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3 years ago

Can I get the answers for number 14 plz?

Mathematics

2 answers:

nataly862011 [7]3 years ago

8 0

A.
-1:
$2(2)^{-1} \\ 2( \frac{1}{2}) \\ \frac{2}{2} \\ 1$
(-1,1)

0:
$2(2)^{0} \\ 2(1) \\ 2$
(0, 2)

1:
$2(2)^{1} \\ 2(2) \\ 4$
(1, 4)

2:
$2(2)^{2} \\ 2(4) \\ 8$
(2, 8)

3:
$2(2) ^{3} \\ 2(8) \\ 16$
(3,16)

b.
To graph the equation, simply go through the points (-2, 0.5), (-1, 1), (0,2), (1,4), (2,8), and (3,16). Make sure you never go below 0 on the x-axis, because there's an asymptote there.

Hope this helps!

attashe74 [19]3 years ago

7 0

Ive gotta say, amazing job brooke....^

You might be interested in

Here's the data (sorted) of the ages of 91 women who won the Oscar for Best Actress in a Leading Role:

Andreyy89

Answer:

B. 21, 28, 33, 41, 81

Step-by-step explanation:

Data : 21, 22, 22, 24, 24, 24, 24, 25, 26, 26, 26, 26, 26, 26, 26, 26, 26, 27, 27, 27, 27, 28, 28, 28, 28, 29, 29, 29, 29, 29, 30, 30, 30, 30, 30, 30, 31, 31, 31, 32, 32, 33, 33, 33, 33, 33, 33, 33, 34, 34, 34, 34, 34, 35, 35, 35, 35, 35, 37, 37, 37, 37, 38, 38, 38, 39, 39, 39, 41, 41, 41, 41, 42, 42, 44, 45, 45, 45, 47, 49, 49, 54, 60, 60, 61, 61, 61, 62, 62, 74, 81

Five number summary includes :

1. Minimum vale

2. First Quartile

3. Median : mid value of the given data is 46th term = 33

4.Third Quartile

5.Maximum value

Solution :

1. Minimum vale : smallest value of data = 21

2. First Quartile: median of the data (first number to the actual median) i.e. from 1st term to 46th term

Median for even number of terms : $\frac{\frac{n}{2}th+(\frac{n}{2}+1)th}{2}$

n = 46

so, median = $\frac{\frac{46}{2}th+(\frac{46}{2}+1)th}{2}$

= $\frac{23th+24th}{2}$

= $\frac{28+28}{2}$

= $\frac{56}{2}$

= $28$

Thus first quartile = 28

3. Median : mid value of the given data(ages of 91 women) is 46th term = 33

4.Third Quartile :median of the data (46th number[actual median] to the 91th)

Median for even number of terms : $\frac{\frac{n}{2}th+(\frac{n}{2}+1)th}{2}$

n = 46

so, median = $\frac{\frac{46}{2}th+(\frac{46}{2}+1)th}{2}$

= $\frac{23th+24th}{2}$

= $\frac{41+41}{2}$

= $\frac{82}{2}$

= $41$

Thus third quartile = 41

5.Maximum value : highest value of data = 81

Thus the five number summary is 21, 28, 33, 41, 81

Hence Option B is correct

8 0

3 years ago

I need help with this problem from the calculus portion on my ACT prep guide

LenaWriter [7]

Given a series, the ratio test implies finding the following limit:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=r$

If r<1 then the series converges, if r>1 the series diverges and if r=1 the test is inconclusive and we can't assure if the series converges or diverges. So let's see the terms in this limit:

$\begin{gathered} a_n=\frac{2^n}{n5^{n+1}} \\ a_{n+1}=\frac{2^{n+1}}{(n+1)5^{n+2}} \end{gathered}$

Then the limit is:

$\lim _{n\to\infty}\lvert\frac{a_{n+1}}{a_n}\rvert=\lim _{n\to\infty}\lvert\frac{n5^{n+1}}{2^n}\cdot\frac{2^{n+1}}{\mleft(n+1\mright)5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert$

We can simplify the expressions inside the absolute value:

$\begin{gathered} \lim _{n\to\infty}\lvert\frac{2^{n+1}}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^{n+1}}{5^{n+2}}\rvert=\lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert \\ \lim _{n\to\infty}\lvert\frac{2^n\cdot2}{2^n}\cdot\frac{n}{n+1}\cdot\frac{5^n\cdot5}{5^n\cdot5\cdot5}\rvert=\lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert \\ \lim _{n\to\infty}\lvert2\cdot\frac{n}{n+1}\cdot\frac{1}{5}\rvert=\lim _{n\to\infty}\lvert\frac{2}{5}\cdot\frac{n}{n+1}\rvert \end{gathered}$

Since none of the terms inside the absolute value can be negative we can write this with out it: