Fill in the boxes using the whole numbers 1 through 9 to make the biggest 3
1 answer:
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Laura's age is 9 years.
Solution:
Let x be the age of April.
Laura's age = 2 years more than half of April's age
<u>Convert statement into algebraic expression:</u>
Half of April's age =
2 years more than half of April's age =
Combined age of April and Laura = 23
⇒ April's age + Laura's age = 23
To add the fractions make the denominators same.
Multiplying 2 on both numerator and denominator of unlike terms, we get
Denominators are same, now add the fractions.
Do cross multiplication.
Aprils's age = 14 years
Laura's age =
= 7 + 2
Laura's age = 9
Hence Laura's age is 9 years.
The given exponential function can be used to determine the
population of the fish at different times.
Response:
1. 200
2. 0.05
3. Please find attached the graph of the exponential function and the table
4. Range; [0, ∞)
Domain: [200, ∞)
5. 346 fishes (approximate value)
<h3>How is the population of the fish obtained from the exponential function?</h3>The given function is f(x) = 200·(1.05)ˣ
1. The initial population is given by the value of the function when x = 0,
which is given as follows;
At x = 0, f(0) = 200·(1.05)⁰ = 200
- The initial population of the fish is <u>200</u>
2. The rate of growth of the fish from the exponential function is given as follows;
The growth factor = 1.05
- The growth rate =<u> 0.05</u>
3. Please find attached the graph of the exponential function created
with MS Excel.
The asymptote of the exponential function is the line, <u>y = 0</u>
The table of values is presented as follows;
4. The domain is the possible x-values which is; 0 ≤ x < ∞
The range is the possible y-values which is; 200 ≤ f(x) < ∞
Which gives;
- Domain; [0, ∞)
- Range: [200, ∞)
5. The population of the fish after 11 years is f(11) = 200·(1.05)¹¹
f(11) = 200 × (1.05)¹¹ ≈ 342
- The population of the fish after 11 years is <u>346 (fishes)</u>
Learn more about exponential function here:
