Refer to our Texas Go Math Grade 6 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 6 Lesson 11.2 Answer Key Evaluating Expressions.

## Texas Go Math Grade 6 Lesson 11.2 Answer Key Evaluating Expressions

**Essential Question**

How can you use the order of operations to evaluate algebraic expressions?

**Your Turn**

Evaluate each expression for the given value of the variable.

Question 1.

4x; x = 8 ____

Answer:

Solution to this example is given below

4x; x = 8

4(8) Substitute 8 for x

32 Multiply

When x = 8, 4x = 32

32 Final solution

Question 2.

6.5 – n;n = 1.8 ___

Answer:

Given expression:

6.5 – n

Substitute n = 1.8 in the given expression:

= 6.5 – 1.8

Evaluate:

= 4.7

6.5 – n = 4.7

Question 3.

\(\frac{m}{6}\);m = 18 ____

Answer:

Solution to this example is giver below

\(\frac{m}{6}\); m = 18

\(\frac{18}{6}\) Substitute 18 for m

3 Divide

When m = 18, \(\frac{m}{6}\) = 3

3 Final solution

**Your Turn**

Evaluate each expression for n = 5.

Question 4.

3(n + 1) ____

Answer:

Solution to this example is given below

3(n + 1); n = 5

3(5 + 1) Substitute 5 for n

3(6) Add inside the parentheses

Multiply

When n = 5, 3(n + 1) = 18

18 Final solution

Question 5.

4(n – 4) + 14 ____

Answer:

Given expression:

4(n – 4) – 14

Substitute n = 5 in the given expression:

= 4(5 – 4) + 14

Simplify the expression in the parentheses:

= 4(1) + 14

Expand the parentheses:

= 4 + 14

Evaluate:

= 18

4(n – 4) + 14 = 18

Question 6.

6n + n^{2} ____

Answer:

Solution to this example is given below

6n + n^{2}; n = 5

6(5) + 5^{2} Substitute 5 for n

6(5) + 25 Evaluate exponents

30 + 25 Multiply

55 Add

When n = 5, 6n + n^{2} = 55

55 Final solution

Evaluate each expression for a = 3, b = 4, and c = -6.

Question 7.

ab – c ____

Answer:

Solution to this example is given below

ab – c; a = 3,b = 4, c = 6

3(4) – 6 Substitute 3 for a, 4 for b, and 6 for c

12 – 6 Multiply

6 Subtract

When a = 3, b = 4, and c = 6 ab – c = 6

6 Final solution

Question 8.

bc + 5a ___

Answer:

Solution to this example is given below

bc + 5a; a = 3, b = 4, c = 6

4(6) + 5(3) Substitute 3 for a, 4 for b, and 6 for c

24 + 15 Multiply

39 Subtract

When a = 3, b = 4, and c = 6 bc + 5a = 39

39 Final solution

Question 9.

a^{2} – (b + c) ____

Answer:

Evaluate the expression by substituting the given values.

= 3^{2} – (4 + (-6)) substitute the values to the given expression

= 9 – (-2) evaluate the exponent and subtract the numbers insde the parenthesis

= 9 + 2 add the numbers

= 11 value of the expression

The value of the expression is 11.

**Your Turn**

Question 10.

The expression 6x^{2} gives the surface area of a cube, and the expression x^{3} gives the volume of a cube, where x is the length of one side of the cube. Find the surface area and the volume of a cube with a side length of 2 m.

S = ____ m^{2} ; V = ___ m^{3}

Answer:

Given expression:

Surface Area = 6x^{2}

Substitute x = 2 in the given expression:

Surface Area = 6(2)^{2}

Evaluate:

Surface Area = 6(4) = 24

Surface area of the given cube is 24 square meters.

Given expression:

Volume = x^{3}

Substitute x = 2 in the given expression:

Volume = 2^{3}

Evaluate:

Volume = 8

Volume of the given cube is 8 cubic meters.

