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:
4 1/6
Step-by-step explanation:
4 x 6 = 24
1/6 x 6 = 6/6 = 1
24 + 1 = 25 :)
Only this function is lineer.
Because it is arithmetic.
The answer is area= 113.05 Circumference= 37.68
To find out if the points lie on the line y = 1/3x + 9, plug in the x value of the coordinate into the equation and see if the y value results in the same number
A. (0,9)
y = 1/3x + 9 Plug in 0 for "x"
y = 1/3(0) + 9
y = 9 This point is on the line
B. (-9,6)
y = 1/3x + 9 Plug in -9 for "x"
y = 1/3(-9) + 9
y = -9/3 + 9
y = -3 + 9
y = 6 This point is on the line
C. (3,10)
y = 1/3x + 9 Plug in 3 for "x"
y = 1/3(3) + 9
y = 3/3 + 9
y = 1 + 9
y = 10 This point is on the line
D. (12, -5)
y = 1/3x + 9 Plug in 12 for "x"
y = 1/3(12) + 9
y = 12/3 + 9
y = 4 + 9
y = 13 This point is not on the line
Your answer is A, B, and C