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kolbaska11 [484]
4 years ago
15

The graph of a logarithmic function is shown below.

Mathematics
1 answer:
Dmitrij [34]4 years ago
3 0

Domain means all the x-values include in the graph.

In this case the graph starts including x-values at 0 and all x-values that are greater then 0.

In other words x can be zero or anything larger

Technically you would show this like so:

x ≥ 0

You use the inequality symbol with the bar underneath because you want to show that 0 is included in the domain. It faces away from the zero and towards the x because x can be larger then 0 not smaller.

I guess in this case you wouldn't be including 0 so your answer is:

B. x > 0

Hope this helped!

~Just a girl in love with Shawn Mendes

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I need help on this question
schepotkina [342]

Answer:

  0 4 9 1

             _____________

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 − 0      

   9 3    

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Step-by-step explanation:

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Find a power series representation for the function. (Assume a>0. Give your power series representation centered at x=0 .)
melamori03 [73]

Answer:

Step-by-step explanation:

Given that:

f_x = \dfrac{x^2}{a^7-x^7}

= \dfrac{x^2}{a^7(1-\dfrac{x^7}{a^7})}

= \dfrac{x^2}{a^7}\Big(1-\dfrac{x^7}{a^7} \Big)^{-1}

since  \Big((1-x)^{-1}= 1+x+x^2+x^3+...=\sum \limits ^{\infty}_{n=0}x^n\Big)

Then, it implies that:

\implies  \dfrac{x^2}{a^7} \sum \limits ^{\infty}_{n=0} \Big(\Big(\dfrac{x}{a} \Big)^{^7} \Big)^n

= \dfrac{x^2}{a^7} \sum \limits ^{\infty}_{n=0} \Big(\dfrac{x}{a} \Big)^{^{7n}}

= \dfrac{x^2}{a^7} \sum \limits ^{\infty}_{n=0} \Big(\dfrac{x^{7n}}{a^{7n}} \Big)}

\mathbf{=  \sum \limits ^{\infty}_{n=0} \dfrac{x^{7n+2}}{a^{7n+7}} }}

7 0
3 years ago
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