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2 years ago

Will mark as brainliest for correct answer. The picture looks strange for some reason by the way.

Mathematics

1 answer:

alina1380 [7]2 years ago

4 0

The third one because of the equations amplitude.

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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

2 years ago

The general solution to the second-order differential equation y′′+10y=0 is in the form y(x)=c1cosβx+c2sinβx. Find the value of

allsm [11]

Answer:β=√10 or 3.16 (rounded to 2 decimal places)

Step-by-step explanation:

To find the value of β :

we will differentiate the y(x) equation twice to get a second order differential equation.
We compare our second order differential equation with the Second order differential equation specified in the problem to get the value of β

y(x)=c1cosβx+c2sinβx

we use the derivative of a sum rule to differentiate since we have an addition sign in our equation.

Also when differentiating Cosβx and Sinβx we should note that this involves function of a function. so we will differentiate βx in each case and multiply with the differential of c1cosx and c2sinx respectively.

lastly the differential of sinx= cosx and for cosx = -sinx.

Knowing all these we can proceed to solving the problem.

y=c1cosβx+c2sinβx

y'= β×c1×-sinβx+β×c2×cosβx

y'=-c1βsinβx+c2βcosβx

y''=β×-c1β×cosβx + (β×c2β×-sinβx)

y''= -c1β²cosβx -c2β²sinβx

factorize -β²

y''= -β²(c1cosβx +c2sinβx)

y(x)=c1cosβx+c2sinβx

therefore y'' = -β²y

y''+β²y=0

now we compare this with the second order D.E provided in the question

y''+10y=0

this means that β²y=10y

β²=10

B=√10 or 3.16(2 d.p)

5 0

2 years ago

(88÷1×2×5−9)+3×8 what does this make

cupoosta [38]

Answer: 6992

Step-by-step explanation: (88 / 1 = 88) (88 * 2 = 176) (176 * 5 = 880) (880 - 9 = 871) (871 + 3 = 874) (874 * 8 = 6992)

5 0

2 years ago

Given z=f(x,y),x=x(u,v),y=y(u,v), with x(4,1)=5 and y(4,1)=2, calculate zv(4,1) in terms of some of the values given in the tabl

Yuliya22 [10]

$z=f(x(u,v),y(u,v))\implies z_v=f_xx_v+f_yy_v$

$z_v(4,1)=f_x(5)x_v(4,1)+f_y(2)y_v(4,1)$

No table = incomplete answer

3 0

2 years ago

Can any one help me at this q

Lisa [10]

Answer:

-14.4

Step-by-step explanation:

the reason being is because 6 degrees celsius is equivalent to 42.8 degrees 14 degrees celsius is equivalent to 57.2 degrees, that stated subtract subtract 57.2 from 42.8 and get -14.4

7 0

2 years ago