QUESTION IMAGE
Question
7 - m2 - tc - lesson 13
- -3(4)
- -2(6)
- 2(-6)
- 6(-2)
- refer to the expression 0 - 3 - 3 to complete parts (a)-(c).
a. write the expression as a repeated addition expression.
b. write the expression as a multiplication expression.
c. evaluate the expression.
Step1: Solve -3(4)
Multiply -3 and 4. $-3\times4=-12$
Step2: Solve -2(6)
Multiply -2 and 6. $-2\times6 = -12$
Step3: Solve 2(-6)
Multiply 2 and -6. $2\times(-6)=-12$
Step4: Solve 6(-2)
Multiply 6 and -2. $6\times(-2)=-12$
Step5: Solve part (a) of question 10
Rewrite subtraction as addition of the opposite. $0 - 3-3=0+(- 3)+(-3)$
Step6: Solve part (b) of question 10
Since we have two - 3's being added to 0, we can write it as a multiplication. $0+(-3)+(-3)=0 + 2\times(-3)$
Step7: Solve part (c) of question 10
First, $0-3=-3$, then $-3 - 3=-6$
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- -12
- -12
- -12
- -12
10.
a. $0+(-3)+(-3)$
b. $0 + 2\times(-3)$
c. -6