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3 years ago

Solve this equation 2(y + 2) = 3(y - 7)

Mathematics

1 answer:

aliina [53]3 years ago

8 0

Answer:

$2(y + 2) = 3(y - 7) \\ 2y + 2 \times 2 = 3y - 3 \times 7 \\ 2y + 4 = 3y - 21 \\ y \: terms \: togther \\ 3y - 2y = 4 + 21 \\ y = 25$

$thank \: you$

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3 years ago

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

The standard form of equation of the circle is

${(x + 2)}^{2} + {(y + 3)}^{2} = 4$

<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).

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

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

Now to find the radius of the circle

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.

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

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

${(x + 2)}^{2} + {(y + 3)}^{2} = 4$

Learn more about circle, refer:

brainly.com/question/24810873

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