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3 years ago

Oh lord help q-q I have like no time left-

Mathematics

1 answer:

galina1969 [7]3 years ago

5 0

Answer:

(Triangular pyramid)

Lateral area = 1650 $m^{2}$

Total surface area = 1860.1 $m^{2}$

(Triangular prism)

Volume = 660 $cm^{3}$

Step-by-step explanation:

<u> For the triangular pyramid</u>

Area of triangle = $\frac{1}{2}$ × base × height

Area of the base = $\frac{1}{2}$ × 22 × 19.1 = 210.1

Area of the sides = $\frac{1}{2}$ × 22 × 50 = 550

Number of sides (excluding base) = 3

∴ Total Area = 210.1 + (550 × 3)

= 210.1 + 1650 = 1860.1 $m^{2}$

Lateral Area = (550 × 3) = 1650 $m^{2}$

<u>For the triangular prism</u>

Area of the triangle = $\frac{1}{2}$ × 12 × 5 = 30

Volume = Area x Height

Volume = 30 × 22 = 660 $cm^{3}$

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Crank

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

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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 $\pm-\sqrt{4-y^2-z^2}$

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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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3 years ago

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Answer:

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