2/3.
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
92 ÷ 3 = 30 remainder 2
3/3 + 3/3 = 6/3 = 2
Well, honey, the least common multiple of a monomial like a^3s and s^2 is simply a^3s^2. You just gotta take the highest power of each variable that appears in either monomial, slap 'em together, and there you have it. Math made sassy.
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!