the sum. d^2+d-30/d^2+3d-40 + d^2+14d+48/d^2-2d-48 A. d^2+14d+16 / (d+8)(d-8) B. 2d^2+14d+16/(d+8)(d-8) C. 2d^2+15d+18/(d+8)(d-8)
Question
Updated 4/24/2014 12:41:55 AM
Flagged by yeswey [4/24/2014 12:41:55 AM]
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Original conversation
User: the sum. d^2+d-30/d^2+3d-40 + d^2+14d+48/d^2-2d-48 A. d^2+14d+16 / (d+8)(d-8) B. 2d^2+14d+16/(d+8)(d-8) C. 2d^2+15d+18/(d+8)(d-8)

Question
Updated 4/24/2014 12:41:55 AM
Flagged by yeswey [4/24/2014 12:41:55 AM]
Rating
8
(d^2+d-30)/(d^2+3d-40) + (d^2+14d+48)/(d^2-2d-48)
= (d - 5)(d + 6)/(d + 8)(d - 5) + (d + 6)(d + 8)/(d - 8)(d + 6)
= (d + 6)/(d + 8) + (d + 8)/(d - 8)
= (d + 6)(d - 8)/(d + 8)(d - 8) + (d + 8)(d + 8)/(d - 8)(d + 8)
= (d^2 - 2d - 48)/(d + 8)(d - 8) + (d^2 + 16d + 64)/(d + 8)(d - 8)
= (2d^2 + 14d + 16)/(d + 8)(d - 8)
Confirmed by jeifunk [4/24/2014 12:43:37 AM]

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Updated 67 days ago|11/26/2022 12:24:35 AM
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Added 67 days ago|11/26/2022 12:24:30 AM
Simplify the sum. d^2+d-30/d^2+3d-40 + d^2+14d+48/d^2-2d-48 A. 2d62+15d+18/2d^2+d-88 B. d^2+14d+16 / (d+8)(d-8) C. 2d^2+14d+16/(d+8)(d-8) D. 2d^2+15d+18/(d+8)(d-8)
Weegy: A. 2d62+15d+18/2d^2+d-88 is the answer. (More)
Question
Use the Binomial Theorem to expand the binomial. (x + 2y^2)3 A. x3 - 3x^2y2 + 3xy^4 - y^6 B. x3 + 6^x2y2 + 12xy^4 + 8y^6 C. x3 - 6x^2y2 + 12xy^4 - 8y^6 w D. x3 + 3x^2y^2 + 3xy^4 + y6
Weegy: C. x3 - 6x^2y2 + 12xy^4 - 8y^6 (More)
Question
Updated 6/10/2014 11:44:25 PM
(x + 2y^2)^3
= x^3 + 3x^2(2y^2) + 3x(2y^2)^2 + (2y^2)^3
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Confirmed by jeifunk [6/10/2014 11:54:09 PM]
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Question
Updated 7/30/2014 10:03:25 AM
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Confirmed by andrewpallarca [7/30/2014 10:05:16 AM]
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Weegy: the answer is C (More)
Question
Updated 5/7/2014 9:03:25 PM
v7x(vx-7v7) = xv7-49vx
Confirmed by jeifunk [5/7/2014 9:04:50 PM]
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