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2 years ago

How many time does x go into zero ?

Mathematics

1 answer:

mylen [45]2 years ago

8 0

Answer:

Step-by-step explanation:

Anything divided by 0 is 0. 0 dovided by anything is also 0

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Find the width of a rectangular prism if the volume is 546cm, the height is 7cm and the length is 13cm.

Margaret [11]

7x w x 13=546

(7x13)*w= 546

91w=546

91w/91= 546/91

w= 6

the width is 6cm

7 0

2 years ago

Use the quadratic formula to solve the equation to the exact value or round to two decimal places.

lys-0071 [83]

First set everything to 0.
${x}^{2} + x - 5$
Giving you A= 1 , B= 1 and C= -5
Then input everything into the quadratic formula and either solve by hand or use a calculator.
Note exact values are found by hand, decimals are found using a calculator.

6 0

2 years ago

75 students in a kindergarten. 40% of them can read. how many can read?

Paul [167]

Answer:

Step-by-step explanation:

75 times .40 = 30

8 0

2 years ago

Read 2 more answers

Could you Show work please

notka56 [123]

Answer:

2784 cm^2

Step-by-step explanation:

I have indicated in the pic there are 4 types of rectangles:

green : 20 x 4 = 80

red : 18 x 4 = 72

yellow : 20 x 18 = 360

blue : 18 x 16 = 288

green has 6 identical surface area so 6 x 80 = 480

red has 4 so 4 x 72 = 288

yellow has 4 so 4 x 360 = 1440

blue has 2 so 2 x 288 = 576

total : 480 + 288 + 1440 + 576 = 2784

7 0

2 years ago

Read 2 more answers

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