Hello!
.
1 liter is equal to 1000 milliliter.
So,
4 L = 4000 mL
.
Max bought 4 liter or 4000 mL of fruit punch. To know how many glasses full of fruit punch can he serve, you just need to divide the amount of fruit punch he bought by the amount of fruit punch a glass can hold.
4 L / 200 mL
The dividend and the divisor have different unit so we must make them have the same unit.
4000 mL / 200 mL
= 20
.
Therefore, Max can make 20 full glasses of fruit punch for the party.
:)
(f∘g)(x) is equivalent to f(g(x)). We solve this problem just as we solve f(x). But since it asks us to find out f(g(x)), in f(x), each time we encounter x, we replace it with g(x).
In the above problem, f(x)=x+3.
Therefore, f(g(x))=g(x)+3.
⇒(f∘g)(x)=2x−7+3
⇒(f∘g)(x)=2x−4
Basically, write the g(x) equation where you see the x in the f(x) equation.
f∘g(x)=(g(x))+3 Replace g(x) with the equation
f∘g(x)=(2x−7)+3
f∘g(x)=2x−7+3 we just took away the parentheses
f∘g(x)=2x−4 Because the −7+3=4
This is it
g∘f(x) would be the other way around
g∘f(x)=2(x+3)−7
now you have to multiply what is inside parentheses by 2 because thats whats directly in front of them.
g∘f(x)=2x+6−7
Next, +6−7=−1
g∘f(x)=2x−1
Its a lts easier than you think!
Hope this helped
Her profit was 126$ because she spent 24$ on her materials and then charged 5$ per 2 so you get 150$ then subtract the 24
(I think)
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 → ∞
![\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 }](https://tex.z-dn.net/?f=%5Cmathbf%7B%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7Dy_1%5C%5Cy_2%5C%5C%5Cend%7Barray%7D%5Cright%5D%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bc%7D0%5C%5C0%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5C%20%5C%20%20so%20%5C%20stable%20%5C%20at%20%5C%20node%20%5C%20%5Cinfty%20%7D)
