63
Rkfj
(11S+3)
click on the clocks that are at the times 2 3 5 7 11
B = bigger numberS = smaller numberYou told us:B = 3SB - S = 2,184Substitute 3S in place of B in the second statement:3S - S = 2,1842S = 2,184S = 1,092Their sum is B+S, and we know that B=3S.B + S = 3S + S = 4S = 4,368.
8.3 times or 8.333333333333333 (8 and 15 .3's)
Well 3 20's, 3 10's and 3 5's would be $105.
25+11=36: Let f and s represent the first and second numbers respectively. The statement of the problem yields two equations: f + s =36 and f = 3 + 2s. Substituting the function given in the second equation for f into the first equation yields 3 + 2s + s = 36, or (subtracting 3 from each side and merging the s terms, 3s = 33 or s = 11. Then f + 11 = 36 (substituting the value for s into the first equation), or f = 25.
11s + 3
=7(x+s)
11 (interior angles total 18 x 90)
click on the clocks that are at the times 2 3 5 7 11
It is: 1*11 = 11 which is a prime number
The sum is 50 and 31/40 The difference is 1/40
111 + 11 + 1 = 123
You record the various speed on the same distance and sum it. Then divide that sum with how many times you record it. Like for x meters : 1. a m/s 2. b m/s 3. c m/s . . n. y m/s The average would be : (a+b+c +...y)/n
11 times
8.3 times or 8.333333333333333 (8 and 15 .3's)
B = bigger numberS = smaller numberYou told us:B = 3SB - S = 2,184Substitute 3S in place of B in the second statement:3S - S = 2,1842S = 2,184S = 1,092Their sum is B+S, and we know that B=3S.B + S = 3S + S = 4S = 4,368.
Well 3 20's, 3 10's and 3 5's would be $105.