Frogs+Fleas+and+Painted+Cubes

Frogs Fleas and Painted Cubes 
 * Alya Shaiful**

**Notes**
x^3 = volume <for cube**
 * x^2 = area < for square

**Problem 5.1**

 * A1. For an edge of two centimeters (2cm) in the large cube, the number of centimeter cubes painted on three faces would be eights (8). For a number of centimeter cubes that are painted in two faces, one face, or unpainted at all, they are all zero (0).

A2.** Large Cube || Number of centimeter Cubes || Cubes painted on 3 faces || Cubes painted on 2 faces || Cubes painted on 1 face || Cubes painted on 0 face ||
 * Edge Length of
 * 2 || 8 || 8 || 0 || 0 || 0 ||
 * 3 || 27 || 8 || 12 || 6 || 1 ||
 * 4 || 64 || 8 || 24 || 24 || 8 ||
 * 5 || 125 || 8 || 36 || 54 || 27 ||
 * 6 || 216 || 8 || 48 || 96 || 64 ||


 * B. The relationship between the edge length of the large cube and the number of centimeter cubes are not quadratic because it is cubed. In a square relationship, it would be 2^2=4, but in the cube relationship, it would be 2^3=8. In this relationship, the first and second differences are not constant, but the third difference is constant. You could calculate the relationship by V=x^3.

C. The relationship between the edge length of the large cube and the number of cubes painted on three faces is linear because as x increases by one (1), y increases by zero (0) and remains constant of eight (8). The relationship between the edge and the number of cubes painted on one (1) face is quadratic because the first difference is not constant, but the second difference is constant. The relationship between the edges of the large cube and the number of cubes painted on zero (0) is not quadratic because the first and second differences are not constant but the third one is constant. The number of cubes painted on two (2) faces is linear because as x increases by one (1), y increases by twelve (12).**

**Follow-up 5.1**
Large Cube || Number of centimeter Cubes || Cubes painted on 3 faces || Cubes painted on 2 faces || Cubes painted on 1 face || Cubes painted on 0 faces ||
 * 1a.**
 * Edge Length of
 * 7 || 343 || 8 || 60 || 150 || 125 ||
 * 8 || 512 || 8 || 72 || 216 || 216 ||
 * 9 || 729 || 8 || 84 || 294 || 343 ||

.**
 * 1b.
 * [[image:3_face_graph.JPG width="364" height="440" caption="3 face painted on"]] ||
 * 3 face painted on ||


 * [[image:Number_of_cubes_total_graph.JPG width="341" height="438" caption="Number of Centimeter Cubes "]] ||
 * Number of Centimeter Cubes ||


 * 2. I would assemble a large cube with no unpainted face showing by stackign the npainted cubes int he middle, and you embed the unpainted cubes in the painted cubes.

Lukas Grimm January 29, 2008 / Day 48 4.3 Putting It All Together Journal

A. Make a table for each quadratic equation below. Use Integer values of x from -5 to 5. Add columns to your tables showing first and second differences.

y=2x(x+3) y=3x-x2 y=(x-2)^2 y=x^2+5x+6 B. Consider the patterns of change in the y values and in the first and second differences for the four equations. In what ways are the patterns similar for the four equations? In what ways are they different?

The patterns are similar in the sense that they are quadratic just like the equations except the 2nd differences which are different because they are constant.

Follow Up 1 The Patters of differences for quadratic functions are not like the patterns of differences for the linear and exponential functions you studied in other units. a. Make a table of (x,y) values for each equation below. Include columns for the first and second differences. y=x+2 y=2x y=2^x y= x^2 b. For each table, look at the pattern of change in the y value as the x value increases by 1. How are the patterns similar in the four table? How are they diffirent? The patterns of change are all similar in the sense that they increase each time x increases by 1 but they are different because they all increase differently. The top two increase linear, and the bottom left increases exponentially while the bottom right increases quadratically.

c. How are the patterns of change in the tables reflected in the equations? The patterns of change are written as numbers with x's which are used for substitution. If the number 2 is written next to an x in superscrypt, (bottom, left) it means that the pattern of change is 2 to the power of x.

2. You have senn that a parabola is a symmetric shape with either a maximum point or minimum point. Describe the graph of each equation in Problem 4.3 and question1 above. Be sure to consider maximum or minimum points, x-intercepts, and lines of symmetry. Use a calculator to check the descriptions.

y=2x(x+3)-Upside down parabola, minimum point -4, x-intercepts (0,0);(-3,0) y=3x-x^2-normal parabola, maximum point 2, x-intercepts (0,0);(3,0) y=(x-2)^2-Upside down parabola, minimum 0, x-intercepts (2,0) y=x^2+5x+6-Upside down parabola, minimum 0, x-intercepts (-3,0);(-2,0) y=x+2-diagonal line going to the top right corner, x-intercepts none y=2x-diagonal line going to the top right corner, x-intercpets (0,0) y=2^x-curve with doubling steepness, x-intercepts (0,0) y=x^2-curve with increasing but not doubling steepness, x-intercepts (0.0)

Homework: 4.2 + follow up 4.3 + follow up ACE

Alina Kepple**

**Journal**
Frog: H= -16t²+12t+0.2 Flea: H=-16t²+8t Basketball PLayer: H= -16t²+16t+6.5 Tables** B. What is the Maximum height reached by each jumper, and when is the maximum height reached? The maximum height that was jumped by the frog was 2.44 at 0.4 seconds. The maximum height that was jumped by the flea was 0.96 at 0.2-0.3 secounds. The maximum height that was jumped by the basketball player was 10.5 at 0.5 seconds. C. HOw long does each jump last? Explain how you found your answer. The frog's jump lasted for only 0.7 seconds. Because in the table anything beyond 0.7 seconds becomes a negative hieght meaning the frog would be going under ground. The flea's jump only lasted for 0.5 seconds. Because same with the frog anything beyond 0.5 seconds means there was negative numbers meaning the flea would be going under grond. The basketball player's jump tkaes a whole 1 secound because at 0 he started his jump and at 1 secound he was back at the high he was at originally. D. What do the constant tearms 0.2 and 6.5 tell youHeya! asl? about the frog and the basketball player? The constant tearms of 0.2 and 6.5 means thats where the frog and basketball player started their jump. Problem 4.2 Follow up 1. For each jumper, describe the pattern of change in the height over time, and explain how the pattern is reflected in the table and graph. The pattern I see for all the equations is that they all become a parabola when it becomes a graph. I could tell this by the table as the numbers grow up to hit a high peek than it falls back down to 0 or the starting number it was originally on. 2. Mr. Jain is a jewelry maker. He would like to increase his profit by raising the rpice of his jade earrings. However, he knows that if he raises the price too high, he wont sell as many earrings and his profit will decrease. Using records of past sales, a business consultant developed the equation p=50s-s² to predict the monthly profit, p, for a given sales price, s. a. Make a table and a graph for this equation.** B. What do the equation, the table, and the graph suggest about the relationship between price and profit? Well it's obviouse that the original price was 50 dollars and as the sales price goes up so does the profit income until it hits a peak of the half price mark or 25$ than from then and on the profit actually decrease all the way down to 0. . What price is going to bring the biggest profit? The price that brings the biggest profit is at 25$ the profit will be 625. D. How does this equation compare with the equations in problem 4.2? The way that these quainots compare to each other is that they are both qudratic equations.
 * A. Use your calculator to make tables and graphs of these three equations. Since a jump doesn't take much time, look at heights for time values between 0 second and 1 second. In your tables, use intercals of 0.1 second.
 * Time (seconds) || Frog (height) || Flea (height) || Basketball (height) ||
 * 0 || .2 || 0 || 6.5 ||
 * 0.1 || 1.24 || 0.64 || 7.94 ||
 * 0.2 || 1.96 || 0.96 || 9.06 ||
 * 0.3 || 2.36 || 0.96 || 9.86 ||
 * 0.4 || 2.44 || 0. 64 || 10.34 ||
 * 0.5 || 2.2 || 0 || 10.5 ||
 * 0.6 || 1.64 || -0.96 || 10.34 ||
 * 0.7 || .76 || -2.24 || 9.86 ||
 * 0.8 || -.44 || -3.84 || 9.06 ||
 * 0.9 || --1.96 || -5.76 || 7.94 ||
 * 1 || -3.8 || -8 || 6.5 ||
 * Graphs:[[image:graph_4.2.jpg width="465" height="1336" caption="graph_4.2.jpg"]]
 * Sales price || Profit income ||
 * 1 || 49 ||
 * 5 || 225 ||
 * 10 || 400 ||
 * 15 || 525 ||
 * 20 || 600 ||
 * 25 || 625 ||
 * 30 || 600 ||
 * 35 || 525 ||
 * 40 || 400 ||
 * 45 || 225 ||
 * 50 || 0 ||
 * Graph:

Megan Nguyen**

**Journal**
The height of the ball increases, reaches the maximum point, then decreases. It's maximum point is 64ft. From +2 seconds, its height decreases at the same rate it increased. This height of the ball vs. time is a quadratic relationship. The values are symmetrical, after +2 seconds the values mimic each other from before 2 seconds. To observe the pattern of change, I will find the difference between each 4-value as the x-value increases by 1.** (2, 64) is when the ball starts traveling downwards being pulled my gravity. The pattern of change is also symmetrical with the y-value and the differences with the 'line of symmetry' being (2, 64).
 * A.Describe how the height of the ball changes over this 4-second time period
 * Time || Height || To get the next y-value || Difference between the changes ||
 * 0.00 || 0 || +15 || -2 ||
 * 0.25 || 15 || +13 || -2 ||
 * 0.50 || 28 || +11 || -2 ||
 * 0.75 || 39 || +9 || -2 ||
 * 1.00 || 48 || +7 || -2 ||
 * 1.25 || 55 || +5 || -2 ||
 * 1.50 || 60 || +3 || -2 ||
 * 1.75 || 63 || +1 || -2 ||
 * 2.00 || 64 || -1 || -2 ||
 * 2.25 || 63 || -3 || -2 ||
 * 2.50 || 60 || -5 || -2 ||
 * 2.75 || 55 || -7 || -2 ||
 * 3.00 || 48 || -9 || -2 ||
 * 3.25 || 39 || -11 || -2 ||
 * 3.50 || 28 || -13 || -2 ||
 * 3.75 || 15 || -15 || -2 ||
 * 4.00 || 0 ||  ||   ||
 * The 'difference between each change' column proves that the pattern of change is as x increases by 0.25, the rate of change, decreases by 2 each time. So, the difference between each x-value of 0.25 isn't constant, the rate of decrease, decreases by -2 each time the x-value increases by 0.25.

