" Solue it ":(i)3(t-3)=5(2t+1)

Solve the following equations : 3 (t-3) = 5 (2t - 1)

If 3(t-3) = 5(2t + 1) , then t = ?

If u=sin^(-1)(x-y),x=3t,y=4t^(3) , then what is the derivative of u with respect to t? (A) 3(1-t^2) (B) 3(1-t^2)^(-1/2) (C) 5(1-t^2)^(-1/2) (D) 5(1-t^2)

If A=[[3,5],[2,1]] Show that (A^(T))^(-1)=(A^(-1))^(T) .

If T_(n)=sin^(n)theta+cos^(n)theta ,prove that (i)(T_(3)-T_(5))/(T_(1))=(T_(5)-T_(7))/(T_(3)) (ii) 2T_(6)-3T_(4)+1=0 (iii) 6T_(10)-15T_(8)+10T_(6)-1=0

Solve the equation : (3t+1)/16-(2t-3)/7=(t+3)/8+(3t-1)/14

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)

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)+at_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)

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)+at_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)

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)+at_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)