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sammy [17]
3 years ago
9

A flagpole stands perpendicular to the ground. It is supported by a 20-foot cable that runs from the top of the flagpole to the

ground at a point that is 12 feet from the base of the pole. Find the height of the flagpole. The Pythagorean theorem is a2 b2
Mathematics
1 answer:
3241004551 [841]3 years ago
7 0

Answer:

h  =  16 feet

Step-by-step explanation:

Pythagoras´ theorem establishes, that in a right triangle square of the hypothenuse (c²) is the sum of leg ( b²)  plus  leg  (a²)

Then

c²  =  a²  +  b²

where according to the problem statement

(20)²  =  h²  +  (12)²           (  h is the height of the flagpole )

h²  =  400 -  144

h²  =  256

h  =  √ 256

h  =  16 feet

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y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\

\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\

This shows that y' = y is true when y = e^x

-----------------------

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