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Is the anser
Answer is in the file below
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120/4=30. Remember its widthxlength. if you forget ask your teacher or look online for more help. Good luck.
Answer:
The set


Step-by-step explanation:
We separate the left and right side of the equation, to create two new equations:
(1) 
(2). 
and then to solve Tanisha's original equation, we find the solution to the system that we got above.
The first step in solving the above system of equations is to take the value for
from equation(1), and substitute it into equation(2) which gives us:

which is the same equation that Tanisha set out to solve.
let's keep in mind that i² = -1, and let's use the conjugate of the denominator.
