ositadima1:

Try this baby, lets c

Same method, I'll do it later.

doubleDx:

Same method, I'll do it later.

Ok, man

2nioshine: d truth is that i actually tried same method as u did bt i got stock somewere after those identity stuff(factorizatn)...must confes av been seen as d bst n ma sku...bt i off my cap 4 u tnks.@ositadime i apreciate u also

You are welcome bro. Maybe you made a mistake when you were collecting like terms. It happens all the time!

doubleDx:

Lemme expand 2nioshine's solution so you can understand better.

x^4 - 1
Can be written as:
(x^2)^2 - 1^2

Lets x^2 = P

P^2 - 1^2
Since a^2 - b^2 = (a - b)(a + b) - difference of two squares
P^2 - 1^2
= (P - 1)(P + 1)
Substituting P = x^2 yields:

= (x^2 - 1)(x^2 +1)
= (x^2 - 1^2)(x^2 +1)
Appylying the difference of two squares to (x^2 -1^2) gives : (x -1)(x + 1)

:. x^4 - 1 = (x -1)(x + 1)(x^2 +1)

Kudos!
@All how about solving this:

If f(x) = a(x^3) + 2(x^2) + bx + c, and f(0) = f(1) = 4, and f(-1) = 8, Find: a, b, and c.

ositadima1:

Ok, man

a+b+c = 8 ...equation 1
a^2+b^2+c^2 =16 ...equation 2
a^3+ b^3+c^3 = 64 ...equation 3

Before we proceed. We have to create an expression for a^3 + b^3 in terms of (a + b)

Using addition and substraction method to factorize the polynomial a^3 + a^3 yields:

a^3 + b^3 can be expressed as : (a + b)^3 - 3ab (a + b)

Also a^2 + y^2 can be expressed as : (a + b)^2 - 2ab

Now, having created an expressing for the above.

From equation 1, (a + b) = (8 - c) ....equation 4

Substituting the expression for a^2 + b^2 in equation 2 yields :
(a + b)^2 - 2ab + c^2 = 16 .... equation 5

Substituting the expression for a^3 + b^3 in equation 3 yields:
(a + b)^3 - 3ab(a + b) + c^3 = 64 .... equation 6

Since we already know the expression for a + b = (8 - c) in terms of c. We can now find an expression for ab from equation 5 so we can substitute back in equation 6 to give us the values of c.

Computing ab in terms of c, subtitute a + b = (8 - c) in equation 5:

(8 - c)^2 - 2ab + c^2 = 16
2ab = (8 - c)( 8 - c) - 16 + c^2
2ab = 64 -16c + c^2 - 16 +c^2
2ab = 2c^2 - 16c + 48
:. ab = c^2 - 8c + 24

Having known the expression for ab interms of c as written above. We can now substitute both expression (i.e the for (a + b) and ab) in terms of c in equation 6. a + b = (8 - c) and ab = c^2 - 8c + 24

(a + b)^3 - 3ab (a + b) + c^3 = 64
substituting yields :
(8 - c)^3 - 3 ( c^2 - 8c + 24) (8 - c) + c^3 = 64
(8 - c)(64 -16c + c^2) - 3 ( c^2 - 8c + 24) (8 - c) + c^3 = 64
Expanding and collecting like terms yields:

3c^3 - 24c^2 - 72c - 64 = 64
Simplifying further =>
c^3 - 8c^2 - 24c = 128/3

^The above polynomial has no perfect factor and cannot be factorize, so I guess the answer should be a little complex. I'll compute it for the values of c later.

Hardfact:
Kudos!
@All how about solving this:

If f(x) = a(x^3) + 2(x^2) + bx + c, and f(0) = f(1) = 4, and f(-1) = 8, Find: a, b, and c.

f(0) = 0 + 0+ 0+ c = 4 thus c = 4
f(1) = a + 2 + b + 4 = 4 thus a+b = -2 A
f(-1) = -a + 2 - b + 4 = 8 thus -a-b = 2 B

Hmmm

The equation has a fault somewhere as the simultaneous equations A & B are similar

ositadima1:


Back to this question

Each term => (n+7)^3
where n=1 to n=200-7=193

Meaning we have 193 terms from 8^3 to 200^3.

Summation of (n+7)^3 for n=1 to n=193

Sn= (1+7)^3+(2+7)^3+(3+7)^3+...+(193+7)^3

Let`s expand (n+7)^3
(n+7)^3=(n+7)(n+7)(n+7
=n^3+21n^2+147n+343

Let's see the summation of each term above

S= n^3= 1^3+2^3+3^3+...+nth= (1/4)(n^2)(n+1)^2


S=n^2= 1^2+2^2+3^2+...+nth= (1/6)(n)(n+1)(2n+1)

S=n =1+2+3+...+nth= (1/2)(n)(n+1)

S=1+1+1+...+nth= n

Now substituting back

Sn= (1/4)(n^2)[(n+1)^2]+21(1/6)(n)(n+1)(2n+1)+147(1/2)(n)(n+1)+343(n)

So, if u now carefully substitute n=193

Sn=404009216

Actually that equation can give u sum to any number of terms u want, hope this was helpful

tnx bros i gat it

biolabee: f(0) = 0 + 0+ 0+ c = 4 thus c = 4 f(1) = a + 2 + b + 4 = 4 thus a+b = -2 A f(-1) = -a + 2 - b + 4 = 8 thus -a-b = 2 B Hmmm The equation has a fault somewhere as the simultaneous equations A & B are similar

yep i also got c=4 but it seems the values of f(1) and f(-1) generates inconsistent simultaneous eqns

biolabee:

f(0) = 0 + 0+ 0+ c = 4 thus c = 4
f(1) = a + 2 + b + 4 = 4 thus a+b = -2 A
f(-1) = -a + 2 - b + 4 = 8 thus -a-b = 2 B

Hmmm

The equation has a fault somewhere as the simultaneous equations A & B are similar

Thanks for the effort. Lifted it from somewhere. Anyone else?

guys please help me with this,i v been stuck on it for a while naw.......
Prove that sin A-sin B=2cos(A+B)/2.sin (A-B)/2.......hope u'll understand it like dis......

1000 naira GLO recharge card for anyone who can prove the below mathematical expression.

Given that Q = xyz
Prove that the CURL of GRAD Q = 0

digitalwale: 1000 naira GLO recharge card for anyone who can prove the below mathematical expression.

Given that Q = xyz
Prove that the CURL of GRAD Q = 0

Use the concept of DOROGO and arrive at the result.

digitalwale: 1000 naira GLO recharge card for anyone who can prove the below mathematical expression.

Given that Q = xyz
Prove that the CURL of GRAD Q = 0

Game on, let me chop first...

digitalwale: 1000 naira GLO recharge card for anyone who can prove the below mathematical expression.

Given that Q = xyz
Prove that the CURL of GRAD Q = 0

The scalar Q=xyz
grad=(d/dx, d/dy, d/dz)

Where d/dx, d/dy, d/dz are partial derivatives

grad(xyz)=[d(xyz)/dx, d(xyz)/dy, d(xyz)/dz]

grad(xyz)=[yz, xz, xy]

We now have a vector...[yz, xz, xy]

curl[yz, xz, xy]= grad x [yz, xz, xy]

where x is a vector cross

curl[yz, xz, xy]= [d/dx, d/dy, d/dz] x [yz, xz, xy]
Matrix form for d curl

curl[yz, xz, xy]=
| i j k |
|d/dx d/dy d/dz|
| yz xz xy |
Breaking down d matrix...

curl[yz, xz, xy]= [d(xy)/dy-d(xz)/dz]i-[d(xy)/dx-d(yz)/dz]j+[d(xz)/dx-d(yz)/dy]k

curl[yz, xz, xy]=[(x-x)i-(y-y)j+(z-z)k]

curl[yz, xz, xy]=[(0)i-(0)j+(0)k]

curl[yz, xz, xy]=[0]=0

^^^Na phone I use type oo, no light for my area.

kasbeats: guys please help me with this,i v been stuck on it for a while naw.......
Prove that sin A-sin B=2cos(A+B)/2.sin (A-B)/2.......hope u'll understand it like dis......

before attempting this question, one has to know the basic rules of sine addition, and the derivation of the factor formulae
NOW;

Sin(P+Q) = SinPCosQ + CosPSinQ......................(1)
Sin(P-Q) = SinPCosQ - CosPSinQ......................(2)

Subtracting eqn(2) from eqn(1),

Sin(P+Q) - Sin(P-Q) = CosPSinQ + CospSinQ

Sin(P+Q) - Sin(P-Q) = 2CosPSinQ ....................(3)

Now we have to do a little trick by letting;

P+Q = A ............................................(4)
P-Q = B ............................................(5)

adding eqns (4) and (5) we get;

2P = A + B
P = (A+B)/2 ........................................(6)

subtracting eqn(5) from eqn(4) we get;

2Q = A - B
Q = (A-B)/2 ........................................(7)

By appropriate substitution for A,B,P and Q in equation(3) we get ;

SinA - SinB = 2[Cos(A+B)/2*Sin(A-B)/2]

Q. E. D

nice one guys...

Nice work people!

kasbeats: guys please help me with this,i v been stuck on it for a while naw.......
Prove that sin A-sin B=2cos(A+B)/2.sin (A-B)/2.......hope u'll understand it like dis......

