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3 years ago

Frorn first to last, when did the presidents serve?

Mathematics

1 answer:

navik [9.2K]3 years ago

8 0

Step-by-step explanation:

you got this i believe in you

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if an external point of a circle is at a distance equal to the diameter of the circle from the center of the circle, the length

jeka57 [31]

Let the radius of the circle be r. Then the line from the external point through the center of the circle which extends to the far point on the circle has length 3r .By the tangent - secant theorem

t^2 = 3r * r = 3r^2 ( where t is the length of the tangent).

So t = √(3r^2) = √3r answer.

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3 years ago

Dr. Jon planned to reward his class with candy because everyone got a A on the test. He surveyed every 4th student that entered

USPshnik [31]

This is an example of a systematic sample.

Dr. Jon developed a system in order to get a random sample. His plan was to survey the 4th student that entered his classroom.

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3 years ago

Can someone help me find the volume and the surface area of this shape.

mash [69]

Answer:

Got my points deducted by a loser name PoeticAesthetics

Step-by-step explanation:

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2 years ago

Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{

SVETLANKA909090 [29]

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

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

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

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Note 1: A more general solution is $y = Ce^x$ for some constant C.
Note 2: It might be tempting to say the general solution is $y = e^x+C$ , but that is not the case because $y = e^x+C \to y' = e^x+0 = e^x$ and we can see that $y' = y$ would only be true for C = 0, so that is why $y = e^x+C$ does not work.

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3 years ago

Ramapo College has a total of 25,000 students enrolling this year. 42% of

salantis [7]

Answer: im sorry im not sure but you can find the answer if you look it up

Step-by-step explanation:

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2 years ago

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