Is done by (3/3) plus (3/3) = 2
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∙ 6y ago2/3.
First, observe that 5 = 3 + 2. That uses 1 of the 3s. How do we get 2 from what's left? Add 3 and 3 to get 6, then divide that by the remaining 3 to get a 2. Done!
6 because 6/3 = 2
20 divided by 3 = 6 and 2/3
Four years ago the son was 8, and the father 32. Now the son is 12, and the dad 36. To solve this problem, you should use the substitution method. The current age of the father(F) and the son(S) are represented by the equation F=3S. The mans age minus four years(F-4) is four times the sons minus four years(S-4), so the equation is (F-4)=4x(S-4). The next step is to simplify equation 2. (F-4)=4x(S-4) becomes F-4=4S-16, and then F=4S-12. Now you substitute equation one into equation two. F=3S, F=4S-12 3S=4S-12 Simplify to find the current age of the son: 0=1S-12 12 = S
3/3 + 3/3 = 6/3 = 2
The LCM of 3s and s^2 is 3s^2
33/3 = 11
-2/2-2/2 = -2
no because 2/3s is 2/3s of 1 therefore 1 is more then anything of it
2/3.
3. (2 (1s), 2 (2s) 6 (2p), 2 (3s) 1 (3p))
5x7=35, then 3-1=2, so 35+2=37
The are two electrons in the 3s orbital of magnesium (Mg.)
-7 = 3s - 1 +1 +1 (add 1 to both sides to get the variable alone) ___ ____ -6 = 3s __ ___ (-6 and 3s both divided by 3) 3 3 -2 = s
3 x (22)2 + 2 = 50
First, observe that 5 = 3 + 2. That uses 1 of the 3s. How do we get 2 from what's left? Add 3 and 3 to get 6, then divide that by the remaining 3 to get a 2. Done!