22/2 *2/2
Not sure about five, but you can make 42 with 6 2s with this method: 2[(22)! - (2/2) - 2] = 42
2s + 16 = 4s - 6 Subtract 2s from both sides: 16 = 2s - 6 Add 6 to both sides: 22 = 2s divide both sides by 2: s = 11
(22/2+2)x2
2 + 2 + 2 + 2/2 = 6 - 1 = 5
22/2 *2/2
Not sure about five, but you can make 42 with 6 2s with this method: 2[(22)! - (2/2) - 2] = 42
7-2s+4+5s ..... Okay first you combine like terms(add the like terms)..so its going to be 11+3s I did 7+4=11 then -2s+5s=3s So the answer can be 11+3s or 3s+11
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
2s + 16 = 4s - 6 Subtract 2s from both sides: 16 = 2s - 6 Add 6 to both sides: 22 = 2s divide both sides by 2: s = 11
(22/2+2)x2
2 + 2 + 2 + 2/2 = 6 - 1 = 5
22 + 22 - 2 = 42
2x2x2+2/2 = 9
You can use three 2s to make 26 as follows (2x2)! + 2 = 26 So you can simply make 26 with five 2s as (2x2!) + 2 - 2 + 2 = 26
L = S + 11; 2S + 3(S + 11) = 123, ie 2S + 3S + 33 = 123ie 5S = 90 ie S = 18 and L = 29
Nitrogen has five electron orbitals: one 2s orbital and three 2p orbitals.