HELP!!

the area A of the circle in the figure
can be represented by A(r)=pi,r^2 where r is the radius
what are the resonable domain and range for each function? explain

the problme before this made me make a equation of B(r)=5^2-pi,r^2 im not sure if there asking this or A(r)=pi,r^2 please answer both just in case tho

the figure is a circle in a 5x5 square the circle shows radius as r

Well, look at you juggling equations! Let's break it down.

For A(r) = πr^2, this represents the area of the circle with radius r. The domain for this function is all positive real numbers since radius can't be negative or imaginary. However, since your circle is within a 5x5 square, the reasonable domain for this specific situation would be 0 ≤ r ≤ 2.5 (half the length of the square).

As for the range, it represents the possible values for the area. Since the area can never be negative, the reasonable range for A(r) is A(r) ≥ 0.

Now onto your B(r) = 25 - πr^2 equation. This seems like you're trying to find the area of the remaining portion of the square when the circle is subtracted. The domain and range for this equation are actually the same as A(r) since it's still related to the circle's radius.

Remember, don't go clowning around with negative or imaginary radii, and keep your areas grounded in reality!

To determine the reasonable domain and range for each function, let's break down the problem.

First, we'll start with the function A(r) = πr^2. This function represents the area of a circle given its radius, which is denoted by r.

Domain: The domain of a function refers to the possible values that the input (in this case, r) can take. Since we're dealing with the area of a circle, the radius must be a positive real number. Therefore, the reasonable domain for A(r) = πr^2 is all positive real numbers.

Range: The range of a function represents the possible values that the output (in this case, the area A) can take. Since the area of a circle can't be negative (as it represents a physical quantity), the range of A(r) = πr^2 will be all positive real numbers.

Now let's move on to the equation B(r) = 5^2 - πr^2.

Domain: The domain of B(r) would be the same as that of A(r), as both functions depend on the radius of the circle. Thus, the domain of B(r) is all positive real numbers.

Range: To determine the range of B(r), we need to consider the equation itself. B(r) = 5^2 - πr^2 means that we subtract the area of a circle with radius r from the area of a square with side 5 units. The area of the square is always positive, so the minimum value B(r) can have is 0. However, since the area of a circle can't exceed the area of a square with the same side length, the maximum value of B(r) would be the area of the square, which is 5^2 = 25. Therefore, the range of B(r) is from 0 to 25.

To summarize:
- The reasonable domain for both A(r) = πr^2 and B(r) = 5^2 - πr^2 is all positive real numbers.
- The range of A(r) = πr^2 is all positive real numbers.
- The range of B(r) = 5^2 - πr^2 is from 0 to 25.