Mathematics, 30.08.2019 00:00 adeanbates

Arrange the circles (represented by their equations in general form) in ascending order of their radius lengths. tiles x2 + y2 − 2x + 2y − 1 = 0 x2 + y2 − 4x + 4y − 10 = 0 x2 + y2 − 8x − 6y − 20 = 0 4x2 + 4y2 + 16x + 24y − 40 = 0 5x2 + 5y2 − 20x + 30y + 40 = 0 2x2 + 2y2 − 28x − 32y − 8 = 0 x2 + y2 + 12x − 2y − 9 = 0

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Answer from: keoni18

We will proceed to convert the equations into standard format to determine the solution.

we know that

The Standard Form Equation of a Circle is equal to

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

where

(h,k) is the center of the circle

r is the radius of the circle

Case N

$x^{2}+y^{2}-2x+ 2y- 1= 0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}-2x)+ (y^{2}+ 2y)=1$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}- 2x+1)+ (y^{2}+ 2y+1)=1+1+1$

Rewrite as perfect squares

$(x-1)^{2}+(y+1)^{2}=3$

$(x-1)^{2}+ (y+1)^{2}=\sqrt{3}^{2}$

Case N

$x^{2}+ y^{2}-4x + 4y- 10= 0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2} - 4x)+ (y^{2}+ 4y)=10$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2} - 4x+4)+ (y^{2}+ 4y+4)=10+4+4$

Rewrite as perfect squares

$(x-2)^{2}+ (y+2)^{2}=18$

$(x-2)^{2}+ (y+2)^{2}=\sqrt{18}^{2}$

Case N

$x^{2}+ y^{2}-8x - 6y-20= 0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2}- 8x)+ (y^{2} - 6y)=20$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}- 8x+16)+ (y^{2}- 6y+9)=20+16+9$

Rewrite as perfect squares

$(x-4)^{2}+ (y-3)^{2}=45$

$(x-4)^{2}+ (y-3)^{2}=\sqrt{45}^{2}$

Case N

$4x^{2}+4y^{2}+16x +24y- 40= 0$

Simplify divide by both sides

$x^{2}+ y^{2}+4x+6y- 10= 0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2} +4x)+ (y^{2} + 6y)=10$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}+4x+4)+(y^{2} + 6y+9)=10+4+9$

Rewrite as perfect squares

$(x+2)^{2}+ (y+3)^{2}=23$

$(x+2)^{2}+ (y+3)^{2}=\sqrt{23}^{2}$

Case N

$5x^{2}+ 5y^{2}-20x +30y+ 40= 0$

Simplify divide by both sides

$x^{2}+ y^{2}-4x +6y + 8= 0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2} -4x)+ (y^{2}+ 6y)=-8$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2}-4x+4)+ (y^{2}+ 6y+9)=-8+4+9$

Rewrite as perfect squares

$(x-2)^{2} + (y+3)^{2}=5$

$(x-2)^{2} + (y+3)^{2}=\sqrt{5}^{2}$

Case N

$2x^{2} + 2y^{2}-28x -32y-8= 0$

Simplify divide by both sides

$x^{2} + y^{2} -14x-16y-4= 0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2} -14x)+ (y^{2} -16y)=4$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2} -14x+49)+ (y^{2} -16y+64)=4+49+64$

Rewrite as perfect squares

$(x-7)^{2} + (y-8)^{2}=117$

$(x-7)^{2} + (y-8)^{2}=\sqrt{117}^{2}$

Case N

$x^{2}+ y^{2}+12x - 2y- 9 = 0$

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$(x^{2} +12x)+ (y^{2} - 2y)=9$

Complete the square twice. Remember to balance the equation by adding the same constants to each side.

$(x^{2} +12x+36)+ (y^{2} - 2y+1)=9+36+1$

Rewrite as perfect squares

$(x+6)^{2} + (y-1)^{2}=46$

$(x+6)^{2} + (y-1)^{2}=\sqrt{46}^{2}$

the circles in ascending order of their radius lengths is

$x^{2}+ y^{2}-2x + 2y- 1=0$

$(x-1)^{2}+ (y+1)^{2}=\sqrt{3}^{2}$

$5x^{2}+ 5y^{2}-20x +30y+40=0$

$(x-2)^{2}+(y+3)^{2}=\sqrt{5}^{2}$

$x^{2}+y^{2}-4x+4y- 10=0$

$(x-2)^{2}+(y+2)^{2}=\sqrt{18}^{2}$

$4x^{2}+4y^{2}+16x+24y-40=0$

$(x+2)^{2}+(y+3)^{2}=\sqrt{23}^{2}$

$x^{2}+y^{2}-8x- 6y-20= 0$

$(x-4)^{2}+(y-3)^{2}=\sqrt{45}^{2}$

$x^{2}+ y^{2}+12x- 2y-9= 0$

$(x+6)^{2}+(y-1)^{2}=\sqrt{46}^{2}$

$2x^{2}+2y^{2}-28x-32y-8=0$

$(x-7)^{2}+(y-8)^{2}=\sqrt{117}^{2}$

Answer from: Quest

i get you, what you should do is think of the positive things that would come out of it. it will make yours parents proud and raise your grade. you should always try and if you can't focus tell the teacher.

Answer from: Quest

the area of a triangle is determined by the formula: base x height * 1/2

you'll need to assign a base to the triangle. based on the vertices given, the simplest base would be the line segment from (-2,1) to (2,1). the distance between any two numbers is found by just subtracting them. here, you should subtract the x values to find the distance of this side.

2 - (-2) = 4.

so the distance (aka the length of the base) is 4. the height of the triangle is the distance between the base and the highest point on the triangle which is (3,4). we should subtract the y values this time to find the height/distance.

4 - 1 = 3

so the height is 3.

base = 4

height = 3

hope this if needed hmu

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