![\bf f(x)=0.35x^2-2.1x+15.8\qquad \begin{cases} x=\textit{year since 1995}\\ f(x)=\textit{population amount} \end{cases}](https://tex.z-dn.net/?f=%5Cbf%20f%28x%29%3D0.35x%5E2-2.1x%2B15.8%5Cqquad%20%0A%5Cbegin%7Bcases%7D%0Ax%3D%5Ctextit%7Byear%20since%201995%7D%5C%5C%0Af%28x%29%3D%5Ctextit%7Bpopulation%20amount%7D%0A%5Cend%7Bcases%7D)
the equation is a quadratic one, and it has a positive coefficient on the leading term, meaning, is opening upwards, so it has a "burrow" for the vertex.
the minimum or lowest point for a quadratic opening upwards is, well, the vertex point :), the "x" value is the year, the "y" or f(x) value is the population, we're asked for the year, or the x-coordinate of the vertex
well
Subtract 4 from both sides, solve using quadratic formula
ax^2+bx+c
(-b(+or-) Square Root of b^2 - 4ac)/2a
9x^2+9x-4=0
-9(+or-)Square root of 9^2-4(9)(-4)/2(9)
Solve^
Answer: She is correct because each ratio is equivalent to 5:1, except for the last one, which is 1:5
5/1 = 5:1
10/2 = 5:1
25/5 = 5:1
10/50 = 1:5
The <em>first three</em> elements of the <em>recursive</em> series are 6, 10, 8. (Correct choice: B)
<h3>How to generate values from a recursive function</h3>
In this question we have a kind of <em>recursive</em> function known as Fibonacci's function, where a value of the series is generated from at least <em>immediately previous</em> elements. In this case, we need to find the <em>first</em> three elements from the <em>fifth</em> and <em>fourth</em> elements of the series:
![a_{n-2} = a_{n-1} - a_{n} + 4](https://tex.z-dn.net/?f=a_%7Bn-2%7D%20%3D%20a_%7Bn-1%7D%20-%20a_%7Bn%7D%20%2B%204)
a₄ = a₅ - a₆ + 4
a₄ = - 2 - 0 + 4
a₄ = 2
a₃ = a₄ - a₅ + 4
a₃ = 2 - (- 2) + 4
a₃ = 8
a₂ = a₃ - a₄ + 4
a₂ = 8 - 2 + 4
a₂ = 10
a₁ = a₂ - a₃ + 4
a₁ = 10 - 8 + 4
a₁ = 6
The <em>first three</em> elements of the <em>recursive</em> series are 6, 10, 8. (Correct choice: B)
To learn more on recursive series: brainly.com/question/8972906
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