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Luba_88 [7]
2 years ago
15

\Suppose that a person has an average heart rate of 88.6 beats/min. (Express your answers to problems in this section to the cor

rect number of significant figures and proper units.) (a) How many beats does he or she have in 5.0 y
Mathematics
1 answer:
vaieri [72.5K]2 years ago
8 0

Answer:

He or she has 232,840,800 beats in five minutes.

Step-by-step explanation:

This question is solved by proportions.

In this question:

A year has 365 days.

Each day has 24 hours.

Each hour has 60 minutes.

Minutes in five years:

5*365*24*60 = 2,628,000 minutes.

How many beats in five minutes?

88.6 beats/min, 2,628,000 minutes. So the number of bears is:

88.6*2628000 = 232,840,800

He or she has 232,840,800 beats in five minutes.

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You would need 16apples
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Expand (x – 4)^5 using the Binomial Theorem and Pascal’s triangle
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5th \: row = 1 \:  \: \:  5 \: \:   \: 10 \:  \:  \: 10 \:  \: \:  5 \: \:   \: 1

(x - 4) {}^{5}  = x {}^{5}  + 5x {}^{4} ( - 4) {}^{1}  + 10x {}^{3} ( - 4) {}^{2}  + 10x {}^{2} ( - 4) {}^{3}  + 5x( - 4) {}^{4}  + ( - 4) {}^{5}

(x - 4)  {}^{5}  = x {}^{5}  - 20x {}^{4}  + 160x {}^{3}  - 640x {}^{2}  + 1280x  - 1024

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2 years ago
Elizabeth company has 10,000 units of their new product in warehouse. They sold 10,000 in pre-order that will need to be shipped
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Answer:

m = 300w OR w = m/300

Step-by-step explanation:

They start with 10k, but that's immediately sold, so they're back at a flat 0. Every week they gain 1000 but lose 700 for a net profit of positive 300.

Therefore, the maximum capacity would be 300 multiplied by the number of weeks. If we know the maximum, then the answer would be m (maximum) divided by 300.

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Let y 00 + by0 + 2y = 0 be the equation of a damped vibrating spring with mass m = 1, damping coefficient b > 0, and spring c
stira [4]

Answer:

Step-by-step explanation:

Given that:    

The equation of the damped vibrating spring is y" + by' +2y = 0

(a) To convert this 2nd order equation to a system of two first-order equations;

let y₁ = y

y'₁ = y' = y₂

So;

y'₂ = y"₁ = -2y₁ -by₂

Thus; the system of the two first-order equation is:

y₁' = y₂

y₂' = -2y₁ - by₂

(b)

The eigenvalue of the system in terms of b is:

\left|\begin{array}{cc}- \lambda &1&-2\ & -b- \lambda \end{array}\right|=0

-\lambda(-b - \lambda) + 2 = 0 \ \\ \\\lambda^2 +\lambda b + 2 = 0

\lambda = \dfrac{-b \pm \sqrt{b^2 - 8}}{2}

\lambda_1 = \dfrac{-b + \sqrt{b^2 -8}}{2} ;  \ \lambda _2 = \dfrac{-b - \sqrt{b^2 -8}}{2}

(c)

Suppose b > 2\sqrt{2}, then  λ₂ < 0 and λ₁ < 0. Thus, the node is stable at equilibrium.

(d)

From λ² + λb + 2 = 0

If b = 3; we get

\lambda^2 + 3\lambda + 2 = 0 \\ \\ (\lambda + 1) ( \lambda + 2 ) = 0\\ \\ \lambda = -1 \ or   \  \lambda = -2 \\ \\

Now, the eigenvector relating to λ = -1 be:

v = \left[\begin{array}{ccc}+1&1\\-2&-2\\\end{array}\right] \left[\begin{array}{c}v_1\\v_2\\\end{array}\right] = \left[\begin{array}{c}0\\0\\\end{array}\right]

\sim v = \left[\begin{array}{ccc}1&1\\0&0\\\end{array}\right] \left[\begin{array}{c}v_1\\v_2\\\end{array}\right] = \left[\begin{array}{c}0\\0\\\end{array}\right]

Let v₂ = 1, v₁ = -1

v = \left[\begin{array}{c}-1\\1\\\end{array}\right]

Let Eigenvector relating to  λ = -2 be:

m = \left[\begin{array}{ccc}2&1\\-2&-1\\\end{array}\right] \left[\begin{array}{c}m_1\\m_2\\\end{array}\right] = \left[\begin{array}{c}0\\0\\\end{array}\right]

\sim v = \left[\begin{array}{ccc}2&1\\0&0\\\end{array}\right] \left[\begin{array}{c}m_1\\m_2\\\end{array}\right] = \left[\begin{array}{c}0\\0\\\end{array}\right]

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∴

\left[\begin{array}{c}y_1\\y_2\\\end{array}\right]= C_1 e^{-t}  \left[\begin{array}{c}-1\\1\\\end{array}\right] + C_2e^{-2t}  \left[\begin{array}{c}-1/2\\1\\\end{array}\right]

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\mathbf{ \left[\begin{array}{c}y_1\\y_2\\\end{array}\right]=  \left[\begin{array}{c}0\\0\\\end{array}\right] \ \  so \ stable \ at \ node \ \infty }

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Explanation:

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