enVisionmath 2.0: Grade 6, Volume 1 Chapter 1 Algebra: Understand Numerical And Algebraic Expressions Guided Practice 7

Chapter 1 Algebra: Understand Numerical And Algebraic Expressions

Guided Practice 7

Page 57 Exercise 1 Answer

Formulas are useful because it generalize a relationship between two or more quantitites.

Result

Because it generalize a relationship between quantities

Page 57 Exercise 2 Answer

It is important to define each variable used in formulas because a formula uses symbols to relate between quantities.

Result

Because formula uses symbols to relate between quantities

Read And Learn More: enVisionmath 2.0 Grade 6 Volume 1 Solutions

Page 57 Exercise 3 Answer

Step 1 → Identify the values in the formulas

d = distance

r = rate = 400 miles per hour

t = time = 5 hours

Step 2 → Substitute the values in the formula and evaluate

d = rt

d = 400 × 5

d = 2000

The plane will travel 2000 miles in 5 hours.

Result

2000 miles

Page 57 Exercise 4 Answer

Step 1 → Identify the values in the formulas

d = distance = 510 miles

r = rate = 68 miles per hour

t = time

Step 2 → Substitute the values in the formula and evaluate

d = rt

510 = (68)t Substitute the value

510 = 68t Simplify

\(\frac{510}{68}=\frac{68 t}{68}\) Divide both sides by 68

t = 7.5

It will take 7.5 hours for a car travelling at a rate of 68 miles per hour to go 510 miles.

Result

7.5 hours

Page 57 Exercise 5 Answer

Step 1 → Identify the values in the formulas

V = volume

s = length = width = height = 8 cm

Step 2 → Substitute the values in the formula and evaluate

V = \(s^3\)

V = \((8)^3\) Substitute the value

V = 512

The Volume of the cube is 512\(cm^3\)

Result

512\(cm^3\)

Page 57 Exercise 6 Answer

Step 1 → Identify the values in the formulas

A = Total surface area

s = length = width = height = 8 cm

Step 2 → Substitute the values in the formula and evaluate

A = \(6s^2\)

A = \(6(8)^2\) Substitute the value

A = 6 × 64 Simplify

A = 384\(cm^2\)

The Total surface area of the cube is 384\(cm^2\)

Result

384 \(cm^2\)

Page 57 Exercise 7 Answer

Step 1 → Identify the values in the formulas

g = number of gallons of gasoline used

m = miles per gallon = 16

d = distance traveled = 296

Step 2 → Substitute the values in the formula and evaluate

g = \(\frac{d}{m}\)

g = \(\frac{296}{16}\) Substitute the value

g = 18.5

Myra will need 18.5 gallons of gasoline to travel 296 miles.

Result

18.5 gallons

Page 58 Exercise 8 Answer

Step 1 → Identify the values in the formulas

F = Fahrenheit

C = Celsius = 26°

Step 2 → Substitute the values in the formula and evaluate

F = (C × 1.8) + 32

F = (26 × 1.8) + 32 Substitute the value

F = 46.8 + 32 Simplify

F = 78.8°

The temperature in Fahrenheit is 78.8°

Result

78.8°

Page 58 Exercise 9 Answer

Step 1 → Identify the values in the formulas

F = Fahrenheit = 45°F

C = Celsius = 13°C

Step 2 → Substitute the values in the formula and evaluate

F = (C × 1.8) + 32

45 = (13 × 1.8) + 32 Substitute the value

45 = 23.4 + 32 Simplify

45 ≠ 55.4

Since Left hand side is not equal to Right hand side.

So, the thermometer cannot show as both 45°F and 13°C

Result

NO

Page 58 Exercise 10 Answer

Step 1 → Identify the values in the formulas

A = Average grade

X = test score = 78

Y = test score = 90

Z = test score = 81

Step 2 → Substitute the values in the formula and evaluate

A = \(\frac{X+Y+Z}{3}\)

A = \(\frac{78+90+81}{3}\) Substitute the value

A = \(\frac{249}{3}\) Simplify

A = 83

Jules average test grade is 83

Result

83

Page 58 Exercise 11 Answer

Step 1 → Identify the values in the formulas

d = density

m = mass of the object = 65 grams

v = volume = 8 cubic meters

Step 2 → Substitute the values in the formula and evaluate

d = \(\frac{m}{v}\)

d = \(\frac{65}{8}\) Substitute the value

d = 8.125 gram per cubic meter

Density of the object is 8.125 gram per cubic meter

Result

8.125 gram per cubic meter

Page 58 Exercise 12 Answer

\(7\left(3^2+5\right)-\left(\frac{81}{9}\right)\) Evaluate

= \(7(9+5)-\left(\frac{81}{9}\right)\) \(3^2\) = 9

= 7(14) − 9 Evaluate inside the parentheses

= 98 − 9 Multiply

= 89 Subtract

Result

89

Page 58 Exercise 13 Answer

Step 1→Identify the values in the formulas

I = Interest = $696

p = Principal loan amount = $5800

r = rate = 4% = 0.04

t = time

Step 2 → Substitute the values in the formula and evaluate

I = prt

696 = 5800 ⋅ 0.04 ⋅ t Substitute the value

696 = 232t Simplify

\(\frac{696}{232}=\frac{232 t}{232}\) Divide both sides by 232

t = 3

Janie will take 3 years to pay off the loan

Result

3 years

Page 58 Exercise 14 Answer

Option 1 → 4.50 for each hour worked.

Option 2 → Final payment of $350

Step 1 → Identify the values in the formulas

p = total payment after 15 days

h = hours worked each day=6 hours

Step 2 → Substitute the values in the formula and evaluate

p = 15 × 4.50h

p = 15 × 4.50(6) Substitute the value

p = 15 × 27 Simplify

p = 405

Jeremiah should choose Option 1 as that he will get more payment than Option 2.

Result

Option 1

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