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Re: Nairaland Mathematics Clinic by Nobody: 6:27pm On Nov 14, 2015
Soneh, my boss has solved the rest. Cheers.
Re: Nairaland Mathematics Clinic by agentofchange1(m): 9:18pm On Nov 14, 2015
sorry, Had flat ba3

cont .Q4

Re: Nairaland Mathematics Clinic by agentofchange1(m): 9:57pm On Nov 14, 2015
Am just a baby sir , hoping to be like you when i grow up wink
Karmanaut:
Soneh, my boss has solved the rest.
Cheers.
Re: Nairaland Mathematics Clinic by seuntosyn: 11:32pm On Nov 14, 2015
Help solve this 6log(n+4)=2log8. What's n
Re: Nairaland Mathematics Clinic by Nobody: 6:39am On Nov 15, 2015
seuntosyn:
Help solve this 6log(n+4)=2log8. What's n
n=-2.

First divide both sides by 6
Log(n+4)=log(8 )/3
Log(n+4)=1/3log(8 )
Recall the law of logarithms that says a*log(b) = log(b)a
So you can rewrite the RHS as log(8 )1/3
Log(n+4)=log(2) [The cube root of 8 is 2]
Log cancels log.
n+4 = 2
n=-2.

Attached is the graph.

Re: Nairaland Mathematics Clinic by LORDDICE(m): 9:23am On Nov 15, 2015
Karmanaut:
n=-2.
First divide both sides by 6 Log(n+4)=log(8 )/3 Log(n+4)=1/3log(8 ) Recall the law of logarithms that says a*log(b) = log(b)a So you can rewrite the RHS as log(8 )1/3 Log(n+4)=log(2) [The cube root of 8 is 2] Log cancels log. n+4 = 2 n=-2.
Attached is the graph.

please what software do u use to plot these graphs
Re: Nairaland Mathematics Clinic by Nobody: 9:59am On Nov 15, 2015
LORDDICE:



please what software do u use to plot these graphs
If I'm on Mobile, I use Grapher.
If I'm with my system then Geogebra or Mathematica.
If I'm with neither I use my trusty Voyager 200.
Re: Nairaland Mathematics Clinic by MathsChic(f): 11:00am On Nov 15, 2015
agentofchange1:
Try...


for. 0<x<2π

Let m = 2015
Let I1 = sinm(x)/(sinm(x)+cosm(x))

∫sinm(x)/(sinm(x)+cosm(x)) dx, 0<x<2pi

I1 is a periodic function, and if shifted by pi/2 will give
sinm(x-pi/2)/(sinm(x-pi/2)+cosm(x-pi/2)), and should give the same result when integrated between 0<x<2pi

Let's call it I2
So, I2 = cosm(x)/(cosm(x)+sinm(x))

∫I1, 0<x<2pi =M
∫I2, 0<x<2pi =M

therefore, 2M = ∫I1 + ∫I2
2M = ∫((cosm(x)+sinm(x))/(cosm(x)+sinm(x)), 0<x<2pi
2M = ∫dx, 0<x<2pi
2M = [x],0,2pi
2M = 2pi

therefore M = pi.

∫I1 = pi smiley

m can be any value, answer will still be pi.

Re: Nairaland Mathematics Clinic by ayokunlei(m): 12:01pm On Nov 15, 2015
agentofchange1:
Am just a baby sir , hoping to be like you when i grow up wink

boss, I understand HW u got the value of A and B, HW did u you get 'A-B-D = 2
thanks.
Re: Nairaland Mathematics Clinic by ladokuntlad(m): 12:13pm On Nov 15, 2015
ayokunlei:


boss, I understand HW u got the value of A and B, HW did u you get 'A-B-D = 2
thanks.
comparing coefficients
compare constants at the LHS with constants at the RHS gives that
Re: Nairaland Mathematics Clinic by Nobody: 12:18pm On Nov 15, 2015
agentofchange1:
This man is sure on point Nice piece sir, planning/thinking of doing mathematical statistics or statistical modelling or statistical simulations in either Staford , Princeton or Cambridge, any requites to note before embarking on any ? currently and undergrad, with statistics major ( loves anything that involves calculus or D.Es, in solving real problems. tnx in advance

