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wel
2 years ago
10

If one row in an echelon form of an augmented matrix is left bracket Start 1 By 5 Matrix 1st Row 1st Column 0 2nd Column 0 3rd C

olumn 0 4st Column 5 5st Column 0 EndMatrix right bracket 0 0 0 5 0 ​, then the associated linear system is inconsistent.Is this statement true or​ false?
Mathematics
1 answer:
Step2247 [10]2 years ago
8 0

Answer:

Inconsistent - TRUE

Step-by-step explanation:

A system of linear equation is said to be inconsistent if it has no solution. that is from the graph, the lines does not intersect compared to a consistent equations that has solutions and intersect on the graph.

From the written augmented matrix = ( 0)

                                                               ( 0)

                                                               ( 0)

                                                               ( 5)

                                                               ( 0)

from the second row of the matrix, it is evident that it has no solutions hence INCOSISTENT

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Answer:

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Step-by-step explanation:

The third degree Taylor polynomial for the cosine function centered at a = \frac{\pi}{2} is:

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86^{\circ} = \frac{86^{\circ}}{180^{\circ}}\times \pi

86^{\circ} = \frac{43}{90}\pi\,rad

Then, the cosine of 86º is:

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\cos 86^{\circ} \approx 0.06976

The cosine of 86º is approximately 0.06976.

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The zero product property is best applied to factorable quadratic equations in this case.

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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)
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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

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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}

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Substitute the value in 1

Cardiac output:F=\frac{3}{54.49}

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Hence Cardiac output:F=0.055 L\s

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