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Sav [38]
3 years ago
12

At the city museum, child admission is

Mathematics
1 answer:
Irina18 [472]3 years ago
5 0
You have to multiply ......... and use 186 and divide. i think.
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Determine whether the integral is convergent or divergent. ∫[infinity] 2 e^−1/x / x^2 dx : O Convergent O divergent If it is con
monitta

Let f(x)=e^{-1/x}. Then f'(x)=\frac1{x^2}e^{-1/x}>0 for all x\ge2, so f is strictly increasing. As x\to\infty, e^{-1/x}\to e^0=1, so f is bounded above by 1. This is to say,

e^{-1/x}

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We have, by setting y=-\frac1x,

\displaystyle\int_2^\infty\frac{e^{-1/x}}{x^2}\,\mathrm dx=\int_{-1/2}^0e^y\,\mathrm dy=e^0-e^{-1/2}=1-\frac1{\sqrt e}

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3 years ago
−|−37| = 37 true or false
Anvisha [2.4K]

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Step-by-step explanation:

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