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2 years ago

Which of the following is(are) the solution(s) to |11x +3 = 6 ?

Mathematics

1 answer:

AnnZ [28]2 years ago

3 0

Hey there!

11x + 3 = 6

SUBTRACT 3 to BOTH SIDES

11x + 3 - 3 = 6 - 3

CANCEL out: 3 - 3 because it gives you 0

KEEP: 6 - 3 because it helps you solve for your x-value

11x = 6 - 3

11x = 3

DIVIDE 11 to BOTH SIDES

11x/11 = 3/11

CANCEL out: 11/11 because that gives you 1

KEEP: 3/11 because it simplifies for the x-value

SOLVE

x = 3/11

Therefore, your answer should bed x = 3/11

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

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Answer:

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step-by-step explanation:

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2 years ago

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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.

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

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

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

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 $\dfrac{\frac{\sqrt3}2}{\frac{7\pi^2}9}=\dfrac{9\sqrt3}{14\pi^2}$ .

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2 years ago

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1 year ago

Read 2 more answers

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