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 Module 13 Answer Key Inequalities and Relationships.
Texas Go Math Grade 6 Module 13 Answer Key Inequalities and Relationships
Texas Go Math Grade 6 Module 13 Are You Ready? Answer Key
Write an integer to represent each situation.
Question 1.
a loss of $75
Answer:
Because here we have loss, we will have -, so, the required integer is 75.
Question 2.
a football player’s gain of 9 yards
Answer:
Here, because we have gain, we will have +. so, the required integer is 9.
Question 3.
spending $1,200 on a flat screen TV __________
Answer:
Because here we have spending, we will have -, so the required integer is -1.200.
Question 4.
a climb of 2,400 feet _________
Answer:
Because here we have climbing, we will have +, so the required integer is 2.400.
Find the product or quotient.
Question 5.
6 × 9 __________
Answer:
Here, the signs of the integers are the same, so, the product will be positive. We have that result is 54.
Question 6.
15 ÷ (-5) ________
Answer:
Here, the signs of integers are different, so the quotient will be negative.
We have that result is -3.
Question 7.
-8 × 6 _______
Answer:
Here, the signs of integers are different, so the product will be negative.
So, the result is -48.
Question 8.
-100 ÷ (10) _________
Answer:
Here, the signs of integers are different, so the quotient will be negative.
We have that result is -10.
Question 9.
3 × (-7) ________
Answer:
Here, the signs of integers are different, so the product will be negative.
So, the result is -21.
Question 10.
-64 ÷ 8 ________
Answer:
Here, the signs of integers are different, so the quotient will be negative.
We have that result is -8.
Question 11.
-8 × (-2) __________
Answer:
Here, the signs of the integers are the same, so, the product will be positive We have that result is 16.
Question 12.
32 ÷ 2 ___________
Answer:
Here, the signs of the integers are the same, so, the quotient will be positive. We have that result is 16.
Solve.
Question 13.
9p = 108 __________
Answer:
Here we have that p is multiplied by 9, so, we will multiply both sides by the reciprocal, \(\frac{1}{9}\), in order to isolate variable. After that we will simplify and get the following:
\(\frac{1}{9}\) ∙ 9p = 108 ∙ \(\frac{1}{9}\)
p = \(\frac{108 \cdot 1}{9}\)
p = 12
Question 14.
\(\frac{3}{5}\)n = 21 _________
Answer:
3 5
Here we have that n is multiplied by \(\frac{3}{5}\), so, we will multiply both sides by the reciprocal, \(\frac{3}{5}\), in order to isolate the variable.
After that we will simplify and get the following:
\(\frac{5}{3}\) ∙ \(\frac{3}{5}\)n = 21 ∙ \(\frac{5}{3}\)
n = \(\frac{21 \cdot 5}{3}\)
n = 35
Question 15.
\(\frac{4}{7}\)k = 84 __________
Answer:
\(\frac{7}{4}\) ∙ \(\frac{4}{7}\)k = 84 ∙ \(\frac{7}{4}\)
k = \(\frac{84 \cdot 7}{4}\)
k = 147
Question 16.
\(\frac{3}{20}\) e = 24 ___________
Answer:
\(\frac{20}{3}\) ∙ \(\frac{3}{20}\)e = 24 ∙ \(\frac{20}{3}\)
e = \(\frac{24 \cdot 20}{3}\)
e = 160
Texas Go Math Grade 6 Module 13 Reading Start-Up Answer Key
Visualize Vocabulary
Use the ✓ words to complete the graphic.
Understand Vocabulary
Match the term on the left to the correct expression on the right.
Answer:
1 – A. Solution of an inequality is a value or values that make the inequality true. ConcLusion ¡s that A matches to this term.
2 – C. Coefficient is the number that is multiplied by the variable in an algebraic expression. Conclusion is that C matches to this term.
3 – B. Constant is a specific number whose value does not change. Conclusion is that B matches this term.
Active Reading
Two-Panel Flip Chart Create a two-panel flip chart to help you understand the concepts in this module. Label one flap “Adding and Subtracting Inequalities.” Label the other flap “Multiplying and Dividing InequalitiesTM As you study each lesson, write important ideas under the appropriate flap.
