Suppose that y varies directly as the square root of x, and that y = 49 when x = 49. What is y when x = 128? Round your answer to two decimal places if necessary

Respuesta :

Since y varies directly with square root x, so

[tex]y=k\sqrt{x}[/tex]

k is the constant of variation

We will find it using the initial values of x and y

Since the initial values are:

y = 49

x = 49

Substitute them in the rule to find k

[tex]\begin{gathered} 49=k\sqrt{49} \\ 49=k(7) \\ 49=7k \\ \frac{49}{7}=\frac{7k}{7} \\ 7=k \end{gathered}[/tex]

The value of k is 7, so the equation will be

[tex]y=7\sqrt{x}[/tex]

We need to find the value of y at x = 128, so

Substitute x in the rule by 128

[tex]\begin{gathered} y=7\sqrt{128} \\ y=56\sqrt{2}=79.1959595 \end{gathered}[/tex]

Round it to two decimal places, so

y = 79.20