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### Theory:
Suppose we want to locate the number $$5.6$$$\overline{34}$ on the number line up to five decimal places.
That is, we have to represent $$5.6343434...$$ on the following number line.
But it is as asked to represent up to $$5$$ decimal places. Thus, we need to locate until $$5.63434$$.
Step 1: The range of the number $$5.63434$$ is $$5$$ and $$6$$.
Step 2: Look for the range of the number on the number line and magnify that range ($$5-6$$) alone. That is, divide the portion into $$10$$ parts.
Step 3: Now let us look for the number with the first decimal point on the number line and find the range of that. That is, $$5.6 -5.7$$
Step 4: Again magnify the range and divide the portion into $$10$$ parts. Now look for the number with two decimal point on the number line. That is $$5.63-5.64$$.
Step 5: Again magnify the range and divide the portion into $$10$$ parts. Now look for the number with three decimal point on the number line $$5.634 - 5.645$$. Repeating the process of successive magnification, we need to magnify further $$5.6343 - 5.6344$$. In this magnification, we can locate $$5.63434$$.
Thus, we located the number $$5.63434$$ on the number line by the process of successive magnification. |
# 6.6 Parallel and Perpendicular Lines Parallel Lines:
6.6 Parallel and Perpendicular Lines
1.
Write the definition and draw a diagram for each:
Parallel Lines:
Perpendicular Lines:
2.
The following pairs of lines are parallel to each other. With your group, write an equation in slopea. intercept form for each line.
b. c.
Line 1: ___________________
Line 2: ___________________
Line 1: __________________
Line 2: __________________
Line 1: _________________
Line 2: _________________
3.
Examine the slopes within each example above. Discuss with your group a pattern you see between the slopes of parallel lines. If two equations are in slope-intercept form, how can you tell if the lines are parallel?
4.
Given two equations. In the space provided state if the two lines are parallel or not parallel. (Hint: Make sure they’re in the correct form to see the slope!) y
3 x
5 y
3 x
7 _______________________ y
1
2 x
2 y y
3
4 x
x
4
7
7 y
1
2 x
2 y y
3
3
7
4
4 x x
_______________________
_______________________
_______________________
5.
Work in your group to create your own equation of a line that would be parallel to each of the given equations. y
2 x
5 y
2
5 x
__________________________________
___________________________________________________
2 y
3 x
7 ___________________________________
6.
The following pairs of lines are perpendicular to each other. With your group, write an equation in slope-intercept form of each line. a. b. c.
Line 1: ________________
Line 2: ________________
Line 1: _________________ Line 1: _________________
Line 2: _________________ Line 2: _________________
7.
In your groups discuss the relationship between the slopes of two lines that are perpendicular to each other. Pay special attention to the slope and the two things that change . Develop a rule that communicates how to find the slope of a line that is perpendicular to a line with an equation given in slope-intercept form.
8.
Below are four equations. What would the slope be of a line that is perpendicular to these equations?
Equation y
2 x
5
Perpendicular slope
_____________ y
1
3 x y
3
4 x
7
_____________
_____________
3 y
7
2 x _____________
9.
Given the two equations below, write an equation of a line perpendicular to the given equations and a line parallel to the given equations.
Equation: y
2 x
5
Parallel: ____________________
Perpendicular: ____________________ y
2
3 x
7
_____________________
_____________________
10.
Tell whether the lines for each pair of equations are parallel, perpendicular, or neither.
(Remember: what form does the equation need to be in to see the slopes?) a.
y
3 x
3
2
2 y x
2
12 b. y
5 x
1
2
10 y x
3
2
15 c. y
4 x
3
4
3 x y
2
6 |
# PROBABILITY for Class 9 MATHS
## INTRODUCTION
Hello bachcho, aaj hum padhenge probability ke bare mai. Probability means possibility. Probability tell us how often some event will happen after many repeated trials.Probability is an essential tool in applied mathematics and mathematical modeling. You use probability in daily life to make decisions when you don't know for sure what the outcome will be. Probability can range between 0 and 1, where 0 probability means the event to be an impossible one and probability of 1 indicates a certain event.
## Explanation
Hum apni daily life mai kahi tarah ki statement dekhte hai aur jinke bare mai hum sure nahi hote hai jaise ki aaj baarish hogi, ya aaj share market ka price kam hoga. Iss tarah se hum apni apni prediction karte hai. Toh ye chapter yani probability kahi na kahi hamari practical life se connected hai.
## PROBABILITY
A probability is a number that reflects the chance or likelihood that a particular event will occur.
## Probability — A Theoretical Approach
Toh baccho probability ko samajhne ke liye ek situation lete hai. Suppose kare ki humne randomly ek coin ko toss kara hai. Jab bhi hum coin ki baat karte hai toh iska matlab hai ki ek fair coin (unbiased). By the phrase ‘random toss’, we mean that the coin is allowed to fall freely without any bias or interference. Hum jante hai ki coin ke 2 possible outcomes hai one is head and another one is tail . We can reasonably assume that each outcome, head or tail, is as likely to occur as the other. We refer to this by saying that the outcomes head and tail, are equally likely.
For another example of equally likely outcomes, Suppose kare ki hum ek dice ko throw kar rahe hai. For us, a die will always mean a fair die. For us, a die will always mean a fair die.
They are 1,2,3,4,5,6.
Each number has the same possibility of showing up. So the equally likely outcomes of throwing a die are 1,2,3,4,5 and 6.
However, in this chapter, from now on, we will assume that all the experiments have equally likely outcomes. In Class IX, we will define the experimental or empirical probability P(E) of an event E as
P(E) = Number of trials in which the event happenedTotal number of trials
P(E) = Number of outcomes favourable to E Number of all possible outcomes of the experiment
Where we assume that the outcomes of the experiment are equally likely. The event E , representing ‘not E’, is called the complement of the event E. We also say that E and ¯¯¯E are complementary events.
The probability of an event which is impossible to occur is 0. Such an event is called an impossible event.
The probability of an event which is sure (or certain) to occur is 1. Such an event is called a sure event or a certain event.
0P(E)1 Aapne apni life mai kabhi na kabhi to cards zarur hi khele honge. Toh aaj hum cards ke bare mai hi thoda janege. There is 52 cards in a deck.
There are two colors of cards in each deck:
(i) RED
(ii) BLACK
There are 26 red cards and 26 black cards.
The types of cards in a deck are:
• Hearts
• Diamonds
• Clubs
• Suits in a deck of cards are the representations of red and black color on the cards.
HEARTS
DIAMONDS
CLUBS
Each card can be categorized into 4 suits constituting 13 cards each. There is one more categorization of a deck of cards:
• Face cards
• Number cards
• Aces
• Face Cards
These cards are also known as court cards. They are Kings, Queens, and Jacks in all 4 suits.
Number Cards
All the cards from 2 to 10 in any suit are called the number cards. These cards have numbers on them along with each suit being equal to the number on number cards.
Aces
There are 4 Aces in every deck, 1 of every suit.
## Example 1
A spinner is coloured by 3 different colours-yellow, blue and red in 12 equal sectors. After spinning the wheel, what is the probability that
(i) wheel stops at yellow colour.
(ii) wheel stops at red colour.
(iii) wheel stops at blue colour.
(i) P(wheel stops at yellow colour)=n(Y)n(S)=412=13
(ii)P(wheel stops at red colour)=n(R)n(S)=312=14
(iii) P(wheel stops at blue colour)=n(B)n(S)=512
or P(blue)=1(13+14)=1712=512
## Example 2
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, find the number of blue balls in the bag.
Number of red balls in a bag = 5
Suppose the number of blue balls in the bag = x RB5x–––––
Total number of balls in the beg=5+x
P(red ball)=55+x
P(blue ball)=x5+x
Now, P(blue ball) = 2 P(red ball) (given)
x5+x=2×55+x
x=10
Therefore, the number of blue balls in the bag are 10.
## Example 3
A die is thrown 250 times and the outcomes are noted as given below :
Outcomes 1 2 3 4 5 6Frequency 60 50 40 20 30 50
If a die is thrown at random, find the probability of getting.
(i)1 (ii)2 (iii) 3 (iv) 4 (v) 5 (vi) 6
Total number of trials = 250
In a random throw of a dia, let E1,E2,E3,E4,E5andE6 be the events of getting 1, 2, 3, 4, 5 and 6, respectively.
(i) P (getting 1)=P(E1)=number of times 1 appeartotal number of trials=6025=625
(ii) P (getting 2)=P(E2)=number of times 2 appearstotal number of trials=5025=15
(iii) P (getting 3)=P(E3)=number of times 3 appearstotal number of trials=40250=425
(iv) P (getting 4)=P(E4)=number of times 4 appearstotal number of trials=20250=225
(v) P (getting 5)=P(E5)=number of times 5 appearstotal number of trials=30250=325
(iv) P (getting 6)=P(E6)=number of times 6 appearstotal number of trials=50250=15
## Example 4
1000 families with 2 children were selected randomly and the following data were recorded.
Number of boys in a family 0 1 2Number of families 140 560 300
If a family is chosen at random, find the probability that it has (i) no boy (ii) one boy (iii) two boys (iv) at least one boy (v) at most one boy.
(i) No of families having no boy child = 140
P(getting a family with no child)=number of families with no boy childtotal number of families
=1401000=750
(ii) Number of families having one boy child = 560
P(getting a family with one boy child)=5601000=1425
(iii) Number of families having two boy children = 300
P(getting a family with two boy children)=3001000=310
(iv) Number of families having at least one boy child = 560 + 300 = 860
P(getting a families having at least one boy child)=8601000=0.86
(v) Number of families having at most one boy child = 140 + 560 = 700
P(getting a family with at most one boy child)=7001000=0.7
## Basic Probability Rules
• Probability Rule One (For any event A,0P(A)1)
• Probability Rule Two (The sum of the probabilities of all possible outcomes is 1)
• Probability Rule Three (The Complement Rule)
• Probabilities Involving Multiple Events.
• Probability Rule Four (Addition Rule for Disjoint Events)
## Summary
In this chapter, you have studied the following points :
1. The difference between experimental probability and theoretical probability.
2. The theoretical (classical) probability of an event E, written as P(E), is defined as
P(E) = Number of outcomes favourable to ENumber of all possible outcomes of the experiment
where we assume that the outcomes of the experiment are equally likely.
3. The probability of a sure event (or certain event) is 1
4. The probability of an impossible event is 0.
5. The probability of an event E is a number P(E) such that
0P(E)1
6. An event having only one outcome is called an elementary event. The sum of the probabilities of all the elementary events of an experiment is 1.
7. For any event E,P(E)+P(E)=1, where E stands for ‘not E’. E and E are called complementary events.
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## Engage NY Eureka Math 5th Grade Module 4 Lesson 20 Answer Key
### Eureka Math Grade 5 Module 4 Lesson 20 Problem Set Answer Key
Question 1.
Convert. Show your work. Express your answer as a mixed number. (Draw a tape diagram if it helps you.) The first one is done for you.
a. 2 $$\frac{2}{3}$$ yd = 8 ft
2 $$\frac{2}{3}$$ yd = 2 $$\frac{2}{3}$$ × 1 yd
= 2 $$\frac{2}{3}$$ × 3 ft
= $$\frac{8}{3}$$ × 3 ft
= $$\frac{24}{3}$$ ft
= 8 ft
b. 1$$\frac{1}{2}$$ qt = $$\frac{3}{8}$$ gal
1$$\frac{1}{2}$$ × 1 qt
= 1 $$\frac{1}{2}$$ × $$\frac{1}{4}$$ gal
= $$\frac{3}{2}$$ × $$\frac{1}{4}$$ gal
= $$\frac{3}{8}$$ gal.
c. 4 $$\frac{2}{3}$$ ft = ______________ in
4 $$\frac{2}{3}$$ × 1 ft
= $$\frac{14}{3}$$ × 12 in
= $$\frac{168}{3}$$ in
= 56 in.
d. 9 $$\frac{1}{2}$$ pt = ______________ qt
9 $$\frac{1}{2}$$ × 1 pt
= $$\frac{19}{2}$$ × $$\frac{1}{2}$$ qt
= $$\frac{19}{4}$$ qt
= 4 $$\frac{3}{4}$$ qt.
e. 3 $$\frac{3}{5}$$ hr = ______________ min
3 $$\frac{3}{5}$$ × 1 hr
= $$\frac{18}{5}$$ × 60 min
= $$\frac{1080}{5}$$ min
= 216 mins.
f. 3 $$\frac{2}{3}$$ ft = ______________ yd
3 $$\frac{2}{3}$$ × 1 ft
= $$\frac{11}{3}$$ × $$\frac{1}{3}$$ yd
= $$\frac{11}{9}$$
= 1 $$\frac{2}{9}$$ yd.
Question 2.
Three dump trucks are carrying topsoil to a construction site. Truck A carries 3,545 lb, Truck B carries 1,758 lb, and Truck C carries 3,697 lb. How many tons of topsoil are the 3 trucks carrying altogether?
The 3 trucks carrying altogether are 4.5 tons.
Explanation:
Given that there are three dump trucks are carrying topsoil to a construction site and Truck A carries 3,545 lb, Truck B carries 1,758 lb, and Truck C carries 3,697 lb, so the total weight carried altogether is 3,545 + 1,758 + 3,697 = 9,000 lb. As each ton is 2000 pounds, so altogether the trucks are carrying is 9000 × $$\frac{1}{2000}$$ which is 4.5 tons.
Question 3.
Melissa buys 3$$\frac{3}{4}$$ gallons of iced tea. Denita buys 7 quarts more than Melissa. How much tea do they buy altogether? Express your answer in quarts.
The total tea they bought is 37 quarts.
Explanation:
Given that Melissa buys 3$$\frac{3}{4}$$ gallons of iced tea, so total iced tea for Melissa is $$\frac{15}{4}$$ which is 3.75. And Denita buys 7 quarts more than Melissa, so the total iced tea for Denita is, as 1 quart is 0.25 gallon and for 7 quarts it will be 7 × 0.25 which is 1.75 gallon. so the total iced tea for Denita is 1.75 + 3.75 which is 5.5. Then the total tea they bought is 3.75 + 5.5 = 9.25 gallon which is 9.25 × 4 = 37 quarts.
Question 4.
Marvin buys a hose that is 27$$\frac{3}{4}$$ feet long. He already owns a hose at home that is $$\frac{2}{3}$$ the length of the new hose. How many total yards of hose does Marvin have now?
The total yards of hose does Marvin have now is 15 $$\frac{5}{12}$$ yd.
Explanation:
Given that Marvin buys a hose that is 27$$\frac{3}{4}$$ feet long and he owns a hose at home that is $$\frac{2}{3}$$ the length of the new hose, so $$\frac{2}{3}$$ of 27$$\frac{3}{4}$$ which is
$$\frac{2}{3}$$ × $$\frac{111}{4}$$
= $$\frac{222}{12}$$
= 18 $$\frac{1}{2}$$,
So the total yards of hose does Marvin have now is
27$$\frac{3}{4}$$ + 18$$\frac{1}{2}$$
= $$\frac{111}{4}$$ + $$\frac{37}{2}$$
= $$\frac{185}{4}$$
= 46 $$\frac{1}{4}$$.
So total in yards, it will be 46 $$\frac{1}{4}$$ × 1 yd
= $$\frac{185}{4}$$ × $$\frac{1}{3}$$ yd
= $$\frac{185}{12}$$
=15 $$\frac{5}{12}$$ yd.
### Eureka Math Grade 5 Module 4 Lesson 20 Exit Ticket Answer Key
a. 2$$\frac{1}{6}$$ ft = ______________ in
26 in.
Explanation:
2$$\frac{1}{6}$$ ft = $$\frac{13}{6}$$ × 1 ft
= $$\frac{13}{6}$$ × 12 in
= $$\frac{156}{6}$$ in
= 26 in.
b. 3$$\frac{3}{4}$$ ft = ______________ yd
45 in.
Explanation:
3$$\frac{3}{4}$$ ft = $$\frac{15}{4}$$ ft × 1 ft
= $$\frac{15}{4}$$ × 12 in
= $$\frac{180}{4}$$ in
= 45 in.
c. 2$$\frac{1}{2}$$c = ______________ pt
1 $$\frac{1}{4}$$ pt.
Explanation:
2$$\frac{1}{2}$$c = $$\frac{5}{2}$$ × 1 c
= $$\frac{5}{2}$$ × $$\frac{1}{2}$$ pt
= $$\frac{5}{4}$$ pt
= 1 $$\frac{1}{4}$$ pt.
d. 3$$\frac{2}{3}$$ years = ______________ months
44 months.
Explanation:
3$$\frac{2}{3}$$ years = $$\frac{11}{3}$$ × 1 year
= $$\frac{11}{3}$$ × 12 months
= 44 months.
### Eureka Math Grade 5 Module 4 Lesson 20 Homework Answer Key
Question 1.
Convert. Show your work. Express your answer as a mixed number. The first one is done for you.
2 $$\frac{2}{3}$$ yd = 8 ft
2 $$\frac{2}{3}$$ yd = 2 $$\frac{2}{3}$$ × 1 yd
= 2 $$\frac{2}{3}$$ × 3 ft
= $$\frac{8}{3}$$ × 3 ft
= $$\frac{24}{3}$$ ft
= 8 ft
b. 1 $$\frac{1}{4}$$ ft = $$\frac{5}{12}$$ yd
1 $$\frac{1}{4}$$ ft = 1 $$\frac{1}{4}$$ × 1 ft
= 1 $$\frac{1}{4}$$ × $$\frac{1}{3}$$ yd
= $$\frac{5}{4}$$ × $$\frac{1}{3}$$ yd
= $$\frac{5}{12}$$ yd.
c. 3$$\frac{5}{6}$$ ft = ______________ in
46 in.
Explanation:
3$$\frac{5}{6}$$ ft = 3$$\frac{5}{6}$$ ft × 1 ft
= $$\frac{23}{6}$$ × 12 in
= 46 in.
d. 7 $$\frac{1}{2}$$ pt = ______________ qt
3 $$\frac{3}{4}$$ qt.
Explanation:
7 $$\frac{1}{2}$$ pt = 7 $$\frac{1}{2}$$ pt × 1 pt
= $$\frac{15}{2}$$ × $$\frac{1}{2}$$ qt
= $$\frac{15}{4}$$ qt
= 3 $$\frac{3}{4}$$ qt.
e. 4$$\frac{3}{10}$$ hr = ______________ min
258 mins.
Explanation:
4$$\frac{3}{10}$$ hr = 4$$\frac{3}{10}$$ × 1 hr
= $$\frac{43}{10}$$ × 60 mins
= 258 mins
f. 33 months = ______________ years
2 $$\frac{3}{4}$$ years.
Explanation:
33 × $$\frac{1}{12}$$
= $$\frac{33}{12}$$
= $$\frac{11}{4}$$
= 2 $$\frac{3}{4}$$ years.
Question 2.
Four members of a track team run a relay race in 165 seconds. How many minutes did it take them to run the race?
The number of minutes did it take them to run the race is 2 $$\frac{3}{4}$$ mins.
Explanation:
Given that there are four members of a track team run a relay race in 165 seconds, so the number of minutes did it take them to run the race is 165 × $$\frac{1}{60}$$
= $$\frac{11}{4}$$
= 2 $$\frac{3}{4}$$ mins.
Question 3.
Horace buys 2$$\frac{3}{4}$$ pounds of blueberries for a pie. He needs 48 ounces of blueberries for the pie. How many more pounds of blueberries does he need to buy?
Horace need more blueberries to buy is 0.25 pounds.
Explanation:
Given that Horace buys 2$$\frac{3}{4}$$ pounds of blueberries for a pie and he needs 48 ounces of blueberries for the pie, so the number of pounds of blueberries does he need to buy is, as 1 pound is 16 ounces and 1 ton is 2,200 pounds which is 32,000 ounces. As Horace needs 48 ounces of blueberries for the pie and we need to convert into ounce, so Horace need more blueberries to buy is 3 – 2$$\frac{3}{4}$$ which is 3 – $$\frac{11}{4}$$ = $$\frac{1}{4}$$ = 0.25 pounds.
Question 4.
Tiffany is sending a package that may not exceed 16 pounds. The package contains books that weigh a total of 9$$\frac{3}{8}$$ pounds. The other items to be sent weigh $$\frac{3}{5}$$ the weight of the books. Will Tiffany be able to send the package?
The total package is 15 pounds.
Explanation:
Given that Tiffany is sending a package that may not exceed 16 pounds and the package contains books that weigh a total of 9$$\frac{3}{8}$$ pounds and the other items to be sent weigh $$\frac{3}{5}$$ the weight of the books. Let the book package be X and let the other package be Y. So from the above problem,
X= 9$$\frac{3}{8}$$ which is $$\frac{75}{8}$$ and
Y = $$\frac{3}{5}$$ X
= $$\frac{3}{5}$$ × $$\frac{75}{8}$$
= $$\frac{225}{40}$$
= 5 $$\frac{5}{8}$$.
So the total package is
X + Y = 9$$\frac{3}{8}$$ + 5 $$\frac{5}{8}$$
= $$\frac{75}{8}$$ + $$\frac{225}{40}$$
= $$\frac{600}{40}$$
= 15 pounds. |
# ACT Math : Acute / Obtuse Isosceles Triangles
## Example Questions
← Previous 1 3
### Example Question #212 : Plane Geometry
What is the area of an isosceles triangle with a vertex of degrees and two sides equal to ?
Explanation:
Based on the description of your triangle, you can draw the following figure:
You can do this because you know:
1. The two equivalent sides are given.
2. Since a triangle is degrees, you have only or degrees left for the two angles of equal size. Therefore, those two angles must be degrees and degrees.
Now, based on the properties of an isosceles triangle, you can draw the following as well:
Based on your standard reference triangle, you know:
Therefore, is .
This means that is and the total base of the triangle is .
Now, the area of the triangle is:
or
### Example Question #213 : Plane Geometry
An isosceles triangle has a height of and a base of . What is its area?
Explanation:
Use the formula for area of a triangle:
### Example Question #214 : Plane Geometry
An isosceles triangle has a base length of and a height that is twice its base length. What is the area of this triangle?
Explanation:
1. Find the height of the triangle:
2. Use the formula for area of a triangle:
### Example Question #215 : Plane Geometry
The height of an isosceles triangle, dropped from the vertex to its base, is one fourth the length of the base. If the area of this triangle is , what is its perimeter?
Explanation:
Based on the description of this question, you can draw your triangle as such. We can do this thanks to the nature of an isosceles triangle:
Now, you know that the area of a triangle is defined as:
So, for our data, we can say:
Solving for , we get:
Thus, .
Now, for our little triangle on the right, we can draw:
Using the Pythagorean Theorem, we know that the other side is:
This can be simplified to:
Now, we know that this side is the "equal" side of the isosceles triangle. Therefore, we can know that the total perimeter is:
### Example Question #216 : Plane Geometry
The base of an isosceles triangle is five times the length of its correlative height. If the area of this triangle is , what is its perimeter?
Explanation:
Based on the description of this question, you can draw your triangle as such. We can do this thanks to the nature of an isosceles triangle:
Now, you know that the area of a triangle is defined as:
So, for our data, we can say:
Solving for , we get:
Thus, .
Now, for our little triangle on the right, we can draw:
Using the Pythagorean Theorem, we know that the other side is:
This can be simplified to:
Now, we know that this side is the "equal" side of the isosceles triangle. Therefore, we can know that the total perimeter is:
### Example Question #1 : How To Find The Perimeter Of An Acute / Obtuse Isosceles Triangle
What is the area of an isosceles triangle with a vertex of degrees and two sides equal to units?
Explanation:
Based on the description of your triangle, you can draw the following figure:
You can do this because you know:
1. The two equivalent sides are given.
2. Since a triangle is degrees, you have only or degrees left for the two angles of equal size. Therefore, those two angles must be degrees and degrees.
Now, based on the properties of an isosceles triangle, you can draw the following as well:
Based on your standard reference triangle, you know:
Therefore, is .
This means that is , and the total base of the triangle is .
Now, the area of the triangle is:
or
### Example Question #218 : Plane Geometry
What is the perimeter of an isosceles triangle with a vertex of degrees and two sides equal to
Explanation:
Based on the description of your triangle, you can draw the following figure:
You can do this because you know:
1. The two equivalent sides are given.
2. Since a triangle is degrees, you have only or degrees left for the two angles of equal size. Therefore, those two angles must be degrees and degrees.
Now, based on the properties of an isosceles triangle, you can draw the following as well:
Based on your standard reference triangle, you know:
Therefore, is .
This means that is and the total base of the triangle is .
Therefore, the perimeter of the triangle is:
### Example Question #219 : Plane Geometry
Triangle A and Triangle B are similar isosceles triangles. Triangle A's sides measure , and . Two of the angles in Triangle A each measure . Triangle B's sides measure , and . What is the measure of the smallest angle in Triangle B?
Explanation:
Because the interior angles of a triangle add up to , and two of Triangle A's interior angles measure , we must simply add the two given angles and subtract from to find the missing angle:
Therefore, the missing angle (and the smallest) from Triangle A measures . If the two triangles are similar, their interior angles must be congruent, meaning that the smallest angle is Triangle B is also .
The side measurements presented in the question are not needed to find the answer!
### Example Question #220 : Plane Geometry
Triangle A and Triangle B are similar isosceles triangles. Triangle A has a base of and a height of . Triangle B has a base of . What is the length of Triangle B's two congruent sides?
Explanation:
We must first find the length of the congruent sides in Triangle A. We do this by setting up a right triangle with the base and the height, and using the Pythagorean Theorem to solve for the missing side (). Because the height line cuts the base in half, however, we must use for the length of the base's side in the equation instead of . This is illustrated in the figure below:
Using the base of and the height of , we use the Pythagorean Theorem to solve for :
Therefore, the two congruent sides of Triangle A measure ; however, the question asks for the two congruent sides of Triangle B. In similar triangles, the ratio of the corresponding sides must be equal. We know that the base of Triangle A is and the base of Triangle B is . We then set up a cross-multiplication using the ratio of the two bases and the ratio of to the side we're trying to find (), as follows:
Therefore, the length of the congruent sides of Triangle B is .
### Example Question #221 : Plane Geometry
Isosceles triangles and share common side is an obtuse triangle with sides is also an obtuse isosceles triangle, where . What is the measure of ? |
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Hint: To determine the limit of $\cot x$ as x approaches ${{\pi }^{-}}$, we are going to first write $\cot x=\dfrac{\cos x}{\sin x}$ and then as x approaches to ${{\pi }^{-}}$ so we are going to take “h” which lies between 0 and 1 and this value of h is very less than 1. Then we write in place of $\left( x-h \right)$ in place of x in $\dfrac{\cos x}{\sin x}$ and then put the limit h tending to 0. And then simplify.
In the above problem, it is given that $\cot x$ as x approaches ${{\pi }^{-}}$ so writing this limit expression in the mathematical form we get,
$\Rightarrow \displaystyle \lim_{x\to {{\pi }^{-}}}\cot x$
Now, we know from the trigonometric properties that:
$\cot x=\dfrac{\cos x}{\sin x}$
So, using the above relation in the above limit we get,
$\Rightarrow \displaystyle \lim_{x\to {{\pi }^{-}}}\dfrac{\cos x}{\sin x}$
It is given that x approaches ${{\pi }^{-}}$ so this means that x value is a subtraction of $\pi$ with some number. Let us assume that a small number is “h”. This number “h” lies between 0 and 1 in the following way:
$0 < h << 1$
So, we can write $x=\pi -h$ in the above limit where h is approaching 0.
$\Rightarrow \displaystyle \lim_{h \to 0}\dfrac{\cos \left( \pi -h \right)}{\sin \left( \pi -h \right)}$
We know the trigonometric identities of sine and cosine as follows:
\begin{align} & \cos \left( A-B \right)=\cos A\cos B+\sin A\sin B; \\ & \sin \left( A-B \right)=\sin A\cos B-\cos A\sin B \\ \end{align}
Using the above trigonometric identities in the above limit we get,
$\Rightarrow \displaystyle \lim_{h \to 0}\dfrac{\cos \pi \cosh +\sin \pi \sinh }{\sin \pi \cosh -\cos \pi \sinh }$
We know the values of sine of $\pi$ and cosine of $\pi$ as follows:
\begin{align} & \sin \pi =0; \\ & \cos \pi =-1 \\ \end{align}
Substituting the above values in the limit expression we get,
\begin{align} & \Rightarrow \displaystyle \lim_{h \to 0}\dfrac{\left( -1 \right)\cosh +\left( 0 \right)\sinh }{\left( 0 \right)\cosh -\left( -1 \right)\sinh } \\ & \Rightarrow \displaystyle \lim_{h \to 0}\dfrac{-\cosh }{\sinh } \\ \end{align}
Applying the value of limit by putting h as 0 in the above fraction we get,
$-\dfrac{\cos 0}{\sin 0}$
We know that the value of $\sin 0=0\And \cos 0=1$ so substituting these values in the above w eget,
$-\dfrac{1}{0}$
And we know that $\dfrac{1}{0}$ is not defined or infinity.
$-\infty$
Hence, the evaluation of the above limit is $-\infty$.
Note: The possible mistake in the above problem is that in the last two steps you might forget to put a negative sign in front of the infinity because you might think what difference does it make if we remove this negative sign in front of the infinity.
\begin{align} & -\dfrac{1}{0} \\ & =\infty \\ \end{align}
This is the wrong answer because positive and negative infinities are completely different. As one is pointing in a positive direction and the other is in a negative direction so make sure you won’t make this mistake. |
# A hemisphere of maximum possible diameter is placed over a cuboidal block of side 7 cm.
Question:
(i) A hemisphere of maximum possible diameter is placed over a cuboidal block of side 7 cm. Find the surface area of the solid so formed.
(ii) A cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that the hemisphere can have? Find the cost of painting the total surface area of the solid so formed, at the rate of 5 per 100 sq cm. [Use π">π= 3.14]
Solution:
(i) Disclaimer: It is written cuboid in the question but it should be cube.
From the figure, it can be observed that the maximum diameter possible for such hemisphere is equal to the cube’s edge, i.e., 7 cm.
Radius $(r)$ of hemispherical part $=\frac{7}{2}=3.5 \mathrm{~cm}$
Total surface area of solid = Surface area of cubical part + CSA of hemispherical part
− Area of base of hemispherical part
$=6$ (Edge) $^{2}+2 \pi r^{2}-\pi r^{2}=6$ (Edge) $^{2}+\pi r^{2}$
Total surface area of solid $=6(7)^{2}+\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}$
$=294+38.5=332.5 \mathrm{~cm}^{2}$
(ii)
We have,
the edge of the cubical block, $a=10 \mathrm{~cm}$
The largest diameter of the hemisphere $=a=10 \mathrm{~cm}$
Also, the radius of the hemisphere, $r=\frac{10}{2}=5 \mathrm{~cm}$
Now,
Total surface area of the solid = TSA of cube + CSA of hemisphere-Area of circle
$=6 a^{2}+2 \pi r^{2}-\pi r^{2}$
$=6 a^{2}+\pi r^{2}$
$=6 \times 10 \times 10+3.14 \times 5 \times 5$
$=600+78.5$
$=678.5 \mathrm{~cm}^{2}$
As, the rate of painting the solid $=₹ 5$ per $100 \mathrm{~cm}^{2}$
So, the cost of painting the solid $=678.5 \times \frac{5}{100} \approx ₹ 33.92$
hence, the cost of painting the total surface area of the solid is ₹33.92. |
## Constant of Proportionality Calculator
Introducing our Constant of Proportionality Calculator, a helpful tool for quickly determining the constant value that relates two proportional quantities, facilitating mathematical and scientific calculations with ease
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# Constant of Proportionality Calculator: Everything You Need to Know
Understanding the consistent of proportionality is critical for both college students and experts. This concept is essential in mathematics and diverse applications. In this weblog, we can explore the consistent of proportionality formula, the way to find the consistent of proportionality from a table, its utility in seventh grade math, interpreting it from a graph, and more. Let's dive in!
## What is the Constant of Proportionality?
The constant of proportionality is a value that describes the ratio between two directly proportional quantities. It is denoted by the symbol $$k$$ and is used in the equation of the form $$y = kx$$.
Here, $$y$$ and $$x$$ are variables, and $$k$$ represents the constant rate of change between them.
## Constant of Proportionality Formula
$k = \frac{y}{x}$
This formula is used to calculate $$k$$ when you know the values of $$y$$ and $$x$$.
## Importance of the Constant of Proportionality
Understanding and calculating the steady of proportionality is important for several motives:
• Simplifies Complex Equations: It allows in simplifying and solving complex mathematical equations and inequalities.
• Real-lifestyles Applications: It is used in diverse fields which include physics, engineering, economics, and extra to describe relationships among quantities.
• Foundation for Further Studies: Mastering this idea is essential for college students because it bureaucracy the inspiration for superior subjects in arithmetic and science.
## How to Find the Constant of Proportionality from a Table
Finding the steady of proportionality from a desk includes a few steps:
1. Identify the Variables: Determine which columns represent $$x$$ and $$y$$.
2. Select Pairs of Values: Choose pairs of $$x$$ and $$y$$ values.
3. Calculate $$k$$: Use the formula $$k = \frac{y}{x}$$ for each pair to ensure $$k$$ is consistent.
Example
Consider a table showing the gap traveled (y) and time taken (x):
Time (hours) Distance (miles)
1 60
2 120
3 180
### Using the formula, we calculate:
$k = \frac{60}{1} = 60$
$k = \frac{120}{2} = 60$
$k = \frac{180}{3} = 60$
Here, $$k$$ is 60 miles per hour, indicating a constant speed.
## Constant of Proportionality in 7th Grade
In 7th grade, college students frequently come upon problems related to the steady of proportionality. These issues assist college students understand direct relationships between variables and exercise using the formulation $$y = kx$$.
If a recipe calls for 3 cups of flour to make 6 cookies, find the constant of proportionality.
$k = \frac{6 \text{ cookies}}{3 \text{ cups of flour}} = 2$
This means there are 2 cookies per cup of flour.
## Constant of Proportionality Graph
A graph can visually constitute the consistent of proportionality. When graphed, an instantaneous proportional dating will form a instantly line passing thru the starting place.
### How to Find the Constant of Proportionality on a Graph
To discover
π
ok from a graph:
To find $$k$$ from a graph:
1. Plot the Points: Plot the given data points on a graph.
2. Draw the Line: Draw a line through the points. Ensure it passes through the origin.
3. Calculate the Slope: The slope of the line (rise over run) represents the constant of proportionality.
### Example
Consider a graph with the following points: (1, 2), (2, 4), and (3, 6).
The rise between (1, 2) and (2, 4) is 2, and the run is 1.
Thus, the slope $$k$$ is $$\frac{2}{1} = 2$$.
The constant of proportionality $$k$$ is 2.
## Using a Constant of Proportionality Table
A constant of proportionality table helps organize values to identify the proportional relationship easily.
### Example
$$x$$ $$y$$
1 2
2 4
3 6
By using the formula $$k = \frac{y}{x}$$:
$k = \frac{2}{1} = 2$
$k = \frac{4}{2} = 2$
$k = \frac{6}{3} = 2$
Consistent $$k$$ values confirm the direct proportional relationship.
## Constant of Proportionality Example
Consider a real-world constant of proportionality example: the relationship between the cost and the number of items bought.
### Example
If 5 notebooks cost $15, the constant of proportionality is: $k = \frac{15 \text{ dollars}}{5 \text{ notebooks}} = 3$ This means each notebook costs$3.
## Direct Proportion Calculator $$y = kx$$
A direct proportion calculator $$y = kx$$ is a tool that simplifies finding the constant of proportionality. By inputting values of $$x$$ and $$y$$, the calculator instantly provides $$k$$.
### How to Use
1. Enter Values: Input the values of $$x$$ and $$y$$.
2. Calculate: Click the calculate button.
3. Result: The calculator displays the constant of proportionality $$k$$.
### Benefits
• Accuracy: Ensures precise calculations.
• Speed: Saves time by providing instant results.
• Convenience: Easy for students, teachers, and professionals to use.
## Conclusion
The consistent of proportionality is a critical mathematical concept with vast packages in education and numerous fields. Whether you're gaining knowledge of approximately it in seventh grade, studying a graph, or using a constant of proportionality calculator, knowledge this concept helps simplify many problems. With this knowledge, you may hopefully tackle proportional relationships and enhance your mathematical talent.
#### References:
1- Proportionality, Hyperbolic coordinates, Inverse proportionality
What is the constant of proportionality if the two variables are the same?
If the two variables are the same, the constant of proportionality $$k$$ is 1. This is because the ratio of the variables $$\frac{y}{x}$$ will always equal 1 if $$y = x$$.
Can the constant of proportionality be negative?
Yes, the constant of proportionality can be negative. This occurs when the variables $$x$$ and $$y$$ are inversely related. For example, if $$y$$ decreases as $$x$$ increases, $$k$$ will be negative.
How can I calculate the constant of proportionality?
You can calculate the constant of proportionality using the formula:
$k = \frac{y}{x}$
where $$y$$ and $$x$$ are the values of the two variables.
Is the constant of proportionality the same as a slope?
Yes, the constant of proportionality is the same as the slope in a linear relationship. Both represent the rate of change between two variables in the equation $$y = kx$$.
Is proportion a type of ratio?
Yes, proportion is a type of ratio that represents the equality of two ratios. In a direct proportion, the ratio of two variables remains constant.
Can there be two constants of proportionality?
No, in a given linear relationship between two variables, there can only be one constant of proportionality $$k$$. If the relationship is truly proportional, $$k$$ will be the same for all pairs of $$x$$ and $$y$$.
What is the constant equation?
The constant equation refers to the equation of a direct proportion:
$y = kx$
where $$k$$ is the constant of proportionality.
What is meant by force constant?
The force constant, often denoted by $$k$$, refers to a constant used in Hooke's Law in physics:
$F = kx$
where $$F$$ is the force applied to a spring, $$k$$ is the spring constant, and $$x$$ is the displacement of the spring from its equilibrium position. The force constant $$k$$ measures the stiffness of the spring. |
Simplifying Expressions Lesson
starstarstarstarstarstarstarstarstarstar
by Felix Gabathuler
| 35 Questions
Today we are going to make sure that everyone is on the same page when thinking about Algebra! Yay! Scroll through this document and complete the each section as directed. Watch the videos, fill in the blanks, and really concentrate!
You should take notes on what you learn today! I will be checking to see how convincing your work is in your notebooks!
Algebra is the language through which we describe patterns. It's like a way of writing really long patterns that go on forever as something much smaller. As opposed to having to do something over and over again, algebra gives you a simple way to express that repetitive process. It's also seen as a "gatekeeper" subject.
In simple addition, what you did in elementary school, we learned to add all the numbers together to get a sum. In Pre-Algebra, numbers are sometimes attached to variables and we need to make sure that the variables are alike before we add the numbers. Can you figure out the pattern below?
The pattern above increases by 4 every time you increase position by 1. We're going to get into the patterns and how to figure them out later. First, we've got to cover the basics. Watch the video below to understand what a variable is and how we can combine like terms.
1
1
After watching the video, write down how you would explain the concept of a variable to someone who has never heard about that before. Use 1 example.
Now, Let's solidify our combining like terms practice with a few extra questions! Make sure to write your answers properly, doodloodloodloo... (pinkie finger waving), by always writing the term with the variable first. Complete the following questions that have blue bubbles next to them.
2
3
4
5
6
7
8
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10
11
2
1
3
1
4
1
5
1
6
1
7
1
8
1
9
1
10
1
11
1
12
1
Are the following expressions equivalent? Explain your reasoning.
13
1
Write an expression in simplest form that represents the area of the banner.
14
1
Draw a diagram that shows how the expression can represent the area of a figure. Then simplify the expression.
Now, watch this short video on distribution, which will help you combine like terms in bigger expressions
15
1
How would you explain "distribution" to someone who doesn't know? Can you come up with a real world example?
16
1
Simplify the following expression using distribution: 2(y - 8)
-2y + 16
2y - 16
8y - 2
2y + 16
17
1
Simplify the following expression using distribution: -5(2x + 1)
-10x + 5
10x - 5
-10x - 5
10x + 5
18
1
Which of the following choices could be the non-simplified version of: 3x - 5 There might be more than one answer!
3(x-4)
3(x - 1) - 2
2 + 3x - 7
9x - 2(3x + 3) + 1
3x - 6x - 5
Things are a little different when you have more than one parentheses in your expression. Check out this video to learn how to distribute using a negative number and what happen when you are dealing with a really big expression!
Let's practice a few more distribution problems! Only answer the questions with the blue bubbles, keep things proper, and simplify!
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20
21
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24
25
26
19
1
20
1
21
1
22
1
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1
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1
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1
26
1
Sometimes you will see expressions that have more than one pair of parentheses! Instead of crying, watch this short video to see how you can simplify those.
27
1
In your own words, explain the steps to simplify the expression: 3(2x + 1) + 4(5x + 3) . You do not need to solve the problem. Just write what you would do first, second, third, etc.
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28
1
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Extension!
You apply gold foil to a piece of red poster board to make the design shown below.
Write an expression in simplest form that represents the area of the gold foil. |
All the solutions provided in McGraw Hill My Math Grade 5 Answer Key PDF Chapter 10 Review will give you a clear idea of the concepts.
Vocabulary Check
Read each clue. Fill the corresponding section of the crossword puzzle to answer each clue. Use the words In the word bank.
dividend divisor fraction Inverse operations
quotient mixed number scaling unit fraction
Across
1. The result of a division problem.
2. The number that divides the dividend.
3. A number that is being divided.
4. The process of resizing a number when you multiply by a fraction that is greater than or less than 1.
5. Operations that undo each other, such as multiplication and division.
Down
6. A fraction with a numerator of 1.
7. A number that has a whole number part and a fraction part.
8. A number that represents part of a whole or part of a set.
1. the quotient is the number which is generated when we perform division operations on two numbers. Basically, it is the result of the division method.
Dividend = Quotient Ă— Divisor + Remainder
If the remainder is equal to 0, then;
Dividend = Quotient × Divisor
Therefore,
Quotient = Dividend Ă· Divisor
2. In division, we divide a number by any other number to get another number as a result. So, the number which is getting divided here is called the dividend. The number which divides a given number is the divisor.
The operation of division in the form of Dividend Ă· Divisor = Quotient
The above expression can also be written as: Divisor = Dividend Ă· Quotient
– The divisor divides the number into parts.
– The divisor can divide the dividend either completely or partially. When divided completely, the remainder is zero and when divided partially, the remainder is a non-zero integer.
– Divisor could be a positive or negative number.
– A number that divides an integer exactly, leaving no remainder, is also termed the divisor.
– The divisor 1 and -1 can divide every integer, present in the number line.
3. In division, we divide a number by any other number to get another number as a result. So, the number which is getting divided here is called the dividend.
4. Scaling is the factor which is used to represent the object size. The size of the object can be shown by increasing or decreasing its original size. In general, the represented size of the object increased for the small object whereas it decreased for the bigger object. Scaling is used for better viewing of an object.
5. Inverse means the opposite effect of an action or a step. In mathematics, we have operations such as addition(+), subtraction(-), multiplication(x), division(÷), squaring, square root, and logarithms.
– When we use two operations together, it is possible to have an inverse impact on the result due to the operations used. The process in which the effect of one operation is inversed by another operation is termed as inverse operations.
6. All the fractions with 1 as the numerator are known as unit fractions. In these fractions, we take only one part of the whole which is divided equally into a finite number of parts. Unit denotes one. Therefore, they are known as unit fractions. Examples: 1/2, 1/3, 1/4…
7. A mixed number is a combination of a whole number and a proper fraction.
8. A fraction represents a numerical value, which defines the parts of a whole.
Concept Check
Estimate each product. Draw a bar diagram if necessary.
Question 9.
$$\frac{3}{4}$$ Ă— 23
______________
The above-given equation:
3/4 x 23
The estimation of 23 is 24.
3/4 x 24
Question 10.
$$\frac{1}{5}$$ Ă— 22
______________
The above-given:
1/5 x 22
The estimation of 22 is 20.
Multiply. Write in the simplest form.
Question 11.
$$\frac{1}{3}$$ Ă— 21
______________
Simplifying a fraction means reducing a fraction to its simplest form. A fraction is in its simplest form if its numerator and denominator have no common factors other than 1.
To simplify fractions, write the factors of the numerator and denominator and mark the common factors in them. Then, divide the numerator and denominator by the common factors until they have no common factor except 1. The fraction thus obtained is in its simplest form.
The above-given equation: 1/3 x 21
Therefore, $$\frac{1}{3}$$ Ă— 21 = 7.
Question 12.
26 Ă— $$\frac{1}{2}$$
______________
Simplifying a fraction means reducing a fraction to its simplest form. A fraction is in its simplest form if its numerator and denominator have no common factors other than 1.
To simplify fractions, write the factors of the numerator and denominator and mark the common factors in them. Then, divide the numerator and denominator by the common factors until they have no common factor except 1. The fraction thus obtained is in its simplest form.
The above-given equation: 26 x 1/2
Therefore, 26 Ă— $$\frac{1}{2}$$ = 13
Question 13.
$$\frac{1}{5}$$ Ă— $$\frac{3}{8}$$
______________
Simplifying a fraction means reducing a fraction to its simplest form. A fraction is in its simplest form if its numerator and denominator have no common factors other than 1.
To simplify fractions, write the factors of the numerator and denominator and mark the common factors in them. Then, divide the numerator and denominator by the common factors until they have no common factor except 1. The fraction thus obtained is in its simplest form.
The above-given equation: 1/5 x 3/8
Therefore, $$\frac{1}{5}$$ Ă— $$\frac{3}{8}$$ = 3/40
Question 14.
$$\frac{1}{2}$$ Ă— $$\frac{7}{8}$$
______________
Simplifying a fraction means reducing a fraction to its simplest form. A fraction is in its simplest form if its numerator and denominator have no common factors other than 1.
To simplify fractions, write the factors of the numerator and denominator and mark the common factors in them. Then, divide the numerator and denominator by the common factors until they have no common factor except 1. The fraction thus obtained is in its simplest form.
The above-given equation: 1/2 x 7/8
Therefore, $$\frac{1}{2}$$ Ă— $$\frac{7}{8}$$ = 7/16.
Question 15.
3$$\frac{1}{4}$$ Ă— $$\frac{3}{5}$$
______________
Simplifying a fraction means reducing a fraction to its simplest form. A fraction is in its simplest form if its numerator and denominator have no common factors other than 1.
To simplify fractions, write the factors of the numerator and denominator and mark the common factors in them. Then, divide the numerator and denominator by the common factors until they have no common factor except 1. The fraction thus obtained is in its simplest form.
The above-given equation: 3 1/4 x 3/5
3 1/4 is a mixed fraction. So convert it into an improper fraction.
Therefore, 3$$\frac{1}{4}$$ Ă— $$\frac{3}{5}$$ = 1 19/20
Question 16.
1$$\frac{1}{8}$$ Ă— 3$$\frac{2}{3}$$
______________
Simplifying a fraction means reducing a fraction to its simplest form. A fraction is in its simplest form if its numerator and denominator have no common factors other than 1.
To simplify fractions, write the factors of the numerator and denominator and mark the common factors in them. Then, divide the numerator and denominator by the common factors until they have no common factor except 1. The fraction thus obtained is in its simplest form.
The above-given equation: 1 1/8 x 3 2/3
both the fractions are mixed fractions. Convert them into improper fractions.
Find each quotient. Use a model. Check using multiplication.
Question 17.
2 Ă· $$\frac{1}{6}$$ = ______________
Check ____ Ă— $$\frac{1}{6}$$ = ____ or 2
The above-given equation:
2 Ă· 1/6
This can be simplified as:
The model can be represented as:
We need to check the answer:
12 x 1/6 = 2
Question 18.
$$\frac{1}{3}$$ Ă· 5 = ______________
Check ____ Ă— 5 = ____ or $$\frac{1}{3}$$
The above-given equation:
1/3 Ă· 5
This can be simplified as:
we need to check the answer:
1/15 x 5 = 5/15 = 1/3
Problem Solving
Question 19.
Ken is working on a project for social studies. He has a piece of poster board that needs to be divided equally into $$\frac{1}{3}$$-foot sections. The poster board is 4 feet wide. How many sections will Ken have
on the poster board?
The above-given:
The number of sections that need to be divided equally each = 1/3 foot
The length of feet = 4
The number of sections Ken has on the poster board = B
B = 4 Ă· 1/3
Therefore, he has 12 sections.
Question 20.
Samantha measured the dimensions of a rectangular poster frame. It was $$\frac{3}{4}$$– yard long and $$\frac{1}{2}$$– yard wide. What is the area of the frame?
The length of a rectangular poster frame = 3/4
The width of a rectangular poster frame = 1/2
The area of the frame = A
The area of the rectangle = length x width
A = 3/4 x 1/2
A = 3 x 1/Â 4 x 2
A = 3/8 square yards.
Therefore, the area of the frame is 3/8 square yards.
Question 21.
At Middle Avenue School, $$\frac{1}{30}$$– of the students are on the track team. The track coach offered to buy pizza or subs for everyone on the team. Of the students for whom the coach bought food, $$\frac{3}{5}$$ ordered pizza. What fraction of the students at Middle Avenue School ate pizza?
The above-given:
The number of students on the track team = 1/30
The number of students who ordered pizza = 3/5
The fraction of students who ate pizza = F
F = 3/5 of 1/30 students
F = 3/5 x 1/30
F = 3/150
F = 1/50
Therefore, 1/50 is the required fraction of who ate pizza.
Question 22.
Saraid is dividing $$\frac{1}{4}$$– pound of cashews into 5 plastic bags. What fraction of a pound of cashews will be in each plastic bag? Draw a model to help you solve.
The above-given:
The number of pounds of cashews Saraid is dividing = 1/4
The number of plastic bags = 5
The fraction of a pound of cashews in each plastic bag = C
C = 1/4 Ă· 5
Therefore, each plastic bag have 1/20 pound cashews.
Test Practice
Question 23.
The Great Rope Company sells a rope that is 5 feet long. Juan wants to make sections that are $$\frac{1}{3}$$ foot long. How many sections will Juan be able to make? Use the model.
A. 18 sections
B. 15 sections
C. 6 sections
D. 3 sections
The above-given:
The length of rope = 5
The number of sections Juan wants to make each = 1/3
The number of sections he can make = J
J = 5 Ă· 1/3
J = 5/1/1/3
J = 5/1 x 3/1
J = 5 x 3
J = 15
Therefore, he can make 15 sections.
Reflect
Use what you learned about fraction operations to complete the graphic organizer. |
# Solve an equation by combining like terms
## Presentation on theme: "Solve an equation by combining like terms"— Presentation transcript:
Solve an equation by combining like terms
EXAMPLE 1 Solve an equation by combining like terms Solve 8x – 3x – 10 = 20. 8x – 3x – 10 = 20 Write original equation. 5x – 10 = 20 Combine like terms. 5x – = Add 10 to each side. 5x = 30 Simplify. = 30 5 5x Divide each side by 5. x = 6 Simplify.
EXAMPLE 2 Solve an equation using the distributive property Solve 7x + 2(x + 6) = 39. SOLUTION When solving an equation, you may feel comfortable doing some steps mentally. Method 2 shows a solution where some steps are done mentally.
EXAMPLE 2 METHOD 1 METHOD 2 Do Some Steps Mentally Show All Steps 7x + 2(x + 6) = 39 7x + 2(x + 6) = 39 7x + 2x + 12 = 39 7x + 2x + 12 = 39 9x + 12 = 39 9x + 12 = 39 9x + 12 – 12 = 39 – 12 9x = 27 9x = 27 x = 3 = 9x 9 27 x = 3
EXAMPLE 3 Standardized Test Practice SOLUTION In Step 2, the distributive property is used to simplify the left side of the equation. Because –4(x – 3) = –4x + 12, Step 2 should be 5x – 4x + 12 = 17. ANSWER The correct answer is D. A C D B
GUIDED PRACTICE for Examples 1, 2, and 3 Solve the equation. Check your solution. 9d – 2d + 4 = 32 1. 4 ANSWER
EXAMPLE 2 GUIDED PRACTICE for Examples 1, 2, and 3 Solve the equation. Check your solution. 2w + 3(w + 4) = 27 2. 3 ANSWER
EXAMPLE 2 GUIDED PRACTICE for Examples 1, 2, and 3 Solve the equation. Check your solution. 6x – 2(x – 5) = 46 3. 9 ANSWER |
# How do you differentiate y=e^(-x/4)?
Mar 4, 2018
$\frac{\mathrm{dy}}{\mathrm{dx}} = - {e}^{- \frac{x}{4}} / 4$
#### Explanation:
In order to differentiate a function of a function, we use the chain rule. In other words, if $y = f \left(u\right)$ and $u = f \left(x\right)$, then
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}}$
In this case, let $u = - \frac{x}{4}$ and $y = {e}^{u}$. Then
$\frac{\mathrm{dy}}{\mathrm{du}} = {e}^{u} = {e}^{- \frac{x}{4}}$;
$\frac{\mathrm{du}}{\mathrm{dx}} = - \frac{1}{4}$
and, importantly,
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dx}} = {e}^{- \frac{x}{4}} \left(- \frac{1}{4}\right) = - {e}^{- \frac{x}{4}} / 4$ |
Question Video: Finding the Tension in a String Attaching a Sphere to a Wall- and the Normal Reaction with the Wall | Nagwa Question Video: Finding the Tension in a String Attaching a Sphere to a Wall- and the Normal Reaction with the Wall | Nagwa
# Question Video: Finding the Tension in a String Attaching a Sphere to a Wall- and the Normal Reaction with the Wall Mathematics • Second Year of Secondary School
## Join Nagwa Classes
A light string is tied from one end to a point on the surface of a homogeneous sphere, and the other end is attached to a point on a vertical smooth wall. The sphere is resting against the wall and weighs 33 N, and the string inclines to the vertical by 30Β°. Find the tension π in the string and the reaction π
of the wall.
04:34
### Video Transcript
A light string is tied from one end to a point on the surface of a homogeneous sphere, and the other end is attached to a point on a vertical smooth wall. The sphere is resting against the wall and weighs 33 newtons, and the string inclines to the vertical by 30 degrees. Find the tension π in the string and the reaction π
of the wall.
In this question, we have a sphere resting against a smooth vertical wall. And the weight of the sphere is 33 newtons. This means that this force acts vertically downwards. The reaction force π
acts perpendicular to the wall. Therefore, it acts horizontally to the right. We have a third force π, which is the tension in the string. The reaction force and the weight of the sphere act at right angles to one another. And we are also told that the string inclines to the vertical by 30 degrees. Since angles in a triangle sum to 180 degrees, the third angle in our triangle is 60 degrees.
Since there are three forces acting at a point, we can solve this problem using Lamiβs theorem. This states that when three forces acting at a point are in equilibrium, then each force is proportional to the sine of the angle between the other two forces. This can be written as follows. π΄ over sin πΌ is equal to π΅ over sin π½ which is equal to πΆ over sin πΎ, where π΄, π΅, and πΆ are three forces acting at a point. πΌ is the angle between forces π΅ and πΆ; π½, the angle between forces π΄ and πΆ; and πΎ, the angle between the forces π΄ and π΅.
In this question, we already know that the angle between the 33-newton force and π
is 90 degrees. By adding 90 degrees to 60 degrees, the angle between the 33-newton force and the tension π is 150 degrees. And finally, since angles at a point sum to 360 degrees, the angle between the tension and reaction forces is 120 degrees.
We now have three forces and their corresponding angles that we can substitute into Lamiβs theorem. This gives us π over sin of 90 degrees is equal to π
over sin of 150 degrees which is equal to 33 over sin of 120 degrees. The sin of 90 degrees is equal to one, the sin of 150 degrees is one-half, and the sin of 120 degrees is root three over two. Dividing by a half is the same as multiplying by two. Therefore, π is equal to two π
.
The third part of our equation is 33 divided by root three over two. Multiplying the numerator and denominator by two, this is equivalent to 66 over root three. We can then rationalize the denominator by multiplying the numerator and denominator by root three. This gives us 66 root three over three, which in turn is equal to 22 root three. Simplifying all three terms in Lamiβs theorem gives us π is equal to two π
which is equal to 22 root three.
The tension in the string is therefore equal to 22 root three newtons. As two π
is equal to 22 root three, π
is equal to 11 root three. The reaction force of the wall is 11 root three newtons. We have now calculated the values of the missing forces in the question.
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# How many times a wheel of radius 28 cm must rotate to 352 m? {Take pie=22/7}
Given:
The radius of a wheel is $28\ cm$.
To do:
We have to find the number of times the wheel must rotate to cover 352 m.
Solution:
Circumference of a circle of radius $r=2 \pi r$
Distance covered in 1 revolution $=2 \times \frac{22}{7} \times 28$
$=2 \times 22 \times 4$
$= 44 \times 4$
$= 176\ m$
Number of revolutions taken by the wheel to cover 352 m $=\frac{352}{176}$
$=2$
Hence, the wheel must rotate 2 times to cover 352 m.
Tutorialspoint
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Updated on: 10-Oct-2022
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# Domino Addition: Understanding the Part/Part/Whole Relationship
29 teachers like this lesson
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## Objective
Students will be able to correctly identify the parts and the whole for an addition number sentence, using a part/part/whole model.
#### Big Idea
The big idea of this lesson is the understanding that addition can be represented as parts of a whole and that we can use addition sentences to represent those parts.
## Activator and Materials
10 minutes
I begin this lesson by asking 3 girls and 4 boys to stand in front of the class.
What do you know about the number of students in front of the room? After the students return to their seats, I have the class turn and talk about the things they noticed. If necessary, I help by guiding those students who aren't sure where to begin to look at the number of girls and number of boys.
Think about how many students there are in all. How do you know?
Now I'm moving students to think about this more abstractly by asking students to think about how they could represent the information about the parts of the whole group of students who stood in front of the room.
I begin by demonstrating the part/part/whole model, drawing 3 squares to represent the girls in one part and 4 squares in the other part to represent the boys. I ask the students to turn and talk, thinking about if there is a number sentence they could use to represent the part, part, and the whole group of girls and boys at the front of the room. I am asking them to think of quantities (boys, girls, altogether) and relationships between those quantities and make sense of this using a representation, or model (MP1 & MP4).
Part-whole instruction is drawing students' attention specifically to the parts and the relationship of the parts to the whole. Students need many of these experiences to develop the flexibility of understanding that allows them to understand that 10 can be composed as 5 + 5, but it can also be composed of 6 + 1. With this understanding, students will be able to manipulate numbers, with understanding, when adding and subtracting, and later when multiplying and dividing.
## Develop the Concept
20 minutes
Using a Part/Part/Whole Mat, have the students build the part 6 cubes with one color and the part 3 cubes with another color. Have the students work in partners to place the cubes in the correct spot on the model. Have the students then write a number sentence to represent the model. Can they find the whole?
At this time, introduce the vocabulary for this lesson:
Explain to the students that the two parts in an addition number sentence are called the addends. The whole in an addition problem is called the sum. I don't expect my students to know these words, but I will now begin to use them, in context and with specificity so that the students bring them into their own vocabulary.
The next step needs to done thoughtfully. It may seem obvious to us as adults, but it is more abstract for 2nd graders. I have the students switch the two parts on their part/part/whole mat. Looking at the part/part/whole model, have them turn and talk to discuss what has changed and what has stayed the same. Students should recognize that the addends have changed order, but the amount in all has stayed the same.
I teach my students a chant to help them remember: “No matter the order the addends are in, the sum remains the same. Cha Cha!”
Explain to the students that, when we change the order of the addends, we call it a turn around fact and that this relationship is that they are related facts.
The students should build several more examples using their part/part/whole mat and their connecting cubes. Have them practice writing addition number sentences for each model they create on their part/part/whole mat. I want to ensure that my students can demonstrate and explain their understanding (2.NBT.B.9). As the students are working, I circulate around the classroom, stopping to ask questions about the numbers that the students are building. Some of the questions I may ask are: "What are your two addends? How did you know which addend to build first? What if your partner really wanted to build the other addend first, how would that change your sum?"
## Practice the Concept
20 minutes
The materials for this lesson include dominos, which may not be a math manipulative that is already in your classroom. I love them! Give each student a pile of 8 dominoes and the domino addition worksheet. I have found domino sets that have up to 18 dots on each side. You can differentiate based on the need of your students with these as well as the basic dominoes. Those who need more of a challenge can receive dominoes with more dots.
The students should use the halves of the dominoes to help them create a part/part/whole model. Once they have drawn the model, students will then write the addition number sentence that represents the model.
## Summarizer
10 minutes
When students finish their independent work doing domino addition, they work in partners to share a few of their domino number sentences. The purpose of the partner discussion is that they discuss the parts and whole of the model. The partner share is also a means to occupy earlier finishers. Ideally, all students should have time to share with a partner, because this prepares them for the class share. As you circulate during independent work, you'll have identified those students who are struggling, and can work with them individually or in a small group. This way they too will have time to discuss, and make sense, of their work. Then come together as a class and choose a few students share their understandings. |
Teacher resources and professional development across the curriculum
Teacher professional development and classroom resources across the curriculum
A B C
Solutions for Session 10, Part C
See solutions for Problems: C1 | C2 | C3 | C4
Problem C1
a.
Solution:
1 The greatest possible score is 8 + 8 + 8, or 24. 2 The lowest possible score is 5 + 5 + 5, or 15. 3 There are 10 different score combinations possible, but since a score of 21 can be gotten two different ways, there are only nine different scores possible: 8 + 8 + 8 = 24 8 + 8 + 7 = 23 8 + 8 + 5 = 21 8 + 7 + 7 = 22 8 + 7 + 5 = 20 8 + 5 + 5 = 18 7 + 7 + 7 = 21 7 + 7 + 5 = 19 7 + 5 + 5 = 17 5 + 5 + 5 = 15 4 John and Mary Beth both got 21, because that's the only score that can be gotten two different ways.
b.
The number and operations content of this problem involves the ability to use the commutative and associative laws to find a sum with three or more addends. Students will also develop flexibility with numbers.
c.
Students need to use logical thinking and to be able to recognize that the order in which they add three whole numbers doesn't affect the sum. They also need to be able to compare sums of numbers in an organized way.
d.
If students are struggling, ask them to start by writing out the sums. It is important to have an organized system -- for example, add all the sums that start with the 8s (first with three 8s, then two 8s, then one 8) and then repeat for the 7s and 5s.
e.
To get students to think beyond this problem, you can turn it around. Make up several scores that players said they got, and have students decide if these scores are possible.
Problem C2
a.
Solution:
1.
Five-sixths is greater. You know this because five parts out of six is more than five parts out of eight -- the pieces are larger.
2.
Three-fourths is greater. Two pies are each missing one piece. The pie with four pieces has less missing, so it is bigger.
3.
Two-thirds is greater. If you change the 2/3 to 6/9, then you have two pies each missing three pieces. The pie with nine pieces has less missing, so it is bigger.
4.
The order is 2/5, 1/2, 5/8, 2/3, 3/4, 5/6, 9/10, and 5/3. People will explain their reasoning in different ways, but here's one example:
• Two-fifths is the only fraction less than 1/2, so it is first. (Here you used 1/2 as the benchmark.) • Five-thirds is the only fraction greater than 1, so it is last. (Here you used 1 as the benchmark.) • The others are ordered according to the procedures explained above.
b.
The number and operations content of this problem involves understanding the concept of fractions (the roles of the numerator and denominator; understanding that the larger the denominator is, the smaller the piece is; etc.), how to compare and order fractions, equivalent fractions, and benchmarks.
c.
Students will probably want to convert to decimals or get common denominators for all the fractions. The procedures discussed above will help students learn to reason about fractions so that they can avoid tedious computation. Students will also need to use logical thinking.
d.
If students are struggling, you can ask the following questions to try to elicit the following answers:
• Why not just get a common denominator for all the fractions and then order them? (Because this is much more time-consuming and does not help to give us an intuitive feeling about the magnitude of fractions.) • Why not convert all the fractions to decimals and then order the decimals? (Because this is much more time-consuming and loses the fractions -- and thus our understanding of the fractions -- entirely.)
Essentially, students need to think about the benchmarks. They should be able to look at a fraction and know if that fraction is between 0 and 1/2, between 1/2 and 1, or greater than 1, and which of those benchmarks it's closer to. In addition, if the denominators are the same, they should know to compare numerators, and vice versa.
e.
There are several ways to help students think beyond this problem. The simplest is to change the fractions. You could also increase the number of fractions, use larger numbers, or include some decimals in the list. You could also have students put the numbers on a number line.
Problem C3
a.
Solution:
1 The smallest number of blocks you could remove is 4 -- 3 red blocks and 1 green block. To have 3 red blocks for every 4 green blocks, you must have a multiple of 3 + 4, or 7 (e.g., 7, 14, 21) blocks remaining in the box. This could be 3 red and 4 green, 6 red and 8 green, 9 red and 12 green, and so forth. However, since there are only nine of each color in the box, the most you could have is 6 red and 8 green blocks. 2 The smallest number of blocks you could remove is 3 (green blocks). To have 3 red blocks for every 2 green blocks, you must have a multiple of 3 + 2, or 5 (e.g., 5, 10, 15) blocks remaining in the box. This could be 3 red and 2 green, 6 red and 4 green, 9 red and 6 green, and so forth. However, since there are only nine of each color in the box, the most you could have is 9 red and 6 green blocks.
b.
The number and operations content of this problem is ratio and proportional reasoning, patterns, and developing mathematical language (such as understanding what "for every" means).
c.
Students need to use logical thinking. They also need to understand the inverse relationship between the number of parts and the size of the parts. Students need to be able to find equivalent fractions and compare fractions. They need to understand that when they're doing part-part relationships, they must identify the whole. For example, in the block problem, to have 3 red for every 4 green blocks, they need to be aware that the total will be 7 or a multiple of 7.
d.
If students are struggling, have them use actual blocks. They can make two piles and then notice the relationships that are evident.
e.
To help students think beyond the problem, ask them to solve a more complicated version of it, using nine red blocks, nine blue blocks, and nine green blocks. Tell them that there must be 3 red blocks for every 7 non-red blocks and 3 blue blocks for every 7 non-blue blocks. (There are two possible answers: 6 red, 6 blue, and 8 green; and 3 red, 3 blue, and 4 green.)
