<h3><em>This is the range of y , Ry=R−{0}</em></h3>

solve for "k", to find k or the "constant of variation"
then plug k's value back to

now.... what is "p" when q = 5? well, just set "q" to 5 on the right-hand-side, and simplify, to see what "p" is
Answer:
I think it is 4 white and 3 blue
Step-by-step explanation:
Because they have 3 cups of white then 6 cups of white so i think they multiplied it. Then they have 4 cups of blue then 8 cups of blue. So if you multiply it, it will give you the answer. I hope this is right and it helped and can you mark me brainliest!!!!!!!!
If Erica earned a total of $15450 last year from both the jobs then he earns $2840 from college if she earned 1250 more than four times the amount from college from store.
Given Total amount earned=$15450,Amount earned from store is 1250 more than 4 times earned from college.
Amount from store forms an equation.
let the amount earned from college is x.
According to question:
Amount earned from store=4x+1250
Amount earned from college=x
Total amount earned=4x+1250+x
5x+1250=15450
5x=15450-1250
5x=14200
x=14200/5
x=2840
Put the value of x in 4x+1250 to get amount earned from store=4(2840)+1250=$12610.
Hence the amount earned by Erica from college is $2840.
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Answer:

Step-by-step explanation:
Consider the revenue function given by
. We want to find the values of each of the variables such that the gradient( i.e the first partial derivatives of the function) is 0. Then, we have the following (the explicit calculations of both derivatives are omitted).


From the first equation, we get,
.If we replace that in the second equation, we get

From where we get that
. If we replace that in the first equation, we get

So, the critical point is
. We must check that it is a maximum. To do so, we will use the Hessian criteria. To do so, we must calculate the second derivatives and the crossed derivatives and check if the criteria is fulfilled in order for it to be a maximum. We get that


We have the following matrix,
.
Recall that the Hessian criteria says that, for the point to be a maximum, the determinant of the whole matrix should be positive and the element of the matrix that is in the upper left corner should be negative. Note that the determinant of the matrix is
and that -10<0. Hence, the criteria is fulfilled and the critical point is a maximum