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3 years ago

Which equation is not equivalent to the other equations?

Mathematics

1 answer:

Colt1911 [192]3 years ago

8 0

I think the answer is C. X*4=-12+3

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FOR AN EXAM DUE TODAY!!! Simplify the polynomial. 3x – 2x2 + 4 – x2 + x + -

Mashutka [201]

Answer:

${ \tt{ \blue{ - {3x}^{2} + 4x + 4 }}}$

Step-by-step explanation:

${ \tt{3x - {2x}^{2} + 4 - {x}^{2} + x }} \\$

» Collect second degree terms (x² coefficients) together and first degree terms (x coefficients) and constants

$= { \tt{( - 2 - 1) {x}^{2} + (3 + 1)x + 4 }} \\ = { \boxed{ \tt{ \: - {3x}^{2} + 4x + 4}}}$

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Vugfybj Bukhara Hii diss did deism ska she did she dke end skip and sjj?

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Answer:

Step-by-step explanation:

The answer is 4

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3 years ago

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Four people enter their names into a drawing. The winner receives one of two different prizes that are randomly selected. The tr

galina1969 [7]

A) (-4,2π/3)
b) (4,4π/3)
c) (-2,π/3)
d) (2,5π/3)

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3 years ago

Use series to verify that <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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love history [14]

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