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Natasha_Volkova [10]
3 years ago
8

20,I am not sure how to do it

Mathematics
1 answer:
professor190 [17]3 years ago
5 0
Given:
f(x) = x^4 - x

g(x) = x^4 (even)
h(x) = -x (odd)

f(x) as a sum of an even function and an odd function

f(x) = g(x) + h(x)

s(x) = 1 / (x^4 - x)

s(x) as a sum of an even function and an odd function

s(x) = 1 / [g(x) + h(x)]
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If y varies directly as x and y=12 when x=c what is y in terms of c when x=8
tatiyna

"y varies directly as x" means y = kx

you do not know what k is yet, other than that it is some constant number.

Now plug in the given values, y=12 when x=4, in the equation

12 = 4k

=> k = 3

Now rewrite the equation with the k value you just found

y = 3k

Substitute 18 in place of x

y = 3(18)

y = 54

8 0
3 years ago
The base of an isosceles triangle is parallel to the x-axis. The y-axis passes through an end point of the base. If the endpoint
AnnyKZ [126]
The first x coordinate is given by 0 and the second one is given as j. So the length of the base is j. (Sometimes in math they give letters rather than numbers. And since you know it is an isosceles then the other vertex should be at x-coordinate half j or j/2.
5 0
3 years ago
What would the new point be If you rotated the point (3,-5) 270 clockwise.
timurjin [86]

Answer:

see below

Step-by-step explanation:

When you rotate a point 270 degrees clockwise (x, -y) becomes (-y, -x) so the answer would be (-5, -3).

5 0
3 years ago
Drag a figure into each blank space so that the two figures in the row match the description.
LUCKY_DIMON [66]
What’s the description?
7 0
3 years ago
Evaluate using integration by parts ​
PolarNik [594]

Rather than carrying out IBP several times, let's establish a more general result. Let

I(n)=\displaystyle\int x^ne^x\,\mathrm dx

One round of IBP, setting

u=x^n\implies\mathrm du=nx^{n-1}\,\mathrm dx

\mathrm dv=e^x\,\mathrm dx\implies v=e^x

gives

\displaystyle I(n)=x^ne^x-n\int x^{n-1}e^x\,\mathrm dx

I(n)=x^ne^x-nI(n-1)

This is called a power-reduction formula. We could try solving for I(n) explicitly, but no need. n=5 is small enough to just expand I(5) as much as we need to.

I(5)=x^5e^x-5I(4)

I(5)=x^5e^x-5(x^4e^x-4I(3))=(x-5)x^4e^x+20I(3)

I(5)=(x-5)x^4e^x+20(x^3e^x-3I(2))=(x^2-5x+20)x^3e^x-60I(2)

I(5)=(x^2-5x+20)x^3e^x-60(x^2e^x-2I(1))=(x^3-5x^2+20x-60)x^2e^x+120I(1)

I(5)=(x^3-5x^2+20x-60)x^2e^x+120(xe^x-I(0))

Finally,

I(0)=\displaystyle\int e^x\,\mathrm dx=e^x+C

so we end up with

I(5)=(x^4-5x^3+20x^2-60x+120)xe^x-120e^x+C

I(5)=(x^5-5x^4+20x^3-60x^2+120x-120)e^x+C

and the antiderivative is

\displaystyle\int2x^5e^x\,\mathrm dx=(2x^5-10x^4+40x^3-120x^2+240x-240)e^x+C

8 0
3 years ago
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