3 years ago

A three dimensional object has?

Mathematics

1 answer:

Neko [114]3 years ago

5 0

Answer:

Step-by-step explanation:

length

width

height

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Calculate the potential energy associated with 1 m^3 of water at 607 feet tall taking the mass of 1 m^3 of water to be 1000 kg

Lady bird [3.3K]

Answer:

1813.137 KJ

Step-by-step explanation:

potential energy of the body = mgH

where m is mass in Kg , g= 9.81 m/sec^2 and H= height in m

here m= 1000 kg, g= 9.81 m/s^2 and H= 607 feet = 607×0.305= 185.135 m

hence the potential energy p= 1000×9.81×185.135= 1813137.2 J

= 1813.137 KJ

hence the potential energy associated with 1 m^3 of water at 607 feet tall taking the mass as 1000 kg is = 1813.137 KJ

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4 years ago

Which parabola has the graph shown? I think the answer is C. however, I'm not 100% sure.

NemiM [27]

$x=m(y-k)^2+h\\\\(h,\ k)-the\ vertex$

We have the vertex (-4, -1), therefore we have:

$x=(y-(-1))^2-4\ \ \ |+4\\\\(x+4)=(y+1)^2$

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4 years ago

Slope=1÷5; y intercep = -5

sertanlavr [38]

The equation is
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4 years ago

What is the greatest common factor of the terms in the polynomial 4x^4-32x^3-60x^2?

jonny [76]

Answer:

$\mathrm{Factor}\:4x^4-32x^3-60x^2:\quad 4x^2\left(x^2-8x-15\right)\\$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{b+c}=a^ba^c\\x^3=xx^2\\x^4=x^2x^2\\=4x^2x^2-32xx^2-60x^2\\\mathrm{Rewrite\:}60\mathrm{\:as\:}4\cdot \:15\\\mathrm{Rewrite\:}32\mathrm{\:as\:}4\cdot \:8\\=4x^2x^2-4\cdot \:8xx^2-4\cdot \:15x^2\\\mathrm{Factor\:out\:common\:term\:}4x^2\\=4x^2\left(x^2-8x-15\right)$

Hope this helps and have a great day!!!

$Sofia$

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3 years ago

Find the measure of the indicated angel.

schepotkina [342]

Answer:

Step-by-step explanation:

44+32=76

180-76=104

104+36=140

180-140=40

D)40

8 0

3 years ago