SpicyMimi: Wooooow!!! Am soooo thrilled by what am seeing here....is it that u solve and practice Maths everyday Cause this is really beyond me!
Keepit up guys!

Thank you!

ControG: @ Richie nd Doubledx I must commend ur work here...u guys 're doing a great Job here..I enjoy reading stuffs on dis page...alot. m not a maths guru neither am I a learner..I read Economics & Statistics and I am professional in my field..but when I see guys like dis..I luv 2 b a learner.kip it up!

Richiez:

Thanks alot bro, we're just following the saying; 'a candle looses nothing by lighting another candle'

common we've been doing alot of solving, does anyone know the definition of mathematics.

common we've been doing alot of solving, does anyone know the definition of mathematics...

Maths Gurus in d house pls kindly help me out on dis.
The data below indicates the mass disribution score by 50 students in a test:15,16,44,13,17,15,47,27,36,29,38,64,39,46,42,45,29,28,16,12,48,15,26,39,52,37,36,12,27,57,42,56,71,15,23,28,82,29,36,36,82,58,28,62,24,51,37,14,18,38.
Construct a histogram for d table. My problem is how do I group it? I can do d chart myself but to group it I dnt knw how to go abt it.


Tenks

happy weekend everyone

ositadima1:

^^^ Hi

Hiiii...

ositadima1: brachistochrone problem: Suppose that there is a
rollercoaster. There is point 1 and point 2 . Point 1 is at the higher place when compared to the point 2, so the rollercoaster rolls down from
point 1 to point 2. Assuming no friction, we want to
build a rollercoaster path from point 1 to point 2, so
that the rollercoaster will reach point 2 from point 1 in
shortest time. Find the curve dat would help us achieve this?

This question is tricky, emmm, my fellow engineers lets tackle this one.

time to travel from point 1 to the other point 2 is given by the integral

t=∫[ds/v], we know v=ds/dt

s is the path length,
v is the speed at any point on the path

based on the principle of conservation of energy: kinetic energy = potential energy at any point on the path
so, (0.5)*m*v^2=m*g*y

where, m is the mass of the rollercoaster
g is acceleration due to gravity
y is the vertical displacement

manipulating for v, v=(2*g*y)^0.5
we know dat, s^2=x^2+y^2
yep the path s at any point has the coordinates x, y

making s the subject and the infinitesimal value of s at any point in the x-y plane is
ds=(dx^2+dy^2)^0.5
dx={[1+(dy/dx)^2]*dx^2}^0.5
dx=(1+(dy/dx)^2)^0.5*dx

plugging this in our time equation we get
t=∫[{1+(dy/dx)^2}^0.5*dx/(2*g*y)^0.5]
t=∫[1+(dy/dx)^2/(2*g*y)]^0.5*dx

Now the wahala starts,

lets say, t=∫fdx
where f=[1+(dy/dx)^2/(2*g*y)]^0.5 this equation is importance for us, why?...... well because we are goint vary it.
for simplicity sake lets mae y'=dy/dx

because f(y, dy/dx) does not depend on x explicitly we are going to apply Beltrami identity (a contrived form of Euler-Lagrange differential equation)
[f-y'∂f/∂y']=c

computing
using chain rule and nothing dat we are differentiating with y'... so y is constant here.
∂f/∂y'=y'(1+y'^2)^-0.5*(2gy)^-0.5

substituting back into
[f-y'∂f/∂y']=c

(1+y'^2)^0.5/(2*g*y)^0.5-y'[y'(1+y'^2)^-0.5*(2gy)^-0.5]=c
(1+y'^2)^0.5/(2*g*y)^0.5-y'^2*(1+y'^2)^-0.5*(2gy)^-0.5]=c

(2*g*y)^-0.5[(1+y'^2)^0.5-y'^2*(1+y'^2)^-0.5]=c

(2*g*y)^-0.5*(1+y'^2)^-0.5=c

squaring both sides and rearranging

[1+y'^2]y=1/[2gc^2]=k^2

We have [1+(dy/dx)^2]y=k^2

Solve for x and y (i go show the workings for this later, make i go chop first)

x(1/2)*k^2*(θ-sinθ)
y=(1/2)*k^2*(1-cosθ)

...equation of a cycloid

Oboy my head they spin like electron on 4f energy level oo. Guys for the books, go to www.4shared.com register, it's easy and straight forward, and run a search, you will get plenty plenty books for there, all free. I salute, ...military style.

