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| 10 Theory slides |
| 10 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
Ramsha is studying Egyptian history for class.She learned that ancient Egyptians had an interesting way to represent fractions. They used unit fractions, which are fractions of the form N1, to represent all other fractions. The examples of the Egyptian fractions was found in the image of the Eye of Horus. Each part of the eye represents a different fraction.
The least common multiple (LCM) of two whole numbers a and b is the smallest whole number that is a multiple of both a and b. It is denoted as LCM(a,b). The least common multiple of a and b is the smallest whole number that is divisible by both a and b. Some examples can be seen in the table below.
Numbers | Multiples of Numbers | Common Multiples | Least Common Multiple |
---|---|---|---|
2 and 3 | Multiples of 2:Multiples of 3: 2,4,6,8,10,12,… 3,6,9,12,15,…
|
6,12,18,24,… | LCM(2,3)=6 |
8 and 12 | Multiples of 8:Multiples of 12: 8,16,24,32,40,48,… 12,24,36,48,…
|
24,48,72,96,… | LCM(8,12)=24 |
A special procedure exists for finding the LCM(a,b) of a pair of numeric expressions. The LCM of two relatively prime numbers is always equal to their product.
Coprimes | LCM |
---|---|
3 and 5 | 15 |
5 and 4 | 20 |
4 and 9 | 36 |
Polynomials can also have a least common multiple. The LCM of two or more polynomials is the smallest multiple of both polynomials. In other words, the LCM is the smallest expression that can be evenly divided by each of the given polynomials.
Polynomials | Factor | LCM | Explanation |
---|---|---|---|
4x3and6xy2
|
22⋅x3and2⋅3⋅x⋅y2
|
12x3y2 | 12x3y2 is the smallest expression that is divisible by both 4x3 and 6xy2. |
3x3+3x2andx2+5x+4
|
3⋅x2⋅(x+1)and(x+1)(x+4)
|
3x2(x+1)(x+4) | 3x2(x+1)(x+4) is the smallest expression that is divisible by both 3x3+3x2 and x2+5x+4. |
Finding the LCM of polynomials requires identifying the factors with the highest power that appear in each polynomial.
Split into factors
Factor out 12y
a2+2ab+b2=(a+b)2
Both polynomials are now written in factored form. Now, write all the missing related factors to identify the factor with the highest power.
Standard Form | Factored Form | All Related Factors | |
---|---|---|---|
Polynomail I | 12x2y+48xy+48y | 22⋅3⋅y⋅(x+2)2 | 22⋅31⋅x0⋅y1⋅(x+2)2⋅(x−8)1 |
Polynomail II | 3x3y−18x2y−48xy | 3⋅x⋅y⋅(x+2)⋅(x−8) | 20⋅31⋅x1⋅y1⋅(x+2)1⋅(x−8)1 |
What is the least common multiple of the given polynomials?
Split into factors
Factor out 6x
Split into prime factors
Split into factors
Factor out 3x
am⋅bm=(a⋅b)m
Split into factors
Write as a power
a2−2ab+b2=(a−b)2
Given Form | Factored Form | All Related Factors | |
---|---|---|---|
Heichi | 6x2y−6x | 2⋅3⋅x⋅(xy−1) | 21⋅31⋅x1⋅(xy−1)1 |
Ramsha | 3x3y2−6x2y+3x | 3⋅x⋅(xy−1)2 | 20⋅31⋅x1⋅(xy−1)2 |
The least common denominator (LCD) of two fractions is the least common multiple (LCM) of the denominators of the fractions. In other words, the least common denominator is the smallest of all the common denominators. Some examples are provided in the table below.
Fractions | Denominators | Multiples of Denominators | Common Denominators | LCM of Denominators (LCD) |
---|---|---|---|---|
32 and 21 | 3 and 2 | Multiples of 3:Multiples of 2: 3,6,9,12,15,… 2,4,6,8,10,12,…
|
6, 12 | 6 |
65 and 41 | 6 and 4 | Multiples of 6:Multiples of 4: 6,12,18,24,30,… 4,8,12,16,20,24,…
|
12, 24 | 12 |
41 and 25 | 4 and 2 | Multiples of 4:Multiples of 2: 4,8,12,… 2,4,6,8,10,12,…
|
4, 8, 12 | 4 |
When adding and subtracting rational expressions, the same rules apply as when adding and subtracting fractions.