Richiez has given a solution to your question. Lemme add another method =>

sin A - sin B = 2cos(A+B )/2 . sin (A-B )/2

Lets take this from the right hand side:

2cos(A+B )/2 . sin (A-B )/2 , simplifying yields =>
2cos(A/2+B/2 ) . sin(A/2 - B/2)

Lets put P = A/2 and Q = B/2
2cos(A/2+B/2 ) . sin(A/2 - B/2) becomes :
2cos(P +Q ) . sin( P - Q)

Since sin (P - Q) = sinP cosQ - cosP sinQ
&
cos (P + Q) = cosP cos Q - sinP sin Q

:. RHS => 2cos( P + Q ) . sin (P - Q) =>

=>2{cosP cosQ - sinP sinQ } . {sinP cosQ - cos
P sinQ }

Expanding:

=> 2[sinP cosP cos^2Q - sinQ cosQ cos^2P - sin
Q cosQ sin^2P + sinP cosP sin^2Q]

Collecting like terms yields:

=> 2[sinP cosP cos^2Q + sinP cosP sin^2Q - sinQ cosQ cos^2P - sin
Q cosQ sin^2P]

Factorising:
=> 2[ sinP cosP (sin^2Q + cos^2Q) - sinQ cosQ (sin^2P + cos^2P) ]

Since Sin^2 x + Cos^2 x = 1 substituting yields =>

= 2[sinP cosP (1) - sinQ cosQ (1)]
= 2sinP cosP - 2sinQ cosQ

Replacing P = A/2 and Q = B/2

= 2sinA/2 cosA/2 - 2sinB/2 cosB/2

^The above is an expression for the difference of two half angle for sin A and Sin B respectively. Because : Sin A = Sin(A/2 + A/2) = 2sinA/2 cos A/2 and also sin B = sin (B/2 + B/2 ) = 2sinB/2 cos B/2

:.RHS = Sin A - Sin B

Hence => sin A - sin B = 2cos(A+B )/2 . sin (A-B )/2

1. Solve the following simultaneous equations and evaluate (2x^2 - y)/(x - 7).

x + y = 9
3x - ³√y = 1

2. Find the second derivative of sin^3(x) cos(3x)

Thanks.

Ghettoguru: 1. Solve the following simultaneous equations and evaluate (2x^2 - y)/(x - 7).

x + y = 9
3x - ³√y = 1

2. Find the second derivative of sin^3(x) cos(3x)

Thanks.

Question 1

x + y = 9 .....(1)
3x - ³√y = 1 ......(2)

from equation (1)

y = (9 - x) ..... (3)

from equation (2)
³√y = (3x - 1)
cubing both sides to get rid of the cube root yields:
(³√y)^3 = (3x - 1)^3
y = (3x - 1)(3x - 1)(3x - 1) . Expanding gives :

y = 27x^3 - 27x^2 + 9x - 1 .....(4)
Substituting (3) in (4) for the values of x gives =>
9 - x = 27x^3 - 27x^2 + 9x -1
:. 27x^3 - 27x^2 + 10x - 10 = 0

Factorising the polynomial yields :

(x - 1)(27x^2 + 10) = 0
:. x = 1 or 27x^2 = -10
The real value of x = 1, the other two are complex numbers => (1/3)i√(10/3) and -(1/3)i√(10/3)

Substitute x = 1 in equation (3)
y = 9 - 1
y = 8

Evaluating : (2x^2 - y)/
(x - 7) gives =>

= (2(1)^2 - 8 )/
(1 - 7).
= -6/-6
= 1

:.(2x^2 - y)/(x - 7) => 1

I'll check question 2 later in the day. Sleepy now!

Ghettoguru:

2. Find the second derivative of sin^3(x) cos(3x)

Thanks.

Using producr derivate d(vu) = vdu + udv

d\First level
d = -3 sin^3x sin 3x + 3 cos x sin ^2 x cos 3x

Second level (wey una dey see all dese kain question sef )

d2x = -9cos3x sin3x-9 sin3x cosx sin^2x -9 cosx sin^2 x sin3x+3 cos 3x(2 cos^2 x - sin^2 x sin x)

I don try

Pls this integration has been given me prob. Question 1: integrate 1/cos(x-a)cos(x-b) with respect to x. Question2: Pls, I need explanation on complex analysis with some examples and their solutions. Thanks !

I cant even moderate the thread as its way above me everything thing looks correct here , no spam
Ride on guys!

God really bless Some people here wit super brain ohh.. Did u guys study Mathematics as ur field in d university?

Ghettoguru: 1. Solve the following simultaneous equations and evaluate (2x^2 - y)/(x - 7).

x + y = 9
3x - ³√y = 1

2. Find the second derivative of sin^3(x) cos(3x)

Thanks.

