3 years ago

PLEASE HELP ASAP I WILL MARK BRAINLIEST

1 answer:

4 0

Answer:

I think it's parallel to the base through the vertex

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kondaur [170]

Answer:

$\int \dfrac{du}{u}=\log|u|+C$

$\int u^n du=\dfrac{u^{n+1}}{n+1}+C$

$\int \dfrac{du}{u\sqrt{u^2-a^2}}=\dfrac{1}{a}\csc^{-1}(\dfrac{x}{a})+C$

Step-by-step explanation:

$\int \dfrac{du}{u}=\log|u|+C$ $[\because \int \dfrac{dx}{x}=\log |x|+C]$

$\int u^n du=\dfrac{u^{n+1}}{n+1}+C$ $[\because \int x^n dx=\dfrac{x^{n+1}}{n+1}+C]$

$\int \dfrac{du}{u\sqrt{u^2-a^2}}=\dfrac{1}{a}\csc^{-1}(\dfrac{x}{a})+C$ $[\because \dfrac{adx}{x\sqrt{x^2-a^2}}=\csc^{-1}(\dfrac{x}{a})+C]$

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matrenka [14]

Answer:

Step-by-step explanation:

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tester [92]

Answer:

Step-by-step explanation:

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Bumek [7]

I think it is 60 years old?

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Katena32 [7]

Tomorrow night at ten because 48 hours is two days. So two days from last night at ten.

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