Yes i guess you just observed
the tone of his posts is annoying to say d least.. why are you even answering d guy

johnpaul1101: a question which goes like this: an equilateral triangle with the lengths of its sides given in terms of a and b, the length are as follows
4acm
3bcm
(a+b+3)cm
find a and b and hence find the length of the sides of the triangle?
Gudluck

The sides of an equilateral triangle are equal so basically,

4a = 3b .....eqn (1)


And also a+b+3 = 3b
a = 2b - 3 .....eqn (2)

Substitute eqn (2) in eqn (1)
4(2b - 3) = 3b
8b - 12 = 3b
5b = 12
b = 2.4
Substitute b in eqn (2)
a = 2(2.4) - 3
a = 1.8.

Now the length of each of the sides is,

3b = 3*(2.4)
= 7.2cm
or 4a =4 *(1.8 ) = 7.2 cm

wow, this is gr8, it didnt even took u up to 5 minutes 2 solve it

mayb u can still help me with this: a cyclist cycles for x hours at 12km/h and y hours at 18km/h. Altogether she cycles 78km in 5 hours, find x and y

johnpaul1101: mayb u can still help me with this: a cyclist cycles for x hours at 12km/h and y hours at 18km/h. Altogether she cycles 78km in 5 hours, find x and y

x + y = 5
12x + 18y = 78

Solve with substitution method

x = 2 y = 3

Even if u use guess work u go get am

johnpaul1101: thanks a lot guru, but i still have more of those simple, jss, questions dat is confusing me

Bring them on

johnpaul1101: mayb u can still help me with this: a cyclist cycles for x hours at 12km/h and y hours at 18km/h. Altogether she cycles 78km in 5 hours, find x and y

Total time is xhrs + yhrs = 5hrs eqn (1)
And x = 5 -y

Speed = distance/time
total distance therefore 12x + 18y = 78 km
...eqn (2)

Substitute x in eqn (2)
12(5 - y)+ 18y = 78
60 -12y + 18y = 78
6y = 18
y = 3

Substitute y in eqn (1)
x = 5 - y
x = 5 - 3
x = 2

biolabee:

x + y = 5
12x + 18y = 78

Solve with substitution method

x = 2 y = 3

Even if u use guess work u go get am

Thanks jor, the guy's questions are too boring

Nice work people. Greetings!

my questions might luk 2 boring, but its always gud 2 go back 2 jss calculations, nd dats wat i'm trying 2 do but i keep getting confused and frustrated, plz tell me rechiez, how can i b like u?

johnpaul1101: my questions might luk 2 boring, but its always gud 2 go back 2 jss calculations, nd dats wat i'm trying 2 do but i keep getting confused and frustrated, plz tell me rechiez, how can i b like u?

well, basically i'd say it's a gift to be good at logical reasoning just like it is a gift to be able to sing very well.
However, in the introductory part of this thread, i gave the following tips and i hope this helps you out;

1. you have to develop a great interest for maths, and have that strong desire to become a maths guru.

2. Realize that there's nothing magic about mathematics.

3. make friends with people who appreciate mathematics.

4. practise as many maths questions as possible, starting from the very simple ones to more complex ones. develop the urge to solve them on your own, but you can always seek for assistance whenever you get hooked.

5. do not forget any new thing you have learnt, keep reviewing them.

6. learn to memorize some basic maths formulas as this will build your confidence.

GOOD LUCK

doubleDx: Nice work people. Greetings!

greetings bro i wana make 5 people the maths generals in this thread...you are on my list....lol

Richiez:

well, basically i'd say it's a gift to be good at logical reasoning just like it is a gift to be able to sing very well.
However, in the introductory part of this thread, i gave the following tips and i hope this helps you out;

1. you have to develop a great interest for maths, and have that strong desire to become a maths guru.

2. Realize that there's nothing magic about mathematics.

3. make friends with people who appreciate mathematics.

4. practise as many maths questions as possible, starting from the very simple ones to more complex ones. develop the urge to solve them on your own, but you can always seek for assistance whenever you get hooked.

5. do not forget any new thing you have learnt, keep reviewing them.

6. learn to memorize some basic maths formulas as this will build your confidence.

GOOD LUCK

thanks guy, i'll try 2 keep 2 this standards

Ghettoguru: Thanks guys. Help with these questions please.

