AboutScotto Expertise Any kind of mathematics (calculus, analysis, game theory, linear approximation, finite differences, linear regression, linear programming, numerical analysis, probability, statistics, etc.).
I also have answered some questions in
Physics (mass, momentum, falling bodies),
Chemistry (charge, reactions, symbols, molecules), and
Biology.
Experience Experience in the area: I have tutored students in all areas of mathematics for over 20 years.
Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors.
Awards and Honors: I have passed Actuarial tests 100, 110, and 135.
Publications Maybe not a publication, but I have respond to well oveer 3000 questions on the PC.
That's around 2,000 in basic math and 1,000 in advanced math.
Education/Credentials I aquired well over 40 hours of upper division courses. This was well over the number that were required.
I graduated with honors in both my BS and MS degree from Oregon State University.
I was allowed to jump into a few junior level courses my sophomore year.
Awards and Honors I have been nominated as the expert of the month several times.
All of my scores right now are at least a 9.8 average (out of 10).
Past/Present Clients My past clients have been students at OSU, students at the college in South Seattle,
referals from a company, friends and aquantenances, people from my church, and people like you.
Question How can calculate kg into meter i. e. we are giving guage, size? OR
what is meter of 1 Kg oply film having 100 Micron & 1000mm Width?
Answer If I understand the question, we have 1 kg of oply film.
The film is 100 microns thick and 1000 mm wide.
The question is how much film would this make?
What is needed to get the solution of this problem is as follows.
1. What is an 'oply'? (a mistake? an o-ply? ...)
2. What is the density of the film?
The film width is 1000mm, which is 1 m.
The film thickness is 100 microns,
which is 0.1 mm, which is 0.0001 m.
This gives us a cross sectional area of (1)(0.0001) mē,
which is 0.0001 mē.
Multiply this by the length L of the strip gives the volume
0.0001 L m^3.
Multiply this amount by the density D of kg/m^3 and
the answer is 0.0001 L D kg.
When D is known, this equation can be set to 1 kg and L can be solved for. The solution is
