Lesson+Summaries

On this page, an assigned student will write a daily summary of the big ideas developed in class along with any other information s/he wants to share, such as the At Home Extension.

Here is a sample format I'd like you to refer to when writing the summary:


 * = ** The following ideas were the focus of today's class :** =
 * idea 1
 * idea 2
 * idea ......
 * The way we developed idea 1 (2, 3, etc) was ...
 * An important thing to remember about idea 1(2, 3, etc) is ...
 * Idea 3 (idea developed late in the class session) is something that we'll revisit in later class periods but we got a start on this idea by discussing ...

=
Please try and follow the guidelines above when writing your summaries. Avoid writing a linear "Here's what we did first, then second, and so on". The intent is that you begin to think more broadly about the ideas and find more connections; beginning to think like a teacher! So try it out. And please don't erase these general instructions. Thanks. ======

Thursday, April 19 (Summary by Ashley)

To start class, we worked on Check your understanding on page 514 in the coursepack. We worked with groups on problems a. iii-v and b. ii and iv.

During our discussion of these problems, we focused on finding the minimum and maximum of a parabola. To do this, you have to look at what kind of parabola the equation will graph. If the parabola is centered on the y-axis, then all you will have to do is plug in x=0 (ex. y=5x^2+3). If the parabola is not centered on the y-axis, then you will have to first find the x-intercepts by plugging in y=0. After you have found the 2 x-intercepts, you will know that the min/max is directly in between those two points (since parabolas are symmetrical). Once you find the midpoint of the x-intercepts, you plug that value into x and solve. The midpoint will give you the (x,y) point (ex. y=-x^2+2x.

After our discussion on the check your understanding, we moved on to page 515 and worked with the quadratic formula. The quadratic formula shows you the x values when y=0, or the x-intercepts. The formula will give you two x values (assuming that the values are real.. meaning that you do not take the square root of a negative number). The quadratic formula is found on page 515.

====To end our class, we were divided into 4 groups and assigned to make 2 questions on an area that we studied for a review for the final. Dr. B will email these questions soon. ====

===**Study Session for final exam is Sunday at 2 pm in our classroom. Rood Hall will be locked, so we will all have to meet at the door near Everett Tower at 2 so Dr. B can let us in! **===

Expanded Form: abx²±adx±cbx±cd
We went over worksheet 8.3 To find the differences put in lists and go to delta lists under 2nd stat aand then ops to find the change in y values. When the function is quadratic the 2nd difference in the y values will the constant.

The 2nd difference in a quadratic function is double the coefficient. ex. 3x^2+6x

The second difference would be 6

To find b you take you take [f(1)-f(0)]-a f(0)=c f(1)=a+b+c f(1)-f(0)=b+a

Homework: Complete investgation #1 on p.511, p.518 1,4-8,15,17,26

April 12th: Today we re-discussed the meaning of the quadratic equation y=ax^2+bx+c and the particular functions of each group.

A major topic of today's discussion was part of the homework; in which a punkin-chunkin scenario was given, and the initial velocity of three different directions was given. The horizontal velocity was 102 ft/s and the vertical velocity was 92 ft/s. Through the Pythagorean Theorem, the hypotenuse velocity was 150 ft/s. The question offered to the class which velocity was used in the equation h(sub 0)+v(sub 0)t-16t^2? In other words, which of these three velocities was our "Initial Velocity? At first there was some debate that the 92 ft/s vertical velocity was the correct answer because until this point, most questions had involved vertical velocity. However, we decided that because the amount of horizontal velocity an object has can affect the time it takes for an object to reach the ground again; the combination of the two velocities of 150 ft/s was in fact the correct initial velocity.

We also discussed exponential properties (specifically how exponents are multiplied and divided.) It would be a good idea to review any of the this material you are unfamiliar with as the final is about 25% cumulative.

Keep checking your e-mails for a study session to be held by Dr. Browning the weekend before exams.

Homework for the weekend is to finish your Voice-Thread activity if you have not. NOTE: some people had issues watching the videos, email Dr. Browning immediately if you have problems. Watch your own, and the two groups following your own and put comments. Consider the rubric when commenting on videos.

