vera_ewing
  • vera_ewing
Math question
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
vera_ewing
  • vera_ewing
vera_ewing
  • vera_ewing
vera_ewing
  • vera_ewing

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

vera_ewing
  • vera_ewing
vera_ewing
  • vera_ewing
LynFran
  • LynFran
|dw:1443315132541:dw|
LynFran
  • LynFran
to find the translation we add \[J\left(\begin{matrix}-3 \\ 5\end{matrix}\right)+\left(\begin{matrix}1 \\ 8\end{matrix}\right)\]=\[\left(\begin{matrix}-2 \\ 13\end{matrix}\right)\] we do this for ever point \[k \left(\begin{matrix}4 \\ 4\end{matrix}\right)+\left(\begin{matrix}1 \\ 8\end{matrix}\right)=\left(\begin{matrix}5 \\ 12\end{matrix}\right)\]\[L \left(\begin{matrix}3 \\ -2\end{matrix}\right)+\left(\begin{matrix}1 \\ 8\end{matrix}\right)=\left(\begin{matrix}4 \\ 6\end{matrix}\right)\]
LynFran
  • LynFran
|dw:1443316254771:dw|
LynFran
  • LynFran
\[J'\left(\begin{matrix}-2 \\ 13\end{matrix}\right)+\left(\begin{matrix}7 \\ 4\end{matrix}\right)=\left(\begin{matrix}5 \\ 17\end{matrix}\right)\]\[K'\left(\begin{matrix}5 \\ 12\end{matrix}\right)+\left(\begin{matrix}7 \\ 4\end{matrix}\right)=\left(\begin{matrix}12 \\ 16\end{matrix}\right)\]\[L'\left(\begin{matrix}4 \\ 6\end{matrix}\right)+\left(\begin{matrix}7 \\ 4\end{matrix}\right)=\left(\begin{matrix}11 \\ 10\end{matrix}\right)\]
vera_ewing
  • vera_ewing
Wait so which one would it be?
LynFran
  • LynFran
http://assets.openstudy.com/updates/attachments/56074b58e4b0a953837026ba-vera_ewing-1443318652152-screenshot20150926at6.49.39pm.png
vera_ewing
  • vera_ewing
So A?
LynFran
  • LynFran
shd be \[J'\left(\begin{matrix}-2 \\ 13\end{matrix}\right)+\left(\begin{matrix}7 \\ -4\end{matrix}\right)=J''\left(\begin{matrix}5 \\ 9\end{matrix}\right)\]\[K'\left(\begin{matrix}5 \\ 12\end{matrix}\right)+\left(\begin{matrix}7 \\ -4\end{matrix}\right)=K''\left(\begin{matrix}12 \\ 8\end{matrix}\right)\]\[L'\left(\begin{matrix}4 \\ 6\end{matrix}\right)+\left(\begin{matrix}7 \\ -4\end{matrix}\right)=L''\left(\begin{matrix}11 \\ 2\end{matrix}\right)\]
LynFran
  • LynFran
yes A
vera_ewing
  • vera_ewing
Thanks!
LynFran
  • LynFran
welcome the translation to move JKL to J''K''L'' would be
LynFran
  • LynFran
\[J''\left(\begin{matrix}5 \\ 9\end{matrix}\right)- J \left(\begin{matrix}-3 \\ 5\end{matrix}\right)=\left(\begin{matrix}8 \\ 4\end{matrix}\right)\]\[K''\left(\begin{matrix}12 \\ 8\end{matrix}\right)-K \left(\begin{matrix}4 \\ 4\end{matrix}\right)=\left(\begin{matrix}8 \\ 4\end{matrix}\right)\]\[L''\left(\begin{matrix}11 \\ 2\end{matrix}\right)-L \left(\begin{matrix}3 \\ -2\end{matrix}\right)=\left(\begin{matrix}8 \\ 4\end{matrix}\right)\] so 8 units to the right and 4 units upward

Looking for something else?

Not the answer you are looking for? Search for more explanations.