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likoan [24]
2 years ago
8

I NEED HELP ASAP!!!PLEASE EXPLAIN THE ANSWER

Mathematics
1 answer:
Dimas [21]2 years ago
5 0

Answer:

Below

Step-by-step explanation:

To find surface area, you can just find the area of each side and add them up!

11 ft x 9 ft x 2 = 198 ft^2 (this is the surface area of the front and back)

12 ft x 9 ft x 2 = 216 ft^2 (this is the surface area of the sides)

12 ft x 11 ft x 2 = 264 ft^2 (this is the surface area of the top and bottom)

Surface area = 198 + 216 + 264

                      = 678 ft^2

Hope this helps! Best of luck <3

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Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant below the line y=5 and betw
vfiekz [6]

First, complete the square in the equation for the second circle to determine its center and radius:

<em>x</em> ² - 10<em>x</em> + <em>y</em> ² = 0

<em>x</em> ² - 10<em>x</em> + 25 + <em>y </em>² = 25

(<em>x</em> - 5)² + <em>y</em> ² = 5²

So the second circle is centered at (5, 0) with radius 5, while the first circle is centered at the origin with radius √100 = 10.

Now convert each equation into polar coordinates, using

<em>x</em> = <em>r</em> cos(<em>θ</em>)

<em>y</em> = <em>r</em> sin(<em>θ</em>)

Then

<em>x</em> ² + <em>y</em> ² = 100   →   <em>r </em>² = 100   →   <em>r</em> = 10

<em>x</em> ² - 10<em>x</em> + <em>y</em> ² = 0   →   <em>r </em>² - 10 <em>r</em> cos(<em>θ</em>) = 0   →   <em>r</em> = 10 cos(<em>θ</em>)

<em>y</em> = 5   →   <em>r</em> sin(<em>θ</em>) = 5   →   <em>r</em> = 5 csc(<em>θ</em>)

See the attached graphic for a plot of the circles and line as well as the bounded region between them. The second circle is tangent to the larger one at the point (10, 0), and is also tangent to <em>y</em> = 5 at the point (0, 5).

Split up the region at 3 angles <em>θ</em>₁, <em>θ</em>₂, and <em>θ</em>₃, which denote the angles <em>θ</em> at which the curves intersect. They are

<em>θ</em>₁ = 0 … … … by solving 10 = 10 cos(<em>θ</em>)

<em>θ</em>₂ = <em>π</em>/6 … … by solving 10 = 5 csc(<em>θ</em>)

<em>θ</em>₃ = 5<em>π</em>/6  … the second solution to 10 = 5 csc(<em>θ</em>)

Then the area of the region is given by a sum of integrals:

\displaystyle \frac12\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}\left(10^2-(10\cos(\theta))^2\right)\,\mathrm d\theta+\int_{\frac\pi6}^{\frac{5\pi}6}\left((5\csc(\theta))^2-(10\cos(\theta))^2\right)\,\mathrm d\theta\right)

=\displaystyle 50\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\} \sin^2(\theta)\,\mathrm d\theta+\frac12\int_{\frac\pi6}^{\frac{5\pi}6}\left(25\csc^2(\theta) - 100\cos^2(\theta)\right)\,\mathrm d\theta

To compute the integrals, use the following identities:

sin²(<em>θ</em>) = (1 - cos(2<em>θ</em>)) / 2

cos²(<em>θ</em>) = (1 + cos(2<em>θ</em>)) / 2

and recall that

d(cot(<em>θ</em>))/d<em>θ</em> = -csc²(<em>θ</em>)

You should end up with an area of

=\displaystyle25\left(\left\{\int_0^{\frac\pi6}+\int_{\frac{5\pi}6}^{2\pi}\right\}(1-\cos(2\theta))\,\mathrm d\theta-\int_{\frac\pi6}^{\frac{5\pi}6}(1+\cos(2\theta))\,\mathrm d\theta\right)+\frac{25}2\int_{\frac\pi6}^{\frac{5\pi}6}\csc^2(\theta)\,\mathrm d\theta

=\boxed{25\sqrt3+\dfrac{125\pi}3}

We can verify this geometrically:

• the area of the larger circle is 100<em>π</em>

• the area of the smaller circle is 25<em>π</em>

• the area of the circular segment, i.e. the part of the larger circle that is bounded below by the line <em>y</em> = 5, has area 100<em>π</em>/3 - 25√3

Hence the area of the region of interest is

100<em>π</em> - 25<em>π</em> - (100<em>π</em>/3 - 25√3) = 125<em>π</em>/3 + 25√3

as expected.

3 0
2 years ago
A cube has a volume of 0.215ft³. What is te length of each side of the cube?
enyata [817]

Answer:

length ≈ 0.599 feet

Step-by-step explanation:

volume = length³

0.215ft³ = length³

cube root both sides to isolate the length

∛(0.215ft³) = length

length ≈ 0.599 feet

6 0
2 years ago
12
AveGali [126]

The rise/run of AC and CE in the similar triangles are the same, the true statement is: B. slope of AC = slope of CE.

<h3>What is the Slope of the Sides of Similar Triangles?</h3>

On a coordinate plane, the corresponding sides of two triangles are always the same because the ratio of the rise over run is always the same.

Triangles ABC and CDE are similar triangles, therefore the rise over run of AC and CE, which is the slope, will be the same.

Thus, slope of AC = slope of CE.

Learn more about slope on:

brainly.com/question/3493733

7 0
2 years ago
What is the slope of the line represented by the equation y=- 1/2x + 1/4
Andreyy89

Answer:

-1/2x

Step-by-step explanation:

-1/2x represents the slope because it shows that the line is going down 1/2 every x. 1/4 would be where the line started or the y-intercept.

8 0
3 years ago
Write the standard equation of the circle if the center is (-1,2), and a point on the circle is (2,6).
ladessa [460]

Answer:

  • (x + 1)² + (y - 2)² = 25

Step-by-step explanation:

<u>The center is given:</u>

  • (-1, 2)

<u>Find the radius, the distance from the center to a point (2, 6)</u>

  • r² = (2 - (-1))² + (6 - 2)² = 3² + 4² = 25
  • r = √25 = 5

<u>Use circle formula:</u>

  • (x - h)² + (y - k)² = r²
  • (x - (-1))² + (y - 2)² = 5²
  • (x + 1)² + (y - 2)² = 25

8 0
3 years ago
Read 2 more answers
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