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## College Algebra (6th Edition)
$8(8)=64$
See p.340: Strategy for Solving Problems lnvolving Maximizing or Minimizing Quadratic Functions 1. $\quad$Read the problem carefully and decide which quantity is to be maximized or minimized. 2. $\quad$Use the conditions of the problem to express the quantity as a function in one variable. 3. $\quad$Rewrite the function in the form $f(x)=ax^{2}+bx+c$. 4. $\quad$Calculate $-\displaystyle \frac{b}{2a}$. If $a>0, f$ has a minimum at $x=-\displaystyle \frac{b}{2a}$. This minimum value is $f$($-\displaystyle \frac{b}{2a}$). If $a<0, f$ has a maximum at $x=-2ab$. This maximum value is $f$($-\displaystyle \frac{b}{2a}$).. 5. $\quad$Answer the question posed in the problem. ---------------------------- 1. Product is to be maximized 2. Let x be one of the numbers. Then, the other number is 16-x Their product, $f(x)=x(16-x)$ is a quadratic function. 3.$\quad f(x)=16x-x^{2}$ $f(x)=-x^{2}+16x$ $4.\qquad a=-1, b=16, c=0$ $-\displaystyle \frac{b}{2a}=-\frac{16}{2(-1)}=8$ 5. The product is maximum when $x=8.$ The other number is $16-8=8.$ The maximum product is f(8)$=8(8)=64$
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Radical equations are equations that have variables stunk inside a radical. We will show you how to solve this type of equations in this lesson.
#### Lessons
3-step approach for solving radical equations:
1) Isolate the “square root” term.
2) i. Square both sides of the equation to get rid of the “square root”.
ii. (optional) If the resulting equation still has a “square root” term, repeat steps 1 and 2.
3) Solve the equation, and REMEMBER to substitute answers back into the original equation to identify any EXTRANEOUS ROOTS,
• 1.
Solving Radical Equations Involving One Square Root Term
Solve:
a)
$3\sqrt {x - 2} + x = 0$
(hint: no solutions)
b)
$x - 2 = 2\sqrt {x + 1}$
(hint: one solutions)
c)
$x + \sqrt {2{x^2} - 7} = 3$
(hint: two solutions)
d)
$\sqrt {x + 9} - 10 = - 8$
(hint: one solutions)
• 2.
Solving Radical Equations Involving More than One Square Root Term
Solve:
a)
$\sqrt{3x+4}-\sqrt{x+1}=3$
(hint: one solutions)
b)
$\sqrt{x-2}+3=\sqrt{2x+5}$
(hint: two solutions)
c)
$\sqrt{x-4}+\sqrt{x}=4$
(hint: one solutions)
d)
$\sqrt{5x+20}=\sqrt{x+8}+\sqrt{x+3}$
(hint: two solutions)
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chapter 6 solution
# chapter 6 solution - Basic 1 To solve this problem we must...
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Basic 1. To solve this problem, we must find the PV of each cash flow and add them. To find the PV of a lump sum, we use: PV = FV / (1 + r) t [email protected]% = \$950 / 1.10 + \$1,040 / 1.10 2 + \$1,130 / 1.10 3 + \$1,075 / 1.10 4 = \$3,306.37 [email protected]% = \$950 / 1.18 + \$1,040 / 1.18 2 + \$1,130 / 1.18 3 + \$1,075 / 1.18 4 = \$2,794.22 [email protected]% = \$950 / 1.24 + \$1,040 / 1.24 2 + \$1,130 / 1.24 3 + \$1,075 / 1.24 4 = \$2,489.88 2. To find the PVA, we use the equation: PVA = C ({1 – [1/(1 + r) ] t } / r ) At a 5 percent interest rate: PVA = \$6,000{[1 – (1/1.05) 9 ] / .05 } = \$42,646.93 PVA = \$8,000{[1 – (1/1.05) 6 ] / .05 } = \$40,605.54
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And at a 15 percent interest rate: [email protected]%: PVA = \$6,000{[1 – (1/1.15) 9 ] / .15 } = \$28,629.50 [email protected]%: PVA = \$8,000{[1 – (1/1.15) 6 ] / .15 } = \$30,275.86 Notice that the PV of cash flow X has a greater PV at a 5 percent interest rate, but a lower PV at a 15 percent interest rate. The reason is that X has greater total cash flows. At a lower interest rate, the total cash flow is more important since the cost of waiting (the interest rate) is not as great. At a higher interest rate, Y is more valuable since it has larger cash flows. At the higher interest rate, these bigger cash flows early are more important since the cost of waiting (the interest rate) is so much greater. 3. To solve this problem, we must find the FV of each cash flow and add them. To find the FV of a lump sum, we use: FV = PV(1 + r) t = \$940(1.08) 3 + \$1,090(1.08) 2 + \$1,340(1.08) + \$1,405 = \$5,307.71 [email protected]% = \$940(1.11) 3 + \$1,090(1.11) 2 + \$1,340(1.11) + \$1,405 = \$5,520.96 [email protected]% = \$940(1.24) 3 + \$1,090(1.24) 2 + \$1,340(1.24) + \$1,405 = \$6,534.81 Notice we are finding the value at Year 4, the cash flow at Year 4 is simply added to the FV of the other cash flows. In other words, we do not need to compound this cash flow. 4. To find the PVA, we use the equation: PVA = C ({1 – [1/(1 + r) ] t } / r ) PVA = \$5,300{[1 – (1/1.07) 15 ] / .07} = \$48,271.94 PVA = \$5,300{[1 – (1/1.07) 40 ] / .07} = \$70,658.06 PVA = \$5,300{[1 – (1/1.07) 75 ] / .07} = \$75,240.70 To find the PV of a perpetuity, we use the equation: PV = C / r PV = \$5,300 / .07 = \$75,714.29 Notice that as the length of the annuity payments increases, the present value of the annuity approaches the present value of the perpetuity. The present value of the 75 year annuity and the present value of the perpetuity imply that the value today of all perpetuity payments beyond 75 years is only \$473.59.
5. Here we have the PVA, the length of the annuity, and the interest rate. We want to calculate the annuity payment. Using the PVA equation: PVA = C ({1 – [1/(1 + r) ] t } / r ) PVA = \$34,000 = \$ C {[1 – (1/1.0765) 15 ] / .0765} We can now solve this equation for the annuity payment. Doing so, we get: C = \$34,000 / 8.74548 = \$3,887.72 6. To find the PVA, we use the equation: PVA = C ({1 – [1/(1 + r) ] t } / r ) PVA = \$73,000{[1 – (1/1.085) 8 ] / .085} = \$411,660.36 7. Here we need to find the FVA. The equation to find the FVA is: FVA = C {[(1 + r) t – 1] / r } FVA for 20 years = \$4,000[(1.112 20 – 1) / .112] = \$262,781.16 FVA for 40 years = \$4,000[(1.112 40 – 1) / .112] = \$2,459,072.63 Notice that because of exponential growth, doubling the number of periods does not merely double the FVA. 8. Here we have the FVA, the length of the annuity, and the interest rate. We want to calculate the annuity payment. Using the FVA equation: FVA = C {[(1 + r ) t – 1] / r } \$90,000 = \$ C [(1.068 10 – 1) / .068] We can now solve this equation for the annuity payment. Doing so, we get: C = \$90,000 / 13.68662 = \$6,575.77 9. Here we have the PVA, the length of the annuity, and the interest rate. We want to calculate the annuity payment. Using the PVA equation: PVA = C ({1 – [1/(1 + r) ] t } / r ) \$50,000 = C {[1 – (1/1.075) 7 ] / .075} We can now solve this equation for the annuity payment. Doing so, we get: C = \$50,000 / 5.29660 = \$9,440.02 10. This cash flow is a perpetuity. To find the PV of a perpetuity, we use the equation:
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# Complement Probability|Definition & Meaning
## Definition
The Probability of taking the place of one event if and only if the other event does not occur, then we say that the two occurrences are complementary. Furthermore, the outcomes in which the event in question does not occur might be considered its complement. Take A to be happening. For the opposite of A, write A’ or Ac. As a pair, A and A’ are synchronous occurrences.
## What Are Complementary Events?
Two events are complementary if and only if their occurrence is dependent on the absence of the other. When two occurrences are mutually exclusive and exhaustive, we say they are complementary.
Figure 1: Mutually Exclusive Events
In the diagram above, the event A and the event B are mutually exclusive, meaning that their elements do not match. There is no intersection between both events.
Probabilities of mutually exclusive occurrences must add up to 1. Events can only occur in a complementary manner if there are precisely two possible outcomes. This article will provide some background information about complimentary occurrences and some illustrations.
In the Probability field, there are two probable outcomes for an event to be complementary. Take, for instance, the difference between passing and failing a test. One way to think of an event is the collection of results that an experiment produces. As a result, occurrences will consistently constitute a portion of the sample space.
### Properties of Complementary Events
For two occurrences to be considered complimentary to one another, they need to exhibit specific characteristics. These are presented in the following order:
Complementary events are exhaustive and mutually exclusive, which implies events that are complementary to one another cannot coincide. This indicates that two occurrences that are complementary cannot occur at the exact moment. In other words, complementary events do not co-occur.
Events that complement one another are thorough. This suggests that an event and its complement must fill up the sample area for it to be considered valid. Thus, S = A ∪ A’
Figure 2: Representation of S = A ∪ A’
The diagram shown above shows two sets of probability that represents the complement probability concept. The shaded area shows the entire sample space.
### Rules of Complementary Events
According to the concept of complementary events, the total of the probabilities that an event will occur and the probabilities that the event’s complement will also occur will always equal 1. Let’s say there’s an event denoted by the letter A, and let’s call the likelihood of its happening P(A). Therefore, the chance that A will not occur is denoted by the P(A) symbol. Following is a mathematical expression that can be used to describe this rule.
1. P (A) + P (A’) = 1
2. P (A) = 1 – P (A’)
3. P (A’) = 1 – P (A) (A)
These three mathematical assertions are all identical to one another.
### Examples of Complementary Events
Let us consider that we roll two dice. Let’s call this occurrence B when you obtain two different digits. Find P (B).
Because it is a more straightforward method, we will apply the rule of complimentary events to solve this example.
Let us call the occurrence described by B’ the one in which the digits on both dice are the same or do not differ. The formula for determining the sample space for B’ is as follows: S’ = [(1, 1), [(2, 2), [(3, 3), [(4, 4), [(5, 5), and [6, 6]].
When two dice are rolled, there are a total of 36 possible results. The number of positive outcomes equals six.
P(B’) = 6 / 36 = 1 / 6
According to the rule of events that occur in complementary ways,
P(A) = 1 – P(A’).
P(B) = 1 – (1 / 6) = 5 / 6
Figure 3: Representation of P(A) = 1 – P(A’)
Therefore, rolling two dice results in a five out of six Probability of getting two different numbers.
### Odds in a Complementary Event
The odds don’t need to be even. The sum of the two probabilities, namely the probability that an event occurs plus the probability that it does not, must equal 1. This is the most crucial thing to keep in mind. The complementary activity of rolling a die to check if you receive six results in only two possible outcomes: either receiving a six (with a chance of one in six) or not getting a six (with a chance of five in six). If there is a one-in-a-million chance of you winning the lottery, then there is a 999,999 to one chance that you will not win. The two Probability are equivalent to one another:
## Things To Remember
1. Complementary occurrences are those in which the occurrence of one event is contingent upon and is only possible in the absence of another event.
2. Events that are complementary to one another cannot co-occur and exhaust one another.
3. The sample space comprises an event and its associated complement when taken together.
4. The rule that governs complementary events states that P(A) + P(A’) = 1.
## Some Examples of Complement Probability Problems
### Example 1
A jar contains red balls and blue balls. If the probability of picking a red ball is 2/5, find the probability of blue ball being picked up?
### Solution
P (B) = P (not picking a red ball)
= 1 – P (R)
= 1 – (2/5)
= 3/5
The probability of picking a red or blue ball is a complementary event. It is defined as below:
P (B) = P (R’) The compliment of R’ is the set of R
P (R’) = P (B)
Thus, the following can be brought up:
P (A) + P (A’) = 1
### Example 2
There are 8 flowers in a pot of which 2 are red, 3 are black, 1 is blue, 2 are purple, and 2 are pink. The primary color event selection of flowers is X. Find P (X’).
### Solution
X = {2 red, 1 blue}
Total number of flowers: 8
The favorable outcomes are three.
P (X) = 3/10
The rule of complementary events says: P(A’) = 1 – P(A)
P(X’) = 1 – (3 / 10) = 7 / 10
All mathematical drawings and images were created with GeoGebra.
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# Permutation and Combination - Important Concepts and Notes
By Abhinav Gupta|Updated : August 21st, 2021
Permutation and combination are the ways to represent a group of objects by selecting them in a set and forming subsets. It defines the various ways to arrange a certain group of data. When we select the data or objects from a certain group, it is said to be permutations, whereas the order in which they are represented is called combination. Both concepts are very important in Mathematics.
Permutation and combination are the ways to represent a group of objects by selecting them in a set and forming subsets. It defines the various ways to arrange a certain group of data. When we select the data or objects from a certain group, it is said to be permutations, whereas the order in which they are represented is called combination. Both concepts are very important in Mathematics.
## Permutation and Combination: Notes & Important Questions
Permutation and combination
Permutations and combinations are the basic ways of counting from a given set, generally without replacement, to form subsets.
Permutation
Permutation of an object means the arrangement of the object in some sequence or order.
Theorem 1
The total number of permutations of a set of n objects taken r at a time is given by
P( n,r) = n (n- 1)(n- 2)….….(n- r+1) (n ≥ r)
Proof:
The number of permutations of a set of `n’objects taken r at a time is equivalent to the number of ways in which r positions can be filled up by those n objects . When first object is filled then we have (n-1) choices to fill up the second position. Similarly, there are (n- 2) choices fill up the third position and so on.
Therefore, P(n,r) = n (n- 1) (n- 2) …….. (n – r +1)
=n(n−1)(n−2)…….(n−r+1)(n−r)…3.2.1(n−r)……3.2.1n(n−1)(n−2)…….(n−r+1)(n−r)…3.2.1(n−r)……3.2.1
= n!(n−r)!n!(n−r)!
Permutation of objects not all different
The permutation of objects taken all at a time when P of the objects are of the first kind, q of them are of the second kind, of them are of the third kind and the rest all are different.
The total number of permutations =n!p!q!r!n!p!q!r!
Circular permutation
Circular Permutation: The number of ways to arrange distinct objects along a fixed line
The total number of permutation of a set of n objects arranged in a circle is P = (n -1)!
Permutation of repeated things
The permutation of the n objects taken r at a time when each occurs a number of times and it is given by P = nr
Example 1
How many numbers of three digits can be formed from the integers 2,3, 4,5,6? How many of them will be divisible by 5?
Soln:
For the three digits numbers, there are 5 ways to fill in the 1st place, there are 4 ways to fill in the 2nd place and there are 3 ways to fill in the 3rd place. By the basic principle of counting, number of three digits numbers = 5 * 4 * 3 = 60.
Again, for three-digit numbers which are divisible by 5, the number in the unit place must be 5. So, the unit place can be filled up in 1 way. After filling up the unit place 4 numbers are left. Ten’s place can be filled up in 4 ways and hundredths place can be filled up in 3 ways. Then by the basic principle of counting, no.of 3 digits numbers which are divisible by 5 = 1 * 4 * 3 = 12.
Example 2
How many numbers of at least three different digits can be formed the integers 1, 2, 34, 5, 6,?
Soln
Numbers formed should be of at least 3 digits means they may be of 3 digits, 4 digits, 5 digits or 6 digits.
There are 6 choices for the digit in the units place. There are 5 and 4 choices for digits in ten and hundred’s place respectively.
So, the total number of ways by which 3 digits numbers can be formed = 6.5.4 = 120
Similarly, the total no.of ways by which 4 digits numbers can be formed = 6.5.4.3 = 360.
the total no. of ways by which 5 digits numbers can be formed = 6.5.4.3.2 = 720.
The total no.of ways by which 4 digits numbers can be formed = 6.5.4.3.2.1 = 720.
So, total no.of ways by which the numbers of at least 3 digits can be formed = 120 + 360 + 720 + 720 = 1920
Example 3
In how many ways can four boys and three girls be seated in a row containing seven seats
1. if they may sit anywhere
2. if the boys and girls must alternate
3. if all three girls are together?
a.
Soln:
If the boys and girls may sit anywhere, then there are 7 persons and 7 seats. 7 persons in 7 seats can be arranged in P(7,7) ways.
= 7!(7−7)!7!(7−7)! = 7!0!7!0! = 7∗6∗5∗4∗3∗2∗117∗6∗5∗4∗3∗2∗11 = 5,040 ways.
b.
Soln:
If the boys and girls must sit alternately, there are 4 seats for boys and 3 for girls.
Here, for boys n = 4, r = 4
4 boys in 4 seats can be arranged in P(4,4) ways
= 4!(4−4)!4!(4−4)! = 4∗3∗2∗114∗3∗2∗11 = 24 ways.
Again. For girls n = 3, r = 3
3 girls in 3 seats can be arranged in P(n,r) i.e. P(3,3) ways.
= 3!0!3!0! = 3∗2∗113∗2∗11 = 6ways.
So, total no.of arrangement = 24 * 6 = 144 ways.
c.
Suppose 3 girls = 1 object, then total number of student (n) = 4 + 1= 5.
Then the permutation of 5 objects taken 5 at a time.
= P(5,5) = 5!(5−5)!5!(5−5)! = 5∗4∗3∗2∗115∗4∗3∗2∗11 = 120.
We know, 3 girls can be arranged themselves in P(3,3) different ways,
i.e. P(3,3) = 3!(3−3)!3!(3−3)! = 3 * 2 * 1 = 6 different ways.
Therefore, required of arrangements = 120 * 6 = 720.
Examples 4
In how many ways can eight people be seated in a round table if two people insisting sitting next to each other?
a.
Total number of objects = 4 + 4 = 8.
If they may sit anywhere, then it is the circular arrangements of 8 objects taken 8 at a time. So, the total number of permutation.
= (8 – 1)! = 7! = 7 * 6 * 5 * 4 * 3 * 2 * 1 = 5040.
b.
Art and Science students have to sit alternately in a round table. So, there are 4 seats for Art students and 4 for Science students.
4 art students at a round table can be arranged in 4 – 1! ways.
= 3! = 3 * 2 * 1 = 6 ways.
Again. 4 science students can be arranged in P(4,4) ways, i.e. 4! Ways = 4 * 3 * 2 * 1 = 24 ways.
So, total no.of arrangement = 6 * 24 = 144 ways.
Combinations
The combination means a collection of an object without regarding the order of arrangement. The total number of combinations of n objects taken r at a time C(n,r ) is given by C(n,r) = n!(n−r)!r!n!(n−r)!r!
Examples 5
From 4 mathematician, 6 statisticians and 5 economists, how many committees consisting of 3 men and 2 women are possible?
Soln:
2 members can be selected from 4 mathematicians in C(4,2) in different ways.
2 members can be selected from 6 statisticians in C(6,2) in different ways.
2 members can be selected from 5 economics in C(5,2) in different ways.
Therefore, total number of committees = C(4,2) * C(6,2) * C(5,2) = 6 * 15 * 10 = 900.
Examples 6
1. If C(20, r+ 5) = C (20, 2r -7) find C( 15,r)
2. if C(n, 10) + C(n,9) = C( 20,10) find n and C(n, 17)
3. solve for n the equation C(n+ 2,4) = 6 C(n, 2)
a.
Given C(20,r + 5) = C(20,2r – 7)
Then r + 5 = 2r – 7 [if C(n,r) = C(n,r’) then r = r’]
Or, r = 12.
So, C(15,r) = C(15,12) = 15!(15−12)!.12!15!(15−12)!.12! = 455.
b.
Given. C(n,10) + C(n,9) = C(20,10)
Or, C(n + 1,10) = C(20,10) [if C(n,r)+ C(n,r – 1) = C(n + 1, r)]
So, n + 1 = 20
So, n = 19.
Again, C(n,17) = C(19,17) = 19!(19−17)!.17!19!(19−17)!.17! = 171.
c.
C(n + 2,4) = 6 C(n,2)
Or, (n+2)!(n−2)!.4!(n+2)!(n−2)!.4! = 6. n!(n−2)!.2!n!(n−2)!.2!à(n+2).(n+1).n!(n−2)!.4!(n+2).(n+1).n!(n−2)!.4! = 6n!(n−2)!.26n!(n−2)!.2.
Or, (n+2)(n+1)4∗3∗2∗1(n+2)(n+1)4∗3∗2∗1 = 3
Or, n2 + 3n + 2 = 72
Or, n2 + 3n – 70 = 0
Or, (n + 10)(n – 7) = 0
Either, n = - 10 or, n = 7.
n = - 10 is not possible.
So, n = 7.
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# Prove that the result of the ratio does not depend on x. A = (log_5_x^2 + log_5_x^3)/(log_4_x^2 + log_4_x^3)
sciencesolve | Certified Educator
Using the logarithmic identities, you may convert the base of logarithm into a new base, such that:
`log_a b = (ln b)/(ln a)`
Reasoning by analogy yields:
`log_5 x^2 = (ln x^2)/(ln 5)`
`log_5 x^3 = (ln x^3)/(ln 5)`
`log_4 x^2 = (ln x^2)/(ln 4)`
`log_4 x^3 = (ln x^3)/(ln 4)`
Replacing the converted logarithms in fraction yields:
`A = ((ln x^2)/(ln 5) + (ln x^3)/(ln 5))/((ln x^2)/(ln 4) + (ln x^3)/(ln 4))`
`A = ((ln x^2 + ln x^3)/(ln 5))/((ln x^2 + ln x^3)/(ln 4))`
Reducing duplicate factors yields:
`A = (1/(ln 5))/(1/(ln 4)) => A = ln 4/ln 5 => A = log_5 4`
Hence, evaluating the given expression yields `A = log_5 4` and it does not contain x.
giorgiana1976 | Student
We notice that the numerator is a sum of logarithms that have matching bases.
We'll use the rule of product:
log a + log b = log (a*b)
log_5_x^2 + log_5_x^3 = log_5_(x^2*x^3)
log_5_(x^2*x^3) = log_5_x^(2+3)
log_5_x^2 + log_5_x^3 = log_5_x^5
We'll use the power rule of logarithms:
log_5_x^5 = 5*log_5_x (1)
We also notice that the denominator is a sum of logarithms that have matching bases.
log_4_x^2 + log_4_x^3 = log_4_x^5
log_4_x^2 + log_4_x^3 = 5*log_4_x (2)
We'll substitute both numerator and denominator by (1) and (2):
A = 5*log_5_x/5*log_4_x
We'll simplify:
A = log_5_x/log_4_x
We'll transform the base of the numerator, namely 5, into the base 4.
log_4_x = (log_5_x)*(log_4_5)
We'll re-write A:
A = log_5_x/(log_5_x)*(log_4_5)
We'll simplify:
A = 1/log_4_5
A = log_5_4
As we can notice, the result is a constant and it's not depending on the variable x.
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# Circle Graphs / Pie Charts / Pie Graphs
In these lessons, we will learn:
• how to construct circle graphs or pie charts
• how to read or use circle graphs or pie charts
Related Topics:
More Statistics Lessons
Constructing Circle Graphs or Pie Charts
A pie chart (also called a Pie Graph or Circle Graph) makes use of sectors in a circle. The angle of a sector is proportional to the frequency of the data.
The formula to determine the angle of a sector in a circle graph is:
Study the following steps of constructing a circle graph or pie chart:
Step 1: Calculate the angle of each sector, using the formula
Step 2: Draw a circle using a pair of compasses
Step 3: Use a protractor to draw the angle for each sector.
Step 4: Label the circle graph and all its sectors.
Example:
In a school, there are 750 students in Year1, 420 students in Year 2 and 630 students in Year 3. Draw a circle graph to represent the numbers of students in these groups.
Solution:
Total number of students = 750 + 420 + 630 = 1,800.
Draw the circle, measure in each sector. Label each sector and the pie chart.
How to construct a circle graph or pie chart from a table of percentages or fractions?
Steps to making a circle graph
1. Determine what percent of the data is from each category.
2. Determine what percent of 360 degrees represents each category.
3. Divide up the circle. How to make a Pie Graph using a protractor and compass?
This video shows how to construct a circle graph from the given data.
How to construct a circle graph given the percentage of each sector?
How to Draw a Pie Chart?
This video shows how to draw a pie chart by working out angles from a table.
Example: The table shows 18 people's favorite color. Display the information as a pie chart.
### Reading or Using Circle Graphs
We could also use a given circle graph to answer some questions about the data.
Example:
The following pie chart shows a survey of the numbers of cars, buses and motorcycles that passes a particular junction. There were 150 buses in the survey.
a) What fraction of the vehicles were motorcycles?
b) What percentage of vehicles passing by the junction were cars?
c) Calculate the total number of vehicles in the survey.
d) How many cars were in the survey?
Solution:
a) Fraction of motorcycles
b) To convert the angle of a sector into a percentage, we use the formula:
Percentage
Percentage of cars
c) Let x be the total number of vehicles
The total number of vehicles is 1,800
d) Number of cars
How to read Pie Graphs (Circle Graphs)?
Example: For the past year, a travel agency has collected data about the number of individual tickets that it sells for its signature product: a Mediterranean Cruise. The monthly data on ticket sales is shown below. What are the best and worst months for cruise sales? How to answer questions based on given circle graphs?
Use the circle graph below to determine what percent of people prefer either soccer or football.
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# 2. Linear Systems¶
There’s definitely, definitely, definitely no logic
To human behaviour
—“Human Behaviour” by Björk
I’m as excited as you are about making a meteor fly, so let’s take the complexity of our systems up a notch. We’ll set multiple objects in motion in this section and see what we can learn about their behavior.
# 2.1 Systems of Equations¶
## Testing Solutions¶
In the last chapter, you applied linear equations as a mathematical model for the path of a meteor. As the meteor’s $$x$$-coordinate increased, its $$y$$-coordinate increased, and we related the two using an equation like the one below.
$y = 0.5x + 50$
If you substituted $$10$$ for $$x$$, you could evaluate the right-hand side to find, or solve for, $$y$$.
$y = 0.5 \cdot 10 + 50$
$y = 55$
You could use the same approach to find every coordinate pair along the meteor’s path. Substituting the coordinate pair $$(10,55)$$ into the equation $$y=0.5x+50$$ makes both sides equal, so we call it a solution to the equation.
How about the coordinate pair $$(10,60)$$?
$60=0.5 \cdot 10 + 50$
$60 \ne 55$
No luck. The symbol $$\ne$$ means “does not equal.” $$(10,60)$$ is not a solution to this equation.
OK, so we can map out all the points along an object’s path. But what if there were multiple objects moving along multiple paths?
## Solving Equations¶
In the film Gravity, a Space Shuttle is destroyed by a barrage of space debris, forcing the survivors to find another way home to Earth before the debris completes another orbit and threatens them again. Let’s try to help out the astronauts by constructing a simulation of the Shuttle-debris system.
A Space Shuttle and debris in Earth’s orbit.
$$\mapsto$$ Outer space is an empty, infinite, two-dimensional surface. The Space Shuttle is a point moving along the straight path described by the equation $$y=0.5x+50$$. The debris is also a point moving along a straight path, this one described by the equation $$y= -0.5x+150$$.
The coordinates of both objects are ordered pairs of floating point numbers. The Space Shuttle’s initial position is $$(0,150)$$. The debris’ initial position is $$(0,50)$$.
First, paint the sky. Next, update the position of each object. Then, select a pencil color & weight. Finally, draw a point at each object’s position.
Example 2.1 A collision
At the beginning of the film, Mission Control warns the astronauts that danger is coming. In real life, the United States Space Surveillance Network tracks objects in Earth’s orbit using a variety of sensing equipment. The team records the types of objects, when they were launched, their orbits, and their sizes. NASA uses this data to construct models that predict collisions like the one in Gravity.
While we don’t have a giant network of sensors and computers, we do have all the tools we need to figure out where the paths of our simulated Shuttle and debris cross. Let’s combine the equations describing each object’s path into a system of equations.
$$y = 0.5x + 50$$
$$y = -0.5x + 150$$
A system of equations is a set of related equations. There are infinitely many possible $$(x,y)$$ coordinate pairs, but only one solves both of these equations. For example, the point $$(50,75)$$ solves the equation for the Shuttle’s path.
$$75 = 0.5 \cdot 50 + 50$$
$$75 = 25 + 50$$
$$75 = 75$$
But does it also solve the equation for the debris’ path?
$$75 = -0.5 \cdot 50 + 150$$
$$150 = -25 + 150$$
$$150 \ne 125$$
No luck. We could try guessing and checking other coordinate pairs, but we might be at it for a while. There are many good techniques for solving systems of linear equations and we need a few additional ideas to apply them. Let’s start with a simpler case.
Math that uses letters as placeholders for numbers is known as algebra. Those placeholders, or variables, can show up anywhere we establish a relationship. Take the equation below as an example.
$2 + 3 = 5$
What if I wrote the following equation instead?
$x + 3 = 5$
We’ll use this example as a starting point for finding “unknown” values like $$x$$, also called solving an equation. Given $$x+3 =5$$, I want to come up with an algorithm that solves for $$x$$. Let’s simplify again.
We know the following equation relates $$2$$, $$3$$, and $$5$$.
$2 + 3 = 5$
Figure 2.1 Three hops to the right
You could express the same relationship another way by rearranging the equation a bit.
$2 = 5 - 3$
Figure 2.2 Three hops to the left
At this point you might think, “OK, the numbers $$2$$ and $$5$$ are $$3$$ units apart. But what does that have to do with finding unknowns?” Everything, as it turns out.
We can think of the original equation $$2+3=5$$ as running “forward” from the starting point $$2$$. In this view, the second equation $$2=5-3$$ runs “backward”. Running an operation backward is known as inverting the operation. And something interesting happens when we apply an operation and its inverse together.
$2 + 3 - 3 = 2$
Figure 2.3 Hopping back and forth
We’re right back where we started! Operations and their inverses undo each other; it’s like nothing happened at all. This fact is key to solving equations.
In math, the $$=$$ symbol is an ironclad statement of equality. The left-hand side must equal the right-hand side, now and always. If you make a change to the left, you’d better do the same to the right. Let’s take the original equation one more time and invert the $$+3$$ operation.
$2 + 3 - 3 = 5 - 3$
Or simplified:
$2 = 2$
Now, let’s apply identical reasoning to the equation with the variable $$x$$.
$x + 3 = 5$
$x + 3 - 3 = 5 - 3$
$x = 2$
OK, let’s see that process one more time with a different example.
$x - 2 = 10$
$x - 2 + 2 = 10 + 2$
$x = 12$
Exercise 2.1 Given the equation $$x-5=10$$, solve for $$x$$. Then, describe the algorithm you used to solve for $$x$$ in plain English.
Muḥammad ibn Mūsā al-Khwārizmī was a mathematician and scientist. Muḥammad’s book The Compendious Book on Calculation by Completion and Balancing laid the foundation for modern algebra. He studied a range of problems in economics and natural science.
Operator Precedence
Inverting addition and subtraction seems to work just fine, but what happens when you include other operations? Let’s take the path of our Shuttle.
$y = 0.5x + 50$
How would you determine the Shuttle’s $$x$$-coordinate when its $$y$$-coordinate is $$150$$?
$150 = 0.5x + 50$
It seems like we could invert this equation to solve for $$x$$, but I’m not certain of how to proceed now that multiplication is in the mix. Figuring this out requires a brief interlude to discuss the order, or precedence, of mathematical operations.
If you come across an expression like $$2 \cdot 3 + 4$$, the mathematical community has agreed that you should multiply before adding. You can compute the value produced in this example as follows.
$$2 \cdot 3 + 4$$
$$6 + 4$$
$$10$$
We’ve seen the computation run forward, so let’s go backward. The last operation applied was $$+4$$, so let’s invert that first.
$$10 = 2 \cdot 3 + 4$$
$$10 - 4 = 2 \cdot 3 + 4 - 4$$
$$6 = 2 \cdot 3$$
You can view the multiplication as $$2$$ multiplying $$3$$ or as $$3$$ multiplying $$2$$. I’ll go with the former and undo multiplication by $$2$$.
$$\frac{6}{2} = \frac{2 \cdot 3}{2}$$
$$3 = 3$$
How about we substitute one of the operands for a variable and solve for it?
$$10 = 2x + 4$$
$$10 - 4 = 2x + 4 - 4$$
$$6 = 2x$$
$$\frac{6}{2} = \frac{2x}{2}$$
$$3 = x$$
It’s been quite a journey, but I think we’re ready to plot a course for our Shuttle.
$$150 = 0.5x + 50$$
$$150 - 50 = 0.5x + 50 - 50$$
$$100 = 0.5x$$
$$\frac{100}{0.5} = \frac{0.5x}{0.5}$$
$$200 = x$$
Exercise 2.2 Find the $$x$$-coordinate of the Shuttle when its $$y$$-coordinate is $$180$$. Then, describe the algorithm you used to solve for $$x$$ in plain English.
The order of mathematical operations is bundled up nice and neat in the acronym PEMDAS.
PEMDAS
Parentheses
Exponents
Multiplication
Division
Subtraction
Let’s focus on MDAS for now as they are key to linear relationships. Multiplication and division have the same precedence. Consider the example below.
$3 \cdot 10 \div 5$
You could compute $$3 \cdot 10 = 30$$, then divide $$30 \div 5$$ to produce $$6$$. Or you could start by dividing $$10 \div 5 = 2$$ before multiplying $$3 \cdot 2$$ to produce $$6$$. There are multiple pathways to the correct answer!
There is a similar story for addition and subtraction.
$3 + 10 - 5$
You could compute $$3 + 10 = 13$$, then subtract $$13 - 5$$ to produce $$8$$. Or you could start by subtracting $$10 - 5 = 5$$ and add $$3 + 5 = 8$$. Once again, multiple pathways!
Exercise 2.3 Evaluate the expression $$10 \cdot 5 \div 2 + 5$$.
Exercise 2.4 Evaluate the expression $$10 ÷ 5 - 2 \cdot 5$$.
## Solving Systems¶
We’re on a tight schedule to be of any help to NASA. Let’s figure out the coordinates $$(x,y)$$ where the paths of the Shuttle and debris intersect. Recall the system of equations describing the scenario.
$$y = 0.5x + 50$$
$$y = -0.5x + 150$$
There are infinitely many points along each object’s path, and the variables $$x$$ and $$y$$ are placeholders for all of them. The Shuttle and debris were moving before they collided, and they continued moving afterward. Those variables $$x$$ and $$y$$ that solved one equation at a time must now solve both equations simultaneously.
Substitution
Algebraic techniques help us solve systems of equations because we really do mean that the $$y$$ in the first equation is the very same $$y$$ in the second equation. Consider the following simplified case for a moment.
$$a = 5$$
$$b = 5$$
$$a = b$$
The same transitive property of equality applies to systems of equations.
$$y = 0.5x + 50$$
$$y = -0.5x + 150$$
$$0.5x + 50 = -0.5x + 150$$
Now we have one equation with one unknown. Let’s solve it for $$x$$!
$$0.5x + 50 = -0.5x + 150$$
$$0.5x + 0.5x + 50 = -0.5x + 0.5x + 150$$
$$x + 50 = 150$$
$$x + 50 - 50 = 150 - 50$$
$$x = 100$$
And now that we’ve found $$x$$, we can substitute the value back into one of the original equations to find $$y$$.
$$y = 0.5 \cdot 100 + 50$$
$$y = 50 + 50$$
$$y = 100$$
We also could have gone with the other equation.
$$y = -0.5 \cdot 100 + 150$$
$$y = -50 + 150$$
$$y = 100$$
Example 2.2 Predicting collision
As you may expect, there’s more than one way to solve this problem. Let’s examine a different algorithm that combines equations to find solutions.
Elimination and Back Substitution
First, I’m going to determine the value of $$y$$ by eliminating the variable $$x$$ from an equation. Unlike the substitution algorithm, which rewrote one variable in terms of another, I’ll eliminate $$x$$ by adding the top equation to the bottom equation. As usual, let’s motivate this idea by considering a simpler case. Take the equations below.
$$20 = 4 \cdot 5$$
$$2x = x + x$$
I could add the two left-hand sides to produce $$2x + 20$$. On the right-hand side, I would have $$x + x + 4 \cdot 5$$. But do these combined expressions equal each other? We’ll follow this one step-by-step.
$$2x + 20 = x + x + 4 \cdot 5$$
$$2x + 20 = x + x + 20$$
$$2x + 20 = 2x + 20$$
Those $$=$$ symbols mean that the expressions on the left- and right-hand sides are equal. When we add the same thing to each side of an equation, we maintain equality. You may hear this called the additive property of equality when you’re out at parties. Now, let’s use this property to eliminate the variable $$x$$ and solve for $$y$$.
$$y = 0.5x + 50$$
$$y = -0.5x + 150$$
$$y + y = -0.5x + 0.5x + 150 + 50$$
$$2y = 200$$
$$\frac{2y}{2} = \frac{200}{2}$$
$$y = 100$$
The $$y$$-value we just found resulted from combining information stored in separate equations. This is the $$y$$-value both equations share. But what about $$x$$? Well, we know that $$y = 100$$, so let’s substitute that part of our solution back into the system.
$$100 = 0.5x + 50$$
$$100 - 50 = 0.5x + 50 - 50$$
$$50 = 0.5x$$
$$\frac{50}{0.5} = \frac{0.5x}{0.5}$$
$$100 = x$$
You’ve now seen two of many possible algorithms for solving systems of linear equations. Practice with them a bit before we build up the logical foundations you need to explore systems of linear inequalities.
Exercise 2.5 Solve the system of equations below using Substitution. Then, solve it using Elimination and Back Substitution. Describe how each algorithm works in plain English.
$$y = 0.25x + 100$$
$$y = -0.75x + 300$$
# 2.2 Logic¶
In the novel 1984, the main character is brainwashed into accepting that $$2 + 2 = 5$$ is true. The scene depicts the ultimate flex of power by an oppressive government. Let’s lay some logical foundations to ensure we always maintain our grasp on the truth.
## Boolean Algebra¶
You just tested a few different coordinate pairs $$(x,y)$$ to determine whether or not they solved an equation. The test could only have gone one of two ways: success or failure, yes or no, true or false. There is an entire branch of algebra called Boolean algebra dedicated to studying these two truth values, which we usually write as $$1$$ (true) and $$0$$ (false). There are not infinitely many truth values like there are numbers; there are only $$1$$ and $$0$$.
A truth value can correspond to a situation in the real world. For example, I could claim, “The sun is up”. This claim happens to be false in my neck of the woods as I write this sentence. I can express this idea using the variable $$s$$ to represent sunniness.
$s = 0$
Even though the sun has already set on this particular day, the sky above me is still momentarily deep blue. I can claim “The sky is blue” and express this blueness, $$b$$, matter-of-factly.
$b = 1$
OK, we have variables with assigned values. But what can we actually do with them?
Let’s begin by combining s and b using our first logical operation: conjunction, also known as AND. The expression s∧b means “$$s$$ is true AND $$b$$ is true”. The sky above my front porch is blue, but the sun is not up, so the combined statement is false. We could write this concisely as $$s \land b = 0$$.
There is a special set of diagrams called logic gates that depict the results of applying logical operations. Each logic gate has a distinctive shape. Below is the diagram for the AND logic gate.
Figure 2.4 The AND logic gate
One of the variables $$s$$ and $$b$$ is true, and we can test for such a condition using the disjunction operation, also known as OR. A disjunction is true if at least one of its operands is true. I could claim “The sun is up OR the sky is blue” and that would be true because $$b = 1$$. We could express idea this as $$s \lor b = 1$$.
Like AND, OR also has its own logic gate.
Figure 2.5 The OR logic gate
The following truth table organizes all of the facts we’ve established about the view of the sky from my front porch.
Table 2.1 A truth table’s view of my piece of sky
$$s$$
$$b$$
$$s \land b$$
$$s \lor b$$
$$0$$
$$1$$
$$0$$
$$1$$
The final logical operation we’ll discuss is negation, also known as NOT. The NOT operation simply flips a truth value from $$1$$ to $$0$$ or from $$0$$ to $$1$$. Let’s take the variable $$s$$ and negate it using the NOT operator, $$\lnot$$.
$$s = 0$$
$$\lnot s = 1$$
NOT is a unary operation, meaning we only apply it to a single truth value at a time. AND and OR are both binary operations, meaning we have to provide a pair of operands.
Not to be left out, NOT also has its own logic gate.
Figure 2.6 The NOT logic gate
And that’s all you need to get started with logic! You can compose logical operations just like you do arithmetic operations. For example, let’s figure out how to cross the street safely using logic. I’ll define the variables $$l$$ and $$r$$ to represent vehicle traffic from the left and traffic from the right, respectively.
If you were trying to cross a busy street, you would want to avoid vehicles. In logical terms, you would check to see that both $$l = 0$$ and $$r = 0$$. “No vehicles on the left? No vehicles on the right? OK, let’s go!” This condition is easily expressed by combining operations $$\lnot l \land \lnot r = 1$$.
Exercise 2.6 Complete the following truth table for two boolean variables $$x$$ and $$y$$.
$$x$$
$$y$$
$$x \land y$$
$$x \lor y$$
$$0$$
$$0$$
$$\,$$
$$\,$$
$$0$$
$$1$$
$$\,$$
$$\,$$
$$1$$
$$0$$
$$\,$$
$$\,$$
$$1$$
$$1$$
$$\,$$
$$\,$$
Exercise 2.7 Compute the value of the expression $$\lnot (1 \land 1)$$.
Exercise 2.8 Compute the value of the expression $$(1 \land 0) \lor (1 \lor 0)$$.
Exercise 2.9 Rewrite the following logic circuit as an equivalent logical expression.
Exercise 2.10 Complete the following truth table for two boolean variables $$x$$ and $$y$$. What do you notice?
$$x$$
$$y$$
$$\lnot (x \land y)$$
$$\lnot x \lor \lnot y$$
$$\lnot (x \lor y)$$
$$\lnot x \land \lnot y$$
$$0$$
$$0$$
$$\,$$
$$\,$$
$$\,$$
$$\,$$
$$0$$
$$1$$
$$\,$$
$$\,$$
$$\,$$
$$\,$$
$$1$$
$$0$$
$$\,$$
$$\,$$
$$\,$$
$$\,$$
$$1$$
$$1$$
$$\,$$
$$\,$$
$$\,$$
$$\,$$
Mary Everest Boole
Mary Everest Boole was a mathematician and teacher. Mary wrote multiple books on teaching mathematics and edited her husband George’s book on algebraic logic. She was an innovative teacher who advocated for a playful, cooperative approach to learning mathematics.
## Branching¶
Logic is a big deal for computation, from the way the machines are physically built to the way we program them. Conditional statements let us test conditions and make decisions while a program executes. For example, let’s revisit the collision scene from Gravity.
Example 2.3 Collision revisited
The simulation works fine, but it doesn’t really convey the full drama of the situation. Let’s revise our system a bit to account for the additional debris generated upon collision.
A Space Shuttle and debris in Earth’s orbit.
$$\mapsto$$ Outer space is an empty, infinite, two-dimensional surface. The Space Shuttle is a point moving along the straight path described by the equation $$y = 0.5x + 50$$. The debris is also a point moving along a straight path, this one described by the equation $$y = -0.5x + 150$$. After colliding, each object leaves a trail of smaller debris along its path.
The coordinates of both objects are ordered pairs of floating point numbers. The Space Shuttle’s initial position is $$(0,50)$$. The debris’ initial position is $$(0,150)$$.
First, test for collision and set the alpha value for the sky. Then, paint the sky. Next, update the position of each object. After that, select a pencil color & weight. Finally, draw a point at each object’s position.
Lucky for us, you already solved this system and know that the two objects collide at $$(100,100)$$. As the sketch continues running, the Shuttle and debris continue moving to the right. We can test for this using an if statement.
if (condition) {
// things to do if condition is true
}
If statements present a logical crossroads in a program. The first line of the if statement is called a header. The header begins with if, defines the condition to test in between a pair of parentheses (), then opens a pair of {}. If the condition is true, JavaScript will execute the set of statements between the {}, also known as the if statement’s body.
if (condition) {
thingOne() // this is part of the body
thingTwo() // this too!
}
thingThree() // this is not
In the Gravity example, we’re testing whether or not the value of the variable x is greater than 100. JavaScript has the following relational operators that work as you would expect for numbers.
Table 2.2 JavaScript’s relational operators
$$Math$$
Code
English
$$=$$
===
Equal to
$$\ne$$
!==
Not equal to
$$>$$
>
Greater than
$$<$$
<
Less than
$$\ge$$
>=
Greater than or equal to
$$\le$$
<=
Less than or equal to
Applying a relational operator produces a boolean value. For example, the expression 2 + 2 === 5 produces the boolean value false because, well, math. 2 + 2 === 4, on the other hand, produces the value true. These sorts of logical expressions are called boolean expressions.
if (x > 100) {
// things to do if x > 100
}
Example 2.4 A more impactful collision
Exercise 2.11 Duplicate your original collision sketch and add a few effects. How should the visual appearance of the Shuttle and debris change after impact?
You can also test multiple conditions together. Thinking back to the meteor sketch, how about we make the meteor glow red as it passes over the middle half of the canvas? In other words, when $$x \ge 50$$ AND $$x \le 150$$.
if (x >= 50 && x <= 150) {
stroke('tomato')
}
&& is one of JavaScript’s boolean operators along with || and !. These JavaScript operators function identically to the logical operators you just studied.
Recall our earlier $$2 + 2$$ example from the novel 1984. Notice you can test for the same condition using different boolean operators.
if (2 + 2 !== 4) {
resist()
}
or
if (!(2 + 2 === 4)) {
resist()
}
Example 2.5 A colorful meteor
Reading through this code, it isn’t immediately clear that ghostwhite is meant to be the default stroke color. You could make this more explicit by adding an else clause to your conditional statement. Here is an example of the syntax.
if (condition) {
// things to do if condition is true
} else {
// things to do if condition is false
}
The conditional statement now has two distinct pathways, or branches, that may be followed depending on the truth value of the condition.
Figure 2.7 Flow of execution with two branches
You could reorganize your sketch to reflect this structure like so.
At this point, you might say, “Branches seem useful, but what if I want more than two in my program?” Say no more! You can chain conditionals together using an else if statement (a combination of “else” and “if”). In a chained conditional, conditions are tested in the order they are written, and only the first branch whose condition is true will execute.
if (condition1) {
thingOne()
} else if (condition2) {
thingTwo()
} else {
thingThree()
}
Example 2.6 A multicolor meteor
Exercise 2.12 Change the previous example so that the following conditions are tested in this order. Can you explain what happened?
if (x >= 50) {
stroke('tomato')
} else if (x >= 150) {
stroke('crimson')
} else {
stroke('ghostwhite')
}
Exercise 2.13 Duplicate one of your previous sketches and modify its behavior using conditional statements. Use at least one relational and one other boolean operator.
# 2.3 Iteration¶
The logical building blocks you assembled in the last two sections let you create branches and make decisions in your programs. In this section, we’ll use many of the same building blocks to form another type of control flow: repetition.
## while statements¶
In Exercise 1.18, you drew a constellation by calling the point() function repeatedly with different arguments. Let’s revisit this exercise using the constellation Orion as a starting point. We’ll focus on Orion’s Belt, which consists of the stars Alnitak, Alnilam, and Mintaka.
Orion’s Belt
$$\mapsto$$ Outer space is an empty, infinite, two-dimensional surface. Stars are points of light within this plane.
Star coordinates are ordered pairs of floating point numbers.
Table 2.3 The “Orion’s Belt” data set
Star Name
$$x$$
$$y$$
Alnitak
$$25$$
$$100$$
Alnilam
$$100$$
$$100$$
Mintaka
$$175$$
$$100$$
First, paint the night sky. Then, select a pencil color & weight. Finally, draw a point at each star’s position.
Example 2.7 Orion’s Belt
The stars Alnitak and Mintaka are actually star systems; each system is made up of multiple stars orbiting one another. I’ll use this fact as my creative license to adjust Orion’s Belt a little. For starters, how about we draw Orion’s Belt with all of the major stars in each system? Let’s keep things simple by assuming all of the stars are the same size and are aligned horizontally in the sky.
Table 2.4 Expanded “Orion’s Belt” data set
Star Name
$$x$$
$$y$$
Alnitak Aa
25
100
Alnitak Ab
50
100
Alnitak B
75
100
Alnilam
100
100
δ Ori Aa1
125
100
δ Ori Aa2
150
100
δ Ori Ab
175
100
The visual result looks good, but the code worries me a little. Notice that I wrote seven nearly identical copies of the same statement to draw the stars. This approach works fine to get started but imagine writing a program that needs to repeat an instruction dozens of times. Or millions of times. Writing each variant by hand would be tedious and error prone. JavaScript’s while statement makes repetition, or iteration, simple.
while (condition) {
// this is the loop body
// statements in here repeat while condition is true
thingOne()
thingTwo()
}
while statements are structured similarly to if statements; they have a header with a condition and a body with code that might be executed. Each statement in the body executes in order, from top to bottom, repeatedly, until the condition in the header is false. Iterative control structures like this are commonly known as loops.
Figure 2.8 Flow of execution in a while loop
Let’s use a while loop to simplify our sketch of Orion’s Belt. Reviewing the previous example, the only difference between the stars is their $$x$$-coordinates. We know where our $$x$$-coordinates start, where they stop, and how much space is between them. This is all the information we need to simplify our work by using a loop.
Example 2.9 Orion’s Belt with iteration
Not too shabby! Given an initial value for x, the while statement draws a point, then increments x by 25, and repeats this process until x is greater than 175.
We could change one line of code from the previous example to pack twice as many stars in the same region of sky.
Example 2.10 Bedazzling Orion’s Belt
Exercise 2.14 Modify the sketch above to draw a row of stars in a different arrangement. What is the initial value of your loop variable x? What condition do you test to end the loop’s execution? How much do you increment x by during each iteration?
You might say, “OK, but what if I turned my head a little? Could I draw the stars along a vertical line instead?” Sure! In this case, you could keep the value of x constant while varying y.
Example 2.11 Switching axes from $$x$$ to $$y$$
## Infinite Loops¶
while statements are powerful tools that should be used with care. Consider the example below.
while (true) {
thingOne()
thingTwo()
}
true is a keyword in JavaScript that corresponds to a boolean value of, you guessed it, true. If a while statement’s condition is always true, then it will continue looping forever, thus creating an infinite loop. Unintended infinite loops are bad news. Let’s examine a slightly modified version of the loop from the previous sketch.
let y = 25
while (y <= 175) {
point(100, y)
}
y += 25 // oops
Notice that I incremented y outside of the loop body. This means that y isn’t incremented after each iteration. Instead, the value of y will always be 25, which is always less than or equal to 175, so the condition 25 <= 175 is always true. The loop never stops executing! If you ran this code, your web browser might give you a friendly notification to stop the sketch before it grinds your computer to a halt.
Exercise 2.15 The code snippet below is meant to draw a horizontal row of points across the canvas. Instead, it creates an infinite loop. Identify the error and fix it. Explain the problem and your solution in plain English.
let x = 0
let y = 100
while (x < 200) {
point(x, y)
x += 10
}
Exercise 2.16 Create a sketch that uses a while statement to draw points along a diagonal line. Review the code snippet below as a hint.
let x = 0
while (x < 200) {
// >> compute y here
point(x, y)
x += 10
}
Exercise 2.17 In Chapter 1, we defined an algorithm for multiplying two integers $$a \cdot b$$ as repeated addition $$b + b + ... + b$$. Fill in the missing code below to express the same algorithm in JavaScript using a while loop.
let a = 5
let b = 3
let product = 0
let i = 0
while (i < a) {
// >> compute product here
i += 1
}
Exercise 2.18 Review the algorithms for integer division and exponentiation, then implement them in JavaScript using a while loop. The subtraction assignment -= and multiplication assignment *= operators might be helpful.
# 2.4 Systems of Inequalities¶
We began this chapter by analyzing a single linear equation in slope-intercept form: $$y = mx + b$$. You could substitute any real number for $$x$$ and compute the corresponding value of $$y$$, making the ordered pair $$(x,y)$$ a solution to the equation. Let’s review a concrete example.
Given the following linear equation
$y = 0.5x + 100$
compute the value of $$y$$ when $$x = 50$$.
$$y = 0.5 \cdot 50 + 100$$
$$y = 25 + 100$$
$$y = 125$$
One solution to this equation is located at $$(50,125)$$. What if we tried $$(50,126)$$ instead?
$$126 = 0.5 \cdot 50 + 100$$
$$126 = 25 + 100$$
$$126 \ne 125$$
It turns out $$(50,126)$$ is not a solution to this particular equation, but there are infinitely many other solutions. For example, we could find solutions to the left and right of $$(50,125)$$ at $$x = 49$$ and $$x = 51$$.
$$y = 0.5 \cdot 49 + 100$$
$$y = 24.5 + 100$$
$$y = 124.5$$
$$y = 0.5 \cdot 51 + 100$$
$$y = 25.5 + 100$$
$$y = 125.5$$
You could use a while loop to quickly compute and visualize all of the solutions in the interval $$0 \le x \le 200$$.
Example 2.12 Visualizing solutions to a linear equation
Mathematicians often express a change in the value of a variable with the Greek letter $$\Delta$$, pronounced “delta”. Using this notation, we can express a change in the variable $$x$$ as $$\Delta x$$. I declare the variable dx on line 11 and use it to increment x on line 15 as a nod to our standard mathematical notation.
## Testing Solutions¶
OK, the solutions to a linear equation generate a line. I wonder what shapes other linear relationships make. Let’s take the previous example and swap out the $$=$$ symbol for a $$>$$.
$y > 0.5x + 100$
You can test solutions to this linear inequality just as you did with linear equations. For example, let’s see if the coordinate pair $$(50,125)$$ produces a truth value of 1 when we substitute the values into our inequality.
$$125 > 0.5 \cdot 50 + 100$$
$$125 > 25 + 100$$
$$125 > 125$$
Uh oh. We evaluated the right-hand side of the inequality and produced a value of $$125$$. But that resulted in a false statement; $$125$$ is not greater than itself. We can conclude that $$(50,125)$$ isn’t a solution to this inequality. How about we move along the $$y-axis$$ a little to $$(50,126)$$?
$$126 > 0.5 \cdot 50 + 100$$
$$126 > 25 + 200$$
$$126 > 125$$
Success! Let’s go a little further along the $$y$$-axis to $$(50,127)$$.
$$127 > 0.5 \cdot 50 + 100$$
$$127 > 25 + 200$$
$$127 > 125$$
Interesting. Let’s try one more coordinate pair $$(50,128)$$ to see if this pattern holds.
$$128 > 0.5 \cdot 50 + 100$$
$$128 > 25 + 200$$
$$128 > 125$$
When $$x = 50$$, we can pair it with any $$y > 125$$ to solve the inequality $$y > 0.5x + 100$$. You could automate this sort of test with a while loop.
Example 2.13 Testing solutions to a linear inequality
The sketch above fixes the value of x at 50 and tests solutions to the inequality for all y values in the interval $$0 \le y < 200$$. Solutions along this column are colored black while other points are colored ghostwhite.
This is the first example we’ve seen of a while loop that includes an if statement in its body. You can put (almost) whatever code you want in the body of a while loop: function calls, arithmetic operations, if statements, and even other while loops. This last option opens up many interesting possibilities.
## Nested Loops¶
You just tested hundreds of possible solutions to the inequality $$y > 0.5x + 100$$ when $$x$$ was fixed at $$50$$. Let’s fully automate the process of testing solutions by iterating over the canvas’ $$x$$-axis just like we’re doing with its $$y$$-axis.
A quick note about the algorithm we are about to run: it is very inefficient and would normally grind your computer to a halt. By default, p5 executes each statement you place in the body of the draw() function in order, from top to bottom, repeatedly, about $$60$$ times per second. Behind the scenes, you can imagine that the code you write in draw() is executing in the body of a while statement.
setup() // all of your code bundled up
while (true) {
draw() // all of your code bundled up
}
Testing individual solutions to a linear inequality requires computing once on each $$(x,y)$$ coordinate pair before drawing a point on the canvas–that’s roughly $$200 \cdot 200 = 40000$$ operations! The algorithm is slow and produces the same results every time, so there is no need to repeat it.
In the next example and those that follow, I will call the noLoop() function once in setup() like so.
function setup() {
createCanvas(400, 400)
noLoop()
}
By calling noLoop(), you change p5’s behavior so that draw() only executes a single time. You can imagine p5 running the following code instead.
setup() // all of your code bundled up
draw() // all of your code bundled up
Now, we can compute once on each $$(x,y)$$ coordinate pair, draw the corresponding point, and produce a single image. This adjustment should keep your computer happy, but it may still take a moment for the result to appear.
Example 2.14 Visualizing a linear inequality
The control structure you just created is called a nested loop. The outer loop increments the variable x while the inner loop increments the variable y and tests for solutions along each column of your canvas. And the result of all that computation? It turns out the set of $$(x,y)$$ coordinate pairs that solve our inequality, known as the solution set, forms a triangle-shaped region with the edges of the canvas. Neat! Can we make a rectangle?
Example 2.15 The rectangle inequality
Exercise 2.19 Modify the previous example to draw a new five-sided shape with your solution set. Closed shapes made by connecting three or more straight lines are known as polygons, and a five-sided polygon is known as a pentagon.
Exercise 2.20Ellsworth Kelly’s “Austin” is a serene little chapel located on the campus of the University of Texas at Austin. Its walls feature a series of fourteen black and white marble panels that look suspiciously like linear inequalities. Use Kelly’s panels as inspiration for your own series of abstract images. How many images will you include in your series? What colors will you use? What shapes will you create?
Drawing with linear inequalities opens up a dizzying number of creative possibilities. But what if you just wanted to draw a rectangle in the middle of your canvas? Meeting this challenge requires expanding our modeling toolkit yet again.
## Solving Systems¶
When you solved your first system of linear equations, you found the point $$(x,y)$$ where two lines intersected. In other words, you found the only ordered pair $$(x,y)$$ that solved both equations simultaneously. We’ll follow a very similar train of thought to solve systems of linear inequalities.
For starters, let’s try consider the system $$y < x + 150$$ and $$y > 75$$. We can test possible solutions $$(x,y)$$ against both inequalities using the && operator.
Example 2.16 A system of linear inequalities
OK, but how would we draw that rectangle in the middle of the canvas? You can think of a rectangle as the set of points between a pair of $$x$$-values and a pair of $$y$$-values. For example, we could take all of the points where $$x > 75$$ AND $$x < 125$$ AND $$y > 100$$ AND $$y < 160$$.
Example 2.17 A more constrained system
At this point you might say, “This is great! But how do I draw multiple shapes?” Simple: define multiple systems of inequalities. I’d like to frame this part of the discussion by studying the work of another abstract painter, Piet Mondrian.
Figure 2.9 “Composition II in Red, Blue, and Yellow” Courtesy Wikimedia Foundation
The painting “Composition II”.
$$\mapsto$$ Each colored region is the solution set to a system of linear inequalities.
The boundaries for each system of inequalities are either vertical or horizontal lines.
Table 2.5 The “Composition II” data set
Name
Left $$x$$
Right $$x$$
Bottom $$y$$
Top $$y$$
Blue corner
0
35
155
200
Yellow corner
188
200
183
200
Red corner
40
200
0
150
Stripe 1
35
40
0
200
Stripe 2
0
35
60
72
Stripe 3
0
200
150
155
Stripe 4
183
188
155
200
Stripe 5
188
200
173
183
First, prime the canvas by painting it ghostwhite. Next, select the paint color by testing whether a point solves a system of inequalities. Finally, paint the point. Repeat for every point on the canvas.
Example 2.18 “Composition II”
Exercise 2.21 Spend a few minutes exploring WikiArt and find an abstract painting or painter who inspires you. Hilma af Klint and Mark Rothko are personal favorites of mine. Then, create a new sketch that uses systems of inequalities to draw your own abstraction. Is your sketch completely abstract or is it based on an object, place, emotion, etc.? What do you like most about your sketch? What was challenging about creating it?
Exercise 2.22 Visualize integer multiplication by building on your solution to Exercise 2.17. Use the starter code below to snap (very tiny) blocks together.
The noStroke() function removes edges p5 draws around text; doing so can make it easier to read some fonts. The fill() function sets the interior color of text. Note that the text() function takes three arguments: the text to be displayed, and the $$x$$- and $$y$$-coordinates where the text should appear.
The control structures you just studied make it possible to construct many useful computations. In the next chapter, we’ll raise the level of abstraction by bundling computations into functions you define. We’ll also have fun drawing with the many built-in shapes that p5 provides.
|
# How do you solve 2x+2y = 2 and 4x+3y=7 using matrices?
Jan 28, 2017
The answer is $\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}4 \\ - 3\end{matrix}\right)$
#### Explanation:
The equations are
$2 x + 2 y = 2$
$4 x + 3 y = 7$
In matrix form , we have
$\left(\begin{matrix}2 & 2 \\ 4 & 3\end{matrix}\right) \cdot \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}2 \\ 7\end{matrix}\right)$
Let $A = \left(\begin{matrix}2 & 2 \\ 4 & 3\end{matrix}\right)$
We must find the inverse matrix ${A}^{-} 1$
The determinant of matrix $A$ is
$\det A = | \left(2 , 2\right) , \left(4 , 3\right) | = 6 - 8 = - 2$
As $\det A \ne 0$, the matrix is invertible
${A}^{-} 1 = \frac{1}{\det} A \cdot \left(\left(3 , - 2\right) \left(- 4 , 2\right)\right)$
$= - \frac{1}{2} \left(\begin{matrix}3 & - 2 \\ - 4 & 2\end{matrix}\right) = \left(\begin{matrix}- \frac{3}{2} & 1 \\ 2 & - 1\end{matrix}\right)$
Verification
${\forall}^{-} 1 = \left(\begin{matrix}2 & 2 \\ 4 & 3\end{matrix}\right) \cdot \left(\begin{matrix}- \frac{3}{2} & 1 \\ 2 & - 1\end{matrix}\right) = \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) = I$
Therefore,
$\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}- \frac{3}{2} & 1 \\ 2 & - 1\end{matrix}\right) \left(\begin{matrix}2 \\ 7\end{matrix}\right) = \left(\begin{matrix}4 \\ - 3\end{matrix}\right)$
|
# What is the limit of [3 + 4/x - 5/x^2 + [x-1]/[x^3+1] as x goes to infinity?
Oct 24, 2015
3
#### Explanation:
Every term in the expression $3 + \frac{4}{x} - \frac{5}{{x}^{2}} + \frac{x - 1}{{x}^{3} + 1}$ goes to zero as $x \rightarrow \infty$ except the 3.
The reason is that the other terms are rational functions where the denominator has a higher degree than the numerator.
Jul 27, 2016
If we actually took the limit, we can separate each term. Since each of these functions have existent limits...
$\textcolor{b l u e}{{\lim}_{x \to \infty} 3 + \frac{4}{x} - \frac{5}{x} ^ 2 + \frac{x - 1}{{x}^{3} + 1}}$
$= {\lim}_{x \to \infty} 3 + {\lim}_{x \to \infty} \frac{4}{x} - {\lim}_{x \to \infty} \frac{5}{x} ^ 2 + {\lim}_{x \to \infty} \frac{x - 1}{{x}^{3} + 1}$
As $x \to \infty$ for $\frac{x - 1}{{x}^{3} + 1}$, the $- 1$ and $+ 1$ become insignificant, so this is equivalent to ${\lim}_{x \to \infty} \frac{1}{{x}^{2}}$:
$\implies 3 + {\cancel{\frac{4}{\infty}}}^{0} - {\cancel{\frac{5}{\infty}}}^{0} + {\cancel{{\lim}_{x \to \infty} \frac{1}{{x}^{2}}}}^{\frac{1}{\infty} \to 0}$
$= \textcolor{b l u e}{3}$
|
# Ten Frames Part 2: Addition and subtraction
Last week’s focus was on using ten frames to help with students’ number sense and conceptual development of number bonds for amounts 1-10. This post will feature ways to use ten frames to enhance students’ understanding of addition and subtraction. Look for freebies and a video!
There are many addition and subtraction strategies to help students memorize the basic facts such as these below. The ten frame is a very good tool for students of all grade levels to make these strategies more concrete and visual. I will focus on some of these today.
• add or take away 1 (or 2)
• doubles, near doubles
• facts of 10
• make a ten
• add or sub. tens and ones
Doubles and near doubles (doubles +1, -1, +2, or -2): If the doubles are memorized, then problems near doubles can be solved strategically.
• Show a doubles fact on a single ten frame (for up to 5 + 5). Use a double ten-frame template for 6 + 6 and beyond.
• With the same doubles fact showing, show a near doubles problem. This should help students see that the answer is just one or two more or less.
• Repeat with other examples.
• Help student identify what a doubles + 1 more (or less) problem looks like. They often have a misconception there should be a 1 in the problem. Make sure they can explain where the “1” does come from. Examples: 7 + 8, 10+11, 24+25, 15 +16, etc.
• For subtraction, start with the doubles problem showing and turn over the 2-color counters or remove them.
Facts of 10: These are important to grasp for higher level addition / subtraction problems as well as rounding concepts.
• Place counters on the ten frame. Determine how many more are needed to fill in the ten frame. This also helps with missing addends. Example: 3 + ___ = 10. Ask, “What goes with 3 to make 10?”
• Using 2-color counters, fill the 10 frame with 1 color. Then turn over some to reveal a number bond of 10 (such as 4 and 6).
Shake and Spill
• Play “Shake and Spill” with 10 two-color counters. Click on these links for Shake and Spill Directions and a Shake and Spill recording page. Basically, the student puts 10 of these counters in a cup, shakes it, and spills it out (gently). Count how many red and how many yellow. Repeat 10 or more times. Keep track of the spills on a recording sheet. Which combination came up most often? Which combination never came up? What is really nice to observe is if a student spills counters and sees 6 are red, do they know automatically there are 4 yellow, or do they still have to count them?
• Since number bonds enable a student to see addition and subtraction problems, the second bullet above will serve subtraction problems very well. Start with 10, turn over 7 to the yellow side. How many counters are red?
Make a Ten: This strategy builds on the above (facts of 10) to help with problems with sums between 10 and 20. Students should readily be able to solve a problem such as 10 + 4 mentally first.
• Use 2 ten frames (see Ten Frames part 1 for a link for templates)
• Let’s say the problem was 8 + 5. Place 8 counters on one ten frame, place 5 on the other.
• Move counters from one ten frame to fill up the other. 8 + 5 is the same as 10 + 3. The problem 10 + 3 should be a mental math problem. Students will need to see that counters were not added, but shifted from one ten frame to the other.
• Repeated practice with this concrete activity helps children think more deeply about the relationship of numbers.
Continued practice with these strategies:
1. During your daily math meeting, flash ten frame dot cards to students in which they must use the above strategies. Use it as a # Talk sessions so students can verbally explain how they solved it.
2. Try this from NMCT Illuminations sight (National Council for Teachers of Mathematics): Interactive ten frame
3. Watch this video of a teacher modeling the Make-a-Ten strategy: Make a ten video using ten frames
4. Learning stations:
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## Calculate real and complex roots of quadratic equations
### Discriminant of a reduced form and factoring quadratic equations
#### Real roots of the quadratic equation
• x1 = 3
• x2 = −1
x² − 2x − 3 = (x − 3)(x + 1)
x² − 2x − 3 = 0
x² − 2x − 3 = 0
#### Discriminant of the reduced form of the quadratic equation
D = q − p²÷4 = (−3) − (−2)²÷4 = −4 < 0
#### The minimum of the quadratic functionf (x) = x² − 2x − 3
f (x) = −4 for x = 1 and f (x) > −4 for x ≠ 1
• To find real and complex roots of a quadratic equation with real coefficients a, b and c:ax² + bx + c = 0 (1)use the following formula:x1,2 = (−b ± b² − 4ac ) ÷ 2a (2)
• Divide the equation (1) by a:x² + px + qa = 0 (2)where:p = b ÷ a (3)q = c ÷ a (4)(2) is called the reduced form of a quadratic equation.
• To find the roots of the reduced form (2) use the following formula:x1,2 = (−p÷2 ± p²÷4 − q ) (5)
• Note: Since the equation (2) is the equation (1) divided by a, the solutions x1,2 found by formulas (2) and (5) are identical.
• The discriminant is equal to the expression under the square root in formula (5) multiplied by −1:D = q − p²÷4 (6)
• Depending on the sign of the discriminant (6) the equations (1) and (2) have:
• D < 0: Two real solitions, i.e. two real roots;
• D = 0: One real solition with two equal (i.e. identical) real roots;
• D > 0: No real solutions, but two conjigate complex roots.
• x1 + x2 = −b ÷ a = −p
• x1 × x2 = c ÷ a = q
• Solve the first derivative f'(x) of a quadratic function f(x) for zero, to find a specific x0, where f(x0) is the extreme of f(x):f(x) = ax² + bx + c ⇛ f'(x) = 2ax + b2ax + b = 0 ⇛ x0 = −b ÷ (2a)Use the second derivative to find whether f(x0) corresponds to the minimum or the maximum value of f(x).f''(x) = 2aFor any quadratic function, the second derivative does not depend on x. The sign of the second derivative of a quadratic equation is the same as the sign of the coefficient a. If a is greater than 0 then f(x0) corresponds to the minimum value of f(x), and if a is less than 0 then f(x0) corresponds to the maximum value of f(x).
• Example:f(x) = −x² + 2x + 3 ⇛ f'(x) = −2x + 2f'(x) = −2x + 2 = 0x = 1 The quadratic function reaches its extreme value at x = 1f''(x) = −2 < 0 The quadratic function reaches its maximum value at x = 1f(1) = −1² + 2×1 + 3 = 4 The maximum value of the quadratic function is equal to 4 at x = 1
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#### Gravels, Substances and Oils
CaribSea, Marine, Arag-Alive, Natural Reef weighs 1 153.3 kg/m³ (71.99817 lb/ft³) with specific gravity of 1.1533 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Nitrobenzol [C6H5NO2] weighs 1 203.7 kg/m³ (75.14454 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ]
Volume to weightweight to volume and cost conversions for Peanut oil with temperature in the range of 10°C (50°F) to 140°C (284°F)
#### Weights and Measurements
kilogray is a derived metric (SI) measurement unit of absorbed radiation dose of ionizing radiation, e.g
Temperature is one of the seven SI base quantities and used as a measure of thermal energy.
Å/h² to fur/s² conversion table, Å/h² to fur/s² unit converter or convert between all units of acceleration measurement.
#### Calculators
Electricity cost calculator per hours, days, weeks, months and years
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Trying to Score Better Scores in Grade 5 Maths? Utilize Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions and make the most out of them. Begin your practice right away using the Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions and understand the concepts. Attain more knowledge using the Big Ideas Math Book 5th Grade Answer Key Chapter 9 Multiply Fractions and cross-check the Solutions from it after your practice sessions.
## Big Ideas Math Book 5th Grade Answer Key Chapter 9 Multiply Fractions
Apply Maths in your Real-Time and get the Tips & Tricks to Solve Various Problems using Big Ideas Math Book 5th Grade Answer Key Chapter 9 Multiply Fractions. Try to utilize the resource available Big Ideas Math Answers Grade 5 Chapter 9 Multiply Fractions and make your preparation way more effective. You just need to tap on the quick links available in order to access them and learn all the topics.
Lesson: 1 Multiply Whole Numbers by Fractions
Lesson: 2 Use Models to Multiply Fractions by Whole Numbers
Lesson: 3 Multiply Fractions and Whole Numbers
Lesson: 4 Use Models to Multiply Fractions
Lesson: 5 Multiply Fractions
Lesson: 6 Find Areas of Rectangles
Lesson: 7 Multiply Mixed Numbers
Lesson: 8 Compare Factors and Products
Chapter: 9 – Multiply Fractions
### Lesson 9.1 Multiply Whole Numbers by Fractions
Explore and Grow
Write any proper fraction that is not a unit fraction. Draw a model to represent your fraction. Draw a model to find a multiple of your fraction?
The proper fraction that is not a unit fraction is: $$\frac{5}{8}$$
Let the proper fraction be multiplied by 5
So,
We have to find the value of 5 × $$\frac{5}{8}$$
Now,
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
5 × $$\frac{5}{8}$$
= $$\frac{5}{1}$$ × $$\frac{5}{8}$$
= $$\frac{5 × 5}{1 × 8}$$
= $$\frac{25}{8}$$
Hence, from the above,
We can conclude that the multiple of your proper fraction is: $$\frac{25}{8}$$
Reasoning
How can you use a model to multiply a whole number by a fraction? Explain.
We can multiply a whole number by a fraction using the properties of multiplication. They are:
A) a = $$\frac{a}{1}$$
B) a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
Hence, by using the above properties, we can multiply the whole number by a fraction.
Think and Grow: Multiply Whole Numbers by Fractions
Example
Find 3 × $$\frac{2}{5}$$
Show and Grow
Multiply.
Question 1.
2 × $$\frac{3}{4}$$ = ______
2 × $$\frac{3}{4}$$ = $$\frac{6}{4}$$
Explanation:
The given numbers are: 2 and $$\frac{3}{4}$$
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
2 × $$\frac{3}{4}$$
= $$\frac{2}{1}$$ × $$\frac{3}{4}$$
= $$\frac{2 × 3}{1 × 4}$$
= $$\frac{6}{4}$$
Hence,
2 × $$\frac{3}{4}$$ = $$\frac{6}{4}$$
Question 2.
4 × $$\frac{5}{8}$$ = ____
4 × $$\frac{5}{8}$$ = $$\frac{20}{8}$$
Explanation:
The given numbers are: 4 and $$\frac{5}{8}$$
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
4 × $$\frac{5}{8}$$
= $$\frac{4}{1}$$ × $$\frac{5}{8}$$
= $$\frac{4 × 5}{1 × 8}$$
= $$\frac{20}{8}$$
Hence,
4 × $$\frac{5}{8}$$ = $$\frac{20}{8}$$
Apply and Grow: Practice
Multiply.
Question 3.
5 × $$\frac{7}{10}$$ = ______
5 × $$\frac{7}{10}$$ = $$\frac{35}{10}$$
Explanation:
The given numbers are: 5 and $$\frac{7}{10}$$
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
5 × $$\frac{7}{10}$$
= $$\frac{5}{1}$$ × $$\frac{7}{10}$$
= $$\frac{5 × 7}{1 × 10}$$
= $$\frac{35}{10}$$
Hence,
5 × $$\frac{7}{10}$$ = $$\frac{35}{10}$$
Question 4.
8 × $$\frac{2}{3}$$ = ______
8 × $$\frac{2}{3}$$ = $$\frac{16}{3}$$
Explanation:
The given numbers are: 8 and $$\frac{2}{3}$$
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
8 × $$\frac{2}{3}$$
= $$\frac{8}{1}$$ × $$\frac{2}{3}$$
= $$\frac{8 × 2}{1 × 3}$$
= $$\frac{16}{3}$$
Hence,
8 × $$\frac{2}{3}$$ = $$\frac{16}{3}$$
Question 5.
7 × $$\frac{5}{6}$$ = ______
7 × $$\frac{5}{6}$$ = $$\frac{35}{6}$$
Explanation:
The given numbers are: 7 and $$\frac{5}{6}$$
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
7 × $$\frac{5}{6}$$
= $$\frac{7}{1}$$ × $$\frac{5}{6}$$
= $$\frac{7 × 5}{1 × 6}$$
= $$\frac{35}{6}$$
Hence,
7 × $$\frac{5}{6}$$ = $$\frac{35}{6}$$
Question 6.
9 × $$\frac{1}{2}$$ = ______
9 × $$\frac{1}{2}$$ = $$\frac{9}{2}$$
Explanation:
The given numbers are: 9 and $$\frac{1}{2}$$
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
9 × $$\frac{1}{2}$$
= $$\frac{9}{1}$$ × $$\frac{1}{2}$$
= $$\frac{9 × 1}{1 × 2}$$
= $$\frac{9}{2}$$
Hence,
9 × $$\frac{1}{2}$$ = $$\frac{9}{2}$$
Question 7.
6 × $$\frac{3}{100}$$ = ______
6 × $$\frac{3}{100}$$ = $$\frac{18}{100}$$
Explanation:
The given numbers are: 6 and $$\frac{3}{100}$$
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
6 × $$\frac{3}{100}$$
= $$\frac{6}{1}$$ × $$\frac{3}{100}$$
= $$\frac{6 × 3}{1 × 100}$$
= $$\frac{18}{100}$$
Hence,
6 × $$\frac{3}{100}$$ = $$\frac{18}{100}$$
Question 8.
15 × $$\frac{4}{7}$$ = ______
15 × $$\frac{4}{7}$$ = $$\frac{60}{7}$$
Explanation:
The given numbers are: 15 and $$\frac{4}{7}$$
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
15 × $$\frac{4}{7}$$
= $$\frac{15}{1}$$ × $$\frac{4}{7}$$
= $$\frac{15 × 4}{1 × 7}$$
= $$\frac{60}{7}$$
Hence,
15 × $$\frac{4}{7}$$ = $$\frac{60}{7}$$
Question 9.
10 × $$\frac{5}{3}$$ = ______
10 × $$\frac{5}{3}$$ = $$\frac{50}{3}$$
Explanation:
The given numbers are: 10 and $$\frac{5}{3}$$
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
10 × $$\frac{5}{3}$$
= $$\frac{10}{1}$$ × $$\frac{5}{3}$$
= $$\frac{10 × 5}{1 × 3}$$
= $$\frac{50}{3}$$
Hence,
10 × $$\frac{5}{3}$$ = $$\frac{50}{3}$$
Question 10.
4 × $$\frac{5}{2}$$ = ______
4 × $$\frac{5}{2}$$ = 10
Explanation:
The given numbers are: 4 and $$\frac{5}{2}$$
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
4 × $$\frac{5}{2}$$
= $$\frac{4}{1}$$ × $$\frac{5}{2}$$
= $$\frac{4 × 5}{1 × 2}$$
= $$\frac{20}{2}$$
= 10
Hence,
4 × $$\frac{5}{2}$$ = 10
Question 11.
3 × $$\frac{11}{8}$$ = ______
3 × $$\frac{11}{8}$$ = $$\frac{33}{8}$$
Explanation:
The given numbers are: 3 and $$\frac{11}{8}$$
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
3 × $$\frac{11}{8}$$
= $$\frac{3}{1}$$ × $$\frac{11}{8}$$
= $$\frac{11 × 3}{1 × 8}$$
= $$\frac{33}{8}$$
Hence,
3 × $$\frac{11}{8}$$ = $$\frac{33}{8}$$
Find the unknown number.
Question 12.
The missing number is: 3
Explanation:
The given fractions are: $$\frac{3}{20}$$ and $$\frac{9}{20}$$
Let the missing number be X
So,
X × $$\frac{3}{20}$$ = $$\frac{9}{20}$$
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = $$\frac{9}{20}$$ ÷ $$\frac{3}{20}$$
We know that,
$$\frac{a}{b}$$ ÷ $$\frac{x}{y}$$ = $$\frac{a}{b}$$ × $$\frac{y}{x}$$
So,
$$\frac{9}{20}$$ ÷ $$\frac{3}{20}$$
= $$\frac{9}{20}$$ × $$\frac{20}{3}$$
= $$\frac{9 × 20}{20 × 3}$$
= 3
Hence,
The missing number is: 3
Question 13.
The missing number is: 6
Explanation:
The given fractions are: $$\frac{4}{9}$$ and $$\frac{24}{9}$$
Let the missing number be X
So,
X × $$\frac{4}{9}$$ = $$\frac{24}{9}$$
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = $$\frac{24}{9}$$ ÷ $$\frac{4}{9}$$
We know that,
$$\frac{a}{b}$$ ÷ $$\frac{x}{y}$$ = $$\frac{a}{b}$$ × $$\frac{y}{x}$$
So,
$$\frac{24}{9}$$ ÷ $$\frac{4}{9}$$
= $$\frac{24}{9}$$ × $$\frac{9}{4}$$
= $$\frac{9 × 24}{9 × 4}$$
= 6
Hence,
The missing number is: 6
Question 14.
The missing number is: 10
Explanation:
The given fractions are: $$\frac{5}{12}$$ and $$\frac{50}{12}$$
Let the missing number be X
So,
X × $$\frac{5}{12}$$ = $$\frac{50}{12}$$
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = $$\frac{50}{12}$$ ÷ $$\frac{5}{12}$$
We know that,
$$\frac{a}{b}$$ ÷ $$\frac{x}{y}$$ = $$\frac{a}{b}$$ × $$\frac{y}{x}$$
So,
$$\frac{50}{12}$$ ÷ $$\frac{5}{12}$$
= $$\frac{50}{12}$$ × $$\frac{12}{5}$$
= $$\frac{12 × 50}{12 × 5}$$
= 10
Hence,
The missing number is: 10
Question 15.
A recipe calls for $$\frac{3}{4}$$ cup of dried rice noodles. You make 44 batches of the recipe. How many cups of dried rice noodles do you use?
The number of cups of dried rice noodles you used is: 33
Explanation:
It is given that a recipe calls for $$\frac{3}{4}$$ cup of dried rice noodles.
It is also given that you make 44 batches of the recipe
So,
The number of cups of dried rice noodles you used = ( The number of batches of the recipe ) × ( The number of dried rice noodles for each batch )
= 44 × $$\frac{3}{4}$$
Now,
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
44 × $$\frac{3}{4}$$
= $$\frac{44}{1}$$ × $$\frac{3}{4}$$
= $$\frac{44 × 3}{1 × 4}$$
= $$\frac{33}{1}$$
= 33
Hence, from the above,
We can conclude that the number of cups of dried noodles you used is: 33 cups
Question 16.
YOU BE THE TEACHER
Your 5 × $$\frac{3}{5}$$ friend says that 5 is equal to $$\frac{3}{5}$$ + $$\frac{3}{5}$$ + $$\frac{3}{5}$$ + $$\frac{3}{5}$$ + $$\frac{3}{5}$$. Is your friend correct? Explain.
Explanation:
It is given that your 5 × $$\frac{3}{5}$$ friend says that 5 is equal to $$\frac{3}{5}$$ + $$\frac{3}{5}$$ + $$\frac{3}{5}$$ + $$\frac{3}{5}$$ + $$\frac{3}{5}$$.
Now,
We know that,
n × a = a + a + a + a + …………….. + n times
Where,
‘n’ is the multiple of a
Here,
‘a’ will be the whole number or the fraction
So,
According to the above property,
5 × $$\frac{3}{5}$$ = $$\frac{3}{5}$$ + $$\frac{3}{5}$$ + $$\frac{3}{5}$$ +$$\frac{3}{5}$$ + $$\frac{3}{5}$$
Hence, from the above,
We can conclude that your friend is correct.
Question 17.
Patterns
Describe and complete the pattern.
The completed pattern is:
From the above pattern,
We can observe that
The whole number is constant i.e., 9
The fraction part is changing i.e., the numerator part is increasing by 1 till 4 and the denominator is constant
So,
Now,
9 × $$\frac{1}{4}$$
= $$\frac{9}{1}$$ × $$\frac{1}{4}$$
= $$\frac{9 × 1}{1 × 4}$$
= $$\frac{9}{4}$$
9 × $$\frac{2}{4}$$
= $$\frac{9}{1}$$ × $$\frac{2}{4}$$
= $$\frac{9 × 2}{1 × 4}$$
= $$\frac{18}{4}$$
So,
The remaining two multiplication equations will also be solved in the same way.
Think and Grow: Modeling Real Life
Example
Your goal is to make a waterslide that is at least 10 meters long. You make the waterslide using 10 plastic mats that are each $$\frac{3}{2}$$ meters long. Do you reach your goal?
Find the length of the waterslide by multiplying the number of mats by the length of each mat.
So,
Show and Grow
Question 18.
An excavator is moving 4 piles of dirt that are the same size. Each pile requires $$\frac{3}{4}$$ hour to move. Can the excavator move all of the piles in 2 hours?
No, the excavator can’t move all of the piles in 2 hours
Explanation:
It is given that an excavator is moving 4 piles of dirt that are of the same size and each pile requires $$\frac{3}{4}$$ hour to move.
So,
We know that,
1 hour = 60 minutes
So,
$$\frac{3}{4}$$ of 1 hour = $$\frac{3}{4}$$ × 60
= $$\frac{3}{4}$$ × $$\frac{60}{1}$$
= $$\frac{3 × 60}{4 × 1}$$
= 45 minutes
So,
The time is taken to move 4 piles of dirt = ( The time is taken to move each pile ) × ( The total number of piles )
= 45 × 4
=180 minutes
It is given that you have to move the piles of dirt in 2 hours but the time taken is 4 hours.
Hence, from the above,
We can conclude that the excavator can’t move all the piles in 2 hours.
Question 19.
You walk dogs $$\frac{5}{4}$$ miles two times each day. How far do you walk the dogs in 1 week?
The number of miles you walk the dogs in 1 week is: $$\frac{35}{2}$$
Explanation:
It is given that you walk dogs $$\frac{5}{4}$$ miles two times each day
So,
The total number of miles you walk dogs in 1 day = 2 × $$\frac{5}{4}$$
= $$\frac{5}{4}$$ × $$\frac{2}{1}$$
= $$\frac{5 × 2}{4 × 1}$$
= $$\frac{5}{2}$$ miles
We know that,
1 week = 7 days
So,
The total number of miles you walk dogs in 1 week = ( The number of miles you walk dogs in 1 day ) × 7
= $$\frac{5}{2}$$ × 7
= $$\frac{5}{2}$$ × $$\frac{7}{1}$$
= $$\frac{5 × 7}{2 × 1}$$
= $$\frac{35}{2}$$
Hence, from the above,
We can conclude that the numebr of miles you walk dogs in 1 week is: $$\frac{35}{2}$$ miles
Question 20.
DIG DEEPER
You have10 feet of string. You need $$\frac{5}{3}$$ feet of string to make 1 necklace. You make 5 necklaces. Do you have enough string to make another necklace? Explain.
Yes, we have enough string to make another necklace
Explanation:
It is given that you have 10 feet of string and you need $$\frac{5}{3}$$ feet of string to make 1 necklace. and you make 5 necklaces.
So,
The length of the string required for 5 necklaces = ( The length of string required for 1 necklace ) × ( The total number of necklaces )
= $$\frac{5}{3}$$ × 5
= $$\frac{5}{3}$$ × $$\frac{5}{1}$$
= $$\frac{5 × 5}{3 × 1}$$
= $$\frac{25}{3}$$ feet
We can write 10 feet as $$\frac{30}{3}$$ feet
So,
When we compare the total length of the string and the length of the string to make 5 necklaces,
We can observe that the length of the string to make 5 necklaces is less than the total length of the string.
Hence, from the above,
We can conclude that we have enough string to make another necklace.
### Multiply Whole Numbers by Fractions Homework & Practice 9.1
Multiply
Question 1.
5 × $$\frac{2}{3}$$ = ______
5 × $$\frac{2}{3}$$ = $$\frac{10}{3}$$
Explanation:
The given numbers are: 5 and $$\frac{2}{3}$$
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
5 × $$\frac{2}{3}$$
= $$\frac{5}{1}$$ × $$\frac{2}{3}$$
= $$\frac{2 × 5}{1 × 3}$$
= $$\frac{10}{3}$$
Hence,
5 × $$\frac{2}{3}$$ = $$\frac{10}{3}$$
Question 2.
9 × $$\frac{7}{8}$$ = ______
9 × $$\frac{7}{8}$$ = $$\frac{63}{8}$$
Explanation:
The given numbers are: 9 and $$\frac{7}{8}$$
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
9 × $$\frac{7}{8}$$
= $$\frac{9}{1}$$ × $$\frac{7}{8}$$
= $$\frac{9 × 7}{1 × 8}$$
= $$\frac{63}{8}$$
Hence,
9 × $$\frac{7}{8}$$ = $$\frac{63}{8}$$
Question 3.
4 × $$\frac{11}{12}$$ = ______
4 × $$\frac{11}{12}$$ = $$\frac{44}{12}$$
Explanation:
The given numbers are: 4 and $$\frac{11}{12}$$
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
4 × $$\frac{11}{12}$$
= $$\frac{4}{1}$$ × $$\frac{11}{12}$$
= $$\frac{4 × 11}{1 × 12}$$
= $$\frac{44}{12}$$
Hence,
4 × $$\frac{11}{12}$$ = $$\frac{44}{12}$$
Question 4.
8 × $$\frac{35}{100}$$ = ______
8 × $$\frac{35}{100}$$ = $$\frac{280}{100}$$
Explanation:
The given numbers are: 8 and $$\frac{35}{100}$$
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
8 × $$\frac{35}{100}$$
= $$\frac{8}{1}$$ × $$\frac{35}{100}$$
= $$\frac{8 × 35}{1 × 100}$$
= $$\frac{280}{100}$$
Hence,
8 × $$\frac{35}{100}$$ = $$\frac{280}{100}$$
Question 5.
3 × $$\frac{1}{2}$$ = ______
3 × $$\frac{1}{2}$$ = $$\frac{3}{2}$$
Explanation:
The given numbers are: 3 and $$\frac{1}{2}$$
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
3 × $$\frac{1}{2}$$
= $$\frac{3}{1}$$ × $$\frac{1}{2}$$
= $$\frac{1 × 3}{1 × 2}$$
= $$\frac{3}{2}$$
Hence,
3 × $$\frac{1}{2}$$ = $$\frac{3}{2}$$
Question 6.
7 × $$\frac{2}{5}$$ = ______
7 × $$\frac{2}{5}$$ = $$\frac{14}{5}$$
Explanation:
The given numbers are: 7 and $$\frac{2}{5}$$
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
7 × $$\frac{2}{5}$$
= $$\frac{7}{1}$$ × $$\frac{2}{5}$$
= $$\frac{2 × 7}{1 × 5}$$
= $$\frac{14}{5}$$
Hence,
7 × $$\frac{2}{5}$$ = $$\frac{14}{5}$$
Question 7.
6 × $$\frac{7}{4}$$ = ______
6 × $$\frac{7}{4}$$ = $$\frac{42}{4}$$
Explanation:
The given numbers are: 6 and $$\frac{7}{4}$$
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
6 × $$\frac{7}{4}$$
= $$\frac{6}{1}$$ × $$\frac{7}{4}$$
= $$\frac{6 × 7}{1 × 4}$$
= $$\frac{42}{4}$$
Hence,
6 × $$\frac{7}{4}$$ = $$\frac{42}{4}$$
Question 8.
12 × $$\frac{8}{7}$$ = ______
12 × $$\frac{8}{7}$$ = $$\frac{96}{7}$$
Explanation:
The given numbers are: 12 and $$\frac{8}{7}$$
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
12 × $$\frac{8}{7}$$
= $$\frac{12}{1}$$ × $$\frac{8}{7}$$
= $$\frac{12 × 8}{1 × 7}$$
= $$\frac{96}{7}$$
Hence,
12 × $$\frac{8}{7}$$ = $$\frac{96}{7}$$
Question 9.
25 × $$\frac{10}{9}$$ = ______
25 × $$\frac{10}{9}$$ = $$\frac{250}{9}$$
Explanation:
The given numbers are: 25 and $$\frac{10}{9}$$
We know that,
a = $$\frac{a}{1}$$
a × $$\frac{a}{b}$$ = $$\frac{a × a}{b × 1}$$
So,
25 × $$\frac{10}{9}$$
= $$\frac{25}{1}$$ × $$\frac{10}{9}$$
= $$\frac{25 × 10}{1 × 9}$$
= $$\frac{250}{9}$$
Hence,
25 × $$\frac{10}{9}$$ = $$\frac{250}{9}$$
Find the unknown number.
Question 10.
The missing number is: 5
Explanation:
The given fractions are: $$\frac{5}{7}$$ and $$\frac{25}{7}$$
Let the missing number be X
So,
X × $$\frac{5}{7}$$ = $$\frac{25}{7}$$
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = $$\frac{25}{7}$$ ÷ $$\frac{5}{7}$$
We know that,
$$\frac{a}{b}$$ ÷ $$\frac{x}{y}$$ = $$\frac{a}{b}$$ × $$\frac{y}{x}$$
So,
$$\frac{25}{7}$$ ÷ $$\frac{5}{7}$$
= $$\frac{25}{7}$$ × $$\frac{7}{5}$$
= $$\frac{7 × 25}{7 × 5}$$
= 5
Hence,
The missing number is: 5
Question 11.
The missing number is: 7
Explanation:
The given fractions are: $$\frac{9}{10}$$ and $$\frac{63}{10}$$
Let the missing number be X
So,
X × $$\frac{9}{10}$$ = $$\frac{63}{10}$$
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = $$\frac{63}{10}$$ ÷ $$\frac{9}{10}$$
We know that,
$$\frac{a}{b}$$ ÷ $$\frac{x}{y}$$ = $$\frac{a}{b}$$ × $$\frac{y}{x}$$
So,
$$\frac{63}{10}$$ ÷ $$\frac{9}{10}$$
= $$\frac{63}{10}$$ × $$\frac{10}{9}$$
= $$\frac{63 × 10}{10 × 9}$$
= 7
Hence,
The missing number is: 7
Question 12.
The missing number is: 9
Explanation:
The given fractions are: $$\frac{3}{5}$$ and $$\frac{27}{5}$$
Let the missing number be X
So,
X × $$\frac{3}{5}$$ = $$\frac{27}{5}$$
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = $$\frac{27}{5}$$ ÷ $$\frac{3}{5}$$
We know that,
$$\frac{a}{b}$$ ÷ $$\frac{x}{y}$$ = $$\frac{a}{b}$$ × $$\frac{y}{x}$$
So,
$$\frac{27}{5}$$ ÷ $$\frac{3}{5}$$
= $$\frac{27}{5}$$ × $$\frac{5}{3}$$
= $$\frac{5 × 27}{5 × 3}$$
= 9
Hence,
The missing number is: 9
Question 13.
You make 5 servings of pancakes. You top each serving with $$\frac{1}{4}$$ cup of strawberries. How many cups of strawberries do you use?
The number of cups of strawberries you used is: 20 cups
Explanation:
It is given that you make 5 servings of pancakes and you top each serving with $$\frac{1}{4}$$ cup of strawberries.
So,
The number of cups of strawberries = $$\frac{The number of servings of pancakes}{Each serving in pancake}$$
= 5 ÷ $$\frac{1}{4}$$
= $$\frac{5}{1}$$ ÷ $$\frac{1}{4}$$
= $$\frac{5}{1}$$ × $$\frac{4}{1}$$
= $$\frac{5 × 4}{1 × 1}$$
= 20 cups
hence, from the above,
We can conclude that the number of cups of strawberries is: 20 cups
Question 14.
Which One Doesn’tBelong?
Which one does not belong with the other three?
4 × $$\frac{3}{8}$$
3 × $$\frac{1}{8}$$
$$\frac{3}{8}$$ + $$\frac{3}{8}$$ + $$\frac{3}{8}$$ + $$\frac{3}{8}$$
1 + $$\frac{1}{2}$$
Let the expresseons be named A., B., C., D.
So,
A) 4 × $$\frac{3}{8}$$
B) 3 × $$\frac{1}{8}$$
C) $$\frac{3}{8}$$ + $$\frac{3}{8}$$ + $$\frac{3}{8}$$ + $$\frac{3}{8}$$
D) 1 + $$\frac{1}{2}$$
So, from the above four expressions,
Expressions A), B), C) are in the normal form and C) is the expanded form of expression A)
Hence, from the above,
We can conclude that expression D) does not belong to the other three.
Question 15.
Modeling Real Life
You complete $$\frac{5}{2}$$ inches of weaving each day for 5 days. The weaving needs to be at least 11 inches long. Is your weaving complete?
Explanation:
It is given that you complete $$\frac{5}{2}$$ inches of weaving each day for 5 days.
So,
The total number of inches you completed in 5 days = ( The number of inches you weaved each day ) × 5
= $$\frac{5}{2}$$ × 5
= $$\frac{5}{2}$$ × $$\frac{5}{1}$$
= $$\frac{5 × 5}{2 × 1}$$
= $$\frac{25}{2}$$
It is also given that the weaving needs to be atleast 11 inches long.