Nice one. Getting into Stanford,Princeton,MIT,Harvard and the likes is very difficult but doable. I will advise you to start preparing your grad application even though you are still an undergraduate student. You will have to take GRE math subject test(this test is very challenging) as well as the GRE general test and get decent scores to be accepted. The secret of acing these standardized test is practice, also get involved in research now(only if you knew the advantage) ask revered professors in your department that you will like to collaborate with them in their research. If you can get a paper or two published in quality journals(emphasis quality not mushroom journals),this will give you an advantage. Also research some of your professors and find the ones who publish in quality journals, try to be close to them so they can give you a very strong letter of recommendation. Do not stop developing yourself like some one will always say when preparation intersects with opportunity people call it luck.

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Re: Nairaland Mathematics Clinic by Nobody: 12:22pm On Nov 15, 2015
[quote author=Karmanaut post=40033285]
n=-2.

First divide both sides by 6
Log(n+4)=log(8 )/3
Log(n+4)=1/3log(8 )
Recall the law of logarithms that says a*log(b) = log(b)a
So you can rewrite the RHS as log(8 )1/3
Log(n+4)=log(2) [The cube root of 8 is 2]
Log cancels log.
n+4 = 2
n=-2.


I am not comfortable with the 'log cancels log' thingy, albeit i know what you mean. I thought they should be a better explanation for that. grin
Re: Nairaland Mathematics Clinic by ladokuntlad(m): 12:49pm On Nov 15, 2015
[quote author=masperano post=40042450][/quote]
log(n+4)=1/3log8
log(n+4)=log2
log(n+4)-log2=0
log(n+4/2)=0
(n+4)/2=100
(n+4)/2=1
n+4=2
n=2-4
n=-2


Is this a better explanation?
Re: Nairaland Mathematics Clinic by MathsChic(f): 12:50pm On Nov 15, 2015
ladokuntlad:
my desert for today
Problem 2
(√2+√11+√13)(√2+√11-√13)(√2-√11+√13)(-√2+√11+√13)
=-(√2+√11+√13)(√2+√11-√13)(√2-√11+√13)(√2-√11-√13)

=-(√2+√11+√13)(√2+√11-√13)(√2-√11+√13)(√2-√11-√13)

=-((√2+√11)2-(√13)2))((√2-√11)2-(√13)2)
=-(2+11+2√22-13)(2+11-2√22-13)
=-(2√22)(-2√22)
=4.22
=88 cheesy
Re: Nairaland Mathematics Clinic by stlaibrowne(m): 12:52pm On Nov 15, 2015
MathsChic:

Problem 2
(√2+√11+√13)(√2+√11-√13)(√2-√11+√13)(-√2+√11+√13)
=-(√2+√11+√13)(√2+√11-√13)(√2-√11+√13)(√2-√11-√13)

=-(√2+√11+√13)(√2+√11-√13)(√2-√11+√13)(√2-√11-√13)

=-((√2+√11)2-(√13)2))((√2-√11)2-(√13)2)
=-(2+11+2√22-13)(2+11-2√22-13)
=-(2√22)(-2√22)
=4.22
=88 cheesy
what's the problem here I don't get it. And secondly this is a primary school sure.
Re: Nairaland Mathematics Clinic by ladokuntlad(m): 12:54pm On Nov 15, 2015
MathsChic:

Problem 2
(√2+√11+√13)(√2+√11-√13)(√2-√11+√13)(-√2+√11+√13)
=-(√2+√11+√13)(√2+√11-√13)(√2-√11+√13)(√2-√11-√13)

=-(√2+√11+√13)(√2+√11-√13)(√2-√11+√13)(√2-√11-√13)

=-((√2+√11)2-(√13)2))((√2-√11)2-(√13)2)
=-(2+11+2√22-13)(2+11-2√22-13)
=-(2√22)(-2√22)
=4.22
=88 cheesy
taaah
u ar too big for dat question naaa
leave it for babies

attempt odas
Re: Nairaland Mathematics Clinic by ayokunlei(m): 12:56pm On Nov 15, 2015
ladokuntlad:
comparing coefficients compare constants at the LHS with constants at the RHS gives that
yes, i did that here too. A +B + C = 0
how do we arrive at A -B - D= 2. agentofchange1
Re: Nairaland Mathematics Clinic by Nobody: 12:59pm On Nov 15, 2015
ladokuntlad:

log(n+4)=1/3log8
log(n+4)=log2
log(n+4)-log2=0
log(n+4/2)=0
(n+4)/2=100
(n+4)/2=1
n+4=2
n=2-4
n=-2


Is this a better explanation?


