Thanks guys. Help with these questions please.

1. Differentiate sin(4x)cosec^2(x) +8cos^2(x) with respect to x.

2. Find the value(s) of p, q and r in the following equations:

p - q - r = -4
p^2 - q^2 - r^2 = -12
p^3 - q^3 - r^3 = -34

doubleDx:

(sinB+sin3B+sin5B )/( cosB+cos3B+cos5B )

To simplify this, you need a know some basic trig identities and expressions like sin2B = 2sinBcosB, cos2B = cos^2B - sin^2B or 1 - 2sin^2B or 2cos^2B - 1, cos^B + sin^B = 1

Lets simplify the numerator =>
sinB + sin3B + sin5B

sin 3B = sin (2B + B )
= sin 2B cos B + cos 2B sin B
Subsituting the expressions for sin2B and cos2B

Sin3B = cosB ( 2sinBcosB ) + sin B (cos^2B - sin^2B )
= sin B (1- 2sin^2 B ) + 2sin B (1 - sin^2 B )
sin 3B = 3sin B - 4 sin^3 B

cos3B
= cos (2B + B ) = cos 2B cos B - sin2B sin B
= cos B(1 - 4sin^2 B )

Now lets simplify sin 5B =>

Sin (2B + 3B ) = sin2B cos3B + cos 2Bsin3B
Subsituting the expressions for sin2B, sin3B, cos2B and cos3B

Sin5B = 2sinBcosB.cosB(1 - 4sin^2 B ) + (1 - 2sin^2B )( 3sinB - 4sin^3B )
Expanding and collecting like terms =>

Sin 5B = 16sin^5B - 20sin^3B + 5sinB

Now, the numerator sinB + sin3B + sin5B =>
= sinB + 3sin B - 4 sin^3B + 16sin^5B - 20sin^3B + 5sinB
Numerator = 16sin^5B - 24sin^3B + 9sinB

Lets take the denominator cosB+ cos3B + cos5B =>

We already know cos3B = cos B ( 1 - 4sin^3B ) if we simplify it in terms of cos B,

cos 3B = 4cos^3B - 3cos B

Cos5B = cos (2B + 3B )
= cos2B cos3B - sin2B sin 3B
Subsituting the expressions for sin2B, sin3B, cos2B and cos3B
cos5B = (2cos^2B -1 )(4cos^3B - 3cosB ) - 2sinBcosB (3sinB - 4sin^3B )

Expanding and collecting like terms,

Cos5B = 16cos^5B - 20cos^3B + 5cosB

:. The denominator cosB+ cos3B + cos5B =>

= cosB + 4cos^3B - 3cosB + 16cos^5B - 20cos^3B + 5cosB

Collecting like terms=>

Denominator = 16cos^5B -16cos^3B + 3cosB

Therefore the fraction (sinB+sin3B+sin5B )/( cosB+cos3B+cos5B ) =>

(16sin^5B - 24sin^3B + 9sinB )/ (16cos^5B -16cos^3B + 3cosB )

= sinB/cosB [ (16sin^4B - 24sin^2B +9)/(6cos^4B -16cos^2B + 3) ]

= tanB [ (16sin^4B - 24sin^2B +9)/(6cos^4B -16cos^2B + 3) ]

The above expression could still be simplified further either in terms of sinB only or cos B only. But I'll stop there.

I hope that helps.

..tank u so much Double...i'm really feeling u....kindly help wit dis***(a) Find d distance of d point(4,5) from d line x+y=7...(b) A line L is perpendicular to AB where A is(-1,4) and B is (4,-1), L divides AB in d ratio 2:3. Find d equation of L?...God bless

Hate Maths!! My head cannot just get it. I'm a beast with words though. **Just came to show off*

Ghettoguru:

1. Differentiate sin(4x)cosec^2(x) +8cos^2(x) with respect to x.

2. Find the value(s) of p, q and r in the following equations:

p - q - r = -4
p^2 - q^2 - r^2 = -12
p^3 - q^3 - r^3 = -34

I'm kind of busy now, so maybe other gurus will help with the second question or when I'm free, I'll take care of it.

Question 2

To differentiate the funtion lets simplify it first =>
sin(4x)cosec^2(x) + 8cos^2(x)
= (sin 4x/sin^2x) + 8cos^2 x

Let split the function into two parts (sin 4x/sin^2x) and 8cos^2 x

To differentiate the first part, lets use quotient rule =>

u = sin 4x, v= sin^2 x
du/dx = 4cos4x and dv/dx = 2sin x cos x = sin 2x

f'(x) = [v(du/dx) - u(dv/dx)]/v²

So the derivative of sin 4x/sin^2x =>

[sin^2 x ( 4cos 4x) - sin 4x (sin 2x)]/sin^4 x
= [4cos 4x sin^2x - sin 4x sin 2x]/sin^4 x

Now lets differentiate the second part 8cos^2 x and add it to the derivative of the first part => using chain rule,

the derivative of 8cos^2 x
= 16cosx (-sin x)
= -16cos x sin x
= - 8(2cos x sin x)
= - 8sin 2x

Adding the derivative of the first and second part of the function yields =>

= { [4cos 4x sin^2x - sin 4x sin 2x]/sin^4 x} - 8sin 2x

= cosec^4 x [4cos 4x sin^2x - sin 4x sin 2x] - 8sin 2x
= 4cosec^2 x cos 4x - cosec^4 x sin 4x sin 2x - 8sin 2x

^ the above could be simplified in terms of cos x or sin x or both, but since there is no instruction to go further, I'll stop there.

:. The derivative of sin(4x)cosec^2(x) + 8cos^2(x) is =>

= 4cosec^2 x cos 4x - cosec^4 x sin 4x sin 2x - 8sin 2x

Fetus: ..tank u so much Double...i'm really feeling u....kindly help wit dis***(a) Find d distance of d point(4,5) from d line x+y=7...(b) A line L is perpendicular to AB where A is(-1,4) and B is (4,-1), L divides AB in d ratio 2:3. Find d equation of L?...God bless

You are welcome bro!

I'll check out the questions later, I'm kinda busy now.

@Doubled, okay thanks. Let me re-check question 1.

Yvete: Hate Maths!! My head cannot just get it. I'm a beast with words though. **Just came to show off*

Excuse me ma, what exactly are u showing off.

gurus pls hlp solve ds quesion,wat will b d value of a,b nd c, in ds binomial expansion (a+bx)^c,if d coefficient of 2nd,3rd nd 4th term are -12,60 nd -160 respectively.

Abiola4eva: gurus pls hlp solve ds quesion,wat will b d value of a,b nd c, in ds binomial expansion (a+bx)^c,if d coefficient of 2nd,3rd nd 4th term are -12,60 nd -160 respectively.

Abiola4eva: gurus pls hlp solve ds quesion,wat will b d value of a,b nd c, in ds binomial expansion (a+bx)^c,if d coefficient of 2nd,3rd nd 4th term are -12,60 nd -160 respectively.

(a + bx)^ c = (nC0)a^c + (nc1)a^c-1 x + (nc1)a^c-2 x2 + (nc1)a^c-3 x3 + ....


hence the equations

c * a^(c-1) = -12 ----(1)
c(c-1)/2 * a^(c-2) = 60 --------(2)
c(c-1)(c-2)/6 * a^(c-3) = -120 --------(3)

This is really a madt equation
But I can hazard the following guess

a is a negative number
c is postive

The rest I leave to my gurus

PS I think we have had enough of trig differentiation and identities
if u read through what has been done so far you should be able to go some distance
Then where you get stuc, bring to the thread

Fetus: ..tank u so much Double...i'm really feeling u....kindly help wit dis***(a) Find d distance of d point(4,5) from d line x+y=7...(b) A line L is perpendicular to AB where A is(-1,4) and B is (4,-1), L divides AB in d ratio 2:3. Find d equation of L?...God bless

....someone kindly assist...and solve did..

Fetus: ..tank u so much Double...i'm really feeling u....kindly help wit dis***(a) Find d distance of d point(4,5) from d line x+y=7...(b) A line L is perpendicular to AB where A is(-1,4) and B is (4,-1), L divides AB in d ratio 2:3. Find d equation of L?...God bless

....someone kindly assist...and solve dis

Abiola4eva: gurus pls hlp solve ds quesion,wat will b d value of a,b nd c, in ds binomial expansion (a+bx)^c,if d coefficient of 2nd,3rd nd 4th term are -12,60 nd -160 respectively.

You made me sweat some...
Based on the binomial theorem each term of (a+bx)^c can be expressed as;

c!/[(c-k)!k!]*a^(c-k)*(bx)^k

Where k= 0, 1, 2, ..., c
I am going to use "*" as multiplied by.
Now,

2nd term : c!/[(c-1)!1!]*a^(c-1)*(bx)^1

3rd term : c!/[(c-2)!2!]*a^(c-2)*(bx)^2

4th term : c!/[(c-3)!3!]*a^(c-3)*(bx)^3

Sifting out the coefficients

c!/[(c-1)!1!]*a^(c-1)*(b)^1=-12......eq1
c!/[(c-2)!2!]*a^(c-2)*(b)^2=60......eq2
c!/[(c-3)!3!]*a^(c-3)*(b)^3=-160......eq3

Now combining equation 1 and 2

c!/[(c-2)!2!]*a^(c-2)*(b)^2=-5[c!/[(c-1)!1!]*a^(c-1)*(b)^1]
***remember -5x-12=60 where -12 is c!/[(c-1)!1!]*a^(c-1)*(b)^1***

Now combining equation 2 and 3

c!/[(c-3)!3!]*a^(c-3)*(b)^3=-(8/3)[c!/[(c-2)!2!]*a^(c-2)*(b)^2]
***-(8/3)x60=-160***

Simplifying;

(c-1)!/(c-2)!2!*b=-5*a
>>(c-1)(c-2)!/(c-2)!2!*b=-5*a ***(c-1)!=(c-1)(c-2)!***
>>(c-1)/2*b=-5*a

Next,
(c-2)!2!/(c-3)!3!*b=-(8/3)*a
>>(c-2)(c-3)!2!/(c-3)!3!*b=-(8/3)*a
>>(c-2)/3*b=-(8/3)*a

we devide this two results to eliminate a and b (akara and bread)

[(c-1)/2*b=-5*a]/[(c-2)/3*b=-(8/3)*a]
>>5c-10=4c-4
>>c=6 yo yo!!!! Uh uh

Now

2nd term : 6!/5!*a^5*b=-12
3rd term : 6!/4!2!*a^4*b^2=60
4th term : 6!/3!3!*a^3*b^3=-160

4th term looks promising
6!/3!3!*a^3*b^3=-160
>>a^3*b^3=-160/(6!/3!3!)
>>a^3*b^3=-8
>>(ab)^3=-8
>>ab=-2

feed it back b=-2/a,

6!/5!*a^5*(-2/a)=-12
>>6*5!/5!*a^4=6
>>a=1
>>b=-2

so... a=1, b=-2, c=6

if u re truely a maths guru as u claim u are, then i believe u can solve this: half of A's money plus one-fifth of B's makes #100. Two-thirds of A's money plus two-fifths of B's makes #150. How much money has each? This is simultaneous equation, show ur workings

Fetus: ....someone kindly assist...and solve did..

i used a graphical method for the two equations

you will then apply pythagoras theorem
dist 2.12

equation
slope of the first line = -1
slope of the new line is then -1/old slope = 1
height from x axis = 6
thus intercept = 3

y = x + 3

johnpaul1101: if u re truely a maths guru as u claim u are, then i believe u can solve this: half of A's money plus one-fifth of B's makes #100. Two-thirds of A's money plus two-fifths of B's makes #150. How much money has each? This is simultaneous equation, show ur workings

reads, looks again, raises d finger and movess on

Miss Odunnu, my post was hiden for no apparent reason. Its painful, I took time to solve a math problem only for it to be wiped off by Pyguru. I don't understand why.

