Solve : `(3t-2)/4 - (2t+3)/3 = 2/3 - t`

solve the equation . (3 t - 2)/(4 ) - (2 t + 3)/(3) = 2/3 - t

If x=t^(2)andy=t^(3) , then (d^(2)y)/(dx^(2)) is equal to: a) (3)/(2) b) (3)/(2)t c) (3)/(2t) d) (3)/(4t)

Simplify t(3t+2)-(5t -4) +7

Prove that: tan^-1(t) +tan^-1 ((2t)/(1-t^2))=tan^-1( (3t-t^3)/(1-3t^2)), if - 1/sqrt(3),tlt 1/sqrt(3)

x=f(t) satisfies (d^2x)/dt^2=2t+3 and for t=0, x=0, dx/dt=0 , then f(t) is given by (A) t^3+t^2/2+t (B) (2t^3)/3+(3t^2)/2+t (C) t^3/3+(3t^2)/2 (D) none of these

The domain of the funciton f(x) given by 3^(x) + 3^(f) = "min" (2t^(3) - 15t^(2) + 36t - 25, 2 + |sin t| , 2 le t le 4) is

The domain of the funciton f(x) given by 3^(x) + 3^(f) = "min" (2t^(3) - 15t^(2) + 36+ - 25, 2 + |sin t| , 2 le t le 4) is

If points (t ,\ 2t),\ (-2,\ 6) and (3,\ 1) are collinear, then t= (a) 3/4 (b) 4/3 (c) 5/3 (d) 3/5

The vertices of a triangle are [a t_1t_2,a(t_1 +t_2)] , [a t_2t_3,a(t_2 +t_3)] , [a t_3t_1,a(t_3 +t_1)] Then the orthocenter of the triangle is: (a) (-a, a(t_1+t_2+t_3)-at_1t_2t_3) (b) (-a, a(t_1+t_2+t_3) + a(t_1t_2t_3) (c) (a, a(t_1+t_2+t_3)+at_1t_2t_3) (d) (a, a(t_1+t_2+t_3)-at_1t_2t_3)

Area of the triangle formed by the threepoints 't_1'. 't_2' and 't_3' on y^2=4ax is K|(t_1-t_2) (t_2-t_3)(t_3-t_1)| then K=