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anonymous

  • 5 years ago

what is the solution in interval notation? -5/3x le -10

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  1. anonymous
    • 5 years ago
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    -5/3x≤-10 multiply each side by -3/5 (don't forget to switch the sign!) x≥6 in interval notation [6, infty)

  2. anonymous
    • 5 years ago
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    its -5 over 3 x \[\le\] -10

  3. anonymous
    • 5 years ago
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    how do i write it in interval notation?

  4. anonymous
    • 5 years ago
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    ooooh got it.

  5. anonymous
    • 5 years ago
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    with 6,infinitie)

  6. anonymous
    • 5 years ago
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    \[[6, \infty)\]

  7. anonymous
    • 5 years ago
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    thank u

  8. anonymous
    • 5 years ago
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    -5/3x ≤ -10 3x ≤ 1/2 x ≤ 1/6 which is to say that x is an element of (epsilon) (0,1/6] (parentheses because you can't have x=0, and a bracket because the expression equals -10 if x = 1/6). TBates, if you plug in 6, or anything greater, into the original inequality you don't even come close to approaching -10. In your method, you forgot that x is in the denominator -- not the numerator. If you graph -5/3x, you have a rational expression with asymptotes. You can use that to find the values at which the function is less than -10: in this circumstance, any value between 0 and 1/6 gives -10 or less. Anything less than 0 gives a value higher than -10, and it certainly cannot equal zero.

  9. anonymous
    • 5 years ago
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    but that answer is still correct...

  10. anonymous
    • 5 years ago
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    Quantum you forgot that the problem had 3 in the denominator, the answer I gave previously is correct.

  11. anonymous
    • 5 years ago
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    How it was written the x is not in parenthesis and therefor not a part of the denominator.

  12. anonymous
    • 5 years ago
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    So it was 5/3 * x? If so, then yeah, you're right. I thought it was 5/(3x).

  13. anonymous
    • 5 years ago
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    sorry, that was my fault. it was hard to write it on here and show the 3 was not part of the x

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