Algebra 2 help?

Question

Thank you so much, would you happen to know the answer to these?
1. A radio active isotope decays according to the exponential decay equation where t is in days.

Round to the thousandths place.

For the half life: The half life is the solution (t) of the equation : a2=ae^−6.345t

2. Zeus Industries bought a computer for $2538. It is expected to depreciate at a rate of 20% per year. What will the value of the computer be in 4 years?


Round to the nearest penny. Do not type the "$" sign in your answer

3. A radio active isotope decays according to the exponential decay equation where t is in days.

Round to the thousandths place.

For the half life: The half life is the solution (t) of the equation : a2=ae^−7.571t


4. An anthropologist finds bone that her instruments measure it as 0.177% of the amount of Carbon-14 the bones would have contained when the person was alive. How long ago did the person die?
The half life of carbon 14 is 5,730 years. Round your answer to the nearest thousand.

5. The annual Gross Domestic Product (GDP) of a country is the value of all of the goods and services produced in the country during a year. During the period 1985-1999, the Gross Domestic Product of the United States grew about 3.2% per year, measured in 1996 dollars. In 1985, the GDP was $577 billion. I what year did/or will the GDP equal $4.0 trillion?


6. The Mendes family bought a new house 8 years ago for $126,000. The house is now worth $156,000. Assuming a steady rate of growth, what was the yearly rate of appreciation? Round your answer to the nearest tenth of a percent (1.2% etc)

Answer 1

1

1/2 = e^-6.345t for half life, not a2=ae^-6.345t

ln(.5) = -6.345t

t = ln.5/-6.345 = 0.1092 days or a little over 2 1/2 hours

odds are though 6.345 should have been 0.06345

634.5% decay rate looks far too rapid

Use 0.06345 and t = 10.92 days or almost 11 days for half life

2

2538 today, 20% depreciation each year, after 4 years?

each year's value is 80% of the previous year

year value

0 2538

1 2030.4

2 1624.32

3 1299.456

4 1039.5648 = 1039.56 to the nearest penny

4 .174 left with half life of 5730 years

how old is it?

5730= ln,5/k where k= the decay rate

k=ln.5/5730 = -0.000121

.1771% = .001771

.001771 = e^-.000121t

t = ln.001771/.000121 = 14.636 years ago

but you probably meant 17.71% = 0.1771

then

0.1771 = e^-.000121t

t = 14,306 years old bone

3

the equation should read

a =2ae^-7.571t

2a =ae^-7.571t is for doubling time, not half time

You probably also may have missed the decimal point on the decay rate. It probably should be .07571 not 7.571

but ignore that decay rate problem and

1 = 2e^-7.571t with t measured in days

divide by 2

1/2 = e^-7.571t

take natural logs of both sides

ln(1/2) = -7.571t

t = ln(1/2)/-7.571 =approximately 0.091553 days

which makes sense as 7.571 means 757.1% decay rate. It's half gone in a 1/10 of a day

but try .07571 as the decay rate, which is 7.571% per day, instead of 757.1% per diem

1/2 = e^-.07571t

ln(1/2) =- ,07571t

t = ln(.5)/-.07571 = about 9.1553 days or 9 days and almost 4 hours.

if you had used the given equation with a2 or 2a on the left side, you end up with a negative time for the half life

6

126,000 to 156,000 in 8 years

at steady growth rate

156 = 126e^8r

8r = ln(156/126)/8 = about 0.0267 = 2.67% growth rate