The idea for transitions for an equation like this is
where m = slope
h = how much left or right (note the negative)
k = how much up or down
moving 13 down would mean
k = -13
so
D. would be your answer
Answer:
(A)74
Explanation:
Let her average score on the last two tests = x.
The six scores will now be: 82,78,87,73, x and x.
Since the average score for her first six chemistry tests = 78
We then solve for x.
Her average score for the last two tests is 74.
Answer:
- y = 81-x
- the domain of P(x) is [0, 81]
- P is maximized at (x, y) = (54, 27)
Step-by-step explanation:
<u>Given</u>
- x plus y equals 81
- x and y are non-negative
<u>Find</u>
- P equals x squared y is maximized
<u>Solution</u>
a. Solve x plus y equals 81 for y.
y equals 81 minus x
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b. Substitute the result from part a into the equation P equals x squared y for the variable that is to be maximized.
P equals x squared left parenthesis 81 minus x right parenthesis
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c. Find the domain of the function P found in part b.
left bracket 0 comma 81 right bracket
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d. Find dP/dx. Solve the equation dP/dx = 0.
P = 81x² -x³
dP/dx = 162x -3x² = 3x(54 -x) = 0
The zero product rule tells us the solutions to this equation are x=0 and x=54, the values of x that make the factors be zero. x=0 is an extraneous solution for this problem so ...
P is maximized at (x, y) = (54, 27).
Answer:
the question is not complete