Note: I didn't solve this entirely on my own, but I simplified some points here and there. Just saying...

estilo: Maths Gurus in d house pls kindly help me out on dis.
The data below indicates the mass disribution score by 50 students in a test:15,16,44,13,17,15,47,27,36,29,38,64,39,46,42,45,29,28,16,12,48,15,26,39,52,37,36,12,27,57,42,56,71,15,23,28,82,29,36,36,82,58,28,62,24,51,37,14,18,38.
Construct a histogram for d table. My problem is how do I group it? I can do d chart myself but to group it I dnt knw how to go abt it.


Tenks

Estilo, the easiest way is to first make a data range/frequency table. Make the table with two columns. Now, for data range, since you have a wide data you can make it 0-10, 11-20, 21-30 and so on. For the frequency part you put in tally every time you have a number falling in a particular range, for simplisity when you have four tallies then you cross the fift to group it in chunks of five. You may want to cross out any number you tallied to avoid counting it multiple times.

I hail all the gurus and able generals on this thread, God bless y'all. Please I need help with this:

1. Evaluate ∫{xy (2y÷x - 3) + 2x^2 - y^2}dx/(x - y)(6x^3 - 4x^2 - 3x^2y + 2xy +16x - 8y)

2. Find the equation of a curve whose gradient to its tangent at a point D (π/4, 4) is given by the equation y' = 8cos^2(x) sin(2x).

ositadima1:

Estilo, the easiest way is to first make a data range/frequency table. Make the table with two columns. Now, for data range, since you have a wide data you can make it 0-10, 11-20, 21-30 and so on. For the frequency part you put in tally every time you have a number falling in a particular range, for simplisity when you have four tallies then you cross the fift to group it in chunks of five. You may want to cross out any number you tallied to avoid counting it multiple times.

Tenks for the attempt. Pls am a mass comm student nd I dnt knw anytin abt calculation,but statistics is compulsory for me dis semester. Why am saying dis is for you to come down to my level.

Since 12 is d least nomba, I used 10-20,20-30..... And as per d frequency am aving problem dere pls explain to me in a simpler way.

Tenks

ositadima1:

time to travel from point 1 to the other point 2 is given by the integral

t=∫[ds/v], we know v=ds/dt

s is the path length,
v is the speed at any point on the path

based on the principle of conservation of energy: kinetic energy = potential energy at any point on the path
so, (0.5)*m*v^2=m*g*y

where, m is the mass of the rollercoaster
g is acceleration due to gravity
y is the vertical displacement

manipulating for v, v=(2*g*y)^0.5
we know dat, s^2=x^2+y^2
yep the path s at any point has the coordinates x, y

making s the subject and the infinitesimal value of s at any point in the x-y plane is
ds=(dx^2+dy^2)^0.5
dx={[1+(dy/dx)^2]*dx^2}^0.5
dx=(1+(dy/dx)^2)^0.5*dx

plugging this in our time equation we get
t=∫[{1+(dy/dx)^2}^0.5*dx/(2*g*y)^0.5]
t=∫[1+(dy/dx)^2/(2*g*y)]^0.5*dx

