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Christian C. How is it different to take the flux integral of a vector field over a full spherical region rather than a hemispherical region? Projecting both shapes results in identical shapes and both normals are the same, resulting in identical triple integrals when the integral is reduced to a simpler form us… Read more »ing the differentiation theorem (I believe it is called? Please don't hesitate to correct me...) Since both integral sets end up being the same, by my logic their numerical results should also be identical which is clearly glaringly wrong. How then do we get the flow of a full sphere? Thanks a million, hope this isn't a waste of your time!!!
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Joe W. In Questions 9-10 of the exercises for the Squeeze Theorem, they multiply the left and right elements in the inequality by an ABSOLUTE VALUE, but they don't mulitply the middle element by an ABSOLUTE VALUE. For example, in Question 9, it reads:
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- ABSOULTE VALUE (x^3) <= x^3sin 5/x <= ABSOULTE VALUE (x^3)
Why can you multiply the left and right elements by the absolute value but not the middle?Classhive April Thank you for bringing this to our attention. You haven't reached the part of our course that explains this. We changed the questions so that all three parts of the inequality have to be multiplied by the absolute value. This should fix your issue.01/28/2011
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Ashley W. Is anyone else taking Thinkwell's Trigonometry course? Or is anyone on here good at Trig and have the time to answer a couple of questions? I was doing fine in the course but I have hit a road block and cannot understand a concept.
~Ashley :) ~ -
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Vicki G. Lesson 3.3.1, Determining Components and Degree -
question #7: shows the correct answer as 6, which is the degree without simplifying, but the other questions # 10 and 11 show the correct degree after simplifying. Why is #7 correct answer not the degree after simplifying? I think Thinkwell didn't think so well on this one. -
Andrew St. John For the pre-calculus class i took all you would need was a good Algebra foundation and some geometry.
01/10/2011
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Alex B. if anybody is taking college algebra on thinkwell can you pleaseeee tell me where the homework link is? i cant find it anywhere
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Melissa T. Diving into calculus at the start of a new term....YAY!!! :-) LOVE Prof. Burger!
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Aliesha V. Ummm...I'm kinda new to this whole Classhve thing. Any help? :)
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Kevin C. Hi this is Kevin and it is a little bit troubling trying to browse through all these sites.
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Jade M. what are the discussions all about. I am missing all grades in them
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Sue C. Hello everyone this is my first semester her at WGU and I am a lil nervous. And right now math is kicking my caboose...Cynthia Balderson Aloha Sue! I am finally at "that" semester where I have to face the math! I've been loving the Thinkwell Pre-Algebra course, but came on here because I obtained the wrong answer on the chapter review of Integers, #25. If you have done it, could you tell me what answer you get? I… Read more »039;ve worked it twice and I don't get 3!
09/21/2010
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Ken F. anyone have any suggestions for what math to take (not ness. from Thinkwell) after BC Calc
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Chad F. I need a little bit of help with the squeeze theorem. I fully understand the theorem but watching the lecture and then working through the example problems, I don't know where they are getting the information in which to sqeeze the answer of the existing problem. Ex. Lim as x approaches 1 of… Read more » |X-1|sin squared times x +1. It just seems like some of the numbers are pulled out of the thin air. HELP...math helper It sounds like the part of the process that you are having trouble with is the algebraic manipulation of the chosen known inequality into an inequality involving the expression from the original limit. And in a way, those steps used to manipulate the inequality are pulled out of thin air. In other w… Read more »ords, there many be different ways to manipulate the inequality, not just one set of corresponding steps. However, there is usually one method that is easier than all others.
For example, suppose you are asked to find a limit of sqrt[x](2(sin x) - 1). Since the expression in the original limit involves sine, start with some known inequality involving sine, such as -1 <= sin[x] <= 1. Now manipulate this inequality (using rules of algebra) until you get sqrt[x](2(sin x) - 1) in the middle.
-1 <= sin[x] <= 1
-2 <= 2sin[x] <= 2 Multiply by 2.
-3 <= 2sin[x] - 1 <= 1 Subtract 1
-3sqrt[x] <= sqrt[x](2sin[x] - 1) <= sqrt[x] Multiply by sqrt[x].
Now take the limits of the two outer functions, -3sqrt[x] and sqrt[x], and if they are equal to the same number, then you can use the Squeeze Theorem to assume that the limit of sqrt[x](2sin[x] - 1) is also equal to that number.08/17/2010
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Christina K. Is anyone else having a problem with the "notes" pages? None of them will load for me today, it just pulls up a black screen.Classhive April Hey Christina! Where exactly are you having this problem? I checked and was able to pull up 2 different notes you had posted previously with no problems. Let me know if this is still happening!
08/12/2010
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Christina K. I found a cool site for graphing ellipses, just click on where it says "click here to start" http://www.analyzemath.com/EllipseEq/EllipseEq.html
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Classhive April It's because when we tallied the winners for the contest, we removed any negative thumbs-down votes, which increased everyone's rep points scores. We did this because several people including Christina were targeted by a group of members who were giving thumbs-down to everyone regardless o… Read more »f their posts. While we removed the points for the contest, I don't believe we removed the points from your total on the site as it would be a huge amount of work to do this for every member and code it into the site.
07/06/2010
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