The height of the dome of the building is given by
[tex] f(x)=2\sqrt{-x^2+10x} [/tex]. The quantity in the square root should be greater than or equal to 0. We need
[tex] -x^2+10x\geq 0\\
x(10-x)\geq 0\\
0\leq x\leq 10 [/tex].
So the domain is [tex] 0\leq x\leq 10 [/tex]
Answer:
0 ≤ x ≤ 10
Step-by-step explanation:
We cannot take the square root of a negative number. This means we cannot use negative numbers in this equation; this means the lower end is 0.
We can only use x values until the value of the negative square is greater than the value of 10 times the x.
10(1) = 10, which is greater than -(1(1)) = -1.
10(2) = 20, which is greater than -(2(2)) = -4.
10(3) = 30, which is greater than -(3(3)) = -9.
10(4) = 40, which is greater than -(4(4)) = -16.
10(5) = 50, which is greater than -(5(5)) = -25.
10(6) = 60, which is greater than -(6(6)) = -36.
10(7) = 70, which is greater than -(7(7)) = -49.
10(8) = 80, which is greater than -(8(8)) = -64.
10(9) = 90, which is greater than -(9(9)) = -81.
10(10) = 100, which is equal to -(10(10)) = -100. This means this is the cutoff point for this function.
