Diodes question (I Got the answer but I just need more clarification)
R1, D2, and R3 are dropping a total of 10V. Since D2 is ideal, it drops 0V. This means that the current flowing must be:

\${10\text{V} \over {10\text{k}\Omega + 5\text{k}\Omega}} \approx 667\mu\text{A}\$

The voltage dropped by R3 must therefore be:

\$667\mu\text{A} \cdot 5\text{k}\Omega \approx 3.34\text{V}\$

The voltage at V must then be:

\$3.34\text{V} + - 5\text{V} = -1.66\text{V}\$