So,
11 inches can also be written as $$\frac{22}{2}$$
So,
When we compare the total number of inches needed for weaving and the total number of inches you weaved in 5 days, we can observe that
You weaved more than the amount of the weaving.
Hence, from the above,
We can conclude that your weaving is completed.
Question 16.
DIG DEEPER!
You spend $$\frac{7}{2}$$ hours playing drums each day for 2 days. Your friend spends $$\frac{5}{4}$$ hours playing drums each day for 6 days. Who spends more time playing drums? How much more?
Your friend spend more time playing drums than you
The amount you played drums more than your friend is: $$\frac{1}{2}$$ hours
Explanation:
It is given that you spend $$\frac{7}{2}$$ hours each day playing drums for 2 days
So,
The time you played drums for 2 days = The time you played drums for each day × 2
= $$\frac{7}{2}$$ × 2
= $$\frac{7}{2}$$ × $$\frac{2}{1}$$
= $$\frac{7 × 2}{2 × 1}$$
= $$\frac{14}{2}$$
= $$\frac{28}{4}$$
= 7 hours
It is also given that Your friend spends $$\frac{5}{4}$$ hours playing drums each day for 6 days.
So,
The time your friend played drums for 6 days = The time your friend played drums for each day × 6
= $$\frac{5}{4}$$ × 6
= $$\frac{5}{4}$$ × $$\frac{6}{1}$$
= $$\frac{5 × 6}{4 × 1}$$
= $$\frac{30}{4}$$ hours
Now,
When we compare the time played drums by you and your friends, your friend played more time than you
So,
The amount of time more your friend played than you = $$\frac{30}{4}$$ – $$\frac{28}{4}$$
= $$\frac{30 – 28}{4}$$
= $$\frac{2}{4}$$
= $$\frac{1}{2}$$ hour
Hence, from the above,
We can conclude that
Your friend spend more time playing drums than you
The amount you played drums more than your friend is: $$\frac{1}{2}$$ hours
Review & Refresh
Find the product.
Question 17.
0.6 × 0.4 = 0.24
Explanation:
The given decimal numbers are: 0.6 and 0.4
the representation of the decimal numbers in the fraction form is: $$\frac{6}{10}$$ and $$\frac{4}{10}$$
Now,
$$\frac{6}{10}$$ × $$\frac{4}{10}$$
= $$\frac{6 × 4}{10 × 10}$$
= $$\frac{24}{100}$$
So,
The representation of $$\frac{24}{100}$$ in the decimal form is: 0.24
Hence, 0.6 × 0.4 = 0.24
Question 18.
2.37 × 1.9 = 4.503
Explanation:
The given decimal numbers are: 2.37 and 1.9
the representation of the decimal numbers in the fraction form is: $$\frac{237}{100}$$ and $$\frac{19}{10}$$
Now,
$$\frac{237}{100}$$ × $$\frac{19}{10}$$
= $$\frac{237 × 19}{100 × 10}$$
= $$\frac{4,503}{1000}$$
So,
The representation of $$\frac{4503}{1000}$$ in the decimal form is: 4.503
Hence, 2.37 × 1.9 = 4.503
Question 19.
52.8 × 0.75 = 39.6
Explanation:
The given decimal numbers are: 52.8 and 0.75
the representation of the decimal numbers in the fraction form is: $$\frac{6}{10}$$ and $$\frac{4}{10}$$
Now,
$$\frac{528}{10}$$ × $$\frac{75}{100}$$
= $$\frac{528 × 75}{10 × 100}$$
= $$\frac{3960}{1000}$$
So,
The representation of $$\frac{3960}{1000}$$ in the decimal form is: 39.6
Hence, 52.8 × 0.75 = 39.6
### Lesson 9.2 Use Models to Multiply Fractions by Whole Numbers
You need to give water to $$\frac{2}{3}$$ of the dogs at a shelter. There are 12 dogs at the shelter. How many dogs need water? Draw a model to support your answer.
The number of dogs that need water is: 8 dogs
Explanation:
It is given that you need to give water to $$\frac{2}{3}$$ of the dogs at a shelter and there are 12 dogs at the shelter
So,
The number of dogs that need water = ( The fraction of dogs that need water ) × ( The total number of dogs )
= $$\frac{2}{3}$$ × 12
We know that,
$$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
= $$\frac{a × x}{b × 1}$$
So,
$$\frac{2}{3}$$ × 12
= $$\frac{2}{3}$$ × $$\frac{12}{1}$$
= $$\frac{2 × 12}{3 × 1}$$
= $$\frac{8}{1}$$
= 8
Hence, from the above,
We can conclude that there are 8 dogs that need water
Reasoning
How can you use a model to multiply a fraction by a whole number? Explain.
We can multiply a fraction by a whole number by using the following multiplication properties. They are:
A) $$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
B) x = $$\frac{x}{1}$$
Think and Grow: Multiply Fractions by Whole Numbers
You can use models to multiply a fraction by a whole number.
Example
Find $$\frac{3}{4}$$ of 8.
Example
Find $$\frac{5}{8}$$ × 4.
Show and Grow
Question 1.
Find $$\frac{2}{5}$$ of 10.
$$\frac{2}{5}$$ × 10 = 4
Explanation:
The given numbers are: $$\frac{2}{5}$$ and 10
We know that,
$$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
= $$\frac{a × x}{b × 1}$$
So,
$$\frac{2}{5}$$ × 10 = $$\frac{2}{5}$$ × $$\frac{10}{1}$$
= $$\frac{2 × 10}{5 × 1}$$
= $$\frac{4}{1}$$
= 4
Hence,
$$\frac{2}{5}$$ × 10 = 4
Question 2.
Find $$\frac{7}{12}$$ × 6.
$$\frac{7}{12}$$ × 6 = $$\frac{7}{2}$$
Explanation:
The given numbers are: $$\frac{7}{12}$$ and 6
We know that,
$$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
= $$\frac{a × x}{b × 1}$$
So,
$$\frac{7}{12}$$ × 6 = $$\frac{7}{12}$$ × $$\frac{6}{1}$$
= $$\frac{7 × 6}{12 × 1}$$
= $$\frac{7}{2}$$
Hence,
$$\frac{7}{12}$$ × 6 = $$\frac{7}{2}$$
Apply and Grow: Practice
Multiply. Use a model to help.
Question 3.
$$\frac{5}{6}$$ of 12
$$\frac{5}{6}$$ × 12 = 10
Explanation:
The given numbers are: $$\frac{5}{6}$$ and 12
We know that,
$$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
= $$\frac{a × x}{b × 1}$$
So,
$$\frac{5}{6}$$ × 12 = $$\frac{5}{6}$$ × $$\frac{12}{1}$$
= $$\frac{5 × 12}{6 × 1}$$
= $$\frac{10}{1}$$
= 10
Hence,
$$\frac{5}{6}$$ × 12 = 10
Question 4.
$$\frac{2}{3}$$ × 9
$$\frac{2}{3}$$ × 9 = 6
Explanation:
The given numbers are: $$\frac{2}{3}$$ and 9
We know that,
$$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
= $$\frac{a × x}{b × 1}$$
So,
$$\frac{2}{3}$$ × 9 = $$\frac{2}{3}$$ × $$\frac{9}{1}$$
= $$\frac{2 × 9}{3 × 1}$$
= $$\frac{6}{1}$$
= 6
Hence,
$$\frac{2}{3}$$ × 9 = 6
Question 5.
$$\frac{1}{5}$$ × 10
$$\frac{1}{5}$$ × 10 = 2
Explanation:
The given numbers are: $$\frac{1}{5}$$ and 10
We know that,
$$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
= $$\frac{a × x}{b × 1}$$
So,
$$\frac{1}{5}$$ × 10 = $$\frac{1}{5}$$ × $$\frac{10}{1}$$
= $$\frac{1 × 10}{5 × 1}$$
= $$\frac{2}{1}$$
= 2
Hence,
$$\frac{1}{5}$$ × 10 = 2
Question 6.
$$\frac{3}{5}$$ of 5
$$\frac{3}{5}$$ × 5 = 3
Explanation:
The given numbers are: $$\frac{3}{5}$$ and 5
We know that,
$$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
= $$\frac{a × x}{b × 1}$$
So,
$$\frac{3}{5}$$ × 5 = $$\frac{3}{5}$$ × $$\frac{5}{1}$$
= $$\frac{3 × 5}{5 × 1}$$
= $$\frac{3}{1}$$
= 3
Hence,
$$\frac{3}{5}$$ × 5 = 3
Question 7.
$$\frac{1}{6}$$ of 3
$$\frac{1}{6}$$ × 3 = $$\frac{1}{2}$$
Explanation:
The given numbers are: $$\frac{1}{6}$$ and 3
We know that,
$$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
= $$\frac{a × x}{b × 1}$$
So,
$$\frac{1}{6}$$ × 3 = $$\frac{1}{6}$$ × $$\frac{3}{1}$$
= $$\frac{1 × 3}{6 × 1}$$
= $$\frac{1}{2}$$
Hence,
$$\frac{1}{6}$$ × 3 = $$\frac{1}{2}$$
Question 8.
$$\frac{3}{8}$$ × 4
$$\frac{3}{8}$$ × 4 = $$\frac{3}{2}$$
Explanation:
The given numbers are: $$\frac{3}{8}$$ and 4
We know that,
$$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
= $$\frac{a × x}{b × 1}$$
So,
$$\frac{3}{8}$$ × 4 = $$\frac{3}{8}$$ × $$\frac{4}{1}$$
= $$\frac{3 × 4}{8 × 1}$$
= $$\frac{3}{2}$$
Hence,
$$\frac{3}{8}$$ × 4 = $$\frac{3}{2}$$
Question 9.
You have 25 beads. You use $$\frac{2}{5}$$ of the beads to make a bracelet. How many beads do you use?
Explanation:
It is given that you have 25 beads and you use $$\frac{2}{5}$$ of the beads to make a bracelet.
So,
The number of beads you used = ( The fraction of beads used to make 1 bracelet ) × ( The total number of beads )
= $$\frac{2}{5}$$ × 25
We know that,
$$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
= $$\frac{a × x}{b × 1}$$
So,
$$\frac{2}{5}$$ × 25 = $$\frac{2}{5}$$ × $$\frac{25}{1}$$
= $$\frac{2 × 25}{5 × 1}$$
= $$\frac{10}{1}$$
Hence, from the above,
We can conclude that the number of beads you used to make bracelets are: 10 beads
Question 10.
Writing
Write and solve a real-life problem for the expression.
$$\frac{3}{4}$$ × 20
Suppose there are 20 passengers in a bus and they occupied the place $$\frac{3}{4}$$ of the bus.
Hence,
The total number of seats in the bus = ( The fraction of the place occupied by the passengers ) × ( The total number of passengers )
= $$\frac{3}{4}$$ × 20
Now,
We know that,
$$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
= $$\frac{a × x}{b × 1}$$
So,
$$\frac{3}{4}$$ × 20 = $$\frac{3}{4}$$ × $$\frac{20}{1}$$
= $$\frac{3 × 20}{4 × 1}$$
= $$\frac{15}{1}$$
= 15 seats
Hence, from the above,
We can conclude that there are 15 seats in the bus.
Question 11.
YOU BE THE TEACHER
Descartes finds $$\frac{2}{3}$$ × 6. Is he correct? Explain.
No, Descartes is not correct
Explanation:
According to Descartes,
The given numbers are: $$\frac{2}{3}$$ and 6
So,
According to Descartes, 6 is divided into 3 equal parts
So,
6 ÷ 3 = 2 equal parts
So,
According to Descartes, these 2 equal parts also divided into 2 parts.
So,
The total number of parts = 2 equal parts + 2 equal parts = 4 parts
So,
The product of $$\frac{2}{3}$$ and 6 is 4
Now,
We know that,
$$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
= $$\frac{a × x}{b × 1}$$
So,
$$\frac{2}{3}$$ × 6 = $$\frac{2}{3}$$ × $$\frac{6}{1}$$
= $$\frac{6 × 2}{3 × 1}$$
= $$\frac{4}{1}$$
= 4
Hence, from the above,
We can conclude that Descartes is not correct.
Think and Grow: Modeling Real Life
Example
A recipe calls for 2 cups of rice. You only have $$\frac{3}{4}$$ of that amount. How much more rice do you need?
Find the number of cups of rice that you have by finding $$\frac{3}{4}$$ of 2.
So,
You have $$\frac{3}{2}$$ cups of rice.
Subtract the amount of rice you have from the amount of rice you need.
2 – $$\frac{3}{2}$$ = $$\frac{2}{1}$$ – $$\frac{3}{2}$$
= $$\frac{(2 × 2) – 3}{2}$$
= $$\frac{1}{2}$$
So,
you need $$\frac{1}{2}$$ cup more of rice.
Show and Grow
Question 12.
You have 12 tokens. You use $$\frac{3}{4}$$ of them to play a pinball game. How many tokens do you have left?
The number of tokens you have left is: 3 tokens
Explanation:
It is given that you have 12 tokens and you use $$\frac{3}{4}$$ of them to play a pinball game.
So,
The number of tokens used to play a pinball game = ( The fraction of tokens used to play a pinball game ) × ( The total number of tokens )
= $$\frac{3}{4}$$ × 12
= $$\frac{3}{4}$$ × $$\frac{12}{1}$$
= $$\frac{3 × 12}{4 × 1}$$
= $$\frac{9}{1}$$
= 9 tokens
So,
The number of tokens left = ( The total number of tokens ) – ( The number of tokens used to play the pinball )
= 12 – 9
= 3 tokens
Hence, from the above,
We can conclude that the number of tokens left are: 3 tokens
Question 13.
A male lion sleeps $$\frac{5}{6}$$ × 6 hours of each day. How many hours does the lion sleep in 1 week?
The number of hours the lion sleeps in 1 week is: 35 hours
Explanation:
It is given that a male lion sleeps $$\frac{5}{6}$$ × 6 hours of each day
So,
The number of hours the lion sleeps in 1 day = $$\frac{5}{6}$$ × 6 hours
= $$\frac{5}{6}$$ × $$\frac{6}{1}$$
= $$\frac{5 × 6}{6 × 1}$$
= $$\frac{5}{1}$$
= 5 hours
We know that,
1 week = 7 days
So,
The total number of hours the lion sleeps in 1 week = ( The number of hours the lion sleeps in 1 day ) × ( The number od fays in 1 week )
= 5 × 7
= 35 hours
Hence, from the above,
We can conclude that the lion sleeps for 35 hours in 1 week
Question 14.
DIG DEEPER!
In a class of 20 students, $$\frac{1}{10}$$ of the students are 10 years old, $$\frac{4}{5}$$of the students are 11 years old, and the rest are 12 years old. How many more 11-year-olds than 12-year-olds are in the class?
The number of 11-year-olds more than 12-year-olds is: 14
Explanation:
It is given that in a class of 20 students, $$\frac{1}{10}$$ of the students are 10 years old, $$\frac{4}{5}$$of the students are 11 years old, and the rest are 12 years old.
So,
The number of 10-year-olds = ( The fraction of 10-year-olds ) × ( The total number of students )
= $$\frac{1}{10}$$ × 20
= $$\frac{1 × 20}{10 × 1}$$
= 2 10-year-olds
The number of 11-year-olds = ( The fraction of 11-year-olds ) × ( The total number of students )
= $$\frac{4}{5}$$ × 20
= $$\frac{4}{5}$$ × $$\frac{20}{1}$$
= $$\frac{4 × 20}{5 × 1}$$
= $$\frac{16}{1}$$
= 16 11-year-olds
So,
The number of 12-year-olds = ( The total number of students ) – ( The number of 10-year-olds + The number of 11-year-olds )
= 20 – ( 16 + 2 )
= 2 12-year-olds
So,
The number of 11-year-olds more than 12-year-olds = ( The number of 11-year-olds ) – ( The numebr of 12-year-olds )
= 16 – 2
= 14 students
Hence, from the above,
We can conclude that there are 14 students who are 11-years-old more than 12-years-old.
### Use Models to Multiply Fractions by Whole Numbers Homework & Practice 9.2
Multiply. Use a model to help.
Question 1.
$$\frac{2}{3}$$ × 6
$$\frac{2}{3}$$ × 6 = 4
Explanation:
The given numbers are: $$\frac{2}{3}$$ and 6
We know that,
$$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
= $$\frac{a × x}{b × 1}$$
So,
$$\frac{2}{3}$$ × 6 = $$\frac{2}{3}$$ × $$\frac{6}{1}$$
= $$\frac{2 × 6}{3 × 1}$$
= $$\frac{4}{1}$$
= 4
Hence,
$$\frac{2}{3}$$ × 6 = 4
Question 2.
$$\frac{3}{5}$$ of 10
$$\frac{3}{5}$$ × 10 = 6
Explanation:
The given numbers are: $$\frac{3}{5}$$ and 10
We know that,
$$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
= $$\frac{a × x}{b × 1}$$
So,
$$\frac{3}{5}$$ × 10 = $$\frac{3}{5}$$ × $$\frac{10}{1}$$
= $$\frac{3 × 10}{5 × 1}$$
= $$\frac{6}{1}$$
= 6
Hence,
$$\frac{3}{5}$$ × 10 = 6
Question 3.
$$\frac{1}{2}$$ of 4
$$\frac{1}{2}$$ × 4 = 2
Explanation:
The given numbers are: $$\frac{1}{2}$$ and 4
We know that,
$$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
= $$\frac{a × x}{b × 1}$$
So,
$$\frac{1}{2}$$ × 4 = $$\frac{1}{2}$$ × $$\frac{4}{1}$$
= $$\frac{1 × 4}{2 × 1}$$
= $$\frac{2}{1}$$
= 2
Hence,
$$\frac{1}{2}$$ × 4 = 2
Question 4.
$$\frac{1}{4}$$ × 12
$$\frac{1}{4}$$ × 12 = 3
Explanation:
The given numbers are: $$\frac{1}{4}$$ and 12
We know that,
$$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
= $$\frac{a × x}{b × 1}$$
So,
$$\frac{1}{4}$$ × 12 = $$\frac{1}{4}$$ × $$\frac{12}{1}$$
= $$\frac{1 × 12}{4 × 1}$$
= $$\frac{3}{1}$$
= 3
Hence,
$$\frac{1}{12}$$ × 4 = 3
Question 5.
$$\frac{5}{6}$$ × 3
$$\frac{5}{6}$$ × 3 = $$\frac{5}{2}$$
Explanation:
The given numbers are: $$\frac{5}{6}$$ and 3
We know that,
$$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
= $$\frac{a × x}{b × 1}$$
So,
$$\frac{5}{6}$$ × 3 = $$\frac{5}{6}$$ × $$\frac{3}{1}$$
= $$\frac{5 × 3}{6 × 1}$$
= $$\frac{5}{2}$$
Hence,
$$\frac{5}{6}$$ × 3 = $$\frac{5}{2}$$
Question 6.
$$\frac{3}{4}$$ of 2
$$\frac{3}{4}$$ × 2 = $$\frac{3}{2}$$
Explanation:
The given numbers are: $$\frac{3}{4}$$ and 2
We know that,
$$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
= $$\frac{a × x}{b × 1}$$
So,
$$\frac{3}{4}$$ × 2 = $$\frac{3}{4}$$ × $$\frac{2}{1}$$
= $$\frac{3 × 2}{4 × 1}$$
= $$\frac{3}{2}$$
Hence,
$$\frac{3}{4}$$ × 2 = $$\frac{3}{2}$$
Question 7.
You have 27 foam balls. You use $$\frac{1}{3}$$ of the balls for a model. How many balls do you use?
The number of balls you used is: 9 balls
Explanation:
It is given that there are 27 foam balls and you use $$\frac{1}{3}$$ of the balls for a model.
So,
The number of balls you used = ( The fraction of balls you used ) × ( The total number of balls )
= $$\frac{1}{3}$$ × 27
Now,
We know that,
$$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
= $$\frac{a × x}{b × 1}$$
So,
$$\frac{1}{3}$$ × 27 = $$\frac{1}{3}$$ × $$\frac{27}{1}$$
= $$\frac{1 × 27}{3 × 1}$$
= $$\frac{9}{1}$$
= 9 balls
Hence, from the above,
We can conclude that the number of foam balls you used are: 27
Question 8.
An object that weighs 1 pound on Earth weighs about $$\frac{1}{15}$$ pound on Pluto. A man weighs 240 pounds on Earth. How many pounds does he weigh on Pluto?
The number of pounds the man weigh in Pluto is: 16 pounds
Explanation:
It is given that an object that weighs 1 pound on Earth weighs about $$\frac{1}{15}$$ pound on Pluto.
It is also given that a man weighs 240 pounds on Earth.
So,
The weight of a man on pluto = ( The fraction of weight of a man on pluto when compared to earth ) × ( The weight of the man on earth )
= $$\frac{1}{15}$$ × 240
Now,
We know that,
$$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
= $$\frac{a × x}{b × 1}$$
So,
$$\frac{1}{15}$$ × 240 = $$\frac{1}{15}$$ × $$\frac{240}{1}$$
= $$\frac{1 × 240}{15 × 1}$$
= $$\frac{16}{1}$$
= 16 pounds
Hence, from the above,
We can conclude that the weight of the man on pluto is: 16 pounds
Question 9.
Structure
Write a multiplication equation represented by the model.
Question 10.
DIG DEEPER!
Find each missing number.
Let the expression be named A), B), C), and D)
So,
The missing number of A is: 12
The missing number of B is: 6
The missing number of C is: 2
The missing number of D is: 5
Explanation:
Let the expressions be named as A), B), C), and D)
So,
A) The given fractions are: $$\frac{1}{3}$$ and 4
Let the missing number be X
So,
X × $$\frac{1}{3}$$ = 4
When multiplication goes from left to right or from right to left, it will become the division
When division goes from left to right or from right to left, it will become multiplication.
So,
X = 4 ÷ $$\frac{1}{3}$$
We know that,
$$\frac{a}{b}$$ ÷ $$\frac{x}{y}$$ = $$\frac{a}{b}$$ × $$\frac{y}{x}$$
So,
$$\frac{4}{1}$$ ÷ $$\frac{1}{3}$$
= $$\frac{4}{1}$$ × $$\frac{3}{1}$$
= $$\frac{4 × 3}{1 × 1}$$
= 12
Hence, the missing number is 12
B) The given numbers are: $$\frac{2}{5}$$ and 15
We know that,
$$\frac{a}{b}$$ × x = $$\frac{a}{b}$$ × $$\frac{x}{1}$$
= $$\frac{a × x}{b × 1}$$
So,
$$\frac{2}{5}$$ × 15 = $$\frac{2}{5}$$ × $$\frac{15}{1}$$
= $$\frac{2 × 15}{5 × 1}$$
= $$\frac{6}{1}$$
= 6
Hence,
$$\frac{2}{5}$$ × 15 = 6
Like the above two expressions, the remaining two expressions can also be solved.
hence, from the above,
We can conclude that
The missing number of A is: 12
The missing number of B is: 6
The missing number of C is: 2
The missing number of D is: 5
Question 11.
Modeling Real Life
You have 28 craft sticks. You use $$\frac{4}{7}$$ of them for a project. How many craft sticks do you have left?
The number of craft sticks you have left is: 12 craft sticks
Explanation:
It is given that you have 28 craft sticks and you used $$\frac{4}{7}$$ of them for a project.
So,
The number of craft sticks you used = ( The fraction of craft sticks you used ) × ( The total number of craft sticks )
= $$\frac{4}{7}$$ × 28
= $$\frac{4}{7}$$ × $$\frac{28}{1}$$
= $$\frac{4 × 28}{7 × 1}$$
= $$\frac{16}{1}$$
= 16
So,
The number of craft sticks you left = ( The total number of craft sticks ) – ( The number of craft sticks you used )
= 28 – 16
= 12 craft sticks
Hence, from the above,
We can conclude that the number of craft sticks you left is: 12 craft sticks
Question 12.
Modeling Real Life
A mother otter spends $$\frac{1}{3}$$ of each day feeding her baby. How many hours does the mother otter spend feeding her baby in 1 week?
The number of hours the mother otter spend feeding her baby in 1 week is: 56 hours
Explanation:
It is given that a mother otter spends $$\frac{1}{3}$$ of each day feeding her baby.
We know that,
1 day = 24 hours
So,
The number of hours the mother otter spends feeding her baby in 1 day = ( The fraction of time the mother otter spends in feeding milk ) × 24
= $$\frac{1}{3}$$ × 24
= $$\frac{1}{3}$$ × $$\frac{24}{1}$$
= $$\frac{1 × 24}{3 × 1}$$
= $$\frac{8}{1}$$
= 8 hours
Now,
We know that,
1 week = 7 days
So,
The number of hours the mother otter spends feeding her milk in 1 week = ( The number of hours the mother otter feeding her milk in 1 day ) × 7
= 8 × 7
= 56 hours
Hence, from the above,
We can conclude that the mother otter spends feeding her milk for 56 hours in 1 week
Review & Refresh
Estimate the quotient.
Question 13.
5,692 ÷ 5
5,692 ÷ 5 =1,138 R 2
Explanation:
By using the partial quotients method,
5,692 ÷ 5 = ( 5,000 + 600 + 90 ) ÷ 5
= ( 5,000 ÷ 5 ) + ( 600 ÷ 5 ) + ( 90 ÷ 5 )
= 1,000 + 120 + 18
= 1,138 R 2
Hence, 5,692 ÷ 5 = 1,138 R 2
Question 14.
309 ÷ 12
309 ÷ 12 = 25 R 9
Explanation:
By using the partial quotients method,
309 ÷ 12 = ( 240 + 60 ) ÷ 12
= ( 240 ÷ 12 ) + ( 60 ÷ 12 )
= 20 + 5
= 25 R 9
Hence, 309 ÷ 12 = 25 R 2
Question 15.
2,987 ÷ 53
2,987 ÷ 53 = 2,987 ÷ ( 50 + 3 )
= ( 2,985 ÷ 50 ) + ( 2,985 ÷ 3 )
= 995 R 2 + 59.7
= 1,054.7 R 2
### Lesson 9.3 Multiply Fractions and Whole Numbers
Explore and Grow
Use models to help you complete the table. What do you notice about each expression and its product?
The completed table is:
From the completed table,
We can observe the multiplication of the whole number and the fraction.
The product of a whole number and a fraction may be a whole number or a fraction.
Construct Arguments
Explain how to multiply fractions and whole numbers without using models.
We can multiply the whole numbers and fractions by using the properties of the multiplication. They are:
A) $$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
B) p = $$\frac{p}{1}$$
Think and Grow: Multiply Fractions and Whole Numbers
Key Idea
You can find the product of a fraction and a whole number by multiplying the numerator and the whole number. Then write the result over the denominator.
Example
Find 2 × $$\frac{5}{6}$$.
Multiply the numerator and the whole number.
Example
Find $$\frac{5}{6}$$ × 2
Multiply the numerator and the whole number.
Show and Grow
Multiply.
Question 1.
3 × $$\frac{5}{8}$$ = _______
3 × $$\frac{5}{8}$$ = $$\frac{15}{8}$$
Explanation:
The given numbers are: 3 and $$\frac{5}{8}$$
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
3 × $$\frac{5}{8}$$ = $$\frac{3}{1}$$ × $$\frac{5}{8}$$
= $$\frac{5 × 3}{8 × 1}$$
= $$\frac{15}{8}$$
Hence,
3 × $$\frac{5}{8}$$ = $$\frac{15}{8}$$
Question 2.
6 × $$\frac{4}{9}$$ = _______
6 × $$\frac{4}{9}$$ = $$\frac{8}{3}$$
Explanation:
The given numbers are: 6 and $$\frac{4}{9}$$
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
6 × $$\frac{4}{9}$$ = $$\frac{6}{1}$$ × $$\frac{4}{9}$$
= $$\frac{6 × 4}{9 × 1}$$
= $$\frac{24}{9}$$
= $$\frac{8}{3}$$
Hence,
6 × $$\frac{4}{9}$$ = $$\frac{8}{3}$$
Question 3.
$$\frac{2}{5}$$ × 15 = _______
15 × $$\frac{2}{5}$$ = 6
Explanation:
The given numbers are: 15 and $$\frac{2}{5}$$
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
15× $$\frac{2}{5}$$ = $$\frac{15}{1}$$ × $$\frac{2}{5}$$
= $$\frac{15 × 2}{5 × 1}$$
= $$\frac{6}{1}$$
= 6
Hence,
15 × $$\frac{2}{5}$$ = 6
Apply and Grow: Practice
Multiply.
Question 4.
$$\frac{3}{5}$$ × 2 = _______
2 × $$\frac{3}{5}$$ = $$\frac{6}{5}$$
Explanation:
The given numbers are: 2 and $$\frac{3}{5}$$
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
2 × $$\frac{3}{5}$$ = $$\frac{2}{1}$$ × $$\frac{3}{5}$$
= $$\frac{2 × 3}{5 × 1}$$
= $$\frac{6}{5}$$
Hence,
2 × $$\frac{3}{5}$$ = $$\frac{6}{5}$$
Question 5.
5 × $$\frac{2}{9}$$ = _______
5 × $$\frac{2}{9}$$ = $$\frac{10}{9}$$
Explanation:
The given numbers are: 5 and $$\frac{2}{9}$$
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
5 × $$\frac{2}{9}$$ = $$\frac{5}{1}$$ × $$\frac{2}{9}$$
= $$\frac{5 × 2}{9 × 1}$$
= $$\frac{10}{9}$$
Hence,
5 × $$\frac{2}{9}$$ = $$\frac{10}{9}$$
Question 6.
$$\frac{5}{6}$$ × 4 = _______
4 × $$\frac{5}{6}$$ = $$\frac{10}{3}$$
Explanation:
The given numbers are: 4 and $$\frac{5}{6}$$
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
4 × $$\frac{5}{6}$$ = $$\frac{4}{1}$$ × $$\frac{5}{6}$$
= $$\frac{5 × 4}{6 × 1}$$
= $$\frac{20}{6}$$
= $$\frac{10}{3}$$
Hence,
4 × $$\frac{5}{6}$$ = $$\frac{10}{3}$$
Question 7.
8 × $$\frac{3}{10}$$ = _______
8 × $$\frac{3}{10}$$ = $$\frac{12}{5}$$
Explanation:
The given numbers are: 8 and $$\frac{3}{10}$$
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
8 × $$\frac{3}{10}$$ = $$\frac{8}{1}$$ × $$\frac{3}{10}$$
= $$\frac{8 × 3}{10 × 1}$$
= $$\frac{24}{10}$$
= $$\frac{12}{5}$$
Hence,
8 × $$\frac{3}{10}$$ = $$\frac{12}{5}$$
Question 8.
$$\frac{1}{5}$$ × 7 = _______
7 × $$\frac{1}{5}$$ = $$\frac{7}{5}$$
Explanation:
The given numbers are: 7 and $$\frac{1}{5}$$
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
7 × $$\frac{1}{5}$$ = $$\frac{7}{1}$$ × $$\frac{1}{5}$$
= $$\frac{7 × 1}{5 × 1}$$
= $$\frac{7}{5}$$
Hence,
7 × $$\frac{1}{5}$$ = $$\frac{7}{5}$$
Question 9.
9 × $$\frac{5}{12}$$ = _______
9 × $$\frac{5}{12}$$ = $$\frac{15}{4}$$
Explanation:
The given numbers are: 9 and $$\frac{5}{12}$$
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
9 × $$\frac{5}{12}$$ = $$\frac{9}{1}$$ × $$\frac{5}{12}$$
= $$\frac{9 × 5}{12 × 1}$$
= $$\frac{45}{12}$$
= $$\frac{15}{4}$$
Hence,
9 × $$\frac{5}{12}$$ = $$\frac{15}{4}$$
Question 10.
15 × $$\frac{5}{8}$$ = _______
15 × $$\frac{5}{8}$$ = $$\frac{75}{8}$$
Explanation:
The given numbers are: 15 and $$\frac{5}{8}$$
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
15 × $$\frac{5}{8}$$ = $$\frac{15}{1}$$ × $$\frac{5}{8}$$
= $$\frac{15 × 5}{8 × 1}$$
= $$\frac{75}{8}$$
Hence,
15 × $$\frac{5}{8}$$ = $$\frac{75}{8}$$
Question 11.
$$\frac{3}{4}$$ × 20 = _______
20 × $$\frac{3}{4}$$ = 15
Explanation:
The given numbers are: 20 and $$\frac{3}{4}$$
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
20 × $$\frac{3}{4}$$ = $$\frac{20}{1}$$ × $$\frac{3}{4}$$
= $$\frac{20 × 3}{4 × 1}$$
= $$\frac{60}{4}$$
= $$\frac{15}{1}$$
= 15
Hence,
20 × $$\frac{3}{4}$$ = 15
Question 12.
$$\frac{7}{9}$$ × 5 = _______
5 × $$\frac{7}{9}$$ = $$\frac{35}{9}$$
Explanation:
The given numbers are: 5 and $$\frac{7}{9}$$
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
5 × $$\frac{7}{9}$$ = $$\frac{5}{1}$$ × $$\frac{7}{9}$$
= $$\frac{5 × 7}{9 × 1}$$
= $$\frac{35}{9}$$
Hence,
5 × $$\frac{7}{9}$$ = $$\frac{35}{9}$$
Question 13.
One-tenth of the 50 states in the United States of America have a mockingbird as their state bird. How many states have a mockingbird as their state bird?
The number of states that have mockingbird as their state bird is: 5 states
Explanation:
It is given that one-tenth of the 50 states in the United States of America have a mockingbird as their state bird
So,
The number of states that have a mockingbird as their state bird = ( The fraction of the states that have a mockingbird as their state bird ) × ( The total number of states in the United States of America )
= $$\frac{1}{10}$$ × 50
Now,
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
50 × $$\frac{1}{10}$$ = $$\frac{50}{1}$$ × $$\frac{1}{10}$$
= $$\frac{50 × 1}{10 × 1}$$
= $$\frac{50}{10}$$
= $$\frac{5}{1}$$
= 5 states
Hence, from the above,
We can conclude that there are 5 states that have mockingbird as their state bird
Question 14.
Writing
Explain why 9 × $$\frac{2}{3}$$ is equivalent to $$\frac{2}{3}$$ × 9.
By using the Commutative property of multiplication,
a × b = b × a
So,
By using the above property,
In 9 × $$\frac{2}{3}$$,
‘a’ is: 9
‘b’ is: $$\frac{2}{3}$$
Hence, from the above,
We can conclude that 9 × “$$\frac{2}{3}$$” is equivalent to “$$\frac{2}{3}$$ × 9″ by using the Commutative property of multiplication.
Question 15.
Reasoning
Without calculating, determine which product is greater. Explain.
$$\frac{1}{8}$$ × 24
$$\frac{7}{8}$$ × 24
The product of “$$\frac{7}{8}$$ × 24″ is greater than “$$\frac{1}{8}$$ × 24″
Explanation:
The given products are: $$\frac{1}{8}$$ × 24 and $$\frac{7}{8}$$ × 24
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So, from the above fractions,
We can observe that except numerator, the denominator and the whole number is all the same.
So,
To find which product is greater, we just have to compare the numerators of the two fractions without calculating the value of the product.
In the comparison of the numerators,
We can observe that,
1 < 7
So,
First fraction numerator < Second fraction numerator
Hence, from the above,
We can conclude that the product of “$$\frac{7}{8}$$ × 24″ is greater than “$$\frac{1}{8}$$ × 24″
Think and Grow: Modeling Real Life
Example
Newton buys 27 songs. Two-thirds of them are classical songs. Descartes buys 16 songs. Seven-eighths of them are classical songs. Who buys more classical songs? How many more?
Multiply $$\frac{2}{3}$$ by 27 to find the number of classical songs Newton buys. Multiply $$\frac{7}{8}$$ by 16 to find the number Descartes buys.
So, Newton buys more classical songs.
Subtract the products to find how many more.
18 – 14 = 4
Hence,
Newton buys 4 more classical songs than Descartes.
Show and Grow
Question 16.
You take 48 pictures on a walking tour. Five-twelfths of them is of buildings. Your friend takes 45 pictures. Six-fifteenths of them are of buildings. Who takes more pictures of buildings? How many more?
You take more pictures
You take 2 pictures more than your friend
Explanation:
It is given that you take 48 pictures on a walking tour and five-twelfths if them is of buildings.
So,
The number of buildings taken by you = ( The total number of pictures taken by you ) × ( The fraction of pictures that is of buildings )
= 48 × $$\frac{5}{12}$$
= $$\frac{48}{1}$$ × $$\frac{5}{12}$$
= $$\frac{48 × 5}{1 × 12}$$
= 20 pictures
It is also given that your friend taken 45 pictures and six-fifteenth of them is buildings
So,
The number of buildings taken by your friend = ( The total number of pictures taken by your friend ) × ( The fraction of pictures that is of buildings )
= 45 × $$\frac{6}{15}$$
= $$\frac{45}{1}$$ × $$\frac{6}{15}$$
= $$\frac{45 × 6}{1 × 15}$$
= 18 pictures
So,
In the comparison of the pictures of buildings,
You take more pictures than your friend.
Now,
The number of pictures more taken by you than your friend = ( The number of pictures taken by you ) – ( The number of pictures taken by your friend )
= 20 – 18
= 2 pictures
Hence, from the above,
We can conclude that
You take more pictures
You take 2 pictures more than your friend
Question 17.
You have 72 rocks in your rock collection. Five-eighths of them are sedimentary, one-sixth of them are igneous, and the rest are metamorphic. How many of your rocks are metamorphic?
The number of rocks that are metamorphic is: 15 metamorphous rocks
Explanation:
It is given that you have 72 rocks in your rock collection and five-eighths of them are sedimentary, one-sixth of them are igneous, and the rest are metamorphic.
So,
The number of rocks that are sedimentary = ( The total number of rocks ) × ( The fraction of rocks that are sedimentary )
= 72 × $$\frac{5}{8}$$
= $$\frac{72}{1}$$ × $$\frac{5}{8}$$
= $$\frac{72 × 5}{1 × 8}$$
= 45 sedimentary rocks
The number of rocks that are igneous = ( The total number of rocks ) × ( The fraction of rocks that are igneous )
= 72 × $$\frac{1}{6}$$
= $$\frac{72}{1}$$ × $$\frac{1}{6}$$
= $$\frac{72 × 1}{1 × 6}$$
= 12 igneous rocks
So,
The number of metamorphous rocks = ( The total number of rocks ) – ( The number of sedimentary rocks + The number of igneous rocks)
= 72 – ( 45 + 12 )
= 15 metamorphous rocks
Hence, from the above,
We can conclude that there are 15 metamorphous rocks
Question 18.
DIG DEEPER!
Each day, you spend $$\frac{3}{4}$$ hour reading and $$\frac{1}{2}$$ hour writing in a journal. How many total hours do you spend reading and writing in 1 week? Describe two ways to solve the problem.
The number of hours you spend on reading and writing in 1 week is:
Explanation:
It is given that, each day you spent $$\frac{3}{4}$$ hour reading and $$\frac{1}{2}$$ hour writing in a journal.
So,
The number of hours you spent on reading and writing a journal in 1 day = ( The number of hours spent on reading ) + ( The number of hours spent on writing )
= $$\frac{3}{4}$$ + $$\frac{1}{2}$$
Multiply $$\frac{1}{2}$$ with $$\frac{2}{2}$$
So,
$$\frac{3}{4}$$ + $$\frac{2}{4}$$
= $$\frac{3 + 2}{4}$$
= $$\frac{5}{4}$$ hours
We know that,
1 day = 24 hours
1 week = 7 days
So,
1 week = 7 × 24 hours
So,
The number of hours you spent on reading and writing in 1 week = ( The number of hours you spent on reading and writing in 1 day ) × ( The number of hours in 1 week )
= $$\frac{5}{4}$$ × 7 × 24
= $$\frac{5}{4}$$ × $$\frac{7}{1}$$ × $$\frac{24}{1}$$
= $$\frac{5 × 7 × 24}{4 × 1}$$
= 5 × 7 × 6
= 210 hours
Hence, from the above,
We can conclude that the number of hours you spent on reading and writing a journal in 1 week is: 210 hours
### Multiply Fractions and Whole Numbers Homework & Practice 9.3
Multiply.
Question 1.
$$\frac{5}{6}$$ × 3 = _______
3 × $$\frac{5}{6}$$ = $$\frac{5}{2}$$
Explanation:
The given numbers are: 3 and $$\frac{5}{6}$$
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
3 × $$\frac{5}{6}$$ = $$\frac{3}{1}$$ × $$\frac{5}{6}$$
= $$\frac{3 × 5}{6 × 1}$$
= $$\frac{5}{6}$$
= $$\frac{5}{2}$$
Hence,
3 × $$\frac{5}{6}$$ = $$\frac{5}{2}$$
Question 2.
$$\frac{2}{3}$$ × 6 = _______
6 × $$\frac{2}{3}$$ = 4
Explanation:
The given numbers are: 6 and $$\frac{2}{3}$$
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
6 × $$\frac{2}{3}$$ = $$\frac{6}{1}$$ × $$\frac{2}{3}$$
= $$\frac{6 × 2}{3 × 1}$$
= $$\frac{12}{3}$$
= $$\frac{4}{1}$$
= 4
Hence,
6 × $$\frac{2}{3}$$ = 4
Question 3.
7 × $$\frac{1}{8}$$ = _______
7 × $$\frac{1}{8}$$ = $$\frac{7}{8}$$
Explanation:
The given numbers are: 7 and $$\frac{1}{8}$$
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
7 × $$\frac{1}{8}$$ = $$\frac{7}{1}$$ × $$\frac{1}{8}$$
= $$\frac{7 × 1}{8 × 1}$$
= $$\frac{7}{8}$$
Hence,
7 × $$\frac{1}{8}$$ = $$\frac{7}{8}$$
Question 4.
2 × $$\frac{1}{2}$$ = _______
2 × $$\frac{1}{2}$$ = 1
Explanation:
The given numbers are: 2 and $$\frac{1}{2}$$
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
2 × $$\frac{1}{2}$$ = $$\frac{2}{1}$$ × $$\frac{1}{2}$$
= $$\frac{2 × 1}{2 × 1}$$
= $$\frac{2}{2}$$
= 1
Hence,
2 × $$\frac{1}{2}$$ = 1
Question 5.
$$\frac{4}{5}$$ × 9 = _______
9 × $$\frac{4}{5}$$ = $$\frac{36}{5}$$
Explanation:
The given numbers are: 9 and $$\frac{4}{5}$$
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
9 × $$\frac{4}{5}$$ = $$\frac{9}{1}$$ × $$\frac{4}{5}$$
= $$\frac{9 × 4}{5 × 1}$$
= $$\frac{36}{5}$$
Hence,
9 × $$\frac{4}{5}$$ = $$\frac{36}{5}$$
Question 6.
4 × $$\frac{5}{12}$$ = _______
4 × $$\frac{5}{12}$$ = $$\frac{5}{3}$$
Explanation:
The given numbers are: 4 and $$\frac{5}{12}$$
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
4 × $$\frac{5}{12}$$ = $$\frac{4}{1}$$ × $$\frac{5}{12}$$
= $$\frac{4 × 5}{12 × 1}$$
= $$\frac{20}{12}$$
= $$\frac{5}{3}$$
Hence,
4 × $$\frac{5}{12}$$ = $$\frac{5}{3}$$
Question 7.
$$\frac{1}{4}$$ × 24 = _______
24 × $$\frac{1}{24}$$ = 6
Explanation:
The given numbers are: 24 and $$\frac{1}{4}$$
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
24 × $$\frac{1}{4}$$ = $$\frac{24}{1}$$ × $$\frac{1}{4}$$
= $$\frac{24 × 1}{4 × 1}$$
= $$\frac{24}{4}$$
= 6
Hence,
24 × $$\frac{1}{4}$$ = 6
Question 8.
16 × $$\frac{3}{8}$$ = _______
16 × $$\frac{3}{8}$$ = 6
Explanation:
The given numbers are: 16 and $$\frac{3}{8}$$
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
16 × $$\frac{3}{8}$$ = $$\frac{16}{1}$$ × $$\frac{3}{8}$$
= $$\frac{16 × 3}{8 × 1}$$
= $$\frac{48}{8}$$
= 6
Hence,
16 × $$\frac{3}{8}$$ = 6
Question 9.
$$\frac{7}{10}$$ × 25 = _______
25 × $$\frac{7}{10}$$ = $$\frac{35}{2}$$
Explanation:
The given numbers are: 25 and $$\frac{7}{10}$$
We know that,
$$\frac{a}{b}$$ × p = $$\frac{a}{b}$$ × $$\frac{p}{1}$$
= $$\frac{a × p}{ b × 1}$$
p = $$\frac{p}{1}$$
So,
25 × $$\frac{7}{10}$$ = $$\frac{25}{1}$$ × $$\frac{7}{10}$$
= $$\frac{7 × 25}{10 × 1}$$
= $$\frac{175}{10}$$
= $$\frac{35}{2}$$
Hence,
25 × $$\frac{7}{10}$$ = $$\frac{35}{2}$$
Question 10.
You spend $$\frac{3}{4}$$ hour jumping rope every week for 8 weeks. How many hours do you jump rope altogether?
The number of hours you jump rope altogether in 8 weeks is: 6 hours
Explanation:
It is given that you spent $$\frac{3}{4}$$ hour jumping rope every week for 8 weeks.
So,
The number of hours you spent jumping rope in 8 weeks = ( The number of hours you jump rope in 1 week ) × ( The total number of weeks )
= $$\frac{3}{4}$$ × 8
= $$\frac{3}{4}$$ × $$\frac{8}{1}$$
= $$\frac{3 × 8}{4 × 1}$$
= 6 hours
hence, from the above,
We can conclude that you spent 6 hours on jumping rope in 8 weeks.
Question 11.
Logic
Your friend finds 25 items that are either insects or flowers. She says that $$\frac{1}{6}$$ of the items are insects. Can this be true? Explain.
No, this is not true
Explanation:
It is given that your friend finds 25 items that are either insects or flowers and she says that $$\frac{1}{6}$$ of the items are insects.
But,
The number of either insects or flowers will always be a whole number.
But, from the given information,
25 will not be divided by $$\frac{1}{6}$$
So,
The statement “$$\frac{1}{6}$$ of the items are insects” is false.
Question 12.
Open-Ended
Write two different pairs of fractions that could represent the insects and flowers your friend finds in Exercise 11.
The different pairs of fractions that could represent the insects and flowers are:
$$\frac{1}{5}$$ and $$\frac{1}{25}$$
Explanation:
These different pairs of fractions have to divide 25
So,
We have to take the fractions the multiples of 5 i.e., 5 and 25
Hence, from the above,
We can conclude that
The different pairs of fractions that could represent the insects and flowers are:
$$\frac{1}{5}$$ and $$\frac{1}{25}$$
Question 13.
Modeling Real Life
Newton bakes 56 treats. Five-eighths of them contains peanut butter. Descartes bakes 120 treats. Five-sixths of them contain peanut butter. Who bakes more peanut butter treats? How many more?
Descartes bakes more peanut butter treats
The number of peanut butter treats Descartes made more than Descartes is: 65
Explanation:
It is given that Newton bakes 56 treats and five-eights of them contains peanut butter
So,
The number of peanut butter treats made by Newton = $$\frac{5}{8}$$ × 56
= $$\frac{5}{8}$$ × $$\frac{56}{1}$$
= $$\frac{5 × 56}{8 × 1}$$
= 35 peanut butter treats
It is also given that Descartes bakes 120 treats and five-sixths of them contain peanut butter
So,
the number of peanut butter treats made by Descartes = $$\frac{5}{6}$$ × 120
= $$\frac{5}{6}$$ × $$\frac{120}{1}$$
= $$\frac{5 × 120}{6 × 1}$$
= 100 peanut butter treats
So,
In the comparison of peanut butter treats,
Descartes bakes more
Now,
The number of peanut butter treats baked by Descartes more than Newton = 100 – 35
= 65 peanut butter treats
Hence, from the above,
We can conclude that
Descartes bakes more peanut butter treats
The number of peanut butter treats Descartes made more than Descartes is: 65
Question 14.
Modeling Real Life
Your class conducts an egg-dropping experiment with 60 eggs. Three-fifths of the eggs break open, one-sixth of the eggs crack, and the rest do not break at all. How many of the eggs do not crack or break open?
The number of eggs that do not break is:
Explanation:
It is given that your class conducts an egg-dropping experiment with 60 eggs and three-fifths of the eggs break open, one-sixth of the eggs crack and the rest do not break.
So,
The number of eggs that break open = 60 × $$\frac{3}{5}$$
= $$\frac{60}{1}$$ × $$\frac{3}{5}$$
= 36
The number of eggs that cracked = 60 × $$\frac{1}{6}$$
= $$\frac{60}{1}$$ × $$\frac{1}{6}$$
= 10
So,
The number of eggs that do not breaked = ( The total number of eggs ) – ( The number of eggs that break open + The number of eggs that cracked )
= 60 – ( 36 + 10 )
= 16
Hence, from the above,
We can conclude that the number of eggs that do not break is: 16 eggs
Review & Refresh
Question 15.
5$$\frac{5}{8}$$ + 6$$\frac{3}{4}$$ = _______
5$$\frac{5}{8}$$ + 6$$\frac{3}{4}$$ = $$\frac{99}{8}$$
Explanation:
The give mixed fractions are: 6$$\frac{3}{4}$$ and 5$$\frac{5}{8}$$
The representation of mixed fractions in the improper form is: $$\frac{45}{8}$$ and $$\frac{27}{4}$$
We have to make the denominators equal.
So,
Multiply $$\frac{27}{4}$$ with $$\frac{2}{2}$$
So,
$$\frac{45}{8}$$ + $$\frac{54}{8}$$ = $$\frac{45 + 54}{8}$$
= $$\frac{99}{8}$$
Hence,
5$$\frac{5}{8}$$ + 6$$\frac{3}{4}$$ = $$\frac{99}{8}$$
Question 16.
1$$\frac{5}{6}$$ + 8$$\frac{1}{12}$$ = ________
1$$\frac{5}{6}$$ + 8$$\frac{1}{12}$$ = $$\frac{119}{12}$$
Explanation:
The give mixed fractions are: 1$$\frac{5}{6}$$ and 8$$\frac{1}{12}$$
The representation of mixed fractions in the improper form is: $$\frac{11}{6}$$ and $$\frac{97}{12}$$
We have to make the denominators equal.
So,
Multiply $$\frac{11}{6}$$ with $$\frac{2}{2}$$
So,
$$\frac{97}{12}$$ + $$\frac{22}{12}$$ = $$\frac{97 + 22}{12}$$
= $$\frac{119}{12}$$
Hence,
1$$\frac{5}{6}$$ + 8$$\frac{1}{12}$$ = $$\frac{119}{12}$$
Question 17.
3$$\frac{1}{2}$$ + $$\frac{3}{5}$$ + 2$$\frac{7}{10}$$ = _______
3$$\frac{1}{2}$$ + 2$$\frac{7}{10}$$ + $$\frac{3}{5}$$ = $$\frac{68}{10}$$
Explanation:
The give mixed fractions are: 3$$\frac{1}{2}$$, $$\frac{3}{5}$$ and 2$$\frac{7}{10}$$
The representation of mixed fractions in the improper form is: $$\frac{7}{2}$$ and $$\frac{27}{10}$$
We have to make the denominators equal.
So,
Multiply $$\frac{7}{2}$$ with $$\frac{5}{5}$$
Multiply $$\frac{3}{5}$$ with $$\frac{2}{2}$$
So,
$$\frac{27}{10}$$ + $$\frac{35}{10}$$ + $$\frac{6}{10}$$ = $$\frac{27 + 35 + 6}{10}$$
= $$\frac{68}{10}$$
Hence,
3$$\frac{1}{2}$$ + 2$$\frac{7}{10}$$ + $$\frac{3}{5}$$ = $$\frac{68}{10}$$
### Lesson 9.4 Use Models to Multiply Fractions
Explore and Grow
Fold a sheet of paper in half. Shade $$\frac{1}{4}$$ of either half. What fraction of the entire sheet of paper did you shade? Draw a model to support your answer.
The fraction of the entire sheet of paper you shaded is: $$\frac{1}{8}$$
Explanation:
Take a full sheet of paper and fold in half
So,
The number of parts of full sheet of paper = $$\frac{1}{2}$$
Now,
Shade $$\frac{1}{4}$$ of the half of the paper
So,,
The fraction of the paper you shaded = $$\frac{1}{4}$$ × $$\frac{1}{2}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$ = $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{1}{4}$$ × $$\frac{1}{2}$$
= $$\frac{1 × 1}{4 × 2}$$
= $$\frac{1}{8}$$
Hence, from the above,
We can conclude that the $$\frac{1}{8}$$ part of the full sheet is shaded
Reasoning
The multiplication expression your model represents is: $$\frac{1}{8}$$
We can obtain the multiplication expression by using the following multiplication properties. They are:
A) $$\frac{a}{b}$$ × $$\frac{p}{q}$$ = $$\frac{a × p}{b × q}$$
B) a = $$\frac{a}{1}$$
Think and Grow: Use Models to Multiply Fractions
You can use models to multiply a fraction by a fraction.
Example
Find $$\frac{1}{2}$$ × $$\frac{1}{3}$$.
Show and Grow
Multiply. Use a model to help.
Question 1.
$$\frac{1}{3}$$ × $$\frac{1}{4}$$ = _______
$$\frac{1}{3}$$ × $$\frac{1}{4}$$ = $$\frac{1}{12}$$
Explanation:
The given fractions are: $$\frac{1}{4}$$ and $$\frac{1}{3}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$ = $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{1}{4}$$ × $$\frac{1}{3}$$
= $$\frac{1 × 1}{4 × 3}$$
= $$\frac{1}{12}$$
Hence,
$$\frac{1}{4}$$ × $$\frac{1}{3}$$ = $$\frac{1}{12}$$
Question 2.
$$\frac{2}{3}$$ × $$\frac{1}{2}$$ = _______
$$\frac{2}{3}$$ × $$\frac{1}{2}$$ = $$\frac{1}{3}$$
Explanation:
The given fractions are: $$\frac{2}{3}$$ and $$\frac{1}{2}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$ = $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{1}{2}$$ × $$\frac{2}{3}$$
= $$\frac{1 × 2}{2 × 3}$$
= $$\frac{1}{3}$$
Hence,
$$\frac{1}{2}$$ × $$\frac{2}{3}$$ = $$\frac{1}{3}$$
Apply and Grow: Practice
Multiply. Use a model to help.
Question 3.
$$\frac{1}{2}$$ × $$\frac{1}{6}$$ = _______
$$\frac{1}{2}$$ × $$\frac{1}{6}$$ = $$\frac{1}{12}$$
Explanation:
The given fractions are: $$\frac{1}{2}$$ and $$\frac{1}{6}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$ = $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{1}{2}$$ × $$\frac{1}{6}$$
= $$\frac{1 × 1}{2 × 6}$$
= $$\frac{1}{12}$$
Hence,
$$\frac{1}{2}$$ × $$\frac{1}{6}$$ = $$\frac{1}{12}$$
Question 4.
$$\frac{1}{5}$$ × $$\frac{1}{8}$$ = _______
$$\frac{1}{5}$$ × $$\frac{1}{8}$$ = $$\frac{1}{40}$$
Explanation:
The given fractions are: $$\frac{1}{5}$$ and $$\frac{1}{8}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$ = $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{1}{5}$$ × $$\frac{1}{8}$$
= $$\frac{1 × 1}{5 × 8}$$
= $$\frac{1}{40}$$
Hence,
$$\frac{1}{5}$$ × $$\frac{1}{8}$$ = $$\frac{1}{40}$$
Question 5.
$$\frac{1}{4}$$ × $$\frac{1}{6}$$ = _______
$$\frac{1}{4}$$ × $$\frac{1}{6}$$ = $$\frac{1}{24}$$
Explanation:
The given fractions are: $$\frac{1}{4}$$ and $$\frac{1}{6}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$ = $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{1}{4}$$ × $$\frac{1}{6}$$
= $$\frac{1 × 1}{4 × 6}$$
= $$\frac{1}{24}$$
Hence,
$$\frac{1}{4}$$ × $$\frac{1}{6}$$ = $$\frac{1}{24}$$
Question 6.
$$\frac{2}{3}$$ × $$\frac{1}{3}$$ = _______
$$\frac{2}{3}$$ × $$\frac{1}{3}$$ = $$\frac{2}{9}$$
Explanation:
The given fractions are: $$\frac{2}{3}$$ and $$\frac{1}{3}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$ = $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{2}{3}$$ × $$\frac{1}{3}$$
= $$\frac{2 × 1}{3 × 3}$$
= $$\frac{2}{9}$$
Hence,
$$\frac{2}{3}$$ × $$\frac{1}{3}$$ = $$\frac{2}{9}$$
Write a multiplication equation represented by the model.
Question 7.
The multiplication equation represented by the model is:
$$\frac{1}{2}$$ × $$\frac{1}{8}$$
Explanation:
The given model is:
From the above model,
The number of rows is: 2
The number of columns is: 8
So,
The value of 1 row = $$\frac{1}{2}$$
The value of 1 column = $$\frac{1}{8}$$
So,
Rows × Columns = $$\frac{1}{8}$$ × $$\frac{1}{2}$$
Hence, from the above,
We can conclude that the multiplication equation represented by the above model is:
$$\frac{1}{8}$$ × $$\frac{1}{2}$$
Question 8.
The multiplication equation represented by the model is:
$$\frac{1}{4}$$ × $$\frac{1}{5}$$
Explanation:
The given model is:
From the above model,
The number of rows is: 4
The number of columns is: 5
So,
The value of 1 row = $$\frac{1}{4}$$
The value of 1 column = $$\frac{1}{5}$$
So,
Rows × Columns = $$\frac{1}{4}$$ × $$\frac{1}{5}$$
Hence, from the above,
We can conclude that the multiplication equation represented by the above model is:
$$\frac{1}{4}$$ × $$\frac{1}{5}$$
Question 9.
One-fifth of the students in your school have tried skating. Of those students, $$\frac{1}{7}$$ have tried ice skating. What fraction of students in your school have tried ice skating?
The fraction of students that have tried ice skating is: $$\frac{1}{35}$$
Explanation:
It is given that one-fifth of the students in your school have tried skating and of these students, $$\frac{1}{7}$$ have tried ice skating
So,
The fraction of students that have tried ice skating = ( The fraction of students that have tried skating ) × ( The fraction of students that tried ice skating out of the total students )
= $$\frac{1}{5}$$ × $$\frac{1}{7}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$ = $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{1}{5}$$ × $$\frac{1}{7}$$
= $$\frac{1 × 1}{5 × 7}$$
= $$\frac{1}{35}$$
Hence, from the above,
We can conclude that $$\frac{1}{35}$$ of the total students tried ice skating.
Question 10.
DIG DEEPER!
Are both Newton and Descartes correct? Explain.
Answer: Yes, both Newton and Descartes are correct.
Explanation:
The given models of Newton and Descartes are:
From the model of Newton,
The number of rows is: 3
The number of columns is: 5
From the model of Descartes,
The number of rows is: 5
The number of columns is: 3
From the above models,
We can observe that the Descartes model is obtained by reversing the rows and columns of Newton i.e., the rows of Newton’s model become the columns of Descartes’s model and the columns of Newton’s model becomes the rows of Descartes’ model.
So,
The multiplication equation represented by Newton is:
$$\frac{1}{3}$$ × $$\frac{1}{5}$$
The multiplication equation represented by Descartes is:
$$\frac{1}{5}$$ × $$\frac{1}{3}$$
Hence, from the above,
We can conclude that Newton and Descartes are correct
Think and Grow: Modeling Real Life
Example
A recipe calls for $$\frac{3}{4}$$ teaspoon of cinnamon. You4want to halve the recipe. What fraction of a teaspoon of cinnamon do you need?
Because you want to halve the recipe, multiply $$\frac{1}{2}$$ by $$\frac{3}{4}$$ to find how many teaspoons of cinnamon you need.
So,
You need $$\frac{3}{8}$$ teaspoon of cinnamon.
Show and Grow
Question 11.
The mass of mango is $$\frac{2}{5}$$ kilogram. The mass of guava is $$\frac{1}{4}$$ as much as the mango. What is the mass of the guava?
The mass of guava is: $$\frac{1}{10}$$ kilograms
Explanation:
It is given that the mass of mango is $$\frac{2}{5}$$ kilogram and the mass of guava is $$\frac{1}{4}$$ as much as the mango.
So,
The mass of guava = ( The mass of mango ) × ( The fraction of mass of guava in the mass of mango )
= $$\frac{2}{5}$$ × $$\frac{1}{4}$$
= $$\frac{1 × 2}{4 × 5}$$
= $$\frac{2}{20}$$
= $$\frac{1}{10}$$ kilograms
Hnce, from the above,
We can conclude that the mass of guava is: $$\frac{1}{10}$$ kilogram
Question 12.
A giant panda spends $$\frac{2}{3}$$ of 1-day eating and foraging. It spends $$\frac{3}{4}$$ of that time for eating bamboo. What fraction of 1 day does the panda spend eating bamboo?
The fraction of 1 day the panda spent on eating bamboo is: $$\frac{1}{2}$$
Explanation:
It is given that a giant panda spends $$\frac{2}{3}$$ of 1-day eating and foraging and it spends $$\frac{3}{4}$$ of that time for eating bamboo.
So,
The fraction of 1 day the panda spent on eating bamboo = ( The fraction of time the panda spent on eating ) × ( The fraction of time the panda spent time on eating bamboo out of the total time )
= $$\frac{2}{3}$$ × $$\frac{3}{4}$$
= $$\frac{3 × 2}{4 × 3}$$
= $$\frac{6}{12}$$
= $$\frac{1}{2}$$ hour
Hence, from the above,
We can conclude that the time spent by the panda on eating bamboo is: $$\frac{1}{2}$$ hour
Question 13.
DIG DEEPER!
You have a half-gallon carton of milk that you use only for cereal. You use the same amount each day for 5 days. There is $$\frac{3}{8}$$ of the carton left. How many cups of milk do you use each day? Explain.
### Use Models to Multiply Fractions Homework & Practice 9.4
Multiply. Use a model to help.
Question 1.
$$\frac{1}{3}$$ × $$\frac{1}{7}$$ = _______
$$\frac{1}{3}$$ × $$\frac{1}{7}$$ = $$\frac{1}{21}$$
Explanation:
The given fractions are: $$\frac{1}{7}$$ and $$\frac{1}{3}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$ = $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{1}{7}$$ × $$\frac{1}{3}$$
= $$\frac{1 × 1}{7 × 3}$$
= $$\frac{1}{21}$$
Hence,
$$\frac{1}{7}$$ × $$\frac{1}{3}$$ = $$\frac{1}{21}$$
Question 2.
$$\frac{1}{2}$$ × $$\frac{1}{9}$$ = _______
$$\frac{1}{2}$$ × $$\frac{1}{9}$$ = $$\frac{1}{18}$$
Explanation:
The given fractions are: $$\frac{1}{2}$$ and $$\frac{1}{9}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$ = $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{1}{2}$$ × $$\frac{1}{9}$$
= $$\frac{1 × 1}{2 × 9}$$
= $$\frac{1}{18}$$
Hence,
$$\frac{1}{2}$$ × $$\frac{1}{9}$$ = $$\frac{1}{18}$$
Question 3.
$$\frac{2}{5}$$ × $$\frac{1}{6}$$ = _______
$$\frac{2}{5}$$ × $$\frac{1}{6}$$ = $$\frac{1}{15}$$
Explanation:
The given fractions are: $$\frac{2}{5}$$ and $$\frac{1}{6}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$ = $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{2}{5}$$ × $$\frac{1}{6}$$
= $$\frac{2 × 1}{5 × 6}$$
= $$\frac{1}{15}$$
Hence,
$$\frac{2}{5}$$ × $$\frac{1}{6}$$ = $$\frac{1}{15}$$
Question 4.
$$\frac{3}{4}$$ × $$\frac{2}{7}$$ = _______
$$\frac{3}{4}$$ × $$\frac{2}{7}$$ = $$\frac{3}{14}$$
Explanation:
The given fractions are: $$\frac{3}{4}$$ and $$\frac{2}{7}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$ = $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{3}{4}$$ × $$\frac{2}{7}$$
= $$\frac{3 × 2}{4 × 7}$$
= $$\frac{3}{14}$$
Hence,
$$\frac{3}{4}$$ × $$\frac{2}{7}$$ = $$\frac{3}{14}$$
Write a multiplication equation represented by the model.
Question 5.
The multiplication equation represented by the model is:
$$\frac{1}{2}$$ × $$\frac{1}{5}$$
Explanation:
The given model is:
From the above model,
The number of rows is: 2
The number of columns is: 5
So,
The value of 1 row = $$\frac{1}{2}$$
The value of 1 column = $$\frac{1}{5}$$
So,
Rows × Columns = $$\frac{1}{2}$$ × $$\frac{1}{5}$$
Hence, from the above,
We can conclude that the multiplication equation represented by the above model is:
$$\frac{1}{2}$$ × $$\frac{1}{5}$$
Question 6.
The multiplication equation represented by the model is:
$$\frac{1}{3}$$ × $$\frac{1}{6}$$
Explanation:
The given model is:
From the above model,
The number of rows is: 3
The number of columns is: 6
So,
The value of 1 row = $$\frac{1}{3}$$
The value of 1 column = $$\frac{1}{6}$$
So,
Rows × Columns = $$\frac{1}{3}$$ × $$\frac{1}{6}$$
Hence, from the above,
We can conclude that the multiplication equation represented by the above model is:
$$\frac{1}{3}$$ × $$\frac{1}{6}$$
Question 7.
One-sixth of the animals at a zoo are birds. Of the birds, $$\frac{1}{3}$$ are female. What fraction of the animals at the zoo are female birds?
The fraction of the animals at the zoo that is female birds is: $$\frac{1}{18}$$
Explanation:
It is given that one-sixth of the animals at the zoo are birds and of the birds, $$\frac{1}{3}$$ is female.
So,
The fraction of the animals at the zoo are female birds = ( The fraction of animals in the zoo that is birds ) × ( The fraction of the animals in the zoo that is female birds )
= $$\frac{1}{3}$$ × $$\frac{1}{6}$$
$$\frac{1 × 1}{3 × 6}$$
= $$\frac{1}{18}$$
Hence, from the above,
We can conclude that the number of animals that are female birds at the zoo is: $$\frac{1}{18}$$
Question 8.
Writing
Write and solve a real-life problem for the expression.
$$\frac{2}{3}$$ × $$\frac{3}{7}$$
Suppose in a school, there are boys ad girls.
In school, $$\frac{2}{3}$$ of the boys and $$\frac{3}{7}$$ of the girls passed the PET test
So,
The fraction of the students that passed the PET test out of the total number of students = ( The fraction of the boys that passed the PET test ) × ( The fraction of the girls that passed the PET test )
= $$\frac{2}{3}$$ × $$\frac{3}{7}$$
= $$\frac{3 × 2}{7 × 3}$$
= $$\frac{6}{21}$$
= $$\frac{2}{7}$$
Hence, from the above,
We can conclude that the number of students who passed the PET test is: $$\frac{2}{7}$$
Question 9.
Modeling Real Life
A Gouldian finch is $$\frac{1}{2}$$ the length of the sun conure. How long is the Gouldian finch?
The length of Gouldian finch is: $$\frac{11}{24}$$ feet
Explanation:
It is given that a Gouldian finch is $$\frac{1}{2}$$ the length of the sun conure.
From the given figure,
The length of the sun conure = $$\frac{11}{12}$$ feet
So,
The length of Gouldian finch = ( The length of Sun Conure ) ÷ 2
= $$\frac{11}{12}$$ ÷ 2
= $$\frac{11}{12}$$ ÷ $$\frac{2}{1}$$
= $$\frac{11}{12}$$ × $$\frac{1}{2}$$
= $$\frac{11 × 1}{12 × 2}$$
= $$\frac{11}{24}$$ feet
Hence, from the above,
We can conclude that the length of the Gouldian finch is: $$\frac{11}{24}$$ feet
Question 10.
DIG DEEPER!
A recipe calls for $$\frac{2}{3}$$ cup of chopped walnuts. You chop 4 walnuts and get $$\frac{1}{4}$$ of the amount you need. How much more of a cup of chopped walnuts do you need? All of your walnuts are the same size. How many more walnuts should you chop? Explain.
The cup of chopped walnuts more you need is: $$\frac{1}{3}$$
The number of walnuts you should chop is: 6
Explanation:
It is given that a recipe calls for $$\frac{2}{3}$$ cup of chopped walnuts.
It is also given that you chop 4 walnuts and get $$\frac{1}{4}$$ of the amount you need.
So,
The cup more of chopped walnuts you need = 1- ( The cup of chopped walnuts you need for recipe )
= 1 – $$\frac{2}{3}$$
= $$\frac{3}{3}$$ – $$\frac{2}{3}$$
= $$\frac{3 – 2}{3}$$
= $$\frac{1}{3}$$ more cup
Now,
The number of walnuts you need more = ( The number of walnuts you needed to get $$\frac{1}{4}$$ of cup of chopped walnuts ) ÷ ( The amount of chopped walnuts you need for recipe )
= 4 ÷ $$\frac{2}{3}$$
= $$\frac{4}{1}$$ ÷ $$\frac{2}{3}$$
= $$\frac{4}{1}$$ × $$\frac{3}{2}$$
= $$\frac{4 × 3}{1 × 2}$$
= $$\frac{12}{2}$$
= 6
Hence, from the above,
We can conclude that
The cup of chopped walnuts more you need is: $$\frac{1}{3}$$
The number of walnuts you should chop is: 6
Review & Refresh
Subtract.
Question 11.
5 – 1$$\frac{3}{4}$$ = ______
5 – 1$$\frac{3}{4}$$ = $$\frac{13}{4}$$
Explanation:
The given numbers are: 5 and 1$$\frac{3}{4}$$
The representation of 1$$\frac{3}{4}$$ in the improper fraction form is: $$\frac{7}{4}$$
So,
5 – $$\frac{7}{4}$$ = $$\frac{20}{4}$$ – $$\frac{7}{4}$$
= $$\frac{20 – 7}{4}$$
= $$\frac{13}{4}$$
Hence,
5 – 1$$\frac{3}{4}$$ = $$\frac{13}{4}$$
Question 12.
13$$\frac{1}{4}$$ – 7$$\frac{5}{8}$$ = ________
13$$\frac{1}{4}$$ – 7$$\frac{5}{8}$$ = $$\frac{45}{8}$$
Explanation:
The given mixed fractions are: 13$$\frac{1}{4}$$ and 7$$\frac{5}{8}$$
The representation of 13$$\frac{1}{4}$$ in the improper fraction form is: $$\frac{53}{4}$$
The representation of 13$$\frac{1}{4}$$ in the improper fraction form is: $$\frac{61}{8}$$
To subtract the fractions, we have to make the denominators equal.
So,
Multiply $$\frac{53}{4}$$ by $$\frac{2}{2}$$
So,
$$\frac{53}{4}$$ = $$\frac{106}{8}$$
So,
$$\frac{106}{8}$$ – $$\frac{61}{8}$$
= $$\frac{106 – 61}{8}$$
= $$\frac{45}{8}$$
Hence,
13$$\frac{1}{4}$$ – 13$$\frac{1}{4}$$ = $$\frac{45}{8}$$
Question 13.
12$$\frac{7}{10}$$ – 5$$\frac{3}{10}$$ – 1$$\frac{1}{5}$$ = ________
12$$\frac{7}{10}$$ – 5$$\frac{3}{10}$$ – 1$$\frac{1}{5}$$ = $$\frac{62}{10}$$
Explanation:
The given mixed fractions are: 12$$\frac{7}{10}$$, 5$$\frac{3}{10}$$ and 1$$\frac{1}{5}$$
The representation of 12$$\frac{7}{10}$$ in the improper fraction form is: $$\frac{127}{10}$$
The representation of 5$$\frac{3}{10}$$ in the improper fraction form is: $$\frac{53}{10}$$
The representation of 1$$\frac{1}{5}$$ in the improper fraction form is: $$\frac{6}{5}$$
To subtract the fractions, we have to make the denominators equal.
So,
Multiply $$\frac{6}{5}$$ by $$\frac{2}{2}$$
So,
$$\frac{6}{5}$$ = $$\frac{12}{10}$$
So,
$$\frac{127}{10}$$ – $$\frac{53}{10}$$ – $$\frac{12}{10}$$
= $$\frac{127 – 53 – 12}{8}$$
= $$\frac{62}{10}$$
Hence,
12$$\frac{7}{10}$$ – 5$$\frac{3}{10}$$ – 1$$\frac{1}{5}$$ = $$\frac{62}{10}$$
### Lesson 9.5 Multiply Fractions
Explore and Grow
Use models to help you complete the table. What do you notice about each expression and its product?
The completed table is:
From the given table,
We can observe the product of fractions is also a product.
Construct Arguments
Explain how to multiply two fractions without using a model.
We can multiply the fractions by using the properties of multiplication. They are:
A) $$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
B) a = $$\frac{a}{1}$$
Think and Grow: Multiply Fractions
Key Idea
You can find the product of a fraction and a fraction by multiplying the numerators and multiplying the denominators.
Example
Find $$\frac{1}{2}$$ × $$\frac{3}{2}$$.
Multiply the numerators and multiply the denominators.
Example
Find $$\frac{5}{6}$$ × $$\frac{3}{5}$$.
Multiply the numerators and multiply the denominators.
Show and Grow
Multiply.
Question 1.
$$\frac{1}{2}$$ × $$\frac{4}{3}$$ = _______
$$\frac{1}{2}$$ × $$\frac{4}{3}$$ = $$\frac{2}{3}$$
Explanation:
The given fractions are: $$\frac{1}{2}$$ and $$\frac{4}{3}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{1}{2}$$ × $$\frac{4}{3}$$
= $$\frac{1 × 4}{2 × 3}$$
= $$\frac{4}{6}$$
= $$\frac{2}{3}$$
Hence,
$$\frac{1}{2}$$ × $$\frac{4}{3}$$ = $$\frac{2}{3}$$
Question 2.
$$\frac{2}{5}$$ × $$\frac{2}{3}$$ = _______
$$\frac{2}{5}$$ × $$\frac{2}{3}$$ = $$\frac{4}{15}$$
Explanation:
The given fractions are: $$\frac{2}{5}$$ and $$\frac{2}{3}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{2}{5}$$ × $$\frac{2}{3}$$
= $$\frac{2 × 2}{5 × 3}$$
= $$\frac{4}{15}$$
Hence,
$$\frac{2}{5}$$ × $$\frac{2}{3}$$ = $$\frac{4}{15}$$
Question 3.
$$\frac{3}{4}$$ × $$\frac{5}{8}$$ = _______
$$\frac{3}{4}$$ × $$\frac{5}{8}$$ = $$\frac{15}{32}$$
Explanation:
The given fractions are: $$\frac{3}{4}$$ and $$\frac{5}{8}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{3}{4}$$ × $$\frac{5}{8}$$
= $$\frac{3 × 5}{4 × 8}$$
= $$\frac{15}{32}$$
Hence,
$$\frac{3}{4}$$ × $$\frac{5}{8}$$ = $$\frac{15}{32}$$
Apply and Grow: Practice
Multiply.
Question 4.
$$\frac{1}{4}$$ × $$\frac{1}{4}$$ = _______
$$\frac{1}{4}$$ × $$\frac{1}{4}$$ = $$\frac{1}{16}$$
Explanation:
The given fractions are: $$\frac{1}{4}$$ and $$\frac{1}{4}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{1}{4}$$ × $$\frac{1}{4}$$
= $$\frac{1 × 1}{4 × 4}$$
= $$\frac{1}{16}$$
Hence,
$$\frac{1}{4}$$ × $$\frac{1}{4}$$ = $$\frac{1}{16}$$
Question 5.
$$\frac{5}{6}$$ × $$\frac{7}{10}$$ = _______
$$\frac{5}{6}$$ × $$\frac{7}{10}$$ = $$\frac{7}{12}$$
Explanation:
The given fractions are: $$\frac{5}{6}$$ and $$\frac{7}{10}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{5}{6}$$ × $$\frac{7}{10}$$
= $$\frac{5 × 7}{6 × 10}$$
= $$\frac{35}{60}$$
= $$\frac{7}{12}$$
Hence,
$$\frac{5}{6}$$ × $$\frac{7}{10}$$ = $$\frac{7}{12}$$
Question 6.
$$\frac{6}{9}$$ × $$\frac{8}{2}$$ = _______
$$\frac{6}{9}$$ × $$\frac{8}{2}$$ = $$\frac{8}{3}$$
Explanation:
The given fractions are: $$\frac{6}{9}$$ and $$\frac{8}{2}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{6}{9}$$ × $$\frac{8}{2}$$
= $$\frac{6 × 8}{2 × 9}$$
= $$\frac{48}{18}$$
= $$\frac{8}{3}$$
Hence,
$$\frac{6}{9}$$ × $$\frac{8}{2}$$ = $$\frac{8}{3}$$
Question 7.
$$\frac{21}{100}$$ × $$\frac{3}{5}$$ = _______
$$\frac{21}{100}$$ × $$\frac{3}{5}$$ = $$\frac{63}{500}$$
Explanation:
The given fractions are: $$\frac{21}{100}$$ and $$\frac{3}{5}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{21}{100}$$ × $$\frac{3}{5}$$
= $$\frac{21 × 3}{100 × 5}$$
= $$\frac{63}{500}$$
Hence,
$$\frac{21}{100}$$ × $$\frac{3}{5}$$ = $$\frac{63}{500}$$
Question 8.
$$\frac{1}{12}$$ × $$\frac{9}{4}$$ = _______
$$\frac{1}{12}$$ × $$\frac{9}{4}$$ = $$\frac{3}{16}$$
Explanation:
The given fractions are: $$\frac{1}{12}$$ and $$\frac{9}{4}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{1}{12}$$ × $$\frac{9}{4}$$
= $$\frac{1 × 9}{12 × 4}$$
= $$\frac{9}{48}$$
= $$\frac{3}{16}$$
Hence,
$$\frac{1}{12}$$ × $$\frac{9}{4}$$ = $$\frac{3}{16}$$
Question 9.
$$\frac{4}{7}$$ × $$\frac{8}{8}$$ = _______
$$\frac{4}{7}$$ × $$\frac{8}{8}$$ = $$\frac{4}{7}$$
Explanation:
The given fractions are: $$\frac{4}{7}$$ and $$\frac{8}{8}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{4}{7}$$ × $$\frac{8}{8}$$
= $$\frac{8 × 4}{7 × 8}$$
= $$\frac{4}{7}$$
Hence,
$$\frac{4}{7}$$ × $$\frac{8}{8}$$ = $$\frac{4}{7}$$
Evaluate
Question 10.
$$\left(\frac{1}{2} \times \frac{7}{8}\right)$$ × 2 = _______
$$\left(\frac{1}{2} \times \frac{7}{8}\right)$$ × 2 = $$\frac{7}{8}$$
Explanation:
The given fractions are: $$\frac{7}{8}$$ and $$\frac{1}{2}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
( $$\frac{1}{2}$$ × $$\frac{7}{8}$$ ) × 2
= ( $$\frac{1 × 7}{2 × 8}$$ ) × $$\frac{2}{1}$$
= $$\frac{7}{16}$$ × $$\frac{2}{1}$$
= $$\frac{7 × 2}{16 × 1}$$
= $$\frac{7}{8}$$
Hence,
$$\frac{4}{7}$$ × $$\frac{8}{8}$$ = $$\frac{4}{7}$$
Question 11.
$$\left(\frac{7}{6}-\frac{5}{6}\right)$$ × $$\frac{2}{3}$$ = _______
$$\left(\frac{7}{6}-\frac{5}{6}\right)$$ × $$\frac{2}{3}$$ = $$\frac{2}{9}$$
Explanation:
The given fractions are: $$\frac{7}{6}$$ , $$\frac{5}{6}$$, and $$\frac{2}{3}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
( $$\frac{7}{6}$$ – $$\frac{5}{6}$$ ) × $$\frac{2}{3}$$
= ( $$\frac{7 – 5}{6}$$ ) × $$\frac{2}{3}$$
= $$\frac{2}{6}$$ × $$\frac{2}{3}$$
= $$\frac{2 × 2}{6 × 3}$$
= $$\frac{4}{18}$$
= $$\frac{2}{9}$$
Hence,
$$\left(\frac{7}{6}-\frac{5}{6}\right)$$ × $$\frac{2}{3}$$ = $$\frac{2}{9}$$
Question 12.
$$\frac{9}{10}$$ × $$\left(\frac{4}{9}+\frac{1}{3}\right)$$ = _________
$$\frac{9}{10}$$ × $$\left(\frac{4}{9}+\frac{1}{3}\right)$$ = $$\frac{7}{10}$$
Explanation:
The given fractions are: $$\frac{9}{10}$$, $$\frac{4}{9}$$ and $$\frac{1}{3}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{9}{10}$$ × ( $$\frac{4}{9}$$ + $$\frac{1}{3}$$ )
= ( $$\frac{4 + 3}{9}$$ ) × $$\frac{9}{10}$$
= $$\frac{7}{9}$$ × $$\frac{9}{10}$$
= $$\frac{7 × 9}{9 × 10}$$
= $$\frac{7}{10}$$
Hence,
$$\frac{9}{10}$$ × $$\left(\frac{4}{9}+\frac{1}{3}\right)$$ = $$\frac{7}{10}$$
Question 13.
At a school, $$\frac{3}{4}$$ of the students play a sport. Of the students that play a sport, $$\frac{1}{5}$$ play baseball. What fraction of the students at the school play baseball?
The fraction of the students at the school that play baseball is: $$\frac{3}{20}$$
Explanation:
It is given that at a school, $$\frac{3}{4}$$ of the students play a sport and of the students that play a sport, $$\frac{1}{5}$$ play baseball.
So,
The fraction of the students at the school that play baseall = ( The fraction of the students that play a sport ) × ( The fraction of the students that play baseball out of thetotal number of students )
= $$\frac{3}{4}$$ × $$\frac{1}{5}$$
= $$\frac{3 × 1}{4 × 5}$$
= $$\frac{3}{20}$$
Hence, from the above,
We can conclude that the fraction of the students that play baseball is: $$\frac{3}{20}$$
Question 14.
Reasoning
Descartes says he can find the product of a whole number and a fraction the same way he finds the product of two fractions. Explain why his reasoning makes sense.
The product of a whole number and a fraction follows the same procedure as that of the product of the two fractions because of the following properties of multiplication:
A) $$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
B) a= $$\frac{a}{1}$$
Hence, from the above two properties of multiplication,
We can conclude that the reasoning of Descartes makes sense
Question 15.
Writing
Explain how multiplying fractions is different than adding and subtracting fractions.
In the multiplication of the fractions, we multiply numerators and denominators.
Example:
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
In addition of the fractions, we add only numerators making the denominators equal
Example:
$$\frac{a}{b}$$ + $$\frac{p}{b}$$
= $$\frac{a + p}{b}$$
In subtraction of the fractions, we subtract only numerators making the denominators equal
Example:
$$\frac{a}{b}$$ – $$\frac{p}{b}$$
= $$\frac{a – p}{b}$$
Think and Grow: Modeling Real Life
Example
A tourist is walking from the Empire State Building to Times Square. She is $$\frac{2}{3}$$ of the way there. What fraction of a mile does she have left to walk?
Find the distance she has walked. Because she has walked $$\frac{2}{3}$$ of $$\frac{3}{4}$$ mile, multiply $$\frac{2}{3}$$ by $$\frac{3}{4}$$.
Show and Grow
Question 16.
At a zoo, $$\frac{3}{5}$$ of the animals are mammals. Of the mammals, $$\frac{5}{12}$$ are primates.What fraction of the animals at the zoo are not primates?
The fractions of the animals at the zoo that are not primates is: $$\frac{3}{4}$$
Explanation:
It is given that at the zoo, $$\frac{3}{5}$$ of the animals are mammals and of the animals, $$\frac{5}{12}$$ are primates
So,
The fraction of the animals that are primates =( The fraction of the animals that are primates ) × ( The fraction of the animals that are primates )
= $$\frac{3}{5}$$ × $$\frac{5}{12}$$
= $$\frac{3 × 5}{5 × 12}$$
= $$\frac{15}{60}$$
= $$\frac{1}{4}$$
Now,
Let the total number of animals that are mammals =1
So,
The fraction of the animals that are not primates = ( The total number of mammals ) – ( The fraction of the animals that are primates )
= 1 – $$\frac{1}{4}$$
= $$\frac{4}{4}$$ – $$\frac{1}{4}$$
= $$\frac{4 – 1}{4}$$
= $$\frac{3}{4}$$
Hence, from the above,
We can conclude that the fraction of the animals that are not primates is: $$\frac{3}{4}$$
Question 17.
You have an album of 216 trading cards. One page contains $$\frac{1}{24}$$ of the cards. On that page, $$\frac{2}{3}$$ of the cards are epic. You only have one page with any epic cards. How many epic cards do you have?
The number of epic cards is: 6 epic cards
Explanation:
It is given that you have an album of 216 trading cards
It is also given that one page contains $$\frac{1}{24}$$ of the cards. On that page, $$\frac{2}{3}$$ of the cards are epic.
So,
The fraction of the epic cards = ( The total number of cards on that page ) × ( The fraction of the epic cards that is on that page )
= $$\frac{1}{24}$$ × $$\frac{2}{3}$$
= $$\frac{2 × 1}{3 × 24}$$
= $$\frac{1}{36}$$
So,
The number of epic cards = ( The total number of cards ) × ( The fraction of the epic cards )
= 216 × $$\frac{1}{36}$$
= $$\frac{216}{1}$$ × $$\frac{}{36}$$
= $$\frac{216}{36}$$
= 6 epic cards
Hence, from the above,
We can conclude that there are 6 epic cards
Question 18.
DIG DEEPER!
In a class, $$\frac{2}{5}$$ of the students play basketball and $$\frac{7}{10}$$ play soccer.Of the students who play basketball, $$\frac{2}{3}$$ also play soccer. There are 30 students in the class. How many students play soccer but do not play basketball?
The number of students that play soccer but do not play basketball is: 4 students
Explanation:
It is given that in a class, $$\frac{2}{5}$$ of the students play basketball and $$\frac{7}{10}$$ play soccer.Of the students who play basketball, $$\frac{2}{3}$$ also play soccer.
So,
The number of students who play basketball and soccer = ( The number of students who play basketball ) × ( The number of students playing soccer who are also playing basketball )
= $$\frac{2}{5}$$ × $$\frac{2}{3}$$
= $$\frac{4}{15}$$
Now,
The number of students who only play soccer but not basketball = ( The number of students who play soccer only ) – ( The number of students who play both basketball and soccer )
= $$\frac{2}{5}$$ – $$\frac{4}{15}$$
In subtraction, the denominators must be equal.
So,
Multiply $$\frac{2}{5}$$ by $$\frac{3}{3}$$
So,
$$\frac{6}{15}$$ – $$\frac{4}{15}$$
= $$\frac{4}{15}$$
It is also given that the total number of students are: 30
So,
The number of students who play soccer only = ( The fraction of the students who play soccer only ) × ( The total number of students )
= $$\frac{2}{15}$$ × 30
= $$\frac{2}{15}$$ × $$\frac{30}{1}$$
= 4 students
Hence, from the above,
We can conclude that there are 4 students who play soccer only.
### Multiply Fractions Homework & Practice 9.5
Multiply.
$$\frac{1}{4}$$ × $$\frac{1}{5}$$ = _______
$$\frac{1}{4}$$ × $$\frac{1}{5}$$ = $$\frac{1}{20}$$
Explanation:
The given fractions are: $$\frac{1}{4}$$ and $$\frac{1}{5}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{1}{4}$$ × $$\frac{1}{5}$$
= $$\frac{1 × 1}{4 × 5}$$
= $$\frac{1}{20}$$
Hence,
$$\frac{1}{4}$$ × $$\frac{1}{5}$$ = $$\frac{1}{20}$$
Question 2.
$$\frac{2}{7}$$ × $$\frac{1}{2}$$ = _______
$$\frac{1}{2}$$ × $$\frac{2}{7}$$ = $$\frac{1}{7}$$
Explanation:
The given fractions are: $$\frac{1}{2}$$ and $$\frac{2}{7}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{1}{2}$$ × $$\frac{2}{7}$$
= $$\frac{1 × 2}{2 × 7}$$
= $$\frac{2}{14}$$
= $$\frac{1}{7}$$
Hence,
$$\frac{1}{2}$$ × $$\frac{2}{7}$$ = $$\frac{1}{7}$$
Question 3.
$$\frac{9}{10}$$ × $$\frac{2}{3}$$ = _______
$$\frac{9}{10}$$ × $$\frac{2}{3}$$ = $$\frac{3}{5}$$
Explanation:
The given fractions are: $$\frac{9}{10}$$ and $$\frac{2}{3}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{9}{10}$$ × $$\frac{2}{3}$$
= $$\frac{9 × 2}{10 × 3}$$
= $$\frac{18}{30}$$
= $$\frac{3}{5}$$
Hence,
$$\frac{9}{10}$$ × $$\frac{2}{3}$$ = $$\frac{3}{5}$$
Question 4.
$$\frac{5}{8}$$ × $$\frac{5}{6}$$ = _______
$$\frac{5}{8}$$ × $$\frac{5}{6}$$ = $$\frac{25}{48}$$
Explanation:
The given fractions are: $$\frac{5}{8}$$ and $$\frac{5}{6}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{5}{8}$$ × $$\frac{5}{6}$$
= $$\frac{5 × 5}{8 × 6}$$
= $$\frac{25}{48}$$
Hence,
$$\frac{5}{8}$$ × $$\frac{5}{6}$$ = $$\frac{25}{48}$$
Question 5.
$$\frac{9}{7}$$ × $$\frac{3}{4}$$ = _______
$$\frac{9}{7}$$ × $$\frac{3}{4}$$ = $$\frac{27}{28}$$
Explanation:
The given fractions are: $$\frac{9}{7}$$ and $$\frac{3}{4}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{9}{7}$$ × $$\frac{3}{4}$$
= $$\frac{9 × 3}{7 × 4}$$
= $$\frac{27}{28}$$
Hence,
$$\frac{9}{7}$$ × $$\frac{3}{4}$$ = $$\frac{27}{28}$$
Question 6.
$$\frac{11}{100}$$ × $$\frac{2}{5}$$ = _______
$$\frac{11}{100}$$ × $$\frac{2}{5}$$ = $$\frac{11}{250}$$
Explanation:
The given fractions are: $$\frac{11}{100}$$ and $$\frac{2}{5}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{11}{100}$$ × $$\frac{2}{5}$$
= $$\frac{11 × 2}{100 × 5}$$
= $$\frac{22}{500}$$
= $$\frac{11}{250}$$
Hence,
$$\frac{11}{100}$$ × $$\frac{2}{5}$$ = $$\frac{11}{250}$$
Question 7.
$$\frac{7}{20}$$ × $$\frac{6}{2}$$ = _______
$$\frac{7}{20}$$ × $$\frac{6}{2}$$ = $$\frac{21}{20}$$
Explanation:
The given fractions are: $$\frac{7}{20}$$ and $$\frac{6}{2}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{7}{20}$$ × $$\frac{6}{2}$$
= $$\frac{7 × 6}{20 × 2}$$
= $$\frac{42}{40}$$
= $$\frac{21}{20}$$
Hence,
$$\frac{7}{20}$$ × $$\frac{6}{2}$$ = $$\frac{21}{20}$$
Question 8.
$$\frac{15}{16}$$ × $$\frac{1}{3}$$ = _______
$$\frac{15}{16}$$ × $$\frac{1}{3}$$ = $$\frac{5}{16}$$
Explanation:
The given fractions are: $$\frac{15}{16}$$ and $$\frac{1}{3}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{15}{16}$$ × $$\frac{1}{3}$$
= $$\frac{15 × 1}{16 × 3}$$
= $$\frac{15}{48}$$
= $$\frac{5}{16}$$
Hence,
$$\frac{15}{16}$$ × $$\frac{1}{3}$$ = $$\frac{5}{16}$$
Question 9.
$$\frac{5}{12}$$ × $$\frac{3}{10}$$ = _______
$$\frac{5}{12}$$ × $$\frac{3}{10}$$ = $$\frac{1}{8}$$
Explanation:
The given fractions are: $$\frac{5}{12}$$ and $$\frac{3}{10}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
$$\frac{5}{12}$$ × $$\frac{3}{10}$$
= $$\frac{5 × 3}{12 × 10}$$
= $$\frac{15}{120}$$
= $$\frac{1}{8}$$
Hence,
$$\frac{5}{12}$$ × $$\frac{3}{10}$$ = $$\frac{1}{8}$$
Evaluate.
Question 10.
3 × $$\frac{3}{10}$$ = _______
3 × $$\frac{3}{10}$$ = $$\frac{9}{10}$$
Explanation:
The given numbers are: 3 and $$\frac{3}{10}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
3 × $$\frac{3}{10}$$
= $$\frac{3}{1}$$ × $$\frac{3}{10}$$
= $$\frac{3 × 3}{10 × 1}$$
= $$\frac{9}{10}$$
Hence,
3 × $$\frac{3}{10}$$ = $$\frac{9}{10}$$
Question 11.
$$\left(\frac{1}{3}+\frac{1}{3}\right)$$ × $$\frac{4}{5}$$ = _______
$$\left(\frac{1}{3}+\frac{1}{3}\right)$$ × $$\frac{4}{5}$$ = $$\frac{8}{15}$$
Explanation:
The given fractions are: $$\frac{1}{3}$$, $$\frac{1}{3}$$ and $$\frac{4}{5}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
( $$\frac{1}{3}$$ + $$\frac{1}{3}$$ ) × $$\frac{4}{5}$$
= ( $$\frac{1 + 1}{3}$$ ) × $$\frac{4}{5}$$
= $$\frac{2}{3}$$ × $$\frac{4}{5}$$
= $$\frac{4 × 2}{3 × 5}$$
= $$\frac{8}{15}$$
Hence,
$$\left(\frac{1}{3}+\frac{1}{3}\right)$$ × $$\frac{4}{5}$$ = $$\frac{8}{15}$$
Question 12.
$$\frac{6}{7}$$ × $$\left(\frac{3}{4}-\frac{5}{12}\right)$$ = _______
$$\frac{6}{7}$$ × $$\left(\frac{3}{4}-\frac{5}{12}\right)$$ = $$\frac{2}{7}$$
Explanation:
The given fractions are: $$\frac{6}{7}$$, $$\frac{3}{4}$$ and $$\frac{5}{12}$$
We know that,
$$\frac{a}{b}$$ × $$\frac{p}{q}$$
= $$\frac{a × p}{b × q}$$
a = $$\frac{a}{1}$$
So,
( $$\frac{3}{4}$$ – $$\frac{5}{12}$$ ) × $$\frac{6}{7}$$
= ( $$\frac{9}{12}$$ – $$\frac{5}{12}$$ ) × $$\frac{6}{7}$$
= $$\frac{4}{12}$$ × $$\frac{6}{7}$$
= $$\frac{4 × 6}{12 × 7}$$
= $$\frac{2}{7}$$
Hence,
$$\frac{6}{7}$$ × $$\left(\frac{3}{4}-\frac{5}{12}\right)$$ = $$\frac{2}{7}$$
Question 13.
A pancake recipe calls for $$\frac{1}{3}$$ cup of water. You want to halve the recipe. What fraction of a cup of water do you need?
The fraction of a cup of water you need is: $$\frac{1}{6}$$
Explanation:
It is given that a pancake recipe calls for $$\frac{1}{3}$$ cup of water and you want to halve the recipe.
So,
The fraction of a cup of water you need for half of the recipe = ( The cup of water you need for a recipe ) × $$\frac{1}{2}$$
= $$\frac{1}{3}$$ × $$\frac{1}{2}$$
= $$\frac{1 × 1}{3 × 2}$$
= $$\frac{1}{6}$$
Hence, from the above,
We can conclude that the fraction of water you need for half of the recipe is: $$\frac{1}{6}$$
Question 14.
Number Sense
Which is greater, $$\frac{3}{4}$$ × $$\frac{1}{5}$$ or $$\frac{3}{4}$$ × $$\frac{1}{8}$$? Explain.
$$\frac{3}{4}$$ × $$\frac{1}{5}$$” is greater than “$$\frac{3}{4}$$ × $$\frac{1}{8}$$
Explanation:
The given fractions are: $$\frac{3}{4}$$, $$\frac{1}{5}$$ and $$\frac{1}{8}$$
Now,
$$\frac{3}{4}$$ × $$\frac{1}{5}$$
= $$\frac{3 × 1}{4 × 5}$$
= $$\frac{3}{20}$$
Now,
$$\frac{3}{4}$$ × $$\frac{1}{8}$$
= $$\frac{3 × 1}{4 × 8}$$
= $$\frac{3}{32}$$
So,
For the comparison of the products, equate the numerators.
Multiply $$\frac{3}{20}$$ by $$\frac{32}{32}$$
Multiply $$\frac{3}{32}$$ by $$\frac{20}{20}$$
So,
$$\frac{3}{20}$$ = $$\frac{96}{640}$$
$$\frac{3}{32}$$ = $$\frac{60}{640}$$
So,
By comparison of the products,
We can observe that
96 > 60
Hence, from the above,
We can conclude that “$$\frac{3}{4}$$ × $$\frac{1}{5}$$” is greater than “$$\frac{3}{4}$$ × $$\frac{1}{8}$$”
Question 15.
Reasoning
Is $$\frac{17}{24}$$ × $$\frac{7}{8}$$ equal to $$\frac{17}{8}$$ × $$\frac{7}{24}$$? Explain.
$$\frac{17}{24}$$ × $$\frac{7}{8}$$ is equal to $$\frac{17}{8}$$ × $$\frac{7}{24}$$
Explanation:
By the commutative property of multiplication,
a × b = b × a
So,
By the above property of multiplication,
We can interchange the numbers
Hence, from the above,
We can conclude that $$\frac{17}{24}$$ × $$\frac{7}{8}$$ is equal to $$\frac{17}{8}$$ × $$\frac{7}{24}$$
Question 16.
Number Sense
In which equations does k = $$\frac{5}{6}$$?
Let the multiplication equations be represented A), B), C) and D)
The given expressions are:
So,
The product of the given expressions are:
A) $$\frac{1}{2}$$ × $$\frac{5}{3}$$ = $$\frac{5}{6}$$
B) $$\frac{1}{10}$$ × k = $$\frac{1}{12}$$
C) $$\frac{1}{3}$$ × k = $$\frac{2}{9}$$
D) $$\frac{5}{4}$$ × $$\frac{2}{3}$$ = k
Hence, from the above expressions,
We can conclude that equation A) fits the value of k perfectly.
Question 17.
Modeling Real Life
At a town hall meeting, $$\frac{37}{50}$$ of the members are present. Of those who are present, $$\frac{1}{2}$$ vote in favor of a new park. What fraction of the members do not vote in favor of the new park?
The fraction of the members that do not vote in favor of the new park is: $$\frac{13}{50}$$
Explanation:
It is given that at a town hall meeting, $$\frac{37}{50}$$ of the members are present. Of those who are present, $$\frac{1}{2}$$ vote in favor of a new park.
So,
The fraction of the members that do not vote in favor of the new park = ( The total number of members at the town hall meeting ) – ( The fraction of the members that vote in the favor of a new park )
= 1 – $$\frac{37}{50}$$
= $$\frac{50}{50}$$ – $$\frac{37}{50}$$
= $$\frac{50 – 37}{50}$$
= $$\frac{13}{50}$$
Hence, from the above,
We can conclude that the number of members that do not vote in the favor of a new park is: $$\frac{13}{50}$$
Question 18.
Modeling Real Life
There are 50 U.S. states. Seven twenty-fifths of the states share a land border with Canada or Mexico.Of those states, $$\frac{2}{7}$$ share a border with Mexico. How many states share a border with Canada?
The number of states that share a border with Canada is: 4 states
Explanation:
It is given that there are 50 U.S. states. Seven twenty-fifths of the states share a land border with Canada or Mexico and of those states, $$\frac{2}{7}$$ share a border with Mexico.
So,
The number of states that share a border with Canada or Mexico = $$\frac{7}{25}$$ × 50
= $$\frac{7}{25}$$ × $$\frac{50}{1}$$
= 14 states
So,
The number of states that share a border with Canada = $$\frac{2}{7}$$ × 14
= $$\frac{2}{7}$$ × $$\frac{14}{1}$$
= 4 states
Hence, from the above,
We can conclude that there are 4 states that share a border with Canada
Review & Refresh
Question 19.
15.67 + 4 + 6.5 = _____
15.67 + 4 + 6.5 = 26.17
Explanation:
The given numbers are: 15.67, 4, and 6.5
The representation of the numbers in the fraction form is: $$\frac{1567}{100}$$, $$\frac{65}{10}$$ and 4
As the highest number in the denominator is 100, make all the denominators hundred
So,
The representation of 4 with 100 in the denominator is: $$\frac{400}{100}$$
The representation of 6.5 with 100 in the denominator is: $$\frac{650}{100}$$
So,
$$\frac{1567}{100}$$ + $$\frac{400}{100}$$ + $$\frac{650}{100}$$
= $$\frac{ 1567 + 400 + 650 }{100}$$
= $$\frac{2,617}{100}$$
= 26.17
Hence, 15.67 + 4 + 6.5 = 26.17
Question 20.
20.7 – 9.54 + 25.81 = _______
20.7 – 9.54 + 25.81 = 36.97
Explanation:
The given numbers are: 20.7, 9.54, and 25.81
The representation of the numbers in the fraction form is: $$\frac{2581}{100}$$, $$\frac{207}{10}$$ and $$\frac{954}{100}$$
As the highest number in the denominator is 100, make all the denominators hundred
So,
The representation of 20.7 with 100 in the denominator is: $$\frac{2070}{100}$$
So,
$$\frac{2581}{100}$$ + $$\frac{2070}{100}$$ – $$\frac{954}{100}$$
= $$\frac{ 2581 + 2070 – 954 }{100}$$
= $$\frac{3,697}{100}$$
= 36.97
Hence, 20.7 – 9.54 + 25.81 = 36.97
### Lesson 9.6 Find Areas of Rectangles
Explore and Grow
Draw and cut out a rectangle that has any two of the side lengths below.
$$\frac{1}{2}$$ ft
$$\frac{1}{3}$$ ft
$$\frac{1}{4}$$ ft
Use several copies of your rectangle to create a unit square. What is the area (in square feet) of each small rectangle? Explain your reasoning.
Let there are two small rectangles with the following lengths:
$$\frac{1}{2}$$ ft and $$\frac{1}{3}$$ ft
$$\frac{1}{2}$$ ft and $$\frac{1}{4}$$ ft
So,
The area of the unit square with lengths $$\frac{1}{2}$$ ft and $$\frac{1}{3}$$ ft is:
$$\frac{1}{2}$$ ft × $$\frac{1}{3}$$ ft
= $$\frac{1 × 1}{2 × 3}$$ ft
= $$\frac{1}{6}$$ ft
The area of the unit square with lengths $$\frac{1}{2}$$ ft and $$\frac{1}{4}$$ ft is:
$$\frac{1}{2}$$ ft × $$\frac{1}{4}$$ ft
= $$\frac{1 × 1}{2 × 4}$$ ft
= $$\frac{1}{8}$$ ft
Hence, from the above,
We can conclude that the areas ofthe smaller rectngles with unit lengths is: $$\frac{1}{6}$$ ft and $$\frac{1}{8}$$ ft
Reasoning
How can you use a rectangle with unit fraction side lengths to find the area of the rectangle below? Explain your reasoning.
The given rectangle is:
To find the area of the rectangle by the unit fractions, multiply the values of the length and the breadth
So,
$$\frac{3}{4}$$ × $$\frac{3}{2}$$
= 3 ($$\frac{1}{4}$$ ) × 3 ( $$\frac{1}{2}$$ )
= 9 ( $$\frac{1}{4}$$ × $$\frac{1}{2}$$ )
= 9 × $$\frac{1 × 1}{4 × 2}$$
= $$\frac{9}{1}$$ × $$\frac{1}{8}$$
= $$\frac{9}{8}$$
The unit fractions are the fractions that contain the value 1 in the numerator.
Here,
$$\frac{1}{4}$$ and $$\frac{1}{2}$$ are the unit fractions
Think and Grow: Find Areas of Rectangles
One way to find the area of a rectangle with fractional side lengths is to fill it with smaller rectangles.
Example
Find the area of the rectangle.
Show and Grow
Question 1.
Find the area of the shaded region.
The area of the shaded region is: $$\frac{3}{12}$$
Explanation:
The given figure is:
From the given figure,
The area of the shaded region = $$\frac{3}{4}$$ × $$\frac{1}{3}$$
= $$\frac{3 × 1}{4 × 3}$$
= $$\frac{3}{12}$$
Hence, from the above,
We can conclude that the area of the shaded region is: $$\frac{3}{12}$$
Apply and Grow: Practice
Find the area of the shaded region.
Question 2.
The area of the shaded region is: $$\frac{2}{12}$$
Explanation:
The given figure is:
From the given figure,
The area of the shaded region = $$\frac{2}{4}$$ × $$\frac{1}{3}$$
= $$\frac{2 × 1}{4 × 3}$$
= $$\frac{2}{12}$$
Hence, from the above,
We can conclude that the area of the shaded region is: $$\frac{2}{12}$$
Question 3.
The area of the shaded region is: $$\frac{3}{16}$$
Explanation:
The given figure is:
From the given figure,
The area of the shaded region = $$\frac{3}{4}$$ × $$\frac{1}{4}$$
= $$\frac{3 × 1}{4 × 4}$$
= $$\frac{3}{16}$$
Hence, from the above,
We can conclude that the area of the shaded region is: $$\frac{3}{16}$$
Use rectangles with unit fraction side lengths to find the area of the rectangle.
Question 4.
The area of the rectangle is: $$\frac{6}{10}$$
Explanation:
The given figure is:
From the given figure,
The area of the rectangle = $$\frac{3}{2}$$ × $$\frac{2}{5}$$
= $$\frac{3 × 2}{2 × 5}$$
= $$\frac{6}{10}$$
Hence, from the above,
We can conclude that the area of the shaded region is: $$\frac{6}{10}$$
Question 5.
The area of the rectangle is: $$\frac{10}{18}$$
Explanation:
The given figure is:
From the given figure,
The area of the rectangle = $$\frac{5}{6}$$ × $$\frac{2}{3}$$
= $$\frac{5 × 2}{6 × 3}$$
= $$\frac{10}{18}$$
Hence, from the above,
We can conclude that the area of the shaded region is: $$\frac{10}{18}$$
Question 6.
Find the area of a rectangle with side lengths of $$\frac{5}{8}$$ and $$\frac{4}{3}$$?
The area of a rectangle is: $$\frac{20}{24}$$
Explanation:
The given side lengths of a rectangle are: $$\frac{5}{8}$$ and $$\frac{4}{3}$$
So,
The area of the rectangle = $$\frac{5}{8}$$ × $$\frac{4}{3}$$
= $$\frac{5 × 4}{8 × 3}$$
= $$\frac{20}{24}$$
Hence, from the above,
We can conclude that the area of the rectangle is: $$\frac{20}{24}$$
Question 7.
Find the area of a rectangle with side lengths of $$\frac{7}{9}$$ and $$\frac{1}{2}$$?
The area of a rectangle is: $$\frac{7}{18}$$
Explanation:
The given side lengths of a rectangle are: $$\frac{7}{9}$$ and $$\frac{1}{2}$$
So,
The area of the rectangle = $$\frac{7}{9}$$ × $$\frac{1}{2}$$
= $$\frac{7 × 1}{9 × 2}$$
= $$\frac{7}{18}$$
Hence, from the above,
We can conclude that the area of the rectangle is: $$\frac{7}{18}$$
Question 8.
Reasoning
Can you find the area of a rectangle with fractional side lengths the same way you find the area of a rectangle with whole-number side lengths? Explain.
Yes, we can find the area of a rectangle with fractional side lengths the same way you find the area of a rectangle with whole-number side lengths.
Example:
Let the fractional side lengths be: $$\frac{1}{5}$$ and $$\frac{1}{9}$$
Let the whole number side lengths be: 3 and 4
So,
The area of the rectangle with the fractional side lengths = $$\frac{1}{5}$$ × $$\frac{1}{9}$$
= $$\frac{1 × 1}{9 × 5}$$
= $$\frac{1}{45}$$
The area of the rectangle with the whole number side lengths = 3 × 4
= $$\frac{3}{1}$$ × $$\frac{4}{1}$$
= $$\frac{3 × 4}{1 × 1}$$
= $$\frac{12}{1}$$
= 12
Hence, from the above,
We can conclude that we can find the area of a rectangle with fractional side lengths the same way you find the area of a rectangle with whole-number side lengths.
Question 9.
YOU BE THE TEACHER
Our friend says she can find the area of a square given only one fractional side length. Is your friend correct? Explain.
Yes, she can find the area of a square by only one fractional side length
Explanation:
Let the unit fractional side length of the square be: $$\frac{1}{2}$$
We know that,
The length of all the sides in a square are equal.
So,
By using this property, we can find the area of the square by taking the fractional unit side length of 1 side as the same for all the sides
So,
The area of the square = $$\frac{1}{2}$$ × $$\frac{1}{2}$$
= $$\frac{1 × 1}{2 × 2}$$
= $$\frac{1}{4}$$
Hence, from the above,
We can conclude that she can find the area of a square by only one fractional side length
Think and Grow: Modeling Real Life
Example
A zoo needs an outdoor enclosure with an area of at least $$\frac{3}{10}$$ square kilometer10for a camel. Is the rectangular enclosure shown large enough for a camel?
Find the area of the enclosure by multiplying the length and the width.
So, the enclosure is large enough for a camel.
Show and Grow
Question 10.
The area of a square dog kennel is 4 square yards. Will the square mat fit in the kennel?
Yes, the square mat will fit in the kennel
Explanation:
It is given that the area of a square dog kennel is 4 square yards and the side of the square mat is $$\frac{5}{3}$$ yd
We know that,
The length of all the sides in a square are equal.
So,
By using this property,
The area of the square mat = $$\frac{5}{3}$$ × $$\frac{5}{3}$$
= $$\frac{5 × 5}{3 × 3}$$
= $$\frac{25}{9}$$ yd
Now,
The area of a square dog kennel can be written as: $$\frac{36}{9}$$ yd
So,
When we compare the 2 areas,
We can get 25 < 36
Hence, from the above,
We can conclude that the square mat will fit in the kennel
Question 11.
DIG DEEPER!
The side lengths of each chalk art square are $$\frac{11}{4}$$ meters. The side lengths of the square zone around each chalk square are an additional $$\frac{3}{4}$$ meter. How many square meters of concrete is used to create the chalk walk shown?
The amount of the concrete used to create the chalk walk is: $$\frac{196}{16}$$ square meters
Explanation:
It is given that the side lengths of each chalk art square are $$\frac{11}{4}$$ meters. The side lengths of the square zone around each chalk square are an additional $$\frac{3}{4}$$ meter.
So,
The total side length of the chalk walk = $$\frac{11}{4}$$ + $$\frac{3}{4}$$
= $$\frac{11 + 3}{4}$$
= $$\frac{14}{4}$$ meters
We know that,
The length of all sides in the square are equal.
So,
By using this property,
The amount of concrete used for the chalk walk = $$\frac{14}{4}$$ × $$\frac{14}{4}$$
= $$\frac{14 × 14}{4 × 4}$$
= $$\frac{196}{16}$$ square meters
Hence, from the above,
We can conclude that the amount of the concrete used to create the chalk walk is: $$\frac{196}{16}$$ square meters
### Find Areas of Rectangles Homework & Practice 9.6
Find the area of the shaded region.
Question 1.
The area of the shaded region is: $$\frac{1}{12}$$
Explanation:
The given figure is:
From the given figure,
The area of the shaded region = $$\frac{1}{2}$$ × $$\frac{1}{6}$$
= $$\frac{1 × 1}{2 × 6}$$
= $$\frac{1}{12}$$
Hence, from the above,
We can conclude that the area of the shaded region is: $$\frac{1}{12}$$
Question 2.
The area of the shaded region is: $$\frac{8}{25}$$
Explanation:
The given figure is:
From the given figure,
The area of the shaded region = $$\frac{2}{5}$$ × $$\frac{4}{5}$$
= $$\frac{2 × 4}{5 × 5}$$
= $$\frac{8}{25}$$
Hence, from the above,
We can conclude that the area of the shaded region is: $$\frac{8}{25}$$
Use rectangles with unit fraction side lengths to find the area of the rectangle.
Question 3.
The area of the rectangle is: $$\frac{2}{24}$$
Explanation:
The given figure is:
From the given figure,
The area of the rectangle = $$\frac{2}{3}$$ × $$\frac{1}{8}$$
= $$\frac{1 × 2}{3 × 8}$$
= $$\frac{2}{24}$$
Hence, from the above,
We can conclude that the area of the shaded region is: $$\frac{2}{24}$$
Question 4.
The area of the rectangle is: $$\frac{35}{90}$$
Explanation:
The given figure is:
From the given figure,
The area of the rectangle = $$\frac{7}{10}$$ × $$\frac{5}{9}$$
= $$\frac{7 × 5}{10 × 9}$$
= $$\frac{35}{90}$$
Hence, from the above,
We can conclude that the area of the shaded region is: $$\frac{35}{90}$$
Question 5.
Find the area of a rectangle with a side length of $$\frac{3}{4}$$ and $$\frac{5}{12}$$.
The area of a rectangle is: $$\frac{15}{48}$$
Explanation:
The given side lengths of a rectangle are: $$\frac{5}{12}$$ and $$\frac{3}{4}$$
So,
The area of the rectangle = $$\frac{5}{12}$$ × $$\frac{3}{4}$$
= $$\frac{5 × 3}{12 × 4}$$
= $$\frac{15}{48}$$
Hence, from the above,
We can conclude that the area of the rectangle is: $$\frac{15}{48}$$
Question 6.
Find the area of a square with side lengths of $$\frac{9}{16}$$
The area of a square is: $$\frac{81}{256}$$
Explanation:
We know that,
The length of all the sides of the square is equal.
So,
The given side lengths of a square are: $$\frac{9}{16}$$ and $$\frac{9}{16}$$
So,
The area of the square = $$\frac{9}{16}$$ × $$\frac{9}{16}$$
= $$\frac{9 × 9}{16 × 16}$$
= $$\frac{81}{256}$$
Hence, from the above,
We can conclude that the area of the rectangle is: $$\frac{81}{256}$$
Question 7.
Open-Ended
The area of a rectangle is $$\frac{16}{24}$$. What are the possible side lengths of the rectangle?
The possible side lengths for the given area of a rectangle are:
A) $$\frac{4}{2}$$ and $$\frac{4}{12}$$
B) $$\frac{1}{3}$$ and $$\frac{4}{8}$$
Explanation:
The given area of a rectangle is: $$\frac{16}{24}$$
So,
To find the possible side lengths of a rectangle, find the factors for the numerator and denominator of $$\frac{16}{24}$$
So,
The factors of 16 are: 1, 2, 4, 8, 16
The factors of 24 are: 1, 2, 3, 4, 6, 8, 12, 24
So,
The possible side lengths of a rectangle are:
A) $$\frac{4}{2}$$ and $$\frac{4}{12}$$
B) $$\frac{1}{3}$$ and $$\frac{4}{8}$$
There are so many possible side lengths of a rectangle like above
Hence, from the above,
We can conclude that the possible side lengths of a rectangle are:
A) $$\frac{4}{2}$$ and $$\frac{4}{12}$$
B) $$\frac{1}{3}$$ and $$\frac{4}{8}$$
Question 8.
Structure
Write an expression that represents the area of the shaded rectangle.
The expression that represents the area of the shaded region is: $$\frac{1}{3}$$ × $$\frac{1}{3}$$
Explanation:
The number of shaded columns is: 3
The number of shaded rows is: 3
The total number of rows is: 1
The total number of columns are: 1
So,
The expression representing the shaded region = $$\frac{The total number of rows}{The number of shaded rows}$$ × $$\frac{The total number of rows}{The number of shaded rows}$$
= $$\frac{1}{3}$$ × $$\frac{1}{3}$$
Hence, from the above,
We can conclude that the expression representing the shaded region of a rectangle is: $$\frac{1}{3}$$ × $$\frac{1}{3}$$
Question 9.
Modeling Real Life
The area of a square table is 9 square feet. Will the board game fit on the table?
Yes, the board game fit on the table
Explanation:
The given area of a square table is 9 square feet
So,
The representation of the area of the square table in the fraction form is: $$\frac{36}{4}$$
Now,
The given board game is:
The area of the board game can find out by multiplying its side lengths
So,
The area of the board game = $$\frac{3}{2}$$ × $$\frac{5}{2}$$
= $$\frac{3 × 5}{2 × 2}$$
= $$\frac{15}{4}$$
So,
Compare the area of the square table and the area of the board game,
We get
Area of the square table > Area of the board game
Hence, from the above,
We can conclude that the board game fit on the table
Review & Refresh
Question 10.
365 ÷ 14 = _______
365 ÷ 14 = 26 R 1
Explanation:
By using the partial quotients method,
365 ÷ 14 = ( 280 ÷ 84 ) ÷ 14
= ( 280 ÷ 14 ) + ( 84 ÷ 14 )
= 20 + 6
= 26 R 1
Hence, 365 ÷ 14 = 26 R 1
Question 11.
282 ÷ 27 = ______
282 ÷ 27 = 10 R 12
Explanation:
By using the partial quotients method,
282 ÷ 27 = 270 ÷ 27
= 10 R 12
Hence, 270 ÷ 27 = 10 R 12
Question 12.
601 ÷ 72 = _____
601 ÷ 72 = 8 R 25
Explanation:
By using the partial quotients method,
601 ÷ 72 = 576 ÷ 72
= 8 R 25
Hence, 601 ÷ 72 = 8 R 25
### Lesson 9.7 Multiply Mixed Numbers
Explore and Grow
Find the area of the rectangle. Explain how you found your answer.
The area of the rectangle is: $$\frac{35}{4}$$
Explanation:
The given rectangle is:
From the above rectangle,
The side lengths are: 2$$\frac{1}{2}$$ and 3$$\frac{1}{2}$$
The representation of the side lengths in the improper form is: $$\frac{5}{2}$$ and $$\frac{7}{2}$$
So,
The area of rectangle = $$\frac{5}{2}$$ × $$\frac{7}{2}$$
= $$\frac{5 × 7}{2 × 2}$$
= $$\frac{35}{4}$$
Hence,
2$$\frac{1}{2}$$ × 3$$\frac{1}{2}$$ = $$\frac{35}{4}$$
Structure
Find the area using a different method. Explain how you found your answer.
The given mixed fractions of a rectangle are: 2$$\frac{1}{2}$$ and 3$$\frac{1}{2}$$
To convert a mixed fraction into an improper fraction, we have to add the whole number part and the fractional part in the mixed fraction.
So,
2$$\frac{1}{2}$$ = 2 + $$\frac{1}{2}$$
= $$\frac{4}{2}$$ + $$\frac{1}{2}$$
= $$\frac{5}{2}$$
3$$\frac{1}{2}$$ = 3 + $$\frac{1}{2}$$
= $$\frac{6}{2}$$ + $$\frac{1}{2}$$
= $$\frac{7}{2}$$
So,
$$\frac{5}{2}$$ × $$\frac{7}{2}$$ = $$\frac{35}{4}$$
Think and Grow: Multiply Mixed Numbers
You can use a model to find the product of two mixed numbers. You can also write the mixed numbers as improper fractions and then multiply.
Example
Find 1$$\frac{1}{2}$$ × 2$$\frac{3}{4}$$.
One Way: Use an area model.
Step 1: Write each mixed number as a sum.
1$$\frac{1}{2}$$ = 1 + $$\frac{1}{2}$$ 2$$\frac{3}{4}$$ = 2 + $$\frac{3}{4}$$
Step 2: Draw an area model that represents the product of the sums.
Step 3: Find the sum of the areas of the sections.
Another Way: Write each mixed number as an improper fraction, then multiply.
Show and Grow
Multiply.
Question 1.
2$$\frac{1}{2}$$ × 1$$\frac{1}{2}$$ = ________
2$$\frac{1}{2}$$ × 1$$\frac{1}{2}$$ = $$\frac{15}{4}$$
Explanation:
The given fractions are: 2$$\frac{1}{2}$$ and 1$$\frac{1}{2}$$
The representation of the fractions in the improper form is: $$\frac{5}{2}$$ and $$\frac{3}{2}$$
So,
$$\frac{5}{2}$$ × $$\frac{3}{2}$$
= $$\frac{5 × 3}{2 × 2}$$
= $$\frac{15}{4}$$
Hence,
2$$\frac{1}{2}$$ × 1$$\frac{1}{2}$$ = $$\frac{15}{4}$$
Question 2.
3$$\frac{1}{4}$$ × 2$$\frac{2}{3}$$ = ______
3$$\frac{1}{4}$$ × 2$$\frac{2}{3}$$ = $$\frac{104}{12}$$
Explanation:
The given fractions are: 3$$\frac{1}{4}$$ and 2$$\frac{2}{3}$$
The representation of the fractions in the improper form is: $$\frac{13}{4}$$ and $$\frac{8}{3}$$
So,
$$\frac{13}{4}$$ × $$\frac{8}{3}$$
= $$\frac{13 × 8}{4 × 3}$$
= $$\frac{104}{12}$$
Hence,
3$$\frac{1}{4}$$ × 2$$\frac{2}{3}$$ = $$\frac{104}{12}$$
Apply and Grow: Practice
Multiply.
Question 3.
1$$\frac{3}{4}$$ × 2$$\frac{1}{6}$$ = ______
1$$\frac{3}{4}$$ × 2$$\frac{1}{6}$$ = $$\frac{91}{24}$$
Explanation:
The given fractions are: 1$$\frac{3}{4}$$ and 2$$\frac{1}{6}$$
The representation of the fractions in the improper form is: $$\frac{7}{4}$$ and $$\frac{13}{6}$$
So,
$$\frac{7}{4}$$ × $$\frac{13}{6}$$
= $$\frac{13 × 7}{4 × 6}$$
= $$\frac{91}{24}$$
Hence,
1$$\frac{3}{4}$$ × 2$$\frac{1}{6}$$ = $$\frac{91}{24}$$
Question 4.
4$$\frac{1}{3}$$ × 1$$\frac{5}{6}$$ = ______
4$$\frac{1}{3}$$ × 1$$\frac{5}{6}$$ = $$\frac{143}{18}$$
Explanation:
The given fractions are: 4$$\frac{1}{3}$$ and 1$$\frac{5}{6}$$
The representation of the fractions in the improper form is: $$\frac{13}{3}$$ and $$\frac{11}{6}$$
So,
$$\frac{13}{3}$$ × $$\frac{11}{6}$$
= $$\frac{13 × 11}{6 × 3}$$
= $$\frac{143}{18}$$
Hence,
4$$\frac{1}{3}$$ × 1$$\frac{5}{6}$$ = $$\frac{143}{18}$$
Question 5.
3$$\frac{2}{5}$$ × 1$$\frac{9}{10}$$ = ______
3$$\frac{2}{5}$$ × 1$$\frac{9}{10}$$ = $$\frac{323}{50}$$
Explanation:
The given fractions are: 3$$\frac{2}{5}$$ and 1$$\frac{9}{10}$$
The representation of the fractions in the improper form is: $$\frac{17}{5}$$ and $$\frac{19}{10}$$
So,
$$\frac{17}{5}$$ × $$\frac{19}{10}$$
= $$\frac{17 × 19}{5 × 10}$$
= $$\frac{323}{50}$$
Hence,
3$$\frac{2}{5}$$ × 1$$\frac{9}{10}$$ = $$\frac{323}{50}$$
Question 6.
2$$\frac{3}{8}$$ × 3$$\frac{1}{2}$$ = ______
2$$\frac{3}{8}$$ × 3$$\frac{1}{2}$$ = $$\frac{133}{16}$$
Explanation:
The given fractions are: 2$$\frac{3}{8}$$ and 3$$\frac{1}{2}$$
The representation of the fractions in the improper form is: $$\frac{19}{8}$$ and $$\frac{7}{2}$$
So,
$$\frac{19}{8}$$ × $$\frac{7}{2}$$
= $$\frac{19 × 7}{8 × 2}$$
= $$\frac{133}{16}$$
Hence,
2$$\frac{3}{8}$$ × 3$$\frac{1}{2}$$ = $$\frac{133}{16}$$
Evaluate.
Question 7.
5$$\frac{1}{4}$$ × $$\frac{2}{5}$$ × 6$$\frac{1}{12}$$ = _________
5$$\frac{1}{4}$$ × $$\frac{2}{5}$$ × 6$$\frac{1}{12}$$ = $$\frac{3,066}{240}$$
Explanation:
The given fractions are: 5$$\frac{1}{4}$$, $$\frac{2}{5}$$ and 6$$\frac{1}{12}$$
The representation of the fractions in the improper form is: $$\frac{21}{4}$$ and $$\frac{73}{12}$$
So,
$$\frac{21}{4}$$ × $$\frac{73}{12}$$ × $$\frac{2}{5}$$
= $$\frac{21 × 73 × 2}{4 × 12 × 5}$$
= $$\frac{3,066}{240}$$
Hence,
5$$\frac{1}{4}$$ × $$\frac{2}{5}$$ × 6$$\frac{1}{12}$$ = $$\frac{3,066}{240}$$
Question 8.
3$$\frac{2}{3}$$ × (10$$\frac{7}{8}$$ – 2$$\frac{1}{4}$$) = _________
3$$\frac{2}{3}$$ × (10$$\frac{7}{8}$$ – 2$$\frac{1}{4}$$) = $$\frac{759}{24}$$
Explanation:
The given fractions are: 3$$\frac{2}{3}$$, 10$$\frac{7}{8}$$ and 2$$\frac{1}{4}$$
The representation of the fractions in the improper form is: $$\frac{11}{3}$$, $$\frac{87}{8}$$ and $$\frac{9}{4}$$
So,
3$$\frac{2}{3}$$ × (10$$\frac{7}{8}$$ – 2$$\frac{1}{4}$$)
= $$\frac{11}{3}$$ × ( $$\frac{87}{8}$$ – $$\frac{9}{4}$$ )
= $$\frac{11}{3}$$ × ( $$\frac{87 – 18}{8}$$ )
= $$\frac{11}{3}$$ × $$\frac{69}{8}$$
= $$\frac{11 × 69}{8 × 3}$$
= $$\frac{759}{24}$$
Hence,
3$$\frac{2}{3}$$ × (10$$\frac{7}{8}$$ – 2$$\frac{1}{4}$$) = $$\frac{759}{24}$$
Question 9.
YOU BE THE TEACHER
Your friend uses the model to find 4$$\frac{1}{2}$$ × 3$$\frac{2}{3}$$. Is your friend correct? Explain.
No, your friend is not correct
Explanation:
The given fractions are: 4$$\frac{1}{2}$$ and 3$$\frac{2}{3}$$
The representation of the side lengths in the improper form is: $$\frac{9}{2}$$ and $$\frac{11}{3}$$
So,
The area of rectangle = $$\frac{9}{2}$$ × $$\frac{11}{3}$$
= $$\frac{9 × 11}{3 × 2}$$
= $$\frac{99}{6}$$
= $$\frac{33}{2}$$
= 15$$\frac{3}{2}$$
4$$\frac{1}{2}$$ × 3$$\frac{2}{3}$$ = 12$$\frac{1}{2}$$
Hence, from the above,
We can conclude that your friend is not correct.
Question 10.
Logic
Find the missing numbers.
The missing numbers are: 1 and $$\frac{8}{5}$$
Explanation:
According to the separation method,
To convert the mixed number into the improper fraction, we can add whole numbers and the fractions separately
Now,
Let the missing numbers be p and q
So,
p + 5 = 6
So,
p = 6 – 5 = 1
Now,
$$\frac{1}{4}$$ + $$\frac{1}{q}$$ = $$\frac{7}{8}$$
So,
$$\frac{1}{q}$$ = $$\frac{7}{8}$$ – $$\frac{1}{4}$$
= $$\frac{7}{8}$$ – $$\frac{2}{8}$$
= $$\frac{7 – 2}{8}$$
= $$\frac{5}{8}$$
So,
q= $$\frac{8}{5}$$
Hence, from the above,
We can conclude that the missing numbers are: 1 and $$\frac{8}{5}$$
Think and Grow: Modeling Real Life
Example
A construction crew is paving 15 miles of a highway. The crew paves 4$$\frac{2}{10}$$ miles each month. Does the crew finish paving the highway in 3$$\frac{1}{2}$$ months?
Find the length of the highway the crew paves by multiplying the number of months by the number of miles they pave each month. Write each mixed number as an improper fraction, then multiply.
Compare the length the crew paves to the amount that needs to be paved.
The Crew can finish paving the highway in 3$$\frac{1}{2}$$ months.
Show and Grow
Question 11.
You have 3 cups of strawberries. You want to make 1$$\frac{1}{2}$$ batches of the recipe. Do you have enough strawberries?
Yes, you have enough strawberries
Explanation:
The given recipe for a strawberry smoothie is:
It is given that you want to make 1$$\frac{1}{2}$$ batches of the recipe
From the table,
The number of cups of strawberry is: 1$$\frac{3}{4}$$
Now,
The representation of 1$$\frac{1}{2}$$ in the iproper fraction is: $$\frac{3}{2}$$
The representation of 1$$\frac{3}{4}$$ in the mixed form is: $$\frac{7}{4}$$
So,
The number of strawberries for the recipe = $$\frac{3}{2}$$ × $$\frac{7}{4}$$
= $$\frac{3 × 7}{4 × 2}$$
= $$\frac{21}{8}$$
It is also given that you have 3 cups of strawberries
So,
We can write 3 in the fraction form and as the multiples of 8 is: $$\frac{24}{8}$$
So,
When we compare the number of cups we obtained and given,
We can say that we have enough strawberries to make the recipe
Hence, from the above,
We can conclude that we have enough strawberries
Question 12.
On Monday,you roller-skate 6$$\frac{1}{4}$$ miles. On Tuesday, you skate 1$$\frac{2}{5}$$ times as far as you did on Monday. How many total miles do you roller-skate on Monday and Tuesday combined?
The number of miles you do roller-skate on Monday and Tuesday combined is: 15 miles
Explanation:
It is given that on Monday,you roller-skate 6$$\frac{1}{4}$$ miles. On Tuesday, you skate 1$$\frac{2}{5}$$ times as far as you did on Monday.
So,
The number of miles you do roller-skate on Tuesday = ( The number of miles you do roller-skate on Monday ) × 1$$\frac{2}{5}$$
= 6$$\frac{1}{4}$$ × 1$$\frac{2}{5}$$
= $$\frac{25}{4}$$ × $$\frac{7}{5}$$
= $$\frac{25 × 7}{5 × 4}$$
= $$\frac{175}{20}$$
Now,
The number of miles you do roller-skate on Monday and Tuesday combined = 6$$\frac{1}{4}$$ + $$\frac{175}{20}$$
= $$\frac{25}{4}$$ + $$\frac{175}{20}$$
= $$\frac{125}{20}$$ + $$\frac{175}{20}$$
= $$\frac{175 + 125}{20}$$
= $$\frac{300}{20}$$
= 15 miles
Hence, from the above,
We can conclude that the number of miles you do roller- skate on Monday and Tuesday combined is: 15 miles
Question 13.
DIG DEEPER!
An artist paints a rectangular mural. The mural is 4$$\frac{1}{3}$$ feet wide. The length is 2$$\frac{1}{4}$$ times the width. What is the area of the mural?
The area of the mural is: $$\frac{117}{12}$$ square feet
Explanation:
It is given that an artist paints a rectangular mural. The mural is 4$$\frac{1}{3}$$ feet wide. The length is 2$$\frac{1}{4}$$ times the width.
So,
The area of the mural = ( The width of the mural ) × ( The length of the mural )
= 4$$\frac{1}{3}$$ × 2$$\frac{1}{4}$$
= $$\frac{13}{3}$$ × $$\frac{9}{4}$$
= $$\frac{13 × 9}{4 × 3}$$
= $$\frac{117}{12}$$
Hence, from the above,
We can conclude that the area of the mural is: $$\frac{117}{12}$$ square feet
### Multiply Mixed Numbers Homework & Practice 9.7
Multiply.
Question 1.
1$$\frac{1}{2}$$ × 1$$\frac{1}{8}$$ = ______
1$$\frac{1}{2}$$ × 1$$\frac{1}{8}$$ = $$\frac{18}{16}$$
Explanation:
The given fractions are: 1$$\frac{1}{2}$$ and 1$$\frac{1}{8}$$
The representation of the fractions in the improper form is: $$\frac{3}{2}$$ and $$\frac{9}{8}$$
So,
$$\frac{9}{8}$$ × $$\frac{3}{2}$$
= $$\frac{9 × 3}{8 × 2}$$
= $$\frac{18}{16}$$
Hence,
1$$\frac{1}{2}$$ × 1$$\frac{1}{8}$$ = $$\frac{18}{16}$$
Question 2.
1$$\frac{5}{6}$$ × 2$$\frac{1}{4}$$ = ______
1$$\frac{5}{6}$$ × 2$$\frac{1}{4}$$ = $$\frac{99}{24}$$
Explanation:
The given fractions are: 1$$\frac{5}{6}$$ and 2$$\frac{1}{4}$$
The representation of the fractions in the improper form is: $$\frac{11}{6}$$ and $$\frac{9}{4}$$
So,
$$\frac{9}{4}$$ × $$\frac{11}{6}$$
= $$\frac{9 × 11}{4 × 6}$$
= $$\frac{99}{24}$$
Hence,
1$$\frac{5}{6}$$ × 2$$\frac{1}{4}$$ = $$\frac{99}{24}$$
Evaluate.
Question 3.
2$$\frac{3}{8}$$ × 2$$\frac{3}{4}$$ = ______
2$$\frac{3}{8}$$ × 2$$\frac{3}{4}$$ = $$\frac{209}{32}$$
Explanation:
The given fractions are: 2$$\frac{3}{8}$$ and 2$$\frac{3}{4}$$
The representation of the fractions in the improper form is: $$\frac{19}{8}$$ and $$\frac{11}{4}$$
So,
$$\frac{19}{8}$$ × $$\frac{11}{4}$$
= $$\frac{19 × 11}{4 × 8}$$
= $$\frac{209}{32}$$
Hence,
2$$\frac{3}{8}$$ × 2$$\frac{3}{4}$$ = $$\frac{209}{32}$$
Question 4.
4$$\frac{1}{6}$$ × 3$$\frac{2}{7}$$ = ______
4$$\frac{1}{6}$$ × 3$$\frac{2}{7}$$ = $$\frac{575}{42}$$
Explanation:
The given fractions are: 4$$\frac{1}{6}$$ and 3$$\frac{2}{7}$$
The representation of the fractions in the improper form is: $$\frac{25}{6}$$ and $$\frac{23}{7}$$
So,
$$\frac{23}{7}$$ × $$\frac{25}{6}$$
= $$\frac{23 × 25}{7 × 6}$$
= $$\frac{575}{42}$$
Hence,
4$$\frac{1}{6}$$ × 3$$\frac{2}{7}$$ = $$\frac{575}{42}$$
Question 5.
2$$\frac{1}{3}$$ × 3$$\frac{9}{10}$$ × 5$$\frac{1}{5}$$ = ______
2$$\frac{1}{3}$$× 3$$\frac{9}{10}$$ ×5$$\frac{1}{5}$$ = $$\frac{7,098}{150}$$
Explanation:
The given fractions are: 2$$\frac{1}{3}$$, 3$$\frac{9}{10}$$ and 5$$\frac{1}{5}$$
The representation of the fractions in the improper form is: $$\frac{7}{3}$$, $$\frac{39}{10}$$ and $$\frac{26}{5}$$
So,
$$\frac{7}{3}$$ × $$\frac{39}{10}$$ ×$$\frac{26}{5}$$
= $$\frac{7 × 39 × 26}{3 × 10 × 5}$$
= $$\frac{7,098}{150}$$
Hence,
2$$\frac{1}{3}$$× 3$$\frac{9}{10}$$ ×5$$\frac{1}{5}$$ = $$\frac{7,098}{150}$$
Question 6.
(1$$\frac{7}{8}$$ + 4$$\frac{4}{5}$$) × 2$$\frac{1}{12}$$ = _______
(1$$\frac{7}{8}$$ + 4$$\frac{4}{5}$$) × 2$$\frac{1}{12}$$ = $$\frac{6,675}{480}$$
Explanation:
The given fractions are: 1$$\frac{7}{8}$$, 4$$\frac{4}{5}$$ and 2$$\frac{1}{12}$$
The representation of the fractions in the improper form is: $$\frac{15}{8}$$, $$\frac{24}{5}$$ and $$\frac{25}{12}$$
So,
(1$$\frac{7}{8}$$ + 4$$\frac{4}{5}$$) × $$\frac{25}{12}$$
= ( $$\frac{15}{8}$$ + $$\frac{24}{5}$$ ) × $$\frac{25}{12}$$
= ( $$\frac{75}{40}$$ + $$\frac{192}{40}$$ ) × $$\frac{25}{12}$$
= $$\frac{75 + 192}{40}$$ × $$\frac{25}{12}$$
= $$\frac{267}{40}$$ × $$\frac{25}{12}$$
= $$\frac{267 × 25}{40× 12 }$$
= $$\frac{6,675}{480}$$
Hence,
(1$$\frac{7}{8}$$ + 4$$\frac{4}{5}$$) × 2$$\frac{1}{12}$$ = $$\frac{6,675}{480}$$
Question 7.
Structure
Find the missing numbers.
The missing numbers are: 5 and $$\frac{3}{10}$$
Explanation:
Let the missing numbers be p and q
By using the partial products method,
( p × 4 ) + ( p × $$\frac{1}{7}$$ ) = 20 + $$\frac{5}{7}$$
p × ( 4 + $$\frac{1}{7}$$ ) = 20 + $$\frac{5}{7}$$
p × $$\frac{29}{7}$$ = $$\frac{145}{7}$$
So,
p = $$\frac{145}{7}$$ ÷ $$\frac{29}{7}$$
= $$\frac{145}{7}$$ × $$\frac{7}{29}$$
= $$\frac{145 × 7}{7 × 29}$$
= 5
Now,
(q × 4 ) + ( q × $$\frac{1}{7}$$ ) = $$\frac{12}{10}$$ + $$\frac{3}{70}$$
q × ( 4 + $$\frac{1}{7}$$ ) = $$\frac{12}{10}$$ + $$\frac{3}{70}$$
q × $$\frac{29}{7}$$ = $$\frac{84 + 3}{70}$$
q × $$\frac{29}{7}$$ = $$\frac{87}{70}$$
So,
q = $$\frac{87}{70}$$ ÷ $$\frac{29}{7}$$
= $$\frac{87}{70}$$ × $$\frac{7}{29}$$
= $$\frac{87 × 7}{70 × 29}$$
= $$\frac{3}{10}$$
Hence, from the above,
We can conclude that the missing numbers are: 5 and $$\frac{3}{10}$$
Question 8.
YOU BE THE TEACHER
Your friend finds 1$$\frac{11}{12}$$ × 2$$\frac{3}{8}$$ . Is your friend correct? Explain.
Explanation:
The given mixed numbers are: 1$$\frac{11}{12}$$ and 2$$\frac{3}{8}$$
The representation of the mixed numbers in the fraction form is: $$\frac{23}{12}$$ and $$\frac{19}{8}$$
So,
$$\frac{23}{12}$$ × $$\frac{19}{8}$$
= $$\frac{23 × 19}{8 × 12}$$
= $$\frac{437}{96}$$
= 4$$\frac{53}{96}$$
Hence, from the above,
We can conclude that your friend is correct
Question 9.
Modeling Real Life
Your friend earns 7$$\frac{1}{2}$$ dollars each hour. Will she earn enough money to buy a $35 toy after working 4$$\frac{3}{4}$$ hours? Answer: she can’t earn enough money to buy$35 toy after working 4$$\frac{3}{4}$$ hours
Explanation:
It is given that your friend earns 7$$\frac{1}{2}$$ dollars each hour.
It is given that will she earn enough money to buy a $35 toy after working 4$$\frac{3}{4}$$ hours So, The number of dollars she can earn = 7$$\frac{1}{2}$$ × 4$$\frac{3}{4}$$ = $$\frac{15}{2}$$ × $$\frac{19}{4}$$ = $$\frac{15 × 19}{4 × 2}$$ = $$\frac{285}{8}$$ Now, The representation of$35 as the multiple of 8 is: $$\frac{280}{8}$$
So,
By comparing the given money with he money she earned,
We can say that she can’t buy the toy.
Hence, from the above,
We can conclude that she can’t earn enough money to buy a $35 toy after working 4$$\frac{3}{4}$$ hours Question 10. Modeling Real Life One class collects 8$$\frac{1}{4}$$ pounds of recyclable materials. Another class collects 1$$\frac{1}{2}$$ times as many pounds as the first class. How many pounds of recyclable materials do the two classes collect altogether? Answer: The number of pounds of recyclable materials the two classes collects together is: $$\frac{231}{8}$$ pounds Explanation: It is given that one class collects 8$$\frac{1}{4}$$ pounds of recyclable materials. Another class collects 1$$\frac{1}{2}$$ times as many pounds as the first class. So, The number of pounds of recyclable materials collected by another class = 8$$\frac{1}{4}$$ × 1$$\frac{1}{2}$$ = $$\frac{33}{4}$$ × $$\frac{3}{2}$$ = $$\frac{99}{8}$$ pounds So, The number of pouds of recyclable materials collected by two classes together = 8$$\frac{1}{4}$$ + $$\frac{99}{8}$$ = $$\frac{33}{4}$$ + $$\frac{99}{8}$$ = $$\frac{132 + 99}{8}$$ = $$\frac{231}{8}$$ pounds Hence, from the above, We can conclude that the number of pounds of recyclable materials the two classes collects together is: $$\frac{231}{8}$$ pounds Review & Refresh Find the product. Question 11. 6 × 5.7 = ______ Answer: 6 × 5.7 = 34.2 Explanation: By using the partial products method, 6 × 5.7 = 6 × ( 5 + 0.7 ) = ( 6 × 5 ) + ( 6 × 0.7 ) = 30 + 4.2 = 34.2 Hence, 6 × 5.7 = 34.2 Question 12. 0.84 × 9 = ______ Answer: 0.84 × 9 = 73.17 Explanation: By using the partial quotients method, 0.84 × 9 = ( 0.81 + 0.03 ) × 9 = ( 0.81 × 9 ) + ( 0.03 × 9 ) = 72.9 + 0.27 = 73.17 Hence, 0.84 × 9 = 73.17 ### Lesson 9.8 Compare Factors and Products Explore and Grow Without calculating, order the rectangles by area from least to greatest. Explain your reasoning. Answer: Let the given rectangles be named A), B), C), and D) So, The side lengths of A are: 1 and 1$$\frac{1}{2}$$ The side lengths of B are: 1$$\frac{1}{3}$$ and $$\frac{11}{12}$$ The side lengths of C are: $$\frac{3}{2}$$ and $$\frac{11}{12}$$ The side lengths of D are: $$\frac{11}{10}$$ and $$\frac{3}{2}$$ Now, The area of A is: $$\frac{1}{2}$$ The area of B is: $$\frac{44}{36}$$ The area of C is: $$\frac{33}{24}$$ The area of D is: $$\frac{33}{20}$$ For comparison make the denominators of the areas of the four rectangles equal. So, The area of A is: $$\frac{18}{36}$$ The area of B is: $$\frac{44}{36}$$ So, by comparing these 2 areas, We can say that B is greater Now, The area of C is: $$\frac{165}{120}$$ The area of D is: $$\frac{198}{120}$$ So, by comparing these 2 areas, We can say that D is greater Hence, from the above, We can conclude that the order of rectangles by areas from the greatest to the least is: B > D> A > C Construct Arguments Explain your strategy to your partner. Compare your strategies. Answer: The strategy you followed is: A) Write the side lengths of the rectangles B) Find the areas of the four rectangles C) Make the denominators of all the four areas of rectangles equal D) Compare the numerators of all the four areas of the four rectangles Think and Grow: Compare Factors and Products Key Idea When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number. Example Without calculating, tell whether the product 3$$\frac{1}{8}$$ × $$\frac{5}{6}$$ is less than, greater than, or equal to or each of its factors. Show and Grow Without calculating, tell whether the product is less than, greater than, or equal to each of its factors. Question 1. 8 × $$\frac{3}{10}$$ Answer: The value of 8 × $$\frac{3}{10}$$ is less than 8 Explanation: The given numbers are: 8 and $$\frac{3}{10}$$ We know that, When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number. So, $$\frac{3}{10}$$ is less than 1 So, 8 × $$\frac{3}{10}$$ is less than 8 Hence, from the above, We can conclude that the value of 8 × $$\frac{3}{10}$$ is less than 8 Question 2. $$\frac{4}{4}$$ × 5$$\frac{2}{3}$$ Answer: The value of $$\frac{4}{4}$$ × 5$$\frac{2}{3}$$ is greater than $$\frac{4}{4}$$ Explanation: The given fractions are: $$\frac{4}{4}$$ and 5$$\frac{2}{3}$$ We know that, When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number. We know that, $$\frac{4}{4}$$ is 1 Now, The representation of 5$$\frac{2}{3}$$ in the improper fraction form is: $$\frac{17}{3}$$ So, $$\frac{17}{3}$$ is greater than 1 So, $$\frac{4}{4}$$ × 5$$\frac{2}{3}$$ is greater than 1 Hence, from the above, We can conclude that the value of $$\frac{4}{4}$$ × 5$$\frac{2}{3}$$ is greater than $$\frac{4}{4}$$ Question 3. $$\frac{4}{3}$$ × $$\frac{1}{6}$$ Answer: The value of $$\frac{4}{3}$$ × $$\frac{1}{6}$$ is greater than $$\frac{1}{6}$$ or less than $$\frac{4}{3}$$ Explanation: The given fractions are: $$\frac{4}{3}$$ and $$\frac{1}{6}$$ We know that, When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number. Now, $$\frac{4}{3}$$ is greater than 1 $$\frac{1}{6}$$ is less than 1 So, $$\frac{4}{3}$$ × $$\frac{1}{6}$$ is less than $$\frac{4}{3}$$ Hence, from the above, We can conclude that the value of $$\frac{4}{3}$$ × $$\frac{1}{6}$$ is greater than $$\frac{1}{6}$$ or less than $$\frac{4}{3}$$ Apply and Grow Without calculating, tell whether the product is less than, greater than, or equal to each of its factors. Question 4. $$\frac{1}{4}$$ × $$\frac{1}{12}$$ Answer: The value of $$\frac{1}{4}$$ × $$\frac{1}{12}$$ is less than 1 Explanation: The given fractions are: latex]\frac{1}{4}[/latex] and $$\frac{1}{12}$$ We know that, When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number. So, $$\frac{1}{4}$$ is less than 1 $$\frac{1}{12}$$ is less than 1 So, $$\frac{1}{4}$$ × $$\frac{1}{12}$$ is less than 1 Hence, from the above, We can conclude that the value of $$\frac{1}{4}$$ × $$\frac{1}{12}$$ is less than 1 Question 5. 3$$\frac{4}{5}$$ × 6$$\frac{7}{8}$$ Answer: The value of 3$$\frac{4}{5}$$ × 6$$\frac{7}{8}$$ is greater than 1 Explanation: The given fractions are: 3$$\frac{4}{5}$$ and 6$$\frac{7}{8}$$ The representation of 3$$\frac{4}{5}$$ and 6$$\frac{7}{8}$$ in the improper fractions form is: $$\frac{19}{5}$$ and $$\frac{55}{8}$$ We know that, When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number. So, $$\frac{19}{5}$$ is greater than 1 $$\frac{55}{8}$$ is greater than 1 So, 3$$\frac{4}{5}$$ × 6$$\frac{7}{8}$$ is greater than 1 Hence, from the above, We can conclude that the value of 3$$\frac{4}{5}$$ × 6$$\frac{7}{8}$$ is greater than 1 Question 6. $$\frac{1}{6}$$ × $$\frac{10}{10}$$ Answer: The value of $$\frac{1}{6}$$ × $$\frac{10}{10}$$ is less than 1 Explanation: The given fractions are: $$\frac{1}{6}$$ and $$\frac{10}{10}$$ We know that, When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number. We know that, $$\frac{10}{10}$$ is 1 So, $$\frac{1}{6}$$ is less than 1 So, $$\frac{1}{6}$$ × $$\frac{10}{10}$$ is less than 1 Hence, from the above, We can conclude that the value of $$\frac{1}{6}$$ × $$\frac{10}{10}$$ is less than 1 Question 7. $$\frac{2}{3}$$ × 5 Answer: The value of $$\frac{2}{3}$$ × 5 is less than 5 Explanation: The given fractions are: $$\frac{2}{3}$$ and 5 We know that, When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number. So, $$\frac{2}{3}$$ is less than 1 So, $$\frac{2}{3}$$ × 5 is less than 5 Hence, from the above, We can conclude that the value of $$\frac{2}{3}$$ × 5 is less than 5 Question 8. $$\frac{7}{10}$$ × 4$$\frac{8}{9}$$ Answer: The value of $$\frac{7}{10}$$ × 4$$\frac{8}{9}$$ is greater than 1 Explanation: The given fractions are: $$\frac{7}{10}$$ and 4$$\frac{8}{9}$$ The representation of 4$$\frac{8}{9}$$ in the improper fraction form is: $$\frac{44}{9}$$ We know that, When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number. So, $$\frac{7}{10}$$ is less than 1 $$\frac{44}{9}$$ is greater than 1 So, $$\frac{7}{10}$$ × 4$$\frac{8}{9}$$ is greater than 1 Hence, from the above, We can conclude that the value of $$\frac{7}{10}$$ × 4$$\frac{8}{9}$$ is greater than 1 Question 9. $$\frac{9}{2}$$ × 1$$\frac{3}{4}$$ Answer: The value of $$\frac{9}{2}$$ × 1$$\frac{3}{4}$$ is greater than 1 Explanation: The given fractions are: $$\frac{9}{2}$$ × 1$$\frac{3}{4}$$ The representation of 1$$\frac{3}{4}$$ in the improper fraction form is: $$\frac{7}{4}$$ We know that, When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number. So, $$\frac{9}{2}$$ is greater than 1 $$\frac{7}{4}$$ is greater than 1 So, $$\frac{9}{2}$$ × 1$$\frac{3}{4}$$ is greater than 1 Hence, from the above, We can conclude that the value of $$\frac{9}{2}$$ × 1$$\frac{3}{4}$$ is greater than 1 Without calculating, order the products from least to greatest. Question 10. Answer: The order of products from the least to the greatest is: $$\frac{5}{6}$$ × $$\frac{1}{3}$$ < $$\frac{5}{6}$$ × $$\frac{7}{7}$$ < $$\frac{5}{6}$$ × 1$$\frac{8}{9}$$ Explanation: The given products are: A) $$\frac{5}{6}$$ × $$\frac{1}{3}$$ B) $$\frac{5}{6}$$ × $$\frac{7}{7}$$ C) $$\frac{5}{6}$$ × 1$$\frac{8}{9}$$ So, In A), $$\frac{5}{6}$$ and $$\frac{1}{3}$$ are less than 1 So, $$\frac{5}{6}$$ × $$\frac{1}{3}$$ is less than 1 In B), $$\frac{5}{6}$$ is less than 1 $$\frac{7}{7}$$ is 1 So, $$\frac{5}{6}$$ × $$\frac{7}{7}$$ is less than 1 In C), $$\frac{5}{6}$$ is less than1 The representation of 1$$\frac{8}{9}$$ in the improper fraction form is: $$\frac{17}{9}$$ So, $$\frac{17}{9}$$ is greater than 1 So, $$\frac{5}{6}$$ × 1$$\frac{8}{9}$$ is greater than 1 Hence, from the above, We can conclude that the order of the products from the least to the greatest is: $$\frac{5}{6}$$ × $$\frac{1}{3}$$ < $$\frac{5}{6}$$ × $$\frac{7}{7}$$ < $$\frac{5}{6}$$ × 1$$\frac{8}{9}$$ Question 11. Answer: The order of products from the least to the greatest is: $$\frac{1}{6}$$ × $$\frac{1}{4}$$ < $$\frac{1}{10}$$ × $$\frac{1}{4}$$ < 5$$\frac{7}{10}$$ × $$\frac{1}{4}$$ Explanation: The given products are: A) $$\frac{1}{6}$$ × $$\frac{1}{4}$$ B) $$\frac{1}{10}$$ × $$\frac{1}{4}$$ C) $$\frac{1}{4}$$ × 5$$\frac{7}{10}$$ So, In A), $$\frac{1}{6}$$ and $$\frac{1}{4}$$ are less than 1 So, $$\frac{1}{6}$$ × $$\frac{1}{4}$$ is less than 1 In B), $$\frac{1}{10}$$ is less than 1 $$\frac{1}{4}$$ is less than 1 So, $$\frac{1}{10}$$ × $$\frac{1}{4}$$ is less than 1 In C), $$\frac{1}{4}$$ is less than1 The representation of 5$$\frac{7}{10}$$ in the improper fraction form is: $$\frac{57}{10}$$ So, $$\frac{57}{10}$$ is greater than 1 So, $$\frac{1}{4}$$ × 5$$\frac{7}{10}$$ is greater than 1 Hence, from the above, We can conclude that the order of the products from the least to the greatest is: $$\frac{1}{6}$$ × $$\frac{1}{4}$$ < $$\frac{1}{10}$$ × $$\frac{1}{4}$$ < 5$$\frac{7}{10}$$ × $$\frac{1}{4}$$ Question 12. YOU BE THE TEACHER Your friend says that $$\frac{1}{2}$$ × 8 is half as much as 8. Is your friend correct? Explain. Answer: Yes, your friend is correct Explanation: The given numbers are: $$\frac{1}{2}$$ and 8 So, $$\frac{1}{2}$$ × 8 = $$\frac{1}{2}$$ × $$\frac{8}{1}$$ = $$\frac{1 × 8}{2 × 1}$$ = 4 It is also given that according to your friend, $$\frac{1}{2}$$ × 8 is half as much as 8. So, 8 ÷ 2 = 4 Hence, from the above, We can conclude that your friend is correct. Question 13. DIG DEEPER! Without calculating, tell whether the product is less than, greater than, or equal to 3$$\frac{3}{4}$$. Explain. $$\left(\frac{1}{2} \times 3 \frac{3}{4}\right)$$ × $$\frac{2}{7}$$ Answer: $$\left(\frac{1}{2} \times 3 \frac{3}{4}\right)$$ × $$\frac{2}{7}$$ is less than 3$$\frac{3}{4}$$ Explanation: The representation of 3$$\frac{3}{4}$$ in the improper fraction form is: $$\frac{15}{4}$$ Now, $$\left(\frac{1}{2} \times 3 \frac{3}{4}\right)$$ × $$\frac{2}{7}$$ = ( $$\frac{1}{2}$$ × $$\frac{15}{4}$$ ) × $$\frac{2}{7}$$ = $$\frac{15}{8}$$ × $$\frac{2}{7}$$ = $$\frac{2 × 15}{7 × 8}$$ = $$\frac{30}{56}$$ = $$\frac{15}{28}$$ Now, $$\frac{15}{4}$$ is multiplied by $$\frac{7}{7}$$ So, $$\frac{15}{4}$$ = $$\frac{105}{28}$$ So, When we compare $$\frac{105}{28}$$ and $$\frac{15}{28}$$ We will get $$\frac{105}{28}$$ is greater Hence, from the above, We can conclude that $$\left(\frac{1}{2} \times 3 \frac{3}{4}\right)$$ × $$\frac{2}{7}$$ is less than 3$$\frac{3}{4}$$ Think and Grow: Modeling Real Life Example Men’s shot put competitions use a shot with a mass of 7$$\frac{1}{4}$$ kilograms. The mass of a bowling ball is $$\frac{7}{8}$$ as much as the mass of the shot. Is the mass of the bowling ball less than, greater than, or equal to the mass of the shot? Write an expression to represent the mass of the bowling ball. So, the mass of the bowling ball is greater than the mass of the shot. Show and Grow Question 14. You practice playing the keyboard for 3$$\frac{1}{2}$$ hours. Your friend practices playing the keyboard for $$\frac{5}{4}$$ as many hours as you. Does your friend practice for fewer hours, more hours, or the same number of hours as you? Answer: You practice more hours than your friend Explanation: It is given that you practice playing the keyboard for 3$$\frac{1}{2}$$ hours. Your friend practices playing the keyboard for $$\frac{5}{4}$$ as many hours as you. So, Now, The representation of 3$$\frac{1}{2}$$ in the improper fraction form is: $$\frac{7}{2}$$ So, The number of hours you practice playing the keyboard is: $$\frac{7}{2}$$ hours The number of hours your friend practice playing the keyboard is: $$\frac{5}{4}$$ hours So, For comparison, make the denominators equal. So, Multiply $$\frac{7}{2}$$ with $$\frac{2}{2}$$ So, $$\frac{7}{2}$$ = $$\frac{14}{4}$$ By comparing the timings , We can say that $$\frac{14}{4}$$ > $$\frac{5}{4}$$ Hence, from the above, We can conclude that you practice more hours than your friend Question 15. The original price of a telescope is$99. The sale price is $$\frac{4}{5}$$ of the original price. An astrologist buys the telescope at its sale price and uses a half-off-coupon. What fraction of the original price does the astrologist pay for the telescope?
The fraction of the original price the astrologist pays for the telescope is: $$\frac{4}{10}$$
Explanation:
It is given that the original price of a telescope is \$99. The sale price is $$\frac{4}{5}$$ of the original price. and an astrologist buys the telescope at its sale price and uses a half-off-coupon.
So,
The cost of the telescope that an astrologist bought = ( The $$\frac{4}{5}$$th of the original price ) × ( Half-off -coupon on the $$\frac{4}{5}$$th of the original price )
= $$\frac{4}{5}$$ × $$\frac{1}{2}$$
= $$\frac{4 × 1}{5 × 2}$$
= $$\frac{4}{10}$$
Hence, from the above,
We can conclude that the fraction of the original price the astrologist pays for the telescope is: $$\frac{4}{10}$$
Question 16.
The Abraj Al-Bait Clock Tower is $$\frac{6}{10}$$ kilometer tall. Zifeng Tower is $$\frac{3}{4}$$ as tall as the clock tower. Is Zifeng Tower shorter than, taller than, or the same height as the Abraj Al-Bait Clock Tower? What is the height of each tower in meters?
The Abraj al-Bait Clock tower is taller than the Zifeng tower
The height of Abraj al-Bait Clock tower is: $$\frac{24}{40}$$
The height of Zifeng tower is: $$\frac{18}{40}$$
Explanation:
It is given that the Abraj Al-Bait Clock Tower is $$\frac{6}{10}$$ kilometer tall and Zifeng Tower is $$\frac{3}{4}$$ as tall as the clock tower.
So,
The height of Abraj al-Bait Clock tower is: $$\frac{6}{10}$$ kilometer
Now,
The height of Zifeng tower is = $$\frac{6}{10}$$ × $$\frac{3}{4}$$
= $$\frac{6 × 3}{10 × 4}$$
= $$\frac{18}{40}$$
Now, for comparison, make the denomonators of both the towers equal.
So,
$$\frac{6}{10}$$ is multiplied by $$\frac{4}{4}$$
So,
$$\frac{6}{10}$$ = $$\frac{24}{40}$$
By comparing the heights of the two towers,
We can say that
$$\frac{24}{40}$$ > $$\frac{18}{40}$$
Hence, from the above,
We can conclude that
The Abraj al-Bait Clock tower is taller than the Zifeng tower
The height of Abraj al-Bait Clock tower is: $$\frac{24}{40}$$
The height of Zifeng tower is: $$\frac{18}{40}$$
### Compare Factors and Products Homework & Practice 9.8
Without calculating, tell whether the product is less than, greater than, or equal to each of its factors.
Question 1.
1$$\frac{3}{4}$$ × 6
The value of 1$$\frac{3}{4}$$ × 6 is greater than 6
Explanation:
The given numbers are: 1$$\frac{3}{4}$$ and 6
The representation of 1$$\frac{3}{4}$$ in the improper fraction form is: $$\frac{7}{4}$$
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
$$\frac{7}{4}$$ is greater than 1
So,
1$$\frac{3}{4}$$ × 6 is greater than 6
Hence, from the above,
We can conclude that the value of 1$$\frac{3}{4}$$ × 6 is greater than 6
Question 2.
$$\frac{5}{12}$$ × $$\frac{1}{6}$$
The value of $$\frac{5}{12}$$ × $$\frac{1}{6}$$ is less than 1
Explanation;
The given fractions are: $$\frac{5}{12}$$ and $$\frac{1}{6}$$
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
$$\frac{5}{12}$$ is less than 1
$$\frac{1}{6}$$ is less than 1
So,
$$\frac{5}{12}$$ × $$\frac{1}{6}$$ is less than 1
Hence, from the above,
We can conclude that the value of $$\frac{5}{12}$$ × $$\frac{1}{6}$$ is less than 1
Question 3.
$$\frac{2}{7}$$ × $$\frac{5}{5}$$
The value of $$\frac{2}{7}$$ × $$\frac{5}{5}$$ is less than 1
Explanation:
The given fractions are: $$\frac{2}{7}$$ and $$\frac{5}{5}$$
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
$$\frac{2}{7}$$ is less than 1
We know that
$$\frac{5}{5}$$ is 1
So,
$$\frac{2}{7}$$ × $$\frac{5}{5}$$ is less than 1
Hence, from the above,
We can conclude that the value of $$\frac{2}{7}$$ × $$\frac{5}{5}$$ is less than 1
Question 4.
3$$\frac{4}{5}$$ × 2$$\frac{9}{10}$$
The value of 3$$\frac{4}{5}$$ × 2$$\frac{9}{10}$$ is greater than 1
Explanation:
the given mixed fractions are: 3$$\frac{4}{5}$$ and 2$$\frac{9}{10}$$
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
The representation of the mixed numbers in the fraction form is: $$\frac{19}{5}$$ and $$\frac{29}{10}$$
So,
$$\frac{19}{5}$$ is greater than 1
$$\frac{29}{10}$$ is greater than
So,
3$$\frac{4}{5}$$ × 2$$\frac{9}{10}$$ is greater than 1
Hence, from the above,
We can conclude that the value of 3$$\frac{4}{5}$$ × 2$$\frac{9}{10}$$ is greater than 1
Question 5.
8 × $$\frac{2}{3}$$
The value of 8 is greater than $$\frac{2}{3}$$
Explanation:
The given numbers are: 8 and $$\frac{2}{3}$$
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
$$\frac{2}{3}$$ is less than 1
So,
8 is greater than $$\frac{2}{3}$$
Hence, from the above,
We can conclude that the value of 8 is greater than $$\frac{2}{3}$$
Question 6.
1$$\frac{7}{8}$$ × $$\frac{1}{4}$$
The value of 1$$\frac{7}{8}$$ × $$\frac{1}{4}$$ is less than 1
Explanation:
The given fractions are: 1$$\frac{7}{8}$$ and $$\frac{1}{4}$$
The representation of 1$$\frac{7}{8}$$ in the fraction form is: $$\frac{15}{8}$$
We know that,
When you multiply a number by a fraction greater than 1, the product is greater than the number. When you multiply a number by a fraction less than 1, the product is less than the number.
So,
$$\frac{15}{8}$$ is greater than 1
$$\frac{1}{4}$$ is less than 1
So,
1$$\frac{7}{8}$$ × $$\frac{1}{4}$$ is less than 1
Hence, from the above,
We can conclude that the value of 1$$\frac{7}{8}$$ × $$\frac{1}{4}$$ is less than 1
Without calculating, order the products from least to greatest.
Question 7.
The order of products from the least to the greatest is:
$$\frac{1}{3}$$ × $$\frac{4}{5}$$ < $$\frac{1}{3}$$ × $$\frac{6}{6}$$ < 8$$\frac{2}{9}$$ × $$\frac{1}{3}$$
Explanation:
The given products are:
A) $$\frac{1}{3}$$ × $$\frac{4}{5}$$
B) $$\frac{1}{3}$$ × $$\frac{6}{6}$$
C) 8$$\frac{2}{9}$$ × $$\frac{1}{3}$$
So,
In A),
$$\frac{1}{3}$$ and $$\frac{4}{5}$$ are less than 1
So,
$$\frac{1}{3}$$ × $$\frac{4}{5}$$ is less than 1
In B),
$$\frac{1}{3}$$ is less than 1
$$\frac{6}{6}$$ is 1
So,
$$\frac{1}{3}$$ × $$\frac{6}{6}$$ is less than 1
In C),
$$\frac{1}{3}$$ is less than1
The representation of 8$$\frac{2}{9}$$ in the improper fraction form is: $$\frac{74}{9}$$
So,
$$\frac{74}{9}$$ is greater than 1
So,
$$\frac{1}{3}$$ × 8$$\frac{2}{9}$$ is greater than 1
Hence, from the above,
We can conclude that the order of the products from the least to the greatest is:
$$\frac{1}{3}$$ × $$\frac{4}{5}$$ < $$\frac{1}{3}$$ × $$\frac{6}{6}$$ < 8$$\frac{2}{9}$$ × $$\frac{1}{3}$$
Question 8.
The order of products from the least to the greatest is:
$$\frac{1}{12}$$ × $$\frac{3}{5}$$ < 4 × $$\frac{3}{5}$$ < 2$$\frac{1}{2}$$ × $$\frac{3}{5}$$
Explanation:
The given products are:
A) $$\frac{1}{12}$$ × $$\frac{3}{5}$$
B) 4 × $$\frac{3}{5}$$
C) 2$$\frac{1}{2}$$ × $$\frac{3}{5}$$
So,
In A),
$$\frac{1}{12}$$ and $$\frac{3}{5}$$ are less than 1
So,
$$\frac{1}{12}$$ × $$\frac{3}{5}$$ is less than 1
In B),
$$\frac{3}{5}$$ is less than 1
So,
4 × $$\frac{3}{5}$$ is less than 4
In C),
$$\frac{3}{5}$$ is less than1
The representation of 2$$\frac{1}{2}$$ in the improper fraction form is: $$\frac{5}{2}$$
So,
$$\frac{5}{2}$$ is greater than 1
So,
$$\frac{3}{5}$$ × 2$$\frac{1}{2}$$ is greater than 1
Hence, from the above,
We can conclude that the order of the products from the least to the greatest is:
$$\frac{1}{12}$$ × $$\frac{3}{5}$$ < 4 × $$\frac{3}{5}$$ < 2$$\frac{1}{2}$$ × $$\frac{3}{5}$$
Question 9.
Logic
Without calculating, use >, <, or = to make the statement true. Explain.
The value of 3$$\frac{1}{3}$$ × $$\frac{1}{6}$$ is less than 3$$\frac{1}{2}$$
Explanation:
The given fractions are: 3$$\frac{1}{3}$$, $$\frac{1}{6}$$ and 3$$\frac{1}{2}$$
The representations of the 3$$\frac{1}{3}$$ and 3$$\frac{1}{2}$$ in the fraction form is: $$\frac{7}{2}$$ and $$\frac{10}{3}$$
So,
3$$\frac{1}{3}$$ × $$\frac{1}{6}$$
= $$\frac{10}{3}$$ × $$\frac{1}{6}$$
= $$\frac{10}{18}$$
To compare the fractions, equate the denominators
So,
Multiply $$\frac{7}{2}$$ by $$\frac{9}{9}$$
= $$\frac{63}{18}$$
So,
By comparison,
We can say that
$$\frac{10}{18}$$ < $$\frac{63}{18}$$
Hence, from the above,
We can conclude that the value of 3$$\frac{1}{3}$$ × $$\frac{1}{6}$$ is less than 3$$\frac{1}{2}$$
Question 10.
Logic
Without calculating, determine which number makes the statement true.
Let the missing number be x.
The options for x are given as:
A) 1 B) $$\frac{1}{2}$$ and C) 1$$\frac{4}{5}$$
So,
From the given three fractions,
The value of x is: 1$$\frac{4}{5}$$
Explanation:
The given multiplication equation is:
x × 1$$\frac{7}{8}$$ is greater than 1$$\frac{7}{8}$$
So,
To find the value of x, there are 3 options. They are:
A) 1 B) $$\frac{1}{2}$$ and C) 1$$\frac{4}{5}$$
Now,
Let x = 1$$\frac{4}{5}$$
So,
1$$\frac{4}{5}$$ × 1$$\frac{7}{8}$$
= $$\frac{9}{5}$$ × $$\frac{15}{8}$$
= $$\frac{27}{8}$$
So, by comparison,
We will get,
$$\frac{27}{8}$$ is greater than $$\frac{15}{8}$$
Hence, from the above,
We can conclude that the missing number is: 1$$\frac{4}{5}$$
Question 11.
Reasoning
Why does multiplying by a fraction greater than one result in a product greater than the original number?
In the given fraction, if the numerator is greater than the denominator, then the given fraction is greater than 1
So,
When the whole number is multiplied by the fraction which is greater than 1, the result in a product will be greater than the original number i.e, the whole number
Question 12.
Modeling Real Life
You snowboard 1$$\frac{7}{8}$$ miles. Your friend snowboards 3$$\frac{2}{3}$$ times as far as you. Does your friend snowboard fewer miles, more miles, or the same number of miles as you?
The number of miles you snowboarded is less than your friend
Explanation:
It is given that you snowboard 1$$\frac{7}{8}$$ miles. Your friend snowboards 3$$\frac{2}{3}$$ times as far as you.
So,
The number of miles you snowboard is: 1$$\frac{7}{8}$$ miles
So,
The number of miles your friend snowboard is = 1$$\frac{7}{8}$$ × 3$$\frac{2}{3}$$
= $$\frac{15}{8}$$ × $$\frac{11}{3}$$
= $$\frac{15 × 11}{3 × 8}$$
= $$\frac{165}{24}$$ miles
So,
For comparison we have to make the denominators equal.
So,
Multiply $$\frac{15}{8}$$ by $$\frac{3}{3}$$
So,
$$\frac{15}{8}$$ = $$\frac{45}{24}$$
So,
By comparing, we will get
$$\frac{45}{24}$$ is less than $$\frac{165}{24}$$ miles
Hence, from the above,
We can conclude that the number of miles you snowboarded is less than your friend
Question 13.
Modeling Real Life
A pet owner has three dogs. The youngest dog weighs $$\frac{1}{4}$$ as much as the second oldest dog. The oldest dog weighs 1$$\frac{1}{4}$$ as much as the second oldest. The second oldest weighs 20 pounds. Which dog weighs the most? the least?
The dog that weighs the most is: The oldest dog
The dog that weighs the least is: The youngest dog
Explanation:
It is given that a pet owner has three dogs. The youngest dog weighs $$\frac{1}{4}$$ as much as the second oldest dog. The oldest dog weighs 1$$\frac{1}{4}$$ as much as the second oldest. The second oldest weighs 20 pounds.
So,
The weight of the second oldest dog is: 20 pounds
Now,
The weight of the oldest dog = 1$$\frac{1}{4}$$ × ( The weight of the oldest second dog )
= 1$$\frac{1}{4}$$ × 20
= $$\frac{5}{4}$$ × 20
= $$\frac{5 × 20}{4 × 1}$$
= 25 pounds
Now,
The weight of the youngest dog = $$\frac{1}{4}$$ × ( The weight of the second oldest dog )
= $$\frac{1}{4}$$ × 20
= $$\frac{1 × 20}{4}$$
= 5 pounds
Hence, from the above,
We can conclude that
The dog that weighs the most is: The oldest dog
The dog that weighs the least is: The youngest dog
Review & Refresh
Compare.
Question 14.
40.5 is greater than 40.13
Explanation:
The given decimal numbers are: 40.5 and 40.13
For comparison, compare the place value of the given digits
If there are more than 2 digits after the decimal, first compare the tenths position and after that only compare the hundredths position
If we can get the result in the tenths position only, then there is no need for further comparison
Hence, from the above,
We can conclude that 40.5 is greater than 40.13
Question 15.
13.90 is equal to 13.9
Explanation:
The given decimal numbers are: 13.90 and 13.9
For comparison, compare the place value of the given digits
If there are more than 2 digits after the decimal, first compare the tenths position and after that only compare the hundredths position
If we can get the result in the tenths position only, then there is no need for further comparison
Hence, from the above,
We can conclude that 13.90 is equal to 13.9
Question 16.
32.006 is less than 32.06
Explanation:
The given decimal numbers are: 32.006 and 32.06
For comparison, compare the place value of the given digits
If there are more than 2 digits after the decimal, first compare the tenths position and after that only compare the hundredths position
If we can get the result in the tenths position only, then there is no need for further comparison
Hence, from the above,
We can conclude that 32.006 is less than 32.06
Question 1.
You see a rock formation at a national park. The formation has layers that formed millions of years ago when particles settled in the water and became rock. You make a model of the rock formation using $$\frac{3}{16}$$ -inch foam sheets.
a. The three types of sedimentary rocks are limestone, sandstone, and shale. Use the number of foam sheets to find the height of each sedimentary rock layer.
The height of limestone is: $$\frac{9}{4}$$ inches
The height of sandstone is: $$\frac{3}{2}$$ inches
The height of shale is: $$\frac{15}{4}$$ inches
Explanation:
It is given that the height of each foam sheet is: $$\frac{3}{16}$$ -inches
From the table,
The number of foam sheets for Limestone is: 12
The number of foam sheets for sandstone is: 8
The number of foam sheets for shale is: 20
So,
The height of limestone is: ( The height of each foam sheet ) × ( The total number of foam sheets )
= $$\frac{3}{16}$$ × 12
= $$\frac{3}{16}$$ × $$\frac{12}{1}$$
= $$\frac{9}{4}$$ inches
The height of sandstone is: ( The height of each foam sheet ) × ( The total number of foam sheets )
= $$\frac{3}{16}$$ × 8
= $$\frac{3}{16}$$ × $$\frac{8}{1}$$
= $$\frac{3}{2}$$ inches
The height of shale is: ( The height of each foam sheet ) × ( The total number of foam sheets )
= $$\frac{3}{16}$$ × 20
= $$\frac{3}{16}$$ × $$\frac{20}{1}$$
= $$\frac{15}{4}$$ inches
Hence, from the above,
We can conclude that
The height of limestone is: $$\frac{9}{4}$$ inches
The height of sandstone is: $$\frac{3}{2}$$ inches
The height of shale is: $$\frac{15}{4}$$ inches
b. What is the combined height of the sedimentary rock layers?
The combined height of the sedimentary layers is: $$\frac{15}{2}$$ inches
Explanation:
The combined height of the sedimentary rock layers = The height of limestone + The height of sandstone + The height of the shale
= $$\frac{9}{4}$$ + $$\frac{3}{2}$$ + $$\frac{15}{4}$$
= $$\frac{24}{4}$$ + $$\frac{3}{2}$$
= $$\frac{12}{2}$$ + $$\frac{3}{2}$$
= $$\frac{15}{2}$$ inches
Hence, from the above,
We can conclude that the combined height of the sedimentary layers is: $$\frac{15}{2}$$ inches
c. Will you use more foam sheets for the granite layers of the shale layers? Explain.
We use more foam sheets for the granite layers
Explanation:
From the above,
The height of the shale layers is: $$\frac{15}{4}$$ inches
From the table,
The height of the granite layers is: 3 inches
When we compare the two values,
We can observe that the height of shale layers is greater than the granite layers
Hence, from the above,
We can conclude that we will use more foam sheets for granite layers
d. The height of the topsoil layer is 1$$\frac{1}{4}$$ times the height of the sandstone layer. How many foam sheets do you use in the topsoil layer?
The height of the topsoil layer is: $$\frac{15}{8}$$ inches
Explanation:
From the above,
The height of the sandstone layer is: $$\frac{3}{2}$$ inches
So,
The height of topsoil layer = 1$$\frac{1}{4}$$ × $$\frac{3}{2}$$
= $$\frac{5}{4}$$ × $$\frac{3}{2}$$
= $$\frac{15}{8}$$ inches
Hence, from the above,
We can conclude that the height of the topsoil layer is: $$\frac{15}{8}$$ inches
e. On your model, 1 inch represents 40 feet. What is the actual height of the rock formation?
The actual height of the rock formation is: 495 feet
Explanation:
From the above,
The height of the sedimentary layers is: $$\frac{15}{2}$$ inches
The height of the topsoil layer is: $$\frac{15}{8}$$ inches
The height of the granite layer is: 3 inches
So,
The combined height of the rock formation = $$\frac{15}{2}$$ + $$\frac{15}{8}$$ + 3
= $$\frac{99}{8}$$ inches
But, it is given that,
1 inch = 40 feet
So,
The combined height of the rock formation in feet = $$\frac{99}{8}$$ inches × 40
= 495 feet
Hence, from the above,
We can conclude that the height of the rock formation in feet is: 495 feet
f. Why do you think the rock formation has layers?
The rock formation has layers because of the tectonic plates.
### Multiply Fractions Activity
Fraction Connection: Multiplication
Directions:
1. Players take turns rolling three dice.
3. The first player to get four in a row, horizontally, vertically, or diagonally, wins?
### Multiply Fractions Chapter Practice
9.1 Multiply Whole Numbers by Fractions
Multiply.
Question 1.
5 × $$\frac{1}{2}$$ = _______
5 × $$\frac{1}{2}$$ = $$\frac{5}{2}$$
Explanation:
The given numbers are: 5 and $$\frac{1}{2}$$
So,
5 × $$\frac{1}{2}$$
= $$\frac{5}{1}$$ × $$\frac{1}{2}$$
= $$\frac{5 × 1}{1 × 2}$$
= $$\frac{5}{2}$$
Hence,
5 × $$\frac{1}{2}$$ = $$\frac{5}{2}$$
Question 2.
2 × $$\frac{7}{10}$$ = _______
2 × $$\frac{7}{10}$$ = $$\frac{14}{10}$$
Explanation:
The given numbers are: 2 and $$\frac{7}{10}$$
So,
2 × $$\frac{7}{10}$$
= $$\frac{2}{1}$$ × $$\frac{7}{10}$$
= $$\frac{2 × 7}{1 × 10}$$
= $$\frac{14}{10}$$
Hence,
2 × $$\frac{7}{10}$$ = $$\frac{14}{10}$$
Question 3.
9 × $$\frac{5}{8}$$ = _______
9 × $$\frac{5}{8}$$ = $$\frac{45}{8}$$
Explanation:
The given numbers are: 9 and $$\frac{5}{8}$$
So,
9 × $$\frac{5}{8}$$
= $$\frac{9}{1}$$ × $$\frac{5}{8}$$
= $$\frac{5 × 9}{1 × 8}$$
= $$\frac{45}{8}$$
Hence,
9 × $$\frac{5}{8}$$ = $$\frac{45}{8}$$
Question 4.
6 × $$\frac{71}{100}$$ = _______
6 × $$\frac{71}{100}$$ = $$\frac{426}{100}$$
Explanation:
The given numbers are: 6 and $$\frac{71}{100}$$
So,
6 × $$\frac{71}{100}$$
= $$\frac{6}{1}$$ × $$\frac{71}{100}$$
= $$\frac{6 × 71}{1 × 100}$$
= $$\frac{426}{100}$$
Hence,
6 × $$\frac{71}{100}$$ = $$\frac{426}{100}$$
Question 5.
4 × $$\frac{8}{5}$$ = _______
4 × $$\frac{8}{5}$$ = $$\frac{32}{5}$$
Explanation:
The given numbers are: 4 and $$\frac{8}{5}$$
So,
4 × $$\frac{8}{5}$$
= $$\frac{4}{1}$$ × $$\frac{8}{5}$$
= $$\frac{4 × 8}{1 × 5}$$
= $$\frac{32}{5}$$
Hence,
4 × $$\frac{8}{5}$$ = $$\frac{32}{5}$$
Question 6.
7 × $$\frac{5}{3}$$ = _______
7 × $$\frac{5}{3}$$ = $$\frac{35}{3}$$
Explanation:
The given numbers are: 7 and $$\frac{5}{3}$$
So,
7 × $$\frac{5}{3}$$
= $$\frac{7}{1}$$ × $$\frac{5}{3}$$
= $$\frac{5 × 7}{1 × 3}$$
= $$\frac{35}{3}$$
Hence,
7 × $$\frac{5}{3}$$ = $$\frac{35}{3}$$
9.2 Use Models to Multiply Fractions by Whole Numbers
Multiply. Use a model to help.
Question 7.
$$\frac{2}{5}$$ of 20
20 × $$\frac{2}{5}$$ = 8
Explanation:
The given numbers are: 20 and $$\frac{2}{5}$$
So,
20 × $$\frac{2}{5}$$
= $$\frac{20}{1}$$ × $$\frac{2}{5}$$
= $$\frac{20 × 2}{1 × 5}$$
= $$\frac{8}{1}$$
= 8
Hence,
20 × $$\frac{2}{5}$$ = 8
Question 8.
$$\frac{1}{6}$$ × 12
12 × $$\frac{1}{6}$$ = 2
Explanation:
The given numbers are: 12 and $$\frac{1}{6}$$
So,
12 × $$\frac{1}{6}$$
= $$\frac{12}{1}$$ × $$\frac{1}{6}$$
= $$\frac{12 × 1}{1 × 6}$$
= $$\frac{2}{1}$$
= 2
Hence,
12 × $$\frac{1}{6}$$ = 2
Question 9.
$$\frac{1}{3}$$ × 6
6 × $$\frac{1}{3}$$ = 2
Explanation:
The given numbers are: 6 and $$\frac{1}{3}$$
So,
6 × $$\frac{1}{3}$$
= $$\frac{6}{1}$$ × $$\frac{1}{3}$$
= $$\frac{6 × 1}{1 × 3}$$
= $$\frac{2}{1}$$
= 2
Hence,
6 × $$\frac{1}{3}$$ = 2
Question 10.
$$\frac{5}{16}$$ of 8
8 × $$\frac{5}{16}$$ = $$\frac{5}{2}$$
Explanation:
The given numbers are: 8 and $$\frac{5}{16}$$
So,
8 × $$\frac{5}{16}$$
= $$\frac{8}{1}$$ × $$\frac{5}{16}$$
= $$\frac{5 × 8}{1 × 16}$$
= $$\frac{5}{2}$$
Hence,
8 × $$\frac{5}{16}$$ = $$\frac{5}{2}$$
Question 11.
Modeling Real Life
You have 24 apples. You use $$\frac{1}{4}$$ of them to make a single serving of applesauce. How many apples do you not use?
The number of apples you do not use is: 18 apples
Explanation:
It is given that you have 24 apples and you use $$\frac{1}{4}$$ of them to make a single serving of applesauce.
So,
The number of apples you used = ( The total number of apples ) × ( The fraction of apples you used to make a single serving of applesauce )
= 24 × $$\frac{1}{4}$$
= $$\frac{24}{1}$$ × $$\frac{1}{4}$$
= $$\frac{6}{1}$$
= 6 apples
Now,
The number of apples that do not use = ( The total number of apples ) – ( The number of apples you used )
= 24 – 6
= 18 apples
Hence, from the above,
We can conclude that the number of apples that do not use is: 18 apples
9.3 Multiply Fractions and Whole Numbers
Multiply.
Question 12.
$$\frac{3}{5}$$ × 15 = _______
15 × $$\frac{3}{5}$$ = 9
Explanation:
The given numbers are: 15 and $$\frac{3}{5}$$
So,
15 × $$\frac{3}{5}$$
= $$\frac{15}{1}$$ × $$\frac{3}{5}$$
= $$\frac{15 × 3}{1 × 5}$$
= $$\frac{9}{1}$$
= 9
Hence,
15 × $$\frac{3}{5}$$ = 9
Question 13.
$$\frac{9}{10}$$ × 30 = _______
30 × $$\frac{9}{10}$$ = 27
Explanation:
The given numbers are: 30 and $$\frac{9}{10}$$
So,
30 × $$\frac{9}{10}$$
= $$\frac{30}{1}$$ × $$\frac{9}{10}$$
= $$\frac{30 × 9}{1 × 10}$$
= $$\frac{27}{1}$$
= 27
Hence,
30 × $$\frac{9}{10}$$ = 27
Question 14.
48 × $$\frac{3}{4}$$ = _______
48 × $$\frac{3}{4}$$ = 36
Explanation:
The given numbers are: 48 and $$\frac{3}{4}$$
So,
48 × $$\frac{3}{4}$$
= $$\frac{48}{1}$$ × $$\frac{3}{4}$$
= $$\frac{48 × 3}{1 × 4}$$
= $$\frac{36}{1}$$
= 36
Hence,
48 × $$\frac{3}{4}$$ = 36
Question 15.
11 × $$\frac{5}{9}$$ = _______
11 × $$\frac{5}{9}$$ = $$\frac{55}{9}$$
Explanation:
The given numbers are: 11 and $$\frac{5}{9}$$
So,
11 × $$\frac{5}{9}$$
= $$\frac{11}{1}$$ × $$\frac{5}{9}$$
= $$\frac{5 × 11}{1 × 9}$$
= $$\frac{55}{9}$$
Hence,
11 × $$\frac{5}{9}$$ = $$\frac{55}{9}$$
Question 16.
$$\frac{1}{6}$$ × 19 = _______
19 × $$\frac{1}{6}$$ = $$\frac{19}{6}$$
Explanation:
The given numbers are: 19 and $$\frac{1}{6}$$
So,
19 × $$\frac{1}{6}$$
= $$\frac{19}{1}$$ × $$\frac{1}{6}$$
= $$\frac{19 × 1}{1 × 6}$$
= $$\frac{19}{6}$$
Hence,
19 × $$\frac{1}{6}$$ = $$\frac{19}{6}$$
Question 17.
7 × $$\frac{13}{50}$$ = _______
7 × $$\frac{13}{50}$$ = $$\frac{91}{50}$$
Explanation:
The given numbers are: 7 and $$\frac{13}{50}$$
So,
7 × $$\frac{13}{50}$$
= $$\frac{7}{1}$$ × $$\frac{13}{50}$$
= $$\frac{7 × 13}{1 × 50}$$
= $$\frac{91}{50}$$
Hence,
7 × $$\frac{13}{50}$$ = $$\frac{91}{50}$$
9.4 Use Models to Multiply Fractions
Multiply. Use a model to help.
Question 18.
$$\frac{1}{2}$$ × $$\frac{1}{10}$$ = _______
$$\frac{1}{2}$$ × $$\frac{1}{10}$$ = $$\frac{1}{20}$$
Explanation:
The given numbers are: $$\frac{1}{2}$$ and $$\frac{1}{10}$$
So,
$$\frac{1}{2}$$ × $$\frac{1}{10}$$
= $$\frac{1 × 1}{2 × 10}$$
= $$\frac{1}{20}$$
Hence,
$$\frac{1}{2}$$ × $$\frac{1}{10}$$ = $$\frac{1}{20}$$
Question 19.
$$\frac{1}{5}$$ × $$\frac{1}{9}$$ = _______
$$\frac{1}{5}$$ × $$\frac{1}{9}$$ = $$\frac{1}{45}$$
Explanation:
The given numbers are: $$\frac{1}{5}$$ and $$\frac{1}{9}$$
So,
$$\frac{1}{5}$$ × $$\frac{1}{9}$$
= $$\frac{1 × 1}{5 × 9}$$
= $$\frac{1}{45}$$
Hence,
$$\frac{1}{5}$$ × $$\frac{1}{9}$$ = $$\frac{1}{45}$$
Question 20.
$$\frac{1}{6}$$ × $$\frac{1}{7}$$ = _______
$$\frac{1}{6}$$ × $$\frac{1}{7}$$ = $$\frac{1}{42}$$
Explanation:
The given numbers are: $$\frac{1}{6}$$ and $$\frac{1}{7}$$
So,
$$\frac{1}{6}$$ × $$\frac{1}{7}$$
= $$\frac{1 × 1}{6 × 7}$$
= $$\frac{1}{42}$$
Hence,
$$\frac{1}{6}$$ × $$\frac{1}{7}$$ = $$\frac{1}{42}$$
Question 21.
$$\frac{1}{3}$$ × $$\frac{1}{8}$$ = _______
$$\frac{1}{3}$$ × $$\frac{1}{8}$$ = $$\frac{1}{24}$$
Explanation:
The given numbers are: $$\frac{1}{3}$$ and $$\frac{1}{8}$$
So,
$$\frac{1}{3}$$ × $$\frac{1}{8}$$
= $$\frac{1 × 1}{3 × 8}$$
= $$\frac{1}{24}$$
Hence,
$$\frac{1}{3}$$ × $$\frac{1}{8}$$ = $$\frac{1}{24}$$
Question 22.
$$\frac{2}{5}$$ × $$\frac{1}{3}$$ = _______
$$\frac{2}{5}$$ × $$\frac{1}{3}$$ = $$\frac{2}{15}$$
Explanation:
The given numbers are: $$\frac{2}{5}$$ and $$\frac{1}{3}$$
So,
$$\frac{2}{5}$$ × $$\frac{1}{3}$$
= $$\frac{2 × 1}{5 × 3}$$
= $$\frac{2}{15}$$
Hence,
$$\frac{2}{5}$$ × $$\frac{1}{3}$$ = $$\frac{2}{15}$$
Question 23.
$$\frac{2}{3}$$ × $$\frac{3}{5}$$ = _______
$$\frac{2}{3}$$ × $$\frac{3}{5}$$ = $$\frac{2}{5}$$
Explanation:
The given numbers are: $$\frac{2}{3}$$ and $$\frac{3}{5}$$
So,
$$\frac{2}{3}$$ × $$\frac{3}{5}$$
= $$\frac{2 × 3}{3 × 5}$$
= $$\frac{2}{5}$$
Hence,
$$\frac{2}{3}$$ × $$\frac{3}{5}$$ = $$\frac{2}{5}$$
9.5 Multiply Fractions
Evaluate.
Question 24.
$$\frac{1}{3}$$ × $$\frac{1}{8}$$ = _______
$$\frac{1}{3}$$ × $$\frac{1}{8}$$ = $$\frac{1}{24}$$
Explanation:
The given numbers are: $$\frac{1}{3}$$ and $$\frac{1}{8}$$
So,
$$\frac{1}{3}$$ × $$\frac{1}{8}$$
= $$\frac{1 × 1}{3 × 8}$$
= $$\frac{1}{24}$$
Hence,
$$\frac{1}{3}$$ × $$\frac{1}{8}$$ = $$\frac{1}{24}$$
Question 25.
$$\frac{5}{6}$$ × $$\frac{1}{4}$$ = _______
$$\frac{5}{6}$$ × $$\frac{1}{4}$$ = $$\frac{5}{24}$$
Explanation:
The given numbers are: $$\frac{5}{6}$$ and $$\frac{1}{4}$$
So,
$$\frac{5}{6}$$ × $$\frac{1}{4}$$
= $$\frac{5 × 1}{6 × 4}$$
= $$\frac{5}{24}$$
Hence,
$$\frac{5}{6}$$ × $$\frac{1}{4}$$ = $$\frac{5}{24}$$
Question 26.
$$\frac{7}{2}$$ × $$\frac{2}{5}$$ = _______
$$\frac{7}{2}$$ × $$\frac{2}{5}$$ = $$\frac{7}{5}$$
Explanation:
The given numbers are: $$\frac{7}{2}$$ and $$\frac{2}{5}$$
So,
$$\frac{7}{2}$$ × $$\frac{2}{5}$$
= $$\frac{7 × 2}{2 × 5}$$
= $$\frac{7}{5}$$
Hence,
$$\frac{7}{2}$$ × $$\frac{2}{5}$$ = $$\frac{7}{5}$$
Question 27.
$$\frac{9}{10}$$ × $$\frac{3}{7}$$ = _______
$$\frac{9}{10}$$ × $$\frac{3}{7}$$ = $$\frac{27}{70}$$
Explanation:
The given numbers are: $$\frac{9}{10}$$ and $$\frac{3}{7}$$
So,
$$\frac{9}{10}$$ × $$\frac{3}{7}$$
= $$\frac{9 × 3}{7 × 10}$$
= $$\frac{27}{70}$$
Hence,
$$\frac{9}{10}$$ × $$\frac{3}{7}$$ = $$\frac{27}{70}$$
Question 28.
$$\frac{4}{5}$$ × $$\frac{13}{100}$$ = _______
$$\frac{4}{5}$$ × $$\frac{13}{100}$$ = $$\frac{52}{500}$$
Explanation:
The given numbers are: $$\frac{4}{5}$$ and $$\frac{13}{100}$$
So,
$$\frac{4}{5}$$ × $$\frac{13}{100}$$
= $$\frac{4 × 13}{5 × 100}$$
= $$\frac{52}{500}$$
Hence,
$$\frac{4}{5}$$ × $$\frac{13}{100}$$ = $$\frac{52}{500}$$
Question 29.
$$\frac{11}{25}$$ × $$\frac{3}{4}$$ = _______
$$\frac{11}{25}$$ × $$\frac{3}{4}$$ = $$\frac{33}{100}$$
Explanation:
The given numbers are: $$\frac{11}{25}$$ and $$\frac{3}{4}$$
So,
$$\frac{11}{25}$$ × $$\frac{3}{4}$$
= $$\frac{11 × 3}{25 × 4}$$
= $$\frac{33}{100}$$
Hence,
$$\frac{11}{25}$$ × $$\frac{3}{4}$$ = $$\frac{33}{100}$$
Question 30.
