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The growth of bacteria makes it necessary to time-date some food products so that they will be sold and consumed before the bacteria count is too high. Suppose for a certain product, the number of bacteria present is given by

[tex]\[ f(t) = 500e^{0.1t} \][/tex]

where [tex]\( t \)[/tex] is time in days and the value of [tex]\( f(t) \)[/tex] is in millions. Find the number of bacteria present at each time:

(a) 3 days
(b) 4 days
(c) 1 week

(a) Approximately how many million bacteria are present in 3 days? (Round to the nearest integer as needed.)

Sagot :

GinnyAnswer
To solve the problem of determining the number of bacteria present at various times, we need to use the given growth function:

[tex]\[ f(t) = 500 \cdot e^{0.1t} \][/tex]

where [tex]\( f(t) \)[/tex] is the number of bacteria in millions, and [tex]\( t \)[/tex] is the time in days. Let's evaluate this function at three specific points: 3 days, 4 days, and 1 week (7 days).

### Step-by-Step Solution

#### (a) Number of bacteria present at 3 days

1. Substitute [tex]\( t = 3 \)[/tex] into the function:

[tex]\[ f(3) = 500 \cdot e^{0.1 \cdot 3} \][/tex]

2. Calculate the exponent:

[tex]\[ 0.1 \cdot 3 = 0.3 \][/tex]

3. Compute the value:

[tex]\[ e^{0.3} \approx 1.34986 \][/tex] (using a calculator or exponential table)

4. Multiply by 500:

[tex]\[ f(3) = 500 \cdot 1.34986 \approx 674.929 \][/tex]

So, the number of bacteria present at 3 days is approximately 674.929 million.

5. Round to the nearest integer:

[tex]\[ 674.929 \approx 675 \][/tex]

Therefore, approximately 675 million bacteria are present in 3 days.

#### (b) Number of bacteria present at 4 days

1. Substitute [tex]\( t = 4 \)[/tex] into the function:

[tex]\[ f(4) = 500 \cdot e^{0.1 \cdot 4} \][/tex]

2. Calculate the exponent:

[tex]\[ 0.1 \cdot 4 = 0.4 \][/tex]

3. Compute the value:

[tex]\[ e^{0.4} \approx 1.49182 \][/tex]

4. Multiply by 500:

[tex]\[ f(4) = 500 \cdot 1.49182 \approx 745.912 \][/tex]

So, the number of bacteria present at 4 days is approximately 745.912 million.

5. Round to the nearest integer:

[tex]\[ 745.912 \approx 746 \][/tex]

Therefore, approximately 746 million bacteria are present in 4 days.

#### (c) Number of bacteria present at 1 week (7 days)

1. Substitute [tex]\( t = 7 \)[/tex] into the function:

[tex]\[ f(7) = 500 \cdot e^{0.1 \cdot 7} \][/tex]

2. Calculate the exponent:

[tex]\[ 0.1 \cdot 7 = 0.7 \][/tex]

3. Compute the value:

[tex]\[ e^{0.7} \approx 2.01375 \][/tex]

4. Multiply by 500:

[tex]\[ f(7) = 500 \cdot 2.01375 \approx 1006.876 \][/tex]

So, the number of bacteria present at 7 days is approximately 1006.876 million.

5. Round to the nearest integer:

[tex]\[ 1006.876 \approx 1007 \][/tex]

Therefore, approximately 1007 million bacteria are present in 1 week.

### Conclusion
- (a) Approximately 675 million bacteria are present in 3 days.
- (b) Approximately 746 million bacteria are present in 4 days.
- (c) Approximately 1007 million bacteria are present in 1 week.
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