I need help with these three math problems: 8. Determine the type of function shown. \( m(x) = \frac{13}{6} \) A. Linear B. Quadratic C. None of these D. Constant 6. Does the following equation determine y as a function of x? \( y^2 = x + 3 \) A. No B. Yes C. Only when x = 1 D. Sometimes 16. Solve the system. \( y = \frac{1}{7}x - 4 \) \( x = 7y + 1 \) A. No solution B. (7, –3) C. (–13, –2) D. There are an infinite number of solutions

8. The function \( m(x) = \frac{13}{6} \) is a function that gives the same output regardless of the input x. Since the output does not depend on the input, and there is no variable attached to x, it is a constant function. Therefore, the correct answer is:

D. Constant

6. The equation \( y^2 = x + 3 \) does not define y as a function of x because for each x there could be two different y values (one positive and one negative) that satisfy the equation. This is because when we solve for y we get \( y = \pm\sqrt{x + 3} \), and a function must assign exactly one output for each input in its domain. Hence, the correct answer is:

A. No

16. To solve the system of equations, we can substitute the second equation into the first. Given:

\( y = \frac{1}{7}x - 4 \) \( x = 7y + 1 \)

Substitute the value of x from the second equation into the first:

\( y = \frac{1}{7}(7y + 1) - 4 \)

Multiply both sides of the equation:

\( y = y + \frac{1}{7} - 4 \)

Now, subtract y from both sides:

\( 0 = \frac{1}{7} - 4 \)

This does not hold true for any y, indicating that there is no solution that satisfies both equations. Therefore, the correct answer is:

A. No solution