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Viefleur [7K]
1 year ago
8

Find the distance between the pair of coordinates: (-4, -3) and (-8, 7). A 10.77 B 3.74 C 04 D 12.65 Question​

Mathematics
1 answer:
kenny6666 [7]1 year ago
6 0

~~~~~~~~~~~~\textit{distance between 2 points} \\\\ (\stackrel{x_1}{-4}~,~\stackrel{y_1}{-3})\qquad (\stackrel{x_2}{-8}~,~\stackrel{y_2}{7})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\\\ d=\sqrt{[-8 - (-4)]^2 + [7 - (-3)]^2}\implies d=\sqrt{(-8+4)^2 + (7+3)^2} \\\\\\ d=\sqrt{(-4)^2 + 10^2}\implies d=\sqrt{116}\implies d\approx 10.77

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Evaluate the summation from n equals 1 to infinity of the quotient of 3 and 2 raised the quantity n minus 1 power
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Answer:

3*∑ (1/2)^n = 6

Step-by-step explanation:

We have the summation from n = 1 to n = ∞, for:

∑ 3/(2)^(n - 1)

We know that the summation between k = 0, and k = N - 1 for:

∑ r^k = (1 - r^N)/(1 - r)

if we have the summation between k = 0 and k = ∞ - 1 = ∞

(here we used that ∞ is really big, then ∞ - 1 = ∞)

In the numerator we will have the term r^∞, if 0 < r < 1, then r^∞ = 0.

Then if we assume that 0 < r < 1 we can write:

∑ r^k =  1/(1 - r)

In our case, we can rewrite our summation as:

3*∑ (1/2)^n

for n = 0 to  n = ∞

You can see that i changed the limits for n, this does not really matter because we are summing between a number and infinity, the only thing you need to take care is that now the power is n, instead of (n - 1)

Then we have r = (1/2) which is clearly smaller than 1.

then (1/2)

Then this summation is equal to:

3*∑ (1/2)^n = 3*( 1/(1 - 1/2)) = 3*( 1/( 2/2 - 1/2)) = 3*( 1/(1/2)) = 3*2 = 6

3*∑ (1/2)^n = 6

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Step-by-step explanation:

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