<span>Simplifying
(6a + -8b)(6a + 8b) = 0
Multiply (6a + -8b) * (6a + 8b)
(6a * (6a + 8b) + -8b * (6a + 8b)) = 0
((6a * 6a + 8b * 6a) + -8b * (6a + 8b)) = 0
Reorder the terms:
((48ab + 36a2) + -8b * (6a + 8b)) = 0
((48ab + 36a2) + -8b * (6a + 8b)) = 0
(48ab + 36a2 + (6a * -8b + 8b * -8b)) = 0
(48ab + 36a2 + (-48ab + -64b2)) = 0
Reorder the terms:
(48ab + -48ab + 36a2 + -64b2) = 0
Combine like terms: 48ab + -48ab = 0
(0 + 36a2 + -64b2) = 0
(36a2 + -64b2) = 0
Solving
36a2 + -64b2 = 0
Solving for variable 'a'.
Move all terms containing a to the left, all other terms to the right.
Add '64b2' to each side of the equation.
36a2 + -64b2 + 64b2 = 0 + 64b2
Combine like terms: -64b2 + 64b2 = 0
36a2 + 0 = 0 + 64b2
36a2 = 0 + 64b2
Remove the zero:
36a2 = 64b2
Divide each side by '36'.
a2 = 1.777777778b2
Simplifying
a2 = 1.777777778b2
Take the square root of each side:
a = {-1.333333333b, 1.333333333b}</span>
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
Learn more about half life here
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Well is just their sum, thus
Answer:
dependent
Step-by-step explanation:
Answer:
use symolab
Step-by-step explanation:
