Frogs+Fleas+and+Painted+Cubes

Problem 5.2 Exploring Painted-Cube Patterns
A. The large cube with sides n long, has a total amount of cubes, expressed by the equation y=n^3. B. On the large cube, there are smaller cubes painted on 3, 2, 1 and no sides. 1. The amount of small cubes painted on 3 sides can be expressed in the equation y=8. 2. The amount of small cubes painted on 2 sides can be expressed in the equation y=12(n-2). 3. The amount of small cubes painted on 1 side can be expressed in the equation y=6(n-2)^2. 4. The amount of small cubes painted on no sides can be expressed in the equation y=(n-2)^3. Follow-Up 1. a.

b. The equations for these graphs and the graphs in 5.1 follow-up are the same, therefore the graphs are the same. 2. a. For the large cube, y=6(n^2) represents the total surface area in comparison to the side length.

b. For the large cube, T=6(n-2)^2+2(12(n-2))+(3x8) represents the total surface area in comparison to the side length. c. The surface-area and total number of painted faces are the same thing. This can be represented by S=T.

3. a. X 1 2 3 4 5 X 1 2 3 4 5

X 1 2 3 4 5 X^2 1 4 9 16 25

X 1 2 3 4 5 X^3 1 8 27 64 125

b. In the first table, both columns increase at a steady rate of 1, this relationship is linear. In the second table, the second column increases at a slowly increasing rate, this relationship is quadratic. In the third table, the second column increases at a rapidly increasing rate, this relationship is cubic. c. The first table is similar to the amount of cubes painted on 2 faces because both are linear. The second table is similar to the amount of cubes painted on 1 face because both are quadratic. The third table is similar to the total amount of cubes faces because both are cubic.

4. The first table is similar to y=2x because both are linear. The first table is similar to y=(x-1)^2 because both are quadratic. The first table is similar to y=x^3-2 because both are cubic.

Homework Assigned
ACE: 15-21 odds and 23 or 25 Complete Mathmatical Reflections 5, pg. 84

Problem 5.1 Analyzing Cube Puzzles
= = =Notes:= To cube something is to add on the third dimension (depth). =Problem:= A.1: A cube with edge lengths of 2cm built from 1cm³ cubes will have each 1cm³ components painted on 3 sides. A.2: A cube with edge lengths of 3cm built from 1cm³ cubes will have 8 mini cubes painted on 3 sides, 12 mini cubes painted on 2 sides, 6 mini cubes painted on 1 side, and 1 mini cube which is not painted. A cube with edge lengths of 4cm built from 1cm³ cubes will have 8 mini cubes painted on 3 sides, 24 mini cubes painted on 2 sides, 24 mini cubes painted on 1 side, and 8 mini cubes which are not painted. A cube with edge lengths of 5cm built from 1cm³ cubes will have 8 mini cubes painted on 3 sides, 36 mini cubes painted on 2 sides, 54 mini cubes painted on 1 side, and 27 mini cubes which are not painted. A cube with edge lengths of 6cm built from 1cm³ cubes will have 8 mini cubes painted on 3 sides, 48 mini cubes painted on 2 sides, 96 mini cubes painted on 1 side, and 64 mini cubes which are not painted.

