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Setler [38]
3 years ago
10

What is the solution to the system of equations? y = A system of equations. Y equals StartFraction 2 over 3 EndFraction x plus 3

. X equals negative 2.X + 3 x = –2
Mathematics
2 answers:
Sauron [17]3 years ago
5 0

Answer:

The solution to the given system of equations is (-2,\frac{5}{3})

Therefore the values of x and y are x=-2 and y=\frac{5}{3}

Step-by-step explanation:

Given equations can be written as

y=\frac{2}{3}x+3\hfill (1)

x=-2x+3x=-2\hfill (2)

Solving equation(2) we get

x=-2

Substitute x=-2 in equation (1) we get

y=\frac{2}{3}(-2)+3

=-\frac{4}{3}+3

=\frac{-4+9}{3}

=\frac{5}{3}

Therefore the values of x and y are  x=-2 and y=\frac{5}{3}

The solution to the given system of equations is (-2,\frac{5}{3})

eimsori [14]3 years ago
3 0

Answer:

It’s B and E

Step-by-step explanation:

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3 years ago
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3 years ago
Find the radius of convergence, r, of the series. ? n2xn 7 · 14 · 21 · ? · (7n) n = 1
defon
I'm guessing the series is supposed to be

\displaystyle\sum_{n=1}^\infty\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}

By the ratio test, the series converges if the following limit is less than 1.

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7\cdot14\cdot21\cdot\cdots\cdot(7n)\cdot(7(n+1))}}{\frac{n^2x^n}{7\cdot14\cdot21\cdot\cdots\cdot(7n)}}\right|

The first n terms in the numerator's denominator cancel with the denominator's denominator:

\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2x^{n+1}}{7(n+1)}}{n^2x^n}\right|

|x^n| also cancels out and the remaining factor of |x| can be pulled out of the limit (as it doesn't depend on n).

\displaystyle|x|\lim_{n\to\infty}\left|\frac{\frac{(n+1)^2}{7(n+1)}}{n^2}\right|=|x|\lim_{n\to\infty}\frac{|n+1|}{7n^2}=0

which means the series converges everywhere (independently of x), and so the radius of convergence is infinite.
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