Question

given a unit circle, what is the value of y at the indicated point? hint: the equation for the unit circle is x² + y² = 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 \(x=\frac{5}{8}\) gives \((\frac{5}{8})^{2}+y^{2}=1\).

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 is in the upper - half of the unit circle (assuming the standard position, as the \(x\) - coordinate is positive and no indication of negative \(y\) from the problem's visual context), we take the positive value.

Answer:

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