Say+it+With+Symbols

Welcome to the Advanced Eighth Grade Math (section 8A) class wiki at the American International School of Dhaka.

Student resources: For posting graphs, use the following website: Tip: For your urgent graphing needs at any time, just point your browser to fooplot.com/sin(x) fooplot.com/x^2+2*x+1 fooplot.com/yourfunctionhere
 * fooplot.com**

Our Current unit is:

Say it With Symbols
journal_homework_record_Math_8_say_it_with_symbols 0708.doc

D.S

Notes:
PEMDAS is the order of solving an equation. It is Parenthesis, exponents, multiplication, division, addition, and subtraction.

1.1 Adding and Multiplying
B I found out that every time it goes down in the table it goes up by 100. Also each time it goes right in the table it goes up by 160. C: The # 100 tells us about the starting price or fee. The 10 tells us about the adult price and the 8 tells us about the price of the child ’ s ticket. D: You must multiply the child by 8, the adults by 10, and then add 100. That is the order you must do it in.
 * X || 20 || 40 || 60 || 80 || 100 ||
 * 10 || 360 || 520 || 680 || 840 || 1000 ||
 * 20 || 460 || 620 || 780 || 940 || 1100 ||
 * 30 || 560 || 720 || 880 || 1040 || 1200 ||
 * 40 || 660 || 820 || 980 || 1140 || 1300 ||

1.1 F/U 1a. P= 2.5V-500 82.5x 300= 1750 P= 2.5 (300)-500250

1b. P= 2.5V-500 2.5x 600= 3500 P= 2.5 (600)-5003000 1c. I used multiplication first than i used subtraction 1.d Yes they do (*) 2a. 600-500R 500x 1.25-600 500x1.25= 625 625-600=25 25=x 2.b 600-500R 500x 1.75-600 500x1.75=875 875-600=275 275= x 2c. I used multiplication first then i used subtraction (*)

2d. Yes it does match with my calculations 3a. 600-500R R=1.5 / 500x1.5-600 1.5x500= 750 750-600=150 X=150 3b. 600-500R 500x1.1-600 500x1.1=550 550-600=-50 -50=X 3c. I used the same operation as questions 1 and 2. I used multiplication first and then subtraction. 4a. P=2.5(600-500R)-500 2.5(600-500R(1.5))-500 500x1.5=750 750-600=150 150x2.5=375 375-500=125 X=125 P=2.5(600-500R)-500 2.5(600-500(1.1) -500 500x1.1=550 550-600=-50 -50x2.5-125 125-500=-375 X=--375 4b. The two equations are very similar but you just have to go through a different process. 5. You would use a acronym called PEMDAS. It stands for Parenthesis, exponents, multiplication, division, addition, and subtraction. For 40- 5x + 7y you would use multiplication, addition, and then subtraction.

Homework:
Collected: Nothing Assigned: ACE: 23-2, 28, 29, 31 1.2 and 1.3

Problem 1.2 Dividing J.K

Feb 24, 08
Note:

Fractions have different ways of writing equations.

A1 Concession will be 250 dollar

A2 Concession per person sould be 0.83 dollar

B

C 250: 0.50 dollar 350: 1.07 dollar 425: 1.32 dollar
 * Visitor || Average Profit ||
 * 100 || - 2.5 ||
 * 200 || 0 ||
 * 300 || 0.83 ||
 * 400 || 1.25 ||
 * 500 || 1.5 ||
 * 600 || 1.66 ||
 * 700 || 1.78 ||
 * 800 || 1.87 ||

D You first multiply the visitors number by 2.50 than subtract this number by 500. When you are done, you take the number and divide it by the number of visitors.

E I agree with both of equations because 1/V (2.50V - 500) and (2.50V -500) / V are same as 2.50 V - 500 / V

1.2 Follow up

1 a When X is 25, the answer would be 7 dollars per person. b I entered (100 + 3X) / X

2 a when X is 10, the answer would be 0.69 dollars per person b I entered 3X/ (4X+3)

3 there are always different ways of making equations for fractions but the thing that you should do before dividing is making brackets for both denominator and numinator.

