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| | 9 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Dominika and her friend Zosia live relatively close to each other. Each morning they leave their homes to go to school, and at some point, they meet and continue the trip together. Usually, when walking separately, they walk at different speeds.
Dominika, a budding citizen scientist, is curious about the changes in temperature where she lives in Phoenix, Arizona. Since she walks to school everyday, the temperature affects her walk. Dominika decides to measure the temperature at three different times on the same day. Here are the measurements she recorded.
| Time | Temperature |
|---|---|
| 16:00 | 94^(∘)F |
| 19:00 | 88^(∘)F |
| 21:00 | 78^(∘)F |
When was the temperature decreasing the fastest: between 16:00 and 19:00 or between 19:00 and 21:00?
| Time Interval | Δ t | Δ T |
|---|---|---|
| 16:00- 19:00 | 3 hours | 94-88= - 6^(∘)F |
| 19:00- 21:00 | 2 hours | 88-78= - 10^(∘)F |
Now, substitute the obtained values of Δ T and Δ t into the formula to find the rate of change of each time interval.
| Time Interval | Δ t | Δ T | Rate of Change |
|---|---|---|---|
| 16:00- 19:00 | 3 hours | - 6^(∘)F | - 6/3=- 2^(∘)F per hour |
| 19:00- 21:00 | 2 hours | - 10^(∘)F | - 10/2=- 5^(∘)F per hour |
As can be seen, the temperature decreased at an average rate of 2^(∘)F in the first time interval and 5^(∘)F in the second time interval. Therefore, the temperature decreased the fastest between 19:00 and 21:00.
Along the walk to school, Dominika passes her father's 24-hour cafe. Running the cafe requires close attention to customer behavior. Her father notices that it is typical to serve 1 customer each night from 2:00AM till 4:00AM. Then, starting from 6:00AM, the cafe slowly begins to fill with regulars.
Let P represent the number of people in the cafe and t represent the time in hours since midnight.
I.& f(12)=20 [0.3em] II.& f(14)-f(12)=20 [0.3em] III.& f(14)-f(12)/2=20 [0.3em] IV.& P=f(12)+20
From 2:00AM till 4:00AM there is usually 1 customer in the cafe. If t is a function of P, written as t=f(P), then the values of P are inputs and the values of t are outputs. Here, the input is 1, the customer, and the outputs are the hours 2:00AM, 3:00AM, and 4:00AM, labeled 2, 3, and 4, respectively. This relation can be visualized with a mapping diagram.
In the diagram, it can be observed that there would be more than one output t associated with the single input value of P=1. This fact contradicts the definition of a function. Therefore, t cannot be a function of P. t=f(P) *
Note that it does not matter that different input values are assigned to the same output value. The important thing is that they are not assigned to more than one output. Therefore, P is a function of t. P=f(t) ✓ To interpret the meaning of f(3), recall that the input t represents the time in hours since midnight, while the output P represents the number of people in the cafe.
The expression f(3) represents the corresponding value of P when t=3. In this situation, t=3 represents 3:00AM and f(3) is the number of people in the cafe at that time.
| Equation | Meaning |
|---|---|
| f(12)=20 | At 12:00PM, there are 20 customers in the cafe. |
| f(14)-f(12)=20 | The difference between the number of customers at 12:00PM and the number of customers at 2:00PM is 20 people. |
| f(14)-f(12)/2=20 | The average change of the number of customers per hour between 12:00PM and 2:00PM is 20 people. |
| P=f(12)+20 | The number of people at the cafe equals the number of customers at 12:00PM plus 20 more customers. |
The second equation can be rephrased as the number of customers at 2:00PM increased by 20 since 12:00PM, which coincides with what Mr. Beckett observed. Therefore, this equation best represents the described situation.
While walking to school, Zosia was thinking about a story her auntie just told her. In the windy city crazy winds, flurries of snow, and a max capacity of 61 500 cheering fans have been a part of the legacy of Soldier Field, where the Chicago Bears have played on the gridiron each Sunday for decades. The average cost of ticket to a game was $45 at some point in time.
