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    Calculus Early Transcendentals
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    Exam 4: Integrals
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    Find the Absolute Minimum Value(s) Of\(y=2 x^{2}-20 x+9\) On the Interval [0, 6]
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Find the Absolute Minimum Value(s) Of y=2x2−20x+9y=2 x^{2}-20 x+9y=2x2−20x+9 On the Interval [0, 6]

Question 90

Question 90

Short Answer

Find the absolute minimum value(s) of y=2x2−20x+9y=2 x^{2}-20 x+9y=2x2−20x+9 on the interval [0, 6].

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