Problem C4
a.
Solution:
1 The sum of 3 + 4 + 5 = 12. Yes, because 12 = 4 • 3. 2 The sum of 7 + 8 + 9 = 24. Yes, because 24 = 8 • 3. 3 The sum of any three consecutive integers is three times the middle number.
b.
The number and operations content of this problem is number theory, addition of whole numbers, fluent computation, the idea of average, and patterns.
c.
Students need to be able to use logical thinking. They need to be able to divide or multiply by 3, add three numbers, and understand what is meant by "three consecutive integers."
d.
If students are struggling, encourage them to make an organized list in which they list the options one by one. They can also use manipulatives, such as Unifix cubes, to help make the patterns more evident. Once they have the numbers represented by stacks, they can take the top cube off the tallest stack and put it on the smallest to see that they are all the same height.
e.
To help students think beyond this problem, ask them to think about five consecutive numbers. (In this case, the sum will be five times the middle number.) |
# 7.3 Unit circle (Page 5/11)
Page 5 / 11
Likewise, there will be an angle in the fourth quadrant with the same cosine as the original angle. The angle with the same cosine will share the same x -value but will have the opposite y -value. Therefore, its sine value will be the opposite of the original angle’s sine value.
As shown in [link] , angle $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ has the same sine value as angle $\text{\hspace{0.17em}}t;$ the cosine values are opposites. Angle $\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$ has the same cosine value as angle $\text{\hspace{0.17em}}t;$ the sine values are opposites.
$\begin{array}{ccc}\mathrm{sin}\left(t\right)=\mathrm{sin}\left(\alpha \right)\hfill & \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}& \mathrm{cos}\left(t\right)=-\mathrm{cos}\left(\alpha \right)\hfill \\ \mathrm{sin}\left(t\right)=-\mathrm{sin}\left(\beta \right)\hfill & \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}& \mathrm{cos}\left(t\right)=\mathrm{cos}\left(\beta \right)\hfill \end{array}$
Recall that an angle’s reference angle is the acute angle, $\text{\hspace{0.17em}}t,$ formed by the terminal side of the angle $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and the horizontal axis. A reference angle is always an angle between $\text{\hspace{0.17em}}0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}90°,$ or $\text{\hspace{0.17em}}0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\frac{\pi }{2}\text{\hspace{0.17em}}$ radians. As we can see from [link] , for any angle in quadrants II, III, or IV, there is a reference angle in quadrant I.
Given an angle between $\text{\hspace{0.17em}}0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}2\pi ,$ find its reference angle.
1. An angle in the first quadrant is its own reference angle.
2. For an angle in the second or third quadrant, the reference angle is $\text{\hspace{0.17em}}|\pi -t|\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}|180°-t|.$
3. For an angle in the fourth quadrant, the reference angle is $\text{\hspace{0.17em}}2\pi -t\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}360°-t.$
4. If an angle is less than $\text{\hspace{0.17em}}0\text{\hspace{0.17em}}$ or greater than $\text{\hspace{0.17em}}2\pi ,$ add or subtract $\text{\hspace{0.17em}}2\pi \text{\hspace{0.17em}}$ as many times as needed to find an equivalent angle between $\text{\hspace{0.17em}}0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}2\pi .$
## Finding a reference angle
Find the reference angle of $\text{\hspace{0.17em}}225°\text{\hspace{0.17em}}$ as shown in [link] .
Because $\text{\hspace{0.17em}}225°\text{\hspace{0.17em}}$ is in the third quadrant, the reference angle is
$|\left(180°-225°\right)|=|-45°|=45°$
Find the reference angle of $\text{\hspace{0.17em}}\frac{5\pi }{3}.$
$\frac{\pi }{3}$
## Using reference angles
Now let’s take a moment to reconsider the Ferris wheel introduced at the beginning of this section. Suppose a rider snaps a photograph while stopped twenty feet above ground level. The rider then rotates three-quarters of the way around the circle. What is the rider’s new elevation? To answer questions such as this one, we need to evaluate the sine or cosine functions at angles that are greater than 90 degrees or at a negative angle. Reference angles make it possible to evaluate trigonometric functions for angles outside the first quadrant. They can also be used to find $\text{\hspace{0.17em}}\left(x,y\right)\text{\hspace{0.17em}}$ coordinates for those angles. We will use the reference angle of the angle of rotation combined with the quadrant in which the terminal side of the angle lies.
## Using reference angles to evaluate trigonometric functions
We can find the cosine and sine of any angle in any quadrant if we know the cosine or sine of its reference angle. The absolute values of the cosine and sine of an angle are the same as those of the reference angle. The sign depends on the quadrant of the original angle. The cosine will be positive or negative depending on the sign of the x -values in that quadrant. The sine will be positive or negative depending on the sign of the y -values in that quadrant.
## Using reference angles to find cosine and sine
Angles have cosines and sines with the same absolute value as their reference angles. The sign (positive or negative) can be determined from the quadrant of the angle.
Given an angle in standard position, find the reference angle, and the cosine and sine of the original angle.
1. Measure the angle between the terminal side of the given angle and the horizontal axis. That is the reference angle.
2. Determine the values of the cosine and sine of the reference angle.
3. Give the cosine the same sign as the x -values in the quadrant of the original angle.
4. Give the sine the same sign as the y -values in the quadrant of the original angle.
bsc F. y algebra and trigonometry pepper 2
given that x= 3/5 find sin 3x
4
DB
remove any signs and collect terms of -2(8a-3b-c)
-16a+6b+2c
Will
Joeval
(x2-2x+8)-4(x2-3x+5)
sorry
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
(X2-2X+8)-4(X2-3X+5)=0 ?
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
Y
master
master
Soo sorry (5±Root11* i)/3
master
Mukhtar
explain and give four example of hyperbolic function
What is the correct rational algebraic expression of the given "a fraction whose denominator is 10 more than the numerator y?
y/y+10
Mr
Find nth derivative of eax sin (bx + c).
Find area common to the parabola y2 = 4ax and x2 = 4ay.
Anurag
A rectangular garden is 25ft wide. if its area is 1125ft, what is the length of the garden
to find the length I divide the area by the wide wich means 1125ft/25ft=45
Miranda
thanks
Jhovie
What do you call a relation where each element in the domain is related to only one value in the range by some rules?
A banana.
Yaona
given 4cot thither +3=0and 0°<thither <180° use a sketch to determine the value of the following a)cos thither
what are you up to?
nothing up todat yet
Miranda
hi
jai
hello
jai
Miranda Drice
jai
aap konsi country se ho
jai
which language is that
Miranda
I am living in india
jai
good
Miranda
what is the formula for calculating algebraic
I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it
Miranda
state and prove Cayley hamilton therom
hello
Propessor
hi
Miranda
the Cayley hamilton Theorem state if A is a square matrix and if f(x) is its characterics polynomial then f(x)=0 in another ways evey square matrix is a root of its chatacteristics polynomial.
Miranda
hi
jai
hi Miranda
jai
thanks
Propessor
welcome
jai
What is algebra
algebra is a branch of the mathematics to calculate expressions follow.
Miranda
Miranda Drice would you mind teaching me mathematics? I think you are really good at math. I'm not good at it. In fact I hate it. 😅😅😅
Jeffrey
lolll who told you I'm good at it
Miranda
something seems to wispher me to my ear that u are good at it. lol
Jeffrey
lolllll if you say so
Miranda
but seriously, Im really bad at math. And I hate it. But you see, I downloaded this app two months ago hoping to master it.
Jeffrey
which grade are you in though
Miranda
oh woww I understand
Miranda
Jeffrey
Jeffrey
Miranda
how come you finished in college and you don't like math though
Miranda
gotta practice, holmie
Steve
if you never use it you won't be able to appreciate it
Steve
I don't know why. But Im trying to like it.
Jeffrey
yes steve. you're right
Jeffrey
so you better
Miranda
what is the solution of the given equation?
which equation
Miranda
I dont know. lol
Jeffrey
Miranda
Jeffrey |
### Two Dice
Find all the numbers that can be made by adding the dots on two dice.
### Biscuit Decorations
Andrew decorated 20 biscuits to take to a party. He lined them up and put icing on every second biscuit and different decorations on other biscuits. How many biscuits weren't decorated?
### Writing Digits
Lee was writing all the counting numbers from 1 to 20. She stopped for a rest after writing seventeen digits. What was the last number she wrote?
# Count the Digits
## Count the Digits
We can do all sorts of things with numbers - add, subtract, multiply, divide ...
Most of us start with counting when we are very little. We usually count things, objects, people etc. In this activity we are going to count the number of digits that are the same.
Rule 1 - The starting number has to have just three different digits chosen from $1, 2, 3, 4$.
Rule 2 - The starting number must have four digits - so thousands, hundreds, tens and ones.
For example, we could choose $2124$ or $1124$.
So when we've got our starting number we'll do some counting. Here is a worked example.
Starting Number
We then count in order the number of $1$s, then the number of $2$s, then $3$s and lastly $4$s, and write it down as shown here.
So the first count gave one $1$, one $3$ and two $4$s.
You may see that this has continued so the third line shows that the line above had three $1$s, one $2$, one $3$ and one $4$.
The fourth line counts the line above giving four $1$s, one $2$, two $3$s and one $4$.
And so it goes on until ... until when?
Your challenge is to start with other four digit numbers which satisfy the two rules and work on it in the way I did.
Tell us what you notice.
What happens if you have five digits in the starting number?
Got a solution? Then click on Submit a Solution above.
### Why do this problem?
This activity, in line with the theme for this month, offers an 'action' to perform on a group of numbers which pupils can continue and explore. Or, you could think of the writing down of the 'description' of a sequence as an action performed on that sequence. It might particularly appeal to those pupils who enjoy number work but who are perhaps not used to succeeding in this area.
### Possible approach
Introduce the task by taking an example and work it through with the group/class of pupils, emphasising how careful we have to be with the simple act of counting.
Give them plenty of time to explore their own choice of numbers before bringing them together to share findings. Depending on the age and experience of the learners, you may like to give them a separate sheet of paper simply to note down anything they notice as they work. Encourage them not to rub out as they go along so they have a record of their thoughts, to some extent.
A whole-group discussion could focus on what they notice and what other questions they might have as a result of working on this task. Some children might be keen to try to explain their findings. Do encourage them in this, even if you are not sure of the reasons yourself. Admitting your possible uncertainty will spur them on!
### Key questions
Tell me about what you see happening.
What will you do now?
Can you make any predictions before you start the next one?
### Possible extension
Change the rules so that only odd numbers are available, for example:
Rule 1 - The starting number must have just three different digits chosen from $1, 3, 5, 7$ Rule 2 - The starting number should have four digits, so thousands, hundreds, tens and ones.
For example, $3155$ or $1135$.
Some children might like to find out about 'Golomb sequences' which are related to this task.
### Possible support
Some pupils may need help in carefully counting the number of occurrences of each digit. It might, therefore, be useful for children to work in pairs so that someone else is always checking the counting. |
The carrots we bought this week were in such a variety of sizes, that our ever ambitious Finny (aged 5) suggested teaching Cian about size order.
• Carrots
• Ruler
Method
Finny started off by explaining what it meant to put the carrots in size order.
Cian always sorts objects in to size by naming them, “Daddy, Mummy, Ioan, Finn, Cian and Pilli”. Sometimes if he has a small object, he names it Soo, after his puppet from Sooty. Finn decided that this was the way to teach Cian.
Finn did very well with his attempts to educate a hungry toddler.
After Cian had finished feeding Soo her carrot, we tried to get him to say the names. Cian (2 years and 2months) is a very reluctant talker.
Finny has always been a big talker, so doesn’t understand Cian’s reluctance to use language. He has made it his personal mission to get him talking!
Finn got to work finding the opposite pairs of size comparisons.
He then added the size comparisons to the carrots.
Finn used a ruler to measure the length of the carrots.
I asked him to work out the difference in size between his smallest carrot (11cm) and largest carrot (23cm). He used the ruler like a number line to count up from 11cm to 23cm. This gave him a difference of 12cm.
Finny worked out that his largest carrot was the same length as his two smallest carrots combined. 12cm + 11cm = 23cm. He set them out in a sum, shown in the final picture below.
DfES Early Learning Goals (2017)
Mathematics
ELG 11 – Numbers:
Children count reliably with numbers from 1 to 20, place them in order and say which number is one more or one less than a given number. Using quantities and objects, they add or subtract two single-digit numbers and count on or back to find the answer.
DfES Outcomes for EYFS and National Curriculum (2013)
Numeracy Year 1 programme of study
Number – measurements
• compare, describe and solve practical problems for lengths and heights [for example, long/short, longer/shorter, tall/short, double/half]
• measure and begin to record lengths and heights
Pupils move from using and comparing different types of quantities and measures using non-standard units, including discrete (for example, counting) measurement, to using manageable common standard units. In order to become familiar with standard measures, pupils begin to use measuring tools such as a ruler. |
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# Which one of the two is greater? ${{\log }_{2}}3$or ${{\log }_{\dfrac{1}{2}}}5$(a) ${{\log }_{2}}3$(b) ${{\log }_{\dfrac{1}{2}}}5$(c) Both are equal (d) Can’t say
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330.9k+ views
Hint: Convert ${{\log }_{a}}x=y$to $x={{a}^{y}}$and find the appropriate values of ${{a}^{y}}$.
Here, we have to find that which is greater among ${{\log }_{2}}3$or ${{\log }_{\dfrac{1}{2}}}5$.
Taking ${{\log }_{2}}3=A$
We know that, when ${{\log }_{b}}a=n$
Then, $a={{b}^{n}}....\left( i \right)$
Similarly, $3={{2}^{A}}...\left( ii \right)$
Now, we know that ${{\left( 2 \right)}^{1}}=2$and ${{\left( 2 \right)}^{2}}=4$.
Also, ${{2}^{1}}<3<{{2}^{2}}$
From equation $\left( ii \right)$, ${{2}^{1}}<{{2}^{A}}<{{2}^{2}}$
Hence, $1Therefore, we get the approximate value of \[A={{\log }_{2}}3$between $1$ and $2$.
Taking ${{\log }_{\dfrac{1}{2}}}5=B$
From equation $\left( i \right)$,
$\Rightarrow 5={{\left( \dfrac{1}{2} \right)}^{B}}$
We know that $\dfrac{1}{a}={{a}^{-1}}$
Hence, we get $5={{2}^{-B}}....\left( iii \right)$
Now, we know that
${{2}^{2}}=4$
And ${{2}^{3}}=8$
Also, ${{2}^{2}}<5<{{2}^{3}}$
As, $-\left( -a \right)=a$
We can also write it as ${{2}^{-\left( -2 \right)}}<5<{{2}^{-\left( -3 \right)}}$
From equation $\left( iii \right)$, ${{2}^{-\left( -2 \right)}}<{{2}^{-B}}<{{2}^{-\left( -3 \right)}}$
Hence, $-2Therefore, we get the approximate value of \[B={{\log }_{\dfrac{1}{2}}}5$between $-2$and $-3$.
Clearly, as $A$is positive and $B$is negative.
$A>B$
Or, ${{\log }_{2}}3>{{\log }_{\dfrac{1}{2}}}5$
Therefore, option (a) is correct
Note: Here, some students may think that as $\dfrac{1}{2}$is smaller as compared to $2$, so it will require greater power to become $5$as compared to power required by$2$ to become $3$. But in this process, they miss the negative sign in the power that will also be required to convert $\dfrac{1}{2}$to $5$and get the wrong result.
Last updated date: 05th Jun 2023
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# 13.3 Resistance
Page 4 / 4
Equivalent resistance of two parallel resistor, ${R}_{p}$
For $2$ resistors in parallel with resistances ${R}_{1}$ and ${R}_{2}$ , the equivalent resistance is:
${R}_{p}=\frac{{R}_{1}{R}_{2}}{{R}_{1}+{R}_{2}}$
## Equivalent parallel resistance
Consider a circuit consisting of a single cell and three resistors that are connected in parallel.
Using what we know about voltage and current in parallel circuits we can define the equivalent resistance of several resistors in parallel as:
Equivalent resistance in a parallel circuit, ${R}_{p}$
For $n$ resistors in parallel, the equivalent resistance is:
$\frac{1}{{R}_{p}}=\left(\frac{1}{{R}_{1}},+,\frac{1}{{R}_{2}},+,\frac{1}{{R}_{3}},+,\cdots ,+,\frac{1}{{R}_{n}}\right)$
Let us apply this formula to the following circuit.
What is the total resistance in the circuit?
$\begin{array}{ccc}\hfill \frac{1}{{R}_{p}}& =& \left(\frac{1}{{R}_{1}},+,\frac{1}{{R}_{2}},+,\frac{1}{{R}_{3}}\right)\hfill \\ & =& \left(\frac{1}{10\phantom{\rule{0.166667em}{0ex}}\Omega },+,\frac{1}{2\phantom{\rule{0.166667em}{0ex}}\Omega },+,\frac{1}{1\phantom{\rule{0.166667em}{0ex}}\Omega }\right)\hfill \\ & =& \left(\frac{1+5+10}{10}\right)\hfill \\ & =& \left(\frac{16}{10}\right)\hfill \\ \hfill \therefore \phantom{\rule{1.em}{0ex}}{R}_{p}& =& 0,625\phantom{\rule{0.166667em}{0ex}}\Omega \hfill \end{array}$
## Aim:
To determine the effect of multiple resistors on current in a circuit
• Battery
• Resistors
• Light bulb
• Wires
## Method:
1. Construct the following circuits
2. Rank the three circuits in terms of the brightness of the bulb.
## Conclusions:
The brightness of the bulb is an indicator of how much current is flowing. If the bulb gets brighter because of a change then more current is flowing. If the bulb gets dimmer less current is flowing. You will find that the more resistors you have the brighter the bulb.
Why is this the case? Why do more resistors make it easier for charge to flow in the circuit? It is because they are in parallel so there are more paths for charge to take to move. You can think of it like a highway with more lanes, or the tube of marbles splitting into multiple parallel tubes. The more branches there are, the easier it is for charge to flow. You will learn more about the total resistance of parallel resistors later but always remember that more resistors in parallel mean more pathways. In series the pathways come one after the other so it does not make it easier for charge to flow.
Two $8\phantom{\rule{2pt}{0ex}}\mathrm{k}\Omega$ resistors are connected in parallel. Calculate the equivalent resistance.
1. Since the resistors are in parallel we can use:
$\frac{1}{{R}_{p}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}$
2. $\begin{array}{ccc}\hfill \frac{1}{{R}_{p}}& =& \frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}\hfill \\ & =& \frac{1}{8\phantom{\rule{0.166667em}{0ex}}\mathrm{k}\phantom{\rule{0.166667em}{0ex}}\Omega }+\frac{1}{10\phantom{\rule{0.166667em}{0ex}}\mathrm{k}\phantom{\rule{0.166667em}{0ex}}\Omega }\hfill \\ \hfill {R}_{p}& =& \frac{2}{8}\hfill \\ & =& 4\phantom{\rule{0.166667em}{0ex}}\mathrm{k}\phantom{\rule{0.166667em}{0ex}}\Omega \hfill \end{array}$
3. The equivalent resistance of two $8\phantom{\rule{2pt}{0ex}}\mathrm{k}\Omega$ resistors connected in parallel is $4\phantom{\rule{2pt}{0ex}}\mathrm{k}\Omega$ .
Two resistors are connected in parallel. The equivalent resistance is $100\phantom{\rule{2pt}{0ex}}\Omega$ . If one resistor is $150\phantom{\rule{2pt}{0ex}}\Omega$ , calculate the value of the second resistor.
1. Since the resistors are in parallel we can use:
$\frac{1}{{R}_{p}}=\frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}$
We are given the value of ${R}_{p}$ and ${R}_{1}$ .
2. $\begin{array}{ccc}\hfill \frac{1}{{R}_{p}}& =& \frac{1}{{R}_{1}}+\frac{1}{{R}_{2}}\hfill \\ \hfill \therefore \phantom{\rule{1.em}{0ex}}\frac{1}{{R}_{2}}& =& \frac{1}{{R}_{p}}-\frac{1}{{R}_{1}}\hfill \\ & =& \frac{1}{100\phantom{\rule{0.166667em}{0ex}}\Omega }-\frac{1}{150\phantom{\rule{0.166667em}{0ex}}\Omega }\hfill \\ & =& \frac{3-2}{300}\hfill \\ & =& \frac{1}{300}\hfill \\ \hfill {R}_{2}& =& 300\phantom{\rule{0.166667em}{0ex}}\Omega \hfill \end{array}$
3. The second resistor has a resistance of $300\phantom{\rule{2pt}{0ex}}\Omega$ .
## Resistance
1. What is the unit of resistance called and what is its symbol?
2. Explain what happens to the total resistance of a circuit when resistors are added in series?
3. Explain what happens to the total resistance of a circuit when resistors are added in parallel?
4. Why do batteries go flat?
The following presentation summarizes the concepts covered in this chapter.
## Exercises - electric circuits
1. Write definitions for each of the following:
1. resistor
2. coulomb
3. voltmeter
2. Draw a circuit diagram which consists of the following components:
1. 2 batteries in parallel
2. an open switch
3. 2 resistors in parallel
4. an ammeter measuring total current
5. a voltmeter measuring potential difference across one of the parallel resistors
3. Complete the table below:
Quantity Symbol Unit of meaurement Symbol of unit e.g. Distance e.g. d e.g. kilometer e.g. km Resistance Current Potential difference
4. Draw a diagram of a circuit which contains a battery connected to a lightbulb and a resistor all in series.
1. Also include in the diagram where you would place an ammeter if you wanted to measure the current through the lightbulb.
2. Draw where and how you would place a voltmeter in the circuit to measure the potential difference across the resistor.
5. Thandi wants to measure the current through the resistor in the circuit shown below and sets up the circuit as shown below. What is wrong with her circuit setup?
6. [SC 2003/11] The emf of a battery can best be explained as the $\cdots$
1. rate of energy delivered per unit current
2. rate at which charge is delivered
3. rate at which energy is delivered
4. charge per unit of energy delivered by the battery
7. [IEB 2002/11 HG1] Which of the following is the correct definition of the emf of a battery?
1. It is the product of current and the external resistance of the circuit.
2. It is a measure of the cell's ability to conduct an electric current.
3. It is equal to the “lost volts” in the internal resistance of the circuit.
4. It is the power supplied by the battery per unit current passing through the battery.
8. [IEB 2005/11 HG] Three identical light bulbs A, B and C are connected in an electric circuit as shown in the diagram below.
1. How bright is bulb A compared to B and C?
2. How bright are the bulbs after switch S has been opened?
3. How do the currents in bulbs A and B change when switch S is opened?
Current in A Current in B (a) decreases increases (b) decreases decreases (c) increases increases (d) increases decreases
9. [IEB 2004/11 HG1] When a current $I$ is maintained in a conductor for a time of $t$ , how many electrons with charge e pass any cross-section of the conductor per second?
1. It
2. It/e
3. Ite
4. e/It
where we get a research paper on Nano chemistry....?
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
The fundamental frequency of a sonometer wire streached by a load of relative density 's'are n¹ and n² when the load is in air and completly immersed in water respectively then the lation n²/na is
Properties of longitudinal waves |
Holt CA Course 1 10-1 Perimeter The perimeter of a polygon is the sum of the lengths of its sides.
Presentation on theme: "Holt CA Course 1 10-1 Perimeter The perimeter of a polygon is the sum of the lengths of its sides."— Presentation transcript:
Holt CA Course 1 10-1 Perimeter The perimeter of a polygon is the sum of the lengths of its sides.
Holt CA Course 1 10-1 Perimeter Vocabulary perimeter
Holt CA Course 1 10-1 Perimeter
Holt CA Course 1 10-1 Perimeter Additional Example 1: Finding the Perimeter of a Polygon Find the perimeter of the figure. 2.8 + 3.6 + 3.5 + 3 + 4.3 Add all the side lengths. The perimeter is 17.2 in.
Holt CA Course 1 10-1 Perimeter Additional Example 2: Using a Formula to Find Perimeter Find the perimeter P of the parallelogram. P = 2l+ 2w Substitute 12 for l and 3 for w. P = (2 12) + (2 3) P = 24 + 6 P = 30 Multiply. Add. The perimeter is 30 cm. 12 cm 3 cm
Holt CA Course 1 10-1 Perimeter A. What is the length of side a if the perimeter equals 1,471 mm? P = sum of side lengths Use the values you know. 1,471 = 416 + 272 + 201 + 289 + a 1,471 = 1,178 + a 1,471 – 1,178 = 1,178 + a – 1,178 Add the known lengths. Subtract 1,178 from both sides. 293 = a Side a is 293 mm long. Additional Example 3: Finding Unknown Side Lengths and the Perimeter of a Polygon Find each unknown measure.
Holt CA Course 1 10-1 Perimeter Additional Example 3: Finding Unknown Side Lengths and the Perimeter of a Polygon Find each unknown measure. B. What is the perimeter of the polygon? Step 1: Find b. Find the sum of the lengths opposite side b. b = 50 + 22 b = 72 P = 72 + 55 + 50 + 33 + 22 + 22 P = 254 Step 2: Find the perimeter. The perimeter of the polygon is 254 cm.
Holt CA Course 1 10-1 Perimeter Check It Out! Example 1 Find the perimeter of the figure. 3.3 + 3.3 + 4.2 + 4.2 + 3 Add all the side lengths. 3.3 in. 4.2 in. 3 in. The perimeter is 18 in.
Holt CA Course 1 10-1 Perimeter Check It Out! Example 2 Find the perimeter P of the parallelogram. P = 2l + 2w Substitute 15 for l and 2 for w. P = (2 15) + (2 2) P = 30 + 4 P = 34 Multiply. Add. The perimeter is 34 cm. 15 cm 2 cm
Holt CA Course 1 10-1 Perimeter Check It Out! Example 3 Find each unknown measure. A. What is the length of side a if the perimeter equals 1,302 mm? P = sum of side lengths Use the values you know. 1,302 = 212 + 280 + 250 + 240 + a 1,302 = 982 + a 1,302 – 982 = 982 + a – 982 Add the known lengths. Subtract 982 from both sides. 320 = a Side a is 320 mm long. 212 mm 280 mm a 250 mm 240 mm
Holt CA Course 1 10-1 Perimeter Lesson Quiz: Part II 4. What is the perimeter of a rectangle with length 22 cm and width 8 cm? 5. The width of a rectangle is 12 in. What is the perimeter of the rectangle if the length is 7 in. longer than the width? 60 cm 62 in.
Holt CA Course 1 10-1 Perimeter P = 14 + 12 + 8 + 4 + 6 + 16 P = 60 The perimeter of the polygon is 60 m. Check It Out! Example 3 Find each unknown measure. B. What is the perimeter of the polygon? 12 m 14 m b m 8 m 6 m 4 m 16 m Step 1: Find b. Find the sum of the lengths opposite side b. b = 12 + 4 b = 16 Step 2: Find the perimeter.
Holt CA Course 1 10-1 Perimeter l = 4w Find the length. l = (4 19) l = 76 Substitute 19 for w. Multiply. Additional Example 3: Finding Unknown Side Lengths and the Perimeter of a Polygon C. The width of a rectangle is 19 cm. What is the perimeter of the rectangle if the length is 4 times the width? 4w4w 19 cm
Holt CA Course 1 10-1 Perimeter Additional Example 3C Continued P = 2 l + 2 w Substitute 76 and 19. P = 2(76) + 2(19) P = 152 + 38 P = 190 Multiply. Add. Use the formula for the perimeter of a rectangle. The perimeter of the rectangle is 190 cm. 4w4w 19 cm
Holt CA Course 1 10-1 Perimeter l = 5w Find the length. l = (5 13) l = 65 Substitute 13 for w. Multiply. Check It Out! Example 3 C. The width of a rectangle is 13 cm. What is the perimeter of the rectangle if the length is 5 times the width? 5w5w 13 cm
Holt CA Course 1 10-1 Perimeter Check It Out! Example 3C Continued 65 cm 13 cm P = 2 l + 2 w Substitute 65 and 13. P = 2(65) + 2(13) P = 130 + 26 P = 156 Multiply. Add. Use the formula for the perimeter of a rectangle. The perimeter of the rectangle is 156 cm.
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# Fractions are an essential part of mathematics
In mathematics, there are various types of quantities. All of those numbers are significant. It is critical to understand all of these quantities and their formations. Whole numbers are the first type of quantity. They are used to represent a specific and fixed amount. They can be used in a variety of problems where the subject needs to be represented by a definite quantity. It has been observed that dealing with problems involving whole numbers is relatively simple. A fraction is the other type of quantity. It can be used in a variety of questions involving partial values. Fractions are used in many important areas of mathematics. Other subjects, such as physics and chemistry, make extensive use of them. Many numerical problems in these critical subjects are solved using fractions. There are many different types of fractions, including equivalent fractions, mixed fractions, proper fractions, and many more. An equivalent fraction is one in which the numerator and denominator of two different fractions can be divided by common terms to yield a common value. It is critical to have adequate information about all of these quantities. This is because questions can only be answered once you have a thorough understanding of the various types of fractions. Fractions are also important in calculating the percentage of a given quantity.
A definite percentage of a specific quantity can be easily computed using the proper fractional operations.
This article discusses the various types of fractions and their importance in answering various types of questions. It is critical to developing clarity in this topic because it is a fundamental component of basic mathematics.
Different types of functions and their significance:
• A mixed fraction is a type of fraction. When the value of the numerator greatly exceeds the value of the denominator, the mixed fraction is formed using division. The quotient is at the beginning, and the remainder is at the end with the denominator. To convert the mixed fraction to a normal fraction, multiply the denominator by the quotient and then add the remainder to the result.
• Proper fraction: A fraction in which the numerator value is less than the denominator value. It has been discovered that performing all operations on proper fractions is relatively easier. When this fraction is converted to decimal, the value obtained is less than one, which is a common way to identify the proper fractions.
• Improper fraction: Improper fractions are fractions in which the value of the denominator is less than the value of the numerator. Converting improper fractions to decimals identifies them. There will always be more than one result.
• Like fractions: This term refers to the relationship between two common fractions. When the denominators of two fractions have the same value, the fractions are said to be like fractions. It is very simple to compare fractions and perform addition and subtraction operations on them.
• Unlike fractions: Two fractions with distinct denominators are referred to as, unlike fractions. By taking the lcm of the denominators and then multiplying the numerator and denominator of the two fractions by a common term in both fractions until the value equals the lcm, two, unlike fractions, can be converted into like fractions. After converting the unlike fractions to like fractions, the two fractions can be easily added and subtracted from or compared.
This article discusses the various types of fractions number Fractions are used extensively in chapters such as percentage, simple interest, compound interest, and many others. This is one of the reasons why it is critical to understand fractions. Students can use Cuemath to clarify any questions they have about this chapter. It is an effective online platform for students to clear their doubts about Mathematics and Coding. |
### Linear Motion
Linear Motion
Unit 1
Velocity and Acceleration
Objectives:
• Define the concept of average velocity in a way that shows you
know how to calculate it
• Calculate average velocity and to solve an equation involving
velocity, distance, and time
• Interpret and plot position-time graphs for positive and negative
velocities
• Determine the slope of a curve on a position-time graph/velocitytime graph and calculate the velocity or acceleration
• Distinguish instantaneous from average velocity
• Define average and instantaneous velocity
• Be able to calculate average velocity, given two velocities and the
time interval between them
• Be able to calculate final velocity in the case of uniform acceleration
• Be able to solve problems of motion of objects uniformly
accelerated by gravity
• Calculate distance fallen in an object affected by gravity
• Learn to use an organized strategy for solving motion problems
Speed
• Instantaneous speed-
– Example: Car speedometer
• Average speed-
Velocity
Instantaneous velocityv =
d
t
v = velocity, m/s
d = displacement, m
t = time, s
d
V
t
Average velocityCan you have a negative velocity? Why or why not?
v = d
t
SI Units: m/s
v = average velocity, m/s
d = displacement, m
t = time, s
English: miles/hour
Example 3.1
Suppose a 100 m walk takes 80 seconds. What is the
average velocity?
Example 3.2
In the Summer Olympics the 100-m race was won in 9.83s.
Find the average velocity in m/s.
Known
Unknown
Equation
Example 3.3 (pg. 43)
A. What is the average speed of a cheetah that sprints
100 meters in 4 seconds?
B. How about if the cheetah sprints 50 meters in 2
seconds?
Example 3.4 (pg. 43)
If a car moves with an average speed of 60 km/h for an
hour, it will have traveled a distance of 60 km.
A. How far would it travel if it moved at a rate for 4 h?
B. For 10 h?
Example 3.5 (pg. 43)
In addition to the speedometer on the dashboard of every
car is an odometer which records the distance traveled. If
the initial reading is set at zero at the beginning of a trip
and the reading is 40 km one-half hour later, what has been
Example 3.6 (pg 43)
Would it be possible to attend this average speed and never
go faster than 80km/h?
Position
Draw a Graph of "Speeding Up"
Time
Graphing Velocity
-
Position
Draw a Graph of "Slowing Down"
Time
Graphing Velocity
-
Position
Draw a Graph of “Constant Speed”
Time
Graphing Velocity
-
Comparisons
• http://www.exploratorium.edu/baseball/reactiontime.html reaction time activity
• reaction time of fox sports science episode 3 – Any (approx.
12 minutes)
• http://www.pbs.org/wgbh/nova/barrier/ - faster than sound
information
Acceleration
Average accelerationa = v
t
a=
v=
t=
SI units = m/s2
English: mi/h
instantaneous accelerationa= v
t
v
a
t
Example 3.7
Suppose you plan to attend the Pittsburgh Steelers home-opener.
The distance from Hollidaysburg to Pittsburgh is 90 miles (145 km).
a) If you are able to average 45 mph (20 m/s), how
long will it take you to travel to the game?
b) As you pull into the parking lot, you decrease your
speed from 20 m/s to 0 m/s over a time period of
4.0 seconds. What is your acceleration?
Example 3.8
A car accelerates along a straight road from rest to 28 m/s in 5.0
s. What is the magnitude of its average acceleration?
Known
Unknown
Equation
Example 3.9 (pg. 46)
What is the acceleration of a race car that whizzes past you at a
constant velocity of 400 km/h?
Example 3.10
The velocity of a car increases from 2.0 m/s at 1.0s to 16 m/s at
4.5 s. What is the car’s average acceleration?
Known
Unknown
Equation
Example 3.11
A car goes faster and faster backwards down a long driveway.
We define the forward velocity as positive, so backward velocity
is negative. The car’s velocity changes from –2.0 m/s to –9.0 m/s
in a 2.0s time interval. Find the acceleration.
Known
Unknown
Equation
What is going on in this graph?
What can you tell from this graph?
What can be read from this graph?
Explain the motion shown in the graph
Review!!!!
Aristotle (384 - 322 BC)
Earth-centric perspective
Proposed Two Types of Motion:
Natural Motion
Objects in the universe have a proper place
based on their nature.
ROCK SMOKE FEATHER - .
Heavier objects fall faster than light ones
Violent Motion
Resulted from pushing/pulling forces
Sustained motion required a sustained force
- Pushing and moving a heavy rock
- Bow and arrow
Copernicus (1543)
Sun is the center of the solar system
Galileo (1564-1642)
Provided evidence that Aristotle's views were
inaccurate.
Leaning Tower of Pisa Experiment
Inertia (Inclined Plane Experiments)
Galileo's Labs
Galileo's Inclined Plane Experiment
Physics
Unit 1 - Motion
Name: ______________________
Period: _____
Data Table
Angle of Incline (o)
Distance Traveled (cm)
Acceleration (m/s )
2
-
Acceleration due to gravity
How Fast- Free Fall
• Freefall- - occurs when gravity is the only force that
acts on an object
•
- No air resistance (in a perfect world)
Can take place when:
• - an object is falling
• - an object is rising
• - an object is traveling at an angle
• - an object is momentarily at rest at its
highest point
Freefall Continued
Gravity(g)- force of attraction between 2 masses
Terminal Velocity-constant maximum velocity
reached by a body falling under gravity
through a fluid
g = 10 m/s
g = v
t
2
Example 3.12 (pg. 48-49)
Using the speedometer on Figure 3.7 what would the
6s after it is dropped? 6.5 seconds?
Example 3.13
The time the Demon drop ride at Cedar Point, Ohio is free falling
is 1.5 s. What is the velocity at the end of this time?
Known
Unknown
Equation
Free Fall Lab
Object
Small blue square
Medium blue square
Large blue square
Softball
Acceleration(m/s2 )
How Far
d = 0.5 gt2
Example 3.14 (pg. 50)
A car steps off a ledge and drops to the ground in 0.5
seconds.
A. What is it’s speed on striking the ground?
B. What is its average speed during the 0.5 seconds?
C. How high is the ledge from the ground?
•
•
•
•
•
•
•
•
http://www.fearofphysics.com/Xva/xva.html
http://www.fearofphysics.com/Fall/fall.html
http://www.fearofphysics.com/Roller/roller.html
http://video.msn.com/?mkt=en-us&brand=foxsports&vid=fce2cab5-a3004f78-880c-18be6d3e3fc2&from=Fox%20Sports&tab=g1196965174236 Ben Rothelsberger and getting off a pass- may not work, but google search
http://www2.swgc.mun.ca/physics/physlets/motiona.html - position vs.
time graphs
http://www2.swgc.mun.ca/physics/physlets/motionb.html - velocity vs.
time graphs
http://www2.swgc.mun.ca/physics/physlets/motionc.html - acceleration
graphs |
# RD Sharma Class 6 ex 6.6 Solutions Chapter 6 Fractions
In this chapter, we provide RD Sharma Class 6 ex 6.6 Solutions Chapter 6 Fractions for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 6 ex 6.6 Solutions Chapter 6 Fractions Maths pdf, Now you will get step by step solution to each question.
# Chapter 6: Fractions – Exercise 6.6
### Question: 1
Reduce each of the following fractions to its lowest term (simplest form):
i) 40/75
ii) 42/28
iii) 12/52
iv) 40/72
v) 80/24
vi) 84/56
### Solution:
i) 40/75
Factors of 40 are 1, 2, 4, 5, 8, 10, 20 and 40.
Factors of 75 are 1, 3, 5, 15 and 75.
Common factors of 40 and 75 are 1 and 5.
HCF = 5
Divide both the numerator & denominator by 5.
Therefore, the simplest form obtained is 8/15
ii) 42/28
Factors of 42 are 1, 2, 3, 6, 7, 14, 21, 42
Factors of 28 are 1, 2, 4, 7, 14, 28
Common factors of 42 & 28 are 1, 2 and 4.
HCF = 4
Divide both the numerator & denominator by
Therefore, the simplest form obtained is 32
iii) 12/52
Factors of 12 are 1, 2, 3, 4, 6 and 12.
Factors of 52 are 1, 2, 4, 13, 26 and 52.
Common factors of 12 and 52 are 1, 2 and 4
HCF = 4
Divide both the numerator & denominator by 4.
On solving the above we have,
3/13
Therefore, the simplest form obtained is 3/13
iv) 40/72
Factors of 40 are 1, 2, 4, 5, 8, 10, 20, 40
Factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
Common factors of 40 and 72 are 1, 2, 4 and 8.
HCF = 8
Divide both the numerator & denominator by 8.
On solving the above equation, we have:
5/9
Therefore, the simplest form obtained is 5/9
v) 80/24
Factors of 80 are 1, 2, 4, 5, 8, 10, 16, 20, 40 and 80.
Factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.
Common factors of 80 and 24 are 1, 2 , 4 , 8
HCF = 8
Divide both the numerator & denominator by 4.
On solving the above equation, we have:
10/3
Therefore, the simplest form obtained is 10/3
vi) 84/56
Factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84.
Factors of 56 are 1, 2, 4, 7, 8, 14, 28 and Common factors of 84 & 56 are 1, 2, 4, 7, 14 and 28.
HCF = 28
Divide both the numerator & denominator by 28
On solving the above equation, we have:
3/2
Therefore, the simplest form obtained is 3/2
### Question: 2
Simplify each of the following to its lowest form:
i) 75/80
ii) 52/76
iii) 84/98
iv) 68/17
v) 150/50
vi) 162/108
### Solution:
i) 75/80
Factors of 75 are 1, 3, 5, 15, 25 and 75.
Factors of 80 are 1, 2, 4, 5, 8, 10, 12, 16, 20, 40 and 80.
Common factors of 75 and 80 are 1 and 5.
HCF of 75 and 80 is 5.
Dividing both the numerator and denominator by 5, we get:
Therefore, the simplest form obtained is
ii) 52/76
Factors of 52 are 1, 2, 4, 13, 26 and 52.
Factors of 76 are 1, 2, 4, 19, 38 and 76.
Common factors of 52 and 76 are 1, 2 and 4.
HCF of 52 and 76 is 4.
Dividing both the numerator and denominator by 5, we get:
Therefore, the simplest form obtained is
iii) 84/98
Factors of 84 are 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42 and 84.
Factors of 98 are 1, 2, 7, 14, 49 and 98.
Common factors of 84 and 98 are 1, 2, 7 and 14.
HCF of 84 and 98 is 14
Dividing both the numerator and denominator by 5, we get:
Therefore, the simplest form obtained is
84/98 = 6/7
iv) 68/17
Factors of 68 are 1, 2, 4, 17, 34 and 68.
Factors of 17 are 1 and 17.
Common factor of 68 and 17 is 17
HCF of 68 and 17 is 17
Dividing both the numerator and denominator by 5, we get:
Therefore, the simplest form obtained is
68/17 = 4/1
v) 150/50
Factors of 150 are 1, 2, 3, 5, 6, 10, 15, 25, 50 and 150.
Factors of 50 are 1, 2, 5, 10, 25 and 50.
Common factor of 150 and 50 is 50
HCF of 150 and 50 is 50
Dividing both the numerator and denominator by 50, we get:
Therefore, the simplest form obtained is
150/50 = 3/1
vi) 162/108
Factors of 162 are 1, 2, 3, 6, 9, 18, 27, 54, 81 and 162.
Factors of 108 are 108, 1, 2, 3, 4, 6, 9, 12, 18, 27 and 54.
Common factor of 162 and 108 are 1, 2, 3, 6, 9, 18, 27, 54
HCF of 162 and 108 is 54
Dividing both the numerator and denominator by 54, we get:
Therefore, the simplest form obtained is
162/108 = 3/2
All Chapter RD Sharma Solutions For Class 6 Maths
I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good |
# 4.3 Rules of Probability
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1 4.3 Rules of Probability If a probability distribution is not uniform, to find the probability of a given event, add up the probabilities of all the individual outcomes that make up the event. Example: Suppose you are given the following probability distribution for a sample space S = {s 1, s 2, s 3, s 4, s 5, s 6 } Outcome s 1 s 2 s 3 s 4 s 5 s Probability Supppose E = {s 1, s 4, s 5 }, F = {s 2, s 3 }, and G = {s 2, s 5 }. Calculate the following. P (s 4 ) P (E) P (F G) P (E F ) P (E c ) P (E G) Example: Suppose an experiment has a sample space S = {s 1, s 2, s 3 } where P (s 1 ) + P (s 2 ) = 0.35 and P (s 2 ) + P (s 3 ) = Find the probability distribution for S. 1
2 What happens if you cannot list out all the outcomes and their probalities (or do not want to)? Or worse, what if we don t even know what the specific outcomes in the events are? We can use the following more general rules. Rules of Probability: 1. 0 P (E) 1 for any event E in a sample space S. In particular P ( ) = 0 and P (S) = Union rule for probability: If E and F are ANY two events, then P (E F ) = P (E) + P (F ) P (E F ) Note: If E and F are mutually exclusive, then E F =, which means P (E F ) = 0 and the formula reduces to just P (E F ) = P (E) + P (F ). 3. Complement Principle: P (E c ) = 1 P (E) or P (E) = 1 P (E c ) Example: Let E and F be two events of an experiment with sample space S. Suppose P (E) = 0.5, P (F ) = 0.4, and P (E F ) = 0.1. Compute the following. P (F c ) P (E c ) P (E F ) Example: If P (E c ) = 0.3 and P (F ) = 0.2 with E and F mutually exclusive, what is P (E F )? Example: In a survey of a group of people it was found that the probability someone did not like Dr. Pepper was 0.65, the probability someone liked Dr. Pepper was 0.45, and the probability that someone liked both Dr. Pepper and Coke was What is the probability that someone in this group likes Dr. Pepper or likes Coke? 2
3 Example: An experiment consists of selecting a card at random from a 52-card deck. Find the probability that a red face card is drawn. Find the probability that a face card is not drawn. Find the probability that a diamond or a club is drawn. Find the probability that a spade or a queen is drawn. Find the probability that a 3 or a red card is drawn. Example: The table below gives the number of students of each classification who are majoring and not majoring in business in a class of 110 students. Freshmen Sophomores Juniors Seniors Total Business Non-Business Total A student is randomly selected from this class. What is the probability that... The student is not a junior? The student is a freshman and a non-business major? The student is a business major or a sophomore? The student is a non-business major or is not a junior? 3
4 4.4 Random Variables and Expected Value A random variable is a rule that assigns a number to each outcome of an experiment. We usually denote a random variable by X. Example: A coin is tossed three times and the sequence of heads and tails is observed. List the outcomes of the experiment. Let the random variable X denote the number of tails that occur. What are the possible values of X? Find the probability distribution of X. What is P (1 X 2)? What is P (X > 0)? 4
5 The expected value of a random variable X, denoted E(X), is given by E(X) = x 1 p 1 + x 2 p x n p n where x 1, x 2,..., x n are the values that X may assume, and p 1, p 2,..., p n are the probabilities of each of these values. Example: Find the expected value of the random variable X given below. X Probability Consider the experiment of rolling 2 fair 5-sided dice. Let X be the sum of the numbers rolled. Find the probability distribution of X. What is E(X)? Example: A survey was conducted of families to determine the distribution of families by size. The results are: Family Size Number of Families Let the random variable X be the number of people in a randomly chosen family. Find the probability distribution for X. What is the expected number of people in a randomly chosen family? 5
6 Expected values are often used in games to determine whether the game is fair. A game is considered fair when the expected NET winnings are 0. You are playing a game at a carnival. The game costs \$1. You select a card from a standard deck. If the card is an ace, then you win \$3. If the card is a face card, you win \$2. Otherwise you win nothing. Find the expected net winnings. Example: A raffle is held people buy a ticket for \$3 each. First prize is \$1500. Second prize is \$750. Then, four \$100 consolation prizes will be given. What are the expected net winnings for one person who buys a \$3 ticket. A game consists of rolling a fair 5-sided die. The game costs \$3 to play. If you roll an odd number, you win an amount of money equal to the number rolled. Otherwise you win nothing. What are your expected net winnings? 6
7 A game consists of rolling a pair of fair 6-sided dice. The game costs \$4 to play. If you roll a double, you win \$a. Otherwise, you get nothing. What value of a would make this game fair? Example: A man purchased a \$25,000 life insurance policy from his employer for \$200/yr. (The cost of \$200 is called the premium.) The probability that he lives another year is What is the life insurance company s expected net gain? If the probability that the man lives drops to 0.98, what is the minimum amount of money, \$a, he can expect to pay for his policy? Note: The insurance company will charge at minimum an amount of money so that their expected net gain is 0. They would probably want to charge more than that to have a positive expected net gain. 7
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# What is y=2(x+3)^2+1 in standard form?
Mar 1, 2016
$y = 2 {\left(x + 3\right)}^{2} + 1$ in standard form is $\textcolor{p u r p \le}{y = 2 {x}^{2} + 12 x + 19}$.
#### Explanation:
$y = 2 {\left(x + 3\right)}^{2} + 1$ is a quadratic equation in vertex form. It can be converted into standard form by doing the following:
Remove the parentheses and exponent by simplifying ${\left(x + 3\right)}^{2}$, which is a sum of squares.
${a}^{2} + {b}^{2} = {a}^{2} + 2 a b + {b}^{2}$, where $a = x , \mathmr{and} b = 3$.
$\textcolor{b l u e}{{\left(x + 3\right)}^{2} = {x}^{2} + 2 \cdot x \cdot 3 + {3}^{2}}$
Simplify.
color(blue)((x^2+6x+9)
Rewrite the equation, substituting $\textcolor{b l u e}{\left({x}^{2} + 6 x + 9\right)}$ for ${\left(x + 3\right)}^{2}$.
$y = 2 \textcolor{b l u e}{\left({x}^{2} + 6 x + 9\right)} + 1$
Distribute the $\textcolor{red}{2}$.
$y = \textcolor{red}{2} \cdot {x}^{2} + 6 x \cdot \textcolor{red}{2} + 9 \cdot \textcolor{red}{2} + 1$
Simplify.
$y = \textcolor{p u r p \le}{2 {x}^{2} + 12 x + 18} + 1$
Simplify.
$\textcolor{p u r p \le}{y = 2 {x}^{2} + 12 x + 19}$ |
# How do determine whether these relations are even, odd, or neither: f(x)=2x^2+7? f(x)=4x^3-2x? f(x)=4x^2-4x+4? f(x)=x-(1/x)? f(x)=|x|-x^2+1? f(x)=sin(x)+1?
Aug 5, 2016
Function 1 is even.
Function 2 is odd.
Function 3 is neither.
Function 4 is odd.
Function 5 is even.
Function 6 is neither.
Next time, try and ask separate questions rather than lots of the same at once, people are here to help you, not to do your homework for you.
#### Explanation:
If $f \left(- x\right) = f \left(x\right)$, function is even.
If $f \left(- x\right) = - f \left(x\right)$, function is odd.
$\textcolor{g r e e n}{\text{Function 1}}$
$f \left(- x\right) = 2 {\left(- x\right)}^{2} + 7 = 2 {x}^{2} + 7 = f \left(x\right)$
$\therefore$ function is even
$\textcolor{g r e e n}{\text{Function 2}}$
$f \left(- x\right) = 4 {\left(- x\right)}^{3} - 2 \left(- x\right) = - 4 {x}^{3} + 2 x = - f \left(x\right)$
$\therefore$ function is odd
$\textcolor{g r e e n}{\text{Function 3}}$
$f \left(- x\right) = 4 {\left(- x\right)}^{2} - 4 \left(- x\right) + 4 = 4 {x}^{2} + 4 x + 4 \ne f \left(x\right) \mathmr{and} - f \left(x\right)$
$\therefore$ function is neither odd nor even
$\textcolor{g r e e n}{\text{Function 4}}$
$f \left(- x\right) = \left(- x\right) - \frac{1}{- x} = - x + \frac{1}{x} = - f \left(x\right)$
$\therefore$ function is odd
$\textcolor{g r e e n}{\text{Function 5}}$
$f \left(- x\right) = \left\mid - x \right\mid - {\left(- x\right)}^{2} + 1 = \left\mid x \right\mid - {x}^{2} + 1 = f \left(x\right)$
$\therefore$ function is even.
$\textcolor{g r e e n}{\text{Function 6}}$
$f \left(- x\right) = \sin \left(- x\right) + 1 = - \sin \left(x\right) + 1 \ne f \left(x\right) \mathmr{and} - f \left(x\right)$
$\therefore$ function is neither even nor odd. |
# What Are Examples Of Complementary Events?
## How do you find complementary events?
The rule of complementary events comes from the fact the the probability of something happening, plus the probability of it not happening, equals 100% (in decimal form, that’s 1).
For example, if the odds of it raining is 40%, the odds of it not raining must equal 60%.
And 40% + 60% = 100%..
## What are complements in probability?
The complement of an event is the subset of outcomes in the sample space that are not in the event. A complement is itself an event. The complement of an event A is denoted as A c A^c Ac or A′. By consequence, the sum of the probabilities of an event and its complement is always equal to 1. …
## What is the complementary rule?
The Complement Rule states that the sum of the probabilities of an event and its complement must equal 1, or for the event A, P(A) + P(A’) = 1.
## What is a complementary angle?
Two angles are called complementary when their measures add to 90 degrees. Two angles are called supplementary when their measures add up to 180 degrees.
## What is the complement formula?
The Complement Rule states that the sum of the probabilities of an event and its complement must equal 1, or for the event A, P(A) + P(A’) = 1.
## What is the complement of at least one?
❖ The complement of getting at least one item of a particular type is that you get no items of that type. ❖ “At least one” is equivalent to “one or more.” To find the probability of at least one of something, calculate the probability of none and then subtract that result from 1.
## Can two complementary events occur at the same time?
In probability theory, the complement of an event A is the event not A; this complementary event is often denoted A’ or Ac. … They are mutually exclusive because the two events cannot occur at the same time, and they are exhaustive because the sum of their probabilities must add to 100%.
## What means complementary?
: completing something else or making it better : serving as a complement. —used of two things when each adds something to the other or helps to make the other better. : going together well : working well together. See the full definition for complementary in the English Language Learners Dictionary. complementary.
## What does C stand for in probability?
P(AB) means the probability that events A and B occur. You could write it P(A∩B). The superscript c means “complement” and Ac means all outcomes not in A. |
Distributing variables over the terms in an algebraic expression involves multiplication rules and the rules for exponents. When different variables are multiplied together, you can write them side by side without using any multiplication symbols. If the same variable is multiplied as part of the distribution, then you add the exponents.
The exponent rule says that when multiplying exponents with the same base, you add the exponents:
This sample problem illustrates the exponent rule for multiplying terms with the same base.
1. Distribute the term outside the parentheses over the terms within.
2. Add the exponents.
3. Complete the distribution.
The next example contains negative powers and fractional powers — the rules for exponents remain the same with negative and fractional exponents.
Example:
1. Distribute the term outside the parentheses over the terms within the parentheses.
2. Add the exponents.
3. Complete the distribution.
The exponent zero means the value of the expression is one:
You combine exponents with different signs by using the rules for adding and subtracting signed numbers. Fractional exponents are combined after finding common denominators. Exponents that are improper fractions are left in that form.
Try going through many of the situations that could arise when distributing, such as distributing both a number and a variable, distributing various powers of more than one variable, distributing negatives, rewriting negative exponents as fractional terms, distributing fractional powers, and distributing radicals. This touches on just about anything you’d probably come across.
Combine the variables by using the rules for exponents.
Example 1: Distribute 5x through the expression
Multiply each term by 5x.
Then multiply the numbers and the variables in each term.
Example 2: Combine the variables with the same base using the rules for exponents. The signs of the results follow the rules for multiplying signed numbers.
–6y(5xy – 4x – 3y + 2)
Multiply each term by -6y.
–6y(5xy) – 6y(–4x) – 6y(–3y) – 6y(2)
Do the multiplication in each term. |
# SOLVED PROBLEMS ON AGES
Problem 1 :
If Azhagan is 50 years old and his son is 10 years old, then the simplest ratio between the age of Azhagan to his son is
(A) 10 : 50 (B) 50 : 10 (C) 5 : 1 (D) 1 : 5
Solution :
Ratio between the age of Azhagan to his son is
= 50 : 10
= ⁵⁰⁄₁₀ ¹⁰⁄₁₀
= 5 : 1
The correct answer choice is option (C).
Problem 2 :
Saran is 6 times as old as his son Sankar. After 4 years, he will be 4 times as old as his son. What are their present ages?
(A) 30, 5 (B) 36, 6 (C) 48, 8 (D) 24, 4
Solution :
Let x be the present age of the son.
Then the present age of Saran is 6x.
After 4 years,
age of Saran = 6x + 4
age of sankar = x + 4
Given : After 4 years, Saran will be 4 times as old as his son Sankar.
6x + 4 = 4(x + 4)
6x + 4 = 4x + 16
2x = 12
x = 6
Present age of Saran = 6(6) = 36 years
Present age of Sankar = 6 years
The correct answer choice is option (B).
Problem 3 :
The sum of the present ages of a father and his son is 60 years. Six years ago, father's age was five times the age of the son. After 6 years, what will the the son's age be?
(A) 12 years (B) 14 years (C) 18 years (D) 20 years
Solution :
Let f and s be the present ages of father son.
Given : The sum of the present ages of a father and his son is 60.
f + s = 60 ----(1)
Given : Six years ago, father's age was five times the age of the son.
f - 6 = 5(s - 6)
f - 6 = 5s - 30
f = 5s - 24
Substitute f = 5s - 24 into (1).
5s - 24 + s = 60
6s - 24 = 60
6s = 84
s = 14
The present age of the son is 16 years.
After 6 years, the age of the son will be
= 14 + 6
= 20 years
The correct answer choice is option (D).
Problem 4 :
The age of a man is 4 times the sum of the ages of his two sons. Ten years hence, his age will be double of the sum of the ages his sons. The father's present age is
(A) 50 years (B) 55 years (C) 60 years (D) 65 years
Solution :
Let f be the present age of father and s be the sum of present ages of two sons.
Given : The age of a man is 4 times the sum of the ages of his two sons.
f = 4s ----(1)
Given : Ten years hence, the age of father will be double of the sum of the ages his sons.
f + 10 = 2(s + 10 + 10)
Here, there are two 10's added to s. Because there are two sons.
f + 10 = 2(s + 20)
f + 10 = 2s + 40
f = 2s + 30 ----(2)
From (1) and (2),
4s = 2s + 30
2s = 30
s = 15
Substitute s = 15 into (1).
f = 4(15)
f = 60
The present age of father is 60 years.
The correct answer choice is option (C).
Problem 5 :
At present, the ratio between the ages of Arun and Deepak is 4 : 3. After 6 years, Arun’s age will be 26 years. What is the age of Deepak at present?
(A) 12 years (B) 15 years (C) 19½ years (D) 21 years
Solution :
Ratio between the present ages of Arun and Deepak is
= 4 : 3
From the above ratio,
present age of Arun = 4x
present age of Deepak = 3x
After 6 years,
age of Arun = 4x + 6
Given : After 6 years, Arun’s age will be 26 years.
Then, we have
4x + 6 = 26
4x = 20
x = 5
Age of Deepak at present :
= 3(5)
= 15 years
The correct answer choice is option (B).
Problem 6 :
A is younger than B. Seven times the difference between their ages is equal to sum of their ages. Ratio of B's age after 12 years and A's present age is 5 : 3, then what is A's present age?
(A) 30 years (B) 36 years (C) 42 years (D) 48 years
Solution :
Let a and b be the present ages of A and B.
Given : A is younger than B. Seven times the difference between their ages is equal to sum of their ages.
7(b - a) = a + b
7b - 7a = a + b
6b = 8a
3b = 4a ----(1)
Given : Ratio of B's age after 12 years and A's present age is 5 : 3.
(b + 12) : a = 5 : 3
3(b + 12) = 5a
3b + 36 = 5a
From (1), substitute 3b = 4a.
4a + 36 = 5a
36 = a
The present age of A is 36 years.
The correct answer choice is option (B).
Problem 7 :
If the ratio of the ages of son and father in 2015 and 2023 are 1 : 4 and 3 : 8 respectively, then the sum of the ages of son and father in 2010 is
(A) 40 (B) 30 (C) 35 (D) 45
Solution :
Given : Ratio of the ages of son and father in 2015 is
= 1 : 4
From the above ratio,
age of son in 2015 = x
age of father in 2015 = 4x
Then, we have
age of son in 2023 = x + 8
age of father in 2023 = 4x + 8
Ratio of the ages of son and father in 2023 is
= (x + 8) : (4x + 8) ----(1)
Given : Ratio of the ages of son and father in 2023 is
= 3 : 8 ----(2)
From (1) and (2),
(x + 8) : (4x + 8) = 3 : 8
8(x + 8) = 3(4x + 8)
8x + 64 = 12x + 24
40 = 4x
10 = x
age of son in 2015 = x = 10
age of father in 2015 = 4x = 40
age of son in 2010 = 10 - 5 = 5
age of father in 2010 = 40 - 5 = 35
Sum of the ages of son and father in 2010 is
= 5 + 35
= 40
The correct answer choice is option (A).
Problem 8 :
If the ratio of father's age to son's age is 4 : 1 and the product of their ages is 196, then the ratio of their ages after 5 years will be
(A) 3 : 1 (B) 10 : 3 (C) 11 : 4 (D) 14 : 5
Solution :
Given : Ratio of father's age to son's age is
4 : 1
From the above ratio,
father's age = 4x
son's age = x
Given : Product of their ages is 196.
(4x)(x) = 196
4x2 = 196
x2 = 49
x = 7
At present,
father's age = 4(7) = 28 years
son's age = 7 years
After 5 years,
father's age = 28 + 5 = 33 years
son's age = 7 + 5 = 12 years
Ratio of their ages after 5 years will be
= 33 : 12
= 11 : 4
The correct answer choice is option (C).
Problem 9 :
The ages of Vivek and Sumit are 2 : 3. After 12 years, their ages will be in the ratio 11 : 15. The age of Sumit is
(A) 32 years (B) 42 years (C) 48 years (D) 56 years
Solution :
Given : The ages of Vivek and Sumit are 2 : 3.
From the above ratio,
present age of Vivek = 2x
present age of Sumit = 3x
After 12 years,
age of Vivek = 2x + 12
age of Sumit = 3x + 12
After 12 years, the ratio between their ages will be
= (2x + 12) : (3x + 12) ----(1)
Given : After 12 years, the ratio betweern their will be
= 11 : 15 ----(2)
From (1) and (2),
(2x + 12) : (3x + 12) = 11 : 15
15(2x + 12) = 11(3x + 12)
30x + 180 = 33x + 132
48 = 3x
16 = x
Present age of Sumit :
= 3x
3(16)
= 48 years
The correct answer choice is option (C).
Problem 10 :
The ratio of the ages of the father and the son at present is 19 : 5. After 4 years, the ratio will become 3 : 1. What is the sum of the present ages of the father and the son?
(A) 40 (B) 42 (C) 48 (D) 52
Solution :
Given : The ratio of the ages of the father and the son at present is
= 19 : 5
From the above ratio,
present age of father = 19x
present age of son = 5x
After 4 years,
age of father = 19x + 4
age of son = 5x + 4
After 4 years, the ratio between their ages will be
= (19x + 4) : (5x + 4) ----(1)
Given : After 4 years, the ratio of their ages will become
= 3 : 1 ----(2)
From (1) and (2),
(19x + 4) : (5x + 4) = 3 : 1
1(19x + 4) = 3(5x + 4)
19x + 4 = 15x + 12
4x = 8
x = 2
present age of father = 19x = 19(2) = 38 years
present age of son = 5x = 5(2) = 10 years
Sum of the present ages of the father and the son :
= 38 + 10
= 48
The correct answer choice is option (C).
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# Lesson 1: Using Decimals in a Shopping Context
Let’s use what we know about decimals to make shopping decisions.
## 1.1: Snacks from the Concession Stand
Clare went to a concession stand that sells pretzels for \$3.25, drinks for \$1.85, and bags of popcorn for \$0.99 each. She bought at least one of each item and spent no more than \$10.
Concession stand + spinning merry go round Copyright Owner: m01229 License: CC BY 2.0 Via: Flickr
1. Could Clare have purchased 2 pretzels, 2 drinks, and 2 bags of popcorn? Explain your reasoning.
2. Could she have bought 1 pretzel, 1 drink, and 5 bags of popcorn? Explain your reasoning.
## 1.2: Planning a Dinner Party
You are planning a dinner party with a budget of \$50 and a menu that consists of 1 main dish, 2 side dishes, and 1 dessert. There will be 8 guests at your party. Choose your menu items and decide on the quantities to buy so you stay on budget. If you choose meat, fish, or poultry for your main dish, plan to buy at least 0.5 pound per person. Use the worksheet to record your choices and estimated costs. Then find the estimated total cost and cost per person. See examples in the first two rows. 1. The budget is \$ ___________ per guest.
item quantity
needed
price
estimated
subtotal
(in dollars)
estimated cost
per person
(in dollars)
row 1 ex. main dish: fish 4 pounds \$6.69 per pound$4\boldcdot 7=2828\div 8 = 3.50$row 2 ex. dessert: cupcakes 8 cupcakes \$2.99 per
6 cupcakes
$2\boldcdot 3 = 6$ $6\div 8= 0.75$
row 3 main dish:
row 4 side dish 1:
row 5 side dish 2:
row 6 dessert:
row 7 estimated
total
2. Is your estimated total close to your budget? If so, continue to the next question. If not, revise your menu choices until your estimated total is close to the budget.
3. Calculate the actual costs of the two most expensive items and add them. Show your reasoning.
4. How will you know if your total cost for all menu items will or will not exceed your budget? Is there a way to predict this without adding all the exact costs? Explain your reasoning.
## Summary
We often use decimals when dealing with money. In these situations, sometimes we round and make estimates, and other times we calculate the numbers more precisely.
There are many different ways we can add, subtract, multiply, and divide decimals. When we perform these computations, it is helpful to understand the meanings of the digits in a number and the properties of operations. We will investigate how these understandings help us work with decimals in upcoming lessons. |
## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.1
12th Maths Exercise 1.1 Answers Question 1.
Find the adjoint of the following:
Solution:
Exercise 1.1 Class 12 Maths State Board Question 2.
Find the inverse (if it exists) of the following:
Solution:
For a matrix A, $$\mathrm{A}^{-1}=\frac{1}{|\mathrm{A}|}(\mathrm{adj} \mathrm{A})$$. Where |A| ≠ 0. If |A| = 0 then A is called a singular matrix and so $$\mathrm{A}^{-1}$$ does not exist.
12th Maths Exercise 1.1 Question 3.
If F(α) = $$\left[\begin{array}{ccc}{\cos \alpha} & {0} & {\sin \alpha} \\ {0} & {1} & {0} \\ {-\sin \alpha} & {0} & {\cos \alpha}\end{array}\right]$$ show that $$[\mathrm{F}(\alpha)]^{-1}=\mathrm{F}(-\alpha)$$
Solution:
Let A = F (α)
So $$[\mathrm{F}(\alpha)]^{-1}=\mathrm{A}^{-1}$$
Now
12th Maths Chapter 1 Exercise 1.1 Question 4.
If A = $$\left[\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right]$$ show that A2 – 3A – 7I2 = O2. Hence find A-1.
Solution:
A = $$\left[\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right]$$
To Find A-1
Now we have proved that A2 – 3A – 7I2 = O2
Post multiply by A-1 we get
A – 3I – 7A-1 = O2
12th Maths 1.1 Exercise Question 5.
