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2 years ago

Use the interval notation to express the solution -6x>42

Mathematics

1 answer:

ser-zykov [4K]2 years ago

6 0

divide each term by -6

x<-7

then convert in to interval

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Sarafina had 34 out of 99 hits when she was at bat during the softball season what was her batting average

prisoha [69]

Sarafina's batting average is 2.91

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3 years ago

Write two and twenty five hundredths as a decimal number

denis23 [38]

The answer is 2.25
1) simplify the fraction
25/100 = 1/4
2) then turn 1/4 into a decimal
1/4 = .25
3) then add
2+0.25= 2.25

8 0

3 years ago

Please! #12, Thank you so much!

vovikov84 [41]

A. 168 feet
b. 1 yard = 3 feet
168*3=504 so, 168 feet =504 yards
c. 504 yards*$15=7560
d. $7560

4 0

3 years ago

Read 2 more answers

HI I NEVER KNEW HOW TO DO THIS BUT I DO SO CAN YALL REALLY HELP ME PLEASE

Aliun [14]

Answer:

A. 4/15 and 0.26 repeating

Step-by-step explanation:

As we see in the long division, 4/15 is equal to 0.26666666666666666666666 (or. 0.26 repeating).

4/15

4 doesn't go into 15, so add a zero.

4.0/15

40 goes into 15 twice

40 - 30 is 10. 10 is less than 15 so add a 0.

4.00

100 goes into 15 6 times.

100 - 90 is 10. 10 is less than 15, so add a 0.

The 100 into 15 process repeats, so there's your answer.

8 0

3 years ago

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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

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

$=\displaystyle2\left(\lim_{n\to\infty}\frac1{(n+1)^2}\right)\left(\lim_{n\to\infty}\left(\frac n{n+1}\right)^n\right)=2\cdot0\cdot e=0$

which is less than 1, so this series is absolutely convergent.

6 0

3 years ago

Use the interval notation to express the solution -6x>42​

Use the interval notation to express the solution -6x>42