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anonymous

  • one year ago

f(x)=4x+3 helpp

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  1. whovianchick
    • one year ago
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    What are we helping to do?

  2. anonymous
    • one year ago
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    figure out f(x)=4x+3

  3. anonymous
    • one year ago
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    can you walk me through it

  4. whovianchick
    • one year ago
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    What do you mean figure it out? What are the instructions on the problem?

  5. anonymous
    • one year ago
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    how do you do it

  6. anonymous
    • one year ago
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    f(x) = 4x + 3 is a statement, not a question. What are you trying to do with the function?

  7. anonymous
    • one year ago
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    get like an answer at the end what would i tell my teacher

  8. anonymous
    • one year ago
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    what were you asked? there has to be some instructions

  9. anonymous
    • one year ago
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    evaluate it

  10. Vocaloid
    • one year ago
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    did they give you any other information?

  11. Vocaloid
    • one year ago
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    like, did they give you an x value?

  12. anonymous
    • one year ago
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    ok, now we're getting somewhere. Are you supposed to evaluate it at a specific value of x?

  13. anonymous
    • one year ago
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    lets say x is 2

  14. anonymous
    • one year ago
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    ok, so you would substitute 2 for x f(x) = 4x + 3 f(2) = 4(2) + 3 then simplify that

  15. whovianchick
    • one year ago
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    Don't multiply f and 2 though, that's basically y. It stays f(2), or function of 2.

  16. anonymous
    • one year ago
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    ok

  17. anonymous
    • one year ago
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    11

  18. anonymous
    • one year ago
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    yep

  19. anonymous
    • one year ago
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    can i ask a couple more

  20. anonymous
    • one year ago
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    ok

  21. anonymous
    • one year ago
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    f(x)=2x+4 G(x)=X+1

  22. anonymous
    • one year ago
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    no actually

  23. anonymous
    • one year ago
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    f(x)=5x+3 G(x)=x+4

  24. anonymous
    • one year ago
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    ok what are the instructions for this one?

  25. anonymous
    • one year ago
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    i have to combine them but I'm confused

  26. anonymous
    • one year ago
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    ok, there are lots of different ways to combine functions. It really depends on what you're asked for. Do you have a specific combination, or do you just want me to show you some

  27. anonymous
    • one year ago
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    can you show me some

  28. anonymous
    • one year ago
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    ok let's start with the basic 4 Adding \(f(x) + g(x) = 5x + 3 + x + 4\) \(= 6x + 7\) Subtracting \(f(x) - g(x) = 5x + 3 - (x + 4)\) \(=5x+3-x-4\) \(=4x-1\) \(g(x) - f(x) = x + 4-(5x+3)\) \(=x+4-5x-3\) =\(-4x+1\) Multiplying \(f(x)\cdot g(x)=(5x+3)(x+4)\) \(=5x^2+20x+3x+12\) \(=5x^2+23x+12\) Dividing \[\frac{ f(x) }{ g(x) }=\frac{ 5x+3 }{ x+4 }\] \[\frac{ g(x) }{ f(x) }=\frac{ x+4 }{ 5x+3 }\]

  29. anonymous
    • one year ago
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    Then you can get more complex with something like \(2f(x)-3g(x)=2(5x+3)-3(x+4)\) =\(10x+6-3x-12\) =\(7x-6\)

  30. anonymous
    • one year ago
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    You can also compose functions, which is basically putting one inside the other. To find f(g(x)) replace the x in f(x) with the function from g(x). \(f(g(x))=5(x+4)+3\) \(=5x+20+3\) \(=5x+23\) To find g(f(x)), do the opposite and replace the x in g(x) with f(x) \(g(f(x))=(5x+3)+4\) \(=5x+7\)

  31. anonymous
    • one year ago
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    ok can you walk me through on Subtracting f(x)−g(x)=5x+3−(x+4) =5x+3−x−4 =4x−1 g(x)−f(x)=x+4−(5x+3) =x+4−5x−3 =−4x+1 how'd you get that

  32. anonymous
    • one year ago
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    from the first line to the second you have to distribute. -(x + 4) = -x - 4 for the first one and -(5x + 3) = -5x - 3 then you combine like terms Is that what you meant?

  33. anonymous
    • one year ago
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    yea thank you and what about the others sorry I'm asking so much its just that i need to pass this

  34. anonymous
    • one year ago
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    no problem. for multiplying you usually have to use FOIL or do some distribution. dividing is just writing one on top of the other like a fraction

  35. anonymous
    • one year ago
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    ok

  36. anonymous
    • one year ago
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    Most of this is really knowing how to apply the distributive property and combine like terms

  37. anonymous
    • one year ago
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    ok thank you

  38. anonymous
    • one year ago
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    you're welcome

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