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Mila [183]
3 years ago
6

34% x 79% pls help ill give brainliest

Mathematics
2 answers:
Leya [2.2K]3 years ago
7 0

Answer:

0.2686

Step-by-step explanation:

It's basically just 34 times 79 lol

s2008m [1.1K]3 years ago
4 0

Answer:

0.2686

Step-by-step explanation:

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Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→[infinity]
Fynjy0 [20]

Answer:

\lim_{x\rightarrow \infty}(\frac{5x-3}{5x+2})^{5x+1}=e^{-5}

Step-by-step explanation:

We are given that

\lim_{x\rightarrow \infty}(\frac{5x-3}{5x+2})^{5x+1}

When substitute limit x tends to infinity then it is 1^{\infty} which is indeterminant form

Wheny=\lim_{x\rightarrow \infty} f(x)^{g(x))

and 1^{\infty}

Then use y=e^{\lim_{x\rightarrow \infty} g(x)(f(x)-1)}

We have g(x)=5x+1 and f(x)=\frac{5x-3}{5x+2}

Substitute the values in the formula

y=e^{\lim_{x\rightarrow \infty}(5x+1)(\frac{5x-3}{5x+2}-1)}

y=e^{\lim_{x\rightarrow \infty}(\frac{-5(5x+1)}{5x+2})}

y=e^{\lim_{x\rightarrow \infty}(\frac{-5(1+\frac{1}{5x})}{1+\frac{2}{5x}})}

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\frac{1}{\infty}=0

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3 years ago
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