Math / Algebra

QuestionBetween 2000 and 2020, the population of Mathville could be modeled by the function $m(t)=100 \sqrt[3]{t}$ , where $m(t)$ is the number of people in Mathville, and $t$ is the number of years since 2000. Between those same years, the population of Calcfield could be modeled by the function $c(t)=18 t$ . A. Graph each function on graph paper or a neatly made coordinate grid by hand. Be sure to consider an appropriate domain for the functions as you make your graph. B. Approximately where do the functions intersect? What does this point of intersection represent? C. Write and solve an equation to algebraically confirm where the two functions intersect. Show your work. D. Write 2-3 complete sentences comparing the relative populations of the cities over time.

Studdy Solution

STEP 1

1. $m(t) = 100 \sqrt[3]{t}$ models the population of Mathville.
2. $c(t) = 18t$ models the population of Calcfield.
3. $t$ represents the number of years since 2000, so $t$ ranges from 0 to 20.
4. We need to graph both functions, find their intersection, and compare the populations over time.

STEP 2

1. Graph the functions $m(t)$ and $c(t)$ .
2. Determine the approximate intersection point graphically.
3. Solve algebraically for the intersection point.
4. Compare the populations over time.

STEP 3

Graph the function $m(t) = 100 \sqrt[3]{t}$ on a coordinate grid. Consider the domain $0 \leq t \leq 20$ .
Graph the function $c(t) = 18t$ on the same coordinate grid. Consider the domain $0 \leq t \leq 20$ .

STEP 4

Examine the graphs to find the approximate intersection point. This point represents the year and population where Mathville and Calcfield have the same population.

STEP 5

Set the equations equal to each other to find the exact intersection point: $100 \sqrt[3]{t} = 18t$

STEP 6

Cube both sides to eliminate the cube root: $(100 \sqrt[3]{t})^3 = (18t)^3$ $100^3 \cdot t = 18^3 \cdot t^3$

STEP 7

Divide both sides by $t$ (assuming $t \neq 0$ ): $100^3 = 18^3 \cdot t^2$

STEP 8

Solve for $t^2$ : $t^2 = \frac{100^3}{18^3}$

STEP 9

Take the square root of both sides to solve for $t$ : $t = \sqrt{\frac{100^3}{18^3}}$
Calculate the value of $t$ .

STEP 10

Compare the populations: - Initially, Calcfield's population grows linearly, while Mathville's grows more slowly. - As time progresses, Mathville's population growth accelerates due to the cubic root function. - The intersection point indicates when both populations are equal, after which Mathville's population surpasses Calcfield's.
The approximate intersection point is calculated to be around $t \approx 7.2$ years after 2000, meaning around the year 2007.2. At this point, both cities have approximately the same population.

Was this helpful?

Studdy Solution

STEP 1

1. $m(t) = 100 \sqrt[3]{t}$ models the population of Mathville.
2. $c(t) = 18t$ models the population of Calcfield.
3. $t$ represents the number of years since 2000, so $t$ ranges from 0 to 20.
4. We need to graph both functions, find their intersection, and compare the populations over time.

STEP 2

1. Graph the functions $m(t)$ and $c(t)$ .
2. Determine the approximate intersection point graphically.
3. Solve algebraically for the intersection point.
4. Compare the populations over time.

STEP 3

Graph the function $m(t) = 100 \sqrt[3]{t}$ on a coordinate grid. Consider the domain $0 \leq t \leq 20$ .
Graph the function $c(t) = 18t$ on the same coordinate grid. Consider the domain $0 \leq t \leq 20$ .

STEP 4

Examine the graphs to find the approximate intersection point. This point represents the year and population where Mathville and Calcfield have the same population.

STEP 5

Set the equations equal to each other to find the exact intersection point: $100 \sqrt[3]{t} = 18t$

STEP 6

Cube both sides to eliminate the cube root: $(100 \sqrt[3]{t})^3 = (18t)^3$ $100^3 \cdot t = 18^3 \cdot t^3$

STEP 7

Divide both sides by $t$ (assuming $t \neq 0$ ): $100^3 = 18^3 \cdot t^2$

STEP 8

Solve for $t^2$ : $t^2 = \frac{100^3}{18^3}$

STEP 9

Take the square root of both sides to solve for $t$ : $t = \sqrt{\frac{100^3}{18^3}}$
Calculate the value of $t$ .

STEP 10

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Studdy Solution

STEP 1

STEP 2

1. Graph the functions m(t) m(t) m(t) and c(t) c(t) c(t).2. Determine the approximate intersection point graphically.3. Solve algebraically for the intersection point.4. Compare the populations over time.

STEP 3

Graph the function m(t)=100t3 m(t) = 100 \sqrt[3]{t} m(t)=1003t​ on a coordinate grid. Consider the domain 0≤t≤20 0 \leq t \leq 20 0≤t≤20.Graph the function c(t)=18t c(t) = 18t c(t)=18t on the same coordinate grid. Consider the domain 0≤t≤20 0 \leq t \leq 20 0≤t≤20.

STEP 4

Examine the graphs to find the approximate intersection point. This point represents the year and population where Mathville and Calcfield have the same population.

STEP 5

Set the equations equal to each other to find the exact intersection point: 100t3=18t 100 \sqrt[3]{t} = 18t 1003t​=18t

STEP 6

Cube both sides to eliminate the cube root: (100t3)3=(18t)3 (100 \sqrt[3]{t})^3 = (18t)^3 (1003t​)3=(18t)3 1003⋅t=183⋅t3 100^3 \cdot t = 18^3 \cdot t^3 1003⋅t=183⋅t3

STEP 7

Divide both sides by t t t (assuming t≠0 t \neq 0 t=0): 1003=183⋅t2 100^3 = 18^3 \cdot t^2 1003=183⋅t2

STEP 8

Solve for t2 t^2 t2: t2=1003183 t^2 = \frac{100^3}{18^3} t2=1831003​

STEP 9

Take the square root of both sides to solve for t t t: t=1003183 t = \sqrt{\frac{100^3}{18^3}} t=1831003​​Calculate the value of t t t.

STEP 10

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Graph the functions $m(t)$ and $c(t)$ .
2. Determine the approximate intersection point graphically.
3. Solve algebraically for the intersection point.
4. Compare the populations over time.

Graph the function $m(t) = 100 \sqrt[3]{t}$ on a coordinate grid. Consider the domain $0 \leq t \leq 20$ .
Graph the function $c(t) = 18t$ on the same coordinate grid. Consider the domain $0 \leq t \leq 20$ .

Set the equations equal to each other to find the exact intersection point: $100 \sqrt[3]{t} = 18t$

Cube both sides to eliminate the cube root: $(100 \sqrt[3]{t})^3 = (18t)^3$ $100^3 \cdot t = 18^3 \cdot t^3$

Divide both sides by $t$ (assuming $t \neq 0$ ): $100^3 = 18^3 \cdot t^2$

Solve for $t^2$ : $t^2 = \frac{100^3}{18^3}$

Take the square root of both sides to solve for $t$ : $t = \sqrt{\frac{100^3}{18^3}}$
Calculate the value of $t$ .