If $$\mathbf{A}=\frac{1}{9}\left[\begin{array}{ccc}{-8} & {1} & {4} \\ {4} & {4} & {7} \\ {1} & {-8} & {4}\end{array}\right]$$ prove that A-1 = AT
Solution:
12 Maths Exercise 1.1 Question 6.
If $$\mathbf{A}=\left[\begin{array}{rr}{8} & {-4} \\ {-5} & {3}\end{array}\right]$$, verify that A(adj A) = (adj A)A = |A| I2
Solution:
12th Maths 1st Chapter Exercise 1.1 Question 7.
If $$\mathbf{A}=\left[\begin{array}{ll}{3} & {2} \\ {7} & {5}\end{array}\right]$$, and $$\mathbf{B}=\left[\begin{array}{cc}{-1} & {-3} \\ {5} & {2}\end{array}\right]$$ verify that (AB)-1 = B-1 A-1.
Solution:
12th Maths Application Of Matrices And Determinants Question 8.
If adj (A) = $$\left[\begin{array}{ccc}{2} & {-4} & {2} \\ {-3} & {12} & {-7} \\ {-2} & {0} & {2}\end{array}\right]$$ find A
Solution:
12th Maths Ex 1.1 Question 9.
If adj(A) = $$\left[\begin{array}{ccc}{0} & {-2} & {0} \\ {6} & {2} & {-6} \\ {-3} & {0} & {6}\end{array}\right]$$ find A-1
Solution:
12th Exercise 1.1 Question 10.
Find adj(adj(A)) if adj A = $$\left[\begin{array}{ccc}{1} & {0} & {1} \\ {0} & {2} & {0} \\ {-1} & {0} & {1}\end{array}\right]$$
Solution:
12th Maths Exercise 1.1 Answers In Tamil Medium Question 11.
Solution:
12th Maths 1st Chapter Question 12.
Find the matrix A for which A $$\left[\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right]=\left[\begin{array}{cc}{14} & {7} \\ {7} & {7}\end{array}\right]$$
Solution:
Given A $$\left[\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right]=\left[\begin{array}{cc}{14} & {7} \\ {7} & {7}\end{array}\right]$$
Let $$\mathrm{B}=\left(\begin{array}{cc}{5} & {3} \\ {-1} & {-2}\end{array}\right) \text { and } \mathrm{C}=\left(\begin{array}{cc}{14} & {7} \\ {7} & {7}\end{array}\right)$$
Given AB = C, To find A
Now AB = C
Post multiply by B-1 on both sides
ABB-1 = CB-1 (i.e) A (BB-1) = CB-1
⇒ A(I) = CB-1 (i.e) A = CB-1
To find B-1:
12th Maths 1.1 Question 13.
Given $$\mathbf{A}=\left[\begin{array}{cc}{1} & {-1} \\ {2} & {0}\end{array}\right], \mathbf{B}=\left[\begin{array}{cc}{3} & {-2} \\ {1} & {1}\end{array}\right] \text { and } \mathbf{C}\left[\begin{array}{ll}{1} & {1} \\ {2} & {2}\end{array}\right]$$, find a matrix X such that AXB = C.
Solution:
A × B = C
Pre multiply by A-1 and post multiply by B-1 we get
A-1 A × BB-1 = A-1CB-1 (i.e) X = A-1CB-1
12 Maths Samacheer Kalvi Solutions Question 14.
Solution:
12th Maths Exercise 1.1 5th Sum Question 15.
Decrypt the received encoded message $$\left[\begin{array}{cc}{2} & {-3}\end{array}\right]\left[\begin{array}{ll}{20} & {4}\end{array}\right]$$ with the encryption matrix $$\left[\begin{array}{cc}{-1} & {-1} \\ {2} & {1}\end{array}\right]$$ and the decryption matrix as its inverse, where the system of codes are described by the numbers 1-26 to the letters A- Z respectively, and the number 0 to a blank space.
Solution:
Let the encoding matrix be $$\left[\begin{array}{cc}{-1} & {-1} \\ {2} & {1}\end{array}\right]$$
So the sequence of decoded matrices is [8 5], [12 16].
### Samacheer Kalvi 12th Maths Solutions Chapter 1 Applications of Matrices and Determinants Ex 1.1 Additional Problems
12th Maths Chapter 1 Question 1.
Using elementary transformations find the inverse of the following matrix
Solution:
12th Maths Guide Question 2.
Using elementary transformations find the inverse of the matrix
Solution:
12th Maths Solutions Samacheer Kalvi Question 3.
Using elementary transformation find the inverse of the matrix
Solution:
12 Maths Chapter 1 Exercise 1.1 Question 4.
Using elementary transformations find the inverse of the matrix
Solution:
12th Maths Exercise 1.1 Solutions Question 5.
Using elementary transformation, find the inverse of the following matrix
Solution:
Samacheer Kalvi 12th Maths Guide Question 6.
Solution:
Samacheer Kalvi 12 Maths Solutions Question 7.
Solution:
12 Maths Solutions Samacheer Kalvi Question 8.
Solution:
Question 9.
Solution:
Question 10.
Solution: |
# How to Calculate Masonry Blocks Needed to Build a House
One of the most important aspects of the building process is estimating the materials needed to complete the project.
The reason this step is so important is simple: Too many materials means the project can go over budget, and too little materials means the project can be delayed waiting on deliveries. One of the easiest items to estimate is concrete block. Whether estimating for a foundation or the entire home, it's a fairly straightforward process, but if it's not done properly, it can shut down a work site until more blocks are delivered. .
Determine what kind of blocks will be used. This information can be found on the drawing. The block size helps to determine the thickness of the wall. Blocks come in many sizes, but the most common is 8 inches tall, 8 inches thick and 16 inches long.
Determine the total square footage for each wall. Use simple math to get this information: basic length times height. For this example, the house will be a simple square, 10 feet on each side with 10-foot-high walls. The total square footage for each wall is 10 feet times 10 feet, or 100 square feet.
Add up the square footage for each wall. For this example, each wall is 100 square feet, so the total square footage is 400.
Determine the square footage covered by one block. Multiply the length of the block times the height of the block--16 inches by 8 inches equals 128 square inches. Now divide the total square inches of the block by 144--128 square inches divided by 144 equals 0.88 square feet.
Determine the number of blocks needed to complete the building. Take the total square footage of all the walls and divide it by the number found in step 4. For this example, 400 square feet is divided by 0.88, which equals 454.54. If the answer has a decimal, round up to the next whole number. This number is the number of blocks needed to build the building.
Add 10 percent to the number found in step 5. To do this, multiply the rounded-up number of blocks from step 5 by 1.1. For this example, 455 times 1.1 equals 500.5. As with the previous step, round decimals up to the next whole number, so the total number of blocks is 501. This number will give enough for any breakage or unusable blocks that come in the order.
## Things You Will Need
• Architectural drawings
• Calculator
## Tip
• Don't forget to ask for corner blocks. |
# TURNING EFFECT OF A FORCE
## Turning Effects
The turning effect of a body is called the moment of that force. The turning effect produced depends on both the size of the force and the distance from the pivot. The moment of a force about a point is the product of the force applied and the perpendicular distance from the pivot (or turning point) to the line of action of the force. Hence, Moments of a force = Force × perpendicular distance from pivot.
## The law of moments
The law of moments states that “when a body is in balance or in equilibrium, the sum of the clockwise moments equals the sum of anti-clockwise moments”.
The SI units of the moments of a force is Newton metre (Nm).
Examples
1. A uniform rod of negligible mass balances when a weight of 3 N is at A, weight of 3 N is at B and a weight of W is at C. What is the value of weight W?
1. The following bar is of negligible weight. Determine the value of ‘ x’ if the bar is balanced.
Solution
The distance from the turning point to the line of action can be determined as,
Clockwise moments = 10 × 30 = 300 N cm, Anticlockwise moments = 10 × ‘x’ = 10 x. N cm. Using the principle of moments
Anti-clockwise moments = clockwise moments
10 x = 300, hence x = 30 cm.
1. Study the diagram below and determine the value of X and hence the length of the bar.
Solution
Clockwise moments = 15x N + 5(X × 20) N
Anticlockwise moments = (20 × 10) + (60 × 10) N cm, = 800 N cm.
Anti-clockwise moments = clockwise moments
800 N cm = 15X + 5X + 100
800 n cm = 20X + 100
20X = 700
X = 35 cm.
Therefore, the length of the bar = 40 + 20 + 35 + 20 = 115 cm.
## The lever
A lever is any device which can turn about a pivot or fulcrum .
The applied force is called the effort and is used to overcome the resisting force called the load. We use the law of moments in the operation of levers.
Example
Consider the following diagram. (The bar is of negligible mass). Determine the effort applied.
Solution
Taking moments about O. Then, clockwise moments = effort × 200 cm. Anticlockwise moments = 200 × 30 cm.
Effort = (200 × 30)/ 200 = 30 N.
MEASUREMENT II
MAGNETISM
CELLS AND SIMPLE CIRCUITS
ELECROSTATICS I
RECTILINEAR PROPAGATION AND REFLECTION AT PLANE SURFACES. INTRODUCTION
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# RD Sharma Class 10 Ex 8.2 Solutions Chapter 8 Quadratic Equations
In this chapter, we provide RD Sharma Class 10 Ex 8.2 Solutions Chapter 8 Quadratic Equations for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 8.2 Solutions Chapter 8 Quadratic Equations pdf, Now you will get step by step solution to each question.
# Chapter 8: Quadratic Equations Exercise – 8.2
### Question: 1
The product of two consecutive positive integers is 306. Form the quadratic equation to find the integers, if x denotes the smaller integer.
### Solution:
Given that the smallest integer of 2 consecutive integers is denoted by x
The two integers be x and x + 1
According to the question, the product of the integers is 306
Now, x(x + 1) = 306
= x+ x – 306 = 0
The required quadratic equation of the equation is x+ x – 306 = 0
### Question: 2
John and Jivani together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 128. Form the quadratic equation to find how many marbles they to start with if John had x marbles.
### Solution:
Given that John and Jilani are having the total of 45 marbles.
Let us consider John is having x marbles
Jivani is having (45 – x) marbles.
Number of marbles john had after losing 5 marbles = x – 5
Number of marbles jivani had after losing 5 marbles = (45 – x) – 5 = 40 – x
According to the question the product of the marbles that they are having now is 128
Now, (x – 5)(40 – x) = 128
= 40x – x– 200 = 128
= x2– 45x + 128 + 200 = 0
= x2 -45x + 328 = 0
The required quadratic equation is x2 – 45x + 328 = 0.
### Question: 3
A cottage industry produces a certain number of toys in a day. The cost of production of each toy was found to be 55 minutes the number of articles produced in a day. On a particular day, the total cost of production was Rs. 750. If x denotes the number of toys produced that day, form the quadratic equation to find x.
### Solution:
Given (y) Denotes the number of toys produced in a day.
The cost of production of each toy = (55 – y)
Total cost of production is nothing but the product of number of toys produced in a day and cost of production of each to y = y (55 – y)
According to the question
The total cost of production is Rs.750
= y (55 – y) = 750
= 55y – y2 = 750
= y– 55y + 750 = 0
The required quadratic equation of the given data is y– 55y + 750 = 0.
### Question: 4
The height of the right triangle is 7 cm less than its base. If the hypotenuse is 13cm, form the quadratic equation to find the base of the triangle.
### Solution:
According to the question
The hypotenuse of the triangle = 13 cm
Let the base of the triangle = x cm
So, the height of the triangle = (x-7) cm
Applying Pythagoras theorem in the right angled triangle, we get,
(Base)+ (height)= (hypotenuse)2
x2 +(x – 7)= (13)2
x+ x+ 49 – 14x = 169
2x-14x – 120 = 0
2(x– 7x – 60) = 0
x– 7x – 60 = 0
The required quadratic equation is x– 7x – 60 = 0
### Question: 5
The average speed of the express train is 11 km/hr more than that of the passenger train. The total distance covered by the train is 132 km. Also, time taken by the express train is 1 hour is less than that of the passenger train. Find the quadratic equation of this problem.
### Solution:
Let the average speed of the express train be = x km/hr
Given, the average speed of the express train is 10 km/hr less than that of passenger train = (x – 11) km/hr
We know that:
Time taken for travel = distance travelled/average speed
Time taken for express train = distance travelled/average speed of the express train
= 132/x
Hence time taken by the passenger train = 132/(x – 11)
According to the question, Time taken by the express train is 1 hour less than that of passenger train Time taken by passenger train – time taken by express train = 1 hour
The quadratic equation of the given problem is x-11x – 1452 = 0.
### Question: 6
A train travels 360 km at a uniform speed. If the speed had been 5 km/ hr more, it would have taken 1 hour less for the same journey. Form the quadratic equation to find the speed of the train.
### Solution:
Let the speed of the train be = x km/hr
Distance travelled by the train = 360 km
We know that, Time taken for travel = distance travelled ÷ speed of the train = 360/x
If the speed of the train is increased by 4 km/hr then time taken = 360/(x + 5)
According to the question,
The time of travel is reduced by 1 hour when the speed of the train is increased by 5 km /hr
The required quadratic equation is x+ 5x – 1800 = 0.
All Chapter RD Sharma Solutions For Class10 Maths
I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good |
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This is a Test of Mathematics Solution Subjective 110 (from ISI Entrance). The book, Test of Mathematics at 10+2 Level is Published by East West Press. This problem book is indispensable for the preparation of I.S.I. B.Stat and B.Math Entrance.
Also visit: I.S.I. & C.M.I. Entrance Course of Cheenta
## Problem:
:
Let ABCD be a cyclic quadrilateral with lengths of sides AB = p , BC = q , CD = r, and DA = s . Show that $\frac{AC}{BD} = \frac{ps+qr}{pq+rs}$
## Solution:
Since ABCD is cyclic $\Delta ABE$ is similar to $\Delta CDE$ (since $\angle ABE = \angle DCE$ as they are subtended by the same arc AD, and $\angle AEB = \angle CED$ as vertically opposite angles are equal)
Hence their corresponding sides are proportional.
$\frac{p}{r} = \frac{BE}{CE} \implies BE = CE \times \frac{p}{r}$
$\frac{p}{r} = \frac{AE}{DE} \implies AE = DE \times \frac{p}{r}$
Similarly $\Delta AED$ is similar to $\Delta BEC$
$\frac{s}{q} = \frac{DE}{CE} \implies DE = CE \times \frac{s}{q}$
$\frac{q}{s} = \frac{CE}{DE} \implies CE = DE \times \frac{q}{s}$
Hence
$BE + DE = BD = CE \times { \frac{p}{r} + \frac{s}{q} } = CE \times \frac{pq+rs}{qr}$
$AE + CE = AC = DE \times { \frac{p}{r} + \frac{q}{s} } = DE \times \frac{ps+qr}{sr}$
Hence $\displaystyle { \frac{AC}{BD} = \frac {ps + qr}{pq + rs} \times \frac {qr}{sr} \times \frac {DE}{CE} }$
Finally we note that since $\Delta AED$ is similar to $\Delta BEC$. $\displaystyle {\frac{q}{s} = \frac {CE}{DE} \implies \frac{q}{s} \times \frac {DE}{CE} = 1 }$
This proves that
$\displaystyle { \frac{AC}{BD} = \frac {ps + qr}{pq + rs}}$ |
# How do you write the equation of line passes through (-4,-3), and is perpendicular to 4x + y=7?
Feb 1, 2017
$3 x - y + 14 = 0$
#### Explanation:
Equation of a line that is perpendicular to $A x + B y = C$
is of the form $B x - A y = k$ i.e. reversing te coefficients of $x$ and $y$ and changing sign of one of them.
Hence, the line perpendicular to $4 x + y = 7$ has equation of the form
$x - 4 y = k$ and as it passes through $\left(- 4 , - 3\right)$
we should have $\left(- 4\right) - 4 \times \left(- 3\right) = k$
and therefore $k = - 4 + 12 = 8$
and desired equation is $x - 4 y = 8$ or $3 x - y + 14 = 0$
graph{(x-4y-8)(4x+y-7)=0 [-10, 10, -5, 5]} |
# 5.3 Features of projectile motion (application) (Page 3/3)
Page 3 / 3
## Composition of motion
Problem : The position of a projectile projected from the ground is :
$\begin{array}{l}x=3t\\ y=\left(4t-2{t}^{2}\right)\end{array}$
where “x” and “y” are in meters and “t” in seconds. The position of the projectile is (0,0) at the time of projection. Find the speed with which the projectile hits the ground.
Solution : When the projectile hits the ground, y = 0,
$\begin{array}{l}0=\left(4t-2{t}^{2}\right)\\ ⇒2{t}^{2}-4t=t\left(2t-2\right)=0\\ ⇒t=0,\phantom{\rule{2pt}{0ex}}t=2\phantom{\rule{2pt}{0ex}}s\end{array}$
Here t = 0 corresponds to initial condition. Thus, projectile hits the ground in 2 s. Now velocities in two directions are obtained by differentiating given functions of the coordinates,
$\begin{array}{l}{v}_{x}=\frac{đx}{đt}=3\\ {v}_{y}=\frac{đy}{đt}=4-4t\end{array}$
Now, the velocities for t = 2 s,
$\begin{array}{l}{v}_{x}=\frac{đx}{đt}=3\phantom{\rule{2pt}{0ex}}m/s\\ {v}_{y}=\frac{đy}{đt}=4-4x2=-4\phantom{\rule{2pt}{0ex}}m/s\end{array}$
The resultant velocity of the projectile,
$\begin{array}{l}v=\sqrt{\left({{v}_{x}}^{2}+{{v}_{y}}^{2}\right)}=\sqrt{\left\{{3}^{2}+{\left(-4\right)}^{2}\right\}}=5\phantom{\rule{2pt}{0ex}}m/s\end{array}$
Problem : A projectile, thrown from the foot of a triangle, lands at the edge of its base on the other side of the triangle. The projectile just grazes the vertex as shown in the figure. Prove that :
$\mathrm{tan}\alpha +\mathrm{tan}\beta =\mathrm{tan}\theta$
where “θ” is the angle of projection as measured from the horizontal.
Solution : In order to expand trigonometric ratio on the left side, we drop a perpendicular from the vertex of the triangle “A” to the base line OB to meet at a point C. Let x,y be the coordinate of vertex “A”, then,
$\mathrm{tan}\alpha =\frac{AC}{OC}=\frac{y}{x}$
and
$\mathrm{tan}\beta =\frac{AC}{BC}=\frac{y}{\left(R-x\right)}$
Thus,
$\mathrm{tan}\alpha +\mathrm{tan}\beta =\frac{y}{x}+\frac{y}{\left(R-x\right)}=\frac{yR}{x\left(R-x\right)}$
Intuitively, we know the expression is similar to the expression involved in the equation of projectile motion that contains range of projectile,
$⇒y=x\mathrm{tan}\theta \left(1-\frac{x}{R}\right)$
$⇒\mathrm{tan}\theta =\frac{yR}{x\left(R-x\right)}$
Comparing equations,
$⇒\mathrm{tan}\alpha +\mathrm{tan}\beta =\mathrm{tan}\theta$
## Projectile motion with wind/drag force
Problem : A projectile is projected at angle “θ” from the horizontal at the speed “u”. If an acceleration of “g/2” is applied to the projectile due to wind in horizontal direction, then find the new time of flight, maximum height and horizontal range.
Solution : The acceleration due to wind affects only the motion in horizontal direction. It would, therefore, not affect attributes like time of flight or maximum height that results exclusively from the consideration of motion in vertical direction. The generic expressions of time of flight, maximum height and horizontal range of flight with acceleration are given as under :
$T=\frac{2{u}_{y}}{g}$
$H=\frac{{u}_{y}^{2}}{2g}=\frac{g{T}^{2}}{4}$
$R=\frac{{u}_{x}{u}_{y}}{g}$
The expressions above revalidate the assumption made in the beginning. We can see that it is only the horizontal range that depends on the component of motion in horizontal direction. Now, considering accelerated motion in horizontal direction, we have :
$x=R\prime ={u}_{x}T+\frac{1}{2}{a}_{x}{T}^{2}$
$⇒R\prime ={u}_{x}T+\frac{1}{2}\left(\frac{g}{2}\right){T}^{2}$
$R\prime =R+H$
Problem : A projectile is projected at angle “θ” from the horizontal at the speed “u”. If an acceleration of g/2 is applied to the projectile in horizontal direction and a deceleration of g/2 in vertical direction, then find the new time of flight, maximum height and horizontal range.
Solution : The acceleration due to wind affects only the motion in horizontal direction. It would, therefore, not affect attributes resulting exclusively from the consideration in vertical direction. It is only the horizontal range that will be affected due to acceleration in horizontal direction. On the other hand, deceleration in vertical direction will affect all three attributes.
1: Time of flight
Let us work out the effect on each of the attribute. Considering motion in vertical direction, we have :
$y={u}_{y}T+\frac{1}{2}{a}_{y}{T}^{2}$
For the complete flight, y = 0 and t = T. Also,
${a}_{y}=-\left(g+\frac{g}{2}\right)=-\frac{3g}{2}$
Putting in the equation,
$⇒0={u}_{y}T-\frac{1}{2}X\frac{3g}{2}X{T}^{2}$
Neglecting T = 0,
$⇒T=\frac{4{u}_{y}}{3g}=\frac{4u\mathrm{sin}\theta }{3g}$
2: Maximum height
For maximum height, ${v}_{y}=0$ ,
$0={u}_{y}^{2}-2X\frac{3g}{2}XH$
$H=\frac{{u}_{y}^{2}}{3g}=\frac{{u}^{2}{\mathrm{sin}}^{2}\theta }{3g}$
2: Horizontal range
Now, considering accelerated motion in horizontal direction, we have :
$x=R={u}_{x}T+\frac{1}{2}{a}_{x}{T}^{2}$
$R={u}_{x}\left(\frac{4{u}_{y}}{g}\right)+\frac{1}{2}\left(\frac{g}{2}\right){\left(\frac{4{u}_{y}}{g}\right)}^{2}$
$⇒R=\left(\frac{4{u}_{y}}{g}\right)\left[{u}_{x}+\frac{1}{2}\left(\frac{g}{2}\right)\left(\frac{4{u}_{y}}{g}\right)\right]$
$⇒R=\left(\frac{4{u}_{y}}{g}\right)\left\{{u}_{x}+{u}_{y}\right\}$
$⇒R=\left(\frac{4{u}^{2}\mathrm{sin}\theta }{g}\right)\left[\mathrm{cos}\theta +\mathrm{sin}\theta \right]$
pls explain what is dimension of 1in length and -1 in time ,what's is there difference between them
what are scalars
show that 1w= 10^7ergs^-1
what's lamin's theorems and it's mathematics representative
if the wavelength is double,what is the frequency of the wave
What are the system of units
A stone propelled from a catapult with a speed of 50ms-1 attains a height of 100m. Calculate the time of flight, calculate the angle of projection, calculate the range attained
58asagravitasnal firce
Amar
water boil at 100 and why
what is upper limit of speed
what temperature is 0 k
Riya
0k is the lower limit of the themordynamic scale which is equalt to -273 In celcius scale
Mustapha
How MKS system is the subset of SI system?
which colour has the shortest wavelength in the white light spectrum
if x=a-b, a=5.8cm b=3.22 cm find percentage error in x
x=5.8-3.22 x=2.58
what is the definition of resolution of forces |
# KS2 Maths Quiz - Fractions (Year 6) (Questions)
In KS2 Year Six, children should be quite comfortable when dealing with fractions in Maths. By now they should be well aware of the values of different fractions and know the difference between numerators and denominators. They should also be able to convert improper fractions into mixed numbers and be good at recognising equivalent values for proper and improper fractions.
Recognising a fraction when you see one is easy. Fractions represent part of a whole number. But not all fractions are the same type. Proper fractions have top numbers, or numerators, lower than their bottom numbers, or denominators. Improper fractions are those where the numerators are bigger than the denominators. Another way to write improper fractions is as mixed numbers where units are shown next to a proper fraction. For example, 43 is the equivalent of 113.
Fractions can be tricky to understand. Give your child some help in our extensive Fractions in Maths article.
1. How many 11,000 in a whole one?
[ ] 10 [ ] 100 [ ] 1,000 [ ] 10,000
2. What is the numerator?
[ ] The top number of a fraction [ ] The bottom number of a fraction [ ] The divider between the two numbers of a fraction [ ] The whole number in a mixed fraction
3. What do we call the bottom number of a fraction?
[ ] Numerator [ ] Factor [ ] Multiple [ ] Denominator
4. What would 129 be when converted to a mixed number?
[ ] 33⁄9 [ ] 11⁄3 [ ] 14⁄9 [ ] 21⁄9
5. What would 226 be when converted to an improper fraction?
[ ] 14⁄2 [ ] 15⁄6 [ ] 14⁄6 [ ] 13⁄6
6. Which of these are NOT equivalent to each other?
[ ] 3⁄9 and 1⁄3 [ ] 6⁄8 and 3⁄4 [ ] 4⁄6 and 8⁄12 [ ] 1⁄10 and 2⁄5
7. Which of these statements is FALSE?
[ ] One half is 3 times more than one sixth [ ] One third is twice as much as one sixth [ ] One twentieth is half of one tenth [ ] One hundredth is ten times more than one tenth
8. How do we reduce 520 to an equivalent fraction?
[ ] Divide the numerator and denominator by 2 [ ] Divide the numerator and denominator by the same number [ ] Multiply the numerator and denominator by 2 [ ] Multiply the numerator and denominator by the same number
9. How do we change 310 to an equivalent fraction?
[ ] Divide the numerator by the denominator [ ] Divide the numerator and denominator by the same number [ ] Multiply the numerator by the denominator [ ] Multiply the numerator and denominator by the same number
10. Which is an equivalent of 615?
[ ] 2⁄5 [ ] 3⁄7 [ ] 2⁄10 [ ] 1⁄3
KS2 Maths Quiz - Fractions (Year 6) (Answers)
1. How many 11,000 in a whole one?
[ ] 10 [ ] 100 [x] 1,000 [ ] 10,000
1,0001,000 = 1, just as 100100 or 33 = 1
2. What is the numerator?
[x] The top number of a fraction [ ] The bottom number of a fraction [ ] The divider between the two numbers of a fraction [ ] The whole number in a mixed fraction
The numerator tells us how many of the equal parts there are
3. What do we call the bottom number of a fraction?
[ ] Numerator [ ] Factor [ ] Multiple [x] Denominator
The denominator tells us the number of parts the whole is divided into
4. What would 129 be when converted to a mixed number?
[ ] 33⁄9 [x] 11⁄3 [ ] 14⁄9 [ ] 21⁄9
129 could be written as 139, 113 or 43
5. What would 226 be when converted to an improper fraction?
[ ] 14⁄2 [ ] 15⁄6 [x] 14⁄6 [ ] 13⁄6
146 could also be written as 73
6. Which of these are NOT equivalent to each other?
[ ] 3⁄9 and 1⁄3 [ ] 6⁄8 and 3⁄4 [ ] 4⁄6 and 8⁄12 [x] 1⁄10 and 2⁄5
15 = 210
7. Which of these statements is FALSE?
[ ] One half is 3 times more than one sixth [ ] One third is twice as much as one sixth [ ] One twentieth is half of one tenth [x] One hundredth is ten times more than one tenth
One tenth is ten times more than one hundredth
8. How do we reduce 520 to an equivalent fraction?
[ ] Divide the numerator and denominator by 2 [x] Divide the numerator and denominator by the same number [ ] Multiply the numerator and denominator by 2 [ ] Multiply the numerator and denominator by the same number
We have to find the lowest common multiple of 5 and 20 which is 5 so the equivalent will be 14
9. How do we change 310 to an equivalent fraction?
[ ] Divide the numerator by the denominator [ ] Divide the numerator and denominator by the same number [ ] Multiply the numerator by the denominator [x] Multiply the numerator and denominator by the same number
If we multiply by 10 the equivalent fraction will be 30100
10. Which is an equivalent of 615?
[x] 2⁄5 [ ] 3⁄7 [ ] 2⁄10 [ ] 1⁄3
6 ÷ 3 = 2 and 15 ÷ 3 = 5 so 25 is the same as 615 |
# Pipes and Cisterns Quiz Set 019
### Question 1
Tap M can fill a cistern in 18 mins. And, a tap N can empty it in 11 mins. In how many minutes will the cistern be emptied if both the taps are opened together when the tank is \$13/17\$th already empty?
A
\$6{78/119}\$ mins.
B
\$7{85/118}\$ mins.
C
\$5{68/121}\$ mins.
D
\$9{60/121}\$ mins.
Soln.
Ans: a
1 filled cistern can be emptied in \${18 × 11}/{18 - 11}\$ mins. So \$1 - 13/17\$ = \$4/17\$ filled cistern can be emptied in \${18 × 11}/{18 - 11}\$ × \$4/17\$ = \${792/119}\$, which is same as: \$6{78/119}\$ mins.
### Question 2
A tank is filled in 13 minutes by three taps running together. Tap A is twice as fast as tap B, and tap B is twice as fast as tap C. How much time will tap A take to fill the tank?
A
91 mins.
B
92 mins.
C
90 mins.
D
93 mins.
Soln.
Ans: a
Let the time taken by tap A be x mins. Then 13 minutes work of all the taps should add to 1. So we have, \$13 × 1/x + 13 × 2/x + 13 × 4/x\$ = 1, which is same as \$13 × 7/x\$ = 1. Solving, we get x = 91 mins.
### Question 3
Pipe A can fill a cistern in 30 minutes, while the pipe B can fill it in 20 minutes. They are alternately open for 1 minute. How long will it take the cistern to fill completely?
A
24 mins.
B
25 mins.
C
23 mins.
D
26 mins.
Soln.
Ans: a
Let the total time taken be 2x minutes. Both X and Y run for x mins. So \$(x/20 + x/30)\$ = 1. Solving for x, we get x = 12, which gives 2x = 24.
### Question 4
A tank is filled in 6 minutes by three taps running together. Times taken by the three taps independently are in an AP[Arithmetic Progression], whose first term is a and common difference d. Then, a and d satisfy the relation?
A
a3 - 18a2 - ad2 + 6d2 = 0.
B
a3 - 12a2 + ad2 + 6d2 = 0.
C
a3 - 6a2 - ad2 + 6d2 = 0.
D
a3 - 30a2 + ad2 + 6d2 = 0.
Soln.
Ans: a
Let the times taken by the three taps be a - d, a and a + d. Then 6 minutes work of all the taps should add to 1. So we have, \$6 × 1/{a - d} + 6 × 1/a + 6 × 1/{a + d}\$ = 1, which is same as a3 - 18a2 - ad2 + 6d2 = 0.
### Question 5
A tank is filled in 3 minutes by three taps running together. Times taken by the three taps to independently fill the tank are in an AP[Arithmetic Progression]. If the first tap is a leakage tap and the second tap takes 1 minute to fill the tank, then, the common difference of the AP can be?
A
2.
B
3.
C
1.
D
4.
Soln.
Ans: a
Let the times taken by the three taps be 1 - d, 1 and 1 + d. The time taken by the first tap will be negative because it is a leakage tap. Then 3 minutes work of all the taps should add to 1. So we have, 3 × \$(1/{1 - d} + 1/1 + 1/{1 + d})\$ = 1, which is same as \$2/{1 - d^2} + 1\$ = \${1/3}\$. Solving we get d = ±2. |
### Transcript
Now we can talk about areas of quadrilaterals. In this video, we will review how to find the area of the special quadrilaterals. So first of all the most special quadrilateral is the square. If a square has the side of s, then the area is simply s squared. In fact, even in algebra, raising a number to the second power is called squaring, precisely because this is the way to find the area of a square.
So the word we use in algebra actually comes from the basic geometric fact. For the other special quadrilaterals, the general formula is area equals base times height, but we need to think carefully about this. The formula area equals base times height, works most obviously for a rectangle, in which b & h are both sides of the rectangle. As with triangles, remember that the base needs not be horizontal.
Any side can be the base, and the height must be perpindicular to it. So, for a rectangle, the area is just the product of the two different side lengths. The area A = bh also works for rhombuses and parallelograms, and any side can be a base, but the height has to be perpendicular to the base. So the height will not lie along a side. Instead the height will be what's called an altitude, a line perpendicular to the base, and so we need to find that height.
The length of the altitude is not given, then almost always one can find it from the Pythagorean theorem. So here is a practice problem, pause the video and then we'll talk about this. Okay, so if JN = 1, and NM = 2, then all the way across from J to M has to be 3. And because it's a rhombus, every side has to have a length of 3.
So now look at the right triangle JKN, in which JK, the side of the rhombus is 3, a nd JN were given as 1. We'll you use the Pythagorean theorem in that triangle to find KN. KN squared = (JK)- (JN) squared. 3 squared is 9, so 9- 1=8. Means that KN is the square root of 8, of course we can simplify that down to 2 root 2, and that is the height of the rhombus.
So, now we're ready to apply area equals base times height. We know that the base is 3 and the height Is 2 root 2 ,and we can simply multiply these and get 6 root 2. If this operation with roots are a little bit unfamiliar, I would suggest going back to the Power and Roots module and watching the video, Operations with Roots. With trapezoids, we have to re-think a bit, because there are two bases, two parallel sides.
So, what exactly would we mean by base times height? Well, the height is pretty clear but we have two bases, so what are we gonna do? One way to find the area is to find the average of the bases, and multiply this by the height. So that is the formula for the area of a rhombus, we average the bases, and multiply the height times the average of the base.
Sometimes we can find the area of a trapezoid by subdividing the trapezoid into a central rectangle, and two side right triangles. So this is often what the test will have us do. We have two side right triangles, we can find information about those with the Pythagorean theorem, and that will allow us to solve for everything and find all the areas.
And of course if it's a symmetrical trapezoid those two side right triangles will be congruent, which makes things even easier. Here's a practice problem, pause the video and then we'll talk about this. In trapezoid ABCD, altitudes are drawn. If AF = 5, find the area of the trapezoid. Well first of all, we're gonna look at ABF.
That little triangle we have a leg of 5, an unknown leg, and a hypotenuse of 13. So of course, that's a 5, 12, 13 triangle, and BF equals 12. So we can find that just from knowledge of our Pythagorean triplet, so we don't even need to do a calculation. So BF equals 12, that means that CE also equals 12.
So, we can find the area of the triangle ABF, and that has to be one half 5 times 12. 5 times 6 is 30. We can also find the area of the rectangle, 10 times 12 is a 120. Now that triangle on the left, triangle CED, that's gonna be blank,12, 15. Well of course that's a 3, 4, 5 triangle multiplied by 3, so that's gonna be 9, 12, 15.
And then the area one-half 9 times 12, well that's 9 times 6 which is 54. So the whole area is gonna be triangle plus rectangle plus triangle. 30 + 120 + 54, and that equals 204. In summary, a square has an area of s squared. The rectangle, rhombus, parallelogram have an area of base times height.
We have to be careful in a rhombus or parallelogram, any side can be the base but the other side is not gonna be the height, the height has to be perpendicular to the base, we need to find the altitude. A trapezoid is the average of the bases times the height. And for any slanty shapes, think about subdividing into rectangles and right triangles, and this might even be true, for example, if we were dealing with an irregular quadrilateral.
And expect to find the Pythagorean theorem involved in anything involving a slant. |
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# Data Representation and Interpretation Study Guide: GED Math
By
Updated on Mar 23, 2011
Practice problems for these concepts can be found at:
Data Analysis, Statistics, and Probability Practice Problems: GED Math
### Data Representation and Interpretation
The GED exam will test your ability to analyze graphs and tables. It is important to study each graph or table very carefully before reading the question. This will help you to process the information that is presented. It is extremely important to read all of the information presented, paying special attention to headings and units of measure. Here is an overview of the types of graphs you will encounter:
### Circle Graphs or Pie Charts
Circle graphs or pie charts show how the parts of a whole relate to one another. A pie chart is a circle divided into slices or wedges—like a pizza. Each slice represents a category.
### Line Graphs
Line graphs show how two categories of data or information (sometimes called variables) relate to one another. The data is displayed on a grid and is presented on a scale using a horizontal axis and a vertical axis for the different categories of information the graph is comparing. Usually, the data points are connected together to form a line so that you can see trends in the data, or how the data changes over time. Therefore, often you will see line graphs with units of time or the word time on the horizontal axis.
Line graphs are frequently used to show the results of a scientific experiment. The variable that the scientist is measuring and tracking is often called the dependent variable. It is usually measured on the vertical axis of a graph. The horizontal axis is usually measuring time, so you can see how the data changes over time.
### Bar Graphs
Like pie charts, bar graphs show how different categories of data relate to one another. A bar represents each category. The length of the bar represents the relative frequency of the category—compared to the other categories on the graph.
Both pie charts and bar graphs are used to compare different categories of data. So when you have data to graph, how do you decide which kind of graph to use? Think about what your purpose is. If your purpose is to compare the absolute values of each category, then a bar chart is probably better because the amounts of each category are shown in comparison to each other. If your purpose is to show how each part relates to the whole, a pie chart is probably better. |
M1410207
# M1410207 - (Section 2.7 Nonlinear Inequalities 2.77 SECTION...
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(Section 2.7: Nonlinear Inequalities) 2.77 SECTION 2.7: NONLINEAR INEQUALITIES We solved linear inequalities to find domains, and we discussed intervals in Section 1.4: Notes 1.24 to 1.30 . In this section, we will solve nonlinear inequalities to find domains. Example 1 Let f x ( ) = x 2 9 . We get real outputs x 2 9 0 . There are different ways to solve this inequality; its solution set is the domain of f . Method 1: Sign Chart Method; we are solving x 2 9 0 The key idea here is that we’d rather perform a sign analysis on products of factors as opposed to sums of terms. (For example, the product of a positive real number and a negative real number is guaranteed to be negative; however, there is no such guarantee regarding their sum.) Factoring can be a key tool. x 2 9 0 x + 3 ( ) x 3 ( ) 0 We need to determine where each of the factors on the left side is negative, 0, and positive in value. We ultimately want to know where their product is 0 or positive. The domain of f is: − ∞ , 3 ( 3, ) .
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(Section 2.7: Nonlinear Inequalities) 2.78 Method 2: Parabola Method (for Quadratic Inequalities); we are solving x 2 9 0 The real zeros of x 2 9 are the x -intercepts of the corresponding parabola: x 2 9 = 0 x 2 = 9 x = ± 3 Warning: Here, the ± symbol means “take both the + 3 and the 3 .” See Warning 1 in Section 1.5: Notes 1.46 . The leading coefficient of x 2 9 is positive, so the parabola opens up. This is enough information for us to sketch the parabola to our satisfaction. Given an input x , the y -coordinate of the corresponding point gives the output (or function value). Because of the “ 0 ” in our inequality, we need the values of x , if any, that correspond to the parts of the parabola that lie above or on the x -axis.
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# How do you find the second derivative of ln(X^(1/2))?
Sep 8, 2017
$- \frac{1}{2} \cdot {x}^{- 2} , \mathmr{and} , - \frac{1}{2 {x}^{2}} .$
#### Explanation:
Let, $f \left(x\right) = \ln \left({x}^{\frac{1}{2}}\right) .$
Using the Rule : $\ln \left({x}^{m}\right) = m \ln x ,$ we find,
$f \left(x\right) = \frac{1}{2} \cdot \ln x .$
Now, for a constant $k , \left\{k \cdot F \left(x\right)\right\} ' = k \cdot F ' \left(x\right) ,$
$\therefore f ' \left(x\right) = \frac{1}{2} \left\{\ln x\right\} ' = \frac{1}{2} \cdot \frac{1}{x} = \frac{1}{2} \cdot {x}^{- 1} .$
Recall that, $f ' ' \left(x\right) = \left\{f ' \left(x\right)\right\} ' .$
$f ' ' \left(x\right) = \left\{\frac{1}{2} \cdot {x}^{- 1}\right\} ' = \frac{1}{2} \left\{{x}^{- 1}\right\} ' .$
But, $\left\{{x}^{n}\right\} ' = n \cdot {x}^{n - 1} ,$
$\therefore f ' ' \left(x\right) = \frac{1}{2} \cdot \left(- 1 \cdot {x}^{- 1 - 1}\right) = - \frac{1}{2} \cdot {x}^{- 2} , \mathmr{and} , - \frac{1}{2 {x}^{2}} .$# |
# What two prime numbers make 63?
## What two prime numbers make 63?
So, the prime factorisation of 63 is 3 × 3 × 7 or 32 × 7, where 3 and 7 are the prime numbers.
### What is the multiple of 63?
Hence, the first five multiples of 63 are 63, 126, 189, 252, and 315.
#### What are two prime numbers?
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
IS 63 is a prime factor?
No, 63 is not a prime number. The number 63 is divisible by 1, 3, 7, 9, 21, 63. Since 63 has more than two factors, i.e. 1, 3, 7, 9, 21, 63, it is not a prime number.
What is the prime numbers of 63?
The prime factors of 63 are 3, 7.
## What is the HCF of 105 and 63?
The GCF of 63 and 105 is 21. To calculate the GCF (Greatest Common Factor) of 63 and 105, we need to factor each number (factors of 63 = 1, 3, 7, 9, 21, 63; factors of 105 = 1, 3, 5, 7, 15, 21, 35, 105) and choose the greatest factor that exactly divides both 63 and 105, i.e., 21.
### What is the perfect square of 63?
Interactive Questions
True
Square root of 63 cannot be simplified in its radical form. TrueTrue – Square root of 63 cannot be simplified in its radical form.
Square root of 63 is a perfect square. TrueTrue – Square root of 63 is a perfect square.
#### What is the lowest multiple of 63?
The lowest common multiple is the lowest multiple shared by two or more numbers. For example, common multiples of 4 and 6 are 12, 24 and 36, but the lowest of those is 12; therefore, the lowest common multiple of 4 and 6 is 12.
Why is 11 not a prime number?
Is 11 a Prime Number? The number 11 is divisible only by 1 and the number itself. For a number to be classified as a prime number, it should have exactly two factors. Since 11 has exactly two factors, i.e. 1 and 11, it is a prime number.
What is Coprime number?
In number theory, two integers a and b are coprime, relatively prime or mutually prime if the only positive integer that is a divisor of both of them is 1. Consequently, any prime number that divides one of a or b does not divide the other.
## What is the HCF of 27 and 63?
9
Answer: HCF of 27 and 63 is 9.
### Why is 65 not a prime number?
For a number to be classified as a prime number, it should have exactly two factors. Since 65 has more than two factors, i.e. 1, 5, 13, 65, it is not a prime number.
#### What is 63 as product of prime numbers?
For instance 63, in our case, is a number gotten as a result of multiplying prime numbers. We can therefore write it as a product of its prime factors. 63 = 3 x 3 x 7. The prime numbers that make up 63 are 3 and 7, and 63 as a product of its prime factors is (3x 3 x7) 0.0.
Is 63 a prime number or a composite number?
The number 63 is not a prime number because it is possible to factorize it. In other words, 63 can be divided by 1, by itself and at least by 3 and 7. This is a so called ‘composite number’.
What is the greatest prime factor of 63?
Prime factors of 63 are 3, 7. Prime factorization of 63 in exponential form is: 63 = 3 2 × 7 1
## What are all the multiples of 63?
Multiples of 63 would be 63, 126, 189, 252 and so on. Now, compare the two lists to find the smallest number the two lists have in common, which is the Least Common Multiple of 12 and 63. |
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# ARITHMETIC SERIES PROBLEMS
Real Life Problems Involving Arithmetic Series
REAL LIFE PROBLEMS INVOLVING ARITHMETIC SERIES. Problem 1 : A construction company will be penalized each day of delay in construction for bridge. The penalty will be \$4000 for the first day and will increase by \$10000 for each following day. Based on its budget, the company can afford to pay a maximum of \$ 165000 toward penalty.
Arithmetic Series -- from Wolfram MathWorld
Apr 13, 2021An arithmetic series is the sum of a sequence, , 2,, in which each term is computed from the previous one by adding (or subtracting) a constant . Therefore, for , (1) Unlimited random practice problems and answers with built-in Step-by-step
Arithmetic progression - Wikipedia
An Arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. For instance, the sequence 5, 7, 9, 11, 13, 15, . . . is an arithmetic progression with a common difference of 2. If the initial term of an arithmetic progression is and the common difference of successive members is d, then the nth term of the
Arithmetic Sequence Practice Problems - ChiliMath
More Practice Problems with the Arithmetic Sequence Formula Direction: Read each arithmetic sequence question carefully, then answer with supporting details. Arithmetic Sequence Practice Problems with Answers 1) Tell whether if the sequence is arithmetic or not. Explain why or why not. Sequence A : Sequence B : Solution: Sequence A is an arithmetic sequence since Arithmetic
Arithmetic- Basics Concepts, Operations & Solved Problems
Arithmetic is one among the oldest and elementary branches of mathematics, originating from the Greek word arithmos, which means number. It involves the study of numbers, especially the properties of traditional operations, such as addition, subtraction, division and multiplication.
Arithmetic - Wikipedia
Arithmetic (from the Greek ἀριθμός arithmos, 'number' and τική, tiké [téchne], 'art' or 'craft') is a branch of mathematics that consists of the study of numbers, especially the properties of the traditional operations on them—addition, subtraction, multiplication, division, exponentiation and extraction of roots. Arithmetic is an elementary part of number theory, and number
Arithmetic Sequences and Series - MATHguide
Jul 16, 2020ideo: Arithmetic Series: Deriving the Sum Formula. Usually problems present themselves in either of two ways. Either the first number and the last number of the sequence are known or the first number in the sequence and the number of terms are known. The following two problems will explain how to find a sum of a finite series.
Examples of arithmetic and geometric sequences and series
Most interest problems would start at time = 0, so I would exclude these unless you said something like "let x = \$ in bank at beginning of each year". Then your first input value would be 1. Also, geometric sequences have a domain of only natural numbers (1,2,3,..), and a graph of them would be only points and not a continuous curved line.
Arithmetic Sequences Problems with Solutions
Arithmetic Sequences Problems with Solutions. Arithmetic sequences are used throughout mathematics and applied to engineering, sciences, computer sciences, biology and finance problems. A set of problems and exercises involving arithmetic sequences, along with detailed solutions and answers, are presented.
Arithmetic Progression - GeeksforGeeks
Dec 10, 2018More problems related to Arithmetic Progression Sum of first n terms of a given series 3, 6, 11, . Ratio of mth and nth terms of an A. P. with given ratio of sums |
# JEE Main Maths Limits, Continuity and Differentiability Previous Year Questions With Solutions
In calculus, limit and continuity are important concepts. A limit is a number that a function approaches as the independent variable of the function approaches a given value. A function can either be continuous or discontinuous. Positive Discontinuity, Jump Discontinuity and Infinite Discontinuity are types of discontinuity. This article covers the definition of limit, types of limit, indeterminate form, algebra of limit, standard limits, expansion of some functions, continuity at a point, continuity in a domain, differentiability at a point. The limits, continuity and differentiability questions from the previous years of JEE Main are present on this page, along with the detailed solution for each question. These questions include all the important topics and formulae. About 2-3 questions are asked from this topic in JEE Examination.
## JEE Main Maths Limits, Continuity and Differentiability Previous Year Questions With Solutions
Question 1: Solve $\underset{x\to 1}{\mathop{\lim}}\,\frac{(2x-3)(\sqrt{x}-1)}{2{{x}^{2}}+x-3}$
Solution:
$\underset{x\to 1}{\mathop{\lim }}\,\,\,\frac{(2x-3)\,(\sqrt{x}-1)\times (\sqrt{x}+1)}{(x-1)\,(2x+3)\times (\sqrt{x}+1)}\\=\frac{-1}{5\,.\,2}\\=\frac{-1}{10}$
Question 2: If f(9)=9, f'(9)=4, then $\lim_{x \rightarrow 9}\frac{\sqrt{f(x)}-3}{\sqrt{x}-3}$
Solution:
Applying L – Hospitals rule,
$\underset{x\to 9}{\mathop{\lim }}\,\frac{\frac{1}{2\sqrt{f(x)}}\cdot {f}'(x)}{\frac{1}{2\sqrt{x}}}\\=\frac{\frac{{f}'(9)}{\sqrt{f(9)}}}{\frac{1}{\sqrt{9}}}\\=\frac{\frac{4}{3}}{\frac{1}{3}}\\=4$
Question 3: Solve $\underset{h\to 0}{\mathop{\lim }}\,\frac{{{(a+h)}^{2}}\sin (a+h)-{{a}^{2}}\sin a}{h}$
Solution:
Apply L-Hospitals rule,
$\underset{h\to 0}{\mathop{\lim }}\,\,\frac{{{(a+h)}^{2}}\sin (a+h)-{{a}^{2}}\sin a}{h}$ $\underset{h\to 0}{\mathop{\lim }}\,\,\,\frac{2\,(a+h)\,\sin \,(a+h)+{{(a+h)}^{2}}\cos \,(a+h)}{1}$ $=2a\,\,\sin a+{{a}^{2}}\cos \,\,a$
Question 4: Solve $\underset{x\to \pi /4}{\mathop{\lim }}\,\frac{\sqrt{2}\cos x-1}{\cot x-1}$
Solution:
$\underset{x\to \pi /4}{\mathop{\lim }}\,\,\frac{(\sqrt{2}-\sec x)\,\cos x\,(1+\cot x)}{\cot x\,[2-{{\sec }^{2}}x]} \\=\underset{x\to \pi /4}{\mathop{\lim }}\,\frac{\sin x\,(1+\cot x)}{(\sqrt{2}+\sec x)}\\=\frac{\frac{1}{\sqrt{2}}(2)}{\sqrt{2}+\sqrt{2}}\\=\frac{1}{2}$
Question 5: Solve $\underset{x\to 0}{\mathop{\lim }}\,\left[ \frac{x}{{{\tan }^{-1}}2x} \right]$
Solution:
Let ${{\tan }^{-1}}2x=\theta \\\,\Rightarrow x=\frac{1}{2}\tan \theta \ and \ as \ x\to 0,\,\,\theta \to 0 \\\Rightarrow \,\,\underset{x\to 0}{\mathop{\lim }}\,\,\frac{x}{{{\tan }^{-1}}2x}\\=\underset{\theta \to 0}{\mathop{\lim }}\,\,\frac{\frac{1}{2}\tan \theta }{\theta }\\=\frac{1}{2}$
Question 6: Solve $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{\frac{1}{2}(1-\cos 2x)}}{x}$
Solution:
$\underset{x\to 0}{\mathop{\lim }}\,\,\frac{\sqrt{\tfrac{1}{2}(1-\cos 2x)}}{x}=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{|\,\,\sin x\,\,|}{x}$
So, $\underset{x\to 0+}{\mathop{\lim }}\,\,\frac{|\,\sin x\,|}{x}=1$ and
$\underset{x\to 0-}{\mathop{\lim }}\,\,\frac{|\,\sin x\,|}{x}=-1$
Hence, the limit doesn’t exist.
Question 7: Solve $\underset{x\to 0}{\mathop{\lim }}\,{{\left\{ \tan \left( \frac{\pi }{4}+x \right) \right\}}^{1/x}}$
Solution:
Given limit = $=\underset{x\to 0}{\mathop{\lim }}\,\,\,{{\left( \frac{1+\tan x}{1-\tan x} \right)}^{1/x}} \\=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{{{\{{{(1+\tan x)}^{1/\tan x}}\}}^{(\tan x)/x}}}{{{\{{{(1-\tan x)}^{1/\tan x}}\}}^{(\tan x)/x}}}\\=\frac{e}{{{e}^{-1}}}\\={{e}^{2}}.$
Question 8: Solve $\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{1+5{{x}^{2}}}{1+3{{x}^{2}}} \right)}^{1/{{x}^{2}}}}$
Solution:
$\underset{x\to 0}{\mathop{\lim }}\,\,{{\left( \frac{1+5{{x}^{2}}}{1+3{{x}^{2}}} \right)}^{1/{{x}^{2}}}}\\=\frac{\underset{x\to 0}{\mathop{\lim }}\,\,\,{{\left[ {{(1+5{{x}^{2}})}^{1/5{{x}^{2}}}} \right]}^{5}}}{\underset{x\to 0}{\mathop{\lim }}\,\,\,{{\left[ {{(1+3{{x}^{2}})}^{1/3{{x}^{2}}}} \right]}^{3}}}\\=\frac{{{e}^{5}}}{{{e}^{3}}}\\={{e}^{2}}$ $[\because \,\,\,\underset{x\to 0}{\mathop{\lim }}\,\,{{(1+x)}^{1/x}}=e]$
Question 9: Solve $\underset{x\to 0}{\mathop{\lim }}\,\frac{x\tan 2x-2x\tan x}{{{(1-\cos 2x)}^{2}}}$
Solution:
$\underset{x\to 0}{\mathop{\lim }}\,\,\frac{x\tan 2x-2x\tan x}{{{(1-\cos \,\,2x)}^{2}}}\\=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{x(\tan \,\,2x-2\tan x)}{{{(2\,{{\sin }^{2}}x)}^{2}}}\\=\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{1}{4}\,\frac{x\,(\tan 2x-2\tan x)}{{{\sin }^{4}}x}\\=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{4}\frac{x\left\{ \left( 2x+\frac{1}{3}{{(2x)}^{3}}+\frac{2}{15}\,{{(2x)}^{5}}+… \right)-2\left( x+\frac{{{x}^{3}}}{3}+\frac{2}{15}{{x}^{5}}+… \right) \right\}}{{{x}^{4}}\,{{\left( 1-\frac{{{x}^{2}}}{3\,\,!}+\frac{{{x}^{4}}}{5\,\,!}+…. \right)}^{4}}}\\=\frac{1}{4}\,.\,\left( \frac{8}{3}-\frac{2}{3} \right)\\=\frac{2}{4}\\=\frac{1}{2}.$
Question 10: The function $f(x)=\frac{\log (1+ax)-\log (1-bx)}{x}$ is not defined at x = 0. The value which should be assigned to f at x = 0 so that it is continuous at x = 0, is
Solution:
Since limit of a function is a + b as x → 0, therefore to be continuous at a function, its value must be a + b at x = 0
⇒ f (0) = a + b
Question 11: Evaluate $f(x)=\begin{array}{cc} \{ & \begin{array}{cc} \frac{{{x}^{3}}+{{x}^{2}}-16x+20}{{{(x-2)}^{2}}} & \text{if}\ x\ne 2\\ \\ \ k & \text{if}\ x=2 \end{array} \end{array}$
Solution:
For continuous $\underset{x\to 2}{\mathop{\lim }}\,\,f(x)=f(2)=k \\\Rightarrow \,\,k=\underset{x\to 2}{\mathop{\lim }}\,\frac{{{x}^{3}}+{{x}^{2}}-16x+20}{{{(x-2)}^{2}}} \\=\underset{x\to 2}{\mathop{\lim }}\,\,\frac{({{x}^{2}}-4x+4)\,\,(x+5)}{{{(x-2)}^{2}}}\\=7 .$
Question 12: $f(x)=\begin{array}{cc} \{ & \begin{array}{cc} \frac{{{x}^{2}}-4x+3}{{{x}^{2}}-1} & \text{for}\ x\ne 1 \\ \ 2 & \text{for }x=1 \end{array} \end{array}$, then find the condition for the function to be continuous or discontinuous.
Solution:
$f(x)=\left\{ \frac{{{x}^{2}}-4x+3}{{{x}^{2}}-1} \right\}$,
for x = 1
f (1) = 2, $f(1+)=\underset{x\to 1+}{\mathop{\lim }}\,\,\frac{{{x}^{2}}-4x+3}{{{x}^{2}}-1}\\=\underset{x\to 1+}{\mathop{\lim }}\,\,\frac{(x-3)}{(x+1)}\\=-1 \\f(1-)=\underset{x\to 1-}{\mathop{\lim }}\,\,\frac{{{x}^{2}}-4x+3}{{{x}^{2}}-1}\\=-1\,\,\\\Rightarrow \,\,f(1)\ne f(1-)$
Hence, the function is discontinuous at x = 1.
Question 13: Which of the following functions have a finite number of points of discontinuity in R ([.] represents the greatest integer function)?
A) tanx
B) x[x]
C) |x| / x
D) sin[πx]
Solution:
f (x) = tanx is discontinuous when x = (2n + 1) π / 2, n ∈ Z
f (x) = x[x] is discontinuous when x = k, k ∈ Z
f (x) = sin [nπx] is discontinuous when nπx = k, k ∈ Z
Thus, all the above functions have an infinite number of points of discontinuity. But, if (x) = |x| / x is discontinuous when x = 0 only.
Question 14: The number of values of x ∈ [0, 2] at which f (x) = ∣x − [1 / 2]∣ + |x − 1|+ tanx is not differentiable is
A) 0
B) 1
C) 3
D) None of these
Solution:
∣x − [1 / 2]∣ is continuous everywhere but not differentiable at x = 1 / 2, |x − 1| is continuous everywhere, but not differentiable at x = 1 and tan x is continuous in [0, 2] except at x = π / 2. Hence, f (x) is not differentiable at x = 1 / 2, 1, π / 2.
Question 15: $\underset{x\to \pi /2}{\mathop{\lim }}\,(\sec \theta -\tan \theta )=$
Solution:
$\underset{\theta \to \pi /2}{\mathop{\lim }}\,\,\,\frac{1-\sin \theta }{\cos \theta }=\underset{\theta \to \pi /2}{\mathop{\lim }}\,\,\,\frac{{{\left( \cos \frac{\theta }{2}-\sin \frac{\theta }{2} \right)}^{2}}}{\left( \cos \frac{\theta }{2}-\sin \frac{\theta }{2} \right)\,\left( \cos \frac{\theta }{2}+\sin \frac{\theta }{2} \right)}=0$ |
# How to Write a Quadratic Function from Its Vertex and Another Point
Using the vertex and another point on the quadratic function, its relation can be written. In this article, step-by-step, the method of writing the equation is explained.
## A step-by-step guide to Writing a Quadratic Function from Its Vertex and Another Point
The general form of the quadratic function is as follows: $$f(x)=ax^2+bx+c$$ where $$a$$, $$b$$, and $$c$$ are real numbers and $$a≠0$$.
The general form of the quadratic function can also be written as $$f(x)=a(x – h)^2+k$$, where $$(h, k)$$ is the coordinate of the vertex.
The graph of a quadratic function is a U-shaped curve called a parabola. An important feature of a graph is that it has a “vertex” point.
If $$a>0$$, the parabola opens upwards and the vertex indicates the lowest point of the graph or the minimum value of the quadratic function.
If $$a<0$$, the parabola opens downwards and the vertex indicates the highest point of the graph or the maximum value of the function. The graph is also symmetrical about the vertical line passing through the vertex.
## Writing a Quadratic Function from Its Vertex and Another Point– Example 1:
A quadratic function has vertex $$(0, 2)$$ and passes through $$(2, 3)$$. Write its equation in vertex form.
Solution:
By using the vertex form formula: $$y=a(x – h)^2+k$$
So, we have: $$(0, 2)$$$$y=a(x – 0)^2+2$$$$y=ax^2+2$$
Substitute $$(2, 3)$$ in the obtained equation, then:
$$3=a(2)^2+2$$$$1=4a$$$$a=\frac{1}{4}$$
Therefore, $$y=\frac{1}{4}x^2+2$$
## Writing a Quadratic Function from Its Vertex and Another Point– Example 2:
A quadratic function opening up or down has vertex $$(0, 0)$$ and passes through $$(4, 5)$$. Write its equation in vertex form.
Solution:
Use the vertex form of the quadratic function as $$y=a(x – h)^2+k$$. Put the coordinate of the vertex $$(0, 0)$$ in the vertex form:
$$(0, 0)$$$$y=a(x – 0)^2+0$$$$y=ax^2$$
To find $$a$$, substitute $$(4, 5)$$ in this equation and calculate. Then,
$$(4, 5)$$→$$5=a(4)^2$$→$$5=16a$$→$$a=\frac{5}{16}$$
Therefore, $$y=\frac{5}{16}x^2$$
## Exercises for Writing a Quadratic Function from Its Vertex and Another Point
1. A quadratic function has vertex $$(0, 4)$$ and passes through $$(3, 0)$$. Write its equation in vertex form.
2. A quadratic function opening up or down has vertex $$(0, 0)$$ and passes through $$(1, 7)$$. Write its equation in vertex form.
1. $$\color{blue}{y=-\frac{4}{9}x^2+4}$$
2. $$\color{blue}{y=7x^2}$$
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# What Is 4/13 as a Decimal + Solution With Free Steps
The fraction 4/13 as a decimal is equal to 0.307.
The process of division is one of the four basic mathematical operations. It is used to describe parts of a whole in real life. In mathematics, division can be represented in the form of fractions like p/q, where p represents the numerator and q the denominator. When we evaluate a fraction, we end up with a decimal value.
Here, we are interested more in the types of division that results in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 4/13.
## Solution
First, we convert the fraction components i.e., the numerator and the denominator, and transform them into the division constituents i.e., the Dividend and the Divisor respectively.
This can be seen done as follows:
Dividend = 4
Divisor = 13
Now, we introduce the most important quantity in our process of division, this is the Quotient. The value represents the Solution to our division, and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 4 $\div$ 13
This is when we go through the Long Division solution to our problem.
## 4/13 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 4, and 13 we can see how 4 is Smaller than 13, and to solve this division we require that 4 be Bigger than 13.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. And if it is then we calculate the Multiple of the divisor which is closest to the dividend and subtract it from the Dividend. This produces the Remainder which we then use as the dividend later.
Now, we begin solving for our dividend 4, which after getting multiplied by 10 becomes 40.
We take this 40 and divide it by 13, this can be seen done as follows:
40 $\div$ 13 $\approx$ 3
Where:
13 x 3 = 39
We add 3 to our quotient. This will lead to the generation of a Remainder equal to 40 – 39 = 1, now this means we have to repeat the process by Converting the 1 into 100 (since 10 is smaller than 13) and solving for that.
Note that 1 needs to be multiplied twice by 10 to become 100, so we add 0 to our quotient because of this. Now:
100 $\div$ 13 $\approx$ 7
Where:
13 x 7 = 91
This, therefore, produces another remainder which is equal to 100 – 91 = 9. We now have up to 3 decimal places, so we stop here with a Quotient equal to 0.307 and a final Remainder equal to 9.
Images/mathematical drawings are created with GeoGebra. |
## How to Do Operations with Polynomials
### How can we Define Polynomials?
To understand the proper definition of polynomials, we must break the word into two “poly” and “nomials”. This means a variety of expressions. To be extremely precise, a polynomial is an algebraic expression which contains various terms separated by certain mathematical operations. These terms can be pure variables, pure constants, or even mixed terms such as variables with coefficients. Now, a crucial point to determining a polynomial is that the variable in the expression (suppose $$x$$), should not have a fractional power.
For example, $$2x^{\frac{1}{3}} \ + \ 3x \ + \ 1$$ is not a polynomial as the first term has a fraction power of $$x$$.
### Types of Polynomials
Based on the number of terms, there are 3 different types of polynomials. They are monomials, binomials and trinomials. Let’s discuss more about them:
• Monomial: They consist of just a single term but there is one condition. The term cannot be zero. Ex: $$6x^2, \ 7x^3, 67x$$, etc.
• Binomial: Binomials consist of just two terms and they must have a mathematical operation in between. Ex: $$6x^2 \ + \ 7x^3, \ 8x^2 \ + \ 7y, \ xy^2 \ + \ 3x$$, etc.
• Trinomial: Trinomial consist of three terms and between every two terms there is a mathematical operation. Ex: $$2x^2 \ + \ 3x \ + \ 10, \ 4x^3 \ + \ 3x^2 \ + \ 7, \ 3x^2y \ + \ 4xy \ + \ 6$$, etc.
### What are Like and Unlike Terms?
In algebra, we have the concept of like and unlike terms. So, if in an expression, you find that two or more same variables have the same exponential powers, then they are considered like terms. So, what we can do to simplify the expression is club those like terms by performing basic mathematical operations between them.
Also, in the case of unlike terms, you will find that the exponential powers of two or more same variables are not the same. So, we have no other option other than to leave them as they are.
Some examples of like terms are $$2x^3 \ - \ x^3, \ 7x^2 \ + \ 3x^2$$. Here we can see that the power of $$x$$ is the same in both the examples. Also, some examples of unlike term expressions would be $$7x^2 \ - \ 9x^3, \ -x^3 \ + \ x^2$$.
### Operations with Polynomials
When operating with polynomial expression, apply the below-mentioned steps:
• First, simplify the expression by adding/subtracting the like terms. Also. Don’t forget to apply the PEMDAS rule in case of brackets or parenthesis.
• Also, wherever possible, use the distributive property.
Ex: $$4x^3 \ + \ 3x^3 \ + \ 2x^2 \ - \ x^2 \ + \ 9 \ = \ 7x^3 \ + \ x^2 \ + \ 9$$.
## Operations with Polynomials Practice Quiz
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# Estimate $5290+17986$?
Last updated date: 20th Jun 2024
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Hint: From the question given we have to estimate the $5290+17986$. Here we have to understand that estimation means sum of the given numbers.so the given problem can be done by doing addition of the given numbers.
Addition means sum or adding numbers, the sign of the addition is $+$ (plus).
First, we have to write the larger number from the given numbers
Here the larger number is $17986$
And then we have to write the smaller number down to the larger number in a particular way that the $0$ should be written under $6$ and $9$ should be written under $8$ and $2$ should be written under $9$ and $5$ should be written under $7$ and under $1$ there will be no number but we can write $0$ .
\begin{align} & \Rightarrow 17986 \\ & +\,\,\ 05290 \\ \end{align}
Now add the numbers from left side onwards,
First add $6$ to the $0$ we will get $6$
\begin{align} & \Rightarrow 17986 \\ & +\,\,\ 05290 \\ \end{align}
$6$
Now add $8$to the $9$ we will get $17$, now $7$ will be written down and $1$ will be carry to the next that is above the $9$
\begin{align} & \quad \quad \ 1 \\ & \Rightarrow 17986 \\ & +\,\,\ 05290 \\ \end{align}
$76$
Now add $9$to the $2$ we will get$11$and now add the $1$ which is carried then we will get $12$ ,now $2$ will be written down and $1$ will be carry to the next that is above the $7$
\begin{align} & \quad \ \,\ 1 \\ & \Rightarrow 17986 \\ & +\,\,\ 05290 \\ \end{align}
$276$
Now add $7$to the $5$ we will get$12$and now add the $\ 1$ which is carried then we will get $13$ ,now $3$ will be written down and $1$ will be carry to the next that is above the $1$
\begin{align} & \quad \,1 \\ & \Rightarrow 17986 \\ & +\,\,\ 05290 \\ \end{align}
$3276$
Now add $1$to the $0$ we will get$1$and now add the $1$ which is carried then we will get $2$, now $2$ will be written down
\begin{align} & \Rightarrow 17986 \\ & +\,\,\ 05290 \\ \end{align}
$23276$
Therefore, the estimation is $23276$
Note: Students should be well aware of the basic summation that is addition. Students should be very careful while doing the calculation part for the given question. We should not add the ones place to hundredths place for example in the first step we should not add 0 and 8 we should add 0 and 6. Students must be very careful in this part or the whole solution will be wrong. |
Simplify 3a to the fifth
over 9a squared times a to the fourth over
a to the third. So before we even
worry about the a’s, we can actually simplify
the 3 and the 9. They’re both divisible by 3. So let’s divide the numerator
and the denominator here by 3. So if we divide the numerator
by 3, the 3 becomes a 1. If we divide the denominator
by 3, the 9 becomes a 3. So this reduces
to, or simplifies to 1a to the fifth
times a to the fourth over– or maybe I should
say, a to the fifth over 3a squared times a to the
fourth over a to the third. Now this, if we just
multiply the two expressions, this would be equal to 1a to
the fifth times a to the fourth in the numerator, and we don’t
have to worry about the one, it doesn’t change the value. So it’s a to the fifth times a
to the fourth in the numerator. And then we have 3a– let
me write the 3 like this– and then we have 3
times a squared times a to the third in
the denominator. And now there’s
multiple ways that we can simplify this from here. One sometimes is called
the quotient rule. And that’s just the idea that
if you have a to the x over a to the y, that this is going to
be equal to a to the x minus y. And just to understand
why that works, let’s think about a to
the fifth over a squared. So a to the fifth
is literally a times a times a times a times a. That right there
is a to the fifth. And we have that over a squared. And I’m just thinking about
the a squared right over here, which is literally
just a times a. That is a squared. Now, clearly, both the
numerator and denominator are both divisible by a times a. We can divide them
both by a times a. So we can get rid of– if
we divide the numerator by a twice, by a times a, so
let’s get rid of a times a. And if we divide the denominator
by a times a, we just get a 1. So what are we just left with? We are left with just a
times a times a over 1, which is just a times a times a. But what is this? This is a to the third power,
or a to the 5 minus 2 power. We had 5, we were able to
cancel out 2, that gave us 3. So we could do the
same thing over here. We can apply the quotient rule. And I’ll do two ways
of actually doing this. So let’s apply the
quotient rule with the a to the fifth and the a squared. So let me do it this way. So let’s apply with
these two guys, and then let’s apply
it with these two guys. And of course, we have
the 1/3 out front. So this can be
reduced to 1/3 times– if we apply the
quotient rule with a to the fifth over
a squared, we just did it over here– that
becomes a to the third power. And if we apply it over here
with the a to the fourth over a to the third, that’ll
give us a– let me do it that same blue color. That’ll give us a– that’s
not the same blue color. There we go. This will give us a to
the 4 minus 3 power, or a to the first power. And of course, we can
simplify this as a to the third times
a– well, actually, let me just do it over here. Before I even rewrite it, we
know that a to the third times a to the first is going to
be a to the 3 plus 1 power. We have the same base,
we can add the exponents. We’re multiplying a times
itself three times and then one more time. So that’ll be a to
the fourth power. So this right over here
becomes a to the fourth power. a to the 3 plus 1 power. And then we have to
multiply that by 1/3. So our answer could be
1/3 a to the fourth, or we could equally write
it a to the fourth over 3. Now, the other way
to do this problem would have been to
apply the product, or to add the exponents
in the numerator, and then add the exponents
in the denominator. So let’s do it that way first. If we add the exponents
in the numerator first, we don’t apply the
quotient rule first. We apply it second. We get in the numerator, a to
the fifth times a to the fourth would be a to the ninth power. 5 plus 4. And then in the denominator
we have a squared times a to the third. Add the exponents, because
we’re taking the product with the same base. So it’ll be a to
the fifth power. And of course, we still
have this 3 down here. We have a 1/3, or we could
just write a 3 over here. Now, we could apply the
quotient property of exponents. We could say, look, we have a to
the ninth over a to the fifth. a to the ninth
over a to the fifth is equal to a to
the 9 minus 5 power, or it’s equal to a
to the fourth power. And of course, we still
have the divided by 3. Either way we got
the same answer. |
# How do you find the derivative of y= sqrt((x-1)/(x+1)) ?
Aug 7, 2014
$y ' = \frac{1}{\sqrt{x - 1} {\left(x + 1\right)}^{\frac{3}{2}}}$
Explanation
y=sqrt((x-1)/(x+1)
taking natural log of both sides,
$\ln y = \frac{1}{2} \ln \left(\frac{x - 1}{x + 1}\right)$,
$\ln y = \frac{1}{2} \left(\ln \left(x - 1\right) - \ln \left(x + 1\right)\right)$
differentiating both sides with respect to $x$,
$\frac{1}{y} \cdot y ' = \frac{1}{2} \left(\frac{1}{x - 1} - \frac{1}{x + 1}\right)$
$y ' = \frac{y}{2} \left(\frac{x + 1 - x + 1}{{x}^{2} - 1}\right)$
$y ' = y \left(\frac{1}{{x}^{2} - 1}\right)$
plugging $y$
$y ' = \left(\sqrt{\frac{x - 1}{x + 1}}\right) \left(\frac{1}{{x}^{2} - 1}\right)$
$y ' = \left(\sqrt{\frac{x - 1}{x + 1}}\right) \left(\frac{1}{\left(x - 1\right) \left(x + 1\right)}\right)$
$y ' = \frac{1}{\sqrt{x - 1} {\left(x + 1\right)}^{\frac{3}{2}}}$ |
JEE Exam » JEE Study Material » Mathematics » Adjoint of Matrix
# Adjoint of Matrix
Adjoint of Matrix is one of the most powerful tools in arithmetic. It is also called an adjugate matrix and is utilised in areas of business and science.
A simple technique used to find out a matrix inverse can be defined as the adjoint of a matrix. A matrix is an ordered rectangular array of numbers or functions in algebra. Matrices may be classified according to the number of rows and columns within which the items are placed.
An adjoint matrix can also be referred to as an adjugate matrix. It is utilised in various areas of life and business such as finance, sales, science etc. It shows its vital role in alternative branches like genetic science, economics, social science, and technology.
## What is a Matrix?
A matrix or a plural matrix is a rectangular shaped array consisting of numbers, symbols, or expressions that are organised in rows and columns.
Usually, a matrix is shown as a capital letter in bold fonts, such as A, B, X etc. The elements or the items of the matrix are shown as lowercase letters with a double subscript (e.g., aij, bij, xij).
## The Formula of Adjoint of a Matrix
With the help of the cofactor and transpose of a matrix, we can derive the formula for the adjoint of a matrix. Below are the formulas and steps involved in determining the adjoint matrix for a given matrix.
Adjoint of a Matrix 2 x 2
Let A be the 2 x 2 matrix and is given by:
A=[a14 a15
a24 a25]
Then, the adjoint of this matrix is:
adj A = [A14 A24
A15 A25]
Here,
A14 = Cofactor of a14
A15 = Cofactor of a15
A24 = Cofactor of a24
Alternatively, the adj A can also be calculated by interchanging a14 and a25 and by changing signs of a15 and a24.
## Properties of the Adjoint of a Matrix
Following are a few important and useful properties of adjoint of a matrix.
1. A.adj(A) = adj.(A).A =|A|I
Here, A is a square matrix, I is an identity matrix and |A| is the determination of matrix A
1. Determination of adjoint A is = determination of A power n – 1, where A is invertible n *n square matrix.
|adjA| = |A| n-1
1. adj(adjA) = |A| n-2-.A , where A is n *n invertible square matrix.
## Use of Adjoint of a Matrix
A few uses or functions of the adjoint of a matrix are mentioned below:
• The adjoint of a matrix helps in solving the system of linear equations. It helps us know whether the answer to equations is consistent or inconsistent.
• Programmers use matrices and their inverse matrices to code or code letters. Matrices are often used to encrypt message codes. A message includes a series of binary numbers that are resolved using coding theory for communication. As a result, the concept of matrices is employed to resolve such equations.
• Engineers and physicists develop models of physical structures and execute the precise calculations needed to operate difficult machinery. Fine-tuned matrix transformation computations are employed in physics, networks, aeroplanes and spacecraft, and chemical processes.
• The matrices are essential when applying Kirchhoff’s laws of voltage and current to resolve issues.
• Algebra is employed to explore electrical circuits, quantum physics, and optics. These matrices are crucial in measuring battery power outputs and changing electricity into alternative usable energy by resistors.
• Many IT organisations use matrix information structures to trace user data, run search queries, and maintain databases. In data security, several frameworks are designed to work with matrices. Matrices are utilised in electronic information compression, like handling biometric data in Mauritius’ new identity card.
### Conclusion
The transpose of a compound matrix of the square matrix is named the adjoint of the matrix. Adjoint of the matrix A is denoted by adj A, and can additionally be referred to as adjugate matrix or adjunct matrix. The adjoint of a matrix is generated by getting the transpose of the matrix’s cofactor members. It is one of the best ways to calculate a matrix’s inverse and the most powerful tool in arithmetic.
We have learnt from the adjoint of a matrix, its properties, and examples that most vital issues can be solved with matrices. We have linear equations and different mathematical functions like calculus, optics, and physics that are all done with these instruments.
It has a good variety of applications within the world that have semiconductor diodes, thereby enjoying an important role in arithmetic.
## Frequently asked questions
Get answers to the most common queries related to the JEE Examination Preparation.
## How does one find the adjoint of a matrix?
Ans: To find out the adjoint of a matrix, first, we should verify the cofactor of every element, followed by two more stages. The ...Read full
## What is the Minor?
Ans: The determinant obtained by deleting the row and column in which that element lies is termed as The Minor of an element in a ...Read full
## . What is Cofactor?
Ans: A number obtained by eliminating the row and column of a specific element within the form of a square or rectangle can be ter...Read full
## What is the difference between the adjoint and transpose of a matrix?
Ans: Transposing a matrix merely means flipping the rows and columns, whereas the transpose of the matrix of cofactors is the adjo...Read full
## How to find the inverse of a 3×3 matrix?
Ans: For finding the inverse of a 3×3 matrix, first, calculate the determinant of the matrix, and if the determinant is zero, the...Read full |
# Relations: Best Answers 007EH
Welcome to Lesson Four – Relations, In this lesson, we shall understand what is discrete maths, why do we need it and how do we apply it in our day-to-day activities.
You can also join the Discrete Mathematics Forum to interact and share ideas, also you can navigate to any session using the Table of Content
Don’t forget to leave your comment, suggestions regarding the previous, and current lecture to help us improve.
# Introduction to Binary Relation
Let P and Q be two non-empty sets. A binary relation R is defined to be a subset of P x Q from a set P to Q. If (a, b) ∈ R and R ⊆ P x Q then a is related to b by R i.e., aRb. If sets P and Q are equal, then we say R ⊆ P x P is a relation on P e.g.
1. (i) Let A = {a, b, c}
2. B = {r, s, t}
3. Then R = {(a, r), (b, r), (b, t), (c, s)}
4. is a relation from A to B.
5.
6. (ii) Let A = {1, 2, 3} and B = A
7. R = {(1, 1), (2, 2), (3, 3)}
8. is a relation (equal) on A.
Example1: If a set has n elements, how many relations are there from A to A.
Solution: If a set A has n elements, A x A has n2 elements. So, there are 2n2 relations from A to A.
Example2: If A has m elements and B has n elements. How many relations are there from A to B and vice versa?
Solution: There are m x n elements; hence there are 2m x n relations from A to A.
Example3: If a set A = {1, 2}. Determine all relations from A to A.
Solution: There are 22= 4 elements i.e., {(1, 2), (2, 1), (1, 1), (2, 2)} in A x A. So, there are 24= 16 relations from A to A. i.e.
1. {(1, 2), (2, 1), (1, 1), (2, 2)}, {(1, 2), (2, 1)}, {(1, 2), (1, 1)}, {(1, 2), (2, 2)},
2. {(2, 1), (1, 1)},{(2,1), (2, 2)}, {(1, 1),(2, 2)},{(1, 2), (2, 1), (1, 1)}, {(1, 2), (1, 1),
3. (2, 2)}, {(2,1), (1, 1), (2, 2)}, {(1, 2), (2, 1), (2, 2)}, {(1, 2), (2, 1), (1, 1), (2, 2)} and ∅.
## Domain and Range of Relation
Domain of Relation: The Domain of relation R is the set of elements in P which are related to some elements in Q, or it is the set of all first entries of the ordered pairs in R. It is denoted by DOM (R).
Range of Relation: The range of relation R is the set of elements in Q which are related to some element in P, or it is the set of all second entries of the ordered pairs in R. It is denoted by RAN (R).
Example:
1. Let A = {1, 2, 3, 4}
2. B = {a, b, c, d}
3. R = {(1, a), (1, b), (1, c), (2, b), (2, c), (2, d)}.
Solution:
```DOM (R) = {1, 2}
RAN (R) = {a, b, c, d}
```
## Complement of a Relation
Consider a relation R from a set A to set B. The complement of relation R denoted by R is a relation from A to B such that
``` R = {(a, b): {a, b) ∉ R}.
```
Example:
1. Consider the relation R from X to Y
2. X = {1, 2, 3}
3. Y = {8, 9}
4. R = {(1, 8) (2, 8) (1, 9) (3, 9)}
5. Find the complement relation of R.
Solution:
```X x Y = {(1, 8), (2, 8), (3, 8), (1, 9), (2, 9), (3, 9)}
Now we find the complement relation R from X x Y
R = {(3, 8), (2, 9)}
```
# Representation of Relations
Relations can be represented in many ways. Some of which are as follows:
1. Relation as a Matrix: Let P = [a1,a2,a3,…….am] and Q = [b1,b2,b3……bn] are finite sets, containing m and n number of elements respectively. R is a relation from P to Q. The relation R can be represented by m x n matrix M = [Mij], defined as
```Mij = 0 if (ai,bj) ∉ R
1 if (ai,bj )∈ R
Example ```
1. Let P = {1, 2, 3, 4}, Q = {a, b, c, d}
2. and R = {(1, a), (1, b), (1, c), (2, b), (2, c), (2, d)}.
The matrix of relation R is shown as fig:
2. Relation as a Directed Graph: There is another way of picturing a relation R when R is a relation from a finite set to itself.
Example
1. A = {1, 2, 3, 4}
2. R = {(1, 2) (2, 2) (2, 4) (3, 2) (3, 4) (4, 1) (4, 3)}
3. Relation as an Arrow Diagram: If P and Q are finite sets and R is a relation from P to Q. Relation R can be represented as an arrow diagram as follows.
Draw two ellipses for the sets P and Q. Write down the elements of P and elements of Q column-wise in three ellipses. Then draw an arrow from the first ellipse to the second ellipse if a is related to b and a ∈ P and b ∈ Q.
Example
1. Let P = {1, 2, 3, 4}
2. Q = {a, b, c, d}
3. R = {(1, a), (2, a), (3, a), (1, b), (4, b), (4, c), (4, d)
The arrow diagram of relation R is shown in fig:
4. Relation as a Table: If P and Q are finite sets and R is a relation from P to Q. Relation R can be represented in tabular form.
Make the table which contains rows equivalent to an element of P and columns equivalent to the element of Q. Then place a cross (X) in the boxes which represent relations of elements on set P to set Q.
Example
1. Let P = {1, 2, 3, 4}
2. Q = {x, y, z, k}
3. R = {(1, x), (1, y), (2, z), (3, z), (4, k)}.
The tabular form of relation as shown in fig:
# Composition of Relations
Let A, B, and C be sets, and let R be a relation from A to B and let S be a relation from B to C. That is, R is a subset of A × B and S is a subset of B × C. Then R and S give rise to a relation from A to C indicated by R◦S and defined by:
1. a (R◦S)c if for some b ∈ B we have aRb and bSc.
2. is,
3. R ◦ S = {(a, c)| there exists b ∈ B for which (a, b) ∈ R and (b, c) ∈ S}
The relation R◦S is known the composition of R and S; it is sometimes denoted simply by RS.
Let R is a relation on a set A, that is, R is a relation from a set A to itself. Then R◦R, the composition of R with itself, is always represented. Also, R◦R is sometimes denoted by R2. Similarly, R3 = R2◦R = R◦R◦R, and so on. Thus Rn is defined for all positive n.
Example1: Let X = {4, 5, 6}, Y = {a, b, c} and Z = {l, m, n}. Consider the relation R1 from X to Y and R2 from Y to Z.
``` R1 = {(4, a), (4, b), (5, c), (6, a), (6, c)}
R2 = {(a, l), (a, n), (b, l), (b, m), (c, l), (c, m), (c, n)}
```
Find the composition of relation (i) R1 o R2 (ii) R1o R1-1
Solution:
(i) The composition relation R1 o R2 as shown in fig:
R1 o R2 = {(4, l), (4, n), (4, m), (5, l), (5, m), (5, n), (6, l), (6, m), (6, n)}
(ii) The composition relation R1o R1-1 as shown in fig:
R1o R1-1 = {(4, 4), (5, 5), (5, 6), (6, 4), (6, 5), (4, 6), (6, 6)}
## Composition of Relations and Matrices
There is another way of finding R◦S. Let MR and MS denote respectively the matrix representations of the relations R and S. Then
Example
1. Let P = {2, 3, 4, 5}. Consider the relation R and S on P defined by
2. R = {(2, 2), (2, 3), (2, 4), (2, 5), (3, 4), (3, 5), (4, 5), (5, 3)}
3. S = {(2, 3), (2, 5), (3, 4), (3, 5), (4, 2), (4, 3), (4, 5), (5, 2), (5, 5)}.
4.
5. Find the matrices of the above relations.
6. Use matrices to find the following composition of the relation R and S.
7. (i)RoS (ii)RoR (iii)SoR
Solution: The matrices of the relation R and S are a shown in fig:
(i) To obtain the composition of relation R and S. First multiply MR with MS to obtain the matrix MR x MS as shown in fig:
The non zero entries in the matrix MR x MS tells the elements related in RoS. So,
Hence the composition R o S of the relation R and S is
1. R o S = {(2, 2), (2, 3), (2, 4), (3, 2), (3, 3), (4, 2), (4, 5), (5, 2), (5, 3), (5, 4), (5, 5)}.
(ii) First, multiply the matrix MR by itself, as shown in fig
Hence the composition R o R of the relation R and S is
1. R o R = {(2, 2), (3, 2), (3, 3), (3, 4), (4, 2), (4, 5), (5, 2), (5, 3), (5, 5)}
(iii) Multiply the matrix MS with MR to obtain the matrix MS x MR as shown in fig:
The non-zero entries in matrix MS x MR tells the elements related in S o R.
Hence the composition S o R of the relation S and R is
1. S o R = {(2, 4) , (2, 5), (3, 3), (3, 4), (3, 5), (4, 2), (4, 4), (4, 5), (5, 2), (5, 3), (5, 4), (5, 5)}.
# Types of Relations
1. Reflexive Relation: A relation R on set A is said to be a reflexive if (a, a) ∈ R for every a ∈ A.
Example: If A = {1, 2, 3, 4} then R = {(1, 1) (2, 2), (1, 3), (2, 4), (3, 3), (3, 4), (4, 4)}. Is a relation reflexive?
Solution: The relation is reflexive as for every a ∈ A. (a, a) ∈ R, i.e. (1, 1), (2, 2), (3, 3), (4, 4) ∈ R.
2. Irreflexive Relation: A relation R on set A is said to be irreflexive if (a, a) ∉ R for every a ∈ A.
Example: Let A = {1, 2, 3} and R = {(1, 2), (2, 2), (3, 1), (1, 3)}. Is the relation R reflexive or irreflexive?
Solution: The relation R is not reflexive as for every a ∈ A, (a, a) ∉ R, i.e., (1, 1) and (3, 3) ∉ R. The relation R is not irreflexive as (a, a) ∉ R, for some a ∈ A, i.e., (2, 2) ∈ R.
3. Symmetric Relation: A relation R on set A is said to be symmetric iff (a, b) ∈ R ⟺ (b, a) ∈ R.
Example: Let A = {1, 2, 3} and R = {(1, 1), (2, 2), (1, 2), (2, 1), (2, 3), (3, 2)}. Is a relation R symmetric or not?
Solution: The relation is symmetric as for every (a, b) ∈ R, we have (b, a) ∈ R, i.e., (1, 2), (2, 1), (2, 3), (3, 2) ∈ R but not reflexive because (3, 3) ∉ R.
Example of Symmetric Relation:
1. Relation ⊥r is symmetric since a line a is ⊥r to b, then b is ⊥r to a.
2. Also, Parallel is symmetric, since if a line a is ∥ to b then b is also ∥ to a.
Antisymmetric Relation: A relation R on a set A is antisymmetric iff (a, b) ∈ R and (b, a) ∈ R then a = b.
Example1: Let A = {1, 2, 3} and R = {(1, 1), (2, 2)}. Is the relation R antisymmetric?
Solution: The relation R is antisymmetric as a = b when (a, b) and (b, a) both belong to R.
Example2: Let A = {4, 5, 6} and R = {(4, 4), (4, 5), (5, 4), (5, 6), (4, 6)}. Is the relation R antisymmetric?
Solution: The relation R is not antisymmetric as 4 ≠ 5 but (4, 5) and (5, 4) both belong to R.
5. Asymmetric Relation: A relation R on a set A is called an Asymmetric Relation if for every (a, b) ∈ R implies that (b, a) does not belong to R.
6. Transitive Relations: A Relation R on set A is said to be transitive iff (a, b) ∈ R and (b, c) ∈ R ⟺ (a, c) ∈ R.
Example1: Let A = {1, 2, 3} and R = {(1, 2), (2, 1), (1, 1), (2, 2)}. Is the relation transitive?
Solution: The relation R is transitive as for every (a, b) (b, c) belong to R, we have (a, c) ∈ R i.e, (1, 2) (2, 1) ∈ R ⇒ (1, 1) ∈ R.
#### ⊥r is not transitive since a ⊥r b, b ⊥r c then it is not true that a ⊥r c. Since no line is ∥ to itself, we can have a ∥ b, b ∥ a but a ∦ a.Thus ∥ is not transitive, but it will be transitive in the plane.
7. Identity Relation: Identity relation I on set A is reflexive, transitive and symmetric. So identity relation I is an Equivalence Relation.
Example: A= {1, 2, 3} = {(1, 1), (2, 2), (3, 3)}
8. Void Relation: It is given by R: A →B such that R = ∅ (⊆ A x B) is a null relation. Void Relation R = ∅ is symmetric and transitive but not reflexive.
9. Universal Relation: A relation R: A →B such that R = A x B (⊆ A x B) is a universal relation. Universal Relation from A →B is reflexive, symmetric and transitive. So this is an equivalence relation.
# Closure Properties of Relations
Consider a given set A, and the collection of all relations on A. Let P be a property of such relations, such as being symmetric or being transitive. A relation with property P will be called a P-relation. The P-closure of an arbitrary relation R on A, indicated P (R), is a P-relation such that
1. R ⊆ P (R) ⊆ S
(1) Reflexive and Symmetric Closures: The next theorem tells us how to obtain the reflexive and symmetric closures of a relation easily.
Theorem: Let R be a relation on a set A. Then:
• R ∪ ∆A is the reflexive closure of R
• R ∪ R-1 is the symmetric closure of R.
Example1:
1. Let A = {k, l, m}. Let R is a relation on A defined by
2. R = {(k, k), (k, l), (l, m), (m, k)}.
Find the reflexive closure of R.
Solution: R ∪ ∆ is the smallest relation having reflexive property, Hence,
```RF = R ∪ ∆ = {(k, k), (k, l), (l, l), (l, m), (m, m), (m, k)}.
```
Example2: Consider the relation R on A = {4, 5, 6, 7} defined by
1. R = {(4, 5), (5, 5), (5, 6), (6, 7), (7, 4), (7, 7)}
Find the symmetric closure of R.
Solution: The smallest relation containing R having the symmetric property is R ∪ R-1,i.e.
```RS = R ∪ R-1 = {(4, 5), (5, 4), (5, 5), (5, 6), (6, 5), (6, 7), (7, 6), (7, 4), (4, 7), (7, 7)}.
```
(2)Transitive Closures: Consider a relation R on a set A. The transitive closure R of a relation R of a relation R is the smallest transitive relation containing R.
Recall that R2 = R◦ R and Rn = Rn-1 ◦ R. We define
The following Theorem applies:
Theorem1: R* is the transitive closure of R
Suppose A is a finite set with n elements.
```R* = R ∪R2 ∪.....∪ Rn
```
Theorem 2: Let R be a relation on a set A with n elements. Then
```Transitive (R) = R ∪ R2∪.....∪ Rn
```
Example1: Consider the relation R = {(1, 2), (2, 3), (3, 3)} on A = {1, 2, 3}. Then
R2 = R◦ R = {(1, 3), (2, 3), (3, 3)} and R3 = R2 ◦ R = {(1, 3), (2, 3), (3, 3)} Accordingly,
Transitive (R) = {(1, 2), (2, 3), (3, 3), (1, 3)}
Example2: Let A = {4, 6, 8, 10} and R = {(4, 4), (4, 10), (6, 6), (6, 8), (8, 10)} is a relation on set A. Determine transitive closure of R.
Solution: The matrix of relation R is shown in fig:
Now, find the powers of MR as in fig:
Hence, the transitive closure of MR is MR* as shown in Fig (where MR* is the ORing of a power of MR).
Thus, R* = {(4, 4), (4, 10), (6, 8), (6, 6), (6, 10), (8, 10)}.
# Equivalence Relations
A relation R on a set A is called an equivalence relation if it satisfies following three properties:
1. Relation R is Reflexive, i.e. aRa ∀ a∈A.
2. Relation R is Symmetric, i.e., aRb ⟹ bRa
3. Relation R is transitive, i.e., aRb and bRc ⟹ aRc.
Example: Let A = {1, 2, 3, 4} and R = {(1, 1), (1, 3), (2, 2), (2, 4), (3, 1), (3, 3), (4, 2), (4, 4)}.
Show that R is an Equivalence Relation.
Solution:
Reflexive: Relation R is reflexive as (1, 1), (2, 2), (3, 3) and (4, 4) ∈ R.
Symmetric: Relation R is symmetric because whenever (a, b) ∈ R, (b, a) also belongs to R.
Example: (2, 4) ∈ R ⟹ (4, 2) ∈ R.
Transitive: Relation R is transitive because whenever (a, b) and (b, c) belongs to R, (a, c) also belongs to R.
Example: (3, 1) ∈ R and (1, 3) ∈ R ⟹ (3, 3) ∈ R.
So, as R is reflexive, symmetric and transitive, hence, R is an Equivalence Relation.
#### Note1: If R1and R2 are equivalence relation then R1∩ R2 is also an equivalence relation.
Example: A = {1, 2, 3}
R1 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
R2 = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)}
R1∩ R2 = {(1, 1), (2, 2), (3, 3)}
#### Note2: If R1and R2 are equivalence relation then R1∪ R2 may or may not be an equivalence relation.
Example: A = {1, 2, 3}
R1 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
R2 = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)}
R1∪ R2= {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (2, 3), (3, 2)}
Hence, Reflexive or Symmetric are Equivalence Relation but transitive may or may not be an equivalence relation.
## Inverse Relation
Let R be any relation from set A to set B. The inverse of R denoted by R-1 is the relations from B to A which consist of those ordered pairs which when reversed belong to R that is:
```R-1 = {(b, a): (a, b) ∈ R}
```
Example1: A = {1, 2, 3}
B = {x, y, z}
Solution: R = {(1, y), (1, z), (3, y)
R-1 = {(y, 1), (z, 1), (y, 3)}
Clearly (R-1)-1 = R
#### Note1: Domain and Range of R-1 is equal to range and domain of R.
Example2: R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (3, 2)}
R-1 = {(1, 1), (2, 2), (3, 3), (2, 1), (3, 2), (2, 3)}
#### Note2: If R is an Equivalence Relation then R-1 is always an Equivalence Relation.
Example: Let A = {1, 2, 3}
R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
R-1 = {(1, 1), (2, 2), (3, 3), (2, 1), (1, 2)}
R-1 is a Equivalence Relation.
#### Note3: If R is a Symmetric Relation then R-1=R and vice-versa.
Example: Let A = {1, 2, 3}
R = {(1, 1), (2, 2), (1, 2), (2, 1), (2, 3), (3, 2)}
R-1 = {(1, 1), (2, 2), (2, 1), (1, 2), (3, 2), (2, 3)}
# Partial Order Relations
A relation R on a set A is called a partial order relation if it satisfies the following three properties:
1. Relation R is Reflexive, i.e. aRa ∀ a∈A.
2. Relation R is Antisymmetric, i.e., aRb and bRa ⟹ a = b.
3. Relation R is transitive, i.e., aRb and bRc ⟹ aRc.
Example1: Show whether the relation (x, y) ∈ R, if, x ≥ y defined on the set of +ve integers is a partial order relation.
Solution: Consider the set A = {1, 2, 3, 4} containing four +ve integers. Find the relation for this set such as R = {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3), (1, 1), (2, 2), (3, 3), (4, 4)}.
Reflexive: The relation is reflexive as for every a ∈ A. (a, a) ∈ R, i.e. (1, 1), (2, 2), (3, 3), (4, 4) ∈ R.
Antisymmetric: The relation is antisymmetric as whenever (a, b) and (b, a) ∈ R, we have a = b.
Transitive: The relation is transitive as whenever (a, b) and (b, c) ∈ R, we have (a, c) ∈ R.
Example: (4, 2) ∈ R and (2, 1) ∈ R, implies (4, 1) ∈ R.
As the relation is reflexive, antisymmetric and transitive. Hence, it is a partial order relation.
Example2: Show that the relation ‘Divides’ defined on N is a partial order relation.
Solution:
Reflexive: We have a divides a, ∀ a∈N. Therefore, relation ‘Divides’ is reflexive.
Antisymmetric: Let a, b, c ∈N, such that a divides b. It implies b divides a iff a = b. So, the relation is antisymmetric.
Transitive: Let a, b, c ∈N, such that a divides b and b divides c.
Then a divides c. Hence the relation is transitive. Thus, the relation being reflexive, antisymmetric and transitive, the relation ‘divides’ is a partial order relation.
Example3: (a) The relation ⊆ of a set of inclusion is a partial ordering or any collection of sets since set inclusion has three desired properties:
1. A ⊆ A for any set A.
2. If A ⊆ B and B ⊆ A then B = A.
3. If A ⊆ B and B ⊆ C then A ⊆ C
(b) The relation ≤ on the set R of real no that is Reflexive, Antisymmetric and transitive.
(c) Relation ≤ is a Partial Order Relation.
## n-Ary Relations
By an n-ary relation, we mean a set of ordered n-tuples. For any set S, a subset of the product set Sn is called an n-ary relation on S. In particular, a subset of S3 is called a ternary relation on S.
## Partial Order Set (POSET):
The set A together with a partial order relation R on the set A and is denoted by (A, R) is called a partial orders set or POSET.
## Total Order Relation
Consider the relation R on the set A. If it is also called the case that for all, a, b ∈ A, we have either (a, b) ∈ R or (b, a) ∈ R or a = b, then the relation R is known total order relation on set A.
Example: Show that the relation ‘<‘ (less than) defined on N, the set of +ve integers is neither an equivalence relation nor partially ordered relation but is a total order relation.
Solution:
Reflexive: Let a ∈ N, then a < a
⟹ ‘<‘ is not reflexive.
As, the relation ‘<‘ (less than) is not reflexive, it is neither an equivalence relation nor the partial order relation.
But, as ∀ a, b ∈ N, we have either a < b or b < a or a = b. So, the relation is a total order relation.
## Equivalence Class
Consider, an equivalence relation R on a set A. The equivalence class of an element a ∈ A, is the set of elements of A to which element a is related. It is denoted by [a].
Example: Let R be an equivalence relations on the set A = {4, 5, 6, 7} defined by
R = {(4, 4), (5, 5), (6, 6), (7, 7), (4, 6), (6, 4)}.
Determine its equivalence classes.
Solution: The equivalence classes are as follows:
{4} = {6} = {4, 6}
{5} = {5}
{7} = {7}.
## Circular Relation
Consider a binary relation R on a set A. Relation R is called circular if (a, b) ∈ R and (b, c) ∈ R implies (c, a) ∈ R.
Example: Consider R is an equivalence relation. Show that R is reflexive and circular.
Solution: Reflexive: As, the relation, R is an equivalence relation. So, reflexivity is the property of an equivalence relation. Hence, R is reflexive.
Circular: Let (a, b) ∈ R and (b, c) ∈ R
⇒ (a, c) ∈ R (∵ R is transitive)
⇒ (c, a) ∈ R (∵ R is symmetric)
Thus, R is Circular.
## Compatible Relation
A binary relation R on a set A that is Reflexive and symmetric is called Compatible Relation.
Every Equivalence Relation is compatible, but every compatible relation need not be an equivalence.
Example: Set of a friend is compatible but may not be an equivalence relation.
Friend Friend
a → b, b → c but possible that a and c are not friends.
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Intermediate Algebra 2e
# 9.1Solve Quadratic Equations Using the Square Root Property
Intermediate Algebra 2e9.1 Solve Quadratic Equations Using the Square Root Property
## Learning Objectives
By the end of this section, you will be able to:
• Solve quadratic equations of the form $ax2=kax2=k$ using the Square Root Property
• Solve quadratic equations of the form $a(x–h)2=ka(x–h)2=k$ using the Square Root Property
## Be Prepared 9.1
Before you get started, take this readiness quiz.
Simplify: $128.128.$
If you missed this problem, review Example 8.13.
## Be Prepared 9.2
Simplify: $325325$.
If you missed this problem, review Example 8.50.
## Be Prepared 9.3
Factor: $9x2−12x+49x2−12x+4$.
If you missed this problem, review Example 6.23.
A quadratic equation is an equation of the form ax2 + bx + c = 0, where $a≠0a≠0$. Quadratic equations differ from linear equations by including a quadratic term with the variable raised to the second power of the form ax2. We use different methods to solve quadratic equations than linear equations, because just adding, subtracting, multiplying, and dividing terms will not isolate the variable.
We have seen that some quadratic equations can be solved by factoring. In this chapter, we will learn three other methods to use in case a quadratic equation cannot be factored.
## Solve Quadratic Equations of the form $ax2=kax2=k$ using the Square Root Property
We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation x2 = 9.
$x2=9x2=9$ Put the equation in standard form. $x2−9=0x2−9=0$ Factor the difference of squares. $(x−3)(x+3)=0(x−3)(x+3)=0$ Use the Zero Product Property. $x−3=0x−3=0x−3=0x−3=0$ Solve each equation. $x=3x=−3x=3x=−3$
We can easily use factoring to find the solutions of similar equations, like x2 = 16 and x2 = 25, because 16 and 25 are perfect squares. In each case, we would get two solutions, $x=4,x=−4x=4,x=−4$ and $x=5,x=−5.x=5,x=−5.$
But what happens when we have an equation like x2 = 7? Since 7 is not a perfect square, we cannot solve the equation by factoring.
Previously we learned that since 169 is the square of 13, we can also say that 13 is a square root of 169. Also, (−13)2 = 169, so −13 is also a square root of 169. Therefore, both 13 and −13 are square roots of 169. So, every positive number has two square roots—one positive and one negative. We earlier defined the square root of a number in this way:
$Ifn2=m,thennis a square root ofm.Ifn2=m,thennis a square root ofm.$
Since these equations are all of the form x2 = k, the square root definition tells us the solutions are the two square roots of k. This leads to the Square Root Property.
## Square Root Property
If x2 = k, then
$x=korx=−korx=±k.x=korx=−korx=±k.$
Notice that the Square Root Property gives two solutions to an equation of the form x2 = k, the principal square root of $kk$ and its opposite. We could also write the solution as $x=±k.x=±k.$ We read this as x equals positive or negative the square root of k.
Now we will solve the equation x2 = 9 again, this time using the Square Root Property.
$x2=9x2=9$ Use the Square Root Property. $x=±9x=±9$ $x=±3x=±3$ $Sox=3orx=−3.Sox=3orx=−3.$
What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation x2 = 7.
$x2=7x2=7$ Use the Square Root Property. $x=7,x=−7x=7,x=−7$
We cannot simplify $77$, so we leave the answer as a radical.
## Example 9.1
### How to solve a Quadratic Equation of the form ax2 = k Using the Square Root Property
Solve: $x2−50=0.x2−50=0.$
## Try It 9.1
Solve: $x2−48=0.x2−48=0.$
## Try It 9.2
Solve: $y2−27=0.y2−27=0.$
The steps to take to use the Square Root Property to solve a quadratic equation are listed here.
## How To
### Solve a quadratic equation using the square root property.
1. Step 1. Isolate the quadratic term and make its coefficient one.
2. Step 2. Use Square Root Property.
3. Step 3. Simplify the radical.
4. Step 4. Check the solutions.
In order to use the Square Root Property, the coefficient of the variable term must equal one. In the next example, we must divide both sides of the equation by the coefficient 3 before using the Square Root Property.
## Example 9.2
Solve: $3z2=108.3z2=108.$
## Try It 9.3
Solve: $2x2=98.2x2=98.$
## Try It 9.4
Solve: $5m2=80.5m2=80.$
The Square Root Property states ‘If $x2=kx2=k$,’ What will happen if $k<0?k<0?$ This will be the case in the next example.
## Example 9.3
Solve: $x2+72=0x2+72=0$.
## Try It 9.5
Solve: $c2+12=0.c2+12=0.$
## Try It 9.6
Solve: $q2+24=0.q2+24=0.$
Our method also works when fractions occur in the equation; we solve as any equation with fractions. In the next example, we first isolate the quadratic term, and then make the coefficient equal to one.
## Example 9.4
Solve: $23u2+5=17.23u2+5=17.$
## Try It 9.7
Solve: $12x2+4=24.12x2+4=24.$
## Try It 9.8
Solve: $34y2−3=18.34y2−3=18.$
The solutions to some equations may have fractions inside the radicals. When this happens, we must rationalize the denominator.
## Example 9.5
Solve: $2x2−8=41.2x2−8=41.$
## Try It 9.9
Solve: $5r2−2=34.5r2−2=34.$
## Try It 9.10
Solve: $3t2+6=70.3t2+6=70.$
## Solve Quadratic Equations of the Form a(x − h)2 = k Using the Square Root Property
We can use the Square Root Property to solve an equation of the form a(xh)2 = k as well. Notice that the quadratic term, x, in the original form ax2 = k is replaced with (xh).
The first step, like before, is to isolate the term that has the variable squared. In this case, a binomial is being squared. Once the binomial is isolated, by dividing each side by the coefficient of a, then the Square Root Property can be used on (xh)2.
## Example 9.6
Solve: $4(y−7)2=48.4(y−7)2=48.$
## Try It 9.11
Solve: $3(a−3)2=54.3(a−3)2=54.$
## Try It 9.12
Solve: $2(b+2)2=80.2(b+2)2=80.$
Remember when we take the square root of a fraction, we can take the square root of the numerator and denominator separately.
## Example 9.7
Solve: $(x−13)2=59.(x−13)2=59.$
## Try It 9.13
Solve: $(x−12)2=54.(x−12)2=54.$
## Try It 9.14
Solve: $(y+34)2=716.(y+34)2=716.$
We will start the solution to the next example by isolating the binomial term.
## Example 9.8
Solve: $2(x−2)2+3=57.2(x−2)2+3=57.$
## Try It 9.15
Solve: $5(a−5)2+4=104.5(a−5)2+4=104.$
## Try It 9.16
Solve: $3(b+3)2−8=88.3(b+3)2−8=88.$
Sometimes the solutions are complex numbers.
## Example 9.9
Solve: $(2x−3)2=−12.(2x−3)2=−12.$
## Try It 9.17
Solve: $(3r+4)2=−8.(3r+4)2=−8.$
## Try It 9.18
Solve: $(2t−8)2=−10.(2t−8)2=−10.$
The left sides of the equations in the next two examples do not seem to be of the form a(xh)2. But they are perfect square trinomials, so we will factor to put them in the form we need.
## Example 9.10
Solve: $4n2+4n+1=16.4n2+4n+1=16.$
## Try It 9.19
Solve: $9m2−12m+4=25.9m2−12m+4=25.$
## Try It 9.20
Solve: $16n2+40n+25=4.16n2+40n+25=4.$
## Media
Access this online resource for additional instruction and practice with using the Square Root Property to solve quadratic equations.
## Section 9.1 Exercises
### Practice Makes Perfect
Solve Quadratic Equations of the Form ax2 = k Using the Square Root Property
In the following exercises, solve each equation.
1.
$a 2 = 49 a 2 = 49$
2.
$b 2 = 144 b 2 = 144$
3.
$r 2 − 24 = 0 r 2 − 24 = 0$
4.
$t 2 − 75 = 0 t 2 − 75 = 0$
5.
$u 2 − 300 = 0 u 2 − 300 = 0$
6.
$v 2 − 80 = 0 v 2 − 80 = 0$
7.
$4 m 2 = 36 4 m 2 = 36$
8.
$3 n 2 = 48 3 n 2 = 48$
9.
$4 3 x 2 = 48 4 3 x 2 = 48$
10.
$5 3 y 2 = 60 5 3 y 2 = 60$
11.
$x 2 + 25 = 0 x 2 + 25 = 0$
12.
$y 2 + 64 = 0 y 2 + 64 = 0$
13.
$x 2 + 63 = 0 x 2 + 63 = 0$
14.
$y 2 + 45 = 0 y 2 + 45 = 0$
15.
$4 3 x 2 + 2 = 110 4 3 x 2 + 2 = 110$
16.
$2 3 y 2 − 8 = −2 2 3 y 2 − 8 = −2$
17.
$2 5 a 2 + 3 = 11 2 5 a 2 + 3 = 11$
18.
$3 2 b 2 − 7 = 41 3 2 b 2 − 7 = 41$
19.
$7 p 2 + 10 = 26 7 p 2 + 10 = 26$
20.
$2 q 2 + 5 = 30 2 q 2 + 5 = 30$
21.
$5 y 2 − 7 = 25 5 y 2 − 7 = 25$
22.
$3 x 2 − 8 = 46 3 x 2 − 8 = 46$
Solve Quadratic Equations of the Form a(xh)2 = k Using the Square Root Property
In the following exercises, solve each equation.
23.
$( u − 6 ) 2 = 64 ( u − 6 ) 2 = 64$
24.
$( v + 10 ) 2 = 121 ( v + 10 ) 2 = 121$
25.
$( m − 6 ) 2 = 20 ( m − 6 ) 2 = 20$
26.
$( n + 5 ) 2 = 32 ( n + 5 ) 2 = 32$
27.
$( r − 1 2 ) 2 = 3 4 ( r − 1 2 ) 2 = 3 4$
28.
$( x + 1 5 ) 2 = 7 25 ( x + 1 5 ) 2 = 7 25$
29.
$( y + 2 3 ) 2 = 8 81 ( y + 2 3 ) 2 = 8 81$
30.
$( t − 5 6 ) 2 = 11 25 ( t − 5 6 ) 2 = 11 25$
31.
$( a − 7 ) 2 + 5 = 55 ( a − 7 ) 2 + 5 = 55$
32.
$( b − 1 ) 2 − 9 = 39 ( b − 1 ) 2 − 9 = 39$
33.
$4 ( x + 3 ) 2 − 5 = 27 4 ( x + 3 ) 2 − 5 = 27$
34.
$5 ( x + 3 ) 2 − 7 = 68 5 ( x + 3 ) 2 − 7 = 68$
35.
$( 5 c + 1 ) 2 = −27 ( 5 c + 1 ) 2 = −27$
36.
$( 8 d − 6 ) 2 = −24 ( 8 d − 6 ) 2 = −24$
37.
$( 4 x − 3 ) 2 + 11 = −17 ( 4 x − 3 ) 2 + 11 = −17$
38.
$( 2 y + 1 ) 2 − 5 = −23 ( 2 y + 1 ) 2 − 5 = −23$
39.
$m 2 − 4 m + 4 = 8 m 2 − 4 m + 4 = 8$
40.
$n 2 + 8 n + 16 = 27 n 2 + 8 n + 16 = 27$
41.
$x 2 − 6 x + 9 = 12 x 2 − 6 x + 9 = 12$
42.
$y 2 + 12 y + 36 = 32 y 2 + 12 y + 36 = 32$
43.
$25 x 2 − 30 x + 9 = 36 25 x 2 − 30 x + 9 = 36$
44.
$9 y 2 + 12 y + 4 = 9 9 y 2 + 12 y + 4 = 9$
45.
$36 x 2 − 24 x + 4 = 81 36 x 2 − 24 x + 4 = 81$
46.
$64 x 2 + 144 x + 81 = 25 64 x 2 + 144 x + 81 = 25$
### Mixed Practice
In the following exercises, solve using the Square Root Property.
47.
$2 r 2 = 32 2 r 2 = 32$
48.
$4 t 2 = 16 4 t 2 = 16$
49.
$( a − 4 ) 2 = 28 ( a − 4 ) 2 = 28$
50.
$( b + 7 ) 2 = 8 ( b + 7 ) 2 = 8$
51.
$9 w 2 − 24 w + 16 = 1 9 w 2 − 24 w + 16 = 1$
52.
$4 z 2 + 4 z + 1 = 49 4 z 2 + 4 z + 1 = 49$
53.
$a 2 − 18 = 0 a 2 − 18 = 0$
54.
$b 2 − 108 = 0 b 2 − 108 = 0$
55.
$( p − 1 3 ) 2 = 7 9 ( p − 1 3 ) 2 = 7 9$
56.
$( q − 3 5 ) 2 = 3 4 ( q − 3 5 ) 2 = 3 4$
57.
$m 2 + 12 = 0 m 2 + 12 = 0$
58.
$n 2 + 48 = 0 . n 2 + 48 = 0 .$
59.
$u 2 − 14 u + 49 = 72 u 2 − 14 u + 49 = 72$
60.
$v 2 + 18 v + 81 = 50 v 2 + 18 v + 81 = 50$
61.
$( m − 4 ) 2 + 3 = 15 ( m − 4 ) 2 + 3 = 15$
62.
$( n − 7 ) 2 − 8 = 64 ( n − 7 ) 2 − 8 = 64$
63.
$( x + 5 ) 2 = 4 ( x + 5 ) 2 = 4$
64.
$( y − 4 ) 2 = 64 ( y − 4 ) 2 = 64$
65.
$6 c 2 + 4 = 29 6 c 2 + 4 = 29$
66.
$2 d 2 − 4 = 77 2 d 2 − 4 = 77$
67.
$( x − 6 ) 2 + 7 = 3 ( x − 6 ) 2 + 7 = 3$
68.
$( y − 4 ) 2 + 10 = 9 ( y − 4 ) 2 + 10 = 9$
### Writing Exercises
69.
In your own words, explain the Square Root Property.
70.
In your own words, explain how to use the Square Root Property to solve the quadratic equation $(x+2)2=16(x+2)2=16$.
### Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
Choose how would you respond to the statement “I can solve quadratic equations of the form a times the square of x minus h equals k using the Square Root Property.” “Confidently,” “with some help,” or “No, I don’t get it.”
If most of your checks were:
…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.
…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?
…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.
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# How do you solve 3x + y = 7 and 3x - y = 5?
Jun 7, 2016
$x = 2 \text{ and } y = 1$
#### Explanation:
There are 2 obvious methods which can be used in this example, based on the x and y values in the equations.
ELIMINATION METHOD
The two y terms are additive inverses. This means they have the same value, but opposite signs.
When they are ADDED together, they will make 0.
$\text{ " 3x + y = 7 " A}$
$\text{ " 3x - y = 5 " B}$
$\text{A + B: " 6x " "= 12" (the y's have been eliminated)}$
$\text{ " x" } = 2$
Substitute into A to find the value of y:
$\text{ } 3 \left(2\right) + y = 7$
$\text{ } 6 + y = 7$
$\text{ } y = 1$
EQUATING METHOD
The two x-terms are exactly the same. Make the x term the subject.
$3 x + y = 7 \text{ and } 3 x - y = 5$
$3 x \text{ " = 7 - y " " 3x " } = 5 + y$
But $3 x \text{ is the same as } 3 x$!!
Therefore $\text{ } 7 - y = 5 + y$
solve for y: $\text{ } 7 - 5 = 2 y$
$\text{ " 2 = 2y " } \Rightarrow y = 1$
If $y = 1$, substitute into either equation to find $x$.
$3 x = 7 - y \Rightarrow 3 x = 7 - 1 = 6$
$3 x = 6$
$x = 2$ |
# One-Dimensional Kinematics with Constant Acceleration| University Physics
Learning Goal:
To understand the meaning of the variables that appear in the equations for one-dimensional kinematics with constant acceleration.
Motion with a constant, nonzero acceleration is not uncommon in the world around us. Falling (or thrown) objects and cars starting and stopping approximate this type of motion. It is also the type of motion most frequently involved in introductory kinematics problems.
The kinematic equations for such motion can be written as
$x\left(t\right)=x_i+v_it+\frac{1}{2}at^2$
$v\left(t\right)=v_i+at,$
where the symbols are defined as follows:
• $x\left(t\right)$ is the position of the particle;
• $x_i$ is the initial position of the particle;
• $v\left(t\right)$ is the velocity of the particle;
• $v_i$ is the initial velocity of the particle;
• $a$ is the acceleration of the particle.
In answering the following questions, assume that the acceleration is constant and nonzero: a≠0.
### PART B. The quantity represented by $x_i$ is a function of time (i.e., is not constant).
Recall that $x_i$ represents an initial value, not a variable. It refers to the position of an object at some initial moment.
### PART D. The quantity represented by v is a function of time (i.e., is not constant).
The velocity v always varies with time when the linear acceleration is nonzero.
### PART E. Which of the given equations is not an explicit function of t and is therefore useful when you don’t know or don’t need the time?
ANSWER: $v^2=\left(v_i\right)^2+2a\left(x-x_i\right)$
### PART F. A particle moves with constant acceleration a. The expression vi+at represents the particle’s velocity at what instant in time?
ANSWER: when the time t has passed since the particle’s velocity was $v_i$
More generally, the equations of motion can be written as
$x\left(t\right)=x_i+v_i\Delta t+\frac{1}{2}a\left(\Delta t\right)^2$
and
$v\left(t\right)=v_i+a\Delta t$
Here Δt is the time that has elapsed since the beginning of the particle’s motion, that is, $\Delta t=t-t_i$, where $t$ is the current time and $t_i$ is the time at which we start measuring the particle’s motion. The terms $x_i$ and $v_i$ are, respectively, the position and velocity at $t=t_i$. As you can now see, the equations given at the beginning of this problem correspond to the case $t_i=0$, which is a convenient choice if there is only one particle of interest.
### What is the equation describing the position of particle B?
ANSWER: $x_B\left(t\right)=x_i+0.5v_i\left(t-t_1\right)+a\left(t-t_1\right)^2$
The general equation for the distance traveled by particle B is $x_B\left(t\right)=x_{iB}+v_{iB}\Delta t+\frac{1}{2}a_B\left(\Delta t\right)^2$ or $x_B\left(t\right)=x_{iB}+v_{iB}\left(t-t_1\right)+\frac{1}{2}a_B\left(t-t_1\right)^2$, since $\Delta t=t-t_1$ is a good choice for B. From the information given, deduce the correct values of the constants that go into the equation for $x_B\left(t\right)$ given here, in terms of A’s constants of motion.
$x_B\left(t\right)=x_i+\frac{1}{2}v_i\left(t-t_1\right)+\frac{1}{2}\left(2a\right)\left(t-t_1\right)^2$
$x_B\left(t\right)=x_i+0.5v_i\left(t-t_1\right)+a\left(t-t_1\right)^2$
### PART H. At what time does the velocity of particle B equal that of particle A?
ANSWER: $t=2t_1+\frac{v_i}{2a}$
Particle A’s velocity as a function of time is $v_A\left(t\right)=v_i+at$, and particle B’s velocity as a function of time is $v_B\left(t\right)=0.5v_i+2a\left(t-t_1\right)$.
Once you have expressions for the velocities of A and B as functions of time, set them equal and find the time t at which this happens.
$v_i+at=\frac{1}{2}v_i+2a\left(t-t_1\right)$
$v_i+at=\frac{1}{2}v_i+2at-2at_1$
$2at-at=2at_1+\frac{1}{2}v_i$
$at=2at_1+\frac{1}{2}v_i$
$t=2t_1+\frac{v_i}{2a}$ |
# Preview of Calculus.
## Presentation on theme: "Preview of Calculus."— Presentation transcript:
Preview of Calculus
The Area Problem Integration
The Tangent Problem Differentiation
Average Velocity Velocity (Slope of the Tangent Line)
Infinite Sequence, Infinite Series
Calculus II (semester 101-2)
1 Functions and Limits 1.1 Four Ways to Represent a Function
1.2 Math. Models: A Catalog of Essential Functions 1.3 New Functions from Old Functions 1.4 The Tangent and Velocity Problems 1.5 The Limit of a Function 1.6 Calculating Limits Using the Limit Laws 1.7 The Precise Definition of a Limit 1.8 Continuity
Four Ways to Represent a Function
1.1 Four Ways to Represent a Function We usually consider functions for which the sets D and E are sets of real numbers. The set D is called the domain of the function. The number f (x) is the value of f at x and is read “f of x.” The range of f is the set of all possible values of f (x) as x varies throughout the domain. A symbol that represents an arbitrary number in the domain of a function f is called an independent variable.
Machine diagram for a function f
About Function A symbol that represents a number in the range of f is called a dependent variable. In Example A, for instance, r is the independent variable and A is the dependent variable. It’s helpful to think of a function as a machine. Machine diagram for a function f Arrow diagram for f
The Graph of a Function The graph of a function is a curve in the xy-plane. But the question arises: Which curves in the xy-plane are graphs of functions? This is answered by the following test.
Representations of Functions
There are four possible ways to represent a function: verbally (by a description in words) numerically (by a table of values) visually (by a graph) algebraically (by an explicit formula)
Symmetry If a function f satisfies f (–x) = f (x) for every number x in its domain, then f is called an even function. For instance, the function f (x) = x2 is even because f (–x) = (–x)2 = x2 = f (x) The geometric significance of an even function is that its graph is symmetric with respect to the y-axis (see Figure 19). An even function Figure 19
Symmetry If f satisfies f (–x) = –f (x) for every number x in its domain, then f is called an odd function. For example, the function f (x) = x3 is odd because f (–x) = (–x)3 = –x3 = –f (x) An odd function
Increasing and Decreasing Functions
The graph shown in Figure 22 rises from A to B, falls from B to C, and rises again from C to D. The function f is said to be increasing on the interval [a, b], decreasing on [b, c], and increasing again on [c, d]. Figure 22
Increasing and Decreasing Functions
Notice that if x1 and x2 are any two numbers between a and b with x1 < x2, then f (x1) < f (x2). We use this as the defining property of an increasing function.
A Catalog of Essential Functions
1.2 A Catalog of Essential Functions Linear Models Polynomials Power Functions Rational Functions Algebraic Functions Trigonometric Functions Exponential Functions Logarithmic Functions
Linear Models When we say that y is a linear function of x, we mean that the graph of the function is a line, so we can use the slope-intercept form of the equation of a line to write a formula for the function as y = f (x) = mx + b where m is the slope of the line and b is the y-intercept.
Polynomials A function P is called a polynomial if
P (x) = anxn + an–1xn– a2x2 + a1x + a0 where n is a nonnegative integer and the numbers a0, a1, a2, . . ., an are constants called the coefficients of the polynomial. The domain of any polynomial is If the leading coefficient an 0, then the degree of the polynomial is n. For example, the function is a polynomial of degree 6.
Power Functions A function of the form f(x) = xa, where a is a constant, is called a power function. We consider several cases. (i) a = n, where n is a positive integer The graphs of f(x) = xn for n = 1, 2, 3, 4, and 5 are shown in Figure 11. (These are polynomials with only one term.) We already know the shape of the graphs of y = x (a line through the origin with slope 1) and y = x2 (a parabola).
Power Functions (ii) a = 1/n, where n is a positive integer
The function is a root function. For n = 2 it is the square root function whose domain is [0, ) and whose graph is the upper half of the parabola x = y2. [See Figure 13(a).] Graph of root function Figure 13(a)
The reciprocal function
Power Functions (iii) a = –1 The graph of the reciprocal function f (x) = x –1 = 1/x is shown in Figure 14. Its graph has the equation y = 1/x, or xy = 1, and is a hyperbola with the coordinate axes as its asymptotes. The reciprocal function Figure 14
The reciprocal function
Rational Functions A rational function f is a ratio of two polynomials: where P and Q are polynomials. The domain consists of all values of x such that Q(x) 0. A simple example of a rational function is the function f (x) = 1/x, whose domain is {x | x 0}; this is the reciprocal function graphed in Figure 14. The reciprocal function Figure 14
Algebraic Functions A function f is called an algebraic function if it can be constructed using algebraic operations (such as addition, subtraction, multiplication, division, and taking roots) starting with polynomials. Any rational function is automatically an algebraic function. Here are two more examples:
Trigonometric Functions
In calculus the convention is that radian measure is always used (except when otherwise indicated). For example, when we use the function f (x) = sin x, it is understood that sin x means the sine of the angle whose radian measure is x.
Exponential Functions
The exponential functions are the functions of the form f (x) = ax, where the base a is a positive constant. The graphs of y = 2x and y = (0.5)x are shown in Figure 20. In both cases the domain is ( , ) and the range is (0, ). Figure 20
Logarithmic Functions
The logarithmic functions f (x) = logax, where the base a is a positive constant, are the inverse functions of the exponential functions. Figure 21 shows the graphs of four logarithmic functions with various bases. In each case the domain is (0, ), the range is ( , ), and the function increases slowly when x > 1. Figure 21
New Functions from Old Functions
1.3 New Functions from Old Functions Likewise, if g(x) = f (x – c), where c > 0, then the value of g at x is the same as the value of f at x – c (c units to the left of x). Therefore the graph of y = f (x – c), is just the graph of y = f (x) shifted c units to the right (see Figure 1). Translating the graph of ƒ Figure 1
Transformations of Functions
The graph of y = –f (x) is the graph of y = f (x) reflected about the x-axis because the point (x, y) is replaced by the point (x, –y). (See Figure 2 and the following chart, where the results of other stretching, shrinking, and reflecting transformations are also given.) Stretching and reflecting the graph of f Figure 2
Combinations of Functions
Two functions f and g can be combined to form new functions f + g, f – g, fg, and f/g in a manner similar to the way we add, subtract, multiply, and divide real numbers. The sum and difference functions are defined by (f + g)(x) = f (x) + g (x) (f – g)(x) = f (x) – g (x) If the domain of f is A and the domain of g is B, then the domain of f + g is the intersection A ∩ B because both f (x) and g(x) have to be defined. For example, the domain of is A = [0, ) and the domain of is B = ( , 2], so the domain of is A ∩ B = [0, 2].
Combinations of Functions
In general, given any two functions f and g, we start with a number x in the domain of g and find its image g (x). If this number g (x) is in the domain of f, then we can calculate the value of f (g (x)). The result is a new function h (x) = f (g (x)) obtained by substituting g into f. It is called the composition (or composite) of f and g and is denoted by f g (“f circle g ”).
The Tangent and Velocity Problems
1.4 Limit |
# Video: Differentiating Reciprocal Trigonometric Functions Using the Chain Rule
Find dπ¦/dπ₯ for the function π¦ = β3 csc (4π₯β΅ β 3).
03:06
### Video Transcript
Find dπ¦ by dπ₯ for the function π¦ is equal to negative three times the csc of four π₯ to the fifth power minus three.
The question wants us to find the first derivative of π¦ with respect to π₯. And we can see that π¦ is the composition of functions; itβs a composition of the cosecant function and the polynomial four π₯ to the fifth power minus three. And we know how to find the derivative of the composition of two functions by using the chain rule.
We recall the chain rule tells us, if we have π¦ is a function of π’ and π’ in turn is a function of π₯, then the derivative of π¦ with respect to π₯ is equal to the derivative of π¦ with respect to π’ multiplied by the derivative of π’ with respect to π₯. So, if we set π’ to be our inner function, thatβs the polynomial four π₯ to the fifth power minus three, then this tells us that π¦ is equal to negative three times the csc of π’. Weβve rewritten π¦ to be a function of π’, and π’, in turn, is a function of π₯.
So, we can evaluate this derivative by using the chain rule. We get dπ¦ by dπ₯ is equal to dπ¦ by dπ’ times dπ’ by dπ₯. And we can actually evaluate both of these derivatives. First, dπ¦ by dπ’ is the derivative of π¦ with respect to π’ and π¦ is equal to negative three times the csc of π’. So, dπ¦ by dπ’ is the derivative of negative three times the csc of π’ with respect to π’.
Next, dπ’ by dπ₯ is the derivative of π’ with respect to π₯, and π’ is four π₯ to the fifth power minus three. So, dπ’ by dπ₯ is equal to the derivative of four π₯ to the fifth power minus three with respect to π₯. And we can just evaluate both of these derivatives. First, we know for any constant π, the derivative of π times the csc of π’ with respect to π’ is equal to negative π times the cot of π’ times the csc of π’. So, applying this to our first derivative, we get negative one multiplied by negative three cot of π’ times the csc of π’.
And our second derivative is just the derivative of a polynomial. We can do this by using the power rule for differentiation. We multiply by our exponent of π₯ and reduce this exponent by one. This gives us 20π₯ to the fourth power. So, we found the following expression for dπ¦ by dπ₯. And we can simplify this expression slightly. First, negative one multiplied by negative three is equal to three. Next, three multiplied by 20π₯ to the fourth power is equal to 60π₯ to the fourth power.
This gives us 60π₯ to the fourth power times the cot of π’ times the csc of π’. And we could leave our answer like this. However, remember, weβre trying to find an expression for the derivative of π¦ with respect to π₯. So, we want our answer to be in terms of π₯. And to do this, weβll use our substitution π’ is equal to four π₯ to the fifth power minus three. And using this substitution, we get our final answer.
If π¦ is equal to negative three times the csc of four π₯ to the fifth power minus three, then dπ¦ by dπ₯ is equal to 60π₯ to the fourth power times the cot of four π₯ to the fifth power minus three multiplied by the csc of four π₯ to the fifth power minus three. |
# What is Symmetry?
As I said in the last post, in group theory, you strip things down to a simple collection of values and one operation, with four required properties. The result is a simple structure, which completely captures the concept of symmetry. But mathematically, what is symmetry? And how can something as simple and abstract as a group have anything to do with it?
Let's look at a simple, familiar example: integers and addition. What does symmetry mean in terms of the set of integers and the addition operation?
Suppose I were to invent a strange way of writing integers. You know nothing about how I'm writing them. But you know how addition works. So, can you figure out which number is which? Let's be concrete
about this: here's the set of numbers from -5 to 5, out of order, in my strange notation. {#, @, *, !, , ^, &, %, \$ }. If you give me two of them, I'll tell you their sum. What can you find out?
Well, you can figure out what zero is. With a bit of experimentation, you can see that adding
"&" and ">" gives you ">", and adding "&" and "~" gives you "~"; the only integer for which that could
be true is zero. So you can tell that "&" is how I'm writing zero.
What else can you figure out? With enough experimentation, you can find an ordering. You can find two
values, "@" and "#", where "@" plus "#" = ">"; ">" plus "#" equals "%", and so on - each time you add "#" to something, you get another value, and if you start with "@", and repeatedly add ">", you'll get a
complete enumeration of them: "@", ">", "%", "\$", "!", "&", "#", "*", "~", "^", ">".
So - which symbol represents one?
You can't tell. "@" might be -5, in which case "#" is 1. Or "@" might be 5, in which case "#" is -1. You can tell that "*" must be either +2 or -2, but you can't tell which. You can't tell which numbers are the positives, and which are the negatives. If all you have available to you is the group operation of addition, then there is no way for you to distinguish between the positive and the negative numbers.
Addition of the integers is symmetric: you can change the signs of the numbers
but by everything you can do with the group operator, the change is invisible. I can write an equation in my representation: "%" plus "*" equals "!", where I know that "%" is +3, "*" is -2, and "!" is +1. I can then switch the positive and the negative numbers, so that "%" is -3, "*" is +2, and "!" is -1, and the equation - in fact, any equation which relies on nothing more than the group operator of addition - can tell the difference. If you don't know what the symbols represent, you wouldn't even be able to tell that I'd changed my mind about what the symbols meant!
That's a simple of example of what symmetry means. Symmetry is an immunity to transformation. If something is symmetric, then that means that there is something you can do to it, some transformation you can apply to it, and after the transformation, you won't be able to tell that any transformation was applied. When something is a group, there is a transformation associated with the group operator which is
undetectable within the structure of the group. That "within the group" part is important: with the integers, if you have multiplication, you can distinguish between 1 and -1; 1 is the identity for multiplication, -1 is not. But if all you have is addition, that the transformation is invisible.
Think of the intuitive notion of symmetry: mirroring. What mirror symmetry means is that you can draw a line through an image, and swap what's on the left-hand side of it with what's on the right-hand side of it - and the end result will be indistinguishable from the original image. Addition based groups of numbers captures the fundamental notion of mirror symmetry: it defines a central division (0), and
the fact that swapping the objects on opposite sides of that division has no discernable effect.
For an example of how that can get more interesting that just the integers without changing the fundamental concept, you can look at the following example: given a hexagon, you can reflect it along many different dividing lines without recognizing any difference.
