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I can answer any questions from the standard four semester Calulus sequence. Derivatives, partial derivatives, chain rule, single and multiple integrals, change of variable, sequences and series, vector integration (Green`s Theorem, Stokes, and Gauss) and applications. Pre-Calculus, Linear Algebra and Finite Math questions are also welcome.
Ph.D. in Mathematics and many years teaching undergraduate courses at three state universities.
B.S. , M.S. , Ph.D.
| User | Date | K | C | T | P | Comments |
|---|---|---|---|---|---|---|
| Kenneth | 10/04/09 | 10 | 10 | 6 | 10 | Thanks for the reply and answer! |
| Xue Ting | 08/30/09 | 10 | 10 | 7 | 10 | |
| Anthony | 05/21/09 | 10 | 10 | 10 | 10 | thanks Socrates |
| bina | 03/26/09 | 10 | 10 | 8 | 10 | |
| Sombra | 02/22/09 | 10 | 10 | 10 | 10 | Thank you so much, you made it ..... |
$15 for 10 people for 10 days would work out to .15 per person per day , so that's not right $150 for 10 people for 10 days would work out to $1.50 per person per day , so this is one posible answer
" x + x^2 + x^3 + ....+ x^n approaches x/(1-x)" x + x^2 + x^3 + ....+ x^n + ..... is a geometric series , with sum x/(1-x) , you should have seen this before you studied Mclaurin series. If not
(Fn(x))/x = 1 + 4x + 9x^2 + ....+ (n^2) (x^(n-1)) = (x + 2x^2 + 3x^3 +....+ nx^n)' = ((x)( 1 + 2x + 3x^2 + ...+ nx^n-1))' = ((x) (x + x^2 + x^3 + ....+ x^n)' )' Since x + x^2 + x^3 + ....+
Your work is correct as far as it goes. The left side is cos A + 1 , as you have found . But the right side is sin^2/1-cos A = (1 - cos^2 A )/ (1 - cos A ) Factor the numerator (1 - cos^2 A )
The common denominator is 12ab (2a+3b)/3a = (4b)(2a+3b)/12ab = (8ab + 12b^2)/12ab (a-2b)/4b = (3a)(a-2b)/12ab = (3a^2 - 6ab)/12ab (a^2+b^2)/ab = (12a^2+12b^2)/12ab So (2a+3b)/3a - (a-2b)/4b
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