If the rational expressions have a common denominator, the numerators can be added or subtracted directly.
Q(x)P(x)±Q(x)R(x)=Q(x)P(x)±R(x)
Here, P(x), Q(x), and R(x) are polynomials and Q(x)=0.
If the denominators are different, the expressions have to be manipulated to find a common denominator before they can be added or subtracted. One way of doing this is to multiply both numerator and denominator of one of the rational expressions with the denominator of the other, and vice versa.
Q(x)P(x)±S(x)R(x)=Q(x)S(x)P(x)S(x)±R(x)Q(x)
Denominator | Factored Form |
---|---|
2x−2 | 2(x−1) |
x2−4x+3 | (x−1)(x−3) |
ba=b⋅(x−3)a⋅(x−3)
Multiply parentheses
Add terms
ba=b⋅2a⋅2
Distribute 2
Multiply
Ramsha is very enthusiastic about learning more about Egypt. She finds an Egyptian pen pal, Izabella. Izabella mentions that her favorite dessert is basbousa.
The talk of dessert reminds Ramsha of something she learned in chemistry class. The pH, or acidity, of a person's mouth changes after eating a dessert. The pH level L of a mouth t minutes after eating a dessert is modeled by the following formula.Distribute 6
Subtract fractions
Commutative Property of Addition
t=15
Calculate power
Multiply
Add and subtract terms
Use a calculator
Round to 2 decimal place(s)
In one of her emails, Izabella tells Ramsha about her father's job. He is in charge of pool maintenance work at a local hotel.
Her father gives Isabella the following set of information.
Let t be the time it takes for the first pipe to fill the pool. Then, the fraction t1 represents the part of the pool that the first pipe fills in one hour.
Let t be the time it takes for Pipe I to fill the pool. Since Pipe II takes 2 times as long to fill the pool as Pipe I, the time needed for Pipe II to fill the pool will be 2 times t. Similarly, the time needed for Pipe III will be 1.5 times 2t because Pipe III takes 1.5 times as long to fill the pool as Pipe II.
Pipe | Time Needed to Fill the Pool (hours) |
---|---|
Pipe I | t |
Pipe II | 2⋅t=2t |
Pipe III | 1.5⋅2t=3t |
All of Them | 6 |
Now, think about how much of the pool is filled by each pipe, alone or together, in one hour. For example, it takes all three pipes 6 hours to fill the pool. Therefore, they would fill 61 of the pool in one hour. The filling rates of the other pipes can be determined by this same logic.
Pipe | Time Needed to Fill the Pool (hours) | Filling Rate (per hour) |
---|---|---|
Pipe I | t | t1 |
Pipe II | 2t | 2t1 |
Pipe III | 3t | 3t1 |
All of Them | 6 | 61 |
Denominator | Factored Form |
---|---|
t | t |
2t | 2⋅t |
3t | 3⋅t |
6 | 2⋅3 |
ba=b⋅6a⋅6
ba=b⋅3a⋅3
ba=b⋅2a⋅2
ba=b⋅ta⋅t
Multiply fractions
Add fractions
Izabella is planning a trip that involves a 90-kilometer bus ride and a high-speed train ride. The entire trip is 300 kilometers.
The average speed of the high-speed train is 30 kilometers per hour more than twice the average speed of the bus.
Add fractions
Ramsha shares what she has learned about Egypt with her mother. She sees how enthusiastic Ramsha is about life in Egypt and considers taking out a loan for the family to take a trip to Egypt.
1=aa
Subtract fractions
a/cb=ba⋅c
Substitute values
Add terms
Multiply
Use a calculator
Round to 2 decimal place(s)
Rational expressions are basically fractions that have variables in their numerators or denominators. As with numeric fractions, rational expressions can be added and subtracted. To clarify the connection between fractions and rational expressions, the challenge presented at the beginning of the lesson will be solved.
