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shutvik [7]
3 years ago
9

Obsessive- compulsive disorder is

Advanced Placement (AP)
1 answer:
zaharov [31]3 years ago
6 0
It is an anxiety disorder, commonly characterised with recurring, obsessive compulsions and habits.
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Plz help quickkkkkkkkk
Whitepunk [10]

Answer:

Second option (B)

Explanation:

The inequality, x ≤ 1,200, can he represented in a number line as follows:

Since x is less "than or equal to 1,200", it means 1,200 is included in the possible values of x. On the number line, the circle of the arrow which starts from 1,200 would be shaded (full). The arrow will point towards our left hand side. This also implies that 1,200 is the maximum possible value of x.

The number line in the second option (B) is the represents the inequality correctly.

8 0
3 years ago
Which of the following represents a conflict between the supremacy clause and the Tenth Amendment?
kobusy [5.1K]

Answer:

A motorist sues the national government for damages when she has an accident driving on a poorly kept federal highway in Montana.

Explanation:

States are supposed to adhere to national guidelines based on the supremacy clause.

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3 years ago
This country is known for having the first nurses and hospitals:
Alex_Xolod [135]
It has to do with Florence Nightingale
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3 years ago
Read 2 more answers
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}

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Piaget believed that in each stage, children
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Are there choices to go with the question?
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