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    Calculus Early Transcendentals
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    Exam 4: Integrals
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    Find the Absolute Maximum Value of Y =\(\sqrt{36-x^{2}}\) On the Interval
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Find the Absolute Maximum Value of Y = 36−x2\sqrt{36-x^{2}}36−x2​ On the Interval

Question 94

Question 94

Short Answer

Find the absolute maximum value of y = 36−x2\sqrt{36-x^{2}}36−x2​ on the interval [−9,9][-9,9][−9,9] .

Correct Answer:

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