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Ms-Brains

  • one year ago

PLEASE HELP ME!!! i'll help you with medals

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  1. Ms-Brains
    • one year ago
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  2. Ms-Brains
    • one year ago
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    SHOW WORK SO I KNOW HOW TO SOLVE TOO

  3. anonymous
    • one year ago
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    Substitute 2 in for wherever you see x in the original inequality.

  4. Avengedslipknot
    • one year ago
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    So all you do it substituent for x 2(2)^2-4(2)+1>2(2)- 2^2/4 and solve

  5. anonymous
    • one year ago
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    Your equation then becomes 2(2)^2 - 4(2) + 1 > 2(2) - [(2)^2]/4

  6. Ms-Brains
    • one year ago
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    I don't know how to solve it. :( help me, so I can understand?

  7. anonymous
    • one year ago
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    Yep. And then when you solve if the number on the left isnt greater than the number on the right. It is false. If it is greater. It is true.

  8. anonymous
    • one year ago
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    Sure, tell me where you are getting stuck. So far you are good.

  9. Ms-Brains
    • one year ago
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    Ughhh, this is so frustrating. :(

  10. anonymous
    • one year ago
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    Just treat the > as an = sign. It has no importance until the end when discerning if the answer is true or false.

  11. Ms-Brains
    • one year ago
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    The awnser is False, I know that. The study guide says it. I just do not know how to simplify it all.

  12. anonymous
    • one year ago
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    Im not sure where you are getting stuck. The first step is to just simplify everything on the left side. Then the second step is to simplify everything on the right side. Then you compare the two numbers to see if the inequality makes sense. You have 2(2)^2-4(2)+1 > 2(2)- 2^2/4 So to simplify the left side first. You can simplify 2(2)^2. First simplify (2)^2 to get 4. And then multiply that by 2. To get 8. Your equation now looks like 8 - 4(2) + 1 > 2(2) - (2^2)/4 The next thing you can do is simplify the -4(2) to -8. So now your inequality looks like 8 - 8 + 1 > 2(2) - (2^2)/4 0 + 1 > 4 - (2^2)/4 1 > 4 - (4/4) 1 > 4 - 1 1 > 3 This says 1 is greater than 3, which is false. So the answer is false like you said.

  13. Ms-Brains
    • one year ago
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    Alright, Thanks! I'm going use this in my notes.

  14. anonymous
    • one year ago
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    No problem. If you are confused on anything I am still here to help as well as many others on this site im sure. Don't be afraid to ask questions :D

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