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1 year ago

Can someone please help me?!? 25 points

Mathematics

1 answer:

vladimir1956 [14]1 year ago

4 0

Answer:

See below

Step-by-step explanation:

There are 52 weeks in a year, so divide t by 52

= 120 (1.016)^(t/52)

then the % for one week = 1.016^1/52 = 1.0003 = .03%

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2 years ago

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Nevertheless, the only way for a fraction to equal zero is to have a zero numerator, i.e.

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Moreover, we have

$\displaystyle \lim_{x\to 0} \dfrac{x}{x^3-2x^2+5x} = \lim_{x\to 0} \dfrac{x}{x(x^2-2x+5)} = \lim_{x\to 0} \dfrac{1}{x^2-2x+5} = \dfrac{1}{5}$

So, we can't even extend with continuity this function in such a way that $f(0)=0$

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3 years ago