First set everything to 0.

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.
Answer:
30
Step-by-step explanation:
75 times .40 = 30
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
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:




(c)
Suppose
, then λ₂ < 0 and λ₁ < 0. Thus, the node is stable at equilibrium.
(d)
From λ² + λb + 2 = 0
If b = 3; we get

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]](https://tex.z-dn.net/?f=v%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%2B1%261%5C%5C-2%26-2%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dv_1%5C%5Cv_2%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C%5Cend%7Barray%7D%5Cright%5D)
![\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]](https://tex.z-dn.net/?f=%5Csim%20v%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%261%5C%5C0%260%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dv_1%5C%5Cv_2%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Let v₂ = 1, v₁ = -1
![v = \left[\begin{array}{c}-1\\1\\\end{array}\right]](https://tex.z-dn.net/?f=v%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-1%5C%5C1%5C%5C%5Cend%7Barray%7D%5Cright%5D)
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]](https://tex.z-dn.net/?f=m%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%5C%5C-2%26-1%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dm_1%5C%5Cm_2%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C%5Cend%7Barray%7D%5Cright%5D)
![\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]](https://tex.z-dn.net/?f=%5Csim%20v%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%261%5C%5C0%260%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dm_1%5C%5Cm_2%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Let m₂ = 1, m₁ = -1/2
![m = \left[\begin{array}{c}-1/2 \\1\\\end{array}\right]](https://tex.z-dn.net/?f=m%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-1%2F2%20%5C%5C1%5C%5C%5Cend%7Barray%7D%5Cright%5D)
∴
![\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]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dy_1%5C%5Cy_2%5C%5C%5Cend%7Barray%7D%5Cright%5D%3D%20C_1%20e%5E%7B-t%7D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-1%5C%5C1%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%2B%20C_2e%5E%7B-2t%7D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D-1%2F2%5C%5C1%5C%5C%5Cend%7Barray%7D%5Cright%5D)
So as t → ∞