An oscilloscope shows a sine wave with a peak value of 9.0 V and a period of 8.0 ms…

Step 1. The frequency. We want $f$, so $f = \dfrac{1}{T} = \dfrac{1}{8.0 \times 10^{-3}\ \text{s}} = 125$ Hz.

Step 2. The angular frequency. $\omega = 2\pi f = 2\pi \times 125\ \text{Hz} = 785$ rad/s.

Step 3. The rms value. $V_{rms} = \dfrac{V_m}{\sqrt{2}} = \dfrac{9.0\ \text{V}}{1.414} = 6.36$ V.

Step 4. The voltage. Starting from zero and rising, $v = V_m \sin \omega t$. The angle is $\omega t = 785\ \text{rad/s} \times 1.5 \times 10^{-3}\ \text{s} = 1.178$ rad, which is $67.5^\circ$. So $v = 9.0\ \text{V} \times \sin(1.178) = 9.0\ \text{V} \times 0.924 = 8.31$ V.