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2 years ago

I NEED HELP AHHH-HHHHHHHHH HELP ME ASAP

Mathematics

1 answer:

zvonat [6]2 years ago

4 0

Answer:

1/5, 1/3, 4/6

Step-by-step explanation:

First, find all the common denominators for these numbers. In this case, it's 30.

1/5 = 10/30

4/6 = 20/30

1/5 = 6/30

Now, you can see which ones are greatest and least.

1. 1/5 (least)

2. 1/3 (middle)

3. 4/6 (greatest)

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What is the volume of the cone to the nearest cubic meter with radius of 4m and height of 5m

adoni [48]

Answer :

B) 84 m³

Step by step :

$\boxed{ \bold{volume = \frac{1}{3}\pi {r}^{2} h }}$

$volume = \frac{1}{3} \times 3.14 \times 4 \times 4 \times 5 \\ volume = \frac{1}{3} \times 251.2 \\ volume = 83.73 \: {m}^{3} \\ volume = 84 \: {m}^{3}$

6 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

2 years ago

Answers:

natita [175]

Answer:

y=-3/4x-3

Step-by-step explanation:

3 0

2 years ago

1) Find the circumference and area of each circle:

MA_775_DIABLO [31]

Answer:

a] Radius of circle = 10 feet

Circumference of circle = 94.2 feet (Approx.)

Area of circle = 314 feet² (Approx.)

b] Radius of circle = 6 cm

Circumference of circle = 37.68 cm (Approx.)

Area of circle = 113.04 cm² (Approx.)

Step-by-step explanation:

Given:

a] Radius of circle = 10 feet

b] Radius of circle = 6 cm

Find:

Circumference an area for both circle

Computation:

Circumference of circle = 2πr

Area of circle = πr²

a] Radius of circle = 10 feet

Circumference of circle = 2(3.14)(10)

Circumference of circle = 94.2 feet (Approx.)

Area of circle = (3.14)(10)²

Area of circle = 314 feet² (Approx.)

b] Radius of circle = 6 cm

Circumference of circle = 2(3.14)(6)

Circumference of circle = 37.68 cm (Approx.)

Area of circle = (3.14)(6)²

Area of circle = 113.04 cm² (Approx.)

7 0

3 years ago

HELP ME PLEASE !!!!!!!!!!!

Scrat [10]

The angle that you need to choose is 2 I think

4 0

2 years ago