Chapter 1 Algebra: Understand Numerical And Algebraic Expressions
Homework And Practice 5
Page 35 Exercise 1 Answer
6a + 4 Evaluate
= \(6\left(\frac{1}{3}\right)+4\) Substitute a = \(\frac{1}{3}\)
= 2 + 4 Multiply
= 6 Add
Result
6
Page 35 Exercise 2 Answer
\(5 a-\frac{2}{3}\) Evaluate
= \(5\left(\frac{1}{3}\right)-\frac{2}{3}\) Substitute a = \(\frac{1}{3}\)
= \(\frac{5}{3}\) – \(\frac{2}{3}\) Multiply
= \(\frac{5-2}{3}\) Write over a common denominator
= \(\frac{3}{3}\) Simplify
= 1 Divide
Result
1
Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions
Page 35 Exercise 3 Answer
5d ÷ c + 2 Evaluate
= 5(10) ÷ (5) + 2 Substitute c = 5 and d = 10
= 50 ÷ 5 + 2 Evaluate the parentheses
= 10 + 2 Divide
= 12 Add
Result
12
Page 35 Exercise 4 Answer
\(b^2\) – 9a Evaluate
= \((9)^2-9\left(\frac{1}{3}\right)\) Substitute b = 9 and a = \(\frac{1}{3}\)
= 81 − 3 Evaluate the parentheses
= 78 Subtract
Result
78
Page 35 Exercise 5 Answer
12a + c − b Evaluate
= \(12\left(\frac{1}{3}\right)+(5)-(9)\) Substitute a = \(\frac{1}{3}\), c = 5 and b = 9
= 4 + 5 − 9 Evaluate the parentheses
= 9 − 9 Add
= 0 Subtract
Result
0
Page 35 Exercise 6 Answer
\(\frac{1}{2} d+c^2-b\) Evaluate
= \(\frac{1}{2}(10)+(5)^2-(9)\) Substitute d = 10, c = 5 and b = 9
= 5 + 25 − 9 Evaluate the parentheses
= 30 − 9 Add
= 21 Subtract
Result
21
Page 35 Exercise 7 Answer
\(d^2\) ÷ 2c − b + 3a Evaluate
\((10)^2 \div 2(5)-(9)+3\left(\frac{1}{3}\right)\) Substitute d = 10, c = 5, b = 9 and a = \(\frac{1}{3}\)
= 100 ÷ 10 − 9 + 1 Evaluate the parentheses
= 10 − 9 + 1 Divide
= 1 + 1 Subtract
= 2 Add
Result
2
Page 35 Exercise 8 Answer
3c + \(b^2\) ÷ 27a – d Evaluate
= \(3(5)+(9)^2 \div 27\left(\frac{1}{3}\right)-(10)\) Substitute c = 5, b = 9, a = \(\frac{1}{3}\) and d = 10
= 15 + 81 ÷ 9 − 10 Evaluate the parentheses
= 15 + 9 − 10 Divide
= 24 − 10 Add
= 14 Subtract
Result
14
Page 35 Exercise 9 Answer
28 − \(c^3\) + 6
28 − \(c^3\) + 6
= 28 − \((1)^3\) + 6 Substitute c = 1
= 28 − 1 + 6 Evaluate the power
= 27 + 6 Subtract
= 33 Add
28 − \(c^3\) + 6
= 28 − \((2)^3\) + 6 Substitute c = 2
= 28 − 8 + 6 Evaluate the power
= 20 + 6 Subtract
= 26 Add
28 − \(c^3\) + 6
= 28 − \((3)^3\) + 6 Substitute c = 3
= 28 − 27 + 6 Evaluate the power
= 1 + 6 Subtract
= 7 Add
Result
33 ; 26; 7
Page 35 Exercise 10 Answer
\(\frac{d}{7}\) − 3 + 10
\(\frac{d}{7}\) − 3 + 10
= \(\frac{28}{7}\) − 3 + 10 Substitute d = 28
= 4 − 3 + 10 Divide
= 1 + 10 Subtract
= 11 Add
\(\frac{d}{7}\) − 3 + 10
= \(\frac{49}{7}\) − 3 + 10 Substitute d = 49
= 7 − 3 + 10 Divide
= 4 + 10 Subtract
= 14 Add
\(\frac{d}{7}\) − 3 + 10
= \(\frac{63}{7}\) − 3 + 10 Substitute d = 63
= 9 − 3 + 10 Divide
= 6 + 10 Subtract
= 16 Add
Result
11 ; 14 ; 16
Page 36 Exercise 11 Answer
a) Expression for the amount she earned for h hours sitting one dog and two days sitting the cats
Cost per hour for one Dog × Number of Hours + Cost per day for two Cats × Number of Days
= 7 × h + 15 × 2
= 7h + 30
b) Evaluate the expression for h = 2
7h + 30
= 7(2) + 30 Substitute h = 2
= 14 + 30 Multiply
= $44
Result
a) 7h + 30
b) $44
Page 36 Exercise 12 Answer
One Dog:
Cost per day = $20
Cost for one hour = $7
Cost for two hours = $7 × 2 = $14
Cost for three hours = $7 × 3 = $21
Thus, we can purchase for 2 hours before it would be cheaper than to pay for one day.
Result
2 hours
Page 36 Exercise 13 Answer
65 + 6x Evaluate
= 65 + 6(4) Substitute x = 4
= 65 + 24 Multiply
= 89 Add
Result
It will cost $89 if he surfs for 4 hours.
Page 36 Exercise 14 Answer
Let the number of lessons be x Assume a variable
12x = 20 + 8x Equate both the expressions
12x − 8x = 20 + 8x − 8x Subtract 8x from both sides
4x = 20 Simplify
\(\frac{4x}{4}\) = \(\frac{20}{4}\) Divide each side by 4
x = 5 Simplify
Result
Thus, it will take 5 lessons for Janet′s class to cost the same amount as Rita′s class.
Page 36 Exercise 15 Answer
3g ÷ \(h^2\) + k − n Evaluate
= 3(12) ÷ \((3)^2\) + (10) − (1) Substitute g = 12, h = 3, k = 10 and n = 1
= 36 ÷ 9 + 10 − 1 Evaluate the parentheses
= 4 + 10 − 1 Divide
= 14 – 1 Add
= 13 Subtract
Result
D) 13
Page 36 Exercise 16 Answer
\(\frac{1}{2} x+y^2-4 z \div t\) Evaluate
= \(\frac{1}{2}(10)+(4)^2-4(5) \div(2)\) Substitute x = 10, y = 4, z = 5, and t = 2
= 5 + 16 − 20 ÷ 2 Evaluate the parentheses
= 5 + 16 − 10 Divide
= 21 − 10 Add
= 11 Subtract
Result
C) 11