SBAC High School Concepts and Procedures. Practice it free.

Find two numbers with product $15$ and sum $-8$: $-3$ and $-5$. So $x^2 - 8x + 15 = (x-3)(x-5)$.

Cross-multiply to clear the fraction: $3(5x - 2y) = 2(x + 3y)$, so $15x - 6y = 2x + 6y$. Collect like terms to get $13x = 12y$. Divide both sides by $y$, then by $13$, to obtain $\dfrac{x}{y} = \dfrac{12}{13}\textsf{.}$

For $x \neq -2$, the factor $x+2$ cancels, giving $x-8 = 5$, so $x = 13\textsf{.}$

Set $y = 0$: $-5(x - 11)(x + 4) = 0$. The constant $-5 \neq 0$, so the zeros are $x = 11$ and $x = -4$. Of the choices, $-4$ is an x-coordinate of an x-intercept.

Substitute $y = 5$. From $x - 2y < -3$, $x - 10 < -3$, so $x < 7$. From $x + y > 6$, $x + 5 > 6$, so $x > 1$. The value $x = 4$ satisfies $1 < 4 < 7$.

The equation is in vertex form $f(x) = a(x - h)^2 + k$ with $a = 1 > 0$, so the graph opens upward and the minimum occurs at the vertex. Rewrite $(x + 3)^2$ as $(x - (-3))^2$: the vertex is at $x = -3$. Therefore, $f(x)$ reaches its minimum when $x = -3$.

Because this parabola opens upward, the smallest $y$ is at the vertex. For $y = ax^2 + bx + c$, the $x$-value at the vertex is $x = -\dfrac{b}{2a}$. Here $a = 1$ and $b = -18$, so $x = -\dfrac{-18}{2(1)} = \dfrac{18}{2} = 9$. Therefore, $y$ reaches its minimum when $x = 9$.

The number of staff members must divide both $24$ and $36$. Common factors greater than $1$ include $2$, $3$, $4$, $6$, and $12$. Only $12$ appears among the choices: $24 \div 12 = 2$ photos and $36 \div 12 = 3$ pins per person.