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1 year ago

Let x and y be real numbers satisfying 2/x=y/3=x/y Determine the value of x^3

Mathematics

1 answer:

svet-max [94.6K]1 year ago

8 0

Answer:

Step-by-step explanation:

If x and y be real numbers satisfying 2/x=y/3=x/y, then any two of the equation are equated as shown;

2/x = y/3 ... 1 and;

y/3 = x/y... 2

From equation 1, 2y = 3x ... 3

and from equation 2; y² = 3x ... 4

Equating the left hand side of equation 3 and 4 since their right hand sides are equal, we will have;

2y = y²

2 = y

y = 2

Substituting y = 2 into equation 3 to get the value of x;

2y = 3x

2(2) = 3x

4 = 3x

x = 4/3

The value of x³ will be expressed as (4/3)³ = 4*4*4/3*3*3 = 64/27

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Lemur [1.5K]

Answer:

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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

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Do, integration by parts

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$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-20te^{-0.6t}}{0.6}+\frac{20e^{-0.6t}}{(0.6)^2}]^{10}_{0}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=[\frac{-200e^{-6}}{0.6}+\frac{20e^{-6}}{(0.6)^2}]+\frac{20}{(0.60^2}$

$[\int{20te^{-0.6t}} \, dt]^{10}_0=\frac{20(1-e^{-6}}{(0.6)^2}-\frac{200e^{-6}}{0.6}$

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Hence Cardiac output: $F=0.055 L\s$

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