Graph y = 1 - x 2 and y = e -x on the same set of coordinate axes. use the graph to find the positive solution of the equation 1 - x 2 = e -x to the nearest tenth.

Please, use " ^ " to denote exponentation.  Thus, your  y = 1 - x 2 and y = e -x
should be written    y = 1 - x^2 and y = e ^(-x)

The first function is a parabola with y-intercept 1; it opens downward.  The second is an exponential function with y-inercept 1; the graph is entirely in the first quadrant , and is always decreasing, and the curve is "concave up."

You need to graph both functions on the same set of axes, and then to the best of your ability determine the coordinates of the point where the 2 curves intersect.  Most teachers would allow you to use a calculator for this work.

According to my graphing calculator, the pts. of intersection are (0,1) and (0.7, .5).

40%=2/550=1/230=3/1056=14/25