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zalisa [80]
3 years ago
13

The landscaper pours 200 gallons of herbicide in a pond. The herbicide degrades 10% each week. How much will be in the pond afte

r 1 week?
Mathematics
2 answers:
DaniilM [7]3 years ago
6 0

Answer:

180 gallons

Step-by-step explanation:

In order to calculate you just need to calculate the 10% of the herbicide that is being degraded in the first week, so we do a rule of three to calculate that 10%:

\frac{200}{100} =\frac{x}{10}\\ x=10*\frac{200}{100} \\x=20

So the 10% is 20 gallons, and that is the gallons of herbicide that degraded in the first week, so we just withdraw the 20 gallons degraded from the initial 200 gallons:

200-20=180

So there will still be 180 gallons in the pond after one week.

weeeeeb [17]3 years ago
5 0

Answer:

180 gallons

Step-by-step explanation:

200 gal

10% = .10

200 x .10 = 20

200 - 20 = 180


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The capital formation of the investment function over a given period is the

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  • (a) The capital formation from the end of the second year to the end of the fifth year is approximately <u>298.87</u>.

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(a) The given investment function is presented as follows;

I(t) = 100 \cdot e^{0.1 \cdot t}

(a) The capital formation is given as follows;

\displaystyle Capital = \int\limits {100 \cdot e^{0.1 \cdot t}} \, dt =1000 \cdot  e^{0.1 \cdot t}} + C

From the end of the second year to the end of the fifth year, we have;

The end of the second year can be taken as the beginning of the third year.

Therefore,  for the three years; Year 3, year 4, and year 5, we have;

\displaystyle Capital = \int\limits^5_3 {100 \cdot e^{0.1 \cdot t}} \, dt \approx 298.87

The capital formation from the end of the second year to the end of the fifth year, C ≈ 298.87

(b) When the capital stock exceeds $100,000, we have;

\displaystyle  \mathbf{\left[1000 \cdot  e^{0.1 \cdot t}} + C \right]^t_0} = 100,000

Which gives;

\displaystyle 1000 \cdot  e^{0.1 \cdot t}} - 1000 = 100,000

\displaystyle \mathbf{1000 \cdot  e^{0.1 \cdot t}}} = 100,000 + 1000 = 101,000

\displaystyle e^{0.1 \cdot t}} = 101

\displaystyle t = \frac{ln(101)}{0.1} \approx 46.15

The number of years before the capital stock exceeds $100,000 ≈ <u>46.15 years</u>.

Learn more investment function here:

brainly.com/question/25300925

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