(17-2k)/(k-5)=-3

Show the following equations and check the result. 3(2k+1)-2(k-5)-5(5-2k)=16

Find the value k if the following equations are consistent (k-2)x+(k-1)y=17,(k-1)x+(k-2)y=18 and x+y=5

tan {2tan ^(-1)""((1)/(5))-(pi)/(4)} is equal to - (k )/(17) then find the value of K .

For what value of k the following distribution is a probability distribution? X=x_(i):0123P(X=x_(i)):2k^(4)3k^(2)-5k^(3)2k-3k^(2)3k-1

If the points whose position vectors are 3i-2j-k,2i+3j-4k,-i+j+2k and 4i+5j+lambda k lie on a plane then lambda is equal to (A)-(146)/(17) (B) (146)/(17) (C) -(17)/(146) (D) (17)/(146)

(i) If the points (1,4),(3,-2)and (k,1)B(3k,2k+3)andC(5k-1,5k) are collinear, fine the value of k. (ii) If the points A(k+1,2k),B(3k,2k+3),B(3k,2k+3)andC(5k-1,5k) are collinear, then show that x + y =2.

sum_(k=1)^(5) (1^3 + 2^3 + ..... + k^3)/(1+3+5+..... + (2k-1) )=