3 years ago

What’s the answer for number 7????????????

Mathematics

1 answer:

kodGreya [7K]3 years ago

6 0

Choice C, because

$\frac{3}{4} = \frac{6}{8}$

So it's less than what she originally had (which isn't possible)

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Evaluate lim x→∞ (3x+1)^(4/x), using l'hospital's rule as needed. show all work using proper notation. as you show your work, if

Alborosie

$\displaystyle\lim_{x\to\inty}(3x+1)^{4/x}=\lim_{x\to\infty}e^{\ln(3x+1)^{4/x}}=e^{\lim\limits_{x\to\infty}\ln(3x+1)^{4/x}}$

$\displaystyle\lim_{x\to\infty}\ln(3x+1)^{4/x}=\lim_{x\to\infty}\frac{4\ln(3x+1)}x\stackrel{\mathrm{LHR}}=\lim_{x\to\infty}\frac{4\frac3{3x+1}}1=\lim_{x\to\infty}\frac{12}{3x+1}=0$

$\implies\displaystyle\lim_{x\to\infty}(3x+1)^{4/x}=e^0=1$

4 0

3 years ago

PLZ ANSWER QUICK HURRY PLZZZ WILL MARK BRANIEST

yuradex [85]

Answer:

1=z-9

2=9+z

3=9/z

4=9z

Step-by-step explanation:

7 0

3 years ago