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jewlzme17

  • one year ago

Find the value x such that f(x)=0, if f(x)=(3x-4)/(5)

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  1. random231
    • one year ago
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    just solve (3x-4)/5=0

  2. jdoe0001
    • one year ago
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    \(\bf f(x)={\color{blue}{ y}}=0\qquad {\color{blue}{ f(x)}}=\cfrac{3x-4}{5}\qquad thus \\ \quad \\ {\color{blue}{ 0}}=\cfrac{3x-4}{5}\impliedby \textit{solve for "x"}\)

  3. jewlzme17
    • one year ago
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    I already know all of that I just don't remember how to solve (3x-4)/5

  4. freckles
    • one year ago
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    (3x-4)/5 there is nothing to solve here do you mean you don't remember how to solve (3x-4)/5=0?

  5. freckles
    • one year ago
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    the fraction is zero when the top is 0

  6. freckles
    • one year ago
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    that you only have to solve 3x-4=0

  7. jewlzme17
    • one year ago
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    so the zero cancels out the 5?

  8. dinamix
    • one year ago
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    (3x-4)=0 *5 3x=4 x=

  9. dinamix
    • one year ago
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    @jewlzme17 u should try little

  10. jewlzme17
    • one year ago
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    I know the answer it 4/3 its just I am having trouble understanding how. You guys are all saying different things and it's confusing. Like can someone just break it down barney style without being rude?

  11. freckles
    • one year ago
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    if 3x-4=0 then 3x has to be 4 another way to say this is 3x=4

  12. freckles
    • one year ago
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    can you solve for x now?

  13. freckles
    • one year ago
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    no one has said anything different

  14. freckles
    • one year ago
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    each of us has given you part of how to get there

  15. freckles
    • one year ago
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    @jdoe0001 showed you exactly what f(x)=0 meant @dinamix showed you how to get to 3x=4

  16. freckles
    • one year ago
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    oh and i forgot to add what I said @freckles said f(x)=0 when the numerator=0

  17. jewlzme17
    • one year ago
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    What I'm confused on is what you do with the denominator?

  18. freckles
    • one year ago
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    the bottom is always 5 it has no deciding factor when the fraction is 0

  19. freckles
    • one year ago
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    do you know that 0/anything is 0 (when anything isn't 0)

  20. freckles
    • one year ago
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    \[h(x)=\frac{f(x)}{g(x)}=0 \text{ when } f(x)=0 \text{ when } x \text{ is in the domain of } h \]

  21. freckles
    • one year ago
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    the only that gets to decide a fraction is 0 when the top is 0 on that function's domain of course

  22. jewlzme17
    • one year ago
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    OKay that makes so much more sense. Thank you

  23. freckles
    • one year ago
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    \[\frac{3x-4}{5}=0 \implies 3x-4=0\]

  24. freckles
    • one year ago
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    though you could have also decide to multiply both sides by 5 and that also works

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