Math / Geometry

QuestionMath 19 Models and Word Problems Worksheet
1. A surveyor plants a stake on one side of a road that runs east to west. The surveyor then walks directly across the road, walks 5 meters to the east, and plants a second stake. The distance between the two stakes is 13 meters. If the surveyor continues to walk to the east a further 11 meters, what would the distance between the surveyor and the first stake be? (a) Begin by drawing a picture, labeling the position of all stakes and distances, including unknowns. Which variable are you solving for? Will there be any intermediate variables you will need to solve for before finding the final answer? (b) Use the Pythagorean Theorem to find the width of the road in meters. (c) Using the information you have, apply the Pythagorean Theorem to solve the original problem.
2. I own a ladder that is 20 feet tall and lean it up against a wall at an angle. The distance between the top of the ladder and the ground is triple the distance between the base of the wall and the bottom of the ladder. What is the distance between the bottom of the ladder and the base of the wall? (Round your answer to the nearest tenth of a foot.)
3. Bamboo is one of the fastest growing plants in the world. Suppose I plant some 15 inch bamboo in my garden. Assume for all parts of this problem that bamboo grows at 35 inches per day. (Amazingly, this is a low estimate for its growth rate!) (a) Write a linear model/function for the height of $h(t)$ the bamboo as it relates to time $t$ in days where $t=0$ is the day I planted it. (b) Express the height of the bamboo on the day that it was planted (c) I have a fence at the back of my garden that is 10 ft tall. How long will it be from the planting of the bamboo until it is as tall as my fence? (d) If we instead choose $t=0$ to be the day that the bamboo is the same height as the fence, how will the model change? (e) How would the model change if we expressed the time $t$ in weeks instead of days?
4. The University of Golden Bears has been working on increasing its enrollment numbers since the year 2005. In 2010, the student population was 11,000 students and by 2020 , the student population was 14,000. (a) Write a function of time $t$ representing the population $P$ . Assume the growth of the student population was linear and use $t=0$ at 2005. (b) What would the model be instead if $t=0$ at 2010?

Studdy Solution

STEP 1

1. We are solving multiple problems involving geometry, algebra, and linear modeling.
2. Each problem is independent and requires its own set of steps.
3. For geometry problems, the Pythagorean Theorem will be used.
4. For linear modeling problems, we will assume constant rates of change.

STEP 2

1. Problem 1: Geometry with stakes and road - Draw a diagram and identify variables - Use the Pythagorean Theorem to find the width of the road - Calculate the final distance using the Pythagorean Theorem
2. Problem 2: Ladder against a wall - Set up the relationship using given conditions - Solve for the unknown distance
3. Problem 3: Bamboo growth modeling - Write a linear model for bamboo growth - Evaluate the model at specific times - Adjust the model for different conditions
4. Problem 4: University enrollment modeling - Write a linear function for student population - Adjust the function for different starting times

STEP 3

(a) Draw a diagram of the scenario with the road, stakes, and distances. Label the width of the road as $w$ , the distance between the stakes as 13 meters, and the additional eastward distance as 11 meters. We are solving for the final distance between the surveyor and the first stake. Intermediate variable: width of the road $w$ .

STEP 4

(b) Use the Pythagorean Theorem to find the width of the road. The first stake, second stake, and the point directly across the road form a right triangle.
The equation is: $w^2 + 5^2 = 13^2$

STEP 5

Solve for $w$ : $w^2 + 25 = 169$ $w^2 = 144$ $w = 12$

STEP 6

(c) Calculate the final distance using the Pythagorean Theorem. The surveyor walks an additional 11 meters east, creating a new right triangle with the road width and the total eastward distance.
The equation is: $\text{Distance}^2 = w^2 + (5 + 11)^2$ $\text{Distance}^2 = 12^2 + 16^2$

STEP 7

Solve for the final distance: $\text{Distance}^2 = 144 + 256$ $\text{Distance}^2 = 400$ $\text{Distance} = 20$

STEP 8

Set up the relationship for the ladder problem. Let $x$ be the distance between the bottom of the ladder and the base of the wall. The top of the ladder is 3 times this distance.
The equation is: $x^2 + (3x)^2 = 20^2$

STEP 9

Solve for $x$ : $x^2 + 9x^2 = 400$ $10x^2 = 400$ $x^2 = 40$ $x = \sqrt{40}$ $x \approx 6.3$

STEP 10

(a) Write a linear model for the bamboo height: $h(t) = 15 + 35t$ .

STEP 11

(b) Express the height of the bamboo on the day it was planted: $h(0) = 15$ .

STEP 12

(c) Solve for when the bamboo reaches 10 feet (120 inches): $15 + 35t = 120$ $35t = 105$ $t = 3$

STEP 13

(d) Adjust the model for $t=0$ when bamboo is 10 feet: $h(t) = 120 + 35t$ .

STEP 14

(e) Change time to weeks: $h(t) = 15 + 35(7t) = 15 + 245t$ .

STEP 15

(a) Write a linear function for student population: $P(t) = 11000 + \frac{3000}{10}(t-5)$ $P(t) = 11000 + 300(t-5)$

STEP 16

(b) Adjust the model for $t=0$ at 2010: $P(t) = 11000 + 300t$
The solutions to the problems are:
1. The distance between the surveyor and the first stake is 20 meters.
2. The distance between the bottom of the ladder and the base of the wall is approximately 6.3 feet.
3. (a) The model is $h(t) = 15 + 35t$ . (b) The height on the day it was planted is 15 inches. (c) It takes 3 days for the bamboo to reach 10 feet. (d) The adjusted model is $h(t) = 120 + 35t$ . (e) The model in weeks is $h(t) = 15 + 245t$ .
4. (a) The function is $P(t) = 11000 + 300(t-5)$ . (b) The adjusted function is $P(t) = 11000 + 300t$ .

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Studdy Solution

STEP 1

1. We are solving multiple problems involving geometry, algebra, and linear modeling.
2. Each problem is independent and requires its own set of steps.
3. For geometry problems, the Pythagorean Theorem will be used.
4. For linear modeling problems, we will assume constant rates of change.

STEP 2

STEP 3

STEP 4

(b) Use the Pythagorean Theorem to find the width of the road. The first stake, second stake, and the point directly across the road form a right triangle.
The equation is: $w^2 + 5^2 = 13^2$

STEP 5

Solve for $w$ : $w^2 + 25 = 169$ $w^2 = 144$ $w = 12$

STEP 6

(c) Calculate the final distance using the Pythagorean Theorem. The surveyor walks an additional 11 meters east, creating a new right triangle with the road width and the total eastward distance.
The equation is: $\text{Distance}^2 = w^2 + (5 + 11)^2$ $\text{Distance}^2 = 12^2 + 16^2$

STEP 7

Solve for the final distance: $\text{Distance}^2 = 144 + 256$ $\text{Distance}^2 = 400$ $\text{Distance} = 20$

STEP 8

Set up the relationship for the ladder problem. Let $x$ be the distance between the bottom of the ladder and the base of the wall. The top of the ladder is 3 times this distance.
The equation is: $x^2 + (3x)^2 = 20^2$

STEP 9

Solve for $x$ : $x^2 + 9x^2 = 400$ $10x^2 = 400$ $x^2 = 40$ $x = \sqrt{40}$ $x \approx 6.3$

STEP 10

(a) Write a linear model for the bamboo height: $h(t) = 15 + 35t$ .

STEP 11

(b) Express the height of the bamboo on the day it was planted: $h(0) = 15$ .

STEP 12

(c) Solve for when the bamboo reaches 10 feet (120 inches): $15 + 35t = 120$ $35t = 105$ $t = 3$

STEP 13

(d) Adjust the model for $t=0$ when bamboo is 10 feet: $h(t) = 120 + 35t$ .

STEP 14

(e) Change time to weeks: $h(t) = 15 + 35(7t) = 15 + 245t$ .

STEP 15

(a) Write a linear function for student population: $P(t) = 11000 + \frac{3000}{10}(t-5)$ $P(t) = 11000 + 300(t-5)$

STEP 16

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Studdy Solution

STEP 1

1. We are solving multiple problems involving geometry, algebra, and linear modeling.2. Each problem is independent and requires its own set of steps.3. For geometry problems, the Pythagorean Theorem will be used.4. For linear modeling problems, we will assume constant rates of change.

