Question and answer
Factor p2 - 10p - 24.
Question
Asked 6/20/2013 1:03:40 PM
Updated 10/24/2014 7:53:12 AM
3 Answers/Comments
This conversation has been flagged as incorrect.
Flagged by janezeshun [10/24/2014 7:53:12 AM]
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User: 6x2 + 8x - 8 = 0

User: Solve x2 - 4x - 3 = 0 by completing the square.

User: Factor p2 - 10p - 24.

Weegy: 3-3 = 0
andrewpallarca|Points 22631|

User: Factor p2 - 10p - 24.

Weegy: 3-3 = 0
andrewpallarca|Points 22631|

User: Factor p^2 - 10p - 24.

Weegy: 4x + 3 is the answer
OxTornado007|Points 6879|

User: What is the center of the circle that has a diameter whose endpoints are (1, 3) and (-7, -1)?

Weegy: the center is (1, 2)
ibontxo|Points 356|

Question
Asked 6/20/2013 1:03:40 PM
Updated 10/24/2014 7:53:12 AM
3 Answers/Comments
This conversation has been flagged as incorrect.
Flagged by janezeshun [10/24/2014 7:53:12 AM]
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The solution for x^2 - 4x - 3 = 0 is x = 7^1/2 + 2 , or x = -7^1/2 + 2 .
x^2 - 4x - 3 = 0
( x - 2 )^2 - 4 - 3 = 0
( x - 2 )^2 = 7
x - 2 = 7^1/2
x = 7^1/2 + 2
or
x - 2 = -7^1/2
x = -7^1/2 + 2
Added 10/24/2014 7:50:28 AM
This answer has been confirmed as correct and helpful.
Confirmed by andrewpallarca [10/24/2014 12:12:49 PM]
0
p^2 - 10p - 24
= ( p - 12 ) ( p + 2 )
Added 10/24/2014 7:51:43 AM
This answer has been confirmed as correct and helpful.
Confirmed by andrewpallarca [10/24/2014 12:13:19 PM]
0
The center of the circle that has a diameter whose endpoints are (1, 3) and (-7, -1) is ( - 3 , 1 ) .
Let the center is ( x , y )
x = ( 1 - 7 ) / 2 = - 3
y = ( 3 - 1) / 2 = 1
Added 10/24/2014 7:53:08 AM
This answer has been confirmed as correct and helpful.
Confirmed by andrewpallarca [10/24/2014 12:14:41 PM]
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1 Answer/Comment
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Updated 6/19/2013 10:26:50 AM
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If you can't answer a question, please pass it to another expert. Thanks!
Added 6/19/2013 10:26:50 AM
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Expert Answered
Updated 314 days ago|9/11/2017 5:11:50 AM
1 Answer/Comment
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Added 314 days ago|9/11/2017 5:11:41 AM
This answer has been confirmed as correct and helpful.
Confirmed by jeifunk [9/11/2017 5:54:15 AM]
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Updated 9/18/2015 1:36:44 AM
1 Answer/Comment
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