Answer:
The hourly decay rate is of 1.25%, so the hourly rate of change is of -1.25%.
The function to represent the mass of the sample after t days is 
Step-by-step explanation:
Exponential equation of decay:
The exponential equation for the amount of a substance is given by:
In which A(0) is the initial amount and r is the decay rate, as a decimal.
Hourly rate of change:
Decreases 26% by day. A day has 24 hours. This means that 
; We use this to find r.
![\sqrt[24]{(1-r)^{24}} = \sqrt[24]{0.74}](https://tex.z-dn.net/?f=%5Csqrt%5B24%5D%7B%281-r%29%5E%7B24%7D%7D%20%3D%20%5Csqrt%5B24%5D%7B0.74%7D)
The hourly decay rate is of 1.25%, so the hourly rate of change is of -1.25%.
Starts out with 810 grams of Element X
This means that 
Element X is a radioactive isotope such that its mass decreases by 26% every day.
This means that we use, for this equation, r = 0.26.
The equation is:
The function to represent the mass of the sample after t days is
It would be 6:42 pm I believe.
Answer:
P = 2000 * (1.00325)^(t*4)
(With t in years)
Step-by-step explanation:
The formula that can be used to calculated a compounded interest is:
P = Po * (1 + r/n) ^ (t*n)
Where P is the final value after t years, Po is the inicial value (Po = 2000), r is the annual interest (r = 1.3% = 0.013) and n is a value adjusted with the compound rate (in this case, it is compounded quarterly, so n = 4)
Then, we can write the equation:
P = 2000 * (1 + 0.013/4)^(t*4)
P = 2000 * (1.00325)^(t*4)
Answer:
15
Step-by-step explanation:
answeram I righatjfj
We can set it up like this, where <em>s </em>is the speed of the canoeist:
To make a common denominator between the fractions, we can multiply the whole equation by s(s-5):
![s(s-5)[\frac{18}{s} + \frac{4}{s-5} = 3] \\ 18(s-5)+4s=3s(s-5) \\ 18s - 90+4s=3 s^{2} -15s](https://tex.z-dn.net/?f=s%28s-5%29%5B%5Cfrac%7B18%7D%7Bs%7D%20%2B%20%5Cfrac%7B4%7D%7Bs-5%7D%20%3D%203%5D%20%5C%5C%2018%28s-5%29%2B4s%3D3s%28s-5%29%20%5C%5C%2018s%20-%2090%2B4s%3D3%20s%5E%7B2%7D%20-15s)
If we rearrange this, we can turn it into a quadratic equation and factor:
Technically, either of these solutions would work when plugged into the original equation, but I would use the second solution because it's a little "neater." We have the speed for the first part of the trip (9 mph); now we just need to subtract 5mph to get the speed for the second part of the trip.
The canoeist's speed on the first part of the trip was 9mph, and their speed on the second part was 4mph.