JH-1.3 Working with exponents

2008/02/18 Day 55
Notes: The equation 5x-0.1x^2 is for the height of the first line of the arch from a specific base at the ground. The base in feet is the x while the height is the y. Problem: 5x-0.1(10^2) (5x10)-(0.1x100) 50-10 40ft 30 ft=60ft 3. 50 feet 5x-0.1x^2 50ft=0ft I used the PEMDAS order of operations. These include exponents, multiplication, subtraction. I first did the exponents then the multiplication and lastly subtraction.
 * 1) Use the equation to find the height of the arch at these distances from the left base. Do your calculations without using a calculator.
 * 2) 10 feet 5x-0.1x^2
 * 1) 30 feet 5x-0.1x^2
 * 1) What operations did you perform to calculate your answers for part A? In what order did you perform these operations?
 * 1) Check your answers for part A by using a graphing calculator to help you make a table.

Use this second expression to calculate the heights for the x values given in part A. 10 ft 0.1x(50-x) (0.1x10)(50-x) 1(50-10) 1x40 40 The same applies to other x distances. 2. In what order did you perform the operations? As you can see from my work the first operation was multiplication then I subtracted the x from 50 and lastly I multiplied the two parts. Follow up: 1.a. Estimate the number of visitors 1 month after the roller-coaster ride opened. About 1200 visitors. b. Estimate the number of visitors 5 months after the roller-coaster ride opened. About 2490 visitors. c. Estimate the number of visitors 12 months after the roller-coaster ride opened. About 8920 visitors. 2. What operations did you perform to find your answers in question 1? In what order did your perform the operations. I used multiplication to find the exponent and then multiplied the 1000 with the other part of the equation. 3.Check your answers to part c of question 1 by entering an expression into your calculator and pressing enter. What expression did your enter? 1000*(1.2^x)
 * Base distance in feet(x) || Height of arch in feet(y) ||
 * 0 || 0 ||
 * 10 || 40 ||
 * 20 || 60 ||
 * 30 || 60 ||
 * 40 || 40 ||
 * 50 || 0 ||
 * 1) 1. The expression 5x-0.1x^2 is equivalent to the expression 0.1x(50-x).

Homework:
Collected: none Assigned: 23-27, 28, 29, 31 odds (30**,36)

ZH - 2.1 Tiling Pools**

- Many possible equations for one problem. - There are also many equations for this problem - The number of tiles that make up a border for a pool can be thought in different ways - This can be expressed through equations or diagrams

Problem 2.1- Tiling pools
A) Make sketches on grid paper to help you figure out how many tiles are needed for the borders of square pools with sides of length 1,2,3,4,6, and 10 feet. Record your results in a table.


 * Pool length || Side Tiles ||
 * 1 || 4 ||
 * 2 || 8 ||
 * 3 || 12 ||
 * 4 || 16 ||
 * 5 || 20 ||
 * 6 || 24 ||
 * 7 || 28 ||
 * 8 || 32 ||
 * 9 || 36 ||
 * 10 || 40 ||

Grid Paper.pdf B) Write an equation for the number of tiles, N, needed to form a border for a square pool with sides of length s feet.

There actually is more than one equation for this situation. Each of these involves looking at the border differently. The equation I came up with first is N=4S+4. This equation lets you look at the situation in a different way. The 4S gives you the amount of tiles needed for you to cover up the edges of the pool. The +4 shows you that there are always 4 corner tiles that are not along the edge. C) Try to write at least one more equation for the number of tiles needed for the border of the pool. How could you convince someone that your expressions for the number of tiles are equivalent?

There is another equation that I can think of. This equation is N=2(2S+2). This looks at the border in a different way. It does exactly as my other equation did except it breaks it into halves. The part in the parentheses is what it takes to get half of the border. Then when you multiply it by 2 then you get the full border. I can prove that my equations are equivalent by making a table or a graph of them.

