^{20}} = \left(\begin{matrix}20 \\ 0\end{matrix}\right) {x}^{20} + \left(\begin{matrix}20 \\ 1\end{matrix}\right) {x}^{19} + \ldots + \left(\begin{matrix}20 \\ 20\end{matrix}\right)$
If you just want a single term from this expansion, then it is probably easiest to calculate directly.
For example, the fourth term is:
((20),(3)) x^17 = (20!)/(17! 3!) x^17 = (20xx19xx18)/(3xx2xx1) x^17 = 1140 x^17
If you want to write out every term, then it may be easier to write out Pascal's triangle as far as the $21$st row - hoping that you make no errors on the way. The $21$st row starts $1 , 20 , 190 , 1140 , \ldots$ consisting of the values:
$\left(\begin{matrix}20 \\ 0\end{matrix}\right) , \left(\begin{matrix}20 \\ 1\end{matrix}\right) , \ldots , \left(\begin{matrix}20 \\ 20\end{matrix}\right)$
conveniently providing all the coefficients you want. |
## Fraction Progress
My third grade students have been exploring fractions. For the past month, students have been delving deeper and constructing a better understanding of fractions. Last week, students cut out fraction area circles and matched them to find equivalent fraction pairs.
For the most part, students were able to match the fractions to observe equivalency. Afterwards, students discussed how to find equivalent fractions through different means. Some students made the connection between doubling the numerator and denominator, while others noticed that they could divide to find an equivalent fraction.
Early this week, students started to place fractions on number lines. They used the whiteboard and a Nearpod activity to become more accurate when identifying and labeling fractions on a line.
It was interesting to see how students showcased their understanding as the number line increased from 0-1 to 0-2, and beyond. Giving an option for students to decide which number to use seemed to encourage them to take a risk with showing their understanding.
On Wednesday, students started a fraction task related to computation. Students were asked to color each fraction bar, cut them out and organize the fraction pieces to complete given number sentences. Students had to rearrange the fraction pieces and found that there were leftover pieces, which makes this a more challenging task. You can find more information about this activity here.
This task took around a day to complete. Students struggled at first and they used a lot of trial-and-error. Students compared the fractions bars and switch the pieces around quite a bit before taping down the sum. A few students needed a second attempt to complete this.
On Friday, students used polygon blocks to show their understanding of fractions. Using polygon blocks, students were asked to take one block and label that as 1/4, 1/2, 1/8, or 1/12. They then combined at least three different blocks to find a sum of 3 1/2.
Students used whiteboards and geometry blocks to combine the fraction pieces. I observed students using different strategies to combine and then take away blocks to find the sum of 3 1/2.
Next week, students will investigate the relationship between fractions and decimals.
## Equivalent Fractions Tweak
A few days ago I started gathering resources to supplement a math unit on fractions. The classroom was studying equivalent fractions and I thought there might be a variety of resources available on a few of the blogs that I regularly visit. I generally follow the #mathchat hashtag and find/share ideas that relate to mathematics. While reading a few math blogs on fractions, I came across John Golden’s site that has some amazing ideas that can be used in math classroom. His triangle pattern template sparked my interest.
John provided a template that’s available on his site. I printed out the template and began filling out each triangle with fractions. I ended up with a sheet that looked like this.
So what happened?
First a lot of brainstorming and error checking. Then I decided to have students cut out the triangles and compile equivalent fractions. This is what happened …
Students in fourth grade cut out each triangle and combined them to make equivalent fraction squares. Students worked in collaborative pairs during the project. I observed students using math vocabulary and having constructive conversations with each other to finish the assignment.
Before giving the assignment to a fifth grade class I decided to eliminate two triangles on the sheet above. It was the job of the student to find what triangles were missing and create equivalent fractions to complete the squares. The students were engaged in this activity from start to finish. Some students even wrote the equivalent decimal next to each square.
Overall this project took approximately 45 minutes to complete and it was worth every minute. Students used the terms fraction, improper fraction, mixed number, numerator, denominator, multiplication, division, and pattern throughout the project.
Just as I did, feel free to tweak this project to best meet the needs of your students. |
The unit circle
Define using the coordinate system a circle whose centre is at the origin and whose radius is 1.
From the course 'Analytical Geometry', we remember that the equation of such a circle is
Oriented angles
Select point A on the circumference of the unit circle.
The angle between the radius drawn at point A and the x-axis is πΌ.
Form a right triangle such that the hypotenuse
is the radius of the circle.
The coordinates of point A are (0,8; 0,6). Since one vertex of the triangle is at the origin, the values of the coordinates of the circumferential point A are formed directly as the lengths of the catheters of the triangle. In this case, the lengths of the catheters are 0,8 and 0,6, and the length of the hypotenuse is 1.
Now we could find the angle πΌ using the trigonometric functions sine or cosine
That is, sin (36,9 Β°) = 0,6 and cos (36,9 Β°) = 0,8 We could indicate the coordinates of point A using the angle πΌ A (cos (36,9 Β°), sin (36,9 Β°)) .
Let (x, y) be the coordinates of the circumferential point A. Then the angle πΌ holds
Example 2
Find the coordinates of the circumferential points A, B and C.
The angle corresponding to point A is 30 Β°
The angle corresponding to point B is 130 Β°
The angle corresponding to point C is 240 Β°
Based on the previous example
A (cos (30 Β°), sin (30 Β°)), B (cos (130 Β°), sin (130 Β°)), C (cos (240 Β°), sin (240 Β°))
We calculate the coordinates to one decimal place.
A (0,9;0,5)
B (-0,6;0,8)
C (-0,5;-0,9)
Properties of sine and cosine
It has been defined above that sine and cosine are the coordinates of the circumference point on a unit circle. The values of sine and cosine thus vary between [-1,1]. Sine and cosine get all the values between [-1,1].
For all the values of the angle πΌ
The sign of sine and cosine is determined by which quarter of the coordinate system the perimeter point is located. Sine is the y-coordinate and cosine is the x-coordinate.
Sine
Cosine
Example 3
At what angle sin (πΌ) = 0,5?
Since the y-coordinate is 0,5 at two circumferential points, there must be two angles with a sine value of 0,5. The circumferential points A and B with a y-coordinate of 0,5 are marked below.
An angle equal to the angle πΌ is formed between the negative x-axis and the line segment drawn at point B. We could find this using similarity. Equal angles are marked in the figure below.
So it has to be that
We solve the angle at which the value of the sine is 0,5
At an angle of 30Β°, the value of sine is 0,5. According to the above, also at an angle of 180Β° - 30Β° = 150Β°, the sine should have a value of 0,5.
Example 4
What angle is πΌ
The cosine is the x-coordinate of the circumferential point and the two circumferential points have an x-coordinate value equal to the requested value.
This forms a corresponding triangle below the x-axis so angle πΌ is equal to the angle π½. The angle rotating clockwise is denoted as negative, so π½ = -πΌ.
Then between 0Β° and 360Β° the cosine would have the same values with
angles πΌ and 360Β° - πΌ
Coterminal angles and supplementary angles
Coterminal angles
Supplementary angle
Periodicity
If a multiple of the full angle 360Β° is added to the angle, the circumference point remains unchanged. That is, sine and cosine always has the same value every 360Β°.
where n is an integer. This is called the periodicity of sine and cosine.
Pythagorean trigonometric identity
The radius drawn from the origin to the circumferential point of the unit circle forms a right triangle in which the radius is the hypotenuse. The catheters are x and y. In this case, Pythagorean theorem states
This is also the equation of an origin-centred circle with a radius of 1.
Since cos (πΌ) = x and sin (πΌ) = y, we get
This is Pythagorean trigonometric identity
A full angle is 360Β° and the circumference of the unit circle is 2π
. Let us introduce the unit of an angle, the radian, where 2π
rad = 360 Β°.
All of the above properties of sine and cosine also apply when the unit of an angle is a radian.
Example 5
From [0,2π
], find the angles that have the same value as
The period of sine and cosine is 2π
, that is, every 2π
sine and cosine has the same value. In addition to this, sine has the same value at supplement angles and cosine at coterminal angles.
a) sin (π
/ 4) = sin (π
-π
/ 4), the angle at which the sine has the same value as the angle π
/ 4
b) cos (2π
/ 5) = cos (-2π
/ 5) and according to the periodicity cos (-2π
/ 5) = cos (-2π
/ 5 + 2π
), the angle at which the cosine has the same value as the angle 2π
/ 5 is
c) π
-π
/ 2 = π
/ 2, the angle and the supplement angle are the same. π
/ 2 radians is 90 Β°. Here the value of the sine is 1 and between [0,2π
] the sine gets the value 1 only at this angle.
Example 6
Find the value of the cosine when we know that
We use Pythagorean trigonometric identity and substitute sine
The angle is between [π
/2, π
], so the cosine is negative. So the answer is
Connection between sine and cosine
In the right triangle
This is true for all angles πΌ. By radians
Tangent
The tangent of the oriented angle is sine divided by cosine. The divisor cannot be zero, so the tangent is not defined by angles with a cosine value of 0. The cosine is zero by the angles π
/ 2 and -π
/ 2 and their multiples. The tangent is thus defined when
Period of the tangent is π
Turn on the subtitles if needed |
Review question
# Where do two tangents cross, and where do two normals cross? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R5703
## Solution
The curve $C$ has equation $xy=\dfrac{1}{2}$. The tangents to the curve $C$ at the distinct points $P(p,\dfrac{1}{2p})$ and $Q(q,\dfrac{1}{2q})$, where $p$ and $q$ are positive, intersect at $T$ and the normals to $C$ at these points intersect at $N$. Show that $T$ is the point $\left(\dfrac{2pq}{p+q},\dfrac{1}{p+q}\right).$
We have the points $P(p,\dfrac{1}{2p})$ and $Q(q,\dfrac{1}{2q})$ where, without loss of generality, $0< p< q$ as in the diagram. We can rearrange the equation of the curve to $y = \dfrac{1}{2x}$ and then differentiate to obtain
$\frac{dy}{dx} = -\frac{1}{2x^2}.$
At the point $P$, with coordinates $(p, \dfrac{1}{2p})$, the gradient is therefore $-\dfrac{1}{2p^2}$. Thus the equation of the tangent through $P$ is $\left(y-\frac{1}{2p}\right)=-\frac{1}{2p^2}(x-p),$ that is, $y=\frac{1}{p}-\frac{1}{2p^2}x.$ Similarly, the equation of the tangent through $Q$ is $y=\dfrac{1}{q}-\dfrac{1}{2q^2}x.$
The two tangents intersect when $\frac{1}{p}-\frac{1}{2p^2}x=\frac{1}{q}-\frac{1}{2q^2}x.$ i.e. when $x\left(\frac{1}{2q^2}-\frac{1}{2p^2}\right) = \frac{1}{q}-\frac{1}{p},$ We can simplify the left-hand bracket, using the difference of two squares, to give $\frac{1}{2q^2}-\frac{1}{2p^2}=\frac{1}{2} \left(\left(\frac{1}{q}\right)^2 -\left(\frac{1}{p}\right)^2\right) = \frac{1}{2}\left(\frac{1}{q}-\frac{1}{p}\right)\left(\frac{1}{q} + \frac{1}{p}\right) = \frac{1}{2}\left(\frac{1}{q}-\frac{1}{p}\right)\left(\frac{p+q}{pq} \right).$ Then the equation involving the $x$-coordinate of $T$ simplifies to $\frac{1}{2} x \left(\frac{1}{q}-\frac{1}{p}\right)\left(\frac{p+q}{pq} \right) = \left(\frac{1}{q}-\frac{1}{p}\right)$ and so, dividing through by the bracketed expression on both sides (which is not zero since $p \neq q$) and making $x$ the subject, we see that $x = \frac{2pq}{p+q}.$ The $y$-coordinate of $T$ is then \begin{align*} y&=\dfrac{1}{p}-\dfrac{1}{2p^2}\times\dfrac{2pq}{p+q}\\ &=\dfrac{1}{p}\left(1-\dfrac{q}{p+q}\right)=\dfrac{1}{p+q}, \end{align*}
as required.
In the case $pq=\dfrac{1}{2}$, find the coordinates of $N$.
It would be helpful to find the equations of the normals to the curve at $P$ and $Q$.
We know that (gradient of the normal) $\times$ (gradient of the tangent) at any point on the curve is $-1$.
We can therefore use what we know about the gradients of the tangents to the curve at $P$ and $Q$.
The gradients of the normals through $P$ and $Q$ are $2p^2$ and $2q^2$, respectively. The equation of the normal through $P$ is therefore $\left(y-\frac{1}{2p}\right)=2p^2(x-p),$ which we can rearrange to give $y=2p^2x+\frac{1}{2p}-2p^3.$ Similarly, the normal through $Q$ has equation $y = 2q^2x + \frac{1}{2q} - 2q^3.$
The two normals intersect at $N$, whose $x$ coordinate satisfies $2p^2x+\frac{1}{2p}-2p^3=2q^2x+\frac{1}{2q}-2q^3,$ that is, \begin{align*} 2(p-q)(p+q)x &= \dfrac{1}{2}\left(\dfrac{1}{q}-\dfrac{1}{p}\right)+2(p^3-q^3)\\ &= \dfrac{1}{2}\left(\dfrac{p-q}{pq}\right)+2(p-q)(p^2+pq+q^2) \end{align*}
and so $x = \frac{1}{p+q}\left(\frac{1}{4pq}+p^2+pq+q^2\right). \qquad (p\neq q)$
Using $pq=\dfrac{1}{2}$, this becomes \begin{align*} x&=\dfrac{1}{p+q}\left(\dfrac{(pq)^2}{pq}+p^2+pq+q^2\right)\\ &=\dfrac{1}{p+q}\left(p^2+2pq+q^2\right)\\ &=\dfrac{1}{p+q}(p+q)^2\\ &=p+q. \end{align*}
Substituting back into the equation of the normal through $P$ we find $y=2p^2(p+q)+\frac{1}{2p}-2p^3=2p^2q+\frac{1}{2p}.$ Again, using $pq=\dfrac{1}{2}$ we find $y=2p\left(\frac{1}{2}\right)+q=p+q.$ Thus $N$ has coordinates $(p+q,p+q)$ when $pq=\dfrac{1}{2}$.
Show (in this case) that $T$ and $N$ lie on the line $y=x$
When $pq=\dfrac{1}{2}$, $T$ has coordinates $\left(\dfrac{1}{p+q},\dfrac{1}{p+q}\right)$, so lies on $y=x$. And we have just shown that $N$ has coordinates $(p+q,p+q)$, which also lies in the line $y=x$.
Alternatively, when $pq = \dfrac{1}{2}$, then $P =\left(p, \dfrac{1}{2p}\right)$ and $Q=\left(q, \dfrac{1}{2q}\right) = \left(\dfrac{1}{2p}, p\right)$, which means by symmetry that $T$ and $N$ lie on $y = x$.
…and are such that the product of their distances from the origin is constant.
We need a way to find the distance of a point from the origin. We can use Pythagoras’s theorem to do this.
The point with coordinates $(a,b)$ is at distance $\sqrt{a^2 + b^2}$ from the origin.
The distance of $T$ from the origin is $\vert OT \vert = \sqrt{\frac{2}{(p+q)^2}}=\frac{\sqrt{2}}{p+q}.$ Similarly, $\vert ON \vert = \sqrt{2(p+q)^2}=\sqrt{2}(p+q).$ Thus the product of their distances from the origin is $\vert OT \vert \times \vert ON \vert = \frac{\sqrt{2}}{p+q} \cdot \sqrt{2}(p+q)=2,$ which is a constant. |
# Adding & Subtracting Fractions with Unlike Denominators (Centers)
12 teachers like this lesson
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## Objective
SWBAT add and subtract fractions with unlike denominators.
#### Big Idea
During choice time, students have an opportunity to select the activities that will help to develop their mathematical fluency.
## Introduction
5 minutes
This lesson is designed to help students increase fluency in the skills we have been learning throughout our unit on adding and subtracting fractions. The next topic is about adding and subtracting mixed numbers. Since this is the last instructional day before the assessment, I provide students with choices of problem activities to practice these skills.
I start with a quick introduction to the lesson by sharing the agenda for the day.
• Must Do
• Choice Time
• Conferences
I explain that throughout the class, students will all complete a must do, then they may make choices about how to spend the rest of the time. Throughout the class, I pull students to conference with me about their recent work with fractions.
## Launch
10 minutes
To introduce the choices, I have students come up and play a demonstration round of each game. I also provide the students with a hand out that includes the rules and expectations of each game.
The choice time activities are:
Problem Solving: Showcase What You Know
1. Make a diagram and write an equation to match the story
2. Solve it
3. Include a picture to show what is happening in the problem.
First to 12 Wholes
The focus of this game is for students to practice using benchmark fractions to estimate.
• Player A flips 4 fraction cards.
• Player B estimates the size of each fraction using benchmark fractions. (Explaining his/her thinking out loud to the other player).
• Player B then estimates the sum of the benchmark fractions and records it.
• Player B flips 4 fraction cards and the cycle continues.
• The first player to get to 12 is the winner.
Possible extensions include incorporating subtraction of benchmark fractions, or try estimating the sum of more than 2 fractions.
Fraction War
The focus of this choice to to have students compare fractions.
Draw a giant fraction bar on a sheet of note book paper (like the model on the board)
Divide a set of fraction cards in half.
Flip 2 number cards to create a fraction. The first card flipped is the numerator, the second is the denominator
Determine which player has the largest fraction (use number sense or create common denominators). Each player must explain their way of knowing which was the largest.
If two players have equivalent fractions, play "war".
It is important to model the rules of the game to make sure students are clear of the expectations. With in the structure of each choice, I allow room for students make some of their own agreements with in their choice activities like how to deal with wild cards.
Define Expectations:
Because there will be a combination of group work, independent work, conferences, and partners working on various activities, it is important to make an agreement of expectations for the lesson.
I ask students to share the expectations they have for themselves and their classmates during the lesson. As each student shares, the others promise to meet these expectations.
## Conferences
5 minutes
Throughout the independent practice, I meet with students one-to-one to discuss their recent work with problem solving.
Students previously solved addition and subtraction word problems that involved fractions with unlike denominators.
When I conference with students I ask three major questions:
1. How are you feeling about your work with fractions?
2. Do you have any questions for me?
3. Will you read the story problem that you solved that you are most proud of, and then explain how you solved it. If there were any parts that you got stuck, can you talk about them?
While students explain their thought process, I listen for and provide prompts to assess three criteria.
• Vocabulary Use
• Ability to explain that when making equivalent fractions the numerator and denominator are both multiplied by the same number because n/n is 1 whole. Multiplying by 1 whole doesn't change the value of a fraction.
• A bar diagram is a part-part-whole model, but the "whole" refers to the entire amount of whatever the problem is about, not a whole number. It is appropriate to have a fraction in the "whole" position of a bar diagram.
## Independent Practice
20 minutes
First, students work on the "must do". This independent activity is used to assess a skill learned in a previous topic.
Must do:
1) .67 x .35
2) 32 divided by 1.5
3) 3.2 divided by 15
4) .32 divided by .15
I choose the same digits in the dividend and divisor so students have to focus on the place value and decimal manipulation, and reasonableness of their answers.
Then students move to their choices, while I meet with one student at a time for conferences.
## Group Share
5 minutes
It may seem unconventional to have a group share at the end of a lesson when so many students were working on different things. However, I find that students are more articulate when they have to explain a unique experience rather than describe an event that was shared by everyone.
The group share prompts for this lesson are:
• What choice did you make today?
• What did you learn while you were working?
I record all recap statements that students share. Moving forward, these statements will be revisited to improve student thinking, address misconceptions, and solidify understandings. |
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Area under the Curve Calculator
Top
Area under a curve in general is the area coming under the curve. It can be calculated with respect to coordinates as the area above the axis minus area below the axis.
If function f(x) is given, then area under the curve between the given limits x = a and x = b is given as:
Area under curve = $\int_{a}^{b}$ f(x) dx.
Area under curve calculator (known as area under a curve calculator even as area under graph calculator) is a online tool to get the area under the curve hence called as area under curve calculator online that finds area under curve. You just have to enter the value of function, lower limit and upper limit and get the answer instantly.
You can see a default function with its lower and upper limit given below. When you click on "Calculate Area", the area under the curve is calculated by integrating the function between the two points.
Steps for Area under the Curve Calculator
Step 1 :
Read the problem and observe the given curve function and the limits.
Step 2 :
Use the formula:
Area of the curve = $\int_{a}^{b}$ f(x) dx.
Substitute the given function and the limits and integrate the function to get the answer.
Problems on Area under the Curve Calculator
1. Find the area under the curve y = 5 - x2 and x-axis where x = -3 and x = 3.
Step 1 :
given: The function is y = 5 - x2
Limits are a = -3 and b = 3.
Step 2 :
Area under the curve = $\int_{a}^{b}$ f(x) dx.
Substituting the values of limits in f(x)
$\int_{-3}^{3} (5 - x^{2})$ dx.
= (5x - $\frac{x^{3}}{3})_{-3}^{3}$
= (5(3) - $\frac{3^{3}}{3}$) - (5(-3) - $\frac{(-3)^{3}}{3}$)
= (15 - 9) - (-15 + 9).
= 15 - 9 + 15 -9
= 12.
$\therefore$ The area under the curve y = 5 - x2 = 12 units.
2. Find the area under the curve y = sin x where the point lies between a = $\pi$ and b = 2 $\pi$.
Step 1 :
given: The given function is f(x) = y = sin x
The limits are a = $\pi$ and b = 2 $\pi$.
Step 2 :
Area under the curve = $\int_{a}^{b}$ f(x) dx where a and b are lower limits.
Substituting the values of limits in f(x)
$\int_{\pi}^{2 \pi}$ sin x dx.
= - $(cos x)_{\pi}^{2 \pi}$
= -[cos $2\pi$ - cos $\pi$]
= -[1 - (-1)]
= -2 |
# Excercise 2.5 Linear Equations in One Variable- NCERT Solutions Class 8
## Chapter 2 Ex.2.5 Question 1
Solve the linear equation \begin{align}\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}\end{align}
### Solution
What is known?
Equation
What is unknown?
Value of the variable
Reasoning:
Multiple both sides by the L.C.M of the denominators to get rid of fractional number. Now transpose variables to one side and constant to another side.
Steps:
\begin{align}\frac{x}{2} - \frac{1}{5} = \frac{x}{3} + \frac{1}{4}\end{align}
LCM of the denominators, $$2, 3, 4,$$ and $$5$$, is $$60.$$
Multiplying both sides by $$60,$$ we obtain
\begin{align}60\left( {\frac{x}{2} - \frac{1}{5}} \right) = 60\left( {\frac{x}{3} + \frac{1}{4}} \right)\end{align}
Opening the brackets, we get,
\begin{align}30x - 12 &= 20x + 15 \\30x - 20x &= 15 + 12\\10x &= 27\quad \quad \\\,\,\,\,\,x &= \frac{{27}}{{10}}\end{align}
## Chapter 2 Ex.2.5 Question 2
Solve the linear equation \begin{align}\frac{n}{2} - \frac{{3n}}{4} + \frac{{5n}}{6} = 21\end{align}
### Solution
What is known?
Equation
What is unknown?
Value of the variable
Reasoning:
Multiple both sides by the L.C.M of the denominators to get rid of fractional number. Now transpose variables to one side and constant to another side.
Steps:
$\frac{n}{2}-\frac{3n}{4}+\frac{5n}{6}=21$
LCM of the denominators, $$2, 4,$$ and $$6,$$ is $$12.$$
Multiplying both sides by $$12,$$ we obtain
\begin{align}6n - 9n + 10n &= 2527\\7n &= 252 \\n &= \frac{{252}}{7}\\n &= 36 \\\end{align}
## Chapter 2 Ex.2.5 Question 3
Solve the linear equation \begin{align}x + 7 - \frac{{8x}}{3} = \frac{{17}}{6} - \frac{{5x}}{2}\end{align}
### Solution
What is known?
Equation
What is unknown?
Value of the variable
Reasoning:
Multiple both sides by the L.C.M of the denominators to get rid of fractional number. Now transpose variables to one side and constant to another side.
Steps:
\begin{align}x + 7 - \frac{{8x}}{3} = \frac{{17}}{6} - \frac{{5x}}{2}\end{align}
LCM of the denominators,$$2, 3,$$ and $$6,$$ is $$6.$$
Multiplying both sides by $$6,$$ we obtain
\begin{align}6x + 42 - 16x &= 17 - 15x \\6x - 16x + 15x &= 17 - 42 \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5x &= - 25 \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x &= \frac{{ - 25}}{5} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x &= - 5 \\\end{align}
## Chapter 2 Ex.2.5 Question 4
Solve the linear equation \begin{align}\frac{{x - 5}}{3} = \frac{x - 3}{5}\end{align}
### Solution
What is known?
Equation
What is unknown?
Value of the variable
Reasoning:
Multiple both sides by the L.C.M of the denominatorss to get rid of fractional number. Now transpose variables to one side and constant to another side.
Steps:
\begin{align}\frac{{x - 5}}{3} = \frac{{x - 3}}{5}\end{align}
LCM of the denominators, $$3$$ and $$5,$$ is $$15.$$
Multiplying both sides by $$15,$$ we obtain
\begin{align}5\left( {x - 5} \right)&= 3\left( {x - 3} \right) \end{align}
Opening the brackets we get,
\begin{align}5x - 25 &= 3x - 9 \\5x - 3x &= 25 - 9 \\2x &= 16 \\\,\,x& = \frac{{16}}{2} \\\,\,x &= 8 \\\end{align}
## Chapter 2 Ex.2.5 Question 5
Solve the linear equation\begin{align}\frac{{3t - 2}}{4} - \frac{{2t + 3}}{3} = \frac{2}{3} - t\end{align}
### Solution
What is known?
Equation
What is unknown?
Value of the variable
Reasoning:
Multiple both sides by the L.C.M of the denominators to get rid of fractional number. Now transpose variables to one side and constant to another side.
Steps:
\begin{align}\frac{{3t - 2}}{4} - \frac{{2t + 3}}{3} = \frac{2}{3} - t\end{align}
LCM of the denominators, $$3$$ and $$4,$$ is $$12.$$
Multiplying both sides by $$12,$$ we obtain
\begin{align}3\left( {3t \! - \! 2} \right) \! - \! 4\left( {2t \! + \! 3} \right) &= \! 8 \! - \! 12t \\9t \! - \! 6 \! - \! 8t \! - \! 12 &= \! 8 \! - \! 12t\, \\ \quad\quad\quad\text{(Opening }& \text{the brackets)}\\9t \! - \! 8t \! + \! 12t &= \! 8 \! + \! 6 \! + \! 12 \\13t &= \! 26 \\t &= \! \frac{{26}}{{13}} \\t &= \! 2 \\\end{align}
## Chapter 2 Ex.2.5 Question 6
Solve the linear equation \begin{align}m - \frac{{m - 1}}{2} = 1 - \frac{{m - 2}}{3}\end{align}
### Solution
What is known?
Equation
What is unknown?
Value of the variable
Reasoning:
Multiple both sides by the L.C.M of the denominators to get rid of fractional number. Now transpose variables to one side and constant to another side.
Steps:
LCM of the denominators, $$2$$ and $$3,$$ is $$6.$$
Multiplying both sides by $$6,$$ we obtain
\begin{align}6m \! - \! 3\left( {m\! -\! 1} \right) & \! = \! 6 \!-\! 2\left( {m \! - \! 2} \right)\\6m \! - \! 3m \! + \! 3 & \! = \! 6 \! - \! 2m \! + \! 4 \\\quad\quad\quad\text{(Opening }& \text{the brackets)}\\6m\! -\! 3m \! + \! 2m & \! = \! 6 \! + \! 4 \! - \! 3\\5m & \! = \! 7\\\,\,m & \! = \! \frac{7}{5}\end{align}
## Chapter 2 Ex.2.5 Question 7
Simplify and solve the linear equation $$3(t - 3) = 5(2t + 1)$$
### Solution
What is known?
Equation
What is unknown?
Value of the variable
Reasoning:
First open the brackets then transpose variable to one side and constant to another side.
Steps:
\begin{align}3\left( {t \! - \! 3} \right) & \! = \! 5\left( {2t \! + \! 1} \right) \\3t \! - \! 9 & \! = \! 10t \! + \! 5 \\\quad\quad\quad\text{(Opening }& \text{the brackets)}\\- 9 \! - \! 5 & \! = \! 10t \! - \! 3t \\ - 14 & \! = \! 7t \\t & \! = \! \frac{{ - 14}}{7} \\t & \! = \! - 2 \\\end{align}
## Chapter 2 Ex.2.5 Question 8
Simplify and solve the linear equation \begin{align}15(y \! - \! 4) \! - \! 2(y \! - \! 9) \! + \! 5(y \! + \! 6) \!= \! 0\end{align}
### Solution
What is known?
Equation
What is unknown?
Value of the variable
Reasoning:
First open the brackets then transpose variable to one side and constant to another side.
