# 10 Questions packed into one Question

**Calculus**Level pending

For statements that is correct, type 1. For statements that is incorrect, type 2. For statements that is partially correct, type 3. So the answer is a 10-digit number \[\] 1) \(\sqrt [ 3 ]{ 27 } \quad -\quad \sqrt { 9 } \quad =\quad 0\) \[\] 2) \(\log { 10 } \quad \times \quad \log { 100 } \quad =\quad \log { 1000 } \) \[\] 3) \({ x }^{ 2 }\quad -\quad 2x\quad -\quad 3\quad =\quad 0\) has the factors \(x\quad =\quad -1\) and \(x\quad =\quad 3\) \[\] 4) \({ 999 }^{ 999 }\) is divisible by 3 \[\] 5) Since \(10!\) has 2 trailing zeros, and \(20!\) has 4 trailing zeros, \(10! + 20!\) has 6 trailing zeros \[\] 6) \(sin\quad { 2460 }^{ \circ }\quad =\quad sin\quad { 300 }^{ \circ }\) \[\] 7) If \(f\left( x \right) \quad =\quad \frac { ax\quad +\quad b }{ cx\quad +\quad d } \), then \(f^{ -1 }\left( x \right) \quad =\quad \frac { -dx\quad +\quad b }{ cx\quad -\quad a } \) \[\] 8) \(\sqrt [ 4 ]{ x } \quad \times \quad \sqrt [ 4 ]{ x } \quad =\quad \sqrt { x } \) \[\] 9) \(\lim _{ n\quad \rightarrow \quad 1 }{ \frac { { n }^{ 2 }\quad -\quad 1 }{ n\quad -\quad 1 } } \quad =\quad 2\) \[\] 10) \(\frac { \sqrt [ 10 ]{ 10 } }{ \log _{ 10 }{ 10 } } \quad =\quad \frac { 5 }{ 2 } \pi \)