2. $\tan ( \alpha + B ) = 50 / \alpha = \frac { 10 / \alpha + \tan B } { 1 - 10 / d \tan B }$\\
solving Yan $B$ and maximizing writ $d$ you'll find $d = 10 \sqrt { 5 }$ feet and corresponding to a\\
3!

$$\begin{aligned}
& 4 \text { SEP; } T \in Q \text { | are tangency points } \\
& C K \text { are mid points of } A B , M N \\
& \triangle P A C \cong \triangle P Q T \text { and } \triangle Q M K \cong \triangle Q P S \\
& y _ { 2 } A B = A C = P A \cdot S T / P Q = \frac { P S Q M } { P Q } = M K = 1 / 2 M N \text { QED } \\
& s = ( x + y + z ) \\
& \triangle = \sqrt { ( x + y + z ) ( x + y + z - x - y ) ( x + y + z - y - z ) ( x + y + z - x - z ) } \\
& = \sqrt { ( x + y + z ) n y z } \\
& \triangle = 1 / 2 r ( ( x + y ) + ( y + z ) + ( x + z ) ) = r ( x - y + z ) \\
& = \sqrt { ( x + y + z ) x y z }
\end{aligned}$$

$$\begin{aligned}
x & = \lim _ { n \rightarrow \alpha } \frac { 1 } { 2 n } \log \frac { 2 n ! } { ( n ! ) ^ { 2 } } \\
& = \lim _ { n \rightarrow \alpha } \frac { 1 } { 2 n } \lim _ { n \rightarrow \alpha } \log \frac { \frac { ( n + 1 ) ( n + 2 ) \cdots ( n + n ) } { n ^ { n } } } { \frac { 12 \cdot 3 \cdots } { n ^ { n } } } \\
& = \lim _ { n \rightarrow \alpha } \frac { 1 } { 2 n } ( \log ( 1 + 1 / n ) ( 1 + 2 / n ) ( 1 + 3 / n ) \cdots ( 1 + n / n ) - \log ( 1 / n + 2 / n + \cdots + n / n ) ) \\
& = 1 / 2 \int _ { 0 } ^ { 1 } \ln ( 1 + x ) - \ln x d x = 1 / 2 \left[ \left. \ln \log ( x + 1 ) \right| _ { 0 } ^ { 1 } - \int _ { 0 } ^ { 1 } \frac { x d x } { 1 + x } \right] - [ x \ln x \cdot x ] _ { 0 } ^ { 1 }
\end{aligned}$$

$$= 1 / 2 ( ( \ln 2 + \ln 2 - 1 ) - ( - 1 ) ) = 1 / 2 \ln 4$$

7

$$\text { a) } \begin{aligned}
f ( x ) & = x ^ { 5 } + x - 10 \\
f ^ { \prime } ( x ) & = 4 x ^ { 4 } + 1 > 0
\end{aligned}$$

$$\text { b) } \begin{gathered}
f ( 1 ) = - 8 ; f ( 2 ) = 24 \\
f ( 1 ) f ( 2 ) < 0
\end{gathered}$$

\begin{center}
\includegraphics[max width=\textwidth, alt={}]{99230309-7375-4af3-83ac-74793f77c0ea-2_358_282_700_982}
\end{center}

$$\text { c) let } \begin{aligned}
x = p / q ; & ( p / q ) ^ { 5 } + ( p / q ) = 10 \\
& \Rightarrow p 5 / q = q 10 q - p
\end{aligned}$$

fraction =integer contradictian!\\