6i-3j
to polar coordinates

Let m and n be two constants. Vectors in the form mi + nj can be represented as ordered pairs in the form (m,n). For the vector 6i-3j, it is located at (6,-3) since m = 6 and n = -3

Find the magnitude of the vector

r = magnitude

r = sqrt(m^2+n^2)

r = sqrt(6^2+(-3)^2)

r = sqrt(36+9)

r = sqrt(45)

r = 6.708 approximate

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Find the angle of the vector that it makes with the positive x axis.

theta = arctan(n/m)

theta = arctan(-3/6)

theta = arctan(-1/2)

theta = -26.565 degrees approximately

theta = -26.565+360

theta = 333.435 this is coterminal with the -26.565 degree angle

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See the diagram below.

10.416 m/s

step-by-step explantion:

distance=100m

time= 9.6s

average speed=100/9.6

=10.416 m/s