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Jet001 [13]
3 years ago
6

Find f'(x). anything helps! also, bonus! f(x)= e^(2x-1) + e^2

Mathematics
1 answer:
MissTica3 years ago
7 0
\bf f(x)=e^{-x}(e^{2x}+x^2)\iff f(x)=e^{-x}e^{2x}+e^{-x}x^2
\\\\\\
f(x)=e^{2x-x}+\cfrac{x^2}{e^x}\implies f(x)=e^x+\cfrac{x^2}{e^x}\\\\
-----------------------------\\\\
\cfrac{dy}{dx}\left[ \cfrac{x^2}{e^x} \right]\implies \cfrac{2xe^x-x^2e^x}{(e^x)^2}\implies \cfrac{xe^x(2-x)}{(e^x)(e^x)}\implies \cfrac{x(2-x)}{e^x}\\\\
-----------------------------\\\\
\cfrac{dy}{dx}=e^x+\cfrac{x(2-x)}{e^x}
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Step-by-step explanation:

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Answer:

y - 5 = -3(x + 2)

General Formulas and Concepts:

<u>Algebra I</u>

Point-Slope Form: y - y₁ = m(x - x₁)  

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3 years ago
If a and b are odd integers, then a – b is an even integer.
Readme [11.4K]
So the answer choices are
1. a-b=even
2. a and b are not odd
3. a and b are odd
4. a-b=even
5. a-b=not even

even=not odd
not even=odd so choice 5 is really a-b=odd
basically choice 4 and 1 are the same so we cross one out
so
the problem said that a and b are odd so therefor choice2 is wrong and choice 3 is correct

then both are odd
odd-odd=even because
an even number is represented as 2n where n is an integer
an odd number can be represented as 2n+1 so assume you have 2 odd numbers 2 away from each other so odd and (odd+2)
odd+2-odd=2n+1+2-(2n+1)=2n+1+2-2n-1=2n-2n+1-1+2=2
you are left with odd
using integers
7 and 11
11-7=4
even

so odd-odd=even, it depends on weather you consider 0 odd or even


so the asnwers are:
a and b are odd
a-b is not an even integer
6 0
3 years ago
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