Question

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

Explanation:

Step1: Substitute \(x = \frac{1}{6}\) into the unit - circle equation

The unit - circle equation is \(x^{2}+y^{2}=1\). Substituting \(x=\frac{1}{6}\), we get \((\frac{1}{6})^{2}+y^{2}=1\).

Step2: Simplify the equation

\(\frac{1}{36}+y^{2}=1\). Then \(y^{2}=1 - \frac{1}{36}\).
Using the formula \(a - b=\frac{36a-36b}{36}\), we have \(y^{2}=\frac{36 - 1}{36}=\frac{35}{36}\).

Step3: Solve for \(y\)

Since the point \((\frac{1}{6},y)\) is in the fourth quadrant (from the position of the point in the coordinate system shown, \(y<0\)), \(y =-\sqrt{\frac{35}{36}}\). And \(\sqrt{\frac{35}{36}}=\frac{\sqrt{35}}{\sqrt{36}}=\frac{\sqrt{35}}{6}\).

Answer:

The value in the numerator (the green box) is \(35\) and the value in the denominator (the grey box) is \(6\). So \(y =-\frac{\sqrt{35}}{6}\), the number under the square - root is \(35\) and the denominator is \(6\).