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3 years ago

A circular piece of wood with a diameter of 5 ft is shown below. A 2 ft square is

Mathematics

1 answer:

Pani-rosa [81]3 years ago

4 0

Answer:

16 ft²

Step-by-step explanation:

Since we're told that 2 ft diameter was cut of the board. We use the formula for hollow circle to find the area.

Area of remaining body = area of full body - area of removed body.

Area = πd²/4

Area of remaining body = (π * 5²)/4 - (π * 2²)/4

Area of remaining body = 19.64 - 3.142

Area of remaining body = 16.49 ft²

Therefore, the remaining area is then calculated into 16 ft²

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Viefleur [7K]

Answer:

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Step-by-step explanation:

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3 years ago

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What does the remainder theorem conclude given that f(x)x+3 has a remainder of 11?

Slav-nsk [51]

By the remainder theorem of polynomial division, the complete equation is f(-3) = 11

<h3>How to complete the blanks?</h3>

The equation is given as:

f(x)/x + 3

Set the divisor to 0

x + 3 = 0

Solve for x

x = -3

Given that the quotient has a remainder of 11.

It means that:

f(-3) = 11

Hence, the complete equation is f(-3) = 11

Read more about remainder theorem at:

brainly.com/question/13328536

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3 years ago

The Waters family budget a maximum amount of $150 per week for groceries. Mr. Waters already spent $40. Write and solve an inequ

ivann1987 [24]

150-40 less than or equal to $110 -

They can spend less than or equal to $110

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3 years ago

A mathematician works for t hours per day and solves p problems per hour, where t and p are positive integers and 1

Crazy boy [7]

Answer:

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Step-by-step explanation:

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3 years ago

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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:

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

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

Let m₂ = 1, m₁ = -1/2

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

∴

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

So as t → ∞

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

5 0

3 years ago