Solve for q. |q+8|≥2 Write a compound inequality like 1 < x < 3 or like x < 1 or x > 3. Use integers, proper fractions, or improper fractions in simplest form.

Step-by-step explanation:

Given the inequality  |q+8|≥2, we are to solve for q. Note that the function inside modulus sign will return both negative and positive value.

For the positive function:

q+8≥2

Subtract 8 from both sides of the inequality.

q+8-8≥2-8

q ≥ -6

For the negative function;

-(q+8)≥2

open the parenthesis

-q-8≥2

add 8 to both sides

-q-8+8≥2+8

-q≥10

Multiply both sides by -1. Note that multiplying both sides of an inequality changes the direction of the inequality sign.

q≤-10

Combining both results i.e q ≥ -6 and q≤-10

q ≥ -6 can also be written as -6≤q

Combining -6≤q with q≤-10, this will give -6≤q≤-10.

Hence the result of the inequality function given is  -6≤q≤-10.

first one is a: renaming a fraction. second one is c: de-c

step-by-step explanation:

my explanation is that these were the answers on edg : )

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