Growing_Growing_Growing

Growing, Growing, Growing Journal and Homework Record Looking For Pythagoras Vocabulary Name (heading 2) Date/Day(heading 2) Problem Number and Title (Heading 3) Notes (Heading 3) Journal (Heading 3) Homework (Heading 3) Collected: (normal) Assigned: (normal)

Notes: What pattern would you expect from the experiment?What would the graph look like? I hypothesis that the hot water will eventually cool down as time passes. The shape of the graph will probably look like an exponential decay graph. In an exponential decay graph the y quickly goes down but as x increases the difference becomes smaller. It’s basically the exact opposite of an exponential growth graph. ==|| Time in minutes || Water temperature in Celsius || Room temperature in Celsius || Temperature difference || || 0 || 79 || 24 || 55 || || 5 || 68 || 25.5 || 42.5 || || 10 || 63 || 26 || 37 || || 15 || 58 || 26 || 32 || || 20 || 54 || 26 || 28 ||==
 * Time in minutes ||
 * Water temperature(Celsius) ||
 * Room temperature(Celsius) ||
 * Difference between room and water temperature(Celsius) ||
 * 0 ||
 * 79 ||
 * 24 ||
 * 55 ||
 * 5 ||
 * 68 ||
 * 25.5 ||
 * 42.5 ||
 * 10 ||
 * 63 ||
 * 26 ||
 * 37 ||
 * 15 ||
 * 58 ||
 * 26 ||
 * 32 ||
 * 20 ||
 * 54 ||
 * 26 ||
 * 28 ||
 * 20 ||
 * 54 ||
 * 26 ||
 * 28 ||
 * 28 ||

Journal:
A. Make a graph of your (time, water temperature) data.

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? As you can see in the following graph the fastest or steepest incline throughout the graph is between 5 to 10 minutes. The slowest is between 15 and 20 minutes.

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

E. Compare the shapes of the graphs. The shapes of the graphs look identical because they have the same characteristics. F. Describe the pattern of change (time, temperature difference) data. When did the temperature difference change most rapidly? When did it change most slowly? From the graph above I think the most rapid change is between 0 and 5 minutes. The slowest change is between 15 and 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. I estimate at about 0.8 because if you divide the y’s by the next y the answer revolves around 0.8. I found the answer by multiplying different numbers to get the desired y. 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. D=55times0.77272727(t) The decay factor is 0.7727272. The 55 is the initial or start. The D stands for the temperature difference and the t stands for time.

Follow up: 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? The change in temperature would get smaller and smaller until it reached the room temperature. Some of the factors include the air temperature because that’s what cools the substance down and the thermometer because it might absorb some of the heat. If you add variables like a spoon or some other object it could significantly affect the rate the water cools down at.
 * 1) What factors might affect the rate at which a cup of hot liquid cools?
 * 1) What factors might be introduce errors in the data you collect?

Homework:
Collected: none Assigned:A.C.E:3,5,7,10

Notes:
Exponential Decay is when the base in the equation is a decimal and Exponential Growth is when the base in the equation is a full number.

Journal:
A. Out of the equations Y=1.25ˆ X, Y=1.5ˆ X, Y=1.75ˆ X and Y=2ˆ X, Y=1.25ˆ X is the least curved and Y=2ˆ X is the most curved in an upward direction. The larger the number, the more curved the line and vice-versa.

B. Out of the equations Y=.25ˆ X, Y=.5ˆ X and Y=.75ˆ X, Y= .75ˆ X is the least curved and Y=.25ˆ X is the most curved in a downward direction. The smaller the decimal, the more curved the line is in a downward direction and vice-versa.

C. If the base in an equation is a full number then the line will curve up. The larger the number, the more curved the line and vice-versa. If the base in an equation is a decimal then the line will curve down. The smaller the decimal, the more curved the line is in downward direction.

4.3 Follow-up:
1. Out of the equations Y=2(2ˆ X), Y=3(2ˆ X) and Y=4(2ˆ X), Y=2(2ˆ X) has the lowest Y-intercept and Y=4(2ˆ X) has the highest y-intercept. the larger the coefficient, the higher up the line is on a graph and the rate of increase is larger.

2.I got the same effect as #1 but the line was curved less.

3. I found that if the coefficient is larger and the base is a decimal, the line is higher up on the graph but the line became less curved as the coefficient went up.

Homework:
Homework Collected: None Homework Assigned: A.C.E 3, 5, 7, 10 Problem: 4.3 and 4.4 and both their follow-ups completed

__David Sidibe__
 * December 5, 2007/ Day 37**
 * 4.3: Exploring Exponential Equations**

Notes:
An Exponential equation is an equation that has both a base with the base being squared n number of times depending on what the exponent is. By reading this problem it looks like we will use both Exponential Growth and Exponential Decay and experiment with a (co- efficient) being the starting value and b (exponent) being the growth or decay factor.

=Journal:= A: While observing the data I got from the graph I found two things. One is that all 4 equations have the same starting point, therefore they have the same y-intercept. The second thing is that the slopes are going up at different rates making the equation with the slope of 2^x have a bigger rise ratio then the run ratio. B: I found that this pattern is an exponential decay pattern (opposite of exponential growth) and also the smaller the base, the less of the line is shown and vice-versa. C: Based on A and B I would say that the bigger the base the greater the increase in x over y. So I would say that if the exponent has a bigger base then it will be greater then the equation with a smaller base. = =

4.3 follow up:
1. I found that the equation with the greater co-efficient of b goes up at a faster rate than the other equations with a lower co-efficient of b. 2. I found that the same thing happens from part 1 but the base is the only difference. 3. This time the opposite happened of part 1 and 2 happened. This happens because when you multiply a co-efficient by a decimal you will always get a number lower then the number you multiply it by, therefore making the equation with the lower co-efficient the one that has the biggest slope. 4: The value of a affects the graph of an equation fo the form y=a(b^x) by doubling, tripling, or quadrupling depending on the co-efficient. When you haved solved (b^x) you multiply it by the a (co-efficient) therefore causing either a huge increase or decrease depending on if the co-efficient is a decimal or a number than one. If the co-efficient was one then the co-efficient would have no effect on the equation and you wouldn't have to write down the one. eg. a=4 b=3 x=3 y=a(b^x) is the equation model y=4(3^3) 3x3x3=27 If you had no co-efficient than the equation would stay at 27 but because you have a co-efficient of 4the you would do 27 multiplied by 4 or 4(3^3) or a(b^x). This would make your answer 108, four times more different without the co-efficient so what I am saying is that co-efficient (a) can completely change the answer.

Homework:
Homework Collected: None Homework Assigned: A.C.E 3, 5, 7, 10 Problem: 4.3 and 4.4 and both their follow-ups completed

Journal
__Problem 4.2__ A. The pattern in the amount of active medicine in the dog's blood is divided by 2 at each stage. B. An equation to model the relationship is: m = 20 ÷ 2^h. In the equation, m is the milligrams of active medicine, 20 is the initial number, 2 is the decay factor and h is the number of hours. C. Based on knowledge of exponential relationships, the pattern that is expected if the data started from 40 milligrams is: If the data started with 40 milligrams and 80% remained then the pattern would be divided by 0.8.

__Problem 4.2 Follow-up__ 1.)
 * Time since dose (hours) || Active medicine in blood (milligrams) ||
 * 0 || 60 ||
 * 1 || 48 ||
 * 2 || 38.4 ||
 * 3 || 30.72 ||
 * 4 || 24.576 ||
 * 5 || 19.6608 ||
 * 6 || 15.72864 ||

2.) I replaced the exponent //h// with the numbers 3, 4 and 5: m = 60(0.8^h) m = 60(0.8^3) m = 60(0.512) m = 30.72

m = 60(0.8^h) m = 60(0.8^4) m = 60(0.4096) m = 24.576

m = 60(0.8^h) m = 60(0.8^5) m = 60(0.32768) m = 19.6608

Using some values from the table above, Janelle’s equations provides the accurate amounts of active medicine in blood. The decay factor shown in the table between each stage is the same as the decay factor in the equation (0.8). Also, the starting number in the table and in the equation is 60.

3.) The rate of decay of 20% is equivalent to 0.8 because when you take out 20% (from 100) you are left with 80%. 80% is the same as (0.8).

Homework
Collected: Ace 3: 1 & 3 and Mathematical Reflection 3 Assigned: ACE4: 1, 6, 9 (pages 53 - 58)

Notes:
- Compound Growth: another term for exponential growth. It is when a value is found by using an operation on the previous value, not the original one. It is usually found in an exponential equation's base. -Sam's interest Rate: 6% (Problem: 3.2) -Growth Rate: 1.06 (Problem: 3.2)

Journal:
A.) Sam's total amount of value left: $1250 Sam's increase of value: 4% Table showing Sam's growth for the next 10 years:
 * Number of Years || Sam's Value/Money ($): ||
 * 0. || 1250 ||
 * 1. || 1300 ||
 * 2. || 1352 ||
 * 3. || 1406.08 ||
 * 4. || 1462.32 ||
 * 5. || 1520.81 ||
 * 6. || 1581.64 ||
 * 7. || 1644.91 ||
 * 8. || 1710.71 ||
 * 9. || 1779.13 ||
 * 10. || 1850.30 ||

B.) Maya's baseball card collection's value: $2500 Maya's increase of value: 4% Table showing Maya's growth for the next 10 years compared to Sam's:

NOTE: Some of the values are rounded to the nearest hundreths.
 * Number of Years: || Sam's money ($) || Maya's money ($) ||
 * 0. || 1250 || 2500 ||
 * 1. || 1300 || 2600 ||
 * 2. || 1352 || 2704 ||
 * 3. || 1406.08 || 2812.16 ||
 * 4. || 1462.32 || 2924.64 ||
 * 5. || 1520.81 || 3401.62 ||
 * 6. || 1581.64 || 3523.28 ||
 * 7. || 1644.91 || 3664.21 ||
 * 8. || 1710.71 || 3884.06 ||
 * 9. || 1779.13 || 4039.06 ||
 * 10. || 1850.30 || 4200.99 ||

C.) The initial value of the collection affects the yearly increase because, the bigger it is the more the money is. So, it depends on the initial value to find out the yearly increase, and also how much it should increase or even decrease.

D.) The initial value of each collection affect the growth rate because the bigger it is, the greater affect the growth factor will have on it.

E.) Equation for Sam's value: //V=1250 (1.04^t)//

F.) What is the value (v) if the time (t) is 30 years? //v=1259(1.04^t)// v=1250(1.04^30) v=1250x3.24 v=4,054.25 So, the value after 30 years is $ 4,054.25.

3.3 Follow Up:
1.) Sam's prediction for the value of his aunt's stamp collection: Value=$2400x1.05x1.05x1.05x1.05 a.) The initial value of the above equation is 2400 because that is what the exponential number is being multiplied by. The rate of increase of the above equation is 1.05 because that is the base of the exponent above. The number of years that Sam is assuming is 4 years because that is how many times the rate of increase is being multiplied by. b.) Sam's calculation: $2917.22 After one year, the money would be: v=2400x1.05x1.05x1.05x1.05x1.05 v=2400(1.05^5) v=2400x1.276 v=3,063.07 The money after one more year would be: $3,063.07

2.) Find the growth factor associated with each percent increase. a.) 30%: growth factor=1.30 b.) 15%: growth factor=1.15 c.) 5%: growth factor=1.05 d.) 75%: growth factor=1.75

Homework:
Assigned: Finish 3.2 and 3.3 with follow ups Ace3: 2,4,5,11 Collected: None

__Austin Dunn__

Notes:
Using %: to find the growth factor for a % from 0 – 99.99 just do 1.(%). Compound Growth: growing exponentially ith numbers smaller than 2, usually written as a percentage. Eg.

2500 + (6% of 2500) = 2500 + 0.06(2500) =2500 + 150 =2650

2650 + (6% of 2650) = 2650 + 0.06(2650) = 2650 + 159 =2809… etc

Problem:
A. Year || Coins: Amount Worth ($) || B. Cards: Amount Worth ($) 0 || ||1250 ||2600 1 || ||1300 ||2704 2 || ||1352 ||2812.16 3 || ||1408.08 ||2924.65 4 || ||1462.32 ||3041.42 5 || ||1520.82 ||3163.30 6 || ||1581.65 ||3281.83 7 || ||1644.91 ||3421.42 8 || ||1710.71 ||3558.28 9 || ||1779.14 ||3700.61 10|| ||1850.31 ||3848.64

B. Refer to table

C. The smaller the intial value, the smaller the growth, because then the numbers have less to multiply off.

D. The initial value doesn't affect the growth factor because they are entirely seperate things.

E. V=(1.04^t)1250 is the equation for the growth of the value of Sam's coin collection.

F. Sam's coin collection's value after thirty years is (1.04^30)1250 = 4052.

Follow-Up
1. A. value = $2400x1.05x1.05x1.05x1.05 2400 is the assumed intial value. The assumed growth rate = 5%. The number of years assumed is four. B. In one more year, there will be $3065.08

2. A. 30% = growth factor of 1.30 B. 15% = growth factor of 1.15 C. 5% = growth factor of1.05 D. 75% = growth factor of 1.75

Homework:
Assigned: Finish 3.2 and 3.3 with follow ups Ace3: 2,4,5,11 Collected: None

__Gus Crowards__

Notes:
Sam's coin collection is originally worth 2500. The value increases by 6% every year.

Journal:
A) this is a table for Sam's cion collecion value in the first 10 years: Year 0 1 2 3 4 5 6 Value 2500 2650 2809 2977.54 3156.19 3345.56 3548.3 7 8 9 10 3759.08 3984.62 4223.7 4477.12

B) The growth factor for the value of the coins is 1.06. 4477.12/4223.7=1.06. 4223.7/3759.08=1.06.

C) 1. If the growth per year changes from 6% to 4% then this is the table: Year 0 1 2 3 4 5 6 Value 2500 2600 2704 2812.16 2924.65 3041.63 3163.3 7 8 9 10 3289.83 3421.42 3558.28 3700.61 2. The growth factor for the value of the coins is 1.04. 2704/2600=1.04. 3163.3/3041.63=1.04

D) If the value increased by 5% each year, then the growth factor would be 1.05 because 5% in decimal form is 0.05. 5% of the previous number plus the previous number is the same as multiplying by 1.05.

Follow-Up:
1. 100%+6%=106%=1.06. this is an easy way to think about having percentage growth. 2. 100%+4%=104%=1.04. this is the growth factor for 4% growth. 100%+5%=105%=1.05. this is the growth factor for 5% growth. These are the same answers that I got in C and D. 3. Yes, Sams formula V=2500(1.06^t) gives the same answers as A. t=4. 1.06^4=1.26. 1.26x2500≈3156.19

Homework:
Assigned: Finish 3.2 and 3.3 and follow-ups Ace 3: 2,4,5,11 Collected: None

__Vivian Ly__

**Notes:**
Growth Factor= base of the exponent

Journal:
A) The growth factor for the rabbit population is an estimation of 1.8 The numbers in the parenthases are the estimated growth factors B) There would be a population of around 35,704 rabbits by 10 years. There would be a population of around 120,886,592 rabbits by 25 years. By 50 years there would be an population of around 5.8026355026 times 10^12 rabbits. C) It would have taken around 24 years for the population to reach around the population of 1,000,000 rabbits. D) The equations Rabbit's Population growth each year would be: P= (1.5^n)100
 * Time (Years) || Rabbit Population ||
 * 0 || 100 (100 to 180= 1.8) ||
 * 1 || 180 (180 to 325= 1.80555....) ||
 * 2 || 325 (325 to 580= 1.78461....) ||
 * 3 || 580 (580 to 1050= 1.810....) ||
 * 4 || 1050 ||

2.3 Follow Up
1.a. It would take 2 years for the population to grow from 1 million to 2 million with the growth factor of 1.5. b. It would take 4 years for the population to grow from 1 million to 4 million with the growth factor of 1.5. c. It would take 6 years for the population to grow from 1 million to 6 million with the growth factor of 1.5. d. It would take 8 years for the population to grow from 1 million to 8 million with the growth factor of 1.5. 2.a. It would have taken 30 years for the population to double from the starting point of 100 with the growth factor of 1.2 b. It would have taken 14 years for the population to double from the starting point of 100 with the growth factor of 1.5 c. It would have take 10 years for the population to double from the starting point of 100 with the growth factior of 1.8 d. I observed that the larger the number is, the less years it would take to double the amount of the starting point. 3.a. The equation assumes that the growth factor is 1.2 b. The equation assumes that the initial population is 15 4.a. 50% for the growth factor of 1.5 b. 25% for the growth factor of 1.25 c. 10% for the growth facotr of 1.1

Homework:
Collected: MR#1 and 2 Assigned: ACE3: 1 and 3 Problem 3.1 and it's follow up __Zaraif Hossain__

Notes:
How can I show a pattern of repeated multiplication? I can show a pattern of repeated multiplication through exponents. Example- 2²=2x2=4

Journal:
A) (answer completely)The pattern is that each number is the result of 3x the previous number. B) A=3^d C) In my equation it shows that to get the are of the mold you will need to find 3 to the power of the number of days. You will need to use three because the **growth factor** of this equation is 3.

2.3 follow up:
1) 2) A=25(3^d) 3) On my equation there is a part which says that you have to multiply 25 with the result of 3^d. The part that shows how the numbers occur and ascends is the part which says 3^d. 4) **Growth Factors** of different plans: a. Making ballots cut in half each time=> 2 b. reward plan1=> 2 c. reward plan2=> 3 d. Growth of bread mold=> 3 5) a. 50 mm² of mold b. 3 was the growth factor c. A=50(3^d)=> A=50(3^5)=> A=50x243=> A= 12,150 d. On day3 the mold are reaches 10cm²
 * Day || Mold Area (mm²) ||
 * 0 (start) || 25 ||
 * 1 || 75 ||
 * 2 || 225 ||
 * 3 || 675 ||
 * 4 || 2,025 ||

Homework:
Collected- none Assigned- ACE2; 10-12

Nabihah Rana

Problem 2.2: Listening to the Queen, pg. 19-20
Notes: 16 squares 500 = Square 1 doubles each time r = 2^n-1 x 500

Journal: A.
 * Square || Number of Rubas ||
 * 5 || 8,000 ||
 * 6 || 16,000 ||
 * 7 || 32,000 ||
 * 8 || 64,000 ||
 * 9 || 128,000 ||
 * 10 || 256,000 ||
 * 11 || 512,000 ||
 * 12 || 1,024,000 ||
 * 13 || 2,048,000 ||
 * 14 || 4,096,000 ||
 * 15 || 8,192,000 ||
 * 16 || 16,384,000 ||

B.r = 2^n-1 x 500 C. The peasant should not accept this offer because the amount she will get in the end is way less then what she would get if they used her original plan. This plan was devised to trick her because it starts off with more rubas, but as the plan goes on, it just doubles itself, but it only goes till square 16.
 * Problem 2.2 Follow up, pg. 20**

1. 5 million = Square 1 y = 1million ^ x + 4 million
 * 1) r = 1,000,000n + 4,000,000
 * 2) The relationship between the number of rubas and the square is a linear relationship. I know this be looking at the equation and noticing that it follows the y = mx+b format.
 * 3) I suggest the peasant go for her original plan because on square 16, following this plan, the number of rubas she gets (20 million) is still very less then what she would get from the other plan.

2. r = 20n +10,000,000 [16 square board] r = 50n + 5,000,000 [16 square board] Both the plans I made are linear. These are supposed to be tough offers to choose from because they both start with a pretty high amount. But, one starts off with 10,000,000 and goes up with an exponential value that is less then the other plan (20). The other one starts off with 5,000,000 but goes up by an exponential growth with 50.

3. In the end, the king and queen just give the peasant a big hug and a 50 million ruba offer. Yet the peasant hesitated about this offer too. So, being close to bursting their heads the royal people decided to have her shipped off to jail for her ignorant behavior to the king and queen.

Homework:
Collected: ACE 1: #1-11 odds, pgs. 10-15 Assigned: ACE 2: #1, 5, and 7, pgs. 22-24 Problem 2.1 and 2.2 and their follow ups

__Huei Yi__

Problem 2.1 Getting Costs in Line
· How are the starting values and the size of the growth reflected in the table, the graph, and the equation for an exponential relationship? · · Equation for Plan 1- r=2^(n-1) · Equation for Plan 4- r=5n+15
 * Notes:**
 * Square || Number of rubas ||  ||
 * || Plan 1 || Plan 4 ||
 * 1 || 1 || 20 ||
 * 2 || 2 || 25 ||
 * 3 || 4 || 30 ||
 * 4 || 8 || 35 ||
 * 5 || 16 || 40 ||
 * 6 || 32 || 45 ||

A. Make a table showing the number of rubas on squares 7, 8, 9 and 10 for plans 1 and 4. B. For Plan 4, write an equation for the relationship between the number of the square, //n//, and the number of rubas, //r//. How is this equation similar to the equation for plan 1? How is it different? An equation for the relationship in Plan 4 is r=5n+15 This equation is similar to the equation for plan 1 (r=2^(n-1)) because in both equations, n has to times by something. Both equations are different though because plan 1 (r=2^(n-1)) shows exponential growth, but plan 4 (r=5n+15) shows a linear relationship. C. For plans 1 and 4, how many rubas would be on square 20? How many rubas would be on square 30? Plan 1-On square 20 there would be 524,288 rubas. r=2^ (n-1) r=2^ (20-1) r=2^19 r=524,288) Plan 1-On square 30 there would be 536,870,912 rubas. r=2 (^n-1) r=2^ (30-1) r=2^29 r=536,870,912 Plan 4-On square 20 there would be 115 rubas. r=5n+15 r= (5x20) +15 r=100+15 r=115 Plan 4- On square 30 there would be 165 rubas. r=5n+15 r= (5x30) +15 r=150+15 r=165 D. Write a paragraph explaining why the peasant should or should not accept the king’s new offer. The peasant should not accept the king’s new offer if she wants a lot of money, because even though in the graph it looks like the peasant’s plan has less money and the king’s plan starts off at a lot of rubas, the graph shows only up to square 6. When the peasant’s plan reaches square 64 though, the number of rubas is a lot because it doubles every time. For the king’s offer, plan 4, the number of rubas just adds 5 every time. So in the end, the peasant would get very little if she chose Plan 4.
 * Journal:**
 * Square number || Number of rubas ||  ||
 * || Plan 1 || Plan 4 ||
 * 7 || 64 || 50 ||
 * 8 || 128 || 55 ||
 * 9 || 256 || 60 ||
 * 10 || 512 || 65 ||

1. For plan 1, there would be 16,384 rubas on square 15. How many rubas would there be on square 16? (The next value is the previous number of rubas times 2. So 16,384 (Number of rubas on square 15) x2=Number of rubas on square 16.) There would be 32,768 rubas. 2. For Plan 4, there would be 90 rubas on square 15. How many rubas would there be on square 16? (The next value is the previous number of rubas plus 5. So 90 (Number of rubas on square 15) +5=Number of rubas on square 16.) There would be 95 rubas on square 16. 3a. How does the number of rubas change from one square to the next for plan 1? The number of rubas changes from one square to the next by doubling the previous value. 3b. How does the number of rubas change from one square to the next for plan 1? The number of rubas changes from one square to the next by adding 5 to the previous value. 4. How are your answers to question 3 represented in the equations for the plans? Plan 1- r=2^(n-1) Plan 4- r=5n+15 My answers for question 3a & b could be seen in the equations. For plan 1, n-1 is the previous value. And 2 is how many times //n// has to times by. (Doubling.) For plan 4, 5n is how many times //n// has to times by, and then +15 means, //n// has to add 15. An easier way of doing this is just adding 5 to the previous value, and it still works.
 * Problem 2.1 Follow-Up**

Homework:
Collected: ACE 1: #1-11 odds. Assigned: ACE 2: #1, 5, and 7. Problem 2.1 and Follow-Up, and Problem 2.2 and Follow-Up.

Day 30
1.3 Making a New Offer

Journal
A. In the table below, plan 1 is the reward requested by the peasant, and plan 2 is the king’s new plan. Complete the table to show the number of rubas on squares 1-16 for each pSquare lan. What are the similarities and differences between the patterns under plan 1 and plan 2? Plans 1 and 2 are similar as they are both examples of exponential growth. They are different because in plan 1, each time the previous value is multiplied by 2 while in plan 2, each time the previous value is multiplied by 3. C. Write an equation for plan 2. //r// = 3//^(n//-1) D. Is the king’s new reward greater or less than the peasant’s original plan? The king’s new reward is much less than the original plan. I know this because the new reward is 14,348,907 rubas. In the previous problem, we figured out that the original reward was 9,223,372,036,854,775,808. Therefore, the new reward is less. Follow up 1. (Notice the break on graph placed below) The following graph compares with the graph for plan 1 as both increase slowly at first and then a lot faster. The only differences are the scales and the speed of growth. 2. (Notice the break on graph placed below) The following graph compares with the graphs for plan 1 and 2 as both increase slowly at first and then a lot faster. The only differences are the scales and the speed of growth. 3. Write an equation for plan 3. //r// = 4^//n//-1 is the equation for plan 3. The 3 equations are similar as they all show exponential growth. They are different as they have different growth factors of 2, 3, and 4. 4. Plan 3 is the best for the king while plan 1 is best for the peasant. 5. Devise another possible plan. Under my plan, the king will make a board with 9 squares. He will start with one ruba on the first square and use a growth factor of 5 from one square to the next. So, the pattern of rubas would be 1, 5, 25, 125, and so on.
 * Square Number || 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 || 2187 ||
 * 9 || 256 || 6561 ||
 * 10 || 512 || 19683 ||
 * 11 || 1024 || 59049 ||
 * 12 || 2048 || 177147 ||
 * 13 || 4096 || 531441 ||
 * 14 || 8192 || 1594323 ||
 * 15 || 16384 || 4782969 ||
 * 16 || 32768 || 14348907 ||

Homework
Collected: None Assigned: ACE 1: #10-12, and Math Reflections pg 16

Notes:
Exponential Growth: A pattern of increase in which each new value is found by multiplying the previous number by 2 or greater (the growth factor). eg. 2, 4, 8, 16, 32…

Journal:
A. Make a table showing the number of rubas the king will place on squares 1 through 16 of the chessboard. B. How does the number of rubas change from one square to the next? The number of rubas on a square is double the number of rubas on the previous square. So it doubles between squares. C. How many rubas will be on square 20? On square 30? On square 64? Square 20= 1,048,576 rubas. Square 30= 1,073,741,824 rubas. Square 64= 18,446,744,070,000,000,000 D. What is the first square on which the king will place at least 1 million rubas? At square 21, the king will place at least 1 million rubas. E. If a Montarek ruba had the value of a U.S. penny, what would be the dollar values of the rubas on squares 10, 20, 30, 40, 50, and 60? Square 10= $5.12 Square 20= $5242.88 Square 30= $5368709.12 Square 40= $5497558139 Square 50= $5.629499534e+12 Square 60= $5.764607523e+15
 * Square Number || 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 ||

Follow Up: 1. As the number of squares increases, how does the number of rubas change? What does this pattern of change tell you about the peasants reward? The number of rubas double between squares. 2. Write an equation for the relationship between the number of the square, n, and the number of rubas, r. R=2n-1. 3. If a chessboard had 100 squares, how many rubas would be on square 100? 6.338253001e+29 rubas would be on square 100. 4. The pattern of change in the number of ballots in Problem 1.1 and the pattern of change in the number of rubas in this problem show **exponential growth**. a. How are the patterns of change in these two situations similar? Both patterns of change are similar because they both double. b. Write an equation for the relationship between the number of ballots, b, and the number of cuts, n, in Problem 1.1. b=2n.

Homework
Assigned: ACE1: 1-11 ODDS Collected: None

**1.1 Making Ballots Jeong Kyu Koh November 17, 07 Day 30**
A B: The pattern that I see is that it's growing germatically fast by doubling what ever it had before. look at A C: He would have 1,048,576 ballots for 20 cuts. If he made 30 cuts, he will have 4,294,968 stocks of ballots. D: For 20 cuts, he would have 4195 stacks of ballots and for 30 cuts, he will have 4,294,968 stacks of ballots. E: 12 cuts because 250 X 120 is 3000 and 12 cuts have around 4000 ballots and 11 cuts have around 2000 ballots.
 * cuts || Paper ||
 * 1 || 2 ||
 * 2 || 4 ||
 * 3 || 8 ||
 * 4 || 16 ||
 * 5 || 32 ||
 * 6 || 64 ||
 * 7 || 128 ||
 * 8 || 256 ||
 * 9 || 512 ||
 * 10 || 1024 ||
 * 11 || 2048 ||
 * 12 || 4096 ||
 * 13 || 8192 ||
 * 14 || 16384 ||
 * 15 || 32768 ||
 * 16 || 65536 ||
 * 17 || 131072 ||
 * 18 || 262144 ||
 * 19 || 524288 ||
 * 20 || 1048576 ||
 * 21 || 2097152 ||
 * 22 || 4194304 ||
 * 23 || 8388608 ||
 * 24 || 16777216 ||
 * 25 || 33554432 ||
 * 26 || 67108864 ||
 * 27 || 134217728 ||
 * 28 || 268435456 ||
 * 29 || 536870912 ||
 * 30 || 1073741824 ||

Follow up
1a: 2³ b: 5⁴ c: 1.5⁷ 2a: 128 b: 27 c: 74.088 3a: 32768 b: 59049 c: 332525673 4 5^2 (5²) is multiplying 5 twice and 2⁵ is multiping 2 five times 5 Yes 5² x 5² 5n4 which means 25 x 25 x x = 625 6 You just need to multiply it's number by 5.

Homework
Collected: None Assigned: Ace #1 1~11 odds