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math

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175
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-2i(4 - 3i) + 6i

Guest Jun 7, 2017

#2
+91465
+2

Thanks x^2 you are absolutely correct :)

But sometimes the letter i is used as the basic unit of the comples (imaginary) number system.

If i is the imaginary number then  $$i=\sqrt{-1}$$

Then

$$6i^2-2i\\ =6*-1-2i\\ =-6-2i$$

Melody  Jun 7, 2017
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#1
+1602
+2

SImplifying this should be relatively simple.

 $$-2i(4-3i)+6i$$ Distribute the -2i into the parentheses $$-8i+6i^2+6i$$ Combine the like terms $$-2i+6i^2$$ Rearrange the terms so the term with the highest degree is first $$6i^2-2i$$ You are done!
TheXSquaredFactor  Jun 7, 2017
#2
+91465
+2

Thanks x^2 you are absolutely correct :)

But sometimes the letter i is used as the basic unit of the comples (imaginary) number system.

If i is the imaginary number then  $$i=\sqrt{-1}$$

Then

$$6i^2-2i\\ =6*-1-2i\\ =-6-2i$$

Melody  Jun 7, 2017

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