When some object is symmetrical with respect to a particular transformation, that means that you cannot distinguish between that object before the transformation, and that object after. There are, of course, many different kinds of symmetry beyond the basic mirroring ones that we're all familiar with. A few examples of basic geometric symmetries:
Scale: scale symmetry means that you can change the size of something without altering it.
Think of geometry, where you're interested in the fundamental properties of a shape - the number
of sides, the angles between them, the relative sizes of the sides. If you don't have any way of
measuring size on an absolute basis, then an equilateral triangle with sides 3 inches long and an equilateral triangle with sides 1 inch long can't be distinguished. You can change the scale of
things without creating a detectable difference.
Translation: translational symmetry means you can move an object without detecting any change. If you have a square grid, like graph paper, drawn on an infinite canvas, you can move it the
distance between adjacent lines, and there will be no way to tell that you changed anything.
Rotation: rotational symmetry means you can rotate something without creating a detectable change. For example, as illustrated in the diagram below, if you rotate a hexagon by 60 degrees, without any external markings, you can't tell that it's rotated.
There are, of course, many more, and we'll talk about some of them in later posts.
For a fun exercise, look at the Escher image at the top of this post. It contains numerous kinds of symmetries; several different kinds of reflective symmetries, translational symmetries, rotational symmetries, color-shift symmetries, and more. See how many you can find. I've been able to figure out at least 16, but I'm sure I've missed something.
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It depends what you mean by a "kind" of symmetry. The group is \$D_3\ltimes\mathbb{Z}^2\$ -- a dihedral group (to flip around one of the triangular shapes) along with a rank-2 lattice to slide the triangular shapes among each other, and a semidirect product to combine them together. So what's a "kind"? An element? A subgroup? A normal subgroup? Some sort of equivalence class of subgroups?
[ Addition of the integers is symmetric: you can change the signs of the numbers but by everything you can do with the group operator, the change is invisible. I can write an equation in my representation: "%" plus "*" equals "!", where I know that "%" is +3, "*" is -2, and "!" is +1. I can then switch the positive and the negative numbers, so that "%" is -3, "*" is +2, and "!" is -1, and the equation - in fact, any equation which relies on nothing more than the group operator of addition - can tell the difference. If you don't know what the symbols represent, you wouldn't even be able to tell that I'd changed my mind about what the symbols meant!
That's a simple of example of what symmetry means. ]
How does group theory then capture the language of symmetry? If 2+3=3+2 indicates a symmetry, then we have symmetries outside of group theory. Do you mean to say that commutative or Abelian group theory captures the notion of symmetry?
By Doug Spoonwood (not verified) on 04 Dec 2007 #permalink
A couple of thoughts:
in a group, if x + y = x
then -x + x + y = -x + x
y = 0
So, once you know that you have a group and x + y = x, then you know that y = 0. Nothing further needed (inverses are unique in a group)
Symmetry and group theory: "kind" of symmetry: I think he means a non-trivial automorphism of a group.
How group theory captures symmetry: a group can act on a set and that group action can be thought of as "a symmetry".
example: consider that "devil angel" tiling of a disk. rotational symmetries of the disk can be obtained by letting the group S^1 (think of the set of complex numbers of modulus 1 with the operation of complex multiplication); this can give various rotations of that disk, some of which correspond to the symmetries of the images on the disk.
Symmetries are elegant, and groups are elegant; but fields, whose two operations form a group and a monoid, have always seemed less so to me, for all their utility... But, the most symmetrical geometry (in their group-theoretic classification) is projective geometry, with its points at infinity; and adjoining a related infinity to a field yields a more symmetrical pair of structures (symmetrical in the mirroring sense), each of which is a commutative generalization of an abelian group. (I just thought I'd throw that in, in case it interests anyone:-)
[A couple of thoughts:
in a group, if x + y = x
then -x + x + y = -x + x
y = 0
So, once you know that you have a group and x + y = x, then you know that y = 0. Nothing further needed (inverses are unique in a group)]
Let '+' stand for ordinary multiplication. Consequently {1, -1}, with 1 and -1 representing the regular numbers '1' and '-1', qualifies as a group under '+', with x+1=x. But, since we let 1 equal the regular number 1, it doesn't equal 0.
I think you mean to say that if x+y=x, then y indicates the identity element of the group. It does so, specifically because the identity element gets defined as the element such that x"*"y=x=y"*"x, where "*" indicates any sort of operation that satisfies the group axioms.
[I think he means a non-trivial automorphism of a group.]
I don't see why you think so, since he didn't bring up the automorphism concept, at least not explicitly. I don't recall it myself. I don't think he meant to assume his readers would know that concept.
[How group theory captures symmetry: a group can act on a set and that group action can be thought of as "a symmetry".]
Perhaps, but Mark didn't talk about symmetry in such a sense. He used a much broader concept. The operator 'max' on {0, 1/2, 1} qualifies as logically symmetric or commutative, has an identity of '0', and works associatively. I can think of 0 min 1 as having 1 min 0 as its symmetrical counterpart. This corresponds to changing "the signs of numbers", as for a set {-1, 0, 1} I could write -1 min 1 and apply the transformation to get 1 min -1, and no change in the structure involved happens. But, there exists no inverse for min, so neither of those sets qualify as a group under min.
By Doug Spoonwood (not verified) on 04 Dec 2007 #permalink
"I can then switch the positive and the negative numbers, so that "%" is -3, "*" is +2, and "!" is -1, and the equation - in fact, any equation which relies on nothing more than the group operator of addition - can tell the difference"
I think that should say, "can't tell the difference".
Let '+' stand for ordinary multiplication. Consequently {1, -1}, with 1 and -1 representing the regular numbers '1' and '-1', qualifies as a group under '+', with x+1=x. But, since we let 1 equal the regular number 1, it doesn't equal 0.
I think you mean to say that if x+y=x, then y indicates the identity element of the group. It does so, specifically because the identity element gets defined as the element such that x"*"y=x=y"*"x, where "*" indicates any sort of operation that satisfies the group axioms.
Nod. He was just using "0" as the name of the identity element. That's common, iirc.
If we were talking about multiplication specifically, then yeah, we'd say "1". However, in more general group terms 0 is used. This is also because they use 1 as the name of the identity of the second operation in a field.
By Xanthir, FCD (not verified) on 04 Dec 2007 #permalink
What does 'iirc' mean?
By Doug Spoonwood (not verified) on 04 Dec 2007 #permalink
Yes, by "0" I meant "group identity element"; of course many might have thought I meant the identity with respect to the "addition" operation in a ring.
Sorry for the confusion.
Doug, "iirc" is '"f I recall correctly".
Doug, about your example: I like it. But I don't think that your set with binary operation (a semigroup) qualifies for symmetry. Here is why:
If one makes a multiplication table for abstract elements a, b, c; the first row of which looks like this:
a*a=a, a*b=b, a*c =c
Then it is clear that a < b and a < c (where "<" is the order associated with your max operator)
then b*c = c means that b < c and there is simply no other choice for the ordering.
On the other hand, if one looks at the group {0, 1, 2, 3, 4} with mod 5 addition as the operation, one can find 3 non trivial symmetries obtained by f(x) = kx, where k is either 2, 3, or 4; that is, f(x+y) = f(x)+f(y). That is, if you knew that the elements of the group were {0, 1, 2, 3, 4} and you were given a multiplication table in terms of {a, b, c, d, e} you couldn't tell which was which.
[Then it is clear that a then b*c = c means that b]
I think you meant to write more here, but I don't know what.
I don't get your point, since if we write out a table with for {a, b, c}, with a
By Doug Spoonwood (not verified) on 05 Dec 2007 #permalink
Remember, everyone, you *can't* use less-than signs in your post directly. The MovableType software that ScienceBlogs uses interprets that as an html comment and tries to eat the rest of the line.
If you really need it, use & lt ;, but without the spaces. Use & gt ; as well if you want the other sign. Those will give you < and >.
As well, remember that the Preview feature will convert those html entities in your actual text, which will then give you the same problems. Either don't Preview, or copy the text of your message *before* hitting Preview, then paste it back in to replace the text that you see in the Preview window.
By Xanthir, FCD (not verified) on 05 Dec 2007 #permalink
Oh dear; I botched that post, didn't I? :)
here is what I wanted to say:
Doug, about your example: I like it. But I don't think that your set with binary operation (a semigroup) qualifies for symmetry. Here is why:
If one makes a multiplication table for abstract elements a, b, c; the first row of which looks like this:
a*a=a, a*b=b, a*c =c
Then it is clear that a is "less than either b and c" and hence is in the minimum position. then b*c = c means that b is less than c as well, so the ordering
a "is less than" b and b is "less than" c is given from the "multiplication table".
--------
The rest of the post was correct. :)
[Oh dear; I botched that post, didn't I? :)]
Yeah... I did it too.
[If one makes a multiplication table for abstract elements a, b, c; the first row of which looks like this:
a*a=a, a*b=b, a*c =c]
So we have the ordering a<b<c, with '*' for max. The whole table looks like this
* a b c
a a b c
b b b c
c c c c
We have a symmetry along the diagonal which stretches from the uppermost left space to the lowermost right space.
I admit that additive mod 5 groups will give you more symmetries, but you still end up with a symmetry in the above table. One can also see symmetry in an operator '\$' on natural numbers where commutavity holds in the following manner: write all natural numbers (1, 2, ...} as a sequence of '*' in terms of their cardinality. In other words, for 1 write *, for 2 write **, and for 3 write ***, etc. Consequently, with the '\$' operator we have
*** \$ **=** \$ ***. Visual speaking we have a symmetry around the '\$' sign, by just using commutavity. I don't see how group theory gives us a language to analyze such a symmetry (we can't talk about a rotation, as we have no assumption nor definition of a rotation here... we just have a '\$' operator and the natural numbers {1, 2...}). Maybe I've missed something, but I don't see it and I don't see how Mark's post lets one analyze such simple situations.
By Doug Spoonwood (not verified) on 05 Dec 2007 #permalink
I'd been giving this alot of thought lately, only not just in groups...in rings and fields too. There's no way to make the following statement rigorous; it's just sort of an heuristic, fun thing. You can take the entirety of your knowledge of automorphisms and view it as a theory of how symmetric certain groups,rings,modules,fields are. Or put another way, automorphisms are a good measure of how much certain elements of a structure lack their own identity (pardon the term). In a field with a huge Galois group where an automorphism can send an element to a hundred different other elements and nobody would give a damn, the automorphism group sortof tells you that the element doesn't have any identity or personality of its own....'Just a new perspective on an old concept you probably never considered. Next time you're dabbling with a separable field extension, take some time to appreciate the roots of a high-degree polynomial for who they are themselves-because as far as the field is concerned, they don't have a name or place of their own.
I think the most interesting aspect of symmetry and group theory is that it is the method that plugs the last hole in Bayesian probability theory. It links probability theory with the symmetrical properties of the universe. One of the most often repeated criticism of Bayesian probability theory is that you need to set priors and these priors are subjective and biased. Bayesians counter that there are principles that allow us to pick the priors that are most uninformative and objective. We just need to following the maximum entropy principle. However it turns out that this is not enough. The MaxEnt principle fails in some situations to remove all ambiguity as to which is the most uninformative prior to choose or leaves the prior undefined. As it turns out, it seems that most of the time we can solve the ambiguity by using as prior information the symmetries inherent in the problems.
Humans seem to do this instinctively, when we probabilistically learn new things about and entity from observations, in many situations, we assume that if the entity moves or rotates it doesn't change the scales of effects and what we have learned is still valid. Translational symmetry seems to be a basic property of the universe in many situations.
And see in this unfinished draft paper how it can be used to solve problems of linear line fitting (I bet you didn't know there was controversy around such as simple concept as linear regressions):
http://bayes.wustl.edu/etj/articles/leapz.pdf
By Benoit Essiambre (not verified) on 06 Dec 2007 #permalink
arXiv:0712.0997
Title: On the realization of Symmetries in Quantum Mechanics
Authors: Kai Johannes Keller (1 and 2), Nikolaos A. Papadopoulos (2), Andrés F. Reyes-Lega (3) ((1) II. Inst. f. Theoretische Physik der Universität Hamburg, Germany, (2) Inst. f. Physik (WA THEP) der Johannes Gutenberg-Universität Mainz, Germany, (3) Departamento de FÃsica, Universidad de los Andes, Bogotá, Colombia) |
# Evaluate the definite integral using linearly and subdivision together with the following...
## Question:
Evaluate the definite integral using linearly and subdivision together with the following results. {eq}\int\limits_{-1}^2 x^2 d x = 3; \int\limits_{-1}^0 x^2 d x = \frac{1}{3}; \int\limits_{-1}^{2} x d x = \frac{3}{2}; \int\limits_0^2 x d x = 2 {/eq} Find {eq}\int\limits_{-1}^2 (x^2 + x) d x = {/eq}
## Integration:
Integration is reverse of differentiation.
We will use rule:
{eq}\displaystyle \int \:\left[f\left(x\right)\pm g\left(x\right)\right]dx=\int \:f\left(x\right)dx\pm \int \:g\left(x\right)dx {/eq}
Given integral is
{eq}\displaystyle I=\int _{-1}^2\left(x^2\:+\:x\right)\:dx\:\\ \displaystyle I=\int _{-1}^2x^2\:dx+\int _{-1}^2\:x\:dx\: {/eq}
Given values are
{eq}\displaystyle \int\limits_{-1}^2 x^2 d x = 3; \int\limits_{-1}^0 x^2 d x = \frac{1}{3}; \int\limits_{-1}^{2} x d x = \frac{3}{2}; \int\limits_0^2 x d x = 2 {/eq}
{eq}\displaystyle \therefore\:I=3+\frac{3}{2}\\ \displaystyle I=\frac{9}{2} {/eq} |
# 1. Find the prime factors of the following numbers by division method:(ii) 1756(iii) 1165(iv) 105002
1. Find the prime factors of the following numbers by division method:
(ii) 1756
(iii) 1165
(iv) 10500
2.Write all the prime factors of the following numbers using factor tree:
(i)550
(ii) 1952
(iii)7952
3. Determine the HCF of numbers in each of the following by prime factorisation method:
(i)144, 192
(ii)225, 450
(iii)624, 936
(iv)513, 783
(v) 288, 1375
(vi)875, 1859
(vii)144, 180and192
(viii)84, 120, 138
(ix)106, 159, 265
(x)625, 3125, 15625
4. Determine the HCF of numbers in each of the following by continued division method:
(i)513, 783
(ii) 216, 1176
(iii) 1965, 2096
(iv)935, 1320
(v) 1624, 522, 1276
(vi) 2241, 8217,747
5. Which of the following pair of numbers are co-prime:
(i)15, 190
(ii) 1903, 130
(iii) 1505, 1030
### 2 thoughts on “<br />1. Find the prime factors of the following numbers by division method:<br />(ii) 1756<br />(iii) 1165<br />(iv) 10500<br />2”
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Factors of 18
Factors of 18 are integers that can be divided evenly into 18. There are overall 6 factors of 18 among which 18 is the biggest factor and its positive factors are 1, 2, 3, 6, 9 and 18. The Pair Factors of 18 are (1, 18), (2, 9) and (3, 6) and its Prime Factors are 1, 2, 3, 6, 9, 18.
Factors of 18: 1, 2, 3, 6, 9 and 18
Negative Factors of 18: -1, -2, -3, -6, -9 and -18
Prime Factors of 18: 2, 3
Prime Factorization of 18: 2 × 3 × 3 = 2 × 32
Sum of Factors of 18: 39
Let us explore more about factors of 18 and ways to find them.
What Are the Factors of 18?
Factors of a number are the numbers that divide the given number exactly without any remainder. According to the definition |
# Introduction to Exponential and Logarithmic Functions
Let’s review some background material to help us study exponential and logarithmic functions.
## Exponential Functions
The function f(x) = 2x is called an exponential function because the variable, x, is the exponent. In general, exponential functions are of the form f(x) = ax, where a is a positive constant. There are three kinds of exponential functions:
Both the red and blue curves above are examples of exponential growth because their base is greater than 1. The green and purple curves are examples of exponential decay because their base is between 0 and 1. The yellow curve is a special case, and we won’t consider this further, since it can be classified as a linear model.
## The Natural Exponential
If you study calculus, you’ll find that the most convenient base for the exponential is Euler’s number, which is denoted by the letter e. This gives us the natural exponential function y = ex.
Euler’s number, e, is the number such that lim (eh – 1)/h = 1
h → 0
(ie., as h gets smaller, the function (eh – 1)/h approaches 1) and has a value such that e = 2.71828.
## Logarithmic Functions
The inverse of an exponential function is called a logarithmic function. Therefore, the inverse of f(x) = ax is the logarithmic function with base a, such that y = logx↔ ay = x.
In the figure above, the red line represents an exponential function and the blue line represents its inverse, the logarithmic function. Since the exponential and logarithmic functions are inverse functions, cancellation laws apply to give:
loga(ax) = x for all real numbers x
alogax = x for all x > 0
## The Natural Logarithm
We already stated that e is the most convenient base to work with for exponential functions. The same is true when working with logarithmic functions. The logarithmic function with base e is called the natural logarithm and is denoted by the special notation:
logex = In x
Now, the same cancellation laws can apply for the natural logarithm, such that:
In(ex) = x for all real numbers x
eInx = x for all x > 0
Finally, note that the logarithm and natural logarithm functions are related by the following change base formula:
logax = In x , where a ≠ 1
Ina |
How to Find the Limit of a Function Algebraically – 13 Best Methods
A Quick Summary of Limit
,A real number $l$ is said to be the limit of a function $f(x)$ at $x=a$ if for every +ve $\varepsilon$ there exists a +ve $\delta$ such that
$\left | f\left ( x \right ) - l \right |< \varepsilon$, whenever $0 < \left | x - a \right |< \delta$
and it is expressed as
$\boldsymbol{\lim_{x \to {\color{Red} a}}{\color{Blue} f(x)}={\color{Green} l}}$
Read more: Concept of Limit of a Function
Table of Contents - What you will learn
How to find the limit of a function Algebraically
There are different ways to find the limit of a function algebraically.
In this article, we will know about the 13 best methods to find the limit of a function.
#1. Direct Substitution
In the substitution method we just simply plug in the value of x in the given function f(x) for the limit.
Look at the examples given below:
$\lim_{x \to 3}5x=5\times {\color{Magenta} 3}=15$
Here we just plugged in x=3.
Is not it easy??
Some other examples:
• $\lim _{x \to -1}\left ( 2+x \right )^{10}=\left( 2{\color{Magenta} -1} \right )^{10}=1^{10}=1$
• $\lim _{x \to 1}\left ( x^{2}+x+3 \right )=({\color{Magenta} 1})^{2}+1+3=1+1+3=5$
• $\lim _{x \to -5}\left ( 10 \right )=10$
• $\lim _{x \to 3}\sqrt[3]{24+x}=\sqrt[3]{24+{\color{Magenta}3}}=\sqrt[3]{27}=3$
• $\lim _{x \to -2}\frac{x^{3}-5x+3}{x^{2}+1}=\frac{({\color{Magenta}-2})^{3}-5({\color{Magenta}-2})+3}{({\color{Magenta}-2})^{2}+1}=\frac{-8+10+3}{4+1}=\frac{5}{5}=1$
#2. Limits by Factoring
Evaluate $\lim _{x \to 2}\frac{x^{2}-5x+6}{x^{3}-5x+2}$.
If we plugin x=2, we get
$\lim _{x \to 2}\frac{x^{2}-5x+6}{x^{3}-5x+2}$=$\frac{({\color{Magenta} 2})^{2}-5({\color{Magenta} 2})+6}{({\color{Magenta} 2})^{3}-5({\color{Magenta} 2})+2}$=$\frac{4-10+6}{8-10+2}$=$\frac{0}{0}$, an indeterminate form
Hence the substitution method does not work to find the limit.
So we have to try another method to find the limit.
Now we try the Factor method.
$\: \: \: \lim _{x \to 2}\frac{x^{2}-5x+6}{x^{3}-5x+2}$
=$\lim _{x \to 2}\frac{\left ( x-3 \right ){\color{Magenta} \left ( x-2 \right )}}{{\color{Magenta} \left ( x-2 \right )}\left ( x^{2}+2x-1\right )}$
=$\lim _{x \to 2}\frac{\left ( x-3 \right )}{\left ( x^{2}+2x-1\right )}$
=$\frac{{\color{Magenta}2}-3}{{\color{Magenta} 2}^{2}+2({\color{Magenta} 2})-1}$ (by substituting x=2)
=$\frac{-1}{4+4-1}$
=$\frac{-1}{7}$
=$-\frac{1}{7}$, a finite number.
$\therefore \lim _{x \to 2}\frac{x^{2}-5x+6}{x^{3}-5x+2}=\frac{-1}{7}$
#3 Common Denominator method
Evaluate $\lim _{x \to 0}\frac{\frac{1}{x+2}-\frac{1}{2}}{x}$
First, we try to find the limit by using the direct substitution method.
If we plug in x=0, we get
$\lim _{x \to 0}\frac{\frac{1}{x+2}-\frac{1}{2}}{x}=\frac{\frac{1}{{\color{Magenta}0}+2}-\frac{1}{2}}{{\color{Magenta}0}}=\frac{\frac{1}{2}-\frac{1}{2}}{0}=\frac{0}{0}$, an indeterminate form.
$\therefore$ the direct substitution method did not work here.
Also, we can not factor the function $\frac{1}{x+2}-\frac{1}{2}$.
So we have to use any other method.
In this case, we use the Common Dinomenator method to find the limit.
$\: \: \: \lim_{x \to 0}\frac{\frac{1}{x+2}-\frac{1}{2}}{x}$
=$\lim_{x \to 0}\frac{\frac{{\color{Magenta}2}\times 1}{{\color{Magenta}2}(x+2)}-\frac{1{\color{Magenta}(x+2)}}{2{\color{Magenta}(x+2)}}}{x}$
=$\lim_{x \to 0}\frac{\frac{2-x-2}{2(x+2)}}{x}$
=$\lim_{x \to 0}\frac{\frac{-x}{2(x+2)}}{x}$
=$\lim_{x \to 0}\frac{-{\color{Magenta}x}}{2(x+2)}\times \frac{1}{{\color{Magenta}x}}$
=$\lim_{x \to 0}\frac{-1}{2(x+2)}$
=$\frac{-1}{2({\color{Magenta}0}+2)}$ (by plugging in x=0)
=$-\frac{1}{4}$
#4. Expansion Method – Open up Parenthesis
Evaluate $\lim _{x \to 0}\frac{(x+3)^{2}-9}{x}$
If we plug in x=0, we get
$\lim _{x \to 0}\frac{(x+3)^{2}-9}{x}=\frac{({\color{Magenta}0}+3)^{2}-9}{{\color{Magenta}0}}=\frac{9-9}{0}=\frac{0}{0}$, an indeterminate form.
So direct substitution does not work here.
Also, see that here we can not apply the Factor method and Common Denominator method to find the limit.
Also, see that finding the limit of a function $\frac{(x+3)^{2}-9}{x}$ using the Factor method and Common Denominator method is not possible.
Hence we use the 4th method i.e., Expansion method
$\: \: \: \lim_{x \to 0}\frac{(x+3)^{2}-9}{x}$
=$\lim_{x \to 0}\frac{x^{2}+6x+9-9}{x}$
=$\lim_{x \to 0}\frac{x^{2}+6x}{x}$
=$\lim_{x \to 0}\frac{{\color{Magenta}x}(x+6)}{{\color{Magenta}x}}$
=$\lim_{x \to 0}(x+6)$
=${\color{Magenta}0}+6$ (by plugging in x=0)
=$6$
$\therefore \lim _{x \to 0}\frac{(x+3)^{2}-9}{x}=6$
#5. Limits by Rationalizing – Rationalization Method
Evaluate $\lim_{x \to 8}\frac{x-8}{\sqrt{x+1}-3}$
If we plug in x=8, we get
$\lim_{x \to 8}\frac{x-8}{\sqrt{x+1}-3}=\frac{8-8}{\sqrt{8+1}-3}=\frac{0}{\sqrt{9}-3}=\frac{0}{3-3}=\frac{0}{0}$, an indeterminate form
$\therefore$ the Direct Substitution method has failed.
You can see that we can not factor $\sqrt{x+1}-3$. So the Factor method also fails.
Also, we can not use the Common Numerator method as there is no fraction in the numerator.
The Expansion Method also failed.
So what can we do to find the limit?
In such cases, we use the Rationalization Method to find the limit like this:
$\: \: \: \lim_{x \to 8}\frac{x-8}{\sqrt{x+1}-3}$
=$\lim_{x \to 8}\frac{x-8}{\sqrt{x+1}-3}\times \frac{{\color{Magenta}\sqrt{x+1}+3}}{{\color{Magenta}\sqrt{x+1}+3}}$ (multiplying numerator and denominator by $\sqrt{x+1}+3$, the conjugate of of $\sqrt{x+1}-3$)
=$\lim_{x \to 8}\frac{(x-8)(\sqrt{x+1}+3)}{(\sqrt{x+1})^{2}-(3)^{2}}$ (by using $(a+b)(a-b)=a^{2}-b^{2}$)
=$\lim_{x \to 8}\frac{(x-8)(\sqrt{x+1}+3)}{x+1-9}$
=$\lim_{x \to 8}\frac{{\color{Magenta}(x-8)}(\sqrt{x+1}+3)}{{\color{Magenta}x-8}}$
=$\lim_{x \to 8}\frac{\sqrt{x+1}+3}{1}$
=$\frac{\sqrt{{\color{Magenta} 8}+1}+3}{1}$ (by plugging in x=8)
=$\sqrt{9}+3$
=$3+3$
=$6$, a finite number
$\therefore \lim_{x \to 8}\frac{x-8}{\sqrt{x+1}-3}=6$
#6. Limit of Absolute Value function
Example: Find the limit if it exists
$\lim_{x \to 3}\left (\frac{\left |x-3\right |}{x-3}\right )$.
Solution:
If we plug in x=2, we get
$\lim_{x \to 3}\left (\frac{\left |x-3\right |}{x-3}\right )=\frac{\left |3-3\right |}{3-3}=\frac{0}{0}$, an indeterminate form.
So we can’t find the limit by substitution.
We will find the limit by using the definition of Absolute Value function.
The Absolute Value function is defined by
Using this definition we can write
Now we have the value of $\frac{\left |x-3\right |}{x-3}$ when $x\geq 0$ and $x<0$
So we have to find both the right-hand limit and the left-hand limit.
$\: \: \: \: \: \: \: \lim_{x \to 3+}\frac{\left |x-3\right |}{x-3}=\frac{x-3}{x-3}=1$
and $\lim_{x \to 3-}\frac{\left |x-3\right |}{x-3}=\frac{-(x-3)}{x-3}=-1$
See that
$\lim_{x \to 3+}\frac{\left |x-3\right |}{x-3}=1\neq -1=\lim_{x \to 3-}\frac{\left |x-3\right |}{x-3}$
i.e., Right hand limit $\neq$ Left hand limit
$\therefore \lim_{x \to 3}\left (\frac{\left |x-3\right |}{x-3}\right )$ does not exists.
#7. Limit of the form of sinx/x
Evaluate $\lim_{x \to 0}\frac{\sin 3x}{x}$
Substituting x=0, we get
$\lim_{x \to 0}\frac{\sin 3x}{x}=\frac{\sin (3\times 0)}{0}=\frac{\sin 0}{0}=\frac{0}{0}$, an indeterminate form.
Therefore we can not find the limit using Direct Substitution.
In such types of problems we use the Limit property given below:
Rule: $\lim_{x \to 0}\frac{\sin x}{x}=1$, where x is measured in radian (i.e., in circular measure).
Now first we transform $\frac{\sin 3x}{x}$ in the form of $\frac{\sin x}{x}$ then we find the limit like this:
$\lim_{x \to 0}\frac{\sin 3x}{x}=\lim_{x \to 0}\frac{\sin 3x}{{\color{Magenta} 3}\times x}\times \frac{{\color{Magenta} 3}}{1}=3\lim_{x \to 0}\frac{\sin 3x}{3x}=3\times 1=3$
Therefore $\lim_{x \to 0}\frac{\sin 3x}{x}=3$
Some other examples:
• $\lim_{x \to 0}\frac{\sin 4x}{8x}=\frac{1}{2}\lim_{x \to 0}\frac{\sin 4x}{4x}=\frac{1}{2}(1)=\frac{1}{2}$,
• $\lim_{x \to 0}\frac{x}{\sin x}=\frac{1}{\lim_{x \to 0}\frac{\sin x}{x}}=\frac{1}{1}=1$,
• $\lim_{x \to 0}\frac{x-\sin x}{x}=\lim_{x \to 0}\left ( 1-\frac{\sin x}{x} \right )=1-1=0$,
• $\lim_{x \to 0}\frac{\tan x}{x}=\lim_{x \to 0}\frac{\sin x}{x}\frac{1}{\cos x}=\lim_{x \to 0}\frac{\sin x}{x}\lim_{x \to 0}\frac{1}{\cos x}=1\times \frac{1}{\cos 0}=1(1)=1$,
#8. Limit of the form of (e^x -1)/x
Rule: $\lim_{x \to 0}\frac{e^{x}-1}{x}=1$
Example 1: How to find the limit of a function $\frac{e^{5h}-1}{3h}$ algebraically as x approaches zero.
Solution:
$\: \: \: \lim_{h \to 0}\frac{e^{5h}-1}{3h}$
=$\lim_{h \to 0}\frac{e^{5h}-1}{{\color{Magenta} 5}h}\frac{{\color{Magenta} 5}}{3}$
=$\frac{5}{3}\lim_{h \to 0}\frac{e^{5h}-1}{5h}$
=$\frac{5}{3}\times 1$ ($\because \lim_{x \to 0}\frac{e^{x}-1}{x}=1$)
=$\frac{5}{3}$
$\therefore \lim_{h \to 0}\frac{e^{5h}-1}{3h}=\frac{5}{3}$
Example 2: How do you find the limit of a function $\frac{e^{px}-e^{qx}}{x}$ as x approaches 0.
Solution:
$\: \: \: \lim_{x \to 0}\frac{e^{px}-e^{qx}}{x}$
=$\lim_{x \to 0}\frac{(e^{px}-1)-(e^{qx}-1)}{x}$
=$\lim_{x \to 0}\frac{e^{px}-1}{x}-\lim_{x \to 0}\frac{e^{qx}-1}{x}$
=$\lim_{x \to 0}\frac{e^{px}-1}{{\color{Magenta} p}x}\frac{{\color{Magenta} p}}{1}-\lim_{x \to 0}\frac{e^{qx}-1}{{\color{Magenta} q}x}\frac{{\color{Magenta} q}}{1}$
=$1\left ( \frac{p}{1} \right )-1\left ( \frac{q}{1} \right )$ ($\because \lim_{x \to 0}\frac{e^{x}-1}{x}=1$)
=$p-q$
$\therefore \lim_{x \to 0}\frac{e^{px}-e^{qx}}{x}=p-q$
Example 3:
$\: \: \: \lim_{x \to 0}\frac{e^{\log x}-1}{e^{x-1}-1}$
=$\lim_{x \to 0}\frac{x-1}{e^{x-1}-1}$ ( $\because e^{\log x}=x$ )
=$\lim_{z \to 0}\frac{z}{e^{z}-1}$ (where $x-1=z$; then $z\rightarrow 0$, when$x\rightarrow 1$)
=$\frac{1}{\lim_{z \to 0}\frac{e^{z}-1}{z}}$
=$\frac{1}{1}$ ($\because \lim_{x \to 0}\frac{e^{x}-1}{x}=1$)
=$1$
$\therefore \lim_{x \to 0}\frac{e^{\log x}-1}{e^{x-1}-1}=1$
#9. Limit of the form of (a^x – 1)/x
Rule: $\lim_{x \to 0}\frac{a^{x}-1}{x}=\log_{e}a,\:a>0$
Example 1: $\lim_{x \to 0}\frac{3^{x}-1}{x}=\log_{e}3 \: \left ( \because 3>0 \right )$
Example 2: Show that $\lim_{x \to 0}\frac{5^{x}-4^{x}}{x}=\log_{e}\left ( \frac{5}{4} \right )$
Solution:
$\: \: \: \lim_{x \to 0}\frac{5^{x}-4^{x}}{x}$
=$\lim_{x \to 0}\frac{(5^{x}-1)-(4^{x}-1)}{x}$
=$\lim_{x \to 0}\frac{5^{x}-1}{x}-\lim_{x \to 0}\frac{4^{x}-1}{x}$
=$\log_{e}5-\log_{e}4$ ($\because 5,4>0$ and $\lim_{x \to 0}\frac{a^{x}-1}{x}=\log_{e}a$, when $a>0$)
=$\log_{e}\left ( \frac{5}{4} \right )$
$\therefore \lim_{x \to 0}\frac{5^{x}-4^{x}}{x}=\log_{e}\left ( \frac{5}{4} \right )$
#10. Limit of the form of (x^n – a^n)/(x-a)
Rule: $\lim_{x \to a}\frac{x^{n}-a^{n}}{x-a}=na^{n-1}$, where $n$ is a rational number.
Example 1: Show that $\lim_{x \to -2}\frac{x^{5}+32}{x+2}=80$.
Solution:
$\: \: \: \lim_{x \to -2}\frac{x^{5}+32}{x+2}$
=$\lim_{x \to -2}\frac{x^{5}-(-2)^{5}}{x-(-2)}$
=$5\times (-2)^{5-1}$ ($\because \lim_{x \to a}\frac{x^{n}-a^{n}}{x-a}=na^{n-1}$ and -2 is a rational number)
=$5\times (-2)^{4}$
=$5\times 16$
=$80$
$\therefore \lim_{x \to -2}\frac{x^{5}+32}{x+2}=80$
Example 2: Prove that $\lim_{h \to 0}\frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h}=\frac{1}{3}x^{\frac{-2}{3}}$.
Solution:
$\: \: \: \lim_{h \to 0}\frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h}$
=$\lim_{h \to 0}\frac{(x+h)^{\frac{1}{3}}-x^{\frac{1}{3}}}{h}$
If we put, $x+h=z$,then $z\rightarrow x$, when $h\rightarrow 0$
Therefore the given limit
=$\lim_{z \to x}\frac{z^\frac{1}{3}-x^\frac{1}{3}}{z-x}$ ($\because x+z=h,\: \therefore h=z-x$)
=$\frac{1}{3}x^{\frac{1}{3}-1}$ ($\because \lim_{x \to a}\frac{x^{n}-a^{n}}{x-a}=na^{n-1}$)
=$\frac{1}{3}x^{\frac{-2}{3}}$
$\therefore \lim_{h \to 0}\frac{\sqrt[3]{x+h}-\sqrt[3]{x}}{h}=\frac{1}{3}x^{\frac{-2}{3}}$
#11. Limit of the form of (1/x)log(1+x)
Rule: $\lim_{x \to 0}\frac{1}{x}\log_{e}\left ( 1+x \right )=1$
Example: Show that $\lim_{x \to 0}\frac{\log_{e}\left ( 1+ax \right )}{x}=a$
Solution:
$\: \: \: \lim_{x \to 0}\frac{\log_{e}\left ( 1+ax \right )}{x}$
=$\lim_{x \to 0}\frac{\log_{e}\left ( 1+ax \right )}{{\color{Magenta}a}x}\frac{{\color{Magenta}a}}{1}$
=$a\lim_{x \to 0}\frac{1}{ax}\log_{e}\left ( 1+ax \right )$
=$a\lim_{z \to 0}\frac{1}{z}\log_{e}\left ( 1+z \right )$ ( where $z=ax$; then $z\rightarrow 0$, when $x\rightarrow 0$ )
=$a$ ($\because \lim_{x \to 0}\frac{1}{x}\log_{e}\left ( 1+x \right )=1$)
$\therefore \lim_{x \to 0}\frac{\log_{e}\left ( 1+ax \right )}{x}=a$
#12. Limit of the form of (1+x)^(1/x)
Rule: $\lim_{x \to 0}\left ( 1+x \right )^{\frac{1}{x}}=e$
Example 1: Prove that $\lim_{x \to 0}\left ( 1+ax \right )^{bx}=e^{ab}$.
Solution:
$\: \: \: \lim_{x \to 0}\left ( 1+ax \right )^{bx}$
=$\lim_{x \to 0}\left ( \left ( 1+ax \right )^\frac{1}{{\color{Magenta} ax}} \right )^{{\color{Magenta} ax}\times \frac{b}{x}}$
=$\left (\lim_{x \to 0} \left ( 1+ax \right )^\frac{1}{ax} \right )^{ab}$
=$\left (\lim_{z \to 0} \left ( 1+z \right )^\frac{1}{z} \right )^{ab}$ (where $z=ax$; then $z\rightarrow 0$, when $x\rightarrow 0$)
=$(e)^{ab}$ ($\because \lim_{x \to 0}\left ( 1+ax \right )^{bx}=e^{ab}$)
=$e^{ab}$
$\therefore \lim_{x \to 0}\left ( 1+ax \right )^{bx}=e^{ab}$
Example 2: Prove that $\lim_{x \to 0}\left ( 1+4x \right )^{\frac{x+2}{x}}=e^{8}$
Solution:
$\: \: \: \lim_{x \to 0}\left ( 1+4x \right )^{\frac{x+2}{x}}$
=$\lim_{x \to 0}(1+4x)^{1+\frac{2}{x}}$
=$\lim_{x \to 0}\left ( (1+4x)(1+4x)^{\frac{2}{x}} \right )$
=$\lim_{x \to 0}(1+4x)\times \lim_{x \to 0}(1+4x)^{\frac{1}{4x}\times 8}$
=$(1+4\times 0)\times \left ( \lim_{x \to 0}(1+4x)^{\frac{1}{4x}} \right )^{8}$
=$1\times \left ( \lim_{x \to 0}(1+4x)^{\frac{1}{4x}} \right )^{8}$
=$\left ( \lim_{x \to 0}(1+4x)^{\frac{1}{4x}} \right )^{8}$
=$\left ( \lim_{z \to 0}(1+z)^{\frac{1}{z}} \right )^{8}$ (where $z=ax$; then $z\rightarrow 0$, when $x\rightarrow 0$)
=$(e)^{8}$ ($\because \lim_{x \to 0}\left ( 1+x \right )^{\frac{1}{x}}=e$)
=$e^{8}$
$\therefore \lim_{x \to 0}\left ( 1+4x \right )^{\frac{x+2}{x}}=e^{8}$
#13. Limit of the form of ((1+x)^n-1)/x
Rule: $\lim_{x \to 0}\frac{(1+x)^{n}-1}{x}=n$
Example 1: Prove that $\lim_{x \to 0}\frac{(1+x)^{6}-1}{2x}=3$
Solution:
$\lim_{x \to 0}\frac{(1+x)^{6}-1}{2x}$
=$\lim_{x \to 0}\frac{(1+x)^{6}-1}{x}\left ( \frac{1}{{\color{Magenta} 2}} \right )$
=$\frac{1}{2}\times \lim_{x \to 0}\frac{(1+x)^{6}-1}{x}$
=$\frac{1}{2}\times 6$ ($\because \lim_{x \to 0}\frac{(1+x)^{n}-1}{x}=n$)
=$3$
$\therefore \lim_{x \to 0}\frac{(1+x)^{6}-1}{2x}=3$
We hope after reading this article you understand How to Find the Limit of a Function Algebraically.
Additionally you can read these articles:
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Alabama - Grade 1 - Math - Measurement - Ordering Objects by Length - 17
Description
Order three objects by length; compare the lengths of two objects indirectly by using a third object.
• State - Alabama
• Standard ID - 17
• Subjects - Math Common Core
• Math
• Measurement
More Alabama Topics
Explain that the two digits of a two-digit number represent amounts of tens and ones. a. Identify a bundle of ten ones as a "ten." b. Identify the numbers from 11 to 19 as composed of a ten and one, two, three, four, five, six, seven, eight, or nine ones. c. Identify the numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 as one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones). C. The numbers 10, 20, 30, 40, 50, 60, 70, 80, 90 refer to one, two, three, four, five, six, seven, eight, or nine tens (and 0 ones).
Use addition and subtraction to solve word problems within 20 by using concrete objects, drawings, and equations with a symbol for the unknown number to represent the problem. a. Add to with change unknown to solve word problems within 20. b. Take from with change unknown to solve word problems within 20. c. Put together/take apart with addend unknown to solve word problems within 20. d. Compare quantities, with difference unknown, bigger unknown, and smaller unknown while solving word problems within 20.
Solve word problems that call for addition of three whole numbers whose sum is less than or equal to 20 by using concrete objects, drawings, or equations with a symbol for the unknown number to represent the problem.
Compare pairs of two-digit numbers based on the values of the tens and ones digits, recording the results of comparisons with the symbols >, =, and < and orally with the words "is greater than," "is equal to," and "is less than."
Apply properties of operations as strategies to add and subtract. Examples: If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known (commutative property of addition). To add 2 + 6 + 4, the second and third numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12 (associative property of addition). When adding 0 to a number, the result is the same number (identity property of zero for addition). |
Apply the distributive property to factor out the greatest common factor.
56+32=
To apply the distributive property and factor out the greatest common factor, you need to find the greatest common factor of 56 and 32, and then rewrite the expression using the distributive property.
To find the greatest common factor of 56 and 32, we can list the factors of each number:
Factors of 56: 1, 2, 4, 7, 8, 14, 28, 56
Factors of 32: 1, 2, 4, 8, 16, 32
The greatest common factor (GCF) is the largest number that divides both 56 and 32 evenly. In this case, the GCF is 8.
Using the distributive property, we can rewrite the expression as:
56 + 32 = (8 * 7) + (8 * 4)
Now, we can factor out the GCF of 8:
56 + 32 = 8 * (7 + 4)
Therefore, the expression 56 + 32 can be factored out with the greatest common factor of 8 as 8 * (7 + 4).
To factor out the greatest common factor (GCF) from a given expression, you need to find the largest number or variable that can divide evenly into all the terms. In this case, let's start by identifying the GCF for the numbers 56 and 32.
To do that, we will list the factors of both numbers:
Factors of 56: 1, 2, 4, 7, 8, 14, 28, 56
Factors of 32: 1, 2, 4, 8, 16, 32
The common factors between both numbers are 1, 2, 4, and 8. The greatest common factor is 8.
Now, we can apply the distributive property by factoring out the GCF of 8 from the expression 56 + 32:
56 + 32 = 8 * (7 + 4)
We have factored out the GCF of 8 and rewritten the expression using the distributive property. |
# We Use Algebra Everyday?
Article Summary: When you have limited money and want to go some where with your friends, you have to budget your money to make sure that you have enough for the whole day. Mental algebra is used to determine costs of things, options for purchasing gifts, and of course having money to eat.
We use algebra everyday? That is the million dollar question. It is usually preceded with, "Why do I have to learn this stuff, I am never going to use it?" You are far from correct. Algebra is used everyday, all the time. It is used in problem solving situations when you are trying to determine how long it will take you to get from your home to your friends house.
Let's look at an example: You live five miles from your friend's home. Your parents need to drive you, so how long does it take them to get ready? If your parents drive the short way it will take 15 minutes, if there is lots of traffic, then it will take longer. So what time do you tell your friend you will be there if you leave at 4:00 PM.
Does this sound familiar to you? Sounds like the old train problem you had school. If the train leaves the station at x o'clock! Guess what, you are using algebra when trying to figure out how long it will take to get to your friend's house and it includes a variable "x" for traffic and time.
Let's look at another example: You and some friends are going to build a skate board half pipe. You draw a model to determine how tall and long to make it. You draw various representations of the half pipe from different angles. Then calculate how much wood you will need and what size, so it does not collapse. Then you need to calculate how much material you will need to make the surface of the half pipe smooth. With every one of your calculations you are using algebra. There are lots of variables and you have to use rational numbers to make your calculations - Algebra.
Every time you need to problem solve a situation that involves money, time, distance, perimeter of a fence or skate ramp, volume of something, comparing prices when you shop, rent something - cost versus time, other situations you are using algebra.
Algebra teaches you logical reasoning and problem solving skills when it comes to most every situation in life. You have to logically think your way through something to obtain the best results. For example: I want to jump my bike off the ramp a distance of 15 feet. You measure the height of the ramp and length of the run up distance, along with is the wind with your or against you. These are variables and rational numbers that are used in algebra. By the way you will also decide that you can or can not make the jump, logical reasoning helped make the decision.
When you play sports you have to mentally determine the angle you throw the ball to make an accurate throw. You now the approximate distance, but you have to determine how much force to apply to your throw. It also applies to soccer, when you are kicking the ball to another player or into the goal. The same mental calculations occur as you consider your options (variables and rational numbers) for making a goal. Algebra in action!
When you have limited money and want to go some where with your friends, you have to budget your money to make sure that you have enough for the whole day. Mental algebra is used to determine costs of things, options for purchasing gifts, and of course having money to eat.
Now for the boring stuff, when you finish school and start applying for jobs your possible future employer may give you a test with some math problems on it. They want to know if you can use mathematical skills to solve a problem and the problems will include algebra - problems with variables. Most employers that pay well will not hire you unless you can solve algebra type problems to prove that you have logical and reasoning problem solving skills. |
<meta http-equiv="refresh" content="1; url=/nojavascript/"> Perimeter and Area | CK-12 Foundation
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# 1.10: Perimeter and Area
Created by: CK-12
0 0 0
Pacing
Day 1 Day 2 Day 3 Day 4 Day 5
Triangles and Parallelograms Trapezoids, Rhombi, and Kites
More Trapezoids, Rhombi, and Kites
Start Area of Similar Polygons
Finish Area of Similar Polygons
Quiz 1
Start Circumference and Arc Length
Day 6 Day 7 Day 8 Day 9 Day 10
Finish Circumference and Arc Length
Investigation 10-1
Area of Circles and Sectors
Quiz 2
Start Review of Chapter 10
Review of Chapter 10 Chapter 10 Test
## Triangles and Parallelograms
Goal
This lesson introduces students to the area and perimeter formulas for triangles, parallelograms and rectangles.
Relevant Review
Most of this lesson should be review for students. They have learned about area and perimeter of triangles and rectangles in a previous math class (Math 6, Pre-Algebra, or equivalent).
Notation Note
In this chapter, students need to use square units. If no specific units are given, students can write $\text{units}^2$ or $u^2$.
Teaching Strategies
If students are having a hard time with the formulas for area and perimeter of a rectangle, place Example 3 on a piece of graph paper or transparency. Then, students can could the squares for the area and perimeter and you can generate the formula together.
If you count all the squares, there are 36 squares in the area, or square centimeters (red numbers). Counting around the rectangle (blue numbers), we see there are 26 squares. Therefore, the perimeter of this square is 26 cm.
This technique will also work for squares.
An important note, each problem will have some sort of units. Remind students that the shapes might not always be drawn to scale.
Example 5 is a counterexample for the converse of the Congruent Areas Postulate. Therefore, the converse is false. An additional counterexample would to have them draw all the possible rectangles with an area of $20 \ in^2$. Use graph paper so students will see that each rectangle has 20 squares. Possible answers are: $20 \times 1, 10 \times 2$, and $5 \times 4$.
The Area Addition Postulate encourages students to separate a figure into smaller shapes. Always divide the larger shape into smaller shapes that students know how to find the area of.
To show students the area of a parallelogram, cut out the picture (or draw a similar picture to cut out) of the parallelogram and then cut the side off and move it over so that the parallelogram is transformed into a rectangle. Explain to students that the line that you cut is the height of the parallelogram, which is not a side of the parallelogram. Then, cut this parallelogram along a diagonal to create a triangle. Here, students will see that the area of a triangle is half the area of a parallelogram. You may need to rotate the halves (triangles) so that they overlap perfectly. This will show the students that the triangles are congruent and each is exactly half of the parallelogram.
Create another set of flashcards for the area formulas in this chapter. These flashcards should be double-sided. The blank side should be a sketch of the figure and its name. The flip side should have the formula for its area and the formula for its perimeter. Students should create flashcards as the chapter progresses.
## Trapezoids, Rhombi, and Kites
Goal
This lesson further expands upon area formulas to include trapezoids, rhombi, and kites.
Relevant Review
Students might need a quick review of the definitions of trapezoids, rhombi, and kites. Go over their properties (especially that the diagonals of rhombi and kites are perpendicular) and theorems. Students may know the area formula of a trapezoid from a previous math class.
Review the Pythagorean Theorem and special right triangles. There are several examples and review questions that will use these properties. If students do not remember the special right triangle ratios, they can use the Pythagorean Theorem.
Teaching Strategies
Use the same technique discussed in the previous lesson for the area of a parallelogram and triangle. Cut out two congruent trapezoids and demonstrate the explanation at the beginning of the lesson explaining the area of a trapezoid. Going over this with students (rather than just giving them a handout or reading it) will enable them to understand the formula better. These activities are done best on an overhead projector.
Conveniently, the area formula of the rhombus and kite are the same. Again, you can cut out a rhombus and kite, then cut them on the diagonals and piece each together to form a rectangle. Generate the formula with students. Another way to write the formula of a rhombus is to say that it has 4 congruent triangles, with area $\frac{1}{2} \left (\frac{1}{2} d_1 \right ) \left (\frac{1}{2} d_2 \right ) = \frac{1}{8} d_1d_2$. Multiplying this by 4, we get $\frac{4}{8}d_1d_2 = \frac{1}{2}d_1d_2$. This process is not as easily done with a kite because one of the diagonals is not bisected.
Additional Example: Find two different rhombi that have an area of $48 \ \text{units}^2$.
Solution: The diagonals are used to find the area, so when solving this problem, we are going to be finding the diagonals’ lengths. $\frac{1}{2} d_1d_2 = 48$, so $d_1d_2 = 96$. This means that the product of the diagonals is double the area.
The diagonals can be: 1 and 96, 2 and 48, 3 and 32, 4 and 24, 6 and 16, 8 and 12.
As an extension, you can students draw the rhombi. The diagonals bisect each other, so have the diagonals cut each other in half and then connect the endpoints of the diagonals to form the rhombus. Three examples are below.
## Areas of Similar Polygons
Goal
Students will learn about the relationship between the scale factor of similar polygons and their areas. Students should also be able to apply area ratios to solving problems.
Relevant Review
Review the properties of similar shapes, primarily triangles and quadrilaterals, from Chapter 7. Remind students that the perimeter, sides, diagonals, etc. have the same ratio as the scale factor. The Review Queue reviews similar squares. As an additional question, ask students to find the perimeter of both squares and then reduce the ratio (smaller square = 40, larger square = 100, ratio is 2:5, the same as the ratio of the sides). Ask students why they think the ratio of the sides is the same as the ratio of the perimeters.
Teaching Strategies
Examples 1 and 2 lead students towards the Area of Similar Polygons Theorem. As an additional example (before introducing the Area of Similar Polygons Theorem), ask students to find the area of two more similar shapes. Having students repeat problems like Example 2, they should see a pattern and arrive at the theorem on their own.
Additional Example: Two similar triangles are below. Find their areas and the ratio of the areas. How does the ratio of the areas relate to the scale factor?
Solution: Each half of the isosceles triangles are 3-4-5 triangles. The smaller triangle has a height of 3 and the larger triangle has a height of 9 (because 12 is 3 time 4, so this triangle is three time larger than the smaller triangle). The areas are: $A_{larger \ \Delta} = \frac{1}{2} \cdot 24 \cdot 9 = 108$ and $A_{smaller \ \Delta} = \frac{1}{2} \cdot 8 \cdot 3 = 12$. The ratio of the area is $\frac{12}{108} = \frac{1}{9}$. The ratios of the scale factor and areas relate by squaring the scale factor, $\frac{1}{9} = \left (\frac{1}{3} \right )^2$.
## Circumference and Arc Length
Goal
The purpose of this lesson is to review the circumference formula and then derive a formula for arc length.
Relevant Review
The Review Queue is a necessary review of circles. Students need to be able to apply central angles, find intercepted arcs and inscribed angles. They also need to know that there are $360^\circ$ in a circle.
Teaching Strategies
Students may already know the formula for circumference, but probably do not remember where $\pi$ comes from. Investigation 10-1 is a useful activity so that students can see how $\pi$ was developed and why it is necessary to find the circumference and area of circles. You can decide to make this investigation teacher-led or allow students to work in pairs or groups. From this investigation, we see that the circumference is dependent upon $\pi$.
When introducing arc length, first have students find the circumference of a circle with radius of $6 (12 \pi)$. Then, see what the length of the arc of a semicircle $(6 \pi)$. Students should make the connection that the arc length of the semicircle will be half of the circumference. Then ask students what the arc length of half of the semicircle is $(3 \pi)$. Ask what the corresponding angle measure for this arc length would be $(90^\circ)$. See if students can reduce $\frac{90}{360}$ and if they make the correlation that the measure of this arc is a quarter of the total circumference, just like $90^\circ$ is a quarter of $360^\circ$. Using this same circle, see if students can find the arc length of a $30^\circ$ portion of the circle $\left (\frac{3 \pi}{3} = \pi \right )$. Then, as students what portion of the total circumference $\pi$ is. $\pi$ is $\frac{1}{12}$ of $12 \pi$, just like $30^\circ$ is $\frac{1}{12}$ of $360^\circ$. This should lead students towards the Arc Length Formula.
Students may wonder why it is necessary to leave answer in exact value, in terms of $\pi$, instead of approximate (multiplying by 3.14). This is usually a teacher preference. By using the approximate value for $\pi$, the answer automatically has a rounding error. Rounding the decimal too short will cause a much larger error than using the decimal to the hundred-thousandths place. Whatever your preference, be sure to explain both methods to your students. The review questions request that answers be left in terms of $\pi$, but this can be easily changed, depending on your decision.
## Area of Circles and Sectors
Goal
This lesson reviews the formula for the area of a circle and introduces the formula for the area of a sector and segment of a circle.
Teaching Strategies
If you have access to an LCD display or a computer lab, show students the animation of the area of a circle formula (link is in the FlexBook).
The formula for the area of a sector is very similar to the formula for arc length. Ask students to compare the two formulas. Stress to students that the angle fraction in the sector formula is the same as it is for the arc length formula. Therefore, students do not need to memorize a new formula; they just need to remember the angle fraction for both.
To find the area of the shaded regions (like Example 8), students will need to add or subtract areas of circles, triangles, rectangles, or squares in order to find the correct area. Encourage students to identify the shapes in these types of problems before they begin to solve it. At the end of this lesson, quickly go over problems 23-25, so that students know how to solve the problems that evening. Remind students to use the examples in the lesson to help them with homework problems.
Finding the area of a segment can be quite challenging for students. This text keeps the angles fairly simple, using special right triangle ratios. Depending on your level of student, you may decide to omit this portion of this lesson. If so, skip Example 9 and review questions 26-31.
Solution: The triangle that is inscribed in the circle is a 45-45-90 triangle and its hypotenuse is on the diameter of the circle. Therefore, the hypotenuse is $24 \sqrt{2}$ and the radius is $12 \sqrt{2}$. The area of the shaded region is the area of the circle minus the area of the triangle.
$A_{\bigodot} &= \pi (12\sqrt{2})^2 = \pi \cdot 144 \cdot 2 = 288 \pi\\A_{\Delta} &= \frac{1}{2} \cdot 24 \cdot 24 = 288$
The area of the shaded region is $288 \pi - 288 \approx 616.78 \ \text{units}^2$
Feb 22, 2012
Aug 21, 2014 |
# 007B Sample Midterm 3, Problem 1 Detailed Solution
Divide the interval ${\displaystyle [0,\pi ]}$ into four subintervals of equal length ${\displaystyle {\frac {\pi }{4}}}$ and compute the right-endpoint Riemann sum of ${\displaystyle y=\sin(x).}$
Background Information:
1. The height of each rectangle in the right-hand Riemann sum
is given by choosing the right endpoint of the interval.
Solution:
Step 1:
Let ${\displaystyle f(x)=\sin(x).}$
Each interval has length ${\displaystyle {\frac {\pi }{4}}.}$
Therefore, the right-endpoint Riemann sum of ${\displaystyle f(x)}$ on the interval ${\displaystyle [0,\pi ]}$ is
${\displaystyle {\frac {\pi }{4}}{\bigg (}f{\bigg (}{\frac {\pi }{4}}{\bigg )}+f{\bigg (}{\frac {\pi }{2}}{\bigg )}+f{\bigg (}{\frac {3\pi }{4}}{\bigg )}+f(\pi ){\bigg )}.}$
Step 2:
Thus, the right-endpoint Riemann sum is
${\displaystyle {\begin{array}{rcl}\displaystyle {{\frac {\pi }{4}}{\bigg (}\sin {\bigg (}{\frac {\pi }{4}}{\bigg )}+\sin {\bigg (}{\frac {\pi }{2}}{\bigg )}+\sin {\bigg (}{\frac {3\pi }{4}}{\bigg )}+\sin(\pi ){\bigg )}}&=&\displaystyle {{\frac {\pi }{4}}{\bigg (}{\frac {\sqrt {2}}{2}}+1+{\frac {\sqrt {2}}{2}}+0{\bigg )}}\\&&\\&=&\displaystyle {{\frac {\pi }{4}}({\sqrt {2}}+1).}\\\end{array}}}$
${\displaystyle {\frac {\pi }{4}}({\sqrt {2}}+1)}$ |
Can someone help me with this calculus population growth problem? I think I need to use a basic population growth equation.
Koala bears live on a farms land in Australia. The population's rate of growth is determined by the equation dK/dt = yK, where y is constant. The initial population is 100 and after one year, 120 koalas are present. Use this to find out how long it will take the population to increase from 100 to 300 koalas. Once the population reaches 300, a Koala-eating anaconda begins eating koalas, at a rate of 80 koalas every year. How long will it take until all the koalas are gone?
Aug 22, 2016
to reach 300, you have to observe koalas for 6 (full) years.
Explanation:
1st year: 20 (increase) Total number at the end of year: 120
Rate of increase is 0.20 or 20% or 20/100
2nd year= 1.2120 = 144 (total koalas)
3rd year=1.2
144 = 173 (total koalas)
4th year = 1.2173 = 208 (total koalas)
5th year= 1.2
208 = 250
6th year = 1.2*250 = 300
The first part of the question, you need 6 years to see 300 koalas.
An anaconda moves to the area, consuming 80 koalas per year.
After anaconda's presence:
1st year = (1.2300) - 80 = 280 (living koalas)
2nd year = (1.2
280) - 80 = 256 (living koalas)
3rd year = (1.2256) - 80 = 227 (living koalas)
4th year = (1.2
227) - 80 = 192 (living koalas)
5th year = (1.2192) - 80 = 150 (living koalas)
6th year = (1.2
150) - 80 = 100 (living koalas)
7th year = (1.2*100) - 80 = 40 (living koalas)
Before 8th year, all koalas will be consumed by the anaconda. Keep it simple and 7.5 years after first appearance of the anaconda, all koalas will be eaten up by this anaconda. |
## Force And Laws Of Motion Notes And NCERT Solutions Of Class 9th Chapter-9 Science.
• March 7, 2021
### Frequently Asked Questions on Force and Laws of Motion
Explain why some of the leaves may get detached from a tree if we vigorously shake its branch?
When the branch of the tree is shaken, the branch moves in a to-and-fro motion. However, the inertia of the leaves attached to the branch resists the motion of the branch. Therefore, the leaves that are weakly attached to the branch fall off due to inertia whereas the leaves that are firmly attached to the branch remain attached.
Why do you fall in the forward direction when a moving bus brakes to a stop and fall backwards when it accelerates from rest?
Initially, when the bus accelerates in a forward direction from a state of rest, the passengers experience a force exerted on them in the backward direction due to their inertia opposing the forward motion. Once the bus starts moving, the passengers are in a state of motion in the forward direction. When the brakes are applied, the bus moves towards a position of rest. Now, a force in the forward direction is applied to the passengers because their inertia resists the change in the motion of the bus. This causes the passengers to fall forwards when the brakes are applied.
Explain, why is it difficult for a fireman to hold a hose, which ejects large amounts of water at a high velocity?
For the hose to eject water at high velocities, a force must be applied on the water (which is usually done with the help of a pump or a motor). Now, the water applies an equal and opposite force on the hose. For the fireman to hold this hose, he must apply a force on it to overcome the force applied on the hose by the water. The higher the quantity and velocity of the water coming out of the hose, the greater the force that must be applied by the fireman to hold it steady.
When a carpet is beaten with a stick, dust comes out of it. Explain.
When the carpet is beaten with a stick, the stick exerts a force on the carpet which sets it in motion. The inertia of the dust particles residing on the carpet resists the change in the motion of the carpet. Therefore, the forward motion of the carpet exerts a backward force on the dust particles, setting them in motion in the opposite direction. This is why the dust comes out of the carpet when beaten.
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Common Core: 5th Grade Math : Number & Operations in Base Ten
Example Questions
← Previous 1 3 4 5 6 7 8 9 68 69
Example Question #1 : Understand Place Value: Ccss.Math.Content.5.Nbt.A.1
300 is how many times greater than 3? Using the same logic, 40 is how many times greater than 4?
300 is 10 times greater than 3, and 40 is 10 times greater than 4.
300 is 10 times greater than 3, and 40 is 100 times greater than 4.
300 is 100 times greater than 3, and 40 is 100 times greater than 4.
300 is 30 times greater than 3, and 40 is 4 times greater than 4.
300 is 100 times greater than 3, and 40 is 10 times greater than 4.
300 is 100 times greater than 3, and 40 is 10 times greater than 4.
Explanation:
Example Question #1 : Understand Place Value: Ccss.Math.Content.5.Nbt.A.1
hundreds is equivalent to how many ones?
Explanation:
The hundreds place has a value of times greater than the ones place value.
If we have hundreds, we can multiply by to find how many ones that is equivalent to.
Example Question #1 : Number & Operations In Base Ten
tens is equivalent to how many ones?
Explanation:
The tens place has a value of times greater than the ones place value.
If we have tens, we can multiply by to find how many ones that is equivalent to.
Example Question #1 : Number & Operations In Base Ten
hundreds is equivalent to how many ones?
Explanation:
The hundreds place has a value of times greater than the ones place value.
If we have hundreds, we can multiply by to find how many ones that is equivalent to.
Example Question #101 : How To Find A Ratio
hundreds is equivalent to how many ones?
Explanation:
The hundreds place has a value of times greater than the ones place value.
If we have hundreds, we can multiply by to find how many ones that is equivalent to.
Example Question #1 : Understand Place Value: Ccss.Math.Content.5.Nbt.A.1
tens is equivalent to how many ones?
Explanation:
The tens place has a value of times greater than the ones place value.
If we have tens, we can multiply by to find how many ones that is equivalent to.
Example Question #761 : Ssat Middle Level Quantitative (Math)
hundreds is equivalent to how many ones?
Explanation:
The hundreds place has a value of times greater than the ones place value.
If we have hundreds, we can multiply by to find how many ones that is equivalent to.
Example Question #1 : Number & Operations In Base Ten
hundreds is equivalent to how many ones?
Explanation:
The hundreds place has a value of times greater than the ones place value.
If we have hundreds, we can multiply by to find how many ones that is equivalent to.
Example Question #1 : Number & Operations In Base Ten
hundreds is equivalent to how many ones?
Explanation:
The hundreds place has a value of times greater than the ones place value.
If we have hundreds, we can multiply by to find how many ones that is equivalent to.
Example Question #2 : Understand Place Value: Ccss.Math.Content.5.Nbt.A.1
tens is equivalent to how many ones? |
# The Meaning of Standard Deviation (σ)
The standard deviation reflects the average of the differences between the individual results and their average, but it is not identical to this average, as we shall show. In order to avoid confusion, we denote the average of the results by the letter M. The greater the gap between the individual results and the M, the greater the standard deviation.
In other words, the more widely the results are dispersed around the average (M), the greater the standard deviation. The difference between the results and the average is always measured as a positive value. It makes no difference whether a datum is to the left or to the right of the average.
The standard deviation is not the average of the differences, although in many cases it is very close to the average of the differences, or is perhaps even equal to it. The standard deviation is calculated as follows (as in the previous example):
1. We take the square of the differences between the individual results and the average.
2. We calculate the average of the squares.
3. We calculate the square root of the averages of the squares (taking the square root is designed to eliminate the effect of squaring the differences).
This series of operations results in differences between the standard deviation and the average of the differences.
We will highlight the differences between the standard deviation and the average of the differences with an example concerning the mathematics grades of six different twelfth grade classes (numbered from 1 to 6). Each class has 10 students.
The highest grade is 11, and the lowest is 1.
The average grade in each class is 6.
The distribution of the grades in each of the classes is displayed in the next slide.
For example, in twelfth grade class no. 1 (the first row):
Table 3.18
grade 11 pts. 10 pts. 9 pts. 8 pts. 7 pts. 6 pts. 5 pts. 4 pts. 3 pts. 2 pts. 1 pt. (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) Class no. 1 ☺ ☺☺ ☺☺ ☺☺ ☺☺ ☺ Class no. 2 ☺☺ ☺☺ ☺☺ ☺☺ ☺☺ Class no. 3 ☺☺ ☺☺ ☺ ☺☺ ☺ ☺☺ Class no. 4 ☺☺ ☺ ☺☺ ☺☺ ☺☺ ☺☺ Class no. 5 ☺☺ ☺☺ ☺ ☺☺ ☺☺ ☺ Class no. 6 ☺☺ ☺☺ ☺ ☺☺ ☺☺ ☺ Calculations grade Average (M) Average Difference Standard deviation (σ) (1) (13) (14) (15) Class no. 1 6 pts. 0.8 pts. 1.1 pts. Class no. 2 6 pts. 1.6 pts. 2 pts. Class no. 3 6 pts. 2.4 pts. 2.45 pts. Class no. 4 6 pts. 3.6 pts. 3.63 pts. Class no. 5 6 pts. 4.0 pts. 4.0 pts. Class no. 6 6 pts. 5.0 pts. 5.0 pts.
Symmetric Distribution
In the previous example, the distribution of marks on both sides of the average is symmetrical for each class, i.e. the number of children who received a mark higher than the average by any specific number of points is the same as the number of children who received a mark lower than the average by that same number of points.
When the distribution is very unsymmetrical, large differences between the standard deviation and the average difference are possible.
Looking at the table, we see that in the lower rows, the grades are more widely-dispersed around the average grade than they are in the higher ones. This fact is obviously reflected in the calculation of the average difference (column 14) and the standard deviation (column 15).