9 × $$\left(\frac{2}{9} \times \frac{1}{2}\right)$$ = _______
9 × ( $$\frac{1}{2}$$ × $$\frac{2}{9}$$ ) = 1
Explanation:
The given numbers are: 9, $$\frac{1}{2}$$ and $$\frac{2}{9}$$
So,
9 × ($$\frac{1}{2}$$ × $$\frac{2}{9}$$ )
= 9 × ($$\frac{1 × 2}{2 × 9}$$ )
= 9 × $$\frac{1}{9}$$
= 1
Hence,
9 × ( $$\frac{1}{2}$$ × $$\frac{2}{9}$$ ) = 1
Question 31.
$$\left(\frac{1}{10}+\frac{7}{10}\right)$$ × $$\frac{3}{4}$$ = _______
$$\left(\frac{1}{10}+\frac{7}{10}\right)$$ × $$\frac{3}{4}$$ = $$\frac{3}{5}$$
Explanation:
The given fractions are: $$\frac{1}{10}$$, $$\frac{7}{10}$$ and $$\frac{3}{4}$$
So,
$$\frac{3}{4}$$ × ($$\frac{1}{10}$$ + $$\frac{7}{10}$$ )
= $$\frac{3}{4}$$ × ($$\frac{7 × 1}{10}$$ )
= $$\frac{3}{4}$$ × $$\frac{8}{10}$$
= $$\frac{3 × 8}{4 × 10}$$
= $$\frac{24}{40}$$
= $$\frac{3}{5}$$
Hence,
$$\left(\frac{1}{10}+\frac{7}{10}\right)$$ × $$\frac{3}{4}$$ = $$\frac{3}{5}$$
Question 32.
$$\frac{4}{7}$$ × $$\left(\frac{5}{8}-\frac{1}{4}\right)$$ = _______
$$\frac{4}{7}$$ × $$\left(\frac{5}{8}-\frac{1}{4}\right)$$ = $$\frac{3}{14}$$
Explanation:
The given fractions are: $$\frac{4}{7}$$, $$\frac{5}{8}$$ and $$\frac{1}{4}$$
So,
$$\frac{4}{7}$$ × ($$\frac{5}{8}$$ – $$\frac{1}{4}$$ )
= $$\frac{4}{7}$$ × ($$\frac{5 – 2}{8}$$ )
= $$\frac{4}{7}$$ × $$\frac{3}{8}$$
= $$\frac{3 × 4}{7 × 8}$$
= $$\frac{12}{56}$$
= $$\frac{3}{14}$$
Hence,
$$\frac{4}{7}$$ × $$\left(\frac{5}{8}-\frac{1}{4}\right)$$ = $$\frac{3}{14}$$
9.6 Find Areas of Rectangles
Use rectangles with unit fraction side lengths to find the area of the rectangle.
Question 33.
The area of the rectangle is: $$\frac{1}{4}$$
Explanation:
The given figure is:
From the given figure,
The side lengths of the rectangle are: $$\frac{3}{8}$$ and $$\frac{2}{3}$$
So,
The area of the rectangle = $$\frac{3}{8}$$ × $$\frac{2}{3}$$
= $$\frac{2 × 3}{3 × 8}$$
= $$\frac{1}{4}$$
Hence, from the above,
We can conclude that the area of the rectangle is: $$\frac{1}{4}$$
Question 34.
The area of the rectangle is: $$\frac{35}{48}$$
Explanation:
The given figure is:
From the given figure,
The side lengths of the rectangle are: $$\frac{7}{12}$$ and $$\frac{5}{4}$$
So,
The area of the rectangle = $$\frac{7}{12}$$ × $$\frac{5}{4}$$
= $$\frac{7 × 5}{12 × 4}$$
= $$\frac{35}{48}$$
Hence, from the above,
We can conclude that the area of the rectangle is: $$\frac{35}{48}$$
Question 35.
Find the area of a rectangle with side lengths of $$\frac{2}{9}$$ and $$\frac{1}{10}$$.
The area of the rectangle is: $$\frac{1}{45}$$
Explanation:
The given side lengths of a rectangle are: $$\frac{2}{9}$$ and $$\frac{1}{10}$$
So,
The area of the rectangle = $$\frac{2}{9}$$ × $$\frac{1}{10}$$
= $$\frac{2 × 1}{10 × 9}$$
= $$\frac{1}{45}$$
Hence, from the above,
We can conclude that the area of the rectangle is: $$\frac{1}{45}$$
Question 36.
Find the area of a with a side length of $$\frac{3}{4}$$.
The area of the square is: $$\frac{9}{16}$$
Explanation:
The given side length of a square is: $$\frac{3}{4}$$
We know that,
The length of all sides of a square are equal
So,
the area of the square = $$\frac{3}{4}$$ × $$\frac{3}{4}$$
= $$\frac{3 × 3}{4 × 4}$$
= $$\frac{9}{16}$$
Hence, from the above,
We can conclude that the area of the square is: $$\frac{9}{16}$$
9.7 Multiply Mixed Numbers
Multiply.
Question 37.
1$$\frac{1}{4}$$ × 2$$\frac{1}{4}$$ = _______
1$$\frac{1}{4}$$ × 2$$\frac{1}{4}$$ =$$\frac{45}{16}$$
Explanation:
The given mixed fractions are: 1$$\frac{1}{4}$$ and 2$$\frac{1}{4}$$
The representation of 1$$\frac{1}{4}$$ and 2$$\frac{1}{4}$$ in the improper fraction form is: $$\frac{5}{4}$$ and $$\frac{9}{4}$$
So,
$$\frac{5}{4}$$ × $$\frac{9}{4}$$
= $$\frac{5 × 9}{4 × 4}$$
= $$\frac{45}{16}$$
Hence,
1$$\frac{1}{4}$$ × 2$$\frac{1}{4}$$ =$$\frac{45}{16}$$
Question 38.
3$$\frac{4}{5}$$ × 1$$\frac{1}{2}$$ = _______
3$$\frac{4}{5}$$ × 1$$\frac{1}{2}$$ = $$\frac{57}{10}$$
Explanation:
The given mixed fractions are: 3$$\frac{4}{5}$$ and 1$$\frac{1}{2}$$
The representation of 3$$\frac{4}{5}$$ and 1$$\frac{1}{2}$$ in the improper fraction form is: $$\frac{19}{5}$$ and $$\frac{3}{2}$$
So,
$$\frac{19}{5}$$ × $$\frac{3}{2}$$
= $$\frac{19 × 3}{5 × 2}$$
= $$\frac{57}{10}$$
Hence,
3$$\frac{4}{5}$$ × 1$$\frac{1}{2}$$ = $$\frac{57}{10}$$
Question 39.
5$$\frac{1}{3}$$ × 4$$\frac{7}{8}$$ = _______
5$$\frac{1}{3}$$ × 4$$\frac{7}{8}$$ = $$\frac{624}{24}$$
Explanation:
The given mixed fractions are: 5$$\frac{1}{3}$$ and 4$$\frac{7}{8}$$
The representation of 5$$\frac{1}{3}$$ and 4$$\frac{7}{8}$$ in the improper fraction form is: $$\frac{16}{3}$$ and $$\frac{39}{8}$$
So,
$$\frac{16}{3}$$ × $$\frac{39}{8}$$
= $$\frac{39 × 16}{3 × 8}$$
= $$\frac{624}{24}$$
Hence,
5$$\frac{1}{3}$$ × 4$$\frac{7}{8}$$ = $$\frac{624}{24}$$
Question 40.
4$$\frac{5}{6}$$ × 2$$\frac{9}{10}$$ × $$\frac{1}{8}$$ = _______
4$$\frac{5}{6}$$ × 2$$\frac{9}{10}$$ × $$\frac{1}{8}$$ = $$\frac{841}{480}$$
Explanation:
The given mixed fractions are: 4$$\frac{5}{6}$$, 2$$\frac{9}{10}$$ and $$\frac{1}{8}$$
The representation of 4$$\frac{5}{6}$$ and 2$$\frac{9}{10}$$ in the improper fraction form is: $$\frac{29}{6}$$ and $$\frac{29}{10}$$
So,
$$\frac{29}{6}$$ × $$\frac{29}{10}$$ × $$\frac{1}{8}$$
= $$\frac{29 × 29 × 1}{8 × 10 × 6}$$
= $$\frac{841}{480}$$
Hence,
4$$\frac{5}{6}$$ × 2$$\frac{9}{10}$$ × $$\frac{1}{8}$$ = $$\frac{841}{480}$$
Question 41.
Logic
Find the missing numbers.
The missing numbers are: 2 and 11
Explanation:
Let the missing numbers be: p and q
So,
4$$\frac{1}{p}$$ × 2$$\frac{1}{5}$$ = 9$$\frac{q}{10}$$
So,
To solve the mixed numbers, we can separate the whole numbers and the fraction numbers
So,
$$\frac{1}{p}$$ × 2$$\frac{1}{5}$$ = $$\frac{q}{10}$$
$$\frac{1}{p}$$ × $$\frac{11}{5}$$ = $$\frac{q}{10}$$
$$\frac{1 × 11}{p × 5}$$ = $$\frac{q}{10}$$
$$\frac{11}{5p}$$ = $$\frac{q}{10}$$
By comparing LHS and RHS
q = 11 and p = 2
Hence, from the above,
We can conclude that the missing numbers are: 2 and 11
9.8 Compare Factors and Products
Without calculating, tell whether the product is less than, greater than, or equal to each of its factors.
Question 42.
1$$\frac{1}{2}$$ × 3
The value of 1$$\frac{1}{2}$$ × 3 is greater than 3
Explanation:
The given numbers are: 3 and 1$$\frac{1}{2}$$
The representation of 1$$\frac{1}{2}$$ in the mixed form is: $$\frac{3}{2}$$
So,
$$\frac{3}{2}$$ is greater than 1
So,
$$\frac{3}{2}$$ × 3 is greater than 3
Hence, from the above,
We can conclude that the value of 1$$\frac{1}{2}$$ × 3 is greater than 3
Question 43.
$$\frac{1}{8}$$ × $$\frac{1}{6}$$
The value of $$\frac{1}{8}$$ × $$\frac{1}{6}$$ is less than 1
Explanation:
The given numbers are: $$\frac{1}{8}$$ and $$\frac{1}{6}$$
So,
$$\frac{1}{8}$$ is less than 1
$$\frac{1}{6}$$ is less than 1
So,
$$\frac{1}{8}$$ × $$\frac{1}{6}$$ is less than 1
Hence, from the above,
We can conclude that the value of $$\frac{1}{8}$$ × $$\frac{1}{6}$$ is less than 1
Question 44.
$$\frac{5}{5}$$ × 2$$\frac{4}{7}$$
The value of $$\frac{5}{5}$$ × 2$$\frac{4}{7}$$ is greater than 1
Explanation:
The given numbers are: $$\frac{5}{5}$$ and 2$$\frac{4}{7}$$
The representation of 2$$\frac{4}{7}$$ in the improper fraction form is: $$\frac{18}{7}$$
So,
$$\frac{18}{7}$$ is greater than 1
$$\frac{5}{5}$$ is 1
So,
$$\frac{5}{5}$$ × 2$$\frac{4}{7}$$ is greater than 1
Hence, from the above,
We can conclude that the value of $$\frac{5}{5}$$ × 2$$\frac{4}{7}$$ is greater than 1
Question 45.
$$\frac{3}{4}$$ × 2$$\frac{11}{12}$$
The value of $$\frac{3}{4}$$ × 2$$\frac{11}{12}$$ is greater than 1
Explanation:
The given numbers are: $$\frac{3}{4}$$ and 2$$\frac{11}{12}$$
The representation of 2$$\frac{11}{12}$$ in the improper fraction form is: $$\frac{35}{12}$$
So,
$$\frac{35}{12}$$ is greater than 1
$$\frac{3}{4}$$ is less than 1
So,
$$\frac{3}{4}$$ × 2$$\frac{11}{12}$$ is greater than 1
Hence, from the above,
We can conclude that the value of $$\frac{3}{4}$$ × 2$$\frac{11}{12}$$ is greater than 1
Question 46.
3$$\frac{1}{9}$$ × 2$$\frac{7}{8}$$
The value of 3$$\frac{1}{9}$$ × 2$$\frac{7}{8}$$ is greater than 1
Explanation:
The given mixed fractions are: 3$$\frac{1}{9}$$ and 2$$\frac{7}{8}$$
The representation of 3$$\frac{1}{9}$$ and 2$$\frac{7}{8}$$ in the improper fractions form is: $$\frac{28}{9}$$ and $$\frac{23}{8}$$
So,
$$\frac{28}{9}$$ is greater than 1
$$\frac{23}{8}$$ is greater than 1
So,
3$$\frac{1}{9}$$ × 2$$\frac{7}{8}$$ is greater than 1
Hence, from the above,
We can conclude that the value of 3$$\frac{1}{9}$$ × 2$$\frac{7}{8}$$ is greater than 1
Question 47.
$$\frac{9}{8}$$ × $$\frac{5}{2}$$
The value of $$\frac{9}{8}$$ × $$\frac{5}{2}$$ is greater than 1
Explanation:
The given fractions are: $$\frac{9}{8}$$ and $$\frac{5}{2}$$
So,
$$\frac{9}{8}$$ is greater than 1
So,
$$\frac{9}{8}$$ × $$\frac{5}{2}$$ is greater than 1
Hence, from the above,
We can conclude that the value of $$\frac{9}{8}$$ × $$\frac{5}{2}$$ is greater than 1
Conclusion:
Hope the information provided in the above answer key is beneficial for all the students of grade 5. Follow our Big Ideas Math Answers and make your learning fun and interesting.
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# 6.3 Taylor and maclaurin series
Page 1 / 13
• Describe the procedure for finding a Taylor polynomial of a given order for a function.
• Explain the meaning and significance of Taylor’s theorem with remainder.
• Estimate the remainder for a Taylor series approximation of a given function.
In the previous two sections we discussed how to find power series representations for certain types of functions––specifically, functions related to geometric series. Here we discuss power series representations for other types of functions. In particular, we address the following questions: Which functions can be represented by power series and how do we find such representations? If we can find a power series representation for a particular function $f$ and the series converges on some interval, how do we prove that the series actually converges to $f?$
## Overview of taylor/maclaurin series
Consider a function $f$ that has a power series representation at $x=a.$ Then the series has the form
$\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}={c}_{0}+{c}_{1}\left(x-a\right)+{c}_{2}{\left(x-a\right)}^{2}+\text{⋯}.$
What should the coefficients be? For now, we ignore issues of convergence, but instead focus on what the series should be, if one exists. We return to discuss convergence later in this section. If the series [link] is a representation for $f$ at $x=a,$ we certainly want the series to equal $f\left(a\right)$ at $x=a.$ Evaluating the series at $x=a,$ we see that
$\begin{array}{cc}\hfill \sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}& ={c}_{0}+{c}_{1}\left(a-a\right)+{c}_{2}{\left(a-a\right)}^{2}+\text{⋯}\hfill \\ & ={c}_{0}.\hfill \end{array}$
Thus, the series equals $f\left(a\right)$ if the coefficient ${c}_{0}=f\left(a\right).$ In addition, we would like the first derivative of the power series to equal ${f}^{\prime }\left(a\right)$ at $x=a.$ Differentiating [link] term-by-term, we see that
$\frac{d}{dx}\left(\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)={c}_{1}+2{c}_{2}\left(x-a\right)+3{c}_{3}{\left(x-a\right)}^{2}+\text{⋯}.$
Therefore, at $x=a,$ the derivative is
$\begin{array}{}\\ \\ \hfill \frac{d}{dx}\left(\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)& ={c}_{1}+2{c}_{2}\left(a-a\right)+3{c}_{3}{\left(a-a\right)}^{2}+\text{⋯}\hfill \\ & ={c}_{1}.\hfill \end{array}$
Therefore, the derivative of the series equals ${f}^{\prime }\left(a\right)$ if the coefficient ${c}_{1}={f}^{\prime }\left(a\right).$ Continuing in this way, we look for coefficients c n such that all the derivatives of the power series [link] will agree with all the corresponding derivatives of $f$ at $x=a.$ The second and third derivatives of [link] are given by
$\frac{{d}^{2}}{d{x}^{2}}\left(\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)=2{c}_{2}+3·2{c}_{3}\left(x-a\right)+4·3{c}_{4}{\left(x-a\right)}^{2}+\text{⋯}$
and
$\frac{{d}^{3}}{d{x}^{3}}\left(\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)=3·2{c}_{3}+4·3·2{c}_{4}\left(x-a\right)+5·4·3{c}_{5}{\left(x-a\right)}^{2}+\text{⋯}.$
Therefore, at $x=a,$ the second and third derivatives
$\begin{array}{cc}\hfill \frac{{d}^{2}}{d{x}^{2}}\left(\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)& =2{c}_{2}+3·2{c}_{3}\left(a-a\right)+4·3{c}_{4}{\left(a-a\right)}^{2}+\text{⋯}\hfill \\ & =2{c}_{2}\hfill \end{array}$
and
$\begin{array}{cc}\hfill \frac{{d}^{3}}{d{x}^{3}}\left(\sum _{n=0}^{\infty }{c}_{n}{\left(x-a\right)}^{n}\right)& =3·2{c}_{3}+4·3·2{c}_{4}\left(a-a\right)+5·4·3{c}_{5}{\left(a-a\right)}^{2}+\text{⋯}\hfill \\ & =3·2{c}_{3}\hfill \end{array}$
equal $f\text{″}\left(a\right)$ and $f\text{‴}\left(a\right),$ respectively, if ${c}_{2}=\frac{f\text{″}\left(a\right)}{2}$ and ${c}_{3}=\frac{f\text{‴}\left(a\right)}{3}·2.$ More generally, we see that if $f$ has a power series representation at $x=a,$ then the coefficients should be given by ${c}_{n}=\frac{{f}^{\left(n\right)}\left(a\right)}{n\text{!}}.$ That is, the series should be
$\sum _{n=0}^{\infty }\frac{{f}^{\left(n\right)}\left(a\right)}{n\text{!}}{\left(x-a\right)}^{n}=f\left(a\right)+{f}^{\prime }\left(a\right)\left(x-a\right)+\frac{f\text{″}\left(a\right)}{2\text{!}}{\left(x-a\right)}^{2}+\frac{f\text{‴}\left(a\right)}{3\text{!}}{\left(x-a\right)}^{3}+\text{⋯}.$
This power series for $f$ is known as the Taylor series for $f$ at $a.$ If $x=0,$ then this series is known as the Maclaurin series for $f.$
## Definition
If $f$ has derivatives of all orders at $x=a,$ then the Taylor series for the function $f$ at $a$ is
$\sum _{n=0}^{\infty }\frac{{f}^{\left(n\right)}\left(a\right)}{n\text{!}}{\left(x-a\right)}^{n}=f\left(a\right)+{f}^{\prime }\left(a\right)\left(x-a\right)+\frac{f\text{″}\left(a\right)}{2\text{!}}{\left(x-a\right)}^{2}+\text{⋯}+\frac{{f}^{\left(n\right)}\left(a\right)}{n\text{!}}{\left(x-a\right)}^{n}+\text{⋯}.$
The Taylor series for $f$ at 0 is known as the Maclaurin series for $f.$
Later in this section, we will show examples of finding Taylor series and discuss conditions under which the Taylor series for a function will converge to that function. Here, we state an important result. Recall from [link] that power series representations are unique. Therefore, if a function $f$ has a power series at $a,$ then it must be the Taylor series for $f$ at $a.$
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
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
ya I also want to know the raman spectra
Bhagvanji
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
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
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Leaves accumulate on the forest floor at a rate of 2 g/cm2/yr and also decompose at a rate of 90% per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time 0 there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?
You have a cup of coffee at temperature 70°C, which you let cool 10 minutes before you pour in the same amount of milk at 1°C as in the preceding problem. How does the temperature compare to the previous cup after 10 minutes?
Abdul
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# Is the dividend the greatest number in a division problem?
Contents
Dividends and divisors are the two key ingredients that yield the quotient, or answer, to a division problem. The dividend is the number being divided, while the divisor is the number by which the dividend is divided.
## Is the dividend always bigger than the divisor?
When the divisor is smaller than the dividend, the quotient is more than 1. Another example where the divisor smaller than the dividend. When the divisor is the same size as the dividend, the quotient is 1. When the divisor is larger than the dividend, the quotient is less than 1.
## Which number is the dividend in a division problem?
In division, the amount or number to be divided is called the dividend. Dividend is the whole that is to be divided into parts. Here, for example, 12 candies are to be divided among 3 children. 12 is the dividend.
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## What is the top number in a division problem called?
In each division problem, you will have one number divided by another. The number you are dividing is called the dividend. The number you are “dividing by” is the divisor. The answers to your division problems are called quotients.
## Which number is the dividend?
The number that is being divided (in this case, 15) is called the dividend, and the number that it is being divided by (in this case, 3) is called the divisor. The result of the division is the quotient.
## How do you divide when the dividend is larger than the divisor?
Perform the division.
1. Remember to place a zero in the quotient when the divisor is larger than the dividend.
2. Place the decimal point in your quotient.
3. Check your answer: Multiply the divisor by the quotient to see if you get the dividend.
## Why is the quotient less than the dividend?
For example take a dividend of 10 and divided by 1/2. … If the divisor is negative and the dividend is positive, then the quotient is negative (and hence less than the dividend).
## What is the second number in a division problem called?
where each of the three expressions means “6 divided by 3”, with 2 as the answer. The first number is the dividend (6), and the second number is the divisor (3). The result (or answer) is the quotient, where any left-over amount as whole numbers is called the “remainder”.
## What is the dividend of 24 divided by 3?
What is the Quotient and Remainder of 24 Divided by 3? The quotient (integer division) of 24/3 equals 8; the remainder (“left over”) is 0. 24 is the dividend, and 3 is the divisor.
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## What comes first divisor or dividend?
When using the short-hand symbols “÷” or “/” to indicate division, the dividend appears to the left and the divisor appears to the right.
## What is the first number in a division problem called?
As in all division problems, one number, called the dividend, is divided by another, called the divisor, producing a result called the quotient.
## What is the quotient in dividing 6 by 3?
The quotient is the number obtained by dividing one number by another. For example, if we divide the number 6 by 3, the result so obtained is 2, which is the quotient.
## What is the dividend of 35 divided by 7?
What is the Quotient and Remainder of 35 Divided by 7? The quotient (integer division) of 35/7 equals 5; the remainder (“left over”) is 0. 35 is the dividend, and 7 is the divisor.
## What is a dividend of 50?
What does bank pays you dividend of 50 in Monopoly? it means that you receive \$50 from the bank because the dividend is the ammount of money that the bank pays you for having stock in their company.
## What is the quotient of 24 is divided by 48?
24 / 48 24+24 = 48 24 / 48 = 2 (the slashes are division..)
## Is the dividend on the top or bottom?
The first digit of the dividend (4) is divided by the divisor. The whole number result is placed at the top. Any remainders are ignored at this point. The answer from the first operation is multiplied by the divisor.
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CAT Number System Tricks (advanced level): Finding Last Non-Zero Digit of any factorial
This article is written by a 100-percentiler in CAT 2016. He is a director of a leading test-prep institute and is the best person to learn Mathematics from. In fact, he was the trainer for Wordpandit himself and Wordpandit credits a lot of his learning to this Master Trainer. Dig into this amazing article by this Mathematics Wizard. We will reveal his identity in the coming posts..:)
Let’s get started with this Advanced maths blog on this topic.
Suppose we have to find last non-zero digit of a simple number say 98670000. Obviously, everyone knows the answer is 7. Another perspective to look at it is if we pull all zeros out of the given number then unit digit of remnant (remaining part) will be our answer i.e. we can write
98670000= 10000*(9867)
And unit digit of 9867 is 7
Let us try to apply similar sort of approach to some problems involving factorial.
Question 1: Find last non-zero digit of 18!?
Solution: We will try to solve this problem by first pulling the number of zeros out of given number 18! and then calculating the unit digit of remnant. So we first calculate the number of zeros in 18!.
We know 18!=1×2× 3×4×5×6×7×8×9×10×11×12×13×14×15×16×17×18
Clearly, calculating number of zeros is equivalent to calculating number of 2’s and 5’s
Now we first calculate number of 2’s present in 18! (Maximum power of 2 present in 18!).For that we will move stepwise
Step 1
We first count those numbers which can contribute 2.Clearly we have 9 such numbers i.e. all evens- 2,4,6,8,10,12,14,16,18.We take 21 from each of these numbers i.e. collectively we will have 29 as each of the even number contributed power of 2.
Step 2
Out of the remaining numbers after step 1, again we count those numbers that can contribute 2.
Clearly after step 1 i.e. after pulling 2 from all even numbers we are left with 1,2,3,4,5,6,7,8,9.Out of these, we count those that can contribute 2. We have 4 such numbers i.e. all evens- 2,4,6,8.We take 21 from each of these numbers i.e. collectively we will have 24 as each of the even number contributed power of 2.
Step 3
Out of the remaining numbers after step 2 again we count those numbers that can contribute 2.
Clearly after step 2 i.e. after pulling 2 from all remaining even numbers we are left with 1,2,3,4.Out of these,we count those that now can contribute 2.We have 2 such numbers i.e. all evens- 2 and 4.We take 21 from each of these numbers i.e. collectively we will have 22 as each of the even number contributed power of 2.
Step 4
Out of the remaining numbers after step 3 again we count those numbers that can contribute 2.
Clearly, after step 3 i.e. after pulling 2 from all remaining even numbers we are left with 1 and 2.Out of these, we count those that now can contribute 2.Clearly, we have only have 1 such number.
Adding all four steps we can say maximum power of 2 present in 18! =9+4+2+1=16
In fact we can generalize what we did above for calculating maximum power of any prime p in any n!
Where[ ] represents integral part
So for calculating maximum power of 5 in 18! We apply the above formula directly
Maximum power of 5
= 3+ 0
=3
So, the number of zeros present in 18! will be Minimum of (3,16)=3(Because for making power of 10 we require both 2 and 5 thus minimum power present will determine the number of zeros)
In fact, we can write prime factorization of 18! by calculating maximum power of each prime involved
Maximum power of 3
= 6+2
=8
Maximum power of 7 =
= 2+ 0
=2
Clearly,
Maximum power of 11 = 1
Maximum power of 13 = 1
Maximum power of 17 = 1
Thus prime factorization of 18! is as follows
18! = 2^16. 3^8. 5^3. 7^2. 11^1. 13^1.17^1
So we pull all zeros out of the given number then calculate the unit digit of remnant (remaining part)
2^3. 5^3(2^13. 3^8. 7^2. 11^1. 13^1.17^1)
Now to calculate Unit digit of 2^13. 3^8. 7^2. 11^1. 13^1.17^1, we calculate unit digit of each separately.
Unit digit of 2^13=2
Unit digit of 3^8=1
Unit digit of 7^2=9
Unit digit of 13=3
Unit digit of 17=7
Unit digit of 2^13. 3^8. 7^2. 11^1. 13^1.17^1=2×1×9×1×3×7=8
NOTE: The question we solved was easy as 18! is not a very big number and we were able to factorise it easily. Now the question arises what if the number is big and process of factorization is very lengthy.
Complete List of Articles in this Series
Question 2: Find last non-zero digit of 2000!?
Solution: We know 2000! is product of 1st 2000 natural numbers. We write these numbers in set of 10 numbers each.
We know
2000! =1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x16x17x18x19x20…………………………x2000
= (1x2x3x4x5x6x7x8x9x10)(11x12x13x14x15x16x17x18x19x20)……………………………………………………………
……………………………………………………………(1991x1992x1993x1994x1995x1996x1997x1998x1999x2000)
i.e. we divided 2000! in groups of 10 each
We write 1st 2000 natural numbers in general form as
10a+1
10a+2
10a+3
10a+4
10a+5
10a+6
10a+7
10a+8
10a+9
10a+10(or we can write it as 10a+0)
Where a varies from 0 to 199 to cover all 1st 2000 natural numbers
When a=0 Then above set of numbers reduces to 1,2,3…….10
Similarly when a=1 Then above set of numbers reduces to 11,12,13…….20
Similarly when a=2 Then above set of numbers reduces to 21,22,23…….30
…………………………………………………………………………………………………………………..
Similarly when a= 199 Then above set of numbers reduces to 1991,1992,1993…….2000
Let us look at the following observations:
Observation 1
(10a+1)(10a+3)(10a+7)(10a+9)
= (10a+1)(10a+9)(10a+3)(10a+7)
=(100a2+100a+9) (100a2+100a+21 )
=10000a4+10000a3+900a2+10000a3+10000a2+900a+2100a2+2100a+189
=10000a4+20000a3+13000a2+3000a+189
=1000(Some integer) +3000a+189
=1000(X) +3000a+189 where X is some integer
Observation 2
(10a+2)(10a+4)(10a+6)(10a+8)
= (10a+2)(10a+8)(10a+4)(10a+6)
= (100a2+100a+16) (100a2+100a+24)
=10000a4+10000a3+1600a2+10000a3+10000a2+1600a+2400a2+2400a+384
=1000(Some integer) +4000a+384
=4 [250(Some integer)+1000a+96]
=22[250(Y)+1000a+96]where Y is some integer
Observation 3
5.10.15.20.25.30.35…………………………………………………………………………..1990.1995.2000
=5400(1.2.3.4.5.6.7………………………………………………………………………………….398.399.400)
=5400(400!)
Let us make use of the above three observations we made in calculating last nonzero digit
Now we pull two 5’s from each group of 10 numbers and making use of Observation 3
We can write
2000!== (1.2.3.4.5.6.7.8.9.10)(11.12.13.14.15.16.17.18.19.20)……………………………………………………………
……………………………………………………………(1991.1992.1993.1994.1995.1996.1997.1998.1999.2000)
=(5.10.15.20.25.30.35………………………….……………….…1990.1995.2000)(1.2.3.4.6.7.8.9) (11.12.13.14.16.17.18.19)………………………………………… (1991.1992.1993.1994.1996.1997.1998.1999)
=5400(400!)(1.2.3.4.6.7.8.9)(11.12.13.14.16.17.18.19)…………………………………………
…………………………………………………………………………(1991.1992.1993.1994.1996.1997.1998.1999) …………….. (A)
Notice that after pulling two 5s from each group of 10 numbers we have 5400(400!)as shown above and we are left with 8 numbers in each group which can be written in generalised form as
(10a+1)(10a+2)(10a+3)(10a+4)(10a+6)(10a+7)(10a+8)(10a+9)
Putting different values of a, we will get members of each group e.g.
Putting a=0 we get members of first group (1.2.3.4.6.7.8.9)
Putting a=1 we get members of second group (11.12.13.14.16.17.18.19)
……………………………………………………………………………………………………………..
……………………………………………………………………………………………………………..
Putting a= 199 we get members of last group (1991.1992.1993.1994.1996.1997.1998.1999)
Now using Observation 1 and 2 we can write
(10a+1)(10a+2)(10a+3)(10a+4)(10a+6)(10a+7)(10a+8)(10a+9)
= (10a+1)(10a+3)(10a+7)(10a+9)(10a+2)(10a+4)(10a+6)(10a+8)
= [1000(X) +3000a+189][22[250(Y)+1000a+96]]
=22[1000(X) +3000a+189][250(Y)+1000a+96]
We have pulled 22 from group of these 8 numbers and in totality we have 200 such sets.
Each such set can contribute 22 and thus we have total of 2400 outside.
From observation 3 we had 5400 outside. Thus in totality by combining 2400 and 5400 we have (2400)(5400)= 10400 i.e. we pull all 400 zeros out of the given number then unit digit of remnant (remaining part) will be our answer.
So lets us look at unit digit of[1000(X) +3000a+189][250(Y) +1000a+96]
= Unit digit of[1000(X) +3000a+189].Unit digit of[250(Y) +1000a+96]
= 9.6
= 4 ……………………………………………………………. (B)
In 1 set of 10 numbers the unit digit of remaining numbers after pulling 400 zeros is 4.Clearly we have 200 such sets with us and each such set gives 4 as unit digit. Thus, in totality, we will have 4200.Thus Combining (A) and (B) and after pulling 400 zeros we can say
Last non-zero digit of 2000! =Unit Digit after pulling 400 zeroes of (400! 4200)
Thus our question reduces from a big number to relatively small number 400!
Again we follow the same process and we can directly writ
400! = Unit of digit after pulling 80 zeros of (80! 440)
Again we follow the same process and we can directly write
80! = Unit of digit after pulling 16 zeros of (16! 48)
Now we cannot apply same further as we are left with 16!.But 16! Is a very small number and we can easily calculate its unit digit by writing its prime factorization
We know 16! =1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16+6
Maximum power of 2
= 8+4+ 2 + 1
=15
Maximum power of 3
= 5+1+0
=6
Maximum power of 5=3
Maximum power of 7=2
Maximum power of 11=1
Maximum power of 13=1
16!=2^15. 3^6. 5^3.7^1. 11^1. 13^1
After pulling zeros we can write it as
16!= 2^3. 5^3(2^12. 3^6. 7^1. 11^1. 13^1)
Thus Unit digit of remnant (remaining part) will be our answer
Unit digit of 2^12=6
Unit digit of 3^6=9
Unit digit of 2^12. 3^6. 7^1. 11^1. 13^1=6.9.7.1.3=4
And Unit digit of 4^8=6
Thus 80! = Unit of digit after pulling zeros of (16! 48)=4×6=4
Question 2: Find last non zero digit of 1000!?
Solution: As we did in last problem we can directly write
1000! =1.2.3.4.5.6.7.8.9.10.11.12.13.14.15.16.17.18.19.20………………………….1000
= (1.2.3.4.5.6.7.8.9.10)(11.12.13.14.15.16.17.18.19.20)……………………………………………………………
……………………………………………………………(991.992.993.994.995.996.997.998.999.1000)
i.e. we divided 1000! In groups of 10 each
We can write
1000!
=(5.10.15.20.25.30.35………………………….……………….…990.995.1000)(1.2.3.4.6.7.8.9) (11.12.13.14.16.17.18.19)………………………………………… (991.992.993.994.996.997.998.999)
=5200(200!)(1.2.3.4.6.7.8.9)(11.12.13.14.16.17.18.19)…………………………………………
…………………………………………………………………………………..(991.992.993.994.996.997.998.999) (A)
Notice that after pulling two 5s from each group of 10 numbers we are left with 8 numbers ineach group which can be written in generalized form as
(10a+1)(10a+2)(10a+3)(10a+4)(10a+6)(10a+7)(10a+8)(10a+9)
As we did earlier, we have pulled 22 from the group of these 8 numbers. Thus we have total of 2200 outside by pulling 22 from each group of 8 numbers of this product (1.2.3.4.6.7.8.9)(11.12.13.14.16.17.18.19)…………………………………………
…………………………………………………………………………………..(991.992.993.994.996.997.998.999)
Thus in totality by combining 2200 and 5200 we have (2200)(5200)=10200 i.e. we pull all zeroes out of the given number then unit digit of remnant (remaining part) will be our answer.
In 1 set of 10 numbers the unit digit of remaining numbers after pulling zeros is 4.Clearly we have 100 such sets with us.Thus after pulling zeros we can say last non-zero digit of 1000! =Unit Digit after pulling zeros of (200! 4100)
Thus our question reduces from a big number to relatively small number 200!
Again we follow the same process and we can directly write
200! = Unit of digit after pulling zeros out of (40! 420)
Again we can reduce 40! And following the same process, we can write
40! = Unit of digit after pulling zeros out of (8! 44)
Now we cannot apply it further as we are left with 8! which only has single 5 in it. But 8! Is a very small number and we can easily calculate its unit digit by writing its prime factorization
We know 8! =1.2.3.4.5.6.7.8
Maximum power of 2
= 4+ 2 + 1 +0
=7
= 2+0
=2
Maximum power of 5=1
Maximum power of 7=1
Thus 8! = 2^7.3^2.5^1.7^1
After pulling zeros we can write it as
8! = 2.5(2^6. 3^2.7^1)
Thus Unit digit of remnant (remaining part) will be our answer
Unit digit of 2^6 = 4
Unit digit of 3^2 = 9
Unit digit of 2^6. 3^2. 7^1 = 4.9.7 = 2
And Unit digit of 4^4 = 6
Thus 40! = Unit of digit after pulling zeros out of (8! 44)=2×6=2.
Rest we hope the process is generalised with the help of above three problems and you can calculate last non-zero digit of any factorial.
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# Calculus 2 : L'Hospital's Rule
## Example Questions
← Previous 1 3 4 5 6 7 8 9
### Example Question #3 : L'hospital's Rule
Solve:
Explanation:
Substitution is invalid. In order to solve , rewrite this as an equation.
Take the natural log of both sides to bring down the exponent.
Since is in indeterminate form, , use the L'Hopital Rule.
L'Hopital Rule is as follows:
This indicates that the right hand side of the equation is zero.
Use the term to eliminate the natural log.
### Example Question #21 : Applications Of Derivatives
Evaluate the limit using L'Hopital's Rule.
Undefined
Explanation:
L'Hopital's Rule is used to evaluate complicated limits. The rule has you take the derivative of both the numerator and denominator individually to simplify the function. In the given function we take the derivatives the first time and get
This still cannot be evaluated properly, so we will take the derivative of both the top and bottom individually again. This time we get
.
Now we have only one x, so we can evaluate when x is infinity. Plug in infinity for x and we get
and
So we can simplify the function by remembering that any number divided by infinity gives you zero.
### Example Question #22 : Applications Of Derivatives
Evaluate the limit using L'Hopital's Rule.
Undefined
Explanation:
L'Hopital's Rule is used to evaluate complicated limits. The rule has you take the derivative of both the numerator and denominator individually to simplify the function. In the given function we take the derivatives the first time and get
Since the first set of derivatives eliminates an x term, we can plug in zero for the x term that remains. We do this because the limit approaches zero.
This gives us
.
### Example Question #23 : Applications Of Derivatives
Evaluate the limit using L'Hopital's Rule.
Undefined
Explanation:
L'Hopital's Rule is used to evaluate complicated limits. The rule has you take the derivative of both the numerator and denominator individually to simplify the function. In the given function we take the derivatives the first time and get
This still cannot be evaluated properly, so we will take the derivative of both the top and bottom individually again. This time we get
.
Now we have only one x, so we can evaluate when x is infinity. Plug in infinity for x and we get
### Example Question #24 : Applications Of Derivatives
Calculate the following limit.
Explanation:
To calculate the limit, often times we can just plug in the limit value into the expression. However, in this case if we were to do that we get , which is undefined.
What we can do to fix this is use L'Hopital's rule, which says
.
So, L'Hopital's rule allows us to take the derivative of both the top and the bottom and still obtain the same limit.
.
Plug in to get an answer of .
### Example Question #4 : L'hospital's Rule
Calculate the following limit.
Explanation:
If we plugged in the integration limit to the expression in the problem we would get , which is undefined. Here we use L'Hopital's rule, which is shown below.
This gives us,
.
However, even with this simplified limit, we still get . So what do we do? We do L'Hopital's again!
.
Now if we plug in infitinity, we get 0.
### Example Question #81 : Functions And Graphs
Calculate the following limit.
Explanation:
If we plugged in directly, we would get an indeterminate value of .
We can use L'Hopital's rule to fix this. We take the derivate of the top and bottom and reevaluate the same limit.
.
We still can't evaluate the limit of the new expression, so we do it one more time.
### Example Question #1 : L'hospital's Rule
Find the
.
Does Not Exist
Explanation:
Subbing in zero into will give you , so we can try to use L'hopital's Rule to solve.
First, let's find the derivative of the numerator.
is in the form , which has the derivative , so its derivative is
is in the form , which has the derivative , so its derivative is .
The derivative of is so the derivative of the numerator is .
In the denominator, the derivative of is , and the derivative of is . Thus, the entire denominator's derivative is .
Now we take the
, which gives us
### Example Question #2 : L'hospital's Rule
Evaluate the following limit.
Explanation:
If we plug in 0 into the limit we get , which is indeterminate.
We can use L'Hopital's rule to fix this. We can take the derivative of the top and bottom and reevaluate the limit.
.
Now if we plug in 0, we get 0, so that is our final limit.
### Example Question #22 : New Concepts
Evaluate the following limit
if possible.
Limit does not exist
Explanation:
If we try to directly plug in the limit value into the function, we get
Because the limit is of the form , we can apply L'Hopital's rule to "simplify" the limit to
.
Now if we directly plug in 0 again, we get
.
← Previous 1 3 4 5 6 7 8 9
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# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2016 | Oct-Nov | (P1-9709/11) | Q#8
Question
The functions f and g are defined by
for
for .
i. Find and simplify an expression for and state the range of .
ii. Find an expression for and find the domain of .
Solution
i.
We are given;
for
for .
We are required to find expression for ;
We are also required to find range of .
To find the range of of , we need domain of of .
For a composite function , the domain D of must be chosen so that the whole of the range of is included in the domain of . The function , is then defined as ,
Therefore, first we need to find range of to compare it with domain of .
Let’s find range of .
Finding range of a function :
· Substitute various values of from given domain into the function to see what is happening to y.
· Make sure you look for minimum and maximum values of y by substituting extreme values of from given domain.
We are given that;
for
We can find the range of by substituting some value(s) of domain of into the function.
It is evident that range of function is;
It is evident that domain of must be chosen such that range of is included in domain of which is .
Therefore, entire range of is included in the domain of .
Therefore, domain of is also same as domain of .
Now let’s find range of .
Finding Range of a Function :
· Substitute various values of from given domain into the function to see what is happening to y.
· Make sure you look for minimum and maximum values of y.
We are given that;
We can find the range of by substituting some value(s) of domain of into the function.
It is evident that range of function is;
ii.
We have;
We write it as;
To find the inverse of a given function we need to write it in terms of rather than in terms of .
Interchanging ‘x’ with ‘y’;
We are also required to find the domain of .
Domain and range of a function become range and domain, respectively, of its inverse function .
Domain of a function Range of
Range of a function Domain of
We have found in (i) that range of is;
Therefore;
Range of a function Domain of
Hence;
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Home » Revision Notes for CBSE Class 6 to 12 » Class 10 Maths for Statistics of Chapter 14 Revision Notes
# Class 10 Maths for Statistics of Chapter 14 Revision Notes
## CBSE Class 10 Maths Chapter 14 – Statistics – Free PDF Download
Free PDF download of Class 10 Maths Chapter 14 – Statistics Revision Notes & Short Key-notes prepared by expert Mathematics teachers from latest edition of CBSE(NCERT) books.
### CBSE Class 10 Maths Revision Notes Chapter 14 Statistics
1. Mean of Grouped Data
2. Mode of Grouped Data
3. Median of Grouped Data
4. Graphical Representation of CF
5. Miscellaneous Questions
1. Mean : The mean for grouped data can be found by :
(i) The direct method =
(ii) The assumed mean method
Where a = Provisional mean
(iii) The step deviation method
2. Mode : The mode for the grouped data can be found by using the formula :
= lower limit of the modal class.
= frequency of the modal class.
= frequency of the preceding class of the modal class.
= frequency of the succeeding class of the modal class.
h = size of the class interval.
Modal class – class interval with highest frequency.
3. Median : Median of continuous series is:
(i) term (if number of terms are odd)
(ii) (if number of terms are even]
(iii) The median for the grouped data can be found by using the formula :
= lower limit of the median class.
n = number of observations.
Cf = cumulative frequency of class interval preceding the median class.
f = frequency of median class.
h = class size.
4.Empirical Formula : Mode = 3 median – 2 mean.
5.Cumulative frequency curve or an Ogive :
(i) Ogive is the graphical representation of the cumulative frequency distribution.
(ii) Less than type Ogive :
• Construct a cumulative frequency table.
• Mark the upper class limit on the x-axis.
(iii) More than type Ogive :
• Construct a frequency table.
• Mark the lower class limit on the x-axis.
(iv) To obtain the median of frequency distribution from the graph :
• Locate point of intersection of less than type Ogive and more than type Ogive :
Draw a perpendicular from this point on x-axis.
• The point at which it cuts the x-axis gives us the median.
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# Minecraft Calculus: River Crossing Escape
This is a classic introductory calculus problem, with a Minecraft theme. This particular variation is inspired by example 4 in section 4.5 of Stewart's calculus.
I'll set the scene: You're weaponless. You have half a heart and you're being chased by a skeleton archer. There's a river between you and your cabin. You need to get to your cabin as quickly as possible!
In more "mathy" terms, you want to get from $A$ to $B$. There are three main ways you can do this:
1. $A$ to $B$ directly, diagonally across the river
2. $A$ to $C$ and then $C$ to $B$
3. At slight angle from $A$ to $D$ and then $D$ to $B$
We know that swimming is slower than running, so option 1 isn't the best. Option 2 involves the smallest time in the water, but also the longest distance traveled.
Option 3 is somewhere in between - you spend a bit more time in the water, but the total distance is decreased. The travel time in this case depends on where exactly point $D$ is. To find the optimal path, we must find the position of $D$ which minimizes the travel time. This sounds like calculus!
Let $x$ be the distance between points $C$ and $D$. In math terms, $|CD|=x$.
We first need an expression for the total time traveled in terms of $x$. The basic equation for time traveled at constant speed is
$t=\frac{distance}{speed}$.
The first part of the trip is in the water where we travel from $A$ to $D$. We can use Pythagorean's theorem to get this distance as $|AD|=\sqrt{x^2+w^2}$. Assuming that we can travel at a speed of $v_w$ in the water, the time for this part is
$t_w=\frac{\sqrt{x^2+w^2}}{v_w}$.
The land part of the trip involves traveling what's left over of $l$ after having already traveled $x$, so $|DB|=l-x$. Traveling at $v_l$ on land, the time for this part of the trip is
$t_l=\frac{l-x}{v_l}$.
The total trip time, $T(x)$, is just the sum of these two.
$T(x)=\frac{\sqrt{x^2+w^2}}{v_w}+\frac{l-x}{v_l}$ (1)
To optimize this function with respect to $x$, we need to find where it is stationary and then verify that this point is a minimum. This means we want to find a place where the function isn't changing with small changes in $x$. In other words, we want to find a spot where the derivative is zero.
Taking the derivative gives:
$T'(x)=\frac{x}{v_w\sqrt{x^2+w^2}}-\frac{1}{v_l}$
Setting this equal to zero, we can solve for $x$:
$x=\frac{v_ww}{\sqrt{v_l^2-v_w^2}}$ (2)
Let's plug in some numbers! In Minecraft, you can swim at about 2.2 m/s and sprint at 5.6 m/s. Let's take the river width $w$ to be 7 blocks (1 block = 1 meter). Plugging in, we find that $x=2.9$. At this point, though, we can't tell if this is a maximum or a minimum. One way to find out is to examine the curvature of the function at this point by using the 2nd derivative:
$T''(x)=\frac{1}{v_w\sqrt{x^2+w^2}}(1-\frac{x^2}{x^2+w^2})$
Plugging in $x=2.9$, we see that $T''(2.9)=0.04$ which, being positive, means that $x=2.9$ is a minimum point for $T(x)$. Visually:
So if you want to get to your cabin as quickly as possible, the fastest route is to swim across the river to a point $D$ that is 2.9 meters from point $C$ and then run the rest of the way.
This is actually a general problem. Try replacing the speeds I used with speeds for soul sand, crouching and walking, a boat, etc. and see what happens! Particularly, what happens to (2) if you can travel more quickly in the water than on land?
#### Scratch Work
Finding $T'(x):$ First, rewrite the square root as a power, $T(x)=\frac{(x^2+w^2)^{\frac{1}{2}}}{v_w}+\frac{l-x}{v_l}$
Differentiating, using the chain rule on the first term, $T'(x)=\frac{\frac{1}{2}(x^2+w^2)^{-\frac{1}{2}}(2x)}{v_w}-\frac{1}{v_l}=\frac{x}{v_w\sqrt{x^2+w^2}}-\frac{1}{v_l}$
Finding $T''(x):$ Starting with $T'(x)=\frac{(x^2+w^2)^{-\frac{1}{2}}x}{v_w}+\frac{1}{v_l}$, we differentiate with respect to x. Note that the derivative of $\frac{1}{v_l}$ with respect to $x$ is 0, so we only have to deal with the first term. Using the product rule and chain rule:
$T''(x)=\frac{-\frac{1}{2}(x^2+w^2)^{-\frac{3}{2}}(2x^2)}{v_w}+\frac{(x^2+w^2)^{-\frac{1}{2}}}{v_w}$
Cleaning up with some algebra:
$T''(x)=\frac{-x^2}{v_w(x^2+w^2)^{\frac{3}{2}}}+\frac{1}{v_w\sqrt{x^2+w^2}} =\frac{1}{v_w\sqrt{x^2+w^2}}(1-\frac{x^2}{x^2+w^2})$
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# Partial Fractions: Decomposition (Type-4)
When the quadratic factor in the denominator is repetitive: Let the quadratic factor in the denominator is $$(ax^2 + bx + c)^2$$, which is repetitive, hence we can write two fractions $$\frac{Ax + B}{(ax^2 + bx + c)}$$ and $$\frac{Cx + D}{(ax^2 + bx + c)^2}$$ equal to the main fraction. we can better understand by the given example.
Example: $$\frac{1}{(1 + x) \ (1 + x^2)^2}$$
Solution: According to the method the given equation can also be written as.
$$\frac{1}{(1 + x) \ (1 + x^2)^2} =$$ $$\frac{A}{(1 + x)} + \frac{Bx + C}{(1 + x^2)}$$ $$+ \frac{Dx + E}{(1 + x^2)^2}....(1)$$
$$\frac{1}{(1 + x) \ (1 + x^2)^2} = \frac{A}{(1 + x)} + \frac{Bx + C}{(1 + x^2)} + \frac{Dx + E}{(1 + x^2)^2}....(1)$$
Now taking the LCM of RHS (Right Hand Side) terms.
$$\frac{1}{(1 + x) \ (1 + x^2)^2} =$$ $$\frac{A (1 + x^2)^2 + (1 + x) (1 + x^2) (Bx + C) + (1 + x) (Dx + E)}{(1 + x) \ (1 + x^2)^2}$$
$$\frac{1}{(1 + x) \ (1 + x^2)^2} = \frac{A (1 + x^2)^2 + (1 + x) (1 + x^2) (Bx + C) + (1 + x) (Dx + E)}{(1 + x) \ (1 + x^2)^2}$$
The terms in the denominators on both sides are the same so they will be canceled.
$$1 = A (1 + x^2)^2 +$$ $$(1 + x) (1 + x^2) (Bx + C) +$$ $$(1 + x) (Dx + E)$$
$$1 = A (1 + x^2)^2 + (1 + x) (1 + x^2) (Bx + C) + (1 + x) (Dx + E)$$
Write down the terms of $$x^4$$, $$x^3$$, $$x^2$$, $$x$$, and constants separately as written below.
$$1 = x^4 \ (A + B) + x^3 \ (B + C) +$$ $$x^2 \ (2A + B + C + D) +$$ $$x \ (B + C + D + E)$$ $$+ (A + C + E)$$
$$1 = x^4 \ (A + B) + x^3 \ (B + C) + x^2 \ (2A + B + C + D) + x \ (B + C + D + E) + (A + C + E)$$
After comparing the values of $$x^4$$, $$x^3$$, $$x^2$$, $$x$$, and constants of both sides we get. $$A + B = 0....(2)$$ $$B + C = 0....(3)$$ $$2A + B + C + D = 0....(4)$$ $$B + C + D + E = 0....(5)$$ $$A + C + E = 1....(6)$$ by subtracting the equation (5) from the equation (6). $$A - B - D = 1....(7)$$ Now by adding equation (4) and (7). $$3A + C = 1....(8)$$ Now by subtracting equation (3) from equation (2). $$A - C = 0....(9)$$ After solving the equations we get. $$A = \frac{1}{4}$$ $$B = \frac{-1}{4}$$ $$C = \frac{1}{4}$$ $$D = \frac{-1}{2}$$ $$E = \frac{1}{2}$$ After putting the values of A, B, C, D, and E in equation (1), we get the final equation of decomposition.
$$\frac{1}{(1 + x) \ (1 + x^2)^2} =$$ $$\frac{1}{4 \ (1 + x)} +$$ $$\frac{(1 - x)}{4 \ (1 + x^2)} +$$ $$\frac{(1 - x)}{2 \ (1 + x^2)^2}$$
$$\frac{1}{(1 + x) \ (1 + x^2)^2} = \frac{1}{4 \ (1 + x)} + \frac{(1 - x)}{4 \ (1 + x^2)} + \frac{(1 - x)}{2 \ (1 + x^2)^2}$$
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20FXandersguidetoeverythingpastchapter4
# 20FXandersguidetoeverythingpastchapter4 - The Chapter...
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The Chapter 5.1-5.3 Super Summary Chapter 5.1. The eigenvectors of a matrix are the non-zero vectors such that is some scalar multiple of . If is an eigenvector, then its eigenvalue is the scalar where . The eigenvalues of a matrix is the set of possible eigenvalues for all its eigenvectors. For any matrix and any eigenvalue of , the eigenspace of corresponding to is the set of all eigenvectors of with eigenvalue . All eigenspaces are vector spaces. If is triangular (either upper- or lower-), its eigenvalues are the entries on its main diagonal. This counts for diagonal matrices as well. Chapter 5.2. There’s a new way to find the determinant of a square matrix. Use row reductions to put it in row echelon form, except: Do NOT multiply a row in-place. (You can still add a multiple of a row to another) Keep track of how many times you exchange rows. Once you finish this, multiply the diagonal terms together (if you end up with a zero in a diagonal, your matrix is not invertible and has determinant zero). Then, multiply by , where is the number of rows exchanges you used. The result will be your determinant. Alternatively, you could multiply rows in-place, so long as you keep track of the multiples. For example, if you divide a row by 2 while row-reducing, you should multiply your final result by 2 to get the determinant. If you divide a row by 3 and multiply another by 4, you’ll have to multiply your result by 3 then divide by 4. Remember, you only need to keep track of this if you multiply a row in-place ; adding a multiple of a row to another can be done freely. Also, never multiply a row by zero -- that would mean you’d have to divide your end result by it! The following are equivalent (if one is true, they all are true): The matrix is not invertible. The number is an eigenvalue of Recall that:
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The characteristic equation of is: is an eigenvalue of if and only if it’s a solution of the characteristic equation. The number of times a factor appears in the characteristic equation is the multiplicity of the eigenvalue. For example, if this is your characteristic equation: Then the eigenvalue has multiplicity 4, the eigenvalue has multiplicity 2, and the eigenvalue has multiplicity 1. The dimension of an eigenspace is, at the maximum , the multiplicity of the eigenvalue. It could be anything between this value and 1. For any matrices and , if there exists an invertible matrix such that: Then and are similar (some call them conjugate ). Similar matrices have the same characteristic equation, and the same eigenvalues (but not the same eigenvectors!) Chapter 5.3. Let be an matrix. is called diagonalizable if it’s similar to a diagonal matrix, i.e. if there exists an invertible matrix and an diagonal matrix , such that: The following are all equivalent, for any matrix : is diagonalizable.
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Mister Exam
# sin(8x-(pi/3))=0 equation
The teacher will be very surprised to see your correct solution 😉
v
#### Numerical solution:
Do search numerical solution at [, ]
### The solution
You have entered [src]
/ pi\
sin|8*x - --| = 0
\ 3 /
$$\sin{\left(8 x - \frac{\pi}{3} \right)} = 0$$
Detail solution
Given the equation
$$\sin{\left(8 x - \frac{\pi}{3} \right)} = 0$$
- this is the simplest trigonometric equation
with the change of sign in 0
We get:
$$\sin{\left(8 x - \frac{\pi}{3} \right)} = 0$$
Divide both parts of the equation by -1
The equation is transformed to
$$\cos{\left(8 x + \frac{\pi}{6} \right)} = 0$$
This equation is transformed to
$$8 x + \frac{\pi}{6} = \pi n + \operatorname{acos}{\left(0 \right)}$$
$$8 x + \frac{\pi}{6} = \pi n - \pi + \operatorname{acos}{\left(0 \right)}$$
Or
$$8 x + \frac{\pi}{6} = \pi n + \frac{\pi}{2}$$
$$8 x + \frac{\pi}{6} = \pi n - \frac{\pi}{2}$$
, where n - is a integer
Move
$$\frac{\pi}{6}$$
to right part of the equation
with the opposite sign, in total:
$$8 x = \pi n + \frac{\pi}{3}$$
$$8 x = \pi n - \frac{2 \pi}{3}$$
Divide both parts of the equation by
$$8$$
$$x_{1} = \frac{\pi n}{8} + \frac{\pi}{24}$$
$$x_{2} = \frac{\pi n}{8} - \frac{\pi}{12}$$
The graph
Sum and product of roots [src]
sum
pi pi
-- + --
24 6
$$\frac{\pi}{24} + \frac{\pi}{6}$$
=
5*pi
----
24
$$\frac{5 \pi}{24}$$
product
pi pi
--*--
24 6
$$\frac{\pi}{24} \frac{\pi}{6}$$
=
2
pi
---
144
$$\frac{\pi^{2}}{144}$$
pi^2/144
Rapid solution [src]
pi
x1 = --
24
$$x_{1} = \frac{\pi}{24}$$
pi
x2 = --
6
$$x_{2} = \frac{\pi}{6}$$
x2 = pi/6
x1 = 90.0589894029074
x2 = 95.9494756283883
x3 = 28.012534494509
x4 = 40.1862060271694
x5 = -15.9697626557481
x6 = 0.130899693899575
x7 = 14.2680666350536
x8 = -73.3038285837618
x9 = -34.0339204138894
x10 = -97.6511716490827
x11 = -9.29387826686981
x12 = -79.9797129726402
x13 = -85.870199198121
x14 = 58.2503637853107
x15 = -83.9067037896274
x16 = -35.997415822383
x17 = -98.0438707307815
x18 = 80.2415123604393
x19 = -27.7507351067098
x20 = 6.02138591938044
x21 = 88.0954939944138
x22 = 22.1220482690281
x23 = 36.2592152101822
x24 = 66.1043454192852
x25 = 46.0766922526503
x26 = 86.1319985859202
x27 = 44.1131968441567
x28 = 73.9583270532597
x29 = 48.0401876611439
x30 = -89.7971900151083
x31 = -25.7872396982162
x32 = -91.7606854236019
x33 = 100.269165527074
x34 = -69.7695368484733
x35 = 38.2227106186758
x36 = -59.9520598060052
x37 = -87.8336946066146
x38 = 51.9671784781312
x39 = 7.98488132787406
x40 = -71.733032256967
x41 = -56.025068989018
x42 = 70.0313362362725
x43 = -78.0162175641465
x44 = -39.9244066393703
x45 = 82.2050077689329
x46 = -41.8879020478639
x47 = -61.9155552144988
x48 = -511.556003759538
x49 = -100.007366139275
x50 = -19.8967534727354
x51 = -67.8060414399797
x52 = -49.7418836818384
x53 = 46.8620904160477
x54 = 92.022484811401
x55 = 16.2315620435473
x56 = 62.177354602298
x57 = -3.79609112308767
x58 = -51.705379090332
x59 = -65.8425460314861
x60 = 60.2138591938044
x61 = 78.2780169519457
x62 = -54.4542726622231
x63 = -47.7783882733448
x64 = -7.72308194007491
x65 = 29.9760299030026
x66 = -45.8148928648512
x67 = -217.424391567194
x68 = -14.0062672472545
x69 = -12.0427718387609
x70 = -43.8513974563575
x71 = -1.83259571459405
x72 = 56.2868683768171
x73 = -29.7142305152035
x74 = -17.9332580642417
x75 = 34.2957198016886
x76 = -93.7241808320955
x77 = 26.0490390860154
x78 = -57.9885643975116
x79 = -37.9609112308767
x80 = 93.9859802198946
x81 = 84.1685031774265
x82 = 50.0036830696375
x83 = 42.1497014356631
x84 = -23.8237442897226
x85 = 18.1950574520409
x86 = -21.860248881229
x87 = -5.75958653158129
x88 = 24.0855436775217
x89 = 64.1408500107916
x90 = 4.05789051088682
x91 = 20.1585528605345
x92 = 71.9948316447661
x93 = -76.4454212373516
x94 = -63.8790506229925
x95 = 2.0943951023932
x96 = -81.9432083811338
x97 = 68.0678408277789
x97 = 68.0678408277789
|
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# Solving Equations With The Addition Method
By Kathleen Cantor, 03 Jul 2020
There are two ways of solving an equation: the addition method and the substitution method. Here, we’re going to take a look at the addition method. The addition method of solving equations is an easy and simple method used in algebra. It’s also called the elimination method.
## Understanding the Parts of an Equation
Before we learn how to use the addition method, let's breakdown of the parts of an equation.
This is our example:
3x + 4y = 10
The numbers 3 and 4 are coefficients that multiply with the variables.
x and y are variables that can have different values depending upon the outcome.
## Solving Using the Addition Method
Now how would you solve this equation? Ideally, you could pair up this equation with another one, let’s say 4x + 2y = 20 and solve it. But because both equations have two different variables, there is no way to identify the values of the variables. You would have to tediously guess the value of x and y, plot it in on a graph, and find the point where the two lines intersect, which will then give you the coordinates of x and y. Even then, if you draw the line or plot the points incorrectly, you won’t get the correct values of x and y.
To save us the hassle of going through that process, we use the addition method for solving equations that have one or two variables.
The addition method has a series of steps that you must follow to solve the equation correctly.
### Step 1
Put two given equations on top of each other and label them.
You need to determine what the two equations are, first. After that, we need to put them on top of each other like a standard addition problem. We will use these two equations as an example:
2x + 4y = 10 equation (1)
4x - 2y = 15 equation (2)
### Step 2
Select the variable that needs to be removed.
We now have to select the variable to remove from the equation. Let's choose the y variable.
### Step 3
Select the equation to convert the coefficient of the variable.
We now need to select the equation in which we will change the coefficient of the selected variable. Depending on the given equation, you can choose one or both. We will proceed with equation (2).
### Step 4
Change the coefficient of the equation.
After selecting the equation, remove the variable to change the coefficient into its opposite. We chose equation (2). Since the y coefficient is already negative, we don’t need to convert the sign. We will just multiply both sides of the equation by 2. We will then end up with a new equation (3).
(4x – 2y) x 2 = 15 x 2
8x – 4y = 30 Equation (3)
### Step 5
Add the two equations and substitute the result into the original equation.
Once the new equations are formed, we need to add the two equations and remove the selected variable. It is illustrated below:
2x + 4y = 10 Equation (1)
8x – 4y = 30 Equation (3)
10x = 40
Divide by 10 on both sides to isolate and solve for x.
x = 4
Now that we found the value of x, we need to substitute it into the original equation (1)
2(4) + 4y = 10
8 + 4y = 10
Subtract 8 from both sides
4y = 2
Divide both sides by 4
y = 0.5
Now we have the value of the two variables x = 4 and y = 0.5. We are done solving this equation.
## Conclusion
There you have it! A simple and effective way to learn the addition or elimination method to solve equations. Don’t worry, it doesn’t matter what two equations you get. Follow the steps above, and you will have no problem solving any two-variable equation question presented to you.
See the 1 Comment below.
### One Comment on “Solving Equations With The Addition Method”
1. Keigan says:
What about somthing like 4x = 9.50
### Comment Preview
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From Math Blogs
|
# 2006 AIME I Problems/Problem 7
## Problem
An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $C$ to the area of shaded region $B$ is 11/5. Find the ratio of shaded region $D$ to the area of shaded region $A.$
$[asy] defaultpen(linewidth(0.7)+fontsize(10)); for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray); } pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10); draw(B--A--C); fill((7.3,0)--(7.8,0)--(7.8,6)--(7.3,6)--cycle, white); clip(B--A--C--cycle); for(int i=0; i<9; i=i+1) { draw((i,1)--(i,6)); } label("\mathcal{A}", A+0.2*dir(-17), S); label("\mathcal{B}", A+2.3*dir(-17), S); label("\mathcal{C}", A+4.4*dir(-17), S); label("\mathcal{D}", A+6.5*dir(-17), S);[/asy]$
## Solution 1
Note that the apex of the angle is not on the parallel lines. Set up a coordinate proof.
Let the set of parallel lines be perpendicular to the x-axis, such that they cross it at $0, 1, 2 \ldots$. The base of region $\mathcal{A}$ is on the line $x = 1$. The bigger base of region $\mathcal{D}$ is on the line $x = 7$. Let the top side of the angle be $y = x - s$ and the bottom side be x-axis, as dividing the angle doesn't change the problem.
Since the area of the triangle is equal to $\frac{1}{2}bh$,
$$\frac{\textrm{Region\ }\mathcal{C}}{\textrm{Region\ }\mathcal{B}} = \frac{11}{5} = \frac{\frac 12(5-s)^2 - \frac 12(4-s)^2}{\frac 12(3-s)^2 - \frac12(2-s)^2}$$
Solve this to find that $s = \frac{5}{6}$.
Using the same reasoning as above, we get $\frac{\textrm{Region\ }\mathcal{D}}{\textrm{Region\ }\mathcal{A}} = \frac{\frac 12(7-s)^2 - \frac 12(6-s)^2}{\frac 12(1-s)^2}$, which is $\boxed{408}$.
## Solution 2
Note that the sections between the two transversals can be divided into one small triangle and a number of trapezoids. Let one side length (not on a parallel line) of the small triangle be $x$ and the area of it be $x^2$. Also, let all sections of the line on the same side as the side with length $x$ on a trapezoid be equal to $1$.
Move on to the second-smallest triangle, formed by attaching this triangle with the next trapezoid. Parallel lines give us similar triangles, so we know the proportion of this triangle to the previous triangle is ${\left(\frac{x+1}{x}\right)}^2$. Multiplying, we get $(x+1)^2$ as the area of the triangle, so the area of the trapezoid is $2x+1$. Repeating this process, we get that the area of B is $2x+3$, the area of C is $2x+7$, and the area of D is $2x+11$.
We can now use the given condition that the ratio of C and B is $\frac{11}{5}$.
$\frac{11}{5} = \frac{2x+7}{2x+3}$ gives us $x = \frac{1}{6}$
So now we compute the ratio of D and A, which is $\frac{(2)(\frac{1}{6}) + 11}{(\frac{1}{6})^2} = \boxed{408.}$
Edit: fixed misplaced brackets
## Solution 3 (Bash)
Let the distances from the apex to the parallel lines be $x$ and $y$ and the distance between the intersections be $a,b.$ We know the area ratio means $\frac{(x+4a)(y+4b)-(x+3a)(y+3b)}{(x+2a)(y+2b)-(x+a)(y+b)} =\frac{5}{11}$ which simplifying yields $ab = 3ay+3bx.$ The ratio we seek is $\frac{(x+6a)(y+6b)-(x+5a)(y+5b)}{xy} =\frac{ay+yx+11ab}{xy}.$ We know that $ab = 3ay+3bx$ so the ratio we seed is $\frac{33(ay+yx)}{11xy}.$ Finally note that by similar triangles $\frac{x}{x+a} =\frac{y}{y+b} \implies bx = ya.$ Therefore the ratio we seek is $\frac{66(ay)}{11xy} =\frac{66a}{11x}.$ Finally note that $ab=3ay+3bx \implies ab = 6bx \implies a = 6x$ so the final ratio is $6 \cdot 68 = \boxed{408}.$
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# Criterio de Leibniz
Now Playing:Es alternating series test – Example 1a
Examples
1. Convergencia y el Criterio de Leibniz
Muestra que las siguientes series convergen:
1. $\large \sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}$
2. $\large \sum_{n=4}^{\infty}\frac{n(-1)^{n+2}}{n^3+1}$
3. $\large \sum_{n=2}^{\infty}\frac{3}{e^n(-1)^{n-2}}$
Introducción a sucesiones
Notes
## Alternating Series Test
The alternating series test (also known as the Leibniz test), is type of series test used to determine the convergence of series that alternate. Keep in mind that the test does not tell whether the series diverges. In order to use this test, we first need to know what a converging series and a diverging series is.
## What is a convergent series? What is a divergent series?
A convergent series is an infinite series which sums up to a finite number. For example, the famous convergent series:
This convergent series sum up to $\frac{\pi^{2}}{6}$! How do you get that? It is a long process that requires a lot of calculations, but usually it is sufficient enough to know that the series is convergent by the p-series test.
A divergent series is an infinite series where the sum is infinity. For example, the series
Adding up all the numbers will give you a sum of infinity. If that doesn't convince you, take a look at this. Note that the series could be written as an $N^{th}$ partial sum.
Now if we were to make it an infinite series, then we are going to take the limit as N goes to infinity of both sides. In other words,
Now that we know what a divergent and convergent series is, let's take a look at the alternating series test.
## Alternating Series Test
In order to use the alternating series test, the series must be alternating. In other words, the series are in the form:
or
where $b_{n} \leq 0$. An alternating series is not limited to these two forms because the exponent on the (-1) can vary. Now the alternating series test states that if the two following conditions are met, then the alternating series is convergent:
For the second condition, $b_{n}$ does not have to be strictly decreasing for all $n \leq 1$. As long as the sequence is decreasing at $n$$\infty$, then that will be sufficient enough to show that it is decreasing. Now that we know what the alternating series test is, let us put it to use for the following examples.
Example 1: Show that the series
is convergent.
Before we want to use the alternating series test, we want to make sure that the series is actually alternating. In other words, turn this series into the form:
As you can see, we can turn our series into that form. Notice that:
And so we know that:
Hence, we can go ahead and use the alternating series test. Now remember the two conditions. First we have to show that
Since our $b_{n} = \frac{1}{n^{2}}$, then our limit is
Notice that this limit does to 0, hence
and so our first condition is fulfilled. Now let's take a look at the second condition. We have to make sure that $\frac{1}{n^{2}}$ is decreasing. How do we do this? There are a total of three ways to do this:
• Method 1:
Write out the first few terms We can write out the first few terms of $b_{n}$ and then conclude if the sequence is decreasing. Notice that:
Notice how the numerator never changes, but the denominator is getting bigger and bigger. As the denominator gets bigger, then the numbers itself get smaller. Hence we can conclude that the sequence is decreasing, and condition is fulfilled. Some teachers may not see this method as legitimate for more complicating questions (because it's harder to compare). In that case, look at the other methods.
• Method 2:
Compare the $n^{th}$ term and $(n+1)^{th}$ term of the sequence $b_{n}$.
Notice that the $n^{th}$ term of the sequence $b_{n}$ is:
And the $(n+1)^{th}$ term is
Now comparing the two terms, you should notice that:
The left side is bigger than the right side because the denominator in the right side is bigger, hence the term is actually smaller. So we just concluded that:
for all $n$>1. This means that the $n^{th}$ term is always going to be bigger than the $(n+1)^{th}$ term, which means the sequence is always decreasing. Hence again, the second condition is fulfilled. Again, sometimes it's really hard to compare the two terms. In this case, look at method 3.
• Method 3:
Take the derivative
What were going to do is take the general term ($b_{n}$) and change all the $n$'s to $x$'s, and set it as $f(x)$. In other words,
Now we are going to take the derivative of this function. This will give us:
Notice that for $x$ > 0, the denominator is going to be positive, and so the derivative f'(x) is negative. In other words, for increasing value of $x$, the function is decreasing. Now let's put that into the perspective of $b_{n}$. This means that for increasing values of n, the sequence $b_{n}$ is always decreasing. Thus, we just fulfilled the second condition again.
Since the two conditions are fulfilled, then we can conclude that the series
converges. Now let's take a look at a more interesting alternating series.
Example 2: Consider the alternating harmonic series:
Is it convergent? If it is, then what is the sum of this series?
• What should we do here? You are probably thinking about one of the series convergence tests. You are most likely thinking about using the alternating series test. Again, we need to show that this is in fact an alternating series before we can apply the alternating series test. Recall that an alternating series could be of the form:
• Notice that:
• And so we know that
• Since we know that it is an alternating series, then we can see if the two conditions are fulfilled. For the first condition, we see that:
• So the first condition is fulfilled. Now the second condition states that $b_{n}$ must be a decreasing sequence. Feel free to use any of the methods, but I will be using method two. See that the $n^{th}$ term and $(n+1)^{th}$ term are
• Notice by comparing $n^{th}$ term and $(n+1)^{th}$ term of the sequence, we have:
• Again, this is because the denominator on the right side is bigger, so the term is actually smaller. In other words, we just concluded that:
• For $n$>1. Hence, every term after is smaller than the one before it. So we can conclude that the sequence is decreasing. Since both of the conditions are fulfilled, then the series is convergent. But what is the sum of the series? Fortunately, there is an easy way to find this. First, we must recognize the Maclaurin series:
• Now if we were to set $x$ = -1, then we will see that
• In other words, we just concluded that:
So the sum of this alternating harmonic series is ln(2). If you want to take a look at more examples of using the alternating series test, click on this link.
http://tutorial.math.lamar.edu/Classes/CalcII/AlternatingSeries.aspx
Now here is an interesting question. If the alternating harmonic series is convergent, then what about the harmonic series itself? Is
convergent or divergent? You may realize that this isn't an alternating series, so we are going to have to use another test. Why don't we use the $n^{th}$ term test?
## nth term test for divergence
Recall that the nth term test (also known as the divergence test) states the following:
So if we take the limit and it is anything BUT 0, then we can say that the series diverge. However, if it does equal 0, then it does NOT mean the series converge. It just means the test has failed, and you would have to use something else to test the convergence.
Now using this test for the harmonic series we let:
Then you will see that:
Since we've got 0, then the nth term test has failed and we would have to try something else. This leads to the question, maybe the harmonic series really converges?
## Does 1/n converge
Let us assume that the harmonic series is convergent. Then that means the series must sum up to a finite number. Let's call that sum S. So
Now we are going to play a little trick here. We can say that:
However on the right hand side of the inequality, see that
In other words, we are saying that
This is impossible, so we reached a contradiction. This situation causes us to get a mathematically illogical statement, so then the harmonic series must be divergent. A lot of people get confused by this method, so I have prepared another easier method to show that the harmonic series diverges. This method will require you to know the p-series rules.
Recall that p-series are in the form:
The p-series rule (or p-series test) states that if p>1, then the series converge. Otherwise, the series diverge. Notice that the series that we have is very similar to it.
In fact, our p=1 in this case. We also know that p=1$\ngtr$1, so then we know that the series diverge. Now we kind of went off topic here, but you must have realized that removing the $(-1)^{(n+1)}$ from the series can actually change the convergence or divergence of a series.
## Alternating Series Estimation Theorem
So we learned that it is possible to find the sum of an alternating harmonic series using a complicated formula that we were unfamiliar with. But what if we are dealing with another alternating series? How would we find the sum? Unfortunately there is no good way to find the exact sum of converging alternating series, but there is a way to estimate the sum. It is called the alternating series estimation theorem.
The alternating series estimation theorem states the following:
Suppose that the alternating series
Is convergent and converges to a finite number $S$. Then
where:
$S_{n}$ = the partial sum of n terms (sum of the first n terms)
$R_{n}$ = the remainder (or error term) that we get from subtracting the actual value of the series with the sum of the first n terms. Sometimes it's called the alternating series error.
$b_{n+1}$ = the neglected term.
Note that this theorem only works if the series is alternating. You are probably very confused at what this theorem is saying, so let us use a series as an example.
Example 3: Let's take a look at the alternating harmonic series we used earlier:
• Let's expand the series out. Doing so will give us:
• Now let's say that we want to estimate the sum of this series. What I'm going to do is estimate the sum of the series by summing the first 4 terms. Summing the first 4 terms gives us:
• Since we only look at the first 4 terms, then the next term after it (the $5^{th}$ term) is the neglected term. Recall that the series can be rewritten as
• So $b_{n} = \frac{1}{n}$, and hence
• Now what the theorem says is that
• So using the 2 pieces of information that we have, then
• Our remainder is kind of in the way of our equation, so we can get rid of it and have
• Now instead of having the absolute value, we can rewrite our inequality to be:
• Adding 0.58333 to all sides of the inequality gives us:
• Hence, we just estimate what the sum of the series is! It must be between 0.38333 and 0.78333. Note that you can also do this by summing the first 5 term or the first 11 terms. Usually the questions you will be dealing with will tell you how many terms you need to sum. In fact, as you sum more and more terms and use this theorem, your estimation becomes more and more accurate.
Now let's take a look at a question where we don't know how many terms we need to sum, but we know that our remainder (or error) has to be less than a certain number.
Example 4: Determine the number of terms required to approximate the sum of the series
with an error of less than 0.0001.
• Recall that the theorem states that:
• If the error must be less than 0.0001, then basically we are saying that
• or in a more simple manner,
• Now we can't really do anything with this inequality because there are too many unknowns. So we need to think of a better way to do this. This is where it gets a little tricky because we actually have to think about this.
• Instead of saying that 0.0001 is less than the neglected term $(0.0001 \leq b_{n+1})$, why don't we say that the neglected term is less than 0.0001. In other words, let's say that
• Why? Think about it. We know that
• The error term is less than the neglected term. So if the neglected term is less than 0.0001, then it must be true that the error term is less than 0.0001. Basically we are saying that:
• So this works! Now going back to our inequality
• We know that
• So
• That means
• We can rewrite this inequality to be:
• That means as long as $n$ > 9999, then the error (or remainder) will be less than 0.0001. Let's pick $n$=10000. This means that summing the first 10000 terms will guarantee the error to be less than 0.0001. That is a lot of terms to sum up! If you want to look at more examples, then take a look at this link:
Una serie alternada tiene la forma:
$\large \sum(-1)^nb_n$ o $\large \sum(-1)^{n+1}b_n$
Donde $b^n \geq0$
Cabe recalcar que las series alternadas no están limitadas a estas dos formas solamente, ya que el exponente en el (-1) varía y con ello la manera en la cual se ven estas series.
La prueba de la serie alternada, también llamado el Criterio de Leibniz dice que si las siguientes dos condiciones se cumplen, entonces la serie alternada es convergente.
1. $\lim$n →$\infty$ $b_n=0$
2. La secuencia $b_n$ es una secuencia decreciente.
Para la segunda condición, $b_n$ no tiene que ser decreciente estrictamente para todas las $n\geq 1$. Mientras la secuencia sea decreciente para $n$$\infty$, es suficiente.
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# How do you write the standard form of the equation of the parabola that has the indicated vertex and whose graph passes through the point Vertex (-2,5); Point ( 0,9)?
Jul 20, 2018
$y = {x}^{2} + 4 x + 9$
#### Explanation:
$\text{the equation of a parabola in "color(blue)"vertex form}$ is.
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y = a {\left(x - h\right)}^{2} + k} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
$\text{where "(h,k)" are the coordinates of the vertex and a}$
$\text{is a multiplier}$
$\text{here } \left(h , k\right) = \left(- 2 , 5\right)$
$y = a {\left(x + 2\right)}^{2} + 5$
$\text{to find a substitute "(0,9)" into the equation}$
$9 = 4 a + 5 \Rightarrow 4 a = 4 \Rightarrow a = 1$
$y = {\left(x + 2\right)}^{2} + 5 \leftarrow \textcolor{red}{\text{in vertex form}}$
$\textcolor{w h i t e}{y} = {x}^{2} + 4 x + 4 + 5$
$y = {x}^{2} + 4 x + 9 \leftarrow \textcolor{red}{\text{in standard form}}$
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# Practice Problems : The Remainder and Factor Theorem
1. What number should be added to $\displaystyle 2x^3 - 3x^2 - 8x$ so that the resulting polynomial leaves the remainder $\displaystyle 10$ when divided by $\displaystyle 2x + 1$?
Show/Hide Solution
Let the number to be added be $\displaystyle k$ and the resulting polynomial be $\displaystyle f(x)$. $\displaystyle \therefore \ f(x)=2{{x}^{3}}-3{{x}^{2}}-8x+k$ When $\displaystyle f(x)$ is divided by $\displaystyle 2x+1$, the remainder is 10. $\displaystyle \begin{array}{l}\therefore \ f\left( {-\displaystyle \frac{1}{2}} \right)=10\\\\\ \ 2{{\left( {-\displaystyle \frac{1}{2}} \right)}^{3}}-3{{\left( {-\displaystyle \frac{1}{2}} \right)}^{2}}-8\left( {-\displaystyle \frac{1}{2}} \right)+k=10\\\\\therefore \ -\displaystyle \frac{1}{4}-\displaystyle \frac{3}{4}+4+k=10\\\\\therefore \ k=7\end{array}$ Hence, the number to be added is 10.
2. What number should be subtracted from $\displaystyle 6x^3 + 7x^2 - 9x+12$ so that $\displaystyle 3x - 1$ is the factor of the resulting polynomial?
Show/Hide Solution
Let the number to be subtracted be $\displaystyle k$ and the resulting polynomial be $\displaystyle f(x)$. $\displaystyle \therefore \ f(x)=6x^3 + 7x^2 - 9x+12-k$ Since $\displaystyle 3x - 1$ is the factor of $\displaystyle f(x)$, $\displaystyle \begin{array}{l}\ \ f\left( {\displaystyle \frac{1}{3}} \right)=0\\\\\ \ 6{{\left( {\displaystyle \frac{1}{3}} \right)}^{3}}+7{{\left( {\displaystyle \frac{1}{3}} \right)}^{2}}-9\left( {\displaystyle \frac{1}{3}} \right)+12-k=0\\\\\therefore \ \displaystyle \frac{2}{9}+\displaystyle \frac{7}{9}-3+12-k=0\\\\\therefore \ k=10\end{array}$ Hence, the number to be subtracted is 10.
3. When divided by $\displaystyle x - 3$ the polynomials $\displaystyle x^3 - px^2 + x + 6$ and $\displaystyle 2x^3 - x^2 - (p + 3) x - 6$ leave the same remainder. Find the value of $\displaystyle p$.
Show/Hide Solution
Let $\displaystyle f(x)={{x}^{3}}-p{{x}^{2}}+x+6$ and $\displaystyle g(x)=2x^3 - x^2 - (p + 3) x - 6$ $\displaystyle f(x)$ and $\displaystyle g(x)$ leave the same remainder when divided by $\displaystyle x - 3$, $\displaystyle \begin{array}{l}\therefore \ f(3)=g(3)\\\\\ \ \ {{(3)}^{3}}-p{{(3)}^{2}}+(3)+6=2{{(3)}^{3}}-{{(3)}^{2}}-(p+3)(3)-6\\\\\ \ \ 27-9p+9=54-9-3p-9-6\\\ \ \\\therefore \ 36-9p=30-3p\ \\\\\therefore \ p=1\ \end{array}$
4. Using remainder theorem, find the value of $\displaystyle a$ if the division of $\displaystyle x^3 + 5x^2 - ax + 6$ by $\displaystyle x -1$ leaves the remainder $\displaystyle 2a$.
Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \text{Let}\ f(x)={{x}^{3}}+5{{x}^{2}}-ax+6\\\\\ \ \ \text{When}\ f(x)\ \text{is divided by }x-1,\ \\\\\ \ \ \text{the remainder}\ =2a\\\\\therefore \ f(1)=2a\\\\\ \ \ {{1}^{3}}+5{{(1)}^{2}}-a(1)+6=2a\\\\\ \ \ 1+5-a+6=2a\\\\\therefore \ 3a=12\\\\\ \ \ a=4\ \ \ \end{array}$
5. Find the value of the constants $\displaystyle a$ and $\displaystyle b$, if $\displaystyle x - 2$ and $\displaystyle x + 3$ are both factors of the expression $\displaystyle x^3 + ax^2 + bx - 12$.
Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \text{Let}\ f(x)={{x}^{3}}+a{{x}^{2}}+bx-12\\\\\ \ \ (x-2)\ \text{and}\ (x+3)\ \text{are factors of }f(x).\ \\\\\therefore \ f(2)=0\\\\\ \ \ {{(2)}^{3}}+a{{(2)}^{2}}+b(2)-12=0\\\\\therefore \ \ 8+4a+2b-12=0\\\\\therefore \ 2a+b=2---(1)\\\\\ \ \text{Again}\ f(-3)=0\\\\\ \ \ {{(-3)}^{3}}+a{{(-3)}^{2}}+b(-3)-12=0\\\\\therefore \ \ -27+9a-3b-12=0\\\\\therefore \ \ 3a-b=13---(2)\\\\\ \ \ (1)+(2)\Rightarrow 5a=15\\\\\therefore \ \ a=3\\\\\therefore \ 2\left( 3 \right)+b=2\\\\\therefore \ b=-4\end{array}$
6. If $\displaystyle x + 2$ and $\displaystyle x - 3$ are factors of $\displaystyle x^3 + ax + b$, find the values of $\displaystyle a$ and $\displaystyle b$. With these values of $\displaystyle a$ and $\displaystyle b$, factorise the given expression.
Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \text{Let}\ f(x)={{x}^{3}}+ax+b\\\\\ \ \ (x+2)\ \text{and}\ (x-3)\ \text{are factors of }f(x).\ \\\\\therefore \ f(-2)=0\\\\\ \ \ {{(-2)}^{3}}+a(-2)+b=0\\\\\therefore \ \ -8-2a+b=0\\\\\therefore \ -2a+b=8---(1)\\\\\ \ \text{Again}\ f(3)=0\\\\\ \ \ {{(3)}^{3}}+a(3)+b=0\\\\\therefore \ \ 27+3a+b=0\\\\\therefore \ \ 3a+b=-27---(2)\\\\\ \ \ (2)-(1)\Rightarrow 5a=-35\\\\\therefore \ \ a=-7\\\\\therefore \ \ -2\left( {-7} \right)+b=8\\\\\therefore \ \ b=-6\\\\\therefore \ \ f(x)={{x}^{3}}-7x-6\\\\\ \ \ \ \text{Let}\ {{x}^{3}}-7x-6=(x+2)(x-3)(x+k)\\\\\therefore \ \ {{x}^{3}}-7x-6={{x}^{3}}+\left( {k-1} \right){{x}^{2}}-\left( {k+6} \right)x-6k\\\\\therefore \ k-1=0\Rightarrow k=1\\\\\therefore \ \ f(x)=(x+2)(x-3)(x+1)\\\ \ \end{array}$
7. Given that $\displaystyle x - 2$ is a factor of the expression $\displaystyle x^3 + ax^2 + bx + 6$. When this expression is divided by $\displaystyle x - 3$, it leaves the remainder $3$. Find the values of $\displaystyle a$ and $\displaystyle b$.
Show/Hide Solution
Let $f(x)=x^{3}+a x^{2}+b x+6\\\\$ $x-2$ is a factor of $f(x)$. \begin{aligned} &\\ &\therefore\ f(2)=0\\\\ &2^{3}+a(2)^{2}+2 b+6=0\\\\ &2 a+b=-7 \ldots(1)\\\\ \end{aligned} When $f(x)$ is divided by $x-3$, the remainder is $3$. \begin{aligned} &\\ &\therefore\ f(3) =3 \\\\ &3^{3}+a(3)^{2}+3 b+6 =3 \\\\ &3 a+b =-10 \cdots(2)\\\\ \end{aligned} By equation $(2)$ - equation $(1)$, \begin{aligned} &\\ a=-3\\\\ \end{aligned} Substituting $a=-3$ in equation $(1)$, \begin{aligned} &\\ &2(-3)+b=-7\\\\ &\therefore\ b=-1 \end{aligned}
8. If $\displaystyle x - 2$ is a factor of the expression $\displaystyle 2x^3 + ax^2 + bx - 14$ and when the expression is divided by $\displaystyle x - 3$, it leaves a remainder $\displaystyle 52$, find the values of $\displaystyle a$ and $\displaystyle b$.
Show/Hide Solution
Let $f(x)=2x^{3}+a x^{2}+b x-14\\\\$ $x-2$ is a factor of $f(x)$. \begin{aligned} &\\ &\therefore\ f(2)=0\\\\ &2(2)^{3}+a (2)^{2}+b (2)-14=0\\\\ &2 a+b=-1 \ldots(1)\\\\ \end{aligned} When $f(x)$ is divided by $x-3$, the remainder is $52$. \begin{aligned} &\\ &\therefore\ f(3) =52 \\\\ &2(3)^{3}+a (3)^{2}+b (3)-14 =52 \\\\ &3 a+b =4 \cdots(2)\\\\ \end{aligned} By equation $(2)$ - equation $(1)$, \begin{aligned} &\\ a=5\\\\ \end{aligned} Substituting $a=5$ in equation $(1)$, \begin{aligned} &\\ &2(5)+b=-1\\\\ &\therefore\ b=-11 \end{aligned}
9. If $ax^3 + 3x^2 + bx - 3$ has a factor $2x + 3$ and leaves remainder $-3$ when divided by $x + 2$, find the values of $a$ and $\displaystyle b$. With these values of $a$ and $b$, factorise the given expression.
Show/Hide Solution
Let $f(x)=ax^3 + 3x^2 + bx - 3\\\\$ $2x+3$ is a factor of $f(x)$. \begin{aligned} &\\ &\therefore\ f\left(-\dfrac{3}{2}\right)=0\\\\ &a\left(-\dfrac{3}{2}\right)^3 + 3\left(-\dfrac{3}{2}\right)^2 + b\left(-\dfrac{3}{2}\right) - 3=0\\\\ &-\frac{27}{8}a-\frac{3}{2}b + \frac{15}{4}=0\\\\ &9 a+4b=10 \ldots(1)\\\\ \end{aligned} When $f(x)$ is divided by $x+2$, the remainder is $-3$. \begin{aligned} &\\ &\therefore\ f(-2) =-3 \\\\ &a(-2)^3 + 3(-2)^2 + b(-2) - 3 =-3 \\\\ &4 a+b =6 \\\\ &b =6-4a \cdots(2)\\\\ \end{aligned} Substituting $b =6-4a$ in equation $(1)$, \begin{aligned} &\\ 9 a+4(6-4a)=10\\\\ &a=2\\\\ \end{aligned} Substituting $a=2$ in equation $(2)$, \begin{aligned} &\\ &b =6-4(2)\\\\ &\therefore\ b=-2\\\\ &\therefore\ f(x)=2x^3 + 3x^2 -2x - 3\\\\ \end{aligned} Since $f(x)$ is a cubic polynomial with factor $2x+3$ and leading coefficient $2$, assume that $f(x) = (2x+3)(x^2 + px + q)$. \begin{aligned} &\\ &2x^3 + 3x^2 -2x - 3=(2x+3)(x^2 + px + q)\\\\ &\therefore\ 2x^3 + 3x^2 -2x - 3=2x^3 + (2p+3)x^2 + (3p+2q)x + 3q\\\\ &\text{Equating the respective terms}\\\\ & 3q=-3\\\\ &\therefore\ q=-1\\\\ & 2p+3 = 3\\\\ &\therefore\ p=0\\\\ &\therefore\ f(x) = (2x+3)(x^2 -1)\\\\ &\therefore\ f(x) = (2x+3)(x +1)(x-1) \end{aligned}
10. Given $\displaystyle f (x) = ax^2 + bx + 2$ and $\displaystyle g (x) = bx^2 + ax + 1$. If $\displaystyle x - 2$ is a factor of $\displaystyle f (x)$ but leaves the remainder $\displaystyle -15$ when it divides $\displaystyle g (x)$, find the values of $\displaystyle a$ and $\displaystyle b$. With these values of $\displaystyle a$ and $\displaystyle b$, factorise the expression $\displaystyle f (x) + g (x) + 4x^2 + 7x$.
Show/Hide Solution
\begin{aligned} &f(x)=a x^{2}+b x+2\\\\ &x-2 \text{ is a factor of } f(x)\\\\ &\therefore\ a(2)^{2}+2 b+2=0\\\\ &2 a+b=-1 \ldots(1)\\\\ &g(x)=b x^{2}+a x+1\\\\ \end{aligned} When g(x) is divided by $x-2$, the remainder is $-15$. \begin{aligned} &\\ \therefore g(2)&=-15 \\\\ b(2)^{2}+2 a+1 &=-15 \\\\ a+2 b &=-8 \ldots(2) \\\\ (1)+(2) \Rightarrow 3 a+3 b &=-9 \\\\ a+b &=-3 \ldots(3) \\\\ (1)-(2) \Rightarrow a-b &=7 \ldots(4) \\\\ (3)+(4) \Rightarrow 2 a &=4 \\\\ a &=2 \\\\ (3)-(4) \Rightarrow 2 b &=-10 \\\\ b &=-5\\\\ \end{aligned} \begin{aligned} \text { Let } h(x) &=f(x)+g(x)+4 x^{2}+7 x \\\\ &=2 x^{2}-5 x+2-5 x^{2}+2 x+1+4 x^{2}+7 x \\\\ &=x^{2}-4 x+3 \\\\ &=(x-1)(x-3) \end{aligned}
11. When $\displaystyle x^3 - 2x^2 + px - q$ is divided by $\displaystyle x^2 - 2x - 3$, the remainder is $\displaystyle x - 6$, What are the values of $\displaystyle p$ and $\displaystyle q$ respectively ?
Show/Hide Solution
Let $f(x)=x^{3}-2 x^{2}+p x-q\\\\$. When $f(x)$ is divided by $x^{2}-2 x-3$, the remainder is $x-6\\\\$. Let $Q(x)$ be the quotient when $f(x)$ is divided by $x^{2}-2 x-3$. \begin{aligned} & \\ & \therefore\ f(x)=Q(x)\left(x^{2}-2 x-3\right)+x-6 \\\\ & x^{3}-2 x^{2}+p x-q=Q(x)(x+1)(x-3)+x-6\\\\ & \text{When } x=-1,\\\\ &(-1)^{3}-2(-1)^{2}-p-q=-7 \\\\ & p+q=4 \ldots(1)\\\\ & \text{When } x=3,\\\\ &(3)^{3}-2(3)^{2}+p(3)-q=-3 \\\\ & 3 p-q=-12 \ldots(2)\\\\ & \text{By equation} (1) + \text{ equation } (2)\\\\ & 4 p =-8 \\\\ & p =-2\\\\ &\text{Substituting } p=-2 \text{ in equation } (1),\\\\ & -2+q=4 \\\\ & q=6 \end{aligned}
12. If $\displaystyle x + k$ is a common factor of $\displaystyle x^2 + px + q$ and $\displaystyle x^2 + lx + m$, Find the value of $\displaystyle k$ in terms of $\displaystyle p, q, l$ and $\displaystyle m$.
Show/Hide Solution
\begin{aligned} \text { Let } f(x) &=x^{2}+p x+q \\\\ g(x) &=x^{2}+l x+m \\\\ \end{aligned} $x+k$ is a common factor of $f(x)$ and $g(x)$. \begin{aligned} & \\ & \therefore \quad f(-k)=0 \\\\ & (-k)^{2}+p(-k)+q=0 \\\\ & k^{2}-p k+q=0 \ldots(1) \\\\ & g(-k)=0 \\\\ &(-k)^{2}+L(-k)+m \\\\ & k^{2}-l k+m=0 \ldots(2)\\\\ \end{aligned} By equation (1) - equation (2), \begin{aligned} &\\ & -p k+l k +q-m=0 \\\\ & (l-p) k =m-q \\\\ & k =\dfrac{m-q}{l-p} \end{aligned}
13. When a polynomial $\displaystyle f(x)$ is divided by $\displaystyle x - 3$ and $\displaystyle x + 6$, the respective remainders are $\displaystyle 7$ and $\displaystyle 22$. What is the remainder when $\displaystyle f(x)$ is divided by $\displaystyle (x - 3) (x + 6)$ ?
