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The problem provides two points, \( A(-2, -4) \) and \( B(4, 6) \), and asks for the value of \( k \) such that the point \( (k, 3) \) lies on the continuous curve AB. 

Assuming that the curve is a straight line between points A and B (since no other information about the curve is provided), we can find the equation of the line passing through points A and B. After that, we can substitute \( y(k) = 3 \) to find the corresponding \( k \)-value.

### Step 1: Find the slope of the line
The slope of a line through two points \((x_1, y_1)\) and \((x_2, y_2)\) is given by:

\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

For points \( A(-2, -4) \) and \( B(4, 6) \):

\[
m = \frac{6 - (-4)}{4 - (-2)} = \frac{6 + 4}{4 + 2} = \frac{10}{6} = \frac{5}{3}
\]

### Step 2: Equation of the line
Using the point-slope form of the equation of a line, which is:

\[
y - y_1 = m(x - x_1)
\]

Substitute \( m = \frac{5}{3} \) and point \( A(-2, -4) \):

\[
y - (-4) = \frac{5}{3}(x - (-2))
\]
\[
y + 4 = \frac{5}{3}(x + 2)
\]

Simplify:

\[
y + 4 = \frac{5}{3}x + \frac{10}{3}
\]
\[
y = \frac{5}{3}x + \frac{10}{3} - 4
\]
\[
y = \frac{5}{3}x - \frac{12}{3}
\]
\[
y = \frac{5}{3}x - 4
\]

### Step 3: Find \( k \) such that \( y(k) = 3 \)
Substitute \( y = 3 \) into the equation:

\[
3 = \frac{5}{3}k - 4
\]

Now, solve for \( k \):

\[
3 + 4 = \frac{5}{3}k
\]
\[
7 = \frac{5}{3}k
\]
\[
k = \frac{7 \times 3}{5} = \frac{21}{5}
\]

Thus, \( k = \frac{21}{5} \) or 4.2.

### Final Answer:
The value of \( k \) is \( \frac{21}{5} \) or approximately 4.2.

Would you like more details, or do you have any questions?

Here are 5 related questions for further exploration:
1. How would the process change if the curve between A and B was not a straight line?
2. What is the importance of the slope in determining the equation of a line?
3. How do you find the distance between two points in a plane?
4. Can you express the equation of the line in standard form?
5. How would the answer change if \( y(k) \) were a different value?

**Tip**: Always check your line equation by substituting the original points to confirm it's correct!

Given A (-2, -4), B(4,6), if curve AB is continuous, y(k)=3, k=?

This problem explores finding the value of k such that the point (k,3) lies on the line between A(-2, -4) and B(4, 6). Learn how to calculate the slope and use the equation of a line to solve this problem. Suitable for students in Grades 9-11.

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