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kiruha [24]
3 years ago
9

What is –3 2/3 · (–2 1/4 )?

Mathematics
1 answer:
algol [13]3 years ago
6 0
-3\frac{2}{3}\cdot\left(-2\frac{1}4{}\right)=\\--------------------\\negative\ number\ multiply\ by\ negative\ number\ is\ a\ positive\ number\\--------------------\\-3\frac{2}{3}\cdot\left(-2\frac{1}4{}\right)=3\frac{2}{3}\cdot2\frac{1}{4}=\\------------------------------\\

if you want to multiply mixed numbers you need to turn them into improper fractions

3\frac{2}{3}\cdot2\frac{1}{4}=\frac{3\cdot3+2}{3}\cdot\frac{2\cdot4+1}{4}=\frac{9+2}{3}\cdot\frac{8+1}{4}=\frac{11}{3}\cdot\frac{9}{4}=\frac{11}{1}\cdot\frac{3}{4}=\frac{11\cdot3}{1\cdot4}\\\center\boxed{=\frac{33}{4}=8\frac{1}{4}}

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Use the Ratio Test to determine the convergence or divergence of the series. If the Ratio Test is inconclusive, determine the co
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<h2>A. The series CONVERGES</h2>

Step-by-step explanation:

If \sum a_n is a series, for the series to converge/diverge according to ratio test, the following conditions must be met.

\lim_{n \to \infty} |\frac{a_n_+_1}{a_n}| = \rho

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If \rho > 1, the series diverges

If \rho = 1, the test fails.

Given the series \sum\left\ {\infty} \atop {1} \right \frac{n^2}{5^n}

To test for convergence or divergence using ratio test, we will use the condition above.

a_n = \frac{n^2}{5^n} \\a_n_+_1 = \frac{(n+1)^2}{5^{n+1}}

\frac{a_n_+_1}{a_n} =  \frac{{\frac{(n+1)^2}{5^{n+1}}}}{\frac{n^2}{5^n} }\\\\ \frac{a_n_+_1}{a_n} = {{\frac{(n+1)^2}{5^{n+1}} * \frac{5^n}{n^2}\

\frac{a_n_+_1}{a_n} = {{\frac{(n^2+2n+1)}{5^n*5^1}} * \frac{5^n}{n^2}\\

aₙ₊₁/aₙ =

\lim_{n \to \infty} |\frac{ n^2+2n+1}{5n^2}| \\\\Dividing\ through\ by \ n^2\\\\\lim_{n \to \infty} |\frac{ n^2/n^2+2n/n^2+1/n^2}{5n^2/n^2}|\\\\\lim_{n \to \infty} |\frac{1+2/n+1/n^2}{5}|\\\\

note that any constant dividing infinity is equal to zero

|\frac{1+2/\infty+1/\infty^2}{5}|\\\\

\frac{1+0+0}{5}\\ = 1/5

\rho = 1/5

Since The limit of the sequence given is less than 1, hence the series converges.

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Answer:

27.5

Step-by-step explanation:

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