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kipiarov [429]
2 years ago
13

What is the derivative of 2te^-t

Mathematics
1 answer:
Sveta_85 [38]2 years ago
6 0
(2te^{-t})'=(2t)'e^{-t}+2t(e^{-t})'=2e^{-t}+2t(-e^{-t})=2e^{-t}-2te^{-t}
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Answer:

<u>x = 5</u>

Step-by-step explanation:

See attached image.

8 0
1 year ago
The measure of each interior angle of a regular polygon is 150 how many sides does the polygon have?
Dmitry [639]
The exterior and an interior angles always add up to 180 degrees.
180-150=One exterior angle
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3 0
3 years ago
Write 3x^2-18x-6 in vertex form
Vedmedyk [2.9K]
The standard form of a quadratic equation is \displaystyle{ y=ax^2+bx+c, while the vertex form is:

                      y=a(x-h)^2+k, where (h, k) is the vertex of the parabola.

What we want is to write \displaystyle{ y=3x^2-18x-6 as y=a(x-h)^2+k

First, we note that all the three terms have a factor of 3, so we factorize it and write:

\displaystyle{ y=3(x^2-6x-2).


Second, we notice that x^2-6x are the terms produced by (x-3)^2=x^2-6x+9, without the 9. So we can write:

x^2-6x=(x-3)^2-9, and substituting in \displaystyle{ y=3(x^2-6x-2) we have:

\displaystyle{ y=3(x^2-6x-2)=3[(x-3)^2-9-2]=3[(x-3)^2-11].

Finally, distributing 3 over the two terms in the brackets we have:

y=3[x-3]^2-33.


Answer: y=3(x-3)^2-33
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