Equations with absolute value:

Where k is a positive number; if k is a negative number, the equation is impossible (absolute value is always positive).
How to solve:


Then:
1. |x+7|=12
x+7=12 V -x-7=12
x=5 V -x=19
x=5 V x=-19
{-19, 5}
2. |2x+4|=8
2x+4=8 V -2x-4=8
2x=4 V -2x=12
x=2 V x=-6
3. 3|3k|=27
3×3k=27 V 3×(-3k)=27
9k=27 V -9k=27
k=3 V k=-3
{-3, 3}
4. 5|b+8|=30
5×(b+8)=30 V 5×(-b-8)=30
5b+40=30 V -5b-40=30
5b=-10 V -5b=70
b=-2 V b=-14
{-14, -2}
5. |m+9|=5
m+9=5 V -m-9=5
m+9=5 V m+9=-5
I had homework on this today lol
First, add 2 to both sides. It would then be 5x = 3x + 10. Then subtract 3x from both sides and you would end up with 2x = 10. Divide 2x by 2 and the 10 by 2. The answer is x = 5.
The domain is 2x and the range is -6
Answer:
D
Step-by-step explanation:
![f(x)=(x-1)(x^2+2)^3\\f'(x)=(x-1)*3(x^2+2)^2*2x+(x^2+2)^3*1\\f'(x)=6x(x-1)(x^2+2)^2+(x^2+2)^3\\f'(x)=(x^2+2)^2[6x^2-6x+x^2+2]\\f'(x)=(x^2+2)^2(7x^2-6x+2)\\D](https://tex.z-dn.net/?f=f%28x%29%3D%28x-1%29%28x%5E2%2B2%29%5E3%5C%5Cf%27%28x%29%3D%28x-1%29%2A3%28x%5E2%2B2%29%5E2%2A2x%2B%28x%5E2%2B2%29%5E3%2A1%5C%5Cf%27%28x%29%3D6x%28x-1%29%28x%5E2%2B2%29%5E2%2B%28x%5E2%2B2%29%5E3%5C%5Cf%27%28x%29%3D%28x%5E2%2B2%29%5E2%5B6x%5E2-6x%2Bx%5E2%2B2%5D%5C%5Cf%27%28x%29%3D%28x%5E2%2B2%29%5E2%287x%5E2-6x%2B2%29%5C%5CD)