B. Without making the graph, describe what the graph of these data would look like. Include as many important features as you can. Without looking at a graph, I can determine the two x-intercepts are (0,0) and (4,0) because on the table, their y-values are 0 indicating that it is the x-intercepts. The maximum or the highest point of the graph is (2, 64) which also means that the line of symmetry is at 2 on the x-axis becuase the highest point is always on the line of symmetry. I can tell the shape is an 'upside-down U-shape' parabola because the values are symmetrical and they increase, peak then decrease.

C.Do you think these data represent a quadratic function? Explain why or why not. Yes, this data represents a quadratic equation because how the values increase to the maximum value, it decreases with the same values in descending order, this shows that the values are symmetrical which is a 'trait' of quadratic equations. The y-values increase, peak, then decrease which is how many quadratic relationship are.

Follow-up 1. The height, h, of the ball after t seconds can be described by the equation. a)Graph this equation on your calculator Graph:

b) Does the graph match your description from part B or Problem 4.1? Explain. Yes, my description was correct. The line of symmetry was located on the x-axis at 2. The highest point is (2,64) abd the y-intercepts are (0,0) and (4,0). And, it is a very narrow, upside-down parabola.

c) Use the equation from the graph to figure out when the ball reaches a height of about 58ft. To find this, I first looked on the graph, using the graphing calulator 'trace', it estimated the time when the ball is at 58 ft was x= 1.4 and x= 2.5. I entered this into the equation. y= 58 x=1.4 y= -16t^2+64t 58= (-16x1.4^2)+(64x1.4) 58 = 58.24 It's not exactly 58, but it comes very close. y= 58 x= 2.5 y= -16t^2+64t y= (-16x2.5^2)+(64x2.5) y=60 It isn't close, so we increase the x-value x=2.6 y= -16t^2+64t y= (-16x2.6^2)+(64x2.6) y= 58.24

So, the time when the ball reaches about 58ft is 1.4 seconds and 2.6 seconds.

d) Use the equation to find the height of the ball after 1.6 seconds. x=1.6 y= -16t^2+64t y= (-16x1.6^2)+(64x1.6) y= 61.44

After 1.6 seconds, the ball will be 61.44 feet.

2. In problem 4.1, the initial height of the ball is 0 feet. This is not very realistic becuase it meants that you would have to lie on the ground and release the ball without extending your arm. A more realistic equation for the height of the ball is after t seconds is : h=-16t^2+64t+6 a) Make a table and a graph**
 * Time || Height || Pattern of change to next y-value ||
 * 0.00 || 6 || +15 ||
 * 0.25 || 21 || +13 ||
 * 0.50 || 34 || +11 ||
 * 0.75 || 45 || +9 ||
 * 1.00 || 54 || +7 ||
 * 1.25 || 61 || +5 ||
 * 1.50 || 66 || +3 ||
 * 1.75 || 69 || +1 ||
 * 2.00 || 70 || -1 ||
 * 2.25 || 69 || -3 ||
 * 2.50 || 66 || -5 ||
 * 2.75 || 61 || -7 ||
 * 3.00 || 54 || -9 ||
 * 3.25 || 45 || -11 ||
 * 3.50 || 34 || -13 ||
 * 3.75 || 21 || -15 ||
 * 4.00 || 0 ||  ||
 * Graph:

b)The 6 in the equation above is called a constant term. A constant term is a term that does not contain a variable. How does the constant term affect the table and the graph? What information does the constant term in this equation give? The constant term is the y-intercept of the relationship, so in the graph when x=0, y= (the constant term). In the table, instead of starting off at 0, the 'starting point' of 0, is now the constant term. The constant term in this equation gives the y-intercept, and also, the height at which the ball was thrown from.

c)Compare the graphs of the equations in question 1 and 2. Discuss similarities and differences in the following: i. the maximum height reached by the ball. In question 1, the maximum height reached by the ball is 64, in question 2, the maximum height is 70. This is because instead of being thrown from right off the ground, the height at which the ball was thrown from was 6, therefore increasing the original height by 6. ii. the x-intercepts In question 1, the x-intercepts are 0 and 4. In question 2, the x-intercepts are (-.1) and 4.1. This is because the constant term has shifted the position of the question 2 graph. iii. the pattern of change in the height of the ball over time The patterns of change are exactly the same except they have different values because of the constant term added. The pattern of change was that the rate of decrease, decreased by (-2) each time x increased by .25 .**

**Homework**
Collected: Mathematical Reflection Investigation 3
 * Assigned: ACE 4: 3, 9, 18, 26

Robert Koehlmoos Jan/26/08

Aki Takeuchi**

**Journal**

 * __Problem 2.4__



A. (Done on Labsheet 2.4) y=x² [Graph 3] y=x(4-x) [Graph 2] y=(x+3)(x+3) [Graph 6] y=x(x-4) [Graph 8] y=x(x+4) [Graph 1] y=(x+3)(x-3) [Graph 7] y=(x+2)(x+3) [Graph 4] y=2x(x+4) [Graph 5]

B. (Done on Labsheet 2.4) [Graph 1] x-intercept(s): (-4,0) and (0,0) [Graph 2] x-intercept(s): (0,0) and (4,0) [Graph 3] x-intercept(s): (0,0) [Graph 4] x-intercept(s): (-3,0) and (-2,0) [Graph 5] x-intercept(s): (-4,0) and (0,0) [Graph 6] x-intercept(s): (-3,0) [Graph 7] x-intercept(s): (-3, 0) and (3,0) [Graph 8] x-intercept(s): (0,0) and (4,0) You can find the x-intercepts by looking at the equation in its factored form. You simply have to substitute 0 for y in the euqation.

C. (Done on Labsheet 2.4) <--- In red color pencil [Graph 1] Line of symmetry at: x= -2 [Graph 2] Line of symmetry at: x=2 [Graph 3] Line of symmetry at: x=0 [Graph 4] Line of symmetry at: x= -2.5 [Graph 5] Line of symmetry at: x= -2 [Graph 6] Line of symmetry at: x= -3 [Graph 7] Line of symmetry at: x=0 [Graph 8] Line of symmetry at: x=2

D. (Done on Labsheet 2.4) The graphs are all parabolas, and they are all symmetric. [Graph 1] Minimum point: (-2,-4) [Graph 2] Maximum point: (2,4) [Graph 3] Minimum point: (0,0) [Graph 4] Minimum points: (-3,0) and (-2,0) [Graph 5] Minimum point: (-2,-8) [Graph 6] Minimum point: (-3,0) [Graph 7] Minimum point: (0,9) [Graph 8] Minumum point: (2,-4)

E. From the equation, you can predict many things. You can predict wether the parabola opens up or faces down, by looking at a in the equation. You can also find the x-intercepts as stated in B, by saying that y=0 to find out the x-values.

F. [Graph 1] Factored form: y=x(x+4) Extended form: y=x²+4x

[Graph 2] Factored form: y=x(4-x) Extended form: y= (-x)²+4x

[Graph 3] Factored form: y=x² Extended form: y=x²

[Graph 4] Factored form: y=(x+2)(x+3) Extended form: y=x²+5x+6

[Graph 5] Factored form: y=2x(x+4) Expanded form: y=2x²+8x

[Graph 6] Factored form: y=(x+3)(x+3) Expanded form: y=x²+6x+9

[Graph 7] Factored from: y=(x+3)(x-3) Expanded form: y=x²-9

[Graph 8] Factored form: y=x(x-4) Expanded form: y=x²-4x

__Problem 2.4 Follow-Up__

1. If an equation is in factored form, you can tell that it is a quadratic function by checking for the 2 'x's. 2. If an equation is in expanded form, you can tell that it is a quadratic function by seeing if it fits the formula, y=ax²+bx+c.**

**Homework**
Handed in: Check-Up 1
 * Assigned: (ACE 2) 2, 24, 37

Lukas Grimm 20th January 2008 Day 45 Problem 2.3: Changing Both Dimensions

A. 1. Copy the new rectangle. Label the area of each of the four sections.

2. Write two expressions, one in factored form and one in expanded form, for the area of the new rectangle.

Factored: (x+3)(x+2)

Expanded: x^2+2x+3x+6

3. Use your expressions from part 2 to write two equations for the area, A, of the rectangle. Graph both equations on your calculator. Compare these graphs with graphs you made in Problem 2.2.

