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2 years ago

I’ll give brainliest

Mathematics

2 answers:

adell [148]2 years ago

5 0

Answer:

Area》22.2 ft^2

Step-by-step explanation:

Hope it helps ...

dlinn [17]2 years ago

3 0

Answer:

22.2

Step-by-step explanation:

Hello There!

This is the sector area formula

$A=\frac{0}{360} (\pi r^2)$

where 0 = sector angle

and r = radius

The given sector has a measure of 50 degrees and a radius of 7.13

Knowing this information we plug it into the formula

$A=\frac{50}{360} (\pi 7.13^2)\\\frac{50}{360} =0.1388888889\\7.13^2=50.8369\\50.8369\pi =159.7088316\\159.7088316*0.1388888889=22.18178216$

so we can conclude that the area of the sector is 22.18178216 ft²

finally we round to the nearest tenth

and get that the answer is 22.2

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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: $F=0.055 L\s$

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Answer:

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Step-by-step explanation:

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Answer:

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