2r + 2s = 50 2r - s = 17 therefore 4r - 2s = 34 Add so that you can eliminate one of the variables: 2r + 2s = 50 4r - 2s = 34 ---------------- 6r + 0s = 84 Solve for r: 6r = 84 r = 14 Substitute r into one of the original equations: 2(14) + 2s = 50 28 + 2s = 50 2s = 22 s = 11 Doublecheck with the other original equation: 2(14) - 11 = 28 - 11 = 17
That's a simple example of system of equations. There are quite a number of methods for solving those, but the easiest here would be to calculate one variable from first equation and substitute it into second. 2s + 3t = 28 5s + 6t = 61 Let's calculate s from the first equation: 2s + 3t = 28, 2s = 28 - 3t s = 14 - 3t/2 Substitute s into second equation 5s + 6t = 61, 5(14 - 3t/2) + 6t = 61, 70 - 15t/2 + 6t = 61, 70 - 3t/2 = 61, 3t/2 = 9, t = 6. We can then substitute t back into first equation: 2s + 3t = 28, 2s + 3 * 6 = 28, 2s = 10, s = 5. So the final answer is: s = 5 t = 6
There is only one 2s orbital in an atom.
25 of them
310.
2000
100 2s for the hundreds digit of 200-30010 2s for the tens digit of 220-23010 2s for the ones digit of 200-300________________________________120 '2s'
unknown
1
36
26
42