Steps:
\begin{align}15\left( {y \! - \! 4} \right) \! - \! 2\left( {y \! - \! 9} \right) \! + \! 5\left( {y \! + \! 6} \right)& \! = \! 0 \\15y \! - \! 60 \! - \! 2y \! + \! 18 \! + \! 5y \! + \! 30 & \! = \! 0 \\\qquad\text{(Opening } \text{the brackets)}\\ \\18y \! - \! 12 & \! = \! 0 \\18y & \! = \! 12 \\y & \! = \! \frac{{12}}{{18}} \\ & \! = \! \frac{2}{3} \end{align}
## Chapter 2 Ex.2.5 Question 9
Simplify and solve the linear equation \begin{align}3(5z \!-\! 7) \!-\! 2(9z - 11) \!=\! 4(8z \!-\! 13) \!-\! 17\end{align}
### Solution
What is known?
Equation
What is unknown?
Value of the variable
Reasoning:
First open the brackets then transpose variable to one side and constant to another side.
Steps:
\begin{align}3(5z\! -\!7)\!-\!2(9z\!-\!11)&\!=\!4(8z\!-\!13)\!-\!17\\15z - 21 - 18z + 22& \!= \!32z - 52 - 17 \\\qquad\qquad\qquad\text{(Opening }& \text{the brackets)}\\\\- 3z + 1 &= 32z - 69\\ - 3z - 32z& = - 69 - 1\\ - 35z &= - 70\\z& = \frac{{70}}{{35}}\\z\, &= 2\\\end{align}
## Chapter 2 Ex.2.5 Question 10
Simplify and solve the linear equation \begin{align}0.25(4f - 3) = 0.05(10f - 9)\end{align}
### Solution
What is known?
Equation
What is unknown?
Value of the variable
Reasoning:
First open the brackets then transpose variable to one side and constant to another side.
Steps:
\begin{align}0.25\left( {4f - 3} \right)&= 0.05\left( {10f - 9} \right)\\\frac{1}{4}\left( {4f - 3} \right) &= \frac{1}{{20}}\left( {10f - 9} \right)\end{align}
Multiplying both sides by $$20$$ we obtain
\begin{align}5\left( {4f - 3} \right) &= 10f - 9\\20f - 15 &= 10f - 9\,\\\qquad\qquad\qquad\text{(Opening }& \text{the brackets)}\\\\20f - 10f &= - 9 + 15\\\quad \,\,\,\,\,\,\,10f &= 6\\\quad \,\,\,\,\,\,\,\,\,\,\,\,f &= \frac{3}{5}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, &= 0.6\end{align}
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# NCERT solutions for Class 10 Mathematics Textbook chapter 10 - Circles [Latest edition]
Course
Textbook page
## Chapter 10: Circles
Ex. 10.1Ex. 10.2
#### NCERT solutions for Class 10 Mathematics Textbook Chapter 10 Circles Exercise 10.1 [Page 209]
Ex. 10.1 | Q 1 | Page 209
How many tangents can a circle have?
Ex. 10.1 | Q 2.1 | Page 209
Fill in the blank:
A tangent to a circle intersects it in _______ point (s).
Ex. 10.1 | Q 2.2 | Page 209
Fill in the blank:
A line intersecting a circle in two points is called a __________.
Ex. 10.1 | Q 2.3 | Page 209
Fill in the blank:
A circle can have __________ parallel tangents at the most.
Ex. 10.1 | Q 2.4 | Page 209
Fill in the blank:
The common point of a tangent to a circle and the circle is called ____
Ex. 10.1 | Q 3 | Page 209
A tangent PQ at a point P of a circle of radius 5 cm meets a line through the centre O at a point Q so that OQ = 12 cm. Length PQ is :
• 12 cm.
• 13 cm
• 8.5 cm
• sqrt119 cm test
Ex. 10.1 | Q 4 | Page 209
Draw a circle and two lines parallel to a given line such that one is a tangent and the other, a secant to the circle.
#### NCERT solutions for Class 10 Mathematics Textbook Chapter 10 Circles Exercise 10.2 [Pages 213 - 214]
Ex. 10.2 | Q 1 | Page 213
From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is
• 7 cm
• 12 cm
• 15 cm
• 24.5 cm
Ex. 10.2 | Q 2 | Page 213
In the given figure, if TP and TQ are the two tangents to a circle with centre O so that ∠POQ = 110°, then ∠PTQ is equal to
• 60°
• 70°
• 80°
• 90°
Ex. 10.2 | Q 3 | Page 213
If tangents PA and PB from a point P to a circle with centre O are inclined to each other an angle of 80°, then ∠POA is equal to
• 50°
• 60°
• 70°
• 80°
Ex. 10.2 | Q 4 | Page 214
Prove that the tangents drawn at the ends of a diameter of a circle are parallel.
Ex. 10.2 | Q 5 | Page 214
Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre
Ex. 10.2 | Q 6 | Page 214
The length of a tangent from a point A at distance 5 cm from the centre of the circle is 4 cm. Find the radius of the circle.
Ex. 10.2 | Q 7 | Page 214
Two concentric circles are of radii 5 cm and 3 cm. Find the length of the chord of the larger circle which touches the smaller circle
Ex. 10.2 | Q 8 | Page 214
In the following Fig, a quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points P, Q, R and S respectively. Prove that AB + CD = BC + DA.
Ex. 10.2 | Q 9 | Page 214
In the given figure, XY and X’Y’ are two parallel tangents to a circle with centre O and another tangent AB with point of contact C intersecting XY at A and X’Y’ at B. Prove that ∠AOB=90°
Ex. 10.2 | Q 10 | Page 214
Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segments joining the pointsof contact to the centre.
Ex. 10.2 | Q 11 | Page 214
Prove that the paralleogram circumscribing a circle, is a rhombus
Ex. 10.2 | Q 12 | Page 214
A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see given figure). Find the sides AB and AC.
Ex. 10.2 | Q 13 | Page 214
Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Ex. 10.1Ex. 10.2
## NCERT solutions for Class 10 Mathematics Textbook chapter 10 - Circles
NCERT solutions for Class 10 Mathematics Textbook chapter 10 (Circles) include all questions with solution and detail explanation. This will clear students doubts about any question and improve application skills while preparing for board exams. The detailed, step-by-step solutions will help you understand the concepts better and clear your confusions, if any. Shaalaa.com has the CBSE Class 10 Mathematics Textbook solutions in a manner that help students grasp basic concepts better and faster.
Further, we at Shaalaa.com provide such solutions so that students can prepare for written exams. NCERT textbook solutions can be a core help for self-study and acts as a perfect self-help guidance for students.
Concepts covered in Class 10 Mathematics Textbook chapter 10 Circles are Circles Examples and Solutions, Introduction to Circles, Tangent to a Circle, Number of Tangents from a Point on a Circle, Circles Examples and Solutions, Introduction to Circles, Tangent to a Circle, Number of Tangents from a Point on a Circle.
Using NCERT Class 10 solutions Circles exercise by students are an easy way to prepare for the exams, as they involve solutions arranged chapter-wise also page wise. The questions involved in NCERT Solutions are important questions that can be asked in the final exam. Maximum students of CBSE Class 10 prefer NCERT Textbook Solutions to score more in exam.
Get the free view of chapter 10 Circles Class 10 extra questions for Class 10 Mathematics Textbook and can use Shaalaa.com to keep it handy for your exam preparation
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# Part 1: Introduction to matrices
A matrix is an array of elements. The elements we will see in matrices will usually be numbers or algebraic expressions. An $$m \times n$$ matrix has $$m$$ rows and $$n$$ columns. In some books, you will find matrices written in square brackets [also known as box brackets], but here we will use round brackets (also known as parentheses). Matrices are denoted by bold, capital letters e.g. A.
The order of a matrix tells you how many rows and columns it has. Therefore, $$\begin{pmatrix} 5 & 2 & 4 \\ 1 & 8 & 2 \\\end{pmatrix}$$ is simply a $$2 \times 3$$ matrix.
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# Part 2: Adding and subtracting matrices
You can only add or subtract matrices A and B if they are the same order. We can work out A + B by simply adding the corresponding elements of the matrices. We can work out A − B by subtracting each element of B from the corresponding element of A.
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# Part 3: Multiplying matrices
### Multiplying a matrix by a scalar
To multiply a matrix by a scalar (a single number or algebraic expression of a number), simply multiply each element in the matrix by the scalar. Note that this fits in with our general understanding of multiplication as repeated addition.
<
### Multiplying a matrix by a matrix
Before getting into the detail of multiplying a matrix by another matrix, we’ll take a look at a simple situation to help illustrate the principle behind matrix multiplication:
A football team scores 3 points for a winning a match, 1 point for drawing, and 0 points for losing. Suppose Alton play 11 games, winning 5, drawing 2, and losing 4. They would score $$5 \times 3 + 2 \times 1 + 4 \times 0 = 17$$ points. We can represent this as a matrix multiplication as follows:
$$\begin{pmatrix} 5 & 2 & 4 \\ \end{pmatrix} \times \begin{pmatrix} 3\\ 1\\ 0 \end{pmatrix} = \begin{pmatrix} 17 \end{pmatrix}$$
Suppose Belton also play 11 games, but they win 1, draw 8, and lose 2. They would score $$1 \times 3 + 8 \times 1 + 2 \times 0 = 11$$ points. We can represent both teams’ results and points scores as a matrix multiplication like this:
$$\begin{pmatrix} 5 & 2 & 4 \\ 1 & 8 & 2 \\\end{pmatrix} \times \begin{pmatrix} 3\\ 1\\ 0 \end{pmatrix} = \begin{pmatrix} 17 \\ 11 \\ \end{pmatrix}$$
Now suppose we wanted to see how the total points scored by each time would differ if 4 points were awarded for a win. We can use the following matrix multiplication:
$$\begin{pmatrix} 5 & 2 & 4 \\ 1 & 8 & 2 \\\end{pmatrix} \times \begin{pmatrix} 3 & 4 \\ 1 & 1\\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 17 & 22 \\ 11 & 12 \\ \end{pmatrix}$$
The activity below walks you through each of the above examples, step-by-step. It then allows you to generate random practice questions. First note the following important facts about matrix multiplication:
• It is not always possible to multiply matrices together. It is only possible to find A $$\times$$ B if the number of columns in A is equal to the number of rows in B. In other words, it is only possible to multiply an $$m \times n$$ matrix by an $$n \times p$$ matrix (where $$m$$ and $$p$$ need not be equal. The result will be an $$m \times p$$ matrix.
• In general, matrix multiplication is not commutative. That is, AB is not always equal to BA.
• Indeed, for the reason mentioned above, it may not even be possible to work out BA even though AB exists. For example, a $$1 \times 2$$ matrix multiplied by a $$2 \times 3$$ matrix will result in a $$1 \times 3$$ matrix, but it is not even possible to multiply a $$2 \times 3$$ matrix by a $$1 \times 2$$ matrix (because $$1$$ ≠ $$3$$).
• Matrix multiplication is associative, however. In other words, A(BC) = (AB)C.
<
### Matrices and index notation
We can use index notation with matrices to indicate repeated multiplication. As you might expect:
• A2 = A $$\times$$ A
• A3 = A $$\times$$A $$\times$$ A
and so on.
<
### The identity matrix
Square matrices in the following form are known as identity matrices:
$$\begin{pmatrix} 1 & 0\\ 0 & 1 \\\end{pmatrix}$$, $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$$, $$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$, etc.
It is often clear from context what size matrix we are dealing with, which is why you will often see references to the identity matrix. We denote the identity matrix using the letter I.
Identity matrices are very special. Given any square matrix A, you will find that:
• I $$\times$$ A = A and
• A $$\times$$ I = A,
where I is the identity matrix that is the same order (i.e. size) as A.
You can think of I as the matrix analogue of the number 1. The number 1 is the multiplicative identity: when you multiply any number $$n$$ by 1, your result is $$n$$ i.e. it is unchanged.. Similarly, when you multiply any suitable matrix M, by I (whether before or after), your result is M, i.e. it is unchanged. Note however that M cannot be any old matrix; it must be a square matrix the same order as I.
Activity: Consider the matrix M $$\begin{pmatrix} a & b\\ c & d \\\end{pmatrix}$$, and verify that I $$\times$$ M = M and that M $$\times$$ I = M.
# Part 4: 2 × 2 Matrices and linear transformations
A 2 × 2 matrix can be used to apply a linear transformation to points on a Cartesian grid. A linear transformation in two dimensions has the following properties:
• The origin (0,0) is mapped to the origin (it is invariant) under the transformation
• Straight lines are mapped to straight lines under the transformation
• Parallel lines remain parallel under the transformation
Questions
Which of the following are linear transformations?
1. Translation by any non-zero vector
2. Rotation about the origin by any angle
3. Rotation about point P, by any angle greater than 0º but less than 360º, where P is not (0,0)
4. Reflection in the y-axis
5. Reflection in the line = 0
6. Reflection in the line y = mx where m is a constant
7. Reflection in the line y = mx + c where m and c are constants and c is non-zero
8. Enlargement by any non-zero scale factor, centre of enlargement (0,0)
9. Enlargement by any non-zero scale factor, centre of enlargement P, where P is not (0,0)
10. Enlargement by scale factor 0, centre of enlargement (0,0)
Answers
1. Translation by any non-zero vector is NOT a linear transformation because the origin is not mapped to itself.
2. Rotation about the origin by any angle is a linear transformation.
3. Rotation about point P, by any angle greater than 0º but less than 360º, where P is not (0,0)
4. Reflection in the y-axis is a linear transformation.
5. Reflection in the line = 0 is a linear transformation.
6. Reflection in the line y = mx where m is a constant is a linear transformation.
7. Reflection in the line y = mx + c where m and c are constants and c is non-zero is NOT a linear transformation because the origin is not mapped to itself.
8. Enlargement by any non-zero scale factor, centre of enlargement (0,0) is a linear transformation.
9. Enlargement by any non-zero scale factor, centre of enlargement P, where P is not (0,0) is NOT a linear transformation because the origin is not mapped to itself.
10. Enlargement by scale factor 0, centre of enlargement (0,0) is NOT a linear transformation because straight lines aren’t mapped to straight lines; in fact every point on the grid is mapped to (0,0).
### The effect of a 2 × 2 transformation matrix
To find where the matrix M $$\begin{pmatrix} a & b\\c & d\end{pmatrix}$$ maps the point Q with coordinates $$(x, y)$$, we multiply the matrix M by the position vector representation of Q:
i.e. we do $$\begin{pmatrix} a & b\\c & d\end{pmatrix} \begin{pmatrix} x\\y\end{pmatrix} = \begin{pmatrix} x’\\y’\end{pmatrix}$$, and Q is mapped to $$(x’, y’)$$.
For example, the matrix $$\begin{pmatrix} 2 & 1\\-1 & 3\end{pmatrix}$$ maps $$(1, 1)$$ to $$\begin{pmatrix} 2 & 1\\-1 & 3\end{pmatrix} \begin{pmatrix} 1\\1\end{pmatrix} = \begin{pmatrix} 3\\2\end{pmatrix}$$ or the point $$(3, 2)$$.
In the following applet, we will take a look at the effect of various transformations on the unit square OPQR:
Possible activities
• Click on “Custom” towards the top of the applet in order to apply custom transformations to the unit square. Drag the blue slider fully to the right and tick the box to show the basis vectors. Then vary $$a$$ and see the impact this has on the basis vectors. Then try varying $$b$$, $$c$$, and $$d$$ (one at a time) to see the impact of varying these.
• Demonstrate how the columns of the transformation matrix correspond to the transformations of two sides of the unit square given.
• Drag the blue slider fully to the left. Tick the boxes to show the basis vectors and to transform the gridlines too. Now drag the blue slider to the right. Note that on the transformed grid, the coordinates of the transformed shape are still at (0,0), (1,0), (1,1) and (0,1). The basis vectors in terms of the untransformed grid are however given by $$\begin{pmatrix} a\\c \end{pmatrix}$$ and $$\begin{pmatrix} b\\d \end{pmatrix}$$. This can be seen by unticking the “Transform gridlines too” box while the blue slider is fully dragged to the right.
### Deducing transformation matrices for common transformations
In the applet above, the point P has position vector $$\begin{pmatrix} 1\\0\end{pmatrix}$$ and the point R has position vector $$\begin{pmatrix} 0\\1\end{pmatrix}$$. The transformation matrix $$\begin{pmatrix} a & b\\c & d\end{pmatrix}$$ maps P to $$\begin{pmatrix} a\\c\end{pmatrix}$$ and R to P to $$\begin{pmatrix} b\\d\end{pmatrix}$$.
You can verify these by working out $$\begin{pmatrix} a & b\\c & d\end{pmatrix} \times \begin{pmatrix} 1\\0\end{pmatrix}$$ and $$\begin{pmatrix} a & b\\c & d\end{pmatrix} \times \begin{pmatrix} 0\\1\end{pmatrix}$$respectively.
By visualising the unit square—in particular how a transformation affects the points P and Q—we can work backwards to quickly deduce the matrices representing many common transformations. For example, a rotation 90º anticlockwise about $$(0,0)$$ maps P to P’, with position vector $$\begin{pmatrix} 0\\1\end{pmatrix}$$, and it maps R to R’ with position vector $$\begin{pmatrix} -1\\0\end{pmatrix}$$. Therefore, the matrix representing this transformation is $$\begin{pmatrix} 0 & -1\\1 & 0\end{pmatrix}$$.
### Summary of transformation matrices that you should learn or be able to deduce quickly
Reflection in the $$x$$-axis: $$\begin{pmatrix} 1 & 0\\0 & -1\end{pmatrix}$$
Reflection in the $$y$$-axis: $$\begin{pmatrix} -1 & 0\\0 & 1\end{pmatrix}$$
Reflection in the $$y=x$$: $$\begin{pmatrix} 0 & 1\\1 & 0\end{pmatrix}$$
Reflection in the $$y=-x$$: $$\begin{pmatrix} 0 & -1\\-1 & 0\end{pmatrix}$$
Enlargement by scale factor $$k$$, centre at $$(0,0)$$: $$\begin{pmatrix} k & 0\\0 & k\end{pmatrix}$$
Rotation 90º anticlockwise about $$(0,0)$$: $$\begin{pmatrix} 0 & -1\\1 & 0\end{pmatrix}$$
Rotation 180º $$(0,0)$$: $$\begin{pmatrix} -1 & 0\\0 & -1\end{pmatrix}$$
Rotation 270º anticlockwise about $$(0,0)$$: $$\begin{pmatrix} 0 & 1\\-1 & 0\end{pmatrix}$$
Rotation $$\theta$$º anticlockwise about $$(0,0)$$: $$\begin{pmatrix} \text{cos} \theta & -\text{sin} \theta\\ \text{sin} \theta & \text{cos} \theta \end{pmatrix}$$
Shear in the $$x$$-direction, shear factor $$k$$: $$\begin{pmatrix} 1 & k\\0 & 1\end{pmatrix}$$
Shear in the $$y$$-direction, shear factor $$k$$: $$\begin{pmatrix} 1 & 0\\k & 1\end{pmatrix}$$
# Part 5: Determinants of 2 × 2 matrices
### Calculating the determinant
The determinant of a 2 × 2 matrix M is written det M or |M|.
For a 2 × 2 matrix $$\begin{pmatrix} a & b\\c & d\end{pmatrix}$$, the determinant can be written det$$\begin{pmatrix} a & b\\c & d\end{pmatrix}$$ or $$\begin{vmatrix} a & b\\c & d\end{vmatrix}$$ and is simply equal to $$ad – bc$$.
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Note: Determinants can only be found for square matrices. There is a general method for working out the determinant of an $$n$$ × $$n$$ matrix, described in Part 9 below. At that stage, you can check that the general method applied to a 2 × 2 matrix gives you the determinant $$ad – bc$$.
### What the determinant represents
The absolute value of the determinant of a 2 × 2 matrix M is equal to the area scale factor by which M transforms the areas of shapes. In particular, consider the parallelogram obtained by transforming the unit square. The unit square has area 1, so the parallelogram will have an area of |M|.
If the determinant is negative, it simply indicates a change of orientation. The vertices of the unit square are O, P, Q, and R going anticlockwise. If the vertices of the image O’, P’, Q’, and R’ also run anticlockwise, then the determinant is positive. If these vertices run clockwise i.e. the orientation has changed, this means that the determinant is negative.
• A matrix with determinant of 0 is called a singular matrix.
• Singular matrices map the unit square to a line—or to (0,0) in the case of the zero matrix. In these cases, the transformed “parallelograms” therefore have area 0.
• A matrix with a non-zero determinant is called non-singular.
• # Part 6: Inverses of 2 × 2 matrices
Given two matrices A and B, if AB = I, the identity matrix, then B is the inverse of A. We can denote the inverse of A as A-1 i.e. B = A-1.
A square matrix M has an inverse, denoted M-1, if and only if its |M| ≠ 0. If the determinant of M is 0, then M has no inverse.
Given a matrix M and its inverse, M-1, the following will be true:
• MM-1 = I
• M-1M = I
The inverse of a 2 × 2 matrix $$\textsf{M}$$ = $$\begin{pmatrix} a & b\\c & d\end{pmatrix}$$ can be found (where it exists) as follows:
$$\textsf{M}^{-1}$$ = $$\dfrac{1}{\begin{vmatrix} \textsf{M} \end{vmatrix}} \begin{pmatrix} d & -b\\-c & a\end{pmatrix}$$ = $$\dfrac{1}{ad-bc} \begin{pmatrix} d & -b\\-c & a\end{pmatrix}$$.
Extension 1: Verify that MM-1 = I.
Extension 2: Verify that M-1M = I.
Extension 3: Find the determinant of M-1.
Questions
1. Can you see why (at least for a 2 $$\times$$ 2 matrix), that an inverse can’t exist if the matrix has determinant 0?
2. Show that $$|\textsf{M}^{-1}| = \dfrac{1}{\textsf{|M|}}$$.
Answers
1. $$\textsf{M}^{-1} = \dfrac{1}{\begin{vmatrix} \textsf{M} \end{vmatrix}} \begin{pmatrix} d & -b\\-c & a\end{pmatrix}$$, so if $$\textsf{|M|} = 0$$, the inverse would be $$\dfrac{1}{0} \begin{pmatrix} d & -b\\-c & a\end{pmatrix}$$, which is a problem because $$\dfrac{1}{0}$$ is not defined.
2. If $$\textsf{M}$$ = $$\begin{pmatrix} a & b\\c & d\end{pmatrix}$$, then $$\textsf{M}^{-1}$$ = $$\dfrac{1}{\begin{vmatrix} \textsf{M} \end{vmatrix}} \begin{pmatrix} d & -b\\-c & a\end{pmatrix} = \begin{pmatrix} \dfrac{d}{\textsf{|M|}} & \dfrac{-b}{\textsf{|M|}}\\\dfrac{-c}{\textsf{|M|}} & \dfrac{a}{\textsf{|M|}}\end{pmatrix}$$.
So $$|\textsf{M}^{-1}| = \dfrac{ad}{\textsf{|M|}^{2}} – \dfrac{(-b)(-c)}{\textsf{|M|}^{2}}=\dfrac{ad-bc}{\textsf{|M|}^{2}}=\dfrac{\textsf{|M|}}{\textsf{|M|}^{2}}=\dfrac{1}{\textsf{|M|}}$$.
# Part 7: Invariant points and lines in 2 dimensions
An invariant point under a transformation is a point that maps to itself. As noted in part 4, linear transformations map the origin to the origin, so the origin is always an invariant point under a linear transformation.
An invariant line is a line that maps to itself. To be precise, every point on the invariant line maps to a point on the line itself. Note that the point needn’t map to itself.
A a line of invariant points is a line where every point every point on the line maps to itself. Any line of invariant points is therefore an invariant line, but an invariant line is not necessarily always a line of invariant points.
Use this applet to see invariant points, invariant lines, and lines of invariant points for three examples of linear transformations.
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# Part 8: 3 × 3 matrices and linear transformations
3 × 3 matrices can be used to apply transformations in 3D, just as we used 2 × 2 matrices in 2D. To find where the matrix M $$\begin{pmatrix} a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{pmatrix}$$ maps the point Q with coordinates $$(x, y, z)$$, we multiply the matrix M by the position vector representation of Q:
i.e. we do $$\begin{pmatrix} a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{pmatrix} \begin{pmatrix} x\\y\\z\end{pmatrix} = \begin{pmatrix} x’\\y’\\z’\end{pmatrix}$$, and Q is mapped to $$(x’, y’,z’)$$.
For example, the matrix $$\begin{pmatrix} 2 & 1 & 0\\-1 & 3 & 0\\0 & 0 & 4\end{pmatrix}$$ maps $$(1, 1, 1)$$ to $$\begin{pmatrix} 2 & 1 & 0\\-1 & 3 & 0\\0 & 0 & 4\end{pmatrix} \begin{pmatrix} 1\\1\\1\end{pmatrix} = \begin{pmatrix} 3\\2\\4\end{pmatrix}$$ or the point $$(3, 2, 4)$$.
In the following applet, we will take a look at the effect of various transformations on the unit cube:
### Deducing transformation matrices for common transformations
The transformation matrix $$\begin{pmatrix} a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{pmatrix}$$ maps $$\begin{pmatrix} 1\\0\\0\end{pmatrix}$$ to $$\begin{pmatrix} a_{11}\\a_{21}\\a_{31}\end{pmatrix}$$, $$\begin{pmatrix} 0\\1\\0\end{pmatrix}$$ to $$\begin{pmatrix} a_{12}\\a_{22}\\a_{32}\end{pmatrix}$$, and $$\begin{pmatrix} 0\\0\\1\end{pmatrix}$$ to $$\begin{pmatrix} a_{13}\\a_{23}\\a_{33}\end{pmatrix}$$.
You can verify these by working out $$\begin{pmatrix} a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{pmatrix} \times \begin{pmatrix} 1\\0\\0\end{pmatrix}$$, $$\begin{pmatrix} a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{pmatrix} \times \begin{pmatrix} 0\\1\\0\end{pmatrix}$$, $$\begin{pmatrix} a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{pmatrix} \times \begin{pmatrix} 0\\0\\1\end{pmatrix}$$ and respectively.
By visualising the unit cube—in particular how a transformation affects the points with position vectors $$\begin{pmatrix} 1\\0\\0\end{pmatrix}$$, $$\begin{pmatrix} 0\\1\\0\end{pmatrix}$$, and $$\begin{pmatrix} 0\\0\\1\end{pmatrix}$$—we can work backwards to quickly deduce the matrices representing many common transformations. For example, a rotation 90º anticlockwise about the $$z$$-axis maps $$\begin{pmatrix} 1\\0\\0\end{pmatrix}$$ to $$\begin{pmatrix} 0\\1\\0\end{pmatrix}$$, $$\begin{pmatrix} 0\\1\\0\end{pmatrix}$$ to $$\begin{pmatrix} -1\\0\\0\end{pmatrix}$$, and $$\begin{pmatrix} 0\\0\\1\end{pmatrix}$$ to itself. Therefore, the matrix representing this transformation is $$\begin{pmatrix} 0 & -1 & 0\\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$.
### Summary of transformation matrices that you should learn or be able to deduce quickly
Reflection in $$x=0$$ (the $$y$$-$$z$$-plane): $$\begin{pmatrix} 1 & 0\\0 & -1\end{pmatrix}$$
Reflection in $$y=0$$ (the $$x$$-$$z$$-plane): $$\begin{pmatrix} -1 & 0\\0 & 1\end{pmatrix}$$
Reflection in $$z=0$$ (the $$x$$-$$y$$-plane): $$\begin{pmatrix} 0 & 1\\1 & 0\end{pmatrix}$$
Enlargement by scale factor $$k$$, centre at $$(0,0,0)$$: $$\begin{pmatrix} k & 0\\0 & k\end{pmatrix}$$
Rotation $$\theta$$º anticlockwise about the $$(x$$-axis: $$\begin{pmatrix} 1 & 0 & 0\\ 0 & \text{cos} \theta & -\text{sin} \theta\\ 0 & \text{sin} \theta & \text{cos} \theta \end{pmatrix}$$
Rotation $$\theta$$º anticlockwise about the $$(y$$-axis: $$\begin{pmatrix} \text{cos} \theta & 0& \text{sin} \theta\\ 0 & 1 & 0\\ -\text{sin} \theta & 0 & \text{cos} \theta \end{pmatrix}$$
Rotation $$\theta$$º anticlockwise about the $$(z$$-axis: $$\begin{pmatrix} \text{cos} \theta & -\text{sin} \theta & 0\\ \text{sin} \theta & \text{cos} \theta & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
# Part 9: Determinants of 3 × 3 matrices
### Minors and cofactors
Before we can find the determinant of a 3 × 3 (or larger) square matrix, we need to learn some new terminology.
Each element of a square matrix has a minor. The minor of the element is found by removing the row and column containing that element, and calculating the determinant of the remaining matrix.
