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2 years ago

Can someone do my hw for me

Mathematics

1 answer:

IrinaVladis [17]2 years ago

6 0

Answer:

depends on what the homework is

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Find the area of the region that lies inside the first curve and outside the second curve.

marishachu [46]

Answer:

Step-by-step explanation:

From the given information:

r = 10 cos( θ)

r = 5

We are to find the the area of the region that lies inside the first curve and outside the second curve.

The first thing we need to do is to determine the intersection of the points in these two curves.

To do that :

let equate the two parameters together

So;

10 cos( θ) = 5

cos( θ) = $\dfrac{1}{2}$

$\theta = -\dfrac{\pi}{3}, \ \ \dfrac{\pi}{3}$

Now, the area of the region that lies inside the first curve and outside the second curve can be determined by finding the integral . i.e

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} (10 \ cos \ \theta)^2 d \theta - \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ 5^2 d \theta$

$A = \dfrac{1}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} 100 \ cos^2 \ \theta d \theta - \dfrac{25}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \ \ d \theta$

$A = 50 \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} \dfrac{cos \ 2 \theta +1}{2} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}}$

$A =\dfrac{ 50}{2} \int \limits^{\dfrac{\pi}{3}}_{-\dfrac{\pi}{3}} \begin {pmatrix} {cos \ 2 \theta +1} \end {pmatrix} \ \ d \theta - \dfrac{25}{2} \begin {bmatrix} \dfrac{\pi}{3} - (- \dfrac{\pi}{3} )\end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin2 \theta }{2} + \theta \end {bmatrix}^{\dfrac{\pi}{3}}_{\dfrac{\pi}{3}} \ \ - \dfrac{25}{2} \begin {bmatrix} \dfrac{2 \pi}{3} \end {bmatrix}$

$A =25 \begin {bmatrix} \dfrac{sin (\dfrac{2 \pi}{3} )}{2}+\dfrac{\pi}{3} - \dfrac{ sin (\dfrac{-2\pi}{3}) }{2}-(-\dfrac{\pi}{3}) \end {bmatrix} - \dfrac{25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\dfrac{\sqrt{3}}{2} }{2} +\dfrac{\pi}{3} + \dfrac{\dfrac{\sqrt{3}}{2} }{2} + \dfrac{\pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = 25 \begin{bmatrix} \dfrac{\sqrt{3}}{2 } +\dfrac{2 \pi}{3} \end {bmatrix}- \dfrac{ 25 \pi}{3}$

$A = \dfrac{25 \sqrt{3}}{2 } +\dfrac{25 \pi}{3}$

The diagrammatic expression showing the area of the region that lies inside the first curve and outside the second curve can be seen in the attached file below.

Download docx

7 0

3 years ago

Tyler goes running every 6 days. Steve goes running every 9 days. If they both ran today, in how many days will they run togethe

sergejj [24]

5 runs more. 5 runs more

8 0

3 years ago

Kira John and Ryan sent a total of 112 text messages during the weekend. Kira sent 8 fewer messages than Ryan. John sent 4 times

Softa [21]

Answer:

Kirasent 16 and Ryan sent 96

Step-by-step explanation:

I got this because Kira is x-8 and Ryan is 4x. 4x-x-8=112 is equal to 5x=120. So x=24.Kira has 16 when plugging in x and Ryan has 96 because 4 times 24 is 96.

4 0

3 years ago

How many inches are in 4 meters

Vilka [71]

Answer:

There are 157.48 inches in 4 meters.

5 0

3 years ago

Read 2 more answers

Please answer will give brainly

kotykmax [81]

Answer:

Step-by-step explanation:

The volume of a cube is given by the formula :

a³ (where a is the side length )

So now we have to cube these lengths :

Part A :

(3x²y)³ =

(3x²y)(3x²y)(3x²y) =

(9x^4y²)(3x²y) =

27x^6y³ (This is now fully simplified so our final answer for a)

Part B:

(5y²)³ =

(5y²)(5y²)(5y²) =

(25y^4)(5y²) =

125y^6 (This is now fully simplified so our final answer for b)

Hope this helped and have a good day

5 0

1 year ago