**April 10th:** **First each group solved the quadratic problem and used our working rubric to justify their solutions. Each group was video taped for voice thread.**

**Next we discussed investigation 3** **The general form of a quadratic equation is y=ax^2+bx+c**

**y=ax^2** **a determines the "width" of the parabola. If 01, then the y-values grow quicker,** **the graph "hugs" the y-axis more as a increases. a also determines if the graph is concave up or concave down. If a is positive then the graph is concave up and has a minimum. If a is negative then the graph is concave down and has a maximum.**

**y=ax^2+c** **c moves the parabola up and down on the y-axis, c is the y-intercept. If c is positive the graph will move up the y-axis and if c is negative the graph will move down the y-axis.**

**y=ax^2+bx** **b moves the parabola left or right. If b is positive the parabola moves left and if b is negative the parabola moves right.**

**Homework:** **AHE p485 #17, 20, 22, 30** **Handout- complete EXPLORE and DISCUSS** **Voice Thread- one comment on your video, one comment on the next 2 successive videos**


 * April 5th: **

** How we developed these ideas: **
=== 1. First we discussed the voice thread videos with our groups. After group discussions, we started talking as a class and relating the videos to our rubric and explaining how we scored videos. As we were doing this, we ended up wanting to make some changes to our rubric. In order to have a full rubric, we decided to add in a 5th point; make sure justifications are clear/concise/organized. After this, we added to the general rule point by saying it needed to be explicit, unless stated otherwise in the question. Next we decided to add that it is necessary to use proper units when making connections. As a class, we also discussed adding in restating the question in our explanation. After a lengthy discussion, we came to the decision that it would not be added into the rubric, but if you personally feel it is necessary to add it into your description, feel free to do so. Lastly we numbered each of the points in the "full points" category of our rubric. After having the 5 points numbered in this category, we decided to change the point values of the rubric. The first/full points category will be worth 5 points. Next is the partial credit categories; if you have touched on 4 of the "points" from the first category, you will recieve 4 points, if you've touched on 3 of the points, you will recieve 3, etc. If you get just the answer correct and nothing else, you will recieve one point, and if you have an answer but it is incorrect, you will recieve 0 points according to our rubric. === === Also while discussing the changes in our rubric, we also discussed the definitions of model, equation, and expression. Model= the "equation" you came up with without using the functions on the calculator (linreg/expreg). This is "your model" of the problem. The model only looks at part of the graph that relates to the problem situation, for example: y=2^x. the domain for a problem situation may only begin with x=0 where the equation and graph can represent x from negative to positive infinity. (Dr. Browning will be emailing us a revised copy of our in class rubric). ===

=== 2. We came to conclusions during the walking activity using the radar guns and going out into the hall to experiment with making different types of graphs. We were asked to make 1. a parabola with a maximum value 2. a parabola with a minimum value 3. exponential 4. a vertical line. In order to make 1. you start away from the radar gun, move toward the gun, and then move back away from it. Need to focus on your rate of change as you make the graph. Moving forward and backward is part of it but how do you get the needed curves in the function? This is true for both paraboloas. 2. Start close to the gun, move away from it and then come back towards it. 3. start close to the gun, take small steps at first, and gradually make the steps larger each time. thus changing your rate of change; you continue to increase the rate of change as you make an exponential function that has a growth factor >1 4. With experimenting, we discovered that this is not possible. One person cannot be in all the places on the y-axis when time/x=0. This means that this SLOPE is undefined. ===

INVESTIGATION 3 page 473 all pieces (do what you need to so that you understand the point of the question). Also do page 479.
**Our Working Rubric for Justification**

To receive full points (5) your justification should include:

• Response is clear/concise/organized.
 * Making connections from the context of the problem to the mathematics or from a solution to a property if there is no context. When in a context, be sure to use appropriate units.
 * A complete verification through describing each of the variables and what they mean.
 * A General rule that is explicit. Recursive rules can be included but are not necessary if not stated in the problem set up.
 * Response addresses all components of the model/formula/equation/expression/or rule.

To receive partial credit your justification would include only parts of the full justification. For each criteria not clearly addressed, a point is lost.

No credit will be given without some of the above things mentioned.

April 3rd:
1.experimenting with quadratic equations. 2.Determining how to find portions of our quadratic equations(where we get values for our variables). 3. Parabolas and the rate of change associated with them.

How we developed these ideas.
1.we worked on number 7 from page 467 in the quadratic functions unit to find a form for quadratics. 2. We used values from the table provided in the problem and "plugged them in" in order to solve for our unknown variables in the quadratic equation. 3. We used a radar gun hooked up to our calculators to try and match our movements with sample graphs.

AHE:
Voice thread task (this pink worksheet was handed out in class) pg. 480 2,4,7,10,12,28,30,34. Both are due Thursday the 5th of April.

