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Hitman42 [59]
3 years ago
7

PLS HELP

Mathematics
1 answer:
lapo4ka [179]3 years ago
4 0

Answer:

AG=5\ cm              

Step-by-step explanation:

we know that

The three perpendicular bisectors of the sides of a triangle meet in a single point, called the circumcenter (point G)

The circumcenter is equidistant from the vertices of the triangle

so

AG=CG=BG

<em>Find the length side BG</em>

In the right triangle BGE

Applying the Pythagorean Theorem

BG^2=GE^2+BE^2

substitute the given values

BG^2=4^2+3^2

BG^2=25

BG=5\ cm

therefore

AG=5\ cm

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What is the sum of the series infinity e n-1 -7(3/8)^n?
soldier1979 [14.2K]

Let

S_N = \displaystyle \sum_{n=1}^N -7 \left(\frac38\right)^n = -7\left(\dfrac38 + \dfrac{3^2}{8^2} + \dfrac{3^3}{8^3} + \cdots + \dfrac{3^N}{8^N}\right)

Then

\dfrac38 S_N = -7\left(\dfrac{3^2}{8^2} + \dfrac{3^3}{8^3} + \dfrac{3^4}{8^4} + \cdots + \dfrac{3^{N+1}}{8^{N+1}}\right)

S_N - \dfrac38 S_N = -7 \left(\dfrac38 - \dfrac{3^{N+1}}{8^{N+1}}\right)

\dfrac58 S_N = -\dfrac{21}8 \left(1 - \left(\dfrac38\right)^N\right)

S_N = -\dfrac{21}5 \left(1 - \left(\dfrac38\right)^N\right)

As N\to\infty, the exponential term will converge to zero, so the infinite sum converges to

\displaystyle \sum_{n=1}^\infty -7 \left(\frac38\right)^n = \lim_{n\to\infty} S_N = \boxed{-\dfrac{21}5}

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2 years ago
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<img src="https://tex.z-dn.net/?f=8%20%5Ctimes%7B10%7D%5E%7B5%7D%20is%20%5C%3A%20how%20%5C%3A%20many%20%5C%3A%20times%20%5C%3A%2
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