Camarasaurus has a sale for 25% of camera bags, which is a discount of

Mathematics

1 answer:

rodikova [14]2 years ago

5 0

Answer:

answer is 80

Step-by-step explanation:

25% = 1/4

you'll need four lines in the diagram, each having 20.

I dont know what diagram your having to use, but I hope this helped!

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How do u turn 2 3/4 into a improper faction

n200080 [17]

To turn a mixed number into an improper fraction, you have to multiply the denominator with the whole number and add the numerator to the result. The denominator doesn't change.

The denominator is 4. The whole number is 2. Multiply:

$4 \times 2 = 8$

The numerator is 3. Add that to the product of the denominator and whole number.

$8+3=11$

The denominator stays the same. The denominator is 4. Make it as a fraction(improper).

$Numerator = 11 \\ Denominator = 4 \\ \\ Fraction: \frac{11}{4}$

Your answer is $\frac{11}{4}$

8 0

3 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

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