[quote author=indoorlove][/quote]..newton's appoximation.

smurfy:

Master? Why you dey fall my hand nah? I be your boy o

I used Microsoft Maths, SIR.

...hmmmm,i respect you bro...i will attempt it when i'm less busy...what's ur course,school and level?moreso,can we have a private chat...?

indoorlove: Log3^x+Logx^3=10/3. Guru in the house, help me with above question

.the question is log3(basex)+logx(base3)=10/3.....from the rule of log...loga(baseb)=1/logb(basea),then...log3(basex)=1/logx(base3),substitute that,therefore logx(base3)+1/logx(base3)=10/3....let logx(base3) be p...then p+1/p=10/3...multiply through by 3p to give 3p^2-10p+3=0...then p=3 or 1/3....from logx(base3)=p...let p be 3...then,logx(base3)=3...x=27...let p be 1/3...then logx(base3)=1/3 and x=3^(1/3)...

Calculusf(x):
.the question is log3(basex)+logx(base3)=10/3.....from the rule of log...loga(baseb)=1/logb(basea),then...log3(basex)=1/logx(base3),substitute that,therefore logx(base3)+1/logx(base3)=10/3....let logx(base3) be p...then p+1/p=10/3...multiply through by 3p to give 3p^2-10p+3=0...then p=3 or 1/3....from logx(base3)=p...let p be 3...then,logx(base3)=3...x=27...let p be 1/3...then logx(base3)=1/3 and x=3^(1/3)...

..i 4 fear na , i tought as much. He wanted to write log 3 _(x) + log x_(3) =10/3 nd nt 'log 3^x + log x^3 =10/3'

nice1 @FX...i respect o...

Calculusf(x):
...hmmmm,i respect you bro...i will attempt it when i'm less busy...what's ur course,school and level?moreso,can we have a private chat...?

Course, school, level? Sorry. Classified.

Calculusf(x):
.the question is log3(basex)+logx(base3)=10/3.....from the rule of log...loga(baseb)=1/logb(basea),then...log3(basex)=1/logx(base3),substitute that,therefore logx(base3)+1/logx(base3)=10/3....let logx(base3) be p...then p+1/p=10/3...multiply through by 3p to give 3p^2-10p+3=0...then p=3 or 1/3....from logx(base3)=p...let p be 3...then,logx(base3)=3...x=27...let p be 1/3...then logx(base3)=1/3 and x=3^(1/3)...

i actually meant what u wrote. Thanks for yous swifty response!!

The mean deviation of 5 numbers is 4, when a number is removed, the mean deviation of the remaining numbers is 3 , what is that number ?

[quote author=benbuks]The mean deviation of 5 numbers is 4, when a number is removed, the mean deviation of the remaining numbers is 3 , what is that number ?[/quote

benbuks: ..i 4 fear na , i tought as much. He wanted to write log 3 _(x) + log x_(3) =10/3 nd nt 'log 3^x + log x^3 =10/3'

nice1 @FX...i respect o...

one can also apply change of base

A man has 10 gallons of a 50% sulphuric acid mixture, and 20 gallons of a 20% solution, and 5 gallons of a 10% solution. He wants to use up all the 10% solution and make 15 gallons of 30% solution. How much of each solution should he use?

benbuks: The mean deviation of 5 numbers is 4, when a number is removed, the mean deviation of the remaining numbers is 3 , what is that number ?

ans is " 8 "....hope i'm right

Wow, i'm glad to be back again, feeling the ingenuity of all the mathematicians and maths lovers in the house, I revere oo!

smurfy:

Wow! Outstandingly correct! Now we all know who General Q is, don't we?

Is it me?

benbuks: The mean deviation of 5 numbers is 4, when a number is removed, the mean deviation of the remaining numbers is 3 , what is that number ?

Let the number be denoted as 'n'
:. (4 x 5)-n/(5 - 1) = 3
(20 - n)/4 = 3
n = 20 - 12
n = 8

smurfy: A man has 10 gallons of a 50% sulphuric acid mixture, and 20 gallons of a 20% solution, and 5 gallons of a 10% solution. He wants to use up all the 10% solution and make 15 gallons of 30% solution. How much of each solution should he use?

Let each of the solution be represented with: 'a', 'b' and 'c', Thus, giving an equation;
50a + 20b + 10c = 30(15)
---
c = 5
15 = a + b + c
15 = a + b + 5
10 = a + b
b = 10 - a
---
50a + 20b + 10c = 450
50a + 20b + 10(5) = 450
50a + 20b + 50 = 450
50a + 20b = 440
---
50a + 20b = 440
50a + 20(10 - a) = 440
50a + 200 - 20a = 440
30a = 240
---
a = 8 gal
b = 2 gal
c = 5 gal

Awesome God...all glory be to him who gives life and knowledge,..whose faithfulness only approaches infinity irrespective of our infinitesimal faithfulness..

