34 is halfway between 23 and 45 inclusively.Therefore, 34 x 2 = 68
Let x = one of the integers x + 1 = the other integer x + x + 1 = 45 2x = 44 x = 22 x + 1 = 23
1 x 45 = 45 2 x 45 = 90 3 x 45 = 135 4 x 45 = 180 5 x 45 = 225 6 x 45 = 270 7 x 45 = 315 8 x 45 = 360 9 x 45 = 405 10 x 45 = 450 11 x 45 = 495 12 x 45 = 540
The first ten positive integer multiples of 45 are as follows: 1 x 45 = 45 2 x 45 = 90 3 x 45 = 135 4 x 45 = 180 5 x 45 = 225 6 x 45 = 270 7 x 45 = 315 8 x 45 = 360 9 x 45 = 405 10 x 45 = 450
0.5111
The answer to 23 X 45 = 1035
1800 25 = 52 40 = 23 x 5 45 = 32 x 5 lcm = 23 x 32 x 52 = 1800
34 is halfway between 23 and 45 inclusively.Therefore, 34 x 2 = 68
The answer is nine. Seriously. Here is my work if you don't believe me: 68=8x+23+4x-7x 68=8x-7x+23+4x 68=x+23+4x 68=x+4x+23 68=5x+23 68-23=5x 45=5x x=45/5 x=9
23 = 68 - 45 = (4 x 17) - (5 x 11) + 10
Let x = one of the integers x + 1 = the other integer x + x + 1 = 45 2x = 44 x = 22 x + 1 = 23
23 x 5 = 40 32 x 5 = 45
68 / 45 = 1 r 23 45 / 23 = 1 r 22 23 / 22 = 1 r 1 22 / 1 = 22 r 0 GCF of 45 and 68 is 1. In prime factorisation: 45 = 32 x 51 68 = 22 x 171 There are no common primes, so the greatest common (and only) factor is 1.
23/27 ÷ 45 = 23/27 ÷ 45/1 = 23/27 × 1/45 = (23×1)/(27×45) = 23/1215
22
45 percent of 23 = 10.35
68