First vector orthogonal<span> to ⟨−</span>3<span>,4</span>
Answer: option d. C (0,3), D (0,5).
Justification:
1) The x - coordinates of the vertices A and B are shown in the diagrama, They are both - 4, so the new vertices C and D must be in a line parallel to y = - 4.
2) The y-coordinates of the vertices A and B are also shown in the diagrama. They are equal to 3 and 5 respectively.
3) We can see that the new points C and D must be over a parallel line to y = - 4 and that their distance to the points A and B has to be the same distance of the point R and S to U and T.
That distance is 4, so the line may be y = - 7 or y = 0.
4) If the line is y = 7 the points C and D would have coordinates (-7,3) and (-7,5), but this points are not among the options.
5) If the line is y = 0 the points C and D would have coordinates (0, 3) and (0,5), which is precisely the points of the option d. That is the answer.
The answer to your question is 405
1&15 and 3&5 are factors of 15
Answer:
P(N = n) = 
Step-by-step explanation:
to find out
Find the PMF PN (n)
solution
PN (n)
here N is random variable
and n is the number of times
so here N random variable is denote by the same package that is N (P)
so here
probability of N is
P(N ) = Ф ( N = n) .................. 1
here n is = 1, 2,3, 4,...................... and so on
so that here P(N = n) will be
P(N = n) =
