. For vectors B⃗ =−iˆ−4jˆB→=−i^−4j^ and A⃗ =−3iˆ−2jˆA→=−3i^−2j^ , calculate (a) A⃗ +B⃗ A→+B→ and its magnitude and direction angle, and (b) A⃗ −B⃗ A→−B→ and its magnitude and direction angle.

Step-by-step explanation:

given are two vectors

a) A+B = sum of twovectors = simply adding corrsponding compnents

=

Magnitude of the above vector =

Direction is making an angle of arc tan (-3) with x axis

b) A-B = Difference

=

Magnitude of difference vector =

Direction is making an angle of arctan (1/2) with positive x axis.

(angle = arctan (jcomponent/icomponent)

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!

step-by-step explanation:

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