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zhenek [66]
2 years ago
10

Can someone please help me ?

Mathematics
1 answer:
tatiyna2 years ago
4 0

Answer:

(x-2),(x-1),(x), and(x+1)

Step-by-step explanation:

In this scenario, we are given nothing about the polynomial except its graph. We can start in problems like these by looking at when y=0 and going from there.

In this graph, we can see that y=0 when x=2, 1, 0, and -1. Looking at our options, for y to be 0, there must be a factor where, when we plug x=2 in, y=0. The only option here is (x-2). Similarly, for x=1, we have (x-1), for x=0, we have x, and for x=-1, we have (x+1). The polynomial has 3 turning points, so the polynomial must be of degree 4 or higher.

So far, we have (x-2)(x-1)(x)(x+1). Plugging this in to a graphing calculator, we can see that this fits. To check if x²+2 and x²+1 would work here, it would require guess and check with values/graphing calculators, seeing if they fit. For example, when x=1.5, y is greater than -1 on the graph and our function. However, if we multiplied that by x²+1 or x²+2. when x=1.5, y would balloon downward, and the graph wouldn't fit

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Angelina_Jolie [31]

Answer:

The value of k is  \pm 9 and The value of p is - 23  

Step-by-step explanation:

Given equation as :

81 y - 17  = k² y + 6 + p

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Now for real solution the equation have no solution

∵ The equation has no solution the ,

The coefficient of y must be equal

I.e 81 = k²

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Again , If the coefficient of y is same then the equation is written as

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3 years ago
Need to know the population in 2015
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\bf P=1110.9e^{kt}\quad 
\begin{cases}
P=1200 &\textit{in thousands}\\
t=2 &\textit{in 2002}
\end{cases}\implies 1200=1110.9e^{k2}
\\\\\\
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\\\\\\
\cfrac{ln\left( \frac{1200}{1110.9} \right)}{2}=k\implies 0.0385755\approx k\implies 0.0386\approx k\\\\
-------------------------------\\\\

\bf P=1110.9e^{0.0386t}\qquad 
\begin{cases}
t=14\\
\textit{year 2015}
\end{cases}\implies P=1110.9e^{0.0386\cdot 14}
\\\\\\
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