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Question
this question has two parts. first, answer part a. then, answer part b.
part a
geometry the area of the base of a cylindrical water tank is 12π square feet. the volume of water in the tank is dependent on the height of the water h and is represented by the function v(h) = 12πh.
a. find v⁻¹(h).
a) v⁻¹(h) = \\(\frac{h}{12π}\\)
b) v⁻¹(h) = \\(\frac{12π}{h}\\)
c) v⁻¹(h) = \\(\frac{h}{6π}\\)
d) v⁻¹(h) = \\(\frac{h}{12}\\)
part b
b. what will the height of the water be when the volume reaches 420π cubic feet?
______ ft
Part A
Step1: Recall inverse function steps
To find the inverse of \( V(h) = 12\pi h \), first, replace \( V(h) \) with \( y \), so \( y = 12\pi h \).
Step2: Swap \( y \) and \( h \)
Swap the variables: \( h = 12\pi y \).
Step3: Solve for \( y \)
Divide both sides by \( 12\pi \): \( y=\frac{h}{12\pi} \). Then, replace \( y \) with \( V^{-1}(h) \), so \( V^{-1}(h)=\frac{h}{12\pi} \).
Step1: Use the volume function
We know the volume function is \( V(h) = 12\pi h \), and we are given \( V = 420\pi \).
Step2: Solve for \( h \)
Substitute \( V = 420\pi \) into \( V(h) = 12\pi h \): \( 420\pi=12\pi h \). Divide both sides by \( 12\pi \): \( h=\frac{420\pi}{12\pi} \). The \( \pi \) cancels out, and \( \frac{420}{12} = 35 \).
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A. \( V^{-1}(h)=\frac{h}{12\pi} \)