Soln: 3red,5blue n 7green.Total case=15(A)Both are red P(R)*P(R){without replacement}> 3/15 *2/14=6/210 or 1/35 (B) FROM TREE DIAGRAM 5/15*7/14 + 7/15*5/14=2/6 (C) 5/15*4/14 +7/15*6/14=31/105 (D) 1/35+1/6+31/105=103/210 THEREFORE 1 - 103/210=107/210 (E) 1/35+1/6+31/105=103/210 ...My own idea.

lol! math clinic sorry the name made me laugh. Anyways nice thread,i luv math though am just on the average side.I Prefer it to sme oda
course(s)or subject as the case may be. where u are required to write and write,and after everything u end up with an annoying grade. when u
can just memorize some formula and some procedures in math and get a
good score. less stress . sha i hope to learn alot from here though am late. i will try to contribute in anyway i can. come where are all the ladies hope am not the only one oh! lol

joywendy: lol! math clinic sorry the name made me laugh. Anyways nice thread,i luv math though am just on the average side.I Prefer it to sme oda
course(s)or subject as the case may be. where u are required to write and write,and after everything u end up with an annoying grade. when u
can just memorize some formula and some procedures in math and get a
good score. less stress . sha i hope to learn alot from here though am late. i will try to contribute in anyway i can. come where are all the ladies hope am not the only one oh! lol

welcome Joywendy....ladies here are very few but the ones we have are geniuses ...let's see how well you rock maths! learn, contribute, correct others and allow yourself to be corrected when necessary, enjoy! it's fun

Richiez:
welcome Joywendy....ladies here are very few but the ones we have are
geniuses ...let's see how well you rock maths! learn, contribute,
correct others and allow yourself to be corrected when necessary, enjoy!
it's fun

thanks alot. alrite. Genuises Mmm i fear. Anyways its 4 fun. Thanks alot i wud try later 2 read from the begining

Onase: Pals i need ur view on this question.I ve solved it base on my own idea but i want to be sure....pls help me to solve it>A bag cotains 3 red balls,5 blue balls nd 7 green balls.two balls are pick 1 after the other without a replacment .find the probability dat (a) both are red (b)first blue other is green (c)1 is blue the other is green (d)both are of different colours (e)both are of the same colours.

..total of 15balls
a) pr(RR)=3/15 x 2/14=1/35
b)pr(BG)=5/15 x 7/14=1/6....
C) pr(1B,1G)=Pr(BG)+Pr(GB)=(5/15 x 7/14) +(7/15 x5/14)=1/3.....
d) pr(different colours)= pr(RB)+pr(BR)+pr(BG)+pr(GB)+pr(RG)+pr(GR)=1/7+1/3+1/5=71/105
e) pr(same colour)=pr(RR)+pr(BB)+pr()=1/35 +2/21 +1/5=34/105

d citizen:
I said 0 factorial equal to 1

Oh okay, sorry =>

n! = n(n-1)!
:. at n = 1

1! = 1.(1-1)!
Since 1! = 1

then 1 = 1.0!

:. 0! = 1

jackpot: dear Maths generals, pls help me with this little problem.


[size=16pt]By using the Cartesian-paraboloidal coordinate transformation equations:
(x=uv sin φ, y=uv cos φ, z=¹/₂(u²-v²),
show that the 3-dimensional Laplace's equation
F_xx+F_yy+F_zz=0
reduces to
¹/_(u²+v²) (F_uu+F_vv+ ¹/_u F_u+¹/_v F_v)+¹/_(u²v²)F_φφ = 0.[/size]

doubleDx, d citizen, rhydex 247, Richiez, benbuks and other maths generals

seeing as you guys dey destroy other peoples questions with confim answers, so, you guys know the answers to my questions but una no wan help me sha. Its unfair ooooo

**sobs**

nobody wants to help me


**sobs even more**

jackpot: dear Maths generals, pls help me with this little problem.


[size=14pt]By using the Cartesian-paraboloidal coordinate transformation equations:
(x=uv sin φ, y=uv cos φ, z=¹/₂(u²-v²),
show that the 3-dimensional Laplace's equation
F_xx+F_yy+F_zz=0
reduces to
¹/_(u²+v²) (F_uu+F_vv+ ¹/_u F_u+¹/_v F_v)+¹/_(u²v²)F_φφ = 0.[/size]

jackpot: doubleDx, d citizen, rhydex 247, Richiez, benbuks and other maths generals

seeing as you guys dey destroy other peoples questions with confim answers, so, you guys know the answers to my questions but una no wan help me sha. Its unfair ooooo

**sobs**

nobody wants to help me


**sobs even more**

Lol, you are a general too ma'am. Okay, I'll try it out tomorrow when I'm free! Math Generals & Doctors, if anyone is free right now, kindly help our female guru here....1luv y'all

Jackpot i would ve luv to help but my Engr.Maths is in sku.I never knew the strike would last this long...

Ade received 3/5 of a sum of a money, Nelly 1/3 of the remainder while Austin share is greater than Nelly's share by N3000. how much did Ade receives

shollynoob:
Ade received 3/5 of a sum of a money, Nelly 1/3 of the remainder while Austin share is greater than Nelly's share by N3000. how much did Ade receives

let d total money b x.
ade=3/5x.
nelly=1/3(2/5x)=2/15x.
austin=2/15x+3000.
total money=ade+nelly+austin.
x=3/5x+2/15x+2/15x+3000.
by l.c.m
13x+45000/15=x.
15x-13x=45000.
2x=45000.
x=22500.
recall::ade=3/5x.
ade=3*22500/5
ade=13500.

d citizen:
Find the area bounded by the curve

y=|x-2| along the x-axis between x= 0 and x = 3.

N.b |x - 2| is not a bracket but a modulus sign. Take note ooo

ur answer is 6

Mathz generalz take diz one chil
1^x + 2^x =3^x-1
find x
p.s:i nor wan hear newton raphson method o

@jackport....what an enigma....dnt worry it will be solved asap...

shollynoob:
Ade received 3/5 of a sum of a money, Nelly 1/3 of the remainder while Austin share is greater than Nelly's share by N3000. how much did Ade receives

ade received 13,500..nelly received 3000...austin received 6000....my first contribution to this amazing family.Hope i got it right?

Sir-Tunechi:
Mathz generalz take diz one chil
1^x + 2^x =3^x-1
find x
p.s:i nor wan hear newton raphson method o

we divide thru bt 3^x ; 1/3^x+(2/3)^x=1/3, now let a=1/3^x then (2/3)^x=1/3-a now from d above parametric equations, xLog(1/3)=Loga & xLog(2/3)=Log(1/3-a)...dividin d last eqn by d precedin one gives, 0.369=Log(1/3-a)/Loga or a^0.369=1/3-a...takin cube of both side; a=(1/3-a)^3 approx. let b=1/3-a, then last eqn becoms 1/3-b=b^3 or b^3+b-1/3=0...solve for b, then for a and then for x from a=1/3-b, 1/3^x=a, u 'll find that x=3 approx...sorry i have no calculator on me...

@ Sir tunechi questn.
1^x+2^x=3^(x-1).
solution.
take log of both sides.
log1^x+log2^x=log3^(x-1).
xlog1+xlog2=(x-1)log3.
xlog1+xlog2=xlog3-log3.
recall that log1=0, log2=0.3010 nd log3=0.4771.
Nw we have.
0.3010x=0.4771x-0.4771.
0.4771=0.4771x-0.3010x.
0.4771=0.17612x
x=0.17612/0.4771
x=2.708929145.
Approximately
x=3.
I want to ask wetin newton raphsön method do u @ sir tunechi.

Breakfast questn
1. Solve. (x^2-16)(x-3)^2+9x^2=0.

rhydex 247: @ Sir tunechi questn.
1^x+2^x=3^(x-1).
solution.
take log of both sides.
log1^x+log2^x=log3^(x-1).
xlog1+xlog2=(x-1)log3.
xlog1+xlog2=xlog3-log3.
recall that log1=0, log2=0.3010 nd log3=0.4771.
Nw we have.
0.3010x=0.4771x-0.4771.
0.4771=0.4771x-0.3010x.
0.4771=0.17612x
x=0.17612/0.4771
x=2.708929145.
Approximately
x=3.
I want to ask wetin newton raphsön method do u @ sir tunechi.

@Rhydex...i dont think d Log of a sum is d same as d sum of d log...as an example...Log(2+3) is not equal to Log2+Log3...if u stil insist, here is a mathematical proof...let a & b any two positive numbers then
Log(a+b)=Loga +Logb now from logarithm, Logab=Loga+Logb...so that Log(a+b)=Logab or a+b=ab that is, d sum of any two numbers is always equal to their product which is blatantly false...

rhydex 247: @ Sir tunechi questn.
1^x+2^x=3^(x-1).
solution.
take log of both sides.
log1^x+log2^x=log3^(x-1).
xlog1+xlog2=(x-1)log3.
xlog1+xlog2=xlog3-log3.
recall that log1=0, log2=0.3010 nd log3=0.4771.
Nw we have.
0.3010x=0.4771x-0.4771.
0.4771=0.4771x-0.3010x.
0.4771=0.17612x
x=0.17612/0.4771
x=2.708929145.
Approximately
x=3.
I want to ask wetin newton raphsön method do u @ sir tunechi.

cuz Newton Raphson dy cum wit limit...nd cant b use in diz kind of question. U guys jst do guez work cuz u knw d ansa alreadi

@ laplacian.
I can still recall when sum1 post dis questn.
5^(x+1)=3^(x^2-1).
which d soln goes dis way.
take log of bth sides.
(x+1)log5=(x^2-1)log3.
(x+1)log5-[(x+1)(x-1)log3]=0
(x+1)[log5-(x-1)log3]=0
x+1=0 or log5-xlog3+log3=0.
x=-1 or log15=xlog3.
x=-1 or x=log15/log3.
x=-1 or x=2.465.
Similarly to sir tunechi questn.
1^x+2^x=3^(x-1).
take log of bth sides.
xlog1+xlog2=(x-1)log3
xlog1+xlog2=xlog3-log3.
0.3010x=0.4771x-0.4771.
x=0.4771/0.17612.
x=2.708929145
x=3.
Hence my soln nd my ans still stand. U can check by puttin x=2.708929145 or x=3.

i wud like to ask d generals in dis thread
''y are u all been so adamant towards my questions

Mbahchiboy: to my generals plz help me out
(1)if
coshx=1+x^2/2!+x^4/4!....,
sinhx=x+x^3/3!+x^5/5!.....,
find tanhx?
(2)find x if x^5-32=0
[hint::x must have five root e.g a,b,c,d or e]

MAKE UNA HELP ME OUT O........
**On my knees**

rhydex 247: Breakfast questn
1. Solve. (x^2-16)(x-3)^2+9x^2=0.

@Rhydex...here's a simple approach...rearrang d eqn to take d followin form...
9x^2=(16-x^2)(x-3)^2...inspection shows that the first expression on the right must be a perfect square...16-x^2=y^2.......eqn1 so we get 9x^2=y^2*(x-3)^2 or
3x=xy-3y or xy=3(x+y)........eqn2 now from eqn1 x^2+y^2=16 or (x+y)^2-2xy=16 hence we can substitut for xy from eqn2 to get;
(x+y)^2-6(x+y)-16=0 so that
x+y=8 or -2 then xy=24 or -6 we now sovlv quadratically for x & y;
z^2-(x+y)z+xy=0 or z^2-8z+24=0 & z^2+2z-6=0...solvin the above quadratics gives u d four values of x...thus x=......plz i have no calculator on me now...

rhydex 247: @ laplacian.
I can still recall when sum1 post dis questn.
5^(x+1)=3^(x^2-1).
which d soln goes dis way.
take log of bth sides.
(x+1)log5=(x^2-1)log3.
(x+1)log5-[(x+1)(x-1)log3]=0
(x+1)[log5-(x-1)log3]=0
x+1=0 or log5-xlog3+log3=0.
x=-1 or log15=xlog3.
x=-1 or x=log15/log3.
x=-1 or x=2.465.
Similarly to sir tunechi questn.
1^x+2^x=3^(x-1).
take log of bth sides.
xlog1+xlog2=(x-1)log3
xlog1+xlog2=xlog3-log3.
0.3010x=0.4771x-0.4771.
x=0.4771/0.17612.
x=2.708929145
x=3.
Hence my soln nd my ans still stand. U can check by puttin x=2.708929145 or x=3.

2+3=5
bt log2 + log3 z nt equal 2 log5

Mbahchiboy: MAKE UNA HELP ME OUT O........
**On my knees**

for your first question use Maclaurin's series...the solution is;
tanhx=x-2x^3/3!+4x^5/5!-...+...
(-2)^n*x^(2n+1)/(2n+1)! For
n=0,1,2...
For your second question,
x^5-32=0 may be written as
x^5=32e^(2#i) where #=22/7 and e=2.71828, it follows that
x=2e^(2#ni/5) for n=0,1,2,3,4 they provide all solutions to the equation...for other soutions check preivous pages...

Mbahchiboy: i wud like to ask d generals in dis thread
''y are u all been so adamant towards my questions

almost everybody gave solutions to your questions on the previous pages...unless u have not viewed them....

can you please solve this for me.

!: sola sells her books to customers at the price marked in her shop. titi allows her own customer 8% reduction from the marked price which means that she sells for 1104 naira the same books sola sells for 112o. determine titi marked price for the books that sola sells 56o naira.
a; what is titi selling price to her customers of this goods in (1) above.
b: find sola selling price for goods titi sells 736 naira.

2: a man left his fortune to be divided so that each of his3 sons receive 1/10,each of his 2 daughter receive 1/8.
#5000 was to be paid to her hospital and the rest to his widow. if the widow received #49000, how much did he leave behind.

Sir-Tunechi:
Mathz generalz take diz one chil
1^x + 2^x =3^x-1
find x
p.s:i nor wan hear newton raphson method o

let # denot congruency....then
1+2^x#0 (mod3) or
2^x#-1 (mod3) or 2^x#2 (mod3)
or 2^(x-1)#1 (mod3) by Fermat's Little Theorem, x-1 must divide @(3)=3-1=2, Euler's phi function. So x-1=1 or 2 so that x=2 or 3...but x=2 does not satisfy d given eqn so x=3....

d citizen:
The five star general,

Here is the clue to the analytical solution to the problem: 3^x + 4^x = 5^x

Divide both side of the equation by 4^x

3^x/4^x + 4^x/4^x = 5^x/4^x

0.75^x + 1 = 1.25^x

(1-0.25)^x + 1 = (1 + 0.25)^x

(1 + (-0.25))^x + 1 = (1 +0.25)^x
U will use binomial theorem to do the expansion and limit the expansion to x^2

(1 +x)^n = 1+ nx + n(n-1)x^2/2!. Use the expansion to get the answer to x by quadratic equation.

why limit it to second order?...why not third or fourth order because in those cases u have differnt results

pls hw can i kw hw to solve maths am so weak at it