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bekas [8.4K]
2 years ago
14

I Have attached an image. Please help me asap

Mathematics
2 answers:
alekssr [168]2 years ago
7 0

Answer:

do you have choices? if not x=

11

16

​

Step-by-step explanation: thats the best o could do

Dmitry_Shevchenko [17]2 years ago
6 0

Answer:

x = 16/11

Step-by-step explanation:

Convert to common denominator of 12   for   4   2 and 3 :

[3x (3)  - 6(2x-4) ] /12   =   4 ( 2x+2) / 12  

   now you can multiply both sides by 12 to get :

9x -12x+24 = 8x+8

-3x+24 = 8x + 8

16 = 11x

x = 16/11  

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Solve the equation ƒ - (-4) = 12.<br> ƒ = -8<br> ƒ = 16<br> ƒ = 8<br> ƒ = -9
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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

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