Show/Hide Solution
Let $f(x)$ be a polynomial.By the problem, \begin{aligned} &\\ f(3)&=7 \\\\ f(-6)&=22\\\\ \end{aligned} Let $Q(x)$ be the quotiont and $a x+b$ be the remainder When $f(x)$ is divided by $(x-3)(x+6)$ \begin{aligned} &\\ \therefore f(x) &=Q(x)(x-3)(x+b)+a x+b \\\\ f(3) &=3 a+b=7 \ldots(1) \\\\ f(-6) &=-6 a+b=22 \ldots(2)\\\\ \end{aligned} By equation $(1)$ - equation $(2)$, \begin{aligned} &\\ 9 a &=-15 \\\\ a &=-\dfrac{5}{3}\\\\ \end{aligned} Substituting $a=-\dfrac{5}{3}$ in equation $(1)$, \begin{aligned} &\\ 3\left(-\frac{5}{3}\right)+b=7 \\\\ b=12\\\\ \end{aligned} $\therefore$ When $f(x)$ is divided by $(x-3)(x+6)$, the remainder is $-\dfrac{5}{3} x+12$. $\displaystyle (x-3)(x+6)$ သည္ polynomial of second degree ျဖစ္ပါသည္။ $\displaystyle f(x)$ ကို $\displaystyle (x-3)(x+6)$ ႏွင့္ စားေသာ အႂကြင္းသည္ စားကိန္းေအာက္ တစ္ထပ္ေလ်ာ့ပါမည္။။ ထို႔ေၾကာင့္ အႂကြင္းသည္ $\displaystyle ax + b$ ပံုစံျဖစ္ပါမည္။
14. Given that $\displaystyle f (x) = x^3 + ax^2 + bx + c$. If $\displaystyle f (1) = f(2) = 0$ and $\displaystyle f(4) = f(0)$, find $\displaystyle a, b$ and $\displaystyle c$.
Show/Hide Solution
\begin{aligned} &f(x)=x^{3}+a x^{2}+b x+c \\\\ &f(1)=0 \\\\ &1^{3}+a(1)^{2}+b(1)+c=0 \\\\ &a+b+c=-1 \cdots(1) \\\\ &f(2)=0 \\\\ &2^{3}+a(2)^{2}+b(2)+c=0 \\\\ &4 a+2 b+c=-8 \ldots(2)\\\\ &f(4)=f(0) \\\\ &4^{3}+a(4)^{2}+b(4)+c=c \\\\ &4 a+b=-16 \ldots(3)\\\\ \end{aligned} By equation $(2)$ - equation $(1)$, \begin{aligned} &\\ 3 a+b&=-7 \ldots(4)\\\\ \end{aligned} By equation $(3)$ - equation $(4)$, \begin{aligned} &\\ a=-9\\\\ \end{aligned} Substituting $a=-9$ in equation $(4)$, \begin{aligned} &\\ 3(-9)+b &=-7 \\\\ b &=20\\\\ \end{aligned} Substituting $a=-9$ and $b=20$ in equation $(1)$, \begin{aligned} &\\ -9+20+c &=-1 \\\\ c &=-12 \end{aligned}
15. If $\displaystyle x – 1$ is a factor of $\displaystyle Ax^3 + Bx^2 - 36x + 22$ and $\displaystyle 2^B = 64^A$, find $\displaystyle A$ and $\displaystyle B$.
Show/Hide Solution
\begin{aligned} &\text { Let } f(x)=A x^{3}+B x^{2}-36 x+22. \\\\ &x-1 \text { is a factor of } f(x). \\\\ &\therefore\ A+B-36+22=0 \\\\ &A+B=14 \ldots (1) \\\\ &2^{B}=64^{A}(\text { given }) \\\\ &2^{B}=2^{6 A} \\\\ &\therefore B=6 A \ldots (2)\\\\ \end{aligned} Substituting $B=6 A$ in equation $(1)$, \begin{aligned} &\\ A+6 A &=14 \\\\ 7 A &=14 \\\\ A &=2\\\\ \end{aligned} Substituting $A=2$ in equation $(2)$. \begin{aligned} &\\ B & =6(2)\\\\ &=12\\\\ \end{aligned}
16. When k is subtracted from $\displaystyle 27x^3 - 9x^2 - 6x - 5$, it is exactly divisible by $3x - 1$, find $\displaystyle k$.
Show/Hide Solution
Let $f(x)=27 x^{3}-9 x^{2}-6 x-5-k\\\\$ $f(x)$ is divisible by $3 x-1$, \begin{aligned} &\\ & \therefore\ f\left(\dfrac{1}{3}\right)=0 \\\\ & 27\left(\dfrac{1}{3}\right)^{3}-9\left(\dfrac{1}{3}\right)^{2}-6\left(\dfrac{1}{3}\right)-5-k=0 \\\\ & 1-1-1-2-5-k=0 \\\\ & k=-8 \end{aligned}
17. Given that $f(x)=4x^3-4kx^2-x+k,\ \$ $g(x)=3x^2+(1-3k)x-k$ and $h(x)=f(x)+g(x)$. If $x-2$ is a factor of $h(x)$, find the value of If $k$ and hence solve the equation $h(x)=0$.
Show/Hide Solution
\begin{aligned} & f(x)=4 x^{3}-4 k x^{2}-x+k \\\\ & g(x)=3 x^{2}+(1-3 k) x-k \\\\ & h(x)=f(x)+g(x) \\\\ & \therefore\ h(x)=4 x^{3}+(3-4 k) x^{2}-3 k x \\\\ & x-2 \text { is a factor of } h(x) . \\\\ &\therefore\ h(2)=0 \\\\ & 4(2)^{3}+(3-4 k)(2)^{2}-3 k(2)=0 \\\\ & 32+12-16 k-6 k=0 \\\\ & 44-22 k=0 \\\\ & \therefore\ k=2 \end{aligned}
စာဖတ်သူ၏ အမြင်ကို လေးစားစွာစောင့်မျှော်လျက်!
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# Mathematical Proof and the Principles of Mathematics/Numbers/Natural numbers
Earlier, we constructed the natural numbers as sets, based on the axioms of set theory.
Now, we approach the natural numbers numbers from an arithmetic viewpoint, and we will see how we can derive the familiar notions of addition and multiplication from the successor operation. As we are moving away from set theory and into arithmetic, we will write ${\displaystyle \mathbb {N} }$ instead of ${\displaystyle \omega }$ to denote the natural numbers. Since we are about to introduce arithmetic, but haven't done so yet, we will provisionally in this section use S(n) instead of n + 1 to denote the successor of n.
## Definitions
Definition Addition is defined from 0 and the successor function by the two rules
Def. 1: a + 0 = a Def. 2: a + S(b) = S(a + b)
We will see first how this definition justifies our previous usage of a + 1 for the successor of a:
For every natural number a, one has
S(a) = S(a + 0) [by Def. 1] = a + S(0) [by Def. 2] = a + 1 [by Def. of 1]
## Induction principle
We will now show how all those properties of addition that you know from school can be derived from the definition.
The proofs rely on the induction principle. Recall the theorem:
Induction principle Suppose ${\displaystyle P(x)}$ is a property of natural numbers for which ${\displaystyle P(0)}$ holds, and such that for all natural numbers ${\displaystyle n}$ for which ${\displaystyle P(n)}$ holds, we have that ${\displaystyle P(n+1)}$ also holds. Then ${\displaystyle P(n)}$ holds for every natural number ${\displaystyle n}$ .
When writing a proof using the induction principle, we call the proof of ${\displaystyle P(0)}$ the base case, and the proof of ${\displaystyle P(a+1)}$ from ${\displaystyle P(a)}$ the induction step. In the proof of the induction step, we will use the fact that we can assume ${\displaystyle P(a)}$ and call that the induction hypothesis. Let's see this in action:
Definition [Def. 1] states directly that 0 is a right identity, that is, that for any number a, a + 0 = a.
We prove that 0 is a left identity, that is, that 0 + a = a by induction on the natural number a.
Theorem For every natural number a, 0 + a = a.
For the base case a = 0, 0 + 0 = 0 by definition [Def. 1].
So we proceed to the induction step. We can assume the induction hypothesis, that 0 + a = a. Then
0 + S(a) = S(0 + a) [by Def. 2] = S(a) [by the induction hypothesis]
So we have shown that 0 + S(a) = a from the assumption that 0 + a = a. So, we can conclude with the induction principle that 0 + a = a for every natural number a.
## Associativity
Let us try our hand at a more complex example, the associativity property:
Theorem For any three natural numbers a, b and c, (a + b) + c = a + (b + c).
The first thing to note is that we have three numbers here, a, b and c. So we will fix a and b to be arbitrary numbers and we will apply the induction principle to c.
More precisely, the property ${\displaystyle P(c)}$ is ${\displaystyle (a+b)+c=a+(b+c)}$ .
For the base case c = 0,
(a+b)+0 = a+b = a+(b+0)
Each equation follows by definition [Def. 1]; the first with a + b, the second with b.
Now, for the induction step. We assume the induction hypothesis, namely we assume that for some natural number c,
(a+b)+c = a+(b+c)
Then it follows,
(a + b) + S(c) = S((a + b) + c) [by Def. 2] = S(a + (b + c)) [by the induction hypothesis] = a + S(b + c) [by Def. 2] = a + (b + S(c)) [by Def. 2]
In other words, ${\displaystyle P}$ holds for S(c). Therefore, the induction on c is complete.
## Commutativity
Another important property that you will now from school is commutativity: When adding to numbers, the order doesn't matter.
Theorem For any two natural numbers a and b, a + b = b + a.
Since this is even more involved, we introduce a lemma first.
Lemma For any two natural numbers a and b, S(a) + b = S(a + b).
Let's perform induction on b, starting with the base case b = 0:
S(a) + 0 = S(a) [by Def. 1] = S(a + 0) [by Def. 1]
And now the induction step:
S(a) + S(b) = S(S(a) + b) [by Def. 2] = S(S(a + b) [by the induction hypothesis] = S(a + S(b)) [by Def. 2]
There we go - the induction on b is complete.
### The proof of commutativity
From this brief digression we have earned ourselves a property that we can now use in the proof of commutativity.
We will proceed again by induction on b, starting with the base case:
a + 0 = a [by Def. 1] = 0 + a [by the Right identity theorem]
Note how the base case is just a straightforward consequence of the identity property we showed above.
Now, suppose that for all natural numbers a, we have a + b = b + a. We must show that for all natural numbers a, we have a + S(b) = S(b) + a. We have
a + S(b) = S(a + b) [by Def. 2] = S(b + a) [by the induction hypothesis] = S(b) + a [by the lemma]
This completes the induction on b.
## Multiplication
By now we introduced addition, and we derived all the basic properties we know about addition on the natural numbers. But arithmetic is more than addition, of course. So, just like we derived addition from the successor relation, we will derive multiplication from addition:
Definition Multiplication is derived from 0, addition and the successor function by the two rules
Def. 3: ${\displaystyle a\cdot 0=0}$ Def. 4: ${\displaystyle a\cdot S(b)=a\cdot b+a}$
We can quickly check mentally that those two equations are indeed true in our intuition of the natural numbers.
Since you are familiar now with how to approach problems using induction, we leave the proofs of commutativity and associativity to you.
## Distributivity
Before we let you, though, let us go through one more property that arises from having two different operators present. It is known as distributivity.
Theorem For every three natural numbers a, b and c, ${\displaystyle a\cdot (b+c)=(a\cdot b)+(a\cdot c)}$ .
We prove distributivity by induction on c:
${\displaystyle a\cdot (b+0)}$ = ${\displaystyle a\cdot b}$ [by Def. 1] = ${\displaystyle a\cdot b+0}$ [by Def. 1] = ${\displaystyle a\cdot b+a\cdot 0}$ [by Def. 3]
...and proceed by induction:
${\displaystyle a\cdot (b+S(c))}$ = ${\displaystyle a\cdot S(b+c)}$ [by Def. 2] = ${\displaystyle a\cdot (b+c)+a}$ [by Def. 4] = ${\displaystyle (a\cdot b+a\cdot c)+a}$ [by the induction hypothesis] = ${\displaystyle a\cdot b+(a\cdot c+a)}$ [by associativity of addition] = ${\displaystyle a\cdot b+a\cdot S(c)}$ [by Def. 4]
And this was to be proven.
## Exercises
Now that you have seen various techniques of inductive argument in action, you should try and figure out the remaining properties of multiplication for yourself.
Try to write them up line by line, with the source of any identity you use to the right of the equation, just as it is laid out on this page.
To help you, there is a skeletal outline of the dependencies between the properties in the diagram on the right. It also includes derivations of the properties that we have already seen, as well as the definitions of multiplication and addition.
Good luck!
## References
The classical reference for the induction proofs on this page is Edmund Landau's Foundations of Analysis, Chelsea Pub Co. ISBN 0-8218-2693-X. there you can find a wealth of further properties and exercises as well as an introduction of exponentiation along the same lines.
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Linear Equations and Word Problems
Word problems that lead to simple linear equations
Number problems
Age problems
Mixture problems
Miscellaneous word problems
Word problems that lead to simple linear equations
The general procedure to solve a word problem is:
1. Set the unknown.
2. Write equation from the text of the problem.
3. Solve the equation for the unknown.
Number problems
Example: A number multiplied by 5 and divided by 4 equals twice the number decreased by 15.
Solution: Let x denotes the number, then
Example: Split the number 61 into two parts so that quarter of the first part increased by sixth of the other
part gives 12. What are these parts?
Solution: Let x denotes the first part of the number, then
Age problems
Example: A father is 48 years old and his son is 14.
a) In how many years will the father be three times older then his son?
b) How many years before was the father seven times older then his son?
Solution: After x years a) 48 + x = 3 · (14 + x)
48 + x = 42 + 3x
x - 3x = 42 - 48
-2x = -6 | ¸ (-2)
x = 3 years.
Before x years b) 48 - x = 7 · (14 - x)
48 - x = 98 - 7x
7x - x = 98 - 48
6x = 50
x = 25/3 = 8 years and 4 months.
Miscellaneous word problems
Example: If fresh grapes contain 90% water and dried 12%, how much dry grapes we get from 22 kg of fresh
grapes?
Solution: Fresh grapes contain 90% water and 10% dry substance.
Dry grapes contain 12% water and 88% dry substance.
22 kg of fresh grapes = x kg of dry grapes, so
Example: An amount decreased 20% and then increased 50%, what is the total increase in relation to initial
value.
Solution: An initial amount x decreased by 20%,
The obtained amount increased by 50%,
The difference in relation to the initial value x,
shows increase by 20%.
Example: From a total deducted is 5% for expenses and the remainder is equally divided to three persons.
What was the total if each person gets \$190?
Solution: If x denotes the total then
Example: Into 10 liters of the liquid A poured is 4 liters of the liquid B and 6 liters of the liquid C.
From obtained mixture D poured is out 3 liters, how many liters of the liquid C remains in the mixture D?
Solution: A + B + C = D => 10 l + 4 l + 6 l = 20 l
Intermediate algebra contents
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# What is the first step in constructing an inscribed square?
Contents
## What is the first step in constructing an inscribed square?
1 Mark a point A on the circle. This will become one of the vertices of the square. 2 Draw a diameter line from the point A, through the center and on to cross the circle again, creating point C. 3 Set the compass on A and set the width to a little more than the distance to O.
## What are the steps to construct an inscribed square?
When constructing an inscribed square by hand, which step comes after constructing a circle? Set compass to the diameter of the circle.Set compass to the radius of the circle.Use a straightedge to draw a diameter of the circle.
## When constructing an inscribed square what step comes after?
1 Mark a point A on the circle. This will become one of the vertices of the square. 2 Draw a diameter line from the point A, through the center and on to cross the circle again, creating point C. 3 Set the compass on A and set the width to a little more than the distance to O.
## What are the steps to inscribe a square in a circle?
When constructing an inscribed square by hand, which step comes after constructing a circle? Set compass to the diameter of the circle.Set compass to the radius of the circle.Use a straightedge to draw a diameter of the circle.
Read also : What did slaves sing when they worked?
## What are the steps for constructing an inscribed square?
1 Mark a point A on the circle. This will become one of the vertices of the square. 2 Draw a diameter line from the point A, through the center and on to cross the circle again, creating point C. 3 Set the compass on A and set the width to a little more than the distance to O.
## What is the next step when constructing a square inscribed circle?
For the square construction, the diameter will be used, but for the equilateral triangle construction, the radius will be used. Which step is the same when constructing an inscribed regular hexagon and an inscribed equilateral triangle? Set the compass width to the radius of the circle.
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# Find the equations to the common tangents of the circle ${x^2} + {y^2} + 4ax = 0$ and parabola ${y^2} = 4ax$
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Hint: From the given two equations we first try to find out the center of the circle and its radius. And then we have to find out if the parabola and the circle are touching each other or not. Thus we can find the result by observing the solution.
Here in this problem, we are given two equations, ${x^2} + {y^2} + 4ax = 0$ and ${y^2} = 4ax$. Where one of them is a circle and the other is a parabola. We need to find the equation of common tangents of these two.
So, the equation of the circle is,
${x^2} + {y^2} + 4ax = 0$
Adding $4{a^2}$ to both sides we get,
$\Rightarrow {x^2} + 4ax + 4{a^2} + {y^2} = 4{a^2}$
On applying ${{\text{a}}^{\text{2}}}{\text{ + 2ab + }}{{\text{b}}^{\text{2}}}{\text{ = (a + b}}{{\text{)}}^{\text{2}}}$, we get,
$\Rightarrow {(x + 2a)^2} + {y^2} = 4{a^2}$……(i)
So, we get the circle with center at $( - 2a,0)$and with radius $2a$ units.
And we have the 2nd equation as, ${y^2} = 4ax$ which is passing through the origin.
Now, if we try to plot them altogether, we get,
Now, from the figure, it is clearly visible that, the circle and the parabola are touching each other along the y-axis and it is the common tangent of these two given equations.
So, now, the equation of the common tangent is, $x = 0$
Note: In this given problem we focused on finding where the different figures are meeting each other and if they are at all touching each other or not. In these types of problems, if the figures are touching each other, our job becomes easier as the common tangent is always passing through their point of intersection.
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# Is Dividend always 0 the quotient?
Contents
## When dividend is 0 then the quotient is always?
The result should be equal to the dividend. Properties of division: When zero is divided by a number the quotient is zero.
## Is dividend a quotient?
The number that is being divided (in this case, 15) is called the dividend, and the number that it is being divided by (in this case, 3) is called the divisor. The result of the division is the quotient.
## Is the quotient always less than the dividend?
The divisor 2 is contained 6 times in the dividend 12. The quotient is larger or smaller than the divisor but it is always smaller than the dividend.
## Is the dividend always bigger than the divisor?
Solution: Know that, Now, we can say that the dividend is greater than the divisor. Hence, “dividend is always greater than the divisor” is True.
## Is division by 0 undefined?
Loosely speaking, since division by zero has no meaning (is undefined) in the whole number setting, this remains true as the setting expands to the real or even complex numbers.
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## What is the quotient of 0 and 0?
One of the properties of 0, is that 0 divided by any number is 0. The exception is 00 which is undefined. (the fact that 0 is being divided by a negative number makes no difference. There is no just thing as +0 or -0.)
## Can zero be divided by zero?
So, that means that this is going to be undefined. So zero divided by zero is undefined.
## What is the rule when dividing by 0?
When you divide by zero, the result is undefined.
## Which part of the expression is a quotient?
The answer after we divide one number by another. dividend ÷ divisor = quotient. Example: in 12 ÷ 3 = 4, 4 is the quotient.
## Which is also known as quotient method?
in arithmetic, long division is a standard division algorithm suitable for dividing multi-digit numbers that is simple enough to perform by hand. … As in all division problems, one number, called the dividend, is divided by another, called the divisor, producing a result called the quotient.
## What is quotient fractions?
The quotient of a fraction is the whole number that is obtained when you simplify the fraction. If the simplification of a fraction is not a whole number, the quotient is the decimal form of the fraction.
## Why is the quotient smaller than the dividend?
Another example where the divisor smaller than the dividend. When the divisor is the same size as the dividend, the quotient is 1. When the divisor is larger than the dividend, the quotient is less than 1. Another example where the divisor is larger than the dividend.
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## Why is the quotient greater than the dividend?
When a whole number is divided by a unit fraction the result is each unit being divided into smaller parts. Dividing each unit of 6 into four equal parts results in 24 equal parts of one- fourths, so the qotient will be a whole number greater than the dividend.
## What is the quotient when the dividend is smaller than the divisor?
If n is less than d then, quotient = 0. remainder = n.
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# What is 5/20 as a decimal?
As you can see, in one quick calculation, we’ve converted the fraction 520 into it’s decimal expression, 0.25.
## How do you write 1/5 as a percent?
Convert 1/5 to Percentage by Changing Denominator Our percent fraction is 20/100, which means that 15 as a percentage is 20%.
## What percent is 5 out of 50?
Now we can see that our fraction is 10/100, which means that 5/50 as a percentage is 10%. And there you have it!
## What is percent example?
For example, 1 percent of 1,000 chickens equals 1/100 of 1,000, or 10 chickens; 20 percent of the quantity is 20/100 1,000, or 200.
## How do you find the percent of a whole number?
Finding the percentage For this type of problem, you can simply divide the number that you want to turn into a percentage by the whole. So, using this example, you would divide 2 by 5. This equation would give you 0.4. You would then multiply 0.4 by 100 to get 40, or 40%.
## What is the percentage of 9 50?
Now we can see that our fraction is 18/100, which means that 9/50 as a percentage is 18%. And there you have it! Two different ways to convert 9/50 to a percentage.
## What is the percent of 1 20?
1 in 20 as a common fraction is 1/20. To express it as a decimal fraction divide 1 by 20 to get 0.05. To express the decimal fraction multiply by 100 to get 5%.
## What is 7 over 20 as a percentage?
Therefore, 7/20 in terms of percentage is 35% Now, let us express 7/20 in terms of decimals. To convert any fraction to a decimal, we divide the numerator by the denominator.
## What is the percent of 9 10?
Answer: 9/10 can be written as 90%. Let’s understand the conversion of a fraction to a percentage. Explanation: To convert a fraction to percent, we multiply it by 100.
## What is 5 percent of a number?
5 percent is one half of 10 percent. To calculate 5 percent of a number, simply divide 10 percent of the number by 2. For example, 5 percent of 230 is 23 divided by 2, or 11.5.
## How do you find 1 percent of a number?
To find 1% of something (1/100 of something), divide by 100. Remember how to divide by 100 mentally: Just move the decimal point two places to the left. For example, 1% of 540 is 5.4. And 1% of 8.30 is 0.083.
## How do you work out 20% on a calculator?
To subtract 20 percent, multiply by 80 percent (0.8). To subtract 30 percent, multiply the number by 70 percent (0.7). Before beginning, determine the gross amount you want to use before subtracting a percentage.
## How do you calculate percentage manually?
You divide your percentage by 100. So, 40 percent would be 40 divided by 100. Once you have the decimal version of your percentage, simply multiply it by the given number (in this case, the amount of your paycheck). If your paycheck is \$750, you would multiply 750 by .
## What is 0.05 as a percent?
To change a decimal to a percentage, multiply it by 100. For example 0.05 is equivalent to 5%, because 0.05 × 100 = 5.
## What is 4/20 as a percentage?
Convert 4/20 to Percentage by Changing Denominator Our percent fraction is 20/100, which means that 420 as a percentage is 20%.
## How do you write 19 50 as a decimal?
As you can see, in one quick calculation, we’ve converted the fraction 1950 into it’s decimal expression, 0.38.
## What is 72 out of 123 as a percentage?
Solution and how to convert 72 / 123 into a percentage 0.59 times 100 = 58.54. That’s all there is to it!
## What is a 20th as a percentage?
Now we can see that our fraction is 20/100, which means that 20/100 as a percentage is 20%. We can also work this out in a simpler way by first converting the fraction 20/100 to a decimal.
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# Conditional Probability: Definition & Uses
Coming up next: Ratios & Rates: Definitions & Examples
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Instructor: Joseph Vigil
We know a coin can land on either heads or tails. But what would happen if one coin flip changed the next? In this lesson, we'll look at events that are dependent on each other, and we'll learn how to calculate the probability of two events occurring in a combined manner.
## Probability
What are the chances your favorite baseball team will make it to their league's playoffs? What are the chances that they make the World Series? Or even that they'll win the World Series? The answers to all these questions depend on probability, which is a numerical expression of how likely it is that an event will occur.
Let's flip a coin, for example. It landing will be one of two sample points, or possible outcomes: heads or tails. These two sample points make up our sample space, the complete set of possibilities in an experiment. Since one out of the two sample points is heads, the probability of the coin landing on heads is a 1 out of 2, or 1/2. Likewise, since one out of the two sample points is tails, the probability of the coin landing on tails is 1 out of 2, or 1/2.
That's simple enough. Every time we flip the coin, we have the same sample space. Even if we flipped it a million times, the next flip will still give either heads or tails. But what if the sample space changed every time we ran an experiment?
## Dependent Events
You've got to get to work! You've slept past your alarm, and it's still dark. You need two black dress socks to finish dressing. You have two white socks and two black ones in your drawer. What are your chances of grabbing the right socks?
Let's look at the socks we have now and how that group of socks may change. When you grab your first sock, it can of course be black or white. Let's say you get lucky, and you grab a black sock on your first try. Since 2 out of the 4 socks are white, you have a 2/4, or 1/2, chance of picking a black sock.
Unlike the coin flip, though, the odds of grabbing a certain sock change every time we run this experiment. This is an example of conditional probability, which is the probability of one event happening given that another event has already happened. In other words, the second event is conditional based on the previous event.
Now you have one black sock and two white socks left. The group of socks has changed. So the odds of picking either a white or black sock has also changed. Each time you pick a sock, you change future options. In other words, each selection is dependent on the one before it; so each selection is a dependent event.
Now that there are two white socks and one black sock left, you have a 1/3 chance of picking a black sock. Since you have a 1/2 chance of picking a black sock the first time and a 1/3 chance of picking a black sock the second time, we'll multiply those two probabilities together. 1/2 * 1/3 = 1/6. You have a one out of six chance of picking two black socks right away. Not very good odds. We'd better turn on the lights!
## Notation for Conditional Probability
We have to use a common language so people can be on the same page when talking about probabilities. So, when we write P(A), we mean the probability that some event, A, will happen.
Let's go back to our coin. If A represents the coin landing on heads, then P(A) = 1/2, because there's a 1 out of 2 chance that the coin will land on heads.
If B represents the coin landing on tails, then P(B) also equals 1/2, because there's a 1 out of 2 chance that the coin will land on tails.
When we were dealing with the socks, though, each event was dependent on those before it. So, we write P(B|A), which means the probability that B happens if A has already happened, also known as the conditional probability of B given A.
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# What Is A Perfect Square In Math latest 2023
You are searching about What Is A Perfect Square In Math, today we will share with you article about What Is A Perfect Square In Math was compiled and edited by our team from many sources on the internet. Hope this article on the topic What Is A Perfect Square In Math is useful to you.
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## Geometry for Beginners – How to Use Pythagorean Triples
Welcome to Geometry for Beginners. In this article, we’ll review the Pythagorean theorem, examine the meaning of the phrase “Pythagorean triple”, and discuss how these triples are used. Additionally, we will list the triplets that need to be memorized. Knowing the Pythagorean triples can save you a lot of time and effort when working with right triangles!
In another article on geometry for beginners, we discussed the Pythagorean theorem. This theorem states a relation on right triangles which is ALWAYS TRUE: in all right triangles, the square of the hypotenuse is equal to the sum of the squares of the legs. In symbols, it looks like c^2 = a^2 + b^2. This formula is one of the most important and widely used in all of mathematics, so it is important that students understand its use.
There are two important applications of this famous theorem: (1) to determine if a triangle is a right triangle if the lengths of the 3 sides are given, and (2) to find the length of a missing side of a right triangle if the other two sides are known. This second application sometimes produces a Pythagorean triple – a very special set of three numbers.
A Pythagorean triple is a set of three numbers that share two qualities: (1) they are the sides of a right triangleand (2) they are all integers. The whole quality is particularly important. Since the Pythagorean theorem involves squaring each variable, the process of solving any of the variables involves taking the square root of both sides of the equation. Only a few times “taking a square root” produces an integer value. Usually the missing value will be irrational.
For example: Find the side length of a right triangle with an 8 inch hypotenuse and a 3 inch leg.
The solution. Use the Pythagorean relation and remember that vs is used for the hypotenuse while a and b are the two legs: vs^2 = a^2 + b^2 becomes 8^2 = 3^2 + b^2 or 64 = 9 + b^2 or b^2 = 55. To solve b, take the square root of both sides of the equation. Since 55 is NOT a perfect square, we cannot eliminate the radical sign, so b = sqrt(55). This means that the missing length is a irrational Number. This is a typical result.
This next example is NOT so typical: Find the hypotenuse of a right triangle with 6 inch and 8 inch legs.
The solution. Again, using the Pythagorean theorem, vs^2 = a^2 + b^2 becomes vs^2 = 6^2 + 8^2 or vs^2 = 36 + 64 or vs^2 = 100. Recall that, algebraically, vs has two possible values: +10 and -10; but, geometrically, the length cannot be negative. Thus, the hypotenuse has a length of 10 inches. WOW! All three sides – 6, 8 and 10 – are integers. It is special ! These “special” situations are Pythagorean triples.
Pythagorean triples should be thought of as “families” based on the smallest set of numbers in that family. Since 6, 8 and 10 have a common factor of 2, removing this common factor results in values of 3, 4 and 5. Testing with the Pythagorean theorem, we want to know IF 5^2 is equal to 3^2 + 4^2. Is it? Does 25 = 9 + 16? YES! This means that the sides of 3, 4 and 5 form a right triangle; and since all values are integers, 3, 4, 5 is a Pythagorean triple. So, 3, 4, 5 and its multiples – like 6, 8, 10 (a multiple of 2) or 9, 12, 15 (a multiple of 3) or 15, 20, 25 (a multiple of 5) or 30, 40, 50 (a multiple of 10), etc., are all Pythagorean triples of the family 3, 4, 5.
ATTENTION ALL STUDENTS! Standardized test writers often use Pythagorean relations in their math questions, so it will benefit you to memorize the most commonly used values. However, you should be aware that these same test writers often construct questions to confuse those whose understanding of the concept isn’t quite what it should be.
Example of a “meant to catch you” question: Find the hypotenuse of a right triangle with sides 30 and 50 units. The tricky part is that students see a multiplier of 10 and think they have a triplet 3, 4, 5 with a hypotenuse of 40 units. FAKE! Do you see why this is wrong? You won’t be alone if you don’t see it. Remember that the hypotenuse must be the LONGEST side, so 40 cannot be the hypotenuse. Always THINK carefully before jumping on an answer that seems too easy. (Since the triple doesn’t really work here, you’ll have to do the whole formula to find the missing value.)
Pythagorean Triplets to memorize and recognize:
(1) 3, 4, 5 and all its multiples
(2) 5, 12, 13 and all its multiples
(3) 8, 15, 17 and all its multiples
(4) 7, 24, 25 and all its multiples
Memorizing ALL the multiples would be impossible, but you should learn the most commonly used multipliers: 2, 3, 4, 5, and 10. The time you’ll save in years to come is worth every minute you spend learning these combinations now. !
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## What Is A Inequality In Math latest 2023
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# Use integration formulas to get the exact value of \int_{0}^{0.35}\frac{2}{x^2 - 4}dx
## Question:
Use integration formulas to get the exact value of
{eq}\int_{0}^{0.35}\frac{2}{x^2 - 4}dx {/eq}
## Formula for Integrals:
There are some standard formulas of integration to get the values without substitution and partial fraction method.
The formula for this particular problem is:
• {eq}\int \frac{1}{x^2-a^2}\ dx= \frac{1}{2a}\ln|\frac{x-a}{x+a}|+C {/eq}
Where,
• {eq}C {/eq} is the constant of integration and {eq}a {/eq} does not equal to zero.
## Answer and Explanation:
The given definite integral is:
{eq}I=\displaystyle \int_{0}^{0.35}\frac{2}{x^2 - 4}\ dx {/eq}
Using the standard fromula of integration, we get:
{eq}\begin{align*} I&=2\displaystyle \int_{0}^{0.35}\frac{1}{x^2 - 2^{2}}\ dx \\ &=2 \left [\frac{1}{2(2)}\ln|\frac{x-2}{x+2}| \right ]_{0}^{0.35}\\ &=2\cdot \frac{1}{4} \left [\ln|\frac{x-2}{x+2}| \right ]_{0}^{0.35}\\ &=\frac{1}{2} \left (\ln|\frac{0.35-2}{0.35+2}|-\ln|\frac{0-2}{0+2}| \right )\\ &=\frac{1}{2} \left (\ln|\frac{-1.65}{2.35}|-\ln|-1| \right )\\ &=\frac{1}{2} \left (\ln\frac{1.65}{2.35}-\ln 1 \right )\\ &=\frac{1}{2} \left (-0.3536-0 \right )\\ &=-0.1768 \end{align*} {/eq}
#### Learn more about this topic:
Evaluating Definite Integrals Using the Fundamental Theorem
from AP Calculus AB: Exam Prep
Chapter 16 / Lesson 2
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How to Convert Between base-10, Hexadecimal, and Binary
Step 5: Convert a Decimal Number Into Any Base
Convert a denery number into any base-n
To me, this is the hardest part of converting between bases: from decimal to a different base. However, don't let my math conundrums confound you. The process is very simple and without pitfalls, if not a bit circuitous. We'll try one to get a feel for the process because the process is the same for every base, which is a good thing we don't have to remember a bunch of conversion rules. To prove this point subtly, I'll start off converting a denery number into an odd base, base-7.
Convert 15810 into base7
I. The first step is to take the target base to varying powers and attempt to divide it into the source number to get the largest part.
1. 73 = 343 which is larger than 158 so we can't divide a larger number into a smaller number. Step down an exponent.
2. 72 = 49 which is smaller than 158. 49 goes into 158 three times with a remainder. 3 * 49 = 147.
3. Take the number of times 49 goes into 147 and use it as the first digit of the new base: 3xx7
II. Subtract the product of the multiplication from the dividend of the division. This sounds overly complicated. Put another way, 158 - 147 = 11. We're subtracting (the number of times 72 goes into 158 * 72) from the original base-10 number, or our starting number.
1. 158 - 147 = 11
2. Since there is one seven in 11, place that in the tens place of the new number: 31x7
3. Proceed to Step I.
I'll complete the step here.
1. How many 71's are in 11? One and a remainder. 1 * 7 = 7.
2. 11 - 7 = 4
3. There are no seven's in four: 4 x 1 = 4
4. Place the four in the digits section of the base-7 number: 3147
That is our base-7 number: 15810 = 3147
Convert denery into binary
What is 15810 in base-2? We can figure that out now that we have our rules in place. I'll go through it short form, but not an entirely terse fashion. You will need, however, to recall binary value placement. I'll add up the digits into the binary value at the end.
1. 28 is 256 and is too big. 27 is 128. 128 goes into 158 one time.
2. 158 - 128 = 30
3. 24 = 16. 30 - 16 = 14.
4. 23 = 8. 14 - 8 = 6.
5. 22 = 4. 6 - 4 = 2.
6. 21 = 2. 2 - 2 = 0
Notice I didn't move any numbers into the target base locations this time. In binary, we'll either have it going into a number one or zero times. Look back at the calculations as I add the numbers.
128 + 16 + 8 + 4 + 2
```128 64 32 16 8 4 2 1
1 0 0 1 1 1 1 0
```
So the decimal number 15810 is 100111102.
Convert decimal into hexadecimal
Ready to try your hand at converting decimal to hexadecimal? Good, I thought so.
What is 15810 in base-16? First, let's use some intuition. How many digits should the hexadecimal number be? Two, right? The number 158 is less than FF, which is 255.
For converting from decimal to hexadecimal there is a different way that I learned and prefer and I am going to show it to you now. It is much easier -- in my opinion -- that what you've just done thus far.
The general heuristic is divide by 16, multiple remainder by 16, convert to hex.
158 / 16 = 9.875
Take the remainder (.875) and multiply it by 16 and round up if necessary.
.875 * 16 = 14.
This is the digits place and 14 in hexadecimal is E: 0xXE
Take the number without the remainder from the previous step, which is 9 and divide by 16, but since 9/16 = 0, move the nine over: 0x9E
The number 15810 is 9E16, 0x9E or 9Eh hexadecimal. Isn't that much easier?
Convert denery into octal
Because we've covered that odd base-7 earlier, and it's just a bit of the same thing with using eights instead of sevens I'll leave this section as an exercise for the reader. Convert 15810 into base-8.
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nevdull (author) 2 months ago
Hey guys,
Thanks for the comments. I checked and it looks like a typo as I have it correctly typed in the preceding line. The typo below is now corrected.
Thanks for finding that!
TheW13 months ago
158 base 10 should be equal to 314 in base 7.
3 months ago
I second/confirm/approve of or whatever this statement.
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# Law of Cosines
The Law of Cosines is a theorem which relates the side-lengths and angles of a triangle. It can be derived in several different ways, the most common of which are listed in the "proofs" section below. It can be used to derive the third side given two sides and the included angle. All triangles with two sides and an include angle are congruent by the Side-Angle-Side congruence postulate.
## Theorem
For a triangle with edges of length $a$, $b$ and $c$ opposite angles of measure $A$, $B$ and $C$, respectively, the Law of Cosines states:
$$c^2 = a^2 + b^2 - 2ab\cos C$$
$$b^2 = a^2 + c^2 - 2ac\cos B$$
$$a^2 = b^2 + c^2 - 2bc\cos A$$
In the case that one of the angles has measure $90^\circ$ (is a right angle), the corresponding statement reduces to the Pythagorean Theorem.
## Proof 1
### Acute Triangle
$[asy] pair A,B,C,D,E; C=(30,70); B=(0,0); A=(100,0); D=(30,0); size(100); draw(B--A--C--B); draw(C--D); label("A",A,(1,0)); dot(A); label("B",B,(-1,-1)); dot(B); label("C",C,(0,1)); dot(C); draw(D--(30,4)--(34,4)--(34,0)--D); label("f",(30,35),(1,0)); label("d",(15,0),(0,-1)); label("e",(50,0),(0,-1.5)); [/asy]$
Let $a$, $b$, and $c$ be the side lengths, $C$ is the angle measure opposite side $c$, $f$ is the distance from angle $C$ to side $c$, and $d$ and $e$ are the lengths that $c$ is split into by $f$.
We use the Pythagorean theorem:
$$a^2+b^2-2f^2=d^2+e^2$$
We are trying to get $a^2+b^2-2f^2+2de$ on the LHS, because then the RHS would be $c^2$.
We use the addition rule for cosines and get:
$$\cos{C}=\dfrac{f}{a}\cdot \dfrac{f}{b}-\dfrac{d}{a}\cdot \dfrac{e}{b}=\dfrac{f^2-de}{ab}$$
We multiply by $-2ab$ and get:
$$2de-2f^2=-2ab\cos{C}$$
Now remember our equation?
$$a^2+b^2-2f^2+2de=c^2$$
We replace the $-2f^2+2de$ by $-2ab\cos{C}$ and get:
$$c^2=a^2+b^2-2ab\cos{C}$$
We can use the same argument on the other sides.
### Right Triangle
Since $C=90^{\circ}$, $\cos C=0$, so the expression reduces to the Pythagorean Theorem. You can find several proofs of the Pythagorean Theorem here.
### Obtuse Triangle
The argument for an obtuse triangle is the same as the proof for an acute triangle.
## Proof 2 (Vector Dot Product)
Consider $\triangle{ABC}$. Let $\vec{AB}=\vec{c}, \vec{AC}=\vec{b},\vec{BC}=\vec{a}$.
Because of the identity $|\vec{a}|^2=\vec{a}\cdot\vec{a}$,we can complete our proof as the following.
Draw the diagram. Note that $\vec{c}+\vec{a}=\vec{b}$. Then $\vec{b}-\vec{c}=\vec{a}$ and $\vec{a}\cdot \vec{a}=a^2$. $(\vec{b}-\vec{c})^2=b^2+c^2-2\cdot b\cdot c\cdot \cos{A}=|\vec{a}|^2$. Now, we have finished the proof because the two quantities are equal.
Credits to China High School Math textbook $\emph{Mathematics Vol 5B Textbook}$ by People's Education Press (this textbook is currently discontinued but it has been used for hinting the proof. The proof is done by myself. But the letting and the process of guiding students to verify the identity $|\vec{a}|^2=\vec{a}\cdot\vec{a}$ is written in the textbook.
~hastapasta
## Problems
### Introductory
1. If the sides of a triangle have lengths 2, 3, and 4, what is the radius of the circle circumscribing the triangle?
$\mathrm{(A) \ } 2 \qquad \mathrm{(B) \ } 8/\sqrt{15} \qquad \mathrm{(C) \ } 5/2 \qquad \mathrm{(D) \ } \sqrt{6} \qquad \mathrm{(E) \ } (\sqrt{6} + 1)/2$
(Source)
2. In the quadrilateral $ABCD$, $\angle{ADC}=90^\circ$, $AB=2$, $BD=5$.
(1) Find $\cos{\angle{ADB}}$.
(2) If $DC=2\sqrt{2}$, find $BC$.
(2018 China Gaokao Syllabus I #17)
### Intermediate
A tripod has three legs each of length $5$ feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is $4$ feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let $h$ be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then $h$ can be written in the form $\frac m{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $\lfloor m+\sqrt{n}\rfloor.$ (The notation $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x.$)
(Source)
A tetrahedron $ABCD$ is inscribed in the sphere $S$. Find the locus of points $P$, situated in $S$, such that
$\frac{AP}{PA_{1}} + \frac{BP}{PB_{1}} + \frac{CP}{PC_{1}} + \frac{DP}{PD_{1}} = 4,$
where $A_{1}, B_{1}, C_{1}, D_{1}$ are the other intersection points of $AP, BP, CP, DP$ with $S$.
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# Equilateral triangle inscribed within nested circles
This seemingly trivial question inspired this one: Somewhere within a circle of unit radius is placed a circle of radius $r, 0 < r < 1$, such that this inner circle's center is uniformly distributed within $1-r$ of the outer circle's center. An equilateral triangle with uniformly distributed orientation is inscribed within the inner circle. (That is, all three vertices are on the inner circle's circumference.)
What is the probability (in terms of $r$) that the outer circle's center is inside the triangle?
• To choose a random circle then a random orientation for an inscribed triangle is just like to choose a random triangle (with a tailor-made definition of "random"). For any fixed orientation, it is not difficult to figure which triangles enclose the centre of the big circle. By symmetry orientation does not really matter, so we just need to compute the ratio between the area of an equilateral triangle and the area of a circle. Commented Aug 21, 2015 at 19:19
Let's assign some labels. Let the outer circle have centre $O$, the inner circle have centre $P$, let the triangle be $T$ and let r.v. X be distance $\overline{OP},\;$ so $0\lt X\lt 1-r$. Also, let $E$ be the event that $O$ is inside $T$.
For $O$ to be inside $T$, it must also be inside the inner circle, so we want $X\lt r$.
Since the orientation of $T$ is uniformly distributed, the probability of $O$ being inside it equals the fraction of the circumference of a circle, centre $P$ with radius $X$, that is inside $T$. Referring to the diagram of the inner circle, this equals $\frac{6A}{2\pi} = \frac{3A}{\pi}$. Using the Sine Rule, we see:
$$\sin B = \dfrac{r\sin(\pi/6)}{X} = \dfrac{r}{2X}.$$
Angle $B$ is obtuse, so:
$$A = \pi-\dfrac{\pi}{6}-B = \dfrac{5\pi}{6}-\left(\pi-\sin^{-1}\left(\frac{r}{2X}\right)\right) = \sin^{-1}\left(\frac{r}{2X}\right) - \dfrac{\pi}{6}.$$
Note that if $X\lt r/2,\;$ then $O$ is always inside $T$. For one thing, this means that if $2/3 \lt r\lt 1$ then $P(E)=1,\;$ so we assume hereon that $0 \lt r\lt 2/3$.
Now, in the range $r/2\lt X\lt \min\{r,1-r\},\;$ we have $$P(E) = \dfrac{3A}{\pi} = \dfrac{3}{\pi} \sin^{-1}\left(\dfrac{r}{2X}\right) - \dfrac{1}{2}.$$
The distribution of $X$ is as follows. Since $P$ is uniformly distributed inside a circle (of radius $1-r$), $P(X=x)$ is proportional to the circumference of a circle with radius $X$, which means that $f_X(x) = cx$ for some constant $c$, which we find:
$$1=\int_{0}^{1-r}cx\;dx = \dfrac{c}{2}\left[x^2\right]_{0}^{1-r} = \dfrac{c}{2}(1-r)^2,\; \text{ so } c=\dfrac{2}{(1-r)^2},\; \text{ and } f_X(x) = \dfrac{2x}{(1-r)^2}.$$
So we have,
$$P(E) = P(X\lt r/2) + P(E\;\cap\; r/2 \lt X\lt \min\{r,1-r\}).$$
$$P(X\lt r/2) = \int_{0}^{r/2}f_X(x)\;dx = \dfrac{1}{(1-r)^2} \left[x^2\right]_{0}^{r/2} = \dfrac{r^2}{4(1-r)^2}.$$
For $0\lt r\lt 1/2,\; \min\{r,1-r\} = r,\;$ so in this range,
\begin{eqnarray*} P(E\;\cap\; r/2\lt X\lt r) &=& \int_{x=r/2}^r \left( \dfrac{3}{\pi} \sin^{-1}\left(\dfrac{r}{2x}\right) - \dfrac{1}{2} \right) \dfrac{2x}{(1-r)^2}\;dx \\ &=& \dfrac{1}{(1-r)^2} \left[\dfrac{3x}{4\pi} \left(r\sqrt{4-r^2/x^2} + 4\sin^{-1}\left(\dfrac{r}{2x}\right) \right) - \dfrac{x^2}{2} \right]_{x=r/2}^r \\ && \qquad\qquad\qquad\qquad\qquad\qquad\text{using Wolfram Alpha} \\ &=& \dfrac{r^2}{4\pi(1-r)^2} \left(3\sqrt{3} - \pi \right) \\ && \\ \therefore\quad P(E) &=& \dfrac{3\sqrt{3}r^2}{4\pi(1-r)^2}. \end{eqnarray*}
For $1/2 \lt r \lt 2/3,\;$ we just change the upper limit of integration, but it results in a rather ugly expression for $P(E)$:
\begin{eqnarray*} P(E\cap r/2\lt X\lt r) &=& \int_{x=r/2}^{1-r} \left( \dfrac{3}{\pi} \sin^{-1}\left(\dfrac{r}{2x}\right) - \dfrac{1}{2} \right) \dfrac{2x}{(1-r)^2}\;dx \\ && \\ \therefore\quad P(E) &=& \dfrac{1}{4\pi(1-r)^2} \left[ 3r(1-r)\sqrt{4-\dfrac{r^2}{(1-r)^2}} + 12(1-r)^2\sin^{-1}\left( \dfrac{r}{2(1-r)} \right) -2(1-r)^2\pi \right]. \end{eqnarray*}
• Hunh. I didn't expect that. Let me look at this when I get a chance; I'll accept if it passes muster. Looks reasonable at first glance, though. Commented Aug 22, 2015 at 17:01
• @BrianTung Hi Brian. Yes, it would be good of you to check. I might well have a mistake or misunderstood something. Commented Aug 22, 2015 at 17:11
• I've accepted this because it looks basically right, but for some reason I'm not getting the same value for $r=1/2$ in the two expressions. I haven't double checked yet, though. Commented Aug 26, 2015 at 1:10
• @BrianTung Sorry, I made a mistake in the last expression. Fixed now. I replaced $4(1-r)\sin^{-1}\left( \dfrac{r}{2(1-r)} \right)$ with $12(1-r)^2\sin^{-1}\left( \dfrac{r}{2(1-r)} \right)$. I get now $\left(3\sqrt{3}\right)/\left(4\pi\right)$ for both expressions with $r=1/2$. Commented Aug 26, 2015 at 2:31
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# How to Find the Surface Area of a Cube – A Step-by-Step Guide
Jan 17, 2023
## Introduction
A cube is a three-dimensional shape that has six equal sides and eight vertices. It is also known as a regular hexahedron and is made up of six squares. The surface area of a cube is the total area of all its faces or sides. Calculating the surface area of a cube can be done using geometry, formulas, and math. This article will discuss how to find the surface area of a cube in an easy and informative way.
## Calculating the Surface Area of a Cube Using Geometry
To calculate the surface area of a cube using geometry, you need to measure each side of the cube and then multiply that number by six. For example, if the length of one side of the cube is 4 inches, then the surface area of the cube would be 24 square inches (4 x 6 = 24).
## Exploring the Formula to Determine the Surface Area of a Cube
The formula for finding the surface area of a cube is SA = 6s², where s represents the length of one side of the cube. To use this formula, simply plug in the length of one side of the cube into the equation and solve. For example, if the length of one side of the cube is 4 inches, then the surface area of the cube would be 24 square inches (6 x 4² = 24).
## A Step-by-Step Guide to Finding the Surface Area of a Cube
The following steps will help you calculate the surface area of a cube:
• Measure the length of one side of the cube.
• Multiply the length of one side of the cube by six.
• The result is the surface area of the cube.
It is important to remember that the surface area of a cube is always equal to six times the length of one side squared.
## Utilizing Math to Compute the Surface Area of a Cube
Math can also be used to calculate the surface area of a cube. To do this, use the formula SA = 6s², where s represents the length of one side of the cube. For example, if the length of one side of the cube is 4 inches, then the surface area of the cube would be 24 square inches (6 x 4² = 24).
## An Easy Way to Figure Out the Surface Area of a Cube
There is also an easy way to figure out the surface area of a cube. Simply take the length of one side of the cube and multiply it by itself. Then, multiply that number by six. The result is the surface area of the cube. For example, if the length of one side of the cube is 4 inches, then the surface area of the cube would be 24 square inches (4 x 4 x 6 = 24).
## Understanding the Concept Behind Calculating the Surface Area of a Cube
Before attempting to calculate the surface area of a cube, it is important to understand the concept behind it. Surface area is the total area of all the faces or sides of a three-dimensional shape. When calculating the surface area of a cube, you are simply multiplying the length of one side of the cube by six. This is because a cube has six sides.
## Uncovering the Mathematics Behind Determining the Surface Area of a Cube
The mathematics behind calculating the surface area of a cube is relatively simple. The formula SA = 6s² can be used to determine the surface area of a cube, where s represents the length of one side of the cube. To use this formula, simply plug in the length of one side of the cube into the equation and solve. It is important to remember that the surface area of a cube is always equal to six times the length of one side squared.
## Conclusion
Calculating the surface area of a cube can be done using geometry, formulas, and math. Using the formula SA = 6s², where s represents the length of one side of the cube, is the easiest and most efficient way to calculate the surface area of a cube. Remember that the surface area of a cube is always equal to six times the length of one side squared. With these tips, you should have no problem calculating the surface area of a cube.
#### By Happy Sharer
Hi, I'm Happy Sharer and I love sharing interesting and useful knowledge with others. I have a passion for learning and enjoy explaining complex concepts in a simple way.
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# What is 16/84 as a decimal?
## Solution and how to convert 16 / 84 into a decimal
16 / 84 = 0.19
Fraction conversions explained:
• 16 divided by 84
• Numerator: 16
• Denominator: 84
• Decimal: 0.19
• Percentage: 0.19%
Converting 16/84 to 0.19 starts with defining whether or not the number should be represented by a fraction, decimal, or even a percentage. Both represent numbers between integers, in some cases defining portions of whole numbers The difference between using a fraction or a decimal depends on the situation. Fractions can be used to represent parts of an object like 1/8 of a pizza while decimals represent a comparison of a whole number like \$0.25 USD. After deciding on which representation is best, let's dive into how we can convert fractions to decimals.
16 / 84 as a percentage 16 / 84 as a fraction 16 / 84 as a decimal
0.19% - Convert percentages 16 / 84 16 / 84 = 0.19
## 16/84 is 16 divided by 84
The first step in converting fractions is understanding the equation. A quick trick to convert fractions mentally is recognizing that the equation is already set for us. All we have to do is think back to the classroom and leverage long division. The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. To solve the equation, we must divide the numerator (16) by the denominator (84). Here's 16/84 as our equation:
### Numerator: 16
• Numerators are the top number of the fraction which represent the parts of the equation. Overall, 16 is a big number which means you'll have a significant number of parts to your equation. But having an even numerator makes your mental math a bit easier. Large numerators make converting fractions more complex. So how does our denominator stack up?
### Denominator: 84
• Denominators are located at the bottom of the fraction, representing the total number of parts. 84 is a large number which means you should probably use a calculator. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Have no fear, large two-digit denominators are all bark no bite. Now let's dive into how we convert into decimal format.
## How to convert 16/84 to 0.19
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 84 \enclose{longdiv}{ 16 }$$
To solve, we will use left-to-right long division. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 84 \enclose{longdiv}{ 16.0 }$$
We've hit our first challenge. 16 cannot be divided into 84! So we will have to extend our division problem. Add a decimal point to 16, your numerator, and add an additional zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 84 into 16 + 0 or 160.
### Step 3: Solve for how many whole groups you can divide 84 into 160
$$\require{enclose} 00.1 \\ 84 \enclose{longdiv}{ 16.0 }$$
Now that we've extended the equation, we can divide 84 into 160 and return our first potential solution! Multiply this number by 84, the denominator to get the first part of your answer!
### Step 4: Subtract the remainder
$$\require{enclose} 00.1 \\ 84 \enclose{longdiv}{ 16.0 } \\ \underline{ 84 \phantom{00} } \\ 76 \phantom{0}$$
If you hit a remainder of zero, the equation is done and you have your decimal conversion. If you still have a remainder, continue to the next step.
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals is a necessity. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. Same goes for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But each represent values in everyday life! Without them, we’re stuck rounding and guessing. Here are real life examples:
### When you should convert 16/84 into a decimal
Contracts - Almost all contracts leverage decimal format. If a worker is logging hours, they will log 1.19 hours, not 1 and 16/84 hours. Percentage format is also used in contracts as well.
### When to convert 0.19 to 16/84 as a fraction
Carpentry - To build a table, you must have the right measurements. When you stretch the tape measure across the piece of wood, you won't see 10.6 inches. You'll see a tick mark at 10 and 3/5 inches.
### Practice Decimal Conversion with your Classroom
• If 16/84 = 0.19 what would it be as a percentage?
• What is 1 + 16/84 in decimal form?
• What is 1 - 16/84 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.19 + 1/2?
### Convert more fractions to decimals
From 16 Numerator From 84 Denominator What is 16/74 as a decimal? What is 6/84 as a decimal? What is 16/75 as a decimal? What is 7/84 as a decimal? What is 16/76 as a decimal? What is 8/84 as a decimal? What is 16/77 as a decimal? What is 9/84 as a decimal? What is 16/78 as a decimal? What is 10/84 as a decimal? What is 16/79 as a decimal? What is 11/84 as a decimal? What is 16/80 as a decimal? What is 12/84 as a decimal? What is 16/81 as a decimal? What is 13/84 as a decimal? What is 16/82 as a decimal? What is 14/84 as a decimal? What is 16/83 as a decimal? What is 15/84 as a decimal? What is 16/84 as a decimal? What is 16/84 as a decimal? What is 16/85 as a decimal? What is 17/84 as a decimal? What is 16/86 as a decimal? What is 18/84 as a decimal? What is 16/87 as a decimal? What is 19/84 as a decimal? What is 16/88 as a decimal? What is 20/84 as a decimal? What is 16/89 as a decimal? What is 21/84 as a decimal? What is 16/90 as a decimal? What is 22/84 as a decimal? What is 16/91 as a decimal? What is 23/84 as a decimal? What is 16/92 as a decimal? What is 24/84 as a decimal? What is 16/93 as a decimal? What is 25/84 as a decimal? What is 16/94 as a decimal? What is 26/84 as a decimal?
### Convert similar fractions to percentages
From 16 Numerator From 84 Denominator 17/84 as a percentage 16/85 as a percentage 18/84 as a percentage 16/86 as a percentage 19/84 as a percentage 16/87 as a percentage 20/84 as a percentage 16/88 as a percentage 21/84 as a percentage 16/89 as a percentage 22/84 as a percentage 16/90 as a percentage 23/84 as a percentage 16/91 as a percentage 24/84 as a percentage 16/92 as a percentage 25/84 as a percentage 16/93 as a percentage 26/84 as a percentage 16/94 as a percentage
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Quantitative Aptitude - SSC CGL 2020 - Study24x7
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# Quantitative Aptitude - SSC CGL 2020
Updated on 25 March 2020
SSC Preparation Strategies & S
Updated on 25 March 2020
In this article, I am going to explain some of the quantitative aptitude topics from the SSC CGL 2020 syllabus.
## Simplification
The simplification topic must have a holistic SSC CGL preparation strategy because it includes the application of multiple individual topics in it like fractions, square of a number, algebraic identities, LCM and HCF, etc.
## Basic Rule: “VBODMAS”
This rule is the basis of the simplification question. You must remember the order of simplification through this rule and apply it for the simplification of large and complex numerical expressions.
V is forvirnaculum” or bar represented by( ).
B is for ‘brackets’. Here the Order of precedence for brackets is ( ), { }, [ ].
O is ‘order ’, defines the power or root in the numerical expression.
D is ‘division’.
M is ‘multiplication’.
S is ‘subtraction’.
Example: Solve the numerical expression and find the value of x=?
5(25/5) + 3(27/9) = 5(40/5)-x
Solution: Apply the VBODMAS RULE on the given numerical Expression
5(5) + 3(3) =5(8)-x
25+9=40-x
34=40-x
X=40-34
X= 6
## Interest
The Interest topic of the CGL 2020 quantitative aptitude syllabus includes problems on Simple as well as compound interest.
Remember:
Interest = Amount – Principal
## Basic Formula on Simple Interest:
SI = (PXRXT)/100
Example: A person has borrowed a sum of Rs 25000 from the bank for a period of 3 years. After 3 years he has paid Rs 27250 to the bank. Calculate the rate of simple interest at which the bank has lent the money to the person.
Solution:
Principal (P) =Rs 25000
Amount (A) =Rs 27250
Time (T) = 3 Years
Rate of Interest (R) =?
SI= Amount- Principal
SI= 27250-25000
= Rs 2250
SI= (P*R*T)/100
2250= (25000*R*3)/100
225000= 75000*R
R= 225000/75000
R= 3%
## Basic Formulas on Compound Interest
Amount= P* (1+R/100) T
Example: What will be the compound interest on Rs 10000 for 1 year, if the rate of interest is calculated at 24% per annum?
Solution:
Principal (P) = Rs 10000
Rate of interest(R) = 24%
Time (T) = 1 Year
Compound Interest (CI) =?
Amount = P* (1+R/100) T
= 10000*(1+ 24/100)1
= 10000* (1+0.24)1
= 10000* 1.24
= 12400
Compound Interest= Amount- Principal
= 12400- 10000
= Rs 2400
## Averages
The middle value of a given data set represents the average value of that data set.
Basics:
Average= (Two middle terms of series)/2
Average= Middle Term of the series
## Basic Formula
Average= Sum of all values of observational set Number of values in observational set
Example: A person has received Rs 6434, Rs 7231, Rs 6856, Rs 6927 and Rs 6562 for the 5 months. How much money he will get in the 6th month to make an average of Rs 6500?
Solution: Let the money received in the 6th month= x
Sum of the money received in 5 months
= 6434+ 7231 + 6856 + 6927+ 6562
= 34010
Average of 6 month = (34010+x)/ 6
6500* 6 = 34010+ x
39000 = 34010+ x
x = 39000-34010
= Rs 4990
All The Best To All The SSC CGL 2020 Aspirants !
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# Arithmetic Operations On Functions
In these lessons, we will look at how functions can be added, subtracted, multiplied or divided. You may also want to look at the lessons on composite functions.
The following diagram shows the operations with functions: addition, subtraction, multiplication, and division. Scroll down the page for more examples and solutions on function operations.
Functions Can Be Added
Example:
Add the functions f(x) = x + 2 and g(x) = 5x – 6
Solution:
(f + g)(x)
= f(x) + g(x)
= (x + 2) + (5x – 6)
= 6x – 4
Functions Can Be Subtracted
Example:
Given f(x) = x + 2 and g(x) = 5x – 6, find (f – g)(x)
Solution:
(f – g)(x)
= f(x) – g(x)
= (x + 2) – (5x – 6)
= –4x + 8
Functions Can Be Multiplied
Example:
Add the functions f(x) = x + 2 and g(x) = 5x – 6
Solution:
(f • g)(x)
= f(x) • g(x)
= (x + 2)(5x – 6)
= 5x2 + 4x – 12
Functions Can Be Divided
Example:
Given f(x) = x + 2 and g(x) = 5x – 6, find
Solution:
for g(x) ≠ 0
Functions: Adding And Subtracting
Examples:
1. f(x) = x2 + 3x
g(x) = 8x + 9
(f+g)(x) = f(x) + g(x) = ?
2. f(x) = x2 + 2x + 8
g(x) = 3x2 - x + 7
(f-g)(x) = f(x) - g(x) = ?
Functions: Multiplying And Dividing
Example:
f(x) = (x2 - 1)
g(x) = (x + 1)
(fg)(x) = ?
(f/g)(x) = ?
Combining Functions By Addition Subtraction Multiplication And Division
Example:
f(x) = 3x2 + 4
g(x) = x - 5
(f + g)(x) = ?
(f - g)(x) = ?
(fg)(x) = ?
(f/g)(x) = ?
Sum, Difference, Product And Quotient Of Two Functions
How to find the sum, difference, product and quotient of two functions and determine the domain?
Example:
f(x) = 5x - 4
g(x) = -9x + 6
(f + g)(x) = ?
(f - g)(x) = ?
(fg)(x) = ?
(f/g)(x) = ?
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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# Problems from Dependent and independent events
In Barcelona, $$60\%$$ of the population are brunette, $$70\%$$ have brown eyes, and $$80\%$$ are brunette or have brown eyes.
We choose a person at random. If he or she is brunette: what is the probability to that he or she also has brown eyes? Is being a brunette idenpendent of having brown eyes, or is there a correlation?
See development and solution
### Development:
We are considering two events, $$C =$$ "to be brunette", $$O =$$"to have brown eyes". For the statement, we know that $$P(C)=\dfrac{6}{10}={3}{5}$$, $$P(O)=\dfrac{7}{10}$$, $$P(O\cup C)=\dfrac{8}{10}=\dfrac{4}{5}$$.
They ask us about the probability of having brown eyes, knowing that the person is brunette, this is $$P(O/C)$$.
Applying the formula of the conditional probability $$P(O/C)=\dfrac{P(O\cap C)}{P(C)}$$, yet we still do not know $$P(O\cap C)$$.
As we know the probability of the union, we can use the formula $$P(O\cup C)=P(O)+P(C)-P(O\cap C)$$.
By substituting, $$\dfrac{4}{5}=\dfrac{7}{10}+\dfrac{3}{5}-P(O\cap C)$$$and therefore, $$P(O\cap C)=\dfrac{7}{10}+\dfrac{3}{5}-\dfrac{4}{5}=\dfrac{5}{10}=\dfrac{1}{2}$$$
And so, $$P(O/C)=\dfrac{P(O\cap C)}{P(C)}=\dfrac{\dfrac{1}{2}}{\dfrac{3}{5}}=\dfrac{5}{6}$$\$
To calculate if being a brunette is idenpendent of having brown eyes, we must ask ourselves whether $$P(O\cap C)=P(O)\cdot P(C)$$.
Substituting, $$\dfrac{1}{2}\neq \dfrac{7}{10}\cdot\dfrac{3}{5}=\dfrac{21}{50}$$.
Therefore, two events are dependent.
### Solution:
$$P(O/C)=\dfrac{5}{6}$$. The two events are dependent.
Hide solution and development
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# Chapter 4 Sequences and Mathematical Induction. 4.2 Mathematical Induction.
## Presentation on theme: "Chapter 4 Sequences and Mathematical Induction. 4.2 Mathematical Induction."— Presentation transcript:
Chapter 4 Sequences and Mathematical Induction
4.2 Mathematical Induction
A recent method for proving mathematical arguments. De Morgan is credited with its discovery and and name. The validity of proof by mathematical induction is taken as a axiom. – an axiom or postulate is a proposition that is not proved or demonstrated but considered to be either self-evident, or subject to necessary decision. (Wikipedia) self-evidentdecision
Principle of Mathematical Induction Let P(n) be a property that is defined for integers n, and let a be a fixed integer. Suppose the following two statements are true: 1.P(a) is true. 2.For all integers k ≥ a, if P(k) is true then P(k+1) is true. Then the statement, for all integers n ≥ a, P(a) is true
Example For all integers n ≥ 8, n cents can be obtained using 3 cents and 5 cents. or, for all integers n ≥ 8, P(n) is true, where P(n) is the sentence “n cents can be obtained using 3 cents and 5 cents.”
Example This can be proven by exhaustion if we can continue to fill in the table up to \$1.00. The table shows how to obtain k cents using 3 and 5 coins. We must show how to obtain (k+1) cents. Two cases: – k: 3 + 5, k+1: ? replace 5 with 3 + 3 – k: 3 + 3 + 3, k+1: ? replace 3+3+3 with 5+5
Method of Proof Mathematical Induction Statement: “For all integers n≥a, a property P(n) is true.” (basis step) Show that the property is true for n = a. (inductive step) Show that for all integers k≥a, if the property is true for n=k then it is true for n=k+1. – (inductive hypothesis) suppose that the property is true for n=k, where k is any particular but arbitrarily chosen integer with k≥a. – then, show that the property is true for n = k+1
Example Coins Revisited Proposition 4.2.1: Let P(n) be the property “n cents can be obtained using 3 and 5 cent coins.” Then P(n) is true for all integers n≥8. – Proof: – Show that the property is true for n=8: The property is true b/c 8=3+5.
Example cont. – Show that for all integers k≥8, if the property true for n=k, then property true for n=k+1 (inductive hypothesis) Suppose k cents can be obtained using 3 and 5 cent coins for k≥8. Must show (k+1) cents can be obtained from 3 & 5 coin. – Case (3,5 coin): k+1 can be obtained by replacing the 5 coin with two 3 cent coins. This increments the value by 1 (3+3=6) replaces the 5 cent coin. – Case (3,3,3 coin): k+1 can be obtained by replacing the three 3 coins with two 5 coins. k=b+3+3+3=b+9 and k+1=b+9+1=b+5+5=b+10
Example Formula Prove with mathematical induction Identify P(n) Basis step
Example Formula cont. Inductive step – assume P(k) is true, k>=1 – show that P(k+1) is true by subing k+1 for n – show that left side 1+2+…k+1 = right side (k+1)(k+2)/2 – 1+2+…+k+1 = (1+2+…+k) + k+1 – sub from inductive hypothesis:
Theorem 4.2.2 Sum of the First n Integers – For all integers n≥1,
Example Sum of the First n Integers – Find 2+4+6+…+500 Get in form of Theorem 4.2.2 (1+2+…+n) factor out 2: 2(1+2+3+…+250) sum = 2( n(n+1)/2 ), n = 250 sum = 2( 250(250+1)/2 ) = 62,750 – Find 5+6+7+8+…+50 add first 4 terms 1+2+3+4 to problem then subtract back out after computation with 4.2.2 1+2+3+4+5+6+7+…+50 – (1+2+3+4) (50 (50+1)/ 2) – 10 =1,265
Sum of Geometric Sequence Prove that, for all integers n≥0 and all real numbers r except 1. P(n): Basis: Inductive: (n=k) Inductive Hypothesis: (n=k+1)
Theorem 4.2.3 Sum of Geometric Sequences – For any real number r except 1, and any integer n≥0,
Examples Sum of Geometric Sequences – Find 1 + 3 + 3 2 + … + 3 m-2 Sequence is in geometric series, apply 4.2.3 directly – Find 3 2 + 3 3 +… + 3 m rearrange into proper geometric sequence by factoring out 3 2 from sequence 3 2 (1 + 3 + 3 2 + … + 3 m-2 ) =
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## Kiss those Math Headaches GOODBYE!
### Tutor Tales #1: Color in Geometry
Hi,
This if the first in what I hope will be a long series of brief blogs called Tutor Tales blogs.
The idea is that while I’m tutoring I get ideas or insights on how to help students, and then I write up a short blog entry on that experience, preferably on the day that the event occurred.
I hope that these Tutor Tales will give you examples of approaches to math that help students (or that do not help, depending on what I did), and that they give you a chance to reflect on your own teaching.
For the first Tutor Tale entry, I just noticed how useful it can be to use color in geometry.
The girl I was tutoring had a problem: Find out how many diagonals can be drawn inside a regular, convex nine-sided polygon.
I’ve already noticed that this girl likes color, and she is 17 years old. So I had a hunch that she would be open to trying a color-approach.
We created the non-agon by first drawing a circle, and then marking off nine points on the circle. Then we connected the points sequentially.
To find out how many diagonals we could draw for such a figure, we chose one color for the top point, green, and drew all of the diagonals we could for that point, in green. It turned out that there were 6 diagonals, so we put a big 6 in green at this vertex. Then we tried the next vertex, which we colored pink. We found that we could create 6 additional diagonals from this vertex, and we colored these pink. So we put a big pink 6 by this vertex. We went around the circle in a clockwise way, using a different color for each vertex. All in all we found that the pattern of diagonals was: 6, 6, 5, 4, 3, 2, 1, 0, 0, for a grand total of 27 diagonals.
Here’s the image of the figure we worked on.
One thing to consider, especially if you teach geometry, is how many opportunities there are in geometry to use color to separate different concepts and to relate similar concepts. Check it out and see what you discover.
Happy Teaching!
— Josh
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# 7.1: Logarithms
Difficulty Level: At Grade Created by: CK-12
Name: __________________
## Introduction to Logarithms
1. On day , you have . Every day, you double.
1. How many pennies do you have on day ?
2. How many pennies do you have on day ?
3. On what day do you have ? Before you answer, express this question as an equation, where is the variable you want to solve for.
4. Now, what is ?
2. A radioactive substance is decaying. There is currently of the substance.
1. How much substance will there be after half-lives?
2. How much substance will there be after half-lives?
3. After how many half-lives will there be of the substance left? Before you answer, express this question as an equation, where is the variable you want to solve for.
In both of the problems above, part (d) required you to invert the normal exponential function. Instead of going from time to amount, it asked you to go from amount to time. (This is what an inverse function does—it goes the other way—remember?)
So let’s go ahead and talk formally about an inverse exponential function. Remember that an inverse function goes backward. If turns a into an , then must turn an into a .
So, fill in the following table (on the left) with a bunch of and values for the mysterious inverse function of . Pick values that will make for easy values. See if you can find a few values that make be or negative numbers!
On the right, fill in and values for the inverse function of .
Inverse of Inverse of
Inverse of Inverse of
Now, let’s see if we can get a bit of a handle on this type of function.
In some ways, it’s like a square root. is the inverse of . When you see you are really seeing a mathematical question: “What number, squared, gives me ?”
Now, we have the inverse of (which is quite different from of course). But this new function is also a question: see if you can figure out what it is. That is, try to write a question that will reliably get me from the left-hand column to the right-hand column in the first table above.
Do the same for the second table above.
Now, come up with a word problem of your own, similar to the first two in this exercise, but related to compound interest.
Name: ___________________________
Homework—Logs
asks the question: “ to what power is ?” Based on that, you can answer the following questions:
1.
2.
3.
4.
5.
6.
8. Multiple choice: which of the following is closest to ?
A.
B.
C.
D.
E.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
OK. When I say, , that’s the same thing as saying . Why? Because asks a question: “What squared equals ?” So the equation is providing an answer: “six squared equals .”
You can look at logs in a similar way. If I say I’m asking a question: “ to what power is ?” And I’m answering: “. to the fifth power is .” So saying is the same thing as saying .
Based on this kind of reasoning, rewrite the following logarithm statements as exponent statements.
22.
23.
24.
25.
Now do the same thing backward: rewrite the following exponent statements as logarithm statements.
26.
27.
28.
Finally...you don’t understand a function until you graph it...
29. a. Draw a graph of .Plot at least to draw the graph.
b. What are the domain and range of the graph? What does that tell you about this function?
Name: ___________________________
## Properties of Logarithms
1. Based on numbers , finish this sentence in words: when you take of a number, you find:
2. Based on numbers , write an algebraic generalization about logs.
3. Now, let’s dig more deeply into that one. Rewrite problems so they look like problems : that is, so the thing you are taking the log of is written as a power of .
1. :
2. :
3. :
4. Based on this rewriting, can you explain why your generalization from works?
4. Based on numbers , write an algebraic generalization about logs.
5. Based on numbers write an algebraic generalization about logs.
Name: ___________________________
Homework—Properties of Logarithms
Memorise these three rules
1. In class, we demonstrated the first and third rules above. For instance, for the first rule:
This demonstrates that when you multiply two numbers, their logs add.
Now, you come up with a similar demonstration of the second rule of logs, that shows why when you divide two numbers, their logs subtract.
Now we’re going to practice applying those three rules. Take my word for these two facts. (You don’t have to memorize them, but you will be using them for this homework.)
Now, use those facts to answer the following questions.
2.
(Hint: . So this is . Which rule above helps you rewrite this?)
3. How can you use your calculator to test your answer to ? (I’m assuming here that you can’t find on your calculator, but you can do exponents.) Run the test—did it work?
4.
5.
6.
7.
8.
Simplify, using the property:
9.
10.
Below bracket are different size
Simplify, using the property:
11.
12.
13.
Simplify, using the property:
14.
15.
16.
17. a. Draw a graph of . Plot at least to draw the graph.
b. What are the domain and range of the graph? What does that tell you about this function?
Name: ___________________________
Using the Laws of Logarithms
1. Simplify:
2. Simplify:
3. Simplify:
4. Solve for .
5. Solve for :
6. Solve for :
7. Solve for in terms of :
So What Are Logarithms Good For, Anyway?
1. Compound Interest. Andy invests in a bank that pays out interest, compounded annually. Note that your answers to parts (a) and (c) will be numbers, but your answers to parts (b) and (d) will be formulae.
a. After , how much money does Andy have?
b. After , how much money m does Andy have?
c. After how many years does Andy have exactly ?
d. After how many years does Andy have ?
2. Sound Intensity. Sound is a wave in the air—the loudness of the sound is related to the intensity of the wave. The intensity of a whisper is approximately ; the intensity of a normal conversation is approximately . Assuming that a person starts whispering at time , and gradually raises his voice to a normal conversational level by time , show a possible graph of the intensity of his voice. (*You can’t get the graph exactly, since you only know the beginning and the end, but show the general shape.)
3. That was pretty complicated, wasn’t it? It’s almost impossible to graph or visualize something going from a hundred to a million: the range is too big.
Fortunately, sound volume is usually not measured in intensity, but in loudness, which are defined by the formula: , where is the loudness (measured in decibels), and is the intensity.
a. What is the loudness, in decibels, of a whisper?
b. What is the loudness, in decibels, of a normal conversation?
c. Now do the graph again—showing an evolution from whisper to conversation in —but this time, graph loudness instead of intensity.
d. That was a heck of a lot nicer, wasn’t it? (This one is sort of rhetorical.)
e. The quietest sound a human being can hear is intensity . What is the loudness of that sound?
f. The sound of a jet engine—which is roughly when things get so loud they are painful—is loudness . What is the intensity of that sound?
g. The formula gave above gives loudness as a function of intensity. Write the opposite function, that gives intensity as a function of loudness.
h. If sound is twenty decibels higher than sound , how much more intense is it?
4. Earthquake intensity. When an Earthquake occurs, seismic detectors register the shaking of the ground, and are able to measure the “amplitude” (fancy word for “how big they are”) of the waves. However, just like sound intensity, this amplitude varies so much that it is very difficult to graph or work with. So Earthquakes are measured on the Richter scale which is the of the amplitude .
a. A “microearthquake” is defined as or less on the Richter scale. Microearthquakes are not felt by people, and are detectable only by local seismic detectors. If is the amplitude of an earthquake, write an inequality that must be true for it to be classified as a microearthquake.
b. A “great earthquake” has amplitude of or more. There is generally one great earthquake somewhere in the world each year. If is the measurement of an earthquake on the Richter scale, write an inequality that must be true for it to be classified as a great earthquake.
c. Imagine trying to show, on a graph, the amplitudes of a bunch of earthquakes, ranging from microearthquakes to great earthquakes. (Go on, just imagine it—I’m not going to make you do it.) A lot easier with the Richter scale, ain’t it?
d. Two Earthquakes are measured—the second one has times the amplitude of the first. What is the difference in their measurements on the Richter scale?
5. . In Chemistry, a very important quantity is the concentration of Hydrogen ions, written as —this is related to the acidity of a liquid. In a normal pond, the concentration of Hydrogen ions is around . (In other words, every liter of water has about , or of Hydrogen ions.) Now, acid rain begins to fall on that pond, and the concentration of Hydrogen ions begins to go up, until the concentration is (every liter has of ).
a. How much did the concentration go up by?
b. Acidity is usually not measured as concentration (because the numbers are very unmanageable, as you can see), but as , which is defined as . What is the of the normal pond?
c. What is the of the pond after the acid rain?
6. Based on numbers , write a brief description of what kind of function generally leads scientists to want to use a logarithmic scale.
Name: ___________________________
Homework: What Are Logarithms Good For, Anyway?
1. I invest in a bank that pays interest, compounded annually. So after , I have dollars in the bank. When I come back, I find that my account is worth . How many years has it been? Your answer will not be a number—it will be a formula with a log in it.
2. The of a substance is given by the formula , where is the concentration of Hydrogen ions.
a. If the Hydrogen concentration is , what is the ?
b. If the Hydrogen concentration is , what is the ?
c. What happens to the every time the Hydrogen concentration divides by ?
You may have noticed that all our logarithmic functions use the base . Because this is so common, it is given a special name: the common log. When you see something like with no base written at all, that means the log is . (So is a shorthand way of writing , just like is a shorthand way of writing . With roots, if you don’t see a little number there, you assume a . With logs, you assume a .)
3. In the space below, write the question that asks.
Use the common log to answer the following questions.
4.
5.
6.
7. ( with after it)
8. (use the log button on your calculator)
OK, so the button on your calculator does common logs, that is, logs base .
There is one other button on your calculator. It is called the “natural log,” and it is written (which sort of stands for “natural log” only backward—personally, I blame the French).
means the log to the base . What is ? It’s a long ugly number—kind of like only different—it goes on forever and you can only approximate it, but it is somewhere around . Answer the following questions about the natural log.
9.
10.
11.
12.
13. (*this is the only one that requires the button on your calculator)
Name: _______________
Sample Test: Logarithms
1.
2.
3.
4.
(approximately)
5.
6.
7.
8.
9.
10
11.
12.
13.
14.
15.
16. Rewrite as a logarithm equation (no exponents):
17. Rewrite as an exponent equation (no logs):
For questions , assume that...
18.
19.
20.
21.
22.
23. Graph .
24. What are the domain and range of the graph you drew in ?
25. I invest in a bank that pays interest, compounded annually. So after , I have dollars in the bank. When I come back, I find that my account is worth . How many years has it been? Your answer will be a formula with a log in it.
26. The “loudness” of a sound is given by the formula , where is the loudness (measured in decibels), and is the intensity of the sound wave.
a. If the sound wave intensity is , what is the loudness?
b. If the sound wave intensity is , what is the loudness?
c. If the sound wave intensity is , what is the loudness?
d. What happens to the loudness every time the sound wave intensity multiplies by ?
27. Solve for .
28. solve for .
Extra credit: Solve for (cabin)
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# What is the equation of the parabola that has a vertex at (0, 8) and passes through point (5,-4) ?
Jan 22, 2016
There are an infinite number of parabolic equations that meet the given requirements.
If we restrict the parabola to having a vertical axis of symmetry, then:
$\textcolor{w h i t e}{\text{XXX}} y = - \frac{12}{25} {x}^{2} + 8$
#### Explanation:
For a parabola with a vertical axis of symmetry, the general form of the parabolic equation with vertex at $\left(a , b\right)$ is:
$\textcolor{w h i t e}{\text{XXX}} y = m {\left(x - a\right)}^{2} + b$
Substituting the given vertex values $\left(0 , 8\right)$ for $\left(a , b\right)$ gives
$\textcolor{w h i t e}{\text{XXX}} y = m {\left(x - 0\right)}^{2} + 8$
and if $\left(5 , - 4\right)$ is a solution to this equation, then
$\textcolor{w h i t e}{\text{XXX}} - 4 = m \left({\left(- 5\right)}^{2} - 0\right) + 8 \Rightarrow m = - \frac{12}{25}$
and the parabolic equation is
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l a c k}{y = - \frac{12}{25} {x}^{2} + 8}$
graph{y=-12/25*x^2+8 [-14.21, 14.26, -5.61, 8.63]}
However, (for example) with a horizontal axis of symmetry:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l a c k}{x = \frac{5}{144} {\left(y - 8\right)}^{2}}$
also satisfies the given conditions:
graph{x=5/144(y-8)^2 [-17.96, 39.76, -8.1, 20.78]}
Any other choice for the slope of the axis of symmetry will give you another equation.
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Home | | Maths 10th Std | Surface Area: Frustum of a right circular cone
# Surface Area: Frustum of a right circular cone
When a cone ABC is cut through by a plane parallel to its base, the portion of the cone DECB between the cutting plane and the base is called a frustum of the cone.
Frustum of a right circular cone
In olden days a cone shaped buckets [Fig.7.21(a)] filled with sand / water were used to extinguish fire during fire accidents. Later, it was reshaped to a round shaped bottom [Fig.7.21(b)] to increase its volume.
The shape in [Fig.7.21(c)] resembling a inverted bucket is called as a frustrum of a cone.
The objects which we use in our daily life such as glass, bucket, street cone are examples of frustum of a cone. (Fig.7.22)
## Definition
When a cone ABC is cut through by a plane parallel to its base, the portion of the cone DECB between the cutting plane and the base is called a frustum of the cone.
## Surface area of a frustum
Let R and r be radii of the base and top region of the frustum DECB respectively, h is the height and l is the slant height of the same.
Therefore, C.S.A. = ½ (sum of the perimeters of base and top region) × slant height
= ½ (2πR + 2πr)l
C.S.A. of a frustum = π(R + r)l sq. units
Where,
T.S.A. = C.S.A. + Area of the bottom circular region + Area of the top circular region.
T.S.A. of a frustum = π(R + r)l + πR2 + πr 2 sq. units
Where,
Example 7.13 The slant height of a frustum of a cone is 5 cm and the radii of its ends are 4 cm and 1 cm. Find its curved surface area.
Solution Let l, R and r be the slant height, top radius and bottom radius of the frustum.
Given that, l= 5 cm, R = 4 cm, r = 1 cm
Now, C.S.A. of the frustum = π (R + r) l sq. units
Therefore, C.S.A. = 78.57 cm2
Example 7.14 An industrial metallic bucket is in the shape of the frustum of a right circular cone whose top and bottom diameters are 10 m and 4 m and whose height is 4 m. Find the curved and total surface area of the bucket.
Solution Let h, l, R and r be the height, slant height, outer radius and inner radius of the frustum.
Given that, diameter of the top =10 m; radius of the top R = 5 m.
diameter of the bottom = 4 m; radius of the bottom r = 2 m, height h= 4 m
Therefore, C.S.A. = 110 m2 and T.S.A. = 201.14 m2
Tags : Definition, Formula, Solved Example Problems | Mensuration | Mathematics , 10th Mathematics : UNIT 7 : Mensuration
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
10th Mathematics : UNIT 7 : Mensuration : Surface Area: Frustum of a right circular cone | Definition, Formula, Solved Example Problems | Mensuration | Mathematics
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What Percentage is 38 out of 40? Calculate Your Score with This Easy Guide!
Welcome to Warren Institute! Have you ever wondered what percent 38 out of 40 is? Understanding percentages is crucial in various mathematical applications. In this article, we will delve into the concept of percentages and how to calculate them effectively. By determining the percentage that 38 represents out of a total of 40, we can gain insights into its relative value within the given context. Join us as we explore the process of finding this percentage and learn how it can be applied in real-world scenarios. Let's unravel the mystery behind percentages and empower ourselves with mathematical knowledge!
Understanding Percentages in Mathematics Education
In mathematics education, understanding percentages is crucial for various real-life applications. One common question that arises is how to calculate what percent a certain number is of another.
Calculating Percentage: The Basics
To find out what percent 38 is out of 40, the basic formula is (Part/Whole) x 100. In this case, it would be (38/40) x 100.
Applying the Formula
By substituting the values into the formula, we get (38/40) x 100 = 0.95 x 100 = 95%. Therefore, 38 is 95% of 40.
Interpreting the Result
When we say that 38 is 95% of 40, it means that if we were to divide 40 into 100 equal parts, 38 of those parts would represent 95%.
Significance in Mathematics Education
Understanding percentages not only helps in solving mathematical problems but also in practical scenarios such as calculating discounts, taxes, and interest rates.
Real-World Applications
Being able to calculate percentages accurately is a valuable skill in fields like finance, business, and science. It is essential for making informed decisions based on data analysis and interpretation.
How do you calculate a percentage for "38 out of 40" in Mathematics education?
To calculate the percentage for "38 out of 40" in Mathematics education, you would divide 38 by 40 and then multiply by 100 to get the percentage.
What is the significance of understanding percentages when solving "38 out of 40" in Mathematics education?
Understanding percentages is important in Mathematics education because it allows students to interpret and compare values, such as "38 out of 40." It helps students quantify the relationship between the given numbers and express it as a fraction or a decimal value.
Are there any real-life applications for knowing how to find the percentage of "38 out of 40" in Mathematics education?
Yes, understanding how to find the percentage of "38 out of 40" is important in Mathematics education as it helps students grasp the concept of calculating percentages, which is a fundamental skill used in various real-life applications such as calculating discounts, analyzing data sets, and interpreting statistics.
Can you explain the concept of finding a percentage using the example of "38 out of 40" in Mathematics education?
To find the percentage of 38 out of 40 in Mathematics education, you would divide 38 by 40 and multiply the result by 100 to get the percentage value.
Why is it important for students to grasp the concept of percentages through exercises like "38 out of 40" in Mathematics education?
It is important for students to grasp the concept of percentages through exercises like "38 out of 40" in Mathematics education because it helps them develop a practical understanding of how fractions and decimals relate to real-world situations, fostering numerical fluency and problem-solving skills.
In conclusion, calculating what percent 38 out of 40 is can help students understand the concept of percentages and fractions better in Mathematics education. It is crucial for students to master these foundational skills to succeed in more advanced math concepts.
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# Common Core: 2nd Grade Math : Number Names for Hundreds: CCSS.Math.Content.2.NBT.A.1b
## Example Questions
← Previous 1 3 4
### Example Question #1 : Number Names For Hundreds: Ccss.Math.Content.2.Nbt.A.1b
How many ones are in the number
Three
Two
Zero
Four
One
Zero
Explanation:
There is a in the ones place which means there are no ones in the number .
### Example Question #2 : Number Names For Hundreds: Ccss.Math.Content.2.Nbt.A.1b
How many tens are in the number
Three
Four
Zero
Two
One
Zero
Explanation:
There is a in the tens place which means there are no tens in the number .
### Example Question #3 : Number Names For Hundreds: Ccss.Math.Content.2.Nbt.A.1b
How many hundreds are in the number
Thirteen
Three hundred
Three
Thirty
Zero
Three
Explanation:
There is a in the hundreds place which means there are three hundreds in the number .
### Example Question #4 : Number Names For Hundreds: Ccss.Math.Content.2.Nbt.A.1b
How many ones are in the number
Forty
Two
Zero
Three
Four
Zero
Explanation:
There is a in the ones place which means there are no ones in the number .
### Example Question #5 : Number Names For Hundreds: Ccss.Math.Content.2.Nbt.A.1b
How many tens are in the number
Five
Three
One
Zero
Six
Zero
Explanation:
There is a in the tens place which means there are no tens in the number .
### Example Question #6 : Number Names For Hundreds: Ccss.Math.Content.2.Nbt.A.1b
How many hundreds are in the number
Sixty
Six
Nine
Six hundred
Three
Six
Explanation:
There is a in the hundreds place which means there are six hundreds in the number .
### Example Question #7 : Number Names For Hundreds: Ccss.Math.Content.2.Nbt.A.1b
How many ones are in the number
Seven
Seventy-seven
Zero
Fourteen
Seventy
Zero
Explanation:
There is a in the ones place which means there are no ones in the number .
### Example Question #8 : Number Names For Hundreds: Ccss.Math.Content.2.Nbt.A.1b
How many tens are in the number
Zero
Sixty
Eight hundred
Eight
Eighteen
Zero
Explanation:
There is a in the tens place which means there are no tens in the number .
### Example Question #9 : Number Names For Hundreds: Ccss.Math.Content.2.Nbt.A.1b
How many hundreds are in the number
Nine
Ninteen
Nine hundred
Ninty
Ninty-nine
Nine
Explanation:
There is a in the hundreds place which means there are nine hundreds in the number .
### Example Question #10 : Number Names For Hundreds: Ccss.Math.Content.2.Nbt.A.1b
How many hundreds are in the number
Twelve
Two
Nine
Six
Twenty
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# DAV Class 8 Maths Chapter 2 Worksheet 1 Solutions
The DAV Class 8 Maths Book Solutions and DAV Class 8 Maths Chapter 2 Worksheet 1 Solutions of Cubes and Cube Roots offer comprehensive answers to textbook questions.
## DAV Class 8 Maths Ch 2 WS 1 Solutions
Question 1.
Find the cubes of:
(i) 8
(ii) 13
(iii) 17
(iv) 1.3
(v) 0.06
(vi) 0.4
(vii) $$\frac{2}{3}$$
(viii) – 7
(ix) – 9
(x) – 12
Solution:
(i) 83 = 8 × 8 × 8 = 512
(ii) 133 = 13 × 13 × 13 = 2197
(iii) 173 = 17 × 17 × 17 = 4913
(iv) 1.33 = 1.3 × 1.3 × 1.3 = 2.197
(v) 0.063 = 0.000216
(vi) (0.4)2 = 0.064
(vii) $$\left(\frac{2}{3}\right)^3=\frac{2 \times 2 \times 2}{3 \times 3 \times 3}=\frac{8}{27}$$
(viii) (- 7)3 = (- 7) × (- 7) × (- 7) = – 343
(ix) (- 9)3 = (- 9) × (- 9) × (- 9) = – 729
(x) (- 12)3 = (- 12) × (- 12) × (- 12) = – 1728.
Question 2.
Which of the following numbers are perfect cubes?
(i) 4096
(ii) 108
(iii) 392
(iv) – 27000
(v) $$\frac{-64}{1331}$$
Solution:
(i) 4096
∴ 4096 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 23 × 23 × 23
∴ 4096 is a perfect cube.
(ii) 108
108 = 2 × 2 × 3 × 3 × 3
= 2 × 2 × 33
Here 2 × 2 is left without triplet.
∴ 108 is not a perfect cube.
(iii) 392
392 = 2 × 2 × 2 × 7 × 7
= 23 × 7 × 7
Here 7 × 7 is left without triplet.
∴ 392 is not a perfect cube.
(iv) – 27000 = (- 2) × (- 2) × (- 2) × (- 3) × (- 3) × (- 3) × (- 5) × (- 5) × (- 5)
= (- 2) × (- 3)3 × (- 5)
Here all have triplets.
∴ – 27000 is a perfect cube.
(v) $$\frac{-64}{1331}$$
⇒ – 64 = (- 4) × (- 4) × (- 4)
and 1331 = 11 × 11 × 11
∴ $$\frac{-64}{1331}$$ is a perfect square.
Question 3.
Find the smallest number by which 2560 must be multiplied so that the product is a perfect cube.
Solution:
∴ 2560 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5
Here 5 is left out without triplet.
So for this triplet 5 × 5 is needed
∴ 5 × 5 = 25 is the smallest number by which the given 2560 number must be multiplied to make it a perfect cube.
Question 4.
Find the smallest number by which 8788 be divided so that the quotient is a perfect cube.
Solution:
∴ 8788 = 2 × 2 × 13 × 13 × 13
Here 2 × 2 is left out for making a triplet
∴ 2 × 2 = 4 is the smallest number by which the given number must be divided to make it a perfect cube.
Question 5.
Write True or False for the following statements.
(i) 650 is not a perfect cube.
Solution:
True
(ii) Perfect cubes may end with two zeros.
Solution:
False
(iii) Perfect cubes of odd numbers may not always be odd.
Solution:
False
(iv) Cube of negative numbers are negative.
Solution:
True
(v) For a number to be a perfect cube, it must have prime factors in triplets.
Solution:
False
Question 1.
Using the following matter, fill in the blanks:
(i) 1 = 1 = 13
(ii) 3 + 5 = 8 = 22
(iii) 7 + 19 + 11 = ____ = ____
(iv) 13 + 15 + 17 + 19 = ____ = ____
(v) ____ = 125 = 53
Sol.
(iii) 7 + 9 + 11 = 27 = 33
(iv) 13 + 15 + 17 + 19 = 64 = 43
(u) 21 + 23 + 25 + 27 + 29 = 125 = 53
Question 2.
Using the given matter, fill in the blanks:
(i) 23 – 13 = 1 + 2 × 1 × 3 = 7
(ii) 33 – 23 = 1 + 3 × 2 × 3 = 19
(iii) 43 – 33 = ____ = ____
(iv) 53 – 43 = ____ = ____
(v) 123 – 113 = ____ = ____
(vi) 203 – 193 = ____ = ____
Solution:
(iii) 43 – 33 = 1 + 4 × 3 × 3 = 37
(iv) 53 – 43 = 1 + 5 × 4 × 3 = 61
(v) 123 – 113 = 1 + 12 × 11 × 3 = 397
(vi) 203 – 193 = 1 + 20 × 19 × 3 = 1141
Question 3.
Which of the following are perfect cubes? Also find their cube roots.
(i) 6859
(ii) 2025
(iii) 15625
(iv) 3375
Solution:
(i) ∴ 6859 = 19 × 19 × 19
= $$\sqrt{19^3}$$ = 19
Here 6859 is perfect cube and its cube root is 19.
(ii) 2025 = 3 × 3 × 3 × 3 × 5 × 5
Here 3 and 5 are ungrouped of triplet.
Hence, 2025 is not a perfect cube.
(iii) 15625 = 5 × 5 × 5 × 5 × 5 × 5
= 53 × 53
∴ 15625 is a perfect cube.
Hence, its cube root is 5 × 5 = 25
(iv) 3375 = 3 × 3 × 3 × 5 × 5
= 33 × 53
∴ 3375 is a perfect cube.
Hence, its cube root is 3 × 5 = 15.
Question 4.
Find the cube root of 17576 by estimation method.
Solution:
The following steps are taken to proceed:
(i) Form groups of three, starting from the rightmost digit of 17576.
(ii) 17 576, here 576 has 3 digits whereas 17 has only two digits.
(iii) The digit 6 is at ones place. So we take the one’s place of the required cube root as 6.
(iv) Take the other group, i.e. 17. Cube of 2 is 8 and cube of 3 is 27
∴ 23 < 17 < 33
∴ The ten’s place is 2.
Hence the required cube root is 26.
Question 5.
Find the cube root of 13824 by prime factorisation method.
Solution:
13824 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
= 23 × 23 × 23 × 33
∴ $$\sqrt[3]{13824}$$ = 2 × 2 × 2 × 3 = 24.
### DAV Class 8 Maths Chapter 2 Worksheet 1 Notes
The cube of a number is obtained by multiplying itself three times and is read as, the number raised to the power 3.
e.g. cube of 4 = 43
= 4 × 4 × 4 = 64
cube of 7 = 73
= 7 × 7 × 7 = 343
cube of 9 = 93
= 9 × 9 × 9 = 729
Here 64, 343 and 729 are called perfect cubes.
