anonymous
  • anonymous
4x-4yi=8-12i
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Solving for y?
anonymous
  • anonymous
\[y= \frac{ 8-12i-4x }{ -4i }*\frac{ 4i }{ 4i } = \frac{ 32i+48-16ix }{ 16 }= 2i+3-ix = 3+(2-x)i\]
anonymous
  • anonymous
y= 3+(2-x)i It got cut off.

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anonymous
  • anonymous
solving for y and x
anonymous
  • anonymous
\[x=\frac{ 8-12i+4yi }{ 4 } = 2-3i+yi = 2+(y-3)i\]
anonymous
  • anonymous
thank you so much can you also help me with writing a complex number in standard form square root of 9+square of blank=blank+square root blank(blank)=blank+square root of blank times square root of blank=blank *i know it looks crazy but i swear this is whats in my book
anonymous
  • anonymous
\[\sqrt{9}+_^2= _+_\sqrt( ) = _\sqrt( )\sqrt( )= _\] uh....lol Could you draw what it looks like?
anonymous
  • anonymous
|dw:1370544084858:dw|
anonymous
  • anonymous
So it's the square root of 9 times the square root of blank for the first one? Sorry, could you draw 'blank+square root blank(blank)' It got cut off.
anonymous
  • anonymous
|dw:1370544695313:dw|
anonymous
  • anonymous
sqrt(9)+sqrt()*(1+1/5)?
anonymous
  • anonymous
\[\sqrt{9}+\sqrt{?}=blank \space +\sqrt{?}\left( ? \right)=blank +\sqrt{?}\times \sqrt{?}=blank \space\]
anonymous
  • anonymous
So 'blank' and '?' are different?
anonymous
  • anonymous
when its just blank i put blank when its blank under a square root i leave it blank
anonymous
  • anonymous
if you cant do it i understand
anonymous
  • anonymous
I think I have an idea.
anonymous
  • anonymous
it deals with complex numbers in standard form if that helps
anonymous
  • anonymous
\[sqrt(9)+\sqrt(blank)=blank +\sqrt(blank)\sqrt(blank)\] \[-2blank+\sqrt(blank)+3\] let x = \[\sqrt(blank)\] \[-2x^2+x+3\] Use quadratic formula x= \[\frac{ -1 \pm \sqrt(1-(4*-2*3)) }{ 2(-2) }\]
anonymous
  • anonymous
the part after the first = is blank+a square root and a parenthesis with a blank all under the same square root then it = blank+square root *(times) square root= answer
anonymous
  • anonymous
\[\frac{ 1 }{ 4 }\pm \frac{ \sqrt(-23) }{ -4 } \] x= \[\frac{ 1-i sqrt(23) }{ 4 }\] or \[x=\frac{ 1+i \sqrt(23) }{ 4 }\] x = \[\sqrt(blank)\] \[blank=x^2\]
anonymous
  • anonymous
Right, I skipped the second one and compared the first one with the third since they're all equal anyway.
anonymous
  • anonymous
If foil it out we get \[\frac{ 1 }{ 16 }-\frac{ \sqrt(23) }{ 4 } \pm \frac{ i \sqrt(23) }{ 8 }\]
anonymous
  • anonymous
I'm not 100% sure that's right, but the steps seem to logically make sense.
anonymous
  • anonymous
can you put it in drawing form please
anonymous
  • anonymous
Hang on.
anonymous
  • anonymous
√+?√=blank +?√(?)=blank+?√×?√=blank |dw:1370547228989:dw|
anonymous
  • anonymous
|dw:1370547491785:dw|
anonymous
  • anonymous
where you get theseacond reply from
anonymous
  • anonymous
|dw:1370547692327:dw|
anonymous
  • anonymous
Yeah, I think I got it wrong.
anonymous
  • anonymous
what part of the equation is that
anonymous
  • anonymous
Which part?
anonymous
  • anonymous
yeah
anonymous
  • anonymous
what about the first part
anonymous
  • anonymous
What do you mean?
anonymous
  • anonymous
the square root of 9 + the blank square root what is that
anonymous
  • anonymous
Isn't that what you said?
anonymous
  • anonymous
your confusing me can you draw the whole problem again
anonymous
  • anonymous
\[\sqrt(9)+\sqrt(blank) =blank +\sqrt(blank*blank) = blank +\sqrt(blank) \sqrt(blank)\]=blank
anonymous
  • anonymous
Right?
anonymous
  • anonymous
right
anonymous
  • anonymous
The last one on the right is \[blank +\sqrt(blank)\sqrt(blank)\]
anonymous
  • anonymous
I don't know how to do this problem then. Sorry.
anonymous
  • anonymous
ok can you do 8x+8yi=16+24i question: find the real numbers x and y to make the equation true

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