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eduard
3 years ago
7

Solve for this equation z^2-12z=-13

Mathematics
1 answer:
ASHA 777 [7]3 years ago
5 0

Answer:

z=\sqrt{-1}

Step-by-step explanation:

z^2-12z=-13\\z^2-\frac{-12}{+12} =-13+12\\z^2=-1\\\sqrt{z^{2} } =\sqrt{-1}\\ z=\sqrt{-1}

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Hi guys, can anyone help me with this triple integral? Many thanks:)
Crank

Another triple integral.  We're integrating over the interior of the sphere

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

y^2=4-x^2-z^2

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

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

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.

=(x^2+xy+y^2)z \bigg|_{z=-\sqrt{4-y^2-z^2}}^{z=\sqrt{4-y^2-z^2}}

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  

I gotta go so I'll stop here, sorry.

7 0
3 years ago
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A new outdoor recreation center is being built in Hadleyville. The perimeter of the rectangular playing field is 316 yards. The
algol13

Given :

Perimeter of rectangular filed , P = 316 yards.

The length of the field is 4 yards less than twice its width.

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x = 54 yards.

So, the dimensions are 54 yards and 104.

Hence, this is the required solution.

8 0
3 years ago
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Answer:

l

Step-by-step explanation:

l

4 0
2 years ago
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Alecsey [184]

Answer:

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3 years ago
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