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Mathematics - Algebra

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I'm having a hard time figuring out the solution.

Find the center and vertices.
graph: x2−4y2=16

Standard Form:
Identify which Conic Section:
Characteristics applicable:
Center:
Vertices:
Asymptotes:

Complete the square, identify the graph of the equation, identify the characteristics
of the conic section and graph: 4x2+25y2−32x+150y+189=0

Standard Form:
Identify which Conic Section:
Characteristics applicable:
Center:
Vertices:
Asymptotes:

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Solution Summary

Complete, neat and step-by-step solutions are provided that assists with solving the precalculus probelm. Graph provided in the attached file.

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(1) x^2 - 4y^2 = 16

x^2 /16 - y^2 /4 = 1

Hyperbola

Center is (0, 0)

Vertices are at (4, 0) and ...

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