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exis [7]
2 years ago
12

Can y’all give me the area of each figure

Mathematics
1 answer:
KIM [24]2 years ago
5 0

Answer:

1. 208 in^2

Step-by-step explanation:

1. We can break the shape up into a rectangle in the middle and 2 triangles on either side of said rectangle.

The dimensions of the rectangle are 8 in by 20 in, and we only know one leg of the triangle as well as the hypotenuse.

If we know one leg and the hypotenuse we can use the pythagorean theormed to sovle for the other side and get 6 in.

So we have

(8 * 20) + 2((1/2)(6)(8))

160 + 48

208 in^2

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The volume of the region R bounded by the x-axis is: \mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^3 dr d\theta}

<h3>What is the volume of the solid (R) on the X-axis?</h3>

If the axis of revolution is the boundary of the plane region and the cross-sections are parallel to the line of revolution, we may use the polar coordinate approach to calculate the volume of the solid.

From the given graph:

The given straight line passes through two points (0,0) and (2,8). Thus, the equation of the straight line becomes:

\mathbf{y-y_1 = \dfrac{y_2-y_1}{x_2-x_1}(x-x_1)}

here:

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Suppose we assign (x₁, y₁) = (0, 0) and (x₂, y₂) = (2, 8)  from the graph, we have:

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Now, our region bounded by the three lines are:

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Similarly, the change in polar coordinates is:

  • x = rcosθ,
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where;

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Therefore;

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Then:

\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^2 (rdr d\theta )}

\mathbf{\iint_R(x^2+y^2)dA = \int ^{tan^{-1}(4)}_{0} \int^{\frac{2}{cos \theta}}_{0} \ r^3 dr d\theta}

Learn more about the determining the volume of solids bounded by region R here:

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Are the triangle below similar
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