Answer
\[7\]
Work Step by Step
\[B(y)=\left\{\begin{array}{ll}4-\displaystyle\frac{1}{2}t\;\;\;\;\text{if}\;t<2\\
\\
\sqrt{t+c}\;\;\;\;\text{if}\;t\geq 2\end{array}\right.\]
Consider left hand limit at $t=2$
\[\lim_{t\rightarrow 2^{-}}B(t)=\lim_{t\rightarrow 2}\left(4-\displaystyle\frac{1}{2}t\right)=4-\displaystyle\frac{1}{2}(2)=4-1=3\]
Consider right hand limit at $t=2$
\[\lim_{t\rightarrow 2^{+}}B(t)=\lim_{t\rightarrow 2}\sqrt{t+c}=\sqrt{2+c}\]
Since $\displaystyle\lim_{t\rightarrow 2}B(t)$ exists so $\displaystyle\lim_{t\rightarrow 2^{-}}B(t)= \displaystyle\lim_{t\rightarrow 2^{+}}B(t) $
\[\Rightarrow 3=\sqrt{2+c}\]
\[\Rightarrow 9=2+c\Rightarrow c=7\]
Hence $c=7$
