Find the slope of the line that passes though (-2,5) and (3, 0
2 answers:
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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
The answer is
x = 12.1 <<<<< answer.
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
The answer is
x = 12.1 <<<<< answer.