1. Differentiate sin(4x)cosec^2(x) +8cos^2(x) with respect to x.

2. Find the value(s) of p, q and r in the following equations:

p - q - r = -4
p^2 - q^2 - r^2 = -12
p^3 - q^3 - r^3 = -34

p - q - r = -4
p^2 - q^2 - r^2 = -12
p^3 - q^3 - r^3 = -34

Lets first simplify the equations by dividing through by -1 =>

q + r - p = 4 .....(1)
q^2 + r^2 - p^2 = 12 .....(2)
q^3 + r^3 - p^3 = 34 .......(3)

Remember from the previous equations we solved on this thread that =>

q^2 + r^2 can be rewritten as (q + r)^2 - 2qr
q^3 + r^3 can be rewritten as (p + r)^3 - 3qr (q + r)
From (1) q + r = p + 4 ......(4)

We can now substitute equation (4) into equation (2) to express qr in terms of p. Substituting yields =>

(p + 4)^2 - 2qr - p^2 = 12
2qr = 8p +16 - 12
2qr = 8p + 4
qr = 4p + 2 .....(5)

Next is to substitute the expression for qr = 4p +2 and (q + r) = p + 4 in equation (3) =>
q^3 + r^3 - p^3 = 34 .......(3)
(p + r)^3 - 3qr (q + r) - p^3 = 34
(p + 4)^3 - 3(4p + 2)(p + 4) - p^3 = 34
Expanding and simplifiying=>

(p + 4)(p^2 + 8p + 16) - 3(4p^2 + 18p + 8 ) - p^3 = 34
p^3 + 12p^2 + 48p + 64 - 12p^2 - 54p - 24 - p^3 = 34
-6p = 34 - 40
:. p = 1

You can now substitute p in equations (1) and (2) to determine the values of q and r.

I hope that helps.

Richiez:
greetings bro i wana make 5 people the maths generals in this thread...you are on my list....lol

Lmao thanks for the compliment bruv! You, ositadima, biolabee and other gurus have been doing great jobs here too! My respek brodas !!!

Good work doubledx, ositadima1, biolabee and ghetoguru, nice job!

Good job guys....I am really learning a lot of long-forgotten concepts
Brain don dry finish

Guys bring up more but I don forget all those div, grad n curl o

thanks richiez, it worked, it's really gud 2 go back 2 old school work 2 refresh memory, this is really a site for mathematicians like u nd ur frnds(osita and co), keep it up

johnpaul1101: thanks richiez, it worked, it's really gud 2 go back 2 old school work 2 refresh memory, this is really a site for mathematicians like u nd ur frnds(osita and co), keep it up

It feels good to touch lives positively

in a positive numbr of two digits, the sum of the digits is 15. If the digits are interchanged, the numbr is increased by 9. Find the numbr

johnpaul1101: in a positive numbr of two digits, the sum of the digits is 15. If the digits are interchanged, the numbr is increased by 9. Find the numbr

SOLUTION
let the number be xy
x + y =15 .............(1)

we now know that tens is x and units is y
10y+x = 10x+y + 9
10y-y-10x+x = 9
9y - 9x = 9 ............(2)
dividing through eqn(2) by 9 we get;

y - x = 1 ..............(3)
now, we cansolve eqn(1) and eqn(3) simultaneously

from eqn(3); y=x+1 hence we substitute this for y in eqn(1)
x + x+1 = 15
2x=15-1
2x=14
x=7
since y=x+1
y=7+1 = 8
recall that the number is xy
therefore '78' is that number

@johnpaul, don't feel too bad that we solve your questions too easily, just bring them on....simple or difficult but make sure you're bring them so that it can build your mathematics skills, and not to test us. one love

Richiez:

Mathematical induction is a method of mathematically proving that a statement is true for all natural numbers i.e positive integers
it usually involves 2 basic steps;
1. showing that the statement is true when the lowest natural number is used e.g n=0 or n=1 (preferably)
2. showing that if the statement is true for any natural number n=k, then it is true for the next natural number n=k+1

let us do some examples;