Starting with row 5, column 14 is equal to column 15.
In the first 4 rows column 15 is higher than column 14, but the gap decreases in the lower rows.
The figures in columns no. 14 and 15 are measured in grade points.
For example, in row 3, the average difference (from M) is 2.4 points, and the standard deviation is 2.45 points. |
The eight point circle was first discovered by Louis Brand of Cincinnati in 1944. This theorem can be useful in proving the far more famous nine point circle theorem.
``` D
|
|
P | O
I |
| H
A________________|_________C
X |
| G
F |
|
M | N
|
|
|
|
B
```
Above illustrates quadrilateral ABCD, its perpendicular diagonals, the eight points, and centroid X of ABCD.
The theorem
For a quadrilateral whose diagonals are perpendicular, the midpoints M, N, O, P of each sides, and the feet F, G, H, I of perpendiculars from the midpoints to the opposite sides all lie on a circle centered at the centroid X of the quadrilateral.
To further explain, midpoint M, the midpoint of A and B, is defined as (1/2) (A + B) where "+" is the addition of vectors. Foot F is the perpendicular intersection point of line lAB that passes through A and B, and line lO that passes through point O. The centroid X is defined as (1/4)(A + B + C + D).
The proof
Quadrilateral MNOP is what is known as a Varignon parallelogram. The Varignon parallelogram of a quadrilateral is the parallelogram that forms from the midpoints of the sides of the original quadrilateral. The midpoints form a parallelogram because the new sides are parallel to the original quadrilateral's diagonals. Since it is given that the diagonals are perpendicular, the Varignon parallelogram is rectangular.
The theorem of Thales (pronounced: "tallies") states that for a circle which has PN as a diagonal, O lies on the circle iff angle PON is perpendicular. Since MNOP is a rectangle, both O and M lie on such circle. Since PN is the diagonal of this circle, the center is the midpoint of PN, or the midpoint of (1/2)(A + D) and (1/2)(B + C), which is (1/4)(A + B + C + D), the centroid of quadrilateral ABCD. Since the diagonals of MNOP are equal in length and bisect each other, MO is also a diagonal of the circle that contains points M, N, O, P.
∠ PGN is perpendicular by construction, hence point G also lies on the circle by the theorem of Thales. By reasons of symmetry it follows that all points M, N, O, P, F, G, H, I lie on the same circle.
QED
Source: http://cut-the-knot.com/Curriculum/Geometry/EightPointCircle.html |
Intermediate Algebra Tutorial 8
Intermediate Algebra
Tutorial 8: Introduction to Problem Solving
The sum of a number and 2 is 6 less than twice that number.
Make sure that you read the question carefully several times.
Since we are looking for a number, we will let
x = a number
Step 2: Devise a plan (translate).
Step 3: Carry out the plan (solve).
*Get all the x terms on one side
*Inv. of add 2 is sub. 2
*Inv. of mult. by -1 is div. by -1
If you take the sum of 8 and 2 it is 6 less than twice 8, so this does check.
FINAL ANSWER: The number is 8.
Find 72% of 35.
Make sure that you read the question carefully several times.
Since we are looking for a number, we will let
x = the number
Step 2: Devise a plan (translate).
Step 3: Carry out the plan (solve).
*Multiply
25.2 is 72% of 35.
FINAL ANSWER: The number is 25.2.
A local furniture store is having a terrific sale. They are marking down every price 45%. If the couch you have our eye on is \$440 after the markdown, what was the original price? How much would you save if you bought it at this sale?
Make sure that you read the question carefully several times.
Since we are looking for the price of the couch before the markdown, we will let
x = the price of the couch
Step 2: Devise a plan (translate).
Step 3: Carry out the plan (solve).
*Combine like terms (1 - .45 = .55)
*Inv. of mult. by .55 is div. by .55
If you subtract the 45% markdown form 800 you do get 440.
FINAL ANSWER: The original price is \$800. You save 800 - 440 = \$360.
A rectangular garden has a width that is 8 feet less than twice the length. Find the dimensions if the perimeter is 20 feet.
Make sure that you read the question carefully several times.
We are looking for the length and width of the rectangle. Since width can be written in terms of length, we will let
L = length
Width is 8 feet less than twice the length:
2L - 8 = width
Step 2: Devise a plan (translate).
Step 3: Carry out the plan (solve).
*Mult. ( ) by 2
*Combine like terms
*Inv. of sub. 16 is add 16
*Inv. of mult. by 6 is div. by 6
If length is 6, then width, which is 8 feet less than twice the length, would have to be 4. The perimeter of a rectangle with width of 4 feet and length of 6 feet is 20 feet.
FINAL ANSWER: Width is 4 feet. Length is 6 feet.
Complimentary angles, sum up to be 90 degrees. Find the measure of each angle in the figure below. Note that since the angles make up a right angle, they are complementary to each other.
Make sure that you read the question carefully several times.
We are already given in the figure that
x = one angle
x + 30 = other angle
Step 2: Devise a plan (translate).
Step 3: Carry out the plan (solve).
*Combine like terms
*Inv. of add 30 is sub. 30
*Inv. of mult. by 2 is div. by 2
If x is 30, then x + 30 = 60. 60 and 30 do add up to be 90, so they are complementary angles.
FINAL ANSWER: The two angles are 30 degrees and 60 degrees.
The sum of 3 consecutive odd integers is 57. Find the integers.
Make sure that you read the question carefully several times.
We are looking for 3 ODD consecutive integers, we will let
x = 1st consecutive odd integer
x + 2 = 2nd consecutive odd integer
x + 4 = 3rd consecutive odd integer
Step 2: Devise a plan (translate).
Step 3: Carry out the plan (solve).
*Combine like terms
*Inv. of add 6 is sub. 6
*Inv. of mult. by 3 is div. by 3
If you take the sum of 17, 19 and 21 you do get
FINAL ANSWER: The three consecutive odd integers are 17, 19 and 21.
The cost C to produce x numbers of VCR's is C = 1000 + 100x. The VCR's are sold wholesale for \$150 each, so the revenue is given by R = 150x.
Make sure that you read the question carefully several times.
We are looking for the number of VCR's needed to be sold to break even, we will let
x = the number of VCR's
Step 2: Devise a plan (translate).
Step 3: Carry out the plan (solve).
*Get all the x terms on one side
*Inv. of mult. by 50 is div. by 50
When x is 20 the cost and the revenue both equal 3000. |
Lesson
```1.7
MODELING WITH FUNCTIONS
Who was the
roundest knight at
King Arthur's
Round Table?
Sir Cumference
From Words to Expressions
A number increased by 2 then cut in half
5 decreased by a number then tripled
A number decreased by 7 then doubled
Area of a Circle
We know….
Area=πr²
But….
What happens when we have the
circumference and we have to find area???
Well….
Area of Circle Cont.
C= Circumference
Since C= 2πr we can solve for r to get r=
C/(2π). Then we can substitute to get area: A=
πr²= π(C/2 π))²= πC²/(4π²)= C²/(4π)
So …
Area= C²/(4π)
Example: C= 8 so 8²/(4π)= 5.093
Box Problem
A square of side X is cut out of each corner of
an 8 in. by 15 in. piece of cardboard and the
sides are folded up to form an open-topped
box. How big should the cut-out squares be
in order to produce the box of maximum
volume?
8
x
x
15
Solution
Volume = Length x Width x Height
V = (15-2x) (8-2x) (x)
X = 1.667 inches
Box Problem 2
A square side is cut out of each corner from a
20cm by 8cm piece of cardboard to form an
open-top box. Find the value of x for the box
to have the maximum amount of volume.
Solution
Volume = Length x Width x Height
V = (8-2x) (20-2x) (x)
X = 1.761 centimeters
Box Problem 3
Find the maximum volume.
10
X
X
47
Solution
Volume = Length x Width x Height
V = (10-2x) (47-2x) (x)
V = 526.847 units
``` |
Q:
# Use synthetic division and factor theorem to solve the below equation: 〖-3z〗^3+140-139z+38z^2=0
Accepted Solution
A:
Answer To solve the equation -3z^3 + 38z^2 - 139z + 140 = 0 using synthetic division and the factor theorem, we need to first find a factor of the polynomial. One way to do this is to try different values of z until we find one that makes the polynomial equal to zero. However, since the polynomial has degree three, we know that it has at most three roots, so we can use the rational root theorem to narrow down the possible values of z. The rational root theorem states that if a polynomial with integer coefficients has a rational root p/q, where p and q are integers with no common factors, then p must be a factor of the constant term of the polynomial, and q must be a factor of the leading coefficient of the polynomial. In this case, the constant term is 140, which has factors of ±1, ±2, ±4, ±5, ±7, ±10, ±14, ±20, ±28, ±35, ±70, and ±140, and the leading coefficient is -3, which has factors of ±1 and ±3. Therefore, the possible rational roots of the polynomial are: ±1/1, ±2/1, ±4/1, ±5/1, ±7/1, ±10/1, ±14/1, ±20/1, ±28/1, ±35/1, ±70/1, and ±140/1 ±1/3, ±2/3, ±4/3, ±5/3, ±7/3, ±10/3, ±14/3, ±20/3, ±28/3, ±35/3, ±70/3, and ±140/3 We can use synthetic division to test each of these possible roots. For example, if we test z = 1, we get: 1 | -3 38 -139 140 | -3 35 -104 | -3 35 -104 36 Since the remainder is not zero, z = 1 is not a root of the polynomial. We can continue testing the other possible roots until we find one that works. In this case, we find that z = 4 is a root of the polynomial, so we can use synthetic division to factor the polynomial as: (z - 4)(-3z^2 + 26z - 35) = 0 Therefore, the solutions of the equation are z = 4, z = (13 + √229)/3, and z = (13 - √229)/3. In summary, we can use synthetic division and the factor theorem to solve the equation -3z^3 + 38z^2 - 139z + 140 = 0 by first finding a factor of the polynomial using the rational root theorem, and then using synthetic division to factor the polynomial and find the solutions of the equation. |
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# Expression Evaluation with Products of Fractions
## Evaluating expressions multiplying fractions < 1.
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Expression Evaluation with Products of Fractions
Credit: Oscar
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Jae is creating a model building for a design competition. Each floor will be 312\begin{align*}3\frac{1}{2}\end{align*} inches tall. How can Jae find the total height of his building model. How tall will it be if Jae decides to have 10 floors?
In this concept, you will learn how to evaluate expressions and products of fractions
### Evaluating Expressions with Products of Fractions
An expression is a numerical phrase that combines numbers and operations but no equal sign.
There are two kinds of expressions. Numerical expressions include numbers and operations only. Algebraic expressions include numbers, operations, and variables.
#### Types of Expressions
Includes Examples
numerical expressions numbers, operations
3+4\begin{align*}3+4\end{align*}
34×23\begin{align*}\frac{3}{4} \times \frac{2}{3}\end{align*}
15.68\begin{align*}15.6-8\end{align*}
4(34)\begin{align*}4 \left (\frac{3}{4} \right )\end{align*}
algebraic expressions numbers, operations, variables
3+x\begin{align*}3 + x\end{align*}
34b3\begin{align*}\frac{3}{4} \cdot \frac{b}{3}\end{align*}
15.6q\begin{align*}15.6 - q\end{align*}
c(34)\begin{align*}c \left (\frac{3}{4}\right )\end{align*}
Since a numerical expression includes numbers and operations, perform the operation required to evaluate. To evaluate means to find the value of the expression.
Evaluate the numerical expression.
(14)(34)\begin{align*}\left ( \frac{1}{4} \right ) \left ( \frac{3}{4} \right )\end{align*}
Notice that there are two sets of parentheses. Remember that the parentheses can identify groups and also indicate multiplication.
Evaluate by multiplying and then simplifying or by simplifying first then multiplying.
14×34=316\begin{align*}\frac{1}{4} \times \frac{3}{4} = \frac{3}{16}\end{align*}
The answer is in simplest form.
The value of the expression is 316\begin{align*}\frac{3}{16}\end{align*}.
As you learn about algebra and higher levels of math, you will be working with algebraic expressions.
An algebraic expression is similar to a numeric expressions, except that it uses variables. Variables are letters that represent an unknown number. Sometimes the values of the variables are given.
Evaluate the algebraic expression.
xy\begin{align*}xy\end{align*} when x=34\begin{align*}x = \frac{3}{4}\end{align*}, and y=13\begin{align*}y = \frac{1}{3}\end{align*}
To evaluate this expression, substitute the given values for x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} into the expression. Notice that the expression has x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*} next to each other. When two variables are next to each other the operation is multiplication.
34×13\begin{align*}\frac{3}{4} \times \frac{1}{3}\end{align*}
Then, simplify the expression. Cross simplify the 3s with the greatest common factor (GCF) of 3. Each three becomes a one.
34×13=14×11\begin{align*}\frac{3}{4} \times \frac{1}{3} = \frac{1}{4} \times \frac{1}{1}\end{align*}
Next, multiply to evaluate the expression.
14×11=14\begin{align*}\frac{1}{4} \times \frac{1}{1} = \frac{1}{4}\end{align*}
The value of the expression is 14\begin{align*}\frac{1}{4}\end{align*}.
### Examples
#### Example 1
Earlier, you were given a problem about Jae and his design competition.
Jae is making a model building where each floor is 312\begin{align*}3\frac{1}{2}\end{align*} inches tall. Create an expression that will help determine the total height of the building depending on the number of floors he decides to make.
For Jae's model building, he can create an algebraic expression to determine the height of his model. The total height will be the the height of each floor times the number of floors.
312x,where x is equal to the total number of floors\begin{align*}3\frac{1}{2} x, \text {where } x \text { is equal to the total number of floors}\end{align*}
Find the height of the model when x = 10.
First, substitute the variable with the given value.
312×10\begin{align*}3\frac{1}{2} \times 10\end{align*}
Then, convert the mixed number and whole number into an improper fraction.
312×10=72×101\begin{align*}3\frac{1}{2} \times 10 = \frac {7}{2} \times \frac{10}{1}\end{align*}
Next, simplify the fractions.
72×101=71×51\begin{align*}\frac {7}{2} \times \frac{10}{1}=\frac {7}{1} \times \frac{5}{1}\end{align*}
Finally, multiply the fractions.
71×51=35\begin{align*}\frac {7}{1} \times \frac{5}{1}=35\end{align*}
Jae's 10 floor model will be 35 inches tall.
#### Example 2
Evaluate xy\begin{align*}xy\end{align*} when x\begin{align*}x\end{align*} is 23\begin{align*}\frac{2}{3}\end{align*} and y\begin{align*}y\end{align*} is 812\begin{align*}\frac{8}{12}\end{align*}.
First, substitute the values into the expression.
23×812\begin{align*} \frac{2}{3} \times \frac{8}{12}\end{align*}
Then, simplify the expression.
23×812=13×43\begin{align*} \frac{2}{3} \times \frac{8}{12} = \frac{1}{3} \times \frac{4}{3} \end{align*}
Next, multiply the fractions.
13×43=49\begin{align*} \frac{1}{3} \times \frac{4}{3} = \frac{4}{9}\end{align*}
The answer is 49\begin{align*}\frac{4}{9}\end{align*}.
#### Example 3
Evaluate (47)(2128)\begin{align*}\left ( \frac{4}{7} \right ) \left ( \frac{21}{28} \right )\end{align*}. Answer in simplest form.
First, simplify the fractions.
47×2128=11×37\begin{align*}\frac{4}{7} \times \frac{21}{28} = \frac{1}{1} \times \frac{3}{7} \end{align*}
Then, multiply.
1×37=37\begin{align*}1 \times \frac{3}{7} = \frac{3}{7} \end{align*}
The value of the expression is 37\begin{align*}\frac{3}{7}\end{align*}.
#### Example 4
Evaluate (xy)\begin{align*}(xy)\end{align*} when x\begin{align*}x\end{align*} is 35\begin{align*}\frac{3}{5}\end{align*} and y\begin{align*}y\end{align*} is 1011\begin{align*}\frac{10}{11}\end{align*}. Answer in simplest form.
First, substitute the values into the expression.
35×1011\begin{align*}\frac{3}{5}\times \frac{10}{11}\end{align*}
Then, simplify the fractions.
35×1011=31×211\begin{align*}\frac{3}{5}\times \frac{10}{11}= \frac{3}{1}\times \frac{2}{11}\end{align*}
Next, multiply.
31×211=611\begin{align*}\frac{3}{1}\times \frac{2}{11} = \frac{6}{11}\end{align*}
The value of the expression is 611\begin{align*}\frac{6}{11}\end{align*}.
#### Example 5
Evaluate (59)(4560)\begin{align*}\left ( \frac{5}{9} \right ) \left ( \frac{45}{60} \right )\end{align*}. Answer in simplest form.
First, simplify the fractions.
59×4560=53×14\begin{align*} \frac{5}{9} \times \frac{45}{60} = \frac {5}{3} \times \frac {1}{4}\end{align*}
Next, multiply.
53×14=512\begin{align*} \frac {5}{3} \times \frac {1}{4}= \frac{5}{12}\end{align*}
The value of the expression is 512\begin{align*}\frac{5}{12}\end{align*}.
### Review
Evaluate each expression. Answer in simplest form.
1. Evaluate (xy)\begin{align*}(xy)\end{align*} when x=23\begin{align*}x = \frac{2}{3}\end{align*} and y=610\begin{align*}y = \frac{6}{10}\end{align*}
2. Evaluate (xy)\begin{align*}(xy)\end{align*} when x=13\begin{align*}x = \frac{1}{3}\end{align*} and y=410\begin{align*}y = \frac{4}{10}\end{align*}
3. Evaluate (xy)\begin{align*}(xy)\end{align*} when x=1213\begin{align*}x = \frac{12}{13}\end{align*} and y=26\begin{align*}y = \frac{2}{6}\end{align*}
4. Evaluate (xy)\begin{align*}(xy)\end{align*} when x=13\begin{align*}x = \frac{1}{3}\end{align*} and y=45\begin{align*}y = \frac{4}{5}\end{align*}
5. Evaluate (xy)\begin{align*}(xy)\end{align*} when x=79\begin{align*}x = \frac{7}{9}\end{align*} and y=321\begin{align*}y = \frac{3}{21}\end{align*}
6. Evaluate (xy)\begin{align*}(xy)\end{align*} when x=45\begin{align*}x = \frac{4}{5}\end{align*} and y=1620\begin{align*}y = \frac{16}{20}\end{align*}
7. Evaluate (46)(12)\begin{align*}\left ( \frac{4}{6} \right ) \left ( \frac{1}{2} \right )\end{align*}
8. Evaluate (19)(618)\begin{align*}\left ( \frac{1}{9} \right ) \left ( \frac{6}{18} \right )\end{align*}
9. Evaluate (49)(14)\begin{align*}\left ( \frac{4}{9} \right ) \left ( \frac{1}{4} \right )\end{align*}
10. Evaluate (411)(1112)\begin{align*}\left ( \frac{4}{11} \right ) \left ( \frac{11}{12} \right )\end{align*}
11. Evaluate (910)(56)\begin{align*}\left ( \frac{9}{10} \right ) \left ( \frac{5}{6} \right )\end{align*}
12. Evaluate (89)(36)\begin{align*}\left ( \frac{8}{9} \right ) \left ( \frac{3}{6} \right )\end{align*}
13. Evaluate (1819)(36)\begin{align*}\left ( \frac{18}{19} \right ) \left ( \frac{3}{6} \right )\end{align*}
14. Evaluate (49)(3640)\begin{align*}\left ( \frac{4}{9} \right ) \left ( \frac{36}{40} \right )\end{align*}
15. Evaluate (1214)(78)\begin{align*}\left ( \frac{12}{14} \right ) \left ( \frac{7}{8} \right )\end{align*}
To see the Review answers, open this PDF file and look for section 7.5.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Algebraic Expression An expression that has numbers, operations and variables, but no equals sign.
Numerical expression A numerical expression is a group of numbers and operations used to represent a quantity.
Product The product is the result after two amounts have been multiplied. |
# In a club of 7 members, it was found that after replacing an old member by a new member the average age of the members is the same as it was 5 years a
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In a club of 7 members, it was found that after replacing an old member by a new member the average age of the members is the same as it was 5 years ago. What is the between ages of the replaced and the new member?
1. 37 years
2. 45 years
3. 35 years
4. 47 years
by (53.1k points)
selected
Correct Answer - Option 3 : 35 years
Given
After replacing an old member by a new member the average age of the members is the same as it was 5 years ago.
Formula Used
Average = Sum of observations / Number of observations
Calculation
i) Let the ages of the 7 members at present be a, b, c, d, e, f and g years and the age of the new member be h years
ii) So, the new average age of 7 members = (a + b + c + d + e + f + h) / 7 ----(1)
iii) Their corresponding ages 5 years ago = (a – 5), (b -5), (c- 5), (d – 5), (e -5), (f – 5) and (g -5) years
So, 5 years age let their average ages = [(a + b + c + d + e + f + g)] / 7 = x ----(2)
a + b + c + d + e + f + g = 7x + 35
⇒ a + b + c + d + e + f = 7x + 35 – g (3)
(ii) Put the value of a + b + c + d + e + f from eq. (3) in eq. (1)
The new average age = (7x + 35 – g + h) / 7
Equating this to the average age of 5 years ago in eq. (2)
(7x + 35 – g + h) / 7 = x
⇒ 7x + 35 – g + h = 7x
On solving, we get
g – h = 35
∴ the difference of ages between replaced and new member is 35 years. |
# SAT Math Multiple Choice Question 741: Answer and Explanation
### Test Information
Question: 741
6.
The figure above shows the locations of quadrants I-IV in the xy-plane. Which of the following represents a pair of linear equations that do NOT intersect in quadrant I?
• A. 3x + 5y = 15
y = 4
• B. 5x + 3y = 15
y = 4
• C. 5x - 3y = 15
y = 4
• D. 3x - 5y = 15
y = 4
Explanation:
A
Algebra (graphing lines) MEDIUM-HARD
In quadrant I, both the x- and y-coordinates are positive. Since y = 4 in all four systems, we simply need to find the system for which the x-coordinate of the solution is not positive. We can find the corresponding x-coordinate for each system by just substituting y = 4 and solving for x.
Substitute y = 4 into first equation in (A):
3x + 5(4) = 15
Simplify:
3x + 20 = 15
Subtract 20:
3x = -5
Divide by 3:
x = -5/3
In this case, we don't need to go any further, because the solution to the system in (A) is (-5/3, 4), which is in quadrant II, not quadrant I. |
# Polynomials Class 10th
## Introduction
A polynomial word is made up of two Greek terms “poly” which means many and “nominal” which means terms So, a polynomial means many terms. Polynomials are the combined expression of variables, constants, and exponents (only whole numbers) with arithmetic expressions like addition, subtraction, multiplication, and division (division by only constant terms).
Polynomials can have one variable and more than one variable with a finite number of terms.
Examples
All the above examples are polynomials. A polynomial in one variable is denoted by p(x), g(x), f(a), q(y)…. etc. where the letter written in the bracket shows the variable of the polynomial.
If we take the example p(x) = 5x2 + 6x – 4, there are three terms in it.
Variable = x, Exponent term = x2, constant = 4, coefficient of x2 and x = 5 & 6
Terms – In the polynomial, terms are connected with the arithmetic operations of addition and subtraction.
Examples –
Variable – Variables are unknown values that are denoted by the English alphabet. Like x, y, z, a, t….. etc.
Constants – Constants are the known numbers used in polynomials. It is real numbers. Like 5, 6, -8, -3, 1.4, 6.2, 1/3.…. etc.
Exponent – The number or alphabet which is continuously multiplied by itself can be written in the exponent form. Like x3, y5, z6…… etc.
Coefficient – The multiplication factor of any term is called the coefficient. Like coefficient of x2 in 5x2 is 5.
### Types of Polynomials
Constant Polynomial – The polynomial which contains only a single constant value is called a constant polynomial.
Examples – 3, -8, 2, -9…..etc. 0 is called a zero polynomial.
MonomialsPolynomials having only one term are called monomials.
Examples – 3x, 7x4, 5, t7, – 9z2……etc.
BinomialsPolynomials having two terms are called binomials.
Examples – x + 8, y9 – z21, t8 + 9…….etc.
Trinomials – Polynomials having three terms are called trinomials.
Examples – 6zt – 7x + 8, 8xyz + y3 – t, x4 – 9x + 8……etc.
### Degree of the Polynomials
When the polynomial is in one variable, then the highest power of the variable in the polynomial is called the degree of polynomials.
What to do when more than one variable is present in the polynomial. Here is the example – Find the Degree of 7x2y + x2 + xy + 2
To find the degree, we add the powers (exponents) of variables of each term, then the highest power will be the degree of that polynomial.
In the above example, first term x2y = 2+1 = 3 there is power 1 on y, generally do not write.
Second term x2 = 2
Third term xy = 1+1 = 2
Fourth term 2 = 0 (no variable)
The highest power is 3 so the degree is 3.
• The Degree of a non-zero constant polynomial is zero.
• The degree of a zero polynomial is not defined.
• Polynomials having degree 1 are called linear polynomials. the maximum terms in a linear polynomial are two so the linear polynomial is either a binomial or a monomial.
• Polynomials having degree 2 are called quadratic polynomials. The maximum terms in a quadratic polynomial in one variable are three so the quadratic polynomial can be monomial, binomial, or trinomial.
• Polynomials having degree 3 are called cubic polynomials. The maximum terms in a cubic polynomial in one variable are four so the cubic polynomial can be monomial, binomial, or trinomial.
A polynomial in one variable x of degree n is expressed by the form –
P(x) = anxn + an-1xn-1 +…………+ a1x + a0
Where an ≠ 0, and an, an-1, a1, a0 = constants
• x + ⅟x
• √y + 5
• z-4 – z
### Like Terms and Unlike Terms
Like Terms – In polynomials, if terms have the same algebraic factors, they are called like terms.
Examples –
• 4xy and 7xy
• 2x2z, x2z, and 9zx2
• 3abc, 8acb and 5cba
Unlike Terms – In polynomials, if terms have different algebraic factors, they are called, Unlike terms.
Examples –
• 7st and 6st2
• 3pqr, pqrs and 9p2
• 6sr, s3r, and rst
### Operations on Polynomials
➤ Subtraction of polynomials
➤ Multiplication of polynomials
➤ Division of polynomials
When we add two or more polynomials then only like terms are added. it means terms with the same variables and the same powers are added. unlike terms will not be added, they will remain unchanged. In addition, the degree of the resultant polynomial remains the same.
Example – Add polynomials 3x2 + 4xy + 2y2 and 5y2 – xy + 7x2
Solution – 3x2 + 4xy + 2y2 + 5y2 – xy + 7x2
(3x2 + 7x2) + (4xy – xy) + (2y2 + 5y2)
10x2 + 3xy + 7y2 Ans.
#### Subtraction of Polynomials
Subtraction of the polynomials is the same as the addition of the polynomials. In this, like terms are subtracted and the unlike terms will be unchanged. In this also, the degree of the resultant polynomial will remain the same.
Example – Subtract the polynomial 3xy2 + 4xy – 6x2y + y3 from 8xy2 + 7xy + x2y.
Solution – 8xy2 + 7xy + x2y – (3xy2 + 4xy – 6x2y + y3)
8xy2 + 7xy + x2y – 3xy2 – 4xy + 6x2y – y3
(8xy2 – 3xy2) + (7xy – 4xy) + (x2y + 6x2y) – y3
5xy2 + 3xy + 7x2y – y3 Ans.
#### Multiplication of Polynomials
When two or more polynomials are multiplied then the result is always of a higher degree polynomial. But in two polynomials, if one or both of the polynomials are constant polynomials then the degree will remain the same. In the multiplication of polynomials, powers of the same variables are added by the laws of exponents.
#### Multiplying a Monomial by a Monomial
Multiplying two monomials
Examples – 1) 4x⨯6y = (4⨯6)⨯(x⨯y) = 24xy
2) 3a⨯(-2b) = {3⨯(-2)}⨯(a⨯b) = -6ab
3) 6p2⨯8p = (6⨯8)⨯(p2⨯p) = 48p3
4) x4yz2⨯2xy3z = (1⨯2)⨯(x4⨯x⨯y⨯y3⨯z2⨯z) = 2x5y4z3 Ans.
Multiplying three or more monomials
Examples – 1) 7t⨯2s⨯3r = (7⨯2⨯3)⨯(t⨯s⨯r) = 42tsr Ans.
2) 4pq⨯5p2q2⨯6p3q3
= (4⨯5⨯6)⨯(p⨯p2⨯p3⨯q⨯q2⨯q3)
= 120p6q6 Ans.
3) 2x2y⨯(-4y2z)⨯(-7z2x)⨯2x2yz
= [2x2y⨯(-4y2z)]⨯[(-7z2x)⨯2x2yz]
= (-8x2y3z)⨯(-14x3yz3)
= 112x5y4z4 Ans.
#### Multiplying a Monomial by a Polynomial
Multiplying a monomial by a binomial
Examples – 1) 2a⨯(3b+4)
Using Distributive law,
= 2a⨯3b + 2a⨯4
= 6ab + 8a Ans.
2) 7xy⨯(y2+5)
= 7xy⨯y2 + 7xy⨯5
= 7xy3 + 35xy Ans.
Multiplying a monomial by a trinomial
Examples – 1) 3v⨯(4v2+5v+7)
Using Distributive law,
= 3v⨯4v2 + 3v⨯5v + 3v⨯7
= 12v3 + 15v2 + 21v Ans.
2) x2⨯(9xz+6y2-xyz)
= x2⨯9xz + x2⨯6y2 – x2⨯xyz
= 9x3z + 6x2y2 – x3yz Ans.
#### Multiplying a Polynomial by a Polynomial
Multiplying a binomial by a binomial
Examples – 1) (2x+3y)⨯(3x+2y)
= 2x⨯(3x+2y) + 3y⨯(3x+2y)
= 2x⨯3x + 2x⨯2y + 3y⨯3x + 3y⨯2y
= 6x2 + 4xy + 9xy + 6y2
= 6x2 + 13xy + 6y2 Ans.
2) (p2-6qr)⨯(9q2+7r)
= p2⨯(9q2+7r) – 6qr⨯(9q2+7r)
= p2⨯9q2 + p2⨯7r – 6qr⨯9q2 – 6qr⨯7r
= 9p2q2 + 7p2r – 54q3r – 42qr2 Ans.
Multiplying a binomial by a trinomial
Examples – 1) (a+5)⨯(a2+2a-7)
= a⨯(a2+2a-7) + 5⨯(a2+2a-7)
= a⨯a2+a⨯2a-a⨯7 + 5⨯a2+5⨯2a-5⨯7
= a3+2a2-7a + 5a2+10a-35
= a3+7a2+3a-35 Ans.
2) (x2-3)⨯(xy+yz+zx)
= x2⨯(xy+yz+zx) – 3⨯(xy+yz+zx)
= x2⨯xy+x2⨯yz+x2⨯zx – 3⨯xy – 3⨯yz – 3⨯zx
= x3y + x2yz + x3z – 3xy – 3yz – 3zx Ans.
#### Division of Polynomials
In the division of polynomials, the result is a lesser degree polynomial and if one of the polynomials is a constant polynomial then the degree will remain the same.
#### Dividing a Monomial by a Monomial
For easy Division, we factorize both Dividend and Divisor in this type of Division.
Examples – 1) 6x4 ÷ 3x
= 2⨯3⨯x⨯x⨯x⨯x/3⨯x = 2⨯x⨯x⨯x = 2x3 Ans.
2) 72p2q3r ÷ (- 6p4q)
= 2⨯2⨯2⨯3⨯3⨯p⨯p⨯q⨯q⨯q⨯r/-2⨯3⨯p⨯p⨯p⨯p⨯q
= 2⨯2⨯3⨯q⨯q⨯r/-p⨯p = 12q2r/-p2 Ans.
#### Dividing a Polynomial by a Monomial
Examples – 1) (2xy + 6x) ÷ 2x
= (2xy + 6x)/2x = 2x(y + 3)/2x = (y+3) Ans.
2) (8y3 + 6y2 + 12y) ÷ 2y
= (8y3 + 6y2 + 12y)/2y = 2y(4y2 + 3y+ 6)/2y
= 4y2 + 3y+ 6 Ans.
This can be also solved as = 8y3/2y + 6y2/2y + 12y/2y = 4y2 + 3y+ 6
#### Dividing a Polynomial by a Polynomial
Examples – 1) (7a2 + 14a) ÷ (a + 2)
= (7a2 + 14a)/(a + 2) = 7a(a + 2)/(a + 2) = 7a Ans.
2) (x4 – 5x3 – 24x2) ÷ x(x – 8)
= (x4 – 5x3 – 24x2)/x(x – 8) = x2(x2 – 5x – 24)/x(x – 8)
= x(x2 – 8x + 3x – 24)/(x – 8) [factorization of (x2 – 5x – 24)]
= x{x(x – 8) + 3(x – 8)}/(x – 8)
x(x – 8)(x + 3)/(x – 8) = x(x + 3) Ans.
If the above method does not help you to divide a polynomial by a polynomial, then there is also another method to divide which is called the Division Algorithm.
### Solving Polynomials
To solve a polynomial, we put that polynomial equals zero (0) and find the value of the variable in it. Values of the variable are called zeroes or roots of the polynomial which depend on the degree of the polynomial. if the degree of the polynomial is 1 then there will be one zero or root and if the degree is 2 then there will be two zeroes or roots.
• Solving a linear polynomial
• Solving a cubic polynomial
#### Solving a Linear Polynomial
In the linear polynomial, the degree is always 1 so there will be one and only one zero. if the linear polynomial is p(x) then to find the zero of the polynomial p(x) we have to solve the equation p(x) = 0.
Example – 1) find the zero of the polynomial p(x) = 3x + 2.
Solution – let p(x) = 0
3x + 2 = 0
3x = – 2
x = -2/3
So, x = -2/3 is the zero of the polynomial p(x) = 3x + 2. Ans.
Example – 2) find the zero of the polynomial p(y) = 2y – 6.
Solution – let p(y) = 0
2y – 6 = 0
2y = 6
y = 6/2
y = 3
So, Zero of the polynomial p(y) = 2y – 6 is y = 3. Ans.
Note – Standard form of the linear polynomial is ax + b = 0 where a ≠ 0 so the zero will be x = -b/a. we can also find the zero by comparing it.
In the quadratic polynomial, the degree is 2 so there will be two zeroes.
Example – 1) find the zeroes of the polynomial x2 – 3x.
Solution – let p(x) = x2 – 3x
Now P(x) = 0
x2 – 3x = 0
x(x – 3) = 0
x = 0 and x – 3 = 0
x = 0 and x = 3
Therefore, The two zeroes are x = 0 and x = 3. Ans.
Example – 2) find the zeroes of the polynomial 6x2 + 5x – 6.
Solution – let p(x) = 6x2 + 5x – 6
Now p(x) = 0
6x2 + 5x – 6 = 0
6x2 + 9x – 4x – 6 = 0 (By factorization method) [9−4 = 5 and 9⨯(-4) = -36]
3x(2x+3) – 2(2x+3) = 0
(2x + 3)(3x – 2) = 0
2x + 3 = 0 and 3x – 2 = 0
2x = -3 and 3x = 2
x = -3/2 and x = 2/3
These are the zeroes of the given polynomial. Ans.
#### Solving a Cubic Polynomial
In the cubic polynomial, the degree is 3 so there will be three zeroes. To solve the cubic polynomial first we have to arrange the polynomial in descending order then we solve it by taking common terms or by factorization.
Example – 1) find the zeroes of the polynomial x3 + 2x2 – x – 2.
Solution – let p(x) = x3 + 2x2 – x – 2
Now p(x) = 0
x3 + 2x2 – x – 2 = 0
x2(x + 2) – 1(x + 2) = 0
(x + 2)(x2 – 1) = 0
x + 2 = 0 and x2 – 1 = 0
x = -2 and x2 = 1
x = ±√1
x = ±1
So, the zeroes are x = -2, x = +1, and x = -1. Ans.
Note – We can check the answer by putting the value of each zero in the given polynomial because, at each value of zero, the value of the polynomial is 0.
Example – 2) find the zeroes of the polynomial x3 + 6x2 + 11x + 6.
Solution – let p(x) = x3 + 6x2 + 11x + 6
There is not any common term in it.
So, the factors of the constant term 6 = ±1, ±2, ±3 and ±6
Now at x = +1
p(1) = (1)3 + 6(1)2 + 11(1) + 6
p(1) = 1 + 6 + 11 + 6
p(1) = 24
∵ p(1) ≠ 0 so x = +1 is not the zero of this polynomial.
Here, all the terms in the polynomial are positive so we shall take only negative factors.
Now at x = – 1
p(-1) = (-1)3 + 6(-1)2 + 11(-1) + 6
p(-1) = -1 + 6 – 11 + 6
p(-1) = 0
p(-1) = 0 so x = -1 is the zero of this polynomial.
Now at x = -2
p(-2) = (-2)3 + 6(-2)2 + 11(-2) + 6
p(-2) = – 8 + 6(4) – 22 + 6
p(-2) = – 8 + 24 – 22 + 6
p(-2) = 0
p(-2) = 0 so x = -2 is the zero of this polynomial.
Now at x = -3
p(-3) = (-3)3 + 6(-3)2 + 11(-3) + 6
p(-3) = -27 + 6(9) – 33 + 6
p(-3) = -27 + 54 – 33 + 6
p(-3) = 0
∵ p(-3) = 0 so, x = -3 is the zero of this polynomial.
∵ The degree of the given polynomial is 3 so there will be only three zeroes.
So, the zeroes of the given polynomial are x = -1, x = -2, and x = -3. Ans.
### Division Algorithm
If p(x) and g(x) are two such polynomials that the degree of p(x) is greater than g(x) and g(x) ≠ 0, and we divide p(x) by g(x) then we get two polynomials as quotient q(x) and remainder r(x) such that
p(x) = g(x).q(x) + r(x)
Where r(x) = 0 or the degree of the r(x) is smaller than g(x).
Example – Divide the polynomial p(x) = 2x4 – 3x3 + 3x + 1 by x + 1.
Solution –
Quotient = 2x3 – 5x2 + 5x – 2 Remainder = 3
Steps –
1. First of all, we arrange the dividend and divisor in descending order if they are not arranged.in this question, both are arranged in descending order.
2. Then we divide the first term of the dividend by the first term of the divisor. We divide 2x4 by x and get 2x3. This is the first term of quotient.
3. We multiply the divisor (x+1) by the first term of the quotient 2x3 and subtract the product 2x4 + 2x3 from the dividend. this gives the remainder -5x3 +3x + 1.
4. This remainder -5x3 + 3x + 1 is our new dividend. we repeat step (2) to get the second term of the quotient as – 5x2.
5. Similarly, like step (3) we multiply the second term of the quotient – 5x2 by the divisor (x+1) and subtract the product – 5x3 – 5x2 from the new dividend. We write the like term below the like term and if there is any unlike term then we write that term separately.
This process continues till the remainder is 0 or the degree of the new dividend is less than the degree of the divisor. At the last, the new dividend will be the remainder.
Here, zero of divisor x + 1 is x + 1 = 0 ⇒ x = -1
So, p(x) at x = -1, p(-1) = 2(-1)4 – 3(-1)3 + 3(-1) + 1
p(-1) = 2 – 3(-1) – 3 + 1
p(-1) = 2 + 3 – 3 + 1 = 3 (Remainder)
It means we can find the remainder r(x) of any division by putting the value of the zero of divisor g(x) in dividend p(x).
### Remainder Theorem
Let p(x) be any polynomial of degree one or greater than one and let a be any real number. If p(x) is divided by the linear polynomial (x – a), then the remainder is p(a).
Example – find the remainder when x4 – 4x2 + x3 + 2x + 1 is divided by x – 1.
Solution – let p(x) = x4 – 4x2 + x3 + 2x + 1
Zero of x – 1 is x – 1 = 0 ⇒ x = 1
So, p(x) at x = 1, p(1) = (1)4 – 4(1)2 + (1)3 + 2(1) + 1
p(1) = 1 – 4 + 1 + 2 + 1 = 1
Therefore, Remainder is 1. Ans.
### Factor Theorem
If polynomial p(x) is divided by polynomial g(x) with remainder r(x) = 0 then the polynomial g(x) will be one factor of polynomial p(x) or we can say if g(x) is one factor of p(x) then the remainder r(x) will be zero.
Example – examine whether x – 3 is a factor of polynomial x3 – 3x2 + 4x – 12.
Solution – let p(x) = x3 – 3x2 + 4x – 12
We know if (x – 3) is a factor of p(x) then the remainder will be 0.
the zero of x – 3 is x – 3 = 0 ⇒ x = 3
Now p(x) at x = 3, p(3) = (3)3 – 3(3)2 + 4(3) – 12
p(3) = 27 – 3(9) + 12 – 12
p(3) = 27 – 27 + 12 – 12 = 0
∵ The remainder is 0 so (x – 3) is the factor of polynomial p(x). Ans.
Note – Here, zero is x = 3 so the factor is x – 3. We can find the factor if we know the zero and we can find the zero if we know the factor of any polynomial.
### Relationship Between Zeroes and Coefficients of a Quadratic Polynomial
We know the standard form of a quadratic polynomial is f(x) = ax2 + bx + c.
Let α and β be two zeroes of this polynomial. Then (x – α) and (x – β) will be factors of f(x).
So, for constant k we can write,
f(x) = k(x – α) (x – β)
ax2 + bx + c = k{x2 – (α + β)x + αβ}
ax2 + bx + c = kx2 – k(α + β)x + kαβ
by comparing, a = k, b= – k(α + β), c = kαβ
∵ b= – k(α + β) and c = kαβ
α + β = b/-k and αβ = c/k
∵ a = k
α + β = b/-a or -b/a and αβ = c/a
So, for quadratic polynomial f(x) = ax2 + bx + c
Sum of zeroes (α + β) = -b/a = -(coefficient of x)/(coefficient of x2)
Product of zeroes (αβ) = c/a = (constant term)/(coefficient of x2)
#### Some Examples
Example – 1) find the zeroes of the quadratic polynomial 3x2 + 5x – 2 and verify the relationship between the zeroes and the coefficients.
Solution – let f(x) = 3x2 + 5x – 2
Now f(x) = 0
3x2 + 5x – 2 = 0
3x2 + 6x – x – 2 = 0
3x(x + 2) – 1(x + 2) = 0
(x + 2)(3x – 1) = 0
x + 2 = 0 and 3x – 1 = 0
x = – 2 and x = ⅓
So, the zeroes of polynomial f(x) are x = – 2 and x = ⅓
Now, sum of the zeroes = – 2 + ⅓ = (-6 + 1)/3 = -5/3 = -(coefficient of x)/(coefficient of x2)
Product of zeroes = – 2⨯⅓ = -2/3 = (constant term)/(coefficient of x2)
So, the relationship between the zeroes and the coefficients of the quadratic polynomial is verified.
Example – 2) find a quadratic polynomial, whose sum and product of zeroes are -6 and 5 respectively.
Solution – let α and β be the zeroes of a quadratic polynomial.
For any constant k, the quadratic polynomial will be
k{x2 – (α + β)x + αβ}
In question, the sum of zeroes (α + β) = – 6
Product of zeroes (αβ) = 5
Putting the values,
k{x2 – (- 6)x + 5}
k{x2 + 6x + 5} [Where k = Constant]
So, the required quadratic polynomial is x2 + 6x + 5. Ans.
Example – 3) find all the zeroes of polynomial f(x) = x3 + 13x2 + 32x + 20, if it’s one zero is – 2.
Solution – Here, – 2 is zero so the factor will be (x + 2). It is one factor of f(x).
To find other zeroes we will divide polynomial f(x) by factor (x + 2).
From the division algorithm, quotient x2 + 11x + 10 will be its factor because the remainder is 0.
Now Dividend = Divisor ⨯ Quotient + Remainder
x3 + 13x2 + 32x + 20 = (x + 2)⨯(x2 + 11x + 10) + 0
x3 + 13x2 + 32x + 20 = (x + 2)⨯{ x2 + 10x + x + 10}
x3 + 13x2 + 32x + 20 = (x + 2)⨯{ x(x + 10) + 1(x + 10)}
x3 + 13x2 + 32x + 20 = (x + 2)⨯(x + 10)(x + 1)
To find zeroes, f(x) = 0
(x + 2)(x + 10)(x + 1) = 0
x = -2, x = -10 and x = -1
So, all the zeroes of polynomial f(x) are -2, -10, and -1. Ans.
Polynomials Class 10th in Hindi |
# What Is Matrix in Data Structure With Example?
//
Scott Campbell
What Is Matrix in Data Structure With Example?
A matrix is a two-dimensional data structure that represents a collection of elements arranged in rows and columns. It is an essential concept in computer science and is widely used to solve various computational problems.
## Structure of a Matrix
A matrix consists of rows and columns, forming a grid-like structure. Each element in the matrix is identified by its row and column index. The general representation of a matrix with ‘m’ rows and ‘n’ columns is as follows:
m x n Matrix:
• Rows are numbered from 1 to m.
• Columns are numbered from 1 to n.
• The element at row i and column j is denoted by A[i][j].
For example, consider the following 3×3 matrix:
3 x 3 Matrix:
• A[1][1] = 2, A[1][2] = 5, A[1][3] = -1
• A[2][1] = 0, A[2][2] = 9, A[2][3] = 4
• A[3][1] = -7, A[3][2] = 6, A[3][3] = 8
## Operations on Matrices
In data structures, various operations can be performed on matrices. Some common operations include:
To add two matrices, their corresponding elements are added together. The resulting matrix will have the same dimensions as the input matrices.
For example, let’s consider two matrices:
Matrix A:
• A[1][1] = 2, A[1][2] = 5
• A[2][1] = -3, A[2][2] = 7
Matrix B:
• B[1][1] = 4, B[1][2] = -1
• B[2][1] = 0, B[2][2] = 9
The addition of Matrix A and Matrix B will result in:
Matrix A + B:
• C[1][1] = A[1][1] + B[1][1] = 6
• C[1][2] = A[1][2] + B[1][2] = 4
• C[2][1] = A[2][1] + B[2][1] = -3
• C[2][2] = A[2][2] + B[2][2] = 16
### Multiplication of Matrices
In matrix multiplication, the dot product of rows from the first matrix and columns from the second matrix is calculated to obtain the resulting matrix. The number of columns in the first matrix must be equal to the number of rows in the second matrix.
For example, consider the following matrices:
Matrix A:
• A[1][1] = 2, A[1][2] = -1
• A[2][1] = 3, A[2][2] = 4
Matrix B:
• B[1][1] = -3, B[1][2] = 5
• B[2][1] = 2, B[2][2] = -6
The multiplication of Matrix A and Matrix B will result in:
Matrix A * B:
• C[1][1] = (A[1][1]*B[1][1]) + (A[1][2]*B[2][1]) = -7
• C[1][2] = (A[1][1]*B[1][2]) + (A[1][2]*B[2][2]) = -17
• C[2][1] = (A[2][1]*B[1][1]) + (A[2][2]*B[2][1]) = -6
• C[2]][C]= (A[[22]]*B[[12]]) + (A[[21]]*B[[22]]) = -8
### Transpose of a Matrix
The transpose of a matrix is obtained by interchanging its rows with columns. The resulting matrix will have the dimensions reversed compared to the original matrix.
For example, consider the following matrix:
Matrix A:
• A[1][1] = 2, A[1][2] = -4
• A[2][1] = 5, A[2][2] = 3
The transpose of Matrix A will be:
Transpose of Matrix A:
• A'[1][1] = A[1][1] = 2
• A'[1][2] = A[2][1] = 5
• A'[2][1] = A[1][2] = -4
• A'[2]][C]= A[[22]] =3
## Conclusion
In summary, a matrix is a fundamental data structure in computer science that represents a collection of elements arranged in rows and columns. It allows for efficient manipulation of data and is used in various algorithms and mathematical operations. Understanding matrices and their operations is crucial for solving computational problems efficiently. |
# Solving 1 Divided by Infinity
Instructor: Laura Pennington
Laura received her Master's degree in Pure Mathematics from Michigan State University. She has 15 years of experience teaching collegiate mathematics at various institutions.
In this lesson, we will analyze the expression 1 divided by infinity through the use of limits, tables, and graphs. We will see how to work with the abstract concept of infinity within a mathematical expression.
## Steps to Solve
We want to evaluate 1 divided by infinity. Infinity is a concept, not a number. We know we can approach infinity if we count higher and higher, but we can't ever actually reach it. Because of this, the expression 1/infinity is actually undefined, but that's not the end of the story! Otherwise, that would make for a very short lesson!
As we just said, we can approach infinity, so what we can do is look at what value 1/x approaches as x approaches infinity, or as x gets larger and larger. Let's use the following table to take a look at the value of 1/x for values of x as they approach infinity.
x 1/x
1 1
10 0.1
1,000 0.001
1,000,000 0.000001
1,000,000,000 0.000000001
Notice that as x gets larger and larger, approaching infinity, 1/x gets smaller and smaller and approaches 0. In mathematics, we call this a limit, and because we can't actually find a value for 1 divided by infinity, the limit of 1/x, as x approaches infinity, is as good as we're going to get.
In general, the limit of a function tells us what value the function approaches as x approaches a certain value, and we use the following notation for the limit of f(x) as x approaches a.
In this instance, we are taking the limit of the function 1/x, and x is approaching infinity.
From observing our table, we find that the limit of 1/x, as x approaches infinity, is 0.
## Solution
Though the expression 1 divided by infinity is undefined, we can see what the expression 1/x approaches as x approaches infinity using limits. We found that the limit of 1/x as x approaches infinity is 0.
## Graphs and Limits
In finding the limit of 1/x as x approaches infinity, we used a table of values to observe a pattern to evaluate the limit. This can also be observed graphically. On graphs, limits as x approaches infinity or negative infinity show up as horizontal asymptotes. An asymptote is a line that a graph approaches, but does not touch.
Let's think about our example for a moment. We found the limit of 1/x as x approaches infinity to be 0. Graphically speaking, this says that if we look at the graph of y = 1/x, we will see that as x gets larger and larger, or the further to the right that we go along the x-axis, the closer the graph will get to the line y = 0. However, again, since we can't actually reach infinity to evaluate 1 divided by infinity, the graph won't ever actually reach the line y = 0. It will just approach it. I don't know about you, but I think we just explained an asymptote!
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# Calculators and complex numbers (Part 12)
In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers.
To begin, we recall that the trigonometric form of a complex number $z = a+bi$ is
$z = r(\cos \theta + i \sin \theta)$
where $r = |z| = \sqrt{a^2 + b^2}$ and $\tan \theta = b/a$, with $\theta$ in the appropriate quadrant. As noted before, this is analogous to converting from rectangular coordinates to polar coordinates.
There’s a shorthand notation for the right-hand side ($r e^{i \theta}$) that I’ll justify later in this series.
In the previous post, we made the following definition for $z^q$ if $q$ is a rational number and $-\pi < \theta \le \pi$. (Technically, this is the definition for the principal root.)
Definition. $z^q = r^q (\cos q \theta + i \sin q \theta)$.
As it turns out, one of the usual Laws of Exponents remains true even if complex numbers are permitted.
Theorem. $z^{q_1} z^{q_2} = z^{q_1 + q_2}$
Proof. Using the rule for multiplying complex numbers that are in trigonometric form:
$z^{q_1} z^{q_2} = r^{q_1} (\cos q_1 \theta + i \sin q_1 \theta) \cdot r^{q_2} (\cos q_2\theta + i \sin q_2 \theta)$
$= r^{q_1+q_2} ( \cos [q_1 \theta +q_2\theta] + i \sin [q_1\theta +q_2 \theta])$
$= r^{q_1+q_2} ( \cos [q_1+q_2]\theta + i \sin [q_1+q_2] \theta)$
$= z^{q_1+q_2}$
However, other Laws of Exponents no longer are true. For example, it may not be true that $(zw)^q$ is equal to $z^q w^q$. My experience is that this next example is typically presented in secondary schools at about the time that the number $i$ is first introduced. Let $z = -2$, $w = -3$, and $q = 1/2$. Then
$\sqrt{-2} \cdot \sqrt{-3} = i \sqrt{2} i \sqrt{3} = -\sqrt{6} \ne \sqrt{6} = \sqrt{(-2) \cdot (-3)}$.
Furthermore, the expression $(z^{q_1})^{q_2}$ does not have to equal $z^{q_1 q_2}$ if $z$ is complex. Let $z = -1$, $q_1 = 3$, and $q_2 = 1/2$. Then
$\left[ (-1)^3 \right]^{1/2} = (-1)^{1/2}$
$= [1 (\cos \pi + i \sin \pi)]^{1/2}$
$= \displaystyle 1^{1/2} \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right)$
$= 1(0+1i)$
$= i$.
However,
$(-1)^{3/2} = [1 (\cos \pi + i \sin \pi)]^{3/2}$
$= \displaystyle 1^{3/2} \left( \cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2} \right)$
$= 1(0-1i)$
$= -i$.
All this to say, the usual Laws of Exponents that work for real exponents and positive bases don’t have to work if the base is permitted to be complex… or even negative.
For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.
# Calculators and complex numbers (Part 11)
In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers.
In today’s post, at long last, I can explain one of the unexpected results of the calculator shown in the opening sections of the video below: the different answers for $(-8)^{1/3}$ and $(-8+0i)^{1/3}$.
To begin, we recall that the trigonometric form of a complex number $z = a+bi$ is
$z = r(\cos \theta + i \sin \theta)$
where $r = |z| = \sqrt{a^2 + b^2}$ and $\tan \theta = b/a$, with $\theta$ in the appropriate quadrant. As noted before, this is analogous to converting from rectangular coordinates to polar coordinates.
There’s a shorthand notation for the right-hand side ($r e^{i \theta}$) that I’ll justify later in this series.
In previous posts, we discussed De Moivre’s Theorem:
Theorem. If $n$ is an integer, then $z^n = \left[ r (\cos \theta + i \sin \theta) \right]^n = r^n (\cos n \theta + i \sin n \theta)$.
This motivates the following definition:
Definition. If $q$ is a rational number, then $z^q = r^q (\cos q \theta + i \sin q \theta)$ if $theta$ is chosen to be in the interval $-\pi < \theta \le \pi$.
Technically speaking, this defines the principal value of $z^q$; however, for the purposes of this post, I’ll avoid discussion of branch cuts and other similar concepts from complex analysis. When presenting this to my future secondary teachers, I’ll often break the presentation by asking my students why it’s always possible to choose the angle $\theta$ to be in the range $(-\pi,\pi]$, and why it’s necessary to include exactly one of the two endpoints of this interval. I’ll also point out that this interval really could have been $[0,2\pi)$ or any other interval with length $2\pi$, but we choose $(-\pi,\pi]$ for a very simple reason: tradition.
Using this definition, let’s compute $(-8+0i)^{1/3}$. To begin,
$-8+0i = 8 (\cos \pi + i \sin \pi)$.
So, by definition,
$(-8+0i)^{1/3} = 8^{1/3} \displaystyle \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)$
$= \displaystyle 2 \left( \frac{1}{2} + \frac{\sqrt{3}}{2}i \right)$
$= 1 + i \sqrt{3}$
As noted in an earlier post in this series, this is one of the three solutions of the equation $z^3 = -8$. Using De Moivre’s Theorem, the other two solutions are $z = -2$ and $z = 1 - i \sqrt{3}$.
So, when $(-8+0i)^{1/3}$ is entered into the calculator, the answer $1 + i \sqrt{3}$ is returned.
On the other hand, when $(-8)^{1/3}$ is entered into the calculator, the calculator determines the solution that is a real number (if possible). So the calculator returns $-2$ and not $1 + i \sqrt{3}$.
For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.
# Calculators and complex numbers (Part 10)
In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers.
To begin, we recall that the trigonometric form of a complex number $z = a+bi$ is
$z = r(\cos \theta + i \sin \theta)$
where $r = |z| = \sqrt{a^2 + b^2}$ and $\tan \theta = b/a$, with $\theta$ in the appropriate quadrant. As noted before, this is analogous to converting from rectangular coordinates to polar coordinates.
There’s a shorthand notation for the right-hand side ($r e^{i \theta}$) that I’ll justify later in this series.
Today, I want to share some pedagogical thoughts about this series of posts. I’ll continue with the mathematical development of these ideas tomorrow.
My experience is that most math majors have never seen this particular application of trigonometry to find the $n$th roots of complex numbers… or even are familiar with the idea of expressing a complex number into trigonometric form at all. This personally surprises me, as this was just one of the topics that I had to learn when I took Precalculus (which was called Trig/Analysis when I took it). I really don’t know if I was fortunate to be exposed to these ideas in my secondary curriculum of the 1980’s or if this was simply a standard topic back then. However, at least in Texas, the trigonometric form of complex numbers does not appear to be a standard topic these days.
This certainly isn’t the most important topic in the mathematics secondary curriculum. That said, I really wish that this was included in a standard Pre-AP course in Precalculus to better serve the high school students who are most likely to take more advanced courses in mathematics and science in college. These ideas are simply assumed in, say, Differential Equations, when students are asked to solve
$y^{5} – 32 y = 0$.
The characteristic equation of this differential equation is $r^5 - 32 = 0$, and it’s really hard to find all five complex roots unless De Moivre’s Theorem is employed.
To give another example: In physics, even a cursory look at my old electricity and magnetism text reveals that familiarity with the trigonometric form of complex numbers can only facilitate student understanding of these physical concepts. Ditto for many concepts in electrical engineering. Stated another way, students who aren’t used to thinking of complex numbers in this way may struggle through physics and engineering in ways that could have been avoided with prior mathematical training.
For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.
# Calculators and complex numbers (Part 9)
In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers.
To begin, we recall that the trigonometric form of a complex number $z = a+bi$ is
$z = r(\cos \theta + i \sin \theta)$
where $r = |z| = \sqrt{a^2 + b^2}$ and $\tan \theta = b/a$, with $\theta$ in the appropriate quadrant. As noted before, this is analogous to converting from rectangular coordinates to polar coordinates.
There’s a shorthand notation for the right-hand side ($r e^{i \theta}$) that I’ll justify later in this series.
In the previous three posts, we discussed De Moivre’s Theorem:
Theorem. If $n$ is an integer, then $\left[ r (\cos \theta + i \sin \theta) \right]^n = r^n (\cos n \theta + i \sin n \theta)$.
Yesterday, we used factoring to show that there are three solutions to $z^3 = -27$, namely, $z = -3$ and $z = \displaystyle \frac{3}{2} \pm \frac{3\sqrt{3}}{2} i$. Let’s now use De Moivre’s Theorem to take on the same task. As we’ll see, De Moivre’s Theorem provides a geometrical interpretation of this result that isn’t readily apparent using solely algebra.
Let $z = r(\cos \theta + i \sin \theta)$, so that
$z^3 = -27$
becomes
$r^3 (\cos 3\theta + i \sin 3 \theta) = 27 (\cos \pi + i \sin \pi)$
We now match the corresponding parts. The distances from the original have to match, so that $r^3 = 27$, or $r = 3$. (Notice that there is only one answer because $r$ must be a positive real number.) Also, the angles $3\theta$ and $\pi$ must be coterminal. They do not necessarily have to be equal, but the angles must point in the same direction. Therefore,
$3 \theta = \pi + 2\pi k$, or $\theta = \displaystyle ]frac{\pi}{3} + \frac{2\pi k}{3}$
for any integer $k$. At first blush, it appears that there are infinitely many solutions $z$ since there are infinitely many possible angles $\theta$. However, it turns out that there are only three answers, as expected.
Let’s now plug in numbers for $k$. Let’s use the easiest three numbers, $k = 0, 1, 2$.
If $k = 0$, then $\theta = \displaystyle \frac{\pi}{3}$, so that
$z = 3 \displaystyle \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right)$
$= 3 \displaystyle \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right)$
$= \displaystyle \frac{3}{2} + \frac{3\sqrt{3}}{2} i$.
If $k = 1$, then $\theta = \displaystyle \frac{\pi}{3} + \frac{2\pi}{3} = \pi$, so that
$z = 3 \displaystyle \left( \cos \pi + i \sin \pi \right) = 3(-1+0i) = -3$.
If $k = 2$, then $\theta = \displaystyle \frac{\pi}{3} + \frac{4\pi}{3} = \displaystyle \frac{5\pi}{3}$, so that
$z = 3 \displaystyle \left( \cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3} \right)$
$= 3 \displaystyle \left( \frac{1}{2} - i \frac{\sqrt{3}}{2} \right)$
$= \displaystyle \frac{3}{2} - \frac{3\sqrt{3}}{2} i$.
Not surprisingly, we obtain the same three answers that we did using algebra.
What if we keep increasing the value of $k$? Let’s find out with $k = 3$:
If $k = 3$, then $\theta = \displaystyle \frac{\pi}{3} + \frac{6\pi}{3} = \displaystyle \frac{7\pi}{3}$, so that
$z = 3 \displaystyle \left( \cos \frac{7\pi}{3} + i \sin \frac{7 \pi}{3} \right)$
$= 3 \displaystyle \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right)$
$= \displaystyle \frac{3}{2} + \frac{3\sqrt{3}}{2} i$.
In other words, since $\displaystyle \frac{\pi}{3}$ and $\displaystyle \frac{7\pi}{3}$ are coterminal, we end up with the same answer. This resolves the apparent paradox of having infinitely many possible angles $\theta$ but only three solutions $z$.
Using De Moivre’s Theorem certainly appears to be much more difficult than just factoring! However, this solution provides a geometric interpretation of the three roots that isn’t otherwise apparent. Let’s using the trigonometric form of these three solutions to plot them in the complex plane:
All three points lie a distance of 3 from the origin, and so they lie on the same circle. Also, the angle from the origin increases by $\displaystyle \frac{2\pi}{3}$ as we shift from point to point. This divides the circle (with a total angle of $2\pi$ into three equal parts, and so the points are evenly spaced around the circle. Also, the points form the vertices of an equilateral triangle inscribed within this circle. Again, none of this would have been apparent by strictly factoring the polynomial $z^3 + 27$.
All of the above can be repeated for finding the roots of any equation $z^n = w$. This equation has $n$ roots which lie a distance of $|w|^{1/n}$ from the origin on the points of a regular $n-$gon inscribed in a circle of this radius. The only tricky part is determining the placement of the initial point (i.e., finding the initial value of $\theta$ from which all other points can be determined.
For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.
# Calculators and complex numbers (Part 8)
In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers.
To begin, we recall that the trigonometric form of a complex number $z = a+bi$ is
$z = r(\cos \theta + i \sin \theta)$
where $r = |z| = \sqrt{a^2 + b^2}$ and $\tan \theta = b/a$, with $\theta$ in the appropriate quadrant. As noted before, this is analogous to converting from rectangular coordinates to polar coordinates.
There’s a shorthand notation for the right-hand side ($r e^{i \theta}$) that I’ll justify later in this series.
In the previous three posts, we discussed De Moivre’s Theorem:
Theorem. If $n$ is an integer, then $\left[ r (\cos \theta + i \sin \theta) \right]^n = r^n (\cos n \theta + i \sin n \theta)$.
Let’s now use this theorem to solve an algebraic equation.
Find all $z$ so that $z^3 = -27$.
When I present this to my students, their kneejerk reaction is always to answer “$-3$.” To which I politely point out, “This is a cubic equation. So how many roots does it have?” Of course, they answer “Three.” To which I respond, “OK, so how do we find the other two roots?”
With enough patience, a student will usually volunteer that $z^3 + 27 = 0$ has a known root of $z = -3$, and so the remaining roots can be found using synthetic division by dividing $z+3$ into $z^3 + 27$, yielding
$z^3 + 27 = (z+3)(z^2 - 3z + 9)$
So the other two roots can be found using the quadratic formula:
$z = \displaystyle \frac{3 \pm \sqrt{9 - 36}}{2} = \displaystyle \frac{3}{2} \pm \frac{3\sqrt{3}}{2} i$
So there are indeed three roots. Though I usually won’t take class time to do it, I encourage my students to cube these two answers to confirm that they indeed get $-27$.
As an aside, before moving on to the use of De Moivre’s Theorem, I usually point out to my students that there is a formula for factoring the sum of two cubes:
$a^3 + b^3 = (a+b)(a^2 - ab + b^2)$
And there’s a formula for the difference of two cubes:
$a^3 - b^3 = (a-b)(a^2 + ab + b^2)$
My experience is that even math majors are not familiar with these two formulas. They know the difference of two squares formula, but they’re either not proficient with these two formulas or else they’ve never seen them before. These can be generalized for any odd positive exponent and any positive exponent, respectively.
In tomorrow’s post, I’ll describe how these three roots can be found by using De Moivre’s Theorem.
For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.
# Calculators and complex numbers (Part 7)
In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers.
To begin, we recall that the trigonometric form of a complex number $z = a+bi$ is
$z = r(\cos \theta + i \sin \theta)$
where $r = |z| = \sqrt{a^2 + b^2}$ and $\tan \theta = b/a$, with $\theta$ in the appropriate quadrant. As noted before, this is analogous to converting from rectangular coordinates to polar coordinates.
There’s a shorthand notation for the right-hand side ($r e^{i \theta}$) that I’ll justify later in this series.
This post builds off the previous two posts by completing the proof of De Moivre’s Theorem.
Theorem. If $n$ is an integer, then $\left[ r (\cos \theta + i \sin \theta) \right]^n = r^n (\cos n \theta + i \sin n \theta)$.
The proof has two parts:
1. For $n \ge 0$: proof by induction.
2. For $n < 0$: let $n = -m$, and then use part 1.
In the previous post, I presented how I describe the proof of step 1 to students. Today, I’ll discuss how I present step 2.
My personal opinion is that the proof of step 2 goes easiest when a numerical example is done first. Let $n = -5$. Students can usually volunteer the successive steps of this special case:
$\left[ r (\cos \theta + i \sin \theta) \right]^{-5} = \displaystyle \frac{1}{ \left[ r (\cos \theta + i \sin \theta) \right]^{5} }$
$= \displaystyle \frac{1}{r^5 (\cos 5 \theta + i \sin 5 \theta)}$
At this point, students usually want to multiply by the conjugate of the denominator. There’s nothing wrong with doing that, of course, but it’s more elegant to write the numerator in trigonometric form:
$= \displaystyle \frac{1(\cos 0 + i \sin 0)}{r^5 (\cos 5 \theta + i \sin 5 \theta)}$
$= r^{-5} (\cos [0-5\theta] + i \sin[0-5\theta] )$
$= r^{-5} (\cos [-5\theta] + i \sin[ - 5\theta])$,
which matches what we would have expected. Students are now prepared for the proof, which I try to place alongside the above computation for $n = -5$.
