#7**+3 **

I'll start off with 44.

Apply complex arithmetic rule: (a+bi)(c+di)=(ac-bd)(ad+bc)i

a=4, b=3, c=5, d=-6

Plugging them in, we have:

(4*5-3(-6))+(4(-6)+3*5)i

So, simplifying, we get \(38-9i\)

.tertre Apr 23, 2018

#11**+2 **

41. First add 8 to both sides, to get, |3x+6|=6

Then, we have, |f(x)|=a - f(x)=-a or f(x)=a

So, 3x+6=-6 or 3x+6=6

3x+6=-6: x=-4

3x+6=6: x=0

So, x=-4, or x=0

tertre Apr 23, 2018

#13**+2 **

Same thing as we did in 41.

Add 2 to both sides, to get, \(|2x+1|\geq7\)

|f(x)| greater than or equal to a, f(x) less than or equal to a, or f(x) greater than or equal to a

\(2x+1\leq-7: x\leq-4\)

\(2x+1\geq7: x\geq3\)

Combine the ranges, to get \(x\leq-4 \) or \(x\geq3\)

tertre Apr 23, 2018

#14**+2 **

Ty so so much tertre and Omi! Now I need help understanding 40 and 42 lol

RainbowPanda Apr 23, 2018