If ABCDEF is a regular hexagon , then A vec D + E vec B + F vec C equals
Let ABCDEF be a regular hexagon whose centre is at the origin.(A,B,C,D,E and F are in anticlockwise direction).If the coordinates of A be 1+i then c lies in
A short linear object of length b lies on the axis of a concave mirror of focal length F at a distance u from the pole. The length of the image will be
A C F J ul(?) U B
If f is a continuous function on [a;b] and u(x) and v(x) are differentiable functions of x whose values lie in [a;b] then (d)/(dx){int_(u(x))^(v)(x)f(t)dt}=f(v(x))dv(x)/(dx)-f(u(x))du(x)/(dx)
Assertion : The focal length of the mirrorr is f and distance of the object from the focus is u , the magnification of the mirror is f//u . Reason : Magnification =("Size of the image")/("Size of object" )
Find the number of 6 digit numbers of the form abcdef if the digits satisfy the condition a+b+c+d+e+f=a^(2)+b^(2)+c^(2)+d^(2)+e^(2)+f^(2)