Each element of a square matrix also has a cofactor. The cofactor of the element is either equal to its minor multiplied by 1, or its minor multiplied by -1. In other words, it is either the minor itself or the minor with its sign changed. We decide whether to keep or change the sign as follws: the cofactor of the element in the $$i\text{th}$$ row and $$j\text{th}$$ column is the minor of the element multiplied by $$(-1)^{i+j}$$. In other words, if $$i+j$$ is even, we keep the minor’s sign the same, and if $$i+j$$ is odd, we change its sign. For a 3 × 3 matrix, the following shows how each element’s cofactor is related to its minor:
$$\begin{pmatrix} \text{keep} & \text{change} & \text{keep}\\ \text{change} & \text{keep} & \text{change} \\ \text{keep} & \text{change} & \text{keep} \end{pmatrix}$$
This applet guides you through the process, step-by-step:
### Finding the determinant
To find the determinant of a matrix, we need to find the cofactors of all of the elements in just one row or column This means that we can find the determinant of a 3 × 3 matrix without needing to find the cofactors of all nine elements.
To find the determinant, pick a row or column. For that row or column, multiply each element by its cofactor and note the product, and finally add these products.
Whichever row or column you choose, you should get the same answer, but it will save time if you choose a row or column with zeroes in it, if possible. If you have a 0 element, then when you multiply this by its cofactor, you will get 0. This means that you can save time by not bothering to find that element’s cofactor.
Use this applet to practise finding the determinant. To start with, it would be good to repeat each question using a different row or column to verify that you always get the same answer.
### What the determinant represents
In part 5, we saw that the determinant of a 2 × 2 matrix M is equal to the area scale factor by which M transforms the areas of shapes. The determinant of a 3 × 3 matrix M is equal to the volume scale factor by which M transforms the volume of shapes. (We can also extend this idea to higher dimensions, though it is very hard to visualise object (let alone transformations of such objects) beyond the third dimension!)
Remember, a matrix with determinant is zero is called a singular matrix. Singular 3 × 3 matrices map the unit cube to either a plane, or a line or to the point (0,0,0) in the case of a zero matrix. A matrix with non-zero determinant is called non-singular.
# Part 10: Inverses of 3 × 3 matrices
To find the inverse, M-1, of a 3 × 3 matrix M (if M-1 exists), we first need to find the cofactor matrix of M, which is the matrix made up of the 9 cofactors of each element of M. We first came across cofactors in part 9.
We also need to be able to find the transpose of a matrix. We can obtain the transpose of a matrix by writing its rows as its columns and vice versa. This is equivalent to reflecting its elements along its diagonal (from top-left to bottom-right). Here is an example:
If A $$= \begin{pmatrix} 4 & 5 & -7\\ 2 & -3 & 0 \\ 1 & -6 & 8 \\ \end{pmatrix}$$, the the transpose of A, denoted AT is $$\begin{pmatrix} 4 & 2 & 1\\ 5 & -3 & -6 \\ -7 & 0 & 8 \\ \end{pmatrix}$$.
If M has cofactor matrix C and is non-singular, then M-1$$=\frac{1}{\text{det }\textbf{M}}$$CT. Use this applet to practise finding the inverse of 3 × 3 matrices.
# Part 11: Matrices and simultaneous equations
### Solving simultaneous equations
We can use our knowledge of matrix multiplication and inverse matrices to solve simultaneous equations. For example, consider this pair of simultaneous equations:
$$4x + 5y = 37\\2x + 3y = 19$$
These can be rewritten as a product of a $$\color{red}{\text{coefficient matrix}}$$ and a $$\color{green}{\text{column vector}}$$ $$\color{red}{ \begin{pmatrix} 4 & 5\\ 2 & 3 \\\end{pmatrix}}\color{green}{\begin{pmatrix} x\\ y \\\end{pmatrix}}=\begin{pmatrix} 37\\ 19 \\\end{pmatrix}$$
Remember, solving the pair of simultaneous equations in the above case means finding the values of $$x$$ and $$y$$ that satisfy the equations. We can do this by left-multiplying both sides of the matrix equation by the inverse of $$\color{red}{\begin{pmatrix} 4 & 5\\ 2 & 3 \\\end{pmatrix}}$$, which is $$\color{blue}{\frac{1}{2}\begin{pmatrix} 3 & -5\\ -2 & 4 \\\end{pmatrix}}$$:
$$\color{blue}{\frac{1}{2}\begin{pmatrix} 3 & -5\\ -2 & 4 \\\end{pmatrix}}\color{red}{\begin{pmatrix} 4 & 5\\ 2 & 3 \\\end{pmatrix}}\color{green}{\begin{pmatrix} x\\ y \\\end{pmatrix}}=\color{blue}{ \frac{1}{2}\begin{pmatrix} 3 & -5\\ -2 & 4 \\\end{pmatrix}} \begin{pmatrix} 37\\ 19 \\\end{pmatrix}$$
Since the red and blue are inverses (and therefore multiply to give us the identity matrix), this simplifies to:
$$\color{green}{\begin{pmatrix} x\\ y \\\end{pmatrix}}=\color{blue}{ \frac{1}{2}\begin{pmatrix} 3 & -5\\ -2 & 4 \\\end{pmatrix}} \begin{pmatrix} 37\\ 19 \\\end{pmatrix}$$
Multiplying the right-hand side, we find:
$$\color{green}{\begin{pmatrix} x\\ y \\\end{pmatrix}}=\begin{pmatrix} 8\\ 1 \\\end{pmatrix}$$
### Graphical visualisations
In part 6, we were introduced to the idea that not all matrices have inverses. For the matrix M to have an inverse, we need |M| ≠ 0 i.e. we need M to be non-singular. When the coefficient matrix is non-singular, we can find a unique solution to the set of linear simultaneous equations. A non-singular $$2 \times 2$$ matrix corresponds to a pair of equations of lines that intersect at exactly one point, with this point defining the solution:
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A non-singular $$3 \times 3$$ matrix corresponds to a set of three planes that intersect at exactly one point as shown. It might be easier to appreciate how three planes can intersect at this point by starting with just one plane (the red one), and then introducing the next two planes one at a time:
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Note that you can click and drag in the applet to rotate your view.
#### What if the determinant is 0?
If the coefficient matrix has a determinant of 0, this will either be because the simultaneous equations have no solutions or an infinite number solutions. When there are no solutions, we say that the set of simultaneous equations is inconsistent. (Note that when there are infinite solutions, the equations are consistent, so a coefficient matrix with a determinant of 0 does not necessarily imply that the set of simultaneous equations is inconsistent.) The following applets graphically illustrate the scenarios (in 2D and 3d) in which a coefficient matrix can have a determinant of 0:
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##### 3$$\times$$3 matrices
Click each scenario to see a graphical illustration. Note that you can click and drag in the applet to rotate your view.
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# Evaluate the integrals \int e^{t}\cos (3e^{t}-2)dt
Evaluate the integrals $$\displaystyle\int{e}^{{{t}}}{\cos{{\left({3}{e}^{{{t}}}-{2}\right)}}}{\left.{d}{t}\right.}$$
• Questions are typically answered in as fast as 30 minutes
### Plainmath recommends
• Get a detailed answer even on the hardest topics.
• Ask an expert for a step-by-step guidance to learn to do it yourself.
Nicole Keller
Step 1
We have to evaluate the integral:
$$\displaystyle\int{e}^{{{t}}}{\cos{{\left({3}{e}^{{{t}}}-{2}\right)}}}{\left.{d}{t}\right.}$$
In this case we should use substitution method since derivatives of one function is in the integral.
So let $$\displaystyle{x}={3}{e}^{{{t}}}-{2}$$
differentiating both sides with respect to 't', we get
$$\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={\frac{{{d}{\left({3}{e}^{{{t}}}-{2}\right)}}}{{{\left.{d}{t}\right.}}}}$$
$$\displaystyle={3}{\frac{{{d}{e}^{{{t}}}}}{{{\left.{d}{t}\right.}}}}-{\frac{{{d}{2}}}{{{\left.{d}{t}\right.}}}}$$
$$\displaystyle={3}{e}^{{{t}}}-{0}$$
$$\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{t}\right.}}}}={3}{e}^{{{t}}}$$
$$\displaystyle{\frac{{{\left.{d}{x}\right.}}}{{{3}}}}={e}^{{{t}}}{\left.{d}{t}\right.}$$
Step 2
Substituting above values, we get
$$\displaystyle\int{e}^{{{t}}}{\cos{{\left({3}{e}^{{{t}}}-{2}\right)}}}{\left.{d}{t}\right.}=\int{\cos{{\left({3}{e}^{{{t}}}-{2}\right)}}}{e}^{{{t}}}{\left.{d}{t}\right.}$$
$$\displaystyle={\frac{{{1}}}{{{3}}}}\int{\cos{{x}}}{\left.{d}{x}\right.}$$
$$\displaystyle={\frac{{{1}}}{{{3}}}}{\left({\sin{{x}}}\right)}+{C}$$
Where, C is an arbitrary constant.
Hence, value of given integral is $$\displaystyle{\frac{{{1}}}{{{3}}}}{\sin{{x}}}+{C}$$.
###### Have a similar question?
Prioned
Step 1: Use Integration by Substitution.
Let $$\displaystyle{u}={3}{e}^{{{t}}}-{2},{d}{u}={3}{e}^{{{t}}}{\left.{d}{t}\right.},\ {t}{h}{e}{n}\ {e}^{{{t}}}{\left.{d}{t}\right.}={\frac{{{1}}}{{{3}}}}{d}{u}$$
Step 2: Using u and du above, rewrite $$\displaystyle\int{e}^{{{t}}}{\cos{{\left({3}{e}^{{{t}}}-{2}\right)}}}{\left.{d}{t}\right.}$$.
$$\displaystyle\int{\frac{{{\cos{{u}}}}}{{{3}}}}{d}{u}$$
Step 3: Use Constant Factor Rule: $$\displaystyle\int{c}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}$$.
$$\displaystyle{\frac{{{1}}}{{{3}}}}\int{\cos{{u}}}{d}{u}$$
Step 4: Use Trigonometric Integration: the integral of $$\displaystyle{\cos{{u}}}\ {i}{s}\ {\sin{{u}}}$$.
$$\displaystyle{\frac{{{\sin{{u}}}}}{{{3}}}}$$
Step 5: Substitute $$\displaystyle{u}={3}{e}^{{{t}}}-{2}$$ back into the original integral.
$$\displaystyle{\frac{{{\sin{{\left({3}{e}^{{{t}}}-{2}\right)}}}}}{{{3}}}}$$
$$\displaystyle{\frac{{{\sin{{\left({3}{e}^{{{t}}}-{2}\right)}}}}}{{{3}}}}+{C}$$ |
# Thread: Writing a Fraction in Simplest Form
1. After a week I thought I'd give this another try. I'm using Soroban's formula to reproduce the mathematical joke,
$\displaystyle \frac{1\!\!\!\not{6}}{\not{6}4} = \frac{1}{4}$
While incorrectly solved, it still gives the correct answer. Any hints are appreciated on where I'm going wrong.
Soroban's formula:
We want digits $\displaystyle a, b, c$ so that: $\displaystyle \frac{10a + b}{10b + c} \:= \:\frac{a}{c}$
Solve for $\displaystyle c:\;\;c\:=\:\frac{10ab}{9a + b}$
Step 1
$\displaystyle \frac{10a + b}{10b + c} = \frac{a}{c} = \frac{a}{c}$
$\displaystyle = \frac{10 \cdot 1 + 6}{10 \cdot 4 + 4} = \frac{16}{44}$
Now, $\displaystyle \frac{a = 16}{c = 44}$
Step 2
$\displaystyle c = \frac{10ab}{9a + b} = \frac{10 \cdot 16 \cdot 16}{9 \cdot 16 + 16} = \frac{960}{144}$
$\displaystyle = \frac{\not2 \cdot \not2 \cdot \not2 \cdot \not2 \cdot 2 \cdot 2 \cdot \not3 \cdot 5}{\not2 \cdot \not2 \cdot \not2 \cdot \not2 \cdot \not3 \cdot 3} = \frac{20}{3}$
2. Originally Posted by Euclid Alexandria
After a week I thought I'd give this another try. I'm using Soroban's formula to reproduce the mathematical joke,
$\displaystyle \frac{1\!\!\!\not{6}}{\not{6}4} = \frac{1}{4}$
While incorrectly solved, it still gives the correct answer. Any hints are appreciated on where I'm going wrong.
Soroban's formula:
We want digits $\displaystyle a, b, c$ so that: $\displaystyle \frac{10a + b}{10b + c} \:= \:\frac{a}{c}$
Solve for $\displaystyle c:\;\;c\:=\:\frac{10ab}{9a + b}$
Step 1
$\displaystyle \frac{10a + b}{10b + c} = \frac{a}{c} = \frac{a}{c}$
Error #1
$\displaystyle = \frac{10 \cdot 1 + 6}{10 \cdot 4 + 4} = \frac{16}{44}$
you say b=6 in the numerator, but you change that to b=4 in the denominator!
Now, $\displaystyle \frac{a = 16}{c = 44}$
Step 2
$\displaystyle c = \frac{10ab}{9a + b} = \frac{10 \cdot 16 \cdot 16}{9 \cdot 16 + 16} = \frac{960}{144}$
$\displaystyle = \frac{\not2 \cdot \not2 \cdot \not2 \cdot \not2 \cdot 2 \cdot 2 \cdot \not3 \cdot 5}{\not2 \cdot \not2 \cdot \not2 \cdot \not2 \cdot \not3 \cdot 3} = \frac{20}{3}$
This is not incorrect, however, to produce the mathematical joke you can't have a b or c more than a single-digit number.
3. ## The punchline rears its worn out head
Thanks, Quick. Your tip about the single digit number for c sent me in the right direction. Actually, I see numerous errors now. I also substituted 16 for b in my step 2. And steps 1 and 2 are themselves erroneous. I repeatedly followed Soroban's steps in the order that he presented them in, rather than the order that was intended to be followed.
To solve for c we use
$\displaystyle \frac{10ab}{9a+b}$
where a and b are any single digits we want to test. We want the result of the test to be a whole number, rather than a fraction. For instance, going through various combinations, if
a = 1 and b = 2
$\displaystyle \frac{10 \cdot 1 \cdot 2}{9 \cdot 1 + 2} = \frac{20}{11}$
or if a = 1 and b = 3
$\displaystyle \frac{10 \cdot 1 \cdot 3}{9 \cdot 1 + 3} = \frac{30}{12} = \frac{\not2 \cdot \not3 \cdot 5}{\not2 \cdot 2 \cdot \not3} = \frac{5}{2}$
or if a = 1 and b = 4
$\displaystyle \frac{10 \cdot 1 \cdot 4}{9 \cdot 1 + 4} = \frac{40}{13}$
and so on, then our formula fails the test. However, if a = 1 and b = 6, then
$\displaystyle c = \frac{10 \cdot 1 \cdot 6}{9 \cdot 1 + 6} = \frac{60}{15}$
$\displaystyle = \frac{2 \cdot 2 \cdot \not3 \cdot \not5}{\not3 \cdot \not5} = 4$
and our formula passes the test, because we have a single digit for c.
We then apply our solution for c to the formula
$\displaystyle \frac{10a+b}{10b+c} = \frac{a}{c}$
This formula results in the butt of the joke, because
$\displaystyle \frac{10 \cdot 1 + 6}{10 \cdot 6 + 4} = \frac{16}{64} = \frac{\not2 \cdot \not2 \cdot \not2 \cdot \not2}{\not2 \cdot \not2 \cdot \not2 \cdot \not2 \cdot 2 \cdot 2} = \frac{1}{4}$
Which means that
$\displaystyle \frac{1\not6}{\not64} = \frac{1}{4}$
might be an incorrect method of simplifying the fraction, but it certainly does not give the wrong answer.
Another fraction in Soroban's list,
$\displaystyle \frac{1\!\!\!\not{9}}{\not{9}5} = \frac{1}{5}$
might show an incorrect way of simplifying, but if we solve for c
$\displaystyle \frac{10 \cdot 1 \cdot 9}{9 \cdot 1 + 9} = \frac{90}{18}$
$\displaystyle = \frac{\not2 \cdot 5 \cdot \not3 \cdot \not3}{\not2 \cdot \not3 \cdot \not3} = 5$
and apply the formula
$\displaystyle \frac{10 \cdot 1 + 9}{10 \cdot 9 + 5} = \frac{19}{95}$
$\displaystyle \frac{19}{5 \cdot 19} = \frac{1}{5}$
And so on. Well I really drilled the humor out of the joke, but at least I get the joke now.
Don't all clap at once.
4. good job
(you did do some errors in the begining of your post )
Page 2 of 2 First 12 |
info@mathplay4kids.com
# Expanded Form Of Numbers Game 2nd Grade
Expanded Form of Numbers Game grade 2| what are expanded form of numbers ? | write numbers in expanded form | expanded versus standard form of numbers. This game page deals with numbers writen in their expanded form up one thousand, and the game questions would be related to selecting the standard form of that particular number among the suggested multiple choice answers.
Expanded form is a way of writing numbers that shows the place value of each digit. It is a useful tool for understanding the value of numbers and for doing math operations such as addition and subtraction. To write a number in expanded form, you start by writing the number as a sum of its digits, each one multiplied by its place value. For example, the number 356 can be written in expanded form as 300 + 50 + 6.
The digit 3 stands for 3 hundreds, the digit 5 stands for 50, and the digit 6 stands for 6 ones. Here are some more examples of numbers written in expanded form: 42,000 = 40,000 + 2,000 1,234 = 1,000 + 200 + 30 + 4 9,876 = 9,000 + 800 + 70 + 6 Expanded form can also be used with decimals. For example, the number 0.8 can be written in expanded form as 0.8 = 8 x 0.1. The digit 8 stands for 8 tenths, or 0.8.
Using expanded form can be helpful when doing math operations because it shows the value of each digit in the number. For example, to add the numbers 123 and 456 in expanded form, you can write them as 100 + 20 + 3 and 400 + 50 + 6 and then add them: (100 + 20 + 3) + (400 + 50 + 6) = 500 + 70 + 9 = 579 In this way, expanded form can help you understand the value of numbers and perform math operations more easily. You can also use expanded form as the basis for a math game. One idea is to give students a number and have them write it in expanded form. They can then compare their expanded form to a partner's or to the answer key to see if they got it correct.
You can also have students create their own numbers and write them in expanded form for their partners to solve. Another idea is to use expanded form to play a place value game. Give students a number and have them identify the place value of each digit. For example, in the number 1,234, the digit 1 stands for 1 thousand, the digit 2 stands for 200, the digit 3 stands for 30, and the digit 4 stands for 4 ones. Students can then use this knowledge to solve place value problems or to compare numbers based on their place value. Expanded form can also be used to teach students about rounding numbers.
For example, you can give students a number in expanded form and have them round it to the nearest ten, hundred, or thousand. For example, if the number is 1,234 in expanded form, students could round it to 1,000 or 1,300 depending on the rounding rules. Overall, expanded form is a useful tool for understanding numbers and for doing math operations. By using it as the basis for a math game, you can help students improve their understanding of place value and rounding, and have fun at the same time. |
Need solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 11
The correct option is (b)
Hint:
Use the given formula:
\begin{aligned} &\text { (i) } \lim _{x \rightarrow 0} f(x) g(x)=e \lim _{x \rightarrow 0}(f(x)-1) \cdot g(x)\\ &\text { Where } \lim _{x \rightarrow 0} f(x)=1 \end{aligned}
\begin{aligned} &\text { And } \lim _{x \rightarrow 0} g(x)=0\\ &\text { (ii) } \lim _{x \rightarrow 0}\left\{\frac{\cos x-1}{x}\right\}=-1 \end{aligned}
(iii) A function f(x) is said to be continuous at a point x = a of its domain, if
$\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a)$
Given:
$f(x)=\left\{\begin{array}{l} (\cos x)^{\frac{1}{x}} \\ k \end{array}\right.$ \begin{aligned} , x & \neq 0 \\ , x &=0 \end{aligned}
And f(x) is continuous at x = 0
Solution:
$\lim _{x \rightarrow 0} f(x)=f(0)$
Calculate the value of k
$\lim _{x \rightarrow 0}(\cos x)^{\frac{1}{x}}=k$
Using formula (i)
$\lim _{x \rightarrow 0} f(x)^{g(x)}=e \lim _{x \rightarrow 0}\left\{\frac{\cos x-1}{x}\right\}=k$
Apply formula (ii)
\begin{aligned} &e^{0}=k \\ & k=1 \end{aligned}
So, option (b) is correct. |
Derivatives of Logarithmic Functions Worksheet | Problems & Solutions
# Derivatives of Logarithmic Functions Worksheet
Derivatives of Logarithmic Functions Worksheet
• Page 1
1.
If $g$($l$) = ln (sec 52$l$ + tan 52$l$), then find $g$′ ($l$).
a. b. 52sec 52$l$ c. d. sec 52$l$ e.
#### Solution:
g(l) = ln (sec 52l + tan 52l)
[Write the function.]
g′ (l) = ddl( ln (sec 52l + tan 52l))
[Find g′ (l)]
= 1sec 52l+tan 52lddl(sec 52l+tan 52l)
[Use the Chain Rule.]
= 52sec 52ltan 52l+52sec252lsec 52l+tan 52l
= 52sec 52l(tan 52l+sec 52l)sec 52l+tan 52l = 52sec 52l
[Factor out 52sec 52l and simplify.]
g′ (l) = 52sec 52l
2.
If $g$($q$) = $q$10ln 8$q$, then find $g$′ ($q$).
a. b. 10 c. d. e.
#### Solution:
g(q) = q10ln 8q
[Write the function.]
g ′(q) = ddq(q10ln 8q)
[Find g ′(q).]
= q10ddq(ln 8q)+(ln 8q)ddq(q10)
[Use the Product Rule.]
= q10(1q)+(ln 8q)(10q9)
[Use the Chain Rule.]
= q9(1+10ln 8q)
g ′(q) = q9(1+10ln 8q)
3.
If ($x$) = $x$(ln 8$x$)11, then find ′ ($x$).
a. (ln 8$x$)10($\frac{11}{8}$ + (ln 8$x$)) b. (ln 8$x$)10($\frac{11}{8}$ - (ln 8$x$)) c. (ln 8$x$)10(11 + (ln 8$x$)) d. (10(ln 8$x$))(11 + (ln 8$x$)) e. (ln 8$x$)10(11 - (ln 8$x$))
#### Solution:
(x) = x(ln 8x)11
[Write the function.]
′ (x) = ddx(x(ln 8x)11)
[Find ′ (x)]
= (x)ddx((ln 8x)11)+(ln 8x)11ddx(x)
[Use the Product Rule.]
= 11x(ln 8x)10ddx(ln 8x)+(ln 8x)11(1)
[Use the Chain Rule.]
= 11x(ln 8x)10(1x)+(ln 8x)11
= (ln 8x)10(11 + (ln 8x))
′ (x) = (ln 8x)10(11 + (ln 8x))
4.
If ($k$) = 7(ln $\sqrt{9{k}^{11}+31}\right)$, then find ′ ($k$).
a. ($\frac{693}{2}$) ($\frac{{k}^{10}}{\sqrt{9{k}^{11}+31}}$) b. ($\frac{693}{2}$) ($\frac{{k}^{10}}{9{k}^{11}+31}$) c. ($\frac{{k}^{10}}{9{k}^{11}+31}$) d. 7 ln($\frac{99{k}^{10}}{2\sqrt{9{k}^{11}+31}}$) e. ()
#### Solution:
(k) = 7(ln 9k11+31)
[Write the function.]
′ (k) = ddk(7ln9k11+31)
[Find ′ (k).]
= 7(19k11+31)ddk(9k11+31)
[Use the Chain Rule.]
= 7(19k11+31)(129k11+31)ddk(9k11+31)
[Use the Chain Rule again.]
= 7(9)(11)k102(9k11+31) = (693 / 2) (k109k11+31)
′ (k) = (693 / 2) (k109k11+31)
5.
If $g$ ($x$) = ln ($x$ + $\sqrt{{x}^{2}+8}\right)$, then find $g$′ ($x$).
a. ln(1 + $\frac{x}{\sqrt{{x}^{2}+8}}$) b. $\frac{1}{x}$ + $\frac{x}{\sqrt{{x}^{2}+8}}$ c. $\frac{1}{\sqrt{{x}^{2}+8}}$ d. $\frac{1}{x}$ - $\frac{x}{\sqrt{{x}^{2}+8}}$ e. $\frac{1}{x-\sqrt{{x}^{2}+8}}$
#### Solution:
g(x) = ln (x + x2+8)
[Write the function.]
g′(x) = ddx(ln(x+x2+8))
[Find g′(x).]
= 1x+x2+8ddx(x+x2+8)
[Use the Chain Rule.]
= 1x+x2+8(1 + (2x)2x2+8)
[Use the Chain Rule again.]
= 1x2+8
g′(x) = 1x2+8
6.
If $g$($t$) = , then find $g$′ ($t$).
a. b. c. d. e. 22
#### Solution:
g(t) = (ln (2t+6))11
[Write the function.]
g′ (t) = ddt((ln (2t+6))11)
[Find g′ (t).]
= 11(ln (2t+6))10ddt(ln (2t+6))
[Use the General Power Rule and the Chain Rule.]
= 11(ln (2t+6))10(2(2t+6))
[Use the Chain Rule again.]
g′ (t) = 22(ln (2t+6))10(2t+6)
7.
If ($w$) = ln[cos (2$w$ + 7)], then find ′ ($w$).
a. - tan (2$w$ + 7) b. - 2tan (2$w$ + 7) c. tan (2$w$ + 7) d. 2tan (2$w$ + 7) e. - $\frac{1}{\mathrm{cos}\left(2w+7\right)}$
#### Solution:
(w) = ln[cos (2w + 7)]
[Write the function.]
′ (w) = ddw(ln[cos (2w+7)])
[Find ′ (w).]
= 1cos (2w+7)ddw(cos (2w+7))
[Use the Chain Rule.]
= - 2(sin (2w+7)cos (2w+7))
[Use the Chain Rule again.]
= - 2tan (2w + 7)
′ (w) = - 2tan (2w + 7)
8.
If $g$($u$) = , then find (1440)$g$′ (0).
a. b. 90(ln 10) - 96 c. 96 - 90(ln 10) d. 96 + 90(ln 10) e.
#### Solution:
g(u) = ln (8u+10)9u+12
[Write the function.]
g′ (u) = ddu(ln (8u+10)9u+12)
[Find g′ (u).]
= (9u+12)ddu(ln (8u+10))-(ln (8u+10))ddu(9u+12)(9u+12)2
[Use the Quotient Rule.]
= (9u+12)(88u+10)-(ln (8u+10))(9)(9u+12)2
= 8(9u+12)-(ln (8u+10))(9)(8u+10)(8u+10)(9u+12)2
g′ (u) = 8(9u+12)-9(8u+10)(ln (8u+10))(8u+10)(9u+12)2
g′ (0) = 96-90ln 101440
[Find g′ (0).]
(1440)g′ (0) = 96 - 90(ln 10)
9.
If ($l$) = (ln (9$l$ + 11))(ln (10$l$ + 12)), then find ′ (0).
a. $\left(\frac{1}{12}\right)$ + $\left(\frac{1}{11}\right)$ b. $\frac{1}{132}$ c. $\frac{5}{6}$(ln 11) + $\frac{9}{11}$(ln 12) d. + e. 10(ln 9) + 9(ln 10)
#### Solution:
(l) = (ln (9l + 11))(ln (10l + 12))
[Write the function.]
′ (l) = ddl((ln (9l + 11))(ln (10l + 12)))
[Find ′ (l).]
= ln (9l + 11)ddl(ln (10l + 12)) + ln (10l + 12)ddl(ln (9l + 11))
[Use the Product Rule.]
= (ln (9l + 11))(1010l+12) + (ln (10l + 12))(99l+11)
[Use the Chain Rule.]
= 10(ln (9l + 11))10l+12 + 9(ln (10l + 12))9l+11
′ (0) = 5 / 6(ln 11) + 9 / 11(ln 12)
[Find ′ (0).]
10.
If ($k$) = ln(ln (6$k$ + 7)), then find ′ ($k$).
a. b. c. d. $\frac{6}{\left(6k+7\right)}$ e.
#### Solution:
(k) = ln(ln (6k + 7))
[Write the function.]
′ (k) = ddk(ln(ln (6k+7)))
[Find ′ (k).]
= 1ln (6k+7)ddk(ln (6k+7))
[Use the Chain Rule.]
= 6(6k+7)(ln (6k+7))
′ (k) = 6(6k+7)(ln (6k+7)) |
# How do you integrate int x^3 /sqrt(4 - 2x^2) dx using trigonometric substitution?
Feb 7, 2016
$- \frac{3}{2} \sqrt{1 - {x}^{2} / 2} + \frac{1}{6} \sqrt{1 - {\left(\frac{3}{\sqrt{2}} x - \sqrt{2} {x}^{2}\right)}^{2}} + C$
#### Explanation:
Try the substitution: $x = \sqrt{2} \sin \left(u\right)$
This would mean $\mathrm{dx} = \sqrt{2} \cos \left(u\right) \mathrm{dx}$
Now put this into the integral and we get:
$\int \frac{{2}^{\frac{3}{2}} {\sin}^{3} \left(u\right)}{\sqrt{4 - 4 {\sin}^{2} \left(u\right)}} \sqrt{2} \cos \left(u\right) \mathrm{du}$
Tidying this up a little and using the trig -identity:
${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$, we get:
$\int \frac{4 {\sin}^{3} \left(u\right)}{2 \sqrt{1 - {\sin}^{2} \left(x\right)}} \cos \left(u\right) \mathrm{du}$
$= 2 \int {\sin}^{3} \frac{u}{\sqrt{{\cos}^{2} \left(u\right)}} \cos \left(u\right) \mathrm{du} = 4 \int {\sin}^{3} \left(u\right) \mathrm{du}$
At this point we need another trig identity . The trig identity that we will use is: ${\sin}^{3} x = \frac{3}{4} \sin x - \frac{1}{4} \sin \left(3 x\right)$.
Putting that into the integral we get:
$2 \int \frac{3}{4} \sin u - \frac{1}{4} \sin \left(3 u\right) \mathrm{dx}$
Which can now be integrated to obtain:
$- \frac{3}{2} \cos \left(u\right) + \frac{1}{6} \cos \left(3 u\right) + C$
We obviously have the challenge now of reversing the substituting. Again using the trig identity near the start, this formula can be re - arranged to give:
$- \frac{3}{2} \sqrt{1 - {\sin}^{2} \left(u\right)} + \frac{1}{6} \sqrt{1 - {\sin}^{2} \left(3 u\right)} + C$
At this point we can now rearrange the above trig identity to get:
$\sin \left(3 u\right) = 3 \sin \left(u\right) - 4 {\sin}^{3} \left(u\right)$
We can now rewrite the above expression as:
$- \frac{3}{2} \sqrt{1 - {\sin}^{2} \left(u\right)} + \frac{1}{6} \sqrt{1 - {\left(3 \sin \left(u\right) - 4 {\sin}^{3} \left(u\right)\right)}^{2}} + C$
And finally we can reverse the substitution to obtain:
$- \frac{3}{2} \sqrt{1 - {x}^{2} / 2} + \frac{1}{6} \sqrt{1 - {\left(\frac{3}{\sqrt{2}} x - \sqrt{2} {x}^{2}\right)}^{2}} + C$ |
Subject of a Formula: A formula is a rule or a fact written with mathematical symbols. In mathematics, students find many formulas for various concepts. The main objective of the formula is to solve the problem easily & quickly. We think you all may have an idea about establishing an equation if not, look at our previous article.
The subject of a formula is the part of Changing the subject of a formula concept. Students who are looking to learn about the subject of formula and how to change it can refer to this article without any delay.
It completely helps you to understand what is meant by the subject of the formula and how to make something the subject of a formula. So, jump into the below sections and gain proper knowledge.
## What is the Subject of a Formula?
A linear equation which is stated in variables and literals with mathematical operators is known as a formula. Hence, the variable that we need to calculate & discover as per the given context hints is called the Subject of the Formula.
For instance, let’s assume Newton’s law of motion as an equation ie., v²-u² = 2as
Here, v, u, a, and s are the variables such as final velocity, initial velocity, acceleration, and displacement of the particle respectively.
This equation can be rearranged as:
s = $$\frac { v²-u² } { 2a }$$, s become the subject of the formula.
OR
a = $$\frac { v²-u² }{ 2s }$$, a become the subject of the formula.
### How to make the subject of the formula?
The steps that are involved in making one variable as a subject of the formula are as given.
• First, Isolate the variable by adding or subtracting terms near to the variable, eliminating any fractions, dividing by the coefficient of the variable, and removing a root or power of both sides of the equation.
• Rearrange the equation and the subject variable should be on the left side of the equal sign.
• In case, the same subject variable contains in multiple terms, then perform factorization or any other related operation to make an end result as a single variable left as a subject of the formula on the left side of the equal sign.
### Subject of a Formula Questions | Examples on Change the Subject of the Formula
Example 1:
If z is the subject of a formula, it expresses the product of x and y. Find x as the subject of the formula?
Solution:
Given formula as per the context is z = xy
Now, we have to find out the x as a subject of the formula.
z = xy
Divide y on both sides ie., (÷)y
$$\frac { z} { y }$$ = $$\frac { xy } { y }$$
$$\frac { z } { y }$$ = x
Hence, x = $$\frac { z } { y }$$.
Example 2:
Change the subject of the equation in terms of p: z = x² + 2y +p
Solution:
Given equation is z = x² + 2y +p
We have to change the subject of an equation in terms of p:
z = x² + 2y +p
After applying the steps of how to subject the formula, we get the result as
p = z – x² – 2y
Example 3:
Make ‘x’ the subject of the formula, s = x + bt
Solutions:
Given expression is s = x + bt
x is added to the bt.
Firstly, subtract bt on both sides.
s – bt = x + bt – bt
s – bt = x, Hence, the subject x = s – bt.
### FAQs on Rearrange Formulas or Equations
1. How to find the subject of the formula in x = 6a + 2b?
The variable that expresses in other variables is known as the subject of a formula. In the given formula, the x variable is the subject of the formula.
2. What happens when a positive variable moves to the other side of an equal sign?
When the positive variable moves to the other side of the equal sign it changes to a negative variable or subtracts.
3. What is changing the subject of the formula?
Changing the subject of the formula is nothing but rearranging the formula and getting the different subject as per the given hints in the problem. |
Multiples of 10 are the commodities once an integer is multiplied by 10. Did you understand that 10 is the sum of the initially four consecutive entirety numbers, that is, 1 + 2 + 3 + 4 = 10, and also 10 is the amount of the first three prime numbers, that is, 2 + 3 + 5 = 10? Multiples of 10 are the easiest to memorize. In this mini-leskid, we will calculate the multiples of 10, and also interesting facts around these multiples.
You are watching: What are the multiples of 10
First 5 multiples of 10: 10, 20, 30, 40, 50Prime Factorization of 10: 10 = 2 × 5
Let us explore a tiny bit even more about the multiples of 10 and its properties of it.
1 What are the Multiples of 10? 2 First 20 Multiples of 10 3 Important Notes 4 FAQs on Multiples of 10
## What are the Multiples of 10?
Multiples of 10 are the numbers that can be separated by 10 without any remainder.
To develop a list of multiples of 10,
We first multiply 10 by 1 to acquire the initially multiple of 10 which is 10,Then we multiply 10 by 2 to acquire the second multiple of 10 which is 20,Then we multiply 10 by 3 to obtain the 3rd multiple of 10 which is 30, and also so on.
This list is never-ending, we can generate as many type of multiples of 10 as we desire.
The initially few multiples of 10 are obtained by the outcomes of the times table 10
10 × 1 = 10 10 × 2 = 20 10 × 3 = 30 10 × 4 = 40 10 × 5 = 50
## List of First 20 Multiples of 10
The multiples of 10 are the numbers that are precisely divisible by 10 via zero remainders.
The multiples of 10 are all even numbersThe multiples of 10 finish via 0
Here are the first 10 multiples of 10:
Can you attempt finding the next ten multiples of 10?
Yes, they are 110, 120, 130, 140, 150, 160, 170, 180, 190, 200
To understand also the idea of finding multiples, let us take a few more examples.
Important Notes:
When you multiply a number by 10 you will certainly get the number attached via zero as the outcome. eg: 10 × 2 = 20Multiples of 10 are even numbers that finish through 0If a number is finishing with 0 then it is constantly divisible by 10The prevalent number device is to the base 10
## Multiples of 10 Solved Examples
Example 1: Sam was provided a list of numbers.
Can you aid him find out multiples of 12, multiples of 10, multiples of 8?
Solution: In such cases, the moment table of 12, 10, 8 comes in handy.
Multiples of 12Multiples of 10Multiples of 8
1220, 30, 4032, 40
Example 2:
Harry hregarding deal with the following puzzle that was offered by a friend. Find the number which fits all these statements:
It is a common multiple of 10 and also 8It is between 1 and 50
Solution: List of multiples of 8 are 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, etc;Multiples of 10 are 10, 20, 30, 40, 50, 60, etc. Out of the numbers detailed over only 40 is lying between 0 and also 50 which ends with 0The common multiple of 10 and8, which lies in between 1 & 50 is 40Because of this, 40 is the solution to Harry"s puzzle.
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## Interenergetic Questions
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## FAQs on Multiples of 10
### How many kind of multiples does 10 Have?
The variety of multiples of any type of number is limitless. eg: 10, 20, 30, 40, 50, 60, etc.
See more: Why Are Swamps More Productive Than Streams ? Why Are Swamps More Productive Than Streams
### What is the 10th multiple of 10?
10 × 10 = 100.10th multiple of 10 is 100.
### How carry out you Multiply Multiples of 10?
Multiplication of any type of number through multiples of 10 is way lot easier than regular multiplication.First, we should multiply the digits as it is and affix the trailing zeros at the finish of the number.Similarly, we should do to acquire multiples of 100Eg: 20 × 70=1400, 40 × 3=120, 100 × 5=500, etc;
### What is the Leastern Typical Multiple of 10 and 15?
The least prevalent multiples of two numbers a and also b have the right to be calculated utilizing the formula: (a × b)/GCF(a,b)Here GCF(a,b) is the greatest prevalent determinants of a and also bGCF(10,15) = 5LCM(10,15) = (10 × 15)/GCF(10,15) = 150/5 = 30
### What are the initially 5 multiples of 10?
The initially 5 multiples of 10 are 10, 20, 30, 40, 50 |
# How do you integrate (2-x)/(x^2+5x) using partial fractions?
Nov 19, 2016
The answer is $= \frac{2}{5 x} - \frac{7}{5 \left(x + 5\right)}$
#### Explanation:
Let's go to the decomposition in partial fractions
$\frac{2 - x}{{x}^{2} + 5 x} = \frac{2 - x}{\left(x\right) \left(x + 5\right)}$
$= \frac{A}{x} + \frac{B}{x + 5} = \frac{A \left(x + 5\right) + B x}{x \left(x + 5\right)}$
so, $\left(2 - x\right) = A \left(x + 5\right) + B x$
Let $x = 0$, $\implies$$2 = 5 A$ ; $\implies$ , $A = \frac{2}{5}$
Let $x = - 5$ ; $\implies$ , $7 = - 5 B$ ; $\implies$ , $B = - \frac{7}{5}$
$\frac{2 - x}{{x}^{2} + 5 x} = \frac{\frac{2}{5}}{x} + \frac{- \frac{7}{5}}{x + 5} =$ |
## Introductory Algebra for College Students (7th Edition)
The rectangle has a length of $150$ yards and a width of $50$ yards.
We know that the perimeter is given by the formula: $$P = 2l + 2w$$ where $P$ is the perimeter of the rectangle, $l$ is the measure of its length, and $w$ is the measure of its width. In this problem, we set $w$ as the width. The length is three times the width, so $l = 3w$. We plug in what we have for $w$ and $l$ into the formula for perimeter, where we have $P = 400$: $$400 = 2(3w) + 2w$$ We can now solve for $w$, the width. Let's multiply first: $$400 = 6w + 2w$$ Add the variable terms together: $$400 = 8w$$ Divide both sides by $8$ to solve for $w$: $$w = 50$$ Now that we have the measure for width, we can substitute $50$ for $w$ into the original equation: $$400 = 2l + 2(50)$$ Multiply: $$400 = 2l + 100$$ Subtract $100$ from both sides to isolate the variable on one side and the constant terms on the other: $$300 = 2l$$ Divide both sides by $2$ to solve for $l$: $$l = 150$$ The rectangle has a length of $150$ yards and a width of $50$ yards. |
# How to Get Perimeter: A Comprehensive Guide to Understanding and Calculating Perimeter
## I. Introduction
Perimeter is a fundamental concept in mathematics that is used to measure the distance around an object or shape. It plays a vital role in our everyday lives, from measuring the area of a room before buying furniture or determining the length of a fence needed for our homes.
The purpose of this article is to provide a comprehensive guide to understanding and calculating perimeter, allowing readers to apply their newfound knowledge in a practical and useful way.
## II. A Step-by-Step Guide to Finding the Perimeter
Perimeter is the total distance around a two-dimensional shape. It is calculated by adding up all the lengths of the sides of the shape. The formula for finding the perimeter of each shape is unique.
For example, for a rectangle, you can calculate the perimeter by adding together the lengths of all four sides: P = 2(l + w), where P represents perimeter, l represents the length, and w represents the width.
Similarly, for triangles, the perimeter is the sum of all three sides. For circles, the perimeter is called the circumference and can be calculated using the formula 2πr where π (pi) is a constant value approximated to 3.14159 and r represents the radius of the circle.
It’s important to note that the units should be the same for every side measurement for proper calculation.
To ensure readers understand the concepts presented, here are some example problems and practice questions:
1. Find the perimeter of a rectangle with a length of 10 cm and a width of 5 cm.
Solution: P = 2(l + w) = 2(10+5) = 30 cm
2. Find the perimeter of a right triangle with sides measuring 3 cm, 4 cm, and 5 cm.
Solution: P = 3 + 4 + 5 = 12 cm
3. Find the perimeter of a circle with a radius of 10 cm.
Solution: P = 2πr = 2π(10) ≈ 62.83 cm
## III. The Importance of Understanding Perimeter for Everyday Life
Knowing how to calculate perimeter is essential in many practical situations, including renovating a room, building a fence, or landscaping a garden.
For example, if you want to install new carpet in your bedroom, by measuring and calculating perimeter, you’ll know exactly how much carpet you need.
Conversely, miscalculating perimeter can lead to wasted resources, unnecessary expenses, or even unsafe conditions.
Therefore, understanding the calculations of perimeter helps ensure that you only purchase the right amount of material, or you don’t cut off materials more than necessary.
## IV. The Relationship Between Perimeter and Area
Perimeter and Area are closely related concepts that affect each other. Knowing one of the measurements can allow you to calculate the other. For example, the perimeter of a square is the sum of its four sides, multiplied by each other to get the area.
Understanding this relationship is critical in real-life scenarios such as determining how much paint is needed to cover a wall or how much turf to buy for a soccer field. Once you know the perimeter, can you calculate the area and plan accordingly.
## V. Tips and Tricks for Solving Perimeter Problems
Perimeter problems can be tricky, especially when the shapes are complex. However, there are several tips and tricks that can help you solve them with ease.
One helpful strategy is to break shapes into simpler components, calculating the perimeter of each shape and adding them together. Using visual aids like diagrams or graphs can help simplify complex shapes and make the problem easier to solve.
It’s essential to avoid common mistakes or misconceptions when calculating perimeter, such as forgetting to account for all sides. Double-checking your calculations can help catch errors before you finalize any decision based on such unsatisfactory math calculations.
## VI. Perimeter and Beyond: Advanced Concepts in Geometry
Perimeter is a fundamental concept in geometry, and as such, is part of broader mathematical concepts. Advanced learners may want to explore more advanced topics related to perimeter, such as finding the perimeter of irregular shapes or exploring the connections between perimeter, area, and volume.
Various resources are available, including online math courses, books, and university classes, making it possible to delve deeper into this vital concept.
## VII. Conclusion
Understanding perimeter is essential to performing many everyday tasks. With this comprehensive guide, readers now have a proper foundation for calculating the perimeter of various shapes, including rectangles, triangles, circles, and irregular shapes.
Whether you need to renovate a room, build a fence, or landscape a garden, knowing the perimeter, and its relationship with area and volume can make any project successful and avoid any wastage of resources or late on project delivery.
With practice and additional resources, readers can fully grasp the importance of perimeter in everyday life and pursue advanced concepts in geometry. |
Class 8 Maths Exponents and Powers Introduction
Introduction:
Writing, reading and comapring the smaller numbers are very
easy as compare to larger numbers.
Ex: 15, 100, 1250, 20000 are small number.
But when we talk about larger number, it is very difficult to read,
understand and compare.
Ex: Distance between Sun and Saturn is 1,433,500,000,000 m and distance
between Saturn and Uranus is 1,439,000,000,000 m.
These are very large numbers and also it is very difficult to read,
write, understand and compare. To
make these numbers easy to read, understand and compare,
we use the concept of exponents.
The concept of exponent makes the represenation of
biggere number easier.
Ex: let a number 10,000
We can write 10,000 = 10 * 10 * 10 * 10 = 104
Here the notation 104 is the shot notation of 10 * 10 * 10 * 10 i.e. 10,000
So, 104 is the exponential form of 10,000
Here 10 is called the base and 4 the exponent. The number 104 is read as 10 raised to the power of 4 or
fourth power of 10.
Again, 1,000 = 10 * 10 * 10 * 10 = 103
So, 103 is the exponent form of 1,000
We can use numbers like 10, 100, 100, etc to write a numbers in expanded form.
Ex: 12534 = 1 * 10000 + 2 * 1000 + 5 * 100 + 3 * 10 + 4 * 1
= 1 * 104 + 2 * 103 + 5 * 102 + 3 * 10 + 4
The numbers which have base other than 10 can also be represented in exponent form.
64 = 2 * 2 * 2 * 2 * 2 * 2 = 26
Here, 2 is the base and 6 is the exponent.
So, any number am is in the form of exponent where a is the base and m is the exponent and we read as
“a raised to power m”.
. |
# Lesson 8
Divide to Multiply Non-unit Fractions
## Warm-up: True or False: A Fraction by a Whole Number (10 minutes)
### Narrative
The purpose of this True or False is to elicit the strategies and insights students have for multiplying fractions by whole numbers. Students do not need to find the value of any of the expressions but rather can reason about properties of operations and the relationship between multiplication and division. In this lesson, they will see some ways to find the value of an expression like $$\frac{2}{3} \times 6$$.
### Launch
• Display one statement.
• “Give me a signal when you know whether the statement is true and can explain how you know.”
• 1 minute: quiet think time
### Activity
• Share and record answers and strategy.
• Repeat with each statement.
### Student Facing
Decide if each statement is true or false. Be prepared to explain your reasoning.
• $$2 \times \left(\frac{1}{3} \times 6\right) = \frac{2}{3} \times 6$$
• $$2 \times \left(\frac{1}{3} \times 6\right) = 2 \times (6 \div 3)$$
• $$\frac{2}{3} \times 6 = 2 \times \left(\frac{1}{4} \times6 \right)$$
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
• “How can you explain why $$\frac{2}{3} \times 6 = 2 \times (\frac{1}{4} \times6)$$ is false without finding the value of both sides?” (It can't be true because $$\frac{2}{3} \times 6=2\times\frac{1}{3}\times6$$.)
## Activity 1: Multiply a Whole Number by a Fraction (15 minutes)
### Narrative
The purpose of this activity is for students to relate multiplying a non-unit fraction by a whole number to multiplying a unit fraction by the same whole number. After finding the value of $$\frac{1}{5} \times 3$$ in a way that makes sense to them, they then consider the value of the products $$\frac{2}{5} \times 3$$ and $$\frac{3}{5} \times 3$$. In the synthesis students address how they can use the value of $$\frac{1}{5} \times 3$$ to find the value other expressions.
• Groups of 2
### Activity
• 8 minutes: independent work time
• Monitor for students who:
• draw a diagram
• use division to solve
• recognize a relationship between $$\frac{1}{5}\times3$$$$\frac{2}{5}\times3$$, and $$\frac{3}{5}\times3$$ .
### Student Facing
Find the value of each expression. Explain or show your reasoning. Draw a diagram if it is helpful.
1. $$\frac{1}{5} \times 3$$
2. $$\frac{2}{5} \times 3$$
3. $$\frac{3}{5} \times 3$$
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
• Ask previously selected students to share their solutions.
• Display: $$\frac{1}{5} \times 3$$, $$\frac{2}{5} \times 3$$, $$\frac{3}{5} \times 3$$
• “How are the expressions the same?” (They all have a 3. They all have some fifths and there is a product.)
• “How are the expressions different?” (The number of fifths is different. There is 1 and then 2 and then 3.)
• “How can you use the value of $$\frac{1}{5} \times 3$$ to help find the value of $$\frac{2}{5} \times 3$$?” (I can just double the result because it’s $$\frac{2}{5}$$ instead of $$\frac{1}{5}$$.)
• “What about $$\frac{3}{5} \times 3$$?” (That’s just another $$\frac{1}{5} \times 3$$.)
• Display diagram from student solution or a student generated diagram like it.
• “How does the diagram show $$\frac{1}{5} \times 3$$?” (There is 3 total and $$\frac{1}{5}$$ of it is shaded.)
• Display: $$\frac{2}{5} \times 3$$.
• “How could you adapt the diagram to show $$\frac{2}{5} \times 3$$?” (I could fill in 2 of the fifths in each whole instead of 1.)
• “In the next activity we will study a diagram for $$\frac{2}{5} \times 3$$ more.”
## Activity 2: Match Expressions to Diagrams (20 minutes)
### Narrative
The purpose of this activity is to interpret diagrams in multiple ways, focusing on different multiplication and division expressions. The repeating structure in the diagrams allows for many different ways to find the value and interpret the meaning of the expressions. Encourage students to use words, diagrams, or expressions to explain how the diagram represents each of the expressions.
Monitor for students who:
• can explain that the diagram represents the multiplication expression $$3 \times \frac{2}{5}$$ because it shows 3 groups of $$\frac{2}{5}$$
• can explain that the diagram represents $$2 \times (3 \div 5)$$ because there are 3 wholes divided into 5 equal pieces and 2 of the pieces in each whole are shaded
• can explain how the diagram represents the relationship between $$\frac{6}{5}$$ and $$2 \times (3 \div 5)$$
This activity gives students an opportunity to generalize their learning about fractions, division and multiplication. Students see shaded diagrams in different ways, representing different operations, and begin to see the operations as a convenient way to represent complex calculations (MP8).
Engagement: Provide Access by Recruiting Interest. Provide choice. Invite students to decide which expression to start with.
Supports accessibility for: Visual-Spatial Processing, Conceptual Processing, Attention
• Groups of 2
### Activity
• 5–10 minutes: partner work time
MLR2 Collect and Display
• Circulate, listen for, and collect the language students use to describe how each part of the expression represents each part of the diagram.
• Listen for language described in the narrative.
• Look for notes, labels, and markings on the diagrams that connect the parts of the diagram to the parts of the expressions.
• Record students’ words and phrases on a visual display and update it throughout the lesson.
### Student Facing
Explain how each expression represents the shaded region.
1. $$2 \times (3 \div 5)$$
2. $$\frac{6}{5}$$
3. $$3 \times \frac{2}{5}$$
4. $$3 \times 2 \times \frac{1}{5}$$
### Student Response
For access, consult one of our IM Certified Partners.
If students do not choose any expressions that represent the diagram, ask them to describe the diagram. Write down the words and phrases they use and ask, “Which expressions represent the words you used to describe the diagram?”
### Activity Synthesis
• Display the expression: $$3 \times \frac{2}{5}$$
• “How does the diagram represent the expression?” (It shows 3 groups of $$\frac{2}{5}$$.)
• Display the expression: $$2 \times (3 \div 5)$$
• “How does the diagram represent the expression?”
• Display: $$2 \times (3 \div 5)=\frac{6}{5}$$
• “How do we know this is true?” (We can see both of them in the diagram. $$3\div5$$ is the same as $$\frac{3}{5}$$ and $$2\times\frac{3}{5}=\frac{6}{5}$$
• “Are there any other words, phrases, or diagrams that are important to include on our display?”
• As students share responses, update the display, by adding (or replacing) language, diagrams, or annotations.
• Remind students to borrow language from the display as needed.
## Lesson Synthesis
### Lesson Synthesis
Revisit the chart about the relationship between multiplication and division created in an earlier lesson.
“What would you add to or revise about the relationship between multiplication and division?”
Revise chart as necessary.
## Cool-down: Two Thirds (5 minutes)
### Cool-Down
For access, consult one of our IM Certified Partners.
## Student Section Summary
### Student Facing
In this section, we explored the relationship between multiplication and division. We learned that 1 diagram can represent different multiplication and division expressions. For example, we can interpret this diagram with 4 different expressions:
• $$\frac{3}{4}$$ because each rectangle is divided into 4 equal parts and three of them are shaded.
• $$3 \times \frac{1}{4}$$ because there are 3 parts shaded and each one is $$\frac{1}{4}$$ of the rectangle.
• $$3 \div 4$$ because there are 3 rectangles and each one is divided into 4 equal parts.
• $$\frac14 \times 3$$ because there are 3 rectangles and $$\frac{1}{4}$$ of each one is shaded.
We know that all of these expressions are equal because they all represent the same diagram. We can use any of these expressions to represent and solve this problem:
• Mai ate $$\frac{1}{4}$$ of a 3 pound bag of blueberries. How many pounds of blueberries did Mai eat? |
# SEPARABLE EQUATIONS
In document Elementary Differential Equations with Boundary Value Problems (Page 54-64)
## First Order Equations
### 2.2 SEPARABLE EQUATIONS
A first order differential equation isseparableif it can be written as
h(y)y0 =g(x),
(2.2.1)
where the left side is a product ofy0 and a function ofyand the right side is a function ofx. Rewriting
a separable differential equation in this form is calledseparation of variables. In Section 2.1 we used separation of variables to solve homogeneous linear equations. In this section we’ll apply this method to nonlinear equations.
To see how to solve (2.2.1), let’s first assume thatyis a solution. LetG(x)andH(y)be antiderivatives ofg(x)andh(y); that is,
H0(y) =h(y) and G0(x) =g(x). (2.2.2)
Then, from the chain rule, d dxH(y(x)) =H 0(y(x))y0(x) =h(y)y0(x). Therefore (2.2.1) is equivalent to d dxH(y(x)) = d dxG(x).
Integrating both sides of this equation and combining the constants of integration yields
H(y(x)) =G(x) +c. (2.2.3)
Although we derived this equation on the assumption thatyis a solution of (2.2.1), we can now view it differently: Any differentiable functionythat satisfies (2.2.3) for some constantcis a solution of (2.2.1). To see this, we differentiate both sides of (2.2.3), using the chain rule on the left, to obtain
H0(y(x))y0(x) =G0(x),
which is equivalent to
h(y(x))y0(x) =g(x)
because of (2.2.2).
In conclusion, to solve (2.2.1) it suffices to find functionsG = G(x)andH = H(y)that satisfy (2.2.2). Then any differentiable functiony=y(x)that satisfies (2.2.3) is a solution of (2.2.1).
Example 2.2.1 Solve the equation
y0=x(1 +y2). Solution Separating variables yields
y0 1 +y2 =x. Integrating yields tan−1y= x2 2 +c Therefore y= tan x2 2 +c .
Example 2.2.2
(a) Solve the equation
y0=
−x
y. (2.2.4)
(b) Solve the initial value problem
y0=−xy, y(1) = 1. (2.2.5)
(c) Solve the initial value problem
y0=
−xy, y(1) =−2. (2.2.6)
SOLUTION(a) Separating variables in (2.2.4) yields
yy0 = −x. Integrating yields y2 2 =− x2 2 +c, or, equivalently, x 2+y2= 2c.
The last equation shows thatc must be positive ify is to be a solution of (2.2.4) on an open interval. Therefore we let2c=a2(witha >0) and rewrite the last equation as
x2+y2 =a2. (2.2.7)
This equation has two differentiable solutions foryin terms ofx:
y= pa2x2, a < x < a, (2.2.8) and
y=−pa2x2, a < x < a. (2.2.9) The solution curves defined by (2.2.8) are semicircles above thex-axis and those defined by (2.2.9) are semicircles below thex-axis (Figure2.2.1).
SOLUTION(b) The solution of (2.2.5) is positive whenx= 1; hence, it is of the form (2.2.8). Substituting
x= 1andy= 1into (2.2.7) to satisfy the initial condition yieldsa2= 2; hence, the solution of (2.2.5) is
y=p2−x2, 2< x <2.
SOLUTION(c) The solution of (2.2.6) is negative when x = 1 and is therefore of the form (2.2.9).
Substitutingx = 1andy =−2 into (2.2.7) to satisfy the initial condition yieldsa2 = 5. Hence, the solution of (2.2.6) is
Section 2.2Separable Equations 47 x y 1 2 −1 −2 1 2 −1 −2 (a) (b) Figure 2.2.1 (a)y=√2−x2,2< x <2; (b)y=5x2, 5< x <5
Implicit Solutions of Separable Equations
In Examples 2.2.1and2.2.2we were able to solve the equationH(y) = G(x) +cto obtain explicit formulas for solutions of the given separable differential equations. As we’ll see in the next example, this isn’t always possible. In this situation we must broaden our definition of a solution of a separable equation. The next theorem provides the basis for this modification. We omit the proof, which requires a result from advanced calculus called as theimplicit function theorem.
Theorem 2.2.1 Supposeg=g(x)is continous on(a, b)andh=h(y)are continuous on(c, d).LetG
be an antiderivative ofgon(a, b)and letH be an antiderivative ofhon(c, d).Letx0be an arbitrary point in(a, b),lety0be a point in(c, d)such thath(y0)6= 0,and define
c=H(y0)−G(x0). (2.2.10) Then there’s a functiony=y(x)defined on some open interval(a1, b1),wherea≤a1< x0< b1≤b, such thaty(x0) =y0and
H(y) =G(x) +c (2.2.11)
fora1< x < b1. Thereforeyis a solution of the initial value problem
h(y)y0 =g(x), y(x
0) =x0. (2.2.12)
It’s convenient to say that (2.2.11) withcarbitrary is animplicit solutionofh(y)y0 =g(x). Curves
defined by (2.2.11) are integral curves ofh(y)y0 =g(x). Ifcsatisfies (2.2.10), we’ll say that (2.2.11) is
animplicit solution of the initial value problem(2.2.12). However, keep these points in mind: • For some choices ofcthere may not be any differentiable functionsythat satisfy (2.2.11).
• The functionyin (2.2.11) (not (2.2.11) itself) is a solution ofh(y)y0 =g(x).
Example 2.2.3
(a) Find implicit solutions of
y0 = 2x+ 1
5y4+ 1. (2.2.13)
(b) Find an implicit solution of
y0 = 2x+ 1
5y4+ 1, y(2) = 1. (2.2.14)
SOLUTION(a) Separating variables yields
(5y4+ 1)y0 = 2x+ 1.
Integrating yields the implicit solution
y5+y=x2+x+c. (2.2.15)
of (2.2.13).
SOLUTION(b) Imposing the initial conditiony(2) = 1in (2.2.15) yields1 + 1 = 4 + 2 +c, soc=−4.
Therefore
y5+y=x2+x−4
is an implicit solution of the initial value problem (2.2.14). Although more than one differentiable func- tiony = y(x)satisfies 2.2.13) near x = 1, it can be shown that there’s only one such function that satisfies the initial conditiony(1) = 2.
Figure2.2.2shows a direction field and some integral curves for (2.2.13). Constant Solutions of Separable Equations
An equation of the form
y0 =g(x)p(y)
is separable, since it can be rewritten as
1
p(y)y
0=g(x).
However, the division byp(y)is not legitimate ifp(y) = 0for some values ofy. The next two examples show how to deal with this problem.
Example 2.2.4 Find all solutions of
y0 = 2xy2. (2.2.16)
Solution Here we must divide byp(y) =y2 to separate variables. This isn’t legitimate ifyis a solution of (2.2.16) that equals zero for some value ofx. One such solution can be found by inspection: y ≡0. Now supposeyis a solution of (2.2.16) that isn’t identically zero. Sinceyis continuous there must be an interval on whichyis never zero. Since division byy2is legitimate forxin this interval, we can separate variables in (2.2.16) to obtain
y0
Section 2.2Separable Equations 49 1 1.5 2 2.5 3 3.5 4 −1 −0.5 0 0.5 1 1.5 2 x y
Figure 2.2.2 A direction field and integral curves fory0= 2x+ 1
5y4+ 1
Integrating this yields
−1
y =x 2+c, which is equivalent to
y=−x21+c. (2.2.17)
We’ve now shown that ify is a solution of (2.2.16) that is not identically zero, thenymust be of the form (2.2.17). By substituting (2.2.17) into (2.2.16), you can verify that (2.2.17) is a solution of (2.2.16). Thus, solutions of (2.2.16) arey ≡0and the functions of the form (2.2.17). Note that the solutiony≡0
isn’t of the form (2.2.17) for any value ofc.
Figure2.2.3shows a direction field and some integral curves for (2.2.16)
Example 2.2.5 Find all solutions of
y0 = 1 2x(1−y
2). (2.2.18)
Solution Here we must divide byp(y) = 1−y2 to separate variables. This isn’t legitimate ify is a solution of (2.2.18) that equals±1for some value ofx. Two such solutions can be found by inspection: y≡1andy ≡ −1. Now supposeyis a solution of (2.2.18) such that1−y2isn’t identically zero. Since
1−y2is continuous there must be an interval on which1y2is never zero. Since division by1y2is legitimate forxin this interval, we can separate variables in (2.2.18) to obtain
2y0
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2 −2 −1.5 −1 −0.5 0 0.5 1 1.5 2 y x
Figure 2.2.3 A direction field and integral curves fory0 = 2xy2
A partial fraction expansion on the left yields
1 y−1 − 1 y+ 1 y0 = −x, and integrating yields
ln y−1 y+ 1 =− x2 2 +k; hence, y−1 y+ 1 =eke−x2 /2.
Sincey(x)6=±1forxon the interval under discussion, the quantity(y−1)/(y+ 1)can’t change sign in this interval. Therefore we can rewrite the last equation as
y−1
y+ 1=ce −x2/2
,
wherec=±ek, depending upon the sign of(y1)/(y+ 1)on the interval. Solving foryyields y= 1 +ce
−x2
/2
1−ce−x2/2. (2.2.19)
We’ve now shown that ifyis a solution of (2.2.18) that is not identically equal to±1, thenymust be as in (2.2.19). By substituting (2.2.19) into (2.2.18) you can verify that (2.2.19) is a solution of (2.2.18). Thus, the solutions of (2.2.18) arey ≡1,y ≡ −1and the functions of the form (2.2.19). Note that the
Section 2.2Separable Equations 51
constant solutiony ≡1can be obtained from this formula by takingc= 0; however, the other constant solution,y≡ −1, can’t be obtained in this way.
Figure2.2.4shows a direction field and some integrals for (2.2.18).
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2 −3 −2 −1 0 1 2 3 x y
Figure 2.2.4 A direction field and integral curves fory0 =x(1−y2)
2
Differences Between Linear and Nonlinear Equations
Theorem2.1.2states that ifpandf are continuous on(a, b)then every solution of
y0+p(x)y=f(x)
on(a, b)can be obtained by choosing a value for the constantcin the general solution, and ifx0is any point in(a, b)andy0is arbitrary, then the initial value problem
y0+p(x)y=f(x), y(x0) =y0
has a solution on(a, b).
The not true for nonlinear equations. First, we saw in Examples 2.2.4and 2.2.5 that a nonlinear equation may have solutions that can’t be obtained by choosing a specific value of a constant appearing in a one-parameter family of solutions. Second, it is in general impossible to determine the interval of validity of a solution to an initial value problem for a nonlinear equation by simply examining the equation, since the interval of validity may depend on the initial condition. For instance, in Example2.2.2 we saw that the solution of
dy dx =− x y, y(x0) =y0 is valid on(−a, a), wherea=px2 0+y20.
Example 2.2.6 Solve the initial value problem
y0= 2xy2, y(0) =y0 and determine the interval of validity of the solution.
Solution First supposey06= 0. From Example2.2.4, we know thatymust be of the form
y=−x21+c. (2.2.20)
Imposing the initial condition shows thatc = −1/y0. Substituting this into (2.2.20) and rearranging terms yields the solution
y= y0
1−y0x2.
This is also the solution ify0 = 0. Ify0 <0, the denominator isn’t zero for any value ofx, so the the solution is valid on(−∞,∞). Ify0>0, the solution is valid only on(−1/√y0,1/√y0).
### 2.2 Exercises
In Exercises1–6find all solutions.
1. y0= 3x2+ 2x+ 1
y−2 2. (sinx)(siny) + (cosy)y 0 = 0 3. xy0+y2+y= 0 4. y0ln |y|+x2y= 0 5. (3y3+ 3ycosy+ 1)y0+(2x+ 1)y 1 +x2 = 0 6. x2yy0 = (y21)3/2
In Exercises7–10find all solutions. Also, plot a direction field and some integral curves on the indicated rectangular region. 7. C/G y0=x2(1 +y2); {−1≤x≤1, −1≤y≤1} 8. C/G y0(1 +x2) +xy= 0; {−2x2, 1y 1} 9. C/G y0= (x −1)(y−1)(y−2); {−2≤x≤2, −3≤y≤3} 10. C/G (y−1)2y0 = 2x+ 3; {−2≤x≤2, −2≤y≤5}
In Exercises11and12solve the initial value problem.
11. y0= x2+ 3x+ 2
y−2 , y(1) = 4
12. y0+x(y2+y) = 0, y(2) = 1
In Exercises13-16solve the initial value problem and graph the solution.
Section 2.2Separable Equations 53
14. C/G y0+(y+ 1)(y−1)(y−2)
x+ 1 = 0, y(1) = 0
15. C/G y0+ 2x(y+ 1) = 0, y(0) = 2
16. C/G y0= 2xy(1 +y2), y(0) = 1
In Exercises17–23solve the initial value problem and find the interval of validity of the solution.
17. y0(x2+ 2) + 4x(y2+ 2y+ 1) = 0, y(1) =1 18. y0=2x(y23y+ 2), y(0) = 3 19. y0= 2x 1 + 2y, y(2) = 0 20. y 0= 2yy2, y(0) = 1 21. x+yy0 = 0, y(3) = −4 22. y0+x2(y+ 1)(y2)2= 0, y(4) = 2 23. (x+ 1)(x−2)y0+y= 0, y(1) =3 24. Solvey0 = (1 +y 2)
(1 +x2)explicitly. HINT:Use the identitytan(A+B) =
tanA+ tanB
1−tanAtanB. 25. Solvey0p
1−x2+p
1−y2= 0explicitly. HINT:Use the identitysin(AB) = sinAcosB
cosAsinB.
26. Solvey0= cosx
siny, y(π) = π
2 explicitly. HINT:Use the identitycos(x+π/2) =−sinxand the periodicity of the cosine.
27. Solve the initial value problem
y0 =ay−by2, y(0) =y0. Discuss the behavior of the solution if(a)y0≥0;(b)y0<0. 28. The populationP =P(t)of a species satisfies the logistic equation
P0 =aP(1
−αP)
andP(0) =P0>0. FindP fort >0, and findlimt→∞P(t).
29. An epidemic spreads through a population at a rate proportional to the product of the number of people already infected and the number of people susceptible, but not yet infected. Therefore, if Sdenotes the total population of susceptible people andI =I(t)denotes the number of infected people at timet, then
I0=rI(S
−I),
wherer is a positive constant. Assuming thatI(0) = I0, findI(t) fort > 0, and show that
limt→∞I(t) =S.
30. L The result of Exercise29is discouraging: if any susceptible member of the group is initially infected, then in the long run all susceptible members are infected! On a more hopeful note, suppose the disease spreads according to the model of Exercise29, but there’s a medication that cures the infected population at a rate proportional to the number of infected individuals. Now the equation for the number of infected individuals becomes
I0 =rI(S
−I)−qI (A)
(a) ChooserandSpositive. By plotting direction fields and solutions of (A) on suitable rectan- gular grids
R={0≤t≤T, 0≤I≤d}
in the(t, I)-plane, verify that ifIis any solution of (A) such thatI(0)>0, thenlimt→∞I(t) =
S−q/rifq < rSandlimt→∞I(t) = 0ifq≥rS.
(b) To verify the experimental results of(a), use separation of variables to solve (A) with initial conditionI(0) = I0 >0, and findlimt→∞I(t). HINT:There are three cases to consider:
(i)q < rS;(ii)q > rS;(iii)q=rS.
31. L Consider the differential equation
y0 =ay
−by2−q, (A)
wherea,bare positive constants, andqis an arbitrary constant. Supposeydenotes a solution of this equation that satisfies the initial conditiony(0) =y0.
(a) Chooseaandbpositive andq < a2/4b. By plotting direction fields and solutions of (A) on suitable rectangular grids
R={0≤t≤T, c≤y≤d} (B)
in the(t, y)-plane, discover that there are numbers y1 andy2 withy1 < y2 such that if y0> y1thenlimt→∞y(t) =y2, and ify0< y1theny(t) =−∞for some finite value oft.
(What happens ify0=y1?)
(b) Chooseaandbpositive andq = a2/4b. By plotting direction fields and solutions of (A) on suitable rectangular grids of the form (B), discover that there’s a numbery1such that if y0≥y1thenlimt→∞y(t) =y1, while ify0 < y1theny(t) =−∞for some finite value of
t.
(c) Choose positivea,b andq > a2/4b. By plotting direction fields and solutions of (A) on suitable rectangular grids of the form (B), discover that no matter whaty0is,y(t) =−∞for some finite value oft.
(d) Verify your results experiments analytically. Start by separating variables in (A) to obtain y0
ay−by2q = 1.
To decide what to do next you’ll have to use the quadratic formula. This should lead you to see why there are three cases. Take it from there!
Because of its role in the transition between these three cases,q0=a2/4bis called abifur-
cation valueofq. In general, ifqis a parameter in any differential equation,q0is said to be a bifurcation value ofqif the nature of the solutions of the equation withq < q0is qualitatively different from the nature of the solutions withq > q0.
32. L By plotting direction fields and solutions of y0 =qy−y3,
convince yourself thatq0 = 0is a bifurcation value ofqfor this equation. Explain what makes you draw this conclusion.
33. Suppose a disease spreads according to the model of Exercise29, but there’s a medication that cures the infected population at a constant rate ofqindividuals per unit time, whereq >0. Then the equation for the number of infected individuals becomes
I0 =rI(S
−I)−q.
Section 2.3Existence and Uniqueness of Solutions of Nonlinear Equations 55
34. Assuming thatp6≡0, state conditions under which the linear equation y0+p(x)y=f(x)
is separable. If the equation satisfies these conditions, solve it by separation of variables and by the method developed in Section 2.1.
Solve the equations in Exercises35–38using variation of parameters followed by separation of variables.
35. y0+y= 2xe −x 1 +yex 36. xy 0 −2y= x 6 y+x2 37. y0 −y= (x+ 1)e 4x (y+ex)2 38. y 02y= xe 2x 1−ye−2x
39. Use variation of parameters to show that the solutions of the following equations are of the form y=uy1, whereusatisfies a separable equationu0=g(x)p(u). Findy1andgfor each equation.
(a)xy0+y=h(x)p(xy) (b)xy0 −y=h(x)py x (c)y0+y=h(x)p(exy) (d)xy0+ry=h(x)p(xry) (e)y0+v 0(x) v(x)y=h(x)p(v(x)y)
In document Elementary Differential Equations with Boundary Value Problems (Page 54-64) |
# How do you find the derivative of f(x)=(x+3)/(x-3)?
Using the quotient rule we have that
f'(x)=((x+3)'*(x-3)-(x-3)'*(x+3))/(x-3)^2=> f'(x)=((x-3)-(x+3))/(x-3)^2=> f'(x)=-6/(x-3)^2
Feb 2, 2016
$- \frac{6}{x - 3} ^ 2$
#### Explanation:
differentiate using the$\textcolor{b l u e}{\text{ quotient rule }}$
for a rational function $f \left(x\right) = g \frac{x}{h \left(x\right)}$
then: $f ' \left(x\right) = \frac{h \left(x\right) . g ' \left(x\right) - g \left(x\right) . h ' \left(x\right)}{h \left(x\right)} ^ 2$
applying this to the above function gives :
$\frac{d}{\mathrm{dx}} \left(\frac{x + 3}{x - 3}\right) = \frac{\left(x - 3\right) \frac{d}{\mathrm{dx}} \left(x + 3\right) - \left(x + 3\right) \frac{d}{\mathrm{dx}} \left(x - 3\right)}{x - 3} ^ 2$
$= \frac{\left(x - 3\right) .1 - \left(x + 3\right) .1}{x - 3} ^ 2 = \frac{x - 3 - x - 3}{x - 3} ^ 2$
$= - \frac{6}{x - 3} ^ 2$ |
# Finding Patterns
In this lesson, we will learn different problem solving strategies to find patterns.
The following are some examples of problem solving strategies.
Explore it//Act it/Try it (EAT) method (Advanced)
Finding a Pattern (Basic)
Finding a Pattern (Intermediate)
Finding a Pattern (Advanced)
## Find A Pattern (Advanced)
In this lesson, we will look at some advanced examples of Find a Pattern method of problem solving strategy.
Example:
Each hexagon below is surrounded by 12 dots.
a) Find the number of dots for a pattern with 6 hexagons in the first column.
b) Find the pattern of hexagons with 229 dots.
Solution:
1st column Pattern Total dots 1 12 12 2 12 + 16 28 3 12 + 16 + 21 49 4 12 + 16 + 21 + 26 75 5 12 + 16 + 21 + 26 + 31 106 6 12 + 16 + 21 + 26 + 31 + 36 142 7 12 + 16 + 21 + 26 + 31 + 36 + 41 183 8 12 + 16 + 21 + 26 + 31 + 36 + 41 + 46 229
a) The number of dots for a pattern with 6 hexagons in the first column is 142
b) If there are 229 dots then the pattern has 8 hexagons in the first column.
Example:
Each member of a club shook hands with every other member who came for a meeting. There were a total of 45 handshakes. How many members were present at the meeting?
A B C D E F G H I J A B ● C ● ● D ● ● ● E ● ● ● ● F ● ● ● ● ● G ● ● ● ● ● ● H ● ● ● ● ● ● ● I ● ● ● ● ● ● ● ● J ● ● ● ● ● ● ● ● ● HS 9 8 7 6 5 4 3 2 1
Solution:
Total = 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 45 handshakes
There were 10 members
Example:
In the figure, a pinball is released at A.
How many paths are there for it to drop from A to E?
Solution:
from A to B: 2
B to C: 6
A to C: 2 × 6 = 12
C to D: 70
A to D: 12 × 70 = 840
D to E: 2
A to E: 2 × 840 = 1680
There are 1680 paths from A to E
Example:
A group of businessmen were at a networking meeting. Each businessman exchanged his business card with every other businessman who was present.
a) If there were 16 businessmen, how many business cards were exchanged?
b) If there was a total of 380 business cards exchanged, how many businessmen were at the meeting?
Solution:
a) 15 + 14 + 13 + 12 + 11 + 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 120 exchanges
120 × 2 = 240 business cards
If there were 16 businessmen, 240 business cards were exchanged.
b) 380 ÷ 2 = 190
190 = (19 × 20) ÷ 2 = 19 + 18 + 17 + … + 3 + 2 + 1
If there was a total of 380 business cards exchanged, there were 20 businessmen at the meeting.
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# What is 12/5 as a mixed fraction?
The aim of this question is to learn how to convert simple fractions into mixed fractions.
Fractions can be subdivided into two types, proper and improper. A fraction is said to be a proper fraction if the numerator magnitude is smaller than the denominator magnitude. $\dfrac{ 1 }{ 2 }$ is an example of a proper fraction.
An improper fraction is such a fraction whose numerator value is equal to or greater than that of the denominator. Improper fractions can be converted into mixed fractions. $\dfrac{ 88 }{ 2 }$ is an example of a proper fraction.
A mixed fraction is a type of fraction that has a whole number part and a proper fraction part. $14 \ + \ \dfrac{ 1 }{ 2 }$ is an example of a proper fraction.
Given the fraction:
$\dfrac{ 12 }{ 5 }$
Substituting $12 \ = \ 10 \ + \ 2$ in the above equation:
$\dfrac{ 10 \ + \ 2 }{ 5 }$
Separating the denominator:
$\dfrac{ 10 }{ 5 } \ + \ \dfrac{ 2 }{ 5 }$
Substituting $10 \ = \ ( 2 )( 5 )$ in the above equation:
$\dfrac{ ( 2 )( 5 ) }{ 5 } \ + \ \dfrac{ 2 }{ 5 }$
$2 \times \dfrac{ 5 }{ 5 } \ + \ \dfrac{ 2 }{ 5 }$
$2 \times 1 \ + \ \dfrac{ 2 }{ 5 }$
$2 \ + \ \dfrac{ 2 }{ 5 }$
Which can be written as:
$2 \dfrac{ 2 }{ 5 }$
## Numerical Results
$2 \dfrac{ 2 }{ 5 }$
## Example
Write the mixed fraction of 33/8 and 15/2.
Part (a) – Given the fraction:
$\dfrac{ 33 }{ 8 }$
Substituting $33 \ = \ 32 \ + \ 1$ in the above equation:
$\dfrac{ 32 \ + \ 1 }{ 8 }$
Separating the denominator:
$\dfrac{ 32 }{ 8 } \ + \ \dfrac{ 1 }{ 8 }$
Substituting $32 \ = \ ( 4 )( 8 )$ in the above equation:
$\dfrac{ ( 4 )( 8 ) }{ 8 } \ + \ \dfrac{ 1 }{ 8 }$
$4 \ + \ \dfrac{ 1 }{ 8 }$
Which can be written as:
$4 \dfrac{ 1 }{ 8 }$
Part (b) – Given the fraction:
$\dfrac{ 15 }{ 2 }$
Substituting $15 \ = \ 14 \ + \ 1$ in the above equation:
$\dfrac{ 14 \ + \ 1 }{ 2 }$
Separating the denominator:
$\dfrac{ 14 }{ 2 } \ + \ \dfrac{ 1 }{ 2 }$
Substituting $14 \ = \ ( 7 )( 2 )$ in the above equation:
$\dfrac{ ( 7 )( 2 ) }{ 2 } \ + \ \dfrac{ 1 }{ 2 }$
$7 \ + \ \dfrac{ 1 }{ 2 }$
Which can be written as:
$7 \dfrac{ 1 }{ 2 }$ |
# Find Out the Speed of the Stream.
## Question: The speed of a boat in still water is 11 km/hr. It can go 12 km upstream and return downstream to its original point in 2 hr 45 min. Find out the speed of the stream.
The question is a real-life application of linear equations in two variables.
## Answer: The speed of the stream is 5 km/hr.
Let's explore the water currents.
## Explanation:
Let the speed of the stream be x km/hr
Given that, the speed boat in still water is 11 km/hr.
⇒ speed of the boat upstream = (11 - x) km/hr
⇒ speed of the boat downstream = (11+ x) km/hr
It is mentioned that the boat can go 12 km upstream and return downstream to its original point in 2 hr 45 min.
⇒ One-wayDistance traveled by boat (d) = 12 km
⇒ Tupstream + Tdownstream = 2 hr 45 min = (2 + 3/4) hr = 11/4 hr
⇒ [distance / upstream speed ] + [distance / downstream speed] = 11/4
⇒ [ 12/ (11-x) ] + [ 12/ (11+x) ] = 11/4
⇒ 12 [ 1/ (11-x) + 1/(11+x) ] = 11/4
⇒ 12 [ {11 - x + 11 + x} / {121 - x2} ] = 11/4
⇒ 12 [ {22} / {121 - x2} ] = 11/4
⇒ 12 [ 2 / {121 - x2} ] = 1/4
⇒ 24 / {121 - x2} = 1/4
⇒ 24 (4) = {121 - x2
⇒ 96 = 121 - x2
⇒ x2 = 121 - 96
⇒ x2 = 25
⇒ x = + 5 or -5
As speed to stream can never be negative, we consider the speed of the stream(x) as 5 km/hr. |
# Function Composition: How to Decompose a Function
Calculus > Function Composition
In order to figure out function composition (or to decompose a function), you must be familiar with the eight common function types and with basic function transformations, like what a negative sign does to a function (flips an axis around the origin) or what happens if you add a constant (it shifts to the left that number of units). In calculus, composite functions are usually represented by f(x) and g(x), where f(x) is a function that takes some kind of action on g(x). Being able to split a function into two can be useful in calculus if the original function is too complicated to work with.
## Function Composition Sample Problem
Sample problem 1: Identify the functions in the equation f(g(x)) = -(x – 3)2 + 5
Step 1: Identify the original function(s). The original function is either going to be: linear, polynomial, square (quadratic), absolute value, square root, rational, sine or cosine. In this example, the original function is the square – (x – 3)2 (the square – (x – 3) has been shifted up five units).
Step 2: Write the functions using standard terminology (f(x) and g(x)).
f(g(x)) = -(x – 3)2 + 5, so:
g(x) = – (x – 3)2
f(x) = x + 5
Sample problem 2: Identify the functions in the equation f(g(x)) = (x + 2 / x)3
Step 1: Look for the original function f(x) — see Step 1 of sample problem 1 above. In this example, the original function isn’t an obvious example of a basic function type. However, while the function f(x) = x + 2/x isn’t a basic type, the second function g(x) — x3 is (it’s a cubic polynomial). So:
f(g(x)) = (x + 2 / x)2
f(x) = x + 2 /x
g(x) = x3
Tip: When trying to find composite functions, look for the simplest transformation, usually involving x and a cube, square, simple addition, division, multiplication, subtraction etc.. This simple transformation is either going to be f(x) or g(x).
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If you prefer an online interactive environment to learn R and statistics, this free R Tutorial by Datacamp is a great way to get started. If you're are somewhat comfortable with R and are interested in going deeper into Statistics, try this Statistics with R track. |
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Alignments to Content Standards: 5.NF.A.1
1. To add fractions, we usually first find a common denominator.
1. Find two different common denominators for $\frac{1}{5}$ and $\frac{1}{15}$.
2. Use each common denominator to find the value of $\frac{1}{5}+\frac{1}{15}$. Draw a picture that shows your solution.
2. Find $\frac{3}{4}+\frac{1}{5}$. Draw a picture that shows your solution.
3. Find $\frac{14}{8}+\frac{15}{12}$.
IM Commentary
Part (a) of this task asks students to find and use two different common denominators to add the given fractions. The purpose of this question is to help students realize that they can use any common denominator to find a solution, not just the least common denominator. Part (b) does not ask students to solve the given addition problem in more than one way. Instead, the purpose of this question is to give students an opportunity to choose a denominator and possibly to compare their solution method with another student who chose a different denominator. The two most likely student choices are the least common denominator and the product of the two denominators.
Parts (a) and (b) ask students to draw pictures to help them to see why finding a common denominator is an important part of solving the given addition problems. Part (c) does not ask students to draw a picture. Instead, the purpose of part (c) is to give students who are ready to work symbolically a chance to work more efficiently.
The numbers were chosen in each part for a specific reason. Part (a) asks students to add unit fractions, helping to emphasize the need for a common denominator. Additionally, one denominator is a multiple of the other, so students should recognize it as a common denominator for both fractions. Part (b) asks students to add a unit fraction and a non-unit fraction. Part (c) asks students to add fractions where one of the addends is greater than 1. It also gives students a chance to write one of the fractions with a smaller denominator before adding, making the arithmetic that follows easier.
Solutions
1. The denominators of the fractions in the first problem are 5 and 15. 5 divides evenly into 15, meaning that 15 is a multiple of 5, so 15 itself is a common denominator for the fractions $\frac{1}{5}$ and $\frac{1}{15}$. The picture below shows this:
Here is a picture showing the fractions when they are both written in terms of fifteenths.
Any multiple of 15 is also a common multiple of 5 and 15. This means that $2\times15=30$ is a common multiple and is a common denominator for the fractions $\frac{1}{5}$ and $\frac{1}{15}$. Thus, 15 and 30 are two different common denominators for the fractions $\frac{1}{5}$ and $\frac{1}{15}$.
2. The first common denominator that we identified in part (i) was 15. Here is a picture that represents $\frac{1}{5}+\frac{1}{15}$:
This is how we can write the process of finding this common denominator and adding using symbols:
\begin{align} \frac{1}{5}+\frac{1}{15}&=\frac{1\times3}{5\times3}+\frac{1}{15}\\ &=\frac{3}{15}+\frac{1}{15}\\ &=\frac{3+1}{15}\\ &=\frac{4}{15} \end{align}
The picture shows that after we convert $\frac{1}{5}$ to $\frac{3}{15}$ we can add $\frac{1}{15}$ and we are left with $\frac{4}{15}$, as we found symbolically.
The second common denominator that we identified in part (i) was 30. Here is a picture that represents $\frac{1}{5}+\frac{1}{15}$:
This is how we can write the process of finding this common denominator and adding using symbols:
\begin{align} \frac{1}{5}+\frac{1}{15}&=\frac{1\times6}{5\times6}+\frac{1\times2}{15\times2}\\ &=\frac{6}{30}+\frac{2}{30}\\ &=\frac{6+2}{30}\\ &=\frac{8}{30}\\ &=\frac{4\times2}{15\times2}\\ &=\frac{4}{15}\\ \end{align}
The picture shows that after we convert $\frac{1}{5}$ to $\frac{6}{30}$ and $\frac{1}{15}$ to $\frac{2}{30}$ we can add the thirtieths and are left with $\frac{8}{30}$. This solution is equivalent to $\frac{4\times 2}{15\times 2}\frac{4}{15}$, as we found above. Thus, we get the same answer using the two different common denominators that we identified in part (i), as we would expect.
3. In order to find the solution to the addition problem $\frac{3}{4}+\frac{1}{5}$, we can first find a common denominator. 20 is a multiple of 4 and 5 because $4\times5=20$, meaning that 20 is a common denominator for the fractions $\frac{3}{4}$ and $\frac{1}{5}$. Here is a picture that shows these fractions when they are both written in terms of twentieths:
Here is a picture that represents $\frac{3}{4}+\frac{1}{5}$:
This is how we can write the process of finding a common denominator and adding using symbols:
\begin{align} \frac{3}{4}+\frac{1}{5} &=\frac{3\times5}{4\times5}+\frac{1\times4}{5\times4}\\ &=\frac{15}{20}+\frac{4}{20}\\ &=\frac{15+4}{20}\\ &=\frac{19}{20} \end{align}
The picture shows that when we convert $\frac{3}{4}$ to $\frac{15}{20}$ and $\frac{1}{5}$ to $\frac{4}{20}$ and add twentieths, we have $\frac{19}{20}$, as we found symbolically.
