Acar leave p and drives 3km north to q. from q it's drives 15km on a bearing north 45degree east to r. find the distance and bearing of r from p

17.25 km N37.9°E

Step-by-step explanation:

The way to solve this is to determine the vertical and horizontal distances traveled.

The bearing means 45° east of north.

Vertically, the car travels a distance of:

y = 3 + 15 cos 45°

y = 3 + 7.5√2

y ≈ 13.6

Horizontally, the car travels a distance of:

x = 15 sin 45°

x = 7.5√2

x ≈ 10.6

The total distance we find with Pythagorean theorem:

d = √(x² + y²)

d = √(10.6² + 13.6²)

d ≈ 17.25

The bearing from north we find with tangent:

tan θ = x / y

tan θ = 10.6 / 13.6

θ ≈ 37.9°

the equation for this situation is y=122+45. to solve the equation you would add 45 and 122 to get a total of 167lbs. in conclusion, grace's weight eight weeks ago was 167lbs.  

55% are girls and 45% are boys : ) hope this

step-by-step explanation: