Question

a town is expanding the size of a square athletic field. the equation shown determines the building cost, y, in dollars, for every foot, x, added to the width of the athletic field.
y = (10.25)x²
which equation shows an equivalent equation rearranged to determine x in terms of y?
〇 x = 10.5√y
〇 x = (10.25)/√y
〇 x = √y / 10.1
〇 x = 10.25√y

Explanation:

Step1: Start with the given equation

We have the equation \( y = 0.0125x^{2} \). Our goal is to solve for \( x \) in terms of \( y \).

Step2: Isolate \( x^{2} \)

First, we divide both sides of the equation by \( 0.0125 \) to get \( x^{2} \) by itself. So, \( \frac{y}{0.0125}=x^{2} \).

Step3: Solve for \( x \)

To solve for \( x \), we take the square root of both sides. Remember that when we take the square root, we consider the positive root since \( x \) represents a length (number of feet added, which can't be negative in this context). So, \( x = \sqrt{\frac{y}{0.0125}} \). We can rewrite \( \sqrt{\frac{y}{0.0125}} \) as \( \frac{\sqrt{y}}{\sqrt{0.0125}} \). Calculating \( \frac{1}{\sqrt{0.0125}} \), we know that \( \sqrt{0.0125}=\sqrt{\frac{1}{80}}\approx0.1118 \), and \( \frac{1}{0.1118}\approx8.944 \), but more accurately, let's rationalize or simplify. Wait, actually, \( 0.0125=\frac{1}{80} \), so \( \frac{1}{0.0125} = 80 \), so \( \sqrt{\frac{y}{0.0125}}=\sqrt{80y} \)? Wait, no, wait. Wait, original equation: \( y = 0.0125x^{2} \), so \( x^{2}=\frac{y}{0.0125} \), then \( x=\sqrt{\frac{y}{0.0125}} \). Let's compute \( \frac{1}{0.0125} \): \( 0.0125=\frac{1}{80} \), so \( \frac{1}{0.0125}=80 \)? Wait, no, \( 0.0125\times80 = 1 \), yes. Wait, but let's check the options. Wait, maybe there was a typo in the original problem, maybe the coefficient is \( 0.0125 \) or maybe \( 0.0125 \) is a typo for \( 0.0125 \) or maybe \( 0.0125 \) is \( \frac{1}{80} \), but looking at the options, one of the options is \( x = \frac{\sqrt{y}}{0.11} \) (maybe the \( 0.11 \) is a typo for \( \sqrt{0.0125} \) or maybe the original coefficient was different). Wait, alternatively, let's re-express the steps. Starting with \( y = 0.0125x^{2} \), divide both sides by \( 0.0125 \): \( x^{2}=\frac{y}{0.0125} \), then take square roots: \( x=\sqrt{\frac{y}{0.0125}}=\frac{\sqrt{y}}{\sqrt{0.0125}} \). Now, \( \sqrt{0.0125}\approx0.1118 \approx 0.11 \) (maybe rounded). So \( x=\frac{\sqrt{y}}{0.11} \), which matches the third option (the one with the blue dot, \( x = \frac{\sqrt{y}}{0.11} \) assuming the \( 0.11 \) is the approximate value of \( \sqrt{0.0125} \)). Let's verify: if we start with \( y = 0.0125x^{2} \), solve for \( x \):

1. Divide both sides by \( 0.0125 \): \( x^{2}=\frac{y}{0.0125} \)
2. Take square root: \( x = \sqrt{\frac{y}{0.0125}}=\frac{\sqrt{y}}{\sqrt{0.0125}} \)
3. Calculate \( \sqrt{0.0125}\approx0.1118\approx0.11 \), so \( x\approx\frac{\sqrt{y}}{0.11} \), which is the third option.

Answer:

The correct option is the one with \( x = \frac{\sqrt{y}}{0.11} \) (the option with the blue dot, assuming the \( 0.11 \) is the approximate value of \( \sqrt{0.0125} \)). If we follow the steps:

1. Start with \( y = 0.0125x^{2} \)
2. Divide both sides by \( 0.0125 \): \( x^{2}=\frac{y}{0.0125} \)
3. Take square root: \( x = \sqrt{\frac{y}{0.0125}}=\frac{\sqrt{y}}{\sqrt{0.0125}} \approx \frac{\sqrt{y}}{0.11} \)

So the answer is the option \( x = \frac{\sqrt{y}}{0.11} \) (the third option, marked with the blue dot).