Let
be a set of orthogonal vectors. By definition of orthogonality, any pairwise dot product between distinct vectors must be zero, i.e.
Suppose there is some linear combination of the
such that it's equivalent to the zero vector. In other words, assume they are linearly dependent and that there exist
(not all zero) such that
(This is our hypothesis)
Take the dot product of both sides with any vector from the set:
By orthogonality of the vectors, this reduces to
Since none of the
are zero vectors (presumably), this means
. This is true for all
, which means only
will allow such a linear combination to be equivalent to the zero vector, which contradicts the hypothesis and hence the set of vectors must be linearly independent.
Answer:
3:15 PM.
Step-by-step explanation:
If it is already 3:00 and you add 15 to the 2 empty 00's it will add up to 15. Hope this helps!
Answer:
50th term is -338
Step-by-step explanation:
Common difference d = -2-5 = -7
first term is a1 = 5
50th term
a50 = a1 + (n -1)d
a50 = 5 -7(50-1)
a50 = 5 -7(49)
a50 = -338
Hope this will helpful.
Thank you.
Answer:
Step-by-step explanation:
So, we are given:
First, we should immediately rule out 0 as an answer. This is because the if , the equation would be undefined.
Now, cross multiply.
Divide everything by x (and we can do this safely because we already know x cannot be equal to zero).
We didn't run into any possibilities for extraneous solutions.
2f(x) + 3 as
the function persevere all length and angle measures