Maths Game

[Helena]

:w:

i know its abit lame, boring, but on other hand educational game which can benefit every member inshAllah..to improve their skills in maths...such as students taking sats in yr 9(if we have lil kiddies around here), GCSE students, A-level students and and and uni students(doubt they need it....me dnt)...and any other brothers sisters.....were maths is used in everyday life....


its a basic game..can be played at easy, hard, difficult levels....ur choice....make it challenging.......

were each members take in turn...only one question at a time please....on subtraction,multiplication,adding, algebra,...well anything relating to maths....

one question,is given at one go....

if its correct...then other members can take there turn inshAllah...

okie dokie....seriously am not a maths person..wont be seeing me around here.....lol.....


and plz......no calculator....esp no cheating...got my eyes on ya......:uuh: :p

does it sound easy and straightforward?...

get rolling.......:D

:sl:

[Helena]

:w:

am back again..not much dice are rolling...:rollseyes .....:X ....

let moi start it inshAllah....

wots....

1) 3*12+119=

[...]

155

2) [(11*34)-74]/ 3

['Abd al-Baari]

salams
i'm taking my ks3 sats this year
3*12+119=155
do i have 2 ask question now?

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['Abd al-Baari]

12*12=144
144/6= 24
24-24= 0

Ans=0

next question
13 squared - 12 squared =

[...]

25

shall we go into logs? :X nah..

[(5^2)^2]/25

[...]

btw ^ = to the power of

['Abd al-Baari]

is it 25?
sooyy you beat me^

[...]

edit read that wrong

['Abd al-Baari]

(2^10)/4=256

next question 25*36/18
people keeb getting answers before me lol anyways how come 1 got 256

['Abd al-Baari]

(6*7)/3 = 14
next question (14*11)-4 /25

[...]

yeh its 256

(25*36)/18 = 50

(2^-5)*32

[...]

(14*11)-4 /25 = 6

['Abd al-Baari]

(2^-5)*32
hard question anyways i got 1

[...]

yeh thats right next?

['Abd al-Baari]

um cant think of questions now
.
.
12*12-6*6-5*5=

['Abd al-Baari]

i'm not anways lol
whats the next question bro?

[~*MaNaL*~]

is it 81????

['Abd al-Baari]

RE: 45/[5*(81/9)]
i got 1

[~*MaNaL*~]

yep now i got 1.

[~*MaNaL*~]

ok so whats the nex one??

['Abd al-Baari]

(25*55)/5

[~*MaNaL*~]

275?????????

['Abd al-Baari]

Sister Manal Whats The Next Question Please?

[The Ruler]

:sl:

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

:w:

[lolwatever]

quadratic formula?

[The Ruler]

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 :confused:

but yeah...you next

:w:

[lolwatever]

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)}

[AmarFaisal]

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:

[...]

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:

lol you either use your head or you lose it! :p

^yes nexxxxt? :D

[lolwatever]

^ 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

[The Ruler]

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

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:

[The Ruler]

Correct...:eek: Please...don't kill yourself...:X

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:

[...]

Oh already answered! so whats next? :p

[The Ruler]

Oh already answered! so whats next? :p

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.

:w:

[lolwatever]

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:

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? ...

[lolwatever]

^ 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..?

[The Ruler]

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

uuh...why didn't you do that before then?

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

Nopesss

And I'm waiting for someone to answer the q I posted above.

:w:

[lolwatever]

uuh...why didn't you do that before then?

:w:

it didnt ask 2 simplify did it :X

n it cant b that confusing.. its like saying 'y didnt u simplifiy log[base 2]2 to tell me the answer ( which is equal to one)... :uhwhat

:w:

[...]

:uhwhat... can't be bothered waiting lol i'm gona give the next Q :p

Find the stationary point of the curve: y = 3x^2 – 6x + 2

[...]

The stationary point is when the gradient of the curve = zero

So firstly you need to find the gradient, and then find the value of x when that is equal to zero..

(have you done diferetiation? you need to use that)

[The Ruler]

(have you done diferetiation? you need to use that)

Uuh Hallaww! Its jee see ess eee here!

:w:

[lolwatever]

:uhwhat... can't be bothered waiting lol i'm gona give the next Q :p

Find the stationary point of the curve: y = 3x^2 – 6x + 2

d/dx (3x^2 -6x + 2) = 6x -6

d/dx = 0 (stationary point)

x = 1

thus.... y = 3-6 +2 = -1

(1,-1) --> stationary point


:D

[lolwatever]

^ if teh above is correct:


A manufacturing company operates its two available machines to polish its metal products. The two machines are equally efficient, although their maintenance costs are different. The company's daily maintenance and operating costs are given in dollars as:

f(x1, y) = 60 - x - 1.2y + 0.05(x^2) + 0.1xy + 0.07(y^2)

where x and y are the hours of operation of the two machines. Find the number of hours to operate each machine in order to minimize daily cost.


:D

(Also requires differentiation, of a different sort;D)

[...]

I haven't covered that :confused: Which exam board did you do?

And yeh the other one is correct

[lolwatever]

^ it's something a lil beyond A-levels... a topic called "functions of several variables"

sorry :hiding:

[afriend]

^ it's something a lil beyond A-levels... a topic called "functions of several variables"

sorry :hiding:

:sl:

So you should!! imsadIt's bad enough I hate maths...

And who's coming up with the next question?

:w:

[lolwatever]

^ u could of asked 2 c the solution atleast!!!!! :raging: :anger:

[afriend]

lol ok whatever man :D

Next Question: Find the height of this scalene triangle:

[Ibn Al Aqwa]

thats quite hard bro, give us a clue :)

[lolwatever]

zero?!