If a plant grew from 6 inches, by 10 inches by, what percent did it increase

ULLIVLY/1/2/3/assessment

zvonat [6]

<h3>The worth after 4 years is $ 680.24</h3>

Solution:

The formula for compound interest, including principal sum, is:

$A = p(1+\frac{r}{n})^{nt}$

Where,

A = the future value of the investment

P = the principal investment amount

r = the annual interest rate (decimal)

n = the number of times that interest is compounded per unit t

t = the time the money is invested

From given,

n = 1 ( since interest is compounded annually)

p = 500

t = 4

$r = 8 \% = \frac{8}{100} = 0.08$

Substituting the values we get,

$A = 500(1+ \frac{0.08}{1})^{1 \times 4}\\\\A = 500(1.08)^4\\\\A = 500 \times 1.36048896\\\\A = 680.24448 \approx 680.24$

Thus the worth after 4 years is $ 680.24

3 0

3 years ago

Rewrite each of the following durations using eighth notes. Write in words and as a fraction: 1 quarter note 1 half note 1 full

LenKa [72]

Answer: one third. one half and 1 whole

Step-by-step explanation:

4 0

3 years ago

Which statements are true select each correct answer

alexandr1967 [171]

The first and the second one are true statements.

4 0

3 years ago

1367324+3469345-75369x3429x0=??

Mnenie [13.5K]

Answer:

it's ans is 4761300

Step-by-step explanation:

by using BODMAS

75369×3429×0

=0

0+1367324+3469345

=4836669

4836669-75369

=4761300

plz mark me as brainliest plz

5 0

3 years ago

Use any of the methods to determine whether the series converges or diverges. Give reasons for your answer.

Aleks [24]

Answer:

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

Step-by-step explanation:

The actual Series is::

$\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$

The method we are going to use is comparison method:

According to comparison method, we have:

$\sum_{n=1}^{inf}a_n\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n$

If series one converges, the second converges and if second diverges series, one diverges

Now Simplify the given series:

Taking"n^2"common from numerator and "n^6"from denominator.

$=\frac{n^2[7-\frac{4}{n}+\frac{3}{n^2}]}{n^6[\frac{12}{n^6}+2]} \\\\=\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{n^4[\frac{12}{n^6}+2]}$

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\ \ \ \ \ \ \ \ \sum_{n=1}^{inf}b_n=\sum_{n=1}^{inf} \frac{1}{n^4}$

Now:

$\sum_{n=1}^{inf}a_n=\sum_{n=1}^{inf}\frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\ \\\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{[7-\frac{4}{n}+\frac{3}{n^2}]}{[\frac{12}{n^6}+2]}\\=\frac{7-\frac{4}{inf}+\frac{3}{inf}}{\frac{12}{inf}+2}\\\\=\frac{7}{2}$

So a_n is finite, so it converges.

Similarly b_n converges according to p-test.

P-test:

General form:

$\sum_{n=1}^{inf}\frac{1}{n^p}$

if p>1 then series converges. In oue case we have:

$\sum_{n=1}^{inf}b_n=\frac{1}{n^4}$

p=4 >1, so b_n also converges.

According to comparison test if both series converges, the final series also converges.

It means $\sum_{n=1}^\inf} = \frac{7n^2-4n+3}{12+2n^6}$ also converges.

5 0

3 years ago

If a plant grew from 6 inches, by 10 inches by, what percent did it increase​

If a plant grew from 6 inches, by 10 inches by, what percent did it increase