Solve this inequality. -35 – 2 >7 A. B. C. D.

At the end of 2006, the population of Riverside was 400 people. The population for this small town can be modeled by the equation below, where t represents the number of years since the end of 2006 and P represents the number of people. $P = 400 ( 1.2 )^ t$ Based on this model, approximately what was the increase in the population of Riverside at the end of 2009 compared to the end of 2006?

(A) 291

(B) 691

(C) 1040

(D) 1440

Answer:

The increase in the population at the end of 2009 is 291 people

Step-by-step explanation:

We are given the equation as $P = 400 ( 1.2 )^ t$

where

P = No of People

t= No of Years

it is given that in the year 2006 the population is 400

this will only happen when we take t= 0

so for

Year value of t

2006- t = 0

2007- t = 1

2008- t = 2

2009 t = 3

No of people in 2009 will be

$P = 400(1.2)^3$

= 400*1.728

P = 691.2

Since the equation represents no of people so it can't be in decimals, Therefore the population will be 691

Increase = P(2009) - P(2006)

= 691 - 400

= 291

The increase in the population at the end of 2009 is 291 people.

4 0

3 years ago

A = 1011 + 337 + 337/2 +1011/10 + 337/5 + ... + 1/2021

egoroff_w [7]

The sum of the given series can be found by simplification of the number

of terms in the series.

A is approximately 2020.022

Reasons:

The given sequence is presented as follows;

A = 1011 + 337 + 337/2 + 1011/10 + 337/5 + ... + 1/2021

Therefore;

$\displaystyle A = \mathbf{1011 + \frac{1011}{3} + \frac{1011}{6} + \frac{1011}{10} + \frac{1011}{15} + ...+\frac{1}{2021}}$

The n + 1 th term of the sequence, 1, 3, 6, 10, 15, ..., 2021 is given as follows;

$\displaystyle a_{n+1} = \mathbf{\frac{n^2 + 3 \cdot n + 2}{2}}$

Therefore, for the last term we have;

$\displaystyle 2043231= \frac{n^2 + 3 \cdot n + 2}{2}$

2 × 2043231 = n² + 3·n + 2

Which gives;

n² + 3·n + 2 - 2 × 2043231 = n² + 3·n - 4086460 = 0

Which gives, the number of terms, n = 2020

$\displaystyle \frac{A}{2} = \mathbf{ 1011 \cdot \left(\frac{1}{2} +\frac{1}{6} + \frac{1}{12}+...+\frac{1}{4086460} \right)}$

$\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2} +\frac{1}{2} - \frac{1}{3} + \frac{1}{3}- \frac{1}{4} +...+\frac{1}{2021}-\frac{1}{2022} \right)$

Which gives;

$\displaystyle \frac{A}{2} = 1011 \cdot \left(1 - \frac{1}{2022} \right)$

$\displaystyle A = 2 \times 1011 \cdot \left(1 - \frac{1}{2022} \right) = \frac{1032231}{511} \approx \mathbf{2020.022}$

A ≈ 2020.022

Learn more about the sum of a series here:

brainly.com/question/190295

8 0

2 years ago

Read 2 more answers

Solve this inequality. -35 – 2 >7 A. B. C. D. > -3