Which expressions of signed decimal numbers have negative quotients? Check all that apply.

2 answers:

7 0

A negative number divided by positive is a negative quotient and vice versa.
So
5.7 / -2.2
-0.01 / 0.05
Are the ones with negative quotients

pav-90 [236]2 years ago

3 0

Answer:

im sorry im late but its b and c

Step-by-step explanation:

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M(5, 7) is the mid-point of the l.ine segment joining A (3,4) to B. Find the coordinates of B.

Nina [5.8K]

Answer:

B(7,10)

Step-by-step explanation:

$M(5,7)=(x ,y)\\A(3,4)=(x_1,y_1)\\B(x_2,y_2)\\\\Endpoint ;\\x = \frac{x_1+x_2}{2}\\ \\5 = \frac{3+x_2}{2}\\ Cross\:Multiply\\5\times 2 =3+x_2\\10=3+x_2\\10-3 =x_2\\7=x_2\\\\\\y = \frac{y_1+y_2}{2} \\\\7 = \frac{4+y_2}{2}\\ Cross\:Multiply\\7\times 2 =4+y_2\\14 =4+y_2\\14-4 =y_2\\10 =y_2\\\\\\B(7,10)$

5 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

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