Px350x8600 Px350x8600 Px350x 2 8600 Px350x28600 Px350x 2 7x8600 Px350x27x8600

Px350x8600 px350x8600 px350x 2 8600 px350x28600

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P(x)=3.50x−8600 P(x)=3.50x−8600 P(x)=3.50x 2 +8600 P(x)=3.50x2+8600 P(x)=3.50x 2 −7x−8600 P(x)=3.50x2−7x−8600 The revenue function R(x)=5.75x R(x)=5.75x. The cost function C(x)=2.25x+8600 C(x)=2.25x+8600 P(x)=R(x)−C(x)=5.75x−(2.25x+8600)=3.50x−8600 P(x)=R(x) −C(x)=5.75x−(2.25x+8600)=3.50x−8600 Question 11 1.67 / 1.67 pts Simplify the expression: ( 16x 2 y 2 3x −2 y 4 ) 12 (16x2y23x−2y4)12 8√xy3√ 8xy3
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43√y 43y 43 xy 43xy Correct! 4x 2 3√y 4x23y 8x3y 3 8x3y3 (16x 2 y 2 3x −2 y 4 ) 12 =(16x 4 3y 2 ) 12 =4x 2 3– √y (16x2y23x−2y4)12=(16x43y2)12=4x23y Question 12 0 / 1.66 pts The following formula describes the quantity Q of a substance in milligrams (mg) as a function of time t, in hours. Q=100(0.97) t . Q=100(0.97)t. Choose the correct statement below. You Answered When t=0, Q=97mg. Q is increasing exponentially at 97% per hour. When t=0, Q=100mg. Q is decreasing linearly at 0.97 mg per hour. When t=0, Q=0.97mg. Q is increasing exponentially at 100% per hour. When t=0, Q=0mg. Q is increasing linearly at 100mg per hour.
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Correct Answer When t=0, Q=100mg. Q is decreasing exponentially at 3% per hour. When t=0 t=0, Q=100(0.97) 0 =100⋅1=100 Q=100(0.97)0=100 1=100 mg. Q Q is decreasing exponentially, at 3% 3% per hour. The base 0.97 0.97 corresponds to a 3% 3% decrease. Question 13 1.67 / 1.67 pts A quantity P is a exponential function of time t, that is P=P 0 a t , P=P0at, where a is a positive constant and P 0 P0 is the quantity at t=0. Suppose that P=100 when t=1,P=75 when t=4 P=100 when t=1,P=75 when t=4 Use this information to find a and P 0 P0 and state the rate of exponential growth or decay. a=0.75, P 0 =133 P0=133 , exponential decay at a rate of 25%. a=1.09, P 0 =110 P0=110 , exponential growth at a rate of 9%. a=1.05, P 0 =125 P0=125 , exponential growth at a rate of 5%.
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Correct! a=0.91, P 0 =110 P0=110 , exponential decay at a rate of 9%. a=0.86, P 0 =137 P0=137 , exponential decay at a rate of 14%. We have two equations and two unknowns. We can eliminate P 0 P0 if we first solve each equation for P 0 P0. This first step gives us two equations for P 0 P0. P 0 = 100a ,P 0 = 75a 4 P0=100a,P0=75a4 We have two expressions that both equal P 0 P0, so we can set them equal to each other, and then solve for a a: Use a=0.91 a=0.91 in either above equation to find P 0 P0. You should get the same answer either way so you can use both as a check. P 0 = 1000.91 =110 P0=1000.91=110 Question 14 1.67 / 1.67 pts Suppose that a certain pain medication is eliminated from the body (by metabolism) according to the model
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Q=150(0.70) t , Q=150(0.70)t, where Q is the quantity of the medication in the body in milligrams (mg) and t in the time in hours since the medication was administered. How much medication was administered (include the units)? What is the hourly decay rate of the medication? Correct! 150 mg of medication was administered. The hourly decay rate is 30%. 150 mg of medication was administered. The hourly decay rate is 70%. 105 mg of medication was administered. The hourly decay rate is 90%. 70 mg of medication was administered. The hourly decay rate is 30%. 0.70 mg of medication was administered. The hourly decay rate is 150%.
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