How do I solve this his math 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?
4 Answers
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Well,
the best thing is to make a drawing my my points
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
hope it' ll help !!
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Draw the lines on a graph. You will see that the polygon has 5 sides, bounded by the x axis, the y axis, the vertical line x = 2, the horizontal line y = 1, and the graph of y = -(1/2)x + 3/2.
5 sides
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5 sides
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3, one for each inequality