The answers will be 14 with how many zeros are present; 140 and 14,000 respectively.
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∙ 9y agolcm(7, 20, 35) = 140 7 = 7 20 = 2^2 x 5 35 = 5 x 7 lcm = 2^2 x 5 x 7 = 140
7 x 23 = (7 x 20) + (7 x 3)
7 x 420 = (7 x 400) + (7 x 20)
x - 7 = 13 Therefore, x = 13 + 7 x = 20
Suppose the answer is x.Then x + 1/2 = 7/20 so that x = 7/20 - 1/2 = 7/20 - 10/20 = -3/20
-x + 7 = 27 Subtract 7 from both sides: -x = 20 or x = -20
lcm(7, 20, 35) = 140 7 = 7 20 = 2^2 x 5 35 = 5 x 7 lcm = 2^2 x 5 x 7 = 140
7 x 420 = 7(400 + 20) = 7 * 400 + 7 * 20.
7 x 23 = (7 x 20) + (7 x 3)
7 x 420 = (7 x 400) + (7 x 20)
1 x 20=20 7 x 1 = 7 20/7 make sure to convert into a mixed number like this: 2 6/7. <-answer
x - 7 = 13 Therefore, x = 13 + 7 x = 20
the answer is 140 .
The only common factor is: 1 20= 2 x 5 x 2 x 17 = 7 x 1The only common factor of 20 & 7 is 1
Suppose the answer is x.Then x + 1/2 = 7/20 so that x = 7/20 - 1/2 = 7/20 - 10/20 = -3/20
Factor them. 7 2 x 2 x 5 = 20 5 x 7 = 35 Combine the highest amount of each factor. 2 x 2 x 5 x 7 = 140, the LCM
Let call them x and y x+y=7 and x*y=20 From the first you get x = 7 -y, put into the second get (7-y)*y=20 -y^2+7*y=20 or y^2-7+20=0 Delta = (-7)^2-4*1*20 = -31 y = (7-i*sqrt(31))/ 2 or y = (7+i*sqrt(31))/ 2 and x = 7-y = (7+i*sqrt(31))/ 2 or x = (7-i*sqrt(31))/ 2