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3 years ago

Solve 5x + x − 3x > −12

Mathematics

1 answer:

Svetllana [295]3 years ago

5 0

X>-4 would be the correct answer

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There are 20 alligators in the swamp. Each year, the number of alligators increases by 25%. There are 25 crocodiles in the swamp

Genrish500 [490]

A)

A=20(1.25^t)

C=25+10t

b)

A(4)=20(1.25^4)≈48.82≈49 alligators to nearest whole alligator

C(4)=25+10(4)=65 crocodiles.

c)

20(1.25^t)=25+10t

1.25^t=1.25+0.5t

t≈7

4 0

3 years ago

Can you please help me figure out 6m=<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B-1%7D%7B3%7D%20" id="TexFormula1" title=" \f

Rainbow [258]

-1/3 divided by six... m=-1/18

8 0

3 years ago

Read 2 more answers

7=2(x+5) hellppppppppp

ExtremeBDS [4]

0 = x because 0 + 5 = 5 plus 2 = 7

4 0

3 years ago

Read 2 more answers

Identify the graph of the equation. What is the angle of rotation for the equation?

borishaifa [10]

Answer:

d. parabola, 0°

Step-by-step explanation:

y² + 8x - 0

y² = -8x

Where x = cos t , y = sin t

Sin² t = -8 Cos t

1 - Cos² t = -8 Cos t

- Cos² t + 8 Cos t + 1 = 0

t = 2лπ ± (3 + √10) , л∈Z

Angle of rotation

5 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