Question 11.

The expression 60m gives the number of seconds in m minutes. How many seconds are there in 7 minutes? ____ seconds

Answer:

Solution to this example is given below

60(m); m = 7

60(7) Substitute 7 for m

420 Multiply

When m = 7, 60(m) = 420

There are 420 seconds in 7 minutes.

420 Final solution

**Texas Go Math Grade 6 Lesson 11.2 Guided Practice Answer Key**

**Evaluate each expression for the given value(s) of the variable(s). (Examples 1 and 2)**

Question 1.

x – 7; x = 23 ________________

Answer:

Solution to this example is given below

x – 7; x = 23

23 – 7 Substitute 23 for x

16 Subtract

When x = 23, x – 7 = 16

16 Final solution

Question 2.

3a – b; a = 4, b = 6 _________

Answer:

Solution to this example is given below

3a – b; a = 4, b = 6

3(4) – 6 Substitute 4 for a, and 6 for b.

12 – 6 Multiply

6 Subtract

When a = 4, and b = 6, 3a – b = 6

6 Final solution

Question 3.

\(\frac{8}{t}\); t = 4 __________________

Answer:

Solution to this example is given below

\(\frac{8}{t}\); t = 4

\(\frac{8}{4}\) Substitute 4 for t

2 Divide

When t = 4, \(\frac{8}{t}\) = 2

2 Final solution

Question 4.

9 + m; m = 1.5 ___________

Answer:

Solution to this example is given below

9 + m; m = 1.5

9 + 1.5 Substitute 1.5 for m

10.5 Add

When m = 1.5, 9 + m = 10.5

10.5 Final solution

Question 5.

\(\frac{1}{2}\)\(\frac{1}{2}\)w + 2; w = \(\frac{1}{9}\) ___________________

Answer:

Solution to this example is given below

\(\frac{1}{2}\)w + 2; w = \(\frac{1}{9}\)

\(\frac{1}{2}\)(\(\frac{1}{2}\)) + 2 Substitute \(\frac{1}{9}\) for w

Substitute \(\frac{1}{18}\) + 2 Multiply

\(\frac{1+36}{18}\) Add

\(\frac{37}{18}\) Add

When w = \(\frac{1}{9}\), \(\frac{1}{2}\)w + 2 = \(\frac{37}{18}\)

\(\frac{37}{18}\) Final solution

Question 6.

5(6.2 + z); z = 3.8 _____________

Answer:

Given expression:

5(6.2 + z)

Substitute the value of z in the given expression:

= 5(6.2 – 3.8)

Simplify the parentheses:

= 5(10)

Expand the parentheses:

= 50

5(6.2 + z) = 50

Question 7.

The table shows the prices for games in Bella’s soccer league. Women’s Soccer league. Her parents and grandmother attended a soccer game. How much did they spend if they all went together in one car? Student tickets (Example 3)

a. Write an expression that represents the cost of one Parking carful of nonstudent soccer fans. Use x as the number of people who rode in the car and attended the game. __________ is an expression that represents the cost of one carful of nonstudent soccer fans.

Answer:

a. There are x Nonstudent people so the cost of their tickets will be 12 × x = 12x plus the parking fee, so the total cost is given by the expression: 12x + 5.

b. Since there are three attendees, evaluate the expression

12x + 5 for x = 3.

12(___) + 5 = _____ + 5 = _____

The family ___ spent to attend the game.

Answer:

Substitute x = 3 in the expression, therefore: 12(3) + 5 = 36 + 5 = $41.

The family spent $41 to attend the game.

Question 8.

Stan wants to add trim all around the edge of a rectangular tablecloth that measures 5 feet long by 7 feet wide. The perimeter of the rectangular tablecloth is twice the length added to twice the width. How much trim does Stan need to buy? (Example 3)

a. Write an expression that represents the perimeter of the rectangular tablecloth. Let l represent the length of the tablecloth and w represent its width. The expression would be ____.