Edge length of large cube
|| ==== ====

Number of cm cubes
|||||||| ==== ====

Number of cm cubes painted on
|| B: The relationship between the edge length of the large cube and the number of cm cubes is the edge length of the large cube cubed equals the number of cm cubes. C: The relationship between the edge length of large cube and the number of cm cubes painted on 3 faces is a constant. The relationship between the edge length of large cube and the number of cm cubes painted on 2 faces is Y=12x-2. It is a linear relationship. The relationship between the edge length of large cube and the number of cm cubes painted on 1 face is Y=6(n-2)^2. It is a quadratic relationship. The relationship between the edge length of large cube and the number of cm cubes painted on 0 faces is Y=(n-2)^3. This is neither a linear, exponential, or quadratic relationship. Follow-up 1.a: (see added rows in table) 1.b: 2: All the mini cubes that are painted on three sides would be the corners, the mini cubes that are painted on one side would be in the middle of a face, and all the mini cubes that are painted on two sides would be on the edges.
 * ||  || ====3 faces==== || ====2 faces==== || ====1 face==== || ====0 faces==== ||
 * 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 ||
 * 7 || 343 || 8 || 60 || 150 || 125 ||
 * 8 || 512 || 8 || 72 || 216 || 216 ||
 * 9 || 729 || 8 || 84 || 294 || 343 ||

Problem 4.3 Putting It All Together
=Notes=
 * Let’s observe the patterns of change for different equations. To do this, we can look at first differences and second differences in the y values as x changes and we can look at the properties of the graphs of those equations.

=Problem= 1.A

X y for 2x(x+3) First diff Second diff



-5 20 -4 8 -12 -3 0 -8 4 -2 -4 -4 4 -1 -4 0 4 0 0 4 4 1 8 8 4 2 20 12 4 3 36 16 4 4 56 20 4 5 80 24 4

X y for 3x-x^2 -5 -40 -4 -28 12 -3 -18 10 -2 -2 -10 8 -2 -1 -4 6 -2 0 0 4 -2 1 2 2 -2 2 2 0 -2 3 0 -2 -2 4 -4 -4 -2 5 -10 -6 -2

X y for y=(x-2)^2 -5 49 -4 36 -13 -3 25 -11 2 -2 16 -9 2 -1 9 -7 2 0 4 -5 2 1 1 -3 2 2 0 -1 2 3 1 1 2 4 4 3 2 5 9 5 2

X y for y=x^2+5x+6 -5 6 -4 2 -4 -3 0 -2 2 -2 0 0 2 -1 2 2 2 0 6 4 2 1 12 6 2 2 20 8 2 3 30 10 2 4 42 12 2 5 56 14 2 B. In the first column, their patterns are similar because they change from positive/negative to negative/positive. They also change by a constant amount each time. But, some of the amounts are different and they can change in either direction. In the second column, the number is always the same, although in one it is negative and for the rest it is positive. They are mostly varying second differences.

4.3 FU

x y for y = x+2 first diff second diff 0 2 1 3 1 2 4 1 0 3 5 1 0 4 6 1 0 5 7 1 0 6 8 1 0 7 9 1 0 8 10 1 0 9 11 1 0 10 12 1 0

x y for y = 2x 0 0 1 2 2 2 4 2 0 3 6 2 0 4 8 2 0 5 10 2 0 6 12 2 0 7 14 2 0 8 16 2 0 9 18 2 0 10 20 2 0

x y for y = 2^x 0 1 1 2 1 2 4 2 1 3 8 4 2 4 16 8 4 5 32 16 8 6 64 32 16 7 128 64 32 8 256 128 64 9 512 256 128 10 1024 512 256

x y for y = x^2 0 0 1 1 1 2 4 3 2 3 9 5 2 4 16 7 2 5 25 9 2 6 36 11 2 7 49 13 2 8 64 15 2 9 81 17 2 10 100 19 2

B. Y value is always rising, always positive. They are all different patterns because of their equations. Two are linear, and one is exponential and one is quadratic.

C. In the first one, the pattern of change reflect the fact that x is by itself, and just added to a constant. In the second one, the pattern is reflected by the fact that x is multiplied by two. In the third, the exponent x applied to the constant # 2 reflects the pattern of change. And in the last one, the pattern of change is quadratic, and rises at a rising, symmetric rate.

