3 years ago

Look at my link please.

Mathematics

1 answer:

il63 [147K]3 years ago

6 0

Answer:

m^2 + 2n-17

Step-by-step explanation:

sorry is it too late

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Nimfa-mama [501]

Nothing crazy here, just a matter of figuring out the limits of integration.

$\displaystyle\iiint_Ex^6e^y\,\mathrm dV=\int_{-5}^5\int_{-5}^5\int_0^{25-y^2}x^6e^y\,\mathrm dz\,\mathrm dy\,\mathrm dx$

$=\displaystyle\int_{-5}^{-5}\int_{-5}^5x^6e^y(25-y^2)\,\mathrm dy\,\mathrm dx$

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goldenfox [79]

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3 years ago

Does the graph represent a function?

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