labodinho: Jackpot scares me off here i must say...


A lady??

hmmm!

shhhhh. . . Easy oooo.

The questions I solve are often very cheap left-overs. You should be giving credit to my fellow hommies and 5-STAR Generals who left them for me to use and shine.

They solve the harder ones. More credits to them

jackpot: shhhhh. . . Easy oooo.

The questions I solve are often very cheap left-overs. You should be giving credit to my fellow hommies and 5-STAR Generals who left them for me to use and shine.

They solve the harder ones. More credits to them

You're such a girl....very funny

Pls guyz help me solve this.its giving me headache
the sum of 3 7/8 and 1 1/3 is less than the difference between 3/8 and 1 2/3 by...........answer

I nac ees ym m464m n1 3h7 d esouh...greetins 2u ma.

Alpha Maximus: you can see your madam in the house and you are greeting her eh? Decode this: 123()_-%. Dgdtz2135^3)5x@ srw(:

..sharp guy.

@benbuks are you an 31337?

wisemania: @benbuks are you an 31337?

...mayb..am

Alpha Maximus: Are you serious? why not just ask what 1 times 1 is?

Alpha Maximus: Are you serious? why not just ask what 1 times 1 is?

your post just showed ur upbring.
Did i tell u i was competing with you.y cant u just live n let others live.arrogance diminishes knowledge,its an advice to you to go see a psychologist its obvious ur nat normal.smellos

Pls guyz help me solve this.its giving me headache
the sum of 3 7/8 and 1 1/3 is less than the difference between 3/8 and 1 2/3 by...........answer

Alpha, warnin 2u. Any level of mathematics is accepted here,be it d simplex, average, complex, hyper advanced...even professors's math..bring it up, dats y its called a "mathematics clinic"...if u can solved any problem posted , solve..if u cnt , leave, nd wait 2 c solution frm anoda general ok..?..u kw say u b ma guy bt u dey fall ma hand bigtime...hope say u get ma gist.?

Solve the simultaneous equations
(10^x)(4^y) = 1
8^x = 10^(y+1)

mickywonder:

your post just showed ur upbring.
Did i tell u i was competing with you.y cant u just live n let others live.arrogance diminishes knowledge,its an advice to you to go see a psychologist its obvious ur nat normal.smellos

Hey, we don't use such words here, please. He was just kidding when he said that. Go back to his older posts and you'll get what I'm saying. ...peace!

mickywonder: Pls guyz help me solve this.its giving me headache
the sum of 3 7/8 and 1 1/3 is less than the difference between 3/8 and 1 2/3 by...........answer

,.i guess dis question is frm "comprehensive mathematics"...first add d first two fractions..then keep ur answer.. Then find d difference in d 2nd two fraction( e.g difference btw 6 n 3 =6-3=3)..then finaly get d difference in both answers...hope my explanation helps..

goodevening every1, plz i need ur help here:
find the value of k given that logk - log(k-2) = log5
(WASSCE JUNE 1995 GENERAL MATHEMATICS THEORY, QUESTION 6b)

smurfy: Solve the simultaneous equations
(10^x)(4^y) = 1
8^x = 10^(y+1)

tak d log of both equations...& solv 4 x and y

...there is no analytic solution to:
3^x+4^x=5^x, 4^x=2x, etc...

Laplacian:
tak d log of both sides...& solv 4 x and y

It's not that simple, my dear. I'd like to see one of you guys solve it

johnpaul1101: goodevening every1, plz i need ur help here:
find the value of k given that logk - log(k-2) = log5
(WASSCE JUNE 1995 GENERAL MATHEMATICS THEORY, QUESTION 6b)

logk - log(k-2) = log5
log [k/(k-2)] = log5
k/(k-2) = 5
k = 5(k-2)
k = 5k - 10
5k - k = 10
4k = 10
k = 5/2

Alpha Maximus: ....my post just showed my UPBRING? Don't you mean UPBRINGING? And you're judging me from just one of my posts? Well, if that's the standard method of judgment in vogue, I will have to assert that you are an intractable illiterate who has origins from the lower rungs of the society!!! How does my post even give the slightest hint of undue competition? The question was so simple , it had to be a joke!!! Maybe you're an adult who is still in pry 5!! And you say I'm the one one that needs a psychologist!! You must be drinking a bicycle/jonezing!!!!!! *flushes him down the toilet*

