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3 years ago

Which list of vaules is ordered from lest to greastest?

Mathematics

2 answers:

Gnesinka [82]3 years ago

7 0

The answer is B. Because the lines mean opposite of

inessss [21]3 years ago

6 0

The answer is B it's ordered from least to greatest

You might be interested in

F(x)=5^x if F(-2) what does this equal

KengaRu [80]

Answer:

$\frac{1}{25}$

Step-by-step explanation:

Using the rule of exponents

$a^{-m}$ = $\frac{1}{a^{m} }$ , hence

f(- 2) = $5^{-2}$ = $\frac{1}{5^{2} }$ = $\frac{1}{25}$

3 0

2 years ago

Last month I had 8 guppies. This month I have 20 guppies. What is the percent of increase in guppies from last month to this mon

ludmilkaskok [199]

Answer: 150% increase

Step-by-step explanation:

percent of increase: new/old × 100%-100%

--------------

new/this month: 20 puppies

old/last month: 8 puppies

new/old×100%-100%

=20/8×100%-100%

=2.5×100%-100%

=250% -100%

=150% increase

Hope this helps!! :)

Please let me know if you have any question

8 0

2 years ago

Read 2 more answers

Consider the series ∑n=1[infinity]2nn!nn. Evaluate the the following limit. If it is infinite, type "infinity" or "inf". If it d

Vikki [24]

I guess the series is

$\displaystyle\sum_{n=1}^\infty\frac{2^nn!}{n^n}$

We have

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{2^{n+1}(n+1)!}{(n+1)^{n+1}}}{\frac{2^nn!}{n^n}}\right|=2\lim_{n\to\infty}\left(\frac n{n+1}\right)^n$

Recall that

$e=\displaystyle\lim_{n\to\infty}\left(1+\frac1n\right)^n$

In our limit, we have

$\dfrac n{n+1}=\dfrac{n+1-1}{n+1}=1-\dfrac1{n+1}$

$\left(\dfrac n{n+1}\right)^n=\dfrac{\left(1-\frac1{n+1}\right)^{n+1}}{1-\frac1{n+1}}$

$\implies\displaystyle2\lim_{n\to\infty}\left(\frac n{n+1}\right)^n=2\frac{\lim\limits_{n\to\infty}\left(1-\frac1{n+1}\right)^{n+1}}{\lim\limits_{n\to\infty}\left(1-\frac1{n+1}\right)}=\frac{2e}1=2e$

which is greater than 1, which means the series is divergent by the ratio test.

On the chance that you meant to write

$\displaystyle\sum_{n=1}^\infty\frac{2^n}{n!n^n}$

we have

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{2^{n+1}}{(n+1)!(n+1)^{n+1}}}{\frac{2^n}{n!n^n}}\right|=2\lim_{n\to\infty}\frac1{(n+1)^2}\left(\frac n{n+1}\right)^2$