Question

given a unit circle, what is the value of y at the indicated point?
hint: the equation for the unit circle is $x^{2}+y^{2}=1$.

Explanation:

Step1: Substitute \(x = \frac{5}{8}\) into the unit - circle equation

The equation of the unit circle is \(x^{2}+y^{2}=1\). Substituting \(x=\frac{5}{8}\), we get \((\frac{5}{8})^{2}+y^{2}=1\).

Step2: Simplify the equation

\(\frac{25}{64}+y^{2}=1\). Then \(y^{2}=1 - \frac{25}{64}\).

Step3: Calculate \(1-\frac{25}{64}\)

\(y^{2}=\frac{64 - 25}{64}=\frac{39}{64}\).

Step4: Solve for \(y\)

\(y=\pm\sqrt{\frac{39}{64}}=\pm\frac{\sqrt{39}}{8}\). Since the point \((\frac{5}{8},y)\) is in the first or second quadrant (assuming the standard position of the unit - circle point), and we are looking for the non - negative value (if we consider the form \(\frac{\sqrt{[?]}}{[?]}\) as in the problem's hint structure), \(y=\frac{\sqrt{39}}{8}\).

Answer:

\(\frac{\sqrt{39}}{8}\)