Alpha Maximus: ....my post just showed my UPBRING? Don't you mean UPBRINGING? And you're judging me from just one of my posts? Well, if that's the standard method of judgment in vogue, I will have to assert that you are an intractable illiterate who has origins from the lower rungs of the society!!! How does my post even give the slightest hint of undue competition? The question was so simple , it had to be a joke!!! Maybe you're an adult who is still in pry 5!! And you say I'm the one one that needs a psychologist!! You must be drinking a bicycle/jonezing!!!!!! *flushes him down the toilet*

lol.there's something wrong.i cant believe a guy is acting this way just because of i corrected him.
For ya info that math question was a jamb past question n if u think its a "1 times 1" question i also think u could have solved it.i forgive u

Alpha Maximus: Oga I sabi your yarn wella!! But that question just dey one kind!! No be pikin wey fit punch all those values for calculator and find answer!!! I no even directly insult the guy, I just ask am whether na joke!! Alright, alright...I hear you

Alpha Maximus: Oga I sabi your yarn wella!! But that question just dey one kind!! No be pikin wey fit punch all those values for calculator and find answer!!! I no even directly insult the guy, I just ask am whether na joke!! Alright, alright...I hear you

mr alpha i now get ur point.u actually thought the question is an easy calculation punch question.why dont u try it and understand why i actually brought it on nairaland.as u can see i dont sound dull neither do i sound like someone in a secoundary sch.pls next time know that no body on earth wud take that as a joke.

jackpot: i hope you meant ¹/_{[(cuberoot2)-1]}

Now, since
¹/_(x-1)=(x²+x+1)/_(x³-1),

we have (upon setting x=cuberoot2=2^1/3) that

¹/_{[(cuberoot2)-1]}

equals

(2^2/3+2^1/3+1)/_(2^3/3-1)

which simply equals

2^2/3+2^1/3+1.

hmmm ...... smart

I-am-Winner:
there are certain tins in maths that I cant forget.. Smtin lyk dis..

Ans is (root 2) + 1

WRONG

jackpot: i hope you meant ¹/_{[(cuberoot2)-1]}

Now, since
¹/_(x-1)=(x²+x+1)/_(x³-1),

we have (upon setting x=cuberoot2=2^1/3) that

¹/_{[(cuberoot2)-1]}

equals

(2^2/3+2^1/3+1)/_(2^3/3-1)

which simply equals

2^2/3+2^1/3+1.

sis u got me wrong.
d ques is not 1/[(cuberoot2)-1] but 1/(cuberoot2) -1

Mr Calculus: sis u got me wrong.
d ques is not 1/[(cuberoot2)-1] but 1/(cuberoot2) -1

& DAT MAKES UR ANSWER WRONG......
WEN I DID IT I GOT
-1/(2^1/3+2^2/3+2)
but i dont know if am right.
plz try again and others plz help out

Alpha Maximus: Finally back with some questions!
Question 1 : The equivalent of Mr Coker's grandson's life lived in days is Mr Coker's son's life lived in weeks, and also, the equivalent of his grandson's life lived in months is Mr Coker's life lived in years. Altogether, the sum of their ages is 120 years. In how many years will the ratio of Mr Coker's son's age to his grandson's age be 1/15 of Mr Coker's age and also how old was Mr Coker when the sum of the past ages of him , his son and grandson was 9 less than the current sum of their ages?

Bro are you a social science student?

To your question:

A. Reading over the questions one will observe that;
Mr cooker is the grandpa
His son is the father
His son's son is the grandson.

So let grandson age = gs old. If gs is x days old then father is x weeks old i.e 365/52 = 7gs
And if gs is y months old then grandpa =12gs
Now, the total age of all the 3 added together is 120yrs which brings the equation: gs + 7gs + 12gs = 120
20gs = 120
gs = 6
.: grandson = 6yrs
father = 7(6) = 42yrs and grandpa = 12(6) = 72yrs

B. 5yrs
C. 66yrs and 6 months old or 67yrs