Differentiating each function, we have for all x, unless otherwise indicated,
![h(x) = 2^x - 1 \implies h'(x) = \ln(2) \, 2^x > 0](https://tex.z-dn.net/?f=h%28x%29%20%3D%202%5Ex%20-%201%20%5Cimplies%20h%27%28x%29%20%3D%20%5Cln%282%29%20%5C%2C%202%5Ex%20%3E%200)
![g(x) = -4 (2^x) \implies g'(x) = -4 \ln(2) \, 2^x < 0](https://tex.z-dn.net/?f=g%28x%29%20%3D%20-4%20%282%5Ex%29%20%5Cimplies%20g%27%28x%29%20%3D%20-4%20%5Cln%282%29%20%5C%2C%202%5Ex%20%3C%200)
![f(x) = -3x+7 \implies f'(x) = -3 < 0](https://tex.z-dn.net/?f=f%28x%29%20%3D%20-3x%2B7%20%5Cimplies%20f%27%28x%29%20%3D%20-3%20%3C%200)
![j(x) = x^2 + 8x + 1 \implies j'(x) = 2x + 8 > 0 \text{ only when } x > -4](https://tex.z-dn.net/?f=j%28x%29%20%3D%20x%5E2%20%2B%208x%20%2B%201%20%5Cimplies%20j%27%28x%29%20%3D%202x%20%2B%208%20%3E%200%20%5Ctext%7B%20only%20when%20%7D%20x%20%3E%20-4)
and only h(x) has a strictly positive derivative. (A)
Look at the axis graph. If both of the coordinates are negative, it would be in quadrant lll. If you go over to the x-axis, you would be in quadrant lV, or 4.
(-2,-3) -> (2,-3)
You change the x-factor's sign.
I hope this helps!
~kaikers
Answer:
6/7
Step-by-step explanation:
45x+34=24x+16
collect like terms, we now have;
45x-24x=16-34
21x=-18
X=18/21
X=6/7
Specify, what are you needing?