During The Testing Of A Certain Brand Of Air Purif

Anonymous

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During the testing of a certain brand of air purifier, the amount of smoke remaining t min after the start of the test was A(t)=-0.00006t5 + 0.000468t4 0.1316t3 + 1.915t2 17.63t + 100 percent of the original amount. PART 1: Compute A(15), A(15). a. A(15)=-3.46022% per minute A(15)=0.39776% per minute in the second power b. A(15)=-2.00166% per minute A(15)=0.39776% per minute in the second power c. A(15)=-2.00166% per minute A(15)=0.41984% per minute in the second power d. A(15)=-1.0175% per minute A(15)=0.572% per minute in the second power e. A(15)=-3.46022% per minute A(15)=0.41984% per minute in the second power PART 2: Compute A(3), A(3). Round answers to the nearest thousandth. A(3)=________________________% per minute A(3)=________________________% per minute in the second power

 

Answer No.1

TinyMary5

From: US
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Votes: 12
A(t)=-0.00006t^5 + 0.000468t^4 0.1316t^3 + 1.915t^2 17.63t + 100 A'(t)=-0.00006*5t^4 + 0.000468*4t^3 0.1316*3t^2 + 1.915*2t 17.63 A'(15) = -0.579%/min A"(t)=-0.00006*5*4t^3 + 0.000468*4*3t^2 0.1316*3*2t + 1.915*2 A"(15) = -10.800%/min^2 A'(3) = -9.667 %/min A"(3) = 1.479 %/min^2

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