A right triangle has one vertex on the graph of y=16-x^2, x>0, at (x,y), another at the origin, and the 3rd pn the positive x-axis t (x, 0). Express the area A of the triangle as a function of x.

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Answer 1

For a triangle, A = 1/2bh

In this case "x" is the base and "16-x²" is the height so just plug them in.