Answer:
![a'(d) = \frac{d}{5} + \frac{3}{5}](https://tex.z-dn.net/?f=a%27%28d%29%20%3D%20%5Cfrac%7Bd%7D%7B5%7D%20%2B%20%5Cfrac%7B3%7D%7B5%7D)
![a(a'(d)) = a'(a(d)) = d](https://tex.z-dn.net/?f=a%28a%27%28d%29%29%20%3D%20a%27%28a%28d%29%29%20%3D%20d)
Step-by-step explanation:
Given
![a(d) = 5d - 3](https://tex.z-dn.net/?f=a%28d%29%20%3D%205d%20-%203)
Solving (a): Write as inverse function
![a(d) = 5d - 3](https://tex.z-dn.net/?f=a%28d%29%20%3D%205d%20-%203)
Represent a(d) as y
![y = 5d - 3](https://tex.z-dn.net/?f=y%20%3D%205d%20-%203)
Swap positions of d and y
![d = 5y - 3](https://tex.z-dn.net/?f=d%20%3D%205y%20-%203)
Make y the subject
![5y = d + 3](https://tex.z-dn.net/?f=5y%20%3D%20d%20%2B%203)
![y = \frac{d}{5} + \frac{3}{5}](https://tex.z-dn.net/?f=y%20%3D%20%5Cfrac%7Bd%7D%7B5%7D%20%2B%20%5Cfrac%7B3%7D%7B5%7D)
Replace y with a'(d)
![a'(d) = \frac{d}{5} + \frac{3}{5}](https://tex.z-dn.net/?f=a%27%28d%29%20%3D%20%5Cfrac%7Bd%7D%7B5%7D%20%2B%20%5Cfrac%7B3%7D%7B5%7D)
Prove that a(d) and a'(d) are inverse functions
and ![a(d) = 5d - 3](https://tex.z-dn.net/?f=a%28d%29%20%3D%205d%20-%203)
To do this, we prove that:
![a(a'(d)) = a'(a(d)) = d](https://tex.z-dn.net/?f=a%28a%27%28d%29%29%20%3D%20a%27%28a%28d%29%29%20%3D%20d)
Solving for ![a(a'(d))](https://tex.z-dn.net/?f=a%28a%27%28d%29%29)
![a(a'(d)) = a(\frac{d}{5} + \frac{3}{5})](https://tex.z-dn.net/?f=a%28a%27%28d%29%29%20%20%3D%20a%28%5Cfrac%7Bd%7D%7B5%7D%20%2B%20%5Cfrac%7B3%7D%7B5%7D%29)
Substitute
for d in ![a(d) = 5d - 3](https://tex.z-dn.net/?f=a%28d%29%20%3D%205d%20-%203)
![a(a'(d)) = 5(\frac{d}{5} + \frac{3}{5}) - 3](https://tex.z-dn.net/?f=a%28a%27%28d%29%29%20%20%3D%205%28%5Cfrac%7Bd%7D%7B5%7D%20%2B%20%5Cfrac%7B3%7D%7B5%7D%29%20-%203)
![a(a'(d)) = \frac{5d}{5} + \frac{15}{5} - 3](https://tex.z-dn.net/?f=a%28a%27%28d%29%29%20%20%3D%20%5Cfrac%7B5d%7D%7B5%7D%20%2B%20%5Cfrac%7B15%7D%7B5%7D%20-%203)
![a(a'(d)) = d + 3 - 3](https://tex.z-dn.net/?f=a%28a%27%28d%29%29%20%20%3D%20d%20%2B%203%20-%203)
![a(a'(d)) = d](https://tex.z-dn.net/?f=a%28a%27%28d%29%29%20%20%3D%20d)
Solving for: ![a'(a(d))](https://tex.z-dn.net/?f=a%27%28a%28d%29%29)
![a'(a(d)) = a'(5d - 3)](https://tex.z-dn.net/?f=a%27%28a%28d%29%29%20%3D%20a%27%285d%20-%203%29)
Substitute 5d - 3 for d in ![a'(d) = \frac{d}{5} + \frac{3}{5}](https://tex.z-dn.net/?f=a%27%28d%29%20%3D%20%5Cfrac%7Bd%7D%7B5%7D%20%2B%20%5Cfrac%7B3%7D%7B5%7D)
![a'(a(d)) = \frac{5d - 3}{5} + \frac{3}{5}](https://tex.z-dn.net/?f=a%27%28a%28d%29%29%20%3D%20%5Cfrac%7B5d%20-%203%7D%7B5%7D%20%2B%20%5Cfrac%7B3%7D%7B5%7D)
Add fractions
![a'(a(d)) = \frac{5d - 3+3}{5}](https://tex.z-dn.net/?f=a%27%28a%28d%29%29%20%3D%20%5Cfrac%7B5d%20-%203%2B3%7D%7B5%7D)
![a'(a(d)) = \frac{5d}{5}](https://tex.z-dn.net/?f=a%27%28a%28d%29%29%20%3D%20%5Cfrac%7B5d%7D%7B5%7D)
![a'(a(d)) = d](https://tex.z-dn.net/?f=a%27%28a%28d%29%29%20%3D%20d)
Hence:
![a(a'(d)) = a'(a(d)) = d](https://tex.z-dn.net/?f=a%28a%27%28d%29%29%20%3D%20a%27%28a%28d%29%29%20%3D%20d)
Answer:
167.19
14.50 x 8 = 116.00
14.50 + 1.75 = 16.25 per hour overtime
16.25 × 3.15 hour overtime = 51.19
116.00 + 51.19 = 167.19
Step-by-step explanation:
Note that if we add each side of the 2 equations, y will cancel out:
-12x-y+(17x+y)=6+4
5x=10
x=2
Substituting x=2 in either of the equations, we find the value of y.
Let's use the first equation:
-12x-y=6
-24-y=6
-y=24+6
-y=30, so y=-30.
Thus, the solution is (2, -30)
Answer: C