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4 years ago

Need help solving 9x2squared-81=0

Mathematics

1 answer:

Ksenya-84 [330]4 years ago

5 0

the answer for x is -3 x= -3. i hope this helped you!

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Drag the titles to the boxes to form correct pairs .not all titles will be used. Match the pairs of equation that represents con

Delvig [45]

Answer:

The concentric circles are

$3x^{2}+3y^{2}+12x-6y-21=0$ and $4x^{2}+4y^{2}+16x-8y-308=0$

$5x^{2}+5y^{2}-30x+20y-10=0$ and $3x^{2}+3y^{2}-18x+12y-81=0$

$4x^{2}+4y^{2}-16x+24y-28=0$ and $2x^{2}+2y^{2}-8x+12y-40=0$

$x^{2}+y^{2}-2x+8y-13=0$ and $5x^{2}+5y^{2}-10x+40y-75=0$

Step-by-step explanation:

we know that

The equation of the circle in standard form is equal to

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

where

(h,k) is the center and r is the radius

Remember that

Concentric circles, are circles that have the same center

Convert each equation in standard form and then compare the centers

The complete answer in the attached document

Part 1) we have

$3x^{2}+3y^{2}+12x-6y-21=0$

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

$(3x^{2}+12x)+(3y^{2}-6y)=21$

Factor the leading coefficient of each expression

$3(x^{2}+4x)+3(y^{2}-2y)=21$

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

$3(x^{2}+4x+4)+3(y^{2}-2y+1)=21+12+3$

$3(x^{2}+4x+4)+3(y^{2}-2y+1)=36$

Rewrite as perfect squares

$3(x+2)^{2}+3(y-1)^{2}=36$

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

therefore

The center is the point (-2,1)

Part 2) we have

$5x^{2}+5y^{2}-30x+20y-10=0$

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

$(5x^{2}-30x)+(5y^{2}+20y)=10$

Factor the leading coefficient of each expression

$5(x^{2}-6x)+5(y^{2}+4y)=10$

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

$5(x^{2}-6x+9)+5(y^{2}+4y+4)=10+45+20$

$5(x^{2}-6x+9)+5(y^{2}+4y+4)=75$

Rewrite as perfect squares

$5(x-3)^{2}+5(y+2)^{2}=75$

$(x-3)^{2}+(y+2)^{2}=15$

therefore

The center is the point (3,-2)

Part 3) we have

$x^{2}+y^{2}-12x-8y-100=0$

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

$(x^{2}-12x)+(y^{2}-8y)=100$

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

$(x^{2}-12x+36)+(y^{2}-8y+16)=100+36+16$

$(x^{2}-12x+36)+(y^{2}-8y+16)=152$

Rewrite as perfect squares

$(x-6)^{2}+(y-4)^{2}=152$

therefore

The center is the point (6,4)

Part 4) we have

$4x^{2}+4y^{2}-16x+24y-28=0$

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

$(4x^{2}-16x)+(4y^{2}+24y)=28$

Factor the leading coefficient of each expression

$4(x^{2}-4x)+4(y^{2}+6y)=28$

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

$4(x^{2}-4x+4)+4(y^{2}+6y+9)=28+16+36$

$4(x^{2}-4x+4)+4(y^{2}+6y+9)=80$

Rewrite as perfect squares

$4(x-2)^{2}+4(y+3)^{2}=80$

$(x-2)^{2}+(y+3)^{2}=20$

therefore

The center is the point (2,-3)

Part 5) we have

$x^{2}+y^{2}-2x+8y-13=0$

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

$(x^{2}-2x)+(y^{2}+8y)=13$

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

$(x^{2}-2x+1)+(y^{2}+8y+16)=13+1+16$

$(x^{2}-2x+1)+(y^{2}+8y+16)=30$

Rewrite as perfect squares

$(x-1)^{2}+(y+4)^{2}=30$

therefore

The center is the point (1,-4)

Part 6) we have

$5x^{2}+5y^{2}-10x+40y-75=0$

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

$(5x^{2}-10x)+(5y^{2}+40y)=75$

Factor the leading coefficient of each expression

$5(x^{2}-2x)+5(y^{2}+8y)=75$

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

$5(x^{2}-2x+1)+5(y^{2}+8y+16)=75+5+80$

$5(x^{2}-2x+1)+5(y^{2}+8y+16)=160$

Rewrite as perfect squares

$5(x-1)^{2}+5(y+4)^{2}=160$

$(x-1)^{2}+(y+4)^{2}=32$

therefore

The center is the point (1,-4)

Part 7) we have

$4x^{2}+4y^{2}+16x-8y-308=0$

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

$(4x^{2}+16x)+(4y^{2}-8y)=308$

Factor the leading coefficient of each expression

$4(x^{2}+4x)+4(y^{2}-2y)=308$

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

$4(x^{2}+4x+4)+4(y^{2}-2y+1)=308+16+4$

$4(x^{2}+4x+4)+4(y^{2}-2y+1)=328$

Rewrite as perfect squares

$4(x+2)^{2}+4(y-1)^{2}=328$

$(x+2)^{2}+(y-1)^{2}=82$

therefore

The center is the point (-2,1)

Part 8) Part 9) and Part 10) in the attached document

Download docx

8 0

4 years ago

The output is eight less than the input

irina [24]

Y=x-8, that is the basic equation or rule you are possibly trying to ask.

6 0

4 years ago

The equation of a circle is (x - 4)² + (y + 3)² = 9. What are the coordinates of the center of this circle? (-4, 3) (4, -3) (4,

bezimeni [28]

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Step-by-step explanation:

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3 years ago

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Maurinko [17]

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Step-by-step explanation:

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3 years ago

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