3.75
21
50 = 15% x ? 50 divided by .15 = 333.33 (infinity)
The smallest number that can be subtracted from 1294, yielding a result that when divided by 9, 11, and 13 leaves a remainder of 6 is 1. To prove this, run the following snippit of a program int i; for (i=1294; i!= 0; i--) if (i % 9 6) break; printf("%u", 1294-i); another way to prove it it to multiply 9 11 and 13, yielding 1287, adding six, yielding 1293, which is one away from 1294.
5n + 15 = 25 you subtract 15 from both sides then you get 10 then divided 5n by 10 you get 2n
Nothing Jk
3.75
21
32 and 1/3 32.333333
50 = 15% x ? 50 divided by .15 = 333.33 (infinity)
The smallest number that can be subtracted from 1294, yielding a result that when divided by 9, 11, and 13 leaves a remainder of 6 is 1. To prove this, run the following snippit of a program int i; for (i=1294; i!= 0; i--) if (i % 9 6) break; printf("%u", 1294-i); another way to prove it it to multiply 9 11 and 13, yielding 1287, adding six, yielding 1293, which is one away from 1294.
0.3333
Absolute Value -1294 is 1294.
5n + 15 = 25 you subtract 15 from both sides then you get 10 then divided 5n by 10 you get 2n
p + 3 + = -3
No it is divisably by 2 1294/2=647
Since there are no brackets or parentheses, the divisions must be evaluated from left to write so that the expression is ((25/5x)/15)/3x = ((5/x)/15)/3x = (5/15x)/3x = (1/3x)/3x = 1/(9x2)