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anonymous
 5 years ago
add or subtract:
1.) (82i) + (6+4i)
2.) (9+5i)  (1+8i) + (9+10i)
anonymous
 5 years ago
add or subtract: 1.) (82i) + (6+4i) 2.) (9+5i)  (1+8i) + (9+10i)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0kathyworld, do you know how to do this? If you were paying attention to the past problems you should know how to do these. These are far easier than the other questions you posted and to solve those problems you needed to know how to add these. You could not have solved the other problems without knowing how to do this. Are you just writing down what people are saying the answers are? I just want you to do good in math and this is not how to do it.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01.) (82i) + (6+4i)=(8+6)+(2+4)i=14+2i 2.) (9+5i)  (1+8i) + (9+10i) = (91+9)+(58+10)i=17+7i. kathy, i know where your problem is. you're not very familiar with the addition, subtract logarithm of imaginaries.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i really don't get the whole "i" process

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes , i get really confuse of how the "i" used

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok. That's a good place to start :) \[i = \sqrt{1}\] Which is obviously not a "real number" because you cannot take the square root of a negative number. So instead, it uses i to mean the square root of negative one which is called an imaginary number. So you can treat it like x like you usually would in algebra.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lol !! oh yes very stupid

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i is just an imaginary number. i know it;s hard for you to change your, maybe deapseated, mind that i has to be a real number...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So solve the problems you posted as if it were x instead of i.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you can regard i as simply an inseparable thing there.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The only difference is if you multiply i by i, which becomes 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0as in just another variable ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0WOW im not surprised some weirdo bully who takes joy in puttin others down is on this site being rude >:O not surprised!!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0right, so you have to classify the given numbers into reals and imaginaries

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No bullying happening here... I'm trying to help. Writing down the answers without knowing why is a bad idea.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think John's purpose is good in helping kathy to solve the problem autonomously, but a little rude, at least in my opinion...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0she didnt mean to you john , she meant to the "someguy"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ah, i made a mistake it's someguy//... i'm sorryyyy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes i did mean the someguy! HAHA sowee if u thought i meant u john! ^^ i just do NOT like bullies at ALL ..pet peeve i guess :P lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0agreed/ back to the problem itself, kathy, do u still have something unclear?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh jeez. Sorry. Thats a funny username. Sorry if I came off as rude. Do you understand how to do the problem now? When you have the standard form a + bi of a complex number ( a complex number is a number with an imaginary part), a is the real part and bi is the imaginary part. When working with complex numbers you can just act as if i is a variable like x except when it is i^2 it "magically" becomes real. Just think of what i means. The square root of 1 times the square root of 1 is just 1!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yesss i understand now , sorry it was just very confusing to me in the beginning

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0except for the fact that i^2=1
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