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 
Answer:
EASY PEASY
Step-by-step explanation:
9²
HOPE IT HELPS YOU
OK...... your answer would be
- 10y^2 - 26y + 1/y
H0P3 It H3LPS :)
The answer is 1.72 and it would be 2 if u estimated
You can start by drawing a number line and labeling it with eighths. ( 1/8, 2/8, etc.) Then you can place a dot on 3/8 and 5/8. From there, it's clear to see that 5/8 is greater than 3/8.