For the first one its 7/2 and the second one is 7/6
and u cant covert them into a proper fraction you can only covert them into a improper fracion
Answer:
51
Step-by-step explanation:
hope that helps
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).
First you are going to solve for f.
6f = -2
you must get f alone so you divide both sides of the equation by 6 which becomes ....
f = -3
now you add the absolute value sign to -3. absolute value is how far away your answer is from 0. -3 is 3 away from zero so f = 3
Answer:
ln|sec θ + tan θ| + C
Step-by-step explanation:
The integrals of basic trig functions are:
∫ sin θ dθ = -cos θ + C
∫ cos θ dθ = sin θ + C
∫ csc θ dθ = -ln|csc θ + cot θ| + C
∫ sec θ dθ = ln|sec θ + tan θ| + C
∫ tan θ dθ = -ln|cos θ| + C
∫ cot θ dθ = ln|sin θ| + C
The integral of sec θ can be proven by multiplying and dividing by sec θ + tan θ, then using ∫ du/u = ln|u| + C.
∫ sec θ dθ
∫ sec θ (sec θ + tan θ) / (sec θ + tan θ) dθ
∫ (sec² θ + sec θ tan θ) / (sec θ + tan θ) dθ
ln|sec θ + tan θ| + C