2. Descriptions of graphs for equations above, in the same order.

Graph 1 – parabola; min at -4.5, x-is at (0,0) and (-3,0). Line of symmetry at x=-1.5

Graph 2 – parabola; max at 2.25, x-is at (0,0) and (3,0), LOS at x=1.5

Graph 3 – parabola; min at 0, x-i at (0,2), LOS at x=2

Graph 4 – parabola; min at -0.25, x-is at (-3,0) and (-2,0), LOS at x=-2.5

Graph 5 – linear; y-I at (0,2), no min or max, no LOS, x-I at -2

Graph 6 – linear; x-I = (0,0), y-I at (0,0), no min/max, no LOS

Graph 7 – exponential; no x-I, y-I = (0,1), no LOS no min/max.

Graph 8 – parabola; min at 0, x-I (0,0), LOS = y axis

=Homework assigned: ACE4: 4-8, 10, 13, 15, 21=

Notes:
Frog: h=-16t^2+12t+0.2 Flea: h=-16t^2+8t Basketball player: h=-16t^2+16t+6.5

The 0.2 and 6.5 are constant terms and are represented as the animal’s height in this particular situation. y=-x^2 is an arch shaped parabola. y=x^2 is a u-shaped parabola.

Journal:
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 seconds and 1 second. In your tables, use intervals of 0.1 second. this is the graphs for the equations(on left). this is the tables for the equations(on left).

B) What is the maximum height reached by each jumper, and when is the maximum height reached? The maximum height reached for the frog was 2.44ft. The Flea reached a height of 11.52 inches or 0.96ft. The basketball player reached a height of 10.5ft. I used the table to find the highest height reached by each animal.

C) How long does each jump last? Explain how you found your answer. The frog lasted about 0.7 seconds. The flea lasted about half a second and the basketball player lasted one second. I used my knowledge of the constant term and used the table to find when the animal reached their initial height.

D) What do the constant terms 0.2 and 6.5 tell you about the frog and the basketball player? The constant terms basically is the height of the animal when it jumps because animals do have a height no matter how small it is.

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 the graph. As you can see from both the graph and the table you can clearly point out the features of these quadratic relationships. In the table the line of symmetry and the x and y intercepts can be easily realted to the various points on the graphs for each equation.

2) Mr. Jain is a jewelry maker. He would like to increase his profit by raising the price of his jade earrings. However, he knows that if he raises the price too high, he won't sell as many earrings and his profit will decrease. Using records of past sales, a business consultant developed the equation p=50s-s^2 to predict the monthly profit,p,for a given sales price,s.

a) Make a table and a graph for this equation. The graph. The table.

b) What do the equation, the table, and the graph suggest about the relationship between price and profit? The equation had two linear factors so it showed that the relationship would be in the form of a quadratic function. The graph is an arch shaped parabola because your subtraction from x. The table also gave a line of symmetry and the features for a quadratic graph.

c) What price will bring the greatest profit? The price should be at twenty five dollars because it has the greatest profit and is the highest point on the graph.

D) How does this equation compare with the equations in Problem 4.2? First of all the equations are in quatratic form but the equations from 4.2 are in expanded form compared to the factored form in this follow up. The parabolas have an arch and they both are dealing with real life situations.

Homework assigned:
A.C.E 4: 4-8,10,13,15,21

Homework collected:
None.

Notes:
Parabolas are quadratic relationships.
 * Time (s) || Height (ft) ||
 * 0.00 || 0 ||
 * 0.25 || 15 ||
 * 0.50 || 28 ||
 * 0.75 || 39 ||
 * 1.00 || 48 ||
 * 1.25 || 55 ||
 * 1.50 || 60 ||
 * 1.75 || 63 ||
 * 2.00 || 64 ||
 * 2.25 || 63 ||
 * 2.50 || 60 ||
 * 2.75 || 55 ||
 * 3.00 || 48 ||
 * 3.25 || 39 ||
 * 3.50 || 28 ||
 * 3.75 || 15 ||
 * 4.00 || 6 ||

Journal:
A. The height of the ball increases after 0-1.75 seconds but when it is 2.00 seconds it stays at 64feet until 2.25 seconds where it starts to decrease. B. A graph of these data would be a parabola. The maximum point is 2.00,64, and the line of symmetry would be x=2. Also the graph would be in a shape of a hill, it would go up, not down. C. Yes I think that these data represents a quadratic function because I can see that the graph makes a parabola, and all parabolas are quadratic relationships.

Problem 4.1 Follow-up 1a) //h// = -16//t//² + 64//t// b) Yes this graph matches my description in Part B above, because it is a parabola that goes up, and also the line of symmetry is at x=2. c) I can use the equation to find out approximately when the ball reaches a height of about 58 feet. //h// = -16//t//² + 64//t// 58 ≈ -16//t//² + 64//t// -16 (1.388²) + 64 x 1.388 ≈ 58 -16 (2.612²) + 64 x 2.612 ≈ 58 The ball reaches about 58 feet when it is 1.388 seconds, or 2.612 seconds. d) I can use the equation to find the height of the ball after 1.6 seconds //h// = -16 (1.6²) + 64 x 1.6 //h// = -40.96 + 102.4 h = 61.44 The height of the ball is 61.44 feet after 1.6 seconds. 2a) //h// = -16//t//² + 64//t// + 6 Table for the equation:

b) The constant term (the 6 in the equation) affects the table, because now the y values start and end at 6, not 0. It affects the graph because the graph is now higher. The constant term tells us where the graph intersects the y-axis. ci) The maximum height reached by the ball Question 1 – 64 feet. Question 2 – 70 feet. The difference between the two heights isn’t that big, it’s only 6 feet. ii) The x-intercepts Question 1 – 0,0 Question 2 – -0.09,0 The x-intercepts aren’t that different either. (By only 0.09) iii) In both graphs in the beginning, as x increases, y increases too, until the middle (maximum point). Then as x increases, y decreases.
 * Time (s) || Height (ft) ||
 * 0.00 || 6 ||
 * 0.25 || 21 ||
 * 0.50 || 34 ||
 * 0.75 || 45 ||
 * 1.00 || 54 ||
 * 1.25 || 61 ||
 * 1.50 || 66 ||
 * 1.75 || 69 ||
 * 2.00 || 70 ||
 * 2.25 || 69 ||
 * 2.50 || 66 ||
 * 2.75 || 61 ||
 * 3.00 || 54 ||
 * 3.25 || 45 ||
 * 3.50 || 34 ||
 * 3.75 || 21 ||
 * 4.00 || 6 ||

Homework:

 * Assigned:** Problem 4.1 & Ace 4: 3,9,18 and 26.
 * Collected:** Mathematical Reflections # 3 & Check-up 2 Redo

Problem 3.2 Exploring Triangular Numbers
Notes: -Triangular Number: A quantity that can be arranged in a triangular pattern. For ex: 1,3,6, 10 etc.

Problem 3.2: A. The two variables in this situation are the # of the triangle (1st, 2nd, 3rd, etc.) and the other one is that # of dots in each figure. The # of the triangle is the independent variable, located in the x-axis while the # of dots is the dependent variable located in the y-axis. B. The table for this situation is located below: C. Each triangular number is being multiplied by one number more than itself and then dividing it in half. For example: if the # of the triangle is 3, then it’s multiplied by 4 (3+1) 12 and then divided in half 6. D. If the # of triangle is 15, then the # of dots is 120. I found it out by doing the same thing I did on C. =15(15+1)/2 =15*16/2 =240/2 =120 E. The equation for the situation above is: D (dots)={n(n+1)}/2 F. Yes, the equation represents a quadratic relationship as the n (x) is being multiplied to itself. And also, if you graph or table the equation, you see that it’s a quadratic relationship.

3.2 Follow Up:

1. Give the dimensions and the area of the rectangle formed by combining two copies of the given figure. a.) the 4th figure: Dimensions- 4,4 Area- 10u b.) the 5th figure: Dimensions- 5,5 Area- 15u c.) the 10th figure: Dimensions- 10,10 Area- 55u d.) the nth figure: Dimensions- n, n Area- n(n+1)/2

2. For each part in question 1, how does the number of squares in the rectangle compare with the number of squares in the original figure? a.) the 4th figure: Area for original figure: 10 Area for real rectangle: 16 Difference of 6u. b.) the 5th figure: Area for original figure: 15 Area for real rectangle: 25 Difference of 10u. c.) the 10th figure: Area for original figure: 55 Area for real rectangle: 100 Difference of 45u. d.) the nth figure: Area for original figure: n(n+1)/2 Area for real rectangle: nxn Difference of nu.

3. Equation for this situation: A= n(n+1)/2 The equation seems to be the same as found in Problem 3.2

4. The area of the 18th triangular number would be: =18(18+1)/2 =18x19/2 =342/2 =171u^2

HOME-WORK:

 * Collected:** Mathematical Reflections # 2
 * Assigned:** Problem 3.1 and 3.2 with follow-ups & Ace (3) 1-12, 15

3.1 Counting Handshakes, Day 46
Notes: Case 1 - Two teams with the same number of players shake hands. 2 people per team: 4 shakes 3 people per team: 9 shakes 4 people per team: 16 shakes 5 people per team: 25 shakes

Case 2 - Two teams shake hands and one team has one more player than the other team. 1 and 2 players: 3 shakes 3 and 4 players: 12 shakes 5 and 6 players: 30 shakes

Case 3 - Players of the same team give each other high fives. 2 players team: 1 high five 3 player team: 3 high fives

Journal: A) 1. According to the pattern in the notes, there will be 100 shakes if both teams have 10 players. 2. If there are 15 people per team, there will be 225 shakes. 3. H = n^2

B) 1. According to the pattern in the notes, there will be 90 shakes if one team has 9 and the other 10. 2. If one team has 14 players and the other 15, there will be 210 shakes. 3. H = n(n-1)

C) 1. There will be 6 high fives with a team of 4 players. 2. A team with 12 players will have 66 high fives. 3. H = n/2 (n-1)

Follow Up: 1a) Case 1: Case 3: b. The pattern for case 1 is that the number of players is multiplied my itself, h=n^2. For case 2, its is the number of players on one team multiplied by the number of players on the other team, h=n(n-1). For case 3, the added amount to the y value increases by 1 more each time, h = n/2 (n-1) c. The patterns are all similar since they all involve the multiplication of both numbers. They are different because the numbers being multiplied keep changing.
 * Number of Players ||
 * || Handshakes ||
 * 1 || 1 ||
 * 2 || 4 ||
 * 3 || 9 ||
 * 4 || 16 ||
 * 5 || 25 ||
 * 6 || 36 ||
 * 7 || 49 ||
 * 8 || 64 ||
 * 9 || 81 ||
 * 10 || 100 ||
 * Number || of players || Number of ||
 * Team 1 || Team 2 || handshakes ||
 * 5 || 6 || 30 ||
 * 6 || 7 || 42 ||
 * 7 || 8 || 56 ||
 * 8 || 9 || 72 ||
 * 9 || 10 || 90 ||
 * Number of players || High fives ||
 * 1 || 0 ||
 * 2 || 1 (+1 from before) ||
 * 3 || 3 (+2) ||
 * 4 || 6 (+3) ||
 * 5 || 10 (+4) ||

2a) (graphs are inserted below) b) The graphs compare since they are all parabolas. The only differences are their location on the grid and the width on the axees.

3) Yes, these equations, tables, and graphs are quadratic. I can tell since the equation has x^2 and the graphs are in the shape of parabolas.

4a) Tyler probably thought out the equation since n is multiplied by one less than it, n-1. Therefore, n(n-1). Asuko thought out his expression because n is multiplied by n, and than 1 is subtracted. So, n^2 - n. b) n
 * n-1 ||

Nabihah Rana January 15th, 2008; Day 43 Problem 2.2: Changing One Dimension, pg. 22-24

Notes: x^2+2x and x(x+2) are the same thing. Journal: A. 1. The dimensions of the added rectangle is x and 2. The area of teh rectangle is x times 2 or 2x. 2. The equation for the whole area of the new rectangle would be: A = x^2 + 2x 3. The length of the new rectangle is x + 2 and the width is x. An equation to get the area would be: A = x(x+2) 4. When the two above equations are graphed, they make the same graph. Both are like a U shape parabola.



B. 1. The new rectangle is 3x greater then the area of the previous square. 2. One equation for the area of the new rectangle is A = x^2 + 3x And the other equation is A = x(x+3) 3. When both the equations are graphed on the calculator, they both come as the same. The shape is like a parabola: U. = =

Follow- Up, pg. 23-24

factored form : A = x(x+4) 2. Area of the unshaded rectangle: expanded form: A = x^2 – 6 factored form: A = x(x-6) 3. (Draw rectangle with that equation and write an equation in expanded form) 4. (Draw rectangle with that equation and write an equation in factored form) 5. If one side of an equation is not written in factored form, I can still tell whether it’s a quadratic equation or not by looking at the other expanded form of the equation.
 * 1) expanded form: A = x^2 + 4x

Vivian Ly 1/20/08 Day: 43 Problem 2.4:Trading Land

Journal: A. Graph 1- y=x(x+4) Graph 2- y=x(4-x) Graph 3- y=x^2 Graph 4- y=(x+3)(x+3) Graph 5- y=2x(x+4) Graph 6- y=(x+2)(x+3) Graph 7- y=(x+3(x-3) Graph 8- y=x(x-4)

B. You can predict the x-intercepts if it matches with the y-intercepts.

C. ?

D.Graph 1: -2,-4 Graph 2: 2,4 Graph 3: 0,0 Graph 4: -2.5,-.5 Graph 5: -2,-8 Graph 6: -3,0 Graph 7: 0,-9 Graph 8: 2,-4

E. x^2 is the basic equation with the symmetrical arc right in the middle of the whole graph. By switching the number around in the brackets will make either an arc or an upside down arc. The number you subtract or add in the equations will make the arc go left or right.

F. I don't know how to put up the pictures... Graph 1: y=x^2+4x Graph 2: y= 4x-x^2 Graph 3: y=x^2 Graph 4: y=x^2+3x+3x+9 Graph 5: y=x^2+3x+2x+6 Graph 6: y=2x^2+4x Graph 7: y=x^2-3x+3x-9 Graph 8: y=x^2-4x

Follow-up 1. You can tell if a factorise equation represents a quadratic form when the equation can make an symmetrical arc on a graph. The equation should have an x^2 and if not you can make it into an expanded form to check. 2. You can tell if an expanded equation represents a quadratic form when the equation has x^2 in it. When it is graphed it should have an symmetrical arc.

Homework assignment: ACE2: #2,34,37 Homework collected: none

Problem 2.1: Trading Land
Notes: Changing one dimension can completely change the area or perimeter of a rectangle. Journal: A: here is the table for 2.1 Side length (m) || Original Square Area || New rectangle length (m) || New rectangle Width (m^2) || New rectangle Area (m^2) || Difference in area (m^2) ||
 * Original Square
 * 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 || n^2 || n+2 || n-2 || (n+2)(n-2) || 4 ||

B. For all the side lengths the difference in the area is always 4 less than the original square making it a bad deal.

C. The expressions are located on the table

2.1 F/U

1a. This is the equation for this graph: n^2

1b. This is the equation for this graph: (n+2)(n-2) or n^2+ 4

1c. the both graphs are in the same shape. They are both in the shape of a porabola that is smiling. So they are both quadratic relationships. The only difference is that one is 1a is much skinnier than 1b.

1d. It should make sense because in a graph all of the side lengths should match up with the area.

