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WINSTONCH [101]
3 years ago
13

What is the equation of the circle center (0, 3) and radius 4?

Mathematics
2 answers:
Oksi-84 [34.3K]3 years ago
5 0

\huge{ \mathfrak{  \underline{ Answer }\:  \:  ✓ }}

The equation of the circle is represented as :

\hookrightarrow (x - h) {}^{2}  + (y - k) {}^{2}  = (r) {}^{2}

where, h is x-coordinate and k is y-coordinate of center of the circle, and "r" is radius of the given circle.

Therefore, the equation of the above circle will look like :

\hookrightarrow(x - 0) {}^{2}  + (y - 3) {}^{2}  =  {4}^{2}

\hookrightarrow {x}^{2}  + (y - 3) {}^{2}  = 16

_____________________________

\mathrm{ ☠ \: TeeNForeveR \:☠ }

Sunny_sXe [5.5K]3 years ago
3 0

Answer:

( x )² + ( y - 3 )² = 16

Step-by-step explanation:

The general format for circular equations is ( x - a )² + ( y - b )² = r²

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Hey, I need help with questions 1 and 2, if anyone can help, please? thanks!
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The correct works are:

  • Blue(s + h) = 2s^2 + 4sh + 2h^2 + 3.
  • \frac{Blue(s + h) - Blue(s)}{h} = 4s + 2h

<h3>Function Notation</h3>

The function is given as:

Blue(s) = 2s^2 + 3

The interpretation when Steven is asked to calculate Blue(s + h) is that:

Steven is asked to find the output of the function Blue, when the input is s + h

So, we have:

Blue(s + h) = 2(s + h)^2 + 3

Evaluate the exponent

Blue(s + h) = 2(s^2 + 2sh + h^2) + 3

Expand the bracket

Blue(s + h) = 2s^2 + 4sh + 2h^2 + 3

So, the correct work is:

Blue(s + h) = 2s^2 + 4sh + 2h^2 + 3

<h3>Simplifying Difference Quotient</h3>

In (a), we have:

Blue(s + h) = 2s^2 + 4sh + 2h^2 + 3

Blue(s) = 2s^2 + 3

The difference quotient is represented as:

\frac{f(x + h) - f(x)}{h}

So, we have:

\frac{Blue(s + h) - Blue(s)}{h} = \frac{2s^2 + 4sh + 2h^2 + 3 - 2s^2 - 3}{h}

Evaluate the like terms

\frac{Blue(s + h) - Blue(s)}{h} = \frac{4sh + 2h^2}{h}

Evaluate the quotient

\frac{Blue(s + h) - Blue(s)}{h} = 4s + 2h

Hence, the correct work is:

\frac{Blue(s + h) - Blue(s)}{h} = 4s + 2h

Read more about function notations at:

brainly.com/question/13136492

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