Equation 1: A=(x+3)(x+2)

Equation 2: A= x^2+2x+3x+6

Graphs: 1 2 B. A square has sides of length x centimetres. One dimension of the square is doubled and then increased by 2 centimetres, and the other dimension is increased by 3 centimetres.

1. Make a sketch to show how the square is transformed into the new rectangle. Label the area of each section of the new rectangle.

The picture shows how the original x by x rectangle is doubled and how the new rectangles are added to it. It also gives the areas of all the rectangles.

2. Write two expressions, one in factored form and one in expanded form, for the area of the new rectangle.

Factored: A=(x+x+2)(x+3)

Expanded: A=x^2+2x+(3x+x)+6

3. Use your expressions from part 2 to write two equations for the area, A, of the rectangle. Graph both equations on your calculator. Compare these graphs with the graphs you made in Problem 2.2.

Equation 1: A=(x+x+2)(x+3)

Equation 2: A=x^2+x^2+3(x+x)+2x+6

Graphs:

C. The rectangle below is divided into four smaller rectangles. Write two expressions, one in factored form and one in expanded form, for the area of the large rectangle.

Factored: x^2

Extended: x^2

Follow Up**


 * 1) **In a-c, write two expressions, one in factored form, and one in extended form, for the area of the entire figure.**

Expanded: x^2+3x+3x+9
 * A Factored: (x+3)(x+3)

B Factored: (x+2)(x+4)

Expanded: x^2+4x+2x+8 C Factored: (x+1x)+(x+1x)

Expanded: x^2+1x+1x

2. One dimension of a square is increased by 1 centimetre, and the other dimension is increased by 3 centimetres. Write two expressions, one in factored and one in expanded form, for the area of the new rectangle.

Factored: (x+1)(x+3)

Expanded: x^2+2x+3x

3. One dimension of a square is doubled, and the other dimension is increased by 4 centimetres. Write two expressions, one in factored form and one in expanded form, for the area of the new rectangle.

Factored: (x*2)(x+4)

Expanded: x^2+x^2+4x*2

4. Do parts a and b for each expression.

1. (x+3)(x+4) 2. (x+5)(x+5) 3. 2x(x+2)**


 * 1) **Draw and label a rectangle whose area is represented by the expression.**


 * 1) **Write an equivalent expression in expanded form.**

b. x^2+3x+4x+12
 * 1 a.

2 a. b. x^2+5x+5x+25

3 a. b. x^2+2x+2x+4

5 a. Copy the diagram below, replacing each question mark with the correct length or area. b. Write two expressions for the area of the rectangle.

Expression 1: (x+3)(x+3)

Expression 2: x^2+6x+9

6 a. Draw a rectangle made up of four smaller rectangles similar to the one in question 5. Make the area of the square in the lower left corner x^2, and make the area of the rectangle in the upper right corner 8. What might the areas of the other two small rectangles be? Find one solution, and use it to label the lengths and areas in your diagram. b. Write two expressions for the area of the large rectangle.

Expression 1: x^2+4x+2x+8

Expression 2: (x+4)(x+2)

7. Do parts a and b for each expression below.

1.x^2+x+5x+5 2.x^2+7x+6 3.x^2+8x+16

a. Draw and label a rectangle whose area is represented by the expression. b. Write an equivalent expression in factored form.

1.a 1.b (x+x)(x+5)

2.a 2.b x(x+7)

3.a 3.b x(x+8)

Homework: Problems: 2.3 Changing both Dimensions 2.4 Looking Back at Parabolas. Ace: 2, 34, 37

Sue Lee**

**Notes:**
This would definitely not a fair trade because then, the area would be much smaller. A perfect square has the largest amount of area in all rectangles so they shouldn't trade this land.**
 * Suppose you own a square piece of land with sides n meters long. You trade your land for a rectangular lot. The length of your new lot is 2 meters longer than the side length of your original lot, and the width of you new lot is 2 meters shorter than the sidelength of the orignal lot. Would this be a fair trade?

**Journal**

 * A.**
 * original square || original square || new rectangle || new rectangle || new rectangle || difference in ||
 * side length(m) || area(m2) || length(m) || width(m) || area(m2) || areas (m2) ||
 * 3 || 9 || 5 || 1 || 5 || 4 ||
 * 4 || 16 || 6 || 2 || 12 || 4 ||
 * 5 || 25 || 7 || 3 || 21 || 4 ||
 * 6 || 36 || 8 || 4 || 32 || 4 ||
 * 7 || 49 || 9 || 5 || 45 || 4 ||
 * n || n2 || n+2 || n-2 || (n+2)(n-2) ||  ||


 * B. For each side length in this table, the area of the new lot is always 4m2 smaller than the original lot. Therefore, none of the side lengths would be a fair trade.

C. Table extended in problem A

Follow Up 1a. Graph on graph paper

b. A=(n+2)(n-2) Graph appears the same as the one i drew in 1a.

c. The shape of my graphs look just the opposite of a parabola graph. It looks like an upside down parabola graph.

d. All of the values for area and side lengths on my table makes sense, but on my graph it doesn't make sense because there can't be negative side lengths.

2a. The tables I made for Ivestigation1 and 2 are not similar because the area increased and then decreased in a table in Ivestigation 1, but in investigation 2, the area kept on increasing as side lengths increased and it didn't decrease. They are also similar because the x values for both investigation increased.

b. The graphs of Investigation 1 and 2 are different because in investigation 1, it was a normal parabola graph, but in investigation 2, the graph was upside down parabola graph. They are also both similar because they all made parabola graphs.

c. The equations of Investigation 1 and 2 are similar because they both made equations of an area. But they are also different because the equation in investigation 2 was an expression.

3. I don't think that the tade offered by US malls Incorporation is a good idea for the family that own the private lot. This is because they will get about 3 times less the area they have right now. (private lot-15625m2, trading lot- 5625m2)**

**Homework:**
Check up 1 ACE 2 (pg 31-39): 10-14
 * Finish Journal 2.1 and 2.2 and their follow ups

Collected: None** = =

= =
 * Rica Duchateau**

**Notes**

 * ¨The difference between the tables of linear, exponential and quadratic relationships is that in linear ones, as x increases, y changes at a constant rate. In exponential ones, as x increases, y multiplies or divides by a certain number each time. In quadratic ones, as x increases, y increases to the maximum value and decreases or vice versa.


 * How can I write a symbol sentence (equation) to describe fixed perimeter problems?

- A //=// lw** = =

= =
 * - A = l (//P//÷2-l)

l**
 * x - l A = _ ||

**Journal**

 * The rectangle below has a perimeter of 20 meters and a side of length //l// meters.

l**
 * 10 - l A = _ ||


 * A. Express the length of each side of the rectangle in terms of //l//. That is, write an expression that contains the variable //l// to represent the length of each side.

- P÷2 - l

B. Write an equation for the area, //A//, of the rectangle in terms of //l//.

- A = l (P÷2 - l)

C. If the length of a side of the rectangle is 6 meters, what is the area?

- A=l(6÷2 - l) ... l=6 ... A=6 (20÷2 - 6) ... A=6 (10 - 6) ... A=6 (4) ... A=24m²

D. Use a calculator to make a table and a graph for your equation. Show x values from 1 to 10 and y values from 0 to 30. Compare your table and graph to those you made in Problem 1.1.**


 * x || y (m²) ||
 * 1 || 9 ||
 * 2 || 16 ||
 * 3 || 21 ||
 * 4 || 24 ||
 * 5 || 25 ||
 * 6 || 24 ||
 * 7 || 21 ||
 * 8 || 16 ||
 * 9 || 9 ||
 * 10 || 0 ||


 * Side Length and Area Relationship

Both the graph and the table are the same with Problem 1.1.

Problem 1.3: Follow-Up

1. Consider rectangles with a perimeter of 60 meters.

a) As in Problem 1.3, draw a rectangle to represent this situation. Label one side l, and label the other sides in terms of l.

l**
 * 30 - l A = ||


 * b) Write an equation for the area, A, in terms of l.

- A = l (60÷2 - l)

c) Make a table for your equation. Then, use your table to estimate the greatest area possible for a rectangle with a perimeter of 60 meters. Give the side lengths of this rectangle.**


 * x || y (m²) ||
 * 1 || 29 ||
 * 2 || 56 ||
 * 3 || 81 ||
 * 4 || 104 ||
 * 5 || 125 ||
 * 6 || 144 ||
 * 7 || 161 ||
 * 8 || 176 ||
 * 9 || 189 ||
 * 10 || 200 ||
 * 11 || 209 ||
 * 12 || 216 ||
 * 13 || 221 ||
 * 14 || 224 ||
 * 15 || 225 ||
 * 16 || 224 ||

Rectangle l 15, w = 15.
 * Greatest area = 225m² for a perimeter of 60m.

d) Use a calculator or data to sketch a graph of the relationship between the length of a side and the area.

Side Length and Area Relationship

e) How can you use your graph to find the maximum area? How does your graph show the side length that corresponds to the maximum area?

- Maximum area is the tip of the parabola in the graph. The x value right below the parabola tip is the side length of the maximum area.

2. An equation for the area of rectangles with a certain fixed perimeter is A = l (35 - l).

a) Draw a rectangle to represent this situation. Label one side l, and label the other sides in terms of l.

l**
 * 35 - l A = _ ||


 * b) If the length of a side of a rectangle with this fixed perimeter is 20 meters, what is the area?

- A=l (35- l) ... A=20 (35 - 20) ... A=20 (15) ... A=300m²

c) Describe two ways you could find the perimeter for the rectangles represented by this equation. What is the perimeter?

- a) A = l (35-l). Multiply the 35 in this equation by 2. - b) Take l. Solve the equation (35 - l), to find the width. Multiply l by 2 and w by 2. Add up. - The perimeter is 70m.

d) Describe the graph of this equation.

- The graph will be the shape of a parabola, with its tip facing up.

e) What is the maximum area for this family of rectangles? What dimensions correspond to this maximum area? Explain how you found your answers.

- 306.25m² Maximum Area, Dimensions 17.5m × 17.5m. To get the answer, I divided 70m by 4.

3. If you know the perimeter of a rectangle, how can you write an equation for the area in terms of the length of a side?

- P= ... A = l (P÷2 - l)

4. Graphs of quadratic functions are called parabolas. Describe the characteristics of the parabolas you have seen so far.

- Parabolas are curved. They don't appear often in nature. It's like half an egg or the St. Louis Arch in Missouri. It increases, tips and then decreases.

5. Study the graphs, tables, and equations for areas of rectangles with fixed perimeters. Which representation is most useful for predicting the maximum area?

- Tables.**

**Homework**
Assigned: Problem 2.2: Changing One Dimension + 2.2 Follow-Up, ACE # 2: 10 - 14 & Check-Up 1
 * Collected: Mathematical Reflections 1

Sue Lee**


 * Notes: None

Problem: A. The shape of the graph looks like a parabola that is upside down. To describe this graph, as the length of a side becomes greater till 20, the graph increases and as the length of a side becomes greater till 40, it dcreases.

B. The greatest area possible for a rectangle with this perimeter is 400 meter squared.

C. The area of the rctangle with a side of length 12 m and 28 m is about 325m. These two rectangles are related because they both have the same area when the length is different.

D. The dimensions of the rectangle with an area of 300 sq m are 10 and 30.

E. The fixed perimeter for the rectangles represented by the graph is 80m. I found this perimeter because if you see where the two x- intercepts are on the same area and add the lengths, they are all 80.**

**Follow Up 1.2**

 * 1. The shape and the special features I observed for the graph in problem 1.2 appears in the table also because it will also make a parabola. The pattern goes up and then down.

2. The fixed perimeter for the rectangles represented in this table is 12. I found this area by seeing the same two spots on the graph and adding them together.

3. The greatest area possible for a rectangle with this perimeter is 36meter squared. The dimensions of these rectangles are 7 and 7.

4. The dimensions would be 4.**

**Homework:**
problem 1.1 follow up and 1.2 follow up ACE1) #3, 5 (pg.12-17)
 * Assigned:

Collected: None

Yoon Sun Lee**

**Notes**
key concept: y=x^2+xc+xd+cd**
 * The general form of quadratic equation is ax^2 + bx + c= 0
 * 1) **factored form: y= (x+c) (x+d)**


 * 1) **expanded form: y=x^2+(c+d)(x)+cd**


 * 1) **when a < 0, graph opens down y==x^2-x+6**


 * 1) **when a > 0, graph opens up y=x^2+x-6**

x=-3x x x =2 intercept (-3), (2,0)**
 * 1) **In factored form: y=x^2-x+6 → ㅛ=ㅌ^2+ㅌ-6**
 * y=(x+3) (x-2)

**Journal**

 * problem 1.1

[| 1.1 graph.tif] A. (please look at the grid paper above.) B.**
 * lengthe of the side (m) || area (m^2) ||
 * 0 || 0 ||
 * 1 || 9 ||
 * 2 || 16 ||
 * 3 || 21 ||
 * 4 || 25 ||
 * 5 || 25 ||
 * 6 || 24 ||
 * 7 || 21 ||
 * 8 || 16 ||
 * 9 || 9 ||
 * 10 || 0 ||

The graph is parabola. It increased really reapidly in the beginning. Agter it reached its maximum point, it started to decrease in a same speed as it increased. D. Thre greatest are that is possible is 25m^2. The dimension of the fencing is 5m x 5m. Both graph and table show the greatest y-value is 25m^2. The x-value, the side length is 5m. To find the other side length, i did 25÷5 and I got 5m for my other side.
 * C. (please look at the graph above.)

Follow up

1. Even if the deimensions are the decimals, I don't think the answer for part D would change. For part D, I wrote the dimension of the biggest area possible 5m x 5m. And I got the area of 25m^2. To check if this is the biggest area possible, I compared with the dimension with the decimals. I worte 5.1m x 4.9m, and I got 24.99m^2. This area is smaller than the are I got it by 5m X 5m. This is a reason why i believe 5m x 5m is a great area that is possible.

2. If the shape of a claim were not restricted to a rectangle, I would use circles. I tried to make a circle. Since the perimeter should be 20m, I did 20m ÷ 3.14 ÷ 2= 3.18, by this i got the radius for the circle. To get the area, I did 3.18 x 3.18 x 3.14 = 31.75m^2. By this I found out that the area of the circle is 6.75m^2 more bigger than the area of the rectangle.**

**Homework**
Assigned: 1.2, ACE1: 3, 5**
 * Collected: none

=**Growing, Growing, Growing**=
 * Journal and Homework Record and Vocabulary

Name (normal) Date/Day(heading 2) Problem Number and Title (Heading 3) Notes (Heading 3) Journal (Heading 3) Homework (Heading 3) Collected: (normal) Assigned: (normal) 1.1 Tarryn 1.2 Yoon Sun 1.3 Sue 2.1 Aki 2.2Alina 2.3 Lucas 3.1 Alya 3.2 Megan 3.3 Lukas 4.1 Robert 4.2 Sajid 4.3 4.4 Rica

Rica Duchateau**

**Notes**
- That the temperature will gradually decrease to become equal to the room temperature.
 * What patterns of change would you expect to find in the temperature of a hot drink as time passes?

- A curved, negative graph line.**
 * What shape would you expect for a graph of (time, drink temperature) data?

**Journal**

 * Test No. ||
 * Time (Minutes) ||
 * Water Temperature (°C) ||
 * Room Temperature (°C) ||
 * Temperature Difference (°C) ||
 * || 0 ||
 * 0 ||
 * 78°C ||
 * 15°C ||
 * 63°C ||
 * || 1 ||
 * 5 ||
 * 70°C ||
 * 16°C ||
 * 54°C ||
 * || 2 ||
 * 10 ||
 * 62°C ||
 * 17°C ||
 * 45°C ||
 * || 3 ||
 * 15 ||
 * 58°C ||
 * 13°C ||
 * 45°C ||
 * || 4 ||
 * 20 ||
 * 54°C ||
 * 16°C ||
 * 38°C ||
 * A. Make a graph of your (time, water temperature) data.

Water Temperature B. Describe the pattern of change in the (time, water temperature) data. When did the water temperature change most rapidly? When did it change most slowly?

- First, the water changed rapidly, but this rate gradually got slower. During 0-10 minutes, the water temperature changed rapidly, while during 10-20 minutes, the water temperature changed slowly.

C. Add a column to your table. In this column, record the difference between the water temperature and the room temperature for each time value.

- See the table.

D. Make a graph of the (time, temperature difference) data.

Temperature Difference E. Compare the shapes of the graphs.

- The graph for A has a slightly curved, negative graph line. The graph for B is also a negative graph line, but the line first decreases linearly (from 0-10 minutes), then levelled (from 10-15 minutes), and then decreased again (from 15-20) minutes.

F. Describe the pattern of change in the (time, temperature difference) data. When did the water temperature change most rapidly? When did it change most slowly?

- First, the temperature decreased, during 0-10 minutes. The temperature levelled during 10-15 minutes. Then it decreased again, during 15-20 minutes.

G. Assume that the relationship between temperature difference and time in this experiment is exponential. Estimate the decay factor for this relationship. Explain how you made your estimate.

- Decay factor: 54/63 = 0.86 First, I took two temperature differences that followed each other. I divided the value for n-1 by the value for n to find the decay factor.

H. Find an equation for the (time, temperature difference) data. Your equation should allow you to predict the temperature difference at the end of any 5-minute interval.

- y 63(0.86^n) (For n, substitute the order of the minutes for numbers by regular numbers, e.g. 0 minutes is 0, 5 is 1, 10 is 2, 15 is 3 & 20 is 4.) Test: y = 63(0.86^2) y = 63(0.7369) y ~ 46°C 46°C is very near to the 45°C on the table.

Test 2: y = 63(0.86^4) y = 63(0.54700816) y = 34.5°C 34.5°C is quite close to the 38°C on the table.**

**Problem 4.4: Follow-Up**

 * 1. What do you think the graph of the (time, temperature difference) data would look like if you had continued the experiment for several more hours?

- I think that the graph line would be a negative with a slight curve.

2. What factors might affect the rate at which a cup of hot liquid cools?

- a) The lower the room temperature, the faster the cooling. - b) The hotter the liquid in the beginning, the longer the time for the liquid to cool.

3. What factors might introduce errors in the data you collect?

- a) Wrong thermometer readings. - b) No equal length time intervals. - c) Adding or removing liquid.**

**Homework**
Assigned: 1. ACE 4: # 3, 5, 7 & 10. (GGG, Pages 53 - 59). 2. Study for the Looking for Pythagoras and Growing, Growing, Growing vocabulary test next class.
 * Collected: None

Alya Shaiful**

**Notes**
y=b^x or y=n(b^x) n being the coefficient Coefficient is the initial value of an equation Base is the growth or decay factor
 * In an exponential equation, the form would be:

To get the growth or decay factor, you could simply write ratios of the y-value over the previous y-value.**

Growth factor:

2/1= 2 4/2 = 2
 * X ||
 * Y ||
 * || 0 ||
 * 1 ||
 * || 1 ||
 * 2 ||
 * || 2 ||
 * 4 ||

2 is the growth factor

Decay factor:

1/2=0.5 0.5/1=0.5
 * X ||
 * Y ||
 * || 0 ||
 * 2 ||
 * || 1 ||
 * 1 ||
 * || 2 ||
 * 1/2 ||

1/2 or 0.5 is the decay factor

Problem 4.3
Graph for Problem A and B
 * [[image:Problem_4.3_alya.png width="520" height="708" caption="Problem 4.3 A and B"]] ||
 * Problem 4.3 A and B ||

A. Investigate the equations: y=1.25^x y=1.5^x y=1.75^x y=2^x

It seems as though the graph gets higher as the base increases. The higher the base, the more rapidly the graph will increase. Also, the graph becomes less curvy when then base gets higher.

B. Investigate the equations: y=0.25^x y=0.5^x y=0.75^x

The graphs seem to decrease slowly as the base increases. The lower the base, the more rapidly the graph will decrease. Also, the graph becomes less curvy as the base increase. The graph seems to get more into the negative zone as the values decrease more.

C. Based on your observations on parts A and B, describe how you would predict the general shape of the graph of y=b^x for a specific value of b.

For any type of y=b^x graph, it will show up as a curved graph. For a value higher than one, the graph increase upwards. With the base that is higher, the graph will look less curvy than a base that has a lower value. However, the graph shows an inverse relationship if the b has a value less than 1 such as 0.5, 0.25, etc. Also, the higher the base, the less curvy it will show.

Follow-up 4.3
Graph for Problem 1, 2, and 3
 * [[image:Followup_4.3_alya.png width="517" height="576" caption="Follow-up 4.3 1,2, and 3"]] ||
 * Follow-up 4.3 1,2, and 3 ||

Questions 1, 2, and 3 will both have the same answers.


 * When there is a coefficient added to an exponential equation, the coefficient will show up as the y-intercept in a graph. The graph will increase exponentially in a rapid rate if the base and coefficient is higher and more than 1, such as y=4(1.5^x). The graph will decrease exponentially in a rapid rate is the coefficient and base is lower and less than 1, such as y=2(0.5^x). All the graphs are curved since they show both exponential growth and decay.

4. Describe the value of a affects the graph of and equation of the form y=a(b^x).

The value of a affects y-intercept of the graph y=a(b^x) because if b^0=1*a, it shows that a is the initial value. Therefore, with a higher value, you either increase faster or decrease slower. Depending on the base of the exponent and value of the coefficient. In an exponential decay, a lower a value helps the graph decrease rapidly. In an exponential growth, a higher a value helps the graph increase rapidly.**

Homework
Collected: None Assigned: ACE 4: 3,5,7,10 Vocabulary Test Next Class (including Looking for Pythagoras vobualary, too)

Sajid Azad

12/3/07 - Day 35
Math 8c

Problem 4.2 - Fighting Fleas
A. How does the amount of active medicine in the dog's blood change from one hour to the next?

- The amount of active medicine, from one hour to the next, decreases by halves.

B. Write an equation to model the relationship between the number of hours since the dose was administered, b, and the milligrams of active medicine, m.

- An equation to model the relationship between the number of hours since the dose was administered and the milligrams of activ e medicine would be: M = 20(.5n)

C. Based on your knowledge of exponential relationships, what pattern would you expect to see in the data if 40 milligrams of the medicine were given to the dog?

- The pattern of change would stay the same, even if 40 mg of medicine is given. The pattern will decrease by halves each time until the medicine eventually fades. The only expected change, however, is for the y-value at x=0 to be 40 instead of 20.

4.2 follow up

1.

2. For the medicine described in question 1, Janelle wrote the equation m= 60(.8^h) to model the relationship between the amount of medicine in the blood and the number of hours since it was administered. Compare the quantitties of active medicine in your table to the quantities given by Janelle's equation for several time values. Explain any similarities or differences you find.
 * Time since dose (hours) ||
 * 0 ||
 * 1 ||
 * 2 ||
 * 3 ||
 * 4 ||
 * 5 ||
 * 6 ||
 * || Active medicine in blood (milligrams) ||
 * 60 ||
 * 48 ||
 * 38.4 ||
 * 30.72 ||
 * 24.576 ||
 * 19.666 ||
 * 16 ||

- M= 60(.8^h) --- M= 60(.8^3)-- 30.72 = 60(.8^3) - M= 60(.8^h) --- M= 60(.8^2)-- 38.4 = 60(.8^2) - M= 60(.8^h) --- M= 60(.8^4)-- 24.576 = 60(.8^4) - Janelle's equation gives results that match the data given in the table. Her equation models the trend of the values decreasing by 20% each time.

3. Janelle's friend Habib was confused by the terms "rate of decay" and "decay factor". He said that since the rate of decay in question 1 is 20%, the decay factor should be .2 and the equation should be m= 60(.2^h). How could you explain to Habib why a rate of decay of 20% is equivalent to a decay factor of .8?

- In this case, we know that the y-values are decreasing by 20% and not EXACTLY by .2. The next data value for y is considered through multiplying the ".2" by the initial y-value, which symbolizes that we want to figure out 20% of that y-value. Afterwards, we subtract this new factor from the original y-value, so this shows that we are now looking for the "change". Once we have our change, we have our answer.

Robert Koehlmoos

Nov/3/07-Day 35 Problem 4.1- Making Smaller Ballots Notes: When things go down at a changing rate that follows a pattern it is called exponential decay. The decimal that each number is compared to the one before it is the decay factor. Problem 3.3 and follow up

A.The sheet of papaer Alejandro started with hadf an area of 64 in2. Copy and complete the table started below to show the area of a ballot after each of the first 10 cuts.


 * Cuts ||
 * Area in2 ||
 * || 0 ||
 * 64 ||
 * || 1 ||
 * 32 ||
 * || 2 ||
 * 16 ||
 * || 3 ||
 * 8 ||
 * || 4 ||
 * 4 ||
 * || 5 ||
 * 2 ||
 * || 6 ||
 * 1 ||
 * || 7 ||
 * 0.5 ||
 * || 8 ||
 * 0.25 ||
 * || 9 ||
 * 0.125 ||
 * || 10 ||
 * 0.0625 ||
 * B.How does the area of a ballot change with each cut?

After each cut the area of a ballot is half of what it was before.

C.How is the pattern of change in the area different from the exponential growth patterns you have seen in this unit? How is it similar?

The pattern of change in the area of the ballot is the same because it dosn't change by the same amount each time, it changes exponentially in fact. But it is different because it gets smaller each time adn changes alot at the start and a little later on.

Follow up 4.1

1. Copy and complete the table below to show the area of a ballot at each stage.**


 * Stage ||
 * Area (cm2) ||
 * || 0 ||
 * 27 ||
 * || 1 ||
 * 9 ||
 * || 2 ||
 * 3 ||
 * || 3 ||
 * 1 ||
 * || 4 ||
 * 0.33333... ||
 * || 5 ||
 * 0.11111... ||
 * || 6 ||
 * 0.0370370370370... ||
 * 2. How does the area of a ballot change at each stage?

After each stage the area of a ballot is a third of what it was a stage ago.

3.Write a equation for the area, A, of a ballot at any stage, n.

A=27(0.3333...^n)

4.Make a graph of the (stage, area) data from your table?

The graph starts out with a large number that goes down rapidly very fast and then goes down very slowly until I can no longer see the diffrence between one stage and the next.**

**Homework:**
Assigned: Probloms 4.1 and 4.2 Ace 1,6,9
 * Collected: MR p.44, Ace 1-11 odds

Lukas Grimm**

**Notes:**
Example: R=20(1.5^t) R equals rent per square meter and 1.5 is the increase in value each year (t).**
 * The growth factor can be used with the starting value to write an equation which can substitue for a table.

**Problem 3.3**

 * A. Suppose the value of the remaining coins increased by 4% each year. Make a table showing the value of the collection each year for the next 10 years.

B. Sam's friend Maya has a baseball card collection worth $2500. Add a column to your table showing the value of Maya's collection each year for a 10-year period if its value increases by 4% each year.**


 * Years ||
 * Value Of Collection (A) USD ||
 * Value of Colleciton (B) USD ||
 * || 0 ||
 * 1250 ||
 * 2500 ||
 * || 1 ||
 * 1300 ||
 * 2600 ||
 * || 2 ||
 * 1352 ||
 * 2704 ||
 * || 3 ||
 * 1406.08 ||
 * 2812.16 ||
 * || 4 ||
 * 1462.32 ||
 * 2924.64 ||
 * || 5 ||
 * 1520.81 ||
 * 3041.63 ||
 * || 6 ||
 * 1581.64 ||
 * 3163.29 ||
 * || 7 ||
 * 1644.91 ||
 * 3289.82 ||
 * || 8 ||
 * 1710.71 ||
 * 3421.42 ||
 * || 9 ||
 * 1779.13 ||
 * 3558.27 ||
 * || 10 ||
 * 1850.30 ||
 * 3700.61 ||


 * C. Compare the values of the collections over the 10-year period. How does the initial value of the collection affect the yearly increase in value?

