Question

In 1999, there were 41,275 shopping centers in a certain country. In 2009, there were 48,199.

A) write an equation expressing the number y of shopping centers in terms of the number x of years after 1999.
B) when will the number of shopping centers reach 80,000?

Answer 1

Answer:

a) $y(x) = 692.4x + 41275$

b) The number of shopping centers will reach 80,000 in 2055.

Step-by-step explanation:

The number of shopping centers in the city may be expressed by the following equation

$y(x) = ax + b$

In which y is the number of shopping centers, x is the number of years, a is the slope(average rate of change) and b is the initial number of shopping centers.

In 1999, there were 41,275 shopping centers in a certain country

This means that $b = 41275$

In 2009, there were 48,199.

2009 is 10 years after 1999. So

$y(10) = 48199$

A) write an equation expressing the number y of shopping centers in terms of the number x of years after 1999.

Applying what we have, we can find a.

$y(x) = ax + b$

$48199 = 10a + 41275$

$10a = 6924$

$a = \frac{6924}{10}$

$a = 692.4$

So

$y(x) = 692.4x + 41275$

B) when will the number of shopping centers reach 80,000?

This is x years after 1999, when y(x) = 80,000. So

$y(x) = 692.4x + 41275$

$692.4x + 41275 = 80000$

$692.4x = 38725$

$x = \frac{38725}{692.4}$

$x = 56$

1999 + 56 = 2055

So the number of shopping centers will reach 80,000 in 2055.