2.1 follow up

1) Make a table and a graph for each equation you wrote in part a of Problem 2.1. Do the table and the graph indicate that the equations are equivalent? Explain.
 * Pool length || Side Tiles ||
 * 1 || 4 ||
 * 2 || 8 ||
 * 3 || 12 ||
 * 4 || 16 ||
 * 5 || 20 ||
 * 6 || 24 ||
 * 7 || 28 ||
 * 8 || 32 ||
 * 9 || 36 ||
 * 10 || 40 ||

2) Is the relationship between the side of the pool and the number of tiles linear, quadratic, exponential, or none of these? Explain your reasoning. This is a linear pattern. There are mainly three things that tell me that this is a linear relationship. I know this because of the table, the graph and the equation. The equation follows the y=mx+b patter. The graph is a straight line, and the table follows a constant patter of adding by 4.

3) a. Write an equation for the area of the pool, A, in terms of the side length, s. The process of finding the pools area is the same as finding the area of a square. This pool is a square. The formula for finding a square is x^2. The side lengths of the pool are represented by the variable s. In this case the equation will be s^2. b. Is the equation you wrote linear, quadratic, and exponential or none of these? Explain. This equation is not a linear equation because it has an exponent. This is now a quadratic relationship. I know this because this power of the exponent is 2. If a number in an equation is taken up to the power of 2 then the equation is a quadratic equation.

4) a. Write an equation for the combined area of the pool and its border, C, in terms of the side length, s. We know how to find the area of a pool. The formula is x squared. In this case the way to find the area of the pool is s squared. If we add this formula to an equation for the border than we can get the total area. The equation can be N= (4S+4)+s^2. b. Is the equation you wrote linear, quadratic, and exponential or none of these? Explain. This equation is not a linear equation anymore because it has an exponent. This is now a quadratic relationship. I know this because this power of the exponent is 2. If a number in an equation is taken up to the power of 2 then the equation is a quadratic equation.

Homework:
Collected: nothing Assigned: Ace2, : 1, 3, 19, 23

PD

2.2 Thinking in Different Ways
Takashi’s expression for the //n// of tiles as the side increases by //s// is: 4s+4 And it can be illustrated like the following: A. Stella wrote another expression: 4(s+1) This expression can be illustrated like this:

B. Jeri’s expression: s+s+s+s+4 This expression can be illustrated like this:

C. Sal’s expression: 2s+2(s+2) This expression can be illustrated like this:

D. Jackie’s expression can be illustrated like this: 4(s+2)-4 This expression can be illustrated like this:

E. Each of the expression from parts A-D are equivalent to Takashi’s equation because only the way of writing the equation and the thinking of each person is different, the overall answer gives us the exact same answer each time. (look at Follow Up 1)

Follow Up:
1. Evaluate each of the five expressions given in the problem for s=10. a.) Takashi’s expression: 4s+4 4(10)+4 40+4 44

b.) Stella’s expression: 4(s+1) 4(10+1) 4(11) 44

c.) Jeri’s expression: s+s+s+s+4 10+10+10+10+4 40+4 44

d.) Sal’s expression: 2s+2(s+2) 2(10)+2(10+2) 2(10)+2(22) 20+24 44

e.) Jackie’s expression: 4(s+2)-4 4(10+2)-4 4(12)-4 48-4 44

Since all of the answers of all of the expressions are the same, I can say that all of these expressions are equivalent.

