Let's go through the questions shown in the image one by one and solve them.

Question 15:

The function $f(x) = \frac{2}{5}x + 10$.

We are asked to find the value of $x$ such that $f(x) = 30$.

Solution: $f(x) = \frac{2}{5}x + 10 = 30$ First, subtract 10 from both sides: $\frac{2}{5}x = 20$ Next, multiply both sides by $\frac{5}{2}$ to solve for $x$: $x = 20 \times \frac{5}{2} = 50$ So, $x = 50$.

Question 16:

The function $f(x) = \frac{1}{4}(x + 4)^2 - 5$ is shown, passing through points $(-10, a)$ and $(a, b)$, where $a$ and $b$ are constants. The question asks for the value of $b$.

Solution:

To find the value of $b$, let's first substitute the point $(-10, a)$ into the function: $f(-10) = \frac{1}{4}((-10 + 4)^2) - 5$ Simplifying: $f(-10) = \frac{1}{4}((-6)^2) - 5 = \frac{1}{4}(36) - 5 = 9 - 5 = 4$ Thus, $b = 4$. The correct answer is A) 4.

Question 17:

We are given the quadratic equation $x^2 - 12x + 23 = 0$, and we are asked to find a solution.

Solution:

We will use the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = -12$, and $c = 23$. Substituting into the quadratic formula: $x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(23)}}{2(1)}$ $x = \frac{12 \pm \sqrt{144 - 92}}{2}$ $x = \frac{12 \pm \sqrt{52}}{2}$ $x = \frac{12 \pm 2\sqrt{13}}{2}$ $x = 6 \pm \sqrt{13}$ So, the solutions are $x = 6 + \sqrt{13}$ and $x = 6 - \sqrt{13}$. The correct answer is D) $6 + \sqrt{13}$.

Question 18:

The points $(-3, 10)$ and $(3, 10)$ are the endpoints of the diameter of a circle. We are asked for the equation of the circle.

Solution:

1. The center of the circle is the midpoint of the diameter, which can be found as: $\text{Midpoint} = \left( \frac{-3 + 3}{2}, \frac{10 + 10}{2} \right) = (0, 10)$
2. The radius is the distance from the center $(0, 10)$ to one of the endpoints, say $(-3, 10)$: $\text{Radius} = \sqrt{(0 - (-3))^2 + (10 - 10)^2} = \sqrt{3^2} = 3$ Thus, the radius is 3.

The equation of a circle is given by: $(x - h)^2 + (y - k)^2 = r^2$ where $(h, k)$ is the center and $r$ is the radius. Substituting $h = 0$, $k = 10$, and $r = 3$: $(x - 0)^2 + (y - 10)^2 = 3^2$ $x^2 + (y - 10)^2 = 9$ The correct answer is A) $x^2 + (y - 10)^2 = 9$.

Would you like further details on any of these solutions?

Here are 5 related questions you could explore:

1. What is the general form of a linear function, and how do you solve for specific values?
2. How do transformations like shifts and scaling affect the graph of a quadratic function?
3. How do you solve quadratic equations using the discriminant to predict the type of roots?
4. What are the properties of a circle, and how do you find the equation of a circle given two points?
5. How do you find the radius of a circle given its endpoints using the distance formula?

Tip: When solving quadratic equations, always check whether factoring is possible before using the quadratic formula—it may save time!