Sam's collection is exactly half of Maya's collection all the time throughout the 10 years. The initial value affects the yearly increase (4%) because the 4% is based on the initial and then current value so depending on the amount of the initial/current value the 4% will increase the collections current worth.

D. How does the initial value of each collection affect the growth factor?

The growth factor is (in this case) always 1.04 and is not changed by the initial value but the initial value affects the amount that it increases by in the first year (the growth factor's value).

E. Write an equation for the value, V, of Sam's $1250 coin collection after t// years.

V=1250(1.04^t)

V=1250(1.04^t) V=1250(1.04^1) V=1250*1.04 V=1300

V=1250(1.04^t) V=1250(1.04^5) V=1250*1.2166529024 V=1520.816128

F. Solve your equation to find the value of Sam's collection after 30 years.

V=1250(1.04^t) V=1250(1.04^30) V=1250*3.2433975100275380436206824984613 V=4216.416763038919465733990239918

3.3 Follow-Up

Sam made the following calculation to predict the value of his aunt's stamp collection several years from now:

value = $2400 x 1.05 x 1.05 x 1.05 x 1.05

A. What initial value, rate of increase in value, and number of years was Sam assuming?

The initial value is $2400, rate of increase is 5% and the number of years were 4.

B. The result of Sam's calculation is $2917.22. If the value continued to increase at this rate, how much would the collection be worth in 1 more year?

The collection would be worth $3063.081.

Find the growth factor associated with each percent increase.

A. 30% = 1.3 B. 15% = 1.15 C. 5% = 1.05 D. 75% = 1.75**

**Homework:**
Assigned: 3.2, 3.3 and follow-ups, Ace 3: 2,4,5,11
 * Collected: ACE 2: 10-12

Megan**

**Notes**
-Earning Interest -How to find the Growth Factor
 * -Is there an easier method?

5% Interest Rate**
 * Year ||
 * Balance ($) ||
 * || 0 ||
 * 1000 ||
 * || 1 ||
 * 1050 ||
 * || 2 ||
 * 1102.5 ||

Example: year 1=n 1050/1000 = 1.05 year 2=n 1102.5/1050 = 1.05
 * value in year n / value in year n-1

There is a pattern here. It shows the ratio, otherwise known as growth factor, which is 1.05**

**Problem 3.2**

 * A. Make a table showing the value of the collection each year for the 10 years after Sam's uncle gave it to him.**
 * Year ||
 * The Value of the Collection ||
 * || 0 ||
 * 2500 ||
 * || 1 ||
 * 2650 ||
 * || 2 ||
 * 2809 ||
 * || 3 ||
 * 2977.54 ||
 * || 4 ||
 * 3156.1924 ||
 * || 5 ||
 * 3345.563944 ||
 * || 6 ||
 * 3546.297781 ||
 * || 7 ||
 * 3759.075647 ||
 * || 8 ||
 * 3984.620186 ||
 * || 9 ||
 * 4223.697398 ||
 * || 10 ||
 * 4477.119241 ||

That is find: value in year n/value in year n-1
 * B. For each year, find the ratio of the value of the coins to the the value for the previous year.

The ratio is the growth factor. Year 1 = n 2650/2500 = 1.06 Year 2 = n 2809/2650 = 1.06 Year 3 = n 2977.54/2809 = 1.06 Year 4 = n 3156.1924/2977.54 = 1.06 Year 5 = n 3345.563944/3156.1924 = 1.06 Year 6 = n 3546.297781/3156.1924 = 1.06 Year 7 = n 3759.075647/3546.297781 = 1.06 Year 8 = n 3984.620186/3759.075647 = 1.06 Year 9 = n 4223.697398/3984.620186 = 1.06 Year 10 = n 4477.119241/4223.697398 = 1.06

The ratio is 1.06 which is the growth factor

C. Suppose the value of the coins increased by 4% each year instead of 6% 1.Make a table showing the vale of the collection each year for the 10 years after Sam's Uncle gave it to him** Year 1 = n 2600/2500 =1.04 Year 2 = n 2704/2600 = 1.04 Year 3 = n 2812.16/2704 Year 4 = n 2924.6464/2812.16 = 1.04 Year 5 = n 3041.632256/2924.6464 = 1.04 Year 6 = n 3163.297546/3041.632256 = 1.04 Year 7 = n 3289.829448/3163.297546 = 1.04 Year 8 = n 3421.422626/3289.829448 = 1.04 Year 9 = n 3558.279531/3421.422626 = 1.04 Year 10 = n 3700.610712/3558.279531 = 1.04
 * Year ||
 * The Value of the Collection ||
 * || 0 ||
 * 2500 ||
 * || 1 ||
 * 2600 ||
 * || 2 ||
 * 2704 ||
 * || 3 ||
 * 2812.16 ||
 * || 4 ||
 * 2924.6464 ||
 * || 5 ||
 * 3041.632256 ||
 * || 6 ||
 * 3163.297546 ||
 * || 7 ||
 * 3289.829448 ||
 * || 8 ||
 * 3421.422626 ||
 * || 9 ||
 * 3558.279531 ||
 * || 10 ||
 * 3700.610712 ||
 * 2. Find the Growth Factor by examining successive values in the table

The growth factor is 1.04

D. What would the growth factor be if the value increased by 5% each year? Explain your answer. The growth factor would be 1.05 becuase when you calculate, you take 5% of the original value, multiply it by .05 because 5%=0.05 and add it to the original value. The original value is 100%, then you add 5% to the 100%. 100%+ 5% = 105% 105% = 1.05 1.05 is the growth factor.**

**Follow-up 3.2**
next year's value = 100% of this year's value + 6% of this year's value = 106% of this year's value = 1.06 x this year's value This helps explain the growth factor because if the growth factor is 1.06, you multiply the 'this year's value' by 1.06 for the next year's value. If you expand on the explanation, it is saying: 1(this year's value) + 0.06 (this year's value) So, it simplifies what the process we went through.
 * 1. How does the reasoning below help explain the growth factor you found for a growth rate of 6%?

2. Use the reasoning from question 1 to find the growth factor if the value increased by 4% each year. Then, find the growth factor if the value increased by 5% each year. Are the growth factors you found the same as those you found in C and D? I use the reasoning from question 1. 4% next year's values = 100% of this year's values + 4% of this year's values = 104% of this year's values = 1.04 x this year's values The growth factor is 1.04 5% next year's values = 100% of this year's values + 5% of this year's values = 105% of this year's values =1.05 x this year's values The growth factor is 1.05

Yes, my answers in this question are the same compared to C and D.

3. Sam wrote this formula for calculating the value of the coins 't' years after he first received them: V 2500(1.06^t) for t 1, 2, 3, ... 10 Does Sam's formula give the same results that are in you table for part A? We have to test the formula out to see. Test Examples t= 1 v = 2500(1.06¹) v = 2500 x 1.06 v = 2050 It matches the table data.

t = 2 v = 2500(1.06²) v = 2500 x 1.1236 v = 2809 It matches the table data.

Yes, the formula produces the same results as the table.**

**Homework:**
Assigned: Problem 3.2 & Follow-up, Problem 3.3 & Follow-up, Ace 3: 2,4,5,11
 * Collected: ACE 2: 10-12

Alya Shaiful**

**Notes**
Doubling (*2) Tripling (*3) Quadrupling (*4) Quintupling (*5)
 * Growth Factor is the base of an exponential relationship.

For example: Y = 2^x 2 is the base since it is doubling**

**Problem 3.1**
325/180 = 1.805... 580/325 = 1.785... 1050/325 = 1.8103...
 * A. Find the growth factor.**
 * Time (Years) ||
 * Rabbit Population ||
 * || 0 ||
 * 100 ||
 * || 1 ||
 * 180 ||
 * || 2 ||
 * 325 ||
 * || 3 ||
 * 580 ||
 * || 4 ||
 * 1050 ||
 * 180/100 = 1.8

Average: 1.8

The growth factor is approximately 1.8 if you average it out.

B. If we follow the growth factor, how many rabbits will there be after 10 years? 20 years? 30 years? 10 years = 35,705 rabbits (at least) Work: (1.8^10) * 100 = 35,704.67227 or 35, 705

20 years = 240,886,592 rabbits (at least) Work: (1.8^20) * 100 = 240,886,592.1 or 240,886,592

30 years = 4551715961 rabbits Work: (1.8^30) * 100 = 4551715961

C. How long will it take the rabbit population to reach to 1 million rabbits? Work: (1.8^15) * 100 = 674,644.0616 or 674,644 rabbits (1.8^16) * 100 = 1,214,395.311 or 1,214,395 rabbits In 15 to 16 years, the rabbit population would reach 1 million rabbits. 16 years being the minimum since the population is closer to 1 million.

D. Write an equation P = 100(1.8^n)

P = 100(1.8^n) P = 100(1.8^1) P = 100 * 1.8 P = 180

P = 100(1.8^n) P = 100(1.8^2) P = 100 * 3.24 P = 324 or 325 approximately

P = 100(1.8^n) P = 100(1.8^3) P = 100 * 5.832 P = 583.2 or 580 approximately**

**Follow-up 3.1**

 * 1a. How many years would the rabbit population reach from 1 million to 2 million with a growth factor of 1.5?

By a growth factor of 1.5, it would take around 1 to 2 years more until the rabbit population reaches from 1 million to 2 million.

