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nikitadnepr [17]
3 years ago
5

Please answer this question

Mathematics
2 answers:
PolarNik [594]3 years ago
6 0

Answer:

9

Step-by-step explanation:

(-1)^4-(-1)^3+(-1)^2+6

1-(-1)+1+6

1+1+1+6

3+6=9

kotegsom [21]3 years ago
6 0

Answer:

Option 2

Step-by-step explanation:

p(t) = t^4-t^3+t^2+6

<u><em>Putting t = -1</em></u>

p(-1) = (-1)⁴-(-1)³+(-1)²+6

p(-1) = 1-(-1)+1+6

p(-1) = 1+1+7

p(-1) = 9

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Answer:7pi\4

Step-by-step explanation:

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Clair collects seashells. At home, she has 15 shells. Every day she brings 3 back to the beach. After 5 days how many seashells
Igoryamba

The equation would be:

y = 3x + 15

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5 0
3 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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