Recall the information Ramsha learned about Egyptian fractions.
ba=b⋅xa⋅x
Add fractions
Cancel out common factors
Simplify quotient
Substitute expressions
Add terms
q1=q+11+q(q+1)1
q1=q+11+q(q+1)1
q+11=q+21+(q+1)(q+2)1
q(q+1)1=q(q+1)+11+q(q+1)(q(q+1)+1)1
By repeating this procedure, replacing every duplicate unit fraction with two unit fractions with larger denominators, all repeating unit fractions will eventually disappear. Therefore, any positive rational number qp can be written as an Egyptian fraction.
Lenses are used in a wide variety of areas, including telescopes, cameras, and magnifying glasses.
We are asked to find a simplified form of the thin-lens equation. When we look at the choices, we see that the variable f is on the left side for each equation, so our first step will be to solve the thin-lens equation for f.
The result corresponds to option D.
We found a simplified equation for variable f in Part A.
f = d_i d_o/d_o+d_i
We can use this equation to find the focal length f of a lens when we know the actual distance d_o of the object from the lens and the distance d_i of the image from the lens. Let's show what we know on a diagram.
We can substitute d_o=50 and d_i=20 into our equation and find the focal length of the lens.
The focal length is about 14.3 centimeters.
A complex fraction is a rational expression that has at least one fraction in its numerator, denominator, or both. We want to simplify the given complex fraction. 1x^2y+ 1xy^2/1x+ 1y To do so, we will combine the expressions in the numerator and the denominator separately. Then, we will multiply the new numerator by the reciprocal of the new denominator. Let's start by simplifying the numerator.
Now let's simplify the denominator.
Next, we can rewrite the complex fraction using the simplified components. 1x^2y+ 1xy^2/1x+ 1y ⇔ y+xx^2y^2/y+xxy Finally, we will multiply the numerator by the reciprocal of the new denominator. y+xx^2y^2/y+xxy ⇔ y+x/x^2y^2 * xy/y+x We can simplify this expression.
Let's analyze the given equation. T=d/s+j+d/s-j The denominators of the rational expressions do not share any common factors. The least common denominator is therefore their product. Let's multiply each fraction by the denominator of the other fraction. We will be able to add them together afterward.
We are given the following information about the round-trip flight between Seattle and Miami.
We are asked to find the total time T of the flight for the given conditions. Let's substitute the values into the simplified formula from Part A and find T.
Therefore, the total time is about 11 hours.
We are given that the amount of the two medications in the bloodstream are determined by A(t) and B(t). A(t) & =5t/3t^2+12t+9 [0.7em] B(t) & =7t/5t^2+20t+15 Let's add these two functions together to determine a function for the total amount of the medication in a person's bloodstream after t hours. A(t) + B(t) = 5t/3t^2+12t+9 + 7t/5t^2 + 20t + 15 Remember, in order to add rational expressions, they must have the same denominator. For this reason, we need to factor the denominators of the rational expressions and find their least common multiple. This will be a candidate for a common denominator.
Factor | |
---|---|
3t^2+12t+9 | 5t^2+20t+15 |
3t^2+ 3t+9t+9 | 5t^2+ 5t+15t+15 |
3t(t+1)+9(t+1) | 5t(t+1)+15(t+1) |
(3t+9)(t+1) | (5t+15)(t+1) |
3(t+3)(t+1) | 5(t+3)(t+1) |
We factored the denominators of the fractions. The least common multiple (LCM) must contain each factor in both denominators. LCM: 3* 5* (t+1)* (t+3) For the fractions to have a denominator as above, let's expand each of them by the factor of LCM their denominators do not have.
Now that the fractions have common denominators, we are able to add them.
We have found the simplified version of the given function!
From Part A, we know that the total amount of medication in LaShay's bloodstream is determined by the following equation.
A(t)+B(t)=46t/15t^2+60t+45
Here, t represents the time in hours after the medication is taken. Since we are asked to find the amount of medication in her system after 6 hours, let's substitute 6 for t into the equation.
We conclude that 6 hours after the medications are taken, the amount in LaShay's bloodstream is approximately 0.29 milligrams.