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ziro4ka [17]
2 years ago
7

Write the equation of the circle centered at (3, 5) and tangent to x = -1. I need help pleasse

Mathematics
1 answer:
Annette [7]2 years ago
7 0

Answer:

(x - 3)² + (y - 5)² = 16

Step-by-step explanation:

the equation of a circle in standard form is

(x - h)² + (y - k)² = r²

where (h, k ) are the coordinates of the centre and r the radius

given (h, k ) = (3, 5 ) we require to find r

r is the distance from the centre to a point on the circle

given x = - 1 is a tangent the r is the distance between the x- coordinates

r = | 3 - (- 1) | = | 3 + 1 | = | 4 | = 4

then equation of circle is

(x - 3)² + (y - 5)² = 4² , that is

(x - 3)² + (y - 5)² = 16

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3 years ago
1. Simplify:<br>4(4y-7y^{2})-9(5y+2)<br><br>2. Simplify:<br>24 – 4(5y – 6z) + 3y – 7z
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Answer:

1)-

How to solve your question

Your question is

4(4−72)−9(5+2)

4(4y-7y^{2})-9(5y+2)4(4y−7y2)−9(5y+2)

Simplify

1

Rearrange terms

4(4−72)−9(5+2)

4({\color{#c92786}{4y-7y^{2}}})-9(5y+2)4(4y−7y2)−9(5y+2)

4(−72+4)−9(5+2)

4({\color{#c92786}{-7y^{2}+4y}})-9(5y+2)4(−7y2+4y)−9(5y+2)

2

Distribute

4(−72+4)−9(5+2)

{\color{#c92786}{4(-7y^{2}+4y)}}-9(5y+2)4(−7y2+4y)−9(5y+2)

−282+16−9(5+2)

{\color{#c92786}{-28y^{2}+16y}}-9(5y+2)−28y2+16y−9(5y+2)

3

Distribute

−282+16−9(5+2)

-28y^{2}+16y{\color{#c92786}{-9(5y+2)}}−28y2+16y−9(5y+2)

−282+16−45−18

-28y^{2}+16y{\color{#c92786}{-45y-18}}−28y2+16y−45y−18

4

Combine like terms

2)

−17y+17z+24

See steps

Step by Step Solution:



STEP1:Equation at the end of step 1

((24 - 4 • (5y - 6z)) + 3y) - 7z

STEP2:

Final result :

-17y + 17z + 24

−282+16−45−18

-28y^{2}+{\color{#c92786}{16y}}{\color{#c92786}{-45y}}-18−28y2+16y−45y−18

−282−29−18

-28y^{2}{\color{#c92786}{-29y}}-18−28y2−29y−18

Solution

−282−29−18

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