Chapter 3 Numeric And Algebraic Expressions
Midpoint
Page 137 Exercise 1 Answer
For the power 43, the base is 4 and the exponent is 3. The exponent means the base will be a factor 3 times when the power is rewritten as repeated multiplication. That is, 43 = 4 × 4 × 4.
Page 137 Exercise 2 Answer
The statement “The GCF of 12 and 28 is 4.” is true.
12 = 2 × 6 = 2 × 2 × 3
28 = 2 × 14 = 2 × 2 × 7
Common factors of 12 and 28 are 2 and 2, so their product, which equals 4, is the GCF of 12 and 28.
Result
True.
Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions
Page 137 Exercise 3 Answer
The statement “The GCF of 18 and 84 is 9.” is false.
18 = 2 × 9 = 2 × 3 × 3
84 = 2 × 42 = 2 × 2 × 21 = 2 × 2 × 3 × 7
Common factors of 18 and 84 are 2 and 3, so their product, which equals 6, is the GCF of 18 and 84.
Result
False.
Page 137 Exercise 4 Answer
The statement “The LCM of 12 and 6 is 6.” is false.
Since 12 is greater than 6, 6 is not the LCM of 6 and 12.
Result
False.
Page 137 Exercise 5 Answer
The statement “The LCM of 4 and 8 is 8.” is true.
Since 8 can be written as a product of 4, 8 is a multiple of 4. Also 8 is the first multiple 8. Thus, the LCM of 4 and 8 is 8.
Result
True.
Page 137 Exercise 6 Answer
0.52 x (20 – 22 x 3) x (\(\frac{2}{5}\) x 25) (Evaluate the power.)
= 0.52 x (20 – 4 x 3) x (\(\frac{2}{5}\) x 25) (Find the product.)
= 0.52 x (20 – 12) x (\(\frac{2}{5}\) x 25) (Find the product.)
= 0.52 x (20 – 12) x (10) (Find the difference.)
= 0.52 ×(8) × (10) (Evaluate the power.)
= 0.25 × (8) × (10) (Find the product.)
= 2 × (10) (Find the product.)
= 20
2[(14 – 3) + 43 ÷ 4] – 2.5 x 12 (Find the difference in the parenthesis.)
= 2[(11) + 43 ÷ 4] – 2.5 x 12 (Evaluate the power.)
= 2[(11) + 64 ÷ 4] – 2.5 x 12 (Find the quotient in the brackets.)
= 2[(11) + 16] – 2.5 x 12 (Find the sum in the brackets.)
= 2[27] – 2.5 x 12 (Find the product.)
= 54 – 2.5 x 12 (Find the product.)
= 54 – 30 (Find the difference.)
= 24
9 + 12 ÷ \(\frac{2}{3}\) – 33 ÷ 3 (Evaluate the power.)
= 9 + 12 ÷ \(\frac{2}{3}\) (Find the quotient.)
= 9 + 18 − 27 ÷ 3 (Find the quotient.)
= 9 + 18 – 9 (Find the sum.)
= 27 – 9 (Find the difference.)
= 18
Result
0.52 x (20 – 22 x 3) x (\(\frac{2}{5}\) x 35) = 20
2[(14−3) + 43 ÷ 4]− 2.5 × 12 = 24
9 + 12 ÷ \(\frac{2}{3}\) ÷ 3 = 18
Page 137 Exercise 7 Answer
First, square the given fraction.
\(\left(\frac{2}{3}\right)^2=\left(\frac{2^2}{3^2}\right)=\left(\frac{4}{9}\right)\)The given fraction \(\left(\frac{2}{3}\right)^2\) is equal to the first option \(\frac{4}{9}\).
The second option the fractions have the same numerators, but the denominators are different, thus the fractions are not equal.
\(\frac{4}{3} \neq \frac{4}{9}\)To compare the third option – a product of two fractions, with a given fraction, find the product and than compare.
\(\frac{1}{3} \times \frac{1}{3}=\frac{1 \times 1}{3 \times 3}=\frac{1}{9} \neq \frac{4}{9}\)The fractions have the same denominators, but the numerators are different, thus the fractions are not equal.
To compare the fourth option – a product of a fraction and a whole number, with a given fraction, find the product and than compare.
\(\frac{1}{9}\) x 4 (Rewrite the whole number as a fraction)
= \(\frac{1}{9} \times \frac{4}{1}\)
= \(\frac{1 \times 4}{9 \times 1}\)
= \(\frac{4}{9}\)
The given fraction \(\left(\frac{2}{3}\right)^2\) is equal to the fourth option \(\frac{1}{9}\) x 4.
To compare the fifth option – a product of two fractions, with a given fraction, find the product and than compare.
\(\frac{2}{3} \times \frac{2}{3}=\frac{2 \times 2}{3 \times 3}=\frac{4}{9}\)The given fraction \(\left(\frac{2}{3}\right)^2\) is equal to the fifth option \(\frac{2}{3}\) x \(\frac{2}{3}\).
