Ajogger ran 3 miles due east of his house. then he ran 5 miles at a heading of 30o east of north (or 30o ne). how far is he from his house after running 8 miles

We want to find how far is the jogger from his house after he runs a total of 8 miles, we will see that the distance is equal to 7 miles.

So we need to define a coordinate axis, the point (0, 0) will be the jogger's house (where he/she starts). North is the positive y-axis and East is the positive x-axis.

Then the jogger starts at (0, 0).

Then he ran 3 miles due East, so the new position is:

(0, 0) + (3mi, 0) = (3mi, 0)

Then he ran 5 miles at 30° East of North (or 60° North of East).

The components are:

Then the new position is:

(3mi, 0) + (2.5mi, 4.33 mi) = (5.5mi, 4.33 mi)

The distance to the jogger's house is just the magnitude of that final vector, which is:

||  (5.5mi, 4.33 mi) || =  √( (5.5mi)^2 + (4.33mi)^2) = 7mi

So the jogger is at 7 miles from his/her house.

If you want to learn more about magnitudes, you can read:

link

let m = monomial

let t = trinomial

so, m times t = t.

monomial times a trinomial = another trinomial.