When you divide exponents, you have to subtract them. So -4-3= -7 as the exponent
When you have a negative exponent, then you have to flip the fraction (all whole numbers are over 1 so flip the 1 to the top and the 5^7 on the bottom)
Real quick so I can sleep. it is because obtuse is over 90 degrees
You add +5k to both sides to eliminate-5k, leaving you with -3k=9 { since you added 5k to both sides } . then you divide both -3k on both sides { to isolate k }, leaving you with k = -3 as your answer.
Answer: a) -24
b) 
c) 4
Step-by-step explanation:
a) To determine the value of (fg)', use the product rule of derivative, i.e.:
(fg)'(x) = f'(x)g(x) + f(x)g'(x)
(fg)'(5) = f'(5)g(5) + f(5)g'(5)
(fg)'(5) = 6(-5) + 3(2)
(fg)'(5) = -24
b) The value is given by the use of the quotient rule of derivative:
![(\frac{f}{g})'(x)=\frac{f'(x)g(x)-f(x)g'(x)}{[g(x)]^2}](https://tex.z-dn.net/?f=%28%5Cfrac%7Bf%7D%7Bg%7D%29%27%28x%29%3D%5Cfrac%7Bf%27%28x%29g%28x%29-f%28x%29g%27%28x%29%7D%7B%5Bg%28x%29%5D%5E2%7D)
![(\frac{f}{g})' (5)=\frac{f'(5)g(5)-f(5)g'(5)}{[g(5)]^2}](https://tex.z-dn.net/?f=%28%5Cfrac%7Bf%7D%7Bg%7D%29%27%20%285%29%3D%5Cfrac%7Bf%27%285%29g%285%29-f%285%29g%27%285%29%7D%7B%5Bg%285%29%5D%5E2%7D)
c) ![(\frac{g}{f})'(5)=\frac{g'(5)f(5)-g(5)f'(5)}{[f(5)]^{2}}](https://tex.z-dn.net/?f=%28%5Cfrac%7Bg%7D%7Bf%7D%29%27%285%29%3D%5Cfrac%7Bg%27%285%29f%285%29-g%285%29f%27%285%29%7D%7B%5Bf%285%29%5D%5E%7B2%7D%7D)
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