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# Tutor profile: Paul G.

Paul G.
Masters Degrees in Applied Statistics and in Mathematical Education

## Questions

### Subject:Geometry

TutorMe
Question:

Why do the three angles of a triangle sum to 180 degrees?

Paul G.

The three angles of a triangle sum to 180 degrees because we accept the Euclidian postulates of parallelism into our system of geometry. One such postulate is, for every point of a line there exists one line through that point parallel to that line. And also, when a transversal intercepts two lines, the lines are parallel if and only if corresponding angles (and alternating interior angles) are congruent. It may be surprising to you that I'm talking so much about parallelism, but it is here that we can answer the question of why! Now take a triangle and draw through any vertex a line parallel to the opposite side. Now the other two sides of the triangle are transversals through two parallel lines. So alternating interior angles are congruent. Therefore, the three angles inside the triangle are the same as the three angles along the line! And any three angles along a line are equal to 180 degrees!

### Subject:Statistics

TutorMe
Question:

What is the P-Value for the outcome of 61 heads when a quarter is flipped 100 times?

Paul G.

A P-Value is a probability of getting what was observed or some value more extreme given the Null Hypothesis is true. Implicitly, the Null Hypothesis here is that a fair coin was flipped in a fair way 100 times. Therefore, outcomes as extreme or more extreme than 61 are 62, 63, 64...100 and also, 39 (since 39 heads are as extreme as 61) 38, 37, 36...0 The values are that are less extreme, of course, are the numbers from 40 to 60. So, on a Ti-84 calculator: 2*binomialcdf(100,.5,39) = .0352, the P-Value for the outcome of 61 heads when a quarter is flipped 100 times.

### Subject:Algebra

TutorMe
Question:

Consider the circles x^2+y^2=1 and (x-4)^2+y^2=4. What is the slope of the equations of the four tangent lines to these two circles?

Paul G.

y=mx+b substituted into x^2+y^2=1 creates x^2+ m^2 x^2+2mxb+b^2=1. Now, if the line is to be tangent, the discriminant has to equal zero. For purposes of solving this problem with the quadratic formula (notice the “b’s” are “different”) , a=m^2+1,b=2mb and c=b^2-1. So, since the discriminate must be zero for the lines to be tangent, 〖(2mb)〗^2-4(m^2+1)(b^2-1)=0. So, 4m^2 b^2-4m^2 b^2+4m^2-4b^2+4=0, which concludes with the nice m^2-b^2+1=0 Apply the same idea to (x-4)^2+y^2=4 with y=mx+b substituted into the y^2: x^2-8x+16+m^2 x^2+2bmx +b^2=4. For purposes of solving this problem with the quadratic formula (notice the “b’s” are “different”) , a = m^2+1, b=2bm+8, and c=b^2+12. So, since the discriminate must be zero for the lines to be tangent, (2bm+8)^2-4(m^2+1)( b^2+12)=0. So, 4b^2 m^2+32mb+64-4m^2 b^2-48m^2-4b^2-48=0 which simplifies to 8mb-12m^2-b^2+4=0. The conclusion from the first paragraph allows us to say 〖b^2=m〗^2+1. So, 8mb-13m^2+3=0. Then, b=(13m^2-3)/8m, and b^2=(169m^4-78m^2+9)/(64m^2 ), so m^2+1=(169m^4-78m^2+9)/(64m^2 ), so 64m^4+64m^2= 169m^4-78m^2+9 and finally we have an equation in quadratic form using have only m, the variable for the slopes, 105m^4-142m^2+9=0 Since it is 4th degree, we know it has four solutions, the four slopes we can visualize in the geometry. m = ±3/√7 and ±1/√15.

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