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

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### **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.

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### **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) \( 6 + \sqrt{13} \).

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### **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) \( x^2 + (y - 10)^2 = 9 \).

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Would you like further details on any of these solutions?

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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!

The user submitted questions from the image including solving a linear function, determining values from a quadratic function graph, solving a quadratic equation, and finding the equation of a circle.

This page provides detailed solutions to problems involving linear functions, quadratic equations, and the equation of a circle. Learn how to solve f(x) = (2/5)x + 10, determine the values of constants in quadratic functions, and find the equation of a circle given endpoints of its diameter. Suitable for high school students in Grades 9-12.

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