Zosia's auntie went to a game against there rivals, only to hear that the owner was thinking of selling the team. If only they could make more money from ticket sales! The amount of money earned from the game her auntie attended is a function of the number of people n who attended it.
M=f(n) Recall that a domain is the set of all inputs for which the function is defined. In this case, the inputs of the function are the values of n. Since n represents the number of people attending the game, it can only have non-negative values. n≥ 0 Also, Soldier Field is said to have a maximum capacity of 61 500. Therefore, n cannot have values greater than 61 500. n≤ 61 500 Additionally, since n is the number of people, it can only have integer values. Gathering all the found information, the domain of the function can now be determined. D={n∈ Z: 0≤ n≤ 61 500}
M≥ 0 In the opposite situation, the stadium would be completely full. This would mean that all tickets were sold, and the earnings would be the maximum possible. In average, a ticket costs $45, and the stadium's total capacity is 61 500, so the product of these values will give the maximum possible earnings. $45* 61 500=$2 767 500 ⇓ M≤ 2 767 500 Finally, because both the price of a ticket and the number of people are integers, M will also be an integer number. With this information, the range of the function can now be summarized. R={M∈ Z:0 ≤ M≤ 2 767 500}
M=45n Now, the cost of organizing the game can be subtracted from the money earned to get the profit. P=45n-100 000 It can be seen that P is a function of n. This can be explicitly indicated by writing the equation for the average profit in function notation. P(n)=45n-100 000
As Dominika was studying temperatures previously, she came across the temperature scales of both Fahrenheit and Celsius. She saw the Kelvin temperature scale — often used in scientific research — for the first time. She finds an online program to make a simple app that converts Kelvins to Fahrenheit and Fahrenheit to Kelvins.
Help Dominika find the needed functions so that she can finish her app.
f(x)=ax+b For this situation, the inputs x will represent a temperature in Kelvins and the outputs f(x) will represent the corresponding temperature in degrees Fahrenheit. It is known that 0 Kelvin equals -459.67^(∘) F. Therefore, the value of the constant b can be calculated by substituting 0 for x and -459.67 for f(x).
x= 0, f(x)= -459.67
Zero Property of Multiplication
Identity Property of Addition
Rearrange equation
Now that the value of b is found, substitute it into the above formula of a linear function. f(x)=ax+ b ⇓ f(x)=ax- 459.67 Using a similar reasoning, the given body temperatures in Kelvin and degrees Fahrenheit can be used to find the value of a. This time, 310.15 will be substituted for x and 98.6 for f(x) into the partial function found previously.
x= 310.15, f(x)= 98.6
LHS+459.67=RHS+459.67
Rearrange equation
.LHS /310.15.=.RHS /310.15.
Finally, the function that converts Kelvins into degrees Fahrenheit can be completed. f(x)=1.8x-459.67
y=1.8 x-459.67 Next, because the inverse of a function reverses x and y, switch x and y places in the function rule. x=1.8 y-459.67 Now, the obtained equation should be solved for y.
LHS+459.67=RHS+459.67
Rewrite 1.8 as 18/10
a/b=.a /2./.b /2.
LHS * 5/9=RHS* 5/9
Rearrange equation
Therefore, the inverse function of f has been found. In the context of this situation, this inverse function converts degrees Fahrenheit into Kelvins. f^(- 1)(x)=5/9(x+459.67) Dominika can now use these two functions to implement the desired feature on her amazing app.
f(x)=x Therefore, substitute f(x) with x and solve the equation using inverse operations.
It can be concluded that 574.6 Kelvins equals 574.6^(∘)F.
Dominika's father, the chef, knows that his daughter and her friends love to stop by his cafe along their walk home from school on Fridays. He is eager to expand his menu and add flor de calabaza quesadillas because it is his daughter's favorite!
