Recent content by jasper10
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Graduate Optimizing Level of Output for Profit-Maximizing Firm
A firm has the following total-cost and demand functions: C = aQ^3 - bQ^2 + cQ + d Q = e - P (d) Find optimizing level of Q. (e) Chooses a,b,c,d and e such that there is only one profit-maximizing level of output Q. I found 2 solutions for question d, but in a very long and messy... -
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Probability - proving independence/dependence
Homework Statement A film is defective either when the level of sensitivity is wrong (defect D1) or when the colours are faulty (defect D2). 2% of all films made have at least one of these two defects. 1% of all films have defect D1 and 0.2% of all films have both defect D1 and D2. Are...- jasper10
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- Probability
- Replies: 5
- Forum: Precalculus Mathematics Homework Help
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Solving 1 + 2cos(2x+ π/3) for Zeroes: Step-by-Step Guide
Ok! thanks, i kind of get it now!- jasper10
- Post #19
- Forum: Calculus and Beyond Homework Help
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Solving 1 + 2cos(2x+ π/3) for Zeroes: Step-by-Step Guide
How does one find the place where cosine is equal to -1/2? and as you said, the solution is in the third quadrant and the first solution (Pi/6) is in the first.- jasper10
- Post #15
- Forum: Calculus and Beyond Homework Help
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Solving 1 + 2cos(2x+ π/3) for Zeroes: Step-by-Step Guide
How do you know this?- jasper10
- Post #13
- Forum: Calculus and Beyond Homework Help
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Solving 1 + 2cos(2x+ π/3) for Zeroes: Step-by-Step Guide
Yes, my question ultimately is How does one determine these other places where cos(y) = -1/2 ?- jasper10
- Post #7
- Forum: Calculus and Beyond Homework Help
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Solving 1 + 2cos(2x+ π/3) for Zeroes: Step-by-Step Guide
Ok Char. Limit, I edited my answer and showed the transition step. What do you mean by "the sum of angles formula" ?- jasper10
- Post #3
- Forum: Calculus and Beyond Homework Help
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Solving 1 + 2cos(2x+ π/3) for Zeroes: Step-by-Step Guide
Homework Statement Find the zeroes of f(x) = 1 + 2cos(2x+ π/3) Range: (0 ; π) The Attempt at a Solution 1 + 2cos(2x + π/3) = 0 hence, cos(2x + π/3) = -1/2 hence, 2x + π/3 = 2π/3 and, x = π/6 which is 30 degrees can someone help me find the other 0 ? I don't know how to...- jasper10
- Thread
- Replies: 18
- Forum: Calculus and Beyond Homework Help
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Need help integrating (4x - 3) / (x^2 +1) for my Maths Baccalaureate exam!
Ok, thanks Mark44 and tiny-tim!- jasper10
- Post #16
- Forum: Calculus and Beyond Homework Help
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Need help integrating (4x - 3) / (x^2 +1) for my Maths Baccalaureate exam!
European baccalaureate..why? :D- jasper10
- Post #15
- Forum: Calculus and Beyond Homework Help
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Need help integrating (4x - 3) / (x^2 +1) for my Maths Baccalaureate exam!
I don't know, nothing seems to work I'm off to bed now, maybe tomorrow i can think more clearly.- jasper10
- Post #11
- Forum: Calculus and Beyond Homework Help
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Need help integrating (4x - 3) / (x^2 +1) for my Maths Baccalaureate exam!
Integration by substitution surely cannot work! it's a vicious circle and keeps getting more complicated whilst integrating- jasper10
- Post #7
- Forum: Calculus and Beyond Homework Help
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Need help integrating (4x - 3) / (x^2 +1) for my Maths Baccalaureate exam!
Ok so i get: ∫4x / (x^2 + 1) dx - ∫3/ (x^2 + 1) dx = 2ln(x^2 + 1) - ? How would you integrate: 3 / (x^2 + 1)? the annoying part is the "x^2" thanks! ps: what do you mean by "trig" substitution? (I have a feeling this type of question isn't on the syllabus, but i found it on a past back...- jasper10
- Post #6
- Forum: Calculus and Beyond Homework Help
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Need help integrating (4x - 3) / (x^2 +1) for my Maths Baccalaureate exam!
Homework Statement Can someone help me integrate (4x - 3) / (x^2 +1) ? The Attempt at a Solution I'm doing my maths baccalaureate in 2 days and came across this question today! I don't know where to start i.e. which method to use: Integration by parts definitely does not work...- jasper10
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- Dx
- Replies: 16
- Forum: Calculus and Beyond Homework Help
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Why Does the Classical Theory of Light Fail to Explain the Threshold Frequency?
What else would be the characteristics of the classical theory? do you know any simple website? wikipedia is far too complicated and redundant.- jasper10
- Post #3
- Forum: Introductory Physics Homework Help