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3 years ago

17 more than twice p

Mathematics

1 answer:

lukranit [14]3 years ago

7 0

Answer:

2p+17

Step-by-step explanation:

twice p = p x 2 or 2p

so, 17 more than 2p (meaning 2p is increased by 17) is equal to 2p+17

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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

SCORPION-xisa [38]

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

$\displaystyle\int\sqrt{13+12x-x^2}\,\mathrm dx$

First complete the square:

$13+12x-x^2=49-(6-x)^2=7^2-(6-x)^2$

Now in the integral, substitute

$6-x=7\sin t\implies\mathrm dx=-7\cos t\,\mathrm dt$

so that

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

Under this change of variables, we have

$7^2-(6-x)^2=7^2-7^2\sin^2t=7^2(1-\sin^2t)=7^2\cos^2t$

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(<em>t</em>) > 0, we can further reduce the integrand to

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$\displaystyle-49\int|\cos t|\cos t\,\mathrm dt=-\frac{49}2\int(1+\cos(2t))\,\mathrm dt=-\frac{49}2\left(t+\frac12\sin(2t)\right)+C$

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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Answer:

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Step-by-step explanation:

complementary - add up to 90 degrees

x+5x=90 degrees

6x=90 degrees

x=90/6

x=15

15*5=75

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Answer:

Step-by-step explanation:

2 : 4

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BSE is square.

Let side be s

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Diagonal be d

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