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3 years ago

Solve the equation 4y – 3 = 1 for y. A.-1 B.0 C.1 D.2

Mathematics

2 answers:

zimovet [89]3 years ago

8 0

Answer:

Step-by-step explanation:

4 times 1 is 4 and 4-3 is 1

Aleks04 [339]3 years ago

8 0

Answer:

c. 1

Step-by-step explanation:

4y-3=1

cancel out the numbers

4y-3=1

+3 +3

4y=4

division,

y=1

hope this helps

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Jason is painting a large circle on one wall of his new apartment. The diameter of the circle will be 8 feet. Approximately how

vampirchik [111]

Answer:

about 50.3 feet squared

Step-by-step explanation:

Area of Circle= pi times radius squared

8/2=4

radius=4

4x4=16

Area=16 pi feet

about 50.3 feet sqared

6 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

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