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Makovka662 [10]
3 years ago
9

Identify the radius of the circle whose equation is (x - 2)2 + (y - 8)2 = 16.

Mathematics
2 answers:
Tanzania [10]3 years ago
8 0
The radius is 4.

That equation is in center-radius form. When that’s the case, the radius is the square root of the number in the right of the equals sign. Square root of 16 is 4.
34kurt3 years ago
4 0
\bf \textit{equation of a circle}\\\\ 
(x- h)^2+(y- k)^2= r^2
\qquad 
center~~(\stackrel{}{ h},\stackrel{}{ k})\qquad \qquad 
radius=\stackrel{}{ r}\\\\
-------------------------------\\\\
(x-\stackrel{h}{2})^2+(y-\stackrel{k}{8})^2=16\implies (x-2)^2+(y-8)^2=\stackrel{r}{4^2}
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konstantin123 [22]

{ \qquad\qquad\huge\underline{{\sf Answer}}}

Here we go ~

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\qquad \sf  \dashrightarrow \:  \cfrac{3{}^{2(x + 1)}  - 3 {}^{2x} }{4 \times 3 {}^{(2x - 1)} }

\qquad \sf  \dashrightarrow \:  \cfrac{3{}^{(2x + 2)}  - 3 {}^{2x} }{4 \times 3 {}^{(2x - 1)} }

here :

  • { \sf {3}^{(2x+2)}=({3}^{2x - 1})\sdot (3³)}

  • { \sf {3}^{(2x)}=({3}^{(2x - 1)})\sdot (3¹)}

\qquad \sf  \dashrightarrow \:  \cfrac{3{}^{(2x  - 1)}(3 {}^{3}   - 3 {}^{1}) }{4 \times 3 {}^{(2x - 1)} }

[ taking { \sf {3}^{(2x - 1)} }common here ]

\qquad \sf  \dashrightarrow \:  \cfrac{27 {}^{}   - 3 {}^{}}{4  }

\qquad \sf  \dashrightarrow \:  \cfrac{24{}^{}   {}^{}}{4  }

\qquad \sf  \dashrightarrow \: 6

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The statement that must be true about square WXYZ are as follows;

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<h3>Properties of a Square.</h3>
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Therefore, the statement that must be true about square WXYZ are as follows;

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learn more on square here: brainly.com/question/15019502

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