The population, P (in thousands), of a town can be modeled by P= 2|t-6|+4, where t=0 represents 1990. During which two years does the town have a population of 8000?

P = 2 * | t - 6 | + 4 = 8

( Remark : 8 mean 8 thousands )

2 * | t - 6 | = 8 - 4

2 * | t - 6 | = 4 Divide both sides with 2

| t - 6 | = 2

| t - 6 | = t - 6

and

| t - 6 | = - ( t - 6 ) also

You must solve two equations:

1.)

t - 6 = 2

t = 2 + 6

t = 8

and

2.)

- ( t - 6 ) = 2

- t + 6 = 2

- t = 2 - 6

- t = - 4 Multiply both sides with - 1

t = 4

So equation:

P = 2 * | t - 6 | + 4 = 8

has the two solutions :

t = 4 yrs

and

t= 8 yrs

Proof:

P = 2 * | t - 6 | + 4

for t = 4 yrs

P = 2 * | 4 - 6 | + 4 =

2 * | - 2 | + 4 =

2 * 2 + 4 =

4 + 4 = 8

P = 2 * | t - 6 | + 4

for t = 8 yrs

P = 2 * | 8 - 6 | + 4 =

2 * | 2 | + 4 =

2 * 2 + 4 =

4 + 4 = 8

1990 + 4 = 1994 yr

1990 + 8 = 1998 yr

Thank you so much!!

Your welcome its no problem

To find the years when the population of the town is 8000, we can plug in this value for P in the given equation and solve for t.

Given equation: P = 2|t-6| + 4

Step 1: Substitute P = 8000 into the equation:
8000 = 2|t-6| + 4

Step 2: Subtract 4 from both sides of the equation:
7996 = 2|t-6|

Step 3: Divide both sides by 2:
3998 = |t-6|

Step 4: Remove the absolute value by considering both positive and negative cases:
Case 1: t - 6 = 3998
Solve for t:
t = 3998 + 6
t = 4004
So one possible year is t = 4004 + 1990 = 6004.

Case 2: t - 6 = -3998
Solve for t:
t = -3998 + 6
t = -3992
So another possible year is t = -3992 + 1990 = -2002.

Therefore, the town has a population of 8000 in the years 6004 and -2002.