Now the wahala starts,

lets say, t=∫fdx
where f=[1+(dy/dx)^2/(2*g*y)]^0.5 this equation is importance for us, why?...... well because we are goint vary it.
for simplicity sake lets mae y'=dy/dx

because f(y, dy/dx) does not depend on x explicitly we are going to apply Beltrami identity (a contrived form of Euler-Lagrange differential equation)
[f-y'∂f/∂y']=c

computing
using chain rule and nothing dat we are differentiating with y'... so y is constant here.
∂f/∂y'=y'(1+y'^2)^-0.5*(2gy)^-0.5

substituting back into
[f-y'∂f/∂y']=c

(1+y'^2)^0.5/(2*g*y)^0.5-y'[y'(1+y'^2)^-0.5*(2gy)^-0.5]=c
(1+y'^2)^0.5/(2*g*y)^0.5-y'^2*(1+y'^2)^-0.5*(2gy)^-0.5]=c

(2*g*y)^-0.5[(1+y'^2)^0.5-y'^2*(1+y'^2)^-0.5]=c

(2*g*y)^-0.5*(1+y'^2)^-0.5=c

squaring both sides and rearranging

[1+y'^2]y=1/[2gc^2]=k^2

We have [1+(dy/dx)^2]y=k^2

Solve for x and y (i go show the workings for this later, make i go chop first)

x(1/2)*k^2*(θ-sinθ)
y=(1/2)*k^2*(1-cosθ)

...equation of a cycloid

Oboy my head they spin like electron on 4f energy level oo. Guys for the books, go to www.4shared.com register, it's easy and straight forward, and run a search, you will get plenty plenty books for there, all free. I salute, ...military style.

Note: I didn't solve this entirely on my own, but I simplified some points here and there. Just saying...

Feeling you.....

Ghettoguru: I hail all the gurus and able generals on this thread, God bless y'all. Please I need help with this:

1. Evaluate ∫{xy (2y÷x - 3) + 2x^2 - y^2}dx/(x - y)(6x^3 - 4x^2 - 3x^2y + 2xy +16x - 8y)

2. Find the equation of a curve whose gradient to its tangent at a point D (π/4, 4) is given by the equation y' = 8cos^2(x) sin(2x).

Question 1

Let's start by simplifying the numerator =>

{xy (2y/x - 3) + 2x^2 - y^2}
= 2y^2 - 3xy + 2x^2 - y^2
= y^2 - 3xy + 2x^2
Factorising yields=>
y^2 - 2xy - xy + 2x^2
y(y - 2x) - x(y - 2x)
(y - x)(y - 2x)

^Substituting back in the question yields=>
= ∫(y - x)(y - 2x)/x/(x - y)(6x^3 - 4x^2 - 3x^2y + 2xy +16x - 8y)
= ∫(y - x)(y - 2x)/x/- (y- x)(6x^3 - 4x^2 - 3x^2y + 2xy +16x - 8y)
= ∫(2x - y)dx/(6x^3 - 4x^2 - 3x^2y + 2xy +16x - 8y)

So from here, I tried checking if the reduced numerator is a factor of the denominator so as to simplify further and it checked out.

Here it goes => let's check again together by substitution. If 2x - y = 0, the y = 2x.

If we substitute y = 2x in the denominator the answer is 0 (zero) too, meaning 2x - y is a factor of (6x^3 - 4x^2 - 3x^2y + 2xy +16x - 8y). Factorising the denominator by initiating the division of polynomial yields =>

= ∫(2x - y)dx/(2x - y)(3x^2 - 2x + 8 )
= ∫(2x - y)dx/(2x - y)(3x^2 - 2x + 8 )
:. The question has been reduced to ∫dx/(3x^2 - 2x + 8 )

Looking at the quadratic denominator, it doesn't have perfect factors and cannot be factorised. Hence we apply the integral formula for a quadratic denominator that cannot be factorised.

i.e => ∫dx/(ax^2 + bx + c) = 2/√(4ac - b^2).arcTan (2ax + b)/√(4ac - b^2)

The denominator is 3x^2 - 2x + 8 which means => a = 3, b = -2 & c = 8. Substituting in the above formula yields =>

∫dx/(3x^2 - 2x + 8 )

= 2/√[(4)(3)(8 ) - (-2)^2].[arcTan (6x - 2)/√[(4)(3)(8 ) - (-2)^2]
= 2/√92 arcTan 2(3x -1)/√92
= 2/2√23 arcTan 2(3x -1)/2√23
= 1/√23 arcTan (3x - 1)/√23

Hence =>

∫{xy (2y÷x - 3) + 2x^2 - y^2}dx/(x - y)(6x^3 - 4x^2 - 3x^2y + 2xy +16x - 8y)

= ∫dx/(3x^2 - 2x + 8 ) =>
= 1/√23 arcTan (3x - 1)/√23

Question 2 coming up later in the day. Take care bruv.