4. In order to solve this addition problem we again find a common denominator for the fractions $\frac{14}{8}$ and $\frac{15}{12}$. 24 is a common multiple for the denominators 8 and 12 because $8\times3=24$ and $12\times2=24$. This means that 24 is a common denominator for the fractions $\frac{14}{8}$ and $\frac{15}{12}$. This is how we can write the process of finding a common denominator and adding using symbols:
\begin{align} \frac{14}{8}+\frac{5}{12} &=\frac{14\times3}{8\times3}+\frac{5\times2}{12\times2}\\ &=\frac{42}{24}+\frac{10}{24}\\ &=\frac{52}{24} \end{align} If we pause for a moment, we will see that $\frac{52}{24}$ is equivalent to $\frac{13}{6}$, although either representation of the sum is correct.
Solution: Alternative approach to part (c)
A student with a robust understanding of fraction equivalence might note that $\frac{14}{8}$ is equivalent to $\frac{7\times 2}{4\times 2}=\frac74$. So:
$$\frac{14}{8}+\frac{5}{12}=\frac74+\frac{5}{12}$$
12 is a common denominator for these two fractions, so:
\begin{align} \frac{14}{8}+\frac{5}{12} &=\frac74+\frac{5}{12}\\ &=\frac{7\times3}{4\times3}+\frac{5}{12}\\ &=\frac{21}{12}+\frac{5}{12}\\ &=\frac{26}{12} \end{align}
Note that $\frac{26}{12}$ is equivalent to $\frac{52}{24}$ as well as $\frac{13}{6}$; any of these answers are acceptable. |
# Parabolas Unit 7.1. Conics When the plane cuts a cone at a right angle 4 types:
## Presentation on theme: "Parabolas Unit 7.1. Conics When the plane cuts a cone at a right angle 4 types:"— Presentation transcript:
Parabolas Unit 7.1
Conics When the plane cuts a cone at a right angle 4 types:
Parabola in every day life Practical Applications Listening Hand blocking Sun
Location of the parabola on Cartesian plot Why study it? Passing to a teammate: football, basketball, volleyball, shooting a basketball
Last year y = ax 2 + bx + c x = input a = determine positive or negative slope c = y intercept Vertex = -b/2x Roots = -b ± √b 2 – 4ac 2a
This Year (because you are smarter and more ambitious) More detailed Locus – set of all points that fulfill a geometric property Focus – locus of points in a plane that are equidistant from a fixed point Directrix – a specific line Axis of symmetry – line perpendicular to the directrix through the focus Vertex – intersection of the parabola and the axis Parabola – the locus of points in a plane
Parabola
Two types Opens vertically (x – h) 2 = 4p(y – k) Opens horizontally (y – k) 2 = 4p(x – h)
Standard form of equations (x – h) 2 = 4p(y – k) a.Vertex ( h,k) b.Focus (h, k + p) c.Axis of symmetry a: (x = h) d. Directrix d: y = k - p (y – k) 2 = 4p(x – h) a.Vertex (h, k) b.Focus (h + p, k) c.Axis of symmetry a: y = k d. Directrix d: x = h - p
Problem 1 Page 428 (x – 3) 2 = 12(y – 7) 1.Does it open horizontally or vertically? 1.Vertically 2. What are h, p and k 1.h = 3, p = 3, k = 7 3. Vertex: (h, k) or (3, 7) 4.Focus: ( h, k + p) = (3, 10) 5.Axis of symmetry: x = h or 3 6.Directrix: y = k – p; 7 – 3 = 4
Unit 7.1 Problems Page 428 Problems 1 - 10
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# Module 4: Fractions
A fraction describes equal parts of a whole:
Using official math vocabulary:
Exercises
The month of April had rainy days and days that were not rainy.
1. What fraction of the days were rainy?
2. What fraction of the days were not rainy?
# Simplifying Fractions
Two fractions are equivalent if they represent the same number. (The same portion of a whole.) To build an equivalent fraction, multiply the numerator and denominator by the same number.
Exercises
3. Write as an equivalent fraction with a denominator of .
4. Write as an equivalent fraction with a denominator of .
Many fractions can be simplified, or reduced. Here are four special cases.
Exercises
Simplify each fraction, if possible.
5.
6.
7.
8.
A fraction is completely reduced, or in simplest form, or in lowest terms, when the numerator and denominator have no common factors other than . To reduce a fraction, divide the numerator and denominator by the same number.
Exercises
Reduce each fraction to simplest form.
9.
10.
# Multiplying Fractions
To multiply fractions, multiply the numerators and multiply the denominators straight across. If possible, simplify your answer.
Exercises
Multiply each pair of numbers. Be sure that each answer is in simplest form.
11.
12.
13.
14.
To find a fraction of a number, multiply.
Exercises
15. To pass his workplace training, Nathan must correctly answer at least of questions. How many questions must he answer correctly to pass the training?
# Dividing Fractions
To divide by a fraction, multiply by the reciprocal of the second number. (Flip the second fraction upside-down.)
Exercises
Divide. Be sure that each answer is in simplest form.
16.
17.
18. Suppose you need to measure cups of flour, but the only scoop you can find is cup. How many scoops of flour will you need?
# Comparing Fractions
If two fractions have the same denominator, we can simply compare their numerators.
If two fractions have different denominators, we can rewrite them with a common denominator and then compare their numerators.
Exercises
19. Cookie recipe A requires cup of sugar, whereas cookie recipe B requires cup of sugar. Which recipe requires more sugar?
To add or subtract two fractions with the same denominator, add or subtract the numerators and keep the common denominator.
Exercises
20. Jack ate of a pizza. Mack ate of the pizza. What fraction of the pizza did they eat together?
21. Tracy ate of a pizza. Stacy ate of the pizza. How much more of the pizza did Tracy eat?
To add or subtract two fractions with different denominators, first write them with a common denominator. Then add or subtract them.
Exercises
A -inch thick sheet of plywood is going to be laid onto a -inch thick sheet of plywood.
22. What is the combined thickness of the two sheets?
23. What is the difference in thickness of the two sheets of plywood?
Jacqueline budgets of her monthly income for food and of her monthly income for rent.
24. What fraction of her monthly income does she budget for these two expenses combined?
25. What fraction more of her monthly income does she budget for her rent than for her food?
# Fractions and Decimals
To write a fraction as a decimal, divide the numerator by the denominator.
A decimal that ends (eventually has a remainder of ) is called a terminating decimal. Fun fact: If the denominator of a fraction has no prime factors other than ‘s and ‘s, the decimal will terminate. Also, the fraction can be built up to have a denominator of , or , or
Exercises
Write each fraction as a decimal.
26.
27.
A decimal that continues a pattern of digits is called a repeating decimal. We can represent the repeating digits by using either an overbar or ellipsis (three dots)…
Exercises
Write each fraction as a decimal.
28.
29.
# Mixed Numbers
A mixed number represents a sum. For example, means .
To write a mixed number as an improper fraction:
1. Multiply the whole number part by the denominator.
2. Add this result to the original numerator to get the new numerator.
3. Keep the same denominator.
Exercises
Rewrite each mixed number as an improper fraction.
30.
31.
To write an improper fraction as a mixed number:
1. Divide the numerator by the denominator to get the whole number part.
2. The remainder after dividing is the new numerator.
3. Keep the same denominator.
Exercises
Rewrite each improper fraction as a mixed number.
32.
33.
Adding or subtracting mixed numbers can be fairly simple or more complicated, depending on the numbers. If adding two mixed numbers would give you an improper fraction as part of your result, you’ll need to carry; if subtracting two mixed numbers would give you a negative fraction as part of your result, you’ll need to borrow.
Exercises
35. Subtract:
Multiplying or dividing mixed numbers is tricky. Change any mixed numbers into improper fractions before doing the calculation, then change the answer back to a mixed number if possible.
Exercises
36. Multiply:
37. cups of water will be divided equally into jars. How much water will go into each jar? |
Lessons
1. Basics
2. Deductive Reasoning
3 - Parallel lines
4 - Congruent Triangles
6 - Inequalities
7 - Similar Polygon
8 - Rt. Triangles
9 - Circles
10 - Constructions
11 - Areas of 2D objects
12 - Areas and Volumes
13 - Coordinates
Areas of Polygons
Objective:
• Learning and using the formulas for the areas of rectangles, parallelograms, triangles, rhombuses, trapezoids, and regular polygons
• Learning the basics of the area of a polygon
Lesson 11-1 Areas of Polygons:
Base: Any side of a rectangle or other parallelogram
Altitude: Any segment perpendicular to the line containing the base from any point on the opposite side
Height: The length of an altitude.
Center of a regular polygon: The center of the circumscribed circle
Radius of a regular polygon: The distance from the center to a vertex
Central angle of a regular polygon: An angle formed by two radii drawn to consecutive vertices
Apothem of a regular polygon: The Perpendicular distance from the center of the polygon to a side
Postulate 17 The area of a square is the square of the length of a side. (A=s2) A=s2 = 22 = 4 Area of the square if 4 Postulate 18 If two figures are congruent, then they have the same area. Postulate 19 The area of a region is the sum of the areas of its non-overlapping parts. Area of ABCD = Area I +Area II Theorem 11-1 The area of a rectangle equals the product of its base and height. (A = bh) Area of ABCD = bh = 10(2) = 20 Theorem 11-2 The area of a parallelogram equals the product of a base and the height to that base (A=bh) Area of ABCD = bh = 20(10) = 200 Theorem 11-3 The area of a triangle equals half the product of a base and the height to that base. (A=1/2 bh) Area of Triangle ABC = 1/2 bh = 1/2(5)(10) = 25 Theorem 11-4 The area of a rhombus equals half the product of its diagonals (A= 1/2D1D2) AC= D1= 10 DB= D2= 5 Area of Rhombus = 1/2 D1D2 = 1/2 (10) (5)= 25 Theorem 11-5 The area of a trapezoid equals half the product of the height and the sum of the bases. (A = 1/2 h (b1 + b2)) Area of ABCD = 1/2 h (b1 + b2) =1/2 (3) (8+10)=27 Theorem 11-6 The area of a regular polygon is equal to half the product of the apothem and the perimeter. (A= 1/2ap) Area of Polygon = 1/2 ap =1/2 (8) (5)=20
Quickie Math Copyright (c) 2000 Team C006354 |
# Rounding
In mathematics, rounding refers to the process of approximating a number to a certain degree of accuracy. Rounding can be useful when dealing with large or complicated numbers, or when we only need an approximate answer. In this lesson, we will learn about the different methods of rounding and how to use them.
## Method of Rounding
There are several methods of rounding, but the two most common are rounding to a certain number of decimal places and rounding to a certain number of significant figures.
1. Rounding to a certain number of decimal places:
When rounding to a certain number of decimal places, we look at the digit to the right of the desired decimal place. If that digit is 5 or greater, we round up the previous digit; otherwise, we round down the previous digit. For example, if we want to round 3.14159 to 2 decimal places, we look at the third decimal place, which is 1. Since 1 is less than 5, we round down the previous digit, which is 4, and get 3.14.
1. Rounding to a certain number of significant figures:
When rounding to a certain number of significant figures, we look at the first non-zero digit from the left. We then count the total number of digits, including the first non-zero digit. If the next digit is 5 or greater, we round up the last digit; otherwise, we leave it unchanged. For example, if we want to round 123.45 to 3 significant figures, we look at the first non-zero digit, which is 1. We count the total number of digits, which is 5. Since the next digit, 4, is less than 5, we leave the last digit, 5, unchanged and get 123.
## Practice Exercise
Now let’s try some practice exercises to reinforce the concept of rounding:
1. Round 4.678 to 2 decimal places.
Answer: 4.68
1. Round 456.789 to 3 significant figures.
Answer: 457
1. Round 0.000045678 to 4 significant figures.
Answer: 0.000046
1. Round 987.654 to 1 decimal place.
Answer: 987.7
## Conclusion
Rounding is an essential skill in mathematics that allows us to estimate and approximate numbers to a certain degree of accuracy. By using the two most common methods of rounding, rounding to a certain number of decimal places and rounding to a certain number of significant figures, we can make our calculations more manageable and meaningful. |
# How do you find the equation of a point on a graph?
## How do you find the equation of a point on a graph?
Correct answer: To determine if a point is on a line you can simply subsitute the x and y coordinates into the equation. Another way to solve the problem would be to graph the line and see if it falls on the line. Plugging in will give which is a true statement, so it is on the line.
## What is a positive slope?
A positive slope means that two variables are positively related—that is, when x increases, so does y, and when x decreases, y decreases also. Graphically, a positive slope means that as a line on the line graph moves from left to right, the line rises.
How do I find slope with two points?
If you know two points on a line, you can use them to write the equation of the line in slope-intercept form. The first step will be to use the points to find the slope of the line. This will give you the value of m that you can plug into y = mx + b. The second step will be to find the y-intercept.
### Which is the equation of a line that has a slope of 4 and passes through Point 1/6 )?
Which is the equation of a line that has a slope of 4 and passes through point (1, 6)? The equation y- 7/2 = 1/2 (x, -4) is written in point-slope form.
### How do you put slope into standard form?
The standard form of such an equation is Ax + By + C = 0 or Ax + By = C. When you rearrange this equation to get y by itself on the left side, it takes the form y = mx +b. This is called slope intercept form because m is equal to the slope of the line, and b is the value of y when x = 0, which makes it the y-intercept.
What does point-slope form look like?
Point-slope is the general form y-y₁=m(x-x₁) for linear equations. It emphasizes the slope of the line and a point on the line (that is not the y-intercept).
#### How do you write slope?
Find the Equation of a Line Given That You Know a Point on the Line And Its Slope. The equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept. If you a point that a line passes through, and its slope, this page will show you how to find the equation of the line.
#### How do you find the points of an equation?
To find points on the line y = mx + b,
1. choose x and solve the equation for y, or.
2. choose y and solve for x.
What are the 3 slope formulas?
There are three major forms of linear equations: point-slope form, standard form, and slope-intercept form. We review all three in this article.
## How do you write in slope intercept form?
To write an equation in slope-intercept form, given a graph of that equation, pick two points on the line and use them to find the slope. This is the value of m in the equation. Next, find the coordinates of the y-intercept–this should be of the form (0, b). The y- coordinate is the value of b in the equation.
## What is the standard form of 200000?
200,000? 200,000 = 2 x 10^5.
How do you turn an equation into Point-slope form?
Thus, to convert to point-slope form, first convert to slope-intercept form, then move the constant term b to the left side of the equation (or isolate x and then divide by the y coefficient). Example: Convert 3x = 4y + 8 to point-slope form.
### How do I find the slope of the line?
The slope of a line characterizes the direction of a line. To find the slope, you divide the difference of the y-coordinates of 2 points on a line by the difference of the x-coordinates of those same 2 points.
### Do parallel lines have the same slope?
In other words, the slopes of parallel lines are equal. Note that two lines are parallel if their slopes are equal and they have different y-intercepts. In other words, perpendicular slopes are negative reciprocals of each other. Here is a quick review of the slope/intercept form of a line.
Is the regression line a good fit?
We use the least squares criterion to pick the regression line. The regression line is sometimes called the “line of best fit” because it is the line that fits best when drawn through the points. It is a line that minimizes the distance of the actual scores from the predicted scores.
#### What is the equation for two points?
Find the Equation of a Line Given That You Know Two Points it Passes Through. The equation of a line is typically written as y=mx+b where m is the slope and b is the y-intercept. If you know two points that a line passes through, this page will show you how to find the equation of the line.
#### What is the distance between 2 points?
For any two points there is exactly one line segment connecting them. The distance between two points is the length of the line segment connecting them. Note that the distance between two points is always positive. Segments that have equal length are called congruent segments.
Can a in standard form be negative?
Standard Form of a Linear Equation A shouldn’t be negative, A and B shouldn’t both be zero, and A, B and C should be integers.
Posted in Interesting |
Trigonometric identities solver
The Best Trigonometric identities solver
Here, we will show you how to work with Trigonometric identities solver. If your child understands the concept of addition, you can start by doing addition drills. For example, you can hand your child a set of counters and ask him or her to add up as many as they can. As your child gets more comfortable, you can ask him or her to keep track of the counters using a tally chart. You can also introduce subtraction by asking your child to count down from 10 by subtracting one number at a time. The main thing is to always keep it fun and make sure you have a good time!
The formula for radius is: The quick and simple way to solve for radius using our online calculator is: R> = (A2 − B2) / (C2 + D2) Where R> is the radius, A, B, C and D are any of the four sides of the rectangle, and A2> - B2> - C2> - D2> are the lengths of those sides. So if we have a square with side length 4cm and want to find its radius value, we would enter formula as 4 cm − 4 cm − 4 cm − 4 cm = 0 cm For example R> = (0cm) / (4 cm + 2cm) = 0.5cm In this case we would know that our square has an area of 1.5cm² and a radius of 0.5cm From here it is easy to calculate the area of a circle as well: (radius)(diameter) = πR>A>² ... where A> is
These are the best hard math problems with answers. The best way to learn math is practice and practice. Most people can do basic math, but some people find it more difficult than others. For these people, there are no shortcuts to learning. They have to practice every day and keep an eye on their progress. The good news is that they can get better with time if they put in the effort.
First determine the y intercept. The y intercept is the value where the line crosses the Y axis. It is sometimes referred to as the "zero" point, or reference point, along the line. The y intercept of an equation can be determined by drawing a vertical line down through the origin of each graph and placing a dot at the intersection of the two lines (Figure 1). When graphing a parabola, the y intercept is placed at the origin. When graphing a line with a slope 1, then both y-intercepts are placed at 0. When graphing a line with a slope >1, then both y-intercepts are moved to positive infinity. In order to solve for x intercept on an equation, first use substitution to solve for one of the variables in terms of another variable. Next substitute back into original equation to find x-intercept. In order to solve for y intercept on an equation, first use substitution to solve for one of the variables in terms of another variable. Next substitute back into original equation to find y-intercept. Example: Solve for x-intercept of y = 4x + 10 Solution: Substitute 4x + 5 = 0 into original problem: y = 4x + 10 => y = 4(x + 5) => y =
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+0
# PLEASE HELP ASAP
0
7
1
+24
Find the number of ordered pairs $$(a,b)$$ of integers such that
$$\frac{a + 2}{a + 5} = \frac{b}{4}$$
Feb 22, 2024
### 1+0 Answers
#1
+1484
0
Here's how to find the number of ordered pairs (a, b) of integers:
Multiply both sides by the common denominator (a + 5) in the first term:
a + 2 = b * (a + 5) / 4
Simplify and expand the expression on the right:
4 * (a + 2) = b * (a + 5)
4a + 8 = ab + 5b
Move all terms containing a to one side:
4a - ab = 5b - 8
a(4 - b) = 5b - 8
Factor out the greatest common factor (GCF) in both sides:
a(b - 4) = 5(b - 8/5)
Analyze the equation:
This equation implies that a and (b - 4) are factors of 5.
Since a and b are integers, the possible factors of 5 are 1, -1, 5, and -5.
Consider each factor combination:
Case 1: a = 1, (b - 4) = 5 - b = 9
Case 2: a = -1, (b - 4) = -5 - b = -1
Case 3: a = 5, (b - 4) = 1 - b = 5
Case 4: a = -5, (b - 4) = -1 - b = -3
Check for solutions:
All four cases lead to valid integer solutions for a and b: (1, 9), (-1, -1), (5, 5), and (-5, -3).
Therefore, there are 4 ordered pairs (a, b) of integers that satisfy the given equation.
Feb 22, 2024 |
# Geometry: The Pythagorean Theorem
## The Pythagorean Theorem
Now that you know how to show that two triangles are similar, you can use CSSTAP to find relationships between the sides of similar triangles. You can even create the theorems necessary to prove one of the most famous theorems in geometry: the Pythagorean Theorem. But before you can tackle the Pythagorean Theorem, you'll need a theorem about altitudes. I'll walk through an explanation of why the theorem is true, but I will not write out a formal proof.
• Theorem 14.1: The altitude drawn to the hypotenuse of a right triangle separates the right triangle into two right triangles that are similar to each other and to the original right triangle.
Figure 14.1 will help clarify what is going on. The altitude drawn to the hypotenuse has to originate at the vertex of the right angle (C) and is perpendicular to the hypotenuse (¯AB). Let's call the point where the altitude and the hypotenuse intersect D. You have three triangles to relate: the original triangle ABC and two new, smaller triangles ACD and CBD. Remember, you have to be careful about the order of the vertices. You have to match up corresponding vertices when representing the similarity of two triangles. You can use the AA Similarity Theorem and the transitive property of similarity (remember, similarity is an equivalence relation) to show that all three triangles are similar.
First, I'll show that ABC and CBD are similar. Both triangles are right triangles, so there's one pair of congruent angles. As for the second angle, notice that B is involved in both triangles. Because B ~= B, you've got a second pair of congruent angles, and by the AA Similarity Theorem, CBD ~= ABC. You need to match up your angles and your vertices: D is the right angle in CBD, which corresponds to C in ABC, B corresponds to itself, and BCD corresponds to A.
You'll use a similar argument to show that ABC and ACD are similar. D is the right angle in ACD, so it corresponds to C in ABC. A corresponds to itself, which leaves ACD and B to correspond.
Because ACD ~ ABC and ABC ~ CBD , the transitive property of ~ shows that ACD ~ CBD. Because CSSTAP, you know that
This should look a bit familiar. From this proportionality, you see that CD is the geometric mean of the lengths of the segments of the hypotenuse! The Pythagorean Theorem is just a couple of algebraic steps away.
Now that you know these three triangles are similar, you can break them apart to get a better handle on the proportionalities involved. You might want to refer to Figure 14.2.
Because ACD ~ ABC, you know that
Because ABC ~ CBD , you know that
• AB/BC = BC/BD.
If you cross-multiply, add the two equations together, and simplify, you will derive the Pythagorean Theorem! I'll write out the details in a two-column proof.
• The Pythagorean Theorem: The square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the lengths of the legs.
Translating the Pythagorean Theorem into the drawing in Figure 14.2, the goal is to prove that (AB)2 = (BC)2 + (AC)2.
Statements Reasons
1. ACD ~ ABC and ABC ~ CBD Previous discussion
2. AB/AC = AC/AD and AB/BC = BC/BD CSSTAP
4. (AB)(AD) = (AC)2 , (AB)(BD) = (BC)2 Means-Extremes property of proportionlity
5. (AB)(AD) + (AB)(BD) = (BC)2 + (AC)2 Addition property of equality
6. (AB)[(AD) + (BD)] = (BC)2 + (AC)2 Algebra (factoring)
7. (AB)[(AB)] = (BC)2 + (AC)2 Substitution (step 3 and 6)
8. (AB)2 = (BC)2 + (AC)2 Algebra
That's the most algebraic proof of the Pythagorean Theorem! The next time you prove the Pythagorean Theorem you will use areas of triangles.
Excerpted from The Complete Idiot's Guide to Geometry © 2004 by Denise Szecsei, Ph.D.. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with Alpha Books, a member of Penguin Group (USA) Inc.
To order this book direct from the publisher, visit the Penguin USA website or call 1-800-253-6476. You can also purchase this book at Amazon.com and Barnes & Noble. |
# 2017 AMC 8 Problems/Problem 7
## Problem
Let $Z$ be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of $Z$?
$\textbf{(A) }11\qquad\textbf{(B) }19\qquad\textbf{(C) }101\qquad\textbf{(D) }111\qquad\textbf{(E) }1111$
## Solution 1 (detailed explanation)
To check, if a number is divisible by 19, take its unit digit and multiply it by 2, then add the result to the rest of the number, and repeat this step until the number is reduced to two digits. If the result is divisible by 19, then the original number is also divisible by 19. Or we could just try to divide the example number the problem gave us by 19.
After we got 19 eliminated, we can see that the other options have a lot of 1's in them. The divisibility rule for 11 is add alternating digits up, then take the difference of them. The example number works like that. If we add variables, ABCDEF to make number ABCABC, we can see that (A+C+B) - (B+A+C) = 0. Which is divisible by 11, so our answer choice is $\boxed{\textbf{(A)}\ 11}$.
by: CHECKMATE2021 (edited by CHECKMATE2021)
## Solution 2
We are given one of the numbers that can represent $Z$, so we can just try out the options to see which one is a factor of $247247$. We get $\boxed{\textbf{(A)}\ 11}$.
~CHECKMATE2021
## Solution 3( similar to solution 1)
Similar to solution 1, let $Z=ABCABC$. To prove it is divisible by 11, we can compute its alternating sum, which is $A-B+C-A+B-C=0$, which is divisible by 11. Therefore, the answer is $\boxed{\textbf{(A)}\ 11}$.
~PEKKA
## Solution 4
We can find that all numbers like $Z$ are divisible by 1001. 1001 is divisible by 11 because when we divide it, we get a whole number. So, the answer is $\boxed{\textbf{(A)}\ 11}$.
~AfterglowBlaziken
## Video Solution (CREATIVE THINKING!!!)
~Education, the Study of Everything
~savannahsolver |
# Search by Topic
#### Resources tagged with Positive-negative numbers similar to Keeping it Safe and Quiet:
Filter by: Content type:
Stage:
Challenge level:
### There are 22 results
Broad Topics > Numbers and the Number System > Positive-negative numbers
### The History of Negative Numbers
##### Stage: 3, 4 and 5
This article -useful for teachers and learners - gives a short account of the history of negative numbers.
### Negative Numbers
##### Stage: 3
A brief history of negative numbers throughout the ages
### Strange Bank Account (part 2)
##### Stage: 3 Challenge Level:
Investigate different ways of making £5 at Charlie's bank.
### Strange Bank Account
##### Stage: 3 Challenge Level:
Imagine a very strange bank account where you are only allowed to do two things...
### Adding and Subtracting Positive and Negative Numbers
##### Stage: 3
How can we help students make sense of addition and subtraction of negative numbers?
### Making Sense of Positives and Negatives
##### Stage: 3
This article suggests some ways of making sense of calculations involving positive and negative numbers.
### Playing Connect Three
##### Stage: 3 Challenge Level:
In this game the winner is the first to complete a row of three. Are some squares easier to land on than others?
### 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?
##### Stage: 1 and 2
In this article for teachers, Liz Woodham describes resources on NRICH that can help primary-aged children get to grips with negative numbers.
### Up, Down, Flying Around
##### Stage: 3 Challenge Level:
Play this game to learn about adding and subtracting positive and negative numbers
### First Connect Three for Two
##### Stage: 2 and 3 Challenge Level:
First Connect Three game for an adult and child. Use the dice numbers and either addition or subtraction to get three numbers in a straight line.
### Being Collaborative - Primary Number
##### Stage: 1 and 2 Challenge Level:
Number problems at primary level to work on with others.
### Vector Racer
##### Stage: 3 and 4 Challenge Level:
The classic vector racing game.
### Tug Harder!
##### Stage: 2 Challenge Level:
In this game, you can add, subtract, multiply or divide the numbers on the dice. Which will you do so that you get to the end of the number line first?
### First Connect Three
##### Stage: 2 and 3 Challenge Level:
The idea of this game is to add or subtract the two numbers on the dice and cover the result on the grid, trying to get a line of three. Are there some numbers that are good to aim for?
### Sea Level
##### Stage: 2 Challenge Level:
The picture shows a lighthouse and many underwater creatures. If you know the markings on the lighthouse are 1m apart, can you work out the distances between some of the different creatures?
### Swimming Pool
##### Stage: 1 and 2 Challenge Level:
In this problem, we're investigating the number of steps we would climb up or down to get out of or into the swimming pool. How could you number the steps below the water?
### Pair Sums
##### Stage: 3 Challenge Level:
Five numbers added together in pairs produce: 0, 2, 4, 4, 6, 8, 9, 11, 13, 15 What are the five numbers?
### 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.
### Consecutive Negative Numbers
##### Stage: 3 Challenge Level:
Do you notice anything about the solutions when you add and/or subtract consecutive negative numbers?
### Weights
##### Stage: 3 Challenge Level:
Different combinations of the weights available allow you to make different totals. Which totals can you make?
### Connect Three
##### Stage: 3 and 4 Challenge Level:
Can you be the first to complete a row of three? |
# Definition of Ellipse
We will discuss the definition of ellipse and how to find the equation of the ellipse whose focus, directrix and eccentricity are given.
An ellipse is the locus of a point P moves on this plane in such a way that its distance from the fixed point S always bears a constant ratio to its perpendicular distance from the fixed line L and if this ratio is less than unity.
An ellipse is the locus of a point in a plane which moves in the plane in such a way that the ratio of its distance from a fixed point (called focus) in the same plane to its distance from a fixed straight line (called directrix) is always constant which is always less than unity.
The constant ratio usually denoted by e (0 < e < 1) and is known as the eccentricity of the ellipse.
If S is the focus, ZZ' is the directrix and P is any point on the ellipse, then by definition
$$\frac{SP}{PM}$$ = e
⇒ SP = e ∙ PM
The fixed point S is called a Focus and the fixed straight line L the corresponding Directrix and the constant ratio is called the Eccentricity of the ellipse.
Solved example to find the equation of the ellipse whose focus, directrix and eccentricity are given:
Determine the equation of the ellipse whose focus is at (-1, 0), directrix is 4x + 3y + 1 = 0 and eccentricity is equal to $$\frac{1}{√5}$$.
