Melody

Can someone answer this: (1-3x)*(2+i), and also explain the FOIL system?

I do not usually teach this method but I shall explain it since you ask.

(1-3x)*(2+i)

The first bracket has 2 terms, they are                  1 and -3x

The second bracket also has 2 terms, they are     2 and   i

F stands for first that is 1*2 = 2

O stands for outside so that is   1*+i = i

I stands for inner                      -3x*2 = -6x

L stands for last                       -3x*i = -3xi

Add them all together and you have

\((1-3x)(2+i)\\=2+i+-6x+-3xi\\ =2+i-6x-3xi\)

i is often used to denote an imaginary number but since this is a low level question I assume that i is a pronumeral the same as x is.  :)

You can do this with the web 2.0calc on the home page.

Just type in 

factor(111111) and press equal

factor(111111)= = (((3*7)*11)*13)*37

I probably cannot help you but what do you think your username is?

I'll see if I can find it, maybe you are spelling it a little differently or not puting in capitals where they belong...

  z^4=81

z=3    is one answer, the others are    z=-3,    z=3i    and     z=-3i

a) 3149810398401392821

factor(3149810398401392821) = 3149810398401392821

b) 3098402938409238409

factor(3098402938409238409) = (47*589783)*111775800209

c) 9348509189323894393

factor(9348509189323894393) = ((17*31)*421)*42135645180779

d) 2309482390481134983

factor( 2309482390481134983) = ((3^3*15901)*86017)*62537737

e) 3294829348201392047

factor(3294829348201392047) = (3881*754249)*1125575263

There you go, web2.0 is a pretty cool calculator :) 

I do not know what you are asking :/

Find the equation of the parabola with coordinates of the vertex (0,0) and equation of the axis y=0 and passing through the point (3,6)

\(y^2=4ax\\ \text{sub in x=3, y=6}\\ 36=4a*3\\ a=3\\ so\\ y^2=12x\)

Express the following in the form a + bi, where a and b are real numbers:      sqrt(i)

\(\sqrt i=a+bi\\ i=(a+bi)^2\\ i=a^2+2abi-b^2\\ i=(a^2-b^2)+2abi\\ so\\ a^2-b^2=0\;\; and\;\ 2ab=1\\ (a-b)(a+b)=0\\ a=\pm b\\ but \;\;since\;\; 2ab=1, \;\;a=b\\ 2a^2=1\\ a=\pm\frac{1}{\sqrt{2}}\\ a=\pm\frac{\sqrt 2}{2}\\ so\\ \sqrt{i}=\pm(\frac{\sqrt 2}{2}+\frac{\sqrt 2i}{2})\)

I wanted to send you that beautiful post that the troll did for you but I don't know where it is.   

I bet Ginger can find it and she will probably read this post.

Maybe she can post the address for us.

That would cheer you up a little I think!

Please Ginger could you do that for us please ???    Pretty please !!!

You could tell your Mum and Dad.

Maybe they will feel so bad that you get even more thant you would have in the first place.  

I am real;ly sorry that you birthday has not been a success.

I hope you enjoy my cake anyway. It won't make you fat  !!