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THE EDGES OF A CUBE ARE EXPANDING A A RATE OF 5 CENTIMETERS PER SECOND.HOW FAST IS THE SURFACE AREA CHANGING WHEN EACH EGDE IS 4.5 CENTIMETERS??
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A=6xr^2 dA/dt=6xd(r^2)/dt=6x2rxdr/dt=12rxdr/dt at r=4.5 and dr/dt=5 dA/dt=12x4.5x5=270 cm^2/sec
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