Answer:
pairs
Step-by-step explanation:
Given
Profit on selling 'x' pairs of shoes is 
Find the derivative of the function to find out the critical points

Thus, for 4 pair of shoes profit is maximize.
In expression form this is:
9 + 2f
Hope this helps!
Answer:
y + 1
Step-by-step explanation:
if y = the total balls
then y + 1 = total balls plus 1
As given by the question
There are given that the point of two-line

Now,
From the condition of a parallel and perpendicular line
If the slopes are equal then the lines are parallel
If the slopes are negative reciprocal then the lines are perpendicular
If the slopes are neither of the above are true then lines are neither
Then,
First, find the slope of both of line
So,
For first-line, from the formula of slope

Now,
For second-line,

The given result of the slope is negative reciprocal because

Hence, the slope of line1 is -1/2, and slope of line2 is 2 and the lines are perpendicular.
Answer:
a. 
b. 
Step-by-step explanation:
The initial value problem is given as:

Applying laplace transformation on the expression 
to get ![L[{y+y'} ]= L[{7 + \delta (t-3)}]](https://tex.z-dn.net/?f=L%5B%7By%2By%27%7D%20%5D%3D%20L%5B%7B7%20%2B%20%5Cdelta%20%28t-3%29%7D%5D)

Taking inverse of Laplace transformation
![y(t) = 7 L^{-1} [ \dfrac{1}{(s+1)}] + L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{(s+1)-s}{s(s+1)}] +L^{-1} [\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7L^{-1} [\dfrac{1}{s}-\dfrac{1}{s+1}] + L^{-1}[\dfrac{e^{-3s}}{s+1}] \\ \\ y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]](https://tex.z-dn.net/?f=y%28t%29%20%3D%207%20L%5E%7B-1%7D%20%5B%20%5Cdfrac%7B1%7D%7B%28s%2B1%29%7D%5D%20%2B%20L%5E%7B-1%7D%20%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D%20%5C%5C%20%5C%5C%20y%28t%29%20%3D%207L%5E%7B-1%7D%20%5B%5Cdfrac%7B%28s%2B1%29-s%7D%7Bs%28s%2B1%29%7D%5D%20%2BL%5E%7B-1%7D%20%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D%20%5C%5C%20%5C%5C%20y%28t%29%20%3D%207L%5E%7B-1%7D%20%5B%5Cdfrac%7B1%7D%7Bs%7D-%5Cdfrac%7B1%7D%7Bs%2B1%7D%5D%20%2B%20L%5E%7B-1%7D%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D%20%5C%5C%20%5C%5C%20y%28t%29%20%3D%207%20%5B1-e%5E%7B-t%7D%20%5D%20%2B%20L%5E%7B-1%7D%20%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D)
![L^{-1}[\dfrac{e^{-3s}}{s+1}]](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D)
![L^{-1}[\dfrac{1}{s+1}] = e^{-t} = f(t) \ then \ by \ second \ shifting \ theorem;](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cdfrac%7B1%7D%7Bs%2B1%7D%5D%20%3D%20e%5E%7B-t%7D%20%20%3D%20f%28t%29%20%5C%20then%20%5C%20by%20%5C%20second%20%5C%20shifting%20%5C%20theorem%3B)
![L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{f(t-3) \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D%20%3D%20%5Cleft%20%5C%7B%20%7B%7Bf%28t-3%29%20%5C%20%5C%20%5C%20t%3E3%7D%20%5Catop%20%7B0%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%20%5C%20t%20%3C3%7D%7D%20%5C%20%5C%20%5C%20%20%5Cright.)
![L^{-1}[\dfrac{e^{-3s}}{s+1}] = \left \{ {{e^{(-t-3)} \ \ \ t>3} \atop {0 \ \ \ \ \ \ \ \ \ t](https://tex.z-dn.net/?f=L%5E%7B-1%7D%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D%20%3D%20%5Cleft%20%5C%7B%20%7B%7Be%5E%7B%28-t-3%29%7D%20%5C%20%5C%20%5C%20t%3E3%7D%20%5Catop%20%7B0%20%5C%20%5C%20%5C%20%5C%20%5C%20%5C%20%20%5C%20%5C%20%20%5C%20t%20%3C3%7D%7D%20%5C%20%5C%20%5C%20%20%5Cright.)

= 
Recall that:
![y(t) = 7 [1-e^{-t} ] + L^{-1} [\dfrac{e^{-3s}}{s+1}]](https://tex.z-dn.net/?f=y%28t%29%20%3D%207%20%5B1-e%5E%7B-t%7D%20%5D%20%2B%20L%5E%7B-1%7D%20%5B%5Cdfrac%7Be%5E%7B-3s%7D%7D%7Bs%2B1%7D%5D)
Then


