steve.muench

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steve.muench last won the day on December 8 2016

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About steve.muench

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  • Gender
    Male
  • Exams
    May 2015
  • Country
    Italy

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  1. Just try typing the link into your browser in a new tab instead of clicking on the link in this page and it will work.
  2. It happens for the new link I posted as well. Going to report the problem to the ibsurvival administrators. Seems like a general problem.
  3. Something seems to have hacked the initial link I provided in this posting. Clicking on the original link I supplied now redirects so some odd survey. Trying again to post the link to the notes: bitly.com/ib-sl-maths-review-notes
  4. Happy they are useful to you. Good luck next year!
  5. Glad the notes are continuing to prove useful to students around the world.
  6. Let me know if you find any errors I can correct, or anything that could be made more clear.
  7. Happy they are proving useful to many students. Good luck at uni.
  8. Glad you all have found the notes useful. At the present time I have not had time to expand them to include Maths High Level topics yet, but maybe I will find the time over the summer or in the fall to do that, as well as to provide additional worked test solutions.
  9. Now that the November 2015 past papers are available for purchase on the http://www.follettibstore.com/main/home I've published my worked solutions for the November 2015 exam. I hope they might be helpful for some of you: https://stevemuench.wordpress.com/2016/04/08/worked-solutions-for-ib-maths-standard-level-november-2015/
  10. May 2014

    Remember the law of "change of bases" for logarithms, which says: log a c log a = ____________ b log b c This lets you re-rexpress any log into a new base "c". If you use "a" for "c" here, you get: log a a 1 log a = ____________ = ______________ b log b log b a a So, in general if you flip-flop the base and the argument of the log function, you get the reciprocal of what you started with. Try using this idea on the left hand side of the equation in the problem, then find a common denominator to add the two fractions together. Then, a consequence of the Pythagorean theorem will let you further simplify, and then you can do one more round of base/argument flip-flopping to get the final answer. Hope this helps.
  11. Your calculator can also compute the definite integral of the absolute value of the function that gives the rate of change. See section 7.28.2 in my notes: http://bitly.com/ib-sl-maths-review-notes
  12. You can try my notes I've published here: https://stevemuench.wordpress.com/2015/04/21/review-notes-for-ib-maths-standard-level/ Its cover page includes links to a two fully explained solutions of two recent past papers. Hope it helps. Good luck.
  13. In case it might prove useful, I've posted worked solutions for the May 2015 Maths Standard exam (TZ2) here: https://stevemuench.wordpress.com/2015/10/06/worked-solutions-for-ib-maths-standard-level-may-2015-tz2/
  14. Now that the May 2015 past papers are available for purchase on the ibo.org website, I've published my worked solutions for the May 2015 Timezone 2 tests my daughter's class did last may. I hope they might come in handy for some of you: https://stevemuench.wordpress.com/2015/10/06/worked-solutions-for-ib-maths-standard-level-may-2015-tz2/
  15. I posted worked solutions to the November exam here: http://bit.ly/ib-sl-maths-nov-2014 Hope this helps.