Which type of transformation is shown below?

Mathematics

2 answers:

Lina20 [59]2 years ago

7 0

Answer:

C.) A dilation. It got bigger so its a dilation

damaskus [11]2 years ago

6 0

Translation is your answer

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What is the equation of this graphed line?

liq [111]

You can put the line into the equation y= MX+c
m = gradient
c = y offset

To work out m you need to divide the change in y by the change in x for two points on the graph:
6--6 = 12 (change in x)
-3--7 = 4 (change in y)

m = 4/12 = 1/3

Then to work out the c variable you need to see where the line crosses the y axis, where this happens is the c variable:
c = -5 because c crosses y at the point -5

Then place it into one equation:
y = 1/3x-5

8 0

3 years ago

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

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