Maths Game

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Re: seriously maths game.....i think soo.......

Sister Manal Whats The Next Question Please?
 
:sl:

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

:w:
 
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)) ?
 
ok heres the next one:

qubed root of {[(18/3)+1)]*(7^2)}
 
What's going on here.:uhwhat ..I still count nos. on my fingers at 28 yrs old. !!!:hiding: I am often surprised to find G1 kids counting in their heads..n u ppl:confused: :confused: !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Maths...definetely not my type:heated:
 
AmarFaisal said:
What's going on here.:uhwhat ..I still count nos. on my fingers at 28 yrs old. !!!:hiding: I am often surprised to find G1 kids counting in their heads..n u ppl:confused: :confused: !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Maths...definetely not my type:heated:
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lol you either use your head or you lose it! :p

^yes nexxxxt? :D
 
^ 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.

:D
 
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Iqram said:
ok next:

Given x=2^k and √4/x=2^c Find c in terms of k

Enjoy :smile: This is surely the end for the next few days...Or just until sis Amamtul Malik comes online ;D
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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!

:w:
 
Iqram said:
Correct...:eek: Please...don't kill yourself...:X
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that was to me right :confused:

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.

:w:
 
Amatul Malik said:
that was to me right :confused:

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.

:w:
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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..?
 

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