If I understand the question, you have to get n alone. in this case, you have to subtract 7 from both sides, resulting in n (is less than) -10
Volume=(pi)(radius^2)(height)
Volume=(pi)(5^2)(12)
V=(pi)(25)(12)
V=(pi)(300)
Answer:
The maximum volume of space that 1,000 BTUs will cool is 400 cubic foot.
Step-by-step explanation:
Given : A total of 1,000 BTUs are needed to cool a space. If 2.5 BTUs are needed to cool 1 cubic foot of space.
To find : What is the maximum volume of space that 1,000 BTUs will cool?
Solution :
Applying unitary method,
2.5 BTU are needed to cool 1 cubic foot of space.
So, 1 BTU are needed to cool
cubic foot of space.
Then,1000 BTU are needed to cool
cubic foot of space.
i.e. 1000 BTU are needed to cool 400 cubic foot of space.
Therefore, the maximum volume of space that 1,000 BTUs will cool is 400 cubic foot.
Answer:
The bounded area is 5 + 5/6 square units. (or 35/6 square units)
Step-by-step explanation:
Suppose we want to find the area bounded by two functions f(x) and g(x) in a given interval (x1, x2)
Such that f(x) > g(x) in the given interval.
This area then can be calculated as the integral between x1 and x2 for f(x) - g(x).
We want to find the area bounded by:
f(x) = y = x^2 + 1
g(x) = y = x
x = -1
x = 2
To find this area, we need to f(x) - g(x) between x = -1 and x = 2
This is:


We know that:



Then our integral is:

The right side is equal to:

The bounded area is 5 + 5/6 square units.