1. prove by mathematical induction 1+2+3+4+.......+n+(n+1) = n(n+1)/2
SOLUTION
note that n represents the number of elements in the series
step 1---we have to show that the statement is true for n=1
1= 1(1+1)/2
1= 2/2
1=1 therefore since L.H.S = R.H.S the statement is true for n=1
we can even go ahead to show that the statement is true for any other small value of n, say n= 4
1+2+3+4 = 4(4+1)/2
10= 4(5)/2
10= 20/2
10= 10 therefore it is also true for n=4
next is step 2
here we shall consider an arbitrary value of n, say n=k and assume it is true
i.e 1+2+3+4+....+k = k(k+1)/2 ................(1)
if the above statement is true, then the final step is to show that it is also true for n= k+1
1+2+3+4+....+k+(k+1) = (k+1)(k+1+1)/2
1+2+3+4+....+k+(k+1) = (k+1)(k+2)/2 ...............(2)

to show that eqn 2 is true, let us add (k+1) to both sides of eqn 1
1+2+3+4+....+k+(k+1) = k(k+1)/2 + (k+1).............(3)
the job here is to make the R.H.S of eqn 3 look like that of eqn 2
taking l.c.m as 2
1+2+3+4+....+k+(k+1) = [k(k+1)+2(k+1)]/2
since (k+1) is a common factor in the R.H.S we factorize
1+2+3+4+....+k+(k+1) = (k+1)(k+2)/2

This is same as eqn 2 above, hence the statement 1+2+3+4+....n+(n+1) = n(n+1)/2 is true for all natural numbers.

feel free to post more exercises.

Impressive!
I love the spirit.
I also expect serious-minded students/undergraduates to ask as many questions as possible here if they want to learn.
Nigerians need to learn to face math challenges that the Asians and develop our IQ.
As someone who has lectured mathematics in a university,I endorse you and I love your spirit of sharing knowledge.
May God bless you.

Evaluate d cube root of (10+6*sqrt of 3)

solomon111: I dread statistics.
That is one area of mathematics,i have not mastered.
As far as am concerned,statisticians are the real mathematicians.

Solomon111 please I beg you to allow Richiez to address the problems as they come.Especially as he was the one that opened the thread.
I have read your solutions and will not want to be an umpire here,but for you to call a subset of mathematics the real mathematicians makes me to wonder.
If you dread statistics,which is basically inference,probability,stochastic methods(probabilistic),quality control,operations research,non-parametric methods,decision theory, will you then be confident with serious analysis? Real analysis,topological spaces,math methods,ord and partial diff equations,linear and abstract algebra,set theory/logic,numerical analysis & optimization theory,measure theory(lebesgue measure & integration),metric spaces,elasticity,complex variables,vector and tensor analysis,etc.
You must be exceptional to find all those cheaper than statistics.

^@Jah man, nice one...

Johncasey: Evaluate d cube root of (10+6*sqrt of 3)

i think binomial theorem might be a useful tool here...wud reply u later

Richiez: @johnpaul, don't feel too bad that we solve your questions too easily, just bring them on....simple or difficult but make sure you're bring them so that it can build your mathematics skills, and not to test us. one love

i cant believe u wud say this richiez, i told u dat i'm preparing for my s.waec, and wen i found out about u and dis site, i was so impressed, because i've had so many problems in maths, those dat i couldnt solve was wat i posted nd fortunately, u nd ur felo mathematicians helped me, i still have many unanswered mathematical question i wud luv u 2 help me with, but no airtime..........moreover......i wouldnt dare.....bsides, wat gives u d feeling dat i'm trying 2 test u?

johnpaul1101: i cant believe u wud say this richiez, i told u dat i'm preparing for my s.waec, and wen i found out about u and dis site, i was so impressed, because i've had so many problems in maths, those dat i couldnt solve was wat i posted nd fortunately, u nd ur felo mathematicians helped me, i still have many unanswered mathematical question i wud luv u 2 help me with, but no airtime..........moreover......i wouldnt dare.....bsides, wat gives u d feeling dat i'm trying 2 test u?

oh i'm very very sorry i got the wrong impression....okay we all are here to help you okay?

johnpaul1101: i cant believe u wud say this richiez, i told u dat i'm preparing for my s.waec, and wen i found out about u and dis site, i was so impressed, because i've had so many problems in maths, those dat i couldnt solve was wat i posted nd fortunately, u nd ur felo mathematicians helped me, i still have many unanswered mathematical question i wud luv u 2 help me with, but no airtime..........moreover......i wouldnt dare.....bsides, wat gives u d feeling dat i'm trying 2 test u?

Sorry buddy, we initially got the wrong impression. Keep your questions coming, there are a lot of gurus on this thread willing to help!

Nice job, the more you solve the better you get at it.

Make x the subject of the formula; STX^3-2WX=10X^2T