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# Tutor profile: Kelsey E.

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Kelsey E.
Licensed Math Teacher with 10 years tutoring experience
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## Questions

### Subject:Pre-Calculus

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Question:

Solve the following inequality: x$$^2$$+3x-10 > 0

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Kelsey E.

One solution to this problem would be to graph the equation x$$^2$$+3x-10 = y and view where the parabola is above the x-axis. To solve this problem without graphing, we first need to identify the zeros of the function -- that is, determine when x$$^2$$+3x-10 = 0. We will begin by factoring the polynomial to get: (x+5)(x-2) = 0 If either factor equals 0 then the product of the two factors will be zero. x+5 = 0 => x = -5 x-2 = 0 => x = 2 So, the zeros of this function are x = 2 and x = -5. These are the points where the value of x$$^2$$+3x-10 will change between a positive and a negative value. There are several ways to test where the value of the polynomial is positive but the easiest way (that can also be applied to more complex polynomials) is to test several values of x. Test one value less than -5, one value between -5 and 2, and one value greater than 2. We'll test -10 first. (-10)$$^2$$+3(-10)-10 = 100-30-10 = 60 Since 60 is a positive number we can know that the polynomial has positive values when x is less than -5. Next we'll test 0. Zero is always a nice, easy number to pick as a test value for problems like this! 0$$^2$$+3(0)-10 = -10 Since -10 is a negative number we can know that the polynomial has a negative value when x is between -5 and 2. Lastly we'll test 3. 3$$^2$$+3(3)-10 = 9+9-10 = 8 Since 8 is a positive number we can know that the polynomial has positive values when x is greater than 2. We have now identified the two ranges where x$$^2$$+3x-10 > 0. The preferred way to write the solution is to write it as the union of the two ranges: (-$$\infty$$,-5) $$\cup$$ (2,$$\infty$$)

### Subject:Trigonometry

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Question:

Let ABC be a triangle with vertices A, B, and C and sides a, b, and c. Assume that side a is opposite angle A, side b is opposite angle B, and side c is opposite angle C. Use the Law of Cosines to find the length of side a if: c = 30; b = 15 and $$\angle$$ B = 20$$^{\circ}$$

Inactive
Kelsey E.

The Law of Cosines is an equation that relates the three sides of a triangle and one of the angles of the triangle. There are three ways to write the Law of Cosines based on which angle of the triangle you are working with. To ensure that our equation only has one unknown (side a), we will use the form of the Law of Cosines that uses angle B since that value is known. b$$^2$$ = a$$^2$$+c$$^2$$ − 2ac* cos B If we substitute the values of our triangle, we get: 15$$^2$$ = a$$^2$$+30$$^2$$ − 2a*30 cos 20$$^{\circ}$$ 225 $$\approx$$ a$$^2$$+900 − 60a*0.94 0 $$\approx$$ a$$^2$$− 56.4a+675 We can plug the coefficients into the quadratic formula to compute two possible values for a: a $$\approx$$ 39.17 or a $$\approx$$ 17.23 Note that it is valid to have two possible values for side a. In this problem there are two possible triangles that meet the requirements. Angle A can be obtuse which causes side a to be approximately 39.17 or angle A can be acute which causes side a to be approximately 17.23.

### Subject:Algebra

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Question:

Vivian and Noelle both leave the park at the same time, but in opposite directions. If Noelle travels 5 mph faster than Vivian and after 8 hours, they are 136 miles apart, how fast in miles per hour is each traveling?

Inactive
Kelsey E.

Let v be the rate (in miles per hour) that Vivian travels. Then v+5 is the rate (in miles per hour) that Noelle travels. Since distance = rate*time, in 8 hours Vivian travels 8v miles and Noelle travels 8(v+5) or 8v+40 miles. Since their total distance is 136 miles we get: 8v+(8v+40) = 136 16v+40 = 136 16v = 96 v = 6 Thus, Vivian travels 6 miles per hour. Since Noelle travels 5 miles per hour faster than Vivian and 6+5 = 11, then Noelle travels 11 miles per hour.

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