ONLINE, ON COURSE


I received the following letters the past few weeks.  They made my day.

Thank you so much, I was struggling and your answer made it simple and understandable. UR GR8.

You are Amazing!

You have helped me in the past and always have accurate answers, I am so grateful you took your time out to help me today, thank you, I appreciate it so much!

Thank you, from the bottom of my heart, you have written this in a way that I totally understand. I will differently (sic) praise you to all my friends and family. God Bless you,

What did I do to deserve these?

Reason for the first:

The area of a triangle is 30 sq in. The base of the triangle measures 2 in more than twice the height of the triangle. Find the measures of the base and height. 

The area of a triangle is (1/2)*base*height.  Remember that?  Therefore, the base*height must equal 60 for the area to be 30.  Let the height= x inches, then the base is 2x+2, two more than twice the height.  Then, 

x*(2x+2)=60, 2x^2+2x=60, and dividing by 2, x^2+x=30.  We can write that as x^2+x-30=0 and factor it as (x+6)(x-5)=0.  That is 0 if x=-6 (not possible for a length) or if x=5.  So, the base is 5 inches and the height is 2(5)+2, or 12 inches.

For the last comment,  I answered the following, taking about a minute in my head, writing it down as I thought.

Write the slope-intercept equation for the line that passes through (-12, 10) and is perpendicular to 4x + 6y = 3. 

One gets the slope first by rewriting the equation as 6y=-4x+3 and dividing by 6 to get y=-(2/3)x+1/2.  The slope is -2/3. The perpendicular line has a negative reciprocal slope.  Turn the fraction over (-3/2) and change the sign (3/2).  That is the slope of the perpendicular line.   Using the point slope formula where we know the slope and a point, x=-12, y=10, y-y1= (3/2) (x-x1).  That is y-10=(3/2)(x+12).  This becomes y-10=(3/2)x+18, and finally y=(3/2)x+28.  Also, y=mx+b, so 10=(3/2)((-12)+b.  That is 10=-18+b, so b=28.  Both methods work; the more ways one knows, the more ways to explain it to students.  One of the ways is likely to stick.

For the one who called me amazing?

Find the accumulated value on an investment of $15,000.00 for 9 years at an interest rate of 11% if the money is compounded 

a) Semi- annually b) Quarterly c) Monthly d) Continuously 

 Here, one uses the formula 

Principal=Starting Principal{1+ rate/compounding per year} raised to (the number of years*compounding per year). P=Po{1+r/t}^nt.  Semi-annual is P=Po{1+(0.11/2)}^18, because it compounds twice a year and there are nine years.  This is $39,322.  For continuously compounding, it is easier, P=Po*e^rt.  e^rt= e^(0.99), because 9*11%=0.99.  Po*e^0.99=$40,368.52.  Continuously compounding gives you more money, although the difference between it and monthly is only $200 less than continuously.  The last formula allows one to prove that the doubling time of money in years is 70 (or 72, which is easier to work with) divided by the interest rate in per cent.  I grew up in the age before calculators, and we had to do this by logs.  On a calculator, it takes about 15 seconds. Dividing 72 by 11 gives a doubling time of about 6 1/2 years, so $15,000 should double once and be well on its way to doing it again.  The answer makes sense.

This is an online math help site.  More than 2000 tutors take part, some of whom have solved one problem, one nearly 70,000.  I’ve solved 2000.  About one in four thanks me.  That’s nice.

Several tutors offer their services for pay, $1 per answer, $2-$5 to show the work.  I do it to relax.  Yes, relax.  This stuff is fun for me, and I have learned the easier the problem for me, the more grateful people tend to be.  I don’t need to hear anything, unless my answer is wrong or not understandable.  I’m there to help.  I don’t know names; I do know I have helped parents help their children.

I’ve learned much.  It has been a great review of my statistics, I now deal with ellipses better, and I understand geometric series better than I ever have before.

I usually want a challenge, so I choose what I want to solve.  I have a big advantage:  I grew up in the era of no Internet, Chemical Rubber Company tables of integrals, no calculators, only log tables to do complex calculations.  In other words, I learned math from first principles, from the ground up.  Yes, it helps to have a genetic ability to do this stuff.  I can’t play the violin, but I can find the vertex of a parabola mentally and write it in three different forms.  Kids need someone to help them understand how to do it, not in their head, but to allow them to understand these and similar problems.

The current list has perhaps 50 problems, and I often work down it until I find a problem I feel like doing.  If interested, I go to the list of unsolved problems.  Last I checked, statistics had about 40,000.  A lot of those are tough, and if I don’t have pen or paper around, I don’t do them.

When I tutor at the community college, I answer algebra questions online while waiting for non-virtual students to ask for help.  I guess I am volunteering, but I am having a lot of fun.  It’s nice to lay out quickly an answer in simple form for a person who is struggling.

The other day at the CC, I was asked to go into the higher level math room to help out.  That was a compliment, because I was felt to be good enough to help out there. I’m the go-to guy for statistics.  The other tutors are really smart, yet all of us at one time or another have trouble with something.  I may struggle at the high levels, but I often find myself pulling stuff out of the air from the past and making sense out of it.  Or better yet, I ask a student where he got a specific term in an equation.  The student looks puzzled then suddenly says, “Oh, wow, I didn’t see that before.  OK, I understand.  Thanks a lot.”  And he leaves.

I hadn’t a clue how to solve the problem, but I think I helped him.

Math is mentally taxing.  After doing about a dozen problems, I take a break.  It helps me later solve troublesome problems.  In the math lab, I have concentrated so deeply that one day when I walked out of the room, I forgot whether it was Tuesday or Friday.

I think the absent-minded professor was probably working overtime on a difficult problem.

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