1x - 39 = 76 1x + -39 = 76 39 = 39 1x + -39 + 39 = 76 + 39 1x + 0 = 115 X = 115 Check 1x -- 39 = 76 If x = 115 1 * 115 -- 39 = 76 115 -- 39 = 76 115 + -39 = 76 76 = 76
The first ten positive integer multiples of 115 are as follows: 115 x 1 = 115 115 x 2 = 230 115 x 3 = 345 115 x 4 = 460 115 x 5 = 575 115 x 6 = 690 115 x 7 = 805 115 x 8 = 920 115 x 9 = 1035 115 x 10 = 1150
5 x 115 = 575
23 x 5 = 115
You can simplify the term by approximation! Here is the approach: Let y = √x = x^(½). Then, dy/dx = ½ * x^(-½) = 1/(2√x) Select 115² to be x. Then, we obtain: y = 115 dy/dx = 1/(2 * 115) = 1 / 230 Therefore, we obtain this equation: y - 115 = 1/230 * (x - 115²) If x = 13144, then we have: y - 115 = 1/230 * (13144 - 115²) y = 1/230 * (13144 - 115²) + 115 ≈ 114.65 OR What you can do is to factor out each term by term and extract the perfect square factor out of the √. For instance: √(13144) = √(4 * 3286) = 2√3286
1x - 39 = 76 1x + -39 = 76 39 = 39 1x + -39 + 39 = 76 + 39 1x + 0 = 115 X = 115 Check 1x -- 39 = 76 If x = 115 1 * 115 -- 39 = 76 115 -- 39 = 76 115 + -39 = 76 76 = 76
The first ten positive integer multiples of 115 are as follows: 115 x 1 = 115 115 x 2 = 230 115 x 3 = 345 115 x 4 = 460 115 x 5 = 575 115 x 6 = 690 115 x 7 = 805 115 x 8 = 920 115 x 9 = 1035 115 x 10 = 1150
x + 7y = 39 So x = 39 - 7y Substitute for x in the second equation: 3(39 - 7y) - 2y = 2 117 - 21y - 2y = 2 115 = 23y y = 5 Substitute this value for y back into the first equation: x + 7*5 = 39 x + 35 = 39 x = 4
1 x 115, 5 x 23, 23 x 5, 115 x 1 = 115
1 x 115, 5 x 23, 23 x 5, 115 x 1 = 115
1 x 115, 5 x 23, 23 x 5, 115 x 1
1 x 115, 5 x 23 | 23 x 5, 115 x 1
From first equation: x = 39 - 7ySubstitute this in second equation: 3(39 - 7y) - 2y = 2ie 117 - 21y - 2y = 2ie 115 = 23yie y = 5, making x = 4.
If the number is x then 135*x = 115 so tat x = 115/135 = 0.851851...
5 x 115 = 575
0 x 39 = 0 1 x 39 = 39 2 x 39 = 78 3 x 39 = 117 4 x 39 = 156 5 x 39 = 195 6 x 39 = 234 7 x 39 = 273 8 x 39 = 312 9 x 39 = 351 10 x 39 = 390
115