Ok, so remember
the legnth of the longest side must be LESS THAN the sum of the measures of the other 2 sides
if no longest side (becasue it has 2 longest sides or 3 equal sides), then it can form a triangle
so the longest side is 36
36 must be less than the sum of 15 and 20
36<15+20
36<35
false
therfor it cannot form a triangle
try it yourself, you cannot connect them that way
So first calculate the time between the plane flied over the Air Force Bases (AFB) and Navy Base Station (NBS):
(time passed Air Force Bases) - (time passed Air Force Bases) = 10:32 - 10:20 = 12 mins
Since we know the distance that plane travel from AFB to NBS is 120 miles and the plane traveled that distance within 12 minutes.
We have
120 miles / 12 minutes = 10 miles / minute
Calculate it into miles per hour (60 minutes):
10 miles/minute x 60 minutes = 600 miles per hour.
ANSWER : 600 mph
Hey there Smarty!
This would be considered to be a (acute angel) which in this case, we would have to make sure that this whole triangle would equal less than 270 because each angle would be less than

°
If we add/multiply

.
So, we would have to know that was ever this would all add up to be, this would have to be less than 270°
![\left[\begin{array}{ccc}\boxed{\boxed{(2.5+5) \\ \\ (2.5*2.5) \\ (2.5-2) \\ (2.5-1) \\}} \\ \\ this \ would \ be \ why \ I \ would \ say \\ that \ this \ would \ be \ the \ answer \end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D%5Cboxed%7B%5Cboxed%7B%282.5%2B5%29%20%5C%5C%20%5C%5C%20%282.5%2A2.5%29%20%5C%5C%20%282.5-2%29%20%5C%5C%20%282.5-1%29%20%5C%5C%7D%7D%20%5C%5C%20%5C%5C%20this%20%5C%20would%20%5C%20be%20%5C%20why%20%5C%20I%20%5C%20would%20%5C%20say%20%5C%5C%20that%20%5C%20this%20%5C%20would%20%5C%20be%20%5C%20the%20%5C%20answer%20%5Cend%7Barray%7D%5Cright%5D%20%20)
I truly hope this helps, and also, it's kind of my
first time doing this I hope this would be helpful.
~Jurgen
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 → ∞