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Phoenix [80]
3 years ago
5

Can someone answer this easy proof

Mathematics
1 answer:
Alja [10]3 years ago
6 0
Answer: in picture

Step-by-Step: See Picture

If this is helpful, please mark me brainliest :)

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Help !!!! Thank you!!!!
castortr0y [4]

Answer:

Option (G)

Step-by-step explanation:

Volume of the real cane = 96 in³

Volume of the model of a can = 12 in³

Volume scale factor = \frac{\text{Volume of the model}}{\text{Volume of the real can}}

                                 = \frac{12}{96}

                                 =\frac{1}{8}

Scale factor of the model = \sqrt[3]{\text{Volume scale factor}}

                                          =\sqrt[3]{\frac{1}{8}}

                                          =\frac{1}{2}

Therefore, scale factor of the model of a can = \frac{1}{2} ≈ 1 : 2

Option (G) will be the correct option.

3 0
3 years ago
PLEASE HELP ME QUICK PLEASE
Vikentia [17]

Answer:

34

Step-by-step explanation:

in the photo illustrated

3 0
3 years ago
How many degrees are in each interior angle of a regular pentagon? 50 72 108 120
Sindrei [870]

The answer 108 degrees.

6 0
3 years ago
Read 2 more answers
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
Students in an art class cut colored paper into equal-sized squares. The paper measures 6 cm by 12 cm and they do not want to wa
solmaris [256]
The greatest possible side length is 24 square piece
7 0
3 years ago
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