Question 2

f(x) = sin^3 x cos 3x

First off, let us simplify cos 3x =>
cos(3x) = cos 2xcos x - sin 2x sin x
= cosx ( 1- 2sin^2 x) - 2sin^2 x cosx
= cos x (1- 4sin^2 x)

:. Inserting cos 3x into the question yields :

f(x) = sin^3x cos x (1- 4sin^2 x)
f(x) = cos x sin^3 x - 4sin^5 x cos x

Now, we can split the function into two (sin^3 x cos x) and (- 4 sin^5 x cos x)

Applying product and chain rule to -4sin^5 x cos x yields :

Put u = - 4sin^5 x, du/dx = - 20sin^4 x cos x
v = cos x, dv/dx = sin x

f'(x) of - 4sin^5 x cos x =>
= udv/dx + vdu/dx
= -4sin^5 x .(-sin x) + cos x (- 20sin^4 x cos x)
= 4sin^6 x - 20sin^4 x cos^2 x
since cos^2 x = (1 - sin^2 x) substituting yields
= 4sin^6 x - 20sin^4 x (1 - sin^2x)
= 4sin^6 x - 20sin^4 x + 20sin^6 x
= 24sin^6 x - 20sin^4 x .....(1)

For the derivative of the second part of f(x) which is cos x sin^3 x

Put u = cos x, du/dx = -sin x
v = sin^3 x, dv/dx = 3sin^2 x cos x

f'(x) for cos x sin^3 x =>
= udv/dx + vdu/dx
f'(x) = cos x . (3sin^2 x cos x) + sin^3 x (-sin x)
= 3sin^2 x cos^2 x - sin^4 x
since cos^2 x = (1 - sin^2 x) substituting yields =>
= 3sin^2 x (1 - sin^2 x) - sin^4 x
= 3sin^2 x - 3sin^4 x - sin^4 x
= 3sin^2 x - 4sin^4 x .... (2)

:.f'(x) for the expression cos x sin^3 x - 4sin^5 x cos x =>

= 3sin^2 x - 4sin^4 x + 24sin^6 x - 20sin^4 x

collecting like times =>

:.The first derivative of the function f(x) =>
f'(x) = 3sin^2 x - 24sin^4 x + 24sin^6 x

Second derivative of f(x) is the derivative of f'(x).

Since f'(x) = 3sin^2 x - 24sin^4 x + 24sin^6 x
f''(x) = 6sin x cos x - 96sin^3 x cos x + 144sin^5 x cos x.

Simplifying =>

f''(x) = 6sin x cos x (1 - 16sin^2 x + 24sin^4 x)

@Ghettoguru, I made some mistakes before. Re-check now, I have corrected them.

Nice job guyZ, keep it up.

[size=15pt]Proof that 12/7 is = 2. Without any reminder or decimal point. [/size]

Answers please

KingOfNairaland: [size=15pt]Proof that 12/7 is = 2. Without any reminder or decimal point. [/size]

Answers please

the roman numeral for 12 is XII and dat of 7 is VII. Therefore: 12/7= XII/VII= X/V( X=10, V=5). X/V=10/5=2ANS.

@DoubleDx thanks, much appreciated. @All, you are doing a great job. Keep them coming!

ositadima1:

The scalar Q=xyz
grad=(d/dx, d/dy, d/dz)

Where d/dx, d/dy, d/dz are partial derivatives

grad(xyz)=[d(xyz)/dx, d(xyz)/dy, d(xyz)/dz]

grad(xyz)=[yz, xz, xy]

We now have a vector...[yz, xz, xy]

curl[yz, xz, xy]= grad x [yz, xz, xy]

where x is a vector cross

curl[yz, xz, xy]= [d/dx, d/dy, d/dz] x [yz, xz, xy]
Matrix form for d curl

curl[yz, xz, xy]=
| i j k |
|d/dx d/dy d/dz|
| yz xz xy |
Breaking down d matrix...

curl[yz, xz, xy]= [d(xy)/dy-d(xz)/dz]i-[d(xy)/dx-d(yz)/dz]j+[d(xz)/dx-d(yz)/dy]k

curl[yz, xz, xy]=[(x-x)i-(y-y)j+(z-z)k]

curl[yz, xz, xy]=[(0)i-(0)j+(0)k]

curl[yz, xz, xy]=[0]=0

This is outstanding! Lemme have your phone number

digitalwale:

This is outstanding! Lemme have your phone number

Can u PM my spoils, Mtn would be nice

ositadima1:

Can u PM my spoils, Mtn would be nice

How can I PM you? I tried going to your profile and clicked to send email but the page that came up only showed a prepared and uneditable message.