Answer:

Here let length of the tablecloth be l and width be w so the expression for perimeter of the tablecloth becomes: 2l + 2w.

b. Evaluate the expression P = 2w + 2l for l = 5 and w = 7.

2(___) + 2(_____) = 14 + _____ = _____

Stan bought __________ of trim to sew onto the tablecloth.

Answer:

Here l = 5 and width be w = 7 so the value of perimeter of the tablecloth becomes: 2(5) + 2(7) = 10 + 14 = 24. The perimeter of the tablecloth is 24 feet. Stan bought 24 feet of trim to sew onto the tablecloth.

Question 9.

Essential Question Follow up How do you know the correct order in which to evaluate algebraic expressions?

Answer:

In any algebraic expression, the simplification of the parentheses has the highest priority followed by division, then multiplication, then addition and finally subtraction. These priorities are used to simplify algebraic expressions.

**Texas Go Math Grade 6 Lesson 11.2 Independent Practice Answer Key**

Question 10.

The table shows ticket prices at the Movie 16 theater. Let a represent the number of adult tickets, c the number of children’s tickets, and s the number of senior citizen tickets.

a. Write an expression for the total cost of tickets.

Answer:

The total cost of the three types of tickets for a adults, c children and s seniors is 8.75a + 6.5c + 6.5s.

b. The Andrews family bought 2 adult tickets, 3 children’s tickets, and 1 senior ticket. Evaluate your expression in part a to find the total cost of the tickets.

Answer:

Expression:

8.75a + 6.5c + 6.5s

Substitute a = 2, b = 3 and c = 1:

= 8.75(2) + 6.5(3) + 6.5(1)

Expand each parentheses:

= 17.5 + 19.5 + 6.5

Evaluate:

= 43.5

Total cost of tickets is $43.50

c. The Spencer family bought 4 adult tickets and 2 children’s tickets. Did they spend the same as the Andrews family? Explain.

Answer:

Expression:

Substitute a = 4, b = 2 and c = 0:

= 8.75(4) + 6.5(2) + 6.5(0)

Expand each parentheses:

= 35 + 13

Evaluate:

= 48

Total cost of tickets is $48. It can be seen that the 2 costs are not equal.

Question 11.

The area of a triangular sail is given by the expression \(\frac{1}{2}\)bh, where b is the length of the base and h is the height. What is the area of a triangular sail in a model sailboat when b = 12 inches and h = 7 inches?

A = ____ in.^{2}

Answer:

Given expression:

\(\frac{1}{2}\)bh

Substitute b = 12 and h = 7:

= \(\frac{1}{2}\)(12)(7)

Simplify:

= 6(7)

Evaluate:

= 42

Area of the triangular sail in a model sailboat is 42 square inches.

Question 12.

Ramon wants to balance his checking account. He has $2,340 in the account. He writes a check for $140. He deposits a check for $268. How much does Ramon have left in his checking account? _________________

Answer:

The expression of his current balance is:

2340 – 140 + 268

The check is a withdrawal from the account so shown with a negative sign.

Evaluate:

= 2468

He will have $2468 in his account.

Question 13.

Look for a Pattern Evaluate the expression 6x – x^{2} for x = 0, 1, 2, 3,4, 5, and 6. Use your results to fill in the table and describe any pattern that you see.

Answer:

x 6x – x^{2}

0 6(0) – 0^{2} = 0

1 6(1) – 1^{2} = 5

2 6(2) – 2^{2} = 8

3 6(3) – 3^{2} = 9

4 6(4) – 4^{2} = 8

5 6(5) – 5^{2} = 5

6 6(6) – 6^{2} = o

The numbers ¡ncrease from 0 to a maximum of 9 and decrease in the same pattern and reach 0 again.

Question 14.

The kinetic energy (in joules) of a moving object can be calculated from the expression \(\frac{1}{2} m v^{2}\), where m is the mass of the object in kilograms and y is its speed in meters per second. Find the kinetic energy of a 0.145-kg baseball that is thrown at a speed of 40 meters per second.