Properties of Cubes of Numbers
1. The cube of any even number is even,
e.g. 43 = 64 (even),
83 = 512 (even)
2. The cube of any odd number is odd,
e.g. (3)3 = 27 (odd),
(9)3 = 729 (odd)
3. The cube of any number multiple of 2 is divisible by 8,
e.g. (6)3 = 216 (divisible by 8)
(8)3 = 512 (divisible by 8)
(12)3 = 1728 (divisible by 8)
4. The cube of a negative number is always negative,
e.g. (- 2) = – 2 × – 2 × – 2 = – 8
(- 5)3 = – 5 × – 5 × – 5 = – 125
5. The cube of a positive number is always positive,
e.g. (3)3= 27 (positive)
(4)3 = 64 (positive)
6. The cube of a rational number is equal to the cube of its numerator divided by the cube of its denominator,
e.g. $$\left(\frac{4}{5}\right)^3=\frac{4^3}{5^3}$$
= $$\frac{64}{125}$$
$$\left(\frac{3}{4}\right)^3=\frac{3^3}{4^3}$$
= $$\frac{27}{64}$$
Example 1.
Is 512 a perfect cube? What is the number whose cube is 512?
Solution:
Resolving 512 into prime factors.
∴ 512 = $$\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2}$$
512 = 23 × 23 × 23
512 = (2 × 2 × 2)3 = (8)
Hence, 512 is a perfect cube and 8 is the number whose cube is 512.
Example 2.
Is 1512 a perfect cube?
Solution:
Resolving 1512 into prime factors.
1512 = $$2 \times 2 \times 2 \times 3 \times 3 \times 3$$ × 7
1512 = 23 × 23 × 7
Here 7 is ungrouped.
Hence, 1512 is not a perfect cube.
Example 3.
Is 53240 a perfect cube? If not, then by which smallest natural number should 53240 be divided so that the quotient is a perfect cube.
Solution:
Resolving 53240 into prime factors.
53240 = $$2 \times 2 \times 2 \times 5 \times \underline{11 \times 11 \times 11}$$
= 23 × 5 × 113
Here 5 is left ungrouped.
So 53240 is not a perfect cube.
∴ 53240 ÷ 5 = 10648
Hence, 10648 is a perfect cube.
Example 4.
Is 68600 a perfect cube? If not, find the smallest number by which 6860C must be multiplied to get a perfect cube.
Solution:
Let us resolve 68600 into prime factors.
∴ 68600 = 2 × 2 × 2 × 5 × 5 × 7 × 7 × 7
= 23 × 5 × 5 × 7
Here 5 × 5 is left without triplet.
∴ 68600 is not a perfect cube.
To make it a perfect cube, we multiply it by 5.
∴ 68600 × 5 = 2 × 2 × 2 × 5 × 5 × 5 × 7 × 7 × 7
= 343000 which is a perfect cube.
Hence, 5 is the number by which the given number must be multiplied 1 to make it a perfect cube.
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# How To Solve Coding And Number Series Quickly
## Concept of Coding And Number Series:-
In this Page How to Solve Coding and Number Series Questions Quickly is given.
Letter-Number Coding is a mark scoring section of the reasoning questions. In competitive examinations reasoning contain a most important part. If you achieve good marks in reasoning you will get a good score and rank in competitive examinations.
## Types of Number Series:
1. Perfect Square Series:
Example$324 = 18^{2} , 361 = 19^{2}, 400 = 20^{2}, 441 = 21^{2}, 484 = 22^{2}$
2. Perfect Cube Series:
Example$8^{3}=512, 9^{3}=729, 10^{3}=1000, 11^{3}=1331$
3. Geometric Series:
Example- $4\times 5=20,20 \times 5=100,100\times 5=500,500 \times 5=2500$
4. Arithmetic Series:
Example- 4,8,12,16,20,24,28
The common difference is constant that is 4.
5. Two Stage type Series:
Example-1, 3, 6, 10, 15…
3 – 1 = 2, 6 – 3 = 3, 10 – 6 = 4, 15 – 10 = 5….
Now, we get an arithmetic sequence 2, 3, 4, 5
6. Mixed Series:
Example-10, 22, 46, 94, 190,?
$10\times2 = 20 +2 = 22,22 \times 2 = 44 + 2 = 46,46 \times 2 = 92 + 2 = 94,94 \times 2 = 188 + 2 = 190,190 \times 2 = 380 + 2 = 382.$
So the missing number is 382.
7. Arithmetic-Geometric Series:
Example- 1, 4, 8, 11, 22, 25, ?
Series $Type +3 , \times2 ({\text{i.e Arithmetic and Geometric Mixing}})(e.g: 1+3=4;4\times 2=8)$
8. Twin/Alternate series:
Example- 3, 4, 8, 10, 13, 16 ? ?
Sol: As we can see, there are two series formed
Series 1 : 3, 8, 13 with a common difference of 5
Series 2 : 4, 10, 16 with a common difference of 6
So, the next two terms of the series should be 18 & 22 respectively.
### How to Solve Coding and Number Series Questions Quickly :
Question 1:
If in a certain language A is coded as 1, B is coded as 2, and so on, how is B I D D I C coded in that code?
Option:
A) 294493
B) 284563
C) 375582
D) 394492
Explanation:
As given the letters are coded as:
A B C D E F G H I
1 2 3 4 5 6 7 8 9
So, in B I D D I C, B is coded as 2, I as 9, D as 4, and C as 3. Thus, B I D D I C is coded as 294493.
Question 2:
If D is coded as 4 and C O V E R is coded as 63, then BASIS will be coded as?
Option:
A) 54
B) 55
C) 49
D) 50
Explanation:
It is clear that the code is arrange in this mannerA = 1. B = 2, C = 3 … so on
so that C O V E R is coded as [C+O+V+E+R]
⇒ 3 + 15 + 22 + 5 + 18 = 63.
Now B A S I S is coded as [ B+A+S+I+S]
⇒2 + 1 + 19 + 9 + 19 = 50.
Question 3:
bcd’ is coded as ‘def’ then ‘True’ is coded as……….
Options:
A) Hynx
B) Vrwx
C) Zbyw
D) Vtwg
Explanation:
b – d (+2)
c – e (+2)
d – f (+2)
+2 letters are considered in this code.
True – Vtwg
Question 4:
In a certain code ‘MONARCHY’ is written as ‘NPOBSDIZ’. How will ‘STANDARD’ be written as in that code?
Options:
A) TUBOEBSE
B) TVBSHBNW
C) WEHJSOLX
D) GHWQMKLJ
Explanation:
Each letter of MONARCHY is replaced by the next letter as per the alphabet
M+ 1 = N
O+ 1 = P
N+ 1 = O
A+ 1 = B
R+ 1 = S
C + 1= D
H+ 1 = I
Y+ 1 = Z
Similarly, STANDARD is written as TUBOEBSE
Question 5:
If the word is KEYBOARD written as LGBFTGYL then, what will be the code for INTERNET ?
A)JPWHUPHW
B)JPHWIUYTP
C)JPWIWTLB
D)None of the above
Explanation:
The 1st, 2nd, 3rd ………. And so on alphabet of the word are moved 1, 2, 3….. Steps forward respectively.
KEYBOARD will be coded as LGBFTGYL
INTERNET will be written as JPWIWTLB
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# Derivative of Inverse Functions
Learn about Derivative of Inverse Functions, its formula along with different examples and ways to calculate it.
Alan Walker-
Published on 2023-05-26
## Introduction to the Derivative of Inverse Functions
In calculus, there are different and easy ways to calculate derivatives of a function by using simple differentiation rules. But sometimes we need to calculate derivatives of an inverse function for which simple derivative rules may not work. Let’s understand how to calculate derivative of an inverse function and the difference between inverse function and implicit differentiation.
## Understanding of the Derivative of inverse functions
The inverse function differentiation is a method to calculate derivative of an inverse function. This method is suitable to calculate the rate of change of a function which is invertible and differentiable. An invertible function is a function whose inverse exists. If an invertible function is differentiable then the inverse of this function is also differentiable.
In inverse function differentiation, we calculate the derivative of inverse. But since there is no specific formula to calculate derivative of such functions. Therefore, the derivative of the original function is used. The relation between a function and its inverse helps to calculate the inverse derivative. By using this differentiation, we can avoid the use of limit definition of derivative.
## Derivative of Inverse Function Formula
The inverse function derivative formula can be obtained by using the relation between a function with its inverse. For this assume an inverse function which is expressed as:
$y=f^{-1}(x)$
Since the function is invertible, so we can write it as,
$x = f(y)$
Now the function becomes an implicit function. We can use implicit differentiation to solve this. Applying derivative,
$\frac{d}{dx}(x)=\frac{d}{dx}[f(y)]$
$1=f'(y)\frac{dy}{dx}$
Or,
$\frac{dy}{dx}=\frac{1}{f'(y)}$
But here $y = f^{-1}(x)$, therefore,
$\frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'[f^{-1}(x)]}$
Or,
$[f^{-1}]'(x)=\frac{1}{f'[f^{-1}(x)]}$
This formula is suitable for inverse trigonometric differentiation which calculates the derivative of inverse trig functions.
## How to calculate derivatives of inverse functions?
The implementation of implicit differentiation is divided into a few steps. These steps assist you to calculate the derivative of an implicit function. These steps are:
1. Write the expression of the function.
2. Identify the independent and dependent variables.
3. Write the inverse function as an explicit function so that it can be differentiated easily.
4. You can also simply use the inverse function differentiation formula.
5. Apply the derivative on both sides of the equation with respect to the independent variable.
6. For example if the left side of the equation contains a power function, use the power rule derivative formula.
7. In the final step, simplify the equation and rearrange it to get dy/dx.
Let’s understand the following examples to find derivative of inverse functions.
### Derivative of inverse function example 1
Find the derivative of inverse trigonometric function tan inverse by using inverse derivative formula.
To calculate the derivative of arctan, suppose that,
$y = \tan^{-1}x$
It can be written as,
$\tan y = x$
Now applying derivative on both sides,
$\frac{d}{dx}(\tan y) = \frac{d}{dx}(x)$
Since the derivative of tan is equal to sec square and the derivative of x is 1.
$\sec^2 y \frac{dy}{dx} = 1$
Rearranging the equation,
$\frac{dy}{dx} = \frac{1}{\sec^2y}$
Since $y = \tan^{-1}x$ then,
$\frac{d}{dx}(\tan^{-1}x) =\frac{1}{\sec^2(\tan^{-1}x)}$
Assume that tan-1x = θ then tan θ = x. Constructing a triangle with angle θ where adjacent side is 1, opposite side is x and then calculating hypotenuse by using Pythagorean Theorem which is √1+x2. Therefore,
$\sec^2(\tan^{-1}x)=\sec^2\theta =(\sqrt{1+x^2})^2 = 1+x^2$
Using the above value in the derivative of tan-1x.
$\frac{d}{dx}(\tan^{-1}x) = \frac{1}{1+x^2}$
### Derivative of inverse function example 2
Find the derivative of inverse trigonometric function sine inverse by using inverse derivative formula.
To calculate the derivative of arcsin, suppose that,
$y=\sin^{-1}x$
It can be written as,
$\sin y = x$
Now applying derivative on both sides,
$\frac{d}{dx}[\sin y] = \frac{d}{dx}(x)$
Since the derivative of sin is equal to cos square and the derivative of x is 1.
$\cos y\frac{dy}{dx}=1$
Rearranging the equation,
$\frac{dy}{dx}=\frac{1}{\cos y}$
Since $y = sin^{-1}x$ then,
$\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sin(\sin^{-1}x)}$
Assume that $\sin^{-1}x = \theta$ then $\sin\theta = x$. Constructing a triangle with angle θ where adjacent side is 1, opposite side is x and then calculating hypotenuse by using Pythagorean Theorem which is √1-x2. Therefore,
$\cos(\sin^{-1}x)=\cos\theta = \sqrt{1-x^2}$
Using the above value in the derivative of sin-1x.
$\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}$
## Calculating derivative of inverse functions by using calculator
The derivative of an inverse function can be also calculated by using an inverse function derivative calculator. It is an online tool that follows the inverse differentiation formula to find derivative. You can find it online by searching for a derivative calculator. For example, to calculate the derivative of cos inverse, the following steps are used by using this calculator.
1. Write the expression of the function in the input box such as, cos^-1 x.
2. Choose the variable to calculate the rate of change, which will be x in this example.
3. Review the input so that there will be no syntax error in the function.
4. Now at the last step, click on the calculate button. By using this step, the inverse function derivative calculator will provide the derivative of cos inverse quickly and accurately which will be -1/√1-x2.
## Comparison between derivative of inverse function and implicit differentiation
The comparison between the derivative of inverse function and implicit differentiation can be easily analysed using the following difference table.
Derivative of inverse function Implicit Differentiation The inverse function differentiation is a method to calculate derivatives of inverse functions. The implicit differentiation is used to calculate derivatives of an implicit function. The inverse derivative of a function is defined as;$[f^{-1}]’(x) = \frac{1}{f’[f^{-1}(x)]}$ There is no specific formula to calculate implicit derivatives. The inverse function derivative uses the relation between a function and its inverse to calculate the derivative. The implicit differentiation can also be used with derivative rules to find rate of change.
## Conclusion
The derivative of an inverse function is a fundamental way to calculate the derivative of a function. But to find an inverse derivative, a function should be both differentiable and invertible. The relation between a function and its inverse is used to find an inverse function derivative.
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# 2006 AMC 10B Problems/Problem 20
## Problem
In rectangle $ABCD$, we have $A=(6,-22)$, $B=(2006,178)$, $D=(8,y)$, for some integer $y$. What is the area of rectangle $ABCD$?
$\mathrm{(A) \ } 4000\qquad \mathrm{(B) \ } 4040\qquad \mathrm{(C) \ } 4400\qquad \mathrm{(D) \ } 40,000\qquad \mathrm{(E) \ } 40,400$
## Solution
### Solution 1
Let the slope of $AB$ be $m_1$ and the slope of $AD$ be $m_2$.
$m_1 = \frac{178-(-22)}{2006-6} = \frac{1}{10}$
$m_2 = \frac{y-(-22)}{8-6} = \frac{y+22}{2}$
Since $AB$ and $AD$ form a right angle:
$m_2 = -\frac{1}{m_1}$
$m_2 = -10$
$\frac{y+22}{2} = -10$
$y = -42$
Using the distance formula:
$AB = \sqrt{ (2006-6)^2 + (178-(-22))^2 } = \sqrt{ (2000)^2 + (200)^2 } = 200\sqrt{101}$
$AD = \sqrt{ (8-6)^2 + (-42-(-22))^2 } = \sqrt{ (2)^2 + (-20)^2 } = 2\sqrt{101}$
Therefore the area of rectangle $ABCD$ is $200\sqrt{101}\cdot2\sqrt{101} = 40,400 \Rightarrow E$
### Solution 2
This solution is the same as Solution 1 up to the point where we find that $y=-42$.
We build right triangles so we can use the Pythagorean Theorem. The triangle with hypotenuse $AB$ has legs $200$ and $2000$, while the triangle with hypotenuse $AD$ has legs $2$ and $20$. Aha! The two triangles are similar, with one triangle having side lengths $100$ times the other!
Let $AD=x$. Then from our reasoning above, we have $AB=100x$. Finally, the area of the rectangle is $100x(x)=100x^2=100(20^2+2^2)=100(400+4)=100(404)=\boxed{40400 \text{ (E)}}$.
## See Also
2006 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 19 Followed byProblem 21 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
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# What is Calculus?
## Primary tabs
Even if you never expect or need to take a course in Calculus, having a strong liberal education should include understanding some of the most basic concepts in science. The cornerstone of today's high tech, big data world is Calculus. It is the language and foundation of modern statistics, engineering, physics, chemistry, medicine, quantitative finance, and the list goes on. So if you have always felt that this subject was complicated then this article is here to answering in the simplest terms: What is Calculus?
Calculus comes in two flavors. The first is Differential Calculus and the second is Integral Calculus. Differential Calculus is finding the slope of curves (mathematicians call curves functions and denote them by $f(x)$) and Integral Calculus is finding the area under curves. Let's talk about Differential Calculus first.
Somewhere in high school you hopefully learned that the slope of a straight line is
$\mbox{Slope}=\frac{\mbox{Rise}}{\mbox{Run}}.$
Differential Calculus is about finding the slope of a function or curve that is not a straight line. We usually stop calling it a slope and start referring to it as a derivative and denote it by $f\,'(x)$ or $df/dx$. In other words,
$\mbox{Slope}=\mbox{Derivative}=f\,'(x)=\frac{df}{dx}.$
So let's explain why this gets complicated. To find the derivative (slope) in Calculus you have to determine the slope as both the rise and the run get smaller and smaller. In other words,
$\mbox{Derivative}=\frac{\mbox{Rise}\to 0}{\mbox{Run}\to 0}.$
If the rise approaches zero more quickly than the run, then the slope is zero. If the run approaches zero more quickly than the rise, then the slope becomes infinitely large. For most functions both the rise and the run approach zero in such a way as to produce some non-zero and non-infinite slope. How we do this is called function limits and is one of the first and most difficult concepts for beginning Calculus students.
Now let's explain Integral Calculus. When we were in Geometry class we had to learn the area of common shapes like rectangles, circles, squares, triangles, etc. In Integral Calculus we find the area underneath a curve. We refer to the area under a curve as an integral and denote it by
$\mbox{Integral}=\int\,f(x)dx.$
The way we do this in practice is to approximate the area under the curve by a bunch of rectangles. As we create rectangles of smaller and smaller width we need to have more and more of them to fill the area. In other words,
$\mbox{Integral}=\left(\mbox{Number of rectangles}\to\infty\right)\times\left(\mbox{Width of each rectangle}\to 0\right).$
As the number of rectangles becomes larger and larger the integral will become larger and larger. However, the rectangles have a smaller and smaller area because their width is getting smaller. The difficulty in Integral Calculus is figuring out whether the integral is infinite, zero or somewhere in between. For most of the functions that describe the real world, there is a sweet spot in this tug of war between infinity and zero that results in a real area that we can calculate.
Hopefully you now have a basic understanding of Calculus and don't need to take a class on it. However, if you do find yourself needing to know more, then we are here to help with lessons, problems and our function wizard to make learning Calculus easier.
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# 3.1 Use a problem-solving strategy (Page 6/8)
Page 6 / 8
According to the National Automobile Dealers Association, the average cost of a car in 2014 was $28,500. This was$1,500 less than 6 times the cost in 1975. What was the average cost of a car in 1975?
$5,000 U.S. Census data shows that the median price of new home in the United States in November 2014 was$280,900. This was $10,700 more than 14 times the price in November 1964. What was the median price of a new home in November 1964?$19,300
## Key concepts
• Problem-Solving Strategy
1. Read the problem. Make sure all the words and ideas are understood.
2. Identify what we are looking for.
3. Name what we are looking for. Choose a variable to represent that quantity.
4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.
• Consecutive Integers
Consecutive integers are integers that immediately follow each other.
$\begin{array}{cccc}\hfill n\hfill & & & {1}^{\text{st}}\phantom{\rule{0.2em}{0ex}}\text{integer}\hfill \\ \hfill n+1\hfill & & & {2}^{\text{nd}}\phantom{\rule{0.2em}{0ex}}\text{integer consecutive integer}\hfill \\ \hfill n+2\hfill & & & {3}^{\text{rd}}\phantom{\rule{0.2em}{0ex}}\text{consecutive integer . . . etc.}\hfill \end{array}$
Consecutive even integers are even integers that immediately follow one another.
$\begin{array}{cccc}\hfill n\hfill & & & {1}^{\text{st}}\phantom{\rule{0.2em}{0ex}}\text{integer}\hfill \\ \hfill n+2\hfill & & & {2}^{\text{nd}}\phantom{\rule{0.2em}{0ex}}\text{integer consecutive integer}\hfill \\ \hfill n+4\hfill & & & {3}^{\text{rd}}\phantom{\rule{0.2em}{0ex}}\text{consecutive integer . . . etc.}\hfill \end{array}$
Consecutive odd integers are odd integers that immediately follow one another.
$\begin{array}{cccc}\hfill n\hfill & & & {1}^{\text{st}}\phantom{\rule{0.2em}{0ex}}\text{integer}\hfill \\ \hfill n+2\hfill & & & {2}^{\text{nd}}\phantom{\rule{0.2em}{0ex}}\text{integer consecutive integer}\hfill \\ \hfill n+4\hfill & & & {3}^{\text{rd}}\phantom{\rule{0.2em}{0ex}}\text{consecutive integer . . . etc.}\hfill \end{array}$
## Practice makes perfect
Use the Approach Word Problems with a Positive Attitude
In the following exercises, prepare the lists described.
List five positive thoughts you can say to yourself that will help you approach word problems with a positive attitude. You may want to copy them on a sheet of paper and put it in the front of your notebook, where you can read them often.
List five negative thoughts that you have said to yourself in the past that will hinder your progress on word problems. You may want to write each one on a small piece of paper and rip it up to symbolically destroy the negative thoughts.
Use a Problem-Solving Strategy for Word Problems
In the following exercises, solve using the problem solving strategy for word problems. Remember to write a complete sentence to answer each question.
Two-thirds of the children in the fourth-grade class are girls. If there are 20 girls, what is the total number of children in the class?
30
Three-fifths of the members of the school choir are women. If there are 24 women, what is the total number of choir members?
Zachary has 25 country music CDs, which is one-fifth of his CD collection. How many CDs does Zachary have?
125
One-fourth of the candies in a bag of M&M’s are red. If there are 23 red candies, how many candies are in the bag?
There are 16 girls in a school club. The number of girls is four more than twice the number of boys. Find the number of boys.
6
There are 18 Cub Scouts in Pack 645. The number of scouts is three more than five times the number of adult leaders. Find the number of adult leaders.
In 10 years, the population of Detroit fell from 950,000 to about 712,500. Find the percent decrease.
how do i set this up
Jenise
25%
Melissa
25 percent
Muzamil
950,000 - 712,500 = 237,500. 237,500 / 950,000 = .25 = 25%
Melissa
I've tried several times it won't let me post the breakdown of how you get 25%.
Melissa
Subtract one from the other to get the difference. Then take that difference and divided by 950000 and you will get .25 aka 25%
Melissa
Finally 👍
Melissa
one way is to set as ratio: 100%/950000 = x% / 712500, which yields that 712500 is 75% of the initial 950000. therefore, the decrease is 25%.
bruce
twenty five percent...
Jeorge
thanks melissa
Jeorge
950000-713500 *100 and then divide by 950000 = 25
Muzamil
Jeannette has $5 and$10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
6t+3
Melissa
6t +3
Bollywood
Tricia got a 6% raise on her weekly salary. The raise was $30 per week. What was her original salary? Iris Reply let us suppose her original salary is 'm'. so, according to the given condition, m*(6/100)=30 m= (30*100)/6 m= 500 hence, her original salary is$500.
Simply
28.50
Toi
thanks
Jeorge
How many pounds of nuts selling for $6 per pound and raisins selling for$3 per pound should Kurt combine to obtain 120 pounds of trail mix that cost him $5 per pound? Valeria Reply Amber wants to put tiles on the backsplash of her kitchen counters. She will need 36 square feet of tiles. She will use basic tiles that cost$8 per square foot and decorator tiles that code $20 per square foot. How many square feet of each tile should she use so that the overal cost of he backsplash will be$10 per square foot?
I need help with maths can someone help me plz.. is there a wats app group?
WY need
Fernando
How did you get $750? Laura Reply if y= 2x+sinx what is dy÷dx formon25 Reply does it teach you how to do algebra if you don't know how Kate Reply Liam borrowed a total of$35,000 to pay for college. He pays his parents 3% interest on the $8,000 he borrowed from them and pays the bank 6.8% on the rest. What average interest rate does he pay on the total$35,000? (Round your answer to the nearest tenth of a percent.)
exact definition of length by bilbao
the definition of length
literal meaning of length
francemichael
exact meaning of length
francemichael
exact meaning of length
francemichael
how many typos can we find...?
5
Joseph
In the LCM Prime Factors exercises, the LCM of 28 and 40 is 280. Not 420!
4x+7y=29,x+3y=11 substitute method of linear equation
substitute method of linear equation
Srinu
Solve one equation for one variable. Using the 2nd equation, x=11-3y. Substitute that for x in first equation. this will find y. then use the value for y to find the value for x.
bruce
I want to learn
Elizebeth
help
Elizebeth
I want to learn. Please teach me?
Wayne
1) Use any equation, and solve for any of the variables. Since the coefficient of x (the number in front of the x) in the second equation is 1 (it actually isn't shown, but 1 * x = x), use that equation. Subtract 3y from both sides (this isolates the x on the left side of the equal sign).
bruce
2) This results in x=11-3y. x is note in terms of y. Use that as the value of x and substitute for all x in the first equation. The first equation becomes 4(11-3y)+7y =29. Note that the only variable left in the first equation is the y. If you have multiple variable, then something is wrong.
bruce
3) Distribute (multiply) the 4 across 11-3y to get 44-12y. Add this to the 7y. So, the equation is now 44-5y=29.
bruce
4) Solve 44-5y=29 for y. Isolate the y by subtracting 44 from birth sides, resulting in -5y=-15. Now, divide birth sides by -5 (since you have -5y). This results in y=3. You now have the value of one variable.
bruce
5) The last step is to take the value of y from Step 4) and substitute into the 2nd equation. Therefore: x+3y=11 becomes x+3(3)=11. Then multiplying, x+9=11. Finally, solve for x by subtracting 9 from both sides. Therefore, x=2.
bruce
6) The ordered pair of (2, 3) is the proposed solution. To check, substitute those values into either equation. If the result is true, then the solution is correct. 4(2)+7(3)=8+21=29. TRUE! Finished.
bruce
|
Algebra II : Adding and Subtracting Logarithms
Example Questions
← Previous 1 3 4 5
Example Question #1 : Adding And Subtracting Logarithms
Simplify the following logarithmic expression:
Possible Answers:
Correct answer:
Explanation:
Each term can be simplified as follows:
Combining these gives the answer:
Example Question #2 : Adding And Subtracting Logarithms
Simplify the expression using logarithmic identities.
Possible Answers:
Cannot be simplified
Correct answer:
Explanation:
The logarithm of a fraction is equal to the logarithm of the numerator minus the logarithm of the denominator.
If we encounter two logarithms with the same base, we can likely combine them. In this case, we can use the reverse of the above identity.
Example Question #3 : Adding And Subtracting Logarithms
Use logarithmic properties to simplify this expression:
Possible Answers:
Correct answer:
Explanation:
Use the sum/product rule to combine the first 2 terms:
Use the difference/quotient rule to combine the remaining terms:
Example Question #4 : Adding And Subtracting Logarithms
Expand the following logarithmic expression into a list of sums or subtractions of logarithms:
Possible Answers:
Correct answer:
Explanation:
One important property of logarithms is that multiplication inside the logarithm is the same thing as addition outside of it. In the same way division is "the same" as subtraction in logarithms. So our expression is the same as
But also, exponents can be moved outside in the same way. is basically , so . This can be reduced even further to our final answer:
Example Question #5 : Adding And Subtracting Logarithms
What is the value of ?
Possible Answers:
Correct answer:
Explanation:
Remember the rules of logarithms:
This means we can simplify it as follows:
The logarithm of anything with the same base is always , so the correct answer is .
Example Question #6 : Adding And Subtracting Logarithms
Which of the following is another way to express
?
Possible Answers:
Correct answer:
Explanation:
Use the rule
therefore
Example Question #7 : Adding And Subtracting Logarithms
Which is another way of expressing
?
Possible Answers:
Correct answer:
Explanation:
Use the rule:
therefore
Example Question #8 : Adding And Subtracting Logarithms
Simplify
Possible Answers:
Correct answer:
Explanation:
This problem can be solved using the properties of logs. When two logs are being subtracted from each other, it is the same thing as dividing two logs together. Remember that to use this rule, the logs must have the same base in this case .
Example Question #9 : Adding And Subtracting Logarithms
Condense this logarithm:
Possible Answers:
Correct answer:
Explanation:
In order to solve this problem you must understand the product property of logarithms and the power property of logarithms . Note that these apply to logs of all bases not just base 10.
first move the constants in front of the logarithmic functions to their proper place using the power rule.
next factor out the logarithmic equation:
change the fractional exponent to a radical
Example Question #10 : Adding And Subtracting Logarithms
Expand
Possible Answers:
Correct answer:
Explanation:
The rule for expanding and dividing logarithms is that you can subtract the terms inside the log. In this case, the question is not asking for an actual number, but just what the expanded version would be. Therefore you separate the terms inside the log by subtracting the denominator from the numerator. Therefore the answer is
← Previous 1 3 4 5
|
# A gas cylinder contains 2.0 mol of gas X and 6.0 mol of gas Y at a total pressure of 2.1 atm. What is is partial pressure of gas Y?
Jul 6, 2017
${P}_{\text{Y" = "1.58 atm}}$
#### Explanation:
Okay, so here is what is going on.
You have some gas cylinder which contains a mixture of 2 different gases: $\text{Gas X" and "gas Y}$. Each gas is exerting its own pressure which is contributing to the total pressure. You are asked to figure out the partial pressure of gas Y. Here is how you go about doing that.
$- - - - - - - - - - - - - - - - - - - - -$
Write out our givens.
Given
• $\text{2.0 moles of gas X}$
• $\text{6.0 moles of gas Y}$
• $\text{Total pressure = 2.1 atm}$
To find the partial pressure of $\text{gas Y}$, you need to use the following formula:
color(white)(aaaaaaaaaaaaaa)color(blue)[P_"Y" = P_"T" * x_"Y"
Where
• $\text{P"_"Y" = "partial pressure of gas Y}$
• $\text{P"_"T" = "total pressure}$
• $\text{x"_"Y" = "mole fraction of gas Y}$
We were given the total pressure of the system, but we need to figure out the mole fraction of gas Y. To calculate it, we use the following equation:
$\textcolor{w h i t e}{a a a a a a a a a a} \textcolor{b l u e}{{x}_{\text{Y" = ("moles of gas Y")/("total moles of gas}}}$
$\textcolor{w h i t e}{a a a a a a a a a a a a a a}$
Plugin and solve for the mole fraction of gas Y
• $\textcolor{b l u e}{{x}_{\text{Y" = ("moles of gas Y")/("total moles of gas}}}$
$\textcolor{w h i t e}{-}$
• $\rightarrow \left(\text{6.0 moles of gas Y")/("2.0 moles of gas X + 6.0 moles of gas Y}\right)$
• $\rightarrow \left(\text{6" cancel"moles")/("8" cancel"moles}\right)$
• $\rightarrow \textcolor{red}{\text{ 0.75}}$
Note: moles over moles cancel out so the mole fraction is unitless
$\textcolor{w h i t e}{a a a a a a a a a a a a a a}$
You now can use the mole fraction to solve for the partial pressure of gas Y by using the first equation which was given out.
• color(blue)[P_"Y" = P_"T" * x_"Y"
• ${P}_{\text{Y" = "(2.1 atm)" * "(0.75)}}$
• ${P}_{\text{Y" = color(magenta)"1.58 atm}}$
Answer: ${P}_{\text{Y" = color(magenta)"1.58 atm}}$
|
# Permutations and Combinations - Formulas, Problems and Video Tutorials
Permutations and Combinations are arrangements or selections of objects out of groups. For example, selecting a team of 11 players out of 20 players.
## Permutations
The arrangements made by taking some or all elements out of a group in a particular manner are called permutations.
For example, in how many way can the letters word ENGLISH be arranged, so that vowels never come together?
The number of permutation of n thing taking r at a time is denoted by $^{n} P_{r}$ and it is defined under :
$^{n} P_{r}\: =\: \frac{n!}{(n-r)!}\: ;$ r ≤ n.
Important notations
n! (Read as n factorial)
Product of first n positive integers is called n factorial.
n! = 1, 2, 3, 4, 5…….n
n! = (n-1)! N
A special case 0 ! = 1
Various Types of Permutations
Case 1
When in a permutation of n thing taken r at a time, a particular thing always occurs.
The required numbers of permutations = $r(^{n-1}P_{r-1})$
Case 2 The number of permutations of n different things taken r at a time, when a particular thing is never taken in each arrangement is $^{n-1}P_{r}$
Case 3 (Permutation of like things) The number of n things taken all at a time, given that $p_{1}$ things are $1^{st}$ alike, $p_{2}$ things are $2^{nd}$ alike, and $p_{r}$ things are $r^{th}$ alike is
$\frac{n!}{p_{1}!p_{2}!......p_{r}!}$
Case 4 (Permutation with repetitions) The number of n different things taken r at a time when each may be repeated any number of times in each arrangements is n.
Case 5 (Circular permutations) Circular permutations are the permutations of things along the circumstance of a circle. We have to consider the relative position of the different things in a circular arrangement.
For example, if there are five letters P, Q, R, S and T, two of the arrangements would be PQRST, TPQRS. These two arrangements are obviously different if the things are to be placed in a straight line. But if the arrangements are written along the circumstance of a circle, then the two arrangements PQRST and TPQRS are one and the same.
As the number of circular permutations depends on the relative position of the objects, we fix the position of one object and then arrange the remaining (n-1)! Ways. Thus the circular arrangements of five letters P, Q, R, S, T will be
(5-1)! = 4! = 4.3.2.1= 24 ways.
Some important result of permutations
1. $np_{n-1}\:=\: np_{n}$
2. $np_{n}\:=\: n$ !
3. $np_{r}\:=\:n\left (n-1 p_{r-1} \right )$
4. $np_{r}\:=\:\left ( n-r+1 \right )\times\: n p_{r-1}$
5. $np_{r}\:=\:n-1p_{r}+ r \left(n-1p _{r-1} \right )$
## Combinations
The groups or selections made by taking some or all elements out of a number of things are called combinations.
For example, find the number of different poker hands in pack 52 playing cards. The number of combinations of n thing r at a time is denoted by image and it is defined as under
$nc_{r}\: =\:\frac{n!}{r!\left ( n-r \right )!}=$
Types of combinations
Case 1
To find the number of ways selecting of ways selecting one or more items out of n given items is
$nc_{1}+nc_{2}+nc_{3}+.....+nc_{n}\: =\: 2^{n}-1$
Case 2 The number of combinations of n items taken r at a time in which given p particular items will always occur is
$n-pc_{r-p}$
Case 3 The number of combinations of n items taken r at a time in which p particular items never occur is image
Some important results of combinations
$n-pc_{r}$
Important result of combination :
1. $nc_{0}\:=\:1\:=\: nc_{n}$
2. $np_{r}\:=\: r!(nc_{r})$
3. $nc_{r}+nc_{r-1}\:=\: n+1c_{r}$
4. $\frac{nc_{r}}{n-r+1}\:=\: \frac{nc_{n-r}}{r}$
5. $n\times n-1c_{r-1}\:=\:(n-r+1)\times nc _{r-1}$
## Permutations and Combinations Examples
Question-1: If $nc_{10}\:=\: nc_{14}$ find the value of n
Solution: $nc_{10}\:=\: nc_{14}\: \Rightarrow n\: =\: (10+14)\: =\: 24$
Question-2: if $nc_{3}\: =\: 220$ then find the value of n.
Solution: $\therefore \frac{n!}{(n-3)!3!}\: =\: 220$
$\frac{n(n-1)(n-2)(n-3)!}{(n-3)!6}\: =\: 220$
n(n-1)(n-2) = 1320 or n(n-1)(n-2) = $11\times 12\times 21$ or n= 12
Question-3: if $nc_{r}+nc_{n+1}\: =\:n+1 c_{x}$ then find the value of x.
Solution: $nc_{r}+nc_{n+1}\: =\:n+1 c_{r+1}$, formula
$nc_{r}+nc_{n+1}\: =\:n+1 c_{x}$ given
by comparing above two statement we can get , x = r+1
Question-4: In how many ways can letter of the word 'APPLE' be arranged?
Solution: there is in all 5 letter .out of these two are p,one is A one is L and one is E.
$\frac{5!}{2!.1!.1!.1!}\:=\: 60$
## Permutations and Combinations Questions from Previous Year Exams
This part will cover the Permutations and Combinations Syllabus of Bank Clerk Exam
## Permutations and Combinations Video Tutorial
Please comment on Permutations and Combinations - Formulas, Problems and Video Tutorials
|
Statistics
# 13.3Facts About the F Distribution
Statistics13.3 Facts About the F Distribution
The following are facts about the F distribution:
• The curve is not symmetrical but skewed to the right.
• There is a different curve for each set of dfs.
• The F statistic is greater than or equal to zero.
• As the degrees of freedom for the numerator and for the denominator get larger, the curve approximates the normal.
• Other uses for the F distribution include comparing two variances and two-way analysis of variance. Two-way analysis is beyond the scope of this chapter.
Figure 13.3
## Example 13.2
### Problem
Let’s return to the slicing tomato exercise in Try It 13.1. The means of the tomato yields under the five mulching conditions are represented by μ1, μ2, μ3, μ4, μ5. We will conduct a hypothesis test to determine if all means are the same or at least one is different. Using a significance level of 5 percent, test the null hypothesis that there is no difference in mean yields among the five groups against the alternative hypothesis that at least one mean is different from the rest.
## Try It 13.2
MRSA, or Staphylococcus aureus, can cause serious bacterial infections in hospital patients. Table 13.6 shows various colony counts from different patients who may or may not have MRSA. The data from the table is plotted in Figure 13.5.
Conc = 0.6Conc = 0.8 Conc = 1.0 Conc = 1.2 Conc = 1.4
9 16 22 30 27
66 93 147 199 168
98 82 120 148 132
Table 13.6
Plot of the data for the different concentrations:
Figure 13.5
Test whether the mean numbers of colonies are the same or are different. Construct the ANOVA table by hand or by using a TI-83, 83+, or 84+ calculator, find the p-value, and state your conclusion. Use a 5 percent significance level.
## Example 13.3
Four sororities took a random sample of sisters regarding their grade means for the past term. The results are shown in Table 13.7.
Sorority 1 Sorority 2 Sorority 3 Sorority 4
2.17 2.63 2.63 3.79
1.85 1.77 3.78 3.45
2.83 3.25 4.00 3.08
1.69 1.86 2.55 2.26
3.33 2.21 2.45 3.18
Table 13.7 Mean Grades for Four Sororities
### Problem
Using a significance level of 1 percent, is there a difference in mean grades among the sororities?
## Try It 13.3
Four sports teams took a random sample of players regarding their GPAs for the last year. The results are shown in Table 13.8.
Basketball Baseball Hockey Lacrosse
3.6 2.1 4.0 2.0
2.9 2.6 2.0 3.6
2.5 3.9 2.6 3.9
3.3 3.1 3.2 2.7
3.8 3.4 3.2 2.5
Table 13.8 GPAs for four sports teams
Use a significance level of 5 percent and determine if there is a difference in GPA among the teams.
## Example 13.4
A fourth-grade class is studying the environment. One of the assignments is to grow bean plants in different soils. Tommy chose to grow his bean plants in soil found outside his classroom mixed with dryer lint. Tara chose to grow her bean plants in potting soil bought at the local nursery. Nick chose to grow his bean plants in soil from his mother’s garden. No chemicals were used on the plants, only water. They were grown inside the classroom next to a large window. Each child grew five plants. At the end of the growing period, each plant was measured, producing the data in inches in Table 13.9.
Tommy’s Plants Tara’s Plants Nick’s Plants
24 25 23
21 31 27
23 23 22
30 20 30
23 28 20
Table 13.9
### Problem
Does it appear that the three soils in which the bean plants were grown produce the same mean height? Test at a 3 percent level of significance.
## Try It 13.4
Another fourth grader also grew bean plants, but in a jelly-like mass. The heights were (in inches) 24, 28, 25, 30, and 32. Do a one-way ANOVA test on the four groups. Are the heights of the bean plants different? Use the same method as shown in Example 13.4.
## Collaborative Exercise
From the class, create four groups of the same size as follows: men under 22, men at least 22, women under 22, women at least 22. Have each member of each group record the number of states in the United States he or she has visited. Run an ANOVA test to determine if the average number of states visited in the four groups are the same. Test at a 1 percent level of significance. Use one of the solution sheets in Appendix E.
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|
Dividing Numbers by A One-Digit Number
# Dividing Numbers by A One-Digit Number Video Lecture | Mental Maths - Class 1
## Mental Maths
30 videos|62 docs|10 tests
## FAQs on Dividing Numbers by A One-Digit Number Video Lecture - Mental Maths - Class 1
1. How do I divide a number by a one-digit number?
Ans. To divide a number by a one-digit number, you can use the long division method. First, write the dividend (the number being divided) on the top and the divisor (the one-digit number) on the outside. Then, proceed to divide as you would with any long division problem, finding the quotient and remainder if necessary.
2. What is the easiest way to divide numbers by a one-digit number?
Ans. One of the easiest ways to divide numbers by a one-digit number is to use mental math. By memorizing the multiplication table and understanding basic division facts, you can quickly determine the quotient without going through the long division process. For example, if you know that 8 divided by 2 is 4, you can easily calculate 80 divided by 2 as 40.
3. Can a one-digit number evenly divide any number?
Ans. No, a one-digit number may not always evenly divide any given number. Whether a one-digit number can evenly divide a number depends on the divisibility rules. For example, if the one-digit number is 2, it can evenly divide any even number. However, if the one-digit number is 3, it can only evenly divide a number if the sum of its digits is divisible by 3.
4. What is the significance of dividing numbers by a one-digit number?
Ans. Dividing numbers by a one-digit number is a fundamental arithmetic operation that helps us understand and solve real-life problems involving sharing, equal distribution, ratios, rates, and proportions. It allows us to divide quantities into smaller parts to determine fair shares or calculate rates of change, among other applications.
5. Are there any shortcuts or tricks to divide numbers by a one-digit number?
Ans. Yes, there are some shortcuts or tricks that can simplify the division process. For example, if the divisor is 9, you can quickly find the quotient by subtracting the sum of the digits in the dividend from 9. Additionally, if the divisor is 5, you can simply divide the last digit of the dividend by 2 and add a decimal point to get the quotient. These shortcuts can save time and mental effort when dividing numbers by a one-digit number.
## Mental Maths
30 videos|62 docs|10 tests
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# How do you simplify ((2/3)^(2/3))^6?
Mar 6, 2017
See the entire simplification process below:
#### Explanation:
First, use this rule of exponents to simplify the exponents on the out parenthesis: ${\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} = {x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}}$
${\left({\left(\frac{2}{3}\right)}^{\textcolor{red}{\frac{2}{3}}}\right)}^{\textcolor{b l u e}{6}} = {\left(\frac{2}{3}\right)}^{\textcolor{red}{\frac{2}{3}} \times \textcolor{b l u e}{6}} = {\left(\frac{2}{3}\right)}^{4}$
We can now complete the simplification as follows:
${\left(\frac{2}{3}\right)}^{4} = {2}^{4} / {3}^{4} = \frac{16}{81}$
|
# Orthogonal Bases and the QR Algorithm
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## Transcription
1 Orthogonal Bases and the QR Algorithm Orthogonal Bases by Peter J Olver University of Minnesota Throughout, we work in the Euclidean vector space V = R n, the space of column vectors with n real entries As inner product, we will only use the dot product v w = v T w and corresponding Euclidean norm v = v v Two vectors v,w V are called orthogonal if their inner product vanishes: v w = In the case of vectors in Euclidean space, orthogonality under the dot product means that they meet at a right angle A particularly important configuration is when V admits a basis consisting of mutually orthogonal elements Definition A basis u,,u n of V is called orthogonal if u i u j = for all i j The basis is called orthonormal if, in addition, each vector has unit length: u i =, for all i =,, n The simplest example of an orthonormal basis is the standard basis e =, e =, e n = Orthogonality follows because e i e j =, for i j, while e i = implies normality Since a basis cannot contain the zero vector, there is an easy way to convert an orthogonal basis to an orthonormal basis Namely, we replace each basis vector with a unit vector pointing in the same direction Lemma If v,,v n is an orthogonal basis of a vector space V, then the normalized vectors u i = v i / v i, i =,, n, form an orthonormal basis Example The vectors v =, v =, v =, are easily seen to form a basis of R Moreover, they are mutually perpendicular, v v = v v = v v =, and so form an orthogonal basis with respect to the standard dot 6// c Peter J Olver
2 product on R When we divide each orthogonal basis vector by its length, the result is the orthonormal basis u = 6 = 6 6, u = =, u = =, 6 satisfying u u = u u = u u = and u = u = u = The appearance of square roots in the elements of an orthonormal basis is fairly typical A useful observation is that any orthogonal collection of nonzero vectors is automatically linearly independent Proposition 4 If v,,v k V are nonzero, mutually orthogonal elements, so v i and v i v j = for all i j, then they are linearly independent Proof : Suppose c v + + c k v k = Let us take the inner product of this equation with any v i Using linearity of the inner product and orthogonality, we compute = c v + + c k v k v i = c v v i + + c k v k v i = c i v i v i = c i v i Therefore, provided v i, we conclude that the coefficient c i = Since this holds for all i =,, k, the linear independence of v,,v k follows QED As a direct corollary, we infer that any collection of nonzero orthogonal vectors forms a basis for its span Theorem Suppose v,,v n V are nonzero, mutually orthogonal elements of an inner product space V Then v,,v n form an orthogonal basis for their span W = span {v,,v n } V, which is therefore a subspace of dimension n = dim W In particular, if dim V = n, then v,,v n form a orthogonal basis for V Computations in Orthogonal Bases What are the advantages of orthogonal and orthonormal bases? Once one has a basis of a vector space, a key issue is how to express other elements as linear combinations of the basis elements that is, to find their coordinates in the prescribed basis In general, this is not so easy, since it requires solving a system of linear equations In high dimensional situations arising in applications, computing the solution may require a considerable, if not infeasible amount of time and effort However, if the basis is orthogonal, or, even better, orthonormal, then the change of basis computation requires almost no work This is the crucial insight underlying the efficacy of both discrete and continuous Fourier analysis, least squares approximations and statistical analysis of large data sets, signal, image and video processing, and a multitude of other applications, both classical and modern 6// c Peter J Olver
3 Theorem 6 Let u,,u n be an orthonormal basis for an inner product space V Then one can write any element v V as a linear combination in which its coordinates v = c u + + c n u n, () c i = v u i, i =,, n, () are explicitly given as inner products Moreover, its norm v = c + + c n = n v u i is the square root of the sum of the squares of its orthonormal basis coordinates i= () Proof : Let us compute the inner product of () with one of the basis vectors Using the orthonormality conditions { i j, u i u j = (4) i = j, and bilinearity of the inner product, we find n v u i = c j u j ; u i = j= To prove formula (), we similarly expand v = v v = n i,j= n j= c j u j u i = c i u i = c i c i c j u i u j = again making use of the orthonormality of the basis elements n i= c i, QED Example 7 Let us rewrite the vector v = (,, ) T in terms of the orthonormal basis u =, u =, u =, constructed in Example Computing the dot products we immediately conclude that v u = 6, v u =, v u = 4, v = 6 u + u + 4 u Needless to say, a direct computation based on solving the associated linear system is more tedious 6// c Peter J Olver
4 While passage from an orthogonal basis to its orthonormal version is elementary one simply divides each basis element by its norm we shall often find it more convenient to work directly with the unnormalized version The next result provides the corresponding formula expressing a vector in terms of an orthogonal, but not necessarily orthonormal basis The proof proceeds exactly as in the orthonormal case, and details are left to the reader Theorem 8 If v,,v n form an orthogonal basis, then the corresponding coordinates of a vector v = a v + + a n v n are given by a i = v v i v i () In this case, its norm can be computed using the formula n n ( ) v v = a i v vi i = (6) v i i= Equation (), along with its orthonormal simplification (), is one of the most useful formulas we shall establish, and applications will appear repeatedly throughout the sequel Example 9 The wavelet basis v =, v =, v =, v 4 =, (7) is an orthogonal basis of R 4 The norms are i= v =, v =, v =, v 4 = Therefore, using (), we can readily express any vector as a linear combination of the wavelet basis vectors For example, 4 v = = v v + v v 4, where the wavelet coordinates are computed directly by v v v = 8 4 =, v v v = 4 4 =, v v v = 6 = v v 4 v 4 = 4 = Finally, we note that 46 = v = v +( ) v + v +( ) v 4 = , in conformity with (6) 6// 4 c Peter J Olver
5 The Gram Schmidt Process Once we become convinced of the utility of orthogonal and orthonormal bases, a natural question arises: How can we construct them? A practical algorithm was first discovered by Pierre Simon Laplace in the eighteenth century Today the algorithm is known as the Gram Schmidt process, after its rediscovery by the nineteenth century mathematicians Jorgen Gram and Erhard Schmidt The Gram Schmidt process is one of the premier algorithms of applied and computational linear algebra We assume that we already know some basis w,,w n of V, where n = dim V Our goal is to use this information to construct an orthogonal basis v,,v n We will construct the orthogonal basis elements one by one Since initially we are not worrying about normality, there are no conditions on the first orthogonal basis element v, and so there is no harm in choosing v = w Note that v, since w appears in the original basis The second basis vector must be orthogonal to the first: v v = Let us try to arrange this by subtracting a suitable multiple of v, and set v = w cv, where c is a scalar to be determined The orthogonality condition = v v = w v cv v = w v c v requires that c = w v / v, and therefore v = w w v v v () Linear independence of v = w and w ensures that v (Check!) Next, we construct v = w c v c v by subtracting suitable multiples of the first two orthogonal basis elements from w We want v to be orthogonal to both v and v Since we already arranged that v v =, this requires and hence = v v = w v c v v, = v v = w v c v v, c = w v v, c = w v v Therefore, the next orthogonal basis vector is given by the formula v = w w v v v w v v v Since v and v are linear combinations of w and w, we must have v, as otherwise this would imply that w,w,w are linearly dependent, and hence could not come from a basis 6// c Peter J Olver
6 Continuing in the same manner, suppose we have already constructed the mutually orthogonal vectors v,,v k as linear combinations of w,,w k The next orthogonal basis element v k will be obtained from w k by subtracting off a suitable linear combination of the previous orthogonal basis elements: v k = w k c v c k v k Since v,,v k are already orthogonal, the orthogonality constraint requires = v k v j = w k v j c j v j v j c j = w k v j v j for j =,, k () In this fashion, we establish the general Gram Schmidt formula v k = w k k j= w k v j v j v j, k =,, n () The Gram Schmidt process () defines an explicit, recursive procedure for constructing the orthogonal basis vectors v,,v n If we are actually after an orthonormal basis u,,u n, we merely normalize the resulting orthogonal basis vectors, setting u k = v k / v k for each k =,, n Example The vectors w =, w =, w =, (4) are readily seen to form a basis of R To construct an orthogonal basis (with respect to the standard dot product) using the Gram Schmidt procedure, we begin by setting The next basis vector is v = w = v = w w v v v = = 4 This will, in fact, be a consequence of the successful completion of the Gram Schmidt process and does not need to be checked in advance If the given vectors were not linearly independent, then eventually one of the Gram Schmidt vectors would vanish, v k =, and the process will break down 6// 6 c Peter J Olver
7 The last orthogonal basis vector is v = w w v v v w v v v = = The reader can easily validate the orthogonality of v,v,v An orthonormal basis is obtained by dividing each vector by its length Since v = 4 7, v =, v = we produce the corresponding orthonormal basis vectors u =, u = , u = () Modifications of the Gram Schmidt Process With the basic Gram Schmidt algorithm now in hand, it is worth looking at a couple of reformulations that have both practical and theoretical advantages The first can be used to directly construct the orthonormal basis vectors u,,u n from the basis w,,w n We begin by replacing each orthogonal basis vector in the basic Gram Schmidt formula () by its normalized version u j = v j / v j The original basis vectors can be expressed in terms of the orthonormal basis via a triangular system w = r u, w = r u + r u, w = r u + r u + r u, w n = r n u + r n u + + r nn u n (6) The coefficients r ij can, in fact, be computed directly from these formulae Indeed, taking the inner product of the equation for w j with the orthonormal basis vector u i for i j, we find, in view of the orthonormality constraints (4), w j u i = r j u + + r jj u j u i = r j u u i + + r jj u n u i = r ij, and hence r ij = w j u i (7) On the other hand, according to (), w j = r j u + + r jj u j = r j + + r j,j + r jj (8) 6// 7 c Peter J Olver
8 The pair of equations (7 8) can be rearranged to devise a recursive procedure to compute the orthonormal basis At stage j, we assume that we have already constructed u,,u j We then compute r ij = w j u i, for each i =,, j (9) We obtain the next orthonormal basis vector u j by computing r jj = w j r j r j,j, u j = w j r j u r j,j u j r jj () Running through the formulae (9 ) for j =,, n leads to the same orthonormal basis u,,u n as the previous version of the Gram Schmidt procedure Example Let us apply the revised algorithm to the vectors w =, w =, w =, of Example To begin, we set r = w =, u = w = r The next step is to compute r = w u = 4, r = w r =, u = w r u = r The final step yields r = r = w u =, r = w u =, w r r = 7, u = w r u r u r = As advertised, the result is the same orthonormal basis vectors that we found in Example When j =, there is nothing to do 6// 8 c Peter J Olver
9 For hand computations, the original version () of the Gram Schmidt process is slightly easier even if one does ultimately want an orthonormal basis since it avoids the square roots that are ubiquitous in the orthonormal version (9 ) On the other hand, for numerical implementation on a computer, the orthonormal version is a bit faster, as it involves fewer arithmetic operations However, in practical, large scale computations, both versions of the Gram Schmidt process suffer from a serious flaw They are subject to numerical instabilities, and so accumulating round-off errors may seriously corrupt the computations, leading to inaccurate, non-orthogonal vectors Fortunately, there is a simple rearrangement of the calculation that obviates this difficulty and leads to the numerically robust algorithm that is most often used in practice The idea is to treat the vectors simultaneously rather than sequentially, making full use of the orthonormal basis vectors as they arise More specifically, the algorithm begins as before we take u = w / w We then subtract off the appropriate multiples of u from all of the remaining basis vectors so as to arrange their orthogonality to u This is accomplished by setting w () k = w k w k u u for k =,, n The second orthonormal basis vector u = w () / w() is then obtained by normalizing We next modify the remaining w (),,w() n to produce vectors w () k = w () k w () k u u, k =,, n, that are orthogonal to both u and u Then u = w () / w() is the next orthonormal basis element, and the process continues The full algorithm starts with the initial basis vectors w j = w () j, j =,, n, and then recursively computes u j = w(j) j w (j) j, w(j+) k = w (j) k w (j) k u j u j, j =,, n, k = j +,, n () (In the final phase, when j = n, the second formula is no longer needed) The result is a numerically stable computation of the same orthonormal basis vectors u,,u n Example Let us apply the stable Gram Schmidt process () to the basis vectors w () = w =, w () = w = The first orthonormal basis vector is u = w () = w () w () u u = 4 w() w () =, w () = w = Next, we compute, w () = w () w () u u = 6// 9 c Peter J Olver
10 The second orthonormal basis vector is u = w () = w () w () u u = w() w () =, u = w() w () = The resulting vectors u,u,u form the desired orthonormal basis Orthogonal Matrices Finally, 6 6 Matrices whose columns form an orthonormal basis of R n relative to the standard Euclidean dot product play a distinguished role Such orthogonal matrices appear in a wide range of applications in geometry, physics, quantum mechanics, crystallography, partial differential equations, symmetry theory, and special functions Rotational motions of bodies in three-dimensional space are described by orthogonal matrices, and hence they lie at the foundations of rigid body mechanics, including satellite and underwater vehicle motions, as well as three-dimensional computer graphics and animation Furthermore, orthogonal matrices are an essential ingredient in one of the most important methods of numerical linear algebra: the QR algorithm for computing eigenvalues of matrices Definition A square matrix Q is called an orthogonal matrix if it satisfies Q T Q = I () The orthogonality condition implies that one can easily invert an orthogonal matrix: Q = Q T () In fact, the two conditions are equivalent, and hence a matrix is orthogonal if and only if its inverse is equal to its transpose The second important characterization of orthogonal matrices relates them directly to orthonormal bases Proposition A matrix Q is orthogonal if and only if its columns form an orthonormal basis with respect to the Euclidean dot product on R n Proof : Let u,,u n be the columns of Q Then u T,,uT n are the rows of the transposed matrix Q T The (i, j) entry of the product Q T Q is given as the product of the i th row of Q T times{ the j th column of Q Thus, the orthogonality requirement () implies, i = j, u i u j = u T i u j = which are precisely the conditions (4) for u, i j,,,u n to form an orthonormal basis QED Warning: Technically, we should be referring to an orthonormal matrix, not an orthogonal matrix But the terminology is so standard throughout mathematics that we have no choice but to adopt it here There is no commonly accepted name for a matrix whose columns form an orthogonal but not orthonormal basis 6// c Peter J Olver
11 u u u θ θ u u = Example A matrix Q = ( ) ( a b,u c = d Figure Orthonormal Bases in R ( ) a b is orthogonal if and only if its columns c d ), form an orthonormal basis of R Equivalently, the requirement ( ) ( ) Q T a c a b Q = = b d c d ( ) a + c ab + cd ab + cd b + d = implies that its entries must satisfy the algebraic equations a + c =, ab + cd =, b + d = ( ), The first and last equations say that the points (a, c ) T and (b, d ) T lie on the unit circle in R, and so a = cos θ, c = sin θ, b = cos ψ, d = sin ψ, for some choice of angles θ, ψ The remaining orthogonality condition is = ab + cd = cos θ cos ψ + sin θ sin ψ = cos(θ ψ), which implies that θ and ψ differ by a right angle: ψ = θ ± π The ± sign leads to two cases: b = sin θ, d = cos θ, or b = sin θ, d = cos θ As a result, every orthogonal matrix has one of two possible forms ( ) ( ) cos θ sin θ cos θ sin θ or, where θ < π () sin θ cos θ sin θ cos θ The corresponding orthonormal bases are illustrated in Figure The former is a right handed basis, and can be obtained from the standard basis e,e by a rotation through angle θ, while the latter has the opposite, reflected orientation 6// c Peter J Olver
12 Example 4 A orthogonal matrix Q = (u u u ) is prescribed by mutually perpendicular vectors of unit length in R For instance, the orthonormal basis constructed in () corresponds to the orthogonal matrix Q = Lemma An orthogonal matrix has determinant det Q = ± Proof : Taking the determinant of (), = det I = det(q T Q) = det Q T det Q = (det Q), which immediately proves the lemma QED An orthogonal matrix is called proper or special if it has determinant + Geometrically, the columns of a proper orthogonal matrix form a right handed basis of R n An improper orthogonal matrix, with determinant, corresponds to a left handed basis that lives in a mirror image world Proposition 6 The product of two orthogonal matrices is also orthogonal Proof : If Q T Q = I = QT Q, then (Q Q )T (Q Q ) = Q T QT Q Q = QT Q = I, and so the product matrix Q Q is also orthogonal QED This property says that the set of all orthogonal matrices forms a group The orthogonal group lies at the foundation of everyday Euclidean geometry, as well as rigid body mechanics, atomic structure and chemistry, computer graphics and animation, and many other areas 4 Eigenvalues of Symmetric Matrices Symmetric matrices play an important role in a broad range of applications, and enjoy a number of important properties not shared by more general matrices Not only are the eigenvalues of a symmetric matrix necessarily real, the eigenvectors always form an orthogonal basis In fact, this is by far the most common way for orthogonal bases to appear as the eigenvector bases of symmetric matrices Let us state this important result, but defer its proof until the end of the section Theorem 4 Let A = A T be a real symmetric n n matrix Then (a) All the eigenvalues of A are real (b) Eigenvectors corresponding to distinct eigenvalues are orthogonal (c) There is an orthonormal basis of R n consisting of n eigenvectors of A In particular, all symmetric matrices are complete 6// c Peter J Olver
13 Example 4 Consider the symmetric matrix A = 4 4 A straightforward computation produces its eigenvalues and eigenvectors: λ = 9, λ =, λ =, v =, v =, v = As the reader can check, the eigenvectors form an orthogonal basis of R An orthonormal basis is provided by the unit eigenvectors u =, u =, u = 6 6 Proof of Theorem 4: First, if A = A T is real, symmetric, then 6 (Av) w = v (Aw) for all v,w C n, (4) where indicates the Euclidean dot product when the vectors are real and, more generally, the Hermitian dot product v w = v T w when they are complex To prove property (a), suppose λ is a complex eigenvalue with complex eigenvector v C n Consider the Hermitian dot product of the complex vectors Av and v: On the other hand, by (4), (Av) v = (λv) v = λ v (Av) v = v (Av) = v (λv) = v T λv = λ v Equating these two expressions, we deduce λ v = λ v Since v, as it is an eigenvector, we conclude that λ = λ, proving that the eigenvalue λ is real To prove (b), suppose Av = λv, Aw = µw, where λ µ are distinct real eigenvalues Then, again by (4), λv w = (Av) w = v (Aw) = v (µw) = µv w, and hence (λ µ)v w = 6// c Peter J Olver
14 Since λ µ, this implies that v w = and hence the eigenvectors v,w are orthogonal Finally, the proof of (c) is easy if all the eigenvalues of A are distinct Part (b) proves that the corresponding eigenvectors are orthogonal, and Proposition 4 proves that they form a basis To obtain an orthonormal basis, we merely divide the eigenvectors by their lengths: u k = v k / v k, as in Lemma To prove (c) in general, we proceed by induction on the size n of the matrix A To start, the case of a matrix is trivial (Why?) Next, suppose A has size n n We know that A has at least one eigenvalue, λ, which is necessarily real Let v be the associated eigenvector Let V = { w R n v w = } denote the orthogonal complement to the eigenspace V λ the set of all vectors orthogonal to the first eigenvector Since dimv = n, we can choose an orthonormal basis y,,y n Now, if w is any vector in V, so is Aw, since, by (4), v (Aw) = (Av ) w = λ v w = Thus, A defines a linear transformation on V represented by an (n ) (n ) matrix with respect to the chosen orthonormal basis y,,y n It is not hard to prove that the representing matrix is symmetric, and so our induction hypothesis then implies that there is an orthonormal basis of V consisting of eigenvectors u,,u n of A Appending the unit eigenvector u = v / v to this collection will complete the orthonormal basis of R n QED The orthonormal eigenvector basis serves to diagonalize the symmetric matrix, resulting in the so-called spectral factorization formula Theorem 4 Let A be a real, symmetric matrix Then there exists an orthogonal matrix Q such that A = Q Λ Q = Q Λ Q T, (4) where Λ = diag (λ,, λ n ) is a real diagonal matrix The eigenvalues of A appear on the diagonal of Λ, while the columns of Q are the corresponding orthonormal eigenvectors Proof : Equation (4) can be rewritten as A Q = Q Λ The k th column of the latter matrix equation reads Av k = λ k v k, where v k is the k th column of Q But this is merely the condition that v k be an eigenvector of A with eigenvalue λ k Theorem 4 serves to complete the proof QED Remark: The term spectrum refers to the eigenvalues of a matrix or, more generally, a linear operator The terminology is motivated by physics The spectral energy lines of atoms, molecules and nuclei are characterized as the eigenvalues of the governing quantum mechanical Schrödinger operator The Spectral Theorem 4 is the finite-dimensional version for the decomposition of quantum mechanical linear operators into their spectral eigenstates The most important subclass of symmetric matrices are the positive definite matrices 6// 4 c Peter J Olver
15 Definition 44 An n n symmetric matrix A is called positive definite if it satisfies the positivity condition x T Ax > for all x R n (4) Theorem 4 A symmetric matrix A = A T is positive definite if and only if all of its eigenvalues are strictly positive Proof : First, if A is positive definite, and v is an eigenvector with (necessarily real) eigenvalue λ, then (4) implies < v T Av = v T (λv) = λ v, (44) which immediately proves that λ > Conversely, suppose A has all positive eigenvalues Let u,,u n be the orthonormal eigenvector basis of R n guaranteed by Theorem 4, with Au j = λ j u j Then, writing x = c u + + c n u n, we find Ax = c λ u + + c n λ n u n Therefore, using the orthonormality of the eigenvectors, for any x, x T Ax = (c u T + + c n ut n ) (c λ u + + c n λ n u n ) = λ c + + λ n c n > since only x = has c = = c n = This proves that A is positive definite QED The Q R Factorization The Gram Schmidt procedure for orthonormalizing bases of R n can be reinterpreted as a matrix factorization This is more subtle than the L U factorization that results from Gaussian Elimination, but is of comparable significance, and is used in a broad range of applications in mathematics, statistics, physics, engineering, and numerical analysis Let w,,w n be a basis of R n, and let u,,u n be the corresponding orthonormal basis that results from any one of the three implementations of the Gram Schmidt process We assemble both sets of column vectors to form nonsingular n n matrices A = (w w w n ), Q = (u u u n ) Since the u i form an orthonormal basis, Q is an orthogonal matrix Moreover, the Gram Schmidt equations (6) can be recast into an equivalent matrix form: r r r n r A = QR, where R = r n () r nn is an upper triangular matrix whose entries are the coefficients in (9 ) Since the Gram Schmidt process works on any basis, the only requirement on the matrix A is that its columns form a basis of R n, and hence A can be any nonsingular matrix We have therefore established the celebrated QR factorization of nonsingular matrices 6// c Peter J Olver
16 start QR Factorization of a Matrix A for j = to n set r jj = a j + + a nj if r jj =, stop; print A has linearly dependent columns else for i = to n next i set a ij = a ij /r jj for k = j + to n set r jk = a j a k + + a nj a nk for i = to n end set a ik = a ik a ij r jk next i next k next j Theorem Any nonsingular matrix A can be factored, A = QR, into the product of an orthogonal matrix Q and an upper triangular matrix R The factorization is unique if all the diagonal entries of R are assumed to be positive We will use the compacted term positive upper triangular to refer to upper triangular matrices with positive entries along the diagonal Example The columns of the matrix A = are the same as the basis vectors considered in Example The orthonormal basis () constructed using the Gram Schmidt algorithm leads to the orthogonal and positive upper triangular matrices Q =, R = 4 () The reader may wish to verify that, indeed, A = QR While any of the three implementations of the Gram Schmidt algorithm will produce the QR factorization of a given matrix A = (w w w n ), the stable version, as encoded in equations (), is the one to use in practical computations, as it is the least likely to fail 6// 6 c Peter J Olver 7
17 due to numerical artifacts produced by round-off errors The accompanying pseudocode program reformulates the algorithm purely in terms of the matrix entries a ij of A During the course of the algorithm, the entries of the matrix A are successively overwritten; the final result is the orthogonal matrix Q appearing in place of A The entries r ij of R must be stored separately Example Let us factor the matrix A = using the numerically stable QR algorithm As in the program, we work directly on the matrix A, gradually changing it into orthogonal form In the first loop, we set r = to be the norm of the first column vector of A We thennormalize the first column by dividing by r ; the resulting matrix is The next entries r = 4, r =, r 4 =, are obtained by taking the dot products of the first column with the other three columns For j =,,, we subtract r j times the first column from the j th 6 4 column; the result is a matrix whose first column is normalized to have unit length, and whose second, third and fourth columns are orthogonal to it In the next loop, we normalize the second column by dividing by its norm r = , and so obtain the matrix 7 7 We then take dot products of the second column with the remaining two columns to produce r = 6 7, r 4 = 7 Subtracting these multiples of the second column from the third and fourth columns, we obtain , which now has its first two columns orthonormalized, and orthogonal to the last two columns We then normalize the third column by dividing 6// 7 c Peter J Olver
18 7 4 6 by r =, and so Finally, we subtract r = times the third column from the fourth column Dividing the resulting fourth column by its norm r 44 = Q = 6 results in the final formulas, for the A = QR factorization, R = 6 Numerical Calculation of Eigenvalues We are now ready to apply these results to the problem of numerically approximating eigenvalues of matrices Before explaining the Q R algorithm, we first review the elementary power method The Power Method We assume, for simplicity, that A is a complete n n matrix, meaning that it has an eigenvector basis v,,v n, with corresponding eigenvalues λ,, λ n It is easy to see that the solution to the linear iterative system 6, v (k+) = Av (k), v () = v, (6) is obtained by multiplying the initial vector v by the successive powers of the coefficient matrix: v (k) = A k v If we write the initial vector in terms of the eigenvector basis then the solution takes the explicit form v = c v + + c n v n, (6) v (k) = A k v = c λ k v + + c n λk n v n (6) Suppose further that A has a single dominant real eigenvalue, λ, that is larger than all others in magnitude, so λ > λ j for all j > (64) This is not a very severe restriction Theorem 4 implies that all symmetric matrices are complete Moreover, perturbations caused by round off and/or numerical inaccuracies will almost inevitably make an incomplete matrix complete 6// 8 c Peter J Olver
19 As its name implies, this eigenvalue will eventually dominate the iteration (6) Indeed, since λ k λ j k for all j > and all k, the first term in the iterative formula (6) will eventually be much larger than the rest, and so, provided c, v (k) c λ k v for k Therefore, the solution to the iterative system (6) will, almost always, end up being a multiple of the dominant eigenvector of the coefficient matrix To compute the corresponding eigenvalue, we note that the i th entry of the iterate v (k) is approximated by v (k) i c λ k v,i, where v,i is the ith entry of the eigenvector v Thus, as long as v,i, we can recover the dominant eigenvalue by taking a ratio between selected components of successive iterates: λ v(k) i, provided that v (k ) v (k ) i (6) i Example 6 Consider the matrix A = 4 As you can check, its eigenvalues and eigenvectors are 9 7 λ =, v =, λ =, v =, λ =, v = Repeatedly multiplying an initial vector v = (,, ) T, say, by A results in the iterates v (k) = A k v listed in the accompanying table The last column indicates the ratio λ (k) = v (k) /v(k ) between the first components of successive iterates (One could equally well use the second or third components) The ratios are converging to the dominant eigenvalue λ =, while the vectors v (k) are converging to a very large multiple of the corresponding eigenvector v = (,, ) T Variants of the power method for finding other eigenvalues include the inverse power method based on iterating the inverse matrix A, with eigenvalues λ,, λ n and the shifted inverse power method, based on (A µ I ), with eigenvalues (λ k µ) The power method only produces the dominant (largest in magnitude) eigenvalue of a matrix A The inverse power method can be used to find the smallest eigenvalue Additional eigenvalues can be found by using the shifted inverse power method, or deflation However, if we need to know all the eigenvalues, such piecemeal methods are too time-consuming to be of much practical value The QR Algorithm The most popular scheme for simultaneously approximating all the eigenvalues of a matrix A is the remarkable Q R algorithm, first proposed in 96 independently by Francis 6// 9 c Peter J Olver
20 and Kublanovskaya The underlying idea is simple, but surprising The first step is to factor the matrix A = A = Q R into a product of an orthogonal matrix Q and a positive (ie, with all positive entries along the diagonal) upper triangular matrix R Next, multiply the two factors together in the wrong order! The result is the new matrix A = R Q We then repeat these two steps Thus, we next factor A = Q R using the Gram Schmidt process, and then multiply the factors in the reverse order to produce A = R Q The complete algorithm can be written as A = Q R, R k Q k = A k+ = Q k+ R k+, k =,,,, (66) where Q k, R k come from the previous step, and the subsequent orthogonal matrix Q k+ and positive upper triangular matrix R k+ are computed by using the numerically stable form of the Gram Schmidt algorithm The astonishing fact is that, for many matrices A, the iterates A k V converge to an upper triangular matrix V whose diagonal entries are the eigenvalues of A Thus, after a sufficient number of iterations, say k, the matrix A k will have very small entries below the diagonal, and one can read off a complete system of (approximate) eigenvalues along its diagonal For each eigenvalue, the computation of the corresponding eigenvector can be done by solving the appropriate homogeneous linear system, or by applying the shifted inverse power method ( ) Example 6 Consider the matrix A = The initial Gram Schmidt fac- torization A = Q R yields Q = ( ), R = These are multiplied in the reverse order to give ( ) 4 A = R Q = ( ) We refactor A = Q R via Gram Schmidt, and then reverse multiply to produce ( ) ( ) Q =, R 4 97 =, 97 ( ) A = R Q = 94 6// c Peter J Olver
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# Math Snap
## Approximately $60 \%$ of mathematics students do their homework on time. In a class of 100 students, what is the mean, variance, and standard deviation if we assume normality and use the normal distribution as an approximation of the binomial distribution? Answer choices rounded to the nearest whole number.
#### STEP 1
Assumptions 1. The probability of a student doing their homework on time, $p$, is 0.60. 2. The number of students, $n$, is 100. 3. We are assuming normality and using the normal distribution as an approximation of the binomial distribution.
#### STEP 2
First, we need to find the mean of the binomial distribution. The mean $\mu$ of a binomial distribution is given by: $\mu = n \cdot p$
#### STEP 3
Plug in the values for $n$ and $p$ to calculate the mean. $\mu = 100 \cdot 0.60$
#### STEP 4
Calculate the mean. $\mu = 60$
#### STEP 5
Next, we need to find the variance of the binomial distribution. The variance $\sigma^2$ of a binomial distribution is given by: $\sigma^2 = n \cdot p \cdot (1 - p)$
#### STEP 6
Plug in the values for $n$, $p$, and $(1 - p)$ to calculate the variance. $\sigma^2 = 100 \cdot 0.60 \cdot (1 - 0.60)$
#### STEP 7
Simplify the expression inside the parentheses. $\sigma^2 = 100 \cdot 0.60 \cdot 0.40$
#### STEP 8
Calculate the variance. $\sigma^2 = 100 \cdot 0.24 = 24$
#### STEP 9
Now, we need to find the standard deviation. The standard deviation $\sigma$ is the square root of the variance. $\sigma = \sqrt{\sigma^2}$
#### STEP 10
Plug in the value of the variance to calculate the standard deviation. $\sigma = \sqrt{24}$
#### STEP 11
Calculate the standard deviation. $\sigma \approx 4.90$
#### STEP 12
Round the standard deviation to the nearest whole number. $\sigma \approx 5$
##### SOLUTION
Summarize the results: - Mean $\mu$ = 60 - Variance $\sigma^2$ = 24 - Standard deviation $\sigma$ = 5
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Chapter 3 (F11)
# Chapter 3 (F11) - Section 3.1 Quadratic Functions A...
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Section 3.1 – Quadratic Functions A quadratic function is one of the form f ( x ) = a x 2 + b x + c Solving quadratic functions using the Quadratic Formula: b 2 – 4ac is called the discriminant. • If b 2 – 4ac > 0 , then the equation will have 2 real solutions • If b 2 – 4ac < 0 , then the equation will have no real solutions (there will be 2 complex sol’ns) • If b 2 – 4ac = 0 , then the equation will have 1 real solution The Quadratic Formula : If a x 2 + b x + c = 0, then x = a ac b b 2 4 2 - ± - Examples: Solve using the quadratic formula: a) 3 x 2 – 7 x + 1 = 0 b) 2 x 2 – 4 x + 3 = 0 The graphs of all quadratic functions are parabolas. Parabolas will either open up or open down , and they will contain a vertex and an axis of symmetry . 1
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Quadratic Form: f ( x ) = a x 2 + b x + c The parabola opens up if a > 0, and opens down if a < 0 The vertex (maximum/minimum) is at ( 29 ) ( , 2 2 a b a b f - - The axis of symmetry is the line x = a b 2 - The y -intercept is at (0, c) The x -intercepts are found by solving f ( x ) = 0. If b 2 – 4ac > 0, then the graph will have 2 x -intercepts If b 2 – 4ac = 0, then the graph will have 1 x -intercept If b 2 – 4ac < 0, then the graph will have 0 x -intercepts Examples : Find the vertex, axis of symmetry, x - and y -intercepts, domain and range. Sketch a graph. a) f ( x ) = x 2 + 6 x + 3 2
b) f ( x ) = –2 x 2 + 8 x – 1 c) f ( x ) = x 2 + 4 x + 5 3
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Applications Involving Maximizing or Minimizing: Quadratic functions will always have either a maximum or a minimum value. Example : A person standing close to the edge on the top of a 160-foot building throws a baseball vertically upward. The quadratic function s ( t ) = –16 t 2 + 64 t + 160 models the ball’s height above the ground, s ( t ), in feet, t seconds after it was thrown. a) After how many seconds does the ball reach its maximum height? b) What is the maximum height? c) How many seconds does it take for the ball to hit the ground? d) Calculate s (0) and describe what it means. 4
Example : Amy has 2500 feet of fencing available to enclose a rectangular garden. Find the dimensions of the garden that will maximize the enclosed area. What is the maximum area? 5
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Section 3.2 – Polynomial Functions and Their Graphs A Polynomial Function of Degree n has the form: f ( x ) = 0 1 1 1 a x a x a x a n n n n + + + + - - (where the a n ’s are real numbers and n is an integer ≥ 0 ) When graphed, polynomial functions are always smooth and continuous. Although the graph of a polynomial may have intervals where it increases or decreases, the graph will
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Chapter 3 (F11) - Section 3.1 Quadratic Functions A...
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LCM of 4 and 16 is the the smallest number amongst all usual multiples of 4 and also 16. The first couple of multiples of 4 and also 16 are (4, 8, 12, 16, 20, 24, 28, . . . ) and also (16, 32, 48, 64, 80, 96, . . . ) respectively. There are 3 commonly used techniques to discover LCM the 4 and 16 - by listing multiples, by division method, and by element factorization.
You are watching: What is the lcm of 4 and 16
1 LCM the 4 and 16 2 List the Methods 3 Solved Examples 4 FAQs
Answer: LCM of 4 and 16 is 16.
Explanation:
The LCM of two non-zero integers, x(4) and y(16), is the smallest optimistic integer m(16) the is divisible by both x(4) and also y(16) without any kind of remainder.
Let's look at the various methods because that finding the LCM the 4 and 16.
By Listing MultiplesBy element Factorization MethodBy division Method
### LCM that 4 and also 16 through Listing Multiples
To calculate the LCM of 4 and 16 by listing the end the common multiples, we can follow the given listed below steps:
Step 1: perform a couple of multiples of 4 (4, 8, 12, 16, 20, 24, 28, . . . ) and 16 (16, 32, 48, 64, 80, 96, . . . . )Step 2: The typical multiples native the multiples the 4 and 16 space 16, 32, . . .Step 3: The smallest typical multiple of 4 and 16 is 16.
∴ The least usual multiple that 4 and also 16 = 16.
### LCM the 4 and also 16 by prime Factorization
Prime administer of 4 and 16 is (2 × 2) = 22 and also (2 × 2 × 2 × 2) = 24 respectively. LCM of 4 and also 16 can be derived by multiplying prime determinants raised to their respective highest power, i.e. 24 = 16.Hence, the LCM the 4 and also 16 by element factorization is 16.
### LCM that 4 and also 16 by division Method
To calculate the LCM the 4 and 16 by the division method, we will certainly divide the numbers(4, 16) by your prime factors (preferably common). The product of these divisors provides the LCM that 4 and also 16.
Step 3: continue the actions until only 1s are left in the last row.
The LCM of 4 and also 16 is the product of every prime numbers on the left, i.e. LCM(4, 16) by division method = 2 × 2 × 2 × 2 = 16.
## FAQs on LCM of 4 and also 16
### What is the LCM of 4 and also 16?
The LCM that 4 and also 16 is 16. To discover the least typical multiple (LCM) that 4 and 16, we require to discover the multiples the 4 and also 16 (multiples that 4 = 4, 8, 12, 16; multiples the 16 = 16, 32, 48, 64) and choose the the smallest multiple that is precisely divisible by 4 and also 16, i.e., 16.
### How to uncover the LCM the 4 and also 16 by prime Factorization?
To uncover the LCM of 4 and also 16 utilizing prime factorization, us will find the prime factors, (4 = 2 × 2) and (16 = 2 × 2 × 2 × 2). LCM that 4 and also 16 is the product of prime factors raised to their respective greatest exponent amongst the numbers 4 and 16.⇒ LCM the 4, 16 = 24 = 16.
### What is the the very least Perfect Square Divisible by 4 and also 16?
The least number divisible through 4 and 16 = LCM(4, 16)LCM the 4 and 16 = 2 × 2 × 2 × 2 ⇒ the very least perfect square divisible by each 4 and also 16 = 16 Therefore, 16 is the forced number.
### If the LCM the 16 and 4 is 16, uncover its GCF.
LCM(16, 4) × GCF(16, 4) = 16 × 4Since the LCM that 16 and also 4 = 16⇒ 16 × GCF(16, 4) = 64Therefore, the greatest typical factor (GCF) = 64/16 = 4.
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### What is the Relation between GCF and LCM of 4, 16?
The complying with equation can be offered to express the relation in between GCF and LCM the 4 and 16, i.e. GCF × LCM = 4 × 16.
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Question: What Is The Multiplicative Property Of Zero?
Why is 1 called the multiplicative identity?
The identity property of 1 says that any number multiplied by 1 keeps its identity.
In other words, any number multiplied by 1 stays the same.
The reason the number stays the same is because multiplying by 1 means we have 1 copy of the number.
For example, 32×1=32..
What is difference between additive identity and multiplicative identity?
Answer: The Multiplicative Identity Axiom states that a number multiplied by 1 is that number. The Additive Inverse Axiom states that the sum of a number and the Additive Inverse of that number is zero. Every real number has a unique additive inverse.
When zero is added to any number?
Identity Law of Addition states that any number added to 0 is equal to itself. Therefore, you can add any number and get the same sum. So you can add 0 to 1, 107, and 1,000,000 and still get the same number that you started with.
Why is the number zero so important?
0 (zero) is a number, and the numerical digit used to represent that number in numerals. It fulfills a central role in mathematics as the additive identity of the integers, real numbers, and many other algebraic structures. As a digit, 0 is used as a placeholder in place value systems.
What are the properties of 0 and 1?
Zero times any number is equal to zero. Which means, multiplying any number by 0 gives 0. Multiplying any number by 1 leaves it unchanged. 1 is called the multiplicative identity hence the property is called multiplicative identity.
What is the multiplicative property of 1?
According to the multiplicative identity property of 1, any number multiplied by 1, gives the same result as the number itself. It is also called the Identity property of multiplication, because the identity of the number remains the same.
What are the properties of zero?
5 Properties of Zero Zero is even (not odd, not neutral) Zero is neither positive nor negative (the only number with this property) Zero is an integer (and must be considered when question limits choices to integers) Zero is a multiple of all numbers (x*0 = 0, so a multiple of any x)More items…•
Is 0 a multiplicative identity?
We observe that the product of any whole number and zero is zero. Multiplicative Identity of Whole Numbers / Identity Property of Whole Numbers: When a number is multiplied by 1, the product is the number itself.
What number is called the multiplicative identity?
The “Multiplicative Identity” is 1, because multiplying a number by 1 leaves it unchanged: a × 1 = 1 × a = a.
What is the multiplicative inverse of 1?
The multiplicative inverse of 1 is in fact 1/1 which is equal to 1.
What is the multiplicative identity of zero?
The multiplication property states that the product of any number and zero is zero. It doesn’t matter what the number is, when you multiply it to zero, you get zero as the answer. So: 2 x 0 = 0.
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Chapter 1: Rational Numbers
Natural numbers are a collection of all positive numbers starting from 1.
Whole numbers are a collection of all natural numbers including 0.
Rational numbers are the numbers that can be written in $\frac{p}{q}$ form, where p and q are integers and q ≠ 0
Closure property of numbers
Whole numbers are closed under addition and multiplication. However, they are not closed under subtraction and division.
This means if a andb are whole numbers, then a + b and a × b are also whole numbers.
Integers are also closed under addition and multiplication. However, they are not closed under subtraction and division.
This means if a and b are integers, then a + b, ab and a × b are also integers.
• Rational numbers are closed under addition, subtraction and multiplication. They are not closed under division.
This means if a and b are rational numbers, then a + b, ab and a × b are also rational numbers.
Commutative property of numbers
• Whole numbers, integers and rational numbers are commutative under addition and multiplication. However, they are not commutative under subtraction and division.
Where a and b are both whole numbers or integers or rational numbers.
Associative property of numbers
• Whole numbers, integers and rational numbers are associative under addition and multiplication. However, none of the number system is associative under subtraction and division.
Where a, b and c are all whole numbers or integers or rational numbers
0 is the additive identity of whole numbers, integers, and rational numbers, since 0 + a = a + 0 = a, where a is a whole number or an integer or a rational number.
1 is the multiplicative identity of whole numbers, integers, and rational numbers, since a × 1 = 1 × a = a, where a is a whole number or an integer or a rational number.
Additive inverse of a number is the number, which when added to a number, gives 0. It is also called the negative of a number.
a and -a are the additive inverse of each other as $a+\left(-a\right)=\left(-a\right)+a=0$
Reciprocal or multiplicative inverse of a number is the number, which when multiplied by the number, gives 1.
a and $\frac{1}{a}$ are the multiplicative inverse of each other as $a×\frac{1}{a}=1$.
Rational numbers are distributive over addition and subtraction.
• For rational numbers a, b, and c:
$a\left(b+c\right)=ab+c\phantom{\rule{0ex}{0ex}}a\left(b+c\right)=ab-ac$
Rational numbers can be represented on the number line in the same way as the fractions are represented.
In between two rational numbers, there exists infinite rational numbers.
Chapter 2: Linear Equations in One Variable
An algebraic equation is an equality involving variables.
• In an equation, the value of expression on the left hand side (LHS) is equal to the value of expression on the right hand side (RHS).
• The equations in which the highest power of variable is 1 are known as linear equations in one variable.
Some linear equations can be solved by transposing terms from one side to the other.
Example: Solve: 2x – 4 = 3x – 2
Solution: Transposing 3x from RHS to LHS, we obtain:
2x – 4 – 3x = –2
Transposing –4 from LHS to RHS, we obtain:
There are certain linear equations in which the denominators of the expressions on both sides are not 1. In such cases, both the sides of the equation are multiplied with the LCM of the denominators on both the sides.
For example: To solve the equation $\frac{3x-1}{4}+2=\frac{2x+3}{6}+7,$ we are first required to reduce it to its simplest form. This can be done as follows:
This is a linear equation and it can be solved further to obtain the value of x.
There are certain equations that can be reduced to linear equations.
For example: To solve the equation $\frac{2-4x}{3x-2}=\frac{3}{2},$ we are first required to reduce it to its simplest form. This can be done by cross-multiplying the terms of both sides as follows:
This is a linear equation in one variable and it can be solved further to obtain the value of x.
Polygons
A simple closed curve made up of line segments only is called a polygon.
• Polygons can be classified according to their number of sides (or vertices).
Number of side/vertices Classification 3 Triangle 4 Quadrilateral 5 Pentagon 6 Hexagon 7 Heptagon . . . . n n - gon
• The line segment connecting two non-consecutive vertices of a polygon are called diagonals.
For polygon ABCD, AC and BD are diagonals and for polygon PQRS, QS and PR are diagonals.
• The polygon, none of whose diagonals lie in its exterior, is called a convex polygon. In the given figure, ABCD is a convex polygon.
• The polygon whose diagonals lie in its exterior is called a concave polygon. PQRS is a concave polygon.
• A polygon, which is both equiangular and equilateral, is called a regular polygon. Otherwise, it is an irregular polygon.
• The sum of all the angles of an n-sided polygon is given by, (n – 2) × 180°.
• The sum of measures of all exterior angles of a polygon is 360°.
A quadrilateral with a pair of parallel sides is called a trapezium.
• A trapezium whose non-parallel sides are equal is called an isosceles trapezium.
A kite is a quadrilateral with exactly two distinct consecutive pairs of sides of equal lengths.
A parallelogram is a quadrilateral whose opposite sides are parallel and equal.
• Its opposite angles are of equal measure.
• The adjacent angles in a parallelogram are supplementary.
• The diagonals of a parallelogram are not equal. However, they bisect each other.
A quadrilateral whose opposite sides are parallel and all sides are of equal lengths is called a rhombus.
• Its opposite angles are of equal measure.
• Its diagonals are perpendicular bisectors of one another.
A rectangle is a parallelogram with equal angles.
• Its each angle is of measure 90°.
• Its diagonals are of equal length and they bisect each other.
A square is a rectangle with equal sides.
• Its diagonals are equal and are perpendicular bisectors of each other.
Chapter 4: Practical Geometry
A unique quadrilateral can be constructed, if any five measurements of the quadrilateral are given.
Construction of a quadrilateral when four sides and a diagonal are given
Example: Construct a quadrilateral WXYZ, where WX = 4.5 cm, XY = 5 cm, YZ = 5.5 cm, ZW = 3 cm, and WY = 6 cm.
Solution:
The steps of constructing quadrilateral WXYZ are as follows:
(1) Draw a line WY of length 6 cm. Draw an arc of radius 4.5 cm with W as centre and another arc of length 5 cm with Y as centre. The intersection of the two arcs will be the point, X. Join WX and XY.
(2) The point, Z, will be on the opposite side of point X with respect to WY. Draw an arc of length 3 cm taking W as centre and another arc of length 5.5 cm taking Y as centre. The intersection of these arcs will be the point, Z. Join WZ and YZ.
Thus, WXYZ is the required quadrilateral.
Construction of a quadrilateral when two diagonals and three sides are given
Example: Construct a quadrilateral PQRS, where PR = 7 cm, QS = 8 cm, PQ = 5 cm, QR = 5 cm, and PS = 5.5 cm.
Solution:
The steps of constructing quadrilateral PQRS are as follows:
(1) Draw a line PR of length 7 cm. Draw an arc of radius 5 cm taking P as centre and an arc of radius 5 cm taking R as centre.The point of intersection of these two arcs will be the point, Q.Join PQ and RQ.
(2) With Q as centre, draw an arc of radius 8 cm. The point, S, will lie on this arc.Then, taking P as centre, draw an arc of radius 5.5 cm. The intersection point of the two arcs will be the point, S.Join PS and RS.
Thus, PQRS is the required quadrilateral.
Construction of a quadrilateral when two adjacent sides and three angles are given:
Example: Construct a quadrilateral ABCD, where AB = 6 cm, AD = 4 cm, A = 90°, B = 105°, and D = 60°.
Solution:
The steps of constructing quadrilateral ABCD are as follows:
(1) Draw a line segment AB of length 6 cm. Make ABX = 105° at B and BAY = 90° at A.
(2) With A as centre, draw an arc of radius 4 cm to cut the ray AY at point D. At D, draw ADZ = 60°. The point of intersection of the rays, BX and DZ, will be the point, C.
Thus, ABCD is the required quadrilateral.
Construction of a quadrilateral when three sides and two included angles are given:
Example: Construct a quadrilateral PQRS with SR = 6.5 cm, PS = 5 cm, QR = 3 cm, R = 120°, and S = 70°.
Solution:
The steps of construction are as follows:
(1) Draw SR = 6.5 cm. Draw SRX = 120° at R and RSY = 70° at S.
(2) With S as centre, draw an arc of radius 5 cm intersecting SY at P. With R as centre, draw an arc of radius 3 cm intersecting RX at Q.
Join PQ to obtain the required quadrilateral PQRS.
Some special quadrilaterals such as square, rhombus, rectangle, kite, and parallelogram can be constructed by using their properties.
• In a square, all the sides are equal and all the angles are equal to 90°. Therefore, a square can be constructed, if one of its sides is given.
• In a rhombus, all the sides are equal. Also, the diagonals of a rhombus are perpendicular bisectors of one another. Therefore, a rhombus can be constructed, if only the measures of two diagonals are given.
• In a rectangle, all angles are of measure 90° and the opposite sides are equal. Therefore, a rectangle can be constructed even if two of its adjacent sides are given.
• In a kite, the adjacent sides are equal. Therefore, if two adjacent sides and a diagonal are given, then the required kite can be constructed.
• In a parallelogram, the opposite sides are equal and parallel. Therefore, a parallelogram can be constructed even if two of its sides and one angle or a diagonal are given.
For example,a parallelogram ABCD with AB = 5 cm, BC = 4 cm, B = 110° can be constructed as follows:
First, draw AB = 5 cm and then construct ABX = 110°. With B as centre, draw an arc of radius 4 cm cutting the ray, BX, to obtain the point, C. Then, draw a line CY parallel to AB. With C as centre, cut an arc of radius 5 cm at CY to obtain the point, D.Join AD to obtain the parallelogram, ABCD.
Chapter 5: Data Handling
The data in an unorganised form is called raw data. In order to draw meaningful inferences from a data, we need to organise the data systematically.
• We can organise a data in the following ways. They are: frequency distribution table, histogram and pie chart
Frequency distribution table
• The number of times a particular entry occurs is called its frequency.
• The difference between the upper and lower class limits is called the width or size of the class interval.
Example: The ages of children (in years) in a group dance competition are given below:
16, 18, 8, 10, 12, 17, 9, 8, 16, 18, 14, 14, 16, 15, 12, 9, 11, 10, 18, 9, 12, 11, 16.
Arrange the following data in a frequency distribution table whose class intervals are 7 – 10, 10 – 13 …
Solution:
Class interval (Age of children) Tally mark Frequency (Number of children) 7 – 10 5 10 – 13 6 13 – 16 4 16 – 19 8
Histogram
• A histogram is a bar graph that is used to represent grouped data. In a histogram, the class intervals are represented on the horizontal axis and the heights of the bars represent frequency. Also, there is no gap between the bars in a histogram.
The above frequency distribution table can be displayed in a histogram as follows:
In a histogram, a broken line can be used along the horizontal axis to indicate that the numbers between 0 to7 are not included.
Pie chart
A pie chart or a circle graph shows the relationship between a whole and its parts.
For example: Consider the given pie chart which shows the favourite colours of the class-VIII students of a school.
In this pie chart, the portion of the sector for the colour red is given by,
Therefore, the sector representing red colour is ${\left(\frac{9}{16}\right)}^{\mathrm{th}}$ part of the circle.
Construction of pie charts
Example: Construct a pie chart for the following data which gives the brands of laptop preferred by the people of a locality.
Brand A : 100 Brand B : 120 Brand C : 180
Solution:
The total number of people is 100 + 180 + 120 = 400.
We can form the following table to find the central angle of each sector:
Brand of laptop Number of people Fraction Central angle A 100 $\frac{100}{400}=\frac{1}{4}$ $\frac{1}{4}×360°=90°$ B 180 $\frac{120}{400}=\frac{3}{10}$ $\frac{3}{10}×360°=108°$ C 120 $\frac{180}{400}=\frac{9}{20}$ $\frac{9}{20}×360°=162°$
To represent the given information using a circle graph, first of all draw a circle with any convenient radius. Let O be the centre of the circle and OX be its radius. Draw the angle of the sector for brand A, which is 90°. Using protractor, draw ∠XOY = 90°. Now, draw the angle of the sectors for brands B and C.
A pie chart is interpreted by analysing it.
Probability is the chance of the occurrence of an event.
• A random experiment is the experiment whose outcome cannot be predicted exactly in advance.
• The outcomes of an experiment having the same chances of occurrence are known as equally-likely outcomes. For example, if we toss a coin, then the possible outcomes are head or tail, and both of them have an equal chance of occurring. So, these are equally-likely outcomes.
• Each outcome of an experiment or collection of outcomes is known as an event.
• When the outcomes of the experiment are equally-likely, the probability of an event is given by:
For example, if two coins are tossed together, then any of the following outcomes may be obtained.
(ii) Head on first coin, tail on second coin (H, T)
(iii) Tail on first coin, head on second coin (T, H)
(iv) Tail on first coin, tail on second coin (T, T)
Now, the probability of getting one head and one tail$=\frac{2}{4}=\frac{1}{2}$
Chapter 6: Squares and Square Roots
If a natural number m can be expressed as n2, where n is also a natural number, then m is a square number. The square numbers are also called perfect squares.
• All the square numbers end in 0, 1, 4, 5, 6, or 9.
• If a number has 1 or 9 at its units place, then its square ends in 1.
• If a square number ends in 6, then the number, whose square it is, will either have 4 or 6 in its units place.
• Square numbers can have even number of zeroes at the end.
If we add two consecutive triangular numbers, then we obtain a square number.
For example, 10 + 15 = 25 = 52
There are 2n non-perfect square numbers between the squares of the numbers, n and (n + 1).
The sum of first n odd natural numbers is n2.
For example, the sum of first 7 odd natural numbers is 49 i.e., 1 + 3 + 5 + 7 + 9 + 11 + 13 = 49 = 72
Square of any odd number can be expressed as the sum of two consecutive positive integers. For example, 92 = 81 = 40 + 41
We can express the product of two consecutive even or odd natural numbers as follows.
