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saw5 [17]
3 years ago
13

Identify the equation of the circle that has its center at (-16, 30) and passes through the origin.

Mathematics
1 answer:
s2008m [1.1K]3 years ago
4 0
The general equation of the circle is:

(x -xo)^2 + (y - yo)^2 = r^2

Where xo,yo are the coordinates of the center and r is the radius of the circle.

Here (xo,yo) is (-16,30) and you can find the radius.

You can find the radius using the equation of the distance (Pythagoras) between the center (-16,30) and any point on the circunference. Here use (0,0)

r^2 = (-16 - 0)^2 + (30 - 0)^2 = 1156

Then, the equation is>

(x - (-16) )^2 + (y - 30)^2 = 1156

=> (x + 16)^2 + (y - 30)^2 = 1156.   <----- answer
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Answer:

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Step-by-step explanation:

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35+62+x=180

97+x=180

Subtract both sides by 97

97+x-97=180-97

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A water tank is a cylinder with radius 40 cm and depth 150 m
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3 years ago
Solve the following systems of equations by substitution method.
Anuta_ua [19.1K]

Answer:

1) x = 3 and y = -2

2) x = 2 and y = 1

3) x = -2 and y = -1

4) x = 3 and y = 2

5) x = 2

6) x = 2 and y = -3

7) x = 4 and y = 3

8) x = -3 and y = -6

Step-by-step explanation:

1. Answer :

y = -2

4x - 3y = 18 ----(1)

Substitute y = -2 in (1).

4x - 3(-2) = 18

4x + 6 = 18

Subtract 6 from both sides.

4x = 12

Divide both sides by 4.

x = 3

So,

x = 3 and y = -2

2. Answer :

y = 6x - 11 ----(1)

2x + 3y = 7 ----(2)

Substitute y = 6x - 11 in (2).

2x + 3(6x - 11) = 7

2x + 18x - 33 = 7

20x - 33 = 7

Add 33 to both sides.

20x = 40

Divide both sides by 20.

x = 2

Substitute x = 2 in (1).

y = 6(2) - 11

= 12 - 11

= 1

So,

x = 2 and y = 1

3. Answer :

x = -3y - 5 ----(1)

4x - 5y = -3 ----(2)

Substitute x = -3y + 5 in (2).

4(-3y - 5) - 5y = -3

-12y - 20 - 5y = -3

-17y - 20 = -3

Add 20 to both sides.

17y = -17

Divide both sides by 17.

y = -1

Substitute y = -1 in (1).

x = -3(-1) - 5

= 3 - 5

= -2

So,

x = -2 and y = -1

4. Answer :

x = 5y - 7 ----(1)

-2x - 3y = -12 ----(2)

Substitute x = 5y - 7 in (2).

-2(5y - 7) - 3y = -12

-10y + 14 - 3y = -12

-13y + 14 = -12

Subtract 14 from both sides.

-13y = -26

Divide both sides by -13.

y = 2

Substitute y = 2 in (1).

x = 5(2) - 7

= 10 - 7

= 3

So,

x = 3 and y = 2

5. Answer :

5x + 3y - 8 = 0 ----(1)

2x - 3y - 6 = 0 ----(2)

Solve for 3y in (1).

5x + 3y - 8 = 0

Subtract 5x from both sides.

3y - 8 = -5x

Add 8 to both sides.

3y = -5x + 8

Substitute 3y = -5x + 8 in (2).

2x - (-5x + 8) - 6 = 0

2x + 5x - 8 - 6 = 0

7x - 14 = 0

Add 14 to both sides.

7x = 14

Divide both sides by 7.

x = 2

6. Answer :

x - 3y - 11 = 0 ----(1)

5x + y - 7 = 0 ----(2)

Solve for x in (1).

x - 3y - 11 = 0

Subtract 3y and 11 to both sides.

x = 3y + 11 ----(3)

Substitute x = 3y + 11 in (2).

5(3y + 11) + y - 7 = 0

15y + 55 + y - 7 = 0

16y + 48 = 0

Subtract 48 from both sides.

16y = -48

Divide both sides by 16.

y = -3

Substitute y = -3 in (3).

x = 3(-3) + 11

= -9 + 11

= 2

So,

x = 2 and y = -3

7. Answer :

2x - 3y = -1 ----(1)

y = x - 1 ----(2)

Substitute y = x - 1 in (1).

2x - 3(x - 1) = -1

2x - 3x + 3 = -1

-x + 3 = -1

Subtract 3 from both sides.

-x = -4

Multiply both sides by -1.

x = 4

Substitute x = x in (3).

y = 4 - 1

= 3

So,

x = 4 and y = 3

8. Answer :

-4x + y = 6 ----(1)

-5x - y = 21 ----(2)

Solve for y in (1).

-4x + y = 6

Add 4x to both sides.

y = 4x + 6 ----(3)

Substitute y = 4x + 6 in (2).

-5x - (4x + 6) = 21

-5x - 4x - 6 = 21

-9x - 6 = 21

Add 6 to both sides.

-9x = 27

Divide both sides by -9.

x = -3

Substitute x = -3 in (3).

y = 4(-3) + 6

= -12 + 6

= -6

So,

x = -3 and y = -6

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Thank You!

Answered by: ms115

6 0
3 years ago
Read 2 more answers
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