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allochka39001 [22]
4 years ago
13

A ball is thrown into the air with a initial velocity of 25 meters per second.The function h(t)+-4.9t^2+25t+6 represents the hei

ght of the t in seconds. ball above the ground in meters,with respect to time.at h(3) then the height of the ball would be
Mathematics
1 answer:
Rainbow [258]4 years ago
6 0

Answer:

36.9 m

Step-by-step explanation:

h(t) = -4.9t² + 25t + 6

At t = 3:

h(3) = -4.9(3)² + 25(3) + 6

h(3) = -44.1 + 75 + 6

h(3) = 36.9

After 3 seconds, the ball will have a height of 36.9 meters.

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Katyanochek1 [597]
<span>1+4=5 2+5=12 3+6=21 8+11=40

Explanation: 1+4=5; 2+5=7+5(the previous result)=12; 3+6=9+12=21;
8+11=19+21=40.

</span>
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3 years ago
The answer I’m soo confused
Arisa [49]

Answer:

The answer would be H for 30

Step-by-step explanation:

7 0
3 years ago
Choose which one you believe doesn't belong.
vodka [1.7K]

Answer:

65

All the other numbers if you add up their digits it is 8.

1 + 7 = 8

2 + 6 = 8

4 + 4 = 8

6 + 5 = 11

26

Is only number divisible by 13.

44 is only divisible by 11

Step-by-step explanation:

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3 years ago
What is the distance between (3,2) and (3,-8)
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The answer is (0,-10)
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3 years ago
Find the surface area of x^2+y^2+z^2=9 that lies above the cone z= sqrt(x^@+y^2)
Mashcka [7]
The cone equation gives

z=\sqrt{x^2+y^2}\implies z^2=x^2+y^2

which means that the intersection of the cone and sphere occurs at

x^2+y^2+(x^2+y^2)=9\implies x^2+y^2=\dfrac92

i.e. along the vertical cylinder of radius \dfrac3{\sqrt2} when z=\dfrac3{\sqrt2}.

We can parameterize the spherical cap in spherical coordinates by

\mathbf r(\theta,\varphi)=\langle3\cos\theta\sin\varphi,3\sin\theta\sin\varphi,3\cos\varphi\right\rangle

where 0\le\theta\le2\pi and 0\le\varphi\le\dfrac\pi4, which follows from the fact that the radius of the sphere is 3 and the height at which the sphere and cone intersect is \dfrac3{\sqrt2}. So the angle between the vertical line through the origin and any line through the origin normal to the sphere along the cone's surface is

\varphi=\cos^{-1}\left(\dfrac{\frac3{\sqrt2}}3\right)=\cos^{-1}\left(\dfrac1{\sqrt2}\right)=\dfrac\pi4

Now the surface area of the cap is given by the surface integral,

\displaystyle\iint_{\text{cap}}\mathrm dS=\int_{\theta=0}^{\theta=2\pi}\int_{\varphi=0}^{\varphi=\pi/4}\|\mathbf r_u\times\mathbf r_v\|\,\mathrm dv\,\mathrm du
=\displaystyle\int_{u=0}^{u=2\pi}\int_{\varphi=0}^{\varphi=\pi/4}9\sin v\,\mathrm dv\,\mathrm du
=-18\pi\cos v\bigg|_{v=0}^{v=\pi/4}
=18\pi\left(1-\dfrac1{\sqrt2}\right)
=9(2-\sqrt2)\pi
3 0
3 years ago
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