Math / Algebra

Question7. The population of a town, $P(t)$ , is modelled by the function $P(t)=6 t^{2}+110 t+3000$ , where $t$ is time in years. Note: $t=0$ represents the year 2000. a) When will the population reach 6000 ? b) What will the population be in 2030?

Studdy Solution

STEP 1

1. The function $P(t) = 6t^2 + 110t + 3000$ models the population of the town over time.
2. $t = 0$ corresponds to the year 2000.
3. To find when the population reaches 6000, we need to solve for $t$ when $P(t) = 6000$ .
4. To find the population in 2030, we need to evaluate $P(t)$ at $t = 30$ .

STEP 2

1. Solve for $t$ when $P(t) = 6000$ .
2. Evaluate $P(t)$ at $t = 30$ .

STEP 3

Set the population function equal to 6000 to find when the population reaches this number:
$P(t) = 6t^2 + 110t + 3000 = 6000$

STEP 4

Subtract 6000 from both sides to set the equation to zero:
$6t^2 + 110t + 3000 - 6000 = 0$ $6t^2 + 110t - 3000 = 0$

STEP 5

Solve the quadratic equation $6t^2 + 110t - 3000 = 0$ using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 6$ , $b = 110$ , and $c = -3000$ .
Calculate the discriminant:
$b^2 - 4ac = 110^2 - 4 \times 6 \times (-3000)$ $= 12100 + 72000$ $= 84100$

STEP 6

Calculate the roots using the quadratic formula:
$t = \frac{-110 \pm \sqrt{84100}}{2 \times 6}$ $t = \frac{-110 \pm 290}{12}$
Calculate the two possible values for $t$ :
$t_1 = \frac{-110 + 290}{12} = \frac{180}{12} = 15$ $t_2 = \frac{-110 - 290}{12} = \frac{-400}{12} = -\frac{100}{3}$
Since $t$ represents time in years and cannot be negative, we discard $t_2$ .

STEP 7

To find the population in 2030, calculate $t$ for the year 2030. Since $t = 0$ corresponds to the year 2000, $t = 30$ corresponds to the year 2030.
Evaluate $P(t)$ at $t = 30$ :
$P(30) = 6(30)^2 + 110(30) + 3000$

STEP 8

Calculate $P(30)$ :
$P(30) = 6(900) + 3300 + 3000$ $= 5400 + 3300 + 3000$ $= 11700$
The population will reach 6000 in the year $2000 + 15 = 2015$ .
The population in 2030 will be 11700.

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Question7. The population of a town, $P(t)$ , is modelled by the function $P(t)=6 t^{2}+110 t+3000$ , where $t$ is time in years. Note: $t=0$ represents the year 2000. a) When will the population reach 6000 ? b) What will the population be in 2030?

Studdy Solution

STEP 1

1. The function $P(t) = 6t^2 + 110t + 3000$ models the population of the town over time.
2. $t = 0$ corresponds to the year 2000.
3. To find when the population reaches 6000, we need to solve for $t$ when $P(t) = 6000$ .
4. To find the population in 2030, we need to evaluate $P(t)$ at $t = 30$ .

STEP 2

1. Solve for $t$ when $P(t) = 6000$ .
2. Evaluate $P(t)$ at $t = 30$ .

STEP 3

Set the population function equal to 6000 to find when the population reaches this number:
$P(t) = 6t^2 + 110t + 3000 = 6000$

STEP 4

Subtract 6000 from both sides to set the equation to zero:
$6t^2 + 110t + 3000 - 6000 = 0$ $6t^2 + 110t - 3000 = 0$

STEP 5

Solve the quadratic equation $6t^2 + 110t - 3000 = 0$ using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ , where $a = 6$ , $b = 110$ , and $c = -3000$ .
Calculate the discriminant:
$b^2 - 4ac = 110^2 - 4 \times 6 \times (-3000)$ $= 12100 + 72000$ $= 84100$

STEP 6

STEP 7

To find the population in 2030, calculate $t$ for the year 2030. Since $t = 0$ corresponds to the year 2000, $t = 30$ corresponds to the year 2030.
Evaluate $P(t)$ at $t = 30$ :
$P(30) = 6(30)^2 + 110(30) + 3000$

STEP 8

Calculate $P(30)$ :
$P(30) = 6(900) + 3300 + 3000$ $= 5400 + 3300 + 3000$ $= 11700$
The population will reach 6000 in the year $2000 + 15 = 2015$ .
The population in 2030 will be 11700.

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Question7. The population of a town, P(t)P(t)P(t), is modelled by the function P(t)=6t2+110t+3000P(t)=6 t^{2}+110 t+3000P(t)=6t2+110t+3000, where ttt is time in years. Note: t=0t=0t=0 represents the year 2000. a) When will the population reach 6000 ? b) What will the population be in 2030?

Studdy Solution

STEP 1

STEP 2

1. Solve for t t t when P(t)=6000 P(t) = 6000 P(t)=6000.2. Evaluate P(t) P(t) P(t) at t=30 t = 30 t=30.

STEP 3

Set the population function equal to 6000 to find when the population reaches this number:P(t)=6t2+110t+3000=6000 P(t) = 6t^2 + 110t + 3000 = 6000 P(t)=6t2+110t+3000=6000

STEP 4

Subtract 6000 from both sides to set the equation to zero:6t2+110t+3000−6000=0 6t^2 + 110t + 3000 - 6000 = 0 6t2+110t+3000−6000=0 6t2+110t−3000=0 6t^2 + 110t - 3000 = 0 6t2+110t−3000=0

STEP 5

STEP 6

STEP 7

To find the population in 2030, calculate t t t for the year 2030. Since t=0 t = 0 t=0 corresponds to the year 2000, t=30 t = 30 t=30 corresponds to the year 2030.Evaluate P(t) P(t) P(t) at t=30 t = 30 t=30:P(30)=6(30)2+110(30)+3000 P(30) = 6(30)^2 + 110(30) + 3000 P(30)=6(30)2+110(30)+3000

STEP 8

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Question7. The population of a town, $P(t)$ , is modelled by the function $P(t)=6 t^{2}+110 t+3000$ , where $t$ is time in years. Note: $t=0$ represents the year 2000. a) When will the population reach 6000 ? b) What will the population be in 2030?

1. Solve for $t$ when $P(t) = 6000$ .
2. Evaluate $P(t)$ at $t = 30$ .

Set the population function equal to 6000 to find when the population reaches this number:
$P(t) = 6t^2 + 110t + 3000 = 6000$

Subtract 6000 from both sides to set the equation to zero:
$6t^2 + 110t + 3000 - 6000 = 0$ $6t^2 + 110t - 3000 = 0$

To find the population in 2030, calculate $t$ for the year 2030. Since $t = 0$ corresponds to the year 2000, $t = 30$ corresponds to the year 2030.
Evaluate $P(t)$ at $t = 30$ :
$P(30) = 6(30)^2 + 110(30) + 3000$