Desperate and quick help needed

**Hallo** · Aug 28th 2007, 08:39 PM

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!

**red_dog** · Aug 28th 2007, 09:02 PM

Originally Posted by Hallo

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
$\displaystyle\frac{3}{16}x=150$

**Hallo** · Aug 28th 2007, 09:04 PM

Originally Posted by red_dog

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.

**red_dog** · Aug 28th 2007, 09:37 PM

$\displaystyle T_f=\frac{9}{5}(-30)+32=-54+32=-22$

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

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