tertre

Welcome! Solving them right now! 

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 \(x\geq3\)

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

Thanks, Gavin! 

Here's a solution.

l=b+16, where l=length and b=breadth

l:b=4:3, so that's 4x:3x

So, x=16

4x=64, and 3x=48

Thus, 64:48=4:3

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 \(38-9i\)

Total Area: 10*9=90u^2

Area of D: 9*3=27u^2

So, \(\frac{27}{90}=\frac{9}{30}=\boxed{\frac{3}{10}}\)

43. \(x\leq -4\) or \(x \geq 3\)

44. Apply complex arithmetic rule, to get 38-9i, (A)

41. |3x+6|-8=-2

3x+6-8=-2 so 3x-2=-2

x=0- one solution

and -4, (B)