Answer to Question #95218 in Algebra for Ruth

Question

0-6[2-12(2-5)+1/2(20)-42+5-3]+0-10

Answer 1

You have the following mathematical operation

$0-6[2-12(2-5)+\frac{1}{2}(20)-42+5-3]+0-10$

We perform the operations between parentheses

$0-6[2-12(-3)+\frac{20}{2}-42+5-3]+0-10\\ 0-6[2+36+10-42+5-3]+0-10$

Now the algebraic sum is made

$2+36+10-42+5-3=8$

$0-10=-10$

Now

$-6[8]-10$ -->$-6[8]=-48$

Continuing

$-48-10=-58$

FInally

Solution $0-6[2-12(2-5)+\frac{1}{2}(20)-42+5-3]+0-10=-58$