Work: 0 year = 1 million X year = 2 million

1 year later: 1,500,000 rabbits (1.5^1) * 1 million = 1,500,000

2 years later: 2,250,000 rabbits (1.5^2) * 1 million = 2,250,000

1b. How long would it take from 1 million to reach 5 million with a growth factor of 1.5?

It will take 3 to 4 years with a growth factor of 1.5 for 1 million to reach 5 million.

Work: 0 year = 1 million X year = 5 million

3 years later: 3,375,000 rabbits (1.5^3) * 1 million = 3,375,000 4 years later: 5,062,500 rabbits (1.5^4) * 1 million = 5,062,500

1c. How long will it take for 1 million to reach 10 million with a growth factor of 1.5?

It will take 5 to 6 years for 1 million to reach 10 million by a growth factor of 1.5.

Work: 0 year = 1 million X year = 10 million

5 years later: 7,593,750 rabbits (1.5^5) * 1 million = 7,593,750

6 years later: 11,390,625 rabbits (1.5^6) * 1 million = 11,390, 625

1d. How long will it take for 1 million to reach 20 million with a growth factor of 1.5?

It will take 7 to 8 years for 1 million to reach 20 million with a growth factor of 1.5.

Work: 0 year = 1 million X year = 20 million

7 years later: 17,085,938 rabbits (1.5^7) * 1 million = 17,085,938

8 years later: 25,628,906 rabbits (1.5^8) * 1 million = 25,628,906

2a. With a growth factor of 1.2, how long will it take to double from any starting population?

It would take around 3 to 4 years.

Work: 1.2 * 2 = 2.4**


 * Year ||
 * Rabbit Population ||
 * || 1 ||
 * 1.2 ||
 * || 2 ||
 * 1.44 ||
 * || 3 ||
 * 1.728 ||
 * || 4 ||
 * 2.0736 ||
 * 2b. With a growth factor of 1.5, how long will it take to double from any starting population?

It would take around 2 to 3 years.

Work: 1.5 * 2 = 3**


 * Year ||
 * Rabbit Population ||
 * || 1 ||
 * 1.5 ||
 * || 2 ||
 * 2.25 ||
 * || 3 ||
 * 3.375 ||


 * 2c. With a growth factor of 1.8, how long will it take to double from any starting population?

It would take around 2 to 3 years.

Work: 1.8 * 2 = 3.6**


 * Year ||
 * Rabbit Population ||
 * || 1 ||
 * 1.8 ||
 * || 2 ||
 * 3.24 ||
 * || 3 ||
 * 5.832 ||


 * 2d. What did you observed in the growth factors of 1.2, 1.5, and 1.8?

I observed that the minimum is either 2 or 3 years because that is when higher values start to double. Also, as it gets higher, the exponential growth gets higher, too.

3a. If the rabbit population has the equation P = 15(1.2^t), what is the growth factor?

The growth factor is 1.2 because that is the base of the exponential equation. Additionally, it is the number that is multiplied several times with the same value.

3b. If the rabbit population has the equation P= 15(1.2^t), what is the initial growth?

The initial population is 15 because it is the coefficient/constant of the equation. Therefore, at 0 years, there would be 15 rabbits.

4a. Find percentage change of 1.5

There would be 50% change. Work:**
 * Year ||
 * Rabbit Population ||
 * || 0 ||
 * 100 ||
 * || 1 ||
 * 150 ||

50/100 = 50%
 * 150-100 = 50

4b. Find the percentage change of 1.25

There would be 25% change.

Work:** 25/100 = 25%
 * Year ||
 * Rabbit Population ||
 * || 0 ||
 * 100 ||
 * || 1 ||
 * 125 ||
 * 125-100 = 25

4c. Find the percentage change of 1.1

There would be 10% change.

Work:**
 * Year ||
 * Rabbit Population ||
 * || 0 ||
 * 100 ||
 * || 1 ||
 * 110 ||

10/100 = 10%**
 * 110-100 = 10

**Homework**
Assigned: Problem 3.1 + follow-up, Ace 3: 1 and 3
 * Collected: Mathematical Reflections pg. 30 (MR #2)

Alina Kepple** Day 31, 24/11/07 Problem 2.2 Listening to the Queen A. Make a table.**
 * Problem 2.2:
 * # squares ||
 * # rubas ||
 * || 1 ||
 * 500 ||
 * || 2 ||
 * 1000 ||
 * || 3 ||
 * 2000 ||
 * || 4 ||
 * 4000 ||
 * || 5 ||
 * 8000 ||
 * || 6 ||
 * 16000 ||
 * || 7 ||
 * 32000 ||
 * || 8 ||
 * 64000 ||
 * || 9 ||
 * 128000 ||
 * || 10 ||
 * 256000 ||
 * || 11 ||
 * 512000 ||
 * || 12 ||
 * 1024000 ||
 * || 13 ||
 * 2048000 ||
 * || 14 ||
 * 4096000 ||
 * || 15 ||
 * 8192000 ||
 * || 16 ||
 * 16384000 ||

B. Write an Equation for the relationship between the number of squares and the number of Rubas. My equation is 500 x 2 (n-1) C. Write a paragrpah on why the peasant should or should not accept this offer. I think that the peasant should not accept this offer because in this original plan (plan1). He might start of small but it gradually becomes bigger. And since the original chess board has 64 squares it also gives time for the coins to go higher.

Problem 2.2 Follow up 4000000=1000000 x n = R 1b. Write about the relationship between squares and rubas. The relationship between the squares and the rubas is a linear relationship. Because the number of squares and rubas grow at a constant rate with each other. Meaning it is a linear relationship. 1c. Would you advice the peasant to take the new offer? I would not advice the peasant to take the new offer. Because like I said before in the original plan she gets more money, because she has more squares in it. 2. Make up two rewards. a. One reward would be a 16 square board and on the first square you get 1 million and each square doubles the amount. Example: 1 million, 2 million, 4 million etc....... b. Another reward would be to start with 5 million on square one on a 16 square board and each square it goes higher by 500 hundred thousand. 3. Come up with an ending. In the end the king and queen could not go back on their word so they went backrupt and became beggers on the street. While in the mean time, the peasant had become the new queen of their country and now is filthy rich. Homework: Collected: ACE 1~11 (odd) Assigned: 2.1 & 2.2 (Follow-Up), ACE 2: 1, 5, 7
 * 1a. Write an equation.

Aki Takeuchi Day 31, 18/11/07 2.1 Getting Costs in Line

Notes:

Equation for Plan 1: r=2^(n-1)

If the values of a variable grow exponentially, how do they change from one stage to the next? They double, triple, quadruple, or increase at a constant multiplying rate.

How are the starting values and the size of the growth reflected in the table, the graph, and the equation for an exponential relationship? If the starting value was 0, then the table and graph would start with x=0. Also, the exponent in the equation would simply be n. However, if the starting value was not 0, then the graph and table would NOT start with 0, and the exponent in the equation would not only be n. (eg. n-1)

Problem 2.1

A.**
 * || Number of Rubas ||
 * Square ||
 * Plan 1 ||
 * Plan 2 ||
 * || 7 ||
 * 64 ||
 * 50 ||
 * || 8 ||
 * 128 ||
 * 55 ||
 * || 9 ||
 * 256 ||
 * 60 ||
 * || 10 ||
 * 512 ||
 * 65 ||
 * 65 ||

B. r=5n+15 This equation is similar to the equation for Plan 1 [r=2^(n-1)] in that they both involve r (rubas) and n (square). However, they are different in that Plan 1 is not a linear relationship, and Plan 4 is. You can tell that Plan 1 is not linear, because it has an exponent.

C. Square 20: ---> Plan 1...r=2^(n-1) r=2^(20-1) =2^19 =524288 ---> Plan 4...r=5n+15 r=5(20)+15 =100+15 =115

Square 30: ---> Plan 1...r=2^(n-1) r=2^(30-1) =2^29 =536870912 ---> Plan 4...r=5n+15 r=5(30)+15 =150+15 =165

D. The peasant should not accept the king's new offer, because she is not going to gain as much money as she wanted. The table and graph which were given to her are quite misleading, because it makes it look like the king's plan is worth more money. For Plan 1, the peasant could get a HUGE amount of money, while in Plan 4, the she can only get 335 rubas.

Problem 2.1 Follow-Up

1. Square 15: 16384, Square 16: 16384×2=32768

2. r=5n+15 r=5(16)+15 =80+15 =95

3a. In Plan 1, the number of rubas double from one square to the next.

b. In Plan 4, the number of rubas start from 20, and it increases by 5 from one square to the next.

4. Plan 1: The description in 3a is reflected in the base of the equation. Plan 2: The description in 3a is reflected in the 5 of "5n" and the "+15".

