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Mathematics, 23.10.2020 17:20 elizabethxoxo3271

PLZ HELP The hyperbolic cross section of a cooling tower is given by the equation 4x2 βˆ’ y2 + 16y βˆ’ 80 = 0. The center of the cooling tower is the same as the center of the hyperbola, and the x-axis represents the ground surface.

The diameter at the center of the tower is

meters. The center of the tower is

meters above the ground.


PLZZZZ HELP

The hyperbolic cross section of a cooling tower is given by the equation 4x2 βˆ’ y2 + 1

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