2. Make a table and a graph for each of the five expressions. Takashi: 4s+4 Stella: 4(s+1) Jeri: s+s+s+s+4 Sal: 2s+2(s+2)
 * Sides || 1 || 2 || 3 || 4 || 5 || 6 || 7 || 8 || 9 || 10 ||
 * # of blocks || 8 || 12 || 16 || 20 || 24 || 28 || 32 || 36 || 40 || 44 ||
 * Sides || 1 || 2 || 3 || 4 || 5 || 6 || 7 || 8 || 9 || 10 ||
 * # of blocks || 8 || 12 || 16 || 20 || 24 || 28 || 32 || 36 || 40 || 44 ||
 * Sides || 1 || 2 || 3 || 4 || 5 || 6 || 7 || 8 || 9 || 10 ||
 * # of blocks || 8 || 12 || 16 || 20 || 24 || 28 || 32 || 36 || 40 || 44 ||

Jackie: 4(s+2)-4
 * Sides || 1 || 2 || 3 || 4 || 5 || 6 || 7 || 8 || 9 || 10 ||
 * # of blocks || 8 || 12 || 16 || 20 || 24 || 28 || 32 || 36 || 40 || 44 ||


 * Sides || 1 || 2 || 3 || 4 || 5 || 6 || 7 || 8 || 9 || 10 ||
 * # of blocks || 8 || 12 || 16 || 20 || 24 || 28 || 32 || 36 || 40 || 44 ||

Graphs for all of the Expressions:

All of these evidences make it clear that all of the expression written by the students are equivalent. The data on the tables are exactly the same, and the graph points are exactly the same as well.

Homework
Collected: MR # 1 Assigned: 2.1 and 2.2 with follow ups Ace (2) 1, 3, 19, 23 Check up: redos

QR

Notes:
There is always more than one equation when dealing with areas of rectangles.

2.3 Diving In
A. 30(50+20) and 30(50)+30(20) are both methods for calculating the area of this pool. 30(50+20) is more efficient because it requires less steps than the other one. B. 30(25+10) and 30(25)+30(10) are both methods for calculating the area of this pool. 30(25+10) is more efficient because it requires less steps than the other one. C. r(25+15) and r(25)+r(15) are both methods for calculating the area of this pool. r(25+15) is more efficient because it requires less steps than the other one. D. r(s+t) and r(s)+r(t) are both methods for calculating the area of this pool. r(s+t) is more efficient because it requires less steps than the other one.

2.3 Follow-up
1a. Triangles i, iii, and v have an area of 5(4+x). b. The equivalent expanded form of 5(4+x) is 5x+4x because they are both describing the area of the same rectangle. one is in factored form the other is in expanded form. c. Triangles i, ii, and iv have an area of 5x+4x d. The equivalent factored form of 5x+4x is 5(4+x) because they are both describing the area of the same rectangle, one is expanded and one is factored. 2a. b. Therefor the expressions 5(x+2) and 5x+10 are equivalent. 3a. 1.5(4+x)'s equivalent expanded form is 1.5x+6. b. x(3+5)'s equivalent expanded form is 3x+5x. 4a. 27+36x's equivalent factored form is 36(.75+x) b 2x+7x's equivalent factored form is x(2+7) 5 . NOT FINISHED Homework (Heading 3) Collected: (normal) Assigned: (normal)
 * Y=5(x+2) ||
 * X || Y ||
 * 0 || 0 ||
 * 1 || 15 ||
 * 2 || 20 ||
 * 3 || 25 ||
 * 4 || 30 ||
 * 5 || 35 ||
 * Y=5x+10 ||
 * X || Y ||
 * 0 || 0 ||
 * 1 || 15 ||
 * 2 || 20 ||
 * 3 || 25 ||
 * 4 || 30 ||
 * 5 || 35 ||

SA

3.1 Walking Together
For each student, write an equation for the amount of money the student will raise if he or she walks x miles. Then write an equation for the total amount the three-person team will raise if they walk x miles. A) Leanne= 16 sponsors to pledge $1 for each mile she walk. $=16X Gilberto= 7 sponsors to pledge $2 for each mile he walks. $=7(2X) Alana= 11 sponsors to pledge $5 plus $0.50 for each mile she walks. $=11(0.5X+5) Total= $=16X+7(2X)+11(0.5X+5) or $= 34(2.5X)+5 B) 1.Excluding the $55, the team will raise $2.50 per mile they walk. 2. Using my answer from part 1, another equation for the total amount the team will raise if they walk X miles is $=34(2.5X) C) The two equations from question A & B are the same because the commutative properties state that the order which the values are added or multiplied do not matter, so therefor the equations are the same.

