Drag and drop the symbols to enter the equation of the circle in standard form with center and radius given. Center (8, 0), radius =
Question
Updated 2/18/2015 3:38:26 PM
This conversation has been flagged as incorrect.
Flagged by yeswey [2/18/2015 3:38:26 PM]
Original conversation
User: Drag and drop the symbols to enter the equation of the circle in standard form with center and radius given. Center (8, 0), radius =

Weegy: In the standard equation of a circle, (x-a)^2 + (y-b)^2 = r^2, the center of the circle is (a, b) and the radius is r.

User: Enter the equation of the line meeting the given conditions. Please put the equation in standard form. Containing C(3, 3) and D(-3, 2)

Weegy: 1+-4=3
jeifunk|Points 42656|

User: Drag and drop the symbols to enter the equation of the circle in standard form with center and radius given. Center (7, 5), radius = 4

Weegy: The equation of the circle with Center (7, 5), radius = 4 in standard form is (x - 7)^2 + (y - 5)^2 = 16.

User: Find the distance between the given points. W(0, 8) and X(0, 12) Distance = 4 4 (13) 10

Weegy: the answer to the equation 3(w+12)<0 is w<-12
shanedunn138|Points 1297|

User: Find the distance between the given points. E(-6, 4) and F(-12, 4) Distance = 6 8 18

User: Enter the equation of the line meeting the given conditions. Please put the equation in standard form. Containing E(4, 3) and F(6, 1)

Question
Updated 2/18/2015 3:38:26 PM
This conversation has been flagged as incorrect.
Flagged by yeswey [2/18/2015 3:38:26 PM]
Rating
3
The equation of the line Containing C(3, 3) and D(-3, 2) in standard form is x -6y = -15
3
The distance between the given points W (0, 8) and X(0, 12) is distance = 4.
Confirmed by sujaysen [2/18/2015 3:36:19 PM]
3
The distance between the given points E(-6, 4) and F(-12, 4) is Distance = 6
Confirmed by sujaysen [2/18/2015 3:36:00 PM]
3
The equation of the line Containing E(4, 3) and F(6, 1) in standard form is x + y = 7

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Enter the equation of the line meeting the given conditions. Please put the equation in standard form. Containing A(5, 3) and perpendicular to a line with slope of -2 User: Drag and drop the symbols to enter the equation of the circle in standard form with center and radius given. Center (-8, -3), radius = 2 User: Find the distance between the given points. A(5, 3) and B(7, -4)
Weegy: -4|-4| = -4 * 4; = -16 User: Find the distance between the given points. R(0, 0) and S(6, 8) Distance = 2 7 (22) 10 Weegy: The answer is zero. The property is identity property of addition. This property states that any number plus 0 equals that number. [smile] User: Indicate the equation of the line meeting the given conditions. Please put the equation in standard form. Containing A(1, 3) and B(0, 2) (More)
Question
Updated 2/19/2015 4:47:20 PM
The equation of the circle in standard form with center (-8, -3), radius = 2 is (x + 8)^2 + (y + 3)^2 = 4.
Confirmed by Andrew. [2/19/2015 6:31:37 PM]
The distance between the given points A(5, 3) and B(7, -4) is sqrt (2^2 + 7^2) = sqrt 53
Confirmed by Andrew. [2/19/2015 6:32:30 PM]
The distance between the given points R(0, 0) and S(6, 8) is Distance = sqrt (6^2 + 8^2) = 10
Confirmed by Andrew. [2/19/2015 6:32:56 PM]
The equation of the line Containing A(1, 3) and B(0, 2) in standard form is x - y = -2
Confirmed by Andrew. [2/19/2015 6:33:27 PM]
Find the distance between the given points. A(5, 3) and B(7, -4) Distance =
Weegy: -4|-4| = -4 * 4; = - 16 User: Please put the equation in standard form. Containing E(4, 3) and F(6, 1) (More)
Question
Updated 2/20/2015 3:41:41 AM
The distance between the given points A (5 ,3) B (7 , -4) is sqrt of 53 .
distance^2 = (5 - 7)^2 + (3 + 4)^2
distance^2 = 4 + 49
distance^2 = 53
distance = sqrt of 53
Confirmed by sujaysen [2/20/2015 7:54:44 AM]
The equation in standar form that containning E (4 , 3), F (6 , 1) is x + y = 7 .
slope = (1 - 3)/(6 - 4) = -2/2 = -1
y - 3 = -1(x - 4)
y - 3 = -x + 4
y = -x + 7
x + y = 7
29,986,070
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