STEP 2

STEP 3

STEP 4

(b) Use the Pythagorean Theorem to find the width of the road. The first stake, second stake, and the point directly across the road form a right triangle.The equation is: w2+52=132 w^2 + 5^2 = 13^2 w2+52=132

STEP 5

Solve for w w w: w2+25=169 w^2 + 25 = 169 w2+25=169 w2=144 w^2 = 144 w2=144 w=12 w = 12 w=12

STEP 6

STEP 7

Solve for the final distance: Distance2=144+256 \text{Distance}^2 = 144 + 256 Distance2=144+256 Distance2=400 \text{Distance}^2 = 400 Distance2=400 Distance=20 \text{Distance} = 20 Distance=20

STEP 8

Set up the relationship for the ladder problem. Let x x x be the distance between the bottom of the ladder and the base of the wall. The top of the ladder is 3 times this distance.The equation is: x2+(3x)2=202 x^2 + (3x)^2 = 20^2 x2+(3x)2=202

STEP 9

Solve for x x x: x2+9x2=400 x^2 + 9x^2 = 400 x2+9x2=400 10x2=400 10x^2 = 400 10x2=400 x2=40 x^2 = 40 x2=40 x=40 x = \sqrt{40} x=40​ x≈6.3 x \approx 6.3 x≈6.3

STEP 10

(a) Write a linear model for the bamboo height: h(t)=15+35t h(t) = 15 + 35t h(t)=15+35t.

STEP 11

(b) Express the height of the bamboo on the day it was planted: h(0)=15 h(0) = 15 h(0)=15.

STEP 12

(c) Solve for when the bamboo reaches 10 feet (120 inches): 15+35t=120 15 + 35t = 120 15+35t=120 35t=105 35t = 105 35t=105 t=3 t = 3 t=3

STEP 13

(d) Adjust the model for t=0 t=0 t=0 when bamboo is 10 feet: h(t)=120+35t h(t) = 120 + 35t h(t)=120+35t.

STEP 14

(e) Change time to weeks: h(t)=15+35(7t)=15+245t h(t) = 15 + 35(7t) = 15 + 245t h(t)=15+35(7t)=15+245t.

STEP 15

(a) Write a linear function for student population: P(t)=11000+300010(t−5) P(t) = 11000 + \frac{3000}{10}(t-5) P(t)=11000+103000​(t−5) P(t)=11000+300(t−5) P(t) = 11000 + 300(t-5) P(t)=11000+300(t−5)

STEP 16

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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. We are solving multiple problems involving geometry, algebra, and linear modeling.
2. Each problem is independent and requires its own set of steps.
3. For geometry problems, the Pythagorean Theorem will be used.
4. For linear modeling problems, we will assume constant rates of change.

(b) Use the Pythagorean Theorem to find the width of the road. The first stake, second stake, and the point directly across the road form a right triangle.
The equation is: $w^2 + 5^2 = 13^2$

Solve for $w$ : $w^2 + 25 = 169$ $w^2 = 144$ $w = 12$

Solve for the final distance: $\text{Distance}^2 = 144 + 256$ $\text{Distance}^2 = 400$ $\text{Distance} = 20$

Set up the relationship for the ladder problem. Let $x$ be the distance between the bottom of the ladder and the base of the wall. The top of the ladder is 3 times this distance.
The equation is: $x^2 + (3x)^2 = 20^2$

Solve for $x$ : $x^2 + 9x^2 = 400$ $10x^2 = 400$ $x^2 = 40$ $x = \sqrt{40}$ $x \approx 6.3$

(a) Write a linear model for the bamboo height: $h(t) = 15 + 35t$ .

(b) Express the height of the bamboo on the day it was planted: $h(0) = 15$ .

(c) Solve for when the bamboo reaches 10 feet (120 inches): $15 + 35t = 120$ $35t = 105$ $t = 3$

(d) Adjust the model for $t=0$ when bamboo is 10 feet: $h(t) = 120 + 35t$ .

(e) Change time to weeks: $h(t) = 15 + 35(7t) = 15 + 245t$ .

(a) Write a linear function for student population: $P(t) = 11000 + \frac{3000}{10}(t-5)$ $P(t) = 11000 + 300(t-5)$