Solution:
Let S (-1, 0) be the focus and ZZ' be the directrix. Let P (x, y) be any point on the ellipse and PM be perpendicular from P on the directrix. Then by definition
SP = e.PM where e = $$\frac{1}{√5}$$.
⇒ SP$$^{2}$$ = e$$^{2}$$ PM$$^{2}$$
⇒ (x + 1)$$^{2}$$ + (y - 0)$$^{2}$$ = $$(\frac{1}{\sqrt{5}})^{2}[\frac{4x + 3y + 1}{\sqrt{4^{2} + 3^{2}}}]$$
⇒ (x + 1)$$^{2}$$ + y$$^{2}$$ = $$\frac{1}{25}$$$$\frac{4x + 3y + 1}{5}$$
⇒ x$$^{2}$$ + 2x + 1 + y$$^{2}$$ = $$\frac{4x + 3y + 1}{125}$$
⇒ 125x$$^{2}$$ + 125y$$^{2}$$ + 250x + 125 = 0, which is the required equation of the ellipse.
The Ellipse
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1. ### Addition of Three 1-Digit Numbers | Add 3 Single Digit Numbers | Steps
Sep 19, 24 01:15 AM
To add three numbers, we add any two numbers first. Then, we add the third number to the sum of the first two numbers. For example, let us add the numbers 3, 4 and 5. We can write the numbers horizont…
2. ### Adding 1-Digit Number | Understand the Concept one Digit Number
Sep 18, 24 03:29 PM
Understand the concept of adding 1-digit number with the help of objects as well as numbers.
3. ### Addition of Numbers using Number Line | Addition Rules on Number Line
Sep 18, 24 02:47 PM
Addition of numbers using number line will help us to learn how a number line can be used for addition. Addition of numbers can be well understood with the help of the number line.
4. ### Counting Before, After and Between Numbers up to 10 | Number Counting
Sep 17, 24 01:47 AM
Counting before, after and between numbers up to 10 improves the child’s counting skills. |
How to Do Long Division How to Do Long Division In arithmetic, long division is a standard procedure suitable for dividing simple or complex multidigit numbers. It breaks down a division problem into a series of easier steps. As in all division problems, one number, called the dividend, is divided by another, called the divisor, producing a result called the quotient. It enables computations involving arbitrarily large numbers to be performed by following a series of simple steps.[1] The abbreviated form of long division is called short division, which is almost always used instead of long division when the divisor has only one digit. Method In English-speaking countries, long division does not use the slash (/) or obelus (รท) signs, instead displaying the dividend, divisor, and (once it is found) quotient in a tableau.The process is begun by dividing the left-most digit of the dividend by the divisor. The quotient (rounded down to an integer) becomes the first digit of the result, and the remainder is calculated (this step is notated as a subtraction). This remainder carries forward when the process is repeated on the following digit of the dividend (notated as 'bringing down' the next digit to the remainder). When all digits have been processed and no remainder is left, the process is complete.n example is shown below, representing the division of 500 by 4 (with a result of 125). Know More About :- Inverse Trig Functions
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125 (Explanations) 4)500 4 (4 × 1 = 4) 10 (5 - 4 = 1) 8 (4 × 2 = 8) 20 (10 - 8 = 2) 20 (4 × 5 = 20) 0 (20 - 20 = 0) In the above example, the first step is to find the shortest sequence of digits starting from the left end of the dividend, 500, that the divisor 4 goes into at least once; this shortest sequence in this example is simply the first digit, 5. The largest number that the divisor 4 can be multiplied by without exceeding 5 is 1, so the digit 1 is put above the 5 to start constructing the quotient. Next, the 1 is multiplied by the divisor 4, to obtain the largest whole number (4 in this case) that is a multiple of the divisor 4 without exceeding the 5; this product of 1 times 4 is 4, so 4 is placed underneath the 5. Next the 4 under the 5 is subtracted from the 5 to get the remainder, 1, which is placed under the 4 under the 5. This remainder 1 is necessarily smaller than the divisor 4. Next the first as-yet unused digit in the dividend, in this case the first digit 0 after the 5, is copied directly underneath itself and next to the remainder 1, to form the number 10. At this point the process is repeated enough times to reach a stopping point: The largest number by which the divisor 4 can be multiplied without exceeding 10 is 2, so 2 is written above the 0 that is next to the 5 — that is, directly above the last digit in the 10. Then the latest entry to the quotient, 2, is multiplied by the divisor 4 to get 8, which is the largest multiple of 4 that does not exceed 10; so 8 is written below 10, and the subtraction 10 minus 8 is performed to get the remainder 2, which is placed below the 8. This remainder 2 is necessarily smaller than the divisor 4. The next digit of the dividend (the last 0 in 500) is copied directly below itself and next to the remainder 2, to form 20. Then the largest number by which the divisor 4 can be multiplied without exceeding 20 is ascertained; this number is 5, so 5 is placed above the last dividend digit that was brought down (i.e., above the rightmost 0 in 500).Then this new quotient digit 5 is multiplied by the divisor 4 to get 20, which is written at the bottom below the existing 20. Then 20 is subtracted from 20, yielding 0, which is written below the 20. We know we are done now because two things are true: Read More About :- What are Rational Numbers
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Thank You
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How to Do Long Division
How to Do Long Division Page : 1/3 Know More About :- Inverse Trig Functions Method Page : 2/3 Read More About :- What are Rational Numbers...
How to Do Long Division
How to Do Long Division Page : 1/3 Know More About :- Inverse Trig Functions Method Page : 2/3 Read More About :- What are Rational Numbers...
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Formulas for Calculating
The Present Value of an Ordinary Annuity
1) Solving the Present Value A friend offers to buy your car if he can pay you \$100 per month for 3 years at an annual interest rate of 7.5% What is the present value of all these payments? The payment per period ('p') is \$100 the total number of periods ('n') is: 12 periods per year for 3 years, equals 12*3 = 36 the interest rate is .075 ÷ 12 = 0.00625 Putting these numbers into the formula: Present Value = 100 * [(1-(1.00625)^-36)/.00625] Present Value = 100 * [(1-0.799075542783109)/.00625] Present Value = 100 * [(0.200924457216891)/.00625] Present Value = 100 * 32.1479131547025 Present Value = 3,214.79
2) Solving the Periodic Payment (Using the data from question 1) A 3 year 7.5% monthly annuity has a present value of \$3,214.79. What was the periodic payment? The total number of periods ('n') is: 12 periods per year for 3 years, equals 12*3 = 36 the interest rate is .075 ÷ 12 = 0.00625 the present value is \$3,214.79 Putting these numbers into the formula: Payment= 3,214.79 ÷ [(1-(1.00625)^-36) / 0.00625] Payment= 3,214.79 ÷ ((1-0.799075542783109) / 0.00625) Payment= 3,214.79 ÷ (0.200924457216891 / 0.00625) Payment= 3,214.79 ÷ 32.1479 Payment= 100.00
3) Solving for Years (Using the data from question 1) A 7.5% annuity with \$100 monthly payments has a present value of \$3,214.79 How many years is this annuity for? the interest rate is .075 ÷ 12 = 0.00625 the present value is \$3,214.79 the monthly payment =100 Putting these numbers into the formula: n=[Log(100) ― Log((100 -(0.00625)*3,214.79))] / Log(1.00625) n=[Log(100) ― Log((100 -(20.0924375)))] / Log(1.00625) n=[Log(100) ― Log((79.9075625))] / Log(1.00625) n=(2 -1.9025878832)/ 0.00270589337592 n=0.0974121168 / 0.00270589337592 n=35.9999834682621 n=36 total number of periods This is a monthly annuity so, dividing 36 by 12 equals 3 years.
4) Solving for Rate The present value of an ordinary annuity formula cannot be solved for rate. The present value calculator solves for rate by using a trial and error process.
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# Question Video: Estimating the Result of an Addition Sentence by Rounding Each Number to the Nearest Ten Mathematics • 4th Grade
Estimate 2,145 + 123 by rounding each number to the tens place.
02:15
### Video Transcript
Estimate 2,145 plus 123 by rounding each number to the tens place.
In this question, we have to estimate the sum of our two numbers by rounding each number to the nearest ten. Let’s start by rounding 2,145 to the nearest ten. The tens digit in 2,145 is a four, worth 40. So we know 2,145 comes between 2,140 and 2,150. So to help us decide whether to round our number down to 2,140 or up to 2,150, we need to look to the ones digit. The ones digit is a five. If the ones digit is worth five or more, we have to round up to 2,150. So the nearest 10 to 2,145 is 2,150. We had to round up.
Let’s round our next number to the nearest ten. The tens digit in 123 is a two, worth two 10s. So we know the two 10s our number is between are 120 and 130. To help us decide whether to round down to 120 or up to 130, we need to look at the ones digit. We know that three is less than five, so we have to round down to 120. Now all we need to do is add together 2,150 and 120, which is 2,270. We estimated 2,145 plus 123 by rounding each number to the tens place. Our estimate is 2,270. |
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# Difference between revisions of "1992 AHSME Problems/Problem 11"
## Problem
$[asy] draw(circle((0,0),18),black+linewidth(.75)); draw(circle((0,0),6),black+linewidth(.75)); draw((-18,0)--(18,0)--(-14,8*sqrt(2))--cycle,black+linewidth(.75)); dot((-18,0));dot((18,0));dot((-14,8*sqrt(2))); MP("A",(-18,0),W);MP("C",(18,0),E);MP("B",(-14,8*sqrt(2)),W); [/asy]$
The ratio of the radii of two concentric circles is $1:3$. If $\overline{AC}$ is a diameter of the larger circle, $\overline{BC}$ is a chord of the larger circle that is tangent to the smaller circle, and $AB=12$, then the radius of the larger circle is
$\text{(A) } 13\quad \text{(B) } 18\quad \text{(C) } 21\quad \text{(D) } 24\quad \text{(E) } 26$
## Solution (Similarity)
We are given that $BC$ is tangent to the smaller circle. Using that, we know where the circle intersects $BC$, it creates a right triangle. We can also point out that since $AC$ is the diameter of the bigger circle and triangle $ABC$ is inscribed the semi-circle, that angle $B$ is a right angle. Therefore, we have $2$ similar triangles. Let's label the center of the smaller circle (which is also the center of the larger circle) as $D$. Let's also label the point where the smaller circle intersects $BC$ as $E$. So $ABC$ is similar to $DEC$. Since $DE$ is the radius of the smaller circle, call the length $x$ and since $DC$ is the radius of the bigger circle, call that length $3x$. The diameter, $AC$ is $6x$. So,
$\frac{AB}{AC} = \frac {DE}{DC} \Rightarrow \frac{12}{6x} = \frac{x}{3x}$
$\fbox{B}$
1992 AHSME (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions |
# It Will Cost Ya!
This week’s math plan is to wrap up our unit on money. Next Monday we will do a #MentalMathMonday focus on money. Here are some ways to practice at home if your child needs some extra support or just so you know what we are working on at school.
How much money? Grab some change and have your child a)count it out coin by coin as needed, then b) when they are ready ask them to add it in their head before telling you how much. a)\$2.00+\$1.00+\$0.25=\$3.25 b)\$3.25
Try it again. This time your child could a)move the coins into amounts that are easier for them to add together (put the quarter alone, then add 2 dimes and a nickel, then the other 2 dimes and count it as \$0.25+\$0.25+\$0.20=\$0.70) b) start from different points and add coins as they are near (this could be counted as dime, dime, quarter nickel, dime, dime) c)group all like coins together then add it up (one quarter, one nickel and 4 dimes). Counting the coins in different ways just helps to solidify their understanding
Add the value of the coins and think about how much more they would need to equal a certain amount (in this example: \$1.00). This one starts at \$0.45 so kids could add on dimes (55,65,75,85,95) then a nickel to make \$1.00 (the answer being 5 dimes and 1 nickel which is \$0.55) or know that \$0.50 would being them to \$0.95 and \$0.05 more would be a \$1.00 (the answer being \$0.50+\$0.05=\$0.55) or knowing compatible numbers to 100 is a great tool (if 45+55=100 then \$0.45 +\$0.55=\$1.00)
Start with an amount (in this example \$1.00 and see what they would need to subtract to end up with a known amount of change (in this example \$0.30). \$1.00-\$0.30 is the same thinking as 100-30. Knowing those compatible numbers to 100 comes in handy again! (Compatible numbers are number pairs that add to a certain number…we’ve already done a lot of practice with 10,20 and 50…now would be a great time to really know the 100 facts!)
Counting money and writing it in different ways…using dollars or cents
Money is similar to time for some kids because we don’t necessarily use actual coins and bills asa often as we might use debit and credit. Give your child real life opportunities to use money to pay for things when you are out, set up a store at home, use real or pretend coins (or you can make your own pretend money if you wanted). If your child isn’t used to telling time or using money these skills won’t happen overnight. Have fun and practice a little each day. Choose one skill to work on at a time.
*There are some students who still need a little support with recognizing coins and knowing their value. Please review toonies, loonies, quarters, dimes, nickels and even pennies at home…I know we don’t use pennies anymore, but they come in handy for math skills! Knowing how many pennies you need to equal the value of a toonie for example is also a nice review of place value understanding.
*You can also do things like how many nickels would you need to make \$0.25 or \$0.75, etc. Or what is the fewest amount of coins you could need to make \$0.60 (and have them draw it out). OR show you all the ways they can think of to make \$1.00.
Happy Practice!
Danielle |
# Square Roots
You already know about squaring, 4² = 16, 7² = 49, and so forth. The opposite, or inverse, of squaring is taking the square root. The symbol for taking the square root, or any root for that matter, , is called a “radical” and is used like this: or
You can take any integer, square it, and end up with a natural number, but it doesn’t work that way with square roots. Take for example. There’s no integer that squares to 5 but there are two ways to handle that.
The first is simply to leave it in radical form and the second is to find the decimal equivalent. For this, you’ll need a calculator (there is a formula for determining a root but that’s a subject for another day). Using a calculator, we get:
≈ 2.236068
Simplify:
Adding and subtracting radicals is similar to adding and subtracting polynomial terms, in that you can only combine like terms. You cannot combine 3x and 2y, so also you cannot combine . You can’t combine 3 and 5x; likewise, you can’t combine .
Simplify:
Don’t assume that expressions with unlike radicals can’t be simplified. It’s possible that, after simplifying the radicals, the expression actually can be simplified. For instance:
Simplify:
Here is an important property of square roots:
Write as the product of two radicals:
Because square roots are flexible with multiplication, you can factor inside a radical. You can then split the roots according to the factors. Likewise, when multiplying radicals, you can multiply the terms within the radicals.
Simplify by writing as one radical:
…which can be further simplified…
Simplify:
The easiest way to factor inside a radical is to take the primes of each term then take out any pairs. For example:
Simplify:
That was pretty simple but, it’s important that you understand the technique to aviod mistakes. In the example above we took the square root of 5² thus wound up with 5 outside the radical. Likewise, we took the square root of x² three times and outside the radical we must take the power of thexs. So the result is x³ NOT 3x. |
Counting Outcomes and Tree Diagrams
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# Counting Outcomes and Tree Diagrams - PowerPoint PPT Presentation
Counting Outcomes and Tree Diagrams. Lesson 10.2. Theoretical probabilities can be determined by finding the ratio of the number of desired outcomes to the number of possible equally likely outcomes.
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## Counting Outcomes and Tree Diagrams
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Presentation Transcript
1. Counting Outcomes and Tree Diagrams Lesson 10.2
2. Theoretical probabilities can be determined by finding the ratio of the number of desired outcomes to the number of possible equally likely outcomes. • In some cases it can be difficult to count the number of possible or desired outcomes. You can make this easier by organizing information and counting outcomes using a tree diagram.
3. Example A • A national advertisement says that every puffed-barley cereal box contains a toy and that the toys are distributed equally. Talya wants to collect a complete set of the different toys from cereal boxes. • a. If there are two different toys, what is the probability that she will find both of them in her first two boxes? • b. If there are three different toys, what is the probability that she will have them all after buying her first three boxes?
4. a. In this tree diagram, the first branching represents the possibilities for the first box and the second branching represents the possibilities for the second box. Thus, the four paths from left to right represent all outcomes for two boxes and two toys. Path 2 and Path 3 contain both toys. If the advertisement is accurate about equal distribution of toys, then the paths are equally likely. So the probability of getting both toys is 2/4, or 0.5.
5. This tree diagram shows all the toy possibilities for three boxes. There are 27 possible paths. You can determine this quickly by counting the number of branches on the far right. Six of the 27 paths contain all three toys, as shown. Because the paths are equally likely, the probability of having all three toys is 6/27, or 0.2222 .
6. The Multiplication Rule • Step 1 On your paper, redraw the tree diagram for Example A, part a. This time, write the probability on each branch. Then find the probability of each path.
7. Step 2 Redraw the tree diagram for Example A, part b. Indicate the probability on each branch, and also write the probability of each path. • What is the sum of the probabilities of all possible paths? • What is the sum of the probabilities of the highlighted paths?
8. Step 3 Suppose the national advertisement mentioned in Example A listed four different toys distributed equally in a huge supply of boxes. Draw only as much of a tree diagram as you need to in order to answer these questions: • a. What would be P(Toy 2) in Talya’s first box? Talya’s second box? Third box? P(any particular toy in any particular box)? • b. In these situations, does the toy she finds in one box influence the probability of there being a particular toy in the next box? • c. One outcome that includes all four toys is Toy 3, followed by Toy 2, followed by Toy 4, followed by Toy 1. What is the probability of this outcome? Another outcome would be the same four toys in a different order. How many such outcomes are there? Are they all equally likely?
9. Step 4 Write a statement explaining how to use the probabilities of a path’s branches to find the probability of the path. • Step 5 What is P(obtaining the complete set in the first four boxes)?
10. Example B • Mr. Roark teaches three classes. Each class has 20 students. His first class has 12 sophomores, his second class has 8 sophomores, and his third class has 10 sophomores. If he randomly chooses one student from each class to participate in a competition, what is the probability that he will select three sophomores?
11. You could consider drawing a tree with 20 branches representing the students in the first class. These would split into 20 branches for the second class, and each of these paths would split into 20 branches for the third class. This would be a tree with 8000 paths! Each class has 20 students. His first class has 12 sophomores, his second class has 8 sophomores, and his third class has 10 sophomores.
12. Instead, you can draw two branches for each stage of the selection process. One branch represents a choice of a sophomore (S) and one represents a choice of a nonsophomore (NS). • This tree shows all eight possible outcomes. However, the outcomes are not equally likely. • For the first class, the probability of choosing a sophomore is 12/20 =0.6, and the probability of choosing a nonsophomore is 1-0.6 =0.4. • Calculate the probabilities for the second and third classes, and represent them on a tree diagram, as shown. Each class has 20 students. His first class has 12 sophomores, his second class has 8 sophomores, and his third class has 10 sophomores.
13. The uppermost path represents a sophomore being chosen from each class. In the investigation you learned to find the probability of a path by multiplying the probabilities of the branches. So the probability of choosing three sophomores is (0.6)(0.4)(0.5), or 0.12.
14. In Example B, the probability of choosing a sophomore in the second class is the same, regardless of whether a sophomore was chosen in the first class. These events are called independent. Events are independent if the occurrence of one has no influence on the probability of the other.
15. Example C • Devon is going to draw three cards, one after the other, from a standard deck. What is the probability that she will draw exactly two hearts?
16. The outcome of each draw can be represented by branches on a tree diagram. Rather than list all of the cards in the deck, classify the result of each draw as a heart (H) or nonheart (NH). Study the tree diagram. There are 52 cards in a standard deck, with 13 cards in each suit. Notice that the denominator of each probability for the second draw is 51 because there are only 51 cards left to choose from.
17. The three highlighted paths show the ways to get exactly two hearts (each has a probability of 0.046 of occurring)
18. For example, the top highlighted path shows getting a heart on the first draw, a heart on the second draw, and a nonheart on the third draw.
19. The probability of any event that can occur along multiple paths is the sum of the probabilities of those paths. Thus, the probability that Devon gets exactly two hearts is the sum of the probabilities of the highlighted paths, which is about 0.046 + 0.046 + 0.046 = 0.138.
20. In a tree diagram, the probability for each simple event along a path may depend on what occurred on the previous branches. In Example C, for instance, the probabilities for the second draw are dependent on the result of the first draw. That is, the events are dependent events.
21. P(A|B) • You can use conditional probability notation to describe both independent and dependentevents. • The probability of event A given event B is denoted with a vertical line: P(A|B) • For example, to denote the probability of drawing a heart on the second draw given that you have already drawn a nonheart, you write P(H2|NH1).
22. What does P(H2|H1) denotes? • The probability of drawing a heart on the second draw given that you have already drawn a heart. • Using the tree diagram from Example C, you can see that P(H2|NH1) =13/51 • and P(H2|H1) =12/51.
23. If events A and B are independent, then the probability of A is the same whether B happens or not. • In this case, P(A|B)= P(A). In Example B, the probability of choosing a sophomore in the second class (S2) is the same whether Mr. Roark chooses a sophomore or a nonsophomore from his first class. This means that P(S2|S1) = P(S2|NS1) = P(S2).
24. This tree could represent any two-stage event with two options at each stage. To find the probability of event C, you must add all paths, or outcomes, that contain C. P(C)= P(A and C)+ P(B and C) = P(A) • P(C|A) + P(B) •P(C|B)
25. The main idea of this lesson is that a tree diagram allows breaking down events into stages that are independent so that the multiplication rule applies: The probability of the event is the product of the probabilities of the branches of its path. If the probability of one of the events depends on which path it’s on, it’s a conditional probability. |
# How do you find a standard form equation for the line with (-4,2) and (6,8)?
Aug 21, 2016
$3 x - 5 y + 22 = 0$
#### Explanation:
The equation of line joining $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is given by
$\frac{y - {y}_{1}}{{y}_{2} - {y}_{1}} = \frac{x - {x}_{1}}{{x}_{2} - {x}_{1}}$.
Hence equation of line joining $\left(- 4 , 2\right)$ and $\left(6 , 8\right)$ is
$\frac{y - 2}{8 - 2} = \frac{x - \left(- 4\right)}{6 - \left(- 4\right)}$ or
$\frac{y - 2}{6} = \frac{x + 4}{10}$ or
$10 y - 20 = 6 x + 24$ or
$6 x - 10 y + 24 + 20 = 0$ or
$6 x - 10 y + 44 = 0$ or
$3 x - 5 y + 22 = 0$ |
# 2001 IMO Problems/Problem 1
## Problem
Consider an acute triangle $\triangle ABC$. Let $P$ be the foot of the altitude of triangle $\triangle ABC$ issuing from the vertex $A$, and let $O$ be the circumcenter of triangle $\triangle ABC$. Assume that $\angle C \geq \angle B+30^{\circ}$. Prove that $\angle A+\angle COP < 90^{\circ}$.
## Solution
Take $D$ on the circumcircle with $AD \parallel BC$. Notice that $\angle CBD = \angle BCA$, so $\angle ABD \ge 30^\circ$. Hence $\angle AOD \ge 60^\circ$. Let $Z$ be the midpoint of $AD$ and $Y$ the midpoint of $BC$. Then $AZ \ge R/2$, where $R$ is the radius of the circumcircle. But $AZ = YP$ (since $AZYP$ is a rectangle).
Now $O$ cannot coincide with $Y$ (otherwise $\angle A$ would be $90^\circ$ and the triangle would not be acute-angled). So $OP > YP \ge R/2$. But $PC = YC - YP < R - YP \le R/2$. So $OP > PC$.
Hence $\angle COP < \angle OCP$. Let $CE$ be a diameter of the circle, so that $\angle OCP = \angle ECB$. But $\angle ECB = \angle EAB$ and $\angle EAB + \angle BAC = \angle EAC = 90^\circ$, since $EC$ is a diameter. Hence $\angle COP + \angle BAC < 90^\circ$.
## Solution 2
Notice that because $\angle{PCO} = 90^\circ - \angle{A}$, it suffices to prove that $\angle{POC} < \angle{PCO}$, or equivalently $PC < PO.$
Suppose on the contrary that $PC > PO$. By the triangle inequality, $2 PC = PC + PC > PC + PO > CO = R$, where $R$ is the circumradius of $ABC$. But the Law of Sines and basic trigonometry gives us that $PC = 2R \sin B \cos C$, so we have $4 \sin B \cos C > 1$. But we also have $4 \sin B \cos C \le 4 \sin B \cos (B + 30^\circ) = 2 (\sin (2B + 30^\circ) - \sin 30^\circ) \le 2 (1 - \frac{1}{2}) = 1$ because $\angle{C} \ge \angle{B} + 30^\circ$, and so we have a contradiction. Hence $PC < PO$ and so $\angle{PCO} + \angle{A} < 90^\circ$, as desired. |
# The High School Gym
Alignments to Content Standards: F-IF.B.6
The school assembly is being held over the lunch hour in the school gym. All the teachers and students are there by noon and the assembly begins. About 45 minutes after the assembly begins, the temperature within the gym remains a steady 77 degrees Fahrenheit for a few minutes. As the students leave after the assembly ends at the end of the hour, the gym begins to slowly cool down.
Let $T$ denote the temperature of the gym in degrees Fahrenheit and $M$ denote the time, in minutes, since noon.
1. Is $M$ a function of $T$? Explain why or why not.
2. Explain why $T$ is a function of $M$, and consider the function $T=g(M)$. Interpret the meaning of $g(0)$ in the context of the problem.
3. Becky says: “The temperature increased 5 degrees in the first half hour after the assembly began.” Which of the following equations best represents this statement? Explain your choice.
1. $g(30)=5$
2. $\frac{g(30)-g(0)}{30}=5$
3. $g(30)-g(0)=5$
4. $T=g(30)-5$
4. Which of these choices below represents the most reasonable value for the quantity $\frac{g(75)-g(60)}{15}$? Explain your choice:
1. 4
2. 0.3
3. 0
4. -0.2
5. -5
## Solution
1. No, $M$ is not a function of $T$. The problem states that the temperature, $T$, remains a steady 77 degrees Fahrenheit for a few minutes, $M$. Thus, we know that there is more than one value of $M$, or more than one minute, associated with the temperature $T=77$. This means that if $M$ were a function of $T$ there would be more than one output M associated with the single input value of $T=77$. By definition of a function each input must be associated with exactly one output. Therefore, $M$ cannot be a function of $T$ because there would be more than one output associated with the input $T=77$.
2. In constrast the previous part, it is the case that for each number of minutes $M$ after noon, there is one and only one temperature $T$ in the room, so $T$ is a function of $M$. The quantity $g(0)$ is the value of $T$ when $M=0$. From the context given in the problem we know that $M$ denotes the time in minutes since noon. Thus, $M=0$ is exactly noon, so $g(0)$ is the temperature in the high school gym in degrees Fahrenheit at exactly noon.
3. Let’s look at the meaning of each equation:
1. $g(30)=5$ says the temperature at 12:30 p.m. is 5 degrees Fahrenheit, which is not what Becky said.
2. $\frac{g(30)-g(0)}{30}=5$ says that the average rate of change of the temperature of the gym between noon and 12:30 p.m. is 5 degrees per minute, which is not what Becky said.
3. $g(0)-g(30)=5$ says that the difference in the temperature of the gym at 12:30 and the temperature of the gym at noon is 5 degrees, which is the same thing as saying the temperature increased by 5 degrees Fahrenheit.
4. $T=g(30)-5$ says that the temperature equals the temperature at 12:30 minus five degrees, which is not what Becky said.
4. As in the previous part, we recognize the quantity $g(75)-g(60)$ as the change in the temperature in the room in the 15 minutes following the students' departure from the gym. We are told that the temperature decreases as the students begin to leave, so we know this quantity must be negative.
Next, we note that the quantity in question, namely $\frac{g(75)-g(60)}{15}$, is precisely the average rate of change of the temperature over these first fifteen minutes, and so is also negative. This eliminates the first three choices for the value of this quantity immediately.
Finally, we note that the last option corresponds to an average decrease of $5$ degrees per minute, which would combine for a 75-degree drop in this 15-minutes span. Since this would leave the gym in a frozen state, we conclude that this was an implausibly fast rate of temperature decrease. This leaves only the value $-0.2$, corresponding to a significantly more plausible temperature drop of 3 degrees over these 15 minutes. |
# Difference between revisions of "2007 AIME I Problems/Problem 8"
## Problem
The polynomial $P(x)$ is cubic. What is the largest value of $k$ for which the polynomials $Q_1(x) = x^2 + (k-29)x - k$ and $Q_2(x) = 2x^2+ (2k-43)x + k$ are both factors of $P(x)$?
## Solution
### Solution 1
We can see that $Q_1$ and $Q_2$ must have a root in common for them to both be factors of the same cubic.
Let this root be $a$.
We then know that $a$ is a root of $Q_{2}(x)-2Q_{1}(x) = 2x^{2}+2kx-43x+k-2x^{2}-2kx+58x+2k = 15x+3k = 0$ , so $x = \frac{-k}{5}$.
