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The quantity y varies directly with the square of x. If y=24 when x=3, find y when x is 4

Sagot :

MrRoyal

Answer:

[tex]y = \frac{384}{9}[/tex]

Step-by-step explanation:

Given

[tex]y\ \alpha\ x^2[/tex] --- direct variation

[tex](x,y) = (3,24)[/tex]

Required

y when x = 4

[tex]y\ \alpha\ x^2[/tex]

Express as an equation

[tex]y = kx^2[/tex]

Substitute: [tex](x,y) = (3,24)[/tex]

[tex]24 = k*3^2[/tex]

[tex]24 = k*9[/tex]

Solve for k

[tex]k = \frac{24}{9}[/tex]

To solve for y when x = 4, we have:

[tex]y = kx^2[/tex]

[tex]y = \frac{24}{9} * 4^2[/tex]

[tex]y = \frac{24}{9} * 16[/tex]

[tex]y = \frac{24 * 16}{9}[/tex]

[tex]y = \frac{384}{9}[/tex]

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