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2 years ago

-3x^2 – 24x – 13 = -13

1 answer:

3 0

$\begin{gathered} -3x^2-24x-13=-13 \\ \text{ Express it as:} \\ 3x^2+24x+26=0 \\ 3x^2+24x=-26 \\ 3(x^2+8x+16)=?-26 \\ 3(x+4)^2=48-26 \\ 3(x+4)^2=22 \end{gathered}$

oh,

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Can someone help me with this

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The center of the circle is (h,k) = (-2,-3)

The radius of the circle is r = 2

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<h3>How to find the center, radius and standrad form of the circle?</h3>

The general form of equation of the circle is

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

Here, (h,k) means centre of the circle.

r means radius of the circle.

given that coordinate points of centre of circle is (-2,-3).

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<h3>How to find the radius of the circle?</h3>

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The distance from a circle's centre to its circumference is its radius.

The distance from a circle's centre (-2,-3) to its circumference (0,-3) is its radius.

using the formula, distance between the two points to obtain radius.

$d = \sqrt{(x1 - x2) {}^{2} + {(y1 - y2)}^{2} } \\ r = \sqrt{ {( - 2 - 0)}^{2} + {( - 3 - ( - 3))}^{2} } \\ r = \sqrt{ {( - 2)}^{2} + {( - 3 + 3)}^{2} } \\ r = \sqrt{ {4}^{2} + 0 } \\ r = \sqrt{4} \\ r = 2$

<h3>How to find the standard form of equation of the circle?</h3>

(h,k) = (-2,-3)

r = 2

subtitue the (h,k) and r values to get the standard form of equation of the circle.

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

${(x - ( - 2))}^{2} + {(y - ( - 3))}^{2} = {r}^{2}$

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Learn more about circle, refer:

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Step-by-step explanation:

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Travka [436]

Answer:

R = 0.8333%/month

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3 years ago

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Answer:

The absolute value makes the number inside it, a positive

Step-by-step explanation:

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