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Likurg_2 [28]
3 years ago
7

Casey has an open book test in his physics class tomorrow and will need to use many formulas. Which of the following is the best

strategy Casey could use to organize his formulas? a. Highlight the formulas in his school’s textbook b. Write the formulas on his hand for easy access c. List all of the formulas on a separate sheet of paper d. Indicate where formulas are located in the book using sticky notes
Advanced Placement (AP)
2 answers:
IRINA_888 [86]3 years ago
3 0
C i think is thr answer
Margaret [11]3 years ago
3 0

c.

List all of the formulas on a separate sheet of paper is the answer.

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When working with HEVs and EVs, you must create a _______ perimeter or buffer zone around the work area that does not have any m
Svetradugi [14.3K]

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36-45 inches

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The regulation for all forms of these types of objects with dangerous voltage and the potential to discharge is generally 36-45 inches, in which a perimeter is to be formed with appropriate precautions and markings with NON conductive materials. . .you know, so electricity doesn't discharge, take the nearest conductor, rebound & amplify, and strike the nearest guy dead.

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8 0
3 years ago
AP CALC QUESTION!! WILL MARK BRAINLIEST (TWO QUESTIONS)
kodGreya [7K]

1. Using washers, the volume is given by the integral

\displaystyle\pi\int_1^{e^2}((2+2)^2-(\ln x+2)^2)\,\mathrm dx

=\boxed{\displaystyle\pi\int_1^{e^2}(20+4\ln x-(\ln x)^2)\,\mathrm dx}

We're using washers whose centers depend on the value of x, hence we integrate with respect to

2. The area of the given region is given by the integral

\displaystyle\int_0^2\sin^{-1}\frac x2\,\mathrm dx

To compute the integral, first consider the substitution u=\frac x2, or 2u=x so that 2\,\mathrm du=\mathrm dx. Then x\to0\implies u\to0 and x\to2\implies u\to1, so the integral is equivalently

\displaystyle2\int_0^1\sin^{-1}u\,\mathrm du

Integrate by parts, taking

f=\sin^{-1}u\implies\mathrm df=\dfrac{\mathrm du}{\sqrt{1-u^2}}

\mathrm dg=\mathrm du\implies g=u

so that

\displaystyle2\int_0^1\sin^{-1}u\,\mathrm du=2\left(u\sin^{-1}u\bigg|_0^1-\int_0^1\frac u{\sqrt{1-u^2}}\,\mathrm du\right)

\sin^{-1}0=0 and \sin^{-1}1=\frac\pi2, so the area is

\displaystyle\pi-2\int_0^1\frac u{\sqrt{1-u^2}}\,\mathrm du

For the remaining integral, substitute w=1-u^2, so that \mathrm dw=-2u\,\mathrm du. Then u\to0\implies w\to1 and u\to1\implies w\to0:

\displaystyle\pi-\int_0^1\frac{\mathrm dw}{\sqrt w}

(notice that the integral is improper)

\displaystyle\pi-\lim_{t\to0^+}2\sqrt w\bigg|_t^1

\displaystyle\pi-2\left(1-\lim_{t\to0^+}\sqrt t\right)=\boxed{\pi-2}

8 0
3 years ago
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