A vase can be modeled using x^2 / 5.5225 - (y-5)^2 / 42.25 = 1...
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A vase can be modeled using x^2 / 5.5225 - (y-5)^2 / 42.25 = 1 and the x-axis, for 0 ? y ? 20, where the measurements are in inches. Using the graph, what is the distance across the base of the vase, and how does it relate to the hyperbola?5.93 inches; distance between the x-intercepts4.50 inches; length of the transverse axis2.97 inches; distance between the intercepts2.35 inches; length of the transverse axis
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