SOLUTION: THE LENGTH OF A RECTANGULAR WAL IS 3M MORE THAN TWICE ITS HEIGHT. THE WIDTH OF A PICTURE ON THE WALL IS 1M LESS THAN THE HEIGHT OF THE WALL. THE PICTURE IS 2M LONGER THAN IT IS WID
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Length-and-distance
-> SOLUTION: THE LENGTH OF A RECTANGULAR WAL IS 3M MORE THAN TWICE ITS HEIGHT. THE WIDTH OF A PICTURE ON THE WALL IS 1M LESS THAN THE HEIGHT OF THE WALL. THE PICTURE IS 2M LONGER THAN IT IS WID
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Question 167994: THE LENGTH OF A RECTANGULAR WAL IS 3M MORE THAN TWICE ITS HEIGHT. THE WIDTH OF A PICTURE ON THE WALL IS 1M LESS THAN THE HEIGHT OF THE WALL. THE PICTURE IS 2M LONGER THAN IT IS WIDE. FIND THE AREA OF THE WALL IF 19 M SQUARED OF THE WALL ARE NOT COVERED BY THE PICTURE? Answer by maali40(13) (Show Source):
You can put this solution on YOUR website! Height of wall = H meters Length of wall =(3+2H)meters Width of picture=(H-1) meters Length of picture=(H+1) meters Area of wall= (3H+2H^2) square meters ----- 3+2H+(3+2H-H-1)(H-1)=19 3+2H+(2+H)(H-1)=19 3+2H+2H+H^2-2-H=19 -------------H^2+3H-18=0 H=(-3+(9-4(-18))^.5)/2 H=3
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