Her father has calculated that there would be a weekly fixed cost of $75 for adding a new dish plus an additional $0.7 for the ingredients per quesadilla. The total cost C for preparing x quesadillas can be represented by a linear function. C(x)=0.7x+75 Now, the chef needs to figure out how much to charge for one quesadilla. Ideally, he wants to earn $0.30 per each quesadilla sold.
| Number of Quesadillas Sold | 10 | 25 | 50 | 75 | 100 | 200 |
|---|---|---|---|---|---|---|
| Total Cost | ||||||
| Cost per Quesadilla | ||||||
| Sales Price per Quesadilla |
Price of One Quesadilla: $4
| Number of Quesadillas Sold | 10 | 25 | 50 | 75 | 100 | 200 |
|---|---|---|---|---|---|---|
| Total Cost | $82 | $92.50 | $110 | $127.50 | $145 | $215 |
| Cost per Quesadilla | $8.20 | $3.70 | $2.20 | $1.70 | $1.45 | $1.08 |
| Sales Price per Quesadilla | $8.50 | $4.00 | $2.50 | $2.00 | $1.75 | $1.38 |
x= 25
Multiply
Add terms
Next, divide the total cost by 25 to calculate the cost of making one quesadilla. Cost per Quesadilla: 92.5/25=$ 3.70 It is given that the chef wants to earn $0.30 from each quesadilla. By adding that value to the cost of making one quesadilla, the price of one quesadilla can be determined. Price of a Quesadilla: 3.70+0.30=$4.00
Average Cost per Quesadilla: 0.7x+75/x Next, add $0.30. This is the profit the chef wants to earn from each quesadilla sold. In doing this, the ideal sales price P of making one quesadilla can be found. P(x)=0.7x+75/x+0.3 Finally, to find the number of quesadillas the chef should sell, substitute P with $1.5 and solve the equation for x.
Therefore, the chef should sell 150 quesadillas each week for the price of one quesadilla to be $1.50.
The cost per quesadilla can be calculated either by dividing the found total cost by 10 or by substituting x with 10 into the expression for the average cost of a quesadilla found in Part B. Cost per Quesadilla: 82/10=$ 8.2 To find the sale price of one quesadilla, add $0.30 to the cost of one quesadilla. Also, the function P(x) from Part B can be used. Price of a Quesadilla: 8.2+0.30=$8.50 Similarly, the total cost, cost per quesadilla, and sales price per quesadilla can be calculated for the rest of the given numbers of sold quesadillas.
| Number of Quesadillas Sold | 10 | 25 | 50 | 75 | 100 | 200 |
|---|---|---|---|---|---|---|
| Total Cost | $82 | $92.50 | $110 | $127.50 | $145 | $215 |
| Cost per Quesadilla | $8.20 | $3.70 | $2.20 | $1.70 | $1.45 | $1.08 |
| Sales Price per Quesadilla | $8.50 | $4.00 | $2.50 | $2.00 | $1.75 | $1.38 |
Let M(x,y) and N(z,w) be two different points.
M(3,y) and N(3,w) In such a case, for the input x=3, the function will have two different outputs, y and w. This contradicts the definition of a function. Therefore, there is not always a function whose graph passes M and N. However, if the points have different x-coordinates, there is always a function whose graph contains those points.
Ax+By=C Next, substitute the coordinates of the two given points, ( 2, 4) and ( - 3, 2), to obtain two equations. 2A+ 4B=C (I) - 3A+ 2B=C (II) To eliminate one variable, for example B, multiply the second equation by 2 so that the coefficients before B match in both equations. - 3A&+2B=C | * 2 &⇕ - 6A&+4B=2C Now, subtract the obtained equation from Equation (I). 2A + 4B &= C ^- -6A + 4B &= 2C 8A &= - C There is not enough information to solve for both A and C. Therefore, choose a convenient value of C, and then use it to solve the equations for A and B. For example, let C be equal to - 8. Use this value to calculate A.
Now, the only unknown value is B. A=1 B= ? C=- 8 By substituting the known values of A and C into either Equation (I) or (II), the value of B can be calculated.
