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anonymous

  • one year ago

can someone give me a algebra function that equals to 13?

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  1. anonymous
    • one year ago
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    idk why but well our math teacher gave us that homework and i have no idea where to start.

  2. anonymous
    • one year ago
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    f(x) = 13 should do the trick

  3. anonymous
    • one year ago
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    well we are supposed to make a function problem for others to solve it...

  4. anonymous
    • one year ago
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    so you want an expression with a variable?

  5. anonymous
    • one year ago
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    and we are learning pretty advanced math, and it could be actually any algebra problem i think...

  6. anonymous
    • one year ago
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    i hope...

  7. anonymous
    • one year ago
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    maybe you are supposed to say "let \(f(x)=2x+1\) solve \(f(x)=13\)" maybe

  8. anonymous
    • one year ago
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    yeah something like that but more complicated... and u could be asking for y inter, roots, anything

  9. LoganH
    • one year ago
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    sorry

  10. anonymous
    • one year ago
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    um... more complicated... the teacher example is like \[((\sqrt{5})^{2})^{2})-(-3)^0+2!+\frac{ 28 }{ 4 }\]

  11. anonymous
    • one year ago
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    for one of her numbers

  12. anonymous
    • one year ago
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    any ideas?

  13. anonymous
    • one year ago
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    well it is suppsoed to be a review of algebra honors i took last year and since we have the same teacher for 8th geo so she decided to give us this thing

  14. anonymous
    • one year ago
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    and i need to make different equations for 13, 4, 2, 0, 4, 0, 2, 2 they have to be different equations idk why.

  15. anonymous
    • one year ago
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    I'm confused. Are you looking for a function whose value is 13 at some point? Or are you looking for a PEMDAS style problem that gives 13 as the answer?

  16. anonymous
    • one year ago
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    it is a non-word problem relating to algebra that is hard to solve.

  17. anonymous
    • one year ago
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    where do the y-intercept/roots come in? The example above doesn't even have a variable. Can you post the exact question you got, maybe that will make it easier to see what it's asking for.

  18. anonymous
    • one year ago
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    well... the problem is that i am supposed to make the question. and that equation up there is one of the teacher's examples.

  19. anonymous
    • one year ago
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  20. anonymous
    • one year ago
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    this is how the teacher did it

  21. anonymous
    • one year ago
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    and so i have me age and bla bla bla bla bla so now i need to make problems out of them

  22. anonymous
    • one year ago
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    ok. I see

  23. anonymous
    • one year ago
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    I never would have understood what you were after without seeing that

  24. anonymous
    • one year ago
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    :P

  25. anonymous
    • one year ago
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    sorry i never noticed that i didnt attach the file :|

  26. anonymous
    • one year ago
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    any ideas?

  27. anonymous
    • one year ago
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    It's hard for someone else to make these up, because we don't really know what you've covered. I don't think they have to be too complex. 2 of them are research style questions and 2 are order of operations type. Maybe do one where you ask for a discriminant of a quadratic equation. Or one that requires some combining of exponents

  28. anonymous
    • one year ago
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    she has one that says number of roots. You can probably do something like sum or product of roots. Or make up a system of equations and have one of the answers be the product of the coordinates.

  29. anonymous
    • one year ago
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    well that is basicially what we've covered. ,ake it pretty hard please :) well i'm not the one doing it :P

  30. anonymous
    • one year ago
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    I wish I could help, Im not very good at math

  31. anonymous
    • one year ago
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    :|

  32. anonymous
    • one year ago
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    @Leong

  33. anonymous
    • one year ago
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    you can... basically just based on her data and makes yours, please check by actually testing and doing it out, not just randomly write it :)

  34. anonymous
    • one year ago
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    yea but i need help making mine cz the numbers are different

  35. anonymous
    • one year ago
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  36. anonymous
    • one year ago
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    okay okay :3 I'll help you :3

  37. anonymous
    • one year ago
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    fhank u!!!!!!!

  38. anonymous
    • one year ago
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    thank*

  39. anonymous
    • one year ago
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    hey wait let me close this and make another one it takes too long to bump and the replies are too long too

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