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cbm

Someone please check. q+12-2(q-22)>0 q+10q-220>0 11q(-220+220)>0 11q>220 q>20

  • one year ago
  • one year ago

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  1. NickR
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    where do you get 10q from?

    • one year ago
  2. cbm
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    12-2, then I distributed by parenthesis

    • one year ago
  3. cbm
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    Is that wrong NickR?

    • one year ago
  4. etemplin
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    you need to distribute first

    • one year ago
  5. theEric
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    \[q+12-2(q-22)>0\]

    • one year ago
  6. theEric
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    That's the start, right?

    • one year ago
  7. theEric
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    @cbm (just making a notification pop up :P )

    • one year ago
  8. jayz657
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    i think you made a mistake there q+12-2(q-22)>0 q+12-2q+44 >0 -q +56 >0 -q > -56 q>56

    • one year ago
  9. jayz657
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    sorry q<56

    • one year ago
  10. theEric
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    I agree with jayz657's work. Any questions, @cbm ?

    • one year ago
  11. cbm
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    Thank ya'll so much! I was reworking it. Yep, once I distributed I got the same answer. The sign at the end changed right? From > to <

    • one year ago
  12. cbm
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    I wish you could give more than one best response

    • one year ago
  13. theEric
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    Yep! Because you multiplied or divided by a negative!

    • one year ago
  14. cbm
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    @theEric thanks!

    • one year ago
  15. theEric
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    My pleasure. I'm glad you got it worked out :) Good luck with any more you have to do!

    • one year ago
  16. cbm
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    Thanks, I'm struggling with one more now trying to set up an inequality.

    • one year ago
  17. theEric
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    You can ask it here or in another question. I have time if you'd like a hand. Algebra like this can be hard at first.

    • one year ago
  18. cbm
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    Ok, it will take me a minute to type it. Thanks for your help. Yes, I'm too old to learn this stuff.

    • one year ago
  19. theEric
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    Haha, no such thing. It might just require some thinking and rethinking, right? Maybe thinking in a different light! I started thinking about counting cookies/blocks again before so the numbers meant something.

    • one year ago
  20. theEric
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    And take your time!

    • one year ago
  21. cbm
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    Your making a rectangular patio that is 10ft wide and 20ft long, using square pavers that have an area of 0.81 sq ft, what is the # of pavers u will need? Write and solve an inequality to find the answer.

    • one year ago
  22. cbm
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    Lol, I'm trying. I thought I had equations but not the fractions

    • one year ago
  23. cbm
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    would it be like (10x20).81=p

    • one year ago
  24. cbm
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    idn

    • one year ago
  25. theEric
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    The first step is always understanding the problem. Then we move on.. So I have a quick question... Each "paver" is .81 square feet, and we need to know how many of the will fit in our 10 foot by 20 foot area?

    • one year ago
  26. cbm
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    Yep, thats the question.

    • one year ago
  27. theEric
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    Alright! Then (10*20) is definately your area of the patio. And you wanna see how many .81 area portions will fit into that. How many times does .81 go into (10*20) is the question to get you to your answer. That's a division thing.

    • one year ago
  28. cbm
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    I got 246,9 so P≥246.9

    • one year ago
  29. theEric
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    That's just one way to look at it. There's another way in which you build the equation and solve.. I got the same number after rounding. Is this supposed to be inequality?

    • one year ago
  30. cbm
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    I'm not sure if I'm writing the inequality right or not (10ftx20ft)/.81=p

    • one year ago
  31. cbm
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    Yes, it's supposed to be an inequality

    • one year ago
  32. theEric
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    That's what I would do to!

    • one year ago
  33. cbm
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    To me it works and gives the answer when you work it.

    • one year ago
  34. theEric
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    I don't see why it would be inequality! There must be something I'm not thinking of.. I thought it would want an exact number...You want so many pavers.... OR! Do you want at least so many pavers? Meaning that many or more?

    • one year ago
  35. cbm
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    Do you want me to create a problem so I can give you a medal? I'm not sure what the medal do.

    • one year ago
  36. cbm
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    I was trying to figure out why it want come out to a whole number also unless it's ≥

    • one year ago
  37. theEric
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    They're just sort of recognition you can count. I'm not worried about it. Thanks though!

    • one year ago
  38. theEric
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    Yeah, I think maybe that's the idea of it.. why they want inequality.. Although you'd think rounding it up to 247 would be good enough! Maybe the question was just not appropriate for this subject!

    • one year ago
  39. cbm
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    Oh, ok thanks! Maybe (10x20)/.81≥p

    • one year ago
  40. theEric
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    I would say \[p\ge 246.0\]

    • one year ago
  41. theEric
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    You would want more pavers rather than less. At least you get your patio done that way :P

    • one year ago
  42. cbm
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    LOL, that's the truth.

    • one year ago
  43. theEric
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    Pavers area things should be greater than or equal to the patio area... to get the patio done :P Yeah!

    • one year ago
  44. theEric
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    Using just that sentence, you can set up the inequality. \[p_{NumberOfPavers}\ge A_{PatioArea}\]

    • one year ago
  45. cbm
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    I've been trying everything to get an even number. I don't understand why it would be a decimal. I tried breaking it up into inches, etc. Thanks again for your help.

    • one year ago
  46. theEric
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    It might just not come out to be an even number. It happens. We could solve it to be a fraction instead of a decimal. With fractions, your always exact (no rounding).

    • one year ago
  47. theEric
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    I have five cookies to share between us. How many can we each have? 2.5. Books don't always make problems come out to nice numbers either..

    • one year ago
  48. theEric
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    So maybe that's the case here.

    • one year ago
  49. cbm
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    true, I hope so.

    • one year ago
  50. theEric
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    Same. Good luck with that! If you solve it as a fraction, note that .81 = 81/100.

    • one year ago
  51. cbm
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    alright, Thanks again!!

    • one year ago
  52. theEric
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    \[\frac{10*20}{.81}=\frac{10*20}{\frac{81}{100}}\] Np!

    • one year ago
  53. cbm
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    :)

    • one year ago
  54. theEric
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    Just in case... From there, it's \[\frac{10*20*100}{81}=\frac{20000}{81}=246\frac{74}{81}\]

    • one year ago
  55. cbm
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    Yep, that works good or at least it looks better. I figured I could write the equation as (p)x0.81≥200. I'm still working it to see if it will work.

    • one year ago
  56. theEric
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    That's great, and it was the other way I would think of the equation to solve!

    • one year ago
  57. theEric
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    And you don't have to change the sign because 0.81 is not negative.|dw:1351404148861:dw|

    • one year ago
  58. theEric
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    All other operations keep the inequality the same direction. Adding (positive number) to both sides: Each side gets the same bump up Subtracting (positive number) from bothe sides: Each side gets the same cut down Multiplying (number greater than 1) to both sides: Each side gets larger proportional to its size Dividing both sides by (number greater than one): Each side gets smaller proportional to its size

    • one year ago
  59. theEric
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    I threw in number specifications, just because adding a negative is like subtracting a positve. Then each side gets cut down the same. Same for multiplication/division. Multiplying by a fraction less than one is like dividing by that fraction's reciprocal\[x*\frac{1}{2}=x\div 2\]and so each side gets proportionally smaller. Just some extra, if you wanna think about it. Take care! It was nice working with you!

    • one year ago
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