Growing_Growing_Growing

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

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

Rica Duchateau**

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

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

**Journal**

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

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

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

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

- See the table.

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

Temperature Difference E. Compare the shapes of the graphs.

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

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

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

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

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

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

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

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

**Problem 4.4: Follow-Up**

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

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

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

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

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

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

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

Alya Shaiful**

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

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

Growth factor:

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

2 is the growth factor

Decay factor:

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

1/2 or 0.5 is the decay factor

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

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

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

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

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

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

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

**Follow-up 4.3**

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

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


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

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

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

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

Sajid Azad**

**12/3/07 - Day 35**

 * Math 8c**

**Problem 4.2 - Fighting Fleas**

 * A. How does the amount of active medicine in the dog's blood change from one hour to the next?

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

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

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

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

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

4.2 follow up

1.**


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

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

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

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

Robert Koehlmoos

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

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


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

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

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

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

Follow up 4.1

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


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

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

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

A=27(0.3333...^n)

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

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

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

Lukas Grimm**

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

**Problem 3.3**

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

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


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


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

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

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

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

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

V=1250(1.04^t)

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

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

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

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

3.3 Follow-Up

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

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

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

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

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

The collection would be worth $3063.081.

Find the growth factor associated with each percent increase.

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

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

Megan**

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

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

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

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

**Problem 3.2**

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

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

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

The ratio is 1.06 which is the growth factor

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

The growth factor is 1.04

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

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

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

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

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

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

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

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

Alya Shaiful**

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

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

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

Average: 1.8

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

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

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

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

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

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

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

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

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

**Follow-up 3.1**

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

It would take around 3 to 4 years.

Work: 1.2 * 2 = 2.4**


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

It would take around 2 to 3 years.

Work: 1.5 * 2 = 3**


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


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

It would take around 2 to 3 years.

Work: 1.8 * 2 = 3.6**


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


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

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

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

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

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

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

4a. Find percentage change of 1.5

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

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

4b. Find the percentage change of 1.25

There would be 25% change.

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

4c. Find the percentage change of 1.1

There would be 10% change.

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

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

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

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

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

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

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

Notes:

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

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

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

Problem 2.1

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

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

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

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

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

Problem 2.1 Follow-Up

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

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

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

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

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

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

__Sue Lee__** November 14th, 2007 - Day 30

1.3 : Making a New Offer

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

__Problem 1.3__

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

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

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

__1.3 Follow Up__

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

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

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

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

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

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

Yoon Sun Lee November 12, 2007/ Day 29

1.2 : Requesting a Reward

Notes

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

Standard form : 16 Journal Problem 1.2

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

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

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

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

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

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

Homework

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

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

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

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

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

__Follow Up__ 1a: 2³ b: 5⁴ c: 1.5ˆ 7

2a: 128 b: 27 c: 74.088

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

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

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

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

21/11/07 Block C Day 29

2.3 Growing mold

Problem:

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

Follow up:

1)

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

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

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

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

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