Can somebody please help me?

1.) If √x = 3, then what is the minimum value of x?

2.) [tex]\frac{6}{2\pi }[/tex]y² ÷ 0.4y/x
Leave your answer in terms of pi.

Respuesta :

Part A

We know that √x = 3

Squaring both sides

x = 3²

⇒ x = 9

Hence, the value of x or the minimum value of x is 9

Part B

([tex]\frac{6}{2\pi }[/tex]*y²) ÷ (0.4*[tex]\frac{y}{x}[/tex]) = ([tex]\frac{6}{2\pi }[/tex]*y²) * [tex]\frac{x}{0.4y}[/tex]

⇒ ([tex]\frac{6}{2\pi }[/tex]*y²) ÷ (0.4*[tex]\frac{y}{x}[/tex]) = [tex]\frac{7.5}{\pi y}[/tex]*yx