Question

which formulas represent linear relationships? select all that apply. the area of a circle, ( a=pi r^{2} ) the perimeter of a regular hexagon, ( p = 6s ) the volume of a sphere, ( v=\frac{4}{3}pi r^{3} ) the perimeter of an isosceles right triangle, ( p=(2+sqrt{2})s )

Explanation:

Step1: Recall the form of a linear relationship

A linear relationship has the form \(y = mx + b\) (where \(b = 0\) in many simple cases of direct proportion), and the highest power of the variable is \(1\).

Step2: Analyze \(A=\pi r^{2}\)

The formula for the area of a circle \(A = \pi r^{2}\) has the variable \(r\) with a power of \(2\). Since the power of \(r\) is not \(1\), it is not a linear relationship.

Step3: Analyze \(P = 6s\)

The formula for the perimeter of a regular hexagon \(P=6s\) can be written as \(P=6s+0\). Here, the variable \(s\) has a power of \(1\), and it is in the form \(y = mx + b\) (\(m = 6\), \(b = 0\)). So it is a linear relationship.

Step4: Analyze \(V=\frac{4}{3}\pi r^{3}\)

The formula for the volume of a sphere \(V=\frac{4}{3}\pi r^{3}\) has the variable \(r\) with a power of \(3\). Since the power of \(r\) is not \(1\), it is not a linear relationship.

Step5: Analyze \(P=(2 + \sqrt{2})s\)

The formula for the perimeter of an isosceles right - triangle \(P=(2+\sqrt{2})s\) can be written as \(P=(2 + \sqrt{2})s+0\). The variable \(s\) has a power of \(1\), and it is in the form \(y=mx + b\) (\(m=(2+\sqrt{2})\), \(b = 0\)). So it is a linear relationship.

Answer:

the perimeter of a regular hexagon, \(P = 6s\); the perimeter of an isosceles right triangle, \(P=(2+\sqrt{2})s\)