Answer:
<em>B.) Kelly got the better deal.</em>
Step-by-step explanation:
Answer:

Step-by-step explanation:
we know that
In the function f(x) the inflection point is at 
In the function g(x) the inflection point is at 
so
the rule of the translation of f(x) to g(x) is equal to

That means-----> the translation is
units to the right and
units down
The equation of g(x) is equal to

therefore

Area of a rectangle is L * W
L = x
W = 5
5x = 20 (Divide by 5 on both sides)
x = 4
L=4
W= 5
The length is 4cm
Hope this helps :)
Answer:
3.) x = 10
Step-by-step explanation:
Let’s plug it into our equation
2(10) + 1 = 21
21 is greater than 15
Answer:
![g(x)=-2\sqrt[3]x](https://tex.z-dn.net/?f=g%28x%29%3D-2%5Csqrt%5B3%5Dx)
or

Step-by-step explanation:
Given
![f(x) = \sqrt[3]x](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%5Csqrt%5B3%5Dx)
Required
Write a rule for g(x)
See attachment for grid
From the attachment, we have:


We can represent g(x) as:

So, we have:
![g(x) = n * \sqrt[3]x](https://tex.z-dn.net/?f=g%28x%29%20%3D%20n%20%2A%20%5Csqrt%5B3%5Dx)
For:

![2 = n * \sqrt[3]{-1}](https://tex.z-dn.net/?f=2%20%3D%20n%20%2A%20%5Csqrt%5B3%5D%7B-1%7D)
This gives:

Solve for n


To confirm this value of n, we make use of:

So, we have:
![-2 = n * \sqrt[3]1](https://tex.z-dn.net/?f=-2%20%3D%20n%20%2A%20%5Csqrt%5B3%5D1)
This gives:

Solve for n


Hence:
![g(x) = n * \sqrt[3]x](https://tex.z-dn.net/?f=g%28x%29%20%3D%20n%20%2A%20%5Csqrt%5B3%5Dx)
![g(x)=-2\sqrt[3]x](https://tex.z-dn.net/?f=g%28x%29%3D-2%5Csqrt%5B3%5Dx)
or:
