A hypothetical temperature scale records the boiling point of water as 58°M and the freezing point as -2°M. What is 2°M in Celius?
Boiling point of water in C is 100
100C = 58M
1M = 50C/29
2M = 50C/29 * 2
= 3.448C
Not fully sure though
I believe it is done this way but check me out.
100 C = 58 M
0 C = -2 M
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100 C divisions = 60 M divisions so
(2 M + 2)*100/60 = 6.667 C
Let's try it. 58 M should calculate to 100 C. That will be
(58+2)*100/60 = 60*100/60 = 100
If you want to go the other way; i.e., from C to M, then
(C*60/100) - 2 = M. So let's see if 100 C ends up at 58 M.
That's (100*60/100) - 2 = 60-2 = 58 M
That checks out to me. Finally, let's convert the 6.667 back to M. That's
6.667 x 60/100 = 4.00 and 4.00 - 2 = 2.0 M Voila!!!
To convert a temperature from the hypothetical scale (°M) to Celsius (°C), we need to establish a conversion formula.
Given that the freezing point is recorded as -2°M in the hypothetical scale and 0°C in Celsius, we can determine the conversion difference between the two scales:
Temperature difference = 0°C - (-2°M) = 0°C + 2°M
Now, let's calculate the temperature difference in Celsius:
0°C + 2°M = 0°C + 2 * (1°C/1°M) = 0°C + 2°C = 2°C
Therefore, 2°M in the hypothetical scale is equivalent to 2°C in Celsius.