Question

the function $d(w)=sqrt{5w}$ gives rope diameter $d$ for weight $w$. which function gives weight when given diameter?
hint: square both sides to undo the square root, then isolate $w$.
skill sheet: solving radical equations steps, p. 3.

a. $w(d)=sqrt{5d}$

b. $w(d)=\frac{d^2}{5}$

c. $w(d)=5d^2$

d. $w(d)=\frac{5}{d^2}$

Explanation:

Step1: Start with the given function

We have \( d(w)=\sqrt{5w} \). We need to solve for \( w \) in terms of \( d \).

Step2: Square both sides to eliminate the square root

Squaring both sides of the equation \( d = \sqrt{5w} \), we get \( d^{2}=(\sqrt{5w})^{2} \). Since \( (\sqrt{x})^{2}=x \) for \( x\geq0 \), this simplifies to \( d^{2}=5w \).

Step3: Isolate \( w \)

To isolate \( w \), we divide both sides of the equation \( d^{2} = 5w \) by 5. So, \( w=\frac{d^{2}}{5} \), which can be written as \( w(d)=\frac{d^{2}}{5} \).

Answer:

B. \( w(d)=\frac{d^{2}}{5} \)