Answer:
1) They are not inverses
2) They are inverses
Step-by-step explanation:
We need to find the composition function between these functions to verify if these functions are inverses. If f[g(x)] and g[f(x)] are equal to x they are inverses.
<u>1)</u>
<u>Let's find f[g(x)] and simplify.</u>
![f[g(x)]=\frac{1}{2}g(x)+\frac{3}{2}](https://tex.z-dn.net/?f=f%5Bg%28x%29%5D%3D%5Cfrac%7B1%7D%7B2%7Dg%28x%29%2B%5Cfrac%7B3%7D%7B2%7D)
As f[g(x)] is not equal to x, these functions are not inverses.
2)
<u>Let's find f[g(x)] and simplify.</u>
![f[g(n)]=\frac{-16+(4n+16)}{4}](https://tex.z-dn.net/?f=f%5Bg%28n%29%5D%3D%5Cfrac%7B-16%2B%284n%2B16%29%7D%7B4%7D)
![f[g(n)]=\frac{-16+4n+16}{4}](https://tex.z-dn.net/?f=f%5Bg%28n%29%5D%3D%5Cfrac%7B-16%2B4n%2B16%7D%7B4%7D)
![f[g(n)]=\frac{4n}{4}](https://tex.z-dn.net/?f=f%5Bg%28n%29%5D%3D%5Cfrac%7B4n%7D%7B4%7D)
![f[g(n)]=n](https://tex.z-dn.net/?f=f%5Bg%28n%29%5D%3Dn)
Now, we need to find the other composition function g[f(x)]
<u>Let's find g[f(x)] and simplify.</u>
![g[f(x)]=4(\frac{-16+n}{4})+16](https://tex.z-dn.net/?f=g%5Bf%28x%29%5D%3D4%28%5Cfrac%7B-16%2Bn%7D%7B4%7D%29%2B16)
![g[f(x)]=-16+n+16](https://tex.z-dn.net/?f=g%5Bf%28x%29%5D%3D-16%2Bn%2B16)
![g[f(x)]=n](https://tex.z-dn.net/?f=g%5Bf%28x%29%5D%3Dn)
Therefore, as f[g(n)] = g[f(n)] = n, both functions are inverses.
I hope it helps you!
The algebraic expression uses the terms to denote symbols
sum of means addition
difference of means subtraction
product of means multiplication
quotient of means division
Here we have the term 6x
which actually means 6 times x or
6 multiplied by x
hence by multiplication we use the word product of
so we have the product of 6 and a number as our right answer
20t² - 13tv - 15v²
20t² - 25tv + 12tv - 15v²
(5t + 3v)(4t - 5v)
Answer:
The 6-minute half-mile time is continuous data; the number of people trying out for the team is discrete data.
Step-by-step explanation: