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Tutor profile: Nina K.

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Nina K.
Learning Center Tutor for 4 years with Bachelors in Mathematics with Minor in Education
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Questions

Subject: Trigonometry

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Question:

Assume the pine tree is at a right angle to the plane Tommy is standing at. Tommy measures his or her distance from the base of the tree, and then uses a clinometer (a small instrument that measures inclination, or angle of elevation) to look at the top of the tree and determine "x" He is standing 71 ft away from a pine tree. the angle of elevation is 35 degrees. How tall is the tree in feet?

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Nina K.
Answer:

Let's draw out a right triangle that connects: Tommy's feet, the bottom of the tree, and top of the tree. If a = the distance between Tommy and the tree, 71 ft x = angle of elevation, 35 degrees b = the height of the tree, We can use the tangent relationship to write the following formula tan(x) = b/a tan(35 degrees) = b/71 b = 71tan(35) b = 49.7 feet The tree is about 49.7 feet tall

Subject: Calculus

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Question:

You are building a fence with width 'x' and length 'y'. You have 500 ft of fencing to use with the only requirement such that inside the fence border, there are four partitions across the length (ie 5x + 2y = 500). What fence dimensions will maximize the total area of the fenced off area? What is the maximum area in ft^{2}?

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Nina K.
Answer:

You are trying to maximize area "A" which is equal to x*y Your restrictions are that 5x + 2y = 500 Another way to write this restriction (through algebraic manipulation) is that y = 250-2.5x By substituting this into the Area problem, we get A = x*y = x*(250-2.5x) = 250x-2.5x^{2} In this case, to maximize A(x) = 250x-2.5x^{2}, we take the fist derivative and set it equal to 0 A'(x) = 250-5x = 0 Therefore x = 50 ft If x = 50, then y = 250 - 2.5(50) = 125 ft A = x*y = (50 ft)*(125ft) = 6250 ft^{2} Therefore, the fence dimensions 50ft by 125 ft will maximize an area of 6250 ft^{2}

Subject: Algebra

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Question:

2x^{2}-9x-5=0

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Nina K.
Answer:

2x^{2}-9x-5=0 (2x-1)(x+5)=0 x = 1/2 or -5

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