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AleksAgata [21]
3 years ago
14

Which expression is equivalent to 2−35,? Choose 1 answer:

Mathematics
2 answers:
sleet_krkn [62]3 years ago
7 0

Answer:

The answer is option A.

Step-by-step explanation:

2 - 35 can be written as 2 + (-35) since in 2+( -35) when the bracket is removed it becomes 2 - 35.

Hope this helps

romanna [79]3 years ago
4 0

Answer:

Your correct answer is option a. 2 + (-35)

Step-by-step explanation:

When you are finding which equation is equivalent to the one you have, the best thing to find is what your equation's answer is.

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Write an equation for when slope =-3 and y-intercept = 36
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Answer:

-3x+36

Step-by-step explanation:

the slope is -3, so just add an x.

the y intercept is 36

put them a together and you get -3x+36

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How many faces doe sthe following shape have?
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The shape consists of 5 faces. One on the front, one on the back, one on the left of the shape, the other on the right, and of course the one on the bottom :)
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Which expression shows 6a + 8b + 10 written as the product of the GCF times a sum?
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3 years ago
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A. 620 is a multiple of 4
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Find the area of the surface. the part of the plane 4x + 4y + z = 16 that lies inside the cylinder x2 + y2 = 9
MAVERICK [17]
As a first step, convert to cylindrical coordinates. The equation x^2+y^2=9=3^2 is a clue to set

x=3r\cos\theta

y=3r\sin\theta

then have 0\le r\le1 and 0\le\theta\le2\pi. Meanwhile, the plane equation tells you to use

4x+4y+z=16\implies z=16-12r\cos\theta-12r\sin\theta

So we parameterize the part of the plane within the cylinder with the vector-valued function

\mathbf s(r,\theta)=(3r\cos\theta,3r\sin\theta,15-12r\cos\theta-12r\sin\theta)

The surface integral giving the area of the surface is


\displaystyle\iint_{\mathcal S}\mathrm dS

where \mathcal S denotes the surface in question, and the surface element \mathrm dS is

\mathrm dS=\left\|\dfrac{\partial\mathbf s}{\partial r}\times\dfrac{\partial\mathbf s}{\partial\theta}\right\|\,\mathrm dr\,\mathrm d\theta

We have

\dfrac{\partial\mathbf s}{\partial r}=(3\cos\theta,3\sin\theta,-12(\cos\theta+\sin\theta))

\dfrac{\partial\mathbf s}{\partial\theta}=(-3r\sin\theta,3r\cos\theta,12r(\sin\theta-\cos\theta))

\implies\mathrm dS=\|(36r,36r,9r\|\,\mathrm dr\,\mathrm d\theta=9\sqrt{33}r\,\mathrm dr\,\mathrm d\theta

So the area is

\displaystyle\iint_{\mathcal S}\mathrm dS=9\sqrt{33}\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1}r\,\mathrm dr\,\mathrm d\theta=18\sqrt{33}\,\pi
4 0
3 years ago
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