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Math Help - Desperate and quick help needed

  1. #1
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    Desperate and quick help needed

    Yes... I need a test that I have to submit relatively soon... I just can't figure these damn answers out. ):

    Anyway.

    Here are the two problems.

    Also... instead of just giving me the answer, can you answer it AND explain how you got to that answer...? Thanks. (:

    The temperature of surface of Mars is recored over a 24-hour period of time. The temperature varied from -30 degrees Celsius to -85 degrees Celsius. Find in Fahrenheit the warmest temperature during the 24-hour time period.

    Possible answers:

    a.) -121 degrees
    b.) -95.4 degrees
    c.) -22 degrees
    d.) -48.7


    And...


    A mining company digs ore from a mountain side. Three-sixteenths of the ore is pure copper. How much of this ore would need to be refined in order to yield 150 pounds of pure copper?

    Possible answers:

    a.) 185 pounds
    b.) 122 pounds
    c.) 7200 pounds
    d.) 800 pounds



    Thanks guys. >_<

    I just really need them soon. T_T; They're confusing. I'm not good with the C/F conversions... and blah. I just plain old don't understand the second.

    Thanks again!
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  2. #2
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by Hallo View Post
    The temperature of surface of Mars is recored over a 24-hour period of time. The temperature varied from -30 degrees Celsius to -85 degrees Celsius. Find in Fahrenheit the warmest temperature during the 24-hour time period.

    Possible answers:

    a.) -121 degrees
    b.) -95.4 degrees
    c.) -22 degrees
    d.) -48.7
    Use the formula conversion:
    \displaystyle T_f=\frac{9}{5}T_c+32
    where T_f=temperature in degrees Fahrenheit
    T_c=remperature in degrees Celsius


    A mining company digs ore from a mountain side. Three-sixteenths of the ore is pure copper. How much of this ore would need to be refined in order to yield 150 pounds of pure copper?

    Possible answers:

    a.) 185 pounds
    b.) 122 pounds
    c.) 7200 pounds
    d.) 800 pounds
    Let x be the quantity of ore.
    Then you have to solve the equation
    \displaystyle\frac{3}{16}x=150
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  3. #3
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    Quote Originally Posted by red_dog View Post
    Use the formula conversion:
    \displaystyle T_f=\frac{9}{5}T_c+32
    where T_f=temperature in degrees Fahrenheit
    T_c=remperature in degrees Celsius




    Let x be the quantity of ore.
    Then you have to solve the equation
    \displaystyle\frac{3}{16}x=150
    Thanks! But I know the equation for the first problem... and blah... I just don't know how to multiply by the fraction. ):

    No idea how to work out the equation. Ahhh... I need to get better at this.
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  4. #4
    MHF Contributor red_dog's Avatar
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    \displaystyle T_f=\frac{9}{5}(-30)+32=-54+32=-22


    \displaystyle\frac{3}{16}x=150\Rightarrow 3x=16\cdot 150\Rightarrow x=16\cdot 50=800
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