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3 years ago

11

Help please me with this ....

1 answer:

Nonamiya [84]3 years ago

5 0

Ext. 13/44 hope it helpa

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raketka [301]

Answer:

y -10 = -1/9 (x-9)

Step-by-step explanation:

Point slope form is

y-y1 = m (x-x1)

y -10 = -1/9 (x-9)

4 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

A magician performing a magic act has asked for 2 volunteers from the audience to help with his routine. If there are 250 people

RUDIKE [14]

The answer is 125. If you were to pair 250 people in 2, that would be 250 divided by 2. And 250 divided by 2 is 125.

5 0

3 years ago

Read 2 more answers

Solve the system of equations.

ycow [4]

The answer is letter c.

7 0

3 years ago

Read 2 more answers

What are the coordinates of the vertex of the graph of

vesna_86 [32]

The vertex is (3,0). The 3 inside the abs val bar tells you the x value of the vertex is 3. No number outside the abs value bar is a +0 which is the y value of the vertex.

4 0

3 years ago