FYI: notation like this: /ab means the square root of ab b/c I can't find anything to represent the square root sign on my computer!!
1. Counterexamples of square roots 2. Using variables within square roots & determining if they can be separated by multiplication (/ab = /a * /b) 3. Raising a base to the 1/2 exponent and raising that to the 2nd power 4. Punkin Chunkin!! (Intro to Quadratic Equations)

The way we developed the ideas were like this:
1. Counterexamples of square roots: Does /2+7 = /2 + /7? - When we did this on our calculators, /2+7 = 3 and /2 + /7 = 4.06. Therefore, /2+7 DOES NOT EQUAL /2 + /7 AND: Does /25+36 = /25 + /36? Since the square root of 25 is 5 and the square root of 36 is 6, we could look at it as Does /25+36 = 5 + 6? Our result was /25+36 (7.81) or /61 DOES NOT EQUAL 5 + 6 = 11

2. Using variables with square roots: We can take /ab and separate them using multiplication: /ab = /a * /b ALSO, /ab = (ab)^1/2 Because it's the definition of 1/2 power = square root ALSO, (a)^1/2 * (b)^1/2 = /a * /b Because this is the power rule by definition

3. Raising a base to the 1/2 exponent and raising that to the 2nd power: (a^m)^n = a^mn FOR EXAMPLE, (2^1/2)^2 = 2^1, and (5^1/2)^2 = 5^1 - This is like our previous example of (w?)^2 = w^1. To solve for ? we multiply the exponents as such: ? * 2 = 1, so ? = 1/2

4. Intro to Quadratic Equations, pg 462 Punkin Chunkin is a good representation of quadratic equations because when pumpkins are shot into the air, they start a a height of 0, gain height and distance, reach a peak and head downward. The shape the data creates on a graph is an "upside down" parabola. To find slope of the line, we revisited distance/time = slope(speed) If we pick any two points on the graphed line and connect them, we can find the slope. There is no constant factor of change because each line looks different.

HOMEWORK: Finish Investigation, just questions & not checking your understanding

= March 27th Lesson Overview: = There were no suggested solutions for p 315 #24d, p 302 #1, or p 348 #9. Look those over again as homework.

Next, we were asked to find a mean growth factor for the table of values.

 * Body Temperature (degrees F) || 50 || 59 || 68 || 77 || 86 || 98.6 ||
 * Brain Activity (%) || 11 || 16 || 24 || 37 || 52 || 100 ||

First, we found values for y=2(1.3)^x. This helped us get an understanding of what we needed to do to find the growth factor of the above table.

 * = x ||= y ||
 * = 0 ||= 2 ||
 * = 1 ||= 2.6 ||
 * = 2 ||= 3.38 ||
 * = 3 ||= 4.39 ||
 * = 4 ||= 5.71 ||
 * = 5 ||= 7.43 ||

=
The difference between x values is 1 while the growth factor of y is 1.3 (each time, the previous y-value is multiplied by 1.3).  You do not look for the difference between y-values because it is an exponential growth. =====

To find the equation for the first table, we have to find the difference in x-values and the factor of change in the y-values.

 * Change in x || 9 || 9 || 9 || 9 || 12.6 ||
 * Factor of Change in y || 1.45 || 1.5 || 1.54 || 1.41 || 1.92 ||

=
1. We'll look at the first two x and y values. From 50 to 59, there are 9 steps, therefore, there are 9 factors of change between 11% and 16% body activity. You have to take the 9th root of 1.45 to get the factor of change for 1 degree which is 1.04. This will be your base number, b. =====

Homework:
Pg. 315 #24d, Pg. 302 #1, Pg 348 #29, Pg. 330 #1, Pg. 325 a-c (CYU: check your understanding), and Pg. 355 Investigation #5 problems 1-5.

Coin (100)~ enter Heads=1; Tails =0
2nd quit 2nd STAT ~ #7 Sum ~List # The sum that you get will be the number of heads you have. Those will be removed so subtract that number fro 100. The amount left will be your new number. Repeat the steps above using your new number. Repeat all steps until you have 5 or less coins left.

=**March 22 Lesson Overview: **=

There was no solutions to Pg 302 #1 (we plan to revisit this problem) and Pg 315 #24d.

We reviewed the idea that the start value is also the coefficient (connect to the specific type of functions we are talking about now) (discussed with pg 338 #1). There were some questions on what 'half-life' meant, as a class we came up with the definition of: how long it will take for half of what you have to go away. The dictionary definition of half-life is the time required for something to fall to half its initial value. So a question is, are the two definitions the same?

Pg 340 #5f: We learned how we could 'zoom-in' on a table to get closer to the exact number (CAREFUL, not really exact solution yet; they are still approximate). We discovered that it might be easier to look at a graph if we have an idea on where the number should be. For this problem we have the radioactive chemical Strontium-90 and we start with 100 grams. Its half-life is around 50 grams and we need to find the year it reaches 50 grams to the nearest tenth. Once we estimate the half-life, we can now zoom-in on the table to find the year.

