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IgorLugansk [536]
3 years ago
15

A chameleon is looking for prey. Let positive numbers represent the elevation of prey above the chameleon and negative numbers r

epresent the elevation of prey below the chameleon. The chameleon spots a fly at 4 \text{ m}4 m4, start text, space, m, end text and a grasshopper at -6\text{ m}−6 mminus, 6, start text, space, m, end text. What does an elevation of 0 \text{ m}0 m0, start text, space, m, end text represent in this situation?
Mathematics
1 answer:
alexira [117]3 years ago
4 0

Answer:

B: The elevation of the chameleon

Step-by-step explanation:

The Fly is <em><u>above</u></em> the chameleon, so its elevation should be represented by a "positive number"(Example of positive number: 16)

Also,

The Grasshopper is <em><u>below </u></em>the chameleon, so its elevation should be represented by a "negative number"(Example of negative number: -23)

ඞHope this helps u because i got dis question from Khan Academyඞ

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1) Use power series to find the series solution to the differential equation y'+2y = 0 PLEASE SHOW ALL YOUR WORK, OR RISK LOSING
iogann1982 [59]

If

y=\displaystyle\sum_{n=0}^\infty a_nx^n

then

y'=\displaystyle\sum_{n=1}^\infty na_nx^{n-1}=\sum_{n=0}^\infty(n+1)a_{n+1}x^n

The ODE in terms of these series is

\displaystyle\sum_{n=0}^\infty(n+1)a_{n+1}x^n+2\sum_{n=0}^\infty a_nx^n=0

\displaystyle\sum_{n=0}^\infty\bigg(a_{n+1}+2a_n\bigg)x^n=0

\implies\begin{cases}a_0=y(0)\\(n+1)a_{n+1}=-2a_n&\text{for }n\ge0\end{cases}

We can solve the recurrence exactly by substitution:

a_{n+1}=-\dfrac2{n+1}a_n=\dfrac{2^2}{(n+1)n}a_{n-1}=-\dfrac{2^3}{(n+1)n(n-1)}a_{n-2}=\cdots=\dfrac{(-2)^{n+1}}{(n+1)!}a_0

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So the ODE has solution

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\boxed{y(x)=a_0e^{-2x}}

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