Pls help I’m really confusedFind AB round to the nearest tenth if necessary
Accepted Solution
A:
Comment By similar triangles it can be shown that AD^2 = AB*AC If you want the proof, Google tangents and secants of a circle.
Find So we want AB
Givens AD = 16 BC = 9
AB = ?? CA = CB + AB CA = 9 + AB
Formula AB * (AB + BC) = AD^2
Sub and Solve AB*(AB + 9) = 16^2 AB*(AB + 9) = 256 Remove the brackets. AB^2 + 9AB = 256 Subtract 256 from both sides. AB^2 + 9AB - 256 = 0
You can only do this either with a graph or the quadratic formula. I'll get the graph for you. You can made these yourself at Desmos.
x = [-b +/- sqrt(b^2 - 4ac)] / (2a) a = 1 b = 9 c = -256
Answer When you substitute these into the quadratic formula, you get x1 = 12.12 and x2 = -21.12 x2 is meaningless. The solution is x = 12.12
Comment But that's not your question. Your question is what is this rounded to the nearest 1/10th? That's a fancy way of saying round to the first decimal place. Since the hundredth place (or second place) is 2, 12.12 rounds to 12.1