5. If R(2.2) and S(5.4) i. Calculate the magnitude of RS ii. Express RS in the form (K, D), where K is the magnitude and Q the bearing. ​

We have:

R = (2, 2)

S = (5, 4)

RS is the segment that connects R with S.

So RS is just the distance between R and S.

Remember that the distance between two points (a, b) and (c, d) is given by:

Then the distance between S and R is

Then:

RS = √13

ii) Now we want to express RS in the form (K, Q)

where K is the magnitude (we already know that it is √13) and Q is the bearing, which is measured with respect to the horizontal axis.

To interpret the bearing, let's look at the image below:

We can use the relation:

Tan(θ) = (opposite cathetus)/(adjacent cathetus)

where the opposite cathetus is just the difference between the y-values:

Opposite cathetus = 4 - 2 = 2

And the adjacent cathetus is the difference between the x-values:

Adjacent cathetus = 5 - 2 = 3

Then we have:

Tan(θ) = 2/3

If we apply the inverse tangent equation to both sides, we get:

Atan(tan(θ)) = Atan(2/3)

θ = Atan(2/3) = 33.69°

The bearing is 33.69°

then:

RS = (√13, 33.69°)

let's add {f(x)=x+1}f(x)=x+1 and {g(x)=2x}g(x)=2x together to make a new function.

f(x)+g(x) =(x+1)+(2x)=x+1+2x=3x+1

let's call this new function hh. so we have:

{h(x)}={f(x)}+{g(x)}{=3x+1}h(x)=f(x)+g(x)=3x+1

we can also evaluate combined functions for particular inputs. let's evaluate function hh above for x=2x=2. below are two ways of doing this.

method 1: substitute x=2x=2 into the combined function hh.

h(x)

h(2)

​    

=3x+1

=3(2)+1

=7

​ since h(x)=f(x)+g(x)h(x)=f(x)+g(x), we can also find h(2)h(2) by finding f(2) +g(2)f(2)+g(2).

first, let's find f(2)f(2):

f(x)

f(2)

​    

=x+1

=2+1

=3

​  

now, let's find g(2)g(2):

g(x)

g(2)

​    

=2x

=2⋅2

=4

​  

so f(2)+g(2)=3+4=\greend7f(2)+g(2)=3+4=7.

notice that substituting x =2x=2 directly into function hh and finding f(2) + g(2)f(2)+g(2) gave us the same answer!