2a. The tables are the same length but they are different because in this problem it takes longer for the y values to get to the line of symmetry.

2b. The graphs from investigation 1 are parabolas just like the ones in this investigation. The main difference from investigation 1 is that this parabola is the opposite way around from the investigation 1 parabola.

2c. In this investigation we use more of expressions than equations. In Investigation 1 we used 2 variables but in this investigation so far we use 1 variable. The equation for this investigation is n^2 but in investigation one it was A=L (half of perimeter-L).

3: The trade offered in “think about this” on page 19 is obviously a very unfair trade. The mall gave an offer of more perimeter, but less land. The farther you get away from a square the less area you get. So the family was getting much more perimeter but much less of an area. That is why it is an unfair deal.

Homework Assigned: ACE2: 10-14, Check up and 2.1 and 2.2 Homework Collected: MMR pg 18

Problem 1.3 Writing a Equation
B a= L(10-L) C the area would be 24 D

Follow up 1b A= L(30-L) 1c The greatest area possible would be 225. 1e when the graph reaches the top, thats the point when the area becomes the greatest number. 2b 300 would be the area 2c don't understand the problem 2d the graph has a shape of parabolas shape. 2e the maximum area would be 306 square meters. 3 the equation would A= L (half of perimeter-L) 4 the charic of parabolas graph is that as much it goes up, it comes down with a same rate as it went up but backwords. 5 I think it's the equation because for some of the numbers, you might have un even number and might go down to decimals. if you do it by equation, all you need to is write numbers and you'll get the answers.
 * x ||  ||   ||   ||
 * Y ||  ||   ||   ||
 * || 0 ||
 * 0 ||
 * || 1 ||
 * 9 ||
 * || 2 ||
 * 16 ||
 * || 3 ||
 * 21 ||
 * || 4 ||
 * 24 ||
 * || 5 ||
 * 25 ||
 * || 6 ||
 * 24 ||
 * || 7 ||
 * 21 ||
 * || 8 ||
 * 16 ||
 * || 9 ||
 * 9 ||
 * || 10 ||
 * 0 ||
 * x || y ||
 * 0 || 0 ||
 * 3 || 81 ||
 * 6 || 144 ||
 * 9 || 189 ||
 * 12 || 216 ||
 * 15 || 225 ||
 * 18 || 216 ||
 * 21 || 189 ||
 * 24 || 144 ||
 * 27 || 81 ||
 * 30 || 0 ||

Notes:
Parabolas: Graphs that rise and then start to fall. In other words the data is a hill like figure when graphed. The highest or the lowest point of the graph can be found in the middle or you could say on the line of symmetry.

Problem 1.2:
Something that I observe right away is that the graph is shaped like a dome or a hill. The figure has a highest point which is in the line of symmetry. I predict that the data on the table will be different. By different I mean that the data won’t just rise or fall. I think that it will be a bit of both. The data for this particular graph will rise, and in the middle it will reach its peak and start to fall.
 * 1) Describe the shape of the graph or any special features you observe.

1. What is the greatest area possible for a rectangle with this perimeter? What are the dimensions of this rectangle? First to find the area of the rectangle I need to find the fixed perimeter. To find the set perimeter I need to find the length of at least on side of the rectangle. The length that sticks out to me right now is the highest point. This length is 20 meters long and the area that goes with it is 400. What multiplied by 20 would equal to 400. 400 divided by 20 is 20. The other length of this rectangle is 20. This also makes sense because the square normally has the largest area.

2. What is the area of the rectangle with a side of length 12 meters? What is the area of the rectangle with a side of length 28 meters? Explain how these two rectangles are related. The area of a rectangle with a side length of 12 is easy to find if we know the fixed perimeter. The fixed perimeter as we know is 80. The formula to find the perimeter of a rectangle is simple, P being perimeter it is P=2(l+w). If we plug in the numbers 12 becomes 24 because 2x12+24. 80 being the total perimeter, 80-24= 56. This means that 56 is the other length. To find the area we need to multiply the length or width. We have to find the half of each of those numbers. We get 12 and 28. 12x28=336.