Proof for $n < 0$. Let $n = -m$. I tell students to imagine that $n$ is $-5$, so that $m$ is equal to (students volunteer) $5$. In other words, $n$ is negative and $m$ is positive. Then
$\left[ r (\cos \theta + i \sin \theta) \right]^{n} = \displaystyle \frac{1}{ \left[ r (\cos \theta + i \sin \theta) \right]^{-n} }$
I again remind everyone that $-n = m$ is positive, and so (like a good MIT freshman) the previous work applies:
$= \displaystyle \frac{1}{ \left[ r (\cos \theta + i \sin \theta) \right]^m }$
The remaining steps mirror the calculation above:
$= \displaystyle \frac{1}{r^m (\cos m \theta + i \sin m \theta)}$
$= \displaystyle \frac{1(\cos 0 + i \sin 0)}{r^m (\cos m \theta + i \sin m \theta)}$
$= r^{-m} (\cos [0-m\theta] + i \sin[0-m\theta] )$
$= r^{-m} (\cos [-m\theta] + i \sin[ - m \theta])$
$= r^{n} (\cos [n\theta] + i \sin[ n \theta])$,
thus ending the proof.
For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures.
# Calculators and complex numbers (Part 6)
In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers.
To begin, we recall that the trigonometric form of a complex number $z = a+bi$ is
$z = r(\cos \theta + i \sin \theta)$
where $r = |z| = \sqrt{a^2 + b^2}$ and $\tan \theta = b/a$, with $\theta$ in the appropriate quadrant. As noted before, this is analogous to converting from rectangular coordinates to polar coordinates.
There’s a shorthand notation for the right-hand side ($r e^{i \theta}$) that I’ll justify later in this series.
In the previous post, I used a numerical example to justify De Moivre’s Theorem:
Theorem. If $n$ is an integer, then $\left[ r (\cos \theta + i \sin \theta) \right]^n = r^n (\cos n \theta + i \sin n \theta)$.
The proof has two parts:
1. For $n \ge 0$: proof by induction.
2. For $n < 0$: let $n = -m$, and then use part 1.
In this post, I describe how I present part 1 to students in class. The next post will cover part 2. As noted before, I typically present this theorem and its proof after a numerical example so that students can guess the statement of the theorem on their own.
Proof for $n \ge 0$.
Base Case: $n = 0$. This is trivial, as the left-hand side is
$\left[ r (\cos \theta + i \sin \theta) \right]^0 = 1$,
while the right-hand side is
$r^0 (\cos 0 \theta + i \sin 0 \theta) = 1(\cos 0 + i \sin 0) = 1(1 + 0i) = 1$.
Assumption. We now assume, for a given integer $latex$n$, that $\left[ r (\cos \theta + i \sin \theta) \right]^n = r^n (\cos n \theta + i \sin n \theta)$. Inductive Step. We now use the above assumption to prove the statement for $n+1$. On the board, I write the left-hand side on the top and the right-hand side on the bottom, leaving plenty of space in between: $\left[ r (\cos \theta + i \sin \theta) \right]^{n+1}$ $\quad$ $\quad$ $\quad$ $\quad$ $\quad$ $= r^{n+1} (\cos [n+1] \theta + i \sin [n+1] \theta)$. All we have to do is fill in the space to transform the left-hand side into the target on the right-hand side. (I like to call the right-hand side “the target,” as it suggests the direction in which the proof should aim.) I also tell the class that we’re more than two-thirds done with the proof, since we’ve finished the first two steps and have made some headway on the third. This usually produces knowing laughter since the hardest part of the proof is creatively converting the left-hand side into the target. The first couple steps of the proof are usually clear to students: $\left[ r (\cos \theta + i \sin \theta) \right]^{n+1} = \left[ r (\cos \theta + i \sin \theta) \right]^n \cdot \left[ r (\cos \theta + i \sin \theta) \right]$ $= \left[ r^n (\cos n \theta + i \sin n \theta) \right] \cdot \left[ r (\cos \theta + i \sin \theta) \right]$ by induction hypothesis. (I’ll also remind students that, as a general rule, when doing a proof by induction, it’s important to actually use the inductive assumption someplace.) At this point, most students want to distribute to get the right answer. This will eventually produce the correct answer using trig identities. However, I again try to encourage them to think like MIT freshmen and use previous work. After all, the distances multiply and the angles add, so the next step can be $=r^n \cdot r (\cos [n \theta + \theta] + i \sin [n \theta + \theta])$ $= r^{n+1} (\cos [n+1]\theta + i \sin [n+1]\theta)$. In tomorrow’s post, I’ll talk about how I present the second part of the proof to my students. For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures. # Calculators and complex numbers (Part 5) In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers. To begin, we recall that the trigonometric form of a complex number $z = a+bi$ is $z = r(\cos \theta + i \sin \theta)$ where $r = |z| = \sqrt{a^2 + b^2}$ and $\tan \theta = b/a$, with $\theta$ in the appropriate quadrant. As noted before, this is analogous to converting from rectangular coordinates to polar coordinates. There’s a shorthand notation for the right-hand side ($r e^{i \theta}$) that I’ll justify later in this series. The trigonometric form of a complex number permits a geometric interpretation of multiplication, given in the following theorem. Theorem. $\left[ r_1 (\cos \theta_1 + i \sin \theta_1) \right] \cdot \left[ r_2 (\cos \theta_2 + i \sin \theta_2) \right] = r_1 r_2 (\cos [\theta_1+\theta_2] + i \sin [\theta_1+\theta_2])$. While this theorem doesn’t seem all that helpful — just multiplying complex numbers seems easier — this theorem will be a great help for the following problem: Compute $(\sqrt{3} + i)^{2014}$. (When teaching this in class, I usually choose the exponent to be the current year.) Let’s discuss the options for evaluating this expression. Method #1: Multiply it out. (Students reflexively wince in pain — or knowing laughter — when I make this suggestion.) Method #2: Use the 2014th row of Pascal’s triangle. (More pain and/or laughter.) Method #3: Use the above theorem. It’s straightforward to write $\sqrt{3} + i$ as $2 \displaystyle \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right)$… for reasons that will become apparent later, I tell my students that I’ll use radians and not degrees for this one. Most students can recognize — and this is important, before I formally prove De Moivre’s Theorem — that they need to multiply $2$ by itself 2014 times and add $\displaystyle \frac{\pi}{6}$ to itself 2014 times. Therefore, $(\sqrt{3} + i)^{2014} = \displaystyle 2^{2014} \left( \cos \frac{2014\pi}{6} + i \sin \frac{2014\pi}{6} \right) = \displaystyle 2^{2014} \left( \cos \frac{1007\pi}{3} + i \sin \frac{1007\pi}{3} \right)$ I then try to coax my students to compute $\displaystyle \cos \frac{1007\pi}{3}$ without a calculator. With some prodding, they’ll recognize that $\displaystyle \frac{1007}{3} = \displaystyle {335}\frac{2}{3}$, and so they can subtract $334\pi$ (not $335\pi$) without changing the values of sine and cosine. Therefore, $(\sqrt{3} + i)^{2014} = \displaystyle 2^{2014} \left( \cos \frac{5\pi}{3} + i \sin \frac{5\pi}{3} \right)$ $= 2^{2014} \left( \frac{1}{2} - i \frac{\sqrt{3}}{2} \right)$ $= 2^{2013} (1-i\sqrt{3})$ By this point, students absolutely believe that the trigonometric form of a complex number serves a useful purpose. Also, this numerical example has prepared students for the formal proof of DeMoivre’s Theorem, which will be the subject of the next two posts. For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures. # Calculators and complex numbers (Part 4) In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers. To begin, we recall that the trigonometric form of a complex number $z = a+bi$ is $z = r(\cos \theta + i \sin \theta)$ where $r = |z| = \sqrt{a^2 + b^2}$ and $\tan \theta = b/a$, with $\theta$ in the appropriate quadrant. As noted before, this is analogous to converting from rectangular coordinates to polar coordinates. In the previous post, I proved the following theorem which provides a geometric interpretation for multiplying complex numbers. Theorem. $\left[ r_1 (\cos \theta_1 + i \sin \theta_1) \right] \cdot \left[ r_2 (\cos \theta_2 + i \sin \theta_2) \right] = r_1 r_2 (\cos [\theta_1+\theta_2] + i \sin [\theta_1+\theta_2])$. Perhaps unsurprisingly, there’s also a theorem for dividing complex numbers. Students can using guess the statement of this theorem. Theorem. $\displaystyle \frac{ r_1 (\cos \theta_1 + i \sin \theta_1) }{ r_2 (\cos \theta_2 + i \sin \theta_2) } = \displaystyle \frac{r_1}{r_2} (\cos [\theta_1-\theta_2] + i \sin [\theta_1-\theta_2])$. Proof. The proof begins by separating the $r_1$ and $r_2$ terms and then multiplying by the conjugate of the denominator: $\displaystyle \frac{ r_1 (\cos \theta_1 + i \sin \theta_1) }{ r_2 (\cos \theta_2 + i \sin \theta_2) }$ $= \displaystyle \frac{r_1}{r_2} \cdot \frac{ \cos \theta_1 + i \sin \theta_1 }{ \cos \theta_2 + i \sin \theta_2 } \cdot \frac{ \cos \theta_2 - i \sin \theta_2 }{ \cos \theta_2 - i \sin \theta_2 }$ $= \displaystyle \frac{r_1}{r_2} \cdot \frac{ (\cos \theta_1 + i \sin \theta_1)( \cos \theta_2 - i \sin \theta_2) } {\cos^2 \theta_2 - i^2 \sin^2 \theta_2}$ $= \displaystyle \frac{r_1}{r_2} \cdot \frac{ (\cos \theta_1 + i \sin \theta_1)( \cos \theta_2 - i \sin \theta_2) } {\cos^2 \theta_2 + \sin^2 \theta_2}$ $= \displaystyle \frac{r_1}{r_2} (\cos \theta_1 + i \sin \theta_1)( \cos \theta_2 - i \sin \theta_2)$ At this juncture in the proof, there are two legitimate ways to proceed. Method #1: Multiply out the right-hand side. After all, this is how we proved the theorem yesterday. For this reason, students naturally gravitate toward this proof, and the proof works after recognizing the trig identities for the sine and cosine of the difference of two angles. However, this isn’t the most elegant proof. Method #2: I break out my old joke about the entrance exam at MIT and the importance of using previous work. I rewrite the right-hand side as $= \displaystyle \frac{r_1}{r_2} (\cos \theta_1 + i \sin \theta_1)( \cos [-\theta_2] + i \sin [-\theta_2])$; this also serves as a reminder about the odd/even identities for sine and cosine, respectively. Then students observe that the right-hand side is just a product of two complex numbers in trigonometric form, and so the angle of the product is found by adding the angles. For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures. # Calculators and complex numbers (Part 3) In this series of posts, I explore properties of complex numbers that explain some surprising answers to exponential and logarithmic problems using a calculator (see video at the bottom of this post). These posts form the basis for a sequence of lectures given to my future secondary teachers. To begin, we recall that the trigonometric form of a complex number $z = a+bi$ is $z = r(\cos \theta + i \sin \theta)$ where $r = |z| = \sqrt{a^2 + b^2}$ and $\tan \theta = b/a$, with $\theta$ in the appropriate quadrant. As noted before, this is analogous to converting from rectangular coordinates to polar coordinates. There’s a shorthand notation for the right-hand side ($r e^{i \theta}$) that I’ll justify later in this series. Why is this important? When students first learn to multiply complex numbers like $1+i$ and $2+i$, they are taught to just distribute (or, using the nomenclature that I don’t like, FOIL it out): $(1+i)(1+2i) = 1 + 2i + i + 2i^2 = 1 + 3i - 2 = -1 + 3i$. The trigonometric form of a complex number permits a geometric interpretation of multiplication, given in the following theorem. Theorem. $\left[ r_1 (\cos \theta_1 + i \sin \theta_1) \right] \cdot \left[ r_2 (\cos \theta_2 + i \sin \theta_2) \right] = r_1 r_2 (\cos [\theta_1+\theta_2] + i \sin [\theta_1+\theta_2])$. Proof. As above, we distribute (except for the $r_1$ and $r_2$ terms): $\left[ r_1 (\cos \theta_1 + i \sin \theta_1) \right] \cdot \left[ r_2 (\cos \theta_2 + i \sin \theta_2) \right]$ $= r_1 r_2 (\cos \theta_1 \cos \theta_2 + i \cos \theta_1 \sin \theta_2 + i \sin \theta_1 \cos \theta_2 + i^2 \sin \theta_1 \sin \theta_2$ $= r_1 r_2 (\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2 + i[ \sin \theta_1 \cos \theta_2 + \cos \theta_1 \sin \theta_2])$ $= r_1 r_2 (\cos [\theta_1+\theta_2] + i \sin [\theta_1+\theta_2])$. When actually doing this in class, the big conceptual jump for students is the last step. So I make a big song-and-dance routine out of this: Cosine of the first times cosine of the second minus sine of the first times sine of the second… where have I seen this before? The idea is for my students to search deep into their mathematical memories until they recall the appropriate trig identity. For the original multiplication problem, we see that $1+i = \sqrt{2} \left( \cos 45^\circ + i \sin 45^\circ \right)$ $1 + 2i = \sqrt{5} \left( \cos[\tan^{-1} 2] + i \sin[\tan^{-1} 2] \right) \approx \sqrt{5} \left( \cos 63.435^\circ + i \sin 63.435^\circ \right)$ Therefore, the product of$1+i$and$1+2i$will be a distance of$\sqrt{2} \cdot \sqrt{5} = \sqrt{10}\$ from the origin, and the angle from the positive real axis will be $45^\circ + \tan^{-1} 2 \approx 45^\circ + 63.435^\circ = 108.435^\circ$. Indeed,
$-1 + 3i \approx \sqrt{10} \left( \cos 108.435^\circ + i \sin 108.435^\circ \right)$.
For completeness, here’s the movie that I use to engage my students when I begin this sequence of lectures. |
## MATHS BITE: Shoelace Theorem
The Shoelace theorem is a useful formula for finding the area of a polygon when we know the coordinates of its vertices. The formula was described by Meister in 1769, and then by Gauss in 1795.
### Formula
Let’s suppose that a polygon P has vertices (a1, b1), (a2, b2), …, (an, bn), in clockwise order. Then the area of P is given by
$$\dfrac{1}{2} |(a_1b_2 + a_2b_3 + \cdots + a_nb_1) - (b_1a_2 + b_2a_3 + \cdots + b_na_1)|$$
The name of this theorem comes from the fact that if you were to list the coordinates in a column and mark the pairs to be multiplied, then the image looks like laced-up shoes.
### Proof
(Note: this proof is taken from artofproblemsolving.)
Let $\Omega$ be the set of points that belong to the polygon. Then
$$A=\int_{\Omega}\alpha,$$
where $\alpha=dx\wedge dy$.
Note that the volume form $\alpha$ is an exact form since $d\omega=\alpha$, where
$$\omega=\frac{x\,dy}{2}-\frac{y\,dx}{2}.\label{omega}$$
Substitute this in to give us
$$\int_{\Omega}\alpha=\int_{\Omega}d\omega.$$
and then use Stokes’ theorem (a key theorem in vector calculus) to obtain
$$\int_{\Omega}d\omega=\int_{\partial\Omega}\omega.$$
where
$\partial \Omega=\bigcup A(i)$
and $A(i)$ is the line segment from $(x_i,y_i)$ to $(x_{i+1},y_{i+1})$, i.e. is the boundary of the polygon.
Next we substitute for $\omega$:
$$\sum_{i=1}^n\int_{A(i)}\omega=\frac{1}{2}\sum_{i=1}^n\int_{A(i)}{x\,dy}-{y\,dx}.$$
Parameterising this expression gives us
$$\frac{1}{2}\sum_{i=1}^n\int_0^1{(x_i+(x_{i+1}-x_i)t)(y_{i+1}-y_i)}-{(y_i+(y_{i+1}-y_i)t)(x_{i+1}-x_i)\,dt}.$$
Then, by integrating this we obtain
$$\frac{1}{2}\sum_{i=1}^n\frac{1}{2}[(x_i+x_{i+1})(y_{i+1}-y_i)- (y_{i}+y_{i+1})(x_{i+1}-x_i)].$$
This then yields, after further manipulation, the shoelace formula:
$$\frac{1}{2}\sum_{i=1}^n(x_iy_{i+1}-x_{i+1}y_i).$$
M x
## Stirling’s Formula
Today I wanted to discuss something I learnt last week in my Probability course: Stirling’s Formula. Stirling’s Formula is an approximation for factorials, and leads to quite accurate results even for small values of n.
The formula can be written in two ways:
or
where the ~ sign means that the two quantities are asymptotic (i.e. their ratios tend to 1 as n tends to infinity).
## Proof of Stirling’s Formula
The following identity arises using integration by parts:
Taking f(x) = log x, we obtain
Next, sum over n, and by recalling that log x + log y = log xy we get the following expression:
where
Next, define
which allows us to rearrange the above expression to:
So as n tends to infinity we get
How do we show that ?
Firstly, note that from (*) it follows that
So, we need to show that
Let’s set
Note that I0=π/2 and I1 = 1. Then for n≥2, we can integrate by parts to see that
And so, we obtain the following two expressions:
In is decreasing in n and In/In-2 → 1, so it follows that I2n/I2n+1 → 1. Therefore,
as required.
Although the end result is satisfying, I find that some steps in this proof are like ‘pulling-a-rabbit-out-of-a-hat’! What do you think? Mx |
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### Section 1.10 : Common Graphs
7. Without using a graphing calculator sketch the graph of $$W\left( x \right) = {{\bf{e}}^{x + 2}} - 3$$.
Hint : The Algebraic transformations that we used to help us graph the first few graphs in this section can be used together to shift the graph of a function both up/down and right/left at the same time.
Show Solution
The Algebraic transformations we were using in the first few problems of this section can be combined to shift a graph up/down and right/left at the same time. If we know the graph of $$g\left( x \right)$$ then the graph of $$g\left( {x + c} \right) + k$$ is simply the graph of $$g\left( x \right)$$ shifted right by $$c$$ units if $$c < 0$$ or shifted left by $$c$$ units if $$c > 0$$ and shifted up by $$k$$ units if $$k > 0$$ or shifted down by $$k$$ units if $$k < 0$$.
So, in our case if $$g\left( x \right) = {{\bf{e}}^x}$$ we can see that,
$W\left( x \right) = {{\bf{e}}^{x + 2}} - 3 = g\left( {x + 2} \right) - 3$
and so the graph we’re being asked to sketch is the graph of $$g\left( x \right) = {{\bf{e}}^x}$$shifted left by 2 units and down by 3 units.
Here is the graph of $$W\left( x \right) = {{\bf{e}}^{x + 2}} - 3$$ and note that to help see the transformation we have also sketched in the graph of $$g\left( x \right) = {{\bf{e}}^x}$$.
In this case the resulting sketch of $$W\left( x \right)$$ that we get by shifting the graph of $$g\left( x \right)$$ is not really the best, as it pretty much cuts off at $$x = 0$$ so in this case we should probably extend the graph of $$W\left( x \right)$$ a little. Here is a better sketch of the graph. |
## Solving Equations Graphically
By this point in time, we have solved first degree equations for years. We did this in elementary school when teachers asked us questions like three times what number plus one is ten. Here, mental math was usually used to find the number to be three.
Later, we learned how to represent this algebraically. This then gave us the equation 3x+1=10. We then learned how solve for x by isolating the variable. Here, we would first subract 1 from both sides to get the equation 3x = 9. Then, we divide both sides by 3 to get the answer of x = 3.
But, why does this work? Is there another way to find the answer x = 3?
Another possible way to find the answers to this is to look at the graph of the equation. What do you think that it will be?
Are you surprised that it is a vertical line at x= 3? You shouldn't be because that is what we already said the answer was. Remember, that one gets a vertical line when the equation is of the form x = some number. Therefore, the answer to the equation occurs at the point where the graph, in this case a vertical line, intersects the x-axis. So, the answer to the equation is 3.
So, what is the point of this? Why does one spend time graphing a simple first degree equation like this? To make connections between different aspects of mathematics and what they are telling us.
What are the connections in this case? Hopefully, we now realize that the graph of first degree equations are linear and that the solution to the equation occurs at the x-intercept. Or, whatever variable you choose to use. So, if one wanted to solve an equation in terms of y-graphically one would need to find the y-intercept.
What is the solution to the equation: 2y + 5 = 17?
Based on the graph, the answer is 6. So, let's check the answer does 2(6) + 5 = 17. Yes, so we found the answer graphically without ever having to use algebraic manipulation.
What would one do if the variable in the equation was something other than x or y? Simple, the variable name does not matter. So, just change the variable to x or y.
Now that you have a better understanding of where the solution to simple one-variable, first degree equations. It is time to solve them using algebraic manipulation.
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# 7.2 Verifying Solutions for Differential Equations
5 min readfebruary 15, 2024
In AP Calculus, one of the fascinating things we learn is how to solve differential equations. In this section, we will focus on verifying solutions to differential equations, a critical skill in both mathematics and real-world problem solving.
## ✅ Verifying Solutions
While actually solving a differential equation may seem daunting, verifying a given solution is a piece of cake! 🍰
Differential equations often have not just one, but infinitely many solutions. These solutions are known as general solutions. Each of these solutions can be tweaked slightly by adding different constants, and yet, they still solve the original differential equation. Imagine a family of curves on a graph, each differing slightly from the others, but all fitting the same overall pattern described by the differential equation.
Image courtesy of Wolfram
This verification process is rooted in understanding derivatives and how they function. When you're given a differential equation and a potential solution, your job is to take the derivative of the proposed solution and see if it fits perfectly into the original equation. It's like having a key and checking to see if it fits the lock.
### 📝 The Verification Process
Let's say you're given a differential equation and a potential solution. How do you verify if this solution is correct? You start by taking the derivative of the proposed solution. Then, you substitute this derivative, along with the original solution, back into the differential equation. If all parts align and the equation holds true, then you've successfully verified the solution. 🥳
For example, consider the differential equation $dy/dx = 3x^2$. If you're given a potential solution $y=x^3$, you would first find the derivative of $y$ , which is $dy/dx = 3x^2$. Then, you substitute this back into the original equation. Since $3x^2 = 3x^2$, the solution is verified.
### 📍 Verifying a Solution Walkthrough Problem
Verify if the function $y = x^2sin(x)$ is a solution to the differential equation $\frac {dy}{dx} = 2xsin(x) + x^2cos(x)$. Let’s go through this! ⬇️
We need to verify if $y = x^2sin(x)$ satisfies the given differential equation. To do this, we'll first find the derivative of $y$ and then check if it matches the right-hand side of the differential equation.
Differentiate $y = x^2sin(x)$. To differentiate, we'll use the product rule since it's a product of two functions which states: if $u = x^2$ and $v=sin(x)$, then $y'=u'v +uv'$.
1. $u’ = 2x$ and $v’ = cos(x)$
2. Multiply based on the product rule: $y’ = (x^2)(cos(x)) + (2x)(sin(x))$
3. $y’ = (x^2)(cos(x)) + (2x)(sin(x))$
Finally, we just need to verify the solution by checking whether $y'=\frac {dy}{dx}$. Since $\frac {dy}{dx} = 2xsin(x) + x^2cos(x)$ and $y’ = (x^2)(cos(x)) + (2x)(sin(x))$, the solution is verified!
## ✏️ Practice Problems
Let’s put your new skills to good use! Try out a couple practice problems:
1. Verify if the function $y = e^{2x}$ is a solution to the differential equation $\frac{dy}{dx} = 2e^{2x}$.
2. Verify if the function $y = 3x^3$ is a solution to the differential equation $\frac{dy}{dx} = 9x^2$.
### ✔️ Step-By-Step Solution: Example 1
First things first, we have to understand the given task.
We need to verify if the function $y = e^{2x}$ satisfies the given differential equation, which is $\frac{dy}{dx} = 2e^{2x}$. To do this, we'll find the derivative of y and check if it matches the right-hand side of the differential equation.
Now, we can differentiate the given function.
Differentiate $y = e^{2x}$ with respect to x. Since it's an exponential function, its derivative is simply itself times the derivative of the exponent:
$\frac {dy}{dx} = e^{2x} * \frac{d}{dx}2x$
Calculate the derivative of 2x:
$\frac{d}{dx} \: 2x = 2$
Therefore,
$\frac {dy}{dx} = e^{2x}*2 = 2e^{2x}$
Last but not least, verify with the differentiate equation!
Since both sides are equal ($\frac{dy}{dx} = 2e^{2x}$), the proposed solution $y = e^{2x}$ successfully verifies as a solution to the differential equation $\frac{dy}{dx} = 2e^{2x}$. ✅
### ✔️ Step-By-Step Solution: Example 2
Go through the same steps!
1. Understand the Task
1. We need to verify if the function $y = 3x^3$ satisfies the given differential equation, which is $\frac{dy}{dx} = 9x^2$. To do this, we'll find the derivative of y and check if it matches the right-hand side of the differential equation.
2. Differentiate the Given Function
1. Differentiate $y = 3x^3$ with respect to x. Since it's a polynomial, we can use the power rule directly: $\frac {dy}{dx} = \frac{d}{dx}(3x^3)$.
1. Apply the power rule: $\frac {dy}{dx} = \frac{d}{dx}(3x^3)$ = $3*3x^{3-1}$ = $9x^2$.
3. Verify with Differential Equation
1. We have found that $\frac{dy}{dx} = 9x^2$ for the function $y = 3x^3$, and this matches the right-hand side of the differential equation, which is also $9x^2$, making $y = 3x^3$ a verified solution to the differential equation.
## ⭐️ Conclusion
Understanding how to verify solutions to differential equations opens up a world of infinite possibilities. It's not just about solving a math problem; it's about exploring a universe of potential solutions, each fitting the equation in its unique way. ✨
# Key Terms to Review (8)
Chain Rule
: The chain rule is a formula used to find the derivative of a composition of two or more functions. It states that the derivative of a composite function is equal to the derivative of the outermost function times the derivative of the innermost function.
Derivative
: A derivative represents the rate at which a function is changing at any given point. It measures how sensitive one quantity is to small changes in another quantity.
Differential Equations
: Differential equations are mathematical equations that involve derivatives. They describe how a function changes over time or in relation to other variables.
Identity
: An identity is an equation that is true for all values of the variable. It represents a fundamental mathematical relationship that holds regardless of the specific values involved.
Implicit Differentiation
: Implicit differentiation is a technique used to differentiate an equation implicitly without explicitly solving for one variable in terms of another.
Initial Conditions
: Initial conditions refer to the values or states of a system at the starting point of a problem or scenario. They are used to determine the specific solution or behavior of the system.
Power Rule
: The power rule is a calculus rule used to find the derivative of a function that is raised to a constant power. It states that if f(x) = x^n, where n is a constant, then the derivative of f(x) with respect to x is equal to n*x^(n-1).
Product Rule
: The product rule is a formula used to find the derivative of the product of two functions. It states that the derivative of the product of two functions is equal to the derivative of the first function times the second function, plus the first function times the derivative of the second function.
# 7.2 Verifying Solutions for Differential Equations
5 min readfebruary 15, 2024
In AP Calculus, one of the fascinating things we learn is how to solve differential equations. In this section, we will focus on verifying solutions to differential equations, a critical skill in both mathematics and real-world problem solving.
## ✅ Verifying Solutions
While actually solving a differential equation may seem daunting, verifying a given solution is a piece of cake! 🍰
Differential equations often have not just one, but infinitely many solutions. These solutions are known as general solutions. Each of these solutions can be tweaked slightly by adding different constants, and yet, they still solve the original differential equation. Imagine a family of curves on a graph, each differing slightly from the others, but all fitting the same overall pattern described by the differential equation.
Image courtesy of Wolfram
This verification process is rooted in understanding derivatives and how they function. When you're given a differential equation and a potential solution, your job is to take the derivative of the proposed solution and see if it fits perfectly into the original equation. It's like having a key and checking to see if it fits the lock.
### 📝 The Verification Process
Let's say you're given a differential equation and a potential solution. How do you verify if this solution is correct? You start by taking the derivative of the proposed solution. Then, you substitute this derivative, along with the original solution, back into the differential equation. If all parts align and the equation holds true, then you've successfully verified the solution. 🥳
For example, consider the differential equation $dy/dx = 3x^2$. If you're given a potential solution $y=x^3$, you would first find the derivative of $y$ , which is $dy/dx = 3x^2$. Then, you substitute this back into the original equation. Since $3x^2 = 3x^2$, the solution is verified.
### 📍 Verifying a Solution Walkthrough Problem
Verify if the function $y = x^2sin(x)$ is a solution to the differential equation $\frac {dy}{dx} = 2xsin(x) + x^2cos(x)$. Let’s go through this! ⬇️
We need to verify if $y = x^2sin(x)$ satisfies the given differential equation. To do this, we'll first find the derivative of $y$ and then check if it matches the right-hand side of the differential equation.
Differentiate $y = x^2sin(x)$. To differentiate, we'll use the product rule since it's a product of two functions which states: if $u = x^2$ and $v=sin(x)$, then $y'=u'v +uv'$.
1. $u’ = 2x$ and $v’ = cos(x)$
2. Multiply based on the product rule: $y’ = (x^2)(cos(x)) + (2x)(sin(x))$
3. $y’ = (x^2)(cos(x)) + (2x)(sin(x))$
Finally, we just need to verify the solution by checking whether $y'=\frac {dy}{dx}$. Since $\frac {dy}{dx} = 2xsin(x) + x^2cos(x)$ and $y’ = (x^2)(cos(x)) + (2x)(sin(x))$, the solution is verified!
## ✏️ Practice Problems
Let’s put your new skills to good use! Try out a couple practice problems:
1. Verify if the function $y = e^{2x}$ is a solution to the differential equation $\frac{dy}{dx} = 2e^{2x}$.
2. Verify if the function $y = 3x^3$ is a solution to the differential equation $\frac{dy}{dx} = 9x^2$.
### ✔️ Step-By-Step Solution: Example 1
First things first, we have to understand the given task.
We need to verify if the function $y = e^{2x}$ satisfies the given differential equation, which is $\frac{dy}{dx} = 2e^{2x}$. To do this, we'll find the derivative of y and check if it matches the right-hand side of the differential equation.
Now, we can differentiate the given function.
Differentiate $y = e^{2x}$ with respect to x. Since it's an exponential function, its derivative is simply itself times the derivative of the exponent:
$\frac {dy}{dx} = e^{2x} * \frac{d}{dx}2x$
Calculate the derivative of 2x:
$\frac{d}{dx} \: 2x = 2$
Therefore,
$\frac {dy}{dx} = e^{2x}*2 = 2e^{2x}$
Last but not least, verify with the differentiate equation!
Since both sides are equal ($\frac{dy}{dx} = 2e^{2x}$), the proposed solution $y = e^{2x}$ successfully verifies as a solution to the differential equation $\frac{dy}{dx} = 2e^{2x}$. ✅
### ✔️ Step-By-Step Solution: Example 2
Go through the same steps!
1. Understand the Task
1. We need to verify if the function $y = 3x^3$ satisfies the given differential equation, which is $\frac{dy}{dx} = 9x^2$. To do this, we'll find the derivative of y and check if it matches the right-hand side of the differential equation.
2. Differentiate the Given Function
1. Differentiate $y = 3x^3$ with respect to x. Since it's a polynomial, we can use the power rule directly: $\frac {dy}{dx} = \frac{d}{dx}(3x^3)$.
1. Apply the power rule: $\frac {dy}{dx} = \frac{d}{dx}(3x^3)$ = $3*3x^{3-1}$ = $9x^2$.
3. Verify with Differential Equation
1. We have found that $\frac{dy}{dx} = 9x^2$ for the function $y = 3x^3$, and this matches the right-hand side of the differential equation, which is also $9x^2$, making $y = 3x^3$ a verified solution to the differential equation.
## ⭐️ Conclusion
Understanding how to verify solutions to differential equations opens up a world of infinite possibilities. It's not just about solving a math problem; it's about exploring a universe of potential solutions, each fitting the equation in its unique way. ✨
# Key Terms to Review (8)
Chain Rule
: The chain rule is a formula used to find the derivative of a composition of two or more functions. It states that the derivative of a composite function is equal to the derivative of the outermost function times the derivative of the innermost function.
Derivative
: A derivative represents the rate at which a function is changing at any given point. It measures how sensitive one quantity is to small changes in another quantity.
Differential Equations
: Differential equations are mathematical equations that involve derivatives. They describe how a function changes over time or in relation to other variables.
Identity
: An identity is an equation that is true for all values of the variable. It represents a fundamental mathematical relationship that holds regardless of the specific values involved.
Implicit Differentiation
: Implicit differentiation is a technique used to differentiate an equation implicitly without explicitly solving for one variable in terms of another.
Initial Conditions
: Initial conditions refer to the values or states of a system at the starting point of a problem or scenario. They are used to determine the specific solution or behavior of the system.
Power Rule
: The power rule is a calculus rule used to find the derivative of a function that is raised to a constant power. It states that if f(x) = x^n, where n is a constant, then the derivative of f(x) with respect to x is equal to n*x^(n-1).
Product Rule
: The product rule is a formula used to find the derivative of the product of two functions. It states that the derivative of the product of two functions is equal to the derivative of the first function times the second function, plus the first function times the derivative of the second function. |
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Chapter 3 – Pair of Linear Equations in Two Variables
Pair of Linear Equations in Two Variables
NCERT Solutions for Class 10th:
Exercise 3.1
1. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.
Ans. Let the present age of Aftab = x
And, present age of his daughter is represented as = y
Seven years ago,
Aftab’s age = x – 7
Age of Aftab’s daughter = y – 7
According to the question,
(x – 7) = 7 (y – 7 )
x – 7 = 7 y – 49
x – 7y = – 49 + 7
x – 7y = – 42 … (i)
x = 7y – 42
Putting y = 5, 6 and 7, we get
x = 7 × 5 – 42 = 35 – 42 = – 7
x = 7 × 6 – 42 = 42 – 42 = 0
x = 7 × 7 – 42 = 49 – 42 = 7
Three years from now,
Aftab’s age = x + 3
Age of Aftab’s daughter = y + 3
According to the question,
(x + 3) = 3 (y + 3)
x + 3 = 3y + 9
x – 3y = 9 – 3
x – 3y = 6 … (ii)
x = 3y + 6
Putting, y = – 2, –1 and 0, we get
x = 3 × – 2 + 6 = – 6 + 6 =0
x = 3 × – 1 + 6 = – 3 + 6 = 3
x = 3 × 0 + 6 = 0 + 6 = 6
Algebraic representation
From equation (i) and (ii)
x – 7y = – 42 … (i)
x – 3y = 6 … (ii)
Graphical representation
2. The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.
Ans. Let cost of one bat be = Rs x
And, cost of one ball be = Rs y
3 bats and 6 balls are for Rs 3900
Therefore, 3x + 6y = 3900 … (i)
Dividing equation by 3, we get
x + 2y = 1300
Subtracting 2y from both side we get
x = 1300 – 2y
Putting y = –1300, 0 and 1300 we get
x = 1300 – 2 (–1300) = 1300 + 2600 = 3900
x = 1300 – 2(0) = 1300 – 0 = 1300
x = 1300 – 2(1300) = 1300 – 2600 = – 1300
Given that she buys another bat and 2 more balls of the same kind for Rs 1300
We get
x + 2y = 1300 … (ii)
Subtracting 2y from both sides we get
x = 1300 – 2y
Putting y = – 1300, 0 and 1300 we get
x = 1300 – 2 (–1300) = 1300 + 2600 = 3900
x = 1300 – 2 (0) = 1300 – 0 = 1300
x = 1300 – 2(1300) = 1300 – 2600 = – 1300
Algebraic representation
3x + 6y = 3900 … (i)
x + 2y = 1300 … (ii)
Graphical representation,
3. The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.
Ans. Let each kg of apples cost = Rs x
And, cost of each kg of grapes = Rs y
Given that the cost of 2 kg of apples and 1kg of grapes on a day was found to be = Rs 160
Therefore,
2 x + y = 160 … (i)
2x = 160 – y
x = (160 – y)/2
Let y = 0 , 80 and 160, we get
x = (160 – ( 0 )/2 = 80
x = (160– 80 )/2 = 40
x = (160 – 2 × 80)/2 = 0
Given that the cost of 4 kg of apples and 2 kg of grapes is Rs 300
Therefore,
4x + 2y = 300 … (ii)
Dividing by 2 we get
2x + y = 150
Subtracting 2x from both the sides, we get
y = 150 – 2x
Putting x = 0 , 50 , 100 we get
y = 150 – 2 × 0 = 150
y = 150 – 2 × 50 = 50
y = 150 – 2 × (100) = – 50
Algebraic representation,
2x + y = 160 … (i)
4x + 2y = 300 … (ii)
Graphical representation,
Exercise 3.2
1. Form the pair of linear equations in the following problems, and find their solutions graphically.
(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
Ans. Let the number of boys be = x
Let the number of girls be = y
Given that total number of students is 10
Therefore x + y = 10
Subtracting y from both the sides we get
x = 10 – y
Putting y = 0, 5, 10 we get
x = 10 – 0 = 10
x = 10 – 5 = 5
x = 10 – 10 = 0
Given: the number of girls is 4 more than the number of boys
Therefore
y = x + 4
Putting x = – 4, 0, 4, and we get
y = – 4 + 4 = 0
y = 0 + 4 = 4
y = 4 + 4 = 8
Graphical representation
Therefore, number of boys = 3 and number of girls = 7.
(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.
Ans. Let the cost of one pencil = Rs x
Let the cost of one pen = Rs y
5 pencils and 7 pens together cost = Rs 50,
Therefore
5x + 7y = 50
Subtracting 7y from both the sides we get
5x = 50 – 7y
Dividing by 5 we get
x = 10 – 7 y /5
Putting value of y = 5 , 10 and 15 we get
x = 10 – 7 × 5/5 = 10 – 7 = 3
x = 10 – 7 × 10/5 = 10 – 14 = – 4
x = 10 – 7 × 15/5 = 10 – 21 = – 11
Given: 7 pencils and 5 pens together cost Rs 46
7x + 5y = 46
Subtracting 7x from both the sides we get
5y = 46 – 7x
Dividing by 5 we get
y = 46/5 – 7x/5
y = 9.2 – 1.4x
Putting x = 0 , 2 and 4 we get
y = 9.2 – 1.4 × 0 = 9.2 – 0 = 9.2
y = 9.2 – 1.4 (2) = 9.2 – 2.8 = 6.4
y = 9.2 – 1.4 (4) = 9.2 – 5.6 = 3.6
Graphical representation
Therefore, cost of one pencil = Rs 3 and cost of one pen = Rs 5.
2. On comparing the ratios a1/a2, b1/b2 and c1/c2, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident.
Ans. (i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0
On comparing these equation with
a1x + b1y + c1 = 0
a2x + b2y + c2= 0
We get
a1 = 5, b1 = – 4, and c1 = 8
a2 =7, b2 = 6 and c2 = – 9
a1/a2 = 5/7,
b1/b2 = – 4/6 and
c1/c2 = 8/–9
Hence, a1/a2 ? b1/b2
Therefore, both the lines intersect at one point.
(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
Comparing these equations with
a1x + b1y + c1 = 0
a2x + b2y + c2= 0
We get
a1 = 9, b1 = 3, and c1 = 12
a2 = 18, b2 = 6 and c2 = 24
a1/a2 = 9/18 = 1/2
b1/b2 = 3/6 = 1/2 and
c1/c2 = 12/24 = 1/2
Hence, a1/a2 = b1/b2 = c1/c2
Therefore, both the lines are coincident
(iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Comparing these equations with
a1x + b1y + c1 = 0
a2x + b2y + c2= 0
We get
a1 = 6, b1 = – 3, and c1 = 10
a2 = 2, b2 = – 1 and c2 = 9
a1/a2 = 6/2 = 3/1
b1/b2 = – 3/–1 = 3/1 and
c1/c2 = 12/24 = 1/2
Hence, a1/a2 = b1/b2 ? c1/c2
Therefore, both lines are parallel
3. On comparing the ratios a1/a2, b1/b2 and c1/c2 find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5 ; 2x – 3y = 7
(ii) 2x – 3y = 8 ; 4x – 6y = 9
(iii) 3/2x + 5/3y = 7 ; 9x – 10y = 14
(iv) 5x – 3y = 11 ; – 10x + 6y = –22
(v) 4/3x + 2y =8 ; 2x + 3y = 12
Ans. (i) 3x + 2y = 5 ; 2x – 3y = 7
a1/a2 = 3/2
b1/b2 = – 2/3 and
c1/c2 = 5/7
Hence, a1/a2 ? b1/b2
These linear equations intersect each other at one point and therefore have only one possible solution. Hence, the pair of linear equations is consistent.
(ii) 2x – 3y = 8; 4x – 6y = 9
a1/a2 = 2/4 = 1/2
b1/b2 = – 3/–6 = 1/2 and
c1/c2 = 8/9
Hence, a1/a2 = b1/b2 ? c1/c2
Therefore, these linear equations are parallel to each other and thus have no possible solution.
Hence, the pair of linear equations is inconsistent.
(iii) 3/2x + 5/3y = 7 ; 9x – 10y = 14
a1/a2 = 3/2/9 = 1/6
b1/b2 = 5/3/–10 = – 1/6 and
c1/c2 = 7/14 = 1/2
Hence, a1/a2 ? b1/b2
Therefore, these linear equations intersect each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.
(iv) 5x – 3y = 11 ; – 10x + 6y = –22
a1/a2 = 5/–10 = – 1/2
b1/b2 = – 3/6 = – 1/2 and
c1/c2 = 11/–22 = – 1/2
Hence, a1/a2 = b1/b2 = c1/c2
Therefore, these linear equations are coincident and have infinite number of possible solutions.
Therefore, the pair of linear equations is consistent.
(v) 4/3x + 2y =8; 2x + 3y = 12
a1/a2 = 4/3/2 = 2/3
b1/b2 = 2 /3 and
c1/c2 = 8/12 = 2/3
Hence, a1/a2 = b1/b2 = c1/c2
Therefore, these linear equations are coincident and have infinite number of possible solutions.
Therefore, the pair of linear equations is consistent.
4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) x + y = 5, 2x + 2y = 10
(ii) x – y = 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
Ans. (i) x + y = 5; 2x + 2y = 10
a1/a2 = 1/2
b1/b2 = 1/2 and
c1/c2 = 5/10 = 1/2
Hence, a1/a2 = b1/b2 = c1/c2
Therefore, these linear equations are coincident and have infinite number of possible solutions.
Therefore, the pair of linear equations is consistent.
x + y = 5
x = 5 – y
And, 2x + 2y = 10
x = 10 – 2y/2
Graphical representation
From the figure, it can be observed that these lines are overlapping each other. Therefore, infinite number of solutions are possible for the given pair of equations.
(ii) x – y = 8, 3x – 3y = 16
a1/a2 = 1/3
b1/b2 = – 1/–3 = 1/3 and
c1/c2 = 8/16 = 1/2
Hence, a1/a2 = b1/b2 ? c1/c2
Therefore, these linear equations are parallel to each other and thus have no possible solution.
Hence, the pair of linear equations is inconsistent.
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
a1/a2 = 2/4 = 1/2
b1/b2 = – 1/2 and
c1/c2 = – 6/–4 = 3/2
Hence, a1/a2 ? b1/b2
Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.
2x + y – 6 = 0
y = 6 – 2x
And, 4x – 2y – 4 = 0
y = 4x – 4/2
Graphical representation
From the figure, it can be observed that these lines are intersecting each other at the only one point i.e., (2,2) which is the solution for the given pair of equations.
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0
a1/a2 = 2/4 = 1/2
b1/b2 = – 2/–4 = 1/2 and
c1/c2 = 2/5
Hence, a1/a2 = b1/b2 ? c1/c2
Therefore, these linear equations are parallel to each other and thus, have no possible solution.
Hence, the pair of linear equations is inconsistent.
5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Ans. Let length of the rectangle be = x m
Let Width of the rectangle be = y m
According to the question,
y – x = 4 … (i)
y + x = 36 … (ii)
y – x = 4
y = x + 4
y + x = 36
Graphical representation
From the figure, it can be seen that these lines intersect each other at only one point i.e., (16, 20).
Therefore, the length and width of the given garden is 20 m and 16 m respectively.
6. Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines
Ans. (i) Intersecting lines:
Condition,
a1/a2 ? b1/b2
The second line such that it is intersecting the given line is
2x + 4y – 6 = 0 as
a1/a2 = 2/2 = 1
b1/b2 = 3/4 and
a1/a2 ? b1/b2
(ii) Parallel lines
Condition,
a1/a2 = b1/b2 ? c1/c2
Hence, the second line can be
4x + 6y – 8 = 0 as
a1/a2 = 2/4 = 1/2
b1/b2 = 3/6 = 1/2 and
c1/c2 = – 8/–8 = 1
and a1/a2 = b1/b2 ? c1/c2
(iii) Coincident lines
Condition,
a1/a2 = b1/b2 = c1/c2
Hence, the second line can be
6x + 9y – 24 = 0 as
a1/a2 = 2/6 = 1/3
b1/b2 = 3/9 = 1/3 and
c1/c2 = – 8/–24 = 1/3
and a1/a2 = b1/b2 = c1/c2
7. Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.
Ans. x – y + 1 = 0
x = y – 1
3x + 2y – 12 = 0
x = 12 – 2y/3
Graphical representation
From the figure, it can be seen that these lines are intersecting each other at point (2, 3) and x-axis at ( – 1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), ( – 1, 0), and (4, 0).
Exercise 3.3
1. Solve the following pair of linear equations by the substitution method.
(i) x + y = 14 ; x – y = 4
(ii) s – t = 3 ; s/3 + t/2 = 6
(iii) 3x – y = 3 ; 9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3 ; 0.4x + 0.5y = 2.3
Ans. (i) x + y = 14 … (i)
x – y = 4 … (ii)
From equation (i), we get
x = 14 – y … (iii)
Putting this value in equation (ii), we get
(14 – y) – y = 4
14 – 2y = 4
10 = 2y
y = 5 … (iv)
Putting this in equation (iii), we get
x = 9
? x = 9 and y = 5
(ii) s – t = 3 … (i)
s/3 + t/2 = 6 … (ii)
From equation (i), we get s = t + 3
Putting this value in equation (ii), we get
t + 3/3 + t/2 = 6
2t + 6 + 3t = 36
5t = 30
t = 30/5 … (iv)
Putting in equation (iii), we get
s = 9
? s = 9, t = 6
(iii) 3x – y = 3 … (i)
9x – 3y = 9 … (ii)
From equation (i), we get
y = 3x – 3 … (iii)
Putting this value in equation (ii), we get
9x – 3(3x – 3) = 9
9x – 9x + 9 = 9
9 = 9
This is always true.
Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by y = 3x – 3
Therefore, one of its possible solutions is x = 1, y = 0.
(iv) 0.2x + 0.3y = 1.3 … (i)
0.4x + 0.5y = 2.3 … (ii)
0.2x + 0.3y = 1.3
Solving equation (i), we get
0.2x = 1.3 – 0.3y
Dividing by 0.2, we get
x = 1.3/0.2 – 0.3/0.2
x = 6.5 – 1.5 y … (iii)
Putting the value in equation (ii), we get
0.4x + 0.5y = 2.3
(6.5 – 1.5y) × 0.4x + 0.5y = 2.3
2.6 – 0.6y + 0.5y = 2.3
–0.1y = 2.3 – 2.6
y = – 0.3/–0.1
y = 3
Putting this value in equation (iii) we get
x = 6.5 – 1.5 y
x = 6.5 – 1.5(3)
x = 6.5 – 4.5
x = 2
? x = 2 and y = 3
(vi) 3/2x – 5/3y = – 2 … (i)
x/3 + y/2 = 13/6 … (ii)
From equation (i), we get
9x – 10y = – 12
x = – 12 + 10y/9 … (iii)
Putting this value in equation (ii), we get
2. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.
Ans. 2x + 3y = 11 … (i)
Subtracting 3y both side we get
2x = 11 – 3y … (ii)
Putting this value in equation second we get
2x – 4y = – 24 … (iii)
11- 3y – 4y = – 24
7y = – 24 – 11
–7y = – 35
y = – 35/–7
y = 5
Putting this value in equation (iii) we get
2x = 11 – 3 × 5
2x = 11 – 15
2x = – 4
Dividing by 2 we get
x = – 2
Putting the value of x and y
y = mx + 3.
5 = – 2m + 3
2m = 3 – 5
m = – 2/2
m = – 1
3. Form the pair of linear equations for the following problems and find their solution by substitution method.
(i) The difference between two numbers is 26 and one number is three times the other.
Find them.
Ans. Let the larger number be = x
Let the smaller number be = y
The difference between the two numbers is 26
x – y = 26
x = 26 + y
Given that one number is three times the other
So x = 3y
Putting the value of x we get
26 + y = 3y
–2y = – 2 6
y = 13
So value of x = 3y
Putting value of y, we get
x = 3 × 13 = 39
Therefore the numbers are 13 and 39.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Ans. Let the first angle be= x
And the second angle be = y
As both angles are supplementary so that sum will 180
x + y = 180
x = 180 – y … (i)
Given; difference is 18 degree
Therefore
x – y = 18
Putting the value of x we get
180 – y – y = 18
– 2y = – 162
y = – 162/–2
y = 81
Putting the value back in equation (i), we get
x = 180 – 81 = 99 Hence, the angles are 99° and 81°.
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats
and 5 balls for Rs 1750. Find the cost of each bat and each ball.
Ans. Let the cost of each bat be = Rs x
Let the cost of each ball be = Rs y
Given: coach of a cricket team buys 7 bats and 6 balls for Rs 3800.
7x + 6y = 3800
6y = 3800 – 7x
Dividing by 6, we get
y = (3800 – 7x)/6 … (i)
Given that she buys 3 bats and 5 balls for Rs 1750 later.
3x + 5y = 1750
Putting the value of y
3x + 5 ((3800 – 7x)/6) = 1750
Multiplying by 6, we get
18x + 19000 – 35x = 10500
–17x =10500 – 19000
–17x = – 8500
x = – 8500/–17
x = 500
Putting this value in equation (i) we get
y = ( 3800 – 7 × 500)/6
y = 300/6
y = 50
Hence the cost of each bat = Rs 500 and the cost of each balls = Rs 50.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km?
How much does a person have to pay for traveling a distance of 25 km?
Ans. Let the fixed charge for taxi = Rs x
And variable cost per km = Rs y
Total cost = fixed charge + variable charge
Given that for a distance of 10 km, the charge paid is Rs 105
x + 10y = 105 … (i)
x = 105 – 10y
Given that for a journey of 15 km, the charge paid is Rs 155
x + 15y = 155
Putting the value of x we get
105 – 10y + 15y = 155
5y = 155 – 105
5y = 50
Dividing by 5, we get
y = 50/5 = 10
Putting this value in equation (i) we get
x = 105 – 10 × 10
x = 5
Cost for traveling a distance of 25 km
= x + 25y
= 5 + 25 × 10
= 5 + 250
=255
A person has to pay Rs 255 for 25 Km.
(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.
Ans. Let the Numerator be = x
Let the Denominator be = y
Fraction = x/y
A fraction becomes 9/11, if 2 is added to both the numerator and the denominator
(x + 2)/y + 2 = 9/11
On Cross multiplying,
11x + 22 = 9y + 18
Subtracting 22 from both sides,
11x = 9y – 4
Dividing by 11, we get
x = 9y – 4/11 … (i)
Given: if 3 is added to both the numerator and the denominator it becomes 5/6.
(x+3)/y +3 = 5/6 … (ii)
On Cross multiplying,
6x + 18 = 5y + 15
Subtracting the value of x, we get
6(9y – 4 )/11 + 18 = 5y + 15
Subtracting 18 from both the sides
6(9y – 4 )/11 = 5y – 3
54 – 24 = 55y – 33
–y = – 9
y = 9
Putting this value of y in equation (i), we get
x = 9y – 4
11 … (i)
x = (81 – 4)/77
x = 77/11
x = 7
Hence our fraction is 7/9.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Ans. Let the present age of Jacob be = x year
And the present Age of his son be = y year
Five years from now,
Jacob’s age will be = x + 5 year
Age of his son will be = y + 5year
Given; the age of Jacob will be three times that of his son
x + 5 = 3(y + 5)
x = 3y + 15 – 5
x = 3y + 10 … (i)
Five years ago,
Jacob’s age= x – 5 year
His son’s age = y – 5 year
Jacob’s age was seven times that of his son
x – 5 = 7(y – 5)
Putting the value of x from equation (i) we get
3y + 10 – 5 = 7y – 35
3y + 5 = 7y – 35
3y – 7y = – 35 – 5
–4y = – 40
y = – 40/–4
y = 10 year
Putting the value of y in equation we get
x = 3 × 10 + 10
x = 40 years
Hence, Present age of Jacob = 40 years and present age of his son = 10 years.
Exercise 3.4
1. Solve the following pair of linear equations by the elimination method and the substitution method:
(i) x + y =5 and 2x –3y = 4
(ii) 3x + 4y = 10 and 2x – 2y = 2
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
(iv) x/2 + 2y/3 = – 1 and x – y/3 = 3
Ans. (i) x + y =5 and 2x –3y = 4
By elimination method
x + y =5 … (i)
2x –3y = 4 … (ii)
Multiplying equation (i) by (ii), we get
2x + 2y = 10 … (iii)
2x – 3y = 4 … (ii)
Subtracting equation (ii) from equation (iii), we get
5y = 6
y = 6/5
Putting the value in equation (i), we get
x = 5 – (6/5) = 19/5
Hence, x = 19/5 and y = 6/5
By substitution method
x + y = 5 … (i)
Subtracting y from both side, we get
x = 5 – y … (iv)
Putting the value of x in equation (ii) we get
2(5 – y) – 3y = 4
–5y = – 6
y = – 6/–5 = 6/5
Putting the value of y in equation (iv) we get
x = 5 – 6/5
x = 19/5
Hence, x = 19/5 and y = 6/5 again
(ii) 3x + 4y = 10 and 2x – 2y = 2
By elimination method
3x + 4y = 10 …. (i)
2x – 2y = 2 … (ii)
Multiplying equation (ii) by 2, we get
4x – 4y = 4 … (iii)
3x + 4y = 10 … (i)
Adding equation (i) and (iii), we get
7x + 0 = 14
Dividing both side by 7, we get
x = 14/7 = 2
Putting in equation (i), we get
3x + 4y = 10
3(2) + 4y = 10
6 + 4y = 10
4y = 10 – 6
4y = 4
y = 4/4 = 1
Hence, answer is x = 2, y = 1
By substitution method
3x + 4y = 10 … (i)
Subtract 3x both side, we get
4y = 10 – 3x
Divide by 4 we get
y = (10 – 3x )/4
Putting this value in equation (ii), we get
2x – 2y = 2 … (i)
2x – 2(10 – 3x )/4) = 2
Multiply by 4 we get
8x – 2(10 – 3x) = 8
8x – 20 + 6x = 8
14x = 28
x = 28/14 = 2
y = (10 – 3x)/4
y = 4/4 = 1
Hence, answer is x = 2, y = 1 again.
(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
By elimination method
3x – 5y – 4 = 0
3x – 5y = 4 …(i)
9x = 2y + 7
9x – 2y = 7 … (ii)
Multiplying equation (i) by 3, we get
9 x – 15 y = 11 … (iii)
9x – 2y = 7 … (ii)
Subtracting equation (ii) from equation (iii), we get
–13y = 5
y = – 5/13
Putting value in equation (i), we get
3x – 5y = 4 … (i)
3x – 5(–5/13) = 4
Multiplying by 13 we get
39x + 25 = 52
39x = 27
x =27/39 = 9/13
Hence our answer is x = 9/13 and y = – 5/13
By substitution method
3x – 5y = 4 … (i)
Adding 5y on both sides we get
3x = 4 + 5y
Dividing by 3 we get
x = (4 + 5y )/3 … (iv)
Putting this value in equation (ii) we get
9x – 2y = 7 … (ii)
9 ((4 + 5y )/3) – 2y = 7
On solve we get
3(4 + 5y ) – 2y = 7
12 + 15y – 2y = 7
13y = – 5
y = – 5/13
(iv) x/2 + 2y/3 = – 1 and x – y/3 = 3
By elimination method
x/2 + 2y/3 = – 1 … (i)
x – y/3 = 3 … (ii)
Multiplying equation (i) by 2, we get
x + 4y/3 = – 2 … (iii)
x – y/3 = 3 … (ii)
Subtracting equation (ii) from equation (iii), we get
5y/3 = – 5
Dividing by 5 and multiplying by 3, we get
y = – 15/5
y = – 3
Putting this value in equation (ii), we get
x – y/3 = 3 … (ii)
x – (–3)/3 = 3
x + 1 = 3
x = 2
Therefore x = 2 and y = – 3.
By substitution method
x – y/3 = 3 … (ii)
Adding y/3 on both sides, we get
x = 3 + y/3 … (iv)
Putting the value in equation (i) we get
x/2 + 2y/3 = – 1 … (i)
(3+ y/3)/2 + 2y/3 = – 1
3/2 + y/6 + 2y/3 = – 1
Multiplying by 6,
9 + y + 4y = – 6
5y = – 15
y = – 3
Therefore x = 2 and y = – 3.
2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:
(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2 if we only add 1 to the denominator. What is the fraction?
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(iv) Meena went to bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Ans. (i) Let the fraction be x/y
According to the question, x + 1/y – 1 = 1
? x – y = – 2 … (i) x/y + 1 = 1/2
? 2x – y = 1 … (ii)
Subtracting equation (i) from equation (ii), we get
x = 3 … (iii)
Putting this value in equation (i), we get
3 – y = – 2
–y = – 5
y = 5
Hence, the fraction is 3/5
(ii) Let the present age of Nuri be = x
Let the present age of Sonu be = y
According to the given information,
(x – 5) = 3(y – 5)
x – 3y = – 10 … (i)
(x + 10y) = 2(y + 10)
x – 2y = 10 … (ii)
Subtracting equation (i) from equation (ii), we get
y = 20 … (iii)
Putting this value in equation (i), we get
x – 60 = – 10
x = 50
Hence, age of Nuri = 50 years and age of Sonu = 20 years.
(iii) Let the unit digit and tens digits of the number be x and y respectively.
Then, number = 10y + x
Number after reversing the digits = 10x + y
According to the question,
x + y = 9 … (i)
9(10y + x) = 2(10x + y)
88y – 11x = 0
–x + 8y =0 … (ii)
Adding equation (i) and (ii), we get
9y = 9
y = 1 … (iii)
Putting the value in equation (i), we get
x = 8
Hence, the number is 10y + x = 10 × 1 + 8 = 18.
(iv) Let the number of Rs 50 notes and Rs 100 notes be x and y respectively.
According to the question,
x + y = 25 … (i)
50x + 100y = 2000 … (ii)
Multiplying equation (i) by 50, we get
50x + 50y = 1250 … (iii)
Subtracting equation (iii) from equation (ii), we get
50y = 750
y = 15
Putting this value in equation (i), we have x = 10
Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs 100.
(v) Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y respectively.
According to the question,
x + 4y = 27 … (i)
x + 2y = 21 … (ii)
Subtracting equation (ii) from equation (i), we get
2y = 6
y = 3 … (iii)
Putting in equation (i), we get
x + 12 =27
x = 15
Hence, fixed charge = Rs 15 and Charge per day = Rs 3.
Exercise 3.5
1. Which of the following pairs of linear equations has unique solution, no solution or infinitely many solutions? In case there is a unique solution, find it by using cross multiplication method.
(i) x – 3y – 3 = 0 ; 3x – 9y – 2 =0
(ii) 2x + y = 5 ; 3x + 2y = 8
(iii) 3x – 5y = 20 ; 6x – 10y = 40
(iv) x – 3y – 7 = 0 ; 3x – 3y – 15= 0
Ans. (i) x – 3y – 3 = 0
3x – 9y – 2 =0
a1/a2 = 1/3
b1/b2 = – 3/-9 = 1/3 and
c1/c2 = – 3/-2 = 3/2
a1/a2 = b1/b2 ? c1/c2
Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations.
(ii) 2x + y = 5
3x + 2y = 8
a1/a2 = 2/3
b1/b2 = 1/2 and
c1/c2 = – 5/–8 = 5/8
a1/a2 ? b1/b2
Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.
By cross-multiplication method,
x/b1c2– b2c1 = y/c1a2 – c2a1 = 1/a1b2 – a2b1
x/-8–(–10) = y/–15 + 16 = 1/4 – 3
x/2 = y/1 = 1
x/2 = 1, y/1 = 1
? x = 2, y = 1.
(iii) 3x – 5y = 20
6x – 10y = 40
a1/a2 = 3/6 = 1/2
b1/b2 = – 5/–10 = 1/2 and
c1/c2 = – 20/–40 = 1/2
a1/a2 = b1/b2 = c1/c2
Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations.
(iv) x – 3y – 7 = 0
3x – 3y – 15= 0
a1/a2 = 1/3
b1/b2 = – 3/–3 = 1 and
c1/c2 = – 7/–15 = 7/15
a1/a2 ? b1/b2
Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.
By cross-multiplication,
x/45 – (21) = y/–21 – (–15) = 1/–3 – (–9)
x/24 = y/–6 = 1/6
x/24 = 1/6 and y/–6 = 1/6
x = 4 and y = – 1
? x = 4, y = – 1.
2. (i) For which values of a and b does the following pair of linear equations have an infinite number of solutions?
2x + 3y =7
(a – b)x + (a + b)y = 3a + b –2
Ans. 2x + 3y –7 = 0
(a – b)x + (a + b)y – (3a + b –2) = 0
a1/a2 = 2/a – b = 1/2
b1/b2 = – 7/a + b and
c1/c2 = – 7/–(3a + b – 2) = 7/(3a + b – 2)
For infinitely many solutions,a1/a2 = b1/b2 = c1/c2
2/a – b = 7/3a + b – 26a + 2b – 4 = 7a – 7b
a – 9b = – 4 … (i)
2/a – b = 3/a+b
2a + 2b = 3a – 3b
a – 5b = 0 … (ii)
Subtracting equation (i) from (ii), we get
4b = 4
b = 1
Putting this value in equation (ii), we get
a – 5 × 1 = 0
a = 5
Hence, a = 5 and b = 1 are the values for which the given equations give infinitely many solutions.
(ii) For which value of k will the following pair of linear equations have no solution?
3x + y = 1
(2k –1)x + (k –1)y = 2k + 1
Ans. 3x + y –1 = 0
(2k –1)x + (k –1)y – (2k + 1) = 0
a1/a2 = 3/2k – 1
b1/b2 = 1/k – 1 and
c1/c2 = – 1/–2k – 1 = 1/2k + 1
For no solutions,
a1/a2 = b1/b2 ? c1/c2
3/2k – 1 = 1/k-1 ? 1/2k + 1
3/2k – 1 = 1/k – 1
3k – 3 = 2k – 1
k = 2
Hence, for k = 2, the given equation has no solution.
3. Solve the following pair of linear equations by the substitution and cross-multiplication methods:
8x + 5y = 9
3x + 2y = 4
Ans. 8x + 5y = 9 … (i)
3x + 2y = 4 … (ii)
From equation (ii), we get
x = 4 – 2y/3 … (iii)
Putting this value in equation (i), we get
8(4 – 2y/3) + 5y = 9
32 – 16y + 15y = 27
–y = – 5
y = 5 … (iv)
Putting this value in equation (ii), we get
3x + 10 = 4
x = – 2
Hence, x = – 2, y = 5
By cross multiplication again, we get
8x + 5y – 9 = 0
3x + 2y – 4 = 0
x/–20 – (–18) = y/–27 – (–32) = 1/16 – 15
x/–2 = y/5 = 1/1
x/–2 = 1 and y/5 = 1
x = – 2 and y = 5
4. orm the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:
(i) A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs 1180 as hostel charges. Find the fixed charges and the cost of food per day.
Ans. Let x be the fixed charge of the food and y be the charge for food per day.
According to the question,
x + 20y = 1000 … (i)
x + 26y = 1180 … (ii)
Subtracting equation (i) from equation (ii), we get
6y = 180
y = 180/6 = 30
Putting this value in equation (i), we get
x + 20 × 30 = 1000
x = 1000 – 600
x = 400
Hence, fixed charge = Rs 400 and charge per day = Rs 30
(ii) A fraction becomes 1/3 when 1 is subtracted from the numerator and it becomes 1/4
when 8 is added to its denominator. Find the fraction.
Ans. Let the fraction be x/y
According to the question,
x–1/y = 1/3
? 3x – y = 3… (i)
x/y + 8 = 1/4
? 4x – y = 8 … (ii)
Subtracting equation (i) from equation (ii), we get
x = 5 … (iii)
Putting this value in equation (i), we get
15 – y = 3
y = 12
Hence, the fraction is 5/12.
(iii) Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
Ans. Let the number of right answers and wrong answers be x and y respectively.
According to the question,
3x – y = 40 … (i)
4x – 2y = 50
? 2x – y = 25 … (ii)
Subtracting equation (ii) from equation (i), we get
x = 15 … (iii)
Putting this value in equation (ii), we get
30 – y = 25
y = 5
Therefore, number of right answers = 15
And number of wrong answers = 5
Total number of questions = 20
(iv) Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
Ans. Let the speed of 1st car and 2nd car be u km/h and v km/h.
Respective speed of both cars while they are travelling in same direction = (u – v) km/h
Respective speed of both cars while they are travelling in opposite directions i.e., travelling towards each other = (u + v) km/h
According to the question,
5(u – v) = 100
? u – v = 20 … (i)
1(u + v) = 100 … (ii)
Adding both the equations, we get
2u = 120
u = 60 km/h … (iii)
Putting this value in equation (ii), we obtain
v = 40 km/h
Hence, speed of one car = 60 km/h and speed of other car = 40 km/h
(v) The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Ans. Let length and breadth of rectangle be x unit and y unit respectively.
