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3 years ago

You can directly attribute your promotion to your impressive professional accomplishments

Advanced Placement (AP)

1 answer:

love history [14]3 years ago

6 0

Answer:

True

Explanation:

Interestingly, in the professional work environment, a person's professional accomplishments like how much they have saved a company, the goals he or she has over-archived, etc can directly attribute an individual's promotion.

This is because companies use professional accomplishments to measure their employee's worth.

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What do egypt and china have in common?

iVinArrow [24]

Egypt and China have the following similarities:
Nile and Yellow river
farmers, servants and slaves were used before (maybe even still today)
writting system
traded

8 0

4 years ago

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 $c=\frac\pi2$ .