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2 years ago

Solve the system: x+4y=1 3x+5y=10

Mathematics

1 answer:

never [62]2 years ago

4 0

Answer:

$x=6.25\\y=-1.25$

Step-by-step explanation:

$3x+5y=10\\x+3y=1\\\\$

Second equation times 3:

$3x+9y=3$

1st equation minus equation above:

$-4y=7\\y=-7/4$

Insert to equation:

$3x+\frac{-35}{4}=10\\12x-35=40\\12x=75\\x=75/12=6.25$

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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

jeka57 [31]

Answer:

<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.

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

If $\rho$ < 1, the series converges absolutely

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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2 years ago

What polynomial must be added to x^2-2x+6 so that the sum is 3x^2+7x?

slavikrds [6]

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2 years ago

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dolphi86 [110]

First isolate the group with h in it
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3 years ago