A workout 4 guys on this tread.
If m:n=2:3, u:v=4:5 and m-u:n+v=2:5, find the ratio m+10u:2n-5v

Johncasey: A workout 4 guys on this tread.
If m:n=2:3, u:v=4:5 and m-u:n+v=2:5, find the ratio m+10u:2n-5v

Solution

To solve this question, we have to express n, u and v in terms of m.

m:n = 2:3
m/n = 2/3
Cross multiplying
2n = 3m......(1)

v:u = 4:5
v/u = 4/5
.: 5v = 4u
v = 4u/5 .....(2)

(m - u) : (n + v) = 2:5
(m - u)/(n + v) = 2/5
Cross multiplying yields =>
5(m - u) = 2( n + v)
5m - 5u = 2n + 2v ..... (3)

Let's substitute equation (1) and (2) in (3) to compute an expression for u in terms of m =>

5m - 5u = 3m + 2(4u/5)
5m - 3m = 5u + (8u/5)
5(2m) = 25u + 8u
10m = 33u
u = 10m/33 .... (4).

Now, substituting u in equation (2) to get an expression for v in terms of m yields=>
v = (4/5)(10m/33)
v = 8m/33 .... (5)

Now, having gotten an expression for n, u and v in terms of m, we can now calculate the ratio (m + 10u) : (2n - 5v) by substituting equation (1), (4) and (5) in the given expression.

= (m + 10u)/(2n - 5v), substituting yields =>
= [m/1 + 10.(10m/33)]/[3m/1 - 5(8m/33)]
= [(33m + 100m)/33]/[99m - 40m]/33
Inverting yields =>
= (133m/33)*(33/59m)
= 133/59
= 133:59

Hence, (m + 10u) : (2n - 5v) = 133:59

Ghettoguru: I hail all the gurus and able generals on this thread, God bless y'all. Please I need help with this:

1. Evaluate ∫{xy (2y÷x - 3) + 2x^2 - y^2}dx/(x - y)(6x^3 - 4x^2 - 3x^2y + 2xy +16x - 8y)

2. Find the equation of a curve whose gradient to its tangent at a point D (π/4, 4) is given by the equation y' = 8cos^2(x) sin(2x).

Question 2

If the gradient to the tangent of the curve is y' = dy/dy then the equation of the curve will be the integral of y' (∫y') which is y.

y' = 8cos^2(x) sin(2x).
Therefore the equation of the curve is=>
y = ∫8cos^2 x sin2x dx
From trig identities, we know that sin2x = 2sin x cos x, substituting back yields=>
y = ∫8cos^2 x (2sin x cos x)dx
y = ∫16cos^3 x sin x dx
Now, let's put u = cos x,
differentiating, du = - sin x dx

Now, y can be rewritten as:
y = -1(∫16cos^3 x (-sin x) dx
substituting u and du yields=>
y = - ∫16u^3du
y = -16u^4/4 + C
y = -4u^4 + C

Remember that u = cos x
:. y = -4cos^4x + C.

From the question the point at which the gradient of the curve is y'= 8cos^3x sin2x is given as D(π/4, 4). Meaning D (π/4, 4) = (x, y) for x = π/4 radians and y = 4.

