povzpetifkzf

2023-02-22

How to find the center of a circle given the three points (-3,5); (3,3); (11,19)?

GreerceViemkpy2

Step 1: Given that:
The given function is
$y=Asi{n}^{2}x$

Given range, $x=0\to x=\pi$
Step 2: Calculation of the average value of the given function:
Average value of a function$\left(\right)=\frac{\int ydx}{\int dx}$
Now, for a given range, the average value will be,

$=\frac{{\int }_{0}^{\pi }A{\mathrm{sin}}^{2}xdx}{{\int }_{0}^{\pi }dx}$
$=\frac{A{\int }_{O}^{\pi }\frac{1-\mathrm{cos}2x}{2}dx}{{\int }_{0}^{\pi }{x}^{O}dx}$
$=\frac{A\left({\int }_{O}^{\pi }\frac{1}{2}dx-\frac{1}{2}{\int }_{O}^{\pi }\mathrm{cos}2xdx\right)}{\left[{x}_{O}^{\pi }\right]}$
Now, $\int \mathrm{cos}nxdx=\frac{1}{n}\mathrm{sin}nx$
$=\frac{A\left(\frac{1}{2}\left({x}_{O}^{\pi }\right)-\frac{1}{2}\left(\frac{1}{2}\mathrm{cos}2{x}_{O}^{\pi }\right)\right)}{\left[{x}_{O}^{\pi }\right]}$
$=A\left[\frac{\frac{1}{2}\left(\pi -O\right)-\frac{1}{4}\left(\mathrm{cos}2\pi -\mathrm{cos}O\right)}{\pi -O}\right]$
Since, $\mathrm{cos}2\pi =\mathrm{cos}O=1=A\left[\frac{\frac{\pi }{2}}{\pi }\right]$
$=\frac{A}{2}$

Hence, the given function's average value in the given range will be
$\frac{A}{2}$
.

resymeguv48h

Equal distances separate the points from the center (a, b). Using the distance squares as an equation,
${\left(a+3\right)}^{2}+{\left(b-5\right)}^{2}={\left(a-3\right)}^{2}+{\left(b-3\right)}^{2}={\left(a-11\right)}^{2}+{\left(b-19\right)}^{2}$,
From the first two,
$3a-b=4$, and from first and the third,
$a+b=16$.
solving, (a, b) = (3, 13)...

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