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