Answer:
a = -36/7
Step-by-step explanation:
55 + 7a = 19
(55 - 55) + 7a = (19 - 55)
7a = -36
7a/7 = -36/7
a = -36/7 the fraction can't be simplified any further
3/16=5/x
times both sides by 16x (aka cross multiply
3x=5*16
3x=80
divide both sides by 3
x=80/3
x=26 and 2/3
the missing number is 80/3 or 26 and 2/3
Double is double, so you can solve
.. 2 = 1*e^(.08t)
and get the correct value of t.
Taking logs,
.. ln(2) = .08t
.. ln(2)/.08 = t ≈ 8.7 . . . . years
Honestly I don’t know the answer
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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