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lyudmila [28]
3 years ago
8

An employee who earned $550 a week working 35 hours had her pay increased by 5 percent. Later, her hours were reduced to 30 per

week, but the new hourly rate of pay was retained. What was her new amount of weekly pay?
Mathematics
1 answer:
Taya2010 [7]3 years ago
3 0

Answer:

  $495

Step-by-step explanation:

After the 5% raise, her weekly pay was ...

  $550 × 1.05 = $577.50

If she works 35 hours for that pay, her hourly rate is

  $577.50/35 = $16.50

Then, working 30 hours, her weekly pay will be ...

  30 × $16.50 = $495.00

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UkoKoshka [18]

Answer:

m = 30, b = 20

Step-by-step explanation:

1. The equation is written in slope-intercept form, so it follows the following equation: y = mx + b, where y = y-coordinate, m = slope, x = x-coordinate, and b = slope.

2. As you can see, 30 is the value that took the place of m, and 20 is the value that took the place of b.

Therefore, m is 30 and b is 20.

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Write an expression for the description: the product of 3 and p subtracted from 6
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6 - 3p

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The word "product" indicates to multiply 3 and p, and "subtracted from" tells us to use subtraction.

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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

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So;

y'₂ = y"₁ = -2y₁ -by₂

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y₁' = y₂

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(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}

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(d)

From λ² + λb + 2 = 0

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\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]

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m = \left[\begin{array}{c}-1/2 \\1\\\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]

So as t → ∞

\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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