Homework: Collected: ACE 1~11 (odd) Assigned: 2.1 & 2.2 (Follow-Up), ACE 2: 1, 5, 7

__Sue Lee__ November 14th, 2007 - Day 30

1.3 : Making a New Offer

A way to check** 1.1 r=2^(n-1) ?
 * Notes:
 * X ||
 * Y ||
 * pattern change ||
 * formula ||
 * || 1 ||
 * 1 ||
 * 3 ⁰ ||
 * 3^(n-1) ||
 * || 2 ||
 * 3 ||
 * 3 exponent 1 ||
 * 3 exponent n-1 ||
 * || 3 ||
 * 9 ||
 * 3 exponent 2 ||
 * 3 exponent n-1 ||
 * || 4 ||
 * 27 ||
 * 3 exponent 3 ||
 * 3 exponent n-1 ||
 * Equations for 1.1 and 1.2

__Problem 1.3__

A**
 * || Number of rubas ||
 * Number of rubas ||
 * || Square ||
 * Plan 1 ||
 * Plan 2 ||
 * || 1 ||
 * 1 ||
 * 1 ||
 * || 2 ||
 * 2 ||
 * 3 ||
 * || 3 ||
 * 4 ||
 * 9 ||
 * || 4 ||
 * 8 ||
 * 27 ||
 * || 5 ||
 * 16 ||
 * 81 ||
 * || 6 ||
 * 32 ||
 * 243 ||
 * || 7 ||
 * 64 ||
 * 729 ||
 * || 8 ||
 * 128 ||
 * 2,187 ||
 * || 9 ||
 * 256 ||
 * 6,561 ||
 * || 10 ||
 * 512 ||
 * 19,683 ||
 * || 11 ||
 * 1,024 ||
 * 59,049 ||
 * || 12 ||
 * 2,048 ||
 * 177,147 ||
 * || 13 ||
 * 4,096 ||
 * 531,441 ||
 * || 14 ||
 * 8,192 ||
 * 1,594,323 ||
 * || 15 ||
 * 16,384 ||
 * 4,782,969 ||
 * || 16 ||
 * 32,768 ||
 * 14,348,907 ||
 * B) The pattern change of plan 1 is similar to plan 2 because they both have exponential growth. The difference is that plan 1 is getting doubled and plan 2 is tripling.

C) Equation for plan 2 - r = 3^(n-1)

D) The total reward under the King's plan is less than the total reward under the peasant's plan. This is because in the 25th square of the peasant's plan, it is already over 14,348,907 rubas. The number of rubas in the 25th square of the peasant's plan is 16,777,216. So obviously, at the 64th square, there will be more rubas in the peasant's reward than in the 16th square in the King's reward.

__1.3 Follow Up__

1) Open file to see graph The graph I made for plan 2 had bigger intervals than in plan 1 since the amount was tripling in plan 2 when the amount was doubling in plan 1.

2) Open file to see graph The graph I made for plan 3 has the biggest interval out of all of the plans since the pattern in the graph is quadroopling.

3) Equation for plan 3 - r = 4 to the power of (n-1) The equations for plan 1, 2, and 3 are all similar, expect for that the base changes depending on wether the pattern is doubling, tripling, or quadrupling. ☞ Plan 1 - r = 2 to the power of (n-1) ☞ Plan 2 - r = 3 to the power of (n-1) ☞ Plan 3 - r = 4 to the power of (n-1)

4) For the peasant, plan 1 is the best because at the end of the 64th square, he will have the most rubas. But for the king, plan 2 is the best because at the end of the 16th square, he will have to give less rubas to the peasant than in plan 1.

5) Answers will vary. Equation for plan 4 - r = 5 to the power of (n-1) I think this growth is exponential because just as the pattern of plans 1, 2, and 3, this plan also has the similar pattern. The pattern of the other plans were doubling, tripling, and quadrupling, but this time, it is multiplying by 5 all the time. I know this has an exponential growth because if I drew a graph out of this equation, it will look similar to the other graphs. The graph will look like first increasing by little, and then, it will increase rapidly by the pattern it has. Generally, the shape of the graph would be curved. This is how I know the growth is exponential.

☞ Plan 4** Home work :** Assigned: Investigation 1) 1.3 & follow up Mathematical Reflections, page 16 ACE 1) # 10 and 12 (problem solving steps for one) Collected: None
 * || *** of Rubas ||
 * || Square ||
 * Plan 4 ||
 * || 1 ||
 * 1 ||
 * || 2 ||
 * 5 ||
 * || 3 ||
 * 25 ||
 * || 4 ||
 * 125 ||
 * || 5 ||
 * 625 ||
 * || 6 ||
 * 3,125 ||
 * || 7 ||
 * 15,625 ||
 * || 8 ||
 * 78,125 ||
 * || 9 ||
 * 390,625 ||
 * || 10 ||
 * 1,953,125 ||
 * || 11 ||
 * 9,765,625 ||
 * || 12 ||
 * 48,828,125 ||
 * || 13 ||
 * 244,140,625 ||
 * || 14 ||
 * 1,220,703,125 ||
 * || 15 ||
 * 6,103,515,625 ||
 * || 16 ||
 * 30,517,578,130 ||

Yoon Sun Lee November 12, 2007/ Day 29

1.2 : Requesting a Reward

Notes

2⁴ 2 = Base ⁴= exponent exponent form : 2x2x2x2

Standard form : 16 Journal Problem 1.2

A.
 * Squares ||
 * Number of Rubas ||
 * || 1 ||
 * 1 ||
 * || 2 ||
 * 2 ||
 * || 3 ||
 * 4 ||
 * || 4 ||
 * 8 ||
 * || 5 ||
 * 16 ||
 * || 6 ||
 * 32 ||
 * || 7 ||
 * 64 ||
 * || 8 ||
 * 128 ||
 * || 9 ||
 * 256 ||
 * || 10 ||
 * 512 ||
 * || 11 ||
 * 1024 ||
 * || 12 ||
 * 2048 ||
 * || 13 ||
 * 4096 ||
 * || 14 ||
 * 8192 ||
 * || 15 ||
 * 16384 ||
 * || 16 ||
 * 32768 ||

B. The number of rubas get twice more bigger than the number of rubas on the previous square.

C. On square 20, there will be 524,288 rubas On square 30, there will be 536,870,912 rubas On square 64, there will be 9,223,372,036,854,775,808 rubas.

D. On square 21, ir will reach one million rubas.

E. The number of rubas on square 10 is 512, then it will worth $5.12. The number of rubas on square 20 is 524,288, then it will worth $5,242.88. The number of rubas on square 30 is 536,870,912, then it will worth $5,368,709.12. The number of rubas on square 40 is 549,755,813,888, then it will worth $5,497,558,138.88. The number of rubas on square 50 is 562,949,953,421,312, then it will worth $5,629,499,534,213.12. The number of rubas on square 60 is 576,260,752,303,423,488, then it will worth $5,762,607,523,034,234.88.

Follow-Up 1, The number of rubas chage rapidly and it has exponential change. In beginning it grows really slowly but all of a sudden it became really huge. This tells that peasant will get a lot of money. 2. The equation is r=2ⁿ 3. If the chessboard had 100 squares, on 100 square there will be 6,338,253,001,141,146,039,913,283,584. 4a. The two problems are similar because they both have exponential growth and change rapidly. 4b. The equation for problem 1.1, it is b=2ⁿ.

Homework

Assigned: Investigation 1) 1.3 & follow up Mathematical Reflections, page 16 ACE 1) # 10 and 12 (problem solving steps for one) Collected: None

__Tarryn__ November 8, 2007 - Day 28 Notes 5ˆ8 = 5⁴ * 5⁴ = 5ˆ5 * 5³ =(5*5*5*5) * (5*5*5*5) When you work out 'the power of' there are many different combinations of the sum. Journal __Problem 1.1__ A, B + C:
 * Cuts ||
 * Ballots ||
 * || 1 ||
 * 2 ||
 * || 2 ||
 * 4 ||
 * || 3 ||
 * 8 ||
 * || 4 ||
 * 16 ||
 * || 5 ||
 * 32 ||
 * || 6 ||
 * 64 ||
 * || 7 ||
 * 128 ||
 * || 8 ||
 * 256 ||
 * || 9 ||
 * 512 ||
 * || 10 ||
 * 1,024 ||
 * || 20 ||
 * 524,288 ||
 * || 30 ||
 * 536,870,912 ||

D: 524,288 / 25 =2097.152'' =174.763'

536,870,912 / 250 = 2,147,483.648'' =178,956.971'

E: 25012= 3,000 sheets 13 cuts (that = 4096 sheets)

__Follow Up__ 1a: 2³ b: 5⁴ c: 1.5ˆ 7

2a: 128 b: 27 c: 74.088

3a: 32,768 b: 59,049 c: 3,325.25673

4: 5² 5*5 25 2ˆ 5 2*2*2*2*2 32 The answer and the method of working is different. You multiply 5*5 and get 25 but the other method you mulitiply 2*itself 5 times and the answer is 32.

6: If 5ˆ 10 9,765,6255 then 5ˆ11 48,828125

Homework Assigned:ACE1: # 1-11 odds Collected: None

21/11/07 Block C Day 29

2.3 Growing mold

Problem:

A: The mold changes by a factor of three every day. B: A: 3^(d-1) is the slope for the given table of mold growth. C: The answer on A should reflect on B because the answer for A is the slope in the equation 3^(d-1) on question B.

Follow up:

1)

Day Mold area (mm^2) 0 (start) 25 1 75 2 225 3 675 4 2025

2) Y=3^(d-1)+25

3) If applied the rule of Y=MX+B, B would indicate the starting point which equals to 25 in the equation Y=3^(d-1)+25 also the rate of cha nge would be represented as X which in the equation Y=3^(d-1)+25 would be 3^(d-1).

4) A: Making ballots had a growth factor of 2 B: reward plan #1 also had a growth factor 2 C: Reward plan #2 also had a growth factor of two but started with 500 and only went up to 16 rather than plan #1 which went up to 64 D: the growth of bread mold had a growth factor of 3

5) A: 50: 3^0=1,X 50=50 B The growth factor on Y is 3 C The area of the mold would be 50X243=12150=mold area after 5 days. D the mold reached 10cm on the 4th day (remember it said reached)

