Isık kirliliğinin careta caretalaraetkisi

Mathematics

2 answers:

ohaa [14]3 years ago

7 0

Answer:

ışık kirliligi sebebiyle caretalar doğduklarında yanlış yöne gider ve bu sebeple nesileri tükene bilir

Step-by-step explanation:

iyi dersler

mina [271]3 years ago

3 0

Answer:

Işık kirliliği, gökyüzünü doğal gece gökyüzünden çok daha parlak hale getirir. Galaksiler ve bulutsular gibi sönük nesnelerden gelen ışığı bastırarak ve sönük yıldızlar için gökyüzü arka planını artırarak astronomi üzerinde önemli etkileri vardır.

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Last one .. worth 20 points please .

777dan777 [17]

Answer:

Should be y=x+2.5

Step-by-step explanation:

8 0

3 years ago

Xe^(-x^2/128) absolute max and absolute min

adoni [48]

$f(x)=xe^{-x^2/128}$
$\implies f'(x)=e^{-x^2/128}+x\left(-\dfrac{2x}{128}\right)e^{-x^2/128}=\left(1-\dfrac1{64}x^2\right)e^{-x^2/128}$

Extrema can occur when the derivative is zero or undefined.

$\left(1-\dfrac1{64}x^2\right)e^{-x^2/128}=0\implies 1-\dfrac1{64}x^2=0\implies x^2=64\implies x=\pm8$

Maxima occur where the first derivative is zero and the second derivative is negative; minima where the second derivative is positive. You have

$f''(x)=-\dfrac1{32}xe^{-x^2/128}+\left(1-\dfrac1{64}x^2\right)\left(-\dfrac{2x}{128}\right)e^{-x^2/128}=\left(-\dfrac3{64}x+\dfrac1{4096}x^3\right)e^{-x^2/128}$

At the critical points, you get

$f''(-8)=\dfrac1{4\sqrt e}>0$
$f''(8)=-\dfrac1{4\sqrt e}$

So you have a minimum at $\left(-8,-\dfrac8{\sqrt e}\right)$ and a maximum at $\left(8,\dfrac8{\sqrt e}\right)$ .

Meanwhile, as $x\to\pm\infty$ , it's clear that $f(x)\to0$ , so these extrema are absolute on the function's domain.