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Aleks04 [339]
3 years ago
11

QUICK ILL GIVE BRAINLIEST

Advanced Placement (AP)
2 answers:
Ganezh [65]3 years ago
8 0

both occur at convergent boundaries

VladimirAG [237]3 years ago
7 0

Answer:

a

Explanation:

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Which NIMS management characteristic may include gathering, analyzing, and assessing weather service data from technical special
weqwewe [10]

Answer: A. Information and Intelligence Management

Choices are:

A. Information and Intelligence Management

B. Integrated Communications

C. Incident Facilities and Locations

D. Management by Objectives

The National Incident Management or NIMS is the first-ever national approach to incident management and response developed in March 2004. It applies to all jurisdictional levels.

One characteristic of NIMS is the Information and Intelligence Management which is the process of managing the process of raw intelligence information. The intelligence cycle includes 5 steps:

1) Planning and Direction

2) Collection

3) Processing

4) All source analysis and production

And 5) Dissemination.


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3 years ago
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If u want points here u go
Morgarella [4.7K]

Answer:

Explanation:

Thx dude

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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
Does anyone do parkour here
stiks02 [169]

Answer:

yessir

Explanation:

i am rad

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What types of food, especially for breakfast, keep your energy levels even all day long?
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All of the above (correct me if I’m wrong) :)
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