Substituting x and y in the equation of the curve to compute the constant C, yields =>
4 = -4cos^4(π/4 ) + C.
C = 4 + [4cos^4(π/4)]
Now, π/4 radians = 45 degrees and cos45 in surd form is (1/√2)

:. C = 4 + [4cos^4(π/4)]
C = 4 + 4(1/√2)^4
C = 4 + 1
C = 5

Therefore the equation of the curve is =>
y = 5 - 4cos^4x

Abeg, una don 4get dis one.
Prove that for any two nos a & b, & for any nEN that
(a+b)^n = a^n + na^(n-1)b + [n(n-1)/2!]a^(n-2)b^(2) + ...+ [n!/(n-k)!k!]a^(n-k)b^k + ... + b^(k)

Nobody is willing to help.

Tenks

abeg gurus help wit dis one
1. prove d existence theorem of laplace transform
2. show L[f"(t)] = s^2F(s) - sf(0) - f'(0)
3. using the methods of contour intergration & convolution theorem, evaluate
L^-1{1/(s+1)(s^2+1)} &
L^-1{1/(s-2)(s-1)^2}

estilo: Nobody is willing to help.

Tenks

The table should be something like this. Re-check coz I checked the data only once.

Mark range: 11 - 20, Frequency : 12 students
Mark range: 21 - 30, Frequency : 11 students
Mark range: 31 - 40, Frequency : 10 students
Mark range: 41 - 50, Frequency : 7 students
Mark range: 51 - 60, Frequency : 5 students
Mark range: 61 - 70, Frequency : 2 students
Mark range: 71 - 80, Frequency : 1 student
Mark range: 81 - 90, Frequency : 2 students

doubleDx:

Question 1

Let's start by simplifying the numerator =>

{xy (2y/x - 3) + 2x^2 - y^2}
= 2y^2 - 3xy + 2x^2 - y^2
= y^2 - 3xy + 2x^2
Factorising yields=>
y^2 - 2xy - xy + 2x^2
y(y - 2x) - x(y - 2x)
(y - x)(y - 2x)

^Substituting back in the question yields=>
= ∫(y - x)(y - 2x)/x/(x - y)(6x^3 - 4x^2 - 3x^2y + 2xy +16x - 8y)
= ∫(y - x)(y - 2x)/x/- (y- x)(6x^3 - 4x^2 - 3x^2y + 2xy +16x - 8y)
= ∫(2x - y)dx/(6x^3 - 4x^2 - 3x^2y + 2xy +16x - 8y)

So from here, I tried checking if the reduced numerator is a factor of the denominator so as to simplify further and it checked out.

Here it goes => let's check again together by substitution. If 2x - y = 0, the y = 2x.

If we substitute y = 2x in the denominator the answer is 0 (zero) too, meaning 2x - y is a factor of (6x^3 - 4x^2 - 3x^2y + 2xy +16x - 8y). Factorising the denominator by initiating the division of polynomial yields =>

= ∫(2x - y)dx/(2x - y)(3x^2 - 2x + 8 )
= ∫(2x - y)dx/(2x - y)(3x^2 - 2x + 8 )
:. The question has been reduced to ∫dx/(3x^2 - 2x + 8 )

Looking at the quadratic denominator, it doesn't have perfect factors and cannot be factorised. Hence we apply the integral formula for a quadratic denominator that cannot be factorised.

i.e => ∫dx/(ax^2 + bx + c) = 2/√(4ac - b^2).arcTan (2ax + b)/√(4ac - b^2)

The denominator is 3x^2 - 2x + 8 which means => a = 3, b = -2 & c = 8. Substituting in the above formula yields =>

∫dx/(3x^2 - 2x + 8 )

= 2/√[(4)(3)(8 ) - (-2)^2].[arcTan (6x - 2)/√[(4)(3)(8 ) - (-2)^2]
= 2/√92 arcTan 2(3x -1)/√92
= 2/2√23 arcTan 2(3x -1)/2√23
= 1/√23 arcTan (3x - 1)/√23

Hence =>

∫{xy (2y÷x - 3) + 2x^2 - y^2}dx/(x - y)(6x^3 - 4x^2 - 3x^2y + 2xy +16x - 8y)

= ∫dx/(3x^2 - 2x + 8 ) =>
= 1/√23 arcTan (3x - 1)/√23

Question 2 coming up later in the day. Take care bruv.

