3 years ago

I need help with exponents

Mathematics

1 answer:

Tanya [424]3 years ago

4 0

How can I help you? Friend me or what every they do on this. Im pretty good at them.

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For what values (cases) of the variables the expression does not exist: a a 2x−4

Natasha2012 [34]

Step-by-step explanation:

When the denominator is zero, the expression is undefined

2x-4=0

2x=4

x=2

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3 years ago

;) yeah Imma just cheat my way through school you guys help a lot

amid [387]

Answer:

Thanks you too

Step-by-step explanation:

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3 years ago

Use l'Hôpital's Rule to find the following limit. ModifyingBelow lim With x right arrow 0StartFraction 3 sine (x )minus 3 x Ove

steposvetlana [31]

Answer:

$\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=-\frac{1}{14}$

Step-by-step explanation:

The limit is:

$\lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{0}{0}$

so, you have an indeterminate result. By using the l'Hôpital's rule you have:

$\lim_{x \to 0} \frac{a(x)}{b(x)}= \lim_{x \to 0} \frac{a'(x)}{b'(x)}$

by replacing, and applying repeatedly you obtain:

$\lim_{x \to 0} \frac{3sinx-3x}{7x^3}= \lim_{x \to 0}\frac{3cosx-3}{21x^2}= \lim_{x \to 0}\frac{-3sinx}{42x}= \lim_{x \to 0}\frac{-3cosx}{42}\\\\ \lim_{x \to 0} \frac{3sinx-3x}{7x^3}=\frac{-3cos0}{42}=-\frac{1}{14}$

hence, the limit of the function is -1/14

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3 years ago

19 books on a shelf. 8 of these books are new. The rest of them are used. What is the ratio of new books to used books

harkovskaia [24]

Answer:

1:8

Step-by-step explanation:

You subtract 19 and 8, and you get 11 then you put them in ration form

Step-by-step explanation:

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16. A newspaper surveyed its subscribers as to which section of the paper they read first. The results are below. Is it reasonab

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Answer:

Step-by-step explanation:n

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3 years ago