Question

Add the additive inverse of $(-12+16-42÷3$ of $7)$ to the product of -2, -3 and -4.

Answer 1

First find the additive inverse of $(-12+16-42÷3of7)$

Additive inverse of a number,$x$ , is the number ,$y$, when you add to $x$ you get $0$.

Find the value of the terms in bracket(use steps of BODMAS):

$-12+16-\left(\frac{42}{3}of7\right)=-12+16-98$

$=16-12-98$

$=16-110$

$=-94$

The additive number is that when added to $-94$ will give zero, hence it is $+94$

Therefore add $94$ to product of $-2,-3,-4$

Product of $-2,-3,-4=(-2)\times (-3)\times (-4)=-24$

$\therefore$ Add $94$ to $-24=94+(-24)=94-24=70$

The answer is $70$