Hello. .
I have solved this.

Please confirm if my answers are right.

1) -5t+10
2) 7.5
3) 41.67

 

First two values are okay but third one RMS seems way off. http://en.wikipedia.org/wiki/Root_mean_square this might help. Break signal as sum of AC and DC. A DC of 7.5 and a sawtooth waveform of maximum amplitude 2.5. http://www.ces.clemson.edu/ge/staff/park/Class/ECE262/Homework/7th/Ch09/SawRMS/SawRMS.html. Then use RMS formula

.
It comes out to be 7.6376.

 

I the RMS over all time of a periodic function is equal to the RMS of one period of the function.

& the RMS for a f'n over all time in the Wiki link that you provided is what I have used. Help!!

 

I the RMS over all time of a periodic function is equal to the RMS of one period of the function.

& the RMS for a f'n over all time in the Wiki link that you provided is what I have used. Help!!

Try this :

1. Integrate the square of the function (that you got) over one period. This gives you mean square value.
2. Under root of this is RMS.

Put the limits from 0 to 1 to get 58.3333

√(58.3333) =7.63

 

The correct answer for RMS value is
\( \sqrt{\int _0^1 (10 - 5t)^2 dt}=5*\sqrt{\frac{7}{3}} \approx 7.63V \)
as pointed out by dalam
http://www.wolframalpha.com/input/?i=sqrt(definite integral from 0 to 1 (10 - 5t)^2 dt)