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MissTica
3 years ago
7

What is the area of the figure below??? Giving first answer brainliest as long as they have an explanation! Worth 10pts!!

Mathematics
1 answer:
ANTONII [103]3 years ago
3 0
<h3>Answer:</h3>

28inch^2

<h3>Step-by-step solution:</h3>

Rectangle:

<em>[</em><em>L</em><em>e</em><em>n</em><em>g</em><em>t</em><em>h</em><em> </em><em>x</em><em> </em><em>B</em><em>r</em><em>e</em><em>a</em><em>d</em><em>t</em><em>h</em><em>]</em>

5 + 3 = 8inch

8 x 2.5 = 20inch^2

Right Triangle:

<em>[</em><em>1</em><em>/</em><em>2</em><em> </em><em>x</em><em> </em><em>B</em><em>a</em><em>s</em><em>e</em><em> </em><em>x</em><em> </em><em>H</em><em>e</em><em>i</em><em>g</em><em>h</em><em>t</em><em>]</em>

1/2 x 5 x 2 = 1/2 x 10

= 5inch^2

Left Triangle:

<em>[</em><em>1</em><em>/</em><em>2</em><em> </em><em>x</em><em> </em><em>B</em><em>a</em><em>s</em><em>e</em><em> </em><em>x</em><em> </em><em>H</em><em>e</em><em>i</em><em>g</em><em>h</em><em>t</em><em>]</em>

1/2 x 3 x 2 = 1/2 x 6

= 3inch^2

20 + 5 + 3 = <u>2</u><u>8</u><u>i</u><u>n</u><u>c</u><u>h</u><u>^</u><u>2</u>

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2 years ago
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jeka94

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Step-by-step explanation:

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3 years ago
Read 2 more answers
How do you do this question?
Ksivusya [100]

Answer:

V = (About) 22.2, Graph = First graph/Graph in the attachment

Step-by-step explanation:

Remember that in all these cases, we have a specified method to use, the washer method, disk method, and the cylindrical shell method. Keep in mind that the washer and disk method are one in the same, but I feel that the disk method is better as it avoids splitting the integral into two, and rewriting the curves. Here we will go with the disk method.

\mathrm{V\:=\:\pi \int _a^b\left(r\right)^2dy\:},\\\mathrm{V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy}

The plus 1 in '1 + 2/x' is shifting this graph up from where it is rotating, but the negative 1 is subtracting the area between the y-axis and the shaded region, so that when it's flipped around, it becomes a washer.

V\:=\:\int _1^3\:\pi \left[\left(1+\frac{2}{y}\right)^2-1\right]dy,\\\\\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx\\=\pi \cdot \int _1^3\left(1+\frac{2}{y}\right)^2-1dy\\\\\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx\\= \pi \left(\int _1^3\left(1+\frac{2}{y}\right)^2dy-\int _1^31dy\right)\\\\

\int _1^3\left(1+\frac{2}{y}\right)^2dy=4\ln \left(3\right)+\frac{14}{3}, \int _1^31dy=2\\\\=> \pi \left(4\ln \left(3\right)+\frac{14}{3}-2\right)\\=> \pi \left(4\ln \left(3\right)+\frac{8}{3}\right)

Our exact solution will be V = π(4In(3) + 8/3). In decimal form it will be about 22.2 however. Try both solution if you like, but it would be better to use 22.2. Your graph will just be a plot under the curve y = 2/x, the first graph.

5 0
3 years ago
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