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jakethekicker

  • 2 years ago

Which of the following is a possible equivalence that can be used in a conversion? ten cubed kL over one L ten squared daL over one L ten to the negative second power L over one cL ten to the negative first power L over one mL

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  1. Claflamme3
    • 2 years ago
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    Let me look this one up

  2. Claflamme3
    • 2 years ago
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    the third one

  3. mr.singh
    • 2 years ago
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    agree with @Claflamme3

  4. theEric
    • 2 years ago
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    Hi! You have your answer, which I agree with, but I want to explain how you'd know that! Firstly, when you convert to a different unit, the value - what the number represents - doesn't change. You're actually multiplying the number by \(1\). Its a fraction, where the top and bottom are equal. I'll derive a simple conversion that I hope you agree with. Even if you're in the stubborn U.S., it's important to know the metric system. Like, 1 meter is one hundred centimters. In math, \(1\ [m]=100\ [cm]\) This would imply that, by dividing by 1 meter,\[\frac{1\ [m]}{1\ [m]}=\frac{100\ [cm]}{1\ [m]}\]\[\qquad\qquad\Downarrow\]\[\frac{\cancel{1\ [m]}}{\cancel{1\ [m]}}=\frac{100\ [cm]}{1\ [m]}\]\[\qquad\qquad\Downarrow\]\[1=\frac{100\ [cm]}{1\ [m]}\] That sort of thing is always good to know for unit conversions. And reversing that on your multiple choices will check to see if they are good! \[\frac{10^3\ [kL]}{[L]}\qquad\rightarrow\qquad 10^3\ [kL]\overset{?}{=}[L]\] \[\frac{10^2\ [daL]}{[L]}\qquad\rightarrow\qquad 10^2\ [daL]\overset{?}{=}[L]\] \[\frac{10^{-2}\ [L]}{[cL]}\qquad\rightarrow\qquad 10^{-2}\ [L]\overset{?}{=}[cL]\] \[\frac{10^{-1}\ [L]}{[mL]}\qquad\rightarrow\qquad 10^{-1}\ [L]\overset{?}{=}[mL]\] And note that \(\Large 10^{-1}=\frac{1}{10^1}\) and that \(\Large 10^{-2}=\frac{1}{10^2}\), just by the meaning of negative exponents. Just for my own fun... \[\frac{10^3\ [kL]}{[L]}\qquad\rightarrow\qquad 10^3\ [kL]\cancel{=}[L]\] \[\frac{10^2\ [daL]}{[L]}\qquad\rightarrow\qquad 10^2\ [daL]\cancel{=}[L]\] \[\frac{10^{-2}\ [L]}{[cL]}\qquad\rightarrow\qquad 10^{-2}\ [L]{=}[cL]\qquad\huge \checkmark \] \[\frac{10^{-1}\ [L]}{[mL]}\qquad\rightarrow\qquad 10^{-1}\ [L]\cancel{=}[mL]\]

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