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slega [8]
3 years ago
9

A square has an area of 289 sq cm. what is the length of one side

Mathematics
1 answer:
MatroZZZ [7]3 years ago
5 0
For this problem you can do it one of two ways. The first way would be simply take the square root because the area of a square is one side squared. The other way using the same rule would be to go up from a reasonable number and square it, if that is not the number you are looking for add 1 to the original number and square it again. if you cannot square and square root in your head, you should be allowed to use a calculator to get the answer or 17 as the length of a side.
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<img src="https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E5_1%20%7B%20%5Cfrac%7BlnR%7D%7B%20R%5E%7B2%7D%20%20%7D%20%5C%2C%20dR%20"
alex41 [277]
Integrate by parts, setting

\begin{matrix}u=\ln R&&\mathrm dv=\dfrac{\mathrm dR}{R^2}\\\mathrm du=\dfrac{\mathrm dR}R&&v=-\dfrac1R\end{matrix}

So the integral is

\displaystyle\int_1^5\frac{\ln R}{R^2}\,\mathrm dR=-\dfrac{\ln R}R\bigg|_{R=1}^{R=5}+\int_1^5\frac{\mathrm dR}{R^2}
\displaystyle\int_1^5\frac{\ln R}{R^2}\,\mathrm dR=-\left(\frac{\ln5}5-\frac{\ln1}1)-\frac1R\bigg|_{R=1}^{R=5}
\displaystyle\int_1^5\frac{\ln R}{R^2}\,\mathrm dR=-\frac{\ln5}5-\left(\frac15-1\right)
\displaystyle\int_1^5\frac{\ln R}{R^2}\,\mathrm dR=\frac15(4-\ln5)
5 0
4 years ago
What’s the sum of 4+4?<br> (Yw)
Olegator [25]

its just 4+4 its 8                                                                                                    

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Determine whether the triangles are simlar. If so, write a similarty statement.
Zolol [24]
Try this option:
6) similar: ∠C=∠E=60; ∠ABC=∠DBE.
7) similar:
ΔLMN: ∠L=75; ∠M=50; ∠N=180-125=55°; ΔPOQ: ∠O=75; ∠Q=55; ∠P=180-130=50°.
3 0
3 years ago
What is the degree of 7?
koban [17]

Answer:

Step-by-step explanation:

7 degrees

5 0
4 years ago
One point or two-point perspective
serious [3.7K]
Two point perspective because of two points on the horizon.
7 0
4 years ago
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