PreCal
posted by Kim .
Perform the operation and write the result in standard form.
(23i)(5i) over 2+3i
Please help!! I do not understand this.

(23i)(5i) over 2+3i
= 5i(23i)/(2+3i)
multiply top and bottom by 23i
= 5i(4  12i + 9i^2)/(49i^2)
= 5i(13  12i)/13
= (65i  60i^2)/13
= (60  65i)/13
or 60/13  5i if by standard form you mean a + bi 
ok, Kim ,I messed up in the arithmetic
but, if you understand what I did, you should be able to fix it yourself.
hint: the error is from
= 5i(4  12i + 9i^2)/(49i^2) to
= 5i(13  12i)/13 
YOU SHOULD KNOW HOW TO DO THIS!
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