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# Tutor profile: Courtney G.

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Courtney G.
Mathematics Tutor, Arts Educator at Root Division
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## Questions

### Subject:Pre-Calculus

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Question:

Does the lim f(g(x)) = f(lim g(x)) as x approaches a when f(x) = 1/x , g(x) = x^3 and a = 0?

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Courtney G.

For the lim f(g(x)) = f(lim g(x)), two conditions must be met: Iim g(x) as x approaches a must exist and the function f must be continuous at the point f(lim g(x)). First, let's see if Iim g(x) exists. lim g(x) = lim x^3, plugging in 0 we get 0. The limit exists. Next let's take our answer 0 and check if f(0) is continuous. f(0) = 1/0 = undefined, therefore not continuous. This means that f(g(x)) does not equal f(lim g(x)) as x approaches a when f(x) = 1/x , g(x) = x^3 and a = 0.

### Subject:Geometry

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Question:

2 lines tangent to a circle, C, intersect the circle at points a & b, where a is not equal to b. The lines intersect each other at a point p, 10m away from the center of the circle. The lines' intersection creates a 60-degree interior angle (angle p = 60 degrees). What is the area of circle C?

Inactive
Courtney G.

We know that the line cp is 10m long and angle p=60 degrees. We need to know the radius of circle C to calculate the area, r = the distance from c to the points where the line intersects the circle, points a and b. Now we can draw two triangles cpa and cpb. We know that these triangles are identical because of the Two Tangents Theorem (line ap=bp, note ac and ab both = r). This means line cp bisects angle p into two 30 degree angles. We can now conclude that our triangles are 30-60-90 triangles because we know that tangent lines form a right angle with the circle's radius (Tangent to a Circle Theorem). Line cp is the hypotenuse which is 10m, meaning the radius (opposite of the 30 degree angle) is 5m. Now we can calculate the area of circle C = πr^2 = π5^2 = 25π meters squared.

### Subject:Calculus

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Question:

Find the derivative of f(x) = cos√x + √x

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Courtney G.

cos√x can be thought of as a composite function, or a function within another function. Let the outer function be denoted as V(u) = cos(u) and the inner function u(x) = √x. From here we can use the chain rule V'(u(x)) * u'(x). We can calculate the simple derivatives V'(u) = -sin(u) and u'(x) = 1/(2√x). Now we are ready to substitute values into the chain rule formula: -sin(u(x))*(1/2√x) = (-sin√x)*(1/2√x), which simplifies to (-sin√x)/(2√x). Next, we can find the derivative of √x, which is a simple application of the power rule: remember that √x=x^1/2, so the derivative of √x = 1/2(x^-1/2) or 1/(2√x) Putting it all together we get out answer: (-sin√x)/(2√x) + 1/(2√x) or (1 - sin√x)/(2√x)

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