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3 years ago

11

Plz help!! Anyone? Would mean a lot

2 answers:

andrezito [222]3 years ago

6 0

12 is A
13 is B
14 is b

omeli [17]3 years ago

4 0

Thank you for changing the prompt now let see

i think the first answer is D.

Im atleast 70% percent sure i about the answer

but i cant see the other question right

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Consider the half of the sphere of radius 2 centered at the origin which lies above the xy-plane (a hemisphere). Suppose the den

Harlamova29_29 [7]

Answer:

25.133 units

Step-by-step explanation:

Since the density ρ = r, our mass is

m = ∫∫∫r³sinθdΦdrdθ. We integrate from θ = 0 to π (since it is a hemisphere), Φ = 0 to 2π and r = 0 to 2 and the maximum values of r = 2 in those directions. So

m =∫∫[∫r³sinθdΦ]drdθ

m = ∫[∫2πr³sinθdθ]dr ∫dФ = 2π

m = ∫2πr³∫sinθdθ]dr

m = 2π∫r³dr ∫sinθdθ = 1

m = 2π × 4 ∫r³dr = 4

m = 8π units

m = 25.133 units

5 0

3 years ago

You are looking at the New York ball drop on New Year’s Eve at a distance of 100 m away from the base of the structure. If the b

Fofino [41]

The question is an illustration of related rates.

The rate of change between you and the ball is 0.01 rad per second

I added an attachment to illustrate the given parameters.

The representations on the attachment are:

$\mathbf{x = 100\ m}$

$\mathbf{\frac{dy}{dt} = 2\ ms^{-1}}$ ---- the rate

$\mathbf{\theta = \frac{\pi}{4}}$

First, we calculate the vertical distance (y) using tangent ratio

$\mathbf{\tan(\theta) = \frac{y}{x}}$

Substitute 100 for x

$\mathbf{y = 100\tan(\theta) }$

$\mathbf{\tan(\theta) = \frac{y}{100}}$

Differentiate both sides with respect to time (t)

$\mathbf{ \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot \frac{dy}{dt}}$

Substitute values for the rates and $\mathbf{\theta }$

$\mathbf{ \sec^2(\pi/4) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

This gives

$\mathbf{ (\sqrt 2)^2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

$\mathbf{ 2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}$

Divide both sides by 2

$\mathbf{ \frac{d\theta}{dt} = \frac{1}{100} }$

$\mathbf{ \frac{d\theta}{dt} = 0.01 }$

Hence, the rate of change between you and the ball is 0.01 rad per second

Read more about related rates at:

brainly.com/question/16981791

8 0

2 years ago

Which applies the power of a product rule to simplify (5t)^3?

frozen [14]

$(5t)^3 = 125t^3$

5 0

3 years ago

Read 2 more answers

20 minute deadline. right answers only.

emmainna [20.7K]

The answer has to be 1

4 0

3 years ago

Rainbow [258]

Answer D is correct

3 0

3 years ago