Answer:
8
Step-by-step explanation:
The given expression is "three less than the quotient of ten and a number, increased by six”
The quotient of ten and a number increased by six is translated into mathematical expression as:

Three less than the quotient of ten and a number, increased by six is

When we substitute n=2, we get:

Divide to get:

Add to get:

Subtract

By looking at it, I can tell the second expression is greater due to a a large postive number that's being added to it. Let's solve.
-12+6=-6-4=-10
-34-3=-37+39=2
So, -34-3+39 is greater.
Answer:
The inverse of function
is ![\mathbf{f^{-1} (x)=\sqrt[5]{x}+7}](https://tex.z-dn.net/?f=%5Cmathbf%7Bf%5E%7B-1%7D%20%28x%29%3D%5Csqrt%5B5%5D%7Bx%7D%2B7%7D)
Option A is correct option.
Step-by-step explanation:
For the function
, Find 
For finding inverse of x,
First let:

Now replace x with y and y with x

Now, solve for y
Taking 5th square root on both sides
![\sqrt[5]{x}=\sqrt[5]{(y+7)^5}\\\sqrt[5]{x}=y+7\\=> y+7=\sqrt[5]{x}\\y=\sqrt[5]{x}-7](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7Bx%7D%3D%5Csqrt%5B5%5D%7B%28y%2B7%29%5E5%7D%5C%5C%5Csqrt%5B5%5D%7Bx%7D%3Dy%2B7%5C%5C%3D%3E%20y%2B7%3D%5Csqrt%5B5%5D%7Bx%7D%5C%5Cy%3D%5Csqrt%5B5%5D%7Bx%7D-7)
Now, replace y with 
![f^{-1} (x)=\sqrt[5]{x}+7](https://tex.z-dn.net/?f=f%5E%7B-1%7D%20%28x%29%3D%5Csqrt%5B5%5D%7Bx%7D%2B7)
So, the inverse of function
is ![\mathbf{f^{-1} (x)=\sqrt[5]{x}+7}](https://tex.z-dn.net/?f=%5Cmathbf%7Bf%5E%7B-1%7D%20%28x%29%3D%5Csqrt%5B5%5D%7Bx%7D%2B7%7D)
Option A is correct option.
If the equation is in the form of y = mx + C, where m and C are real numbers, the slope (which means gradient) of the line is the coefficient of x (I.e. the number to the left of x in simple language), which is m.
In this case, m = 54 and C = -13pie. Both are real numbers, which means that this equation satisfies the form of y = mx + C
m is the coefficient of x, which is 54. 54 is the gradient of this line, which means that for every 54 y-axis units the line increases, the line increases 1 x-axis unit.
Hope this helps! :)