The radius of the circle whose equation is (x + 4)² + (y - 2)² = 36 is
36
18
6

Question

Asked 11/12/2014 7:36:49 AM

Updated 11/12/2014 10:56:02 PM

1 Answer/Comment

This answer has been confirmed as correct and helpful.

s

Question

Asked 11/12/2014 7:36:49 AM

Updated 11/12/2014 10:56:02 PM

1 Answer/Comment

This answer has been confirmed as correct and helpful.

Rating

8

The distance between the points (5, -2) and (-7, 3) is 13 .

Let the distance is d

d^2 = (-7-5)^2 + {3 - (-2)}^2

d^2 = 144 + 25

d^2 = 169

d = 13

Let the distance is d

d^2 = (-7-5)^2 + {3 - (-2)}^2

d^2 = 144 + 25

d^2 = 169

d = 13

Added 11/12/2014 10:45:36 PM

This answer has been confirmed as correct and helpful.

Find the center of the following equation.
x 2 + y 2 + 8 x - 6 y - 15 = 0

Question

Updated 11/10/2014 11:24:59 PM

1 Answer/Comment

x^2 + y^2 + 8x - 6y - 15 = 0

(x + 4)^2 - 16 + (y - 3)^2 - 9 - 15 = 0

(x + 4)^2 + (y - 3)^2 = 40

(-4 , 3) is the center of equation.:x^2 + y^2 + 8x - 6y - 15 = 0 .

(x + 4)^2 - 16 + (y - 3)^2 - 9 - 15 = 0

(x + 4)^2 + (y - 3)^2 = 40

(-4 , 3) is the center of equation.:x^2 + y^2 + 8x - 6y - 15 = 0 .

Added 11/10/2014 11:24:59 PM

This answer has been confirmed as correct and helpful.

Confirmed by jeifunk [11/11/2014 12:51:23 AM]

What is the solution set of 2x(x - 1) = 3?
User: Which of the following constants can be added to x2 - 10x to form a perfect square trinomial?
10
25
100 User: Write the quadratic equation in factored form. Be sure to write the entire equation.
x2 + x - 12 = 0
**Weegy:** (x+4)(x-3)=0 **User:** If the quadratic formula is used to find the solution set of 3x 2 + 4x - 2 = 0, what are the solutions?
**User:** What is the value of b2 - 4ac for the following equation?
2x2 + 3x = -1
0
1
17 **Weegy:** The answer is: b.1 [smile] **User:** What is the solution set of 7x 2 + 3x = 0?
{0, 3/7}
{0, -3/7}
{0, -4/7} **Weegy:** The solution set of 7x^2 + 3x = 0 is: B. {0, -3/7}. **User:** What is the solution set of x 2 + 5x + 1 = 0?
**User:** What is the solution set of (3x - 1)2 = 5?
(More)

Question

Not Answered

Updated 11/11/2014 7:37:43 AM

2 Answers/Comments

(3x - 1)^2 = 5

3x - 1 = ± sqrt 5

3x = 1 ± sqrt 5

x = (1 ± sqrt 5)/3

3x - 1 = ± sqrt 5

3x = 1 ± sqrt 5

x = (1 ± sqrt 5)/3

Added 11/11/2014 7:35:35 AM

This answer has been confirmed as correct and helpful.

x^2 + 5x + 1 = 0

a = 1, b = 5, c = 1

discriminant = b^2 - 4ac = 25 - 4 = 21

x = (-5 ± sqrt 21)/2

a = 1, b = 5, c = 1

discriminant = b^2 - 4ac = 25 - 4 = 21

x = (-5 ± sqrt 21)/2

Added 11/11/2014 7:37:43 AM

This answer has been confirmed as correct and helpful.

Solve (x + 2)(2x + 3) = 6
{-2, -3/2}
{-7/2, 0}
{0}

Question

Not Answered

Updated 11/15/2014 2:38:32 PM

1 Answer/Comment

The solution set for (x + 2)(2x + 3) = 6 is {-7/2, 0}.

(x + 2)(2x + 3) = 6;

2x^2 +3x + 4x + 6 = 6;

2x^2 + 7x + 6 - 6 = 0;

2x^2 + 7x = 0;

x(2x + 7) = 0;

x = 0;

2x + 7 = 0; 2x = -7; x = -7/2

(x + 2)(2x + 3) = 6;

2x^2 +3x + 4x + 6 = 6;

2x^2 + 7x + 6 - 6 = 0;

2x^2 + 7x = 0;

x(2x + 7) = 0;

x = 0;

2x + 7 = 0; 2x = -7; x = -7/2

Added 11/15/2014 2:38:32 PM

This answer has been confirmed as correct and helpful.

What is the midpoint of the line segment whose endpoints are (-8, 12) and (-13, -2)?
(-10.5, 5)
(-10.5, 7)
(-2.5, 7) **Weegy:** The midpoint of the line segment whose endpoints are (-8, 12) and (-13, -2) is (-10.5, 5). ((x1+x2)/2, (y1+y2)/2); = ((-8 - 13)/2, (12 - 2)/2) = (-21/2, 10/2) = (-10.5, 5) **User:** What is the length of the hypotenuse of a right triangle whose legs have lengths of 5 and 12?
13
15
17 **Weegy:** The length of the hypotenuse of a right triangle whose legs have lengths of 5 and 12 is 13. **User:** Find the midpoint of the line segment whose endpoints are (3, 10) and (7, 8).
(2, 1)
(5, 9)
(11, 12) **Weegy:** Find the midpoint of the line segment whose endpoints are (3, 10) and (7, 8) is: (5, 9). M = [(x2 + x1)/2] , [(y2 + y1)/2]; = [(3 + 7)/2] , [(10 + 8)/2]; = (10, 2) , (18, 2); = (5, 9) **User:** What is the equation of a circle whose center is at the origin and whose radius is 16?
x2 + y2 = 4
x2 + y2 = 16
x2 + y2 = 256 (More)

Question

Not Answered

Updated 11/15/2014 2:39:44 PM

1 Answer/Comment

The equation of a circle whose center is at the origin and whose radius is 16 is: x^2 + y^2 = 256

Added 11/8/2014 4:10:30 AM

This answer has been confirmed as correct and helpful.

Confirmed by andrewpallarca [11/15/2014 2:39:54 PM]

The center of a circle is at the origin. An endpoint of a diameter of the circle is at (-3, -4). How long is the diameter of the circle?
5
10
25

Question

Not Answered

Updated 11/9/2014 4:33:59 AM

1 Answer/Comment

The center of a circle is at the origin. An endpoint of a diameter of the circle is at (-3, -4). The diameter of the circle is 10.

Let the radius is r

r^2 = (-3)^2 + (-4)^2

r^2 = 25

r = 5

diameter = 2r = 2*5 = 10

Let the radius is r

r^2 = (-3)^2 + (-4)^2

r^2 = 25

r = 5

diameter = 2r = 2*5 = 10

Added 11/9/2014 4:33:59 AM

This answer has been confirmed as correct and helpful.

Confirmed by andrewpallarca [11/15/2014 2:40:40 PM], Rated good by andrewpallarca

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