What’s the question???...
C² = a² + b² - (2ab * cosC)
<span>c² = 10² + 23² - (2 * 10 * 23 * cos95) </span>
<span>c² = 100 + 529 - (460 * -.08715) </span>
<span>c² = 629 - (-40.1) </span>
<span>c² = 669.1 </span>
<span>c = 25.87 </span>
<span>(Sin C) / C = (Sin A) / A </span>
<span>(Sin 95) / 25.87 = SinA / 10, Remember 0 < A < 85 </span>
<span>(10 * Sin95) / 25.87 = Sin A </span>
<span>A = arcsin ((10 * sin95) / 25.87) </span>
<span>A = 22.65º </span>
<span>B = 180 - A - C </span>
<span>B = 180 - 95 - 22.65 </span>
<span>B = 62.35º </span>
<span>I hope this helps. Have a good day.</span>
Assuming the equation is x^2 + y^2 = 1, then that's a circle, with radius 1, centered on the origin [0,0].
So there are two tangents at x = 0. They are y = 1, and y = -1 (horizontal lines).
There is one tangent at x = 1. It is x = 1 (a vertical line).
There is no tangent at x = 35, because the original equation has no solution at x = 35.
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:
root 16
Step-by-step explanation:
