₦airaland Forum

RECHECK UR QUES@LAZ

i guess everybody has his or her dfficiency it is just a matter of how we deal with it.

hmm........i have never taking dis thing called maths magic really serious but i think i'm beginning to like it.

Redemption camp tinz..................
i salute all the redeeme members on this thread"

smurfy: Yeah. sin¥ = sqrt(1/15) gives ¥ = 14.9 and @ = 28.1.

That's exactly what I got above.

THANKS SMURFY & LAPLACIAN

Mr Calculus: plz guys help me out wit dis......
if 3sin2@=5sin2¥....
find @ & ¥ if(tan@=2tan¥)

plz my generals i need ur help to dis o.@laplacian,rechiez,ben,fx,smurfy,ortarico.e.t.c
i dont know but d answers to d ques are 16.78' & 31.1'.
plz i want d workings not only d answers

smurfy: @ = 28, ¥ = 30

Approximate acute angle values only.

d ans to d ques are 16.78' & 31.1'.
plz i just need d workings

smurfy: @ = 74, ¥ = 19.

not correct sir....
plz recheck ur calculations

plz guys help me out wit dis......
if 3sin2@=5sin2¥....
find @ & ¥ if(tan@=2tan¥)

benbuks: ...beautifully wrong.

THANK U SIR

benbuks: ^^ modified..sori it's ^1/2 ...check it again.

in dat case i got 2/root2 or rather root2

benbuks: Q1 compute the derivative of cosecx wrt.x from first principle

Q2 evaluate lim. as x-->0 of [(1/x +2)^1^2 - (1/x)^1/2 ]

i'm not clear with ur no 2 first bracket,is it ^1^2 or ^1/2?
all dsame i got 2/root2

divide through by ¥x.
¥y/¥x=-[2cos(x+{¥x/2})sin(¥x/2)]/¥x/sinxsin(x+¥x).
¥y/¥x=-cos(x+{¥x/2})sin(¥x/2)/(¥x/2)/sinxsin(x+¥x).
recall::lim x-->0.¥x=0,sin(¥x/2)/(¥x/2)=1,¥y/¥x=dy/dx.
dy/dx=-cos(x+0)/sinxsin(x+0)
dy/dx=-cosx/sin^2x.
dy/dx=-cosx/sinx*1/sinx.
dy/dx=-cotxcosecx

benbuks: Q1 compute the derivative of cosecx wrt.x from first principle ]

to solve d/dx cosecx by first principle::
y=cosecx=1/sinx.
add ¥ to both sides(¥=delta).
y+¥y=1/sin(x+¥x).
¥y=1/sin(x+¥x)-y.
¥y=1/sin(x+¥x)-1/sinx.
¥y=sinx-sin(x+¥x)/sinxsin(x+¥x).
recall::sinx-siny=2cos(x+y/2)sin(x-y/2).
¥y=[2cos(x+x+¥x/2)sin(x-x-¥x/2)]/sinxsin(x+¥x).
¥y=[2cos(2x+¥x/2)sin(-¥x/2)]/sinxsin(x+¥x).
¥y=[2cos(x+{¥x/2})sin(-¥x/2)]/sinxsin(x+¥x).
divide through by ¥x.....
loading.....

Appliedmaths.:
@Boss, please contact me on trustyondstreets@gmial.com // I got your NL message but it couldn't reply.

ALREADY DON SIR

Alpha Maximus: ...CORRECT!!!! You are a true mathematician!!!!!

WONDERFUL

Richiez: Okay jaryeh and smurfy here it is
by Quiz master Alphamaximus


1. The product of 2 numbers is equivalent to the square of of the result of the product of the square root of 4 and 500. The sixth root of one of the numbers is deduced to be equal to the difference between the square root of the other number and 3. What is the square of the reciprocal of the result of the product of the cube root of the larger number and the sixth root of the smaller number?

hmm.............

benbuks: @mr. Calulus .use benbuks10@gmail.com...cnt read d msg u just snt to me..

K SIR

Alim.toheeb:
Pls someone should help me to axe dis S=\/`````tsquare-4t+4.....make t d subject of d formular

t=s+2.....
do u want workings??IF U WANT WORKINGS GO TO MATHS CLINIC

were is factoral and dennice?

aysuccess99: I have lost all hope...the competition is now between oga jaryeh and oga smurfy.

DONT say dat sir dis is like a game u dont know wat will happen next

should there really b bonous questions?

Ortarico: If not for these busy days, I should continue as a full time administrator. My prof, I greet you powerfully. Where is my oga @ the top, 'Mr calculus'?

my general i just dy low key..
watching from d sidelines though

hmm...............

TNX alot@jackpot.
abeg no mind my username oo....
but i no i will still uphold dat name wit a little time......tnx again

jackpot: it doesnt make my answer wrong, since I didn't answer your question

hope you got my drift.
you are totally, entirely, capitally and wholistically wrong since there are serious radicals in your denominator.

You started with a question having only one radical, and your wrong answer had 2 radicals in the denominator!!!

hahaha......na because of fail way i fail am 9 make u dy writ all those vocalbs?....
i dint espect to b right,dats y i posted it.....

hmm.......@jackpot..
first of all u change my ques from -16 to -16+oi....i followed that.
then later to 16e^i(pi) and w^4......i was totally loss by dat move cause u ignored d -ve sign attach to d 16.....
ALL D SAME DIS IS D ANSWER,BUT WAT I WANT IS D SOLUTION::
root2(1+j),,root2(-1+j),,root2(-1-j),,root2(1-j).....
PLZ WEN SOLVIN TRY TO USE SIMPLE TERMS CAUS ME NO TOO GUD 4 MATHS CAUS AM STILL A FRESHER.....
TNX IN ADVANCE@ALL MY GENERALS PLZ HELP ME OUT WIT D SOLUTION.

[quote author=jackpot]in the form a+ib
z=-16+0i
[s]sorry, Mathematicians dont write the imaginary part with j like Physicists and Engineers [/s]

now, for the fourth root, i'll prefer working with the exponential form to avoid the impossible task of typing out square roots here.

z=-16+0i=16e^i(pi)=w⁴

Using the nth root of a complex number formula

w=nth root of |z| times e^i[(Argz+2k pi)/n] for k= 0, 1, 2 ,. . ., n-2, n-1

Thus, the fourth roots are simply

w= 2e^i(pi/4), 2e^i(3

Mr Calculus: sis u got me wrong.
d ques is not 1/[(cuberoot2)-1] but 1/(cuberoot2) -1

& DAT MAKES UR ANSWER WRONG......
WEN I DID IT I GOT
-1/(2^1/3+2^2/3+2)
but i dont know if am right.
plz try again and others plz help out

jackpot: i hope you meant ¹/_{[(cuberoot2)-1]}

Now, since
¹/_(x-1)=(x²+x+1)/_(x³-1),

we have (upon setting x=cuberoot2=2^1/3) that

¹/_{[(cuberoot2)-1]}

equals

(2^2/3+2^1/3+1)/_(2^3/3-1)

which simply equals

2^2/3+2^1/3+1.

sis u got me wrong.
d ques is not 1/[(cuberoot2)-1] but 1/(cuberoot2) -1

I-am-Winner:
there are certain tins in maths that I cant forget.. Smtin lyk dis..

Ans is (root 2) + 1

WRONG