<h3>
Answer: A. 18*sqrt(3)</h3>
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Explanation:
We'll need the tangent rule
tan(angle) = opposite/adjacent
tan(R) = TH/HR
tan(30) = TH/54
sqrt(3)/3 = TH/54 ... use the unit circle
54*sqrt(3)/3 = TH .... multiply both sides by 54
(54/3)*sqrt(3) = TH
18*sqrt(3) = TH
TH = 18*sqrt(3) which points to <u>choice A</u> as the final answer
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An alternative method:
Triangle THR is a 30-60-90 triangle.
Let x be the measure of side TH. This side is opposite the smallest angle R = 30, so we consider this the short leg.
The hypotenuse is twice as long as x, so TR = 2x. This only applies to 30-60-90 triangles.
Now use the pythagorean theorem
a^2 + b^2 = c^2
(TH)^2 + (HR)^2 = (TR)^2
(x)^2 + (54)^2 = (2x)^2
x^2 + 2916 = 4x^2
2916 = 4x^2 - x^2
3x^2 = 2916
x^2 = 2916/3
x^2 = 972
x = sqrt(972)
x = sqrt(324*3)
x = sqrt(324)*sqrt(3)
x = 18*sqrt(3) which is the length of TH.
A slightly similar idea is to use the fact that if y is the long leg and x is the short leg, then y = x*sqrt(3). Plug in y = 54 and isolate x and you should get x = 18*sqrt(3). Again, this trick only works for 30-60-90 triangles.
Answer:

Step-by-step explanation:











Nice question hey there! need help!?
A circle with center
and radius
has equation

If we substitute
in this equation, we have

This equation has two solutions (i.e. the line intersects the circle in two points) if and only if the determinant is greater than zero:

The expression simplifies to

The solutions to the associated equation are

So, the parabola is negative between the two solutions:
