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Hitman42 [59]
3 years ago
11

Check

Mathematics
1 answer:
Westkost [7]3 years ago
3 0

Answer:

I think it's the 3rd one

Step-by-step explanation:

because it makes sense

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Are the functions f(x) = (x^2-1)/(x-1) and g(x)= x+1 equal for all x?
Vinvika [58]
\bf \textit{difference of squares}
\\ \quad \\
(a-b)(a+b) = a^2-b^2\qquad \qquad 
a^2-b^2 = (a-b)(a+b)\\\\
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\bf f(x)=\cfrac{x^2-1}{x-1}\implies f(x)=\cfrac{x^2-1^2}{x-1}\implies f(x)=\cfrac{(\underline{x-1})(x+1)}{\underline{x-1}}
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f(x)=x+1\qquad \qquad \qquad  \qquad  \qquad  g(x)=x+1\\\\
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\textit{they're, kinda, except that, when x = 1}
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g(x)=(1)+1\implies g(x)=2
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f(x)=\cfrac{(1)^2-1}{(1)-1}\implies f(x)=\cfrac{0}{0}\impliedby und efined
7 0
3 years ago
Find the volume of the following solid. The solid between the cylinder ​f(x,y)equals=e Superscript negative xe−x and the region
PilotLPTM [1.2K]

It is hard to comprehend your question. As far as I understand:

f(x,y) = e^(-x)

Find the volume over region R = {(x,y): 0<=x<=ln(6), -6<=y <= 6}.

That is all I understood. It would be easier to understand with a picture or some kind of visual aid.

Anyways, to find the volume between the surface and your rectangular region R, we must evaluate a double integral of f on the region R.

\iint_{R}^{ } e^{-x}dA=\int_{-6}^{6} \int_{0}^{ln(6)} e^{-x}dx dy

Now evaluate,

\int_{0}^{ln(6)}e^{-x}dx

which evaluates to,  5/6 if I did the math correct. Correct me if I am wrong.

Now integrate this w.r.t. y:

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So,

\iint_{R}^{ } e^{-x}dA=\int_{-6}^{6} \int_{0}^{ln(6)} e^{-x}dx dy = 10

7 0
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Mark Brainliest please

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The cosine of an angle is defined as the ratio of the adjacent side to the hypotenuse.

The tangent of an angle is defined as the ratio of the opposite side to the adjacent side.

See the image for Sin J, K & cos j & K & tan J & K

6 0
3 years ago
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Answer: because the other half is used on the base solution

Step-by-step explanation:

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