The temperature at 5:00 a. m. was –6 °F. By 3:00 p. m., the temperature had risen to a high of 36 °F. If the temperature at 8:00 p. m. decreased from the high by latex of the change in temperature in Fahrenheit degrees from 5:00 a. m. to 3:00 p. m., what was the temperature at 8:00 p. m.?

Complete question :

The temperature at 5:00 a.m. was –6 °F. By 3:00 p.m., the temperature had risen to a high of 36 °F. If the temperature at 8:00 p.m. decreased from the high by the change in temperature in Fahrenheit degrees from 5:00 a.m. to 3:00 p.m., what was the temperature at 8:00 p.m.?

-6°F

Step-by-step explanation:

Temperature at 5am (t1) = - 6°F

Temperature at 3pm (t2) = 36°F

Change in temperature, between 5am and 3pm:

(t2 - t1) = (36 - (-6)) = (36 + 6) = 42°F

Hence, temperature at 8:00 pm:

High temperature - change in temperature between (5am - 3pm)

36°F - 42°F = - 6°F

1 number

if we use a graphing calculator, there is only one such number that satisfies the conclusion of the mean value theorem, that is, a number such that

where the right-hand side is the average rate of change of f on the interval [0.5,0.75], which is the slope of the secant line.

in the attached graph, the black line is the secant line which has its slope as the average rate of change of f on [0.5,0.75], while the orange line is the only tangent line to the graph of f in the interval that satisfies the mvt conclusion.

6e14 is the answer.