Let x be the number of times they raise the price on the newspaper. Then the new cost of the newspaper is

Let y be the newspaper they sell, then the income will be

Now, we know that the circulation is of 500, assuming that they sold every newspaper at the original price now the number the will sell will be

Plugging the value of y in the first expression we have that the income will be

Then the income is given by the function

To find the maximum value of this functions (thus the maximum income) we need to take the derivative of the function,

no we equate the derivative to zero and solve for x.

This means that we have an extreme value of the function when x=9. Now we need to find out if this value is a maximum or a minimum. To do this we need to take the second derivative of the function, then

Since the second derivative is negative in the point x=9, we conclude that this value is a maximum of the function.
With this we conclude that the number of times that they should raise the price to maximize the income is 9. This means that they will raise the price of the newspaper (9)($0.05)=$0.45.
Therefore the price to maximize the income is $0.35+$0.45=$0.80.
We can not see the rest of 12
Answer:
There are no choices, but
can be changes in the following ways: multiplying the numerator and the denominator by the same amount, or dividing the numerator and denominator by the same amount.
Hope it helps <3
Answer: 1% red and 3% orange
explanation:
Not sure if its right but did my best!!
Answer:
7.5 30
45 180
Step-by-step explanation:
The given line has x = 1.25 and y = 5.
y/x = 5/1.25 = 4
x is multiplied by 4 to get y since 1.25 * 4 = 5.
That means the equation is
y = 4x
For a proportional function, every value of x must be multiplied by 4 to get y.
x = 7.5
y = 4x = 4(7.5) = 30
7.5 30
We need to find the values for the last line.
11 * 4 = 44
There is no 44
13.75 * 4 = 55
There is no 55.
17.5 * 4 = 70
There is no 70
30 * 4 = 120
There is no 120.
45 * 4 = 180
There is a 180.
45 180
Answer:
7.5 30
45 180