College Algebra Exercises Section 1.2

College Algebra Exercises Section 1.3

As you study these exercises, move your cursor over arrows, equal symbols, light bulbs, and highlighted words. Always check in with the wise old owl and his little apprentice by moving your cursor over them. Study with an active cursor.

The definition that pops up when you move the cursor over a highlighted word is what should pop into your mind when you read, hear, or speak that word.

The material that pops up when you move the cursor over a light bulb is a suggested strategy for solving the problem. You should always formulate a similar strategy when you attempt to solve a problem.

When you move the cursor over an explanation of a step in a solution, the property that pops up is the mathematical justification for taking that action. You must always be able to provide such a justification for every step you take in mathematics.

12. Write an algebraic expression to represent the product of two consecutive natural numbers .

Solution:
Let n be an arbitrary natural number.
Then the next larger natural number is n + 1.
The product of these two consecutive natural numbers is n(n + 1) = n² + n

14. Write an algebraic expression to represent the sum of the squares two consecutive even integers the first of which is 2n.

Solution:
If 2n is the first of two consecutive even integers, then the next larger even integer is 2n + 2.
The squares of these two even integers are (2n)² and (2n + 2)².
The sum of these two squares is (2n)² + (2n + 2)².
A little algebraic simplification is in order.
(2n)² + (2n + 2)² = 4n² + 4n² + 8n + 4 = 8n² + 8n + 4 = 4(2n² + 2n + 1).

18. Write an algebraic expression to represent the sale price of an item that is discounted 20% of its list price L.

Solution:
If the list price is L and the discount is 20% of L, then the discount is 0.2L.
The sale price is then the list price L minus the discount 0.2L.
So the sale price S is given by S = L - 0.2L = 0.8L.

26. Write an algebraic expression or equation to represent the percent change in sales from one month to the next if the sales are S₁ and S₂ respectively.

Solution:
To compute the percent of change we must first compute the amount of change and then determine what percent that change is of S₁.
That amount of change is S₂ - S₁. The question now is: S₂ - S₁ is what percent of S₁ ?
If we let P represent that percent, then P = .

29. The sum of two consecutive natural numbers is 525. Find the numbers.

Solution:
Let n be an arbitrary natural number.
Then the next larger natural number is n + 1.
The sum of these natural numbers is n + (n + 1)
The sum of these natural numbers is 525.
These two representations for the sum of the two numbers must be equal. So we obtain the equation
n + (n + 1) = 525 which may be written as
2n + 1 = 525
2n = 524
n = 262
Then n + 1 = 263.
The desired two natural numbers are 262 and 263.

30. Find three consecutive natural numbers whose sum is 804.

Solution:
Let n be an arbitrary natural number.
Then the next larger natural number is n + 1 and the one following that is n + 2.
Let S be the sum of these three natural numbers.
Then S = 804 and S = n + (n + 1) + (n + 2).
We have two expressions for the same quantity S. Those two expressions must therefore be equal and that gives us the desired model and equation to solve.
n + (n + 1) + (n + 2) = 804
3n + 3 = 804
3n = 801
n = 267
The desired three consecutive natural numbers are 267, 268, and 269.

33. Find two consecutive integers whose product is 5 less than the square of the smaller number.

Solution:
Let n be an arbitrary integer.
Then the next larger integer is n + 1.
The product of these natural numbers is n (n + 1) = n² + n
The sum of these natural numbers is n² - 5.
These two representations for the product of the two numbers must be equal. So we obtain the equation
n² + n = n² - 5
n = - 5
Then n + 1 = - 4.
The desired two integers are - 5 and - 4.

54. Two cars start at an interstate interchange and travel in the same direction at average speeds of 40 miles per hour and 55 miles per hour. How much time must elapse before the two cars are 5 miles apart ?

Solution:
We can use the distance formula d = rt to determine how far each car is from the starting at any given time.
The distance between the two cars is simply the difference of the two distances .
The distance between them is also 5 miles at the time we are trying to find.
Two expressions for the same quantity -- distance between the cars.
That will give us the equation to solve.
Let d₁ = 40t be the distance traveled by the slower car.
Let d₂ = 55t be the distance traveled by the faster car.
The distance between them at any time is D = d₂ - d₁ which is D = 55t - 40t.
Because we are concerned with the time when the cars are 5 miles apart we also know D = 5.
We now have the desired model:
55t - 40t = 5 which we can solve for t
15t = 5
t =
The cars will be 5 miles apart after they have traveled for hour or 20 minutes.

Comment: The solution to this particular problem may be obtained pretty easily by a variety of other analysis. The method presented here is to show a general method for solving these kinds of "distance" problems. For example consider the following problem.

Problem. Two cars start at an interstate interchange and travel in the same direction at average speeds of 54.8 miles per hour and 67.5 miles per hour. How much time must elapse before the two cars are 7 miles apart ? Express your answer so that it is correct to the nearest minute.

Solution:
We can use the distance formula d = rt to determine how far each car is from the starting at any given time.
The distance between the two cars is simply the difference of the two distances .
The distance between them is also 7 miles at the time we are trying to find.
Two expressions for the same quantity -- distance between the cars.
That will give us the equation to solve.
Let d₁ = 54.8t be the distance traveled by the slower car.
Let d₂ = 67.5t be the distance traveled by the faster car.
The distance between them at any time is D = d₂ - d₁ which is D = 67.5t - 54.8t.
Because we are concerned with the time when the cars are 7 miles apart we also know D = 7.
We now have the desired model:
67.5t - 54.8t = 7 which we can solve for t
12.7t = 7
hours
The cars will be 7 miles apart after they have traveled for 0.551181102 hours.
Suppose we want to express this time to the nearest minute. How many decimal places do we need to represent the time so that it is correct to the nearest minute ?
Look at the following table of computations.

From the above calculations it appears clear that we need two decimal place accuracy to represent the time correct to the nearest minute.
Our conclusion is:
The cars will be 7 miles apart after they have traveled for 33 minutes.

76. A trough is 12 feet long, 3 feet deap, and 3 feet wide. Find the depth of the water when the trough contains 70 galons of water.

Solution:
The volume of a rectangular solid is given by Volume = (length)(width)(height)
The volume of water in the tank is V = (12)(3)(h) cu.ft.
The volume of water in the tank is V = 70 gal. = 9.3576 cu.ft.
These two representations for the volume of water must be equal. So we obtain the equation
(12)(3)h = 9.3576
h = (9.3576)/36 = 0.25993 ft = 3.119 in.
When 70 gallons of water are in the trough, the water is 3.119 inches deep.
Remarks:
Notice that we used conversion factors to express gallons as cubic feet and feet as inches.
Here are those conversion factors:
1 cubic foot = 7.481 gallons
So 1 gallon = 0.13368 cubic feet
1 foot = 12 inches
So 1 inch = 0.08333 feet
Many conversion calculators may be found at
http://www.cpdesigncenter.com/