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2 years ago

Please help!! I don't have a whole lot of time!!

Mathematics

1 answer:

Zolol [24]2 years ago

7 0

The midpoint of the line segment is the average of the respective coordinates.

So the x-value of the midpoint if the average of the two individual x-coordinates.

(-3.5 + 4)/2 = 0.5/2 = 0.25

And the y-value of the midpoint if the average of the two individual y-coordinates.

(2 + (-2.5))/2 = -0.5/2 = -0.25

So, the midpoint is (0.25, -0.25) or (1/4, -1/4) or "C".

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Use trigonometric substitution to evaluate the integral 13 + 12x − x2 dx . First, write the expression under the radical in an a

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I don't see a square root sign anywhere, so I'll assume the integral is

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First complete the square:

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Now in the integral, substitute

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so that

$\displaystyle\int\sqrt{13+12x-x^2}\,\mathrm dx=-7\int\sqrt{7^2\cos^2t}\,\cos t\,\mathrm dt=-49\int|\cos t|\cos t\,\mathrm dt$

Under the right conditions, namely that cos(t) > 0, we can further reduce the integrand to

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Expand the sine term as

$\dfrac12\sin(2t)}=\sin t\cos t$

Then

$t=\sin^{-1}\left(\dfrac{6-x}7\right)\implies \sin t=\dfrac{6-x}7$

$t=\sin^{-1}\left(\dfrac{6-x}7\right)\implies \cos t=\sqrt{7^2-(6-x)^2}=\sqrt{13+12x-x^2}$

So the integral is

$\displaystyle-\frac{49}2\left(\sin^{-1}\left(\dfrac{6-x}7\right)+\dfrac{6-x}7\sqrt{13+12x-x^2}\right)+C$

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Please help!! I don't have a whole lot of time!!​

Please help!! I don't have a whole lot of time!!