Question

Solve:
(2-3i)(5i)/ 2+3

Answer 1

complex?

Answer 2

yes

Answer 3

I don't think that's it. You have to use the conjugate of the bottom

Answer 4

oh.

Answer 5

\[\frac{(2 \times 5i)-(3i \times 5i)}{5}= \frac{10i - 15i^2}{5}\]\[i=\sqrt{-1} ... i^2=-1\]\[\frac{10i-15(-1)}{5}= \frac{10i+15}{5} = 2i+3\]

Answer 6

Distibute the top, and then deal with the bottom. There isn't any subtraction sign in the top part

Answer 7

the bottom is 2+3 which is 5
and i already distributed whats in the numerator

Answer 8

2 plus 3 "I"

Answer 9

oh u forgot the i

Answer 10

Sorry :O

Answer 11

Why 2-9?

Answer 12

3i*3i = -9
in the denominator

Answer 13

on the bottom I got 4-9i and then that =13

Answer 14

and why is 20i not -20i

Answer 15

yeah that was my mistake im sorry ... difference between two squares (2-3i)(2+3i)=4-(-9)

Answer 16

wait let me do it again

Answer 17

yes, it is a positive 20, my bad :D

Answer 18

wait no, take back!

Answer 19

\[\frac{10i+15}{2+3i} \times \frac{2-3i}{2-3i} = \frac{20i-30(-1)+30-45i}{4-(-9)}\]\[=\frac{-25i+60}{13}\]

Answer 20

When you multiply the -10 and the 2i should it be -20i

Answer 21

what -10 and 2i?

Answer 22

thers 10i and 2

Answer 23

but the 10 is negative?

Answer 24

no when u distribute at the begining 5i*2 = 10i

Answer 25

I am going to attach my work

Answer 26

kk

Answer 27

can in progress. sorry

Answer 28

*scan

Answer 29

well the solution for \[\frac{(2-3i)(5i)}{2+3i}\]is what i did , last thing

Answer 30

Thanks