Voltage division equation.
The Attempt at a Solution
From a) I know that Rx = 86k ohms.
Part b) and c) befuddle me. I think it is relevant to the transistor being on and off.
In b) I would think that because Rx is so high, Vbe would be too low to activate the transistor, and therefore 10V would flow through the 10kohm and the 866k eq. resistor, both losing 10V along the way.
In c) I would think that Rx is 8600. Therefore Vbe = 4.1V. I don't know where to go from there or even if that is correct.