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4 years ago

ANSWER WITH DETAILS PLS

Mathematics

1 answer:

Radda [10]4 years ago

6 0

Answer:

last one is the answer

(x-a)^2 + (y-b)^2 = r^2

where (a ,b) are the coordinates of the centre and r, the radius

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3 years ago

100 + 100 X (58 divided by 1000)

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You flip a coin, spin the spinner, and flip a coin again. Find the probability of flipping heads, spinning an even number, and t

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3 years ago

Find the average value of f(x+y)=sin(x+y) over

Troyanec [42]

For each given region $D$ , the average value of $f$ over $D$ is the integral of $f$ over $D$ divided by the area of $D$ . In both cases, $D$ is a rectangle so the area is trivial.

a. The area of $D$ is $\dfrac\pi3\cdot\dfrac{5\pi}3=\dfrac{5\pi^2}9$ . The integral of $f$ over $D$ is

$\displaystyle\int_0^{5\pi/3}\int_0^{\pi/3}\sin(x+y)\,\mathrm dx\,\mathrm dy$

$\displaystyle=\int_0^{5\pi/3}\left(\cos y-\cos\left(y+\dfrac\pi3\right)\right)\,\mathrm dy$

$=\sin\dfrac{5\pi}3-\sin\left(\dfrac{5\pi}3+\dfrac\pi3\right) - \sin0+\sin\dfrac\pi3=0$

So the average value of $f$ over this region is 0.

b. The area of $D$ is $\dfrac{2\pi}3\cdot\dfrac{7\pi}6=\dfrac{7\pi^2}9$ . The integral of $f$ is

$\displaystyle\int_0^{7\pi/6}\int_0^{2\pi/3}\sin(x+y)\,\mathrm dx\,\mathrm dy$

$=\displaystyle\int_0^{7\pi/6}\left(\cos y-\cos\left(y+\frac{2\pi}3\right)\right)\,\mathrm dy$

$=\sin\dfrac{7\pi}6-\sin\left(\dfrac{7\pi}6+\dfrac{2\pi}3\right) - \sin0+\sin\dfrac{2\pi}3=\dfrac{\sqrt3}2$

The average vale of $f$ over this region is then $\dfrac{\frac{\sqrt3}2}{\frac{7\pi^2}9}=\dfrac{9\sqrt3}{14\pi^2}$ .

4 0

3 years ago

*ANSWER WITH DETAILS PLS*

ANSWER WITH DETAILS PLS