Result
Expressions \(\frac{2}{3} \times \frac{2}{3}, \frac{1}{9} \times 4, \frac{4}{9}\) are equal to \(\left(\frac{2}{3}\right)^2\).
Page 137 Exercise 8 Answer
Liam bought 2 vintage movie posters which cost $28.50, so the cost of it is the product 2 × 28.50. He bought 2 rock posters which cost $29.75, and 1 rap poster which costs $19.50. The total cost, without the gift card and the \(\frac{1}{2}\) – off coupon, is the sum of the products mentioned before.
Since he applied a $35 gift card to the total purchase, so from the sum mention before we subtract 35. A \(\frac{1}{2}\) – off coupon is applied only to the rap poster so we multiply 1 × 19.50 by \(\frac{1}{2}\).
The numerical expression which then describes the total cost for the posters is:
= [2 × 28.50 + 2 × 29.75 + \(\frac{1}{2}\) (19.50)] − 35
= [2 × 28.50 + 2 × 29.75 + 9.75] − 35
= [57 + 2 × 29.75 + 9.75] − 35
= [57 + 59.5 + 9.75] − 35
= [116.5 + 9.75] − 35
= [126.25] − 35
= 91.25
Result
Liam paid $91.25 for the posters.
Page 137 Exercise 9 Answer
Since Eva volunteers every 3 days and Jin volunteers every 5 days, they will volunteer together on the days which are multiples of both 3 and 5. To find this numbers, first find the least common multiple of 3 and 5. Multiples of the LCM of 3 and 5 are then mulitples of both 3 and 5.
Both 3 and 5 are prime so the LCM is simply their product.
3 x 5 = 15
Since they worked the first day of the month, they will volunteer together on the 15-th day and on the 30-th day.
Result
They will volunteer together on the same day again in 15 days.
Page 138 Exercise 1a Answer
To find the number of gift bags Raoul needs to make, find the greatest common factor of of 72 and 96.
72 = 2 × 36 = 2 × 2 × 18 = 2 × 2 × 2 × 9 = 2 × 2 × 2 × 3 × 3
96 = 2 × 48 = 2 × 2 × 24 = 2 × 2 × 2 × 12 = 2 × 2 × 2 × 2 × 6
= 2 × 2 × 2 × 2 × 2 × 3
The greatest common factor of 72 and 96 is the product of common factors in the prime factorization of 72 and 96.
2 × 2 × 2 × 3 = 24
The GCF of 72 and 96 is 24.
Raoul needs to make 24 gift bags.
To find how many wristbands and how many movie posters go in each gift bag if evenly distributed write 72 and 96 as a product of two numbers whose one factor is 24.
72 = 24 × 3
96 = 24 × 4
In each bag Raoul needs to put 3 wristbands and 4 movie posters.
Result
Raoul needs to make 24 gift bags and in each bag he needs to put 3 wristbands and 4 movie posters.
Page 138 Exercise 1b Answer
To answer the question find the least common multiple of 12 and 10.
2 = 2 × 6 = 2 × 2 × 3
10 = 2 × 5
The LCM of 12 and 10 is the product of the union of factors of 12 and 10.
2 × 2 × 3 × 5 = 60
The LCM of 12 and 10 is 60.
Rewrite 60 as a product of two numbers, one being 12.
60 = 12 × 5
Rewrite 60 as a product of two numbers, one being 10.
60 = 10 × 6
To have an equal number of cups and napkins Monique needs to buy 5 packages of cups and 6 packages of napkins.
Result
Monique needs to buy 5 packages of cups and 6 packages of napkins.
Page 138 Exercise 1c Answer
Monique will have 60 cups and 60 napkins in Part B.
Find the value of each expression to check if they are equal to 60.
21 x 30 (A number raised to the power of one is that number.)
= 2 x 30 (Find the product.)
= 60
102 x 60 (Evaluate the power.)
= 100 x 60 (Find the product.)
= 6000
≠ 60
22 x 15 (Evaluate the power.)
= 4 x 15 (Find the product.)
= 60
4600 (Any number raised to the power of 0 is equal to 1.)
= 1
≠ 60
40 x 60 (Any number raised to the power of 0 is equal to 1.)
= 1 x 60 (Find the product.)
= 60
Result
40 × 60, 22 × 15, and 21 × 30.
Page 138 Exercise 1d Answer
In Part B we calculated that the teachers need to buy 5 packages of cups and 6 packages of napkins. The total cost of cups and napkins is the sum of products 5 × $3.50 and 6 × $4.25.
To answer the question
find the difference between the total cost of supplies and the money teachers have.
(5 x 3.50 + 6 x 4.25) – 25 (Find the products.)
= (17.50 + 25.50) – 25 (Find the sum.)
= 43 – 25 (Find the difference.)
= 18
Result
The teachers need $18 more.