(a + 1) × (a – 1) = a2 – 1
We can find the squares of numbers having more than one digit by making use of the identity,(a + b)2 = a × (a + b) + b × (a + b)
For example (45)2 = (40 + 5)2= (40) × (40 + 5) + 5 × (40 + 5)= 402 + 40 × 5 + 5 × 40 + 52= 1600 + 200 + 200 + 25= 2025
A square number with units digit 5, say (a5), can be written as follows:
(a5)2 = a(a + 1) hundred + 25
For example, 6252 = 390625 = (62 × 63) hundred + 25= 390600 + 25
2m, m2 – 1, and m2+ 1 forms a Pythagorean Triplet, where m is a natural number greater than 1.
Square root is the inverse operation of squaring. The positive square root of a number is denoted by the symbol $\sqrt{}$. Some methods to find the square root of perfect squares are: repeated subtraction method, prime factorisation method and division method.
Example: Find the square root of 36 in repeated subtraction method.
Solution:
36 – 1 = 35, 35 – 3 = 32, 32 – 5 = 27, 27 – 7 = 20, 20 – 9 = 11, 11 – 11= 0
Therefore, from 36, we subtracted successive odd numbers and obtained 0 in the 6th step. Therefore, $\sqrt{36}=6$
Example: Find the smallest number by which 252 can be multiplied to make it a perfect square.
Solution:
We have, 252 = 2 × 2 × 3 × 3 × 7
The number 7 does not occur in pair. Therefore, if we multiply 252 by 7, then it will become a perfect square.
Therefore, 252 × 7 = 2 × 2 × 3 × 3 × 7 × 7
That is, 1764 is a perfect square and $\sqrt{1764}=42$
Example: Find the square root of 1369 by division method.
Solution:
Step1: Firstly, place bars over every pair of digits starting from the digit at ones place. We obtain
Step2: Find the largest number whose square is less than or equal to the number under the extreme left bar.
Take this number as the divisor and the number under the extreme left bar as the dividend. Divide and obtain the remainder.
Step3: Bring down the number under the next bar to the right of the remainder.
Therefore, the new dividend is 469.
Double the divisor and enter it with the blank on its right.
Step 4: Guess the largest possible digit to fill the blank, which becomes the new digit in the quotient, such that when the new digit is multiplied to the new quotient, the product is less than or equal to the dividend.
In this case, 67 × 7 = 469
Therefore, the quotient is 7.
Also, the remainder becomes 0 and no bar is left.
Therefore, $\sqrt{1369}=37$
Square root of a non-perfect square can be estimated to a nearest whole number. Let m be non-perfect square and let n2<m<(n + 1)2.
• If mn2< (n + 1)2m, then $\sqrt{m}$ is approximately equal to n.
• If mn2> (n + 1)2m, then $\sqrt{m}$ is approximately equal to n + 1.
Example: Estimate the value of $\sqrt{180}$ to the nearest whole number.
Solution:
It is known that, 132 = 169 and 142 = 196
169 < 180 < 196
$⇒{13}^{2}<\sqrt{180}<{14}^{2}$
Now, 180 – 132 = 11 and 142 – 180 = 16.
Since, 11 < 16 so 69 is closer to 180 than 196.
Therefore, $\sqrt{180}$ is approximately equal to 13.
Chapter 7: Cubes and Cube Roots
Numbers obtained when any number is multiplied by itself three times are known as cube numbers. For example, 343 is a cube number, since 343 = 7 × 7 × 7
Properties of cubes of numbers
• Cubes of even numbers are even and the cubes of odd numbers are odd.
• If a number has 0 or 1 or 4 or 5 or 6 or 9 in its ones place, then its cube will have the same digit in ones place.
• If a number has 2 in its ones place, then its cube will have 8 in ones place and vice versa.
• If a number has 3 in its ones place, then its cube will have 7 in ones place and vice versa.
Pattern of addition of consecutive odd numbers
13 = 1 = 1
23 = 8 = 3 + 5
33 = 27 = 7 + 9 + 11
43 = 64 = 13 + 15 + 17 + 19 so on
Each prime factor appears three times in the prime factorization of its cube.
Cube root is the inverse operation of finding a cube. The symbol $\sqrt[3]{3}$ denotes cube-root. If a3 = b, then .
Cube root of a perfect cube can be found by prime factorising it.
Example: Find the cube root of 1728
Solution:
Cube root of a perfect cube can be found through estimation.
Example: Find the cube root of 39304 through estimation.
Solution: The steps involved in finding the cube root of a given number are as follows:
Step I: Make groups of three digits starting from the right-most digit of the number.
Step II: The number 304 ends with 4. We know that 4 comes at unit’s place of a number only when it is cube root ends in 4. So, we get 4 at the unit’s place of the cube root.
Step III: Now, consider the other group, that is, 39.
We know that 33 = 27 and 43 = 64.
Also 27 < 39 < 64
Therefore, we take the smallest number amongst 3 and 4, that is, 3 as the ten’s digit of the required cube root.
Thus, we have $\sqrt[3]{39304}=34$Chapter 8: Comparing Quantities
Ratio means comparing two quantities.
Percentages are numerators of fractions with denominator 100.
Increase and decrease percent
•
Discount is the reduction given on the Marked Price (M.P) of an article.
• Discount = Marked Price – Sale price
• Discount = Discount % of Marked Price
Additional expenses made after buying an article are included in the cost price and are known as overhead expenses. These may include expenses such as amount spent on repairs, labour charges, transportation, etc. These expenses are added to the cost price (C.P.) to obtain the real C.P. of an article.
Sales tax is charged on the sale of an item by the government and is added to the bill amount.
Sales Tax (or VAT) = Tax % of bill amount
Interest is the extra money paid by institutions such as banks or post offices on money deposited with them. It is also paid by people when they borrow money from these institutions.
• Amount = Principal + Interest
The interest calculated on the amount of the previous year is known as compound interest.
• Amount (A) when interest is compounded annually is $\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{n},$ Where, P = Principal, R = Rate of interest, n = Time period.
• Amount when interest is compounded half yearly is given by, $\mathrm{A}=\mathrm{P}{\left(1+\frac{\mathrm{R}}{100}\right)}^{2n}$ Where, $\frac{\mathrm{R}}{2}$ = Half-yearly rate and 2n = Number of half years
Example: A sum of Rs 12,000 was kept in a bank for 2 years and 6 months at 5% per annum. Find the compound interest to be paid by the bank when the interest is compounded annually.
Solution:
We have, P = 12000, R = 5 %
For 2 years, the amount is given by,
For the next $\frac{1}{2}$ year, the interest can be calculated as simple interest.
Total compound interest = Rs (1230 + 330.75) = Rs 1560.75
Chapter 9: Algebraic Expressions and Identities
Expressions are formed from variables and constants.
• Terms are added to form expressions.
• Terms themselves can be formed as the product of factors.
An expression that contains only one term is called a monomial. An expression that contains two terms is called a binomial. An expression that contains three terms is called a trinomial. An expression containing one or more terms with non-zero coefficient is called a polynomial.
Like terms are formed by same variables and the powers of these variables are same too.
Addition and subtraction of algebraic expressions
• While performing addition, we write each expression to be added in a separate row. While doing so, we write like terms one below the other and add them.
• Subtraction of a number is same as addition of its additive inverse.
For example, 2x2 + 3yz – 4xy is subtracted from 3x2 + 8xy + 3y2 as:
Multiplication of algebraic expressions
• The multiplication of a monomial by a monomial gives a monomial. While performing multiplication, the coefficients of the two monomials are multiplied and the powers of different variables in the two monomials are multiplied by using the rules of exponents and powers.
For example the expression (–2ab2c) and (3abc2) are multiplied as:
(–2ab2c) × (3abc2) = –6 × (a × a × b2 × b × c × c2) = –6a2b3c3
• While multiplying a monomial by a binomial or trinomial or polynomial, we multiply every term in the polynomial by the monomial by making use of distributive law.
For example, 5a × (2b + c) = (5a × 2b) + (5a × c) = 10ab + 5ac
• While multiplying a polynomial by a binomial (or trinomial), we multiply it term by term. That is, every term of the polynomial is multiplied by every term of the binomial (or trinomial).
For example, (5a c) × (2b + c) = 5a × (2b + c) c × (2b + c) = (5a × 2b) + (5a × c) (c × 2b) (c × c) = 10ab + 5ac 2bc c2
An identity is an equality which is true for all values of the variables in it. However, an equation is true for only certain values of the variables in it. It is not true for all the values of the variable and hence, it is not an identity.
Some standard identities
(i) (a + b)2 = a2 + 2ab + b2
(ii) (a b)2 = a2 – 2ab + b2
(iii) (a + b) (ab) = a2b2
(iv) (x + a) (x + b) = x2 + (a + b) x + ab
Example: Simplify (5x + 2y)2 – (3xy)2.
Solution:
Using identities (i) and (ii), we obtain:
(5x + 2y)2 = (5x)2 + 2 (5x) (2y) + (2y)2 = 25x2 + 20xy + 4y2
(3xy)2 = (3x)2 – 2 (3x) (y) + (y)2 = 9x2 – 6xy + y2
∴ (5x + 2y)2 – (3xy)2 = 25x2 + 20xy + 4y2 – 9x2 + 6xyy2 = 16x2 + 26xy + 3y2
Chapter 10: Visualising Solid Shapes
Three-dimensional objects look differently from different positions.
For example, the following figure (made by joining some cubes) is viewed are given as:
A map depicts the location of a particular object/place in relation to other object(s)/place(s).
• Symbols in a map are used to depict the different objects/places.
• There is no reference or perspectives in a map, i.e., objects that are closer to the observer are shown to be of the same size as those that are further away.
• Maps use scale, which is fixed for a particular map. It reduces the real distances proportionately to the distances on the paper.
For example, the following map shown is the map of a locality.
By looking at this map, some questions can be answered. For example, if the question is which is further west: a market or a temple, then the answer is market.
Three-dimensional objects look differently from different positions.
For example, the following figure (made by joining some cubes) is viewed are given as:
Polyhedron is a solid made up of polygonal regions (called its faces). These faces meet at edges (which are line segments) and the edges meet at the vertices (which are points).
Among the solids shown below, the first two solids are polyhedron.
A polyhedron is a convex polyhedron if the line joining any of its two vertices lies within or on the polyhedron. Otherwise it is called a concave polyhedron.
Among the solids shown below, the first two polyhedrons are convex polyhedrons and the last one is a concave polyhedron.
Regular polyhedrons are polyhedrons whose faces are made up of regular polygons and the same numbers of faces meet at each vertex.
Among the solids shown below, the first polyhedron is a regular polyhedron where as the other one is a concave polyhedron.
A prism is a polyhedron whose base and top are congruent polygons and whose lateral faces are parallelograms in shape. A pyramid is a polyhedron whose base is a polygon (of any number of sides) and whose lateral faces are triangles with a common vertex.
Among the following solids, the first one is a prism and the second one is a pyramid.
For any polyhedron, F + VE = 2, where F is the number of faces, V is the number of vertices and E is the number of edges. This relationship is called Euler’s formula.Chapter 11: Mensuration
Area and perimeter of various plane figures
❖ Shape ❖ Area ❖ Perimeter Rectangle with adjacent sides a and b a × b 2× (a + b) Square with side a a2 4a Circle with radius r πr2 2πr Triangle with base b and its corresponding height h $\frac{1}{2}×b×h$ Sum of the three sides Parallelogram with base b and its corresponding height h b × h Sum of the four sides Trapezium with parallel sides a and b and height h. $\frac{1}{2}×\left(a+b\right)×h$ Sum of the four sides Rhombus with base b and its corresponding height h b × h Sum of the four sides Rhombus with diagonals d1 and d2 $\frac{1}{2}×{d}_{1}×{d}_{2}$ Sum of the four sides Quadrilateral with diagonal d and the altitudes h1 and h2 to this diagonal from the opposite vertices. $\frac{1}{2}×d×\left({h}_{1}+{h}_{2}\right)$ Sum of the four sides Polygon Break the polygon into any type of triangle (s) or any type of quadrilateral (s). Add the areas of these triangle (s) and quadrilateral (s). Sum of all the sides of the polygon
Example: Find the area of the given polygon, where ABCD is a trapezium.
Solution:
Area of ABCED = Area of trapezium ABCD + Area of ∆CDE
$=\left[\frac{1}{2}\left(3+5\right)×3+\frac{1}{2}×5×4\right]{\mathrm{cm}}^{2}$
= (12 + 10) cm2
= 22 cm2
Lateral surface area of a solid is the sum of the areas of its lateral faces.
Total surface area of a solid is the sum of the areas of its all faces.
Volume of a solid is the amount of region occupied by the solid.
Surface area and volumes of various solids
❖ Solid ❖ Lateral Surface Area ❖ Total Surface Area ❖ Volume Cuboid with length l, breadth b and height h 2h × (l + b) 2(lb + bh + hl) l × b × h Cube with edge l 4l2 6l2 l3 Right circular cylinder with radius of circular base r and height h 2πrh 2πr (r + h) πr2h
Capacity is the quantity that a container can hold.
Relationship between common units
1 mL = 1 cm3
1 L = 1000 cm3
1 m3 = 1000000 cm3 = 1000 L
Chapter 12: Exponents and Powers
Very large numbers and very small numbers are difficult to read, understand, and compare. To make this easier, we use exponents by converting many of the large numbers and small numbers into a shorter form.
For example, 10,000,000,000,000 can be written as (10)13.
Here, 10 is called the base and 13 is called the exponent.
For any non-zero integer a${a}^{-m}=\frac{1}{{a}^{m}}$, where m is a positive integer, am is called the multiplicative inverse of am and vice-versa.
Decimal numbers can be written in expanded form using exponents.
For example, 32845.912
$=3×{10}^{4}+2×{10}^{3}+8×{10}^{2}+4×{10}^{1}+5×1+9×\frac{1}{10}+1×\frac{1}{{10}^{2}}+2×\frac{1}{{10}^{3}}\phantom{\rule{0ex}{0ex}}=3×{10}^{4}+2×{10}^{3}+8×{10}^{2}+4×{10}^{1}+5×1+9×{10}^{-1}+1×{10}^{-2}+2×{10}^{-3}$
Laws of exponents
(i) am × an = am + n
(ii)
(iii) (am)n = amn
(iv) am × bm = (ab)m
(v) $\frac{{a}^{m}}{{b}^{m}}={\left(\frac{a}{b}\right)}^{m}$
(vi) a0 = 1 (a 0)
an = 1 implies that n = 0 for any a 0, except at a = 1 or a = –1
• (1)n = 1 for any n
• (–1)p = 1 for any even integer p
Very smaller numbers can be expressed in a simpler way using exponents.
For example the number 0.00000003812 is written in standard form as:
$0.00000003812=\frac{3812}{100000000000}=\frac{3812}{{10}^{11}}=\frac{3.812×{10}^{3}}{{10}^{11}}=\frac{3.812}{{10}^{8}}=3.812×{10}^{-8}$
Here, the decimal in is moved 8 places to the right.
When we have to compare or to add or subtract numbers in standard form, we convert them into numbers with the same exponents.
Example: The thickness of a cloth is 0.251 cm and the thickness of a piece of a paper is 0.0502 cm. Compare the thickness of the cloth to the thickness of the piece of paper.
Solution:
Thus, the thickness of the cloth is 5 times the thickness of the paper.
Chapter 13: Direct and Inverse Proportions
Two quantities, x and y, are said to be in direct proportion, if they increase (decrease) together in such a manner that the ratio of their corresponding values remains constant.
That is, $\frac{x}{y}=k,$
where k is a positive number.
For example, let r be the radius of a circle.
Now, which is a constant
Thus, the circumference of the circle and its radius are in direct proportion.
• If y1, y2 are the values of y corresponding to the values x1, x2 of x respectively, then $\frac{{x}_{1}}{{y}_{1}}=\frac{{x}_{2}}{{y}_{2}}$ is a case of direct proportion.
Two quantities, x and y, are said to be in inverse proportion, if an increase in x causes a proportional decrease in y (and vice-versa) in such a manner that the product of their corresponding values remains constant. That is, xy = k, where k is a positive number.
For example, consider the number of employers and the time taken to finish the work in an organisation. More the number of employers less will be the time taken to finish the work. Therefore, this is a case of inverse proportion.
• If y1, y2 are the values of y corresponding to the values x1, x2 of x respectively, then x1y1 = x2y2 is a case of indirect proportion.
Example: If the cost of 5 movie tickets is Rs 875, then find the cost of 2 tickets.
Solution:
Suppose the cost of 2 tickets and the cost of 7 tickets be Rs x and Rs y respectively. We can form a table as shown below.
Number of tickets 2 5 Cost of tickets (Rs) x 875
As the number of tickets increases, the cost of tickets will also increase in the same ratio. It is a case of direct proportion.
$\frac{2}{x}=\frac{5}{875}\phantom{\rule{0ex}{0ex}}⇒x=\frac{2×875}{5}=2×175=350$
Thus, the cost of 2 tickets is Rs 350.
Chapter 14: Factorisation
Factors of an algebraic term can be numbers or algebraic variables or algebraic expressions.
For example, the factors of 2a2b are 2, a, a, b, since 2a2b = 2 × a × a × b
The factors, 2, a, a, b, are said to be irreducible factors of 2a2b since they cannot be expressed further as a product of factors.
Factorisation of expressions by the method of common factors
This method involves the following steps.
Step 1: Write each term of the expression as a product of irreducible factors.
Step 2: Observe the factors, which are common to the terms and separate them.
Step 3: Combine the remaining factors of each term by making use of distributive law.
Example: Factorize $12{p}^{2}q+8p{q}^{2}+18pq.$
Solution:
We have, 12p2q = 2 × 2 × 3 × p × p × q
8pq2 = 2 × 2 × 2 × p × q × q
18pq = 2 × 3 × 3 × p × q
The common factors are 2, p, and q.
12p2q + 8pq2 + 18pq = 2 × p × q [(2 × 3 × p) + (2 × 2 × q) + (3 × 3)] = 2pq (6p + 4q + 9)
Factorisation by regrouping terms
Sometimes, all terms in a given expression do not have a common factor. However, the terms can be grouped by trial and error method in such a way that all the terms in each group have a common factor. Then, there happens to occur a common factor amongst each group, which leads to the required factorization.
Example: Factorise 2a2b + 2aab.
Solution:
2a2b + 2a ab
= 2a2 + 2abab (The terms, 2a2 and 2a, have common factors, 2 and a.)
= 2a (a + 1) – b (1 + a) (The terms, –b and –ab, have common factors, –1 and b.)
= (a + 1) (2ab)
Factorisation using identities: Some of the expressions can also be factorized by making use of the following identities.
a2 + 2ab + b2 = (a + b)2
a2 – 2ab + b2 = (ab)2
a2b2 = (a + b) (ab)
x2 + (a +b)x + ab = (x + a) (x + b)
Division of any polynomial by a monomial is carried out either by dividing each term of the polynomial by the monomial or by the common factor method.
Division of a polynomial by a polynomial is carried out by factorizing both the polynomials and then cancelling out the common factors.
Example: Divide (35x2 – 70x – 105) by (7x + 7).
Solution:
We have, 35x2 – 70x – 105 = 35 (x2 – 2x – 3) = 35(x2 + x – 3x – 3) = 35[x(x + 1) – 3(x + 1)]= 35(x + 1) (x – 3) and 7x + 7 = 7(x + 1)
$\therefore \frac{35{x}^{2}-70x-105}{7x+7}=\frac{35×\left(x+1\right)×\left(x-3\right)}{7×\left(x+1\right)}=\frac{5×7×\left(x+1\right)×\left(x-3\right)}{7×\left(x+1\right)}=5\left(x-3\right)$
Points to be remembered while carrying out calculations in algebra
• The coefficient, 1, of a term is usually not written. However, while adding or subtracting like terms, we include them also.
• While substituting a negative value, we make use of brackets.
• While multiplying an expression enclosed in a bracket by a constant (or a variable) outside, each term of the expression has to be multiplied by the constant (or the variable).
• While squaring a monomial, the numerical coefficient and each factor have to be squared.
• Before applying an identity, it has to be made sure whether it can be really applied or not.
For example: (x – 5)2x2 – 52. However, (x – 5)2 = x2 – 10x + 25
Chapter 15: Introduction to Graphs
A given data can be represented graphically to make it easier to read and understand.
A data that changes continuously over periods of time can be displayed by a line graph.
For example, consider that the given table shows the profit earned by a shopkeeper in the last five years of his business.
Year 2008 2007 2006 2005 2004 Profit (in Rs) 50,000 60,000 40,000 80,000 90,000
On representing the years on the horizontal axis and the profit earned on the vertical axis,
The given information can be represented by a line graph as
From the above line graph, a lot of information can be interpreted. For example, it can be observed from the line graph that the profit earned by the shopkeeper was maximum in the year 2004 and minimum in the year 2006.
A line graph with a whole unbroken line is called a linear graph.
For locating a point on a graph sheet, the x and y-coordinates of the point are required.
• The horizontal axis is called the x-axis and the vertical axis is called the y-axis.
• The x-coordinate of a point refers to the number of units to be moved to the right from the vertical axis and the y-coordinate of a point refers to the number of units to be moved up from the horizontal axis.
For example, the point (2, 4) can be located by moving 2 units from the left edge and then 4 units from the bottom edge in the graph. This can be shown as
The point O with coordinates (0, 0) is called the origin.
There are many applications of linear graphs. The relation between two variables, one of which is an independent variable (called control variable) and the other is a dependent variable, can be represented through a linear graph.
For example, the relationship between quantity and cost, time and distance can be represented by a linear graph. In this case, quantity and time are independent variable and the cost and the distance are dependent variable as they depend on the quantity and time respectively.
Relationship between principal and simple interest can be represented by a linear graph. In this case, principal is an independent variable and simple interest is a dependent variable.
Example: The amount deposited in a bank and the corresponding simple interest for a year given by the bank are given as follows.
Principal (in Rs) 1500 2000 3000 4000 Simple interest (in Rs) 75 100 150 200
Draw a graph for the given data. From the graph, find the simple interest, if the principal is Rs 1000 and also find the principal whose simple interest is Rs 175.
Solution:
We represent principal on the horizontal axis and simple interest on the vertical axis.
Scale: 1 unit = Rs 1000 on the horizontal axis and 1 unit = Rs 50 on the vertical axis
Plot the points, (1500, 75), (2000, 100), (4000, 200), on a graph paper. Join the four points.
Note that the graph will pass through the origin since the simple interest on the principal of Rs 0 is Rs 0.
From the graph, it can be observed that corresponding to Rs 1000 on the horizontal axis, we obtain the interest as Rs 50 on the vertical axis.
Therefore, for the principal of Rs 1000, the simple interest is Rs 50.
Also, corresponding to Rs 175 on the vertical axis, we obtain Rs 3500 on the horizontal axis.
Therefore, the simple interest of Rs 175 can be earned if the principal is Rs 3500.
Chapter 16: Playing with Numbers
Any two-digit number, ab, made of digits, a and b, can be written in general form as
ab = 10a + b whereas as a three-digit number, abc, made of digits, a, b, and c, can be written as abc = 100a + 10b + c
Some properties of two and three digit numbers
• The sum of a two-digit number and the number formed by reversing the digits is always a multiple of 11.
For example, 28 + 82 = 110 = 11 × 10 is a multiple of 11
• The difference between a two-digit number and the number formed by reversing the digits is always a multiple of 9.
For example, 91 – 19 = 72 = 9 × 8 is a multiple of 9
• The difference between a three-digit number and the number formed by reversing the digits is always a multiple of 99.
For example, 923 – 329 = 594 = 99 × 6 is a multiple of 99
• If abc is a three-digit number, then (abc+ cab + bca) will be a multiple of 37.
For example, 158 + 581 + 815 = 1554 = 37 × 42 is a multiple of 37
We can solve puzzles related to addition and multiplication in which some of the digits are denoted by letters and we have to find the number by which the letters are replaced.
Rules while solving puzzles are as follows.
• Each letter in the puzzle stands for just one digit.
• The first digit of a number cannot be zero.
Example: Find the digits, P and Q, in the given puzzle.
Solution:
The ones digit of the number should be obtained as 0 on multiplying 2 by Q.
Therefore, the possibility for Q is either 0 or 5.
However, Q cannot be 0 because then, the result obtained after multiplication will be 0. Therefore, Q = 5
In the tens column, there will be a carry of 1, which was obtained as tens digit on multiplication of 2 by 5 in the ones column.
Therefore, we should obtain, 1 + P × 5 = 36
P × 5 = 35
P = 7
Therefore, the only possibility for P is 7.
Thus, the puzzle can now be solved as below.
Divisibility rules
• A number is divisible by 10, if its ones digit is 0.
• A number is divisible by 5, if its ones digit is either 0 or 5.
• A number is divisible by 2, if its ones digit is 0, 2, 4, 6, or 8 i.e., if the ones digit is even.
• A number is divisible by 9, if the sum of its digits is divisible by 9.
• A number is divisible by 3, if the sum of its digits is divisible by 3.
Example: If the four-digit number, 835a, is divisible by 9, then what is the value of a?
Solution:
It is known that if a number is divisible by 9, then the sum of its digits is divisible by 9. It is given that the number, 835a, is divisible by 9.
Therefore, 8 + 3 + 5 + a = 16 + a is divisible by 9.
This is possible when 16 + a = 9 or 18 or 27….
However, since a is a digit,
16 + a = 18
a = 2
What are you looking for?
Syllabus
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### Addition Show What You Know
DIRECTIONS 1–2. Count and tell how many.
Draw a set with one more counter. Write how many in each set.
3. Write the number of cubes in each set. Circle the number that
is greater than the other number.
More
Question 1.
Each set – 2, so total 4,
Explanation:
Given one set it has 2 and drawn one more
counter which has 2, So both sets has
2 each making 4 in total.
Question 2.
Each set – 4, so total 8,
Explanation:
Given one set it has 4 and drawn one more
counter which has 4, So both sets has
4 each making 8 in total.
Compare Numbers to 10
Question 3.
Each set 1 has 7, set 2 has 4 as 7 > 4 so circled the number
that is greater than the other number.
Explanation:
Given one set it has 7 and second one
has 4, if we compare 7 is greater than 4, So
circled it.
DIRECTIONS Count and tell how many birds are on the ground.
Count and tell how many birds are flying. Write these numbers to
show a pair of numbers that make ten.
7 birds are on the ground and 3 birds are flying,
Numbers that make sum ten are 7 + 3 = 10,
Explanation:
As shown in the picture 7 birds are on the ground,
3 birds are flying and pair of numbers that make ten
are 7 + 3 = 10.
### Addition Game Pairs That Make 7
DIRECTIONS Play with a partner. The first player rolls the number
cube and writes the number on the yellow boat.
Partners determine what number makes 7 when paired
with the number on the yellow boat. Players take turns rolling
the number cube until that number is rolled. Write the number
beside it on the green boat. Partners continue to roll the number
cube finding pairs of numbers that make 7.
MATERIALS number cube (1–6)
Numbers that make 7 are
2 + 5 = 7,
6 + 1 = 7,
5 + 2 = 7,
4 + 3 = 7,
1 + 6 = 7,
3 + 4 = 7.
Explanation:
Given the first player rolls the number cube and writes the
number on the yellow boat. Partners determine what number makes
7 when paired with the number on the yellow boat.
Players take turns rolling the number cube until that number
is rolled and write the number beside it on the green boat
as shown above, Partners continue to roll the number cube
finding pairs of numbers that make 7 as
2 + 5 = 7,
6 + 1 = 7,
5 + 2 = 7,
4 + 3 = 7,
1 + 6 = 7,
3 + 4 = 7.
DIRECTIONS Shuffle the Vocabulary Cards and place in a pile.
A player takes the top word card from the pile and tells what they
know about the word. The player puts a counter on that word
on the board. Players take turns. The first player to cover all the words
on his or her board says “Bingo.
MATERIALS 2 sets of Vocabulary Cards, 6 two color counters for each player
Both Player 1 and Player 2 cover all the words same time so both
says “Bingo.
Explanation:
Given 2 sets of Vocabulary Cards, 6 two color counters for each player,
Shuffling the Vocabulary Cards and placing in a pile.
A player takes the top word card from the pile and tells
what they know about the word. The player puts a counter
with green color on that word on the board. Players take turns.
The first player to cover all the words on his or her board says “Bingo.
both Player 1 and Player 2 cover all the words same time so both
says “Bingo.
The Write Way
DIRECTIONS Draw and write to show how to find a number pair for 5.
Number pair for 5 are (3, 2), (4, 1) and (5, 0),
Explanation:
To show a number pair for 5 we take numbers as x and y,
So, x + y = 5 and we will take numbers below 5 only,
as if x = 3 then y = 5 – x = 5 -3 = 2,
now if x = 4 then y = 5 – 4 = 1,
now if x = 5 then y = 5 – 0 = 5,
So number pair for 5 is (3, 2), (4, 1) and (5, 0).
Share and Show
DIRECTIONS 1. Listen to the addition word problem. Trace the
number that shows how many children are sitting eating lunch.
Write the number that shows how many children are being added
to the group. Write the number that shows how many children are
having lunch now.
Question 1.
3 children are sitting eating lunch,
1 child being added to the group,
total 4 children are having lunch now.
Explanation:
Tracing the number that shows how many children are
sitting and eating lunch are 3. Now the number 1 shows
children are being added to the group, therefore
total 3 + 1 = 4 children are having lunch now.
DIRECTIONS 2. Listen to the addition word problem. Write the
number that shows how many children are playing with the soccer ball.
Write the number that shows how many children are being added
to the group. Write the number that shows how many children
there are now.
Question 2.
2 children are playing with the soccer ball,
3 children are being added to the group,
Now there are 5 children now,
Explanation:
The number 2 shows children are playing with the soccer ball.
3 shows how many children are being added to the group.
The number that shows how many children are there now are
2 + 3 = 5.
Problem Solving • Applications
DIRECTIONS 3. Two sheep are in a pen. Two sheep are added to
the pen. How can you write the numbers to show the sheep in the
pen and the sheep being added?
4. Write how many sheep are in the pen now.
Question 3.
2 sheep in a pen and 2 sheep are added to pen,
Explanation:
Given two sheep are in a pen, two sheep are added to the pen,
So first we write number 2 and again write 2 number as shown above.
Question 4.
__________
_ _ _ _ _ _ _
___________
Four sheep are in the pen now,
Explanation:
Given two sheep are in a pen, two sheep are added to the pen,
means 2 + 2 = 4, therefore 4 sheep are in the pen now.
HOME ACTIVITY • Show your child a set of four objects.
Have him or her add one object to the set and tell how
many there are now.
Set of four objects and one object make 5 now,
Explanation:
Given my child has a set of four objects and
added one object to the set now there are
4 + 1 = 5 now.
DIRECTIONS 1. There are four red counters in the five frame.
One yellow counter is added. R is for red, and Y is for yellow.
How many are there of each color counter? Write the numbers.
2. Write the number that shows how many counters are in the five frame now.
Question 1.
R – Red-4
Y – Yellow -1,
Explanation:
Given there are four red counters in the five frame.
One yellow counter is added. R is for red and Y is for yellow.
There are 4 Red counters and 1 Yellow counter in the five frame.
Question 2.
__________
_ _ _ _ _ _ _
___________
There are 5 counters are in the five frame now,
Explanation:
Given there are four red counters in the five frame.
One yellow counter is added, So number of counters available
now are 4 + 1 = 5 counters are in the five frame.
DIRECTIONS 1. How many of each color counter? Write the numbers.
2. Count and tell how many balloons. Write the number.
3.
Count and tell how many in each set. Write the numbers.
Compare the numbers. Circle the number that is less.
Lesson Check
Question 1.
2 white color counters, 3 light black counters,
Explanation:
Counting in the frame there are 2 white color counters and
3 light black counters in the frame.
Spiral Review
Question 2.
10 balloons,
Explanation:
Counting the balloons there are 10 in number.
Question 3.
First set 3,
Second set 2,
2 is less than 3, So circled it,
Explanation:
After counting in set 1 there are 3 in set 2 there are 2,
Now on comparing set 1 and set 2 2 is less than 3,
2 < 3, So circled the 2 as shown above.
### Lesson 5.2 Addition: Put Together
Essential Question How can you show addition as putting together?
There were 6 birds and 3 more joined them.
How many birds are there now?
So on adding it will be 6 + 3 = 9 birds now.
Listen and Draw
DIRECTIONS Listen to the addition word problem. Place red and
yellow counters in the ten frame as shown. Trace the numbers and
the symbol to show the sets that are put together. Write the number
that shows how many in all.
yellow – 6 and red – 1,
In all 6 + 1 = 7,
Explanation:
In the ten frame yellow counters are 6 and red
counter is 1 in total there are 6 + 1 = 7 in all.
Share and Show
DIRECTIONS 1. Listen to the addition word problem.
Place red counters in the ten frame as shown. Place yellow
counters to model the sets that are put together. Write the
numbers and trace the symbol. Write the number to show how many in all.
Question 1.
Red 7, Yellow 2, In all – 9,
Explanation:
Red counters – 7,
Yellow counters – 2,
So in all there are 7 + 2 = 9 counters in
the ten frame.
DIRECTIONS 2. Listen to the addition word problem.
Place counters in the ten frame to model the sets that are put together.
How many are there of each color counter? Write the numbers and
trace the symbol. Write the number to show how many in all.
Question 2.
2 red
8 yellow,
In all 10,
Explanation:
Placed in Red counters – 2,
Yellow counters -8,
So in all there are 2 + 8 = 10 counters in
the ten frame.
Problem Solving • Applications
DIRECTIONS 3. Four red apples and two green apples are on the table.
Write the numbers and trace the symbol to show the apples
being put together.
4. Write the number to show how many apples in all.
Question 3.
4 red apples, 2 green apples,
Explanation:
Wrote numbers as 4 red apples, 2 green apples
and traced the symbol to show the apples being put together.
Question 4.
___________
_ _ _ _ _ _ _ _
___________
6 apples in all,
Explanation:
There are four red apples and two green apples,
In all there are 4 + 2 = 6 apples.
HOME ACTIVITY • Show your child two sets of four objects.
Have him or her put the sets of objects together and tell how many in all.
Two sets of four objects put together makes 8 in all,
Explanation:
Given my child has 2 sets of four objects now
put the sets of objects together makes 4 + 4 = 8 in all.
### Addition: Put Together Homework & Practice 5.2
DIRECTIONS Roy has three yellow counters and five red counters.
How many counters does he have in all? 1. Place counters in the
ten frame to model the sets that are put together. Y is for yellow,
and R is for red. Write the numbers and trace the symbol. Write the
number to show how many in all.
Question 1.
Roy has 3 yellow and 5 red counters,
So, 8 counters in all,
Explanation:
Given Roy has three yellow counters and five red counters.
So in all Roy have 3 + 5 = 8 counters
Wrote the numbers and traced the symbols,
wrote the number 8 to show how many are there in all.
DIRECTIONS 1. What numbers show the sets that are put together?
Write the numbers and trace the symbol. 2. Count the dots in the
ten frames. Begin with 6. Write the numbers in order as
you count forward.
3. Paul has a number of counters two less than seven.
Draw the counters in the ten frame. Write the number.
Lesson Check
Question 1.
Numbers 7 and 1,
Explanation:
Numbers 7 and 1 are put together,
Wrote the numbers and traced the symbols.
Spiral Review
Question 2.
Beginning with 6 the numbers in order as we count forward are
6, 7, 8, 9, 10.
Explanation:
Counted the dots in the ten frames, Beginning with 6,
the numbers in order as we count forward are
6, 7, 8, 9, 10.
Question 3.
Paul has 5 number of counters,
Explanation:
Given Paul has a number of counters two less than seven,
means 7 – 2 = 5, Drawn the counters in the ten frame and
wrote the number 5.
### Lesson 5.3 Problem Solving • Act Out Addition Problems
Essential Question How can you solve problems using the strategy act it out?
Problem Solving Strategies
a. Guess (this includes guess and check, guess and improve),
b. Act It Out (act it out and use equipment),
c. Draw (this includes drawing pictures and diagrams),
d. Make a List (this includes making a table),
e. Think (this includes using skills you know already).
Try Another Problem
DIRECTIONS 1. Listen to and act out the addition word problem.
Trace the numbers and the symbols. Write the number that shows
how many children in all.
Question 1.
Numbers are 2, 1, 3 children in all,
Explanation:
Numbers are given as 2, 1,
therefore children in all are 2 + 1 = 3.
Share and Show
DIRECTIONS 2. Listen to and act out the addition word problem.
Trace the numbers and the symbols. Write the number that shows
how many children in all.
Question 2.
Numbers are 3, 2, 5 children in all,
Explanation:
Numbers are given as 3, 2,
therefore children in all are 3 + 2 = 5.
Trace the numbers and the symbols. Write the number that shows
how many puppies there are now.
4. Draw a picture to match this addition sentence.
Question 3.
There are 3 puppies, 1 more puppy is added,
Tell how many puppies are there now?,
Numbers are 3 and 1,
4 puppies are there now in all,
Explanation:
An addition word problem is- There are 3 puppies,
1 more puppy is added, Tell how many puppies are there now ?,
number of puppies are 3 and 1 in total there are 3 + 1 = 4
puppies are there now in all.
Question 4.
Addition Sentence is 1 + 4 = 5,
There is 1 bird flying in the sky again 4 birds joined,
So in all how many birds are flying now?
Explanation:
Drawn a picture of birds flying in the sky according
to match this addition sentence as there is one bird flying
in the sky again 4 birds joined, So in all there are 5 birds
flying.
I will explain my friend about my drawing by showing the
picture there is 1 bird first and 4 birds joined making
in total 5 birds flying.
adding three objects to a set of two objects. Have your child
use toys to act out the word problem.
Three objects to a set of objects,
3 + 2 put together makes 5 in all,
Yes, My child uses toys to act out the
word problem,
Explanation:
Given my child has 3 objects adding to a set of two objects, now
put the sets of objects together makes 3 + 2 = 5 in all.
Yes, My child uses toys to act out the word problem,
Few toys names are Magna-Tiles 32-Piece Solid Colors Set,
Think Fun Zingo etc.
### Problem Solving • Act Out Addition Problems Homework & Practice 5.3
Trace the numbers and the symbols. Write the number that shows
how many children in all.
Question 1.
There are 4 children in the bus, 1 more child is added,
Tell how many children are there now in the bus?,
Numbers are 4, 1, 5 children are there now in all in the bus.
Explanation:
There are 4 children in the bus, 1 more child is added,
Tell how many children are there now in the bus?,
Number of children are 4,1 in total there are 4 + 1 = 5
children are there now in all in the bus.
Question 2.
There are 3 children in the canteen to have meals,
Tell how many children are there now in the canteen?,
Numbers are 3, 2, 5 children are there now in all in the canteen.
Explanation:
There are 3 children in the canteen to have meals,
Number of children are 3,2 in total there are 3 + 2 = 5
children are there now in all in the canteen to have meals.
DIRECTIONS 1. Tell an addition word problem.
Trace the numbers and the symbols. Write the number that
shows how many cats in all. 2. Count and tell how many tigers.
Write the number. 3. Count how many bears. Write the number.
Draw to show a set of counters that has the same number
as the set of bears. Write the number.
Lesson Check
Question 1.
There are 3 cats sitting on the chair in the park,
2 more cats are moving near the chair,
Tell how many cats are there now in the park?,
Numbers are 3, 2, 5 cats are there now in all in the park.
Explanation:
There are 3 cats sitting on the chair in the park,
2 more cats are added near the chair,
Number of cats are 3,2 in total there are 3 + 2 = 5
cats are there now in all in the park.
Spiral Review
Question 2.
6 tigers,
Explanation:
Counting the number of tigers there are 6 in all.
Question 3.
2 bears,
2 bears in one set and 2 more bears in another set,
In total 2 + 2 = 4,
Explanation:
Counting 2 number of bears. Wrote 2,
Drawn to show a set of counters that has the same number
as the set of bears is two more bears. So in total there are
2 + 2 = 4 bears.
### Lesson 5.4 Algebra • Model and Draw Addition Problems
Essential Question How can you use objects and drawings to
When we solve a word problem, look for important words.
These words usually mean we need to add: altogether, total,
plus, sum, make, and, overall.
1. Read the Entire Word Problem. …
2. Think About the Word Problem. …
3. Write on the Word Problem. …
4. Draw a Simple Picture and Label It. …
5.
Estimate the Answer Before Solving. …
6. Check Your Work When Done. …
7. Practice Word Problems Often.
So, objects and drawings helps us to easily solve addition word problems.
Share and Show
DIRECTIONS 1–2. Place cubes as shown. Listen to the addition
word problem.
Model to show the cubes put together. Draw the cube train.
Trace and write to complete the addition sentence.
Question 1.
1 cube and 2 cubes makes 3 cubes in all,
Explanation:
Placed cubes as shown above. Traced and wrote to complete
the addition sentence as 1 + 2 = 3 cubes in all.
Question 2.
1 cube and 3 cubes makes 4 cubes in all,
Explanation:
Placed cubes as shown above. Traced and wrote to complete
the addition sentence as 1 + 3 = 4 cubes in all.
DIRECTIONS 3–4. Place cubes as shown. Listen to the addition
word problem.
Model to show the cubes put together. Draw the cube train.
Trace and write to complete the addition sentence.
Question 3.
4 cubes and 1 cube makes 5 cubes in all,
Explanation:
Placed cubes as shown above. Traced and wrote to complete
the addition sentence as 4 + 1 = 5 cubes in all.
Question 4.
1 cube and 4 cubes makes 5 cubes in all,
Explanation:
Placed cubes as shown above. Traced and wrote to complete
the addition sentence as 1 + 4 = 5 cubes in all.
### Algebra • Model and Draw Addition Problems Homework & Practice 5.4
DIRECTIONS 1–2. Place cubes as shown. B is for blue, and Y is for yellow.
Tell an addition word problem. Model to show the cubes put together.
Draw the cube train. Trace and write the numbers to complete the addition sentence.
Question 1.
2 B- Blue cubes and 1 Y – Yellow cube makes 3 cubes in all,
Explanation:
Placed cubes as shown above. 2 Blue and 1 Yellow cubes
together, Traced and wrote to complete
the addition sentence as 2 + 1 = 3 cubes in all.
Question 2.
1 B- Blue cubes and 3 Y – Yellow cube makes 4 cubes in all,
Explanation:
Placed cubes as shown above. 1 Blue and 3 Yellow cubes
together, Traced and wrote to complete
the addition sentence as 1 + 3 = 4 cubes in all.
DIRECTIONS 1. Look at the cube train. Tell an addition word problem.
Trace and write to complete the addition sentence.
2. How many more counters would you place to model a
way to make 7? Draw the counters.
3. Draw counters to make a set that shows the number.
Lesson Check
Question 1.
A child has 3 cubes with him, his friend gave him 1 more
cube in total how many cubes the child have?
3, 1 in total makes 4 cubes,
Explanation:
Looking at the cube train wrote one addition problem as
A child has 3 cubes with him, his friend gave him 1 more
cube in total how many cubes the child have?
Traced and wrote to complete
the addition sentence as 1 + 3 = 4 cubes in all.
Spiral Review
Question 2.
5 more counters I would place to model a way to
make 7, 2 + 5 =7.
Explanation:
Already there are 2 counters, to make 7 in all we need
7 – 2 = 5 more counters, So 2 + 5 = 7,Drawn 5 more counters.
Question 3.
4 = 2 + 2,
Explanation:
To make a set that shows the number 4 is
2 + 2 = 4.
Concepts and Skills
DIRECTIONS 1. Write the number that shows how many
puppies are sitting.
Write the number that shows how many puppies are
2. Write the numbers and trace the symbol to show
the sets that are put together. (K.OA.A.1)
3.
Circle all the ways that show how many in all. (K.OA.A.1)
Question 1.
2 puppies are sitting,
Explanation:
2 number of puppies are sitting.
1 number of puppy is being added to them.
shown in the picture above.
Question 2.
5 yellow and 3 red numbers respectively,
Explanation:
As shown in the frame the numbers are
5 yellow and 3 red numbers traced the symbol to
the sets that are put together.
Question 3.
THINK SMARTER
1. 1 + 2,
2. 1 plus 2
Explanation:
Given 1 blue cube and 2 orange cubes,
So circled all the ways that showed how many in all
as 1 + 2 and 1 plus 2.
### Lesson 5.5 Algebra • Write Addition Sentences for 10
Essential Question How can you use a drawing to find the number
that makes a ten from a given number?
Counting the given number in the drawing and
checking so that it is a ten.
Share and Show
DIRECTIONS 1–3. Look at the cube train. How many red cubes
do you see? How many blue cubes do you need to add to make 10?
Use blue to color those cubes. Write and trace to show this as
Question 1.
8 red cubes, 2 blue cubes,
addition sentence – 10 cubes in all,
Explanation:
Looking at the cube train, 8 red cubes I can see,
2 blue cubes I need to add to make 10,
Wrote and traced to show this as an addition sentence
as 8 + 2 = 10.
Question 2.
7 red cubes, 3 blue cubes,
addition sentence – 10 cubes in all,
Explanation:
Looking at the cube train, 7 red cubes I can see,
3 blue cubes I need to add to make 10,
Wrote and traced to show this as an addition sentence
as 7 + 3 = 10.
Question 3.
6 red cubes, 4 blue cubes,
addition sentence – 10 cubes in all,
Explanation:
Looking at the cube train, 6 red cubes I can see,
4 blue cubes I need to add to make 10,
Wrote and traced to show this as an addition sentence
as 6 + 4 = 10.
DIRECTIONS 4–6. Look at the cube train. How many red cubes
do you see? How many blue cubes do you need to add to make 10?
Use blue to draw those cubes. Write and trace to show this as an
Question 4.
5 red cubes, 5 blue cubes,
addition sentence – 10 cubes in all,
Explanation:
Looking at the cube train, 5 red cubes I can see,
5 blue cubes I need to add to make 10,
Used blue to draw 5 cubes, Wrote and traced to show this as
an addition sentence as 5 + 5 = 10.
Question 5.
4 red cubes, 6 blue cubes,
addition sentence – 10 cubes in all,
Explanation:
Looking at the cube train, 4 red cubes I can see,
6 blue cubes I need to add to make 10,
Used blue to draw 6 cubes, Wrote and traced to show this as
an addition sentence as 4 + 6 = 10.
Question 6.
3 red cubes, 7 blue cubes,
addition sentence – 10 cubes in all,
Explanation:
Looking at the cube train, 3 red cubes I can see,
7 blue cubes I need to add to make 10,
Used blue to draw 7 cubes, Wrote and traced to show this as
an addition sentence as 3 + 7 = 10.
Problem Solving • Applications
DIRECTIONS 7. Troy has 2 ducks. How many more ducks
does he need to get to have 10 ducks in all? Draw to solve
the problem. Trace and write to show this as an addition sentence.
8. Draw to find the number that makes 10 when put together
with the given number. Trace and write to show this as an
Question 7.
Troy need 8 more ducks,
addition sentence – 2 + 8 = 10 ducks in all,
Explanation:
Given Troy has 2 ducks, Troy need 8 more ducks
to get to have 10 ducks in all, Drawn to solve
the problem, Traced and wrote to show this as an
addition sentence as 2 + 8 = 10.
Question 8.
The number that makes 10 when put together
with the given number 1 is 9,
addition sentence – 1 + 9 = 10,
Explanation:
Drawn the number that makes 10 when put together
with the given number 1 is 9 (10 – 1 = 9). Traced and
wrote to show this as an addition sentence as 1 + 9 = 10.
HOME ACTIVITY
Show your child a number from 1 to 9.
Ask him or her to find the number that makes 10 when put
together with that number. Then have him or her tell a story
to go with the problem.
5 is the number that makes 10 when put
together with that number,
Explanation:
Shown few numbers 1 to 9 to find the number that makes
10 when put together with that number, The child
says 5 number because let us take the number as x,
x + x = 10, 2x = 10, therefore x = 10 ÷ 2 = 5.
Story:
I have 10 cards from 2 pack of cards,
When each pack is put together it makes 10,
So tell me how many number of cards are there in
in each pack of cards?
### Algebra • Write Addition Sentences for 10 Homework & Practice 5.5
DIRECTIONS 1–3. Look at the cube train. How many gray
cubes do you see? How many blue cubes do you need to
add to make 10? Use blue to color those cubes. Write and
trace to show this as an addition sentence.
Question 1.
7 gray cubes, I need 3 blue cubes to make 10,
addition sentence – 7 + 3 = 10,
Explanation:
7 gray cubes I can see, I need 3 blue cubes to
add to make 10, Used blue to color those cubes,
Wrote and traced to show this as an addition sentence as
7 + 3 = 10.
Question 2.
8 gray cubes, I need 2 blue cubes to make 10,
addition sentence – 8 + 2 = 10,
Explanation:
8 gray cubes I can see, I need 2 blue cubes to
add to make 10, Used blue to color those cubes,
Wrote and traced to show this as an addition sentence as
8 + 2 = 10.
Question 3.
9 gray cubes, I need 1 blue cubes to make 10,
addition sentence – 9 + 1 = 10,
Explanation:
9 gray cubes I can see, I need 1 blue cubes to
add to make 10, Used blue to color those cubes,
Wrote and traced to show this as an addition sentence as
9 + 1 = 10.
DIRECTIONS 1. Look at the cube train. How many white cubes
are added to the gray cubes to make 10? Write and trace to show
2. Which number is less? Circle the number.
3. How many cubes are there? Write the number.
Model a cube train that has the same number of cubes.
Draw the cube train. Write how many.
Lesson Check
Question 1.
2 white cubes,
addition sentence – 8 + 2 = 10,
Explanation:
Looking at the cube train, there are 2 white cubes
added to the gray cubes to make 10? Wrote and traced
to show this as an addition sentence as 8 + 2 = 10.
Spiral Review
Question 2.
4 is less, So circled it,
Explanation:
Given numbers are 5 and 4 ,
as 4 is less than 5, 4 < 5,
Therefore circled the number 4.
Question 3.
4 cubes are there,
Added same number of 4 cubes,
Now in total there are 8 cubes in all,
Explanation:
There are 4 cubes, Wrote the number.
Added the same number of cubes.
Drawn the cube train. Now we have
4 + 4 = 8 cubes now in all.
### Lesson 5.6 Algebra • Write Addition Sentences
Essential Question How can you solve addition word problems
To solve addition word problems the main components of
a. The Relationship of the Numbers – Know the problem type and
help to solve for the action in the problem,
b. Differentiate the Numbers – Take just the right numbers so that
it can read the problem without getting bogged down
with the computation,
c. Use Academic Vocabulary – And be consistent in what we are using,
e. Differentiate between the Models and the Strategies – one has
to do with the relationship between the numbers and the other
has to “solve” or compute the problem.
Addition sentence is a number sentence used or
simply an equation used to express addition.
For example:
I have 2 cars, I purchased 3 more,
how many cars do I have now in all?
The addition sentence is 2 + 3 = 5,
So after solving addition word problems,
completing it with a addition sentence.
Share and Show
DIRECTIONS 1. Listen to the addition word problem.
Trace the circle around the set you start with. How many are
being added to the set? How many are there now?
2–3. Listen to the addition word problem. Circle the set you
How many are there now? Write and trace the numbers to
Question 1.
Circled the set which I have started with.
Traced the numbers, 2 numbers and
3 more numbers added, Total 5 are
there now, addition sentence is 2 + 3 = 5,
Explanation:
Circled the set which I have started with.
3 more numbers are added to the set 2 numbers,
there are 5 in all now,
Traced the addition sentence as 2 + 3 = 5.
Question 2.
Circled the set which I have started with.
Traced the numbers, 1 number and
3 more numbers added, Total 4 are there now,
addition sentence is 1 + 3 = 4,
Explanation:
Traced the circle around the set which I have started with.
3 more numbers are added to the set 1 number,
there are 4 in all now,
Traced the addition sentence as 1 + 3 = 4.
Question 3.
Circled the set which I have started with.
Traced the numbers, 3 numbers and 2 more
numbers added, Total 5 are there now,
addition sentence is 3 + 2 = 5,
Explanation:
Traced the circle around the set which I have started with.
2 more numbers are added to the set 3 numbers,
there are 5 in all now,
Traced the addition sentence as 3 + 2 = 5.
How many are there now? Write and trace the numbers to complete
Question 4.
Circled the set which I have started with.
Traced the numbers, 1 number and 4 more numbers added,
Total 5 are there now,
addition sentence is 1 + 4 = 5,
Explanation:
Traced the circle around the set which I have started with.
4 more numbers are added to the set 1 number,
there are 5 in all now,
Traced the addition sentence as 1 + 4 = 5.
Question 5.
Circled the set which I have started with.
Traced the numbers, 3 numbers and 1 more numbers added,
Total 4 are there now,
addition sentence is 3 + 1 = 4,
Explanation:
Traced the circle around the set which I have started with.
1 more number is added to the set 3 numbers,
there are 4 in all now,
Traced the addition sentence as 3 + 1 = 4.
Question 6.
Circled the set which I have started with.
Traced the numbers, 2 numbers and 3 more
numbers are added, Total 5 are there now,
addition sentence is 2 + 3 = 5,
Explanation:
Traced the circle around the set which I have started with.
3 more numbers are added to the set 2 numbers,
there are 5 in all now,
Traced the addition sentence as 2 + 3 = 5.
Problem Solving • Applications
DIRECTIONS 7. Bill catches two fish. Jake catches some fish.
They catch four fish in all. How many fish does Jake catch?
Draw to show the fish. Trace and write to complete the
Question 7.
Jake catches 2 fish,
addition sentence : 2 + 2 = 4,
Explanation:
Given Bill catches two fish. Jake catches some fish.
They catch four fish in all. So number of fishes
Jake catch are 4 – 2 = 2, Drawn to show the fish.
Traced and wrote to complete the addition sentence
as 2 + 2 = 4.
Question 8.
I catch 2 fish, My friend catch 2 fish,
In all 4 fish,
addition sentence 2 + 2 = 4,
Explanation:
I and my friend went for fishing,
I catch 2 fishes by afternoon, My friend also
catch fish by evening, Both catch four fish in all,
So my friend has catch 4 – 2 = 2 fish,
Drawn to show the fish. The drawing tells us
about I had 2 fish plus my friend has 2 fish in all
total both have 4 number of fish as shown,
the addition sentence is 2 + 2 = 4.
HOME ACTIVITY • Have your child show three fingers.
Have him or her show more fingers to make five fingers in all.
Then have him or her tell how many more fingers he or she showed.
3 fingers, He showed 2 more fingers
to make five fingers in all,
Explanation:
My child shows three fingers, asked him or her
Then asked him or her to tell how many more
fingers he or she showed. (So that 3 + 2 = 5 fingers in all).
### Algebra • Write Addition Sentences Homework & Practice 5.6
the sets. Circle the set you start with. How many are being
added to the set? How many are there now?
Write and trace to complete the addition sentence.
Question 1.
3 boats were their in the lake, 2 more joined,
How many boats are there now in all?
Circled the 3 numbers which I started with,
2 more are added, 5 in total,
addition sentence – 3 + 2 = 5,
Explanation:
3 boats were their in the lake, 2 more joined,
How many boats are there now in all?
Circled the set 3 numbers which I started with. 2 more are
added to the set, there are now 5 boats in all,
Wrote and traced to complete the addition sentence as
3 + 2 = 5.
Question 2.
1 helicopter is their on the ground, 3 more joined,
How many helicopters are there now in all?
Circled the 1 number which I started with,
3 more are added, 4 in total,
addition sentence – 1 + 3 = 4,
Explanation:
1 helicopter is their on the ground, 3 more joined,
How many helicopters are there now in all?
Circled the set 1 number which I started with. 3 more are
added to the set, there are now 5 helicopters in all,
Wrote and traced to complete the addition sentence as
1 + 3 = 4.
Question 3.
4 steamers are their in the river, 1 more joined,
How many steamers are there now in all?
Circled the 4 numbers which I started with,
1 more is added, 5 in total,
addition sentence – 4 + 1 = 5,
Explanation:
4 steamers are their in the river, 1 more joined,
How many steamers are there now in all?
Circled the set 4 numbers which I started with. 1 more is
added to the set, there are now 5 steamers in all,
Wrote and traced to complete the addition sentence as
4 + 1 = 5.
Write and trace to complete the addition sentence.
2. How many more counters would you place to model a
way to make 8? Draw the counters.
3. How many paintbrushes are there? Write the number.
Lesson Check
Question 1.
3 planes are their, 2 more are added,
How many in all now?
In all 5 planes now,
addition sentence- 3 + 2 = 5,
Explanation:
3 planes are their, 2 more are added to them,
How many in all now? In all 5 planes now,
Wrote and traced to complete the addition
sentence as 3 + 2 = 5.
Spiral Review
Question 2.
3 more counters,
drawn the counters,
Explanation:
Given 5 counters, more 3 counters we would
place to model a way to make 8 ,(8 – 5 = 3),
So that 3 + 5 = 8, Drawn the counters.
Question 3.
3 paintbrushes are there,
wrote the number,
Explanation:
3 paintbrushes are there,
Wrote the number 3 as shown above.
### Lesson 5.7 Algebra • Write More Addition Sentences
Essential Question How can you solve addition word
problems and complete the addition sentence?
To solve addition word problems the main components of
a. The Relationship of the Numbers – Know the problem type and
help to solve for the action in the problem,
b. Differentiate the Numbers – Take just the right numbers so that
it can read the problem without getting bogged down
with the computation,
c. Use Academic Vocabulary – And be consistent in what we are using,
e. Differentiate between the Models and the Strategies – one has
to do with the relationship between the numbers and the other
has to “solve” or compute the problem.
Addition sentence is a number sentence used or
simply an equation used to express addition.
For example:
I have 4 story books with me, I bought 3 more
story books, how many story books do I have
in all now? So, 4 + 3 = 7 is the addition sentence.
So after solving addition word problems,
Share and Show
DIRECTIONS 1. Listen to the addition word problem.
Circle the set being added. How many are in the set to
2–3. Listen to the addition word problem.
Write and trace to complete the addition sentence.
Question 1.
Circled the set 3 numbers being added.
addition sentence – 4 + 3 = 7,
Explanation:
Circled the set 3 numbers being added,
traced to complete the addition sentence
as 4 + 3 = 7.
Question 2.
Circled the set 7 numbers being added.
addition sentence – 2 + 7 = 9,
Explanation:
Circled the set 7 numbers being added,
Wrote and traced to complete the addition sentence
as 2 + 7 = 9.
Question 3.
Circled the set 4 numbers being added.
addition sentence – 6 + 4 = 10,
Explanation:
Circled the set 4 numbers being added,
Wrote and traced to complete the addition sentence
as 6 + 4 = 10.
DIRECTIONS 4–6. Tell an addition word problem.
Circle the set being added. Draw to show how many are
in the set to start with. Write and trace to complete the
Question 4.
In a class room chart there are 3 butterflies in
the chart, My friend added 5 more to it,
How many are there on the chart in all now?
Circled the set 5 numbers being added.
addition sentence – 3 + 5 = 8,
Explanation:
In a class room chart there are 3 butterflies in
the chart, My friend added 5 more to it,
How many are there on the chart in all now?
Circled the set 5 numbers being added,
Wrote and traced to complete the addition sentence
as 3 + 5 = 8.
Question 5.
In my home there are 6 cockroaches in
the kitchen, 3 more are added,
How many are cockroaches are there in all now?
Circled the set 3 numbers being added.
addition sentence – 6 + 3 = 9,
Explanation:
In my home there are 6 cockroaches in
the kitchen, 3 more are added,
How many are cockroaches are there in all now?
Circled the set 3 numbers being added,
Wrote and traced to complete the addition sentence
as 6 + 3 = 9.
Question 6.
In a park 2 butterflies are flying, 8 more are added,
How many are butterflies are there in all now?
Circled the set 8 numbers being added.
addition sentence – 2 + 8 = 10,
Explanation:
In a park 2 butterflies are flying, 8 more are added,
How many are butterflies are there in all now?
Circled the set 8 numbers being added,
Wrote and traced to complete the addition sentence
as 2 + 8 = 10.
Problem Solving • Applications
DIRECTIONS 7. Tell an addition word problem.
Complete the addition sentence. Draw a picture of real
Question 7.
I had 3 pens with me, My father gave me 2 more,
How many pens do I have now in all?
3 + 2 = 5 addition sentence,
Drawn a picture of objects as shown above,
Explanation:
I had 3 pens with me, My father gave me 2 more,
How many pens do I have now in all?
3 + 2 = 5, Drawn a picture of real objects to show
the problem, Telling my friend about my drawing as
starting see how many pens are there now It is added
with 2, now if we count It’s 5 means I have 5 pens now.
problem such as:
There are some socks in the drawer.
Now there are ten socks in the drawer. How many socks
Yesterday, there were some socks in the drawer,
I added four more socks today, Now there are ten
socks in the drawer, So how many socks were
in the drawer to start with, to make 10 socks in all?.
### Algebra • Write More Addition Sentences Homework & Practice 5.7
DIRECTIONS 1–4. Tell an addition word problem.
Circle the set being added. How many are in the set
Question 1.
There are 5 ducks in the pond, 4 more joined,
how many ducks are there now in all?
Circled the set 4 numbers being added.
4 + 5 = 9 is the addition sentence,
Explanation:
There are 5 ducks in the pond, 4 more joined,
how many ducks are there now in all?
Circled the set 4 numbers being added.
Wrote and traced to complete the
addition sentence as 4 + 5 = 9.
Question 2.
4 honey bees were there near honeycomb,
6 more joined, Now how many honeybees are
there in all?
Circled the set 6 numbers being added.
6 + 4 = 10 is the addition sentence,
Explanation:
4 honey bees were there near honeycomb,
6 more joined, Now how many honeybees are
there in all?
Circled the set 6 numbers being added.
Wrote and traced to complete the
addition sentence as 6 + 4 = 10.
Question 3.
4 planes were flying in the sky,
3 more joined, Now how many planes are
there in all?
Circled the set 3 numbers being added.
3 + 4 = 7 is the addition sentence,
Explanation:
4 planes were flying in the sky,
3 more joined, Now how many planes are
there in all?
Circled the set 3 numbers being added.
Wrote and traced to complete the
addition sentence as 3 + 4 = 7.
Question 4.
3 ducks are swimming in the pond,
5 more joined them, Now how many ducks are
there in all?
Circled the set 5 numbers being added.
5 + 3 = 8 is the addition sentence,
Explanation:
3 ducks are swimming in the pond,
5 more joined them, Now how many ducks are
there in all?
Circled the set 5 numbers being added.
Wrote and traced to complete the
addition sentence as 5 + 3 = 8.
Write and trace to complete the addition sentence.
2. How many more counters would you place to model
a way to make 9? Draw the counters.
3. Count and tell how many trumpets. Write the number.
Lesson Check
Question 1.
3 puppies set were lying on the ground in the park,
6 puppies set joined them, how many puppies are
there now in all?
wrote and traced 6 + 3 = 9 addition sentence,
Explanation:
3 puppies set were lying on the ground in the park,
6 puppies set joined them, how many puppies are
there now in all?
wrote and traced to complete addition sentence
as 6 + 3 = 9.
Spiral Review
Question 2.
1 more counter,
8 + 1 = 9,
Explanation:
Given 8 counters to make 9, I would place 1 counter
(9 – 8 = 1) to model a way, Drawn the counter.
Question 3.
5 number of trumpets,
Explanation:
Counted there are 5 number of trumpets. Wrote the number 5.
### Lesson 5.8 Algebra • Number Pairs to 5
Essential Question How can you model and write
addition sentences for number pairs for sums to 5?
By using models of cubes,
1 cube and 4 cubes makes 5 cubes in all,
2 cubes and 3 cubes makes 5 cubes in all,
Explanation:
By using models of cubes,
The addition sentence for number pairs (1, 4)
for sums to 5 is 1 + 4 = 5 cubes in all.
The addition sentence for number pairs (2, 3)
for sums to 5 is 2 + 3 = 5 cubes in all.
Share and Show
DIRECTIONS Place two colors of cubes on the cube train
to show the number pairs that make 4.
1. Trace the addition sentence to show one of the pairs.
2–3. Complete the addition sentence to show
another number pair.
Color the cube train to match the addition sentence in Exercise 3.
Question 1.
Addition sentence : 4 = 3 + 1 ,
Explanation:
Traced the addition sentence to show one of the pairs
that make 4, 4 = 3 + 1.
Question 2.
Another addition sentence : 4 = 2 + 2 ,
Explanation:
The addition sentence to show another pairs
that make 4, 4 = 2 + 2.
Question 3.
Another addition sentence : 4 = 1 + 3,
Explanation:
The addition sentence to show another pairs
that make 4, 4 = 1 + 3.
Colored the cube train with green 1 cube and
3 blue cubes to match the addition sentence
as 4 = 1 + 3,
DIRECTIONS Place two colors of cubes on the cube train
to show the number pairs that make 5.
4–7. Complete the addition sentence to show a number pair.
Color the cube train to match the addition sentence in Exercise 7.
Question 4.
4 cubes and 1 cube make 5 in all,
Addition sentence : 5 = 4 + 1 ,
Explanation:
4 cubes and 1 cube make 5 in all, So
the addition sentence to show one of the pair
that make 5 is 5 = 4 + 1.
Question 5.
1 cube and 4 cubes make 5 in all,
Addition sentence : 5 = 1 + 4 ,
Explanation:
1 cube and 4 cubes make 5 in all, So
the addition sentence to show one of the pair
that make 5 is 5 = 1 + 4.
Question 6.
2 cubes and 3 cubes make 5 in all,
Addition sentence : 5 = 2 + 3,
Explanation:
2 cubes and 3 cubes make 5 in all, So,
the addition sentence to show pair
that make 5 is 5 = 2 + 3.
Question 7.
3 cubes and 2 cubes make 5 in all,
Addition sentence : 5 = 3 + 2,
Explanation:
3 cubes and 2 cubes make 5 in all, So,
the addition sentence to show pair
that make 5, 5 = 3 + 2.
Colored the cube train with yellow 3 cubes and
2 red cubes to match the addition sentence
as 5 = 3 + 2.
DIRECTIONS 8. Peyton and Ashley have five red apples.
Peyton is holding five of the apples. How many is Ashley
holding? Color the cube train to show the number pair.
9.
Draw to show what you know about a number pair to 5.
Question 8.
Ashley is holding 0 apples,
Addition sentence : 5 = 5 + 0,
Explanation:
Given Peyton and Ashley have five red apples.
Peyton is holding five of the apples. So Ashley is
holding(5 – 5 = 0) 0 apples,
Colored the cube train to show the number pair (5,0),
Completed the addition sentence as 5 = 5 + 0.
Question 9.
Number Pair of 5 : Two numbers when added makes 5 in all,
then the two numbers are called number pair,
Drawn to show about a number pair to 5 using cubes,
Explanation:
Numbers that are pair of 5 are,
5 = 0 + 5,
5 = 5 + 0,
5 = 1 + 4,
5 = 4 + 1,
5 = 2 + 3,
5 = 3 + 2,
Drawn to show about a number pair to 5 using cubes.
HOME ACTIVITY • Have your child tell you the number pairs
for a set of objects up to five. Have him or her tell an
addition sentence for one of the number pairs.
Yes, my child told the number pairs for a set
of objects up to five.
Yes, He or she told an addition sentence for one of the
number pairs.
number pair (2, 3) to make 5 in all,
Addition sentence : 2 + 3 = 5.
### Algebra • Number Pairs to 5 Homework & Practice 5.8
DIRECTIONS 1–3. Look at the number at the beginning
of the addition sentence. Place two colors of cubes on the
cube train to show a number pair for that number.
Complete the addition sentence to show a number pair.
Color the cube train to match the addition sentence.
Question 1.
Addition sentence : 3 = 1 + 2,
Colored the cube train to match the
Explanation:
Placed two colors of 1 cube of blue color,
2 cubes of green color on the cube train to show
a number pair for the number 3,
Completed the addition sentence to show a
number pair as 3 = 1 + 2,
Colored the cube train to match the addition sentence as
shown in the picture.
Question 2.
Addition sentence : 4 = 2 + 2,
Colored the cube train to match the
Explanation:
Placed two colors of 2 cubes of blue color,
2 cubes of green color on the cube train to show
a number pair for the number 4,
Completed the addition sentence to show a
number pair as 4 = 2 + 2,
Colored the cube train to match the addition sentence as
shown in the picture.
Question 3.
Addition sentence : 5 = 2 + 3,
Colored the cube train to match the
Explanation:
Placed two colors 2 cubes of blue color,
3 cubes of green color on the cube train to show
a number pair for the number 5,
Completed the addition sentence to show a
number pair as 5 = 2 + 3,
Colored the cube train to match the addition sentence as
shown in the picture.
DIRECTIONS 1. Complete the addition sentence to
show the numbers that match the cube train.
2. Count the number of turtles in each set.
Circle the set that has the greater number of turtles.
3. How many more counters would you place to model
a way to make 6? Draw the counters.
Lesson Check
Question 1.
Addition sentence : 5 = 3 + 2,
Explanation:
show the numbers that matches the
cube train as 5 = 3 + 2.
Spiral Review
Question 2.
The number of turtles in each set is set 1 – 7,
set 2 – 8, 8 > 7, So circled number 8 turtles,
Explanation:
Counted the number of turtles in each set,
set 1 – 7, set 2 – 8, Now comparing 8 is greater
than 7 turtles, So circled the set 8 number of turtles.
Question 3.
4 more counters,
Drawn the counters,
Explanation:
Given 2 counters to make 6, I need to place
(6 – 2 = 4) 4 more number of counters
to model a way to make 6 as 2 + 4 = 6,
Drawn the counters as shown above in the picture.
### Lesson 5.9 Algebra • Number Pairs for 6 and 7
Essential Question How can you model and write addition
sentences for number pairs for each sum of 6 and 7?
With the help of objects images, I can write addition
sentences for number pairs for each sum of 6 and 7.
Explanation:
With the help of objects images, I can write addition
sentences for number pairs for each sum of 6 and 7.
Example :
Number pair for 6 is 2 + 4,
I had 2 puppies with me, my friend gave me
4 more puppies make in all 6 puppies
Addition sentence : 2 + 4 = 6,
Number pair for 7 is 4 + 3,
My friend have 4 balloons and I have 3
balloons in total make in all 7 balloons,
Addition sentence : 4 + 3 = 7.
Share and Show
DIRECTIONS Place two colors of cubes on the cube
train to show the number pairs that make 6.
1. Trace the addition sentence to show one of the pairs.
2–5. Complete the addition sentence to show a
number pair for 6. Color the cube train to match the
Question 1.
1 cube and 5 cubes make 6 in all,
Addition sentence : 6 = 1 + 5,
Explanation:
1 cube and 5 cubes make 6 in all, So traced
the addition sentence to show one of the pair
that make 6 is 6 = 1 + 5.
Question 2.
3 cubes and 3 cubes make 6 in all,
Addition sentence : 6 = 3 + 3,
Explanation:
3 cubes and 3 cubes make 6 in all,
the addition sentence to show one of the
number pair that make 6 is 6 = 3 + 3.
Question 3.
2 cubes and 4 cubes make 6 in all,
Addition sentence : 6 = 2 + 4,
Explanation:
2 cubes and 4 cubes make 6 in all,
the addition sentence to show one of the
number pair that make 6 is 6 = 2 + 4.
Question 4.
5 cubes and 1 cube make 6 in all,
Addition sentence : 6 = 5 + 1,
Explanation:
5 cubes and 1 cube make 6 in all,
the addition sentence to show one of the
number pair that make 6 is 6 = 5 + 1.
Question 5.
Addition sentence : 6 = 4 + 2,
Colored the cube train to match the
Explanation:
Placed two colors 4 cubes of pink color,
2 cubes of green color on the cube train to show
a number pair for the number 6,
Completed the addition sentence to show a
number pair as 6 = 4 + 2,
Colored the cube train to match the addition sentence as
shown in the picture.
DIRECTIONS Place two colors of cubes on the cube train
to show the number pairs that make 7.
6–10. Complete the addition sentence to show a
number pair for 7. Color the cube train to match the
Question 6.
3 cubes and 4 cubes make 7 in all,
Addition sentence : 7 = 3 + 4,
Explanation:
3 cubes and 4 cubes make 7 in all,
the addition sentence to show one of the
number pair that make 7 is 7 = 3 + 4.
Question 7.
2 cubes and 5 cubes make 7 in all,
Addition sentence : 7 = 2 + 5,
Explanation:
2 cubes and 5 cubes make 7 in all,
the addition sentence to show one of the
number pair that make 7 is 7 = 2 + 5.
Question 8.
1 cube and 6 cubes make 7 in all,
Addition sentence : 7 = 1 + 6,
Explanation:
1 cube and 6 cubes make 7 in all,
the addition sentence to show one of the
number pair that make 7 is 7 = 1 + 6.
Question 9.
4 cubes and 3 cubes make 7 in all,
Addition sentence : 7 = 4 + 3,
Explanation:
4 cubes and 3 cubes make 7 in all,
the addition sentence to show one of the
number pair that make 7 is 7 = 4 + 3.
Question 10.
Addition sentence : 7 = 5 + 2,
Colored the cube train to match the
Explanation:
Placed two colors 5 cubes of brown color,
2 cubes of purple color on the cube train to show
a number pair for the number 7,
Completed the addition sentence to show a
number pair as 7 = 5 + 2,
Colored the cube train to match the addition sentence as
shown in the picture.
Problem Solving • Applications
DIRECTIONS 11. Peter and Grant have six toy cars.
Peter has no cars. How many cars does Grant have?
Color the cube train to show the number pair.
12. Draw to show what you know about a number pair
for 7 when one number is 0. Complete the addition sentence.
Question 11.
Peter is having 0 cars,
Grant have 6 cars,
Addition sentence : 6 = 0 + 6,
Explanation:
Given Peter and Grant have six toy cars.
Peter has no cars. So Grant have (6 = 6-0)
6 toy cars,
Colored the cube train to show the number pair (0, 6),
as Peter nil color and Grant with purple color.
Completed the addition sentence as 6 = 0 + 6.
Question 12.
Number pair for 7 is 7, 0,
Addition sentence is 7 = 7 + 0,
Explanation:
Drawn cubes to show about a number pair for 7
when one number is 0 as 7,0 and 7 number
with green color, The addition sentence is 7 = 7 + 0.
HOME ACTIVITY • Have your child use his or her fingers
on two hands to show a number pair for 6.
Yes, My child has used his or her fingers on two hands to
show a number pair (3,3) for 6 as 3 + 3 = 6.
### Algebra • Number Pairs for 6 and 7 Homework & Practice 5.9
DIRECTIONS 1–2. Look at the number at the beginning
Place two colors of cubes on the cube train to show a
number pair for that number.
Complete the addition sentence to show a number pair.
Color the cube train to match the addition sentence.
Question 1.
Addition sentence : 6 = 4 + 2,
Colored the cube train to match the
Explanation:
Placed two colors 4 cubes of yellow color,
2 cubes of green color on the cube train to show
a number pair for the number 6,
Completed the addition sentence to show a
number pair as 6 = 4 + 2,
Colored the cube train to match the addition sentence
as shown in the picture.
Question 2.
Addition sentence : 7 = 6 + 1,
Colored the cube train to match the
Explanation:
Placed two colors 6 cubes of yellow color,
1 cube of purple color on the cube train to show
a number pair for the number 7,
Completed the addition sentence to show a
number pair as 7 = 6 + 1,
Colored the cube train to match the addition sentence
as shown in the picture.
DIRECTIONS 1. Complete the addition sentence to
show the numbers that match the cube train.
2.
How many more counters would you
place to model a way to make 10? Draw the counters.
3.
Count and tell how many hats. Write the number.
Lesson Check
Question 1.
Addition sentence : 7 = 1 + 6,
Explanation:
show the numbers that matches the
cube train as 7 = 1 + 6.
Spiral Review
Question 2.
2 more counters,
Drawn the counters,
Explanation:
Given 8 counters to make 10, I need to place
(10 – 8 = 2), 2 more number of counters
to model a way to make 10 as 8 + 2 = 10,
Drawn the counters as shown above in the picture.
Question 3.
7 number of hats,
Explanation:
Counted there are 7 number of hats.
Wrote the number 7.
### Lesson 5.10 Algebra • Number Pairs for 8
Essential Question How can you model and write addition
sentences for number pairs for sums of 8?
With the help of objects images, I can write addition
sentences for number pairs for sum of 8.
Explanation:
With the help of objects images, I can write addition
sentences for number pairs for sum of 8.
Example :
Number pair for 8 is 5 + 3,
I have 5 pencils and my grandma gave me
3 more pencils in total I have 8 pencils in all,
Addition sentence : 5 + 3 = 8.
Share and Show
DIRECTIONS Use two colors of cubes to make a cube train
to show the number pairs that make 8.
1. Trace the addition sentence to show one of the pairs.
2–4. Complete the addition sentence to show a number pair for 8.
Color the cube train to match the addition sentence in Exercise 4.
Question 1.
1 cube and 7 cubes make 8 in all,
Addition sentence : 8 = 1 + 7,
Explanation:
1 cube and 7 cubes make 8 in all,
the addition sentence to show one of the
number pair that make 8 is 8 = 1 + 7.
Question 2.
3 cubes and 5 cubes make 8 in all,
Addition sentence : 8 = 3 + 5,
Explanation:
3 cubes and 5 cubes make 8 in all,
the addition sentence to show one of the
number pair that make 8 is 8 = 3 + 5.
Question 3.
4 cubes and 4 cubes make 8 in all,
Addition sentence : 8 = 4 + 4,
Explanation:
4 cubes and 4 cubes make 8 in all,
the addition sentence to show one of the
number pair that make 8 is 8 = 4 + 4.
Question 4.
Addition sentence : 8 = 2 + 6,
Colored the cube train to match the
Explanation:
Placed two colors 2 cubes of blue color,
6 cubes of pink color on the cube train to show
a number pair for the number 8,
Completed the addition sentence to show a
number pair as 8 = 2 + 6,
Colored the cube train to match the addition sentence
as shown in the picture.
DIRECTIONS Use two colors of cubes to make a cube train
to show the number pairs that make 8.
5–7. Complete the addition sentence to show a number pair for 8.
Color the cube train to match the addition sentence in Exercise 7.
Question 5.
7 cubes and 1 cube make 8 in all,
Addition sentence : 8 = 7 + 1,
Explanation:
7 cubes and 1 cube make 8 in all,
the addition sentence to show one of the
number pair that make 8 is 8 = 7 + 1.
Question 6.
6 cubes and 2 cubes make 8 in all,
Addition sentence : 8 = 6 + 2,
Explanation:
6 cubes and 2 cubes make 8 in all,
the addition sentence to show one of the
number pair that make 8 is 8 = 6 + 2.
Question 7.
Addition sentence : 8 = 5 + 3,
Colored the cube train to match the
Explanation:
Placed two colors 5 cubes of blue color,
3 cubes of pink color on the cube train to show
a number pair for the number 8,
Completed the addition sentence to show a
number pair as 8 = 5 + 3,
Colored the cube train to match the addition sentence
as shown in the picture.
Problem Solving • Applications
DIRECTIONS 8. There are eight crayons in a packet.
Eight of the crayons are red. How many are not red?
Draw and color to show how you solved.
9. Draw to show what you know about a different number pair for 8.
Question 8.
0 are not red,
Explanation:
Given there are eight crayons in a packet.
Eight of the crayons are red. So number
that are not red are (8= 8 – 0) 0.
Drawn and colored to show all 8 are red
by using cubes 1 cube means one crayon so I solved,
Completed the addition sentence as 8 = 8 + 0.
Question 9.
0 cube and 8 cubes make 8 in all,
Addition sentence : 8 = 0 + 8,
Explanation:
0 cube and 8 cubes make 8 in all,
means no color cubes and 8 cubes are there
in total also make 8, the addition sentence to
show one of the different number pair that make 8
is 8 = 0 + 8.
HOME ACTIVITY • Have your child tell you the number pairs
for a set of eight objects. Have him or her tell the
addition sentence to match one of the number pairs.
Yes my child tells me about the number pairs
for a set of eight objects. The addition sentence
to match one of the number pairs is 3 and 5,
3 + 5 = 8.
Explanation:
Yes my child tells me about the number pairs
for a set of eight objects.
Example:
My child tells me he had 3 erasers with him,
His friend gave him 5 more, in all he have 8
erasers with him.
The addition sentence to match one of the number pair
is 3 and 5, So 3 + 5 = 8.
### Algebra • Number Pairs for 8 Homework & Practice 5.10
DIRECTIONS Use two colors of cubes to make a cube train
to show the number pairs that make 8.
1–4. Complete the addition sentence to show a number pair for 8.
Color the cube train to match the addition sentence in Exercise 4.
Question 1.
4 cubes and 4 cubes make 8 in all,
Addition sentence : 8 = 4 + 4,
Explanation:
4 cubes and 4 cubes make 8 in all,
the addition sentence to show one of the
number pair that make 8 is 8 = 4 + 4.
Question 2.
7 cubes and 1 cube make 8 in all,
Addition sentence : 8 = 7 + 1,
Explanation:
7 cubes and 1 cube make 8 in all,
the addition sentence to show one of the
number pair that make 8 is 8 = 7 + 1.
Question 3.
2 cubes and 6 cubes make 8 in all,
Addition sentence : 8 = 2 + 6,
Explanation:
2 cubes and 6 cubes make 8 in all,
the addition sentence to show one of the
number pair that make 8 is 8 = 2 + 6.
Question 4.
Addition sentence : 8 = 3 + 5,
Colored the cube train to match the
Explanation:
Placed two colors 3 cubes of green color,
5 cubes of pink color on the cube train to show
a number pair for the number 8,
Completed the addition sentence to show a
number pair as 8 = 3 + 5,
Colored the cube train to match the addition sentence
as shown in the picture.
DIRECTIONS 1. Complete the addition sentence to
show the numbers that match the cube train.
2. Count and tell how many in each set. Write the numbers.
Compare the numbers. Circle the number that is greater.
3. How many more counters would you place in the
five frame to show a way to make 5? Draw the counters.
Lesson Check
Question 1.
Addition sentence : 8 = 1 + 7,
Explanation:
show the numbers that matches the
cube train as 8 = 1 + 7.
Spiral Review
Question 2.
Each set, set 1 has 4 numbers,
set 2 has 3 numbers,
4 is greater than 3, So circled 4 numbers.
Explanation:
Counted each set, Set 1 has 4 numbers,
set 2 has 3 numbers, Wrote the numbers,
Comparing the numbers 4 is greater than 3,
4 > 3, So circled the numbers 4 which is greater.
Question 3.
1 more counter,
Drawn the counter,
Explanation:
Given 4 counters to make 5, I need to place
(5 – 4 = 1), 1 more number of counter
to model a way to make 5 as 4 + 1 = 5,
Drawn the counters as shown above in the picture.
### Lesson 5.11 Algebra • Number Pairs for 9
Essential Question How can you model and write addition sentences
for number pairs for sums of 9?
With the help of objects images, I can write addition
sentences for number pairs for sum of 9.
Explanation:
With the help of objects images, I can write addition
sentences for number pairs for sum of 9.
Example :
Number pair for 9 is 4 + 5,
A shopkeeper has 4 caps and he purchased
5 more caps in total he has 9 caps overall,
Addition sentence : 4 + 5 = 9.
Share and Show
DIRECTIONS Use two colors of cubes to make a cube train
to show the number pairs that make 9.
1. Trace the addition sentence to show one of the pairs.
2–4. Complete the addition sentence to show a number pair for 9.
Color the cube train to match the addition sentence in Exercise 4.
Question 1.
1 cube and 8 cubes make 9 in all,
Addition sentence : 9 = 1 + 8,
Explanation:
1 cube and 8 cubes make 9 in all,
the addition sentence to show one of the
number pair that make 9 is 9 = 1 + 8.
Question 2.
3 cubes and 6 cubes make 9 in all,
Addition sentence : 9 = 3 + 6,
Explanation:
3 cubes and 6 cubes make 9 in all,
the addition sentence to show one of the
number pair that make 9 is 9 = 3 + 6.
Question 3.
2 cubes and 7 cubes make 9 in all,
Addition sentence : 9 = 2 + 7,
Explanation:
2 cubes and 7 cubes make 9 in all,
the addition sentence to show one of the
number pair that make 9 is 9 = 2 + 7.
Question 4.
Addition sentence : 9 = 8 + 1,
Colored the cube train to match the
Explanation:
Placed two colors 8 cubes of brown color,
1 cube of yellow color on the cube train to show
a number pair for the number 9,
Completed the addition sentence to show a
number pair as 9 = 8 + 1,
Colored the cube train to match the addition sentence
as shown in the picture.
DIRECTIONS Use two colors of cubes to make a cube train
to show the number pairs that make 9.
5–8. Complete the addition sentence to show a
number pair for 9.Color the cube train to match the
Question 5.
7 cubes and 2 cubes make 9 in all,
Addition sentence : 9 = 7 + 2,
Explanation:
7 cubes and 2 cubes make 9 in all,
the addition sentence to show one of the
number pair that make 9 is 9 = 7 + 2.
Question 6.
5 cubes and 4 cubes make 9 in all,
Addition sentence : 9 = 5 + 4,
Explanation:
5 cubes and 4 cubes make 9 in all,
the addition sentence to show one of the
number pair that make 9 is 9 = 5 + 4.
Question 7.
0 cube and 9 cubes make 9 in all,
Addition sentence : 9 = 0 + 9,
Explanation:
0 cube and 9 cubes make 9 in all,
the addition sentence to show one of the
number pair that make 9 is 9 = 0 + 9.
Question 8.
Addition sentence : 9 = 6 + 3,
Colored the cube train to match the
Explanation:
Placed two colors 6 cubes of brown color,
3 cubes of yellow color on the cube train to show
a number pair for the number 9,
Completed the addition sentence to show a
number pair as 9 = 6 + 3,
Colored the cube train to match the addition sentence
as shown in the picture.
Problem Solving • Applications
DIRECTIONS 9. Shelby has nine friends. None of them are boys.
How many are girls? Complete the addition sentence to show the number pair.
10. Draw to show what you know about a different number pair for 9.
Question 9.
Shelby have 9 girl friends,
Addition sentence : 9 = 9 + 0,
Explanation:
Given Shelby has nine friends and none of them are boys.
So number of girls are (9 = 9 – 0) = 9,
Completed the addition sentence to show
number pair as 9 = 9 + 0.
Question 10.
7 cubes and 2 cubes make 9 in all,
Addition sentence : 9 = 7 + 2,
Explanation:
7 cubes and 2 cubes make 9 in all,
Drawn to show one of the different number pair
that make 9 is 9 = 7 + 2.
Completed the addition sentence to show
number pair as 9 = 7 + 2.
HOME ACTIVITY • Have your child use his or her fingers on
two hands to show a number pair for 9.
Yes, My child has used his or her fingers on two hands to
show a number pair (4, 5) for 9 as 4 + 5 = 9.
### Algebra • Number Pairs for 9 Homework & Practice 5.11
DIRECTIONS Use two colors of cubes to make a cube train
to show the number pairs that make 9.
1–4. Complete the addition sentence to show a
number pair for 9. Color the cube train to match the
Question 1.
3 cubes and 6 cubes make 9 in all,
Addition sentence : 9 = 3 + 6,
Explanation:
3 cubes and 6 cubes make 9 in all,
the addition sentence to show one of the
number pair that make 9 is 9 = 3 + 6.
Question 2.
9 cubes and 0 cube make 9 in all,
Addition sentence : 9 = 9 + 0,
Explanation:
9 cubes and 0 cube make 9 in all,
the addition sentence to show one of the
number pair that make 9 is 9 = 9 + 0.
Question 3.
1 cube and 8 cubes make 9 in all,
Addition sentence : 9 = 1 + 8,
Explanation:
1 cube and 8 cubes make 9 in all,
the addition sentence to show one of the
number pair that make 9 is 9 = 1 + 8.
Question 4.
Addition sentence : 9 = 2 + 7,
Colored the cube train to match the
Explanation:
Placed two colors 2 cubes of violet color,
7 cubes of orange color on the cube train to show
a number pair for the number 9,
Completed the addition sentence to show a
number pair as 9 = 2 + 7,
Colored the cube train to match the addition sentence
as shown in the picture.
DIRECTIONS 1. Complete the addition sentence to show
the numbers that match the cube train.
2. Count how many birds. Write the number.
3. Count and tell how many in each set. Write the numbers.
Compare the numbers. Circle the number that is less.
Lesson Check
Question 1.
5 cubes and 4 cubes make 9 in all,
Addition sentence : 9 = 5 + 4,
Explanation:
5 cubes and 4 cubes make 9 in all,
the addition sentence to show one of the
number pair that make 9 is 9 = 5 + 4.
Spiral Review
Question 2.
8 number of birds,
Explanation:
Counted there are 8 number of birds.
Wrote the number 8.
Question 3.
Each set, set 1 has 3 numbers,
set 2 has 2 numbers,
2 is less than 3, So circled 2 numbers.
Explanation:
Counted each set, Set 1 has 3 numbers,
set 2 has 2 numbers, Wrote the numbers,
Comparing the numbers 2 is less than 3,
2 < 3, So circled the numbers 2 which is lesser.
### Lesson 5.12 Algebra • Number Pairs for 10
Essential Question How can you model and write addition
sentences for number pairs for sums of 10?
With the help of objects images, I can write addition
sentences for number pairs for sum of 10.
Explanation:
With the help of objects images, I can write addition
sentences for number pairs for sum of 10.
Example :
Number pair for 10 is 8 + 2,
I have 8 oranges with me and my uncle gave me
2 more oranges in total I have 10 oranges in all,
Addition sentence : 8 + 2 = 10.
Share and Show
DIRECTIONS Use two colors of cubes to build a cube train to
show the number pairs that make 10.
1. Trace the addition sentence to show one of the pairs.
2–4. Complete the addition sentence to show a number
pair for 10. Color the cube train to match the
Question 1.
1 cube and 9 cubes make 10 in all,
Addition sentence : 10 = 1 + 9,
Explanation:
1 cube and 9 cubes make 10 in all,
the addition sentence to show one of the
number pair that make 10 is 10 = 1 + 9.
Question 2.
5 cubes and 5 cubes make 10 in all,
Addition sentence : 10 = 5 + 5,
Explanation:
5 cubes and 5 cubes make 10 in all,
the addition sentence to show one of the
number pair that make 10 is 10 = 5 + 5.
Question 3.
3 cubes and 7 cubes make 10 in all,
Addition sentence : 10 = 3 + 7,
Explanation:
3 cubes and 7 cubes make 10 in all,
the addition sentence to show one of the
number pair that make 10 is 10 = 3 + 7.
Question 4.
Addition sentence : 10 = 4 + 6,
Colored the cube train to match the
Explanation:
Placed two colors 4 cubes of green color,
6 cubes of orange color on the cube train to show
a number pair for the number 10,
Completed the addition sentence to show a
number pair as 10 = 4 + 6,
Colored the cube train to match the addition sentence
as shown in the picture.
DIRECTIONS Use two colors of cubes to build a
cube train to show the number pairs that make 10.
5–8. Complete the addition sentence to show a
number pair for 10. Color the cube train to match the
Question 5.
1 cube and 9 cubes make 10 in all,
Addition sentence : 10 = 1 + 9,
Explanation:
1 cube and 9 cubes make 10 in all,
the addition sentence to show one of the
number pair that make 10 is 10 = 1 + 9.
Question 6.
5 cubes and 5 cubes make 10 in all,
Addition sentence : 10 = 5 + 5,
Explanation:
5 cubes and 5 cubes make 10 in all,
the addition sentence to show one of the
number pair that make 10 is 10 = 5 + 5.
Question 7.
2 cubes and 8 cubes make 10 in all,
Addition sentence : 10 = 2 + 8,
Explanation:
2 cubes and 8 cubes make 10 in all,
the addition sentence to show one of the
number pair that make 10 is 10 = 2 + 8.
Question 8.
Addition sentence : 10 = 7 + 3,
Colored the cube train to match the
Explanation:
Placed two colors 7 cubes of green color,
3 cubes of orange color on the cube train to show
a number pair for the number 10,
Completed the addition sentence to show a
number pair as 10 = 7 + 3,
Colored the cube train to match the addition sentence
as shown in the picture.
Problem Solving • Applications
DIRECTIONS 9. There are ten children in the cafeteria.
Ten of them are drinking water. How many children are
drinking milk? Complete the addition sentence to
show the number pair.
10. Draw to show what you know about a different
number pair for 10. Complete the addition sentence.
Question 9.
10 children are drinking milk,
Addition sentence : 10 = 10 + 0,
Explanation:
Given there are ten children in the cafeteria.
Ten of them are drinking water. So children
drinking milk are (10 = 10 -0) 10, Completed
the addition sentence to show the number pair
as 10 = 10 + 0.
Question 10.
8 cubes and 2 cubes make 10 in all,
Addition sentence : 10 = 8 + 2,
Explanation:
8 cubes and 2 cubes make 10 in all,
the addition sentence to show one of the
number pair that make 10 is 10 = 8 + 2.
HOME ACTIVITY • Have your child tell you the number pairs
for a set of ten objects. Have him or her tell the addition sentence
to match one of the number pairs.
Yes, my child told the number pairs for a set
of ten objects.
Yes, He or she told an addition sentence for one of the
number pairs of 5 and 5 makes 10 as 5 + 5 = 10.
Example: My child prepared 5 glass of lemon juice,
I prepared 5 glass of lemon juices for guests
In total makes 10 glasses of lemon juice.
Addition sentence : 5 + 5 = 10.
### Algebra • Number Pairs for 10 Homework & Practice 5.12
DIRECTIONS Use two colors of cubes to build a cube train
to show the number pairs that make 10.
1–4. Complete the addition sentence to show a number pair for 10.
Color the cube train to match the addition sentence in Exercise 4.
Question 1.
6 cubes and 4 cubes make 10 in all,
Addition sentence : 10 = 6 + 4,
Explanation:
6 cubes and 4 cubes make 10 in all,
the addition sentence to show one of the
number pair that make 10 is 10 = 6 + 4.
Question 2.
3 cubes and 7 cubes make 10 in all,
Addition sentence : 10 = 3 + 7,
Explanation:
3 cubes and 7 cubes make 10 in all,
the addition sentence to show one of the
number pair that make 10 is 10 = 3 + 7.
Question 3.
2 cubes and 8 cubes make 10 in all,
Addition sentence : 10 = 2 + 8,
Explanation:
2 cubes and 8 cubes make 10 in all,
the addition sentence to show one of the
number pair that make 10 is 10 = 2 + 8.
Question 4.
Addition sentence : 10 = 9 + 1,
Colored the cube train to match the
Explanation:
Placed two colors 9 cubes of yellow color,
1 cube of orange color on the cube train to show
a number pair for the number 10,
Completed the addition sentence to show a
number pair as 10 = 9 + 1,
Colored the cube train to match the addition sentence
as shown in the picture.
DIRECTIONS 1. Complete the addition sentence to show
the numbers that match the cube train.
2. Count the dots in the ten frames. Trace the number.
Write the numbers in order as you count forward from the
dashed number.
3. Use blue and red to color the cubes to show a way to make 10.
Lesson Check
Question 1.
7 cubes and 3 cubes make 10 in all,
Addition sentence : 10 = 7 + 3,
Explanation:
7 cubes and 3 cubes make 10 in all,
the addition sentence to show one of the
number pair that make 10 is 10 = 7 + 3.
Spiral Review
Question 2.
Counted the dots in the ten frames, Number traced as 1,
Wrote the numbers in order as I count forward from the
dashed number 1 as 1, 2, 3, 4, 5,
Explanation:
Counted the dots in the ten frames,
Traced the given number as 1,
Wrote the numbers in order as I count forward from the
dashed number 1 as 1, 2, 3, 4, 5.
Question 3.
Used 5 blue and 5 red to color cubes to show
a way to make 10,
Explanation:
Used 5 blue and 5 red to color cubes to show
a way to make 10 as 10 = 5 + 5.
DIRECTIONS 1. How many puppies are sitting?
How many puppies are being added to the group?
Write the numbers.
2.
Sonja put 4 red counters in the ten frame.
Then she put 3 yellow counters in the ten frame.
Choose all the ways that show the counters being put together.
3. How many of each color cube is being added?
Trace the numbers and symbols. Write the number that
shows how many cubes in all.
Question 1.
There are 3 puppies are sitting,
1 more puppy is being added
to the group,
Numbers are 3, 1,
Explanation:
There are 3 puppies are sitting,
1 more puppy is being added
to the group, So numbers are 3, 1.
Question 2.
