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Tuesday, June 5, 2012

FINAL EXAM REVIEW

1) Unit 9 Cumulative Review
2) MHS Review Packet (handed out in class)
3) Apart from the two assignments above, AND as I have stated in class --- I feel that the BEST use of your time would be to "thumb thru" the individual chapter reviews in our text book, re-familiarize yourself with some the concepts and do a few problems here and there. All of the answers are in the text book.


TEXT BOOKS SHOULD BE RETURNED ON THE DAY YOU TAKE THE FINAL EXAM!!!

26 comments:

  1. Hi Mr. C i have a question on #12 in the packet is there enough information to solve it

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    Replies
    1. I agree that the wording is a little odd... this is the "slope as a rate of change" concept again. If you recognize that x is the number of hours, then clearly she is being paid $7.50/hr. For some reason she was given a starting amount of $40 that the word problem doesn't explain.

      I think (hope) you'll find the questions on the final more self-explanatory.

      Delete
  2. I think 31 is similar its also confusingly worded

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    1. I'll choose to disagree with you on this one...

      The amount of change you get from a $20 bill is a FUNCTION of the number of lbs of pears you buy at .79/lb.

      Subtract the cost of pears from $20 to calculate your change, right?

      c(x) = 20 - .79x

      Voila!

      Delete
  3. Mr. C
    can you re explain exponential growth and decay to me plz

    ~Danie

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    Replies
    1. exponential growth is when a>1 and b>1 while exponential decay is when a>1 while b<

      Delete
    2. Quite honestly, your time is MUCH BETTER SPENT nailing down slope concepts and factoring. Do as you will.

      This link does better than I can in 10 seconds.

      http://www.regentsprep.org/regents/math/algebra/AE7/ExpDecayL.htm

      ENJOY!

      Delete
  4. hello Mr C! I was just studying when I came across problem 23 in the packet. I am unsure what to do!! please help
    thanks

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    1. IMPORTANT QUESTION (hmmm... why would I say that... what might I know?????)

      TABLES, EQUATIONS, AND GRAPHS, OH MY!

      The DOMAIN is the list of x-values.
      The RANGE is the list of y-values.

      Make a table of values with y=3(4^x) as the function.

      Plug in the 5 x-values (aka the DOMAIN)

      If x is -2, y=3(4^-2) = 3(1/16) = 3/16
      If x is -1, y=3(4^-1) = 3(1/4) = 3/4
      If x is 0, y=3(4^0) = 3(1) = 3
      If x is 1, y=3(4^1) = 3(4) = 12
      If x is 2, y=3(4^2) = 3(16) = 48

      Can you graph the ordered pairs?

      Capeesh??

      Delete
  5. Mr. C I am having a very tough time on how to solve #38, #43 and # 44 on the MHS review packet. Could you please help????

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    1. For #38 (So many ways to do this...)

      you are multiplying fractions, so if you like you can create one big numerator with multiple FACTORS and do the same for the denominator.

      THEN, you could rearrange, such that the 5^4 was over 5^3, which would simplify to 5, yes?

      The 3^-2 would travel to the numerator and become 3^2, yes? THEN you have 3^2 over 3 which simplifies to 3.

      2^5 over 4^3 is 32/64 or 1/2, yes?

      So in the end, you have 5(3)(1/2) = 15/2.
      Capeesh?

      Delete
    2. #43 is hinting that (2x+1) is a factor of 6x^2 - 5x - 4.

      You can confirm this by factoring the quadratic.

      If you don't know how to "box and diamond" a quadratic at this point, it's really too late for me to help. The length will simply be the "other" factor, which in this case is (3x-4).

      Looking at it another way...

      If I had asked you what do you have to multiply (2x+1) by to get 6x^2 - 5x - 4, it should(?) be obvious that you would need a 3x and a -4 to make it work.

      Then just try it and see if the resulting quadratic is obtained... i.e. (2x+1)(3x-4) does in fact equal 6x^2 - 5x - 4.

      YAY!

      Delete
  6. Mr. C, i haveing problems with quite a few things and i think its because i just can't remember how to do everything.the numbers i'm having problems with are 12,13, 18, 20, 23, 36,38, 41, 43, 44, and 49. is it possible to help me? thanks!

    ReplyDelete
    Replies
    1. #12 is answered above
      #13 can be answered with a TOV... plug it into the graphing calculator to check your work
      #18 What's changing? What stays the same (i.e. constant). The cost increases at an hourly rate from a BASE (constant) charge of $10
      f(h)= 8h +10

      The slope = 8.
      #20
      f(x)= 25x + 750
      If you buy 100 books, the cost will be as follows:
      25(100) + 750 = 3,250
      That works out to an average of $32.50 per book, yes?

      If you buy 400 books, the cost will be as follows:
      25(400) + 750 = 10,750
      That works out to an average of $26.88 per book, yes?

      Can you subtract 26.88 from 32.50 to calculate the savings per book?

      Delete
  7. i agree problem 44 is confusing
    for those confused on 23 the domain is the x values and the range is the y values. plug in the domain for x and solve for y and you will get a bunch of points that you can graph.

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    Replies
    1. the thing is, i did, that but part of the graph looks odd that's why i am unsure on if the graph is correct or not.

      Delete
  8. I am having trouble with problems 28, 33 and 43 in the packet.

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    Replies
    1. Mr. C said not to worry about #28. He explained why in class.

      Delete
  9. I'm having trouble with number 49 in the packet

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    1. #49

      6x^4 - 21x^3 - 12x^2

      HUGE FIRST STEP THAT YOU NEED TO RECOGNIZE HERE (THE ONE EVERYONE ALWAYS FORGETS)... LOOK FOR A GCF!!

      AH-HAH! 3x^2 is a factor in each term, so our first step yields:

      3x^2(2x^2 - 7x - 4)

      NOW, you have a quadratic trinomial that may be factorable... box'ed properly, you will be asked to find two numbers that multiply to -8 and sum to -7.

      The fully factored expression is:

      3x^2(2x+1)(x-4)

      Too easy, right?

      Delete
  10. For number 28 how do you figure out how long it takes the tool box to hit the ground

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    1. You meant #48, yes...

      h= height
      t = time in secs
      300 is the starting height

      h=-16t^2 + 300

      The ground is h=0, SO

      0 = -16t^2 + 300 (subtract 300 from both sides)
      -300 = - 16t^2 (divide both sides by -16)
      t^2 = 18.75 (now, take the square root of both sides)
      t = plus-or-minus 4.33

      Throw out the -4.33 since time can't move backwards (aka, -4.33 is an EXTRANEOUS SOLUTION)

      Therefore, it takes a little over 4 seconds for the tool box to hit the ground... RUN FOR IT!!

      Delete
  11. Q. HOW LONG HAVE YOU GUYS HAD THIS PACKET?
    A. about a week

    Q. How many students showed up for extra help after school today?
    A. about ONE

    ReplyDelete
  12. Mr. C-

    I have my textbook still, i will return it in to you tommorrow during school. Sorry for the delay!

    ReplyDelete