Question

The solutions to the inequalities 0≤x≤2, 0≤y≤1, and y≤−1/ 2 x+3/ 2 form a polygonal region in the plane.?
how many sides does this polygon have?

Answer 1

i need help please!

Answer 2

0≤x≤2, 0≤y≤1, forms a rectangle with corners : O(0,0) , A(2,0) , B(2,1) , J(0,1)

and y = -(1/ 2) x + (3/ 2) passes through : C(0, 3/2) and D(3,0) ---> line (L)
and
(L) cuts the rectangle at two points :
E(1,1) so that y = 1 = -(1/2)x+ 3/2 gives: x = 1
and
F(2, f ) so that : f = -(1/ 2) 2 + (3/ 2) = 1/2
therefore :
the polygon has 5 sides

Answer 3

Get it?

Answer 4

helpful but i dont get the part you get 4 sides?

Answer 5

where from?

Answer 6

\(0\le x\le2\):