3 years ago

A triangular prism was sliced parallel to its base. What is the shape of the Cross section shown in the figure?

Mathematics

1 answer:

Nitella [24]3 years ago

8 0

Answer:

it would be a triangle

Step-by-step explanation:

Do good!

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Express the integral as an iterated integral in six different ways, where E is the solid bounded by y=4-x^2-4z^2 and y=0

zmey [24]

Assuming you need the integral expressing the volume of $E$ , the easiest setup is to integrate with respect to $y$ first.

This is done with either

$\displaystyle\iiint_E\mathrm dV=\int_{-2}^2\int_{-2}^2\int_0^{4-x^2-z^2}\mathrm dy\,\mathrm dx\,\mathrm dz$
$\displaystyle\iiint_E\mathrm dV=\int_{-2}^2\int_{-2}^2\int_0^{4-x^2-z^2}\mathrm dy\,\mathrm dz\,\mathrm dx$

Thanks to symmetry, integrating with respect to either $x$ or $z$ first will be nearly identical.

First, with respect to $x$ :

$\displaystyle\iiint_E\mathrm dV=\int_{-2}^2\int_0^4\int_{-\sqrt{4-y-z^2}}^{\sqrt{4-y-z^2}}\mathrm dx\,\mathrm dy\,\mathrm dz$
$\displaystyle\iiint_E\mathrm dV=\int_0^4\int_{-2}^2\int_{-\sqrt{4-y-z^2}}^{\sqrt{4-y-z^2}}\mathrm dx\,\mathrm dz\,\mathrm dy$

Next, with respec to $z$ :

$\displaystyle\iiint_E\mathrm dV=\int_{-2}^2\int_0^4\int_{-\sqrt{4-y-z^2}}^{\sqrt{4-y-x^2}}\mathrm dz\,\mathrm dy\,\mathrm dx$
$\displaystyle\iiint_E\mathrm dV=\int_0^4\int_{-2}^2\int_{-\sqrt{4-y-z^2}}^{\sqrt{4-y-x^2}}\mathrm dz\,\mathrm dx\,\mathrm dy$

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3 years ago

Determine between which two integers the following irrational number lie square root 4 square root 21

Taya2010 [7]

Answer:

Step-by-step explanation:

that what i got

6 0

2 years ago

An apple juice factory produced 444 gallons of apple juice in 8 days. How much apple juice, on average, did the factory produce

mixer [17]

55.5 gallons of apple juice on an average day.