Follow Up (all the variables are 2 for the problems below) 1. 3x+5x and 8x are equivalent because 16 and 16 are the same. 2. 3x+5 and 8x are not equivalent because 11 and 16 aren't the same. 3. 4(x+7) and 4x+7 are not equivalent because 36 and 15 aren't the same. 4. 5(x+2) and 5x+10 are equivalent because 20 and 20 are the same. 5. 12+8x and 4(3+2x) are equivalent because 28 and 28 are the same. 6. 4x+x+2x and 8x are not equivalent because 14 and 16 aren't the same. 7. 7+5x and 5x+7 are equivalent because 17 and 17 are the same. 8. 3x+8 and 8x+3 are not equivalent because 14 and 19 aren't the same. 9. 5x+3x+4x and 4x+5x+3x are equivalent because 24 and 24 are the same. 10. 6+2t and 2(t+3) are equivalent because 10 and 10 are the same. 11. 2(L+2)+2W and 2L + 2W+4 are equivalent because 12 and 12 are the same. 12. 2L+2W+4 and 2(L+W+2) are not equivalent because 12 and 24 aren't the same.

Homework
Collected: None Assigned: ACE3: 1-11 odds, (45-51 odds)

AT - 3.2 Estimating Profit Initials (normal) Date-Day(heading 2) Notes (include the Essential Question)(Heading 3) Problem Number and Title (Heading 3) Homework (Heading 3) Collected: (normal) Assigned: (normal)

TA - 3.3 Finding the Area of a Trapezoid Initials (normal) Date-Day(heading 2) Notes (include the Essential Question)(Heading 3) Problem Number and Title (Heading 3) Homework (Heading 3) Collected: (normal) Assigned: (normal)

NR - 3.4 Writing Quadratic Equations Initials (normal) Date-Day(heading 2) Notes (include the Essential Question)(Heading 3) Problem Number and Title (Heading 3) Homework (Heading 3) Collected: (normal) Assigned: (normal)

GC - 4.1 Comparing Costs Initials (normal) Date-Day(heading 2) Notes (include the Essential Question)(Heading 3) Problem Number and Title (Heading 3) Homework (Heading 3) Collected: (normal) Assigned: (normal)

HK

Notes:
Symbolic Method of solution - Apply any mathematical operatoin to both sides.

Problem 4.2 Solving Linear Equations
A. 100+4x = 25+7x 100+4x-4x = 25+7x-4x 1.Minus 4x from both sides so that it is equal 100 = 25+3x 100-25 = 25+3x-25 2. Minus 25 from both sides so that it is equal 75 = 3x 75/3 = 3x/3 3. Divide by 3 each side so that it is equal 25 = x x = 25 B. Another way I could show this solution beginning with a different step: 100+4x = 25+7x 100+4x-25 = 25+7x-25 75+4x = 7x 75+4x-4x = 7x-4x 75 = 3x 75/3 = 3x/3 25 = x x = 25 C. I can check that x = 25 is the correct solution, by substituting 25 in for all the xs in the equation. If x = 25 100+4x 25+7x 100+4(25) 25+7(25) 100+100 25+175 200 200

Problem 4.2 Follow Up 11x-12 = 30+5x 11x = 42+5x 6x = 42 x = 7 1. 11x-12 = 30+5x 11x-12+12 = 30+5x+12 11x = 42+5x 11x-5x = 42+5x-5x 6x = 42 6x/6 = 42/6 x = 7 2. I can check if x = 7, by substituting 7 for all the xs in the equation. 11x-12 = 30+5x 11(7)-12 = 30+5(7) 77-12 = 30+35 65 = 65 3. I could use a graph and a table to solve the equation. On the graph I could see where the two lines (from the two equations) intercept. Where the intercept is, I could find x and that would be the x for the equation. On a table I could look at what x is, when the y values of the two equations are the same.

