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natita [175]
3 years ago
10

Solve for d: 2d+1/6=6(1/3d+1)

Mathematics
2 answers:
taurus [48]3 years ago
7 0
2d+1/6=2d+6 (I distributed the 6 to what was in the parentheses)
2d-2d=6-1/6 (I subtracted 2d from both sides to get the variables on one side and               1/6 from both sides to get the variables by themselves)
0=35/6 (combined like terms)
0=35/6 is an untrue statement so x doesn't exist

Colt1911 [192]3 years ago
3 0
2d + 1/6 = 6(1/3d + 1)
2d + 1/6 = 2d + 6
2d - 2d = 6 - 1/6
0d = 35/6

Ans: \O
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Hi guys, can anyone help me with this triple integral? Many thanks:)
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x^2+y^2+z^2=2^2

Let's do the outer integral over z.   z stays within the sphere so it goes from -2 to 2.

For the middle integral we have

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x is the inner integral so at this point we conservatively say its zero.  That means y goes from -\sqrt{4-z^2} and +\sqrt{4-z^2}

Similarly the inner integral x goes between \pm-\sqrt{4-y^2-z^2}

So we rewrite the integral

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Let's work on the inner one,

\displaystyle\int_{-\sqrt{4-y^2-z^2}}^{\sqrt{4-y^2-z^2}} (x^2+xy+y^2)dz

There's no z in the integrand, so we treat it as a constant.

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So the middle integral is

\displaystyle\int_{-\sqrt{4-z^2}}^{\sqrt{4-z^2}}2(x^2+xy+y^2)\sqrt{4-y^2-z^2} \ dy  

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sergiy2304 [10]
The answer is 438.6...
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Answer:

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Step-by-step explanation:

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