3. What are the dimensions of the rectangle with an area of 300 square meters? This is an easy question to answer if we look at the graph. First we look at the y-axis which is labeled ‘areas’. We have to look at the area of 300 and look across. The lengths that hit that are 10 and 30. The dimensions of the rectangle therefore are 10x30. 10x30= 300. These shows that 10x30 are the right dimensions.

4. What is the fixed perimeter represented by the graph? Explain how you found the perimeter. Actually I think we answered this question before in his problem. We had tried out man areas and we put in the formula to find the perimeter and each time we got 80.

1.2 follow up:
1. How do the special features you observed appear in the table? The shape appears in the sense of the way the data raises or falls. The data in the table first start to rise. When the length of a side reaches 7 the data falls back down the same exact way.

2. What is the fixed perimeter for the rectangle represented in this table? Explain how you found the perimeter. In this table the fixed perimeter can be found out if we add two opposite numbers such as 1 and 11. 2x1= 2 and 2x11= 22. 22+2=24. The fixed perimeter for this data table is 24.

3. What is the greatest area possible for a rectangle with this perimeter? What are the dimensions of this rectangle? The greatest area is really easy to find on tables. All we have to do is read down the column with all the areas. The largest one is 36; this means that 36 is the largest area. The length of one of the sides in this triangle is 6. As we discovered before the largest area is always a square. So that means that the dimensions will be 6x6.

4. Approximate the dimensions of a rectangle with this fixed perimeter and an area of 16 square meters. If the fixed perimeter is 24 and the area are 16 we have to find a number which multiplied can be 16. Let’s try 4. If four is a side length then what would the total perimeter be? 4x4=16. 16x2= 32. So that doesn’t work. Let’s try the dimensions 2x8. 2x8=16. 2x2=4, 8x2=16. Then the total perimeter would be 20. This means that the dimensions can’t be a whole number. 1.53x10.5= 16. 1.53x2= 3.06, 10.5x2=21. 21+3.06=24.

Homework:
Collected: none Assigned: ACE# 1 => 3,5

Problem 1.1: Staking a Claim Notes
Factored Form: y = (x + c)(x+d) Expanded Form: y = x² + (c+d)x + cd When a < 0 graph opens down Y = -x² - x + 6 When a > 0 graph opens up Y = x² - x +6 Infactored Form: Y =-x² - x + 6 à Y x² - x +6 Y = (x + 3)(x – 2) X -2 ^ x 2 intercepts (-3, 0), (2, 0)

Y = (x + 3 (x + 2) = x² + 3x + 2x + 6 Y = x² + 5x + 6

Journal
Problem 1.1: A.)

B.)
 * Length || Width || Area ||
 * 9 || 1 || 9 ||
 * 8 || 2 || 16 ||
 * 7 || 3 || 21 ||
 * 6 || 4 || 24 ||
 * 5 || 5 || 25 ||

C.)

D.) To enclose the greatest area possible, the dimensions of my fence would be 5 meters x 5 meters. In this case, the area would be 25 meters². It is also the highest point on the graph shown above. Problem 1.1 Follow-up: 1.) This would change the answer to part D because if the numbers were 5.5 x 5.5 rather than 5 x 5, just by adding the extra 0.5 to each end, the area would increase altogether: A 25 || A = 30.25 ||
 * Whole numbers || Integers ||
 * A = 5 x 5
 * A = = = 5.5 x 5.5

2.) To arrange my 20 meters so I could enclose a greater area, I would make the shape of the claim a circle. From the diagrams on page 6 (and below), the circle encloses a greater area whereas the rectangle has a limited and a smaller area.

Homework:
Collected: none Assigned: page 12 - 17 # 3 & 5

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