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# Tutor profile: Greg B.

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Greg B.
Ex-petroleum engineer tutoring math and science
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## Questions

### Subject:Pre-Calculus

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Question:

Find the Foci coordinates for the given hyperbola: 1 - ((x^2)/ 64) = - (y^2) / 36

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Greg B.

First we should try to simplify our equation to match the standard format of a hyperbola: 1 - ((x^2)/ 64) = - (y^2) / 36 1 = (x^2)/64 - (y^2) / 36 (x^2)/64 - (y^2) / 36 = 1 Now we can determine our foci by using the foci equations: c ^ 2 = a ^ 2 + b ^ 2 where a is from (x^2) / (a^2) and b is from (y^2) / (b^2) from our hyperbola formula c ^ 2 = 64 + 36 c ^ 2 = 100 c = + or - sqrt(100) c = + or - 10 From the definition of our focus from a hyperbola, we know that the foci will be in the following coordinates: F ( c , 0) & F` (-c , 0) F ( 10 , 0) & F` ( -10 , 0) (10 , 0) and (-10, 0)

### Subject:Geometry

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Question:

Determine the formula to find the area of a perfect 6-point star ( all exterior angles in a 6 point star are 60 degrees and the length of each line are the same)

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Greg B.

Given this unique shape, we can divide it up into shapes we know then sum up the areas of all those in order to create a formula for the perfect star. First, we can form an equilateral triangle at each point by connecting a line at the base of each point ( /\ becomes ∆ ). We know this triangle must be equilateral, because of combined 3 reasons: The summation of all the angles in a triangle is 180 degree, Angle 1 is 60 degrees, and Angle 2 and Angle 3 are equal ( because the corresponding sides are of the same length). This will give us six equilateral triangles and a hexagon in the center of the star. Now we can also divide up this hexagon again into another 6 equilateral triangles by drawing a line from each point on the hexagon to the center point. Again we can determine these triangles are equilateral as well by the following reasons: 360 degrees / 6 = 60 degrees, in order for the point to be in the exact middle all the lines must be equal lengths ( this also means each angle created from the drawn lines is 60 degrees as well). These triangles will also be the same area as the exterior triangles, as the bases are the same length, giving us 12 equilateral triangles. Now we can simply find the formula for the area of one of the triangles, then multiply it by twelve and come up with our solution: Area of ∆ = (1 / 2 * b * h ) * 12 = 6 * b * h We can further simplify by solving for h using Pythagorean's theorem a^2 + b^2 = c^2 on the equilateral triangle: a = 1/2 (base) ; b = h ; c = base which gives us: (1/2 b) ^2 + h^2 = b ^ 2 1/4 b^2 + h^2 = b^2 h^2 = b^2 - 1/4 b^2 h^2 = 3/4 b^2 h = sqrt ( 3/4 b^2) h = b/2 * sqrt(3) Finally we can plug our value for h into our area equation: 6 * b * h = 6 * b * b / 2 * sqrt(3) = 3 * b^2 * sqrt(3) [ 3* b^2 * sqrt(3) where b = length of line in the star ]

### Subject:Algebra

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Question:

Let's define two functions f(x) and g(x): f(x) = x^2 and g(x) = |x| . Over what intervals is f(x) greater than g(x) and less than g(x)?

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Greg B.

There are a few ways to solve this problem. The first would be to arbitrarily plug in numbers but we could miss one of the intervals in this problem. Instead, we should set f(x) = g(x) , which will give us the intersect points and help us define our intervals. f(x) = g(x) x ^ 2 = |x| x = + or - sqrt( |x| ) Now we simply need to determine what numbers we can plug into the above equation in order to make the left hand side of the equation equal to the right hand side. With the square root, the only numbers that would satisfy this equation are 1, 0, and -1 1 = + or - sqrt ( |1| ); 1 = + or - 1; 1 = +1 Correct 0 = + or - sqrt ( |0| ); 0 = 0 Correct -1 = + or - sqrt ( |-1| ); -1 = + or minus sqrt(1); -1 = + or - 1; -1 = -1 Correct These points will define where our two functions intersect, and therefore where the functions overtake one another. So simply by plugging in a few numbers between our points, we can determine our intervals: f(2) = 2 ^ 2 = 4 ; g(2) = |2| = 2 ; f(x) > g(x) f(0.5) = 0.5 ^ 2 = 0.25 ; g(0.5) = | 0.5 | = 0.5 ; g(x) > f(x) f( -0.5) = (-0.5) ^ 2 = 0.25 ; g(-0.5) = | 0.5 | = 0.5 ; g(x) > f(x) f(-2) = (-2) ^ 2 = 4 ; g( -2) = | -2 | = 2 ; f(x) > g(x) Now we know our can define our intervals: From { - ∞ to -1 } : f(x) > g(x) From {-1 to 0} : g(x) > f(x) From {0 to 1} : g (x) > f(x) From {1 to ∞ } : f(x) > g(x)

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