-50°C equals -58°F
-10°F = -23.3°C(-10°F - 32) multiplied by 5/9 = -23.3°C
-10 °c = 14 °f
-2oc
Temperatures can be MUCH colder than the Zero point on both.
28.4 degrees Fahrenheit.
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
-1 c = 30.2 f
-22 degrees Fahrenheit.
-23.8 degrees Fahrenheit
-3°C = 26.6°F: -3°C multiplied by 1.8 +32 = 26.6°F
C-F will be the most polar because of the four C and F have the largest difference in electronegativity.
A Maclaurin series is centered about zero, while a Taylor series is centered about any point c. M(x) = [f(0)/0!] + [f'(0)/1!]x +[f''(0)/2!](x^2) + [f'''(0)/3!](x^3) + . . . for f(x). T(x) = [f(c)/0!] + [f'(c)/1!](x-c) +[f''(c)/2!]((x-c)^2) + [f'''(c)/3!]((x-c)^3) + . . . for f(x).
Negative 97 Degrees Fahrenheit is negative 71.66667 degrees Celsius.
°C = (°F − 32) × 5⁄9 °C = (-20 − 32) × 5⁄9 °C = -53 × 5⁄9 °C = -29.4 Therefore -20°F is equal to -29.4°C
assume that whatever integers you are using are the variables in this. If you haven't been given integers, assume (for the sake of simplicity) that they are one. a * b * c * d * e * f = x -a * b * c * d * e * f = -x -a * -b * c * d * e * f = x -a * -b * -c * d * e * f = -x see a pattern? any ODD number of negative integers will lead to a negative answer, therefore with the limit being 6, the answer will be 5.
-50°C = -58°F