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2 years ago

Circle has a radius of 4 and the center is (-4,-8) what is the equation of the circle if I double the radius

Mathematics

1 answer:

wariber [46]2 years ago

4 0

9514 1404 393

Answer:

(x +4)^2 +(y +8)^2 = 64

Step-by-step explanation:

The equation of a circle with center (h, k) and radius r is ...

(x -h)^2 +(y -k)^2 = r^2

The circle centered at (h, k) = (-4, -8) with radius 8 has equation ...

(x +4)^2 +(y +8)^2 = 64

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<h3>Answer: c = 7/4</h3>

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Work Shown:

Compute the function value at the endpoints

$f(x) = \sqrt{4-x}\\\\f(-5) = \sqrt{4-(-5)} = 3\\\\f(4) = \sqrt{4-4} = 0\\\\$

With a = -5 and b = 4, we have

$f'(c) = \frac{f(b)-f(a)}{b-a}\\\\f'(c) = \frac{f(4)-f(-5)}{4-(-5)}\\\\f'(c) = \frac{0-3}{9}\\\\f'(c) = -\frac{1}{3}\\\\$

So,

$f(x) = \sqrt{4-x}\\\\f'(x) = -\frac{1}{2\sqrt{4-x}}\\\\f'(c) = -\frac{1}{3}\\\\-\frac{1}{2\sqrt{4-c}} = -\frac{1}{3}\\\\$

Use algebra to solve for c

$-\frac{1}{2\sqrt{4-c}} = -\frac{1}{3}\\\\\frac{1}{2\sqrt{4-c}} = \frac{1}{3}\\\\3 = 2\sqrt{4-c}\\\\2\sqrt{4-c} = 3\\\\\sqrt{4-c} = \frac{3}{2}\\\\4-c = \frac{9}{4}\\\\c = 4-\frac{9}{4}\\\\c = \frac{16-9}{4}\\\\c = \frac{7}{4}\\\\$

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