The "triple product" is defined for vectors. You can't calculate a triple product if you don't know the components of the vectors (or some other information, that allows you to calculate those).
If by "triple dot product" you mean u·v·w, then no, because that would imply the existence of a dot product between a vector and a scalar.
36a2 - 60a + 25 = 36a2 - 30a - 30a + 25 = 6a(6a - 5) - 5(6a - 5) = (6a - 5)(6a - 5) = (6a - 5)2
10a-37=6a+51 -6a -6a subtract 6a from both sides and the 6a's cancel 4a-37=51 +37 +37 4a=88 /4 /4 a=22
15n
6a
-(6a)
B2 has a longer bond length than B2. This is because B2 has a triple bond, which is shorter and stronger than a single bond present in B2.
..
(a-b) (a+b) = a2+b2
If by "triple dot product" you mean u·v·w, then no, because that would imply the existence of a dot product between a vector and a scalar.
Divide by 6a: 6a(a + 3b)
36a2 - 60a + 25 = 36a2 - 30a - 30a + 25 = 6a(6a - 5) - 5(6a - 5) = (6a - 5)(6a - 5) = (6a - 5)2
Suppose you have two sets of n-numbers: {a1, a2, a3, ... , an} and {b1, b2, b3, ... , bn} Then the form for the standard sum of product is a1*b1 + a2+b2 + a3*b3 + ... + an*bn
10a-37=6a+51 -6a -6a subtract 6a from both sides and the 6a's cancel 4a-37=51 +37 +37 4a=88 /4 /4 a=22
15n
6a
18