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2 years ago

The oblique prism below has an isosceles right triangle

Mathematics

1 answer:

Zarrin [17]2 years ago

7 0

The volume of the prism, in cubic units is V = 1/2*x³ + x²

The figure and options are in the figure attached

<h3>What is an Oblique Prism ?</h3>

An oblique prism is a polyhedron figure , with a rectangular base and triangular side faces .

It is given that

The oblique prism below has an isosceles right triangle base.

In the figure attached, the oblique prism is shown.

The volume of the prism is given by

V = b*h

h is the height and b is the Area of the base

It is given that the base is an isosceles right triangle, its area is:

Area of a Triangle = (1/2) base * height

here the base and height is x

b = 1/2*x²

The height of the prism is (x + 2).

Then, the volume is:

V = 1/2*x²*(x + 2)

V = 1/2*x³ + x² cu. units

To know more about Oblique Prism

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Calculus A hard problem help please

dexar [7]

∑x = 1 + 2 + 3 + 4 + 5 + 6 = 21
∑y = 8 + 3 + 0 + 1 + 2 + 1 = 15
∑x^2 = 1 + 4 + 9 + 16 + 25 + 36 = 91
∑y^2 = 64 + 9 + 0 + 1 + 4 + 1 = 79
∑xy = 8 + 6 + 0 + 4 + 10 + 6 = 34

r = (n∑xy - ∑x∑y)/(sqrt(n∑x^2 - (∑x)^2)*sqrt(n∑y^2 - (∑y)^2)) = (6(34) - 21(15))/(sqrt(6(91) - (21)^2)*sqrt(6(79) - (15)^2)) = (204 - 315)/(sqrt(546 - 441)*sqrt(474 - 225)) = -111/(sqrt(105)*sqrt(249)) = -111/(10.25*15.78) = -111/161.7 = -0.68

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3 years ago

Read 2 more answers

Find S8 for the geometric series 3 + -6 + 12 + -24 +…

kirill115 [55]

I guess you are asking to find the sum of the first 8 terms. If so, then:
Sum = a₁(1-rⁿ)/(1-r), where a₁ is the 1st term, r=common ratio and n=number of terms:
the 1st term a₁ =3
common ratio r = - 2 (since -6/3 = - 2, and 12/-6 = - 2, etc.)

Sum = 3[(1- (-2)⁸]/(1-2) = 3(1- 256)/(1/2)
Sum = -1530

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3 years ago

Devon wants to memorize 1/2 of his lines for a play by the end of the week.He has already memorized 1/8 of his lines.what fracti

Ket [755]

Answer:

3/8 left

Step-by-step explanation:

I can't draw a model but I can still give an explanation.

1/2 required to memorize - 1/8 already memorized = 3/8 left

Maybe draw a rectangle with 8 divisions and label the situation.

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3 years ago

Divide.<br><br> 6√3 cis(7π/6) ÷ 3√5 cis(π/3)

Nesterboy [21]

$\dfrac{6\sqrt 3~ \text{cis} \left( \dfrac{7 \pi} 6 \right)}{3 \sqrt 5 ~\text{cis} \left( \dfrac{\pi}{3}\right)}\\\\\\=\dfrac{2\sqrt 3 \cdot e^{i\tfrac{7\pi}{6}}}{\sqrt 5\cdot e^{i \tfrac{\pi}{3}}}\\\\\\=\dfrac{2 \sqrt 3}{\sqrt 5} \cdot \left(e^i \right)^{\tfrac{7\pi}{6} - \tfrac{\pi }{3}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~;\left[\text{cis}~ \theta = e^{i \theta}\right]\\\\\\=\dfrac{2 \sqrt{3}}{\sqrt 5} \cdot \left(e^i\right)^{\tfrac{5\pi}6}\\\\\\=\dfrac{2\sqrt{3}}{\sqrt 5}\cdot e^{i\tfrac{5 \pi}6}\\\\\\$

$=\dfrac{2\sqrt{3}}{\sqrt 5} \left[ \cos \left(\dfrac{5\pi}{6}\right) + i \sin \left(\dfrac{5\pi}{6}\right) \right]~~~~~~~~~~~~~~~~~~;\left[e^{i\theta} = \cos \theta + i \sin \theta} \right]\\\\\\=\dfrac{2\sqrt{3}}{\sqrt 5}\left[ \cos \left(2 \times \dfrac{\pi}2 -\dfrac{\pi}{6} \right) + i\sin \left(2 \times \dfrac{\pi}2 -\dfrac{\pi}{6} \right) \right]\\\\\\=\dfrac{2\sqrt{3}}{\sqrt 5} \left[ -\cos \left(\dfrac{\pi}{6}\right) + i \sin \left(\dfrac{\pi}{6}\right) \right]\\\\\\$

$=\dfrac{2\sqrt{3}}{\sqrt 5}\left( -\dfrac {\sqrt{3}}{2} + i \cdot \dfrac 12 \right)\\\\\\=-\dfrac{3}{\sqrt 5} + i \dfrac{\sqrt 3}{\sqrt 5}$

3 0

2 years ago

Simplify: |5-11|<br> 16<br> -16<br> 6<br> -6

Debora [2.8K]

Subtract 11 from 5 to get -6.

Because the equation is in between two vertical lines, this means the absolute value, which is a positive value, so -6 becomes positive 6.

The answer is 6

4 0

3 years ago