m-((m-1))/(2)=1-((m-2))/(3)

Solve the following linear equations. m -frac(m-1)(2) =1 -frac(m-2)(3)

(m-1)/(m-4)=(2m-3)/(3m-8)

If l : m = 2 (1)/(2) : 1(2)/(3) and m : n = 1 (1)/(4) : 3 (1)/(2) , find l : m : n

In triangle ABC m_(1),m_(2),m_(3) are the lenghts of the medians through A,B and C respectively . If C=(pi)/(2) , then (m_(1)^(2)+m_(2)^(2))/(m_(3)^(2)) =

The lines y=m_(1)x,y=m_(2)xandy=m_(3)x make equal intercepts on the line x+y=1 Then (a) 2(1+m_(1))(1+m_(3))=(1+m_(2))(2+m_(1)+m_(3))(1+m_(1))(1+m_(3))=(1+m_(2))(1+m_(1)+m_(3))(1+m_(1))(1+m_(2))=(1+m_(3))(2+m_(1)+m_(3))(1+m_(1))(1+m_(3))=(1+m_(2))(1+m_(1)+m_(3))2(1+m_(1))(1+m_(3))=(1+m_(2))(1+m_(1)+m_(3))

The value of .^(n)C_(1)+.^(n+1)C_(2)+.^(n+2)C_(3)+"….."+.^(n+m-1)C_(m) is equal to a. .^(m+n)C_(n) - 1 b. .^(m+n)C_(n-1) c. .^(m)C_(1) + ^(m+1)C_(2) + ^(m+2)C_(3) + "…." + ^(m+n-1)C_(n) d. .^(m+n)C_(m) - 1

If m_(1)andm_(2) satisfy the relation "^(m+5)P_(m+1)=11/(2) (m-1) ("^(m+3)P_(m)) then m_(1)+m_(2) is equal to

Find the value of m ( m -1 )/3 - ( m -2 )/4 = 1