For the information given in the statement, we can write the following proportion:
[tex]\begin{gathered} \frac{\text{Width initial photo}}{\text{ Height initial photo}}=\frac{\text{ Width enlarged photo}}{\text{ Height enlarged photo}} \\ \frac{4in}{6in}=\frac{10in}{x\text{ }in} \\ \frac{4}{6}=\frac{10}{x} \end{gathered}[/tex]
Now, we can solve the equation for x:
[tex]\begin{gathered} \text{ Apply cross product} \\ 4\cdot x=10\cdot6 \\ 4x=60 \\ \text{ Divide by 4 from both sides of the equation} \\ \frac{4x}{4}=\frac{60}{4} \\ \boldsymbol{x=15} \end{gathered}[/tex]
Therefore, the height of the enlarged photograph will be 15 inches and, the correct answer is option C.
