High pass filter with given attenuation
In this instance, think of the capacitor as a frequency-dependent resistor.

At high frequencies, you can think of it as a dead short relative to the fixed resistance.  In that case, the input is simply connected to the output, so the gain is 1.

At low frequencies, the capacitor is a open circuit.  Now you only have the two resistors forming a voltage divider with a gain of 1/11.  Whoever said this had a low frequency gain of 1/10 was wrong, unless they consider 1/11 within their error margin.  Note that the gain of the voltage divider is:

&nbsp;&nbsp;gain = R / (10R + R) = R / 11R = 1/11

To find the crossover frequency of a R-C filter, you set the impedance magnitude of the capacitor equal to the resistance:

&nbsp;&nbsp;R = 1 / (2&pi;fC)

When f is in units of Hz, C in Farads, then R is in Ohms.