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stiks02 [169]
3 years ago
14

Solve 2.02W = -3.636

Mathematics
2 answers:
Westkost [7]3 years ago
8 0

Answer:

W = -1.8

Step-by-step explanation:

Step 1: Divide 2.02 on both sides.

W = -1.8

ValentinkaMS [17]3 years ago
7 0

Answer: W=-1.8

Step-by-step explanation:

Divide both sides by 2.02 and you are left with W=-1.8.

Hope this helps!

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The dye dilution method is used to measure cardiac output with 3 mg of dye. The dye concentrations, in mg/L, are modeled by c(t)
Lemur [1.5K]

Answer:

Cardiac output:F=0.055 L\s

Step-by-step explanation:

Given : The dye dilution method is used to measure cardiac output with 3 mg of dye.

To Find : Find the cardiac output.

Solution:

Formula of cardiac output:F=\frac{A}{\int\limits^T_0 {c(t)} \, dt} ---1

A = 3 mg

\int\limits^T_0 {c(t)} \, dt =\int\limits^{10}_0 {20te^{-0.06t}} \, dt

Do, integration by parts

[\int{20te^{-0.6t}} \, dt]^{10}_0=[20t\int{e^{-0.6t} \,dt}-\int[\frac{d[20t]}{dt}\int {e^{-0.6t} \, dt]dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20}{0.6}\int {e^{-0.6t} \,dt]^{10}_0

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}

[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}

[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}

[\int{20te^{-0.6t}} \, dt]^{10}_0\sim {54.49}

Substitute the value in 1

Cardiac output:F=\frac{3}{54.49}

Cardiac output:F=0.055 L\s

Hence Cardiac output:F=0.055 L\s

4 0
3 years ago
A car rental agency rents 190 cars per day at a rate of 29 dollars per day. for each 1 dollar increase in the daily rate, 5 fewe
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The equation that we can create from this situation is:

i = (190 – 5 x) * (29 + x)

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Expanding the equation:

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Taking the 1st derivative:

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So the cars should be rented at:

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The maximum income is:

i = (190 – 5*4.5) * (33.5)

i = 5,611.25 dollars

7 0
3 years ago
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