If `"Delta"_1=|x bb a x b a a x|a n d"Delta"_2=|x b a x|` are the given determinants, then `"Delta"_1=3("Delta"_2)^2` b.`d/(dx)("Delta"_1)=3"Delta"_2` c. `d/(dx)("Delta"_1)=3("Delta"_2)^2` d. `"Delta"_1=3"Delta"2 3//2`

Delta_1 = |[x,b,b], [a,x, b] ,[a, a, x]| and Delta_2 = |[x,b],[a,x]| show that d(Delta_1)/ dx = 3 Delta_2

If Delta_1=|(1,0),(a,b)| and Delta_2=|(1,0),(c,d)| , then Delta_2Delta_1 is equal to :

In a triangle A B C ,(r_1+r_2)/(1+cosC) is equal to (2a b)/(c"Delta") (b) ((a+b))/(c"Delta") (a b c)/(2"Delta") (d) (a b c)/("Delta"^2)

A substance is carried through the following transformations: The equation Delta S_(1) + Delta S_(2) = Delta S_(3) + Delta S_(4) + Delta D_(5)

Delta_(1) and Delta_(2) are respectively areas of in circle and circumcircle of triangle formed by (0,0), (2,0) and (1,sqrt(3)) then which of following is NOT correct (A) Delta_(1)+Delta_(2)=(5 pi)/(3) (B) |Delta_(1)-Delta_(2)|=pi (C) 3Delta_(1)+6Delta_(2)=9 pi (D) [2Delta_(1)+3Delta_(2)]=12 , where [.] is G.I.F.

If p_1,\ p_2,\ p_3 are the altitudes of a triangle from its vertices A ,\ B ,\ C and "Delta" , the area of the triangle A B C , then 1/(p_1)+1/(p_2)-1/(p_3) is equal to s/"Delta" (b) (s-c)/"Delta" (c) (s-b)/"Delta" (d) (s-a)/"Delta"

The base BC of Delta ABC is divided at D, so that BD=(1)/(2)DC. Prove that ar(Delta ABD)=(1)/(3)ar(Delta ABC)

Differentiate the following by delta method : (x-1)(x-2)