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3 years ago

Please someone help me out. I beggggg

Mathematics

1 answer:

bezimeni [28]3 years ago

4 0

Answer:

1. 144 2. 16 3. 1 4. 3x-6

Step-by-step explanation:

So think of this as a function in a function. So you work from the inside to the outside. So for problem 1, we start with f(4)) [you read it "f of 4"] so what is the solution when x = 4, since f(x) means the function of x so f(4) means 'the function of 4' inside f(x).

Since f(x) = 3x then f(4) = 3(4) [notice how you substitute the 4 everywhere you see a letter x]

so f(4) = 12, now you work the next part h(f(4)) since f(4)=12 then h(12)

So take the h(x) function which is h(x) = $x^{2}$ then h(12) = $12^{2}$ so h(12) = 144

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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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3 years ago

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3 years ago

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3 years ago

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3 years ago

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andre [41]

Answer:

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