E = __ joules

Answer:

Given expression:

\(\frac{1}{2}\)mv^{2}

Substitute m = 0.145 and v = 40:

= \(\frac{1}{2}\)(0.145)(40)^{2}

Simplify:

= 0.0725(1600)

Evaluate:

= 116

Kinetic energy of the baseball is 116 Joules.

Question 15.

The area of a square is given by x^{2}, where x is the length of one side. Mary’s original garden was in the shape of a square. She has decided to double the area of her garden. Write an expression that represents the area of Mary’s new garden. Evaluate the expression if the side length of Mary’s original garden was 8 feet.

Answer:

Given expression:

Area = s^{2}

New length of the square garden is 2s^{2}:

Area = 2s^{2} = 2x^{2}

Substitute x = 8:

Area = 2(8)^{2}

Simplify:

Area = 2(64)

Evaluate:

Area = 128

Area of the new garden is 128 square feet

Question 16.

The volume of a pyramid with a square base is given by the expression \(\frac{1}{3} s^{2} h\), where s is the length of a side of the base and h is the height. Find the volume of a pyramid with a square base of side length 24 feet and a height of 30 feet.

Answer:

Solution to this example is given below

\(\frac{1}{3}\)s^{2}h, s = 24, and h = 30

\(\frac{1}{3}\)(24)^{2}(30) Substitute 24 for s, and 30 for h

(576)(30) Evaluate exponents

(576)(10) Divide

5760 Multiply

When s = 24, and h = 30, \(\frac{1}{3}\)s^{2}h = 5760

Volume of the pyramid is 5760 cubic feet.

5760 Final solution

**Texas Go Math Grade 6 Lesson 11.2 H.O.T. Focus On Higher Order Thinking Answer Key**

Question 17.

**Draw Conclusions** Consider the expressions 3x(x – 2) + 2 and 2x^{2} + 3x – 12.

a. Evaluate each expression for x = 2 and for x = 7. Based on your results, do you know whether the two expressions are equivalent? Explain.

Answer:

Check the value of the expression if they are equivalent when x = 2 and x = 7.

3(2)(2 – 2) + 2 = 2(2)^{2} + 3(2) – 12 substitute for the value of x

6(0) + 2 = 2(4) + 6 – 12 perform the indicated operation inside the parentheses

0 + 2 = 8 + 6 – 12 simplify

2 = 2 value of the expression

The expressions are equivalent when the value of x is 2.

3(7)(7 – 2) +2 = 2(1)^{2} + 3(7) – 12 substitute for the value of x

21(5) + 2= 2(49) + 21 – 12 perform the indicated operation inside the parentheses

105 + 2 = 98 + 21 – 12 simplify

107 = 107 value of the expression

The expressions are equivalent when the value of x is 7.

b. Evaluate each expression for x = 1. Based on your results, do you know whether the two expressions are equivalent? Explain.

Answer:

Check the value of the expression if they are equivalent when x = 1.

3(1)(1 – 2) + 2 = 2(1)^{2} + 3(1) – 12 substitute for the value of x

3(-1) + 2 = 2 (1) + 3 – 12 perform the indicated operation inside the parentheses

-3 + 2 = 2 + 3 – 12 simplify

-1 ≠ -7 value of the expression

The expressions are not equivalent when the value of x is 1

Question 18.

**Critique Reasoning** Marjorie evaluated the expression 3x + 2 for x = 5 as shown:

3x + 2 = 35 + 2 = 37

What was Marjorie’s mistake? What is the correct value of 3x + 2 for x = 5?

Answer:

Given expression:

3x + 2

Substitute x = 5:

= 3(5) + 2

Simplify. Marjorie’s mistake was in the incorrect multiplication of 3 and 5:

= 15 + 2

Evaluate:

= 17

3x + 2 = 17