x6 - 27 = (x2 - 3)*(x4 + 3x2 + 9) Danny rocks
If we just assume that the contestant risks exactly how much the clue is originally worth on the Daily Doubles and they risk all their winning and get the clue correct in Final Jeopardy, here is the addition problem (In Dollars):'Single' Jeopardy:(200+400+600+800+1000)x6Double Jeopardy:(400+800+1200+1600+2000)x6Final Jeopardy:All that x2[(200+400+600+800+1000)x6]+[(400+800+1200+1600+2000)x6]x2=[(3000x6)+(6000x6)]x2=(18000+36000)x2=54000x2=$108,000.00
You cant get that many but if you buy the x2 and x6 and the x8 and the x10 wich ='s x6350 and go into a double score zone you can get 1,000,000 stud for each thing you destroy. :) HOW MUCH IS 2 TIMES MULTIPLYER
(x+2)6 (x2 + 4x + 4)(x+2)4 (x2 + 4x + 4)(x2 + 4x + 4)(x+2)2 (x2 + 4x + 4)(x2 + 4x + 4)(x2 + 4x + 4) (x4 + 4x3 + 4x2 + 4x3 + 16x2 + 16x + 4x2 + 16x + 16)(x2 + 4x + 4) (x4 + 8x3 + 24x2 + 32x + 16)(x2 + 4x + 4) (x6 + 4x5 + 4x4 + 8x5 + 32x4 + 32x3 + 24x4 + 96x3 + 96x2 + 32x3 + 128x2 +128x + 16x2 + 64x + 64) (x6 + 12x5 + 60x4 + 160x3 + 240x2 + 192x +64) the coefficeints are 1, 12, 60, 160, 240, and 192
You must have at least 2 numbers to have a LCM. If you mean 30 and 42... LCM = 5 x6 x 7 = 210
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