Answer: i - j - k
Step-by-step explanation:
Taking the cross product between two vectors will give you a third vector that is orthogonal(perpendicular) to both vectors.
<1,1,0> x <1,0,1>
![det(\left[\begin{array}{ccc}i&j&k\\1&1&0\\1&0&1\end{array}\right] )](https://tex.z-dn.net/?f=det%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Di%26j%26k%5C%5C1%261%260%5C%5C1%260%261%5Cend%7Barray%7D%5Cright%5D%20%29)
the determinate of the matrix: <1,-(1),-1>
or: i - j - k
63 is the answer to your problem. Hope this helps!
Answer:
Ok, first in our series we can see two numbers in the Sigma, one bellow 0, and other above, 4.
This means that the value of k will go from 0 to 4, then all the numbers in the sum are:
(-1/2)^0 + (-1/2)^1 + (-1/2)^2 + (-1/2)^3 + (-1/2)^4
So we have 5 terms in our series.
b) to see the sign in each term, we must solve the powers, remember that:
(-1)^n is -1 if n is odd, and is equal to 1 if n is even, so we have:
(-1/2)^0 + (-1/2)^1 + (-1/2)^2 + (-1/2)^3 + (-1/2)^4
= 1 -1/2 + 1/4 - 1/8 + 1/16.
So the sign in each term of the series alternates.
Answer:
D
Step-by-step explanation:
