If a^2+b^2+c^2=ab+bc+ca...then a^3+b^3+c^3 ??
@mickym said: a/(1-a) + b/(1-b)+ c/(1-c) = 1then find1/(1-a) + 1/(1-b) + 1/(1-c)
a/(1-a)+1 + b/(1-b) +1+c/(1-c)+1 = 1+1+1+1
1/1-a +1/1-b +1/1-c =4
cos(2pie/7)+cos(4pie/7) +cos(6pie/7) = ??
3abc
RHS will become zero
ABC is a triangle.The medians CD & BE intersect each other at point O.then area(ode)/area(abc)...
a) 1:3 b) 1:4
c) 1:6 d) 1:12
yar tier1 ka ques hai..agar pehle solve kar rakha ho to link de do..otherwise give proper explaination..
@JS-M said: cos(2pie/7)+cos(4pie/7) +cos(6pie/7) = ??
I always get confused in such q plz give a detailed sol
@wolverine1986 said: ABC is a triangle.The medians CD & BE intersect each other at point O.then area(ode)/area(abc)...a) 1:3 b) 1:4c) 1:6 d) 1:12yar tier1 ka ques hai..agar pehle solve kar rakha ho to link de do..otherwise give proper explaination..
@sumit1234567890 said: @wolverine1986 ans kya hai mera B aa raha hai
1:12 hai
median divides area in half & itself in ratio 2:1 use kar k dekho
@watsnew24 said:bhai xplain kr dena
Multiply & div by sin(pi/7) & solve
use formula for 2 Sin A Cos B
bhaiyo aur koi question???
@sumitrewri said:
yar iska logic btana koi
P,Q,R employed to do work for 5750. P+Q do 19/23 wrok and Q+R do 8/23 work. Wage of Q
@watsnew24 said:0cube on both side& derive the equation by taking x^6 common
yar ye walla btana plz
q was :
x+ 1/x = rt(3)
value of
x^18 + x^12 + x^6 + x^1
@JS-M said: If sinA = b/a, then sqrt{(a-b)/(a+b)} + sqrt{(a+b)/(a-b)}
@JS-M said: If 8tanA = 15, then (sinA-cosA)/(sinA+cosA)
how to do these???
@Brooklyn said:yar ye walla btana plzq was :x+ 1/x = rt(3)value ofx^18 + x^12 + x^6 + x^1
bhai 2-3 page piche hi explain kiya hai agar cube karoge to x^3 + 1/x^3 = 0 aa jayega
@mickym said:Multiply & div by sin(pi/7) & solveuse formula for 2 Sin A Cos B
yar plz elaborate a lil more