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anonymous
 one year ago
Let u = <5, 1>, v = <7, 4>. Find 9u  6v.
anonymous
 one year ago
Let u = <5, 1>, v = <7, 4>. Find 9u  6v.

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jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.19u < scalar multiplication, same for 6v 9u  6v < vector addition do the scalar mutiplication first, to get a resultant vector then "add" them up, or subtract in this case

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You might get annoyed, but I'm new to this topic, so I don't know what to do. Maybe baby steps? :)

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1\(\bf { \begin{array}{llll} u=<5,1>\\ v=<7,4> \end{array}\qquad \begin{array}{llll} 9u\implies <9\cdot 5,9\cdot 1>\implies <45,9>\\ 6v\implies <6\cdot 7,6\cdot 4>\implies <42,24> \end{array} \\ \quad \\ 9u6v\implies <45,9> + <42,24>\\ \quad \\ <45+(42)\quad ,\quad 9+24> }\)

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1well... could have been done using the  I gather

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ohhh, so does it look like I'm distributing 9u to the first one (u), and 6v to v?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And the answer is <87, 33>?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much! It became clear now. :)

jdoe0001
 one year ago
Best ResponseYou've already chosen the best response.1\(\begin{array}{llll} u=<5,1>\\ v=<7,4> \end{array}\qquad \begin{array}{llll} 9u\implies <9\cdot 5,9\cdot 1>\implies <45,9>\\ 6v\implies <6\cdot 7,6\cdot 4>\implies <42,24> \end{array} \\ \quad \\ 9u6v\implies <45,9>  <42,24>\implies <4542\quad ,\quad 9(24)>\) would have done the same thing anyhow
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