Question

ajutooòr,dau coroana!
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Answer 1

a) 3+2×[-3 ^ 11]^2 ÷9^22=

3+2×[3^11]^2 : 3^44 <=> 3+2×3^22÷3^44 <=> 3+

$\frac{2}{3 {}^{22 } }$

b)12÷[-1^22]^3 +(-8)=20

12÷[-1]^3 - 8 <=> 12 ÷ [-1] - 8 <=>(-12) - 8 = -20

C)(-7)^14 ÷7 ^12 -49=0

7^14÷7^12 - 49 <=> 7^2 - 49 <=> 49-49 = 0

D)(-1-3-8)-(-1-6) = -5

(-12)-(-7) <=> -12+7= -5

E) (-2) × (-10) - 5 × 12= -40

1)(-2)×(-10)=20

2)5×12=60

3)20-60=-40

F)(-49)÷(-7)÷(-1)+12÷(-2)÷(-6)+14÷(-1+1)=-6

1)(-1+1)=0

2)(-49)÷(-7)=7

3)7÷(-1)=-7

4)12:(-2)=-6

5)(-6):(-6)=1

6)(-7)+1=-6

7)14:0=0

8)(-6)+0=-6