Bless you sir. Thanks a bunch!

yungryce: abeg gurus help wit dis one
1. prove d existence theorem of laplace transform
2. show L[f"(t)] = s^2F(s) - sf(0) - f'(0)
3. using the methods of contour intergration & convolution theorem, evaluate
L^-1{1/(s+1)(s^2+1)} &
L^-1{1/(s-2)(s-1)^2}

Bros you be engineering student? Make you dey help others make others dey help you too naw. How u see am bro? Abeg no vex o, na so we dey do am here since sha.

plz give me a hint on how to solve this simple question under chords and arcs(circle geometry)
an equilateral triangle of side 20cm is inscribed in a circle. Calculate the distance of a side of the triangle from the centre of the circle?

Ghettoguru:

The table should be something like this. Re-check coz I checked the data only once.

Mark rage: 11 - 20, Frequency : 12 students
Mark rage: 21 - 30, Frequency : 11 students
Mark rage: 31 - 40, Frequency : 10 students
Mark rage: 41 - 50, Frequency : 7 students
Mark rage: 51 - 60, Frequency : 5 students
Mark rage: 61 - 70, Frequency : 2 students
Mark rage: 71 - 80, Frequency : 1 student
Mark rage: 81 - 90, Frequency : 2 students

At last. Tenk u so much and God bless

Nice one

estilo:

At last. Tenk u so much and God bless

Nice one

You are welcome, I hope you crosschecked? I'm glad I could help.

Typo - it's supposed be 'mark range' not 'mark rage'.

johnpaul1101: plz give me a hint on how to solve this simple question under chords and arcs(circle geometry)
an equilateral triangle of side 20cm is inscribed in a circle. Calculate the distance of a side of the triangle from the centre of the circle?

HINT:draw lines from d centre of d circle to d vertices of d triangle.....a straight line from the upper vertex to d middle of d line below it......
Use pythagoras' theorem all thru and u r good

Ghettoguru:

Bros you be engineering student? Make you dey help others make others dey help you too naw. How u see am bro? Abeg no vex o, na so we dey do am here since sha.

ok sir

Hi friends! Hope u all had a great weekend

my maths guru in the house, i hail.
if the root of an equation x^2+Kx+11 are alpha and beta ,alpha^2+beta^2=27.find the possible number of K given alpha^2+beta^2=(alpha+beta)^2 -2*alpha*beta.
help me crack this one.

double j1: my maths guru in the house, i hail.
if the root of an equation x^2+Kx+11 are alpha and beta ,alpha^2+beta^2=27.find the possible number of K given alpha^2+beta^2=(alpha+beta)^2 -2*alpha*beta.
help me crack this one.

^^^
Solution

The roots of the equation x^2 + Kx + 11 are α and β
x^2 + Kx + 11 = 0 ....eqn(1)

Then, x = α or x = β
To form the equation
(x - α)(x - β) = 0
x^2 - xβ - xα + αβ = 0
x^2 - x(α + β) + αβ = 0 .....eqn (2)
x^2 + Kx + 11 = 0 .....eqn (1)
Comparing equations (1) & (2), the coefficient of x can be equated as follows.

Kx = - x (α + β)
(α + β) = -K ...eqn(3)
Equating the constants
αβ = 11 .....eqn(4)

The question says α^2 + β^2 = 27 = (α + β)^2 - 2αβ
(α + β)^2 - 2αβ = 27 ....eqn(5)

Now, substitute the values of (α + β)....eqn(3) and αβ...eqn (4) in eqn(5).
(α + β)^2 - 2αβ = 27
(-K)^2 - 2(11) = 27
K^2 - 22 = 27
K^2 = 27 + 22
K^2 = 49
K = ±√49
K = ±7

Richiez: Hi friends! Hope u all had a great weekend

My abled-general Richiez, I salute you sir.

@Ghetoguru, see how u destroyed that math question. i salute you sir

@everyone, sorry for my disappearing act, i was fixing my modem. i'm back so let's roll