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irinina [24]
3 years ago
15

Choose all that rearrange the formula correctly.If

Mathematics
1 answer:
Sonbull [250]3 years ago
5 0

Answer:

I sure hope none is an answer because none of these work!

Step-by-step explanation:

C = 59 (F - 32)

(C/59) = F - 32

F = C/59 + 32

So no :(

m = x + y + z3

y = m - x - z3

So no :((

s = r^2 - 1

r = sqrt(s+1)

So no :000

A = 12(a+b)

A = 12a + 12b

12b = A - 12a

b = A/12 - a

So no :00(((

m = x + y^2

y = sqrt (m-x)

AND no :|

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the hot air balloon is hotter than me

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
Find the volume of the region between the planes x plus y plus 2 z equals 2 and 4 x plus 4 y plus z equals 8 in the first octant
Alex787 [66]

Find the intercepts for both planes.

Plane 1, <em>x</em> + <em>y</em> + 2<em>z</em> = 2:

y=z=0\implies x=2\implies (2,0,0)

x=z=0\implies y=2\implies(0,2,0)

x=y=0\implies 2z=2\implies z=1\implies(0,0,1)

Plane 2, 4<em>x</em> + 4<em>y</em> + <em>z</em> = 8:

y=z=0\implies4x=8\implies x=2\implies(2,0,0)

x=z=0\implies4y=8\impliesy=2\implies(0,2,0)

x=y=0\implies z=8\implies(0,0,8)

Both planes share the same <em>x</em>- and <em>y</em>-intercepts, but the second plane's <em>z</em>-intercept is higher, so Plane 2 acts as the roof of the bounded region.

Meanwhile, in the (<em>x</em>, <em>y</em>)-plane where <em>z</em> = 0, we see the bounded region projects down to the triangle in the first quadrant with legs <em>x</em> = 0, <em>y</em> = 0, and <em>x</em> + <em>y</em> = 2, or <em>y</em> = 2 - <em>x</em>.

So the volume of the region is

\displaystyle\int_0^2\int_0^{2-x}\int_{\frac{2-x-y}2}^{8-4x-4y}\mathrm dz\,\mathrm dy\,\mathrm dx=\displaystyle\int_0^2\int_0^{2-x}\left(8-4x-4y-\frac{2-x-y}2\right)\,\mathrm dy\,\mathrm dx

=\displaystyle\int_0^2\int_0^{2-x}\left(7-\frac72(x+y)\right)\,\mathrm dy\,\mathrm dx=\int_0^2\left(7(2-x)-\frac72x(2-x)-\frac74(2-x)^2\right)\,\mathrm dx

=\displaystyle\int_0^2\left(7-7x+\frac74 x^2\right)\,\mathrm dx=\boxed{\frac{14}3}

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Help ASAP! And explain!
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