What are expressions that are equivalent to -4/3p

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

Simplify these expressions

Xelga [282]

$\frac{10a}{2} \\ \\ 5a \\ \\$

The answer is: 5a

-----------

$\frac{36b}{6b} \\ \\ \frac{36}{6} \\ \\ 6 \\ \\$

The answer is: 6

------------

$\frac{16c^2}{8c} \\ \\ 2c^{2-1} \\ \\ 2c^1 \\ \\ 2c \\ \\$

The answer is: 2c

------------

$\frac{18ab^2}{2b} \\ \\ 9ab^{2-1} \\ \\ 9ab^1 \\ \\ 9ab \\ \\$

The answer is: 9ab

------------

$\frac{40b^2c^2}{5bc} \\ \\ 8b^{2-1} c^{2-1} \\ \\ 8b^1c^1 \\ \\ 8bc^1 \\ \\ 8bc \\ \\$

The answer is: 8bc

-----------

$\frac{63ab^2c^3}{7b^2c^2} \\ \\ \frac{63ac^3}{7c^2} \\ \\ 9ac^{3-2} \\ \\ 9ac^1 \\ \\ 9ac \\ \\$

The answer is: 9ac

6 0

3 years ago

Which inequality represent the statement: John is at least 5 feet tall.

belka [17]

Answer:

$h\geq 5 \text{ feet}$