Since $x-9=(\sqrt{x}{)}^2-3^2$, the denominator factors as $(\sqrt{x}-3)(\sqrt{x}+3)$. So

Because $x>9$, we have $\sqrt{x}-3\ne 0$, so that factor can be canceled. The expression becomes

Use the function rule $y=\sqrt{3^x-1}$ with $x=2$:

This also fits the pattern shown in the table, since for $x=3$, $y=\sqrt{3^3-1}=\sqrt{27-1}=\sqrt{26}$, and for $x=4$, $y=\sqrt{3^4-1}=\sqrt{81-1}=\sqrt{80}$. So the missing value is $\sqrt{8}$.

Evaluate the two heights at the given times:

so $h(0)>d(0)$. Also,

so $h(4)<d(4)$ because $\sqrt{13}<4$. Since both functions are continuous for $t\ge 0$, one starts above the other and later is below it, so they must be equal at some time between $0$ and $4$. Therefore, the statement that must be true is that the student is correct because $h(0)>d(0)$ and $h(4)<d(4)$.

Let $y=a^x$. Since $a>0$, this substitution is valid, and $a^{-x}=\frac{1}{a^x}=\frac{1}{y}$. The equation becomes

Multiplying both sides by $y$ gives

Because the original equation has solutions $x=r$ and $x=-r$ with $r>0$, the corresponding values of $y$ are $a^r$ and $a^{-r}$, which are the two roots of this quadratic. Using the quadratic formula,

Since $a^r>1$ when $r>0$ and the two roots are reciprocals, $a^r$ must be the larger root. Therefore,

Because the graphs intersect at $(7,4)$, both functions must have value $4$ when $t=7$.

First check the radical function:

So the point is on $y=h(t)$.

Now use the exponential function:

Since $4=2^2$, we have

so the exponents must be equal:

Therefore, the value of $k$ is $5$.