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Re: Nairaland Mathematics Clinic by gpercuxionz(m): 12:03pm On Nov 20, 2017
Good day everyone pls I need help to the question 25. Thanks

Re: Nairaland Mathematics Clinic by Umartins1(m): 3:59pm On Nov 23, 2017
gpercuxionz:
Good day everyone pls I need help to the question 25. Thanks

Let the ogas cross check

1 Like

Re: Nairaland Mathematics Clinic by Angelan97: 10:40pm On Nov 23, 2017
Hi, can someone help me with this one:

To cover a distance of 10km, a girl runs some of the way at 15km/h and walks the rest of the way at 5km/h. Her total time of journey is 1hour. A) if the distance walked is x km find a simple expession for the distance run. B) Derive an equation involving x and use it to find the distance walked?
Re: Nairaland Mathematics Clinic by Umartins1(m): 10:26am On Nov 24, 2017
Angelan97:
Hi, can someone help me with this one:

To cover a distance of 10km, a girl runs some of the way at 15km/h and walks the rest of the way at 5km/h. Her total time of journey is 1hour. A) if the distance walked is x km find a simple expession for the distance run. B) Derive an equation involving x and use it to find the distance walked?


The answer provided is subject to further scrutiny

Re: Nairaland Mathematics Clinic by Umartins1(m): 10:54am On Nov 24, 2017
Check this alternative solution for distance run. I dey argue with one of my guys ontop the one wey correct

Re: Nairaland Mathematics Clinic by meekmill(m): 12:51pm On Dec 01, 2017
Guys help me out

Re: Nairaland Mathematics Clinic by Umartins1(m): 2:03pm On Dec 01, 2017
Here
meekmill:
Guys help me out

1 Like

Re: Nairaland Mathematics Clinic by meekmill(m): 4:02pm On Dec 01, 2017
Umartins1:
Here

Bro thanks very much
Re: Nairaland Mathematics Clinic by personal59: 12:09pm On Dec 07, 2017
please help out guys

Re: Nairaland Mathematics Clinic by GeniusDavid(m): 10:25am On Dec 23, 2017
If 47% of the people in a community voted in a local election and 75% voted in a federal election, what is the least percentage that voted in both?
Re: Nairaland Mathematics Clinic by Umartins1(m): 11:17am On Dec 23, 2017
GeniusDavid:
If 47% of the people in a community voted in a local election and 75% voted in a federal election, what is the least percentage that voted in both?

47-x+75-x+x=100

122-x=100

x = 122-100

x= 22
Re: Nairaland Mathematics Clinic by Redder97(m): 8:51am On Dec 24, 2017
Umartins1:


45-x+75-x+x=100

120-x=100

x = 120-100

x= 20


47% not 45%.
Re: Nairaland Mathematics Clinic by assavin: 2:38pm On Dec 25, 2017
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Re: Nairaland Mathematics Clinic by Crizillion1(m): 8:33am On Dec 26, 2017
add me up on whatsapp group 08131667273
Re: Nairaland Mathematics Clinic by IFEOLUWAKRIZ: 1:16pm On Dec 26, 2017
shocked shocked shocked



I register my presence .

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1 Like 2 Shares

Re: Nairaland Mathematics Clinic by tobillionaire(m): 9:49am On Dec 30, 2017
pls I nid help on d working of this

find d integral of √tanx
Re: Nairaland Mathematics Clinic by chrismato(m): 12:24pm On Dec 30, 2017
Youngsage:

x + y =5 -------- eqn i.
x^x + y^y=31 ----- eqn ii.
From equatns i & ii, take the log of both sides
(xlog^x + ylog^y) = log 31------ eqn iii.
log(x+y) = log 5 ------------ eqn iv.
xylogxy= log 31 -------- eqn v.
log xy = log 5 --------- eqn vi.
(Using elimination method);
xylog5 = log31
xy = log 31/log 5 =6.2 approx. 6.

