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PolarNik [594]
2 years ago
6

I need to find the x pls help and show work

Mathematics
2 answers:
KIM [24]2 years ago
3 0

Answer:

X=63

Step-by-step explanation:

The quadrilateral above are similar and they have a ratio. To find the ratio divide a bigger side in the first shape by a corresponding smaller side in the other shape

36/4 = 9/1

The ratio is 9:1

When a side in the big shape is 9 the other shape is 1

Therefore x/7 =9

x = 7 * 9

=63

liraira [26]2 years ago
3 0

Answer:

63

Step-by-step explanation:

36 / 9 = 4

90 / 9 = 10

126 / 9 = 14

is the ratio 9:1

so 7 * 9 = 63

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Hey there!

4 + (14 - 2)

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4 + 14 - 2

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What exponential expression is equal to <br> 2^-5 * 2^8?
amm1812

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Step-by-step explanation:

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3 years ago
Please help ASAP! BRAINLIEST to best/right answer (: SCATTER PLOT
Serhud [2]
The answer is D. Negative Correlation.
3 0
3 years ago
A statistician studies the relationship between income and the choice of individuals to attend college. He notices that during a
Gnoma [55]

The conclusion is is misleading because the extraneous variable, cost of tuition was not accounted for in the experiment.

<h3>What are the variables in this experiment?</h3>

The independent variable is the variable that the person carrying out an experiment changes or manipulates. This variable is income. The dependent variable is the variable that is being measured in an experiment. This variable is college enrolment.

The extraneous variable is the variable  that is not been researched in the experiment but it has the potential to influence the results of the experiment. This variable is the cost of tuition.

To learn more about independent variables, please check: brainly.com/question/26287880

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6 0
1 year 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
2 years ago
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