Questions: 6x+y=0 Function O Not a function x^2+7y=6 Function O Not a function x^2+y^2=4 Function O Not a function y=sqrt(x-1) Function O Not a function

Transcript text: \begin{tabular}{|c|lc|} \hline $6 x+|y|=0$ & $\bigcirc$ Function & O Not a function \\ \hline$x^{2}+7 y=6$ & Function & O Not a function \\ \hline$x^{2}+y^{2}=4$ & Function & O Not a function \\ \hline$y=\sqrt{x-1}$ & Function & O Not a function \\ \hline \end{tabular}

Solution

Solution Steps

Given Equation: $y = |6x + 1|$

Step 1: Check if y is a function of x

Since the absolute value function passes the Vertical Line Test, y is a function of x.

Final Answer:

Yes, the given equation defines y as a function of x.

Given Equation: $x + by = 6$

Step 1: Solve for y

Rearrange the equation to solve for y: $y = \frac{6 - x}7$ Since for every value of x, there is exactly one corresponding value of y, y is a function of x.

Final Answer:

Yes, the given equation defines y as a function of x.

Given Equation: $x^2 + by^2 = 4$

Step 1: Attempt to solve for y

Rearrange the equation: $by^2 = {c} - {a}{variable_name}^2$ Solving for y gives: $y = \pm\sqrt{\frac{{{c} - {a}{variable_name}^2}}{b}}$ Since for a single value of x, we get two values of y, y is not a function of x based on the Vertical Line Test.

Final Answer:

No, the given equation does not define y as a function of x.

Given Equation: $1\sqrt{x-1} + y = 1$

Step 1: Solve for y

Rearrange the equation to solve for y: $y = 1 - 1\sqrt{x-1}$ Since for every value of x, there is exactly one corresponding value of y, y is a function of x.

Final Answer:

Yes, the given equation defines y as a function of x.

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