₦airaland Forum

Somethings waz omitted....he was the school of engineering president...and he's popularly known as OBAMA...kudos bro...as for jokogbola Nofisat,i doff my hat for you....

Atolagbe1: They asked for cgpa.but as at now u only have gpa,I advice u put nil.

..i kept nil bt is nt goin thru until u put a figure

Fetus: @olsey..plz wat do i put for my cgpa? Cuz i'm a 100l witout a GP..jst finish my first semester's exam..urgent reply, plz....and kindly forward ur number 2 me: talk2fetus@gmail.com...tank u

...rummies abeg ansa me...

@olsey..plz wat do i put for my cgpa? Cuz i'm a 100l witout a GP..jst finish my semester's exam..urgent reply, plz....and kindly forward ur number 2 me: talk2fetus@gmail.com...tank u

@all tank u so much 4 ur opinions...i wud work on it!

@olsey..kindly asist me...i wants to apply for d scholarship bt d application is nt displaying....urgent reply plss...thanks

Kindly help with dis...(1) the sum of the first n terms of a G.P is 127 and the sum of their reciprocal is 127/64.the first term is 1. Find n and the common ratio respectively...(2)find 3 numbers in G.P whose sum is 13 and whose product is -64.....(3) resolve into partial fraction 7x+2/(2x-3)(x+1)^2.......tank u and God bless as u asist me....

Plz...guyz check dis out...find d general solutions of (1) 7sec^2€= 6tan€ + 8...(2) 2cos^2€-3sin2€-2...where ^ and € means power and tetha respectively..i.e 2cos squre tetha= 2cos^2€...tanx alot

Plz, is there no scholarship awards for pure sciences? Like chemistry...

doubleDx: Solution to Question 1b

To prove that √10 = 3[1 + 1/20 + 1(3)/[(20)(40)] + [(1)(3)(5)/[(20)40(60) + ....]

Evaluating the RHS yields =>
= 3[1 + 1/20 + 1(3)/[(20)(40)] + [(1)(3)(5)/[(20)40(60) + ....]
= 3 (1.0541)
= 3.162 approx.

√10 can be rewritten as =>

= √[9(1 + 1/9)]
= 3√(1 + 1/9)
= 3 (1 + 1/9)^(1/2)
Expanding the bolded part yields =>
= 3[1 + (1/2)(1/9) + (1/2)(-1/2)(1/9)^2/2! + (1/2)(-1/2)(-3/2)(1/9)^3/3! + ....]
Simplifying further yields =>
= 3[ 1.0541 approx]
= 3.162 approx

Thus,
√10 = 3(1 + 1/20 + 1(3)/[(20)(40)] + [(1)(3)(5)/[(20)40(60) + ...] = 3[1 + (1/2)(1/9) + (1/2)(-1/2)(1/9)^2/2! + (1/2)(-1/2)(-3/2)(1/9)^3/3! + .... = 3.162 approx

...God bless u man.....tanx alot.....

doubleDx: No probs bruv, check this out =>

2√(x + 4) - x = 1
2√(x + 4) = x + 1

For x = -3 on the LHS

2√(x + 4)
= 2 . [±√(-3 + 4)]
= 2. [±√(1)]
= 2 .(+ 1 ) or 2. (- 1)
= 2 or -2

For x = -3 on the RHS

= x + 1
= -3 + 1
= - 2

It's true for x = -3 that the LHS = RHS with a value of (-2)
Note=> (.) is for multiplication!

..u are nt solving my questions ni?...kindly asist me na

mr byron: it 19 years not 13 years please.... we last won the nations cup in 1994

....d last time we entered final was in 2000,,,dats 13yrs ago...

Kindly help with dis questions....1) write down d first four terms of d bionomial series expansion in ascending power of x of d functions (1-x)^-1/2 and hence show dat root 10= 3(1+1/20+1*3/20*40+1*3*5/20*40*60+........) No(2) find d common ratio of d geometric sequence sin2€, -sin€cos2€, sin2€cos^2(2€),.....prove dat for @<€<₹/2...the series sin2€-sin2€cos2€ + sin2€sin^2(2€) +....has a sum to infinity and show dat d sum to infinity is tan€... Where €,₹ and @ are alpha, pie and teetha respectively.... Tanks

Richiez: 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

....kindly help with dis...simplify tan^2(1/4¥ - 1/2€)....where ¥ and € are pie and teetha respectively....tanks

Richiez: 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

....Yepa! I doff my hat 4 u oooo.....tankz so much man...

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

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..

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

doubleDx: To solve this question, apply the trig identity for tan (x + y) which is => (tan x + tan y)/(1 - tan x tan y)

Since tan (x + y) = 4/3
(tan x + tan y)/(1 - tan x tan y) = 4/3
cross multiplying yields =>
4 (1 - tan x tan y) = 3 (tan x + tan y)
4 - 4tan x tan y = 3tan x + 3tan y

Collecting like terms =>

3tan y + 4tan x tan y = 4 - 3tan x
tan y (3 + 4tan x) = 4 - 3tan x
tan y = (4 - 3tan x)/(3 + 4tan x)
Now remember that tan x = 1/2, substituting yields =>

tan y = [4 - 3(1/2)]/[3 + 4(1/2)]
tan y = [4 - 3/2]/5
tan y = (5/2) .(1/5)
:. tan y = 1/2.

I'm on phone, I will post the solution to question 2 later.

....Tank u so much...God bless u...

Richiez: but for the 2nd part of the question, is it (sinB+sin3B+sin5B)/(cosB+cos3B+cos5B) or sinB+sin3B+(sin5B/cosB)+Cos3B+cos5B

pls reply quickly for prompt solution

(sinB+sin3B+sin5B)/(cosB+cos3B+cos5B)...yea dis one

Can sumone solve dis 4 me....IF Tan(x+y)=4/3 and tanx=1/2, evaluate tany....kindly show d workings.....another is...Simplify sinB +sin3B +sin5B/ cosB + cos3B + cos5B...note B=beta

Salam alaikum to all my muslims brothers and sisters. I'm a muslim whom fell in love with a christian.. we both really each other...my challenges are now that she is really worried about the difference in religion likewise me...my question what is d quran view on this?..need your urgent replies....kindly help us...we are worried! Ma salam

Still passing-by

Can anyone remember this sweet...how often did u take it back then As for me; it was lyk everyday,,,

D last tym i cried 4 real was 2007 wen i lost my parents,,,,thou,it tuk me days 2 cry bt i eventually did wen i saw my 6yrs old sister still believing our parents was still alive.......itz very tragedic...u dont want 2 xperience it

Mascotizer: sup fetus....i am chemistry student, but a stalite though, on be half of all the Final Year Students of Chemistry dept, I welcome to our beloved department.
wish you the best of lucks in our academic pursuit.
I hope we get to meet someday.

sure,,,tank u....can u kindly drop ur 2go id,,,,or number so dat we can talk,,,,i tink i wud nid ur assistance...

Yepa!! Any chemistry student in da house?? I went 2 d skul yesterday nd it was only 2 chemistry student(fresher) dat was around,,,,,can jst 2 student run d programme?

Take u all...i appreciate ur contributions; @1dacrux tinkin of where 2 work is part of my future na...dint u get it! Well, ur oda opinion is noted 2...smiles