Area = xy
According to the question,
(x – 5) (y + 3) = xy – 9
? 3x – 5y – 6 = 0 … (i)
(x + 3) (y + 2) = xy + 67
? 2x – 3y – 61 = 0 … (ii)
By cross multiplication, we get
x/305 – (–18) = y/–12 – (–183) = 1/9 – (–10)
x/323 = y/171 = 1/19
x = 17, y = 9
Hence, the length of the rectangle = 17 units and breadth of the rectangle = 9 units.
Exercise 3.6
1. Solve the following pairs of equations by reducing them to a pair of linear equations:
Ans. (i) 1/2x + 1/3y = 2
1/3x + 1/2y = 13/6
Let 1/x = p and 1/y = q, then the equations changes as below:
p/2 + q/3 = 2
? 3p + 2q – 12 = 0 … (i)
p/3 + q/2 = 13/6
? 2p + 3q – 13 = 0 … (ii)
By cross-multiplication method, we get
p/–26 – (–36) = q/–24 – (–39) = 1/9–4
p/10 = q/15 = 1/5
p/10 = 1/5 and q/15 = 1/5
p = 2 and q = 3
1/x = 2 and 1/y = 3
Hence, x = 1/2 and y = 1/3
2p + 3q = 2 … (i)
4p – 9q = – 1 … (ii)
Multiplying equation (i) by 3, we get
6p + 9q = 6 … (iii)
Adding equation (ii) and (iii), we get
10p = 5
p = 1/2 … (iv)
Putting in equation (i), we get
2 × 1/2 + 3q = 2
3q = 1
q = 1/3
p = 1/vx = 1/2
vx = 2
x = 4
and
q = 1/vy = 1/3
vy = 3
y = 9
Hence, x = 4, y = 9
(iii) 4/x + 3y = 14
3/x – 4y = 23
Putting 1/x = p in the given equations, we get
4p + 3y = 14
? 4p + 3y – 14 = 0
3p – 4y = 23
? 3p – 4y – 23 = 0
By cross-multiplication, we get
p/–69 – 56 = y/–42 – (–92) = 1/–16 – 9
? – p/125 = y/50 = – 1/25
Now,
–p/125 = – 1/25 and y/50 = – 1/25
? p = 5 and y = – 2
Also, p = 1/x = 5
? x = 1/5
So, x = 1/5 and y = – 2 is the solution.
(iv) 5/x – 1 + 1/y – 2 = 2
6/x – 1 – 3/y – 2 = 1
Putting 1/x – 1 = p and 1/y – 2 = q in the given equations, we obtain
5p + q = 2 … (i)
6p – 3q = 1 … (ii)
Now, by multiplying equation (i) by 3 we get
15p + 3q = 6 … (iii)
Now, adding equation (ii) and (iii)
21p = 7
? p = 1/3
Putting this value in equation (ii) we get,
6×1/3 – 3q =1
? 2 – 3q = 1
? – 3q = 1 – 2
? – 3q = – 1
? q = 1/3
Now,
p = 1/x – 1 = 1/3
?1/x – 1 = 1/3
? 3 = x – 1
? x = 4
Also,
q = 1/y – 2 = 1/3
? 1/y – 2 = 1/3
? 3 = y – 2
? y = 5
Hence, x = 4 and y = 5 is the solution.
(v) 7x – 2y/xy = 5
? 7x/xy – 2y/xy = 5
? 7/y – 2/x = 5 … (i)
8x + 7y/xy = 15
? 8x/xy + 7y/xy = 15
? 8/y + 7/x = 15 … (ii)
Putting 1/x = p and 1/y = q in (i) and (ii) we get,
7q – 2p = 5 … (iii)
8q + 7p = 15 … (iv)
Multiplying equation (iii) by 7 and multiplying equation (iv) by 2 we get,
49q – 14p = 35 … (v)
16q + 14p = 30 … (vi)
Now, adding equation (v) and (vi) we get,
49q – 14p + 16q + 14p = 35 + 30
? 65q = 65
? q = 1
Putting the value of q in equation (iv)
8 + 7p = 15
? 7p = 7
? p = 1
Now,
p = 1/x = 1
? 1/x = 1
? x = 1
also, q = 1 = 1/y
? 1/y = 1
? y = 1
Hence, x =1 and y = 1 is the solution.
(vi) 6x + 3y = 6xy
? 6x/xy + 3y/xy = 6
? 6/y + 3/x = 6 … (i)
2x + 4y = 5xy
? 2x/xy + 4y/xy = 5
? 2/y + 4/x = 5 … (ii)
Putting 1/x = p and 1/y = q in (i) and (ii) we get,
6q + 3p – 6 = 0
2q + 4p – 5 = 0
By cross multiplication method, we get
p/–30 – (–12) = q/–24 – (–15) = 1/6 – 24
p/–18 = q/–9 = 1/–18
p/–18 = 1/–18 and q/–9 = 1/–18
p = 1 and q = 1/2
p = 1/x = 1 and q = 1/y = 1/2
x = 1, y = 2
Hence, x = 1 and y = 2
(vii) 10/x+y + 2/x – y = 4
15/x + y – 5/x – y = – 2
Putting 1/x + y = p and 1/x – y = q in the given equations, we get:
10p + 2q = 4
? 10p + 2q – 4 = 0 … (i)
15p – 5q = – 2
? 15p – 5q + 2 = 0 … (ii)
Using cross multiplication, we get
p/4 – 20 = q/–60 – (–20) = 1/–50 – 30
p/–16 = q/–80 = 1/–80
p/–16 = 1/–80 and q/–80 = 1/–80
p = 1/5 and q = 1
p = 1/x + y = 1/5 and q = 1/x – y = 1
x + y = 5 … (iii)
and x – y = 1 … (iv)
Adding equation (iii) and (iv), we get
2x = 6
x = 3 …. (v)
Putting value of x in equation (iii), we get
y = 2
Hence, x = 3 and y = 2
(viii) 1/3x + y + 1/3x – y = 3/4
1/2(3x – y) – 1/2(3x – y) = – 1/8
Putting 1/3x + y = p and 1/3x – y = q in the given equations, we get
p + q = 3/4 … (i)
p/2 – q/2 = – 1/8
p – q = – 1/4 … (ii)
Adding (i) and (ii), we get
2p = 3/4 – 1/4
2p = 1/2
p = 1/4
Putting the value in equation (ii), we get
1/4 – q = – 1/4
q = 1/4 + 1/4 = 1/2
p = 1/3x + y = 1/4
3x + y = 4 … (iii)
q = 1/3x – y = 1/2
3x – y = 2 … (iv)
Adding equations (iii) and (iv), we get
6x = 6
x = 1 … (v)
Putting the value in equation (iii), we get
3(1) + y = 4
y = 1
Hence, x = 1 and y = 1
2. Formulate the following problems as a pair of equations, and hence find their solutions:
(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.
Ans. Let the speed of Ritu in still water and the speed of stream be x km/h and y km/h respectively.
Speed of Ritu while rowing
Upstream = (x – y) km/h
Downstream = (x + y) km/h
According to question,
2(x + y) = 20
? x + y = 10 … (i)
2(x – y) = 4
? x – y = 2 … (ii)
Adding equation (i) and (ii), we get
Putting this equation in (i), we get
y = 4
Hence, Ritu’s speed in still water is 6 km/h and the speed of the current is 4 km/h.
(ii) 2 women and 5 men can together finish an embroidery work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work, and also that taken by 1 man alone.
Ans. Let the number of days taken by a woman and a man be x and y respectively.
Therefore, work done by a woman in 1 day = 1/x
According to the question,
4(2/x + 5/y) = 1
2/x + 5/y = 1/4
3(3/x + 6/y) = 1
3/x + 6/y = 1/3
Putting 1/x = p and 1/y = q in these equations, we get
2p + 5q = 1/4
By cross multiplication, we get
p/–20 – (–18) = q/–9 – (–18) = 1/144 – 180
p/–2 = q/–1 = 1/–36
p/–2 = – 1/36 and q/–1 = 1/–36
p = 1/18 and q = 1/36
p = 1/x = 1/18 and q = 1/y = 1/36
x = 18 and y = 36
Hence, number of days taken by a woman = 18 and number of days taken by a man = 36
(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Ans. Let the speed of train and bus be u km/h and v km/h respectively.
According to the given information,
60/u + 240/v = 4 … (i)
100/u + 200/v = 25/6 … (ii)
Putting 1/u = p and 1/v = q in the equations, we get
60p + 240q = 4 … (iii)
100p + 200q = 25/6
600p + 1200q = 25 … (iv)
Multiplying equation (iii) by 10, we get
600p + 2400q = 40 …. (v)
Subtracting equation (iv) from (v), we get1200q = 15
q = 15/200 = 1/80 … (vi)
Putting equation (iii), we get
60p + 3 = 4
60p = 1
p = 1/60
p = 1/u = 1/60 and q = 1/v = 1/80
u = 60 and v = 80
Hence, speed of train = 60 km/h and speed of bus = 80 km/h.
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# Precalculus : Simplify Expressions With Rational Exponents
## Example Questions
← Previous 1
Simplify
Explanation:
.
### Example Question #1 : Simplify Expressions With Rational Exponents
Simplify the expression.
Explanation:
Using the properties of exponents, we can either choose to subtract the exponents of the corresponding bases or rewrite the expression using negative exponents as such:
Here, we combine the terms with corresponding bases by adding the exponents together to get
Placing the x term (since it has a negative exponent) in the denominator will result in the correct answer. It can be shown that simply subtracting the exponents of corresponding bases will result in the same answer.
### Example Question #1 : Simplify Expressions With Rational Exponents
Simplify the expression .
Explanation:
We proceed as follows
Write as a fraction
The denominator of the fraction is a , so it becomes a square root.
Take the square root.
Raise to the power.
### Example Question #2 : Simplify Expressions With Rational Exponents
What is the value of
Explanation:
Recall that when considering rational exponents, the denominator of the fraction tells us the "root" of the expression.
Thus in this case we are taking the fifth root of .
The fifth root of is , because .
Thus, we have reduced our expression to
### Example Question #4 : Simplify Expressions With Rational Exponents
Simplify the expression:
Explanation:
Simplify the constants:
Subtract the "x" exponents:
This is how the x moves to the denominator.
Finally subtract the "y" exponents:
### Example Question #6 : Simplify Expressions With Rational Exponents
Solve:
Explanation:
To remove the fractional exponents, raise both sides to the second power and simplify:
Now solve for :
### Example Question #3 : Simplify Expressions With Rational Exponents
Solve:
Explanation:
To remove the rational exponent, cube both sides of the equation:
Now simplify both sides of the equation:
### Example Question #6 : Simplify Expressions With Rational Exponents
Simplify and rewrite with positive exponents:
Explanation:
When dividing two exponents with the same base we subtract the exponents:
Negative exponents are dealt with based on the rule
:
### Example Question #7 : Simplify Expressions With Rational Exponents
Simplify the function:
Explanation:
When an exponent is raised to the power of another exponent, just multiply the exponents together.
Simplify: |
# Word problem solver online
This Word problem solver online helps to fast and easily solve any math problems. Our website can solving math problem.
## The Best Word problem solver online
In this blog post, we discuss how Word problem solver online can help students learn Algebra. Geometry is the math of shapes and solids. In a right triangle, the longest side is opposite the right angle and is called the hypotenuse. The other two sides are the short side and the long side. To find x, use the Pythagorean theorem which states that in a right angled triangle, the sum of the squares of the two shorter sides is equal to the square of the length of the hypotenuse. This theorem is represented by the equation: a^2 + b^2 = c^2. To solve for x, plug in the known values for a and b (the two shorter sides) and rearrange the equation to isolate c (the hypotenuse). For example, if a=3 and b=4, then c^2 = 3^2 + 4^2 = 9 + 16 = 25. Therefore, c = 5 and x = 5.
Algebra is the branch of mathematics that deals with the solution of equations. In an equation, the unknown quantity is represented by a letter, usually x. The object of algebra is to find the value of x that will make the equation true. For example, in the equation 2x + 3 = 7, the value of x that makes the equation true is 2. To solve an equation, one must first understand what each term in the equation represents. In the equation 2x + 3 = 7, the term 2x represents twice the value of x; in other words, it represents two times whatever number is assigned to x. The term 3 represents three units, nothing more and nothing less. The equal sign (=) means that what follows on the left-hand side of the sign is equal to what follows on the right-hand side. Therefore, in this equation, 2x + 3 is equal to 7. To solve for x, one must determine what value of x will make 2x + 3 equal to 7. In this case, the answer is 2; therefore, x = 2. |
# A Small Block of Mass 200 G is Kept at the Top of a Frictionless Incline Which is 10 M Long and 3⋅2 M High. How Much Work Was Required (A) to Lift the Block from the Ground and Put It an the Top, - Physics
Sum
A small block of mass 200 g is kept at the top of a frictionless incline which is 10 m long and 3⋅2 m high. How much work was required (a) to lift the block from the ground and put it an the top, (b) to slide the block up the incline? What will be the speed of the block when it reaches the ground if (c) it falls off the incline and drops vertically to the ground (d) it slides down the incline? Take g = 10 m/s2
#### Solution
$\text{ Given },$
$\text{ Mass of the block, m = 200 g = 0 . 2 kg}$
$\text{ Length of the incline, s = 10 m },$
$\text{ Height of the incline, h = 3 . 2 m }$
$\text{ Acceleration due to gravity, g = 10 m/ s}^2$
(a) Work done, W = mgh = 0.2 × 10 × 3.2 =6 .4 J
(b) Work done to slide the block up the incline
$\text{ W }= \left( \text{ mg } \sin \theta \right) \times \text{ s }$
$= \left( 0 . 2 \right) \times 10 \times \left( 3 . 2/10 \right) \times 10$
$= 6 . 4 \text{ J }$
(c) Let final velocity be v when the block falls to the ground vertically.
Change in the kinetic energy = Work done
$\frac{1}{2}\text{mv}^2 - 0 = 6 . 4 \text{ J }$
$\Rightarrow \nu = 8 \text{ m/s }$
(d) Let $\nu$ be the final velocity of the block when it reaches the ground by sliding.
$\frac{1}{2}m \nu^2 - 0 = 6 . 4 \text{ J }$
$\Rightarrow \text{ v = 8 m/s }$
Is there an error in this question or solution?
#### APPEARS IN
HC Verma Class 11, 12 Concepts of Physics 1
Chapter 8 Work and Energy
Q 34 | Page 134 |
# Time and Work Quiz 13
Home > > Tutorial
5 Steps - 3 Clicks
# Time and Work Quiz 13
### Introduction
Time and Work is one of important topic in Quantitative Aptitude Section. In Time and Work Quiz 13 article candidates can find questions with an answer. By solving this questions candidates can improve and maintain, speed, and accuracy in the exams. Time and Work Quiz 13 questions are very useful for different exams such as IBPS PO, Clerk, SSC CGL, SBI PO, NIACL Assistant, NICL AO, IBPS SO, RRB, Railways, Civil Services etc.
### Q1
P can do a piece of work in 20 days. Q is 25 percent more efficient than P. In how many days half the work is completed when both are working simultaneously?
A. $\frac {41}{9}$ B. $\frac {40}{9}$ C. $\frac {39}{9}$ D. $\frac {43}{9}$
B
Q is 25 percent more efficient so he will complete the work in 16 days
$(\frac {1}{20} + \frac {1}{16}) \times t = \frac {1}{2}$
$(\frac {4 + 5}{80}) \times t = \frac {1}{2}$
$t = \frac {1}{2} \times \frac {80}{9}$
$t = \frac {40}{9}$
### Q2
A does half as much work as B does in one sixth of the time. If together they take 20 days to complete the work, then what is the time taken by B to complete the work independently.
A. 80 days B. 100 days C. 120 days D. 140 days
A
Let B take X days to complete the work then in one – sixth of the time i.e. $\frac {x}{6}$ days. Now A do half work as done by B so A will take twice the time i.e. $2 \times \frac {x}{6} = \frac {x}{3}$ to complete the job alone
So $\frac {1}{x} + \frac {3}{x} = \frac {1}{20}, x = 80$ days
### Q3
A contractor undertakes to make a mall in 60 days and he employs 30 men. After 30 days it is found that only $\frac{1}{3}$ of the work is completed. How many extra men should he employ so that the work is completed on time?
A. 20men B. 25men C. 30men D. 40men
C
Let total work is w and it is given that one-third of the work is completed after 30 days. Means
$M \times D = 30 \times 30 = \frac {w}{3}$, so total work = $30 \times 30 \times 3$
$2700 = 30 \times 30 + (30 + p) \times 30$, so we get P = 30 (p = additional men)
### Q4
P and Q were assigned to do a work for an amount of 1200. P alone can do it in 15 days while Q can do it in 12 days. With the help of R they finish the work in 6 days. Find the share if C.
A. 100 B. 120 C. 140 D. 160
B
$\frac {1}{15} + \frac {1}{12} + \frac {1}{C} = \frac {1}{6}$, we got C = 60 (it means C will take 60 days to complete the work alone)
So ratio of work done by P : Q : R = 4 : 5 : 1
So c share = $(\frac {1}{10}) \times 1200 = 120$
### Q5
P does half as much work as Q in three-fourth of the time. If together they take 24 days to complete the work, how much time shall P take to complete the work?
A. 50 days B. 60 days C. 70 days D. 80 days
B
Let Q take x days to complete the work, so P will take $2 \times \frac {3}{4}$ of X day to complete the work i.e. $\frac {3x}{2}$ days
$\frac {1}{x} + \frac {2}{3x} = \frac {1}{24}$, we get x = 40 days, so P will take = $\frac {3}{2}$ of 40 = 60 days |
# What is the maximum value of Q(p,q,r)=2pq+2pr+2qr subject to p+q+r=1?
Apr 7, 2018
The maximum value of $Q$ is $\frac{2}{3}$.
#### Explanation:
We want to maximise $Q \left(p , q , r\right) = 2 p r + 2 p q + 2 q r$ subject to $p + q + r - 1 = 0$.
Let $P \left(p , q , r\right) = q + p + r - 1$. By the method of Lagrange multipliers, the extrema of $Q$ occur where
$\nabla Q = \lambda \nabla P$
$\Rightarrow \left(\begin{matrix}2 q + 2 r \\ 2 p + 2 r \\ 2 p + 2 q\end{matrix}\right) = \lambda \left(\begin{matrix}1 \\ 1 \\ 1\end{matrix}\right)$
So
$2 q + 2 r = \lambda$ $\left(1\right)$
$2 p + 2 r = \lambda$ $\left(2\right)$
$2 p + 2 q = \lambda$ $\left(3\right)$
$\left(1\right) - \left(2\right) \Rightarrow 2 q - 2 p = 0 \Rightarrow p = q$
$\left(1\right) - \left(3\right) \Rightarrow 2 r - 2 p = 0 \Rightarrow p = r$
Since $p + q + r = 1$, it follows that $p = q = r = \frac{1}{3}$
So the maximum value of $Q$ is $2 \left(\frac{1}{3}\right) \left(\frac{1}{3}\right) + 2 \left(\frac{1}{3}\right) \left(\frac{1}{3}\right) + 2 \left(\frac{1}{3}\right) \left(\frac{1}{3}\right) = \frac{2}{3}$
Apr 7, 2018
See below.
#### Explanation:
${\left(p + q + r\right)}^{2} = {p}^{2} + {q}^{2} + {r}^{2} + Q \left(p , q , r\right) = 1$ then
$1 - Q \left(p , q , r\right) = {p}^{2} + {q}^{2} + {r}^{2}$
The solution is at the tangency points between the sphere
$S \left(p , q , r\right) = {p}^{2} + {q}^{2} + {r}^{2}$
with the plane
$\Pi \to p + q + r = 1$
This point is for $p = q = r = \frac{1}{3} \Rightarrow Q \left(p , q , r\right) = \frac{2}{3}$ |
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# Doubling the Cube
## Overview and Objective
In this lesson, students learn about one of the oldest mathematical stories while exploring the volume of a cube, constructible numbers, square root, and cube root numbers.
## Warm-Up
Share this polypad with students and invite them to work on the area of the squares and volume of the cubes. Click here to learn how to share polypads with students and how to view their work. After a few minutes of work time, share some student work with the class. Share the numbers like 1, 4, 9, 16 .. are called square numbers, whereas 1, 8, 27, 64... are cube numbers. Discuss which numbers are both square and cube numbers.
## Main Activity
Share the following legend with students:
According to legend, the city of Delos in ancient Greece was once faced with a terrible plague. An oracle told them that this was a punishment from the Gods, and the plague would go away if they built a new temple that was exactly twice the volume of the existing one. Delians doubled the edges of the temple, but the plague did not stop. What went wrong?
Share this canvas with students and give them some time to let them work on the problem. Students should use the custom polygon to create a cube with the given conditions.
Students are challenged to create additional cubes with twice the area and the volume, as well as the one that Delians built. Students may discover that when Delians doubled the edges, the new temple had an area 4 times bigger and a volume 8 times bigger. Since they were tasked with doubling the volume, they did not stop the plague. Ask them how to fix that historical mistake.
Let's first try doubling the area. Use the pink free polygon to create the new cubic temple with twice the area.After students figure out the side length must be $sqrt{2}$, ask them to create a square with this side length. Some of the students may remember $sqrt{2}$ is the diagonal length of a 1 x 1 square. They can use this idea to create the square, then the net of the cube.
Explain that the tools that Delians use at the time are only a compass and straightedge. However, they could still construct a length of $sqrt{2}$ to double the area of the cube.
The √2 is the length of the hypotenuse of a right triangle with legs of length 1. Therefore it can be constructed using a compass and straightedge, which makes √2 a constructible number.
Now, it is time to create a cube with twice the volume. In algebraic terms, doubling a unit cube requires the construction of a line segment of length $x$, where $x^3$ = 2
in other words, x = $^3sqrt{2}$, the cube root of two.
After students figure out the side length, let them try to create a square with this side length using the pink free polygon tool on the canvas. They will realize that it is not as easy as constructing √2. In fact, it is not possible to construct this number using a compass and a straightedge. $^3sqrt{2}$ is not a constructible number
That's why doubling the volume of a cube, in fact, for that time was an impossible request.
## Closure
You can close the lesson by mentioning that doubling the cube is one of the 3 main geometric problems in history (Geometric Problems of Antiquity) whose solutions were sought using only compass and straightedge:
1. Circle Squaring
2. Doubling a cube
3. Trisecting an angle
More than 2000 years after they were formulated, only in modern times were all three ancient problems proved insoluble using only compass and straightedge. Today, there are ways to draw a line segment with a length of $^3sqrt{2}$ without the restriction of using a compass and a straightedge.
## Support and Extension
For students ready for additional extension in this lesson, consider asking questions like constructing line segments with side lengths of √3, √5,√6.. and so on a blank Polypad canvas with a square grid background.
## Polypads for This Lesson
To assign these to your classes in Mathigon, save a copy to your Mathigon account. Click here to learn how to share Polypads with students and how to view their work. |
# WORD PROBLEMS ON RATIONAL EQUATIONS WORKSHEET
Problem 1 :
Decreasing the reciprocal of a number by 7 results ⁻²⁰⁄₃. What is the number?
Problem 2 :
7 more than three times the reciprocal of a number results ³⁸⁄₅. What is the number?
Problem 3 :
Kevin needs 150 ounces of sugar for making 50 pounds cookies. If Kevin currently has 80 ounces of sugar, how many more ounces of sugar does he need to make 30 pounds cookies?
Problem 4 :
John is able to complete a certain work in 10 days, but Peter is abobe to complete the same work in 8 days. How many days will it take them to complete the work, if they work together?
Problem 5 :
The numerator and denominator of a fraction add up to 11. If 5 be added to both numerator and denominator, the fraction becomes ¾. Find the original fraction.
Problem 6 :
Divide 15 into two parts such that both of them are positive and the difference between their reciprocals is ¼
Let x be the number.
¹⁄ₓ - 7 = ⁻²⁰⁄₃
Multiply both sides by x to get rid of the denominator x on the left side.
x(¹⁄ₓ - 7) = x(⁻²⁰⁄₃)
x(¹⁄ₓ) - x(7) = ⁻²⁰ˣ⁄₃
1 - 7x = ⁻²⁰ˣ⁄₃
Multiply both sides by 3 to get rid of the denominator 5 on the right side.
3(1 - 7x) = 3(⁻²⁰ˣ⁄₃)
3 - 21x = -20x
3 = x
The number is 3.
Let x be the number.
3(¹⁄ₓ) + 7 = ³⁸⁄₅
³⁄ₓ + 7 = ³⁸⁄₅
Multiply both sides by x to get rid of the denominator x on the left side.
x(³⁄ₓ + 7) = x(³⁸⁄₅)
x(³⁄ₓ) + x(7) = ³⁸ˣ⁄₅
3 + 7x = ³⁸ˣ⁄₅
Multiply both sides by 5 to get rid of the denominator 5 on the right side.
5(3 + 7x) = 5(³⁸ˣ⁄₅)
15 + 35x = 38x
Subtract 35x from both sides.
15 = 3x
Divide both sides by 3.
5 = x
The number is 5.
Let x be the additional ounces ofsugar needed to make 30 pounds of cookies.
150 ounces : 50 pounds = (80 + x) ounces : 30 pounds
Write each ratio in the equation above as a fraction.
¹⁵⁰⁄₅₀ = ⁽⁸⁰ ⁺ ˣ⁾⁄₃₀
3 = ⁽⁸⁰ ⁺ ˣ⁾⁄₃₀
Multiply both sides by 30 to get rid of the denominator 30 on the right side.
30(3) = 30(⁽⁸⁰ ⁺ ˣ⁾⁄₃₀)
90 = 80 + x
Subtract 80 from both sides.
10 = x
Kevin needs 10 more ounces of sugar to make 30 pounds.
Givren : John is able to complete a certain work in 10 days and Peter is abobe to complete the same work in 8 days.
Part of the work completed by John in 1 hour :
=
Part of the work completed by Peter in 1 hour :
Let x be the number of days required to complete the work, if both John and Peter work together.
Part of the work completed by both John and Peter in 1 hour :
¹⁄ₓ
¹⁄ₓ
Multiply both sides by x to get rid of the denominator on the left side.
x(¹⁄ₓ) = x( )
1 = x() + x()
1 = ˣ⁄₁₀ + ˣ⁄₈
The least common multiple of (10, 8) = 40.
Multiply both sides by 40 to get rid of the denominators 10 and 8 on the right side.
40(1) = 40(ˣ⁄₁₀ + ˣ⁄₈)
40 = 40(ˣ⁄₁₀) + 40(ˣ⁄₈)
40 = 4x + 5x
40 = 9x
Divide both sides by 9.
⁴⁰⁄₉ = x
4⁴⁄₉ = x
If both John and Peter work together, they will be able to complete the work in 4⁴⁄₉ days.
Let x be the denominator of the fraction.
Since the numerator and denominator add up to 11, the numerator is (11 - x).
Fraction = ⁽¹¹ ⁻ ˣ⁾⁄ₓ
Givren : If 5 is added to both numerator and denominator, the fraction becomes ¾.
⁽¹¹ ⁻ ˣ ⁺ ⁵⁾⁄₍ₓ ₊ ₅₎ = ¾
⁽¹⁶ ⁻ ˣ⁾⁄₍ₓ ₊ ₅₎ = ¾
By cross multiplying,
4(16 - x) = 3(x + 5)
64 - 4x = 3x + 15
64 = 7x + 15
Subtract 15 from both sides.
49 = 7x
Divide both sides by 7.
7 = x
11 - x = 11 - 7 = 4
⁽¹¹ ⁻ ˣ⁾⁄ₓ ⁴⁄₇
The required fraction is ⁴⁄₇.
Let x be one of the parts of 15.
Then the other part of 15 is (15 - x).
Given : Difference between their reciprocals of the parts of 25 is ¼.
¹⁄ₓ - ¹⁄₍₁₅ ₋ ₓ₎ = ¼
Multiply both sides by x to get rid of the denominator x on the left side.
x[¹⁄ₓ - ¹⁄₍₁₅ ₋ ₓ₎] = x(¼)
x(¹⁄ₓ) - x[¹⁄₍₁₅ ₋ ₓ₎] = ˣ⁄₄
1 - ˣ⁄₍₁₅ ₋ ₓ₎ = ˣ⁄₄
Multiply both sides by (15 - x) to get rid of the denominator (15 - x) on the left side.
(15 - x)[1 - ˣ⁄₍₁₅ ₋ ₓ₎] = (15 - x)(ˣ⁄₄)
(15 - x)(1) - (15 - x)[ˣ⁄₍₁₅ ₋ ₓ₎] = (15 - x)(ˣ⁄₄)
15 - x - x = (15 - x)(ˣ⁄₄)
15 - 2x (15 - x)(ˣ⁄₄)
Multiply both sides by 4 to get rid of the denominator 4 on the right side.
4(15 - 2x) = 4(15 - x)(ˣ⁄₄)
60 - 8x = (15 - x)(x)
60 - 8x = 15x - x2
x- 8x + 60 = 15x
Subtract 15x from both sides.
x- 23x + 60 = 0
Factor and solve.
x- 3x - 20x + 60 = 0
x(x - 3) - 20(x - 3) = 0
(x - 3)(x - 20) = 0
x - 3 = 0 or x - 20 = 0
x = 3 or x = 20
If x = 3,
15 - x = 15 - 3
15 - x = 12
If x = 20,
15 - x = 15 - 20
15 - x = -5
When x = 20, the other part (15 - x) is -5, which is negative. So, it can be ignored.
Therefore, the two parts of 15 are 3 and 12.
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# 3.3 Future value annuities
## 3.3 Future value annuities (EMCFZ)
For future value annuities, we regularly save the same amount of money into an account, which earns a certain rate of compound interest, so that we have money for the future.
## Worked example 3: Future value annuities
At the end of each year for $$\text{4}$$ years, Kobus deposits $$\text{R}\,\text{500}$$ into an investment account. If the interest rate on the account is $$\text{10}\%$$ per annum compounded yearly, determine the value of his investment at the end of the $$\text{4}$$ years.
### Write down the given information and the compound interest formula
$A = P{\left(1+i\right)}^{n}$ \begin{align*} P &= \text{500} \\ i &= \text{0,1} \\ n &= 4 \end{align*}
### Draw a timeline
The first deposit in the account earns the highest amount of interest (three interest payments) and the last deposit earns the least interest (no interest payments).
We can summarize this information in the table below:
Deposit No. of interest payments Calculation Accumulated amount Year 1 $$\text{R}\,\text{500}$$ $$\text{3}$$ $$\text{500}(1 + \text{0,1})^{3}$$ $$\text{R}\,\text{665,50}$$ Year 2 $$\text{R}\,\text{500}$$ $$\text{2}$$ $$\text{500}(1 + \text{0,1})^{2}$$ $$\text{R}\,\text{605,00}$$ Year 3 $$\text{R}\,\text{500}$$ $$\text{1}$$ $$\text{500}(1 + \text{0,1})^{1}$$ $$\text{R}\,\text{550,00}$$ Year 4 $$\text{R}\,\text{500}$$ $$\text{0}$$ $$\text{500}(1 + \text{0,1})^{0}$$ $$\text{R}\,\text{500,00}$$ Total $$\text{R}\,\text{2 320,50}$$
### Deriving the formula (EMCG2)
Note: for this section is it important to be familiar with the formulae for the sum of a geometric series (Chapter 1):
\begin{align*} S_{n} = \frac{a\left({r}^{n}-1\right)}{r-1} & \qquad \text{for } r >1 \\ S_{n} = \frac{a\left(1-{r}^{n}\right)}{1-r} & \qquad \text{for } r < 1 \end{align*}
In the worked example above, the total value of Kobus' investment at the end of the four year period is calculated by summing the accumulated amount for each deposit:
$$\begin{array}{c@{\;}l@{\;}l@{\;}l@{\;}l@{\;}} \text{R}\,\text{2 320,50} &= \text{R}\,\text{500,00} \quad + & \text{R}\,\text{550,00} \quad + & \text{R}\,\text{605,00} \quad + & \text{R}\,\text{665,50} \\ &= \text{500}(1 + \text{0,1})^{0} \enspace + & \text{500}(1 + \text{0,1})^{1} \enspace + & \text{500}(1 + \text{0,1})^{2} \enspace + & \text{500}(1 + \text{0,1})^{3} \end{array}$$
We notice that this is a geometric series with a constant ratio $$r = 1 + \text{0,1}$$.
Using the formula for the sum of a geometric series:
\begin{align*} a &= \text{500} \\ r &= \text{1,1} \\ n &= 4 \\ & \\ S_{n} &= \frac{a\left({r}^{n}-1\right)}{r-1} \\ &= \frac{\text{500}\left({\text{1,1}}^{4}-1\right)}{\text{1,1} - 1} \\ &= \text{2 320,50} \end{align*}
We can therefore use the formula for the sum of a geometric series to derive a formula for the future value ($$F$$) of a series of ($$n$$) regular payments of an amount ($$x$$) which are subject to an interest rate ($$i$$):
\begin{align*} a &= x \\ r &= 1 + i \\ & \\ S_{n} &= \frac{a\left({r}^{n}-1\right)}{r-1} \\ \therefore F &= \frac{x\left[(1 + i)^{n}-1\right]}{(1 + i)-1} \\ &= \frac{x\left[(1 + i)^{n}-1\right]}{i} \end{align*}
Future value of payments:
$F = \frac{x\left[(1 + i)^{n}-1\right]}{i}$
If we are given the future value of a series of payments, then we can calculate the value of the payments by making $$x$$ the subject of the above formula.
Payment amount:
$x = \frac{F \times i}{\left[(1 + i)^{n}-1\right]}$
## Worked example 4: Future value annuities
Ciza decides to start saving money for the future. At the end of each month she deposits $$\text{R}\,\text{900}$$ into an account at Harringstone Mutual Bank, which earns $$\text{8,25}\%$$ interest p.a. compounded monthly.
1. Determine the balance of Ciza's account after $$\text{29}$$ years.
2. How much money did Ciza deposit into her account over the $$\text{29}$$ year period?
3. Calculate how much interest she earned over the $$\text{29}$$ year period.
### Write down the given information and the future value formula
$F = \frac{x\left[(1 + i)^{n}-1\right]}{i}$ \begin{align*} x &= \text{900} \\ i &= \frac{\text{0,0825}}{12} \\ n &= 29 \times 12 = \text{348} \end{align*}
### Substitute the known values and use a calculator to determine $$F$$
\begin{align*} F &= \dfrac{\text{900}\left[(1 + \frac{\text{0,0825}}{12})^{\text{348}}-1\right]}{\frac{\text{0,0825}}{12}} \\ &= \text{R}\,\text{1 289 665,06} \end{align*}
Remember: do not round off at any of the interim steps of a calculation as this will affect the accuracy of the final answer.
### Calculate the total value of deposits into the account
Ciza deposited $$\text{R}\,\text{900}$$ each month for $$\text{29}$$ years:
\begin{align*} \text{Total deposits: } &= \text{R}\,\text{900} \times 12 \times 29 \\ &= \text{R}\,\text{313 200} \end{align*}
### Calculate the total interest earned
\begin{align*} \text{Total interest } &= \text{final account balance } - \text{total value of all deposits} \\ &= \text{R}\,\text{1 289 665,06} - \text{R}\,\text{313 200} \\ &= \text{R}\,\text{976 465,06} \end{align*}
temp text
Useful tips for solving problems:
1. Timelines are very useful for summarising the given information in a visual way.
2. When payments are made more than once per annum, we determine the total number of payments ($$n$$) by multiplying the number of years by $$p$$:
Term $$p$$ yearly / annually $$\text{1}$$ half-yearly / bi-annually $$\text{2}$$ quarterly $$\text{4}$$ monthly $$\text{12}$$ weekly $$\text{52}$$ daily $$\text{365}$$
3. If a nominal interest rate $$\left( i^{(m)} \right)$$ is given, then use the following formula to convert it to an effective interest rate:
$1 + i = \left( 1 + \frac{i^{(m)}}{m} \right)^{m}$
## Worked example 5: Calculating the monthly payments
Kosma is planning a trip to Canada to visit her friend in two years' time. She makes an itinerary for her holiday and she expects that the trip will cost $$\text{R}\,\text{25 000}$$. How much must she save at the end of every month if her savings account earns an interest rate of $$\text{10,7}\%$$ per annum compounded monthly?
### Write down the given information and the future value formula
$F = \frac{x\left[(1 + i)^{n}-1\right]}{i}$
To determine the monthly payment amount, we make $$x$$ the subject of the formula:
$x = \frac{F \times i}{\left[(1 + i)^{n}-1\right]}$ \begin{align*} F &= \text{25 000} \\ i &= \frac{\text{0,107}}{12} \\ n &= 2 \times 12 = \text{24} \end{align*}
### Substitute the known values and calculate $$x$$
\begin{align*} x &= \dfrac{\text{25 000} \times \frac{\text{0,107}}{12}}{\left[(1 + \frac{\text{0,107}}{12})^{24}-1\right]} \\ &= \text{R}\,\text{938,80} \end{align*}
Kosma must save $$\text{R}\,\text{938,80}$$ each month so that she can afford her holiday.
## Worked example 6: Determining the value of an investment
Simon starts to save for his retirement. He opens an investment account and immediately deposits $$\text{R}\,\text{800}$$ into the account, which earns $$\text{12,5}\%$$ per annum compounded monthly. Thereafter, he deposits $$\text{R}\,\text{800}$$ at the end of each month for $$\text{20}$$ years. What is the value of his retirement savings at the end of the $$\text{20}$$ year period?
### Write down the given information and the future value formula
$F = \frac{x\left[(1 + i)^{n}-1\right]}{i}$ \begin{align*} x &= \text{800} \\ i &= \frac{\text{0,125}}{12} \\ n &= 1 + (20 \times 12) = \text{241} \end{align*}
Note that we added one extra month to the $$\text{20}$$ years because Simon deposited $$\text{R}\,\text{800}$$ immediately.
### Substitute the known values and calculate $$F$$
\begin{align*} F &= \frac{\text{800}\left[(1 + \frac{\text{0,125}}{12})^{\text{241}}-1\right]}{ \frac{\text{0,125}}{12}} \\ &= \text{R}\,\text{856 415,66} \end{align*}
Simon will have saved $$\text{R}\,\text{856 415,66}$$ for his retirement.
## Future value annuities
Textbook Exercise 3.2
Shelly decides to start saving money for her son's future. At the end of each quarter she deposits $$\text{R}\,\text{500}$$ into an account at Durban Trust Bank, which earns an interest rate of $$\text{5,96}\%$$ per annum compounded quarterly.
Determine the balance of Shelly's account after $$\text{35}$$ years.
Write down the given information and the future value formula:
$F = \frac{x\left[(1 + i)^{n}-1\right]}{i}$ \begin{align*} x &= \text{500} \\ i &= \frac{\text{0,0596}}{4} \\ n &= 35 \times 4 = \text{140} \end{align*}
Substitute the known values and use a calculator to determine $$F$$:
\begin{align*} F &= \dfrac{\text{500}\left[(1 + \frac{\text{0,0596}}{4})^{\text{140}}-1\right]}{\frac{\text{0,0596}}{4}} \\ &= \text{R}\,\text{232 539,41} \end{align*}
How much money did Shelly deposit into her account over the $$\text{35}$$ year period?
Calculate the total value of deposits into the account:
Shelly deposited $$\text{R}\,\text{500}$$ each quarter for $$\text{35}$$ years:
\begin{align*} \text{Total deposits: } &= \text{R}\,\text{500} \times 4 \times 35 \\ &= \text{R}\,\text{70 000} \end{align*}
Calculate how much interest she earned over the $$\text{35}$$ year period.
Calculate the total interest earned:
\begin{align*} \text{Total interest } &= \text{final account balance } - \text{total value of all deposits} \\ &= \text{R}\,\text{232 539,41} - \text{R}\,\text{70 000} \\ &= \text{R}\,\text{162539,41} \end{align*}
Gerald wants to buy a new guitar worth $$\text{R}\,\text{7 400}$$ in a year's time. How much must he deposit at the end of each month into his savings account, which earns a interest rate of $$\text{9,5}\%$$ p.a. compounded monthly?
Write down the given information and the future value formula:
$F = \frac{x\left[(1 + i)^{n}-1\right]}{i}$
To determine the monthly payment amount, we make $$x$$ the subject of the formula:
$x = \frac{F \times i}{\left[(1 + i)^{n}-1\right]}$ \begin{align*} F &= \text{7 400} \\ i &= \frac{\text{0,095}}{12} \\ n &= 1 \times 12 = \text{12} \end{align*}
Substitute the known values and calculate $$x$$:
\begin{align*} x &= \dfrac{\text{7 400} \times \frac{\text{0,095}}{12}}{\left[(1 + \frac{\text{0,095}}{12})^{12}-1\right]} \\ &= \text{R}\,\text{590,27} \end{align*}
Gerald must deposit $$\text{R}\,\text{590,27}$$ each month so that he can afford his guitar.
A young woman named Grace has just started a new job, and wants to save money for the future. She decides to deposit $$\text{R}\,\text{1 100}$$ into a savings account every month. Her money goes into an account at First Mutual Bank, and the account earns $$\text{8,9}\%$$ interest p.a. compounded every month.
How much money will Grace have in her account after $$\text{29}$$ years?
\begin{align*} F & = \frac{x\left[(1+i)^n - 1\right]}{i} \\ \text{Where: } \quad x & = \text{R}\,\text{1 100} \\ i &= \text{0,089} \\ n & = \text{29} \end{align*} \begin{align*} F & = \frac{(\text{1 100}) \left[ \left(1 + \frac{\text{0,089}}{\text{12}}\right) ^{(\text{29} \times \text{12})} - 1 \right]} {\left(\frac{\text{0,089}}{\text{12}} \right)} \\ & = \text{R}\,\text{1 792 400,11} \end{align*}
After $$\text{29}$$ years, Grace will have $$\text{R}\,\text{1 792 400,11}$$ in her account.
How much money did Grace deposit into her account by the end of the $$\text{29}$$ year period?
The total amount of money Grace saves each year is $$\text{1 100} \times \text{12} = \text{R}\,\text{13 200}$$. From that we can determine the total amount she saves by multiplying by the number of years: $$\text{13 200} \times \text{29} = \text{R}\,\text{382 800}$$.
After $$\text{29}$$ years, Grace deposited a total of $$\text{R}\,\text{382 800}$$ into her account.
Ruth decides to save for her retirement so she opens a savings account and immediately deposits $$\text{R}\,\text{450}$$ into the account. Her savings account earns $$\text{12}\%$$ per annum compounded monthly. She then deposits $$\text{R}\,\text{450}$$ at the end of each month for $$\text{35}$$ years. What is the value of her retirement savings at the end of the $$\text{35}$$ year period?
Write down the given information and the future value formula:
$F = \frac{x\left[(1 + i)^{n}-1\right]}{i}$ \begin{align*} x &= \text{450} \\ i &= \frac{\text{0,12}}{12} \\ n &= 1 + (35 \times 12) = \text{421} \end{align*}
Substitute the known values and calculate $$F$$:
\begin{align*} F &= \frac{\text{450}\left[(1 + \frac{\text{0,12}}{12})^{\text{421}}-1\right]}{ \frac{\text{0,12}}{12}} \\ &= \text{R}\,\text{2 923 321,08} \end{align*}
Ruth will have saved $$\text{R}\,\text{2 923 321,08}$$ for her retirement.
Musina MoneyLenders offer a savings account with an interest rate of $$\text{6,13}\%$$ p.a. compounded monthly. Monique wants to save money so that she can buy a house when she retires. She decides to open an account and make regular monthly deposits. Her goal is to end up with $$\text{R}\,\text{750 000}$$ in her account after $$\text{35}$$ years.
How much must Monique deposit into her account each month in order to reach her goal?
\begin{align*} F & = \frac{x\left[(1+i)^n - 1\right]}{i} \\ F & = \text{R}\,\text{750 000} \\ i & = \text{0,0613} \\ n & = \text{35} \end{align*} \begin{align*} \text{750 000} & = \frac{x \left[ \left(1 + \frac{\text{0,0613}}{\text{12}}\right) ^{(\text{35} \times \text{12})} \right]} {\left(\frac{\text{0,0613}}{\text{12}} \right)} \\ \therefore x &= \frac{ \text{750 000} \times \left(\frac{\text{0,0613}}{\text{12}} \right) }{\left[ \left(1 + \frac{\text{0,0613}}{\text{12}}\right) ^{(\text{35} \times \text{12})} \right]} \\ &= \text{510,84927} \ldots \end{align*}
In order to save $$\text{R}\,\text{750 000}$$ in $$\text{35}$$ years, Monique will need to save $$\text{R}\,\text{510,85}$$ in her account every month.
How much money, to the nearest rand, did Monique deposit into her account by the end of the $$\text{35}$$ year period?
The final amount calculated in the question above includes the money Monique deposited into the account plus the interest paid by the bank. The total amount of money Monique put into her account during the $$\text{35}$$ year is the product of $$\text{12}$$ payments per year, $$\text{35}$$ years, and the payment amount itself:
$\text{R}\,\text{510,85} \times 12 \times 35 = \text{R}\,\text{214 557,00}$
After $$\text{35}$$ years, Monique deposited a total of $$\text{R}\,\text{214 557}$$ into her account.
Lerato plans to buy a car in five and a half years' time. She has saved $$\text{R}\,\text{30 000}$$ in a separate investment account which earns $$\text{13}\%$$ per annum compound interest. If she doesn't want to spend more than $$\text{R}\,\text{160 000}$$ on a vehicle and her savings account earns an interest rate of $$\text{11}\%$$ p.a. compounded monthly, how much must she deposit into her savings account each month?
First calculate the accumulated amount for the $$\text{R}\,\text{30 000}$$ in Lerato's investment account:
$A = P(1 + i)^{n}$ \begin{align*} P &= \text{30 000} \\ i &= \text{0,13} \\ n &= \text{5,5} \end{align*} \begin{align*} A &= \text{30 000}(1 + \text{0,13})^{\text{5,5}}\\ &= \text{R}\,\text{58 756,06} \end{align*}
In five and a half years' time, Lerato needs to have saved $$\text{R}\,\text{160 000} - \text{R}\,\text{58 756,06} = \text{R}\,\text{101 243,94}$$.
$x = \frac{F \times i}{\left[(1 + i)^{n}-1\right]}$ \begin{align*} F &= \text{101 243,94} \\ i &= \frac{\text{0,11}}{12} \\ n &= \text{5,5} \times 12 = \text{66} \end{align*}
Substitute the known values and calculate $$x$$:
\begin{align*} x &= \dfrac{\text{101 243,94} \times \frac{\text{0,11}}{12}}{\left[(1 + \frac{\text{0,11}}{12})^{66}-1\right]} \\ &= \text{R}\,\text{1 123,28} \end{align*}
Lerato must deposit $$\text{R}\,\text{1 123,28}$$ each month into her savings account.
Every Monday Harold puts $$\text{R}\,\text{30}$$ into a savings account at the King Bank, which accrues interest of $$\text{6,92}\%$$ p.a. compounded weekly. How long will it take Harold's account to reach a balance of $$\text{R}\,\text{4 397,53}$$. Give the answer as a number of years and days to the nearest integer.
\begin{align*} F & = \frac{x\left[(1+i)^n - 1\right]}{i} \\ F & = \text{R}\,\text{4 397,53} \\ x & = \text{R}\,\text{30} \\ i &= \text{0,0692} \end{align*} \begin{align*} \text{4 397,53} & = \frac{(\text{30}) \left[ \left(1 + \frac{\text{0,0692}}{\text{52}}\right) ^{(n \times \text{52})} - 1 \right]} {\left(\frac{\text{0,0692}}{\text{52}} \right)} \\ \text{4 397,53} & = \frac{(\text{30}) \left[ \left( \text{1,00133} \right) ^{\text{52}n} - 1 \right]} {\text{0,00133} \ldots} \end{align*} \begin{align*} (\text{0,00133} \ldots)(\text{4 397,53}) & = (\text{30}) \left[ \left( \text{1,00133} \ldots \right) ^{\text{52}n} - 1 \right] \\ \frac{\text{5,85209} \ldots}{\text{30}} & = \left[ \left( \text{1,00133} \ldots \right) ^{\text{52}n} - 1 \right] \end{align*} \begin{align*} \text{0,19506} \ldots + 1 & = \left( \text{1,00133} \ldots \right) ^{\text{52}n} \\ \text{1,19506} \ldots & = \left( \text{1,00133} \ldots \right) ^{\text{52}n} \\ \text{Change to logarithmic form: } \quad \text{52}n & = \log_{\text{1,00133} \ldots} (\text{1,19506} \ldots) \\ \text{52}n & = \text{134} \\ n & = \frac{\text{134}}{\text{52}} \\ n & = \text{2,57692} \ldots \end{align*}
To get to the final answer for this question, convert $$\text{2,57692} \ldots$$ years into years and days.
$$\qquad (\text{0,57692} \ldots) \times \frac{\text{365}}{\text{year}} = \text{210,577}$$ days
Harold's investment takes $$\text{2}$$ years and $$\text{211}$$ days to reach the final value of $$\text{R}\,\text{4 397,53}$$.
How much interest will Harold receive from the bank during the period of his investment?
The total amount Harold invests is as follows:
$$\quad \text{30} \times \text{52} \times \text{2,57692} \ldots = \text{R}\,\text{4 020,00}$$
Therefore, the total amount of interest paid by the bank: $$\text{R}\,\text{4 397,53} - \text{R}\,\text{4 020,00} = \text{R}\,\text{377,53}$$.
### Sinking funds (EMCG3)
Vehicles, equipment, machinery and other similar assets, all depreciate in value as a result of usage and age. Businesses often set aside money for replacing outdated equipment or old vehicles in accounts called sinking funds. Regular deposits, and sometimes lump sum deposits, are made into these accounts so that enough money will have accumulated by the time a new machine or vehicle needs to be purchased.
## Worked example 7: Sinking funds
Wellington Courier Company buys a delivery truck for $$\text{R}\,\text{296 000}$$. The value of the truck depreciates on a reducing-balance basis at $$\text{18}\%$$ per annum. The company plans to replace this truck in seven years' time and they expect the price of a new truck to increase annually by $$\text{9}\%$$.
1. Calculate the book value of the delivery truck in seven years' time.
2. Determine the minimum balance of the sinking fund in order for the company to afford a new truck in seven years' time.
3. Calculate the required monthly deposits if the sinking fund earns an interest rate of $$\text{13}\%$$ per annum compounded monthly.
### Determine the book value of the truck in seven years' time
\begin{align*} P &= \text{296 000} \\ i &= \text{0,18} \\ n &= \text{7} \\ & \\ A &= P(1 - i)^{n} \\ &= \text{296 000}(1 - \text{0,18})^{\text{7}} \\ &= \text{R}\,\text{73 788,50} \end{align*}
### Determine the minimum balance of the sinking fund
Calculate the price of a new truck in seven years' time:
\begin{align*} P &= \text{296 000} \\ i &= \text{0,09} \\ n &= \text{7} \\ & \\ A &= P(1 + i)^{n} \\ &= \text{296 000}(1 + \text{0,09})^{\text{7}} \\ &= \text{R}\,\text{541 099,58} \end{align*}
Therefore, the balance of the sinking fund ($$F$$) must be greater than the cost of a new truck in seven years' time minus the money from the sale of the old truck:
\begin{align*} F &= \text{R}\,\text{541 099,58} - \text{R}\,\text{73 788,50} \\ &= \text{R}\,\text{467 311,08} \end{align*}
### Calculate the required monthly payment into the sinking fund
$x = \frac{F \times i}{\left[(1 + i)^{n}-1\right]}$ \begin{align*} F &= \text{467 311,08} \\ i &= \frac{\text{0,13}}{12} \\ n &= \text{7} \times 12 = \text{84} \end{align*}
Substitute the values and calculate $$x$$:
\begin{align*} x &= \dfrac{\text{467 311,08} \times \frac{\text{0,13}}{12}}{\left[(1 + \frac{\text{0,13}}{12})^{84}-1\right]} \\ &= \text{R}\,\text{3 438,77} \end{align*}
Therefore, the company must deposit $$\text{R}\,\text{3 438,77}$$ each month.
## Sinking funds
Textbook Exercise 3.3
Mfethu owns his own delivery business and he will need to replace his truck in $$\text{6}$$ years' time. Mfethu deposits $$\text{R}\,\text{3 100}$$ into a sinking fund each month, which earns $$\text{5,3}\%$$ interest p.a. compounded monthly.
How much money will be in the fund in $$\text{6}$$ years' time, when Mfethu wants to buy the new truck?
\begin{align*} F & = \frac{x\left[(1+i)^n - 1\right]}{i} \\ \text {Where: } \quad & \\ x & = \text{3 100} \\ i & = \text{0,053} \\ n & = \text{6} \end{align*}
Interest is compounded monthly: $$\; i = \text{0,053} \rightarrow \frac{\text{0,053}}{12} \;$$ and $$\; n = \text{6} \rightarrow \text{6} \times (12) \;$$.
\begin{align*} F & = \frac{(\text{3 100}) \left[ \left(1 + \frac{\text{0,053}}{12}\right) ^{(\text{6} \times 12)} - 1 \right]} {\left(\frac{\text{0,053}}{12} \right)} \\ & = \text{R}\,\text{262 094,55} \end{align*}
After $$\text{6}$$ years, Mfethu will have $$\text{R}\,\text{262 094,55}$$ in his sinking fund.
If a new truck costs $$\text{R}\,\text{285 000}$$ in $$\text{6}$$ years' time, will Mfethu have enough money to buy it?
No, Mfethu does not have enough money in his account:
$\text{R}\,\text{285 000} - \text{R}\,\text{262 094,55} = \text{R}\,\text{22 905,45}$
Atlantic Transport Company buys a van for $$\text{R}\,\text{265 000}$$. The value of the van depreciates on a reducing-balance basis at $$\text{17}\%$$ per annum. The company plans to replace this van in five years' time and they expect the price of a new van to increase annually by $$\text{12}\%$$.
Calculate the book value of the van in five years' time.
\begin{align*} P &= \text{265 000} \\ i &= \text{0,17} \\ n &= \text{5} \\ & \\ A &= P(1 - i)^{n} \\ &= \text{265 000}(1 - \text{0,17})^{\text{5}} \\ &= \text{R}\,\text{104 384,58} \end{align*}
Determine the amount of money needed in the sinking fund for the company to be able to afford a new van in five years' time.
\begin{align*} P &= \text{265 000} \\ i &= \text{0,12} \\ n &= \text{5} \\ & \\ A &= P(1 + i)^{n} \\ &= \text{265 000}(1 + \text{0,12})^{\text{5}} \\ &= \text{R}\,\text{467 020,55} \end{align*}
Therefore, the balance of the sinking fund ($$F$$) must be greater than the cost of a new van in five years' time less the money from the sale of the old van:
\begin{align*} F &= \text{R}\,\text{467 020,55} - \text{R}\,\text{104 384,58} \\ &= \text{R}\,\text{362 635,97} \end{align*}
Calculate the required monthly deposits if the sinking fund earns an interest rate of $$\text{11}\%$$ per annum compounded monthly.
Calculate the required monthly payment into the sinking fund:
$x = \frac{F \times i}{\left[(1 + i)^{n}-1\right]}$ \begin{align*} F &= \text{362 635,97} \\ i &= \frac{\text{0,11}}{12} \\ n &= \text{5} \times \text{12} = \text{60} \end{align*}
Substitute the values and calculate $$x$$:
\begin{align*} x &= \dfrac{\text{362 635,97} \times \frac{\text{0,11}}{12}}{\left[(1 + \frac{\text{0,11}}{12})^{60}-1\right]} \\ &= \text{R}\,\text{4 560,42} \end{align*}
Therefore, the company must deposit $$\text{R}\,\text{4 560,42}$$ each month.
Tonya owns Freeman Travel Company and she will need to replace her computer in $$\text{7}$$ years' time. Tonya creates a sinking fund so that she will be able to afford a new computer, which will cost $$\text{R}\,\text{8 450}$$. The sinking fund earns interest at a rate of $$\text{7,67}\%$$ p.a. compounded each quarter.
How much money must Tonya save quarterly so that there will be enough money in the account to buy the new computer?
\begin{align*} F & = \frac{x\left[(1+i)^n - 1\right]}{i} \\ \text{Where: } \quad & \\ F & = \text{8 450} \\ i & = \text{0,0767} \\ n & = \text{7} \end{align*}
Interest is compounded per quarter, therefore $$\; i = \text{0,0767} \rightarrow \frac{\text{0,0767}}{\text{4}}$$ and $$n = \text{7} \rightarrow \text{7} \times \text{4}$$
\begin{align*} \text{8 450} & = \frac{x \left[ \left(1 + \frac{\text{0,0767}}{\text{4}}\right) ^{(\text{7} \times \text{4})} - 1 \right]} {\left(\frac{\text{0,0767}}{\text{4}} \right)} \\ \therefore x & = \frac{\left( \text{8 450} \times \frac{\text{0,0767}}{\text{4}} \right)}{ \left[ \left(1 + \frac{\text{0,0767}}{\text{4}}\right) ^{(\text{7} \times \text{4})} - 1 \right]} \\ &= \text{230,80273} \ldots \end{align*}
Tonya must deposit $$\text{R}\,\text{230,80}$$ into the sinking fund quarterly.
How much interest (to the nearest rand) does the bank pay into the account by the end of the $$\text{7}$$ year period?
Total savings:
$\text{R}\,\text{230,80} \times \text{4} \times \text{7} = \text{R}\,\text{6 462,40}$
Interest earned:
$\quad \text{R}\,\text{8 450} - \text{R}\,\text{6 462,40} = \text{R}\,\text{1 987,60}$
To the nearest rand, the bank paid $$\text{R}\,\text{1 988}$$ into the account. |
## Time, Distance and Work
Theory
The questions in this section can vary from being very easy to surprisingly difficult. This is a conceptual section (especially questions involving clocks) and some of the questions can consume a lot of time. While solving, write down the equations as far as possible to avoid mistakes. The few extra seconds can help you avoid silly mistakes. Also, check if the units of distance, speed and time match up. So if you see yourself adding a unit of distance like m to a unit of speed m/s, you would realize you have missed a term. Choose to apply the concept of relative speed wherever possible as it can greatly reduce the complexity of the problem. Like speed and distance, in time and work while working with terms ensure that you convert all terms to consistent units like man-hours.
Theory
Constant Distance:
Let the distances travelled in each part of the journey be $$d_{1}, d_{2}, d_{3}$$ and so on till $$d_{n}$$ and the speeds in each part be $$s_{1}, s_{2}, s_{3}$$ and so on till $$s_{n}$$.
If $$d_{1} = d_{2} = d_{3} =...= d_{n}$$= d, then the average speed is the harmonic mean of the speeds $$s_{1}, s_{2}, s_{3}$$ and so on till $$s_{n}$$.
Constant Time:
Let the distances travelled in each part of the journey be $$d_{1}, d_{2}, d_{3}$$ and so on till $$d_{n}$$ and the time taken for each part be $$t_{1}, t_{2}, t_{3}$$ and so on till $$t_{n}$$.
If $$t_{1} = t_{2} = t_{3} =...= t_{n}$$= t, then the average speed is the arithmetic mean of the speeds $$s_{1}, s_{2}, s_{3}$$ and so on till $$s_{n}$$.
Tip
• In a journey travelled with different speeds, if the distance covered in each stage is constant, the average speed is the harmonic mean of the different speeds.
• In a journey travelled with different speeds, if the time travelled in each stage is constant, the average speed is the arithmetic mean of the different speeds.
Formula
PIPES & CISTERNS:
Inlet Pipe: A pipe which is used to fill the tank is known as Inlet Pipe.
Outlet Pipe: A pipe which can empty the tank is known as an Outlet Pipe.
• If a pipe can fill a tank in 'x' hours then the part filled per hour= 1/x
• If a pipe can empty a tank in 'y' hours, then the part emptied per hour= 1/y
• If pipe A can fill a tank 'x' hours and pipe can empty a tank in 'y' hours, if they are both active at the same time, then
The part filled per hour =$$\frac{1}{x}-\frac{1}{y}$$(if y>x)
The part emptied per hour =$$\frac{1}{y}-\frac{1}{x}$$(if x>y)
Formula
CLOCKS:
• In a well functioning clock, both hands meet after every 720 / 11 Mins.
• It is because the relative speed of the minute hand with respect to the hour hand = 11/2 degrees per minute.
Formula
BOATS & STREAMS
• If the speed of water is 'W' and speed of a boat in still water is 'B'
1. Speed of the boat downstream is B+W
2. Speed of the boat upstream is B-W
The direction along the stream is called downstream. And, the direction against the stream is called upstream.
• If the speed of the boat downstream is x km/hr and the speed of the boat upstream is y km/hr, then
Speed of boat in still water= $$\frac{x+y}{2}$$km/hr
Rate of stream= $$\frac{x-y}{2}$$km/hr
• While converting the speed in m/s to km/hr, multiply it by 3.6(18/5).
1m/s = 3.6 km/h
• While converting km/hr into m/sec, we multiply by 5/18
Formula
Distance = Speed$$\times$$Time
Speed = $$\frac{Distance}{Time}$$
Time = $$\frac{Distance}{Speed}$$
While covering the Speed in m/s to km/hr. multiply it by 3.6. It is because 1m/s = 3.6 km/hr
If the ratio of the speeds of A and B is a : b, then
• The ratio of the times taken to cover the same distance is 1/a : 1/b or b : a.
• The ratio of distance travelled in equal time intervals is a : b
Average speed= $$\frac{Total Distance Travelled}{Total Time Taken}$$
If a part of a journey is travelled at speed $$S_{1}$$ km/hr in $$T_{1}$$ hours and the remaining part at speed $$S_{2}$$ km/hr in $$T_{2}$$ hours then.
Total distance travelled= $$S_{1}T_{1}$$+$$S_{2}T_{2}$$ km
Average speed=$$\frac{S_{1}T_{1}+S_{2}T_{2}}{T_{1}+T_{2}}$$ km/hr
If $$D_{1}$$ km is travelled at speed of $$S_{1}$$ km/hr, and $$D_{2}$$ km is travelled at speed of $$S_{2}$$ km/hr then
Average Speed= $$\frac{D_{1}+D_{2}}{\frac{D_{1}}{S_{1}}+\frac{D_{2}}{S_{2}}}$$ km/hr
• In a journey travelled at different speeds, if the distance covered in each stage is constant, the average speed is the harmonic mean of the different speeds.
• Suppose a man covers a certain distance st x km/hr and an equal distance at y km/hr
Then the average speed during the whole journey is $$\frac{2xy}{x+y}$$ km/hr
• In a journey travelled with different speeds, if the time travelled in each stage is constant, the average speed is the harmonic mean of the different speeds.
• If a man travelled for a certain time at the speed of x km/hr and travelled for an equal amount of time at the speed of y km/hr then
Then the average speed during the whole journey is $$\frac{x+y}{2}$$ km/hr
Formula
Circular Tracks
If two people are running on a circular track with speeds in the ratio a:b where a and b are co-prime, then
• They will meet at a+b distinct points if they are running in the opposite directions.
• They will meet at |a-b| distinct points if they are running in the same direction.
If two people are running on a circular track having perimeter I, with speeds m and n,
• The time for their first meeting = $$\frac{I}{(m+n)}$$
(when they are running in opposite directions)
• The time for their first meeting = $$\frac{I}{(|m-n|)}$$
(when they are running in the same direction)
If a person P starts from A and heads towards B and another person Q starts from B and heads towards A and they meet after a time 't' then,
t = $$\sqrt{x*y}$$
where x = time taken (after the meeting) by P to reach B and y = time taken (after the meeting) by Q to reach A.
A and B started st a time towards each other. After crossing each other, they took $$T_{1}$$ hrs, $$T_{2}$$ hrs respectively to reach their destinations. If they travel at constant speeds $$S_{1}$$ and $$S_{2}$$ respectively all over the journey, Then
$$\frac{S_{1}}{S_{2}}$$=$$\sqrt{\frac{T_{2}}{T_{1}}}$$
Formula
TRAINS:
Two trains of length $$L_1$$ and $$L_2$$ travelling at speeds of $$S_1$$ and $$S_2$$ cross each other in
• $$\frac{L_1+L_2}{S_1+S_2}$$ if they are going in opposite directions
• $$\frac{L_1+L_2}{S_1-S_2}$$ if they are going in the same direction
Formula
• If X can do a work in 'n' days, the fraction of work X does in a day is $$\frac{1}{n}$$
• If X can do a work in 'x' days, and Y can do a work in 'y' days, the number of days taken by both of them together is $$\frac{x*y}{x+y}$$
• If $$A_1$$ men can do $$B_1$$ work in $$C_1$$ days and $$A_2$$ men can do $$B_2$$ work in $$C_2$$ days, then $$\frac{A_1 C_1}{B_1}$$ =$$\frac{A_2 C_2}{B_2}$$
Formula
• While converting the speed in m/s to km/hr, multiply it by 3.6. It is because 1 m/s = 3.6 km/hr
Tip
• In a well functioning clock, both the hands meet after every $$\frac{720}{11}$$ mins. It is because relative speed of minute hand with respect to hour hand = $$\frac{11}{2}$$ degrees per minute.
Formula
Work:
• If X can do a work in 'n' days, the fraction of work X does in a day us 1/n
• If X can do a work in 'x' days, and Y can do a work in 'Y' days,
The number of days taken by both of them together is $$\frac{x*y}{x+y}$$
• If $$M_{1}$$ men work for $$H_{1}$$ hours per day and worked for $$D_{1}$$ days and completed $$W_{1}$$ work, and if $$M_{2}$$ men work for $$H_{2}$$ hours per day and worked for $$D_{2}$$ days and completed $$W_{1}$$ work, then
$$\frac{M_{1}H_{1}D_{1}}{W_{1}}$$=$$\frac{M_{2}H_{2}D_{2}}{W_{2}}$$ |
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