Apowerboat heads due northwest at 13 m/s relative to the water across a river that flows due north at 5.0 m/s. what is the velocity (both magnitude and direction) of the motorboat relative to the shore?

The boat is travelling at 16.9 m/s at 57 degrees North of West. 

This is a vector problem. 
The '^' symbol denotes a power, ('4^2' is '4 squared') 

The first step is to break down the components of the NorthWest motion (13 m/s). The phrase Northwest means it has an angle of 45 degrees north of west. So both the north and west components will be equal. 
Pythagorean theorem: A^2 + B^2 = C^2 
We have C^2, it is 13^2 which is 169. 
Divide that by two to get the squares of one of the components: 84.5,. 
The square root of 84.5 is 9.19. 

So the components of the Northwest motion at 13 m/s is 
9.19 m/s north and.. 
9.19 m/s west. 

Add the north motion of the boat relative to the water to the motion of the water (north at 5 m/s) 
9.19+5= 14.19 m/s North. 

Our new vector components are 
14.19 m/s North and 
9.19 m/s West. 
Pthagorean theorem again tells us the magnitude of the motion: 
14.19^2 + 9.19^2 = C^2 = 285.81 
Take the square root and get 16.9 m/s 

Now we have a vector triangle. 

Opposite wall: 14.19 
Adjacent Wall: 9.19 
Hypotenuse: 16.9 

Now we need the angle in the bottom right corner: 
Sin(angle)= opposite / hypotenuse = 14.19/16.9 
Sin(angle)= .8396449704 
ArcSin(.8396449704) = 57.1 degrees 

The magnitude and direction: 
16.9 m/s at 57 degrees North of West. 

ArcSin is a reverse trigonometry operation available on most graphing calculators, it appears as "Sin^-1"