Greetings to you all brothers and sisters.. Happy sunday and happy solving.

1love

@Ortarico n sir chides.
Correct, good job..

solomon111: x^x^x=log3
x^y^y=log8.
Find x and y.

...

Ortarico:

Is it me?

Thou sayest

Ortarico:

Let the number be denoted as 'n'
:. (4 x 5)-n/(5 - 1) = 3
(20 - n)/4 = 3
n = 20 - 12
n = 8



Let each of the solution be represented with: 'a', 'b' and 'c', Thus, giving an equation;
50a + 20b + 10c = 30(15)
---
c = 5
15 = a + b + c
15 = a + b + 5
10 = a + b
b = 10 - a
---
50a + 20b + 10c = 450
50a + 20b + 10(5) = 450
50a + 20b + 50 = 450
50a + 20b = 440
---
50a + 20b = 440
50a + 20(10 - a) = 440
50a + 200 - 20a = 440
30a = 240
---
a = 8 gal
b = 2 gal
c = 5 gal

Beautifully wrong.

Where is my big oga na. @ master richiez? .guess in sch nw...pls i want 2 link 1 thread with another., hw do i do that. Pls .?

smurfy:

Beautifully wrong.

find the missing numbers : 2,_10,_40 :-(hint : _using the commutative law of addition

I hail my generals.

benbuks: I hail my generals.

u are the biggest general here sir....please oga mi sir, help us to solve that trig. question, sec@+cot@=5, find cos@, i believe u can help us 2 solve it o.... and i hope u are true with the mourning of NELSON MANDELA MADIBA

aysuccess99: oga mi,DOUBLEDX, i hail o, longest time bro...how's maths life going sir? and hope all is well? happy maths solving sir.

Hey Mr. AYSUCCESS99; I saw your ID in one of my funny Math dreams LOL !

Math life is great! I have been watching from the sidelines lately though... I'm doing good, I hope you are too? Yahweh has been good to me! I'll start contributing on the thread come Jan. 2014 by His grace...

aysuccess99:
u are the biggest general here sir....please oga mi sir, help us to solve that trig. question, sec@ cot@=5, find cos@, i believe u can help us 2 solve it o.... and i hope u are true with the mourning of NELSON MANDELA MADIBA

1/cos@ +cos@/sin@=5
Sin@ cos^2@/cos@sin@=5
5cos@sin@=sin@ cos^2@
Divide all tru by cos@
5sin@=tan@ cos@
*cos@=5sin@-tan@...i tink am half correct sir.

smurfy:

Beautifully wrong.

hahahahahahahahahahahahahahahaha, IDONBILIVIT!

Viewing this topic: Alpha Maximus(m), Ortarico(m)

Thank God you're here, oya attempt it na.

aysuccess99:
u are the biggest general here sir....please oga mi sir, help us to solve that trig. question, sec@+cot@=5, find cos@, i believe u can help us 2 solve it o.... and i hope u are true with the mourning of NELSON MANDELA MADIBA

...haha yes boss
the question is suppose 2b
cosec@ + cot@ =5
=>1/sin@ + cos@/sin@ =5
we have 5sin@=1+ cos@
from pythagorean identities , cos @ = sqrt[1-sin^2 @]
==>5sin@-1 =sqrt[1-sin^2 @]

taking square of both sides and puting sin@ as 'r' we have
25r^2 - 10r+1 =1-r^2
=r(26r-10)= 0
==>r=0 or 5/13
since sin@=r
==> @=0 (when r=0) or 22.61¤(when r=5/13)

hence cos@= 1 (when @=0) or 12/13=0.923 (when @=22.61)

benbuks: ...haha yes boss
the question is suppose 2b
cosec@ + cot@ =5
=>1/sin@ + cos@/sin@ =5
we have 5sin@=1+ cos@
from pythagorean identities , cos @ = sqrt[1-sin^2 @]
==>5sin@-1 =sqrt[1-sin^2 @]

taking square of both sides and puting sin@ as 'r' we have
25r^2 - 10r+1 =1-r^2
=r(26r-10)= 0
==>r=0 or 5/13
since sin@=r
==> @=0 (when r=0) or 22.61¤(when r=5/13)

hence cos@= 1 (when @=0) or 12/13=0.923 (when @=22.61)

wow, great workings bro....wow!!! thank GOD o, i have gain something 2day.....and tanx 4 checking,correcting and answering d question...more wisdom, knowledge and understanding to ur brain, and more battery or charger 2 ur calculator, oh, i have 4goten dat u self gan u are a calculator bro, gud work sir...kudos to u.,