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valentina_108 [34]
3 years ago
9

An argument against cost/benefit analysis is that it requires _____.

Advanced Placement (AP)
2 answers:
puteri [66]3 years ago
8 0

Answer:

the answer is c.

Explanation:

8_murik_8 [283]3 years ago
6 0

The correct answer is placing monetary values on everything

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HEYY I NEED HELP WITH AN AP CALC ASSIGNMENT ASAP
Vladimir79 [104]

CALCULATOR PART

1. The area of R + S is unsigned, meaning you want to find

\displaystyle\int_a^b\left|f(x)-g(x)\right|\,\mathrm dx

where [a,b] is the interval between the leftmost and rightmost intersections of f and g.

First use your calculator to find these intersections:

\cos x=\dfrac{x+1}3\implies x\approx-3.64,x\approx-1.86,x\approx0.889

so that a=-3.64 and b=0.889. Now compute the integral using your calculator:

\displaystyle\int_a^b\left|f(x)-g(x)|\,\mathrm dx\approx1.662

2. The volume, using the washer method, is given by the integral

\displaystyle\pi\int_{-1.86}^{0.889}(|2-g(x)|^2-|2-f(x)|^2)\,\mathrm dx\approx12.078

3. A circle of radius r has area \pi r^2; a semicircle with the same radius thus has area \frac{\pi r^2}2. Each cross section of this solid is a semicircle whose diameter is the vertical distance between f(x) and g(x), or |f(x)-g(x)|. In terms of the diameter d=2r, the area of each semicircle would be \frac{\pi d^2}8. Then the volume of the solid is

\displaystyle\frac\pi8\int_{-3.64}^{-1.86}|f(x)-g(x)|^2\,\mathrm dx\approx0.0425

NON-CALCULATOR PART

4. The mean value theorem says that for a function F continuous on an interval [a,b] and differentiable on (a,b), there is some c\in(a,b) such that

F'(c)=\dfrac{F(b)-F(a)}{b-a}

If this F happens to be an antiderivative of f, then we end up with

f(c)=\displaystyle\frac1{b-a}\int_a^bf(x)\,\mathrm dx

\cos x is continuous and differentiable everywhere, so the MVT applies. We have F'(x)=f(x)=\cos x, so the MVT tells us there is some c\in[0,\pi such that

\cos c=\dfrac{\sin\pi-\sin0}{\pi-0}=0

That is, the average value of f(x) on [0,\pi] is 0. The MVT says there is some c in the interval such that the function takes on the average value itself; this happens for c=\frac\pi2.

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