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3 years ago

Line segment YV of rectangle YVWX measures 24 units.

Mathematics

1 answer:

Anna [14]3 years ago

7 0

Answer: 8√3

<u>Step-by-step explanation:</u>

ΔXVW is a 30°-60°-90° triangle. This special triangle has corresponding sides of lengths: b - b√3 - 2b. So,

VW = b

XW = b√3

XV = 2b

Since, YV = 24, then XW = 24 and we stated above that XW = b√3. So,

24 = b√3

$\frac{24}{\sqrt{3}} = \frac{b\sqrt{3}}{\sqrt{3}}$

$\frac{24}{\sqrt{3}}$ = b

$\frac{24}{\sqrt{3}}(\frac{\sqrt{3}}{\sqrt{3}})$ = b

$\frac{24\sqrt{3}}{3}$ = b

8√3 = b

We are looking for side YX which is congruent (equal) to VW and we stated above that VW = b, So, VW = 8√3

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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.

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

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

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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