Recall, from eqn i, x + y = 5.
So x = 6/y or y = 6/x.
Therefore x=2 when y=3.
ans: x=2, and y=3.
Youngsage:

x + y =5 -------- eqn i.
x^x + y^y=31 ----- eqn ii.
From equatns i & ii, take the log of both sides
(xlog^x + ylog^y) = log 31------ eqn iii.
log(x+y) = log 5 ------------ eqn iv.
xylogxy= log 31 -------- eqn v.
log xy = log 5 --------- eqn vi.
(Using elimination method);
xylog5 = log31
xy = log 31/log 5 =6.2 approx. 6.

Recall, from eqn i, x + y = 5.
So x = 6/y or y = 6/x.
Therefore x=2 when y=3.
ans: x=2, and y=3.
Youngsage:

x + y =5 -------- eqn i.
x^x + y^y=31 ----- eqn ii.
From equatns i & ii, take the log of both sides
(xlog^x + ylog^y) = log 31------ eqn iii.
log(x+y) = log 5 ------------ eqn iv.
xylogxy= log 31 -------- eqn v.
log xy = log 5 --------- eqn vi.
(Using elimination method);
xylog5 = log31
xy = log 31/log 5 =6.2 approx. 6.

Recall, from eqn i, x + y = 5.
So x = 6/y or y = 6/x.
Therefore x=2 when y=3.
ans: x=2, and y=3.
I,disagree with this solution. bcus log31/logo =2.13approx.an not 6 by approx
Re: Nairaland Mathematics Clinic by Mechanics96(m): 3:04pm On Dec 30, 2017
This solution is wrong sir; u don't take the log of x^x+y^y as xlog^x +ylog^y, and neither do you say xlogx+ ylogy. It should be log(x^x+y^y). These two are not the same.

Simple test ; log3+log7 is never equal to log (3+7)

Also log31/log5 is not 6.2 but 2.13...




Let's get something clear on this question. It is only an IQ test, no maths rule supports the solution, yet the solution is crystal clear.
There are many of such questions, the tricks for hacking the solution is thought in numerology.
For example; if the sum of two numbers that have themselves as powers is a prime number then the numbers must be prime or one prime one even. x^x+y^y=31, x and y are both prime, one of them is even.

Not classroom mathematics...
Re: Nairaland Mathematics Clinic by Mechanics96(m): 3:37pm On Dec 31, 2017
tobillionaire:
pls I nid help on d working of this

find d integral of √tanx

Re: Nairaland Mathematics Clinic by Mechanics96(m): 3:45pm On Dec 31, 2017
Cont'd

Re: Nairaland Mathematics Clinic by inspirational93: 11:47pm On Dec 31, 2017
click this link to download free PDF document files and any related courses on any topic
http://viwright.com/4as2
Re: Nairaland Mathematics Clinic by KlarkDevlin: 4:37pm On Jan 03, 2018
Mechanics96:
This solution is wrong sir; u don't take the log of x^x+y^y as xlog^x +ylog^y, and neither do you say xlogx+ ylogy. It should be log(x^x+y^y). These two are not the same.

Simple test ; log3+log7 is never equal to log (3+7)

Also log31/log5 is not 6.2 but 2.13...




Let's get something clear on this question. It is only an IQ test, no maths rule supports the solution, yet the solution is crystal clear.
There are many of such questions, the tricks for hacking the solution is thought in numerology.
For example; if the sum of two numbers that have themselves as powers is a prime number then the numbers must be prime or one prime one even. x^x+y^y=31, x and y are both prime, one of them is even.

Not classroom mathematics...
I also got carried away recently by numerology. If someone is interested, then there is a lot of useful information https://numbervoice.com/ It is very interesting.
Re: Nairaland Mathematics Clinic by jojokings: 6:11pm On Jan 03, 2018
Please I need help regarding my junior sisters maths problem, she has written maths in weac 4 times now yet the highest she got was D7. Who can assist for her to obtain her papers in one sittings. Anyhow that is possible. Am ready for her to secure university admission to study accounting. Preferable if your withing Abuja and environment. Call my number let's discuss. 08036568875
Re: Nairaland Mathematics Clinic by jerome263: 6:41pm On Jan 06, 2018
jojokings:
Please I need help regarding my junior sisters maths problem, she has written maths in weac 4 times now yet the highest she got was D7. Who can assist for her to obtain her papers in one sittings. Anyhow that is possible. Am ready for her to secure university admission to study accounting. Preferable if your withing Abuja and environment. Call my number let's discuss. 08036568875

I have a friend who is well gifted in mathematics.... he is currently studying mathematics in Uniben..........quote me if you need his number....
Re: Nairaland Mathematics Clinic by jojokings: 7:21am On Jan 08, 2018
jerome263:


I have a friend who is well gifted in mathematics.... he is currently studying mathematics in Uniben..........quote me if you need his number....


