Answer
$\overrightarrow{a}+\overrightarrow{b} = <6,3>$
$4\overrightarrow{a}+ 2\overrightarrow{b}=<6,14>$
$| \overrightarrow{a} |=5$
$| \overrightarrow{a}-\overrightarrow{b} |=13$
Work Step by Step
Given vectors $\overrightarrow{a}=<-3,4>$ and $\overrightarrow{b}=<9,-1>$
1) We find $\overrightarrow{a}+\overrightarrow{b}$ by adding the components of $\overrightarrow{a}$ and $\overrightarrow{b}$
$\overrightarrow{a}+\overrightarrow{b} = <-3+9,4-1>=<6,3>$
2) We find $4\overrightarrow{a}+ 2\overrightarrow{b}$ by multiplying the components by their respective coefficients:
$4\overrightarrow{a} = <4(-3),4(4)>=<-12,16>$
$2\overrightarrow{b} = <2(9),2(-1))>=<18,-2>$
and then adding their components:
$4\overrightarrow{a}+ 2\overrightarrow{b}=<(-12+18),(16-2)>=<6,14>$
3)We find $| \overrightarrow{a} |$ with the distance formula:
$| \overrightarrow{a} |=\sqrt{(a_x)^2+(a_y)^2}$
$| \overrightarrow{a} |=\sqrt{(-3)^2+(4)^2}=\sqrt{25}=5$
4) Finally, we find $| \overrightarrow{a}- \overrightarrow{b} |$ by first subtracting the components of $\overrightarrow{b}$ from $\overrightarrow{a}$
$\overrightarrow{a}-\overrightarrow{b} = <-3-9,4-(-1)>=<-12,5>$
and then using the distance formula
$| \overrightarrow{a}-\overrightarrow{b} |=\sqrt{(-12)^2+(5)^2}=\sqrt{144+25}=13$
