Say+it+With+Symbols

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Say it With Symbols
[| journal_homework_record_Math_8_say_it_with_symbols 0708.doc] 

Journal 1.1
A. Number of Children Equation : p = 100+10a+8c
 * Number of Adults / Number of Children || 20 || 40 || 60 || 80 || 100 ||
 * 10 || 360 || 520 || 680 || 840 || 1000 ||
 * 20 || 460 || <span style="font-size: 12pt; font-family: 굴림">620 || <span style="font-size: 12pt; font-family: 굴림">780 || <span style="font-size: 12pt; font-family: 굴림">940 || <span style="font-size: 12pt; font-family: 굴림">1100 ||
 * <span style="font-size: 12pt; font-family: 굴림">30 || <span style="font-size: 12pt; font-family: 굴림">560 || <span style="font-size: 12pt; font-family: 굴림">720 || <span style="font-size: 12pt; font-family: 굴림">880 || <span style="font-size: 12pt; font-family: 굴림">1040 || <span style="font-size: 12pt; font-family: 굴림">1200 ||
 * <span style="font-size: 12pt; font-family: 굴림">40 || <span style="font-size: 12pt; font-family: 굴림">660 || <span style="font-size: 12pt; font-family: 굴림">820 || <span style="font-size: 12pt; font-family: 굴림">890 || <span style="font-size: 12pt; font-family: 굴림">1140 || <span style="font-size: 12pt; font-family: 굴림">1300 ||

Steps: 10 adults and 20 children p = 100+10(10)+8(20) = 100+100+160 = $360 20 adults and 20 children p = 100+10(20)+8(20) = 100+200+160 = $460 30 adults and 20 children p = 100+10(30)+8(20) = 100+300+160 = $560 40 adults and 20 children p = 100+10(40)+8(20) = 100+400+160 = $660 10 adults and 40 children p = 100+10(10)+8(40) p = 100+100+320 p = $520 20 adults and 40 children p = 100+10(20)+8(40) p = 100+200+320 p = $620 30 adults and 40 children p = 100+10(30)+8(40) p = 100+300+320 p = $720 40 adults and 40 children p = 100+10(40)+8(40) p = 100+400+320 p = $820 10 adults and 60 children p = 100+10(10)+8(60) p = 100+100+480 p = $680 20 adults and 60 children p = 100+10(20)+8(60) p = 100+200+480 p = $780 30 adults and 60 children p = 100+10(30)+8(60) p = 100+300+480 p = $880 40 adults and 60 children p = 100+10(40)+8(60) p = 100+400+480 p = $980 10 adults and 80 children p = 100+10(10)+8(80) p = 100+100+640 p = $840 20 adults and 80 children p = 100+10(20)+8(80) 30 adults and 80 children p = 100+10(30)+8(80) p = 100+300+640 p = $1040 40 adults and 80 children p = 100+10(40)+8(80) p = 100+400+640 p = $1140 10 adults and 100 children p = 100+10(10)+8(100) p = 100+100+800 p = $1000 20 adults and 100 children p = 100+10(20)+8(100) p = 100+200+800 p = $1100 30 adults and 100 children p = 100+10(30)+8(100) p = 100+300+800 p = $1200 40 adults and 100 children p = 100+10(40)+8(100) p = 100+400+800 p = $1300

B. The pattern of the table is that as you go down the row, the value increases by $100 in each colum. In the equation p = 100+10a+8c, 10a and 8c creates the pattern of the table.

C. In the equation p = 100+10a+8c, 100 is the constant price that everyone pays for coming in, 10 is the amount of money per adult, and 8 is the amount of money per child. So, if you add all of these, it would add up to the total price.

D. For this situation, to perform the operation of equation p = 100+10a+8c, you first need to multiply the variables, and then add them together.

Follow up

1a. If 300 people visit the park, about $250 concession profit will be made. -> p = 2.5(300)-500 p = 750-500 p = $250

b. If 600 people visit the park, about $1000 concession profit will be made. -> p = 2.5(600)-500 p = 1500-500 p = $250

c. For this equation, I performed the multiplications of the variables first, and then did the subtraction.

d. Yes, the results agree with the results I found by doing the calculations by hand.

2a. If the probability of rain is 25%, about 475 people will visit the park. -> v = 600-500(0.25) v = 600-125 v = 475

b. If the probability of rain is 75%, about 225 people will visit the park. -> v = 600-500(0.75) v = 600-375 v = 225

c. To perform this equation, I did the multiplication first and then did the subtraction.

d. The results agree with the results I found by doing the caluculations by hand.

3a. If the probability of rain is 50%, about $375 profit will be made from the concession stands. -> v = 600-500(0.5) v = 600-250 v = 350 -> p = 2.5(350)-500 p = 875-500 p = $375

b. If the probability of rain is 10%, about $875 profit will be made from the concession stands. -> v = 600-500(0.10) v = 600-50 v = 550 -> p = 2.5(550)-500 p = 1375 - 500 p = $875

c. To perform the operations, in this equation, I did the multiplication first and then did the subtraction.

4a. If the probability of rain is 50%, the profit would be $375, and if the probability of rain is 10%, the profit will be $875. -> p = 2.5[600-500(0.50)]-500 p = 2.5(600-250)500 p = 2.5(350)-500 p = 875-500 p = $375

b. The answers for part a is the exact same with #3.

c. To evaluate 40-5x+7y when x=3 and y=5, I would first do the two multiplications, add, and then subtract. -> b = 40-5(3)+7(5) b = 40-15+35 b = 40-50 b = -10

Homework:
Problem 1.1 and follow up ACE1) #1~5 odds, 18~22, and 24

None**
 * Collected:

=Megan Nguyen=

Journal 1.2
A. 1. If 300 people visited the park, about how much concession profit will be made? a = 2.50V - 500 v = 300 a = 2.50(300) - 500 multiply a = 750 - 500 subtract a = 250 If 300 people visited the part, the concession profit would be $250 2. About how much concession profit will be made per visitor? a = (2.50V-500)/V v = 300 a = (2.50 x 300 - 500) / 300 parentheses - multiply a = (750 - 500)/300 parentheses - subtract a = 250/300 divide a = $.83

About $.83 concession profit will be made per visitor.

B. Copy and complete the table below to show the average per-visitor concession profit for various numbers of visitors. Do your calculations without using your calculator.


 * Visitors || 100 || 200 || 300 || 400 || 500 || 600 || 700 || 800 ||
 * Average Profit ($) || - 2.5 || 0 || .83 || 1.25 || 1.5 || 1.6 || 1.79 || 1.875 ||

v = 100 a = (2.5 x 100 - 500) / 100 a = (250 - 500) / 100 a = -250 / 100 a = -2.5

v = 200 a = (2.5 x 200 - 500) / 200 a = (500 - 500) / 200 a = 0 / 200 a = 0

v = 300 a = (2.5 x 300 - 500) / 300 a = (750 - 500) / 300 a = 250 / 300 a = .833

v = 400 a = (2.5 x 400 - 500) / 400 a = (1000 - 500) / 400 a = 500 / 400 a = 1.25

v = 500 a = (2.5 x 500 - 500) / 500 a = (1250 - 500) / 500 a = 750 / 500 a = 1.5

v = 600 a = (2.5 x 600 - 500) / 600 a = (1500 - 500) / 600 a = 1000 / 600 a = 1.66

v = 700 a = (2.5 x 700 - 500) / 700 a = (1750 - 500) / 700 a = 1250 / 700 a = 1.79

v = 800 a = (2.5 x 800 - 500) / 800 a = (2000 - 500) / 800 a = 1500 / 800 a = 1.875

C. Find the average per-visitor concession profit for 250, 350, and 425 visitor. v = 250 a = (2.5 x 250 - 500) / 250 a = (625 - 500) / 250 a = 125 / 250 a = 0.5 The average profit per 250 customers is $0.5

v = 350 a = (2.5 x 350 - 500) / 350 a = (875 - 500) / 250 a = 375 / 250 a = 1.07 The average per-visitor concession profit for 350 visitors is about $1.07

v = 425 a = (2.5 x 425 - 500) / 425 a = (1062.5 - 500) / 425 a = 562.5 / 425 a = 1.32 The average per-visitor concession profit for 425 visitor is about $1.32

D. What mathematical operation do you need to preform to calculate the average per-visitor profit for a given number of visitors? In what order must you perform the operations? First, you work inside the parentheses. You multiply V by 2.5, then you take that value and subtract it by 500. Then you divide it by V. So you, work in the parentheses. Within the parentheses, you multiply, subtract, then, once you have simplified the equation within the parentheses, you divide.

E. The Water Works business manager claims that the average concession profit per visitor can also be calculated with either of these equations: a = 1/V (2.50V - 500)

a = (2.50V - 500) / V

Do you agree? Explain.

To test whether the equations are equivalent, we test them v = 500 a = 1/V (2.50V-500) a = 1/500 (2.5 x 500 - 500) a = 1/500 (1250 - 500) a = 1/500 x 750 a = 1.5

a = (2.50V - 500) / V a = (2.5 x 500 - 500) / 500 a = (1250 - 500) / 500 a = 750 / 500 a = 1.5 Yes, the two equations can also be used to calculate the average concession profit becuase they produce the same answer as the original equation. They are all equivalent but expressed differently.

Follow-up 1. a. Use the example of calculating the average per-visitor concession profit as a guide to help you evaluate __100 + 3x__ when x = 25. x x = 25 y = (100 + 3x) / x y = (100 + 3 x 25) / 25 y = (100 + 75) / 25 y = 175 / 25 y = 7

b. Check your answer by entering an expression into your graphing calculator and pressing 'ENTER'. What expression did you enter? My calculations are right. I entered (100 + 3 x 25) / 25 = 7

2. a. Use the example of calculating the average per-visitor concession profit as a guide to help you evaluate __3x__ when x = 10. 4x + 3 x = 10 y = (3x) / (4x + 3) y = (3 x 10) / (4 x 10 + 3) y = (30) / (40 + 3) y = 30 / 43 y = .69767 ...

b. Check your answer by entering an expression into your graphing calculator and pressing 'ENTER'. What expression did you enter? My calculations were right. I entered (3 x 10) / (4 x 10 + 3).

3. When evaluation an expression that is in fraction form, in what order should you preform the operations? First you simplify the numerator and denominator acting as the numerator and denominator are in parentheses. So once you have simplified the numerator or denominator by multiplying or adding or subtracting to get a numeral, you then divide the numerator by the denominator.

Homework:
Journal 1.2 and 1.3 follow up ACE1: 23 - 27, 28, 29,31

=February 23rd, 2008, Day 55=

Notes:
None

Journal 1.3

Use the equation to find the height of the arch at these distances from the left base. y = 5x - 0.1x^2

A1) y = 5(10) - 0.1(10^2) y = 50 - 10 y = 40 2) y = 5(30) - 0.1(30^2) y = 150 - 0.1(900) y = 150 - 90 y = 60 3) y = 5(50) - 0.1(50^2) y = 250 - 0.1(2500) y = 250 - 250 y = 0 B. The operations I performed to calculate my answers were multiplication and subtraction. The orders I operated them in were first multiplication, exponents, and then the subtraction.

C.
 * Distance || Height of arc ||
 * 10 || 40 ||
 * 30 || 60 ||
 * 50 || 0 ||

D1) Use this equivalent expression to the first part to find the height of the arch at these distances from the left base for the x values given in part A.

y = 0.1(10)(50-10) y = 1(40) y = 40

y = 0.1(30)(50-30) y = 3(20) y = 60

y = 0.1(50)(50-50) y = 5(0) y = 0

2) I operated the expression in the order of multiplication, subtraction, and then subtraction.

Follow up
The business manager wrote the equation V = 1000(1.2^m) to estimate the daily number of visitors //m// months after the roller coaster ride opened.

1a) Q: Estimate the number of visitors 1 month after the roller-coaster ride opened.

V = 1000(1.2^1) V = 1000(1.2) V = 1200

b) Q: Estimate the number of visitors 5 months after the roller-coaster ride opened.

V = 1000(1.2^5) V = 1000(2.48) V = 2480

c) Q: Estimate the number of visitors 12 months after the roller-coaster ride opened.

V = 1000(1.2^12) V = 1000(10.69) V = 10690

2) I operated the expression in the order of exponent and then multiplying.

3) I entered the expression V = 1000(1.2^m) in to the graphing calculator to check my answer.

Homework:
Journal 1.2 and 1.3 follow up ACE1) #23~27, 28, 29,31 MN

Notes (include the Essential Question):
None

Problem 2.1: A. Make sketches on grid paper to help you figure out how many tiles are needed for the borders of square pools with side length of 1, 2, 3, 4, 6, and 10 feet. Record your results in a table. Sketches:
 * Side Length || Number of Border Tiles ||
 * 1 || 8 ||
 * 2 || 12 ||
 * 3 || 16 ||
 * 4 || 20 ||
 * 6 || 28 ||
 * 10 || 44 ||

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. The equation for the relationship between the nuimber of tiles needed to form a border for a square pool is: N = 4(s+1)

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? An equivalent equation would be N = 4s + 4. I know this is equivalent becuase I used the distributive property. I used to factor of 4 to multiply it to s and 1. We can also test both equations. If the answers are equal, it means the equations are equivalent. s = 7 n = 4s + 4 n = 4 x 7 + 4 n = 28 + 4 n = 32

n = 4(s+1) n = 4(7+1) n = 4 x 8 n = 32

Yes, n = 4s + 4 is equivalent to n = 4(s+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 graph indicate that the equations are equivalent? Graph: Comparing n = 4s + 4 and n = 4(s+1) There is another line plot behind that but it can't be seen because it is covered up by the seconded equation I plotted.
 * Side Length || n = 4(s+1) || n = 4s + 4 ||
 * 1 || 8 || 8 ||
 * 2 || 12 || 12 ||
 * 3 || 16 || 16 ||
 * 4 || 20 || 20 ||
 * 5 || 24 || 24 ||
 * 6 || 28 || 28 ||
 * 7 || 32 || 62 ||
 * 8 || 36 || 36 ||
 * 9 || 40 || 40 ||
 * 10 || 44 || 44 ||
 * 10 || 44 || 44 ||

Yes, the table and graph indicate that the equations are equivalent becuase they produce the same value on the same x-value. The y-value are always the same with the same x-value. On the graph, they plot the same linear graph. Both equations produce the same answers.

2. Is the relationship between the side length of the pool and the number of tiles linear, quadratic, exponential, or none of these? Explain your reasoning. The relationship between the side length of the pool and the number of tiles is linear becuase once it is graphed, it forms a straight line. also, it has a constant rate of change, as x increases by 1, y increases by 4.

3. a. Write an equation for the area of the pool, A, in terms of the side length, s. The relationship between the area of the pool and the side length is : A = s². This is because it is a shape of a square. b. Is this equation you wrote linear, quadratic, exponential, or none of these? Explain. This equation is quadratic because s is raised to the power of two. It is the quadratic function, and the difference between each consecutive y-values increase by 2.

4.a. Write an equation for the combined area of the pool and its border, C, in terms of the side length, s. The relationship between the combined area and the side length is C= 3^2 + 4s + 4 or C = (s+2)^2. b. Is the equation you wrote linear, quadratic, exponential, or none of these? Explain. This relationship is quadratic because s is raised to the power of 2 meaning it is the quadratic function. Also, it follows the basic format of a quadratic relationship, y = ax^2 + bx + c.

Homework
Collected: Mathematical Reflection Investigation 1, Assigned: Problem 1.1, Problem 1.2, ACE 2: 1, 3, 19, 23, Portfolio Reflection

RD

Problem 2.2: Thinking in Different Ways
Takashi thought of the pool's border as being composed of four 1-by-s rectangles, each made from s tiles, and four corner squares, each made from one tile. he wrote the expression 4s+4 to represent the total number of border tiles.


 * A.** Stella wrote the expression 4(s+1) to represent the number of border tiles. Draw a picture that illustrates how Stella might have been thinking about the border of the pool.




 * B.** Jeri wrote the expression s+s+s+s+4 to represent the number of border tiles. Draw a picture that illustrates how Jeri might have been thinking about the border of the pool.




 * C.** Sal wrote the expression 2s+2(s+2) to represent the number of border tiles. Draw a picture that illustrates how Sal might have been thinking about the border of the pool.




 * D.** Jackie wrote the expression 4(s+2)-4 to represent the number of border tiles. Draw a picture that illustrates how Jackie might have been thinking about the border of the pool.



- Each expression in parts A-D is equivalent to Takashi's expression, because when this equation is illustrated, and you separate the parts and put them back together again, you will get the same border as Takashi's.
 * E.** Explain why each expression in parts A-D is equivalent to Takashi's expression.

Problem 2.2: Follow-Up

 * 1.** Evaluate each of the five expressions given in the problem for s=10. Can you conclude from your results that all the expressions are equivalent? Explain your reasoning.

Takashi: 4s+4 -> 4(10)+4 -> 40+4=__44__ Stella: 4(s+1) -> 4(10+1) -> 4(11) -> __44__ Jeri: s+s+s+s+4 -> (10+10+10+10)+4 -> 40+4=__44__ Sal: 2s+2(s+2) -> 2(10)+2(10+2) -> 2(10)+2(12) -> 20+24=__44__ Jackie: 4(s+2)-4 -> 4(10+2)-4 -> 4(12)-4 -> 48-4=__44__

All the expressions are equivalent, because are the results are the same.


 * 2.** Make a table and a graph for each of the five expressions. Do the tables and graphs indicate that the expressions are equivalent? Explain.

Takashi: 4s+4
 * Side Length || Pool Border Tiles ||
 * 6 || 28 ||
 * 7 || 32 ||
 * 8 || 36 ||
 * 9 || 40 ||
 * 10 || 44 ||

Stella: 4(s+1)
 * Side Length || Pool Border Tiles ||
 * 6 || 28 ||
 * 7 || 32 ||
 * 8 || 36 ||
 * 9 || 40 ||
 * 10 || 44 ||

Jeri: s+s+s+s+4
 * Side Length || Pool Border Tiles ||
 * 6 || 28 ||
 * 7 || 32 ||
 * 8 || 36 ||
 * 9 || 40 ||
 * 10 || 44 ||

Sal: 2s+2(s+2)
 * Side Length || Pool Border Tiles ||
 * 6 || 28 ||
 * 7 || 32 ||
 * 8 || 36 ||
 * 9 || 40 ||
 * 10 || 44 ||

Jackie: 4(s+2)-4
 * Side Length || Pool Border Tiles ||
 * 6 || 28 ||
 * 7 || 32 ||
 * 8 || 36 ||
 * 9 || 40 ||
 * 10 || 44 ||

All the results for the tables and the graphs are the same. Therefore, they indicate that the expressions are equivalent.

Homework
Collected: Mathematical Reflections: Investigation 1 - Order of Operations and Say It With Symbols Check-Up 1 redo. Assigned: Problem 2.1 + Follow-Up, Problem 2.2 + Follow-Up, ACE 2: # 1, 3, 19 & 23 and Portfolio Piece Self Assessment. Sue Lee

AK

Investigational 3.3 Finding The Area of a Trapezoid
a. Explain each student's method for finding the area. Tua's method for finding the area in a trapezoid was to devide the trapezoid into two triangles than to use the "triangle method" to find the area of both triangles. Once she had both areas she would plus them together to get the area of the trapezoid. Sam's method for finding the area in a trapezoid was to double the trapezoid, join them together to make a parrallelogram. So when you cut the two sides, you get a rectangle than on the rectangle you times width by height and than you join the two cut of triangles to get a square and for the square you times width by height and plus it with the rectangle than when you get the total area you devide by 2. Carlos's method to find the area of a trapezoid would be to cut the trapezoid into two than join the top bit by the bottom bit to make a parraleelogram. Than you do the exact same as Sam's method but not deviding by two at the end.

b.Write an algebraic expression to describe each method. Tua's algebraic expression would be: __(2 x h) + (b x h)__ 2 2 Sam's a;gebraic expression would be: __(b + a) x h__ 2 Carlos's algebraic expression would be: (b + a) x __h__ 2 C. Show that the expressions you wrote in part B are equivalent. Example: a=1 b=3 h=2 Tua's ( a x h ) + (b x h)= (1 x 2) + (3 x 2) = 2 + 6= 8= 4 2 2 2 2 2 2 2 Sam's (b + a) x h= (3 + 1) x 2 = 4 x 2 = 8 = 4 2 2 2 2 Carlos's (b + a) x h = (3 + 1) x 2= 4 x 2 = 8= 4 2 2 2 2 All answers came to 4. So meaning there are all equvialent.

Problem 3.3 follw up.
a. Natasha lost her drawing, but she had written this expression for finding the area of a trapezoid. 1/2h(b-a)+h(a). Use this expression to decide what Natasha's drawing might have looked like. Make a drawing, and use it to help explain Natasha's method for finding the area. Natasha's drawing:

b. Is Natasha's expression equivalent to the three expressions in part b? Explain. a = 1 b = 3 h = 2 1/2h( b - a ) + h (a) = 1 ( 3 - 1 ) + 2 (1)= 1 (2)+ 2 (1)= 2+2 = 4. Since a, b and c had all the name numerbs and came out with the same answer. It proves that they are all equivalent. 2. A trapezoid has a height of 10 centimeters and a base of 9 centimeters and 15 centimeters. Find the area of the trapezoid using each of the four expressions. Tua's Method: (a x h ) + ( b x h ) = ( 9 x 10 ) + ( 15 x 10 ) = 90 + 150= 45 + 75 = 120. 2 2 2 2 2 2 So area is 120. Sam's Method: (b = a) x h = (15 + 9) x 10 = 24 x 10 = 240 = 120. 2 2 2 2 So total are is 120. Carlos's Method. ( b + a) x h = ( 15 + 9 ) x 10 = 24 x 10 = 240= 120 2 2 2 2 So area is 120. Natasha's Method: 1/2h (b-a)+h(a)= 5(15-9) = 10(9)= 5(6)+ 10(9)= 30+ 90= 120. So the area is 120. So overal the total area is 120.

AK 11-March-2008

Investigation 3.4 Writing Quadratic Equations
In parts A-C, you will explore three ways of thinking about this question: If a league has n team plays each of the other teams twice, how many games are played in all? A.Figure out how many games would be played for leagues with 2,3,4,5, and 6 teams. Record your results in a table similar to the one below. Look for a pattern in your table. Use the pattern to write an expression for the number of games played by a league with n teams. The pattern I see is every number of teams times the ssame number of teams minus one. So the expression would be: Games= n(n-1) B. Suppose a sports reporter wants to attend exactly one game in the schedule of n-team league. 1. How many choices does the reporter have for the home team for the game she attends? n-teams 2.Once she has chosen a home team, how many choices does she have for the visiting team? n-1 teams. 3. Use your answers from part 1 and 2 to write an expression for the total number of games the reporter can choose from. n(n-1) C.Suppose you made a table similar to the one on page 41 to record wins and losses for an n-team league. 1. How many cells would your table have? 25 cells. 2. How many cells in the table would not be used for W or L entries? 5 cells. 3. Use your answers from part 1 and 2 to write an expression for the total number of games player. My expression would be : 25-5=20 So basically total teams times total teams minus one= games played. D. In parts a-c, your wrote expressions for the number of games played by an n-team league. Show that these expressions are equivalent. All the equations I wrote were the same = n(n-1)
 * Number of teams || 2 || 3 || 4 || 5 || 6 ||
 * Number of games || 2 || 6 || 12 || 20 || 30 ||
 * || T1 || T2 || T3 || T4 || T5 ||
 * T1 || - || w || w || L || W ||
 * T2 || L || - || L || L || W ||
 * T3 || w || W || - || L || L ||
 * T4 || w || W || W || - || W ||
 * T5 || L || W || W || W || - ||

Problem 3.4 follow up
1. x(-4x+3)= (-4x)x+(3x) 2.(4x+3)(x+7)=(4+3)x(X+7) 3. (x-4.5)(x+6.5) 4.(2x-5)(5x-2) 5.x²-6x= x(x-6x) 6.12x-3x²= 12x-(3x)x

AS - 4.3 Reasoning with Symbols 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)

AS

**Problem 4.3**
In A-D, use the symbolic method to solve the equation

Symbolic Method: 7x + 15 = 12x + 5 12-7x+15 = 12x-12+5 5x+15-15 = 15-5 5x = 10 5x/5 = 10/5 __x = 2__
 * A. ** 7x + 15 = 12x + 5

Symbolic Method: 14-3x = 1.5x +5 14 - 3x +3 = 3 + 1.5x + 5 14 = 4.5x + 5-5 9 = 4.5x 9/4.5 = 4.5x/4.5 __2 = x__
 * B. ** 14-3x = 1.5x +5

Symbolic Method: -3x + 5 = 2x - 10 -3x +3 + 5 = 2x + 3 - 10 10 + 5 = 5x - 10 + 10 15 = 5x 15/5 = 5x/5 __3 = x__
 * C. ** -3x + 5 = 2x - 10

Symbolic Method: 3 + 5(x+2) = 7x - 1 3 + 5x + 10 = 7x -1 5x + 13 = 7x -1 5-5x + 13 = 7x - 5 - 1 13 = 2x - 1 1- 13= 2x -1 - 1 -14 = 2x -14/-2 = -2x/-2 __7 = x__
 * D. ** 3 + 5(x+2) = 7x - 1


 * E. ** Check your solutions to the equations in A-D

In solution A, both equations have the same value of ** 29 ** when substituted with x = 2.

Work:

y = 7x + 15 y = 7(2) + 15 y = 14 = 15 __y = 29__

y = 12x + 5 y = 12(2) + 5 y = 24 + 5 __y = 29__

In solution B, both equations have the same value of ** 8 ** when substituted with x = 2.

Work:

y = 14 - 3x y = 14 - 3(2) y = 14 - 6 __y = 8__

y = 1..5x + 5 y = 1.5 (2) + 5 y = 3 + 5 __y = 8__

In solution C, both equations have the same value of ** -4 ** when substituted with x = 3.

Work:

y = -3x + 5 y = -3(3) + 5 y = -9 + 5 __y = -4__

y = 2x - 10 y = 2(3) - 10 y = 6 - 10 __y = -4__

In solution D, both equations have the same value of ** 48 ** when substituted with x = 7.

Work:

y = 3 + 5(x + 2) y = 3 + 5(7 + 2) y = 3 + 5(9) y = 3 + 45 __y = 48__

y = 7x - 1 y = 7(7) - 1 y = 49 -1 __y = 48__


 * F. **Look over your work for A-D. Record any general strategies that seem to work well in solving linear equations.

General Strategies: ·<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal"> If they have parentheses, use distributive property so you can multiply and subtract easily. ·<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal"> If the outside of the parentheses have subtraction or addition as they order or operation, when you use the distributive property, you should change the operation similar to the one outside of the parentheses. ·<span style="font-family: 'Times New Roman'; font-style: normal; font-variant: normal; font-weight: normal; font-size: 7pt; line-height: normal; font-size-adjust: none; font-stretch: normal"> If you have 6x - 6, the one that you are subtracting to is a negative so it would be 6x + -6.

**Follow-up 4.3**
In a-c, the given equation is related to the equation y = 7.5x - 23.5. Explain what question the solution to the given equation answers. For example, the solution to 7.5x - 23.5 = 10 answers the question "What is the value of x in the equation y = 7.5x - 23.5 when y = 10?" Then use a table or graph to solve the equation, and explain how you found your solution/


 * 1a. ** 7.5x - 23.5 = 51.5

Question: What is the value of x in the equation y = 7.5x - 23.5 when y = 51.5?

Symbolic Method: 7.5x - 23.5 = 51.5 7.5x - 23.5 + 23.5 = 51.5 + 23.5 7.5x = 78 7.5x/7.5 = 78/7.5 __x = 10__ Table: Explanation: I found the answer by using the symbolic method. Then, I checked it by using a table.
 * x || y ||
 * 2 || -8.5 ||
 * 4 || 6.5 ||
 * 6 || 21.5 ||
 * 8 || 36.5 ||
 * ** 10 ** || ** 51.5 ** ||


 * 1b. **7.5x - 23.5 = 0

Question: What is the value of x in the equation y = 7.5x - 23.5 when y = 0?

Symbolic Method: 7.5x - 23.5 = 0 7.5x - 23.5 + 23.5 = 0 + 23.5 7.5x = 23.5 7.5x/7.5 = 23.5/7.5 __x = 3.133...__

Table: Explanation: I used the symbolic method and use a table to clarify my answer.
 * x || y ||
 * 3.10 || -0.25 ||
 * 3.11 || -0.175 ||
 * 3.12 || -0.1 ||
 * ** 3.13 ** || ** 0 ** ||


 * 1c. **7.5x - 23.5 = -30

Question: What is the value of x in the equation y = 7.5x - 23.5 when y = -30?

Symbolic Method: 7.5x - 23.5 = -30 7.5 - 23.5 + 23.5 = -30 + 23.5 7.5x = -6.5 7.5x/7.5 = -6.5/7.5 x = -0.866... Explanation: I used the symbolic method and generated a table to find x = -0.866... y = -30
 * x || y ||
 * -1 || -31 ||
 * -0.9 || -30.25 ||
 * -0.866... || -30 ||
 * -0.8 || -29.5 ||
 * -0.7 || -28.75 ||


 * 2. **Write a linear equation with a solution of x = 2. Do you think everyone in your class wrote the same equation? Explain.

Explanation: My equation is y = 3x + 8 when y = 14. (3x + 8 = 14). I don't think that everyone in my class would have the same expression. They are many expressions that have x = 2. For example, in problem 4.3, there are 4 equations that have x = 2 such as 7x + 15 = 29. 2 can be multiplied by any number so there are many variations of x = 2.

Work: 3x + 8 = 14 3x + 8 - 8 = 14 - 8 3x = 6 3x/3 = 6/3 x = 2


 * 3. ** The school choir is selling boxes of greeting cards to raise money for a trip. The equation for the profit in dollars, // P //, in terms of the number of boxes sold, // x, // is ** P = 3x - (100 + 2x) **

Explanation: I found my answer by using the symbolic method. Then, I used the equation P = 3x – (100 + 2x) to check my work as I substituted x with 100.
 * 3a. **** How many boxes must the choir sell to make a $200 profit? Explain how you found your answer. **
 * Symbolic Method: **
 * 3x – (100 + 2x) = 200 **
 * 3x – 100 - 2x = 200 **
 * 3x – 2x - 100 = 200 **
 * x - 100 = 200 **
 * x – 100 + 100 = 200 + 100 **
 * x = 300 **
 * Check: **
 * P = 3x – (100 +2x) **
 * P = 3(300) – (100 + 2*300) **
 * P = 3(300) – (100 + 600) **
 * P = 3(300) – (700) **
 * P = 900 – 700 **
 * __ P= 200 __**
 * Explanation: **
 * I found my answer by using the symbolic method. To check my work, I used the equation P = 3x – (100 +2x) and substituted x with 300. **
 * 3b. **** How many boxes must the choir sell to break even? Explain how you found your answer. **
 * Symbolic Method: **
 * 3x – (100 + 2x) = 0 **
 * 3x – 100 – 2x = 0 **
 * x – 100 = 0 **
 * x – 100 + 100 = 0 + 100 **
 * x = 100 **
 * Check: **
 * P = 3x – (100 + 2x) **
 * P = 3(100) – (100 + 2*100) **
 * P = 3(100) – (100 + 200) **
 * P = 3(100) – (300) **
 * P = 300 – 300 **
 * __ P = 0 __**

**Homework**
Collected: ACE 3: 13 - 27 odds, 39 (53-57 odds) Assigned: ACE 4: 3, 4, and 13

Lukas Grimm Day 66/March 30 2008

Notes: none

Stacking Rods Problem 5.1 and Follow Up
 * 1) The length of a longer rod is 5 unit rods because when the longer rods are stacked I can see that there are 2 unit rods on each side but still there is some space in the middle (appr. The size of one unit rod).

(L) || Surface Area (S) || B.
 * Rod Length
 * 1 || 6 ||
 * 2 || 10 ||
 * 3 || 14 ||
 * 4 || 18 ||

C. An appropriate equation would be S=(L-1)4+6

Follow Up 1a. I reckon they will all be equivalent. I think so because equations can look very different but if you solve them and the math is correct, you should get the same answer no matter what the equation looks like. 1b. (L) || Surface Area (S) ||
 * Rod Length
 * 1 || 6 ||
 * 2 || 10 ||
 * 3 || 14 ||
 * 4 || 18 ||

I can use this table to find out if other people’s equations are equivalent by comparing the results that I got. If the surface area matches that of other people’s tables, they must be equivalent.

2a. S=(12-1)4+6 S=(11)4+6 S=44+6 S=50 2b.S=(20-1)4+6 S=(19)4+6 S=76+6 S=82 3. Yes my equation works for any different length of unit rods.

4. It is linear because it always increases by 4 with a y-intercept of 6.

Homework: Assigned:5.1 and Follow Up Ace Collected: Ace 4: 3,4,13** MR4, pg 64 Check up 2 Assessment Check up 3 and 4

MN-1.2 Dividing 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)

SL- 1.3 Working with Exponents 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)

MN- 2.1 Tiling Pools 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)

RD- 2.2 Thinking in Different Ways 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)

RK - 2.3 Diving In 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)

YL -3.1 Walking Together 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)

LC - 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)

AK - 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)

? - 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)

AT- 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)

SA - 4.2 Solving Linear 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)

AS - 4.3 Reasoning with Symbols 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)

RK - 4.4 Solving 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)