Ask question

Ask question

All categories

3 years ago

15

Write a paragraph comparing the distribution of dollars spent on monthly entertainment by 12-year-old girls and boys. Including

your response measures of shape, center, and spread in reference to the context of data presented

1 answer:

Lina20 [59]3 years ago

4 0

Answer:

Veaca dom vetta me lato da la te amo lattea.

Step-by-step explanation:

You might be interested in

Perform the requested operation or operations.

alukav5142 [94]

$f(x) = \frac{x-3}{5}$
$g(x) = 5x+3$
$g(f(x)) = 5(\frac{x-3}{5})+3 = x-3+3 = x$

EDIT:
$f(g(x))=\frac{(5x+3)-3}{5}=\frac{5x+0}{5}=\frac{5x}{5}=x$

4 0

3 years ago

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

goblinko [34]

The answer should be B

8 0

3 years ago

Read 2 more answers

264 divided by 8 what is the answer

zheka24 [161]

Answer:

20.5 or 20 1/2

Step-by-step explanation:

calculator -.-

8 0

3 years ago

Solve the given equation. (Enter your answers as a comma-separated list. Let k be any integer. Round terms to three decimal plac

storchak [24]

Answer:

θ = πk, 1.318+2πk, 4.965+2πk

Step-by-step explanation:

4sin(2θ) − 2sin(θ) = 0

8sin(θ)cos(θ) - 2sin(θ) = 0

2sin(θ)[4cos(θ)-1] = 0

2sin(θ) = 0

sin(θ) = 0

θ = πk

4cos(θ)-1 = 0

4cos(θ) = 1

cos(θ) = 1/4

θ ≈ 1.318+2πk, 4.965+2πk

Therefore, θ = πk, 1.318+2πk, 4.965+2πk

4 0

2 years ago