A room has a wall with an R value of 18 F sq. ft. hr/BTU. The room is 15 feet long and 11 feet wide with walls that are 9 ft high. On a particular day the average outside temperature is 40 degrees F. How much heat is transferred through the walls on that day (in BTUs)?

The heat transferred through the wall that day is  13728 BTUs

Explanation:

Here, we have the area of the wall given as

Area of wall = 2 × Length × Height + 2 × Width × Height

Length = 15 feet

Width = 11 Feet and

Height = 9 feet

Therefore, the area = 2×15×9 + 2×11×9 = 468 ft²

Temperature difference is given by

Average outside temperature - Wall temperature = 40 - 18 = 22 °F

Therefore the heat transferred through the wall that day (24 hours) at 18 sq.ft. hr/BTU is given by;

468 × 22 × 24/18 = 13728 = 13728 BTUs.

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .