If a vector's magnitude is 36 units right and 33 units up, what's the approximate magnitude of a vector that's one-third the length of the first vector? a. 48.8 b. 16.3 c. 12.2 d. 24.4

The correct option is B. 16.3

Step-by-step explanation:

We are given a vector with components

V = 36x +33y

To solve this problem, we find its magnitude

|V| = SQRT (36^2 + 33^2)

|V| = SQRT (2385)

|V| = 48.836

To find the magnitude of the new vector we divide by three the previous quantity

|V2| = 48.836/3 = 16.278

if you explain the the question more correctly i might be able to answer this better

step-by-step explanation:

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.