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just olya [345]
2 years ago
5

Write any algebraic eqaution.

Mathematics
2 answers:
Vesnalui [34]2 years ago
6 0

Answer:

(y4x2 + 2xy – y)/(x – 1) = 12

Step-by-step explanation:

RideAnS [48]2 years ago
6 0

Answer:

x3 + 1

Step-by-step explanation:

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Write an addition expression that equivalent to 65-79. Then evaluate the expression
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Which of the following equations is the result after the first step in solving 3x + 6 = 12?
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7 0
3 years ago
Can someone help me with this
mash [69]

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

#SPJ9

5 0
1 year ago
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