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3 years ago

Which function describes this graph? (CHECK PHOTO FOR GRAPH)

Mathematics

1 answer:

romanna [79]3 years ago

6 0

Answer:

Option A. y = x² + 7x + 10

Step-by-step explanation:

We'll begin calculating the roots of the equation from the graph.

The roots of the equation on the graph is where the curve passes through the x-axis.

The curve passes through the x-axis at –5 and –2

Next, we shall determine the equation. This can be obtained as follow:

x = –5 or x = –2

x + 5 = 0 or x + 2 = 0

(x + 5)(x + 2) = 0

Expand

x(x + 2) + 5(x + 2) = 0

x² + 2x + 5x + 10 = 0

x² + 7x + 10 = 0

y = x² + 7x + 10

Thus, the function that describes the graph is y = x² + 7x + 10

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d1i1m1o1n [39]

Given:

The given expression to find the nth term of the sequence is $d(n)=d(n-1) \cdot (-5)$

The first term of the sequence is $d(1)=8$

We need to determine the third term of the sequence.

Second term:

The second term of the sequence can be determined by substituting n = 2 in the nth term of the sequence.

Thus, we have;

$d(2)=d(2-1) \cdot (-5)$

$d(2)=d(1) \cdot (-5)$

$d(2)=8 \cdot (-5)$

$d(2)=-40$

Thus, the second term of the sequence is -40.

Third term:

The third term of the sequence can be determined by substituting n = 3 in the nth term of the sequence.

Thus, we have;

$d(3)=d(3-1) \cdot (-5)$

$d(3)=-40 \cdot (-5)$

$d(3)=120$

Thus, the third term of the sequence is 120.

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dlinn [17]

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A country's population in 1994 was 182 million. In 2002 it was 186 million. Estimate the population in 2004 using the exponentia

iogann1982 [59]

$\bf =ae^{kt}\qquad 
\begin{cases}
1994\impliedby \textit{year 0, starting point}\\
t=0\qquad P=182
\end{cases}\implies 182=ae^{k0}
\\\\\\
182=a\cdot e^0\implies 182=a\cdot 1\implies 182=a
\\\\\\
thus\qquad P=182e^{kt}\\\\
-------------------------------\\\\$

$\bf P=182e^{kt}\qquad 
\begin{cases}
2002\impliedby \textit{8 years later}\\
t=8\qquad P=186
\end{cases}\implies 186=182e^{k8}
\\\\\\
\cfrac{186}{182}=e^{8k}\implies ln\left( \frac{93}{91} \right)=ln(e^{8k})\implies ln\left( \frac{93}{91} \right)=8k
\\\\\\
\cfrac{ln\left( \frac{93}{91} \right)}{8}=k\implies 0.0027\approx k\implies \boxed{P=182e^{0.0027t}}$

what's the population in 2004? well, from 1994 to 2004 is 10 years later, so t = 10

plug that in, to get P for 2004

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Answer:

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Step-by-step explanation:

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