Ms-Brains
  • Ms-Brains
PLEASE HELP ME!!! i'll help you with medals
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Ms-Brains
  • Ms-Brains
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Ms-Brains
  • Ms-Brains
SHOW WORK SO I KNOW HOW TO SOLVE TOO
anonymous
  • anonymous
Substitute 2 in for wherever you see x in the original inequality.

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Avengedslipknot
  • Avengedslipknot
So all you do it substituent for x 2(2)^2-4(2)+1>2(2)- 2^2/4 and solve
anonymous
  • anonymous
Your equation then becomes 2(2)^2 - 4(2) + 1 > 2(2) - [(2)^2]/4
Ms-Brains
  • Ms-Brains
I don't know how to solve it. :( help me, so I can understand?
anonymous
  • anonymous
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.
anonymous
  • anonymous
Sure, tell me where you are getting stuck. So far you are good.
Ms-Brains
  • Ms-Brains
Ughhh, this is so frustrating. :(
anonymous
  • anonymous
Just treat the > as an = sign. It has no importance until the end when discerning if the answer is true or false.
Ms-Brains
  • Ms-Brains
The awnser is False, I know that. The study guide says it. I just do not know how to simplify it all.
anonymous
  • anonymous
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.
Ms-Brains
  • Ms-Brains
Alright, Thanks! I'm going use this in my notes.
anonymous
  • anonymous
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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