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Old 07-25-2007   #1
Anonymous
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Default pls help algebra problems :(?

1)the sum of four consecutive integers is -106.2)The product of two consectuve even integers is 1683) The sum of four consecutive even integers is -100
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Old 07-25-2007   #2
Lidya
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The product of two consecutive even integers is 168 and the numbers are 12 and 14the sum of four consecutive intigers is -106 and the numbers are -25, -26,-27 and -28The sum of four consecutive even inegers is -100 and the numbers are -22, -24, -26 and -28.
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Old 07-25-2007   #3
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x+x+1+x+2+x+3 = -1064x +6 = -1064x = -112x = -28x(x + 2) = 168x^2 + 2x -168 = 0(x +14)(x - 12)12, 14x+x+2+x+4+x+6 = -1004x + 12 = -1004x = -112x = -28
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Old 07-25-2007   #4
ucla bruin fan!
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A) -25+-26+-27+-28b) 12 * 14c) )-22+-24+-26+-28
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Old 07-25-2007   #5
zohair
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1)the sum of four consecutive integers is -106.x+x+1+x+2+x+3=4x+6=-1064x=-112x=-28ans -28 -27 -26 -252)The product of two consectuve even integers is 1682x(2x+2)=1684x^2+4x-168=0x^+x-42-0x=6, -7ans12 14or-14 -123) The sum of four consecutive even integers is -1002x+2x+2+2x+4+2x+6=1008x+12=1008x=88x=11ans22 24 26 28
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Old 07-25-2007   #6
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(1) x + (x+1) + (x +2) + (x+3) = -1064x + 6 = -106x = -28so the 4 integers are -28, -27, -26, -25(2) (2n)*(2n+2) = 1684n^2 + 4n - 168 = 0solve this quadratic to get n = 6the 2 integers are 12, 14(3) 2p + (2p+2) + (2p+4) + (2p+6) = -1008p + 12 = -100p = -14the 4 integers are -28, -26, -24, -22
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