Collected:
ACE 3: 1-11 odds, (45-51 odds), 13-27 odds, 39 and MR #3

Assigned:
Problem 4.1 & 4.2, ACE 4: 5-9 odds,16

AD

4.3 by Austin Dun
A. 7x+15=12x+5 7x+15-7x=12+5-7x 15=5x+5 15-5=5x+5-5 10=5 10/5=5x/5 2=x

B. 14-3x=1.5x+5 14-3x+3x=1.5x+5+3x 14=4.5x+5 14-5=4.5x+5-5 9=4.5x 9/4.5=4.5x/4.5 2=x

C. -3x+5=2x-10 -3x+5+3x=2x-10+3x 5=5x-10 5+10=5x-10+10 15/5=5x/5 3=x

D. 3+5(x+2)=7x-1 3+5x+10=7x-1 13+5x=7x-1 13+5x-5x=7x-1-5x 13=2x-1 13+1=2x-1+1 14/2=2x/2 7=x

E. A. 14+15=29 24+5=29 B. 14-6=8 Checked. 3+5=8 C. -9+5=-4 6-10=-4 D. 3+(5*9)=3+45=48 49-1=48

F. Strategies on solving linear equations: - Doing the same to both sides. - Bringing things with and without x’s to separate sides. - Expand anything in factored form first.

Follow Up:

1. A. 7.5x-23.5+23.5=51.5+23.5 7.5x/7.5=75/7.5, so x=10 B. 7.5x-23.5+23.5=0+23.5 7.5x/7.5=23.5/7.5, so x=3.133… C. 7.5x-23.5=-30 7.5x-23.5+23.5=-30+23.5 7.5x/7.5=-6.5/7.5, so x=0.866…

2. 7=3x+1. I doubt anyone else wrote this because there are an infinite number of equations where x=2

3. A. To get $200, x=300 I used a graph to find this, but I could set P=200=3x-100-2x P=200=x-100, so x=300 B. 100 boxes must be sold to break even. P=0=3x-100-2x P=0=x-100, so x=100

Homework: ACE4: 5-9 odds, 16 Assigned: ACE4:3, 4, 13**
 * Collected: Nothing

AC

Notes:
Solving a quadratic doesn't have to use graphs and tables, we could solve it by using a symbolic method. The solution to x² + 5x = 0 are called the **roots** of the equation y = x² + 5x. Every quadratic equation may have 0 to 2 solutions.

Problem 4.4 Solving Quadratic Equations
A. x² + 3x = x (x+3)

B. x² + 3x = 0 x² + 3x-3x = 0 - 3x x² = -3x x² ÷ x = -3x ÷ x x = -3 I know I found all the the possible solutions for x² + 3x = 0 already because a quadratic equation could have 0-2 solutions, and the solutions if this equations are 0 and -3.

C. The x-intercepts of y= x² + 3x are 0 and -3. The answer of part B helps me answer the question because the number of x means the x-intercept.

D. The expanded from of x² + 3x is more useful for finding the roots because we could easily find out the roots by using symbolic method.

E 1. y = 4x² - 8x 0 = 4x² - 8x 0+ 8x = 4x² - 8x + 8x 8x = 4x² 8x ÷ 4 = 4x² ÷ 4 2x = x² 2x ÷ x = x² ÷ x 2 = x

y = 4x (x-2) 0 = 4x (x-2) 0 = 4x² - 8x 0 + 8x = 4x² - 8x + 8x 8x = 4x² 8x ÷ 4x = 4x² ÷ 4x 2 = x

2. y = 6x (5-2x) 0 = 6x (5-2x) 0 = 30x - 12x² 0 + 12x² = 30x - 12x² + 12x² 12x² = 30x 12x² ÷ 12x = 30x ÷ 12x x = 2.5

y = 30x - 12x² 0 = 30x - 12x² 0 + 12x² = 30x - 12x² + 12x² 12x² = 30x 12x² ÷ 12x = 30x ÷ 12x x = 2.5