Yea this is probably the way i will teach some one unfamiliar with logarithmic concept.
Re: Nairaland Mathematics Clinic by stlaibrowne(m): 1:04pm On Nov 15, 2015
I just jammed this thread.
Re: Nairaland Mathematics Clinic by ladokuntlad(m): 1:05pm On Nov 15, 2015
ayokunlei:


yes, i did that here too. A +B + C = 0

how do we arrive at A -B - D= 2.
agentofchange1
boss
A-B-d=2 are coefficients for constants
i.e constants on the LHS is 2 which corresponds to A-B-D on the RHS

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Re: Nairaland Mathematics Clinic by Nobody: 1:13pm On Nov 15, 2015
Find the volume of a cylinder with a diameter of 4xsq - 6x + 2 and a height of 2x + 7.
Re: Nairaland Mathematics Clinic by LORDDICE(m): 1:14pm On Nov 15, 2015
Karmanaut:
If I'm on Mobile, I use Grapher. If I'm with my system then Geogebra or Mathematica. If I'm with neither I use my trusty Voyager 200.
u na boss.. d voyager 200... is it a calculator? if it is, huw much is it?
Re: Nairaland Mathematics Clinic by LORDDICE(m): 1:15pm On Nov 15, 2015
stlaibrowne:
I just jammed this thread.

you are very much welcome...
Re: Nairaland Mathematics Clinic by stlaibrowne(m): 1:18pm On Nov 15, 2015
all4naija:
Find the volume of a cylinder with a diameter of 4xsq - 6x + 2 and a height of 2x + 7.

(2x+1)(x+1)(2x+1)(x+1)(2x+7)pie
Re: Nairaland Mathematics Clinic by stlaibrowne(m): 1:19pm On Nov 15, 2015
LORDDICE:



you are very much welcome...

Thank you
Re: Nairaland Mathematics Clinic by Nobody: 1:24pm On Nov 15, 2015
stlaibrowne:


(2x+1)(x+1)(2x+1)(x+1)(2x+7)pie
Please, show how you arrive at that. I don't think that is the correct answer.
Re: Nairaland Mathematics Clinic by LORDDICE(m): 1:24pm On Nov 15, 2015
stlaibrowne:

Thank you

u Also a madmatician ?
Re: Nairaland Mathematics Clinic by ladokuntlad(m): 1:38pm On Nov 15, 2015
stlaibrowne:


(2x+1)(x+1)(2x+1)(x+1)(2x+7)pie
(2x-1)2(x-1)2(2x+7)pie

-1 and not +1
Re: Nairaland Mathematics Clinic by Nobody: 1:40pm On Nov 15, 2015
LORDDICE:

u na boss.. d voyager 200... is it a calculator?
if it is, huw much is it?
Yeah, a programmable calculator.
Looks like this:

Costs $223(Approximately #45, 000) on Amazon.

1 Like

Re: Nairaland Mathematics Clinic by ladokuntlad(m): 1:42pm On Nov 15, 2015
all4naija:
Find the volume of a cylinder with a diameter of 4xsq - 6x + 2 and a height of 2x + 7.
(2x-1)2(x-1)2(2x+7)pie
How?
volume = pieXr2h
r=D/2=(4x2 - 6x + 2)/2=2x2-3+1
h=2x+7
subtitute into the formula
gives the above result
Re: Nairaland Mathematics Clinic by Nobody: 1:47pm On Nov 15, 2015
[quote author=masperano post=40042450][/quote]
Na so them take teach me.
Re: Nairaland Mathematics Clinic by Nobody: 2:14pm On Nov 15, 2015
ladokuntlad:

(2x-1)2(x-1)2(2x+7)pie
How?
volume = pieXr2h
r=D/2=(4x2 - 6x + 2)/2=2x2-3+1
h=2x+7
subtitute into the formula
gives the above result
How is that correct when the radius R is 2x 2 - 3x + 1 ?

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