So how could he be if help?
Re: Nairaland Mathematics Clinic by jerome263: 8:56am On Jan 08, 2018
jojokings:


So how could he be if help?
contact me on WhatsApp let's reason zero 7 zero, six 8 nine 9 six 2 six zero
Re: Nairaland Mathematics Clinic by jojokings: 9:00am On Jan 08, 2018
Please who knows when NABTEB O LEVEL exam will be holding, or anyone that knows about special reliable centers for making ones o level papers clear.?
Re: Nairaland Mathematics Clinic by Red2C(m): 11:31am On Jan 08, 2018
No. 22 pls

Re: Nairaland Mathematics Clinic by lordbeans(m): 6:47am On Jan 11, 2018
Red2C:
No. 22 pls

Ok let me have a crack on this.


We were asked to show that 2y^2-y-1<=0 for all values of x in R (i.e. (-inf, inf)), if y=(2x+3)/(x^2+2x+3) ;

Let's first factorize 2y^2-y-1=(2y+1)(y-1);=> (2y+1)(y-1)<=0; for this to be true, it means (2y+1)<=0 and (y-1)>=0 (We will call this scenario 1) OR (2y+1)>=0 and (y-1)<=0 (scenario 2).

Assessing scenario 1(a) 2y+1<=0; => y<=-1/2;(that is y can be -1,-2,-3,... etc). BUT this is in contradiction to scenario 1 (b) y-1>=0 (i.e. -1-1=-2 CANNOT be greater than 0). This means there exist values for y such that scenario one is not satisfied (there goes scenario 1 off the window). We are then left with scenario 2.

We must now show that substituting y=(2x+3)/(x^2+2x+3) into scenario 2 must satisfy the inequality.

Let's substitute it into scenario 2(a), we obtain 2((2x+3)/(x^2+2x+3))+1 ?>=0 (i.e. is this >=0), => 2(2x+3) ?>= -(x^2+2x+3) , simplifying further we have x^2+6x+9 ?>=0; => (x+3)^2 ?>=0. Any number squared is always positive irrespective of whether the number is negative or positive, hence we can now be confident to remove the question mark in front of the inequality and say yes for any value of x, 2y+1>=0, and this implies that y-1<=0 for y=(2x+3)/(x^2+2x+3); hence the proof of 2y^2-y-1<=0 is complete.

Second part (finding the maximum and minimum of y)

y=(2x+3)/(x^2+2x+3); we look for the stationary or turning points which are usually maximum, minimum or inflexion points. The stationary points are points where there is a sign change of y as increases that is y moves from positive to negative or vice versa; this means y must go through 0. Knowing this characteristic, we would want to seek these points ; one way of achieving this is to find the points where the derivative of y wrt x is 0 (i.e. dy/dx=0).

Differentiating dy/dx using quotient rule gives dy/dx=(-2x^2-6x)/(x^2+2x+3)^2=0;

-2x^2-6x=0; =>x^2+3x=> x(x+3)=0=> x=0, x=-3; Hence the turning points occur at x=0 an -3; but you were asked about the maximum and minimum values not where in the x space they occur. Substituting the values into the function we get y=1 at x=0 and y=-1/2 at x=-3.
all other values of y are bounded between 1 and -1/2, which implies the maximum value of y=1 and the minimum value of y=-1/2.
Re: Nairaland Mathematics Clinic by Aybalance: 4:37pm On Jan 19, 2018
pls i need help on this... ........nP3+nC4=6.find n
Re: Nairaland Mathematics Clinic by lordbeans(m): 1:40pm On Jan 20, 2018
Aybalance:
pls i need help on this... ........nP3+nC4=6.find n

n=3

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