Math worksheet on bear-related algebraic expressions for educational use.

Problem: Solve the tasks related to bears and write algebraic expressions for each scenario.

#### 1. Bear Population
- Task: The population of bears in a national park is currently \( p \) and is increasing by 10% each year. Write an algebraic expression for the population of bears after 1 year.
- Solution:
- If the current population is \( p \), a 10% increase means the population will be multiplied by \( 1 + 0.10 = 1.10 \).
- Therefore, the population after 1 year is:
\[
p \times 1.10 = 1.1p
\]
- Answer: \( 1.1p \)

#### 2. Bear Cubs
- Task: A female bear gives birth to 3 cubs every 2 years. If the number of cubs in the first year is \( x \), write an algebraic expression for the number of cubs after \( n \) years.
- Solution:
- Since the bear gives birth to 3 cubs every 2 years, the number of cubs born in \( n \) years is proportional to \( \frac{n}{2} \).
- The total number of cubs after \( n \) years is the initial number of cubs (\( x \)) plus the additional cubs born over \( n \) years.
- The additional cubs born in \( n \) years are \( 3 \times \frac{n}{2} = \frac{3n}{2} \).
- Therefore, the total number of cubs after \( n \) years is:
\[
x + \frac{3n}{2}
\]
- Answer: \( x + \frac{3n}{2} \)

#### 3. Food Consumption
- Task: A bear consumes 5 pounds of food per day. If the bear has been hibernating for \( d \) days, write an algebraic expression for the total amount of food consumed during hibernation.
- Solution:
- The bear consumes 5 pounds of food per day.
- Over \( d \) days, the total amount of food consumed is:
\[
5 \times d = 5d
\]
- Answer: \( 5d \)

#### 4. Bear Weight
- Task: A bear weighs \( w \) kilograms. If the bear gains 2 kilograms every month for \( m \) months, write an algebraic expression for the bear's weight after \( m \) months.
- Solution:
- The bear starts with a weight of \( w \) kilograms.
- It gains 2 kilograms every month for \( m \) months, so the total weight gained is \( 2m \) kilograms.
- The bear's weight after \( m \) months is:
\[
w + 2m
\]
- Answer: \( w + 2m \)

#### 5. Speed
- Task: A bear is running at a constant speed. If the distance traveled by the bear is \( d \) kilometers and the time taken is \( t \) hours, write an algebraic expression relating the speed, distance, and time.
- Solution:
- Speed is defined as the distance traveled divided by the time taken.
- Therefore, the speed \( s \) is:
\[
s = \frac{d}{t}
\]
- Answer: \( s = \frac{d}{t} \)

#### 6. Bear Tracking
- Task: A group of researchers is tracking the movement of bears in a forest. The position of a bear at time \( t \) is given by the equation \( p = 2t^2 - 5t + 10 \), where \( p \) is the position in kilometers. Find the position of the bear at \( t = 3 \).
- Solution:
- Substitute \( t = 3 \) into the equation \( p = 2t^2 - 5t + 10 \):
\[
p = 2(3)^2 - 5(3) + 10
\]
- Calculate step-by-step:
\[
p = 2(9) - 15 + 10 = 18 - 15 + 10 = 13
\]
- Answer: \( 13 \)

#### 7. Bear Lifespan
- Task: The lifespan of a bear is 15 years longer than twice the age of a cub. If the age of a cub is \( y \) years, write an algebraic expression for the lifespan of a bear.
- Solution:
- Twice the age of a cub is \( 2y \).
- The lifespan of a bear is 15 years more than this, so:
\[
\text{Lifespan} = 2y + 15
\]
- Answer: \( 2y + 15 \)

#### 8. Conservation Area
- Task: A square bear conservation area has an area of 64 square kilometers. Write an algebraic equation to represent the length, \( L \), of one side of the conservation area.
- Solution:
- The area of a square is given by the side length squared: \( \text{Area} = L^2 \).
- Given that the area is 64 square kilometers:
\[
L^2 = 64
\]
- Answer: \( L^2 = 64 \)

#### 9. Fishing
- Task: A bear catches 5 fish every hour. If the bear has been fishing for \( h \) hours, write an algebraic expression for the total number of fish caught.
- Solution:
- The bear catches 5 fish per hour.
- Over \( h \) hours, the total number of fish caught is:
\[
5 \times h = 5h
\]
- Answer: \( 5h \)

#### 10. Migration
- Task: A group of bears migrates from their summer habitat to their winter habitat. If the distance between the two habitats is \( d \) kilometers and the bears travel at a speed of \( s \) kilometers per hour, write an algebraic expression for the time taken to migrate.
- Solution:
- Time is calculated as distance divided by speed.
- Therefore, the time taken to migrate is:
\[
\text{Time} = \frac{d}{s}
\]
- Answer: \( \frac{d}{s} \)

Final Answers:

1. \( \boxed{1.1p} \)
2. \( \boxed{x + \frac{3n}{2}} \)
3. \( \boxed{5d} \)
4. \( \boxed{w + 2m} \)
5. \( \boxed{s = \frac{d}{t}} \)
6. \( \boxed{13} \)
7. \( \boxed{2y + 15} \)
8. \( \boxed{L^2 = 64} \)
9. \( \boxed{5h} \)
10. \( \boxed{\frac{d}{s}} \)