Question

(-56 ÷ -7) - -3 that; (-15 + -45) ÷ 6 one; (-9 · -4) + (-4 · 5) never; 9(-11 - -4) maybe; -3 · -8 · -2 anything; (-4 · 4) - (5 · -5) story

Explanation:

Step1: Solve $(-56 \div -7) - -3$

First calculate division: $\frac{-56}{-7}=8$, then simplify subtraction of negative: $8 + 3 = 11$

Step2: Solve $(-15 + -45) \div 6$

First sum inside parentheses: $-15 + (-45) = -60$, then divide: $\frac{-60}{6} = -10$

Step3: Solve $(-9 \cdot -4) + (-4 \cdot 5)$

Calculate multiplications: $(-9)(-4)=36$, $(-4)(5)=-20$, then add: $36 + (-20) = 16$

Step4: Solve $9(-11 - -4)$

Simplify inside parentheses: $-11 + 4 = -7$, then multiply: $9 \times (-7) = -63$

Step5: Solve $-3 \cdot -8 \cdot -2$

Multiply step-by-step: $(-3)(-8)=24$, then $24 \times (-2) = -48$

Step6: Solve $(-4 \cdot 4) - (5 \cdot -5)$

Calculate multiplications: $(-4)(4)=-16$, $(5)(-5)=-25$, then subtract: $-16 - (-25) = -16 + 25 = 9$

Answer:

1. $(-56 \div -7) - -3 = 11$
2. $(-15 + -45) \div 6 = -10$
3. $(-9 \cdot -4) + (-4 \cdot 5) = 16$
4. $9(-11 - -4) = -63$
5. $-3 \cdot -8 \cdot -2 = -48$
6. $(-4 \cdot 4) - (5 \cdot -5) = 9$