All categories

3 years ago

Determine if the sequence is geometric. If it is, find the term named in the problem.

Mathematics

1 answer:

tino4ka555 [31]3 years ago

4 0

Answer:

120

Step-by-step explanation:

Each new term is twice the previous term, so this is a geometric sequence. The applicable formula is

a(n) = a(1)*4*(n - 1)

and so a(9) = 4*4(9 - 1) = 120

You might be interested in

Here’s another one thank u all for helping me. I really appreciate it!

nikdorinn [45]

Answer:

628 feet..............

6 0

3 years ago

The town of Madison has a population of 25000 The population is increasing by a factor of 1.12 each year.

Fudgin [204]

Answer:

$P(t) = 25,000*1.12^t$

Step-by-step explanation:

The populational growth is exponential with a factor of 1.12 each year. An exponential function has the following general equation:

$y(x) = ab^x$

Where 'a' is the initial population (25,000 people), 'b' is the growth factor (1.12 per year), 'x' is the time elapsed, in years, and 'y(x)' is the population after 'x' years.

Therefore, the function P(t) that models the population in Madison t years from now is:

$P(t) = 25,000*1.12^t$

8 0

3 years ago

Read 2 more answers

It cost Joseph $20.10 to send 134 text messages. How much does it cost to send one text message? Show your work.

Licemer1 [7]

$20.10/134
$0.15 cost for one message

8 0

3 years ago

The ratio of the numerator and denominator of a rational number is 2 : 7 and

Nadya [2.5K]

Answer:

6/21.

Step-by-step explanation:

let the rational number be 2x/7x.

2x + 4 / 7x - 2 = 10 /19

Cross multiply:

19(2x + 4) = 10(7x - 2)

38x + 76 = 70x - 20

96 = 70x - 38x

32x = 96

x = 3.

So the rational number is 2*3/ 7*3

= 6/21.

4 0

3 years ago

Determine whether the improper integral converges or diverges, and find the value of each that converges.

Ksju [112]

Answer:converge at $I=\frac{1}{3}$

Step-by-step explanation:

Given

Improper Integral I is given as

$I=\int^{\infty}_{3}\frac{1}{x^2}dx$

integration of $\frac{1}{x^2}$ is - $\frac{1}{x}$

$I=\left [ -\frac{1}{x}\right ]^{\infty}_3$

substituting value

$I=-\left [ \frac{1}{\infty }-\frac{1}{3}\right ]$

$I=-\left [ 0-\frac{1}{3}\right ]$

$I=\frac{1}{3}$

so the value of integral converges at $\frac{1}{3}$

8 0

4 years ago