^^^Why are they hiding some posts? I think you should let the moderators know about this.

does it mean u cant solve it?@osita dima does it mean u cant solve it?@osita dima

Abiola4eva: gurus pls hlp solve ds quesion,wat will b d value of a,b nd c, in ds binomial expansion (a+bx)^c,if d coefficient of 2nd,3rd nd 4th term are -12,60 nd -160 respectively.

I am doing this a second time...

Based on binomial theorem each term of (a+bx)^c is n!/(n-k)!k!*a^(c-k)*(bx)^k
Where k=0, 1, 2, 3, 4,..., c and "*" stands for multiplied by

2nd term; n!/(n-1)!1!*a^(c-1)*(bx)^1
3rd term; n!/(n-2)!2!*a^(c-2)*(bx)^2
4th term; n!/(n-3)!3!*a^(c-3)*(bx)^3

Sifting out x coefficients;

n!/(n-1)!1!*a^(c-1)*b^1=-12..........eq1
n!/(n-2)!2!*a^(c-2)*b^2=60.........eq2
n!/(n-3)!3!*a^(c-3)*b^3=-160.........eq3

Combining equations 1 and 2 noting dat -5*-12=60 where -12=n!/(n-1)!1!*a^(c-1)*b^1

n!/(n-2)!2!*a^(c-2)*b^2=-5*n!/(n-1)!1!*a^(c-1)*b^1
>>(n-1)!/(n-2)!*b=-5*a
Note (n-1)!=(n-1)(n-2)!
>>(n-1)*b=-5*a.........eq4

Combining equations 2 and 3 noting dat -8/3*60=-160

n!/(n-3)!3!*a^(c-3)*b^3=-(8/3)*n!/(n-2)!2!*a^(c-2)*b^2
>>(n-2)!2!/(n-3)!3!*b=-(8/3)*a
>>(n-2)!/3*b=-(8/3)*a..........eq5

Dividing equation 5 with 4 to eliminate a and b.

>>-5*(n-2)=-8*(n-1)
>>5c-10=4c-4
>>c=6

TOSIN ACCA: Please kindly solve this jamb question for me: b=a +cp and r=ab + 1/2cp^2, express b^2 in terms of a, c and r

Osita are you there?please help me solve this question.

ositadima1: Miss Odunnu, my post was hiden for no apparent reason. Its painful, I took time to solve a math problem only for it to be wiped off by Pyguru. I don't understand why.

I beg you again, use the complaints thread next time you want the attention of any of the MODs. If you just vent off here, very likely none of us will see and take action.
You are lucky now though

[quote author=Richiez]

U can call me 2 get answers to your questions on Mathematics

Odunnu:
I beg you again, use the complaints thread next time you want the attention of any of the MODs. If you just vent off here, very likely none of us will see and take action.
You are lucky now though

Thank you.

TOSIN ACCA: Osita are you there?please help me solve this question.

if this na jamb question i go murd

p is the variable u need to get rid of

b = a + cp
p = (b-a)/c
p2 = (b2 + a2 - 2ab)/c2

also p2 = 2(r-ab)/c

Equate the two equations

So you get

(b2+a2-2ab)/c2 = 2r-2ab/c

(b2+a2-2ab)/c = 2r-2ab

cross multiply

b2 -2ab + 2abc + a2 - 2rc

Use almighty formula and solve

a = 1
b = 2ac - 2a
c = a2- 2rc

Fetus: ..tank u so much Double...i'm really feeling u....kindly help wit dis***(a) Find d distance of d point(4,5) from d line x+y=7...(b) A line L is perpendicular to AB where A is(-1,4) and B is (4,-1), L divides AB in d ratio 2:3. Find d equation of L?...God bless

Question 1
Now we have to rewrite the straight line eqn
x + y = 7
x + y - 7 = 0 ................(1)
For any straight line Ax+By+c=0 and a point (m,n) on same plane, the perpendicular distance between the point and the plane is given by;
d = (lAm+Bn+Cl)/√(A^2 + B^2)
obviously, A=1 , B=1 , C+ -7 , m=4 , n=5
therefore;
d = (l4+5-7l)/√(1^2 +1^2)

d = l2l/√(2) = 1.414units(approx)

Question 2
here, the main task is to find the coordinates of the point on line AB, where the line L cuts it in the ratio 2:3

