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2 years ago

The midpoint M and one endpoint of JK are given. Find the coordinates of the other endpoint. J(7,2) and M(1,-2)

Mathematics

1 answer:

djverab [1.8K]2 years ago

3 0

Step-by-step explanation:

Xj + Xk/2 = Xm

7 + Xk/2 = 1

to get rid of the bracket, multiply all two sides by the denominator.

2(7 + Xk/2) = 1(2)

7 + Xk = 2

Xk = 2 - 7

Xk = -5

Yj + Yk/2 = Ym

2 + Yk/2 = -2

to get rid of the bracket, multiply all two sides by the denominator.

2(2 + Yk/2) = -2(2)

2 + Yk = -4

Yk = -4 - 2

Yk = -6

Therefore the coordinates of point K is (-5,-6)

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$\lim_{n \to \infty} U_n =0$

Given series is convergence by using Leibnitz's rule

Step-by-step explanation:

Step(i):-

Given series is an alternating series

∑ $(-1)^{n} \frac{n^{2} }{n^{3}+3 }$

Let $U_{n} = (-1)^{n} \frac{n^{2} }{n^{3}+3 }$

By using Leibnitz's rule

$U_{n} - U_{n-1} = \frac{n^{2} }{n^{3} +3} - \frac{(n-1)^{2} }{(n-1)^{3}+3 }$

$U_{n} - U_{n-1} = \frac{n^{2}(n-1)^{3} +3)-(n-1)^{2} (n^{3} +3) }{(n^{3} +3)(n-1)^{3} +3)}$

Uₙ-Uₙ₋₁ <0

Step(ii):-

$\lim_{n \to \infty} U_n = \lim_{n \to \infty}\frac{n^{2} }{n^{3}+3 }$

$= \lim_{n \to \infty}\frac{n^{2} }{n^{3}(1+\frac{3}{n^{3} } ) }$

= $= \lim_{n \to \infty}\frac{1 }{n(1+\frac{3}{n^{3} } ) }$

$=\frac{1}{infinite }$

$\lim_{n \to \infty} U_n =0$

∴ Given series is converges

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