Finally, the equation can be completed. x-2.5y=- 8 Note that if another value of C was chosen, the coefficients of A and B would be different. For instance, if C=16, then A=- 2 and B=5. Then, a different equation is obtained. - 2x+5y=16 However, these are equivalent equations of the same line, as the first equation multiplied by - 2 is the same as the second equation. x-2.5y=- 8 | * (- 2) ⇕ - 2x+5y=16 If those equations were rewritten in slope-intercept form, they would be the same. This observation strongly emphasizes an advantage of the slope-intercept form: for every line, there is only one equation that describes it.
(4,- 2) and (4,5) As can be noticed, the points have the same x-coordinates. Therefore, they both lie on the vertical line x=4.
Additionally, by multiplying both sides of this equation by some number, an equivalent equation that also contains these points can be found. 2x=8 or - 3x=- 12 Keep in mind that these equations do not describe a function, as for one input of 4, there are many different outputs.
Finally, the challenge presented at the beginning will be solved. It stated that Dominika and her friend Zosia live relatively close to each other. Each morning they leave their homes to walk to school, and at some point, they meet and continue the trip together. Usually, when walking separately, they walk at different speeds.
Graph:
Dominika: f(x)=5x+a Zosia: g(x)=3x+b Note that a function with a greater slope has a steeper graph. On the given diagram, the blue line is steeper than the purple line. With this information, the functions can now be labeled.
It is given that x represents the time spent walking, and f and g represent each girl's distance from Dominika's home. From the diagram, it can be concluded that a equals 0. This makes sense, since Dominika is going to school from her home, so when she has yet to start her walk, her distance from her house is 0. a= Dominika's distance from her house when she has not started walking to school yet In the case of Zosia, b denotes her distance from Dominika's house when she has not started her walk to school. If she is also starting from her own home, then b is the distance between the houses of the two girls. b= Zosia's distance from Dominika's house when she has not started walking to school yet
Slope off(x): m_1= 5 Slope of g(x): m_2= 3 Recall that the slope gives the number of units a function rises when moving 1 unit to the right. By using this information, the number of units that each function rises when moving 2 units to the right can be determined. f(x):& 5*2=10 units up g(x):& 3*2=6 units up Now, by adding this values to n, the y-coordinates of the points with the x-coordinate of m+2 can be found. f(x):& (m+2,n+10) g(x):& (m+2,n+6) Finally, the difference between the found y-coordinates can be determined. (n+10)-(n+6)=4
A rectangle has a perimeter of 24 cm and an area that can be modeled by the function y=A(x) where x is a side length in centimeter. Determine the range of A(x).
The perimeter of a rectangle can be calculated by the following formula, where w is the width of the rectangle and l is its length. P=2w+2l We know that one side has a length of x cm and the perimeter of the rectangle is 24 centimeters. If we let the width be x cm, we can create an expression for the length in terms of x.
Now we can determine an expression for the area by multiplying the width and the length. A(x)=x(12-x)
To find the range of the function, we will first identify its domain. We will need to analyze what values x can take. Note that x must be greater than 0 for a rectangle to form at all. Similarly, since x is the width, it should be less than 12 to form a rectangle. Therefore, the domain is as follows.
Domain: 0
We will now analyze all the possible y-values. The area must be greater than 0. To determine the largest possible area of the rectangle, we need to graph the function. To do this, let's first make a table of values.
| x | x(12-x) | A(x) |
|---|---|---|
| 1 | 1(12- 1) | 11 |
| 2 | 2(12- 2) | 20 |
| 3 | 3(12- 3) | 27 |
| 4 | 4(12- 4) | 32 |
| 5 | 5(12- 5) | 35 |
| 6 | 6(12- 6) | 36 |
| 7 | 7(12- 7) | 35 |
| 8 | 8(12- 8) | 32 |
| 9 | 9(12- 9) | 27 |
| 10 | 10(12- 10) | 20 |
Now we can draw the function by plotting the points and connecting them with a smooth curve.
As we can see, the function has a maximum value at x=6. For that value, the area of the rectangle is 36 cm^2. With this information the range can be identified. Range: 0