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3 years ago

Help me with this please!!!!

Mathematics

1 answer:

EleoNora [17]3 years ago

7 0

Answer: the volume is 3.75

Step-by-step explanation:

Lengin x width x Height= Volume

2.5 x 1.5 x 1 = 3.75

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Answer number 20. Please answer all parts to the question.

emmasim [6.3K]

1: 7
2: Yes, 20 because it’s isolated from the rest of the points
(I don’t know the other ones, sorryyyyy)

5 0

2 years ago

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: $F=0.055 L\s$

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3 years ago

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tiny-mole [99]

Answer:

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Step-by-step explanation:

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