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x can be 5 and y 5 x can be 7 and y 3 there are too many combinations of x and y. there has to be a logical reason why you would choose 0 as one and 10 for the other.

x can be 5 and y 5 x can be 7 and y 3 there are too many combinations of x and y. there has to be a logical reason why you would choose 0 as one and 10 for the other.

Please explain.

in order to find the enclosed area you first have to find where {x,y} intersect the coordinate plane.

if x=0 then (0,10) and if y=0 (10,0)

then you can tell its 10*10 area.

your solution of {5,5} is true algebraic wise but that isn't what you are being asked for.

x can be 5 and y 5 x can be 7 and y 3 there are too many combinations of x and y. there has to be a logical reason why you would choose 0 as one and 10 for the other.

Please explain.

in order to find the enclosed area you first have to find where {x,y} intersect the coordinate plane.

if x=0 then (0,10) and if y=0 (10,0)

then you can tell its 10*10 area.

your solution of {5,5} is true algebraic wise but that isn't what you are being asked for.

i think you all are missing something the figure formed is symmetrical about the x and y axis so we get 4 triangles in each quadrant that are congurent area of each is 1/2(base * height) =1/2(10*10) =50 as we get 4 triangles total area=4(50) =200

x can be 5 and y 5 x can be 7 and y 3 there are too many combinations of x and y. there has to be a logical reason why you would choose 0 as one and 10 for the other.

Please explain.

in order to find the enclosed area you first have to find where {x,y} intersect the coordinate plane.

if x=0 then (0,10) and if y=0 (10,0)

then you can tell its 10*10 area.

your solution of {5,5} is true algebraic wise but that isn't what you are being asked for.

i think you all are missing something the figure formed is symmetrical about the x and y axis so we get 4 triangles in each quadrant that are congurent area of each is 1/2(base * height) =1/2(10*10) =50 as we get 4 triangles total area=4(50) =200