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cbm

  • 2 years ago

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

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  1. NickR
    • 2 years ago
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    where do you get 10q from?

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

  3. cbm
    • 2 years ago
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    Is that wrong NickR?

  4. etemplin
    • 2 years ago
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    you need to distribute first

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

  6. theEric
    • 2 years ago
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    That's the start, right?

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

  8. jayz657
    • 2 years ago
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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

  9. jayz657
    • 2 years ago
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    sorry q<56

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

  11. cbm
    • 2 years ago
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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 <

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

  13. theEric
    • 2 years ago
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    Yep! Because you multiplied or divided by a negative!

  14. cbm
    • 2 years ago
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    @theEric thanks!

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

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

  17. theEric
    • 2 years ago
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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.

  18. cbm
    • 2 years ago
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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.

  19. theEric
    • 2 years ago
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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.

  20. theEric
    • 2 years ago
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    And take your time!

  21. cbm
    • 2 years ago
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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.

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

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

  24. cbm
    • 2 years ago
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    idn

  25. theEric
    • 2 years ago
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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?

  26. cbm
    • 2 years ago
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    Yep, thats the question.

  27. theEric
    • 2 years ago
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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.

  28. cbm
    • 2 years ago
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    I got 246,9 so P≥246.9

  29. theEric
    • 2 years ago
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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?

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

  31. cbm
    • 2 years ago
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    Yes, it's supposed to be an inequality

  32. theEric
    • 2 years ago
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    That's what I would do to!

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

  34. theEric
    • 2 years ago
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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?

  35. cbm
    • 2 years ago
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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.

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

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

  38. theEric
    • 2 years ago
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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!

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

  40. theEric
    • 2 years ago
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    I would say \[p\ge 246.0\]

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

  42. cbm
    • 2 years ago
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    LOL, that's the truth.

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

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

  45. cbm
    • 2 years ago
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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.

  46. theEric
    • 2 years ago
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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).

  47. theEric
    • 2 years ago
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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..

  48. theEric
    • 2 years ago
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    So maybe that's the case here.

  49. cbm
    • 2 years ago
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    true, I hope so.

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

  51. cbm
    • 2 years ago
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    alright, Thanks again!!

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

  53. cbm
    • 2 years ago
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    :)

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

  55. cbm
    • 2 years ago
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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.

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

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

  58. theEric
    • 2 years ago
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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

  59. theEric
    • 2 years ago
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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!

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