Answer:
The inverse of the function is ![f^{-1}(x) = \sqrt[3]{\frac{x-16}{8}}](https://tex.z-dn.net/?f=f%5E%7B-1%7D%28x%29%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7Bx-16%7D%7B8%7D%7D)
Step-by-step explanation:
Inverse of a function:
Suppose we have a function y = g(x). To find the inverse, we exchange the values of x and y, and then isolate y.
In this question:

Exchanging x and y:



![y = \sqrt[3]{\frac{x-16}{8}}](https://tex.z-dn.net/?f=y%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7Bx-16%7D%7B8%7D%7D)
The inverse of the function is ![f^{-1}(x) = \sqrt[3]{\frac{x-16}{8}}](https://tex.z-dn.net/?f=f%5E%7B-1%7D%28x%29%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7Bx-16%7D%7B8%7D%7D)
Answer:
the approximate area of this circle is 200.96 inches long.
Step-by-step explanation:
To answer this problem we need to remember that the area of a circle is given by the formula:
Area = π
where r is the radius.
and the perimeter is:
Perimeter = 2πr
Now, the problem tells us that the circle is enclosed by a piece of rope that's 50.24 inches long. So the perimeter of the circle is 50.24 inches.
Since we have the value of the perimeter and the value of pi, we are going to substitute these values in the perimeter formula to find r.
Perimeter = 2πr
50.24=2(3.14)r
50.24= 6.28r
50.24/6.28= r
8= r
Thus, the radius of the circle is 8 inches long.
Now, we can use this value to find the area of the circle:
Area = π
Area = π
Area = 3.14 (64)
Area = 200.96
Therefore, the approximate area of this circle is 200.96 inches long.
Answer:
Option A 
Step-by-step explanation:
we know that
The volume of a cone is equal to

In this problem we have

substitute the values


Answer:
Q1 - 2
Q1 - 5
Q3 - 7
maximum 9
minimum 1
Step-by-step explanation:
order the numbers from least to greatest
Answer:
I sure hope none is an answer because none of these work!
Step-by-step explanation:
C = 59 (F - 32)
(C/59) = F - 32
F = C/59 + 32
So no :(
m = x + y + z3
y = m - x - z3
So no :((
s = r^2 - 1
r = sqrt(s+1)
So no :000
A = 12(a+b)
A = 12a + 12b
12b = A - 12a
b = A/12 - a
So no :00(((
m = x + y^2
y = sqrt (m-x)
AND no :|