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3 years ago

I honestly just don't get this question. I'll give 30 points!!!

Mathematics

2 answers:

Svetllana [295]3 years ago

7 0

I wish I could help but but I can't I'm dumb as a box of rocks

ExtremeBDS [4]3 years ago

7 0

sorry about your other question but here is the answer to this one

ok so her is what I think you do...

you have the equation y=3(x+1)^2+4 you use the distributive property to get rid of the prenticies so 3*x=3x and 3*1=3.

Then you have y=3x^2+3^2+4 both of the numbers have a square root because the 2 beside the parinticies so both of those are squared

then u square what you can square so it would them be y=3x^2+9+4 because you can't square a variable and a number it stays the same but because 3^2 can be squared is become 3*3 which equals 9

Then u combine the like terms you have making it y=3x^2+13 because of 9+4=14

then it says that y=x^2 so you put x^2=3x^2+13

after that you subtract -3x^2 from both sides making it -2x^2=13 because +x^2-3^2=-2x^2 and 3x^2-3x^2=0 so that goes away and you keep the +13

that is when you would divide by -2 on both sides making it -x^2=-13/2 which also equals -6.5(we will be using the decimal soon) because -2^2 divided by -2=-x^2 and 13 divided by -2 is -6.5

then u would find the square root of that by-{equation:\/(-x^2)=\/(-6.5)} -->\/(-x^2) which equals x because we don't know what x is it turns into jus x then as we move on =\/(-6.5) which when you use a calculator it equals 2.54950976 but I would just round to the nearest hundreths making it 2.55 and that would be your answer

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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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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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