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Oxana [17]
3 years ago
9

A triangular prism has a volume of 2040 units. The dimensions of the base of the prism are shown below.

Mathematics
1 answer:
Bad White [126]3 years ago
7 0

Answer:

Height = 21.25 units

Step-by-step explanation:

Triangular prism (with triangular base at bottom) will have volume formula as:

Triangular Prism Volume = area of base * height

We need to find area of base (which is the triangle shown).

We can use heron's formula which is:

A=\sqrt{p(p-a)(p-b)(p-c)}

Where

A is area

p is HALF of the perimeter (sum of all sides)

a,b,c are the three sides

So p = (12+16+20)/2=24

Now,

A=\sqrt{p(p-a)(p-b)(p-c)}\\A=\sqrt{24(24-20)(24-16)(24-12)}\\A=\sqrt{(24)(4)(8)(12)} \\A=\sqrt{9216} \\A=96

Thus,

Volume = 96 * height

2040 = 96 * height

height = 2040/96 = 21.25 units

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A container in the shape of a pyramid has a rectangular base with dimensions of 5 inches by 15 inches. The height of the rectang
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Step-by-step explanation:

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The that in one hour it travels 4 miles, and in 2 hours it travels 8 miles have been given so that someone can easily confuse the speed of electric skateboard as a discrete data, but in actual speed of skateboard is a continuous data.

8 0
3 years ago
WHO CAN solve it Please !
Mariulka [41]

Answer:

a) True

<em>      </em>\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha  }   )d\alpha =0<em></em>

Step-by-step explanation:

<u><em>Step(i):-</em></u>

<em>Given that the definite integration</em>

<em>            </em>\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha  }   )d\alpha<em></em>

<em>we know that the trigonometric formula</em>

<em> sin²∝+cos²∝ = 1</em>

<em>            cos²∝ = 1-sin²∝</em>

<u><em>step(ii):-</em></u>

<em>Now the  integration</em>

<em>         </em>\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha  }   )d\alpha = \int\limits^\pi _0 {(\sqrt{cos^{2} \alpha } } \, )d\alpha<em></em>

<em>                                      = </em>\int\limits^\pi _0 {cos\alpha } \, dx<em></em>

<em>Now, Integrating </em>

<em>                                  </em>= ( sin\alpha )_{0} ^{\pi }<em></em>

<em>                                = sin π - sin 0</em>

<em>                               = 0-0</em>

<em>                              = 0</em>

<u><em>Final answer:-</em></u>

<em>      </em>\int\limits^\pi _ {0} \,(\sqrt{1-sin^{2}\alpha  }   )d\alpha =0<em></em>

<em></em>

<em></em>

5 0
3 years ago
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