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Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!
Over 15 years teaching at the college level.
NCTM, NYSMATYC, AMATYC, MAA, NYSUT, AFT.
B.S. in Mathematics from Rensselaer Polytechnic Institute
M.S. (and A.B.D.) in Applied Mathematics from SUNY @ Stony Brook
| User | Date | K | C | T | P | Comments |
|---|---|---|---|---|---|---|
| Warm | 08/14/09 | 10 | 10 | 10 | 10 | Thanks |
| Jason | 07/22/09 | 10 | 10 | 10 | 10 | Thank you |
| shikhin | 07/20/09 | 10 | 10 | 10 | 10 | Thanks a lot....got the point.....I ..... |
| Jason | 07/09/09 | 10 | 10 | 10 | 10 | Thank you very much |
| Attir | 07/09/09 | 10 | 10 | 10 | 10 | very nice i really appriciate your fast ..... |
Hello from the US! :-) Use the sum and difference formulas for tangent: tan(30+a)=[tan(30)+tan(a)]/[1-tan(30)tan(a)]=[sqrt(3)+3tan(a)]/[3-sqrt(3)tan(a)] tan(60-a)=[tan(60)-tan(a)]/[1+tan(60)tan(a)]=[sqrt(3)-tan(a)]/[1+sqrt(3)tan(a)]
Hello, Since (1,6) satisfies y=ax^2+bx+c, we get: 6=a+b+c...since the tangent line at x=1 is horizontal, that means the turning point is at x=1 (and y=6), so we can write the parabola as y=a(x-1)^2
Since the area of a triangle can be written as: A=(1/2)(a)(b)sin(C), where C is the angle between the two sides. So, the maximum area will be when sin(C) is as large as possible. Thus, C=90 degrees
1. What is happening to f(x) or f''(x)??? 2. You want to maximize f'''(x)? So, take the derivative of f'''(x), . which is f''''(x)=24, but f''''=24 and is never zero. So the . maximum and minimum
1. y=x^2-2x has vertex at x=-b/(2a)=2/2=1, so x=1 and y=-1 ==> (1,-1) 2. Basically, if you arrange the equation for a parabola in this form: (x-h)^2=4p(y-k) The vertex is (h,k); the focus is (h, k+p)

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