If the radio active isotope has the half life of 24100 years, then the initial quantity is 2.16 grams
The half life in years = 24100
Consider the quantity of the radio active isotope remaining
y = 
When t = 1000 the y = 1.2
y = C/2 when t = 1599
Substitute the values in the equation
C/2 = 
Cancel the C in both side
1/2 = 
Here we have to apply ln to eliminate the e terms
ln (1/2) = 24100k
k = ln(1/2) / 24100
k = -2.87× 10^-5
To find the initial value we have to substitute the value of k and y in the equation
1.2 = Ce^{1000 × -2.87× 10^-5}
C = 1.2 / e^(-0.0287)
C = 2.16 gram
Hence, the initial quantity of the radioactive isotope is 2.16 gram
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Answer: if it's 18 times 29 then it's 522. If it's 18 to the power of 29 it's 2.52873108 times 10^36 (decimal form)
Step-by-step explanation:
Answer:
The answer is D
Step-by-step explanation:
Since the input is -6, you input it into the equation. That would make it f(-6)=-(-6)-1 which would be equivalent to f(-6)=6-1 , as the negatives at the beginning of the equation cancel each other out and make it positive. Then you solve and get 5, which is your output. Hope I could help :)
A. But you can also read a book and not rely on this
Answer:
x = 1/2 = 0.500
Step-by-step explanation:
Solve : 2x-1 = 0
Add 1 to both sides of the equation :
2x = 1
Divide both sides of the equation by 2:
x = 1/2 = 0.500