F 1. x² + 4.5x = 0 x (x+4.5) = 0 0 = x ² + 4.5x 0 - 4.5x = x² +4.5x - 4.5x -4.5x = x² -4.5x ÷ x = x² ÷ x -4.5 = x

2. x² - 9x = 0 x (x-9) = 0 0 = x² - 9x 0 + 9x = x² - 9x + 9x 9x = x² 9x ÷ x = x² ÷ x 9 = x

3. -x² +10x = 0 x (-x+10) = 0 0 = -x² + 10x 0 - 10x = -x² + 10x -10x -10x = x² -10x ÷ x = x² ÷ x -10 = x

Problem 4.4 Follow-Up
1. F1. x = -4.5 x² + 4.5x = 0 -4.5x² + 4.5 (-4.5) = 20.25 - 20.25 = 0

F2. x = 9 x² - 9x = 0 9² - 9 (9) 81 - 81 = 0

F3. x=10 -x² + 10x -10² + 10 (10) -100 + 100 = 0

2. F1. y = x² + 4.5x F2. y = x² - 9x F3. y = -x² + 10x 3a. i ) (2x+1) (x+5) = 2x² + 10x + x + 5 = 2x² + 11x + 5 ii) (x+2) (x-2) = x² - 2x + 2x -4 = x² -4
 * x || y ||
 * -5 || 2.5 ||
 * -4 || -2 ||
 * -3 || -4.5 ||
 * -2 || -5 ||
 * -1 || -3.5 ||
 * 0 || 0 ||
 * 1 || 5.5 ||
 * x || y ||
 * -1 || 10 ||
 * 0 || 0 ||
 * 1 || -8 ||
 * 2 || -14 ||
 * 3 || -18 ||
 * 4 || -20 ||
 * 5 || -20 ||
 * 6 || -18 ||
 * 7 || -14 ||
 * 8 || -8 ||
 * 9 || 0 ||
 * x || y ||
 * -1 || -11 ||
 * 0 || 0 ||
 * 1 || 9 ||
 * 2 || 16 ||
 * 3 || 21 ||
 * 4 || 24 ||
 * 5 || 25 ||
 * 6 || 24 ||
 * 7 || 21 ||
 * 8 || 16 ||
 * 9 || 9 ||
 * 10 || 0 ||

b. I would use (x+2) (x-2) to predict the x-intercepts of the related graph. The intercepts are -2 and 2. I would choose this equation because it is more easier to find out the x-intercepts.

4. x² + 8x + 12 = (x+8) (x+1.5)

Homework:
Collected: ACE 4 # 3, 4, 13 Assigned: Problem 4.4, MR #4, ACE 4 # 20, 21

VL Day 64

Problem 5.1 A. The dimensions of one longer rod using the unit rod would be 1 x 5.

B.
 * = Number of rods stacked ||= Surface Area (cm²) ||
 * = 1 ||= 22 ||
 * = 2 ||= 36 ||
 * = 3 ||= 50 ||
 * = 4 ||= 64 ||
 * = 5 ||= 78 ||
 * = 6 ||= 92 ||
 * = 7 ||= 100 ||
 * = 8 ||= 120 ||
 * = 9 ||= 134 ||
 * = 10 ||= 148 ||

C. The equation to find the surface areas would be: y=2(5x)+(2 x 5)+2(x+x-1)

Follow-up 1a. All the expressions should be equivalent, because they all give the same 1b.You could check it be entering the expression in the graphing calculator. Usually on a graph you would only see one line, because it is the same. On a table all they x ands y are the same numbers.

2a. The surface area of 12 rods would be 176 cm² 2b. The surface area of 20 rods would be 288 cm²

3. To change the length of each rod on the equation you would just need to change the 5 into another length.

4. This equation is a linear equation, because it goes by 14.

HW: ACE5: 1,3,4