3 years ago

How can I graph that

1 answer:

REY [17]3 years ago

5 0

Slope:2
y intercept: 5
so if your test has like a graph the dot will be at y=5

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Nimfa-mama [501]

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

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

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

$=\displaystyle\left(\int_{-5}^5x^6\,\mathrm dx\right)\left(\int_{-5}^5e^y(25-y^2)\,\mathrm dy\right)=\boxed{\frac{625,000(3+2e^{10})}{7e^5}}$

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lutik1710 [3]

2 x 3 x 2 x 3 units rectangle as shown in the picture would work!

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