Match the polygons formed by the sets of points with their perimeters (rounded to the nearest hundredth). Tiles A(1, 1), B(6, 13
), C(8, 13), D(16, -2), E(1, -2) 38 units F(14, -10), G(16, -10), H(19, -6), I(14, -2), J(11, -6) 25.4 units K(4, 2), L(8, 2), M(12, 5), N(6, 5), O(4, 4) 50 units P(7, 2), Q(12, 2), R(12, 6), S(7, 10), T(4, 6) 19.24 units U(4, -1), V(12, -1), W(20, -7), X(8, -7), Y(4, -4)
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Substitute this into the parabolic equation,
We're told the line intersects
twice, which means the quadratic above has two distinct real solutions. Its discriminant must then be positive, so we know
We can tell from the quadratic equation that has its vertex at the point (3, 6). Also, note that
and
so the furthest to the right that extends is the point (5, 2). The line
passes through this point for
. For any value of
, the line
passes through
either only once, or not at all.
So ; in set notation,