We then know that $\frac{-k}{5}$ is a root of $Q_{1}$ so we get: $\frac{k^{2}}{25}+(k-29)\left(\frac{-k}{5}\right)-k = 0 = k^{2}-5(k-29)(k)-25k = k^{2}-5k^{2}+145k-25k$ or $k^{2}=30k$, so $k=30$ is the highest.
We can trivially check into the original equations to find that $k=30$ produces a root in common, so the answer is $030$.
### Solution 2
Again, let the common root be $a$; let the other two roots be $m$ and $n$. We can write that $(x - a)(x - m) = x^2 + (k - 29)x - k$ and that $2(x - a)(x - n) = 2\left(x^2 + \left(k - \frac{43}{2}\right)x + \frac{k}{2}\right)$.
Therefore, we can write four equations (and we have four variables), $a + m = 29 - k$, $a + n = \frac{43}{2} - k$, $am = -k$, and $an = \frac{k}{2}$.
The first two equations show that $m - n = 29 - \frac{43}{2} = \frac{15}{2}$. The last two equations show that $\frac{m}{n} = -2$. Solving these show that $m = 5$ and that $n = -\frac{5}{2}$. Substituting back into the equations, we eventually find that $k = 30$.
### Solution 3
Since $Q_1(x)$ and $Q_2(x)$ are both factors of $P(x)$, which is cubic, we know the other factors associated with each of $Q_1(x)$ and $Q_2(x)$ must be linear. Let $Q_1(x)R(x) = Q_2(x)S(x) = P(x)$, where $R(x) = ax + b$ and $S(x) = cx + d$. Then we have that $((x^2 + (k-29)x - k))(ax + b) = ((2x^2+ (2k-43)x + k))(cx + d)$. Equating coefficients, we get the following system of equations:
\begin{align} a = 2c \\ b = -d \\ 2c(k - 29) - d = c(2k - 43) + 2d \\ -d(k - 29) - 2ck = d(2k - 43) + ck \end{align}
Using equations $(1)$ and $(2)$ to make substitutions into equation $(3)$, we see that the $k$'s drop out and we're left with $d = -5c$. Substituting this expression for $d$ into equation $(4)$ and solving, we see that $k$ must be $\boxed {30}$.
~ anellipticcurveoverq
### Solution 4
Notice that if the roots of $Q_1(x)$ and $Q_2(x)$ are all distinct, then $P(x)$ would have four distinct roots, which is a contradiction since it's cubic. Thus, $Q_1(x)$ and $Q_2(x)$ must share a root. Let this common value be $r.$
Thus, we see that we have \begin{align*} & r^2 + (k - 29)r - k = 0, \\ & 2r^2 + (2k - 43)r + k = 0. \end{align*} Adding the two equations gives us $$3r^2 + 3k - 72r = 0 \implies r = 0, 24 - k.$$ Now, we have two cases to consider. If $r = 0,$ then we have that $Q_1(r) = 0 = r^2 + (k - 29)r - k \implies k = 0.$ On the other hand, if $r = 24 - k,$ we see that $$Q_1(r) = 0 = (24 - k)^2 + (k - 29)(24 - k) - k \implies k = \boxed{030}.$$ This can easily be checked to see that it does indeed work, and we're done!
~Ilikeapos |
# Trigonometry
Page Contents
## Pythagoras' Theorem
Consider a right angled triangle (drawn in red) with hypotenuse ~a, and shorter sides ~b and ~c as shown.
Draw three other "copies" of the triangle as shown. Now:
Area of red triangle = ~b~c / 2
Area of large square = (~b + ~c)^2 = ~b^2 + ~c^2 + 2~b~c
Area of small square = ~a^2
But the large square is just the sum of the small square and the four triangles. _ So:
~b^2 + ~c^2 + 2~b~c = ~a^2 + 2~b~c
So _ _ _ ~b^2 + ~c^2 = ~a^2
## Trigonometrical Definitions
With reference to the right-angled triangle (right): _ _ _ sine A _ ( #:=#: _ sin A ) _ #:= _ opposite ./ hypotenuse . cosine A _ ( #:=#: _ cos A ) _ #:= _ adjacent ./ hypotenuse . tangent A _ ( #:=#: _ tan A ) _ #:= _ opposite ./ adjacent .
#{Mnemonic}
~some ~officers ~have _ ~curly ~auburn ~hair _ ~to ~offer ~attraction.
~sine = ~opposite ./ ~hypotenuse , _ ~cosine = ~adjacent ./ ~hypotenuse , _ ~tangent = ~opposite ./ ~adjacent .
To define sine and cosine of the general angle, consider a circle of unit radius drawn with centre at the origin. Four Examples are shown below, with the angle &theta. in each quadrant. From our elementary definitions, and considering the first diagram, we can put :
_ _ _ sin &theta. _ #:= _ ~y ./ 1 _ = _ ~y
_ _ _ cos &theta. _ #:= _ ~x ./ 1 _ = _ ~x
_ _ _ tan &theta. _ #:= _ ~y ./ ~x _ = _ sin &theta. ./ cos &theta.
Here is a plot of _ sin &theta. _ and _ cos &theta. _ against _ &theta. _ (in radians):
Finally we define the other trigonometrical quantities :
_ _ _ cosecant &theta. _ [ #:=#: _ cosec &theta. ] _ #:=#: _ 1 ./ sin &theta.
_ _ _ secant &theta. _ [ #:=#: _ sec &theta. ] _ #:= _ 1 ./ cos &theta.
_ _ _ cotangent &theta. _ [ #:=#: _ cotan &theta. ] _ #:= _ 1 ./ tan &theta.
## Compound Angles
Consider the right-anlgled triangles &triangle.ABC and &triangle.ADE [The right-angles being at B and D respectively], where AC = AE = 1, and &angle.BAC = ~a, and &angle.DAE = ~b.
Drop the vertical from E to meet AB at F, and draw the horizontal through D to meet the vertical at G. Drop the vertical from D to meet AB at H. Note that &angle.GED = ~a. We have:
ED = sin(~b) _ so EG = cos(~a) sin(~b)
AD = cos(~b) _ so DH = sin(~a)cos(~b)
EF = EG + GF = EG + DH _ = cos(~a) sin(~b) + sin(~a)cos(~b)
but _ EF = sin(~a + ~b), so
sin(~a + ~b) = cos(~a) sin(~b) + sin(~a)cos(~b)
similarly:
cos(~a + ~b) = AF = AH &minus. FH = AH &minus. GD = cos(~a)cos(~b) &minus. sin(~a)sin(~b)
sin(~a + ~b) _ = _ cos(~a) sin(~b) + sin(~a)cos(~b) cos(~a + ~b) _ = _ cos(~a)cos(~b) &minus. sin(~a)sin(~b)
when ~a = ~b:
sin(2~a) _ = _ 2 cos(~a) sin(~a) cos(2~a) _ = _ cos^2(~a) &minus. sin^2(~a)
## Area of Triangle
By studying the diagrams on the right you should be able to convince yourself that the area of any triangle is half the length of it's base (i.e. any side you wish to choose) times it's height (i.e. the length of the perpendicular from the base to the opposite apex. (See fig. 6). For any triangle (fig. 1) take two copies (fig. 2) to make a rhombus or parallelogram (fig. 3). Move the red triangle shown (fig. 4) over to the left to give the rectangle (fig. 5) whose area is base times height, and is just twice the area of the original triangle.
## Sine Rule
~c : sin ~C _ = _ ~a : sin ~A _ = _ ~b : sin ~B
~{area} &triangle. = &hlf. # ~{height} # ~{base} = &hlf. # ~y # ~b . First consider the acute &angle. ~C sin ~C = ~y / ~a _ => _ ~y = ~a sin ~C _ => _ ~{area} = &hlf. # ~a~b sin ~C &angle. ~B is also acute so by symmetry _ ~{area} = &hlf. # ~a~c sin ~B . &angle. ~A obtuse _ sin ~A = sin ( &pi. - ~A ) = ~y / ~c _ => _ ~y = ~c sin ~A _ => _ ~{area} = &hlf. # ~b~c sin ~A ~a~b sin ~C _ = _ ~b~c sin ~A _ = _ ~c~a sin ~B
## Cosine Rule
~a^2 = ~c^2 + ~b^2 - 2~b~c cos ~A
~a^2 = ~y^2 + (~b + ~x)^2 _ = ~y^2 + ~b^2 + ~x^2 + 2~b~x _ = (~c^2 - ~x^2) + ~b^2 + ~x^2 + 2~b~x _ = ~c^2 + ~b^2 + 2~b~x but _ ~x = ~c cos ( &pi. - ~A ) = - ~c cos ~A ~a^2 = ~c^2 + ~b^2 - 2~b~c cos ~A
~a^2 = ~y^2 + (~b - ~x)^2 _ = ~y^2 + ~b^2 + ~x^2 - 2~b~x _ = (~c^2 - ~x^2) + ~b^2 + ~x^2 - 2~b~x _ = ~c^2 + ~b^2 - 2~b~x but _ ~x = ~c cos ~A ~a^2 = ~c^2 + ~b^2 - 2~b~c cos ~A |
# Rational expression solver
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Simply point your camera at the problem and watch as the app displays the answer on screen. Not only does PhotoMath save you time, but it can also help you to better understand the concepts behind the problem. With its step-by-step solution guide, you can see how PhotoMath arrived at the answer, giving you a valuable learning opportunity. So next time you're stuck on a math word problem, reach for your phone and let PhotoMath do the work for you!
To find the domain and range of a given function, we can use a graph. For example, consider the function f(x) = 2x + 1. We can plot this function on a coordinate plane: As we can see, the function produces valid y-values for all real numbers x. Therefore, the domain of this function is all real numbers. The range of this function is also all real numbers, since the function produces valid y-values for all real numbers x. To find the domain and range of a given function, we simply need to examine its graph and look for any restrictions on the input (domain) or output (range).
Logarithmic functions are a type of math used to calculate an exponent. The log function is the inverse of the exponential function, meaning that it can be used to solve for x when given a number raised to a power. In order to solve logarithmic functions, you need to use a few basic steps. First, identify the base of the logarithm. This is usually either 10 or e. Next, identify the number that is being raised to a power. This number is called the argument. Finally, set up an equation using these two numbers and solve for x. With a little practice, solving logarithmic functions can be easy and even enjoyable!
When it comes to math apps, there is no shortage of options to choose from. However, not all math apps are created equal. Some are more comprehensive than others, some are more user-friendly, and some are just plain more fun to use. So, which is the best app to solve math problems? It really depends on your individual needs and preferences. However, here are three apps that are definitely worth checking out: 1. Photomath is a great option for those who want a comprehensive app that can provide step-by-step solutions to even the most complex math problems. 2. If you're looking for an app that's easy to use and navigate, then Mathway is definitely worth considering. 3. Finally, if you want an app that's both educational and entertaining, then we suggest giving Socratic a try.
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Pemdas equation solver Website with math textbook answers Free homework help online chat Geometry solver Trigonometric solver |
### Solving Equations Containing Variable Exponents
To solve an equation containing a variable exponent, isolate the exponential quantity. Then take a logarithm, to the base of the exponent, of both sides.
Example 1: Solve for x: 3x = 15.
3x = 15
log33x = log315
x = log315
x =
x 2.465
Example 2: Solve for x: 4·52x = 64.
4·52x = 64
52x = 16
log552x = log516
2x = log516
2x =
2x 1.723
x 0.861
### Solving Equations Containing Logarithms
To solve an equation containing a logarithm, use the properties of logarithms to combine the logarithmic expressions into one expression. Then convert to exponential form and evaluate. Check the solution(s) and eliminate any extraneous solutions--recall that we cannot take the logarithm of a negative number.
Example 1: Solve for x: log3(3x) + log3(x - 2) = 2.
log3(3x) + log3(x - 2) = 2
log3(3x(x - 2)) = 2
32 = 3x(x - 2)
9 = 3x2 - 6x
3x2 - 6x - 9 = 0
3(x2 - 2x - 3) = 0
3(x - 3)(x + 1) = 0
x = 3, - 1
Check:
• x = 3: log3(3·3) + log31 = 2 + 0 = 2. x = 3 is a solution.
• x = - 1: log3(3· -1) + log3(- 1 - 2) = log3(- 3) + log3(- 3) does not exist. x = - 1 is not a solution.
Thus, x = 3.
Example 2: Solve for x: 2 log(2x+1)(2x + 4) - log(2x+1)4 = 2.
2 log(2x+1)(2x + 4) - log(2x+1)4 = 2
log(2x+1)(2x + 4)2 - log(2x+1)4 = 2
log(2x+1) = 2
(2x + 1)2 =
(2x + 1)2 =
4x2 +4x + 1 = x2 + 4x + 4
3x2 - 3 = 0
3(x2 - 1) = 0
3(x + 1)(x - 1) = 1
x = 1, - 1
Check:
• x = 1: 2 log36 - log34 = log362 - log34 = log3 = log39 = 2. x = 1 is a solution.
• x = - 1: 2 log-12 - log-14 does not exist (the base cannot be negative).
Thus, x = 1. |
# AP Statistics : How to find the standard deviation for a set of data
## Example Questions
### Example Question #226 : Ap Statistics
A bird watcher observed how many birds came to her bird feeder for four days. These were the results:
Day 1: 15
Day 2: 12
Day 3: 10
Day 4: 13
Which answer is closest to the standard deviation of the number of birds to visit the bird feeder over the four days?
Explanation:
Standard deviation is essentially the average distance from the mean of a group of numbers. There are a number of steps in computing standard deviation, but the steps are not too complicated if you take them one at a time. First, find the mean of the values. Second, subtract the mean from the first value and square the result. Do this for all remaining values. Third, add these results together. Fourth, divide this value by the number of values. Finally, find the square root of the result.
1:
2:
3:
4:
5:
### Example Question #227 : Ap Statistics
Alice recorded the outside temperature at noon each day for one week. These were the results.
Monday: 78
Tuesday: 85
Wednesday: 82
Thursday: 84
Friday: 82
Saturday: 79
Sunday: 80
What is the standard deviation of the temperatures?
Explanation:
There are five steps to finding the standard deviation of a group of values. First, find the mean of the values. Second, subtract the mean from the first value and square the result. Do this for all remaining values. Third, add these results together. Fourth, divide this value by the number of values minus one. Finally, find the square root of the result.
### Example Question #1 : How To Find The Standard Deviation For A Set Of Data
A bird watcher observed how many birds came to her bird feeder for four days. These were the results:
Day 1: 15
Day 2: 12
Day 3: 10
Day 4: 13
What is the variance of the number of birds that visited the bird feeder over the four days?
Explanation:
Variation measures the average difference between values within a group. The process is not complicated but there are four steps that can take time. First, find the mean of the values. Second, subtract the mean from the first value and square the result. Do this for all remaining values. Third, add these results together. Fourth, divide this value by the number of values in the group minus one (in this case, there are four days).
1:
2:
3:
4:
Note that to find the standard deviation, we would simply take one additional step of finding the square root of the variance.
### Example Question #229 : Ap Statistics
Alice recorded the outside temperature at noon each day for one week. These were the results.
Monday: 78
Tuesday: 85
Wednesday: 82
Thursday: 84
Friday: 82
Saturday: 79
Sunday: 80
What is the variance of the temperatures?
Explanation:
There are four steps to finding the variance of values within a group. First, find the mean of the values. Second, subtract the mean from the first value and square the result. Do this for all remaining values. Third, add these results together. Fourth, divide the result by the number of values in the group minus one (in this case, there are seven days, so you must divide by six).
### Example Question #230 : Ap Statistics
The average height of females in a class is inches, with a standard deviation of inches. In the same class, the average height of boys is inches, with a standard deviation of inches. What is the mean height of both males and females?
Explanation:
To find the mean of the whole population, multiply the female's average by the number of females, and then multiply the male's average by the number of males. Sum up these products and divide by the total number of males AND females:
### Example Question #2 : How To Find The Standard Deviation For A Set Of Data
The standard deviation of a population is 7.5. What is the variance of the population? |
# Difference between revisions of "2000 AMC 10 Problems/Problem 7"
## Problem
In rectangle $ABCD$, $AD=1$, $P$ is on $\overline{AB}$, and $\overline{DB}$ and $\overline{DP}$ trisect $\angle ADC$. What is the perimeter of $\triangle BDP$?
$[asy] draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle); draw((0,0)--(1.3,2)); draw((0,0)--(3.4,2)); dot((0,0)); dot((0,2)); dot((3.4,2)); dot((3.4,0)); dot((1.3,2)); label("A",(0,2),NW); label("B",(3.4,2),NE); label("C",(3.4,0),SE); label("D",(0,0),SW); label("P",(1.3,2),N); [/asy]$
$\mathrm{(A)}\ 3+\frac{\sqrt{3}}{3} \qquad\mathrm{(B)}\ 2+\frac{4\sqrt{3}}{3} \qquad\mathrm{(C)}\ 2+2\sqrt{2} \qquad\mathrm{(D)}\ \frac{3+3\sqrt{5}}{2} \qquad\mathrm{(E)}\ 2+\frac{5\sqrt{3}}{3}$
## Solution
$[asy] draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle); draw((0,0)--(1.3,2)); draw((0,0)--(3.4,2)); dot((0,0)); dot((0,2)); dot((3.4,2)); dot((3.4,0)); dot((1.3,2)); label("A",(0,2),NW); label("B",(3.4,2),NE); label("C",(3.4,0),SE); label("D",(0,0),SW); label("P",(1.3,2),N); label("1",(0,1),W); label("2",(1.7,1),SE); label("\frac{\sqrt{3}}{3}",(0.65,2),N); label("\frac{2\sqrt{3}}{3}",(0.85,1),NW); label("\frac{2\sqrt{3}}{3}",(2.35,2),N); [/asy]$
$AD=1$.
Since $\angle ADC$ is trisected, $\angle ADP= \angle PDB= \angle BDC=30^\circ$.
Thus, $PD=\frac{2\sqrt{3}}{3}$
$DB=2$
$BP=\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}$.
Adding, we get $\boxed{\textbf{(B) } 2+\frac{4\sqrt{3}}{3}}$.
## See Also
2000 AMC 10 (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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# 2.3: Permutation Problems
Difficulty Level: Basic Created by: CK-12
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Practice Permutation Problems
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Suppose you want to select items to place on your shelf at home from left to right. If you have 45 items from which to select how many possible arrangements can you make? Can you see why a calculator might be useful to find this answer?
### Watch This
First watch this video to learn about calculating permutations with calculators.
Then watch this video to see some examples.
### Guidance
To calculate permutations (\begin{align*}nPr\end{align*}) on the TI calculator, first enter the \begin{align*}n\end{align*} value, and then press \begin{align*}\boxed{\text{MATH}}\end{align*}. You should see menus across the top of the screen. You want the fourth menu: PRB (arrow right 3 times). The PRB menu should appear as follows:
You will see several options, with \begin{align*}nPr\end{align*} being the second. Press \begin{align*}\boxed{2}\end{align*}, and then enter the \begin{align*}r\end{align*} value. Finally, press \begin{align*}\boxed{\text{ENTER}}\end{align*} to calculate the answer.
#### Example A
Compute \begin{align*}{_9}P_5\end{align*} using your TI calculator.
\begin{align*}\boxed{9} \ \boxed{\text{MATH}} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ (\text{PRB}) \ \boxed{2} \ (\text{nPr}) \ \boxed{5} \ \boxed{\text{ENTER}}\end{align*}
After pressing \begin{align*}\boxed{\text{ENTER}}\end{align*}, you should see the following on your calculator's screen:
Therefore, \begin{align*}{_9}P_5= 15,120\end{align*}.
#### Example B
In how many ways can first and second place be awarded to 10 people? Compute the answer using your TI calculator.
There are 10 people \begin{align*}(n = 10)\end{align*}, and there are 2 prize winners \begin{align*}(r = 2)\end{align*}, so to find the answer, enter the following into your TI calculator:
\begin{align*}\boxed{10} \ \boxed{\text{MATH}} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ (\text{PRB}) \ \boxed{2} \ (\text{nPr}) \ \boxed{2} \ \boxed{\text{ENTER}}\end{align*}
After pressing \begin{align*}\boxed{\text{ENTER}}\end{align*}, you should see the following on your calculator's screen:
Therefore, \begin{align*}{_{10}}P_2= 90\end{align*}, which means that the number of ways that first and second place can be awarded to 10 people is 90.
#### Example C
In how many ways can 3 favorite desserts be listed in order from a menu of 10? Compute the answer using your TI calculator.
There are 10 menu items \begin{align*}(n = 10)\end{align*}, and you are choosing 3 favorite desserts \begin{align*}(r = 3)\end{align*} in order, so to find the answer, enter the following into your TI calculator:
\begin{align*}\boxed{10} \ \boxed{\text{MATH}} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ (\text{PRB}) \ \boxed{2} \ (\text{nPr}) \ \boxed{3} \ \boxed{\text{ENTER}}\end{align*}
After pressing \begin{align*}\boxed{\text{ENTER}}\end{align*}, you should see the following on your calculator's screen:
Therefore, \begin{align*}{_{10}}P_3= 720\end{align*}, which means that the number of ways that 3 favorite desserts can be listed in order from a menu of 10 is 720.
### Guided Practice
Verify that the \begin{align*}nPr\end{align*} option on the PRB menu of the TI calculator works by calculating \begin{align*}{_{18}}P_{18}\end{align*} with both this option and with the factorial function. Note that with the TI calculator, you can find the factorial function by pressing \begin{align*}\boxed{\text{MATH}}\end{align*}, pressing the right arrow 3 times, and pressing \begin{align*}\boxed{4}\end{align*}.
First, lets compute \begin{align*}{_{18}}P_{18}\end{align*} with the factorial function. Remember, the formula to solve permutations like these is:
\begin{align*}{_n}P_r=\frac{n!}{(n-r)!}\end{align*}
This means that \begin{align*}{_{18}}P_{18}\end{align*} can be written as follows:
\begin{align*}{_{18}}P_{18} &= \frac{18!}{(18-18)!}\\ {_{18}}P_{18} &= \frac{18!}{0!}\\ {_{18}}P_{18} &= \frac{18!}{1}\\ {_{18}}P_{18} &= 18!\end{align*}
To find 18!, enter the following into your TI calculator:
\begin{align*}\boxed{1} \ \boxed{8} \ \boxed{\text{MATH}} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ (\text{PRB}) \ \boxed{4} \ (\text{!}) \ \boxed{\text{ENTER}}\end{align*}
After pressing \begin{align*}\boxed{\text{ENTER}}\end{align*}, you should see the following on your calculator's screen:
Now let's compute \begin{align*}{_{18}}P_{18}\end{align*} with the \begin{align*}nPr\end{align*} option on the PRB menu.
\begin{align*}\boxed{1} \ \boxed{8} \ \boxed{\text{MATH}} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ (\text{PRB}) \ \boxed{2} \ (\text{nPr}) \ \boxed{1} \ \boxed{8} \ \boxed{\text{ENTER}}\end{align*}
After pressing \begin{align*}\boxed{\text{ENTER}}\end{align*}, you should see the following on your calculator's screen:
The answers for the 2 methods are the same, so the \begin{align*}nPr\end{align*} option on the PRB menu of the TI calculator does work.
### Practice
1. Enter each of the following sets of keystrokes into your TI calculator to compute the corresponding permutations.
2. \begin{align*}\boxed{1} \ \boxed{2} \ \boxed{\text{MATH}} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ (\text{PRB}) \ \boxed{2} \ (\text{nPr}) \ \boxed{8} \ \boxed{\text{ENTER}}\end{align*}
3. \begin{align*}\boxed{1} \ \boxed{5} \ \boxed{\text{MATH}} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ (\text{PRB}) \ \boxed{2} \ (\text{nPr}) \ \boxed{5} \ \boxed{\text{ENTER}}\end{align*}
4. \begin{align*}\boxed{2} \ \boxed{0} \ \boxed{\text{MATH}} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ (\text{PRB}) \ \boxed{2} \ (\text{nPr}) \ \boxed{7} \ \boxed{\text{ENTER}}\end{align*}
5. \begin{align*}\boxed{1} \ \boxed{1} \ \boxed{\text{MATH}} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ (\text{PRB}) \ \boxed{2} \ (\text{nPr}) \ \boxed{6} \ \boxed{\text{ENTER}}\end{align*}
6. \begin{align*}\boxed{1} \ \boxed{4} \ \boxed{\text{MATH}} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ (\text{PRB}) \ \boxed{2} \ (\text{nPr}) \ \boxed{4} \ \boxed{\text{ENTER}}\end{align*}
7. \begin{align*}\boxed{1} \ \boxed{9} \ \boxed{\text{MATH}} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ (\text{PRB}) \ \boxed{2} \ (\text{nPr}) \ \boxed{3} \ \boxed{\text{ENTER}}\end{align*}
8. \begin{align*}\boxed{2} \ \boxed{2} \ \boxed{\text{MATH}} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ (\text{PRB}) \ \boxed{2} \ (\text{nPr}) \ \boxed{9} \ \boxed{\text{ENTER}}\end{align*}
9. \begin{align*}\boxed{1} \ \boxed{8} \ \boxed{\text{MATH}} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ (\text{PRB}) \ \boxed{2} \ (\text{nPr}) \ \boxed{2} \ \boxed{\text{ENTER}}\end{align*}
10. \begin{align*}\boxed{2} \ \boxed{5} \ \boxed{\text{MATH}} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ (\text{PRB}) \ \boxed{2} \ (\text{nPr}) \ \boxed{3} \ \boxed{\text{ENTER}}\end{align*}
11. \begin{align*}\boxed{1} \ \boxed{6} \ \boxed{\text{MATH}} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ \boxed{\blacktriangleright} \ (\text{PRB}) \ \boxed{2} \ (\text{nPr}) \ \boxed{6} \ \boxed{\text{ENTER}}\end{align*}
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English Spanish
TermDefinition
$n$ value When calculating permutations with the TI calculator, the $n$ value is the number of objects from which you are choosing, and the $r$ value is the number of objects chosen.
$r$ value When calculating permutations with the TI calculator, the $n$ value is the number of objects from which you are choosing, and the $r$ value is the number of objects chosen.
combination Combinations are distinct arrangements of a specified number of objects without regard to order of selection from a specified set.
factorial The factorial of a whole number $n$ is the product of the positive integers from 1 to $n$. The symbol "!" denotes factorial. $n! = 1 \cdot 2 \cdot 3 \cdot 4...\cdot (n-1) \cdot n$.
n value When calculating permutations with the TI calculator, the n value is the number of objects from which you are choosing.
Permutation A permutation is an arrangement of objects where order is important.
r value When calculating permutations with the TI calculator, the r value is the number of objects chosen.
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# Lesson 8Making the GradePractice Understanding
## Learning Focus
Compare sets of data using center, spread, and shape.
How do I use measures of center and spread together to make decisions about data?
How do I compare two or more data sets that are represented by different plots?
## Open Up the Math: Launch, Explore, Discuss
Principal Scanner wants to make his school the best ever. He has offered to buy pizza for the class that shows the best results on the department final at the end of the semester. He tells all the teachers in the department to submit their student test-score data so that he can determine which class did the best. Unfortunately, Principal Scanner underestimated the job he was going to have in determining the best class. To make matters worse, teachers used different representations for the data they submitted for him to evaluate.
So, Principal Scanner has turned the job over to you. You are to compare the two data sets in each problem and give the principal a report, using the sentence frames provided for this purpose. Give each comparison some thoughtful analysis, and use your best statistical language to describe your findings. Somebody’s pizza is depending on you!
Data Set I: Williams’ Class
Data Set II: Lemon’s Class
Data Set III: Croft’s Class
Data Set IV: Anderson’s Class
Data Set V: Hurlea’s Class
Data Set VI: Jones’ Class
### 1.
Compare data distributions between Anderson’s and Williams’ classes.
#### a.
Complete the sentence frames with your findings:
The data from has a greater than because .
The data from has less than because .
The data from is skewed left/right (choose one) while the data from is not because .
#### b.
Which of these two classes do you think did the best? Why?
### 2.
Compare data distributions between Croft’s and Hurlea’s classes.
#### a.
Complete the sentence frames with your findings:
The data from has a greater than because .
The data from has less than because .
The data from contains an outlier while the data from . I know this because .
#### b.
Which of these two classes do you think did the best? Why?
### 3.
Compare data distributions between Jones’ and Williams’ classes.
#### a.
Complete the sentence frames with your findings.
The data from has a greater than because .
The data from has less than because .
The data from is skewed left/right (choose one) while the data from is not because .
#### b.
Which of these two classes do you think did the best? Why?
### 4.
Compare data distributions between Lemon’s class and that of a teacher of your choice.
#### a.
Complete the sentence frames with your findings.
The data from has a greater than because .
The data from has less than because .
#### b.
Which of these two classes do you think did the best? Why?
You now get to make your final recommendation to Principal Scanner. Which class do you choose, and why?
Box plot:
Histogram:
Dot plot:
## Lesson Summary
In this lesson, we compared data distributions using shape, center, and spread. We learned to consider the interquartile range and the median when using a box plot and to consider the mean and the standard deviation when using a histogram. We also discussed the effect of outliers on the median, mean, and standard deviation.
## Retrieval
Sketch a graph for each of the functions. |
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