If y varies directly as the square of x, and y=20 when x=3, find y when x=9.
Question
Updated 3/23/2014 7:18:57 PM
This conversation has been flagged as incorrect.
Flagged by yeswey [3/23/2014 7:12:45 PM]
Original conversation
User: If y varies directly as the square of x, and y=675 when x=25, find y when x=50.

User: If y varies directly as the square of x, and y=20 when x=3, find y when x=9.

Weegy: If y is 5 when x is 2.5 and y varies directly with x. When x = 10, y = ? 5/2.5 = y/10; y = 50/2.5; y = 20;
erikalui|Points 367|

Question
Updated 3/23/2014 7:18:57 PM
This conversation has been flagged as incorrect.
Flagged by yeswey [3/23/2014 7:12:45 PM]
Rating
3
y varies directly as the square of x, the equation is y = kx^2,
y=675 when x=25, so k = y/x^2 = 675/(25)^2 = 1.08
The equation is y = 1.08x^2
when x = 50, y = 1.08*(50)^2 = 2700
Confirmed by jeifunk [3/23/2014 7:31:10 PM]
3
y varies directly as the square of x, the equation is y = kx^2,
y=20 when x=3, so k = y/x^2 = 20/3^2 = 20/9
The equation is y = (20/9)x^2
when x = 9, y = (20/9)*9^2 = 180
Confirmed by jeifunk [3/23/2014 7:31:10 PM]

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