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ikadub [295]
3 years ago
15

The weight of an object on Mars varies directly with its weight on Earth an object that weighs 50 pounds on Mars weighs 150 lbs

on Earth of an object weighs 120 pounds on Earth write and solve a direct variation equation to find how much in object would weigh on
Mathematics
1 answer:
Scorpion4ik [409]3 years ago
3 0
So it would weight 20 pound because they removed th 100 so that means any thing the. is in 100 like 140 would be 40
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Mr. Cromleigh and his mom stayed at a hotel while in Vegas. If the room was 12ft by 8ft by 9ft, what is the volume of the room?
Komok [63]

Answer:

864 ft³

Step-by-step explanation:

Volume = Length * Width * Height

Substitute

Volume = 12 * 8 * 9

Volume = 864 ft³

Hope this helps :)

7 0
4 years ago
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If f(x) = 2x - 7, what is the equation for f(x)<br> URGENT PLEASE ANSWER!<br> *see attachment*
Serhud [2]

Answer:

a

Step-by-step explanation:

inverse ----> y = x

x = 2y - 7

2y = x + 7

y = (x+7)/2

8 0
4 years ago
The tax rate is 3.9%<br> What is the tax on 42$? <br> Round to the nearest hundredth
mafiozo [28]
To find the total price, we use this equation:
42 + 0.039(42)
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1.039(42)
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7 0
3 years ago
Mrs.Jacinto asked carpenter to construct a rectangular bulletin board for her classroom. She told the Carpenter that the board's
Misha Larkins [42]

Answer:

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Step-by-step explanation:

3 0
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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