The difference of c and 5 is simply c - 5. This can't be simplified, unless you assume a specific value for c.
The difference of c and 5 is simply c - 5. This can't be simplified, unless you assume a specific value for c.
The difference of c and 5 is simply c - 5. This can't be simplified, unless you assume a specific value for c.
The difference of c and 5 is simply c - 5. This can't be simplified, unless you assume a specific value for c.
The difference of c and 5 is simply c - 5. This can't be simplified, unless you assume a specific value for c.
22 - 5 = 17 degrees
(3c +15)/(c2 - 5) + c/(c + 5)= (3c + 15)/(c - 5)(c + 5) + c/(c + 5)= (3c + 15)/(c - 5)(c + 5) + [c(c - 5)/(c + 5)(c - 5)(since the common denominator is (c + 5)(c - 5)= (3c + 15)/(c - 5)(c + 5) + (c2 - 5c)/(c + 5)(c - 5)= (3c + 15 + c2 - 5c)/(c - 5)(c + 5)= (c2 - 2c + 15)/(c - 5)(c + 5)= [(c - 5)(c + 3)]/(c - 5)(c + 5) (simplify)= (c + 3)/(c + 5)
The -7 is called the difference. In any subtraction problem: a = b - c a is the difference b and c are terms (technically, b is minuend and c is subtrahend, but these terms are not really used in modern math)
If you are you refering to temperature scales Celcius and Fahrenheit, then the crossing of the two scales happens at -40 degrees. That is because of a 32 degree difference at freezing point of water and the F-scale is 5/9ths of the C-scale. C = (F - 32) * 5/9 and F = (C * 9/5) + 32 so -40 = (-40 * 9/5) + 32 = -40
F to C- Deduct 32, then multiply by 5, then divide by 9 C to F- Multiply by 9, then divide by 5, then add 32 This is how you find Celsius to Fahrenheit and Fahrenheit to Celsius.
5 F = -15 C 5 C = 41 F The difference is 20 C or 36 F.
22 - 5 = 17 degrees
27
In C programming 5 is 5 (obviously), while '5' if 0x35 in ASCII, 0xF5 in EBCDIC.
0.5556
10 degrees
(3c +15)/(c2 - 5) + c/(c + 5)= (3c + 15)/(c - 5)(c + 5) + c/(c + 5)= (3c + 15)/(c - 5)(c + 5) + [c(c - 5)/(c + 5)(c - 5)(since the common denominator is (c + 5)(c - 5)= (3c + 15)/(c - 5)(c + 5) + (c2 - 5c)/(c + 5)(c - 5)= (3c + 15 + c2 - 5c)/(c - 5)(c + 5)= (c2 - 2c + 15)/(c - 5)(c + 5)= [(c - 5)(c + 3)]/(c - 5)(c + 5) (simplify)= (c + 3)/(c + 5)
If the K in the question refer to Kelvin, it is not "in Celsius".The Celsius scale is an interval scale. This means that the difference between 5 deg C and 10 deg C is the same as that between 32 deg C and 37 deg C, or -56 deg C and -51 deg C. However, the choice of the 0 for this scale (the triple point of water) is arbitrary. As a result, 10 deg C is not twice anything that is 5 deg C (other its numerical value).By contrast absolute temperature, measured in terms of thermodynamic activity, is measured in Kelvin. The difference between 5 Kelvin and 10 K is the same as the difference between 5 deg C and 10 deg C, or 32 deg C and 37 deg C. But additionally, 10 K represents double the thermodynamic activity than does 5 K.So a Kelvin degree is the same change as a Celsius degree but 1 K = -272.15 deg C.
8 degrees C
The difference between C and the advanced C is that C is basic. On the other hand, the advanced C is thorough and to the detail.
The -7 is called the difference. In any subtraction problem: a = b - c a is the difference b and c are terms (technically, b is minuend and c is subtrahend, but these terms are not really used in modern math)
16232