1. Given that v(t) = -6 + 12 cos(2π200t + π/4) - 9 cos(2π400t + π/6) volts. The signal is applied to a 14 kΩ resistor. Find the average power dissipated on the resistor in mW. Round your answer off to two decimal places. 2. Pavg = (sqrt(Vdc^2 + vp^2/2))/R 3. I used the above equation just P=Vrms^2/R to calculate the power, this is wrong and I'm not sure what to do.