You also need to brake down this answer farther.
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:
y = 
Step-by-step explanation:
Let the equation of the line is,
y = mx + b
Here m = slope of the line
b = y-intercept
Slope of a line passing through two points
and
is,
m = 
From the graph attached,
Since, the given line passes through (0, -1) and (3, 4),
Slope 'm' = 
m = 
y - intercept 'b' = -1
Therefore, equation of the line will be,
y = 
Answer:
i think 0 slop. wait for the other person to answers this, to be sure of the answer
Step-by-step explanation:
Answer:
8n³ + 10n² - 13n - 15
Step-by-step explanation:
Distribute the factors by multiplying each term in the first factor by each term in the second factor, that is
4n(2n² + 5n + 3) - 5(2n² + 5n + 3) ← distribute both parenthesis
= 8n³ + 20n² + 12n - 10n² - 25n - 15 ← collect like terms
= 8n³ + 10n² - 13n - 15