Let the coordinate of the point be P(x,y)
For the line AB;
A(x1,y1) = A(-1,4)
B(x2,y2) = B(4,-1)

and the ratio m:n = 2:3

x = (mx2 + nx1)/(m+n)
x = [2*4 + 3*(-1)]/(2+3)
x = 5/5 =1

y = (my2 + ny1)/(m+n)
y = [2*(-1) + 3*4)/(2+3)
y =10/5 =2

hence the coordinates of the point p(x,y) = P(1,2)

next is to find the gradient of the line AB
m1 = (y2-y1)/(x2-x1)
= (-1-4)/[4-(-1)]
= -5/5 = -1
let the gradient of the line L perpendicular to AB be m2

for any two perpendicular lines, m1*m2 = -1 or m2= -1/m1
m2= -1/-1 = 1

the equation of a straight line given gradient m2 and p(1,2) can also be written as;
yo-y = m2(xo-x)
hence, for line L
yo-2 = 1(xo-1)
yo =xo-1+2
yo = xo+1
OR simply y=x+1
thanks

interesting we have slightly diff sln.. i used graph sha

johnpaul1101: if u re truely a maths guru as u claim u are, then i believe u can solve this: half of A's money plus one-fifth of B's makes #100. Two-thirds of A's money plus two-fifths of B's makes #150. How much money has each? This is simultaneous equation, show ur workings

Well this question nothing near a threat for a Jss kid, let alone a maths guru.....here's your solution

Let A's money = A
and B's money = B

(1/2)A + (1/5)B = 100 .......(1)
(2/3)A + (2/5)B = 150 .......(2)

now, multiply eqn(1) by 2

A + (2/5)B = 200 ............(3)

next, subtract eqn(2) from eqn(3)
_ A + (2/5)B = 200
(2/3)A + (2/5)B = 150

(1/3)A = 50
A = 50*3 =150
from eqn(3)
A + (2/5)B = 200
(2/5)B = 200-A
therefore B = 5/2(200-A)
= 5/2(200-150)
= 5/2(50)= 125
A's money =#150
B's money =#125

biolabee:

if this na jamb question i go murd

p is the variable u need to get rid of

b = a + cp
p = (b-a)/c
p2 = (b2 + a2 - 2ab)/c2

also p2 = 2(r-ab)/c

Equate the two equations

So you get

(b2+a2-2ab)/c2 = 2r-2ab/c

(b2+a2-2ab)/c = 2r-2ab

cross multiply

b2 -2ab + 2abc + a2 - 2rc

Use almighty formula and solve

a = 1
b = 2ac - 2a
c = a2- 2rc

I have tried solving it yet no solution.I stopped at this point: a-ac+or -sqrt(4a^2c^2-8a^2c+8rc).I guess there is no solution to it.Even the past questions booklet said so.Thanks for your help.I'll be back with a handful of questions tomorrow.God loves you.

ok.. lets see

Richiez:

Well this question nothing near a threat for a Jss kid, let alone a maths guru.....here's your solution

Let A's money = A
and B's money = B

(1/2)A + (1/5)B = 100 .......(1)
(2/3)A + (2/5)B = 150 .......(2)

now, multiply eqn(1) by 2

A + (2/5)B = 200 ............(3)

next, subtract eqn(2) from eqn(3)
_ A + (2/5)B = 200
(2/3)A + (2/5)B = 150

(1/3)A = 50
A = 50*3 =150
from eqn(3)
A + (2/5)B = 200
(2/5)B = 200-A
therefore B = 5/2(200-A)
= 5/2(200-150)
= 5/2(50)= 125
A's money =#150
B's money =#125

thanks a lot guru, but i still have more of those simple, jss, questions dat is confusing me

a question which goes like this: an equilateral triangle with the lengths of its sides given in terms of a and b, the length are as follows
4acm
3bcm
(a+b+3)cm
find a and b and hence find the length of the sides of the triangle?
Gudluck

^^^ What is it with Goodluck? Are you giving folks a test or something? Because it doesn't sound like you are looking for help...Smh! And try to upgrade your questions they are damn too cheap !