All the ways that show the counters
being put together are 4 plus 3,
4 + 3, So circled them,
Explanation:
Given Sonja put 4 red counters in the ten frame
and then she put 3 yellow counters in the ten frame.
All the ways that show the counters being put together
are 4 plus 3 and 4 + 3, So circled them as shown above.
Question 3.
Each orange 2, green 2 color cubes
In all there are 2 + 2 = 4 cubes,
Explanation:
Each of orange 2 color cubes
and green 2 color cubes are being added,
Traced the numbers and symbols,
Wrote the number that shows how many cubes in all
as 2 + 2 = 4.
DIRECTIONS 4. Annabelle has 2 red cubes. She has 2 yellow cubes.
How many cubes does she have? Draw the cubes.
Trace the numbers and symbols. Write how many in all.
5. Look at the cube train. How many red cubes do you see?
How many more cubes do you need to add to make 10?
Draw the cubes. Color them blue. Write and trace to show
6. Write and trace the numbers to complete the addition sentence.
Question 4.
THINK SMARTER+
Annabelle has 4 cubes in all,
Explanation:
Annabelle has 2 red cubes and she has 2 yellow cubes,
In all she has 2 + 2 = 4, Drawn the cubes.
Traced the numbers and symbols,
Wrote how many in all as 2 + 2 = 4.
Question 5.
Red cubes I can see are 7, 3 more cubes are needed,
Addition sentence = 7 + 3 = 10.
Explanation:
Looking at the cube train 7 red cubes I can see,
3 more cubes I do need to add to make 10
as 10 -7 = 3.
Drawn the cubes. Colored them as blue.
Wrote and traced to show addition sentence
as 7 + 3 = 10.
Question 6.
Addition sentence : 3 + 2 = 5.
Explanation:
Wrote and traced the numbers to complete
the addition sentence as 3 + 2 = 5,
DIRECTIONS 7. Trace and write the numbers and trace
the symbols to complete the addition sentence.
8. Nora has 1 green crayon. Gary has some red crayons.
Together they have 5 crayons. Draw to show how many
red crayons Gary has. Complete the number pair.
9. Does this show a number pair for 7? Choose Yes or No.
Question 7.
Addition sentence : 4 + 2 = 6,
Explanation:
Traced and wrote the numbers and trace
the symbols to complete the addition sentence
as 4 + 2 = 6.
Question 8.
THINK SMARTER+
Gary has 4 red crayons,
Number pair 1, 4, 5 = 1 + 4,
Explanation:
Given Nora has 1 green crayon and Gary has some
red crayons.
Together they have 5 crayons. Drawn and showed
how many red crayons Gary has 4 red crayons.
(5 – 1 = 4), Completed the number pair as 1, 4 makes
5 = 1 + 4.
Question 9.
Yes, 4 + 3 makes 7,
Explanation:
Yes, 4 + 3 shows a number pair for 7.
DIRECTIONS 10. Circle all the number pairs for 8.
11. Larry counted out 9 cubes. The cubes were either red or blue.
How many red and blue cubes could he have? Color the cubes
to show the number of red and blue cubes. Write the numbers
12.
Complete the addition sentence to show a number pair for 10.
Question 10.
Circled 2+6 and 1 + 7,
Explanation:
Circled all the number pairs for 8,
2 + 6 and 1 + 7.
Question 11.
Larry can have 4 red, 5 blue cubes,
Addition sentence : 9 = 4 + 5,
Explanation:
Given Larry counted out 9 cubes,
The cubes were either red or blue.
Larry can have 4 red and 5 blue cubes,
Colored the cubes to show the number of red
and blue cubes. Wrote the numbers
to complete the addition sentence as 9 = 4 + 5.
Question 12.
|
# Álgebra: Factorización
1.
$$12{x}^{3}+11{x}^{2}+2x$$
Solución
2.
$$60{h}^{2}+280h+45$$
Solución
3.
$$8{x}^{3}-125$$
Solución
4.
$$-3{x}^{2}+36x-108$$
Solución
5.
$$3{x}^{3}+21{x}^{2}+36x$$
Solución
6.
$$6{x}^{2}y+4xy+2ya$$
Solución
7.
$${x}^{3}+5x+2{x}^{2}+10$$
Solución
# Factoring - Introduction
A polynomial is an expression composed of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents. A common form of polynomials are quadratic expressions, which follows the form:
$$a{x}^{2}+bx+c$$
. For example:
$${x}^{2}+10x+16$$
Other variations might include extra terms, higher-power exponents or negative numbers. Regardless, in many cases, it is possible to turn the polynomial into a simpler form through a process called “factoring”. Let’s try a few simple factoring problems and find out how.
# The Simple Case
The simplest case is where there is no number in front of the
$${x}^{2}$$
, meaning that the cofficient is “1”. Let's use the previous expression as our example:
$${x}^{2}+10x+16$$
To factor this expression, we look for two numbers that multiply to
$$b$$
$$c$$
. In this case, these are
$$8$$
and
$$2$$
, which let us split the second term. This gives us:
$${x}^{2}+8x+2x+16$$
Next, we split the polynomial into two sets of terms, like this:
$$({x}^{2}+8x)+(2x+16)$$
Then we factor out the common terms to give:
$$x(x+8)+2(x+8)$$
Now, we factor out the common term again, which produces:
$$(x+2)(x+8)$$
# More Complex Cases
What happens if we are dealing with a more complex polynomial, such as:
$$12{x}^{3}+11{x}^{2}+2x$$
Here, we have a thrid-degree polynomial with a leading cofficient that is not
$$1$$
. Let's start by taking out the greatest common factor (GCF) of all three terms, which is
$$x$$
. This gives:
$$x(12{x}^{2}+11x+2)$$
Now, we want to do the same thing as above: Split the second term. Start by multiplying the coefficient of the first term,
$$12$$
, by the constant term,
$$2$$
. This gives
$$24$$
. Then, ask the same question as above — what numbers add up to
$$11$$
and multiply to
$$24$$
$$8$$
and
$$3$$
, which gives:
$$x(12{x}^{2}+8x+3x+2)$$
Now, we can split the polynomial in two and find the common term in
$$12{x}^{2} + 8x$$
, which is
$$4x$$
. This lets us remove the factor and gives:
$$x(4x(3x+2)+(3x+2))$$
Once we factor the common term
$$(3x+2)$$
, we arrive at the final answer:
$$x(3x+2)(4x+1)$$
|
FutureStarr
35 40 As a Percentage:
## 35 40 As a Percentage
via GIPHY
The difference in the number of people is 115. The number of men is 115. The number of women is 40. So the difference between the men and the women is 35.
### Number
In this section we learn how to express one number as a percentage of annother. For example, by the end of this section we'll have no trouble showing that: $18 = 45\% \ \text{of} \ 40$ We start by learning the method as well as read through a worked example before working through some exercises. The percentage increase calculator above computes an increase or decrease of a specific percentage of the input number. It basically involves converting a percent into its decimal equivalent, and either subtracting (decrease) or adding (increase) the decimal equivalent from and to 1, respectively. Multiplying the original number by this value will result in either an increase or decrease of the number by the given percent. Refer to the example below for clarification.In calculating 34% of a number, sales tax, credit cards cash back bonus, interest, discounts, interest per annum, dollars, pounds, coupons,34% off, 34% of price or something, we use the formula above to find the answer. The equation for the calculation is very simple and direct. You can also compute other number values by using the calculator above and enter any value you want to compute.percent dollar to pound = 0 pound
; hence the net change is an overall decrease by x percent of x percent (the square of the original percent change when expressed as a decimal number). Thus, in the above example, after an increase and decrease of x = 10 percent, the final amount, $198, was 10% of 10%, or 1%, less than the initial amount of$200. The net change is the same for a decrease of x percent, followed by an increase of x percent; the final amount is p(1 - 0.01x)(1 + 0.01x) = p(1 − (0.01x) The word "percentage" is often a misnomer in the context of sports statistics, when the referenced number is expressed as a decimal proportion, not a percentage: "The Phoenix Suns' Shaquille O'Neal led the NBA with a .609 field goal percentage (FG%) during the 2008–09 season." (O'Neal made 60.9% of his shots, not 0.609%.) Likewise, the winning percentage of a team, the fraction of matches that the club has won, is also usually expressed as a decimal proportion; a team that has a .500 winning percentage has won 50% of their matches. The practice is probably related to the similar way that batting averages are quoted. (Source: en.wikipedia.org)
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|
# 2006 AMC 12B Problems/Problem 16
## Problem
Regular hexagon $ABCDEF$ has vertices $A$ and $C$ at $(0,0)$ and $(7,1)$, respectively. What is its area?
$\mathrm{(A)}\ 20\sqrt {3} \qquad \mathrm{(B)}\ 22\sqrt {3} \qquad \mathrm{(C)}\ 25\sqrt {3} \qquad \mathrm{(D)}\ 27\sqrt {3} \qquad \mathrm{(E)}\ 50$
## Solution
To find the area of the regular hexagon, we only need to calculate the side length. a distance of $\sqrt{7^2+1^2} = \sqrt{50} = 5\sqrt{2}$ apart. Half of this distance is the length of the longer leg of the right triangles. Therefore, the side length of the hexagon is $\frac{5\sqrt{2}}{2}\cdot\frac{1}{\sqrt{3}}\cdot2 = \frac{5\sqrt{6}}{3}$.
The apothem is thus $\frac{1}{2}\cdot\frac{5\sqrt{6}}{3}\cdot\sqrt{3} = \frac{5\sqrt{2}}{2}$, yielding an area of $\frac{1}{2}\cdot10\sqrt{6}\cdot\frac{5\sqrt{2}}{2}=25\sqrt{3} \implies \mathrm{(C)}$.
## See also
2006 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 15 Followed byProblem 17 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 12 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
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To solve this problem, we use trigonometry, specifically the tangent function. Initially, the angle of elevation is 30°, and finally, it becomes 60°. Let’s denote the initial and final distances from the boy to the building as d1 and d2 respectively.
Initial Position (30° angle): Using tan(30°) = 1/√3, the height of the building minus the height of the boy is 30 m − 1.5 m = 28.5 m. So, 1/√3 = 28.5/d1. Solving for d1, we get d1 = 28.5√3.
Final Position (60° angle): Using tan(60°) = √3, we have √3 = 28.5/d2. Solving for d2, we get d2 = 28.5/√3.
The distance walked is d1 − d2 = 28.5√3 − 28.5/√3, which simplifies to approximately 24.75 meters. Therefore, the boy walked about 24.75 meters towards the building.
Let’s discuss in detail
## Trigonometric Motion Analysis
Trigonometry, a branch of mathematics, is not only about solving theoretical problems but also about understanding real-world situations, such as the movement of a person in relation to a structure. The problem at hand involves a boy walking towards a building and the consequent change in the angle of elevation from his eyes to the top of the building. This scenario is a classic example of how trigonometry can be applied to analyze motion and distance in a physical context, demonstrating the practical utility of mathematical concepts in everyday life.
### Motion Towards a Building
The problem presents a 1.5-meter-tall boy standing at a certain distance from a 30-meter-tall building. As the boy walks towards the building, the angle of elevation from his eye level to the top of the building increases from 30° to 60°. The objective is to calculate the distance the boy has walked towards the building. This scenario forms two different right-angled triangles at different instances – one at the initial position (30° angle) and the other at the final position (60° angle).
#### The Role of Tangent in Angle of Elevation
In trigonometry, the tangent of an angle in a right-angled triangle is the ratio of the opposite side (in this case, the difference in height between the boy’s eyes and the top of the building) to the adjacent side (the distance from the boy to the building). By applying the tangent function to the angles of elevation, we can determine the boy’s distances from the building at both the initial and final positions.
Calculating the Initial Distance
Initially, when the angle of elevation is 30°, we use the formula tan(30°) = opposite/adjacent. The opposite side is the height difference between the building and the boy’s eyes, which is 30 m − 1.5 m = 28.5 m. The tangent of 30° is 1/√3. Therefore, 1/√3 = 28.5/d1, where d1 is the initial distance. Solving for d1, we find d1 = 28.5√3 meters.
##### Calculating the Final Distance
At the final position, where the angle of elevation is 60°, we apply the formula tan(60°) = opposite/adjacent. The tangent of 60° is √3. Therefore, √3 = 28.5/d2, where d2 is the final distance. Solving for d2, we find d2 = 28.5/√3 meters.
###### Determining the Distance Walked
The distance the boy walked towards the building is the difference between the initial and final distances, d1 − d2. Substituting the values, we get 28.5√ − 28.5/√3, which simplifies to approximately 24.75 meters. This calculation not only demonstrates the application of trigonometry in analyzing real-world motion but also highlights the importance of understanding mathematical principles to interpret and solve practical problems. The scenario underscores how trigonometry can provide insights into everyday situations, bridging the gap between abstract mathematics and tangible experiences.
Discuss this question in detail or visit to Class 10 Maths Chapter 9 for all questions.
Questions of 10th Maths Exercise 9.1 in Detail
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case? The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 30°. Find the height of the tower. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building. From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower. A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building. Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the canal. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower. As observed from the top of a 75 m high lighthouse from the sea-level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships. A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30°. Find the distance travelled by the balloon during the interval. A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
|
14
Nov 18
## Constructing a projector onto a given subspace
Let $L$ be a subspace of $R^n.$ Let $k=\dim L\ (\leq n)$ and fix some basis $x^{(1)},...,x^{(k)}$ in $L.$ Define the matrix $X=(x^{(1)},...,x^{(k)})$ of size $n\times k$ (the vectors are written as column vectors).
Exercise 1. a) With the above notation, the matrix $(X^TX)^{-1}$ exists. b) The matrix $P=X(X^TX)^{-1}X^T$ exists. c) $P$ is a projector.
Proof. a) The determinant of $A=X^TX$ is not zero by linear independence of the basis vectors, so its inverse $A^{-1}$ exists. We also know that $A$ and its inverse are symmetric:
(1) $A^T=A,$ $(A^{-1})^T=A^{-1}.$
b) To see that $P$ exists just count the dimensions.
c) Let's prove that $P$ is a projector. (1) allows us to make the proof compact. $P$ is idempotent:
$P^2=(XA^{-1}X^T)(XA^{-1}X^T)=XA^{-1}(X^TX)A^{-1}X^T$ $=X(A^{-1}A)A^{-1}X^T=XA^{-1}X^T=P.$
$P$ is symmetric:
$P^T=[XA^{-1}X^T]^T=(X^T)^T(A^{-1})^TX^T=XA^{-1}X^T=P.$
Exercise 2. $P$ projects onto $L:$ $\text{Img}(P)=L.$
Proof. First we show that $\text{Img}(P)\subseteq L.$ Put
(2) $y=A^{-1}X^Tx,$
for any $x\in R^n.$ Then
$Px=XA^{-1}X^Tx=Xy=\sum x^{(j)}y_j\in L.$
This shows that $\text{Img}(P)\subseteq L.$
Let's prove the opposite inclusion. Any element of $L$ is of form $\sum x^{(j)}y_j$ with some $y.$ $N(X)=\{0\}$ because we are dealing with a basis. This fact and the general equation $N(X)\oplus \text{Img}(X^T)=R^n$ imply $\text{Img}(X^T)=R^n.$ Hence for any given $y$ there exists $x$ such that $Ay=X^Tx.$ Then (2) is true and, as above, $Px=\sum x^{(j)}y_j.$ We have proved $\text{Img}(P)\supseteq L.$
|
# Surface Area and Volume of Cones
## Presentation on theme: "Surface Area and Volume of Cones"— Presentation transcript:
Surface Area and Volume of Cones
Goal: Students will find the surface area and volume of cones.
Lateral Areas and Surface Areas of Cones
A cone is like a pyramid, but its base is circular, and the vertex is not in the same plane as the base. The radius of the base is the radius of the cone. The height is the perpendicular distance between the vertex and the center of the base, called the altitude. The slant height, l, is the distance between the vertex and a point on the base edge.
A The lateral surface of a cone consists of all segments that connect the vertex with points on the base edge.
Theorem 12.5 Surface Area of a Right Cone:
Lateral Area: LA = πrl Surface Area: S = πrl + B S = πrl + πr2 where B = Area of the Base l = slant height r = radius of cone
Ex. 1: The radius of the base of a cone is 6 m. Its height is 8 m
Ex.1: The radius of the base of a cone is 6 m. Its height is 8 m. Find its lateral and surface area to the nearest tenth. Ex.2: The radius of the base of a cone is 15 cm. Its height is 20 cm. Find its surface area to the nearest tenth.
Ex.3: A traffic cone can be approximated by a right cone with radius 5.7 inches and height 18 inches. Find the approximate lateral area and surface area of the traffic cone to the nearest tenth.
Volume of Cones Theorem Volume of a Cone: Volume: where h is the height of the cone r is the radius of the cone
Ex.4: Find the volume of the solid.
Ex. 6: Find the volume of the solid. Ex
Ex.6: Find the volume of the solid. Ex.7: Find the volume of the solid shown.
Ex.8: Originally, the pyramid had height 144 meters and volume 2, 226, 450 cubic meters. Find the side length of the square base. Ex.9: The volume of a right cone is 1350π cubic meters and the radius is 18 meters. Find the height of the cone.
Ex.10: Find the surface area of the solid. Round to two decimal places.
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# Question Video: Using Group Models to Understand Multiplication by 0 and 1 Mathematics • 3rd Grade
Matthew started with 12 stars in three groups of four. 3 groups of 4, 3 × 4 = 12. Remove one group at a time. Find the missing numbers. 2 groups of 4, 2 × 4 = ＿. 1 group of 4, 1 × 4 = ＿. 0 groups of 4, 0 × 4 = ＿
01:58
### Video Transcript
Matthew started with 12 stars in three groups of four. Three groups of four. Three times four equals 12. Remove one group at a time. Find the missing numbers. Two groups of four, two times four equals what. One group of four, one times four equals what. Zero groups of four, zero times four equals what.
Matthew started with 12 stars in three groups of four. We have to remove one group of four at a time to find the missing numbers. We know that three times four is 12. So to find two groups of four, we just need to take away one group of four. What is four less than 12? Two times four is eight. Four less than 12 is eight.
If we take away another four, we will have one group of four, which is four. One group of four or one times four is four. And to find zero times four, we just need to take away another group of four. Four take away four is zero. Zero times four equals zero.
Matthew started with 12 stars in three groups of four. And he removed one group at a time. Two groups of four or two times four equals eight. One group of four or one times four equals four. And zero groups of four equal zero.
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### Polynomial PPT
```Warm up
List all POSSIBLE zeros, factor
completely, and give all ACTUAL zeros
1. x 3 - 7x 2 +14 x -8
2. 2x
3
+ 15 x
2
+22 x -15
Warm up
1. List all POSSIBLE zeros, factor completely, and
give all ACTUAL zeros
x3 + x2 - 10x + 8
2. Write a polynomial in standard form with the
given zeros.
0, -1, 2
3. Divide the following using SYNTHETIC
DIVISION
Warm up
1. Check if x + 3 is a factor of
f(x) = 2x 3 + 15 x 2 +22 x -15
2. Check if x - 2 is a factor of
f(x) = x 3 - 7x 2 +14 x -8
3. Check if x - 3 is a factor of
f(x) = 2x 3 + 15 x 2 +22 x -15
Warm up
1. Write a polynomial in standard form with the
given zeros.
a. 5/3, 1 , -1
b. 5, -1, 0
Divide the following using SYNTHETIC DIVISION.
Show all work.
Warm up
1. Write the polynomial in standard form and
give the degree.
2. Divide the following using LONG DIVISION.
Show all work.
3. Divide the following using SYNTHETIC
DIVISION.
Divide 2x3 + 11x2 - 2x – 25 by 2x + 3
Divide 3x3 – 5x2 + 10x – 3 by 3x + 1
Writing a polynomial function from
zeros
Steps
• Convert the zeros to factors.
• Multiply the factors.
• Combine like terms and write with powers of x
in descending order, which is the standard
form of a polynomial function.
Writing Polynomial Function from its
Zeros
Zeros means the value of x at which the value of
y is zero.
Write a polynomial function from its zeros
Suppose the zeros of a polynomial are -4, 1, and
2
Y=?
Y =(x+ 4) (x- 1) (x – 2)
Y = x3 + x2 - 10x + 8
Check: These values of x cause this
polynomial equal to zero.
Write a polynomial in standard form
with the given zeros.
1. 0, -3
2. 5, -1, 0
3. 2, 3, 3
5. 5/3, 1 , -1
6. -1, 3/2
Questions
What are the zeros of the polynomial function y = (x 3) (2x – 5) (x – 6)
Which polynomial function has zeros at -4, 3, and 5?
a. f(x) = (x + 4)(x + 3)(x + 5) b. g(x) = (x + 4)(x - 3)(x - 5)
c. h(x) = (x - 4)(x - 3)(x - 5) d. k(x) = (x - 4)(x + 3)(x + 5)
Write a polynomial function of least degree with
integral coefficients that has the given zeros?
1.) 3, 2, -2
2.) 3, 1, -2
3.) 5, -1, 0
4.) -3, -1/3, 5
5.) 5/3, 1, -1
6.) 2, 5/3, -5
If the given number x = a is a zero of the
polynomial function f(x), then f(a) = 0.
Example:
Verify if the given number is a zero of the
function.
Verify if the given number is a zero of
the function
m 14m 42m 29, x 9
3
2
Rational Zero Theorem
If a polynomial function, written in descending order
of the exponents, has integer coefficients, then any
rational zero must be of the form ± p/ q, where p is a
factor of the constant term and q is a factor of the
Rational Zeros
Steps to finding all rational zeros:
1) List the possible rational zeros by calculating p/q.
2) Use synthetic division to test the polynomial at each of
the possible rational zeros that you found in step 1.
OR Graph the function using your calculator to find a
starting point by identifying just one rational zero. (If
there are no rational zeros, you are done.)
3) Perform synthetic division on your polynomial using
the rational zero from step 2.
4) Repeat step 2 until you reach a quotient that is a
factoring to find the remaining zeros.
Find all the rational zeros of
f ( x) = 2 x 3 + 3 x 2 – 8 x + 3
P: Factors of 3 = ± 1, ± 3
q: Factors of 2 = ± 1, ± 2
The possibilities of p/q are ± 1,
1
±
2
,± 3,±
3
2
How do you find all the rational zeros
of a polynomial function?
1. x 3 - 7x 2 +14 x -8
(1, 2, 4)
2. 2x 3 + 15 x 2 +22 x -15
(-3, -5, ½)
The Factor Theorem
If the function f(x) = 0 at x = a, then (x – a) must be
a factor of the polynomial.
Example: x2-3x-4
f(4) = (4)2-3(4)-4 = 16-12-4 = 0
So (x – 4) must be a factor of x2-3x-4
1. Check if x - 2 is a factor of x 3 7x 2 +14 x -8
2. Check if x +5 is a factor of
F(x) = 2x 3 + 15 x 2 +22 x -15
The Remainder Theorem:
When we divide a polynomial f(x) by x- a the
remainder equals f(a)
Find the remainder when 2x2-5x-1 divided by x-3
Calculate f(3) to find the remainder.
F(3) =
2(3)2-5(3)-1 = 2x9-5x3-1 = 18-15-1 = 2
Remainder is 2.
Find the remainder when x3 + x2 - 10x + 8 is divided
by (x + 2).
End Behavior of the Polynomial
Function
The end behavior of a polynomial function is the behavior of the graph
of f(x) as x approaches positive infinity or negative infinity. The degree
and the leading coefficient of a polynomial function determine the end
behavior of the graph of polynomial.
End Behavior
1. If the degree n of a polynomial is even, then the arms of the graph
are either both up or both down.
2. If the degree n is odd, then one arm of the graph is up and one is
down.
3. If the leading coefficient is positive, the right arm of the graph is up.
4. If the leading coefficient an is negative, the right arm of the graph is
down.
Moe Specific Feasible Cases of End
Behavior
Feasible Cases:
Case 1: If Leading coefficient(an) is positive and the power(n) is even,
both ends of the graph go up.
Case 2: If Leading coefficient(an) is negative and the power(n) is even,
both ends of the graph go down.
Case 3: If Leading coefficient(an) is positive and the power(n) is odd,
the right hand of the graph goes up and the left hand goes down.
Case 4: If Leading coefficient(an) is negative and the power(n) is odd,
the right hand of the graph goes down and the left hand goes up.
End Behavior
A polynomial function is a function of the form
where each of the coefficients (the a’s) is a real
number and all exponents are positive. The
domain is all real numbers. The graph is
continuous and created with smooth curves.
Polynomial function
of odd degree, positive leading coefficient.
Polynomial function
of odd degree, negative leading coefficient
Polynomial function of even degree, positive leading
coefficient.
Polynomial function of even degree, negative leading
coefficient
•
•
•
•
Algebra II
Name: _________________________
Symmetry & End Behavior
Date: __________ ______
Circle whether the function is even, odd or neither. Then describe end behavior of the graph of the polynomial function by filling in the blanks with .
1.
f(x) = -6x3 + 8x
•
•
End Behavior:
the left side goes __________
•
•
the right side goes _________
2.
•
•
End Behavior:
the left side goes __________
•
the right side goes _________
•
3.
•
•
End Behavior:
the left side goes __________
•
the right side goes _________
•
4.
•
•
End Behavior:
the left side goes __________
•
the right side goes _________
•
5.
•
End Behavior:
f(x) = 7x4 - x3 + 7x + 1
f(x) = 5x3 - 5x2 - 7x – 3
f(x) = 2x2 – 1
f(x) = x5 - 6x7 - 4x
Use what you know about end behavior to match the polynomial function with its graph.
9. f ( x) 4 x 6 3x 2 5 x 2
10. f ( x) 2 x3 5x 2
11. f ( x) x 4 1
12.f ( x) 6 x3 1
y
y
A.
B.
x
C.
x
D.
y
x
y
x
List all POSSIBLE zeros, factor
completely, and give all ACTUAL zeros.
• What is the equation for the cubic model?
• Can you predict the value of y for x = 17?
• Steps to Finding a Cubic Regression using a graphing
calculator:
• 1) Go to Stat 1:Edit and enter the data into L1 and L2
• 2) Go to 2nd Y= and turn Plot1 On.
• 3) Go to Stat, Calc, 6:CubicReg, VARS, Y-VARS, Enter,
Enter, Enter.
• 4) Look at your graph by using Zoom 9:ZoomStat
• 5) Look at your equation by going to Y=.
• 6) Calculate a value by using 2nd Trace, 1:Value, X=.
Comparing Model
• Below is 3 scatterplots for the given data. Each
scatterplot is fitted with a different model.
Which model is the best fit?
```
|
Linear Functions
Student: So I have practiced guessing functions and am getting pretty good at it as long as there is one operation. The more complicated functions are harder.
Mentor: Yes, they are. The best way to understand such functions is to study one kind at a time. Let's start with functions of the form:
`Y = ____ * X + ____`
These functions are called linear functions, and are often written as:
`Y = m * X + b `
Where m represents the number multiplied to X and b represents the number added to the result.
Student: What's so important about these?
Mentor: These functions increase or decrease steadily. Look at the following function and table of points from the function:
`Y = 4 * X + 2 `
X Y
0 2
1 6
2 10
3 14
4 18
Now, answer some questions for me. What is the value of the function when X is 0?
Student: 2.
Mentor: Good. What is the change in the value of the function as X increases by 1?
Student: Well, the value of the function goes from 2 to 6 to 10. So at each step the function increases by 4.
Mentor: Now look at your answers: 2 for the starting point, when X is 0 and 4 for the increase. Do those numbers look familiar?
Student: In the original function, Y = 4 * X + 2, m = 4 and b = 2. The same numbers we got for the starting value and the increasing value. Is this a coincidence?
Mentor: No, it is not a coincidence. This always works. Try some.
Student: Here are a few:
• `Y = 10 * X - 1`
Change = 10, start = -1
• `Y = -2 * X + 3`
Change = -2, start = 3
• `Y = 5 * X + 11`
Change = 5, start = 11
Mentor: Good! But before we begin, let's get the terminology right: The change is called the slope and the starting value is called the intercept. We'll learn why these words are used later when we talk about graphs. Can you build a few tables of ordered pairs to further demonstrate these facts about your functions? You may wish to use Simple Plot to plot the ordered pairs from your table.
|
### The perimeter of a rhombus is 74 cm and one of its diagonals is 35 cm. What is the length of other diagonal?
12 cm
Step by Step Explanation:
1. One way to solve this is as follows:
We know that,
a) The sides of a rhombus are equal. Therefore one side =
74 4
= 18.5
b) A diagonal of a rhombus divides the rhombus into 2 equal triangles.
c) The area of a rhombus is
1 2
(Diagonal1 × Diagonal2) ------(1)
2. Taking one of the two triangles formed by the diagonal with length 35 cm.
Area (using Heron's formula) = $\sqrt{ S(S-18.5)(S-18.5)(S-35) }$
Where, S =
2 × 18.5 + 35 2
=
72 2
= 36
Area = $\sqrt{ 36(36-18.5)(36-18.5)(36-35) }$ = 210 ------(2) [The details of this computation are left to the student.]
3. On comparing equation (1) and (2) we get,
1 2
(Diagonal1 × Diagonal2) = 210
⇒
1 2
(35 × Diagonal2) = 210
⇒ Diagonal2 = 2 ×
210 35
= 12 cm
|
Question #e7e98
Feb 16, 2018
$x < 1 \mathmr{and} x > 7$
Explanation:
Given ${x}^{2} - 8 x + 4 > - 3$
Adding 3 on both sides will give us
${x}^{2} - 8 x + 7 > 0$
Solving this equation using the quadratic formula, we get
$\frac{8 \pm \sqrt{{8}^{2} - 4 \left(1 \cdot 7\right)}}{2 \cdot 1}$
This gives us
$\frac{8 \pm \sqrt{64 - 28}}{2}$
Which is $\frac{8 \pm \sqrt{36}}{2}$
Which is $\frac{8 \pm 6}{2}$
Which gives 2 solutions:
$x = 7$ and $x = 1$
Now check whether the inequality is valid for values greater or lesser than the ones we have obtained from the quadratic equation.
Substituting $x = 8$ we can see that the original inequality given in the question is true, so x must be greater than 7
$x > 7$
And taking a value less than 1, suppose 0, also shows that the inequality in the question is true.
So $x < 1$
|
Edit Article
# wikiHow to Find the Slope of an Equation
The slope of a line is a measure of how fast it is changing. This can be for a straight line -- where the slope tells you exactly how far up (positive slope) or down (negative slope) a line goes while it goes how far across. Slope can also be used for a line tangent to a curve. Or, it can be for a curved line when doing Calculus, where slope is also known as the "derivative" of a function. Either way, think of slope simply as the "rate of change" of a graph: if you make the variable "x" bigger, at what rate does "y" change? That is a way to see slope as a cause and an effect event.
### Method 1 Finding the Slope of a Linear Equation
1. 1
Use slope to determine how steep, and in what direction (upward or downward), a line goes. Finding the slope of a line is easy, as long as you have or can setup a linear equation. This method works if and only if:
• There are no exponents on the variables
• There are only two variables, neither of which are fractions (for example, you would not have ${\displaystyle {\frac {1}{x}}}$
• The equation can be simplified to the form ${\displaystyle y=mx+b}$, where m and b are constants (numbers like 3, 10, -12, ${\displaystyle {\frac {4}{3}},{\frac {3}{5}}}$).[1]
2. 2
Find the number in front of the x, usually written as "m," to determine slope. If your equation is already in the right form, ${\displaystyle y=mx+b}$, then simply pick the number in the "m" position (but if there is no number written in front of x then the slope is 1). That is your slope! Note that this number, m, is always multiplied by the variable, in this case an "x." Check the following examples:
• ${\displaystyle y=2x+6}$
• Slope = 2
• ${\displaystyle y=2-x}$
• Slope = -1
• ${\displaystyle y={\frac {3}{8}}x-10}$
• Slope = ${\displaystyle {\frac {3}{8}}}$ [2]
3. 3
Reorganize the equation so one variable is isolated if the slope isn't apparent. You can add, subtract, multiply, and more to isolate a variable, usually the "y." Just remember that, whatever you do to one side of the equal sign (like add 3) you must do to the other side as well. Your final goal is an equation similar to ${\displaystyle y=mx+b}$. For example:
• Find the slope of ${\displaystyle 2y-3=8x+7}$
• Set to the form ${\displaystyle y=mx+b}$:
• ${\displaystyle 2y-3+3=8x+7+3}$
• ${\displaystyle 2y=8x+10}$
• ${\displaystyle {\frac {2y}{2}}={\frac {8x+10}{2}}}$
• ${\displaystyle y=4x+5}$
• Find the slope:
• Slope = M = 4[3]
### Method 2 Finding the Slope with Two Points
1. 1
Use a graph and two points to find slope without the equation handy. If you've got a graph and a line, but no equation, you can still find the slope with ease. All you need are two points on the line, which you plug into the equation ${\displaystyle {\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}}$. While finding the slope, keep in mind the following information to help you check if you're on the right track:
• Positive slopes go higher the further right you go.
• Negative slopes go lower the further right you go.
• Bigger slopes are steeper lines. Small slopes are always more gradual.
• Perfectly horizontal lines have a slope of zero.
• Perfectly vertical lines do not have a slope at all. Their slope is "undefined."[4]
2. 2
Find two points, putting them in simple (x,y) form. Use the graph (or the test question) to find the x and y coordinates of two points on the graph. They can be any two points that the line crosses through. For an example, assume that the line in this method goes through (2,4) and (6,6).[5]
• In each pair, the x coordinate is the first number, the y coordinate comes after the comma.
• Each x coordinate on a line has an associated y coordinate.
3. 3
Label your points x1, y1, x2, y2, keeping each point with it's pair. Continuing our first example, with the points (2,4) and (6,6), label the x and y coordinates of each point. You should end up with:
• x1: 2
• y1: 4
• x2: 6
• y2: 6[6]
4. 4
Plug your points into the "Point-Slope Formula" to get your slope. The following formula is used to find slope using any two points on a straight line: ${\displaystyle {\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}}$. Simply plug in your four points and simplify:
• Original Points: (2,4) and (6,6).
• Plug into Point Slope:
• ${\displaystyle {\frac {6-4}{6-2}}}$
• ${\displaystyle {\frac {2}{4}}={\frac {1}{2}}}$ = Slope
5. 5
Understand how the Point-Slope Formula works. The slope of a line is “Rise over Run:” how much the line goes up divided by how much the line "runs" to the right. The “rise” of the line is the difference between the y-values (remember, the Y-axis goes up and down), and the “run” of the line is the difference between the x-values (and the X-axis goes left and right).
6. 6
Recognize other ways you may be tested to find slope. The equation of the slope is ${\displaystyle {\frac {y_{2}-y_{1}}{x_{2}-x_{1}}}}$. This may also be shown using the Greek letter “Δ”, called “delta”, meaning “difference of”. Slope can also be shown as Δy/Δx, meaning "difference of y / difference of x:" this is the same exact question as "find the slope between
### Method 3 Using Differential Calculus to Find the Slope of a Curve
1. 1
Review how to take a variety of derivatives from common functions. Derivatives give you the rate of change (or slope) at a single point on a line. The line can be curved or straight -- it doesn't matter. Think of it as how much the line is changing at any time, instead of the slope of the entire line. How you take derivatives changes depending on the type of function you have, so review how to take common derivatives before moving on.
2. 2
Understand what questions are asking for a slope using derivatives. You will not always be asked to explicitly find the derivative or slope of a curve. You might also be asked for the "rate of change at point (x,y). You could be asked for an equation for the slope of the graph, which simply means you need to take the derivative. Finally, you may be asked for "the slope of the tangent line at (x,y)." This, once again, just wants the slope of the curve at a specific point, (x,y).
• For this method, consider the question: "What is the slope of the line ${\displaystyle f(x)=2x^{2}+6x}$ at the point (4,2)?"[7]
• The derivative is often written as ${\displaystyle f'(x),y',}$ or ${\displaystyle {\frac {dy}{dx}}}$[8]
3. 3
Take the derivative of your function. You don't even really need you graph, just the function or equation for your graph. For this example, use the function from earlier, ${\displaystyle f(x)=2x^{2}+6x}$. Following the methods outlined here, take the derivative of this simple function.
• Derivative: ${\displaystyle f'(x)=4x+6}$
4. 4
Plug in your point to the derivative equation to get your slope. The differential of a function will tell you the slope of the function at a given point. In other words, f’(x) is slope of the function at any point (x,f(x)) So, for the practice problem:
• What is the slope of the line ${\displaystyle f(x)=2x^{2}+6x}$ at the point (4,2)?
• Derivative of Equation:
• ${\displaystyle f'(x)=4x+6}$
• Plug in Point for x:
• ${\displaystyle f'(x)=4(4)+6}$
• Find the Slope:
• The slope of the ${\displaystyle f(x)=2x^{2}+6x}$ at (4,2) is 22.
5. 5
Check your point against a graph whenever possible. Know that not all points in calculus will have a slope. Calculus gets into complex equations and difficult graphs, and not all points will have a slope, or even exist on every graph. Whenever possible, use a graphing calculator to check the slope of your graph. If you can't, draw the tangent line using your point and the slope (remember -- "rise over run") and note if it looks like it could be correct.
• Tangent lines are just lines with the exact same slope as your point on the curve. To draw one, go up (positive) or down (negative) your slope (in the case of the example, 22 points up). Then move over one and draw a point. Connect the dots, (4,2) and (26,3) for your line.
## Community Q&A
Search
• So is the slope of y=-2 .-2?
No. The slope is zero, because there is no x-term in the equation, meaning that m (the x-coefficient and also the slope) is zero.
• What if the equation is like x+y=0 or x-y=0?
That's no problem. x+y=0 is equal to y=-x. On the other hand, x-y=0 is equivalent to y=x. Both of these lines have a slope of 1.
• What is the slope for the equation y=1?
The graph of y=1 is a straight, horizontal line, meaning that it does not rise or fall as it moves left or right. Its slope is therefore zero.
• What is the slope of x = -2?
The slope is called "undefined." That means in this case that the slope is infinity (because there are an infinite number of y values associated with the only x value).
200 characters left
## Article Info
Categories: Algebra
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# How do you solve the system of linear equations by elimination 12x-7y=-2, 8x+11y=30?
Jun 8, 2018
$x = 1$
$y = 2$
#### Explanation:
Solving By Elimination
$12 x - 7 y = - 2$
$8 x + 11 y = 30$
First, let's have a look at our system and recall what the elimination method is. To eliminate a variable (either $x$ or $y$), we have to multiply the different equations in the system to get the least common multiple. One equation must be positive and the other negative to cancel each other out. The least common multiple of $12$ and $8$ is $24$, so I multiply the first equation of the system by $2$ to get to $24 x$, and the second equation by $- 3$ to get $- 24 x$ to cancel out the positive $24 x$.
$2 \left(12 x - 7 y = - 2\right)$
$- 3 \left(8 x + 11 y = 30\right)$
Distribute. All you have to do is multiply each term in the parentheses by the number on the outside, so:
First equation
$2 \cdot 12 x = 24 x$
$2 \cdot - 7 y = - 14 y$
$2 \cdot - 2 = - 4$
Second equation
$- 3 \cdot 8 x = - 24 x$
$- 3 \cdot 11 y = - 33 y$
$- 3 \cdot 30 = - 90$
Your equations should now look like this:
$24 x - 14 y = - 4$
$- 24 x - 33 y = - 90$
Because $24 x$ and $- 24 x$ cancel out, you're left with $- 14 y$, $- 33 y$, $- 4$, and $- 90$. Combine like terms, so:
$- 47 y = - 94$
Divide by $- 47$ to isolate for $y$.
$y = 2$
Now that you know that $y = 2$, plug it into an equation of the system to solve for $x$:
$12 x - 7 y = - 2$
$12 x - 7 \left(2\right) = - 2$
$12 x - 14 = - 2$
$12 x = 12$
$x = 1$
Now that you know that $x$ is $1$, you can plug both values back into the system to confirm they're correct:
$12 x - 7 y = - 2$
$12 \left(1\right) - 7 \left(2\right) = - 2$
$12 - 14 = - 2$
$- 2 = - 2$
$8 x + 11 y = 30$
$8 \left(1\right) + 11 \left(2\right) = 30$
$8 + 22 = 30$
$30 = 30$
They're correct!
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# Difference between revisions of "2018 AMC 8 Problems/Problem 3"
## Problem 3
Students Arn, Bob, Cyd, Dan, Eve, and Fon are arranged in that order in a circle. They start counting: Arn first, then Bob, and so forth. When the number contains a 7 as a digit (such as 47) or is a multiple of 7 that person leaves the circle and the counting continues. Who is the last one present in the circle?
$\textbf{(A) } \text{Arn}\qquad\textbf{(B) }\text{Bob}\qquad\textbf{(C) }\text{Cyd}\qquad\textbf{(D) }\text{Dan}\qquad \textbf{(E) }\text{Eve}$
## Solution
The five numbers which cause people to leave the circle are $7, 14, 17, 21,$ and $27.$
Arn counts $7$ (assuming they start at $1$) so he leaves first. Then Cyd counts $14$, as there are $7$ numbers to be counted from this point. Then Fon, Bob, Eve, so last one standing is Dan, hence meaning the answer would be $\textbf{(D)}$
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## How Do You Find The Equation Of A Tangent Line?
A tangent line to a function is a line that looks most like the function at a point. In common terms, it just grazes the function.
To find its equation, we need to locate the point where the two meet as well as the slope of the function at that point. Then we can use the slope-intercept form or point-slope form of a line to get the equation.
Find the equation of the tangent line to
at x = 3.
Since this problem is asking for the equation of a line, let’s start with the point-slope form
This requires a point (x1, y1) and slope m. We’ll use the function to get the point and the derivative to get the slope of the tangent line.
Find the point: We are given a point x = 3. To find the corresponding y value, put the x value into the function
Find the slope of the tangent line: We need h′(3) to get the slope of the tangent line. We’ll use the Power Rule to take the derivative,
The slope of the tangent is
Write the equation of the tangent line: Putting the point (3, 10) and the slope 9 into the line yields
If you are asked to write this in slope-intercept form, you’ll need to solve this for y to give
If you graph h(x) and the tangent line together, it should be obvious that your tangent line is correct (ie. tangent).
## How Do You Calculate A Rate From A Function?
In many business problems, we are interested in knowing the rate at which some quantity is changing. If this quantity is modeled by a function, the rate is modeled by the derivative of the function.
In the problem below, we are given a model of revenue for Verizon, a national wireless carrier. To find the rate, we’ll need to take the derivative of the revenue function.
From 2006 through 2011, Verizon Wireless grew steadily. As the number of connections increased, so did the revenue from those connections.A cubic model for the revenue over this period is
where c is the number of connections in millions.
At what rate is the revenue increasing when there are 80,000,000 connections?
Questions that ask about rate or refer to “marginal” are questions about the derivative. In this case, it is referring to the value of the derivative at c = 80. The derivative of the revenue function is
The value of the derivative at 80 million connections is
To find the units on this rate, recall that the derivative is also the slope of the tangent line at c = 80. The slope of this line has vertical units of billions of dollars and horizontal units of millions of connections. This may be simplified to
Since a billion divided by a million is one thousand. So means that an additional connection at this level results in an increase in revenue of 0.31265 thouusand dollars or \$321.65.
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$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 3.2: The Mean Value Theorem
We motivate this section with the following question: Suppose you leave your house and drive to your friend's house in a city 100 miles away, completing the trip in two hours. At any point during the trip do you necessarily have to be going 50 miles per hour?
In answering this question, it is clear that the average speed for the entire trip is 50 mph (i.e. 100 miles in 2 hours), but the question is whether or not your instantaneous speed is ever exactly 50 mph. More simply, does your speedometer ever read exactly 50 mph?. The answer, under some very reasonable assumptions, is "yes."'
Let's now see why this situation is in a calculus text by translating it into mathematical symbols.
First assume that the function $$y = f(t)$$ gives the distance (in miles) traveled from your home at time $$t$$ (in hours) where $$0\le t\le 2$$. In particular, this gives $$f(0)=0$$ and $$f(2)=100$$. The slope of the secant line connecting the starting and ending points $$(0,f(0))$$ and $$(2,f(2))$$ is therefore
$$\frac{\Delta f}{\Delta t} = \frac{f(2)-f(0)}{2-0} = \frac{100-0}{2} = 50 \, \text{mph}.$$
The slope at any point on the graph itself is given by the derivative $$f'(t)$$. So, since the answer to the question above is "yes," this means that at some time during the trip, the derivative takes on the value of 50 mph. Symbolically,
$$f'(c) = \frac{f(2)-f(0)}{2-0} = 50$$
for some time $$0\le c \le 2.$$
How about more generally? Given any function $$y=f(x)$$ and a range $$a\le x\le b$$ does the value of the derivative at some point between $$a$$ and $$b$$ have to match the slope of the secant line connecting the points $$(a,f(a))$$ and $$(b,f(b))$$? Or equivalently, does the equation
$f'(c) = \frac{f(b)-f(a)}{b-a}$
have to hold for some $$a < c < b$$?
Let's look at two functions in an example.
Example $$\PageIndex{1}$$: Comparing average and instantaneous rates of change
Consider functions
$$f_1(x)=\frac{1}{x^2}\quad \text{and} \quad f_2(x) = |x|$$
with $$a=-1$$ and $$b=1$$ as shown in Figure $$\PageIndex{1}$$ (a) and (b), respectively. Both functions have a value of 1 at $$a$$ and $$b$$. Therefore the slope of the secant line connecting the end points is $$0$$ in each case. But if you look at the plots of each, you can see that there are no points on either graph where the tangent lines have slope zero. Therefore we have found that there is no $$c$$ in $$[-1,1]$$ such that
$$f'(c) = \frac{f(1)-f(-1)}{1-(-1)} = 0.$$
Figure $$\PageIndex{1}$$: A graph of $$f_1(x) = 1/x^2$$ and $$f_2(x) = |x|$$ in Example $$\PageIndex{1}$$.
So what went "wrong"'? It may not be surprising to find that the discontinuity of $$f_1$$ and the corner of $$f_2$$ play a role. If our functions had been continuous and differentiable, would we have been able to find that special value $$c$$? This is our motivation for the following theorem.
Theorem $$\PageIndex{1}$$: The Mean Value Theorem of Differentiation
Let $$y=f(x)$$ be continuous function on the closed interval $$[a,b]$$ and differentiable on the open interval $$(a,b)$$. There exists a value $$c$$, $$a < c < b$$, such that
$$f'(c) = \frac{f(b)-f(a)}{b-a}.$$
That is, there is a value $$c$$ in $$(a,b)$$ where the instantaneous rate of change of $$f$$ at $$c$$ is equal to the average rate of change of $$f$$ on $$[a,b]$$.
Note that the reasons that the functions in Example $$\PageIndex{1}$$ fail are indeed that $$f_1$$ has a discontinuity on the interval $$[-1,1]$$ and $$f_2$$ is not differentiable at the origin.
We will give a proof of the Mean Value Theorem below. To do so, we use a fact, called Rolle's Theorem, stated here.
Theorem $$\PageIndex{2}$$: Rolle's Theorem
Let $$f$$ be continuous on $$[a,b]$$ and differentiable on $$(a,b)$$, where $$f(a) = f(b)$$. There is some $$c$$ in $$(a,b)$$ such that $$f'(c) = 0.$$
Consider Figure $$\PageIndex{2}$$ where the graph of a function $$f$$ is given, where $$f(a) = f(b)$$. It should make intuitive sense that if $$f$$ is differentiable (and hence, continuous) that there would be a value $$c$$ in $$(a,b)$$ where $$f'(c)=0$$; that is, there would be a relative maximum or minimum of $$f$$ in $$(a,b)$$. Rolle's Theorem guarantees at least one; there may be more.
FIgure $$\PageIndex{2}$$: A graph of $$f(x) = x^3-5x^2+3x+5$$, where $$f(a) = f(b)$$. Note the existence of $$c$$, where $$a<c<b$$, where $$f'(c)=0$$.
Rolle's Theorem is really just a special case of the Mean Value Theorem. If $$f(a) = f(b)$$, then the average rate of change on $$(a,b)$$ is $$0$$, and the theorem guarantees some $$c$$ where $$f'(c)=0$$. We will prove Rolle's Theorem, then use it to prove the Mean Value Theorem.
Proof of Rolle's Theorem
Let $$f$$ be differentiable on $$(a,b)$$ where $$f(a)=f(b)$$. We consider two cases.
Case 1: Consider the case when $$f$$ is constant on $$[a,b]$$; that is, $$f(x) = f(a) = f(b)$$ for all $$x$$ in $$[a,b]$$. Then $$f'(x) = 0$$ for all $$x$$ in $$[a,b]$$, showing there is at least one value $$c$$ in $$(a,b)$$ where $$f'(c)=0$$.
Case 2: Now assume that $$f$$ is not constant on $$[a,b]$$. The Extreme Value Theorem guarantees that $$f$$ has a maximal and minimal value on $$[a,b]$$, found either at the endpoints or at a critical value in $$(a,b)$$. Since $$f(a)=f(b)$$ and $$f$$ is not constant, it is clear that the maximum and minimum cannot both be found at the endpoints. Assume, without loss of generality, that the maximum of $$f$$ is not found at the endpoints. Therefore there is a $$c$$ in $$(a,b)$$ such that $$f(c)$$ is the maximum value of $$f$$. By Theorem 3.1.2, $$c$$ must be a critical number of $$f$$; since $$f$$ is differentiable, we have that $$f'(c) = 0$$, completing the proof of the theorem.
$$\square$$
We can now prove the Mean Value Theorem.
Proof of the Mean Value Theorem
Define the function
$$g(x) = f(x) - \frac{f(b)-f(a)}{b-a}x.$$
We know $$g$$ is differentiable on $$(a,b)$$ and continuous on $$[a,b]$$ since $$f$$ is. We can show $$g(a)=g(b)$$ (it is actually easier to show $$g(b)-g(a)=0$$, which suffices). We can then apply Rolle's theorem to guarantee the existence of $$c \in (a,b)$$ such that $$g'(c) = 0$$. But note that
$$0= g'(c) = f'(c) - \frac{f(b)-f(a)}{b-a} \ ;$$
hence
$$f'(c) = \frac{f(b)-f(a)}{b-a},$$
which is what we sought to prove.
$$\square$$
Going back to the very beginning of the section, we see that the only assumption we would need about our distance function $$f(t)$$ is that it be continuous and differentiable for $$t$$ from 0 to 2 hours (both reasonable assumptions). By the Mean Value Theorem, we are guaranteed a time during the trip where our instantaneous speed is 50 mph. This fact is used in practice. Some law enforcement agencies monitor traffic speeds while in aircraft. They do not measure speed with radar, but rather by timing individual cars as they pass over lines painted on the highway whose distances apart are known. The officer is able to measure the average speed of a car between the painted lines; if that average speed is greater than the posted speed limit, the officer is assured that the driver exceeded the speed limit at some time.
Note that the Mean Value Theorem is an existence theorem. It states that a special value $$c$$ exists, but it does not give any indication about how to find it. It turns out that when we need the Mean Value Theorem, existence is all we need
Example $$\PageIndex{2}$$: Using the Mean Value Theorem
Consider $$f(x) = x^3+5x+5$$ on $$[-3,3]$$. Find $$c$$ in $$[-3,3]$$ that satisfies the Mean Value Theorem.
Solution
The average rate of change of $$f$$ on $$[-3,3]$$ is:
$$\frac{f(3)-f(-3)}{3-(-3)} = \frac{84}{6} = 14.$$
We want to find $$c$$ such that $$f'(c) = 14$$. We find $$f'(x) = 3x^2+5$$. We set this equal to 14 and solve for $$x$$.
\begin{align*} f'(x) &= 14 \\ 3x^2 +5 &= 14\\ x^2 &= 3\\ x &= \pm \sqrt{3} \approx \pm 1.732 \end{align*}
We have found 2 values $$c$$ in $$[-3,3]$$ where the instantaneous rate of change is equal to the average rate of change; the Mean Value Theorem guaranteed at least one. In Figure $$\PageIndex{3}$$ $$f$$ is graphed with a dashed line representing the average rate of change; the lines tangent to $$f$$ at $$x=\pm \sqrt{3}$$ are also given. Note how these lines are parallel (i.e., have the same slope) as the dashed line.
Figure $$\PageIndex{3}$$: Demonstrating the Mean Value Theorem in Example $$\PageIndex{2}$$.
While the Mean Value Theorem has practical use (for instance, the speed monitoring application mentioned before), it is mostly used to advance other theory. We will use it in the next section to relate the shape of a graph to its derivative.
### Contributors
• Gregory Hartman (Virginia Military Institute). Contributions were made by Troy Siemers and Dimplekumar Chalishajar of VMI and Brian Heinold of Mount Saint Mary's University. This content is copyrighted by a Creative Commons Attribution - Noncommercial (BY-NC) License. http://www.apexcalculus.com/
• Integrated by Justin Marshall.
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Words to Expressions 1
Alignments to Content Standards: 5.OA.A.2
Write an expression that records the calculations described below, but do not evaluate.
Add 2 and 4 and multiply the sum by 3. Next, add 5 to that product and then double the result.
IM Commentary
This problem allows student to see words that can describe the expression from part (c) of "5.OA Watch out for Parentheses." Additionally , the words (add, sum) and (product, multiply) are all strategically used so that the student can see that these words have related meanings.
Solution
Taking the instructions step by step we have:
1. Add 2 and 4 and multiply the sum by 3: $3(2+4)$ or possibly $(2+4)3$.
2. Add 5 to that product gives $5+3(2+4)$ or $3(2+4)+5$ or even $(2+4)3+5$.
3. Then double the result means we need to multiply the entire expression by 2: $$2(5 + 3(2 + 4))$$ or $$(5 + 3(2 + 4))2$$ or $$2(3(2+4)+5)$$ or $$(3(2+4)+5)2$$ or $$2((2+4)3+5)$$ or $$((2+4)3+5)2$$ Additionally, this can be written as $$(5 + 3(2 + 4)) + (5 + 3(2 + 4)).$$
The differing ways the calculations can be expressed also help to illustrate the commutative laws of multiplication and addition.
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# 3 ways to decompose numbers
## Table of contents:
Number decomposition helps younger students understand the arrangement and relationships between the digits of the same number and between the numbers in an operation. You can decompose a number into hundreds, tens, and ones, or by separating the numbers into several addends.
## Steps
### Method 1 of 3: Decompose into Hundreds, Tens, and Ones
#### Step 1. Understand the difference between tens and ones
When you see a number with two digits and no decimal point, the digits represent the tens and the ones. The tens are on the left and the ones on the right.
• The number in the units place can be read exactly as it appears. The only numbers that are considered units are those from 0 to 9 (zero, one, two, three, four, five, six, seven, eight, and nine).
• The number in the tens place is apparently of the same type as the one in the ones place. However, if they are analyzed separately, the ten actually has a 0 to the right, so it is a higher number than the one in the ones place. The numbers that are considered tens are: 10, 20, 30, 40, 50, 60, 70, 80 and 90 (ten, twenty, thirty, forty, fifty, sixty, seventy, eighty and ninety).
#### Step 2. Separate the two-digit number into two parts
A two-digit number will always have a part of ones and another part of tens. To decompose the number, you will have to separate it into those two parts.
• Example: decompose the number 82.
• 8 is the tens digit, so this part of the number can be separated and written as 80.
• The 2 is the ones digit, so this part of the number can be separated and written as 2.
• To write the result, you must express it as follows: 82 = 80 + 2.
• Also, keep in mind that a number written in normal form is expressed in its "standard form", but a decomposed number is expressed in expanded form.
### Based on the example above, "82" would be the standard form and "80 + 2" would be the expanded form
#### Step 3. Enter the hundreds
A three-digit number without a decimal point will always have units, tens, and hundreds. The hundreds are to the left of the number, the tens in the center, and the ones to the right.
• Units and tens work exactly the same as for two-digit numbers.
• The number in the hundreds place is apparently the same as the number in the ones place, but when analyzed separately, a hundred actually has two leading zeros. The numbers that are considered hundreds are: 100, 200, 300, 400, 500, 600, 700, 800 and 900 (one hundred, two hundred, three hundred, four hundred, five hundred, six hundred, seven hundred, eight hundred, nine hundred).
#### Step 4. Separate the three-digit number into three parts
A three-digit number will always have ones, tens, and hundreds. To decompose a number of this size, you will have to separate it into its three parts.
• Example: decompose the number 394.
• The 3 is in the hundreds place, so this part can be separated and written as 300.
• The 9 is in the tens place, so this part of the number can be separated and written as 90.
• The 4 is in the ones place, so this part of the number can be separated and written as 4.
• You must write the final result as follows: 394 = 300 + 90 + 4.
• If you write the number as 394, it will be expressed in its standard form. If you write it as 300 + 90 + 4, it will be expressed in expanded form.
#### Step 5. Apply this decomposition pattern to numbers with infinite digits
You can decompose larger numbers by following the same procedure.
• A digit, located anywhere in a number, can be expressed separately, replacing the digits on the right with zeros. This is applicable to any number, regardless of the number of digits that compose it.
• Example: 5,394,128 = 5,000,000 + 300,000 + 90,000 + 4,000 + 100 + 20 + 8
#### Step 6. Understand how decimals work
You can decompose a decimal number, but each number after the comma must also be expressed separately with a decimal point.
• The tenths place is used when there is a single digit to the right of the decimal point.
• The hundredths place is used when there are two digits to the right of the decimal point.
• The thousandths place is used when there are three digits to the right of the decimal point.
#### Step 7. Separate a decimal number into several parts
If you have a number with digits to the left and to the right of the decimal point, you will need to decompose it by separating both sides.
• Note that all the digits that appear to the left of the decimal point must be decomposed the same as if there were no comma.
• Example: decompose the number 431, 58
• The 4 is in the hundreds place, so it should be expressed separately as: 400
• The 3 is in the tens place, so it should be expressed separately as: 30
• The 1 is in the ones place, so it should be expressed separately as: 1
• The 5 is in the tenths place, so it should be expressed separately as: 0, 5
• The 8 is in the hundredths place, so it should be expressed separately as: 0, 08
• To write the final result, you must express it as follows: 431, 58 = 400 + 30 + 1 + 0, 5 + 0, 08
### Method 2 of 3: Decompose into Multiple Addends
#### Step 1. Understand the concept
When we decompose a number into several addends, what we do is separate it into other numbers (addends) that can be added to obtain the original value.
• When one addend is subtracted from the original number, the result is equal to the second addend.
• When both addends are added, the result is equal to the original number.
#### Step 2. Practice with a simple number
It's easier to understand the concept by practicing with single-digit numbers (numbers that have only units).
### You can use the principles learned in the section "Decomposing into Hundreds, Tenths, and Units" to decompose higher numbers, but since there are so many possible combinations of addends in multi-digit numbers, this method is not very practical for such cases
#### Step 3. Work with all possible combinations of sums
To decompose a number into addends, you just have to write all the possible ways to get the original value using smaller numbers and sums.
• Example: decompose the number 7 into its different addends.
• 7 = 0 + 7
• 7 = 1 + 6
• 7 = 2 + 5
• 7 = 3 + 4
• 7 = 4 + 3
• 7 = 5 + 2
• 7 = 6 + 1
• 7 = 7 + 0
#### Step 4. Use objects if necessary
For some people who are learning this concept, it may be helpful to use visual examples to illustrate the process in a practical way.
• Start by gathering the original number of something. For example, if the number is seven, you can start by collecting seven pieces of candy.
• Separate the candies into two groups, placing one aside. Count the candies that are left in the second group and explain that the seven at the beginning have broken down into two groups of one and six.
• Keep separating candies into two different groups, removing one more from the first group and adding it to the second. Each time you make a move, count the number of candies in both groups.
• You can do this with many different objects, such as small candles, squares of paper, colored clothespins, buckets or buttons.
### Method 3 of 3: Decompose to Solve Features
#### Step 1. Look at a simple sum
You can combine the two decomposition methods to separate this type of operation in different ways.
### This is easy enough for decomposing simple operations, but not as practical when used to decompose long operations
#### Step 2. Decompose the numbers of the operation
Look at the operation and separate the numbers into tens and ones. If necessary, you can separate the units into smaller groups.
• Example: decompose and solve the operation: 31 + 84.
• You can decompose 31 into: 30 + 1.
• You can decompose 84 into: 80 + 4.
#### Step 3. Modify and rewrite the operation more easily
The operation can be rewritten by expressing each decomposed component separately, or by combining certain decomposed components to see the whole operation more clearly.
### Example: 31 + 84 = 30 + 1 + 80 + 4 = 30 + 80 + 5 = 100 + 10 + 5
#### Step 4. Solve the operation
Once you have rewritten the operation to simplify it and make more sense, you just have to add the numbers and find the result.
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# Unit Vector Calculator
In this article, the calculator developed by us will easily calculate the unit vector that is the vector of length 1 unit in any plane. The article will first explain to you the concept of the unit vector in physics, its formulas followed by a few solved examples and generally asked FAQs.
A unit vector is a vector defined that of length equal to 1unit in any plane. When we use a unit vector to describe a spatial direction, we call it a direction vector with respect to that plane. In a Cartesian coordinate system(which we generally work in), the three-unit vectors that form the basis of the 3D space are as follow:
(1, 0, 0) — Directs the vector in the x-direction;
(0, 1, 0) — Directs the vector in the y-direction; and
(0, 0, 1) — Directs the vector in the z-direction.
## What is a unit vector?
Talking about an arbitrary vector, it is possible to calculate what the unit vector is along the same direction in any plane. And to do that, we can apply the following formula to calculate the unit vector ; û = u / |u|
where:
û — Unit vector;
u — Arbitrary vector in the form (x, y, z) or in any dimensional plane; and
|u| — Magnitude of the vector u in one particular direction.
You can calculate the magnitude of a vector using our distance calculator or simply by the equation:
|u| = √(x² + y² + z²)
The unit vector is a useful concept in describing linear transformations of any vector in a plane. For example, the matrix norm describes how much a unit vector is stretched when multiplied by a matrix of a dimensional vector.
### How to calculate the unit vector?
Let's consider an example of a vector u = (1, -3, 2). To calculate the unit vector in the same direction, you have to follow these steps:
1. Write down the x, y, and z components of the vector. In this case, x₁ = 1, y₁ = -3 and z₁ = 2.
2. Calculate the magnitude of the vector u:
|u| = √(x₁² + y₁² + z₁²)
|u| = √(1² + (-3)² + 2²)
|u| = √(1 + 9 + 4)
|u| = √14
|u| = 3.8
3. Now that you know the magnitude of the vector u, you probably want to know how to calculate the unit vector. All you have to do is divide each of the initial vector's components by |u|.
x₂ = x₁ / |u| = 1 / 3.8 = 0.263
y₂ = y₁ / |u| = -3 / 3.8 = 0.789
z₂ = z₁ / |u| = 4 / 3.8 = 1.05
4. Now, write these results in vector form to find the vector û = (0.263,0.789,1.05).
5. The user can verify the result if the magnitude of your unit vector should be equal to 1.
### Solved Examples on Unit Vector
Question 1:
Find the unit vector →aa→for the given vector, 12ii^– 3^jj^– 4 ^kk^.
Solution:
Given
To find the magnitude of the given vector first, →aa→ is :
Let’s use this magnitude to find the unit vector now:
Substituting the values:
a^=a|a|=xi^+yj^+zk^x2+y2+z2
a^=12i^−3j^–4j^1
The unit vector in Bracket form is given as follow:
a^=1213i^−313j^−413k^
Question 2: Find the unit vector →bb→ for the given vector, −2^i+4^j–4^k.−2i^+4j^–4k^.
Solution:
Given:
To find the magnitude of the given vector first,
→b is :|b|= √x2+y2+z2|q|=√−22+(4)2+(−4)2
|bI= √4+16+16
Ib|= √36
|b|= 6
The magnitude to find the unit vector now:
|b|= 6
### FAQs on Unit Vector Calculator
1. How to find a unit vector in the same direction as that of the plane?
To find this, divide the original vector by its magnitude. For example, the vector u = (2, 3) has a magnitude of √(2² + 3²) = √13. Therefore the unit vector that has the same direction is û = (2/√13, 3/√13) = (0.5547, 0.832).
2. Is (1, 1) a unit vector?
No, the length of a unit vector needs to be equal to 1. Calculating the length of (1, 1), we find that √(1² + 1²) = √2 = 1.414, which is not equal to 1. Hence this is not a unit vector.
3. What is the magnitude of a unit vector?
The magnitude, or length, of a unit vector, is 1 units.
4. What is unit vector notation?
The notation you use to denote a unit vector to place a circumflex, or "hat" above the lowercase letter representing the vector.
5. How do you find the unit vector?
To find a unit vector with the same direction as a given vector, simply divide the vector by its magnitude.
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• -2
Newbie
# The sum of the areas of two squares is 157m^2. If the sum of their perimeters is 68m, find the sides of the squares.
• -2
This is the Important question based on Linear Equations in two Variables Chapter of R.S Aggarwal book for
ICSE & CBSE Board.
Question Number 52 Exercise 3E of RS Aggarwal Solution.
In this Question following Points are given.
(i)The sum of the areas of two squares
(ii) the difference of their perimeters
You have to find the sides of squares.
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1. Let the sides of the two squares be x and y.
Their areas will be :
x² and y² respectively.
Now their perimeters will be :
4x and 4y respectively.
We do this since the sides of a square are equal.
From the question we can do the substitution as follows:
Sum of the areas:
x² + y² = 157………..1)
4x + 4y = 68……2)
Divide equation 2 all through by 4 to get :
x + y = 17
We need to solve for x and y.
By substitution we have :
x = 17 – y
Replace this in equation 1 as follows:
(17 – y)² + y ² = 157
289 – 34y + y² + y² = 157
Collecting the like terms together we have :
2y² – 34y + 132 = 0
Divide through by 2 to get :
y² – 17y + 66 = 0
The roots are :
-11 and -6
We expand the equation as follows:
y² – 11y – 6y + 66 = 0
y(y – 11) – 6(y – 11) = 0
(y – 6)(y – 11) = 0
y = 6 or 11
When y is 11 x is (17 – 11) = 6
So the values can either be 6 or 11 for y or vice versa for x.
The sides are thus :
11 meters and 6 meters
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# xplusa into xplus b
x plus a into x plus b Formula :
In this section, we are going to see the formula/expansion for (x + a)(x + b).
That is,
(x + a)(x + b) = x2 + xb + xa + ab
(x + a)(x + b) = x2 + (b + a)x + ab
(x + a)(x + b) = x2 + (a + b)x + ab
## x plus a into x plus b Formula - Example Problems
Problem 1 :
Find the product of :
(x + 2)(x + 3)
Solution :
(x + 2)(x + 3) is in the form of (x + a)(x + b)
Comparing (x + a)(x + b) and (x + 2)(x + 3), we get
x = x
a = 2
b = 3
Write the formula / expansion for (x + a)(x + b).
(x + a)(x + b) = x2 + (a + b)x + ab
Substitute x for x, 2 for a and 3 for b.
(x + 2)(x + 3) = x2 + (2 + 3)x + (2)(3)
(x + 2)(x + 3) = x2 + 5x + 6
So, the product of (x + 2) and (x + 3) is
x2 + 5x + 6
Problem 2 :
Find the product of :
(2p + 1)(2p + 3)
Solution :
(2p + 1)(2p + 3) is in the form of (x + a)(x + b)
Comparing (x + a)(x + b) and (2p + 1)(2p + 3), we get
x = 2p
a = 1
b = 3
Write the formula / expansion for (x + a)(x + b).
(x + a)(x + b) = x2 + (a + b)x + ab
Substitute 2p for x, 1 for a and 2 for b.
(2p + 1)(2p + 3) = (2p)2 + (1 + 3)(2p) + (1)(3)
(2p + 1)(2p + 3) = 4p2 + (4)(2p) + 3
(2p + 1)(2p + 3) = 4p2 + 8p + 3
So, the product of (2p + 1) and (2p + 3) is
4p2 + 8p + 3
## x plus a into x minus b - Formula
(x + a)(x - b) = x2 - xb + xa - ab
(x + a)(x - b) = x2 + xa - xb - ab
(x + a)(x - b) = x2 + (a - b)x - ab
## x minus a into x plus b - Formula
(x - a)(x + b) = x2 + xb - xa - ab
(x - a)(x + b) = x2 - xa + xb - ab
(x - a)(x + b) = x2 - (a - b)x - ab
## x minus a into x minus b - Formula
(x - a)(x - b) = x2 - xb - xa + ab
(x - a)(x - b) = x2 - xa - xb + ab
(x - a)(x - b) = x2 - (a + b)x + ab
After having gone through the stuff given above, we hope that the students would have understood the product of two binomials.
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# 7.2 Electric potential and potential difference (Page 2/12)
Page 2 / 12
## Calculating energy
You have a 12.0-V motorcycle battery that can move 5000 C of charge, and a 12.0-V car battery that can move 60,000 C of charge. How much energy does each deliver? (Assume that the numerical value of each charge is accurate to three significant figures.)
## Strategy
To say we have a 12.0-V battery means that its terminals have a 12.0-V potential difference. When such a battery moves charge, it puts the charge through a potential difference of 12.0 V, and the charge is given a change in potential energy equal to $\text{Δ}U=q\text{Δ}V.$ To find the energy output, we multiply the charge moved by the potential difference.
## Solution
For the motorcycle battery, $q=5000\phantom{\rule{0.2em}{0ex}}\text{C}$ and $\text{Δ}V=12.0\phantom{\rule{0.2em}{0ex}}\text{V}$ . The total energy delivered by the motorcycle battery is
$\text{Δ}{U}_{\text{cycle}}=\left(5000\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(12.0\phantom{\rule{0.2em}{0ex}}\text{V}\right)=\left(5000\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(12.0\phantom{\rule{0.2em}{0ex}}\text{J/C}\right)=\phantom{\rule{0.2em}{0ex}}6.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}$
Similarly, for the car battery, $q=60,000\phantom{\rule{0.2em}{0ex}}\text{C}$ and
$\text{Δ}{U}_{\text{car}}=\left(60,000\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(12.0\phantom{\rule{0.2em}{0ex}}\text{V}\right)=7.20\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{J}\text{.}$
## Significance
Voltage and energy are related, but they are not the same thing. The voltages of the batteries are identical, but the energy supplied by each is quite different. A car battery has a much larger engine to start than a motorcycle. Note also that as a battery is discharged, some of its energy is used internally and its terminal voltage drops, such as when headlights dim because of a depleted car battery. The energy supplied by the battery is still calculated as in this example, but not all of the energy is available for external use.
Check Your Understanding How much energy does a 1.5-V AAA battery have that can move 100 C?
$\text{Δ}U=q\text{Δ}V=\left(100\phantom{\rule{0.2em}{0ex}}\text{C}\right)\left(1.5\phantom{\rule{0.2em}{0ex}}\text{V}\right)=150\phantom{\rule{0.2em}{0ex}}\text{J}$
Note that the energies calculated in the previous example are absolute values. The change in potential energy for the battery is negative, since it loses energy. These batteries, like many electrical systems, actually move negative charge—electrons in particular. The batteries repel electrons from their negative terminals ( A ) through whatever circuitry is involved and attract them to their positive terminals ( B ), as shown in [link] . The change in potential is $\text{Δ}V={V}_{B}-{V}_{A}=+12\phantom{\rule{0.2em}{0ex}}\text{V}$ and the charge q is negative, so that $\text{Δ}U=q\text{Δ}V$ is negative, meaning the potential energy of the battery has decreased when q has moved from A to B .
## How many electrons move through a headlight each second?
When a 12.0-V car battery powers a single 30.0-W headlight, how many electrons pass through it each second?
## Strategy
To find the number of electrons, we must first find the charge that moves in 1.00 s. The charge moved is related to voltage and energy through the equations $\text{Δ}U=q\text{Δ}V.$ A 30.0-W lamp uses 30.0 joules per second. Since the battery loses energy, we have $\text{Δ}U=-30\phantom{\rule{0.2em}{0ex}}\text{J}$ and, since the electrons are going from the negative terminal to the positive, we see that $\text{Δ}V=\text{+12.0}\phantom{\rule{0.2em}{0ex}}\text{V}\text{.}$
## Solution
To find the charge q moved, we solve the equation $\text{Δ}U=q\text{Δ}V:$
$q=\frac{\text{Δ}U}{\text{Δ}V}.$
Entering the values for $\text{Δ}U$ and $\text{Δ}V$ , we get
$q=\frac{-30.0\phantom{\rule{0.2em}{0ex}}\text{J}}{+12.0\phantom{\rule{0.2em}{0ex}}\text{V}}=\frac{-30.0\phantom{\rule{0.2em}{0ex}}\text{J}}{+12.0\phantom{\rule{0.2em}{0ex}}\text{J/C}}=-2.50\phantom{\rule{0.2em}{0ex}}\text{C}\text{.}$
The number of electrons ${n}_{e}$ is the total charge divided by the charge per electron. That is,
${n}_{e}=\frac{-2.50\phantom{\rule{0.2em}{0ex}}\text{C}}{-1.60\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{\phantom{\rule{0.2em}{0ex}}10}^{-19}\phantom{\rule{0.2em}{0ex}}{\text{C/e}}^{-}}=1.56\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{\phantom{\rule{0.2em}{0ex}}10}^{19}\phantom{\rule{0.2em}{0ex}}\text{electrons}\text{.}$
## Significance
This is a very large number. It is no wonder that we do not ordinarily observe individual electrons with so many being present in ordinary systems. In fact, electricity had been in use for many decades before it was determined that the moving charges in many circumstances were negative. Positive charge moving in the opposite direction of negative charge often produces identical effects; this makes it difficult to determine which is moving or whether both are moving.
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Home >> Equality >> Multiply with different numbers >>
## Multiply both sides of equality with different numbers
Add same number Add different number Subtract same number Subtract different number Multiply with same number Multiply with different numbers Divide by same number Divide by different number
Explanation: When the sides of equation i.e. L.H.S. and R.H.S of the equation are multiplied with different numbers, the equality fails to holds.
Let’s understand it with the help of following examples:
Example 1 - Multiply L.H.S. by 2 and R.H.S. by 3 of given equation and check what happens to equality
2 X 10 = 5 X 4
Solution - This proceeds as :
Multiply L.H.S. by 2 and R.H.S. by 3 of given equation and we get;
2 X 10 X 2 = 5 X 4 X 3
Solve L.H.S. and we get;
L.H.S. = 2 X 10 X 2
L.H.S. = 40
Solve R.H.S. and we get
R.H.S. = 5 X 4 X 3
R.H.S. = 60
Since L.H.S. in not equals to R.H.S i.e. 40 is not equal to 60
So the given equation 2 X 10 = 5 X 4 fails to hold equality, when we multiply L.H.S. by 2 and R.H.S. by 3 of given equation and hence we get that
“When the sides of equation i.e. L.H.S. and R.H.S of the equation are multiplied with different numbers, the equality fails to holds.”
Example 2 - Multiply L.H.S. by 5 and R.H.S. by 7 of given equation and check what happens to equality
5 + 2 = 6 + 1
Solution - This proceeds as:
Multiply L.H.S. by 5 and R.H.S. by 7 of given equation and we get;
(5 + 2) X 5 = (6 + 1) X 7
Solve L.H.S. and we get;
L.H.S. = (5 + 2) X 5
L.H.S. = 35
Solve R.H.S. and we get
R.H.S. = (6 + 1) X 7
R.H.S. = 49
Since L.H.S. in not equals to R.H.S i.e. 35 is not equal to 49
So the given equation 5 + 2 = 6 + 1 fails to hold equality, when we multiply L.H.S. by 5 and R.H.S. by 7 of given equation and hence we get that
“When the sides of equation i.e. L.H.S. and R.H.S of the equation are multiplied with different numbers, the equality fails to holds.”
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# 2.3: Subtraction of Rational Numbers
Difficulty Level: At Grade Created by: CK-12
In the previous two lessons, you have learned how to find the opposite of a rational number and to add rational numbers. You can use these two concepts to subtract rational numbers. Suppose you want to find the difference of 9 and 12. Symbolically, it would be 912\begin{align*}9 - 12\end{align*}. Begin by placing a dot at nine and move to the left 12 units.
912=3\begin{align*}9 - 12 = -3\end{align*}
Rule: To subtract a number, add its opposite.
35=3+(5)=2916=9+(16)=7\begin{align*}3 - 5 = 3 + (-5) = -2 && 9 - 16 = 9 + (-16) = -7\end{align*}
A special case of this rule can be written when trying to subtract a negative number.
The Opposite-Opposite Property: For any real numbers a\begin{align*}a\end{align*} and b, a(b)=a+b\begin{align*}b, \ a-(-b) = a + b\end{align*}.
Example 1: Simplify 6(13).\begin{align*}-6 - (-13).\end{align*}
Solution: Using the Opposite-Opposite Property, the double negative is rewritten as a positive.
6(13)=6+13=7\begin{align*}-6 - (-13) = -6 + 13 = 7\end{align*}
Example 2: Simplify 56(118).\begin{align*}\frac{5}{6} - \left ( - \frac{1}{18} \right ).\end{align*}
Solution: Begin by using the Opposite-Opposite Property.
56+118\begin{align*}\frac{5}{6} + \frac{1}{18}\end{align*}
Next, create a common denominator: 5×36×3+118=1518+118.\begin{align*}\frac{5 \times 3}{6 \times 3} + \frac{1}{18} = \frac{15}{18} + \frac{1}{18}.\end{align*}
Add the fractions: 1618.\begin{align*}\frac{16}{18}.\end{align*}
Reduce: 2×2×2×23×3×2=89.\begin{align*}\frac{2 \times 2 \times 2 \times 2}{3 \times 3 \times 2} = \frac{8}{9}.\end{align*}
## Evaluating Change Using a Variable Expression
You have learned how to graph a function by using an algebraic expression to generate a table of values. Using the table of values you can find the change in the dependent values between any two independent values.
In Lesson 1.5, you wrote an expression to represent the pattern of the total cost to the number of CDs purchased. The table is repeated below:
Number of CDsCost ()22444867289610120\begin{align*}&\text{Number of CDs} && 2 && 4 && 6 && 8 && 10\\ &\text{Cost (\)} && 24 && 48 && 72 && 96 && 120\end{align*} To determine the change, you must find the difference between the dependent values and divide it by the difference in the independent values. Example 2: What is the cost of a CD? Solution: We begin by finding the difference between the cost of two values. For example, the change in cost between 4 CDs and 8 CDs. 9648=48\begin{align*}96-48 = 48\end{align*} Next, we find the difference between the number of CDs. Finally, we divide.84=4484=12\begin{align*}&&&8 - 4 = 4\\ &\text{Finally, we divide.}&& \quad \frac{48}{4}=12\end{align*} ## Practice Set Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Subtraction of Rational Numbers (10:22) In 1 – 20, subtract the following rational numbers. Be sure that your answer is in the simplest form. 1. 914\begin{align*}9 - 14\end{align*} 2. 27\begin{align*}2 - 7\end{align*} 3. 218\begin{align*}21 - 8\end{align*} 4. 8(14)\begin{align*}8 - (-14)\end{align*} 5. 11(50)\begin{align*}-11 - (-50)\end{align*} 6. 512918\begin{align*}\frac{5}{12} - \frac{9}{18}\end{align*} 7. 5.41.01\begin{align*}5.4 - 1.01\end{align*} 8. 2314\begin{align*}\frac{2}{3} - \frac{1}{4}\end{align*} 9. 3413\begin{align*}\frac{3}{4} - \frac{1}{3}\end{align*} 10. 14(23)\begin{align*}\frac{1}{4} - \left (- \frac{2}{3} \right )\end{align*} 11. 151197\begin{align*}\frac{15}{11} - \frac{9}{7}\end{align*} 12. 213111\begin{align*}\frac{2}{13} - \frac{1}{11}\end{align*} 13. 78(83)\begin{align*}-\frac{7}{8} - \left (- \frac{8}{3} \right )\end{align*} 14. 727939\begin{align*}\frac{7}{27} - \frac{9}{39}\end{align*} 15. 611322\begin{align*}\frac{6}{11} - \frac{3}{22}\end{align*} 16. 3.121.49\begin{align*}-3.1 - 21.49\end{align*} 17. 1364740\begin{align*}\frac{13}{64} - \frac{7}{40}\end{align*} 18. 11701130\begin{align*}\frac{11}{70} - \frac{11}{30}\end{align*} 19. 68(22)\begin{align*}-68 - (-22)\end{align*} 20. 1312\begin{align*}\frac{1}{3} - \frac{1}{2}\end{align*} 21. Determine the change in y\begin{align*}y\end{align*} between (1, 9) and (5, –14). 22. Consider the equation y=3x+2\begin{align*}y = 3x + 2\end{align*}. Determine the change in y\begin{align*}y\end{align*} between x=3\begin{align*}x = 3\end{align*} and x=7.\begin{align*}x = 7.\end{align*} 23. Consider the equation y=23x+12\begin{align*}y = \frac{2}{3}x + \frac{1}{2}\end{align*}. Determine the change in y\begin{align*}y\end{align*} between x=1\begin{align*}x = 1\end{align*} and x=2\begin{align*}x = 2\end{align*}. 24. True or false? If the statement is false, explain your reasoning. The difference of two numbers is less than each number. 25. True or false? If the statement is false, explain your reasoning. A number minus its opposite is twice the number. 26. KMN stock began the day with a price of4.83 per share. At the closing bell, the price dropped 0.97 per share. What was the closing price of KMN stock? In 27 – 32, evaluate the expression. Assume a=2, b=3,\begin{align*}a=2, \ b= -3,\end{align*} and c=1.5.\begin{align*}c = -1.5.\end{align*} 1. (ab)+c\begin{align*}(a-b)+c\end{align*} 2. |b+c|a\begin{align*}|b+c|- a\end{align*} 3. a(b+c)\begin{align*}a-(b+c)\end{align*} 4. |b|+|c|+a\begin{align*}|b|+ |c|+ a\end{align*} 5. \begin{align*}7b+4a\end{align*} 6. \begin{align*}(c-a)- b\end{align*} Mixed Review 1. Graph the following ordered pairs: \begin{align*}\left \{(0,0),(4,4),(7,1),(3,8) \right \}\end{align*}. Is the relation a function? 2. Evaluate the expression when \begin{align*}m= \left (- \frac{2}{3} \right ): \ \frac{2^3+m}{4}\end{align*}. 3. Translate the following into an algebraic equation: Ricky has twelve more dollars than Stacy. Stacy has 5 less dollars than Aaron. The total of the friends’ money is62.
4. Simplify \begin{align*}\frac{1}{3} + \frac{7}{5}\end{align*}.
5. Simplify \begin{align*}\frac{21}{4} - \frac{2}{3}\end{align*}.
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## What is the application of quadratic equation in real life?
Answer: In daily life we use quadratic formula as for calculating areas, determining a product’s profit or formulating the speed of an object. In addition, quadratic equations refer to an equation that has at least one squared variable.
## What are the applications of quadratic functions?
Throwing a ball, shooting a cannon, diving from a platform and hitting a golf ball are all examples of situations that can be modeled by quadratic functions. In many of these situations you will want to know the highest or lowest point of the parabola, which is known as the vertex.
## What are the 5 examples of quadratic equation?
Examples of Quadratic Equation6x² + 11x – 35 = 0.2x² – 4x – 2 = 0.-4x² – 7x +12 = 0.20x² -15x – 10 = 0.x² -x – 3 = 0.5x² – 2x – 9 = 0.3x² + 4x + 2 = 0.-x² +6x + 18 = 0.
## Why are quadratic equations important?
So why are quadratic functions important? Quadratic functions hold a unique position in the school curriculum. They are functions whose values can be easily calculated from input values, so they are a slight advance on linear functions and provide a significant move away from attachment to straight lines.
## Why do quadratic equations equal zero?
We use the zero product property when we solve quadratic equations. You may have noticed that we always manipulate quadratic equations to ax2+bx+c=0. This is because factoring the equation gives us two expressions that multiply to zero. We can set each factor equal to zero and solve for x.
## Who invented the quadratic equation?
580 BC Pythagoras hates irrational numbers. 400 BC Babylonians solved quadratic equations. 300 BC Euclid developed a geometrical approach and proved that irrational numbers exist. 598-665AD Brahmagupta took the Babylonian method that allowed the use of negative numbers.
## How do quadratic functions work?
A quadratic function is one of the form f(x) = ax2 + bx + c, where a, b, and c are numbers with a not equal to zero. The graph of a quadratic function is a curve called a parabola. Parabolas may open upward or downward and vary in “width” or “steepness”, but they all have the same basic “U” shape.
## What is another name for the standard form of a quadratic function?
Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function. As with the general form, if a>0, the parabola opens upward and the vertex is a minimum. If a<0, the parabola opens downward, and the vertex is a maximum.
## What are the three types of quadratic equations?
Here are the three forms a quadratic equation should be written in:1) Standard form: y = ax2 + bx + c where the a,b, and c are just numbers.2) Factored form: y = (ax + c)(bx + d) again the a,b,c, and d are just numbers.3) Vertex form: y = a(x + b)2 + c again the a, b, and c are just numbers.
You might be interested: Compliance equation
## Can a quadratic always be factored?
If that is a perfect square, then the equation can be factored nicely. If not, then at least you are halfway toward finding the roots using the quadratic formula. You can only factorise easily (without involving surds) if the discriminant is a perfect square.
## How do you simplify quadratic equations?
Completing the squarePut the equation into the form ax 2 + bx = – c.Make sure that a = 1 (if a ≠ 1, multiply through the equation by. before proceeding).Using the value of b from this new equation, add. Find the square root of both sides of the equation.Solve the resulting equation.
## How do you explain quadratic equations?
SummaryQuadratic Equation in Standard Form: ax2 + bx + c = 0.Quadratic Equations can be factored.Quadratic Formula: x = −b ± √(b2 − 4ac) 2a.When the Discriminant (b2−4ac) is: positive, there are 2 real solutions. zero, there is one real solution. negative, there are 2 complex solutions.
## Why is it called a quadratic equation?
We use the word quadratic because “quadra” refers to a square, and the leading term in a quadratic equation is “squared.” This is consistent with calling a degree three polynomial a “cubic” for the leading term represents a cube. The word for an equation with a leading term of x^4 is “quartic.”
### Releated
#### Equation of vertical line
How do you write an equation for a vertical and horizontal line? Horizontal lines go left and right and are in the form of y = b where b represents the y intercept. Vertical lines go up and down and are in the form of x = a where a represents the shared x coordinate […]
#### Bernoulli’s equation example
What does Bernoulli’s equation State? Bernoulli’s principle states the following, Bernoulli’s principle: Within a horizontal flow of fluid, points of higher fluid speed will have less pressure than points of slower fluid speed. Why is Bernoulli’s equation used? The Bernoulli equation is an important expression relating pressure, height and velocity of a fluid at one […]
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# Centroid: Simple Definition, Examples, Finding
Contents:
## Definition & Basic Geometry
The centroid is the average of all points in an object. It is sometimes called the center of area or the center of volume.
## Uniform Density
If a shape has uniform density, the centroid is the same as the center of mass. If this shape is also in a uniform gravitational field, this is also the same as the center of gravity— the point where the whole weight of a body interacts. Exactly where the center is depends on the shape of the body; Surprisingly, the center of gravity doesn’t necessarily have to be inside the body. It might be a point outside, somewhere in space.
## Finding the Centroid of Plane Figures
There’s a hands-on way to find the centroid of a plane figure. Draw it on a piece of card, cut it out, and then find the place where it balances perfectly on a pin or on the tip of a pencil. In theory, this should be simple. But in practice, it takes a little skill to balance the shape. Plus, merely balancing the shape won’t give you information like where the coordinates of the shape are. Some centroids, like circles, rectangles and triangles, are even easier to find:
## 1. Circle
To find the center of the circle: fold the paper in half one way, then another:
## 2. Rectangle
To find the center of the rectangle, fold the paper (diagonally) in half from corner to corner:
## 3. Triangle
Find the place where the triangle’s three medians interact. To find any one median of a triangle, draw a line segment from one vertex (corner) to another point half way on the opposite side.
## Finding the Centroid of Two Dimensional Shapes Using Calculus
Two methods can be used to find the centroid’s location:
## Summation
Let’s say we have a more complicated shape that we can turn into a compound shape (i.e. divide it into smaller primary shapes). If we know how to find the centroids for each of the individual shapes, we can find the compound shape’s centroid using the formula:
Where:
• xi is the distance from the axis to the centroid of the simple shape,
• Ai is the area of the simple shape.
• Σ is summation notation, which basically means to “add them all up.”
The same formula, with yi substituting for xi, gives us the y coordinate of the centroid.
## Integration.
An integral can be used to find the centroid of shape too complicated to be broken down into its primary parts. Integrating is working with infinitesimally small areas; Finding the centroid of parts tell us what the centroid of the whole will be.
If a shape or region is bounded by two functions, f(x) and g(x), the coordinates of the centroid is given by:
where A is the area calculated by the integral
.
## References
Gere, J. & Goodno, B. (2008). Mechanics of Materials. Cengage Learning.
Pytel, A. & Kiusalaas, J. (2009). Engineering Mechanics: Statics. Cengage Learning.
Rajput, R. (2015). A Textbook of Applied Mechanics. Laxmi Publications.
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```7.2 Polynomial Functions and Their Graphs
Objectives:
•Identify and describe the important features of the
graph of a polynomial function
•Use a polynomial function to model real-world data
Exploring End Behavior of f(x) = axn
Graph each function separately. For each function,
1) y = x2
2) y = x4
3) y = 2x2
4) y = 2x4
5) y = x3
6) y = x5
7) y = 2x3
8) y = 2x5
9) y = -x2
10) y = -x4
11) y = -2x2
12) y = -2x4
13) y = -x3
14) y = -x5
15) y = -2x3
16) y = -2x5
a. Is the degree of the function even or odd?
b. Is the leading coefficient positive or negative?
c. Does the graph rise or fall on the left? on the right?
Exploring End Behavior of f(x) = axn
Rise or Fall???
a > 0
left
a < 0
right
left
right
n is even
rise
rise
fall
fall
n is odd
fall
rise
rise
fall
Example 1
Describe the end behavior of each function.
a) V(x) = x3 – 2x2 – 5x + 3
falls on the left and rises on the right
b) R(x) = 1 + x – x2 – x3 + 2x4
rises on the left and the right
Graphs of Polynomial Functions
f(a) is a local maximum if there is an interval around
a such that f(a) > f(x) for all values of x in the
interval, where x = a.
f(a) is a local minimum if there is an interval around
a such that f(a) < f(x) for all values of x in the
interval, where x = a.
Increasing and Decreasing Functions
Let x1 and x2 be numbers in the domain of a function, f.
The function f is increasing over an open interval if for
every x1 < x2 in the interval, f(x1) < f(x2).
The function f is decreasing over an open interval if for
every x1 < x2 in the interval, f(x1) > f(x2).
Example 2
Graph P(x) = -2x3 – x2 + 5x + 6.
a) Approximate any local maxima or minima to the
nearest tenth.
minimum: (-1.1,2.0)
maximum: (0.8,8.3)
b) Find the intervals over which the function is increasing
and decreasing.
increasing: x > -1.1 and x < 0.8
decreasing: x < -1.1 or x > 0.8
Example 3
The table below gives the number of students who
participated in the ACT program during selected years
from 1970 to 1995. The variable x represents the
number of years since 1960, and y represents the
number of participants in thousands.
x
y
a) Find a quartic regression model for the
10
714
number of students who participated in
15
822
the ACT program during the given years
20 836
A(x) = -0.004x4 + 0.44x3 – 17.96x2 + 293.83x - 836
25 739
b) Use the regression model to estimate
30
817
the number of students who participated
in the ACT program in 1985.
35
estimate using model is about 767,000
945
Homework
Lesson 7.2 exercises 29-36
```
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# Difference between revisions of "2015 AMC 8 Problems/Problem 19"
A triangle with vertices as $A=(1,3)$, $B=(5,1)$, and $C=(4,4)$ is plotted on a $6\times5$ grid. What fraction of the grid is covered by the triangle?
$\textbf{(A) }\frac{1}{6} \qquad \textbf{(B) }\frac{1}{5} \qquad \textbf{(C) }\frac{1}{4} \qquad \textbf{(D) }\frac{1}{3} \qquad \textbf{(E) }\frac{1}{2}$
$[asy] draw((1,0)--(1,5),linewidth(.5)); draw((2,0)--(2,5),linewidth(.5)); draw((3,0)--(3,5),linewidth(.5)); draw((4,0)--(4,5),linewidth(.5)); draw((5,0)--(5,5),linewidth(.5)); draw((6,0)--(6,5),linewidth(.5)); draw((0,1)--(6,1),linewidth(.5)); draw((0,2)--(6,2),linewidth(.5)); draw((0,3)--(6,3),linewidth(.5)); draw((0,4)--(6,4),linewidth(.5)); draw((0,5)--(6,5),linewidth(.5)); draw((0,0)--(0,6),EndArrow); draw((0,0)--(7,0),EndArrow); draw((1,3)--(4,4)--(5,1)--cycle); label("y",(0,6),W); label("x",(7,0),S); label("A",(1,3),dir(210)); label("B",(5,1),SE); label("C",(4,4),dir(100)); [/asy]$
## Solution 1
The area of $\triangle ABC$ is equal to half the product of its base and height. By the Pythagorean Theorem, we find its height is $\sqrt{1^2+2^2}=\sqrt{5}$, and its base is $\sqrt{2^2+4^2}=\sqrt{20}$. We multiply these and divide by 2 to find the of the triangle is $\frac{\sqrt{5 \cdot 20}}2=\frac{\sqrt{100}}2=\frac{10}2=5$. Since the grid has an area of $30$, the fraction of the grid covered by the triangle is $\frac 5{30}=\boxed{\textbf{(A) }\frac{1}{6}}$.
### Solution 2
Note angle $\angle ACB$ is right this the area is $\sqrt{1^2+3^2} \times \sqrt{1^2+3^2}\times \dfrac{1}{2}=(10)\times \dfrac{1}{2}=5$ thus the fraction of the total is $\dfrac{5}{30}=\boxed{\textbf{(A)}~\dfrac{1}{6}}$
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Complement and Set Difference
Readings for Session 5 – (Continued)
Complement and Set Difference
Remember that we often work with a specific set of objects when solving problems or discussing issues. We called this set of objects a universal set or universe. For example, in the lead-in problem above, the universal set could be either the set of all U. S. dollars or the set of the \$836 Sam originally had in the checking account.
Complement of a Set: The complement of a set, denoted A', is the set of all elements in the given universal set U that are not in A.
In set- builder notation, A' = {x U : x A}.
The Venn diagram for the complement of set A is shown below where the shaded region represents A'.
Example: For the lead-in example on the previous page, let the universal set U be the \$836 Sam originally has in the checking account and let A be the set of the \$429 of the check. The complement of set A would be the set of the \$407 remaining in the checking account.
Example: Let U = {1, 2, 3, 4, 5, 6} and A = {1, 3, 5}. Then A' = {2, 4, 6}.
Example: U' = The complement of the universe is the empty set.
Example: ' = U The complement of an empty set is the universal set.
Set Difference: The relative complement or set difference of sets A and B, denoted AB, is the set of all elements in A that are not in B.
In set-builder notation, AB = {x U : x A and x B}= A B'.
The Venn diagram for the set difference of sets A and B is shown below where the shaded region represents AB.
Example: For the lead-in example on the previous page, let the universal set U be the set of all U.S. dollars, let set A be the set of \$836 Sam originally has in the checking account, and let B be the set of the \$429 of the check. Then the set difference of A and B would be the \$407 remaining in the checking account.
Example: Let A = {a, b, c, d} and B = {b, d, e}. Then A B = {a, c} and BA = {e}.
Example: Let G = {t, a, n} and H = {n, a, t}. Then G H = .
How should we define the subtraction of whole numbers?
In the lead-in example on an earlier page of this section, the remaining balance was the difference between the cardinalities of the sets for the checking account and the check. This also works for the third example (above)where
n(G) – n(H) = 3 – 3 = 0 = n().
But, with the second example (above) the difference between the cardinalities does not give the expected result, e.g.,
n(A) – n(B) = 4 – 3 = 1 ≠ 2 = n(AB).
In this case, B is not a subset of A. This leads to the set definition for subtraction of whole numbers given on the next page.
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