Question

Trig help! Screenshot is attached

Answer 1

\(\bf \textit{distance between 2 points or vector's magnitude}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
&({\color{red}{ \square }}\quad ,&{\color{blue}{ \square }})\quad &({\color{red}{ \square }}\quad ,&{\color{blue}{ \square }})
\end{array}\qquad
d = \sqrt{({\color{red}{ x_2}}-{\color{red}{ x_1}})^2 + ({\color{blue}{ y_2}}-{\color{blue}{ y_1}})^2}\)

Answer 2

Is it 10?

Answer 3

hmm... what are your points though?

Answer 4

sqrt(-2-4)^2+(-3-5)^2

Answer 5

5 is where I messed up right? It would be positive which one make the answer 2sqrt10?

Answer 6

\(\bf \textit{distance between 2 points or vector's magnitude}\\ \quad \\
\begin{array}{lllll}
&x_1&y_1&x_2&y_2\\
&({\color{red}{ -4}}\quad ,&{\color{blue}{ 5}})\quad &({\color{red}{ -2}}\quad ,&{\color{blue}{ -3}})
\end{array}\qquad
d = \sqrt{({\color{red}{ -2}}-{\color{red}{ (-4)}})^2 + ({\color{blue}{ -3}}-{\color{blue}{ 5}})^2}\)

Answer 7

that is.. using vector "u"

Answer 8

actually.. they aren't labeled =).. anyhow the one on the left

Answer 9

I see where I messed up! Let me see if I can get the 2nd part right! haha

Answer 10

Got it! Sorry I seem to make everything more difficult than it needs to be ;-/! thank you!

Answer 11

yw

Answer 12

@jdoe0001 Could you explain to me how to find lwl?

Answer 13

maybe better if you post anew.. dan815 may know better