Since $f(x)=\sqrt{x+4}$, set $\sqrt{x+4}=6$. Squaring both sides gives $x+4=36$, so $x=32$. This also fits the pattern in the table: when the output increases from $5$ to $6$, the value inside the square root must go from $25$ to $36$, so $x$ goes from $21$ to $32$.

Since $\sqrt{x}=7$, square both sides to get $x=7^2=49$. Because $x$ is positive, this value fits the condition. So the statement that must be true is $x=49$.

For a square, the area is $s^2$, where $s$ is the side length. Since the area is $49$, solve $s^2=49$. Taking the positive square root gives $s=\sqrt{49}=7$. So the side length is $7$ feet.

When a quantity doubles every equal time interval, it grows exponentially. Starting with $200$ cells means the initial value is $200$, and doubling every hour means multiplying by $2$ for each hour. So after $t$ hours, the model is $N=200(2{)}^t$.

Since $(k,3)$ is on the graph of $y=\sqrt{x+5}$, substitute $x=k$ and $y=3$:

Square both sides:

So,