Imcool

oh thanks

cool.

Let's say that z = a+bi and w = c+di

okokok

$zw - 3w - 2iw + 4z = 7iw + 5w + 11iz - 13 + 15i$

$= (a+bi)(c+di) - 3(c+di) - 2i(c+di) + 4(a+bi) = 7i(c+di) + 5(c+di) + 11i(a+bi) - 13 + 15i$

so then...

$= ac+adi+bci-bd - 3c-3di - 2ic-2di^2 + 4a+4bi = 7ic+7di + 5c+5di + 11ia+11bi^2 - 13 + 15i$

This is not the correcr way to do it oh gosh...

$= ac+adi+bci-bd -2di^2 + 4a+4bi = 9ic+15di + 8c + 11ia+11bi^2 - 13 + 15i$

$ac-bd+ 4a =+ 8c +(11b^2-4b-bc-ad+2di+9c+15d+11a+15)i - 13$

yeahhhh..... Im not the one solving this

I don't think 2877 is correct..

Um the file could't be acceced

I want you to do a portion of it to so im going to do half of it.

so first lets simplify this

$x^2 - 5x + 3 = 2x^2 - 11x + 14$

$x^2-6x+11=0$

oh man... this is crazy..

$a=3+\sqrt{2}i,\:b=3-\sqrt{2}i$

this is definently not what i had in mind...

ok so now you square them...

$a=7+(6i)\sqrt{2}$

$b=7+(-6i)\sqrt{2}$

I guess you have to find a quadratic...

@guest, next time, please try to include an explanation.

As you see, if you google, this, ANOTHER PERSON WHO DIDN"T CARE TO GIVE AN EXPLANATION EVEN WHEN THE ASKER SAID SO said the answer was 11. SO WHICH ONE IS RIGHT? Now they don't know.

if at least 3 of them are red... then they have to be positioned like this so that they don't touch:

red ? red ? red

now for the first question mark there are 2 choices, same for the second one. so our answer is 2x2 which is 4

$\boxed{4}$

BRO, DO YOU THINK I CAN SOLVE THIS WITHOUT THE PICTURE!?!?!!

sorry for the shouting, just that ive been answering questions like this  A LOT

so basically

G=M-15

G+M=61, now solve for M

substitute the first equation into the second one

M-15+M=61

2m-15=61

2m=76

m=38

CHECK

38 + 23 = 61 YAY

I got a bit too excited, i was 2x2.5 not 2.5x2.5 im sorry