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kitsune0724

  • 3 years ago

If x is the smallest integer, which of the following equations models the problem? I got X+(x+2)+(x+4)+(x+6)=636. Is this correct? Please and thank you.

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  1. zzr0ck3r
    • 3 years ago
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    there is no smallest integer

  2. zzr0ck3r
    • 3 years ago
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    so they are all true

  3. kitsune0724
    • 3 years ago
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    So it would none of the above?

  4. zzr0ck3r
    • 3 years ago
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    if you are doing if then -> -p ^ q

  5. Directrix
    • 3 years ago
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    @What is the problem being modeled? x appears to be 156 on what was posted but what is the scenario of the problem that calls for the smallest x? @kitsune0724

  6. zzr0ck3r
    • 3 years ago
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    they are all true

  7. kitsune0724
    • 3 years ago
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    The question just says if x is the smallest integer, which of the equations models the problem. The choices are: 4x=636, x(x+2)(x+4)(x+6)=636, x+(x+2)+(x+4)+(x+6)=636, x+(x+2)+(x+4)+(x+6)=636 and none of the above. I got the third choice. @Directix

  8. kitsune0724
    • 3 years ago
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    @Directrix

  9. Directrix
    • 3 years ago
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    @kitsune0724 I assume that we have to solve all the equations for x and then choose as the answer the equation which yields the smallest value of x of all options. Let me check your answer.

  10. Directrix
    • 3 years ago
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    @kitsune0724 In the options, I think you wrote one of them twice: x+(x+2)+(x+4)+(x+6)=636, x+(x+2)+(x+4)+(x+6)=636. What is supposed to be there?

  11. zzr0ck3r
    • 3 years ago
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    how is there a smallest integer? I think this is a question about if then....which is -P v Q thus since P is false the statement is true for what follows...

  12. Directrix
    • 3 years ago
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    Something bothers me about this problem. The question asks the following: " which of the equations **models the problem.** @kitsune0724 --> What problem?

  13. kitsune0724
    • 3 years ago
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    yes. it's suppose to be there. @Directrix

  14. kitsune0724
    • 3 years ago
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    That's all it says in the problem.

  15. zzr0ck3r
    • 3 years ago
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    plewase tell me why you guys are ignoring the smallest integer part? and my explination...

  16. kitsune0724
    • 3 years ago
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    I'm not ignoring your explanation @zzr0ck3r but that's not one of the choices.

  17. zzr0ck3r
    • 3 years ago
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    hehe, they are all true is the choice:)

  18. Directrix
    • 3 years ago
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    @zzr0ck3r -P v Q thus since P is false the statement is true for what follows... I am not commenting on your contributions because I do not follow what they have to do with the problem. And, I STILL do not think we have been given the entire problem. My best guess is that there's a given problem about consecutive integers and we are to select the equation that can be used to solve that problem. Yet, I cannot find out from @kitsune0724 what that problem is.

  19. kitsune0724
    • 3 years ago
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    There's no problem given @Directrix. I'm sorry.

  20. Directrix
    • 3 years ago
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    @kitsune0724 Why did you pick the third choice? And, I know two of the choices cannot be identical. So, there's something faulty about this problem. I cannot help without more information. The choices are: ... SAME x+(x+2)+(x+4)+(x+6)=636, SAME x+(x+2)+(x+4)+(x+6)=636 ....

  21. Directrix
    • 3 years ago
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    @kitsune0724 I'm going to try to get a fresh pair of eyes for this problem. Hold on.

  22. zzr0ck3r
    • 3 years ago
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    @Directrix it says if x is the smallest integer then .... thus P = x is the smallest integer this is never true^^^^^^ thus we have ~P which implies Q is true for any Q. we have 4 Q's

  23. zzr0ck3r
    • 3 years ago
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    all true....

  24. Directrix
    • 3 years ago
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    @saifoo.khan --> We need a fresh pair of eyes on this problem. Would you render your thoughts, please? Thanks.

  25. zzr0ck3r
    • 3 years ago
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    I dont see why you would try and solve a problem that cant be solved..... if x is an irrational rational number what is x^2 <----would you try and solve this?

  26. saifoo.khan
    • 3 years ago
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    I don't really get the question. Sorry. :/ Are you sure it's complete? @kitsune0724

  27. kitsune0724
    • 3 years ago
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    Thank y'all for your help.

  28. Directrix
    • 3 years ago
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    Glad to try to help. It was a mathematical adventure. :)

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