" 10."quad 3x-(y+7)/(11)+2=10;quad 2y+(x+11)/(7)=10

3x-(y+7)/(11)+2=10 and 2y+(x+11)/(7)=10

Solve: 3x - (y+7)/11+2 =10 and 2y+ (x+11)/7 =10

x-y=0.9; 11/(2(x+y))=10.5

x-y=0.9; 11/(2(x+y))=10.5

If A = (3,-4) , B= (7,0) and C = (14 , -7) are the three consecutive vertices of a parallelogram ABCD , then find the slope of the diagonal BD . The following are the steps involved in solving the above problem . Arrange them is sequential order (A) ((x + 7)/(2) , (y + 0)/(2)) = ((3 + 14)/(2) , (-4-7)/(2)) . (B) The slope of BD = (-11-0)/(10-7) = (-11)/(3) . (C) (x+7)/(2) = (17)/(2) and (y+0)/(2) - (-11)/(2) implies x = 10 , y = -11 therefore D = (10 , -11) . (D) Let the fourth vertex be D(x , y) . we know that the diagonals of a parallelogram bisect each other .

If ( 10)^(9) + 2( 11)^(1) ( 10)^(8)+ 3 ( 11)^(2) ( 10)^(7)+"……….." + 10 ( 11)^(9)= k ( 10)^(9) , then k is equal to :

If the lines (1-x)/3=(y-2)/(2a)= (z-3)/2 and (x-1)/(3a) =y-1 = (6-z)/5 are perpendicular then the value of a is a) (-10)/7 b) 10/7 c) (-10)/11 d) 10/11

Solve x + y = 7 and 3x - 2y = 11.

Solve x + y = 7 and 3x - 2y = 11.

Solve by the method of elimination 2x-y=3,3x +y=7 (ii) x-y=5, 3x+ 2y = 25 (iii) (x)/(10 ) +(y)/(5) =14 ,(x)/( 8) + (y)/( 6) =15 (iv) 3(2x+y) =7xy,3 (x+3y ) =11xy (v) (4)/(x) +5y =7,(3)/(x) +4y =5 (vi) 13x +11y = 70,11x +13y =74