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2 years ago

7

What is the domain of the function

1 answer:

mixas84 [53]2 years ago

3 0

Answer:

(a) -∞ < x < ∞

Step-by-step explanation:

Unlike the square root function, the cube root function is defined for all values of its argument. Here, x can take on any value and the function will be defined for that value.

-∞ < x < ∞

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Find the integral <br> ∫√(9+x)/(9-x)

densk [106]

I suppose you mean

$\displaystyle \int \frac{\sqrt{9+x}}{9-x} \, dx$

Substitute y = √(9 + x). Solving for x gives x = y² - 9, so that 9 - x = 18 - y², and we have differential dx = 2y dy. Replacing everything in the integral gives

$\displaystyle \int \frac{2y^2}{18 - y^2} \, dy$

Simplify the integrand by dividing:

$\dfrac{2y^2}{18 - y^2} = -2 + \dfrac{36}{18 - y^2}$

$\implies \displaystyle \int \left(\frac{36}{18-y^2} - 2\right) \, dy$

For the first term of this new integral, we have the partial fraction expansion

$\dfrac1{18 - y^2} = \dfrac1{\sqrt{72}} \left(\dfrac1{\sqrt{18}-y} + \dfrac1{\sqrt{18}+y}\right)$

$\implies \displaystyle \frac{36}{\sqrt{72}} \int \left(\frac1{\sqrt{18}-y} + \frac1{\sqrt{18}+y}\right) \, dy - 2 \int dy$

The rest is trivial:

$\displaystyle \sqrt{18} \int \left(\frac1{\sqrt{18}-y} + \frac1{\sqrt{18}+y}\right) \, dy - 2 \int dy$

$= \displaystyle \sqrt{18} \left(\ln\left|\sqrt{18}+y\right| - \ln\left|\sqrt{18}-y\right|\right) - 2y + C$

$= \displaystyle \sqrt{18} \ln\left|\frac{\sqrt{18}+y}{\sqrt{18}-y}\right| - 2y + C$

$= \boxed{\displaystyle \sqrt{18} \ln\left|\frac{\sqrt{18}+\sqrt{9+x}}{\sqrt{18}-\sqrt{9+x}}\right| - 2\sqrt{9+x} + C}$

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2 years ago

77 meter=_____dm ? <br><br> Can someone help ?

MariettaO [177]

Answer:

<h2>770 dm</h2>

Step-by-step explanation:

1 meter = 10 decimeter. Deci means ten.

77 meters = 770 dm

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3 years ago

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Step-by-step explanation:

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2 years ago

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amm1812

Answer:

It will be neither. Also, thanks for the points friend

Step-by-step explanation:

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The correct answer would be A.

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