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3 years ago

I need the work shown not just the answers please

Mathematics

1 answer:

Yuliya22 [10]3 years ago

3 0

1.) To find the surface area, we want to find the area of each side, then add them all up together:
Area of base=lwA=(5)(5)A=25 cm²
Area of triangle=1/2bhA=(1/2)(5)(12.5)A=31.25 cm²
Since the 4 triangles have the same area, we can multiply the area we found by 4 and then add the area of the square:
SA=(4)(31.25) + 25SA=125+25SA=150 cm²

2.) To find the volume use the formula:

V=(1/3)lwh
V=(1/3)(5)(5)(12.5)
V=104.167 cm^3

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Step-by-step explanation:

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3 years ago

Read 2 more answers

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.

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3 years ago

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If you open a window for 6 hours, you would keep 6,987,000 pounds of CO₂ for entering the atmosphere.

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