how about this...give the name of this: [-b+- square root of (b^2-4ac)]/2a

quadratic formula?

hmm...i think so...i was told completing the square...but i cheked my book...and completing the square was different...and i am like what the

but yeah...you next

yeh sis its also Completing the square, except that it handles harder cases where the answer isn't an 'easy' fraction or straight number.

hmmn me... d/dx(tan(3x)) ?

cos(3x)
______
-sin(3x)

nexttt?

ok heres the next one:

qubed root of {[(18/3)+1)]*(7^2)}

What's going on here. ..I still count nos. on my fingers at 28 yrs old. !!! I am often surprised to find G1 kids counting in their heads..n u ppl !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Maths...definetely not my type

lol you either use your head or you lose it!

^yes nexxxxt?

^ lol

c = log[base 2] (2^(1-k))

i'm assuming that √4/x=2^c u meant sqrt(4) / x not sqrt(4/x) btw.

haraam 3alaik little akhii... Mustn't trouble a sister when she's trying to have some peace.

Gah, anyways, this is what I got... c = 1 - k

Tell me I'm wrong, and I shall shed tears and you shall drown. Yeah, consider that a warning!

Now I wonder how the heck I am to go through with my A-level maths... Let alone double maths. Ayeyooh!

that was to me right

If so, then, what on the sweet creation of dear God did lolwhatever get?

EDIT: remind me... did I mention I was going to kill myself? I thought I threatened to drown you.

Oh already answered! so whats next?

If p is directly proportional to q, what is the equation connecting p and q, given that p = 1.5 when q = 4.5?

copy andpaste from bitesize. Believe it or not, I haven't read it yet.

I just read it.

lol its same as ur answer when u simplify it!
c = log[base 2] (2^(1-k))

because .... c = log[base 2] (2^(1-k)) = (1-k)log[base2]2 = (1-k) * 1 = (1-k)

do u mean u didnt use logs btw? ...

^ so how did u go abt solvin it without logs? i thougth that would prob be the easiest way


here's my workn out:


eq(1) --- x = 2^k

eq(2) --- sqrt(4)/x = 2^c


we'll make them both equations have x as their dependent variables.. and then equate them. so...

eq(1) ---> x = 2^k
eq(2) ---> x = sqrt(4)/2^c

eq(1) = eq(2)

2^k = sqrt(4)/2^c

2^c = sqrt(4) / 2^k


take log(base2) of both sides...

log[base2] 2^c = log[base2] sqrt(4) / 2^k

c = log[base2] sqrt(4)/2^k = log[base2] 2^(1-k) = 1 - k

hence... c = 1-k


ne simpler methods out of curiosity..?