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notka56 [123]
3 years ago
14

A student answered 86 problems on a test correctly and received a grade 98%. How many problems were on the test, if all the prob

lems were worth the same number of points?
Mathematics
1 answer:
irakobra [83]3 years ago
5 0

Answer:

87                

Step-by-step explanation:

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5. A tank initially contains 20 lb of salt dissolved in 200-gallon of water. Starting at time t = 0, a solution containing 0.5 l
sladkih [1.3K]

Answer:

(a) Amount of salt as a function of time

A(t)=100-80e^{-0.02t}

(b) The time at which the amount of salt in the tank reaches 50 lb is 23.5 minutes.

(c) The amount of salt when t approaches to +inf is 100 lb.

Step-by-step explanation:

The rate of change of the amount of salt can be written as

\frac{dA}{dt} =rate\,in\,-\,rate\,out\\\\\frac{dA}{dt}=C_i*q_i-C_o*q_o=C_i*q_i-\frac{A(t)}{V}*q_o\\\\\frac{dA}{dt}=0.5*4-\frac{A(t)}{200}*4\\\\\frac{dA}{dt} =\frac{100-A(t)}{50}=-(\frac{A(t)-100}{50})

Then we can rearrange and integrate

\frac{dA}{dt}= -(\frac{A(t)-100}{50})\\\\\int \frac{dA}{A-100}=-\frac{1}{50}  \int dt\\\\ln(A-100)=-\frac{t}{50}+C_0\\\\ A-100=Ce^{-0.02*t}\\\\\\A(0)=20 \rightarrow 20-100=Ce^0=C\\C=-80\\\\A=100-80e^{-0.02t}

Then we have the model of A(t) like

A(t)=100-80e^{-0.02t}

(b) The time at which the amount of salt reaches 50 lb is

A(t)=100-80e^{-0.02t}=50\\\\e^{-0.02t}=(50-100)/(-80)=0.625\\\\-0.02*t=ln(0.625)=-0.47\\\\t=(-0.47)/(-0.02)=23.5

(c) When t approaches to +infinit, the term e^(-0.02t) approaches to zero, so the amount of salt in the solution approaches to 100 lb.

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3 years ago
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Answer:

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Step-by-step explanation:

y=f(x)=2^3

y=f(x)=8

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Graph the equation y=×^2-9
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