Answer:
each folder costed 0.35
Step-by-step explanation:
8.75 divided by 25 would have given you the answer
I believe the correct expression is: <span>11.50(1.083)^t where t is the time.
Now, we are given that the average price of the ticket is $11.5
The given expression means that this average value is dependent on the variable t. Therefore, the average price of the ticket increases exponentially with the time with the rate of growth equals 1.083
Now, to better understand this, we will get the price of the ticket at different times:
At t = 1: price = </span><span>11.50(1.083)^1 = $12.4545
At t = 2: price = </span><span>11.50(1.083)^2 = $13.4882235
At t = 3: price = </span><span>11.50(1.083)^3 = $14.60774605
We can notice that the price of the ticket increases exponentially as the time increases.
Hope this helps :)</span>
Answer:
(13/10)H
got it on khan academy brainliest pls thx uwu
Remember that 30% in fraction form is 33/100
The amount of health points (H) restored would depend on the amount of the current H so it means it would add 30% of the current which we can write as:
(30/100)H
And since it would add that to the current total we can right the current total as:
(100/100)H
So our equation would be
=(30/100)H + (100/100)H
=(130/100)H
=(13/10)H
400. Just add a zero when you're dealing with tens
Answer:

Step-by-step explanation:

Applying the Laplace transform:
![\mathcal{L}[y'']+5\mathcal{L}[y']+4\mathcal{L}[y']=0](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5By%27%27%5D%2B5%5Cmathcal%7BL%7D%5By%27%5D%2B4%5Cmathcal%7BL%7D%5By%27%5D%3D0)
With the formulas:
![\mathcal{L}[y'']=s^2\mathcal{L}[y]-y(0)s-y'(0)](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5By%27%27%5D%3Ds%5E2%5Cmathcal%7BL%7D%5By%5D-y%280%29s-y%27%280%29)
![\mathcal{L}[y']=s\mathcal{L}[y]-y(0)](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5By%27%5D%3Ds%5Cmathcal%7BL%7D%5By%5D-y%280%29)
![\mathcal{L}[x]=L](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5Bx%5D%3DL)

Solving for 




Apply the inverse Laplace transform with this formula:
![\mathcal{L}^{-1}[\frac1{s-a}]=e^{at}](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5E%7B-1%7D%5B%5Cfrac1%7Bs-a%7D%5D%3De%5E%7Bat%7D)
![y=3\mathcal{L}^{-1}[\frac1{s+4}]=3e^{-4t}](https://tex.z-dn.net/?f=y%3D3%5Cmathcal%7BL%7D%5E%7B-1%7D%5B%5Cfrac1%7Bs%2B4%7D%5D%3D3e%5E%7B-4t%7D)