A new idea we learned in class was about DECAY FACTORS and DECAY RATE. Decay factor is the opposite of growth factor, it shows how something decays (self-explanatory). EX: y=10(2^x) When we create a table, we can see the growth factor is 2. This is a growth because the base,2, is greater than 1. In the equation y= 100(.6)^x, .6 is the decay factor. This is a decay because the base is less than 1. In the first equation, the growth rate is 100%. We extended our definition of growth rate is how much above 100% or one, so in the equation y=10(4^x) the growth rate is 300%. Decay Rate is how much below 100% or one. In our equation y=100(.6)^x our decay rate is 40% since .6 is .4 away from 1. The class noticed decay factor + decay rate =1. You can tell an equation is going to be a decay when the base number is a decimal. Be more precise. If the base was 1.5, does that mean this could involve a rate of decay????

Pg 334 -Check your understanding we are looking at B and H and writing out the properties to write in simplest form. AHE: was sent in an email to wmich addresses. The Project is due April 3rd.

Next week we are going to start working with Radicals & Square Roots, we touched on this a little at the end of class.

= = =**March 20th Lesson Overview:**=

We spent time having a discussion about problem #24 on page 315. The now / next rule for this problem was next=now x 1.5 starting at 20pi. The growth rate for this equation was 50% because we times it by 1.5, which means it is increasing by .5 every time. We then plugged in the data into lists in our calculators and used the exp. regression to see what equation the calculator formulated.

The equation we got was: y=51 x 1.03^x which is close to y=20pi x 1.5^x. In this equation, x is the radius of the circle and y is the circumference. you would use this equation if you know the radius in order to find the circumference. Another equation for this problem was: y= 20pi x 1.5^n. This equation can be used if you know the circle number.

We also went over problem b on pg. 303 we found that: i) linear ii) exponential ii) linear for iii we used the standard y=mx+b formula to find a formula for this problem. we said that b is the y intercept, so b=1.6. We came up with this value because when x=0, y=1.6, so it is our initial value. We then picked 2 points from the data set to find the slope and found that the slope was not consistent.

A.H.E. Go voer pg. 302 #1: If this was a test question could you do it?

We also went over a worksheet about diabetes that introduced us to exponential decay

We punched the information into lists on the calculator and created a scatter plot. We then used the standard formula y= a x b^x to create a formula for this problem. We set a=10 because when x=0, y=10, which would make it our y-intercept or our starting value we then plugged in a set of numbers from the data set to solve the formula: 8.8 =10 x b^3 .88 = b^3 if we take the 3rd root, which we learned in the previous class, we get the solution: .958 = b so, the equation is y = 10 (.958)^x Now / Next rule: Next = Now x .96 starting at 10

We shouldn't have a b value greater than 1 when there is decay or else it would not be getting smaller.

A.H.E: Finish Medicine Handout pg. 338 #1, 5, 6, (12-16 AMAYN), 20, 21, 29 44

Mathematical Practices: Make sense of problems and solving them.


 * March 15th Lesson overview**

General form for exponential equation: y=a*b^x

Pg. 311 a. v=a(b^t) a=700,000

b. 500=b^40 To solve for this take the 40th root from each side. (You need to have 40 plugged into your calculator before you find the x√ button)
 * t || v ||
 * 0 || 700k ||
 * 40 || 350Mil ||

b≈ 1.681

c. 16.81%

d. Put y=700,00∗1.168^x into y= and go to 2nd table and it takes 2.3 years to get to 1,000,000

We then talked about properties of exponents examples below:

2^3 * 2^2 = 2^5 (2*2*2) *(2*2) = 2^5  2*2*2*2*2 = 2^5

a^m * a^n = a^m+n (product of powers)

a^m * b^m = (a*b)^m

(a^m)^n = a^(m*n)

a^m/a^n= a^(m-n)

1= (7^5/7^5) = 7^(5-5) = 7^0

7^5/7^6 = 7^-1 = (7*7*7*7*7) / (7*7*7*7*7*7) = 1/7

(2/7)^-2 = (7/2)^2 = 49/4

We then discussed the affect of "a" on a graph. For example, as a gets bigger (the coefficient get bigger), the slope gets steeper fast. If "a" is negative, the result would be a reflective image of the positive over the x-axis.

the coefficient is also the y-intercept when graphing. For example:

y=4(2^x) y=8(2^x) y=4(2^x) +24

They shift according to the y-intecept of "a" so when x=0, y=4, 8, and 28 (going from top to bottom).


 * For homework we had a handout as well and problem Dr. Bemailed to us. Hope this helps! See you all on Tuesday!! :) **
 * Courtney P. **

March 13th

Answers for the Homework assigned before spring break

Bunny 3.1A NEXT+NOW * 1.8 Starting @100
 * 1) 1. 1.8, 180/100=1.8, 325/180=1.8, 538/325=1.8, 1050/583=1.8
 * 2) 2. P=100(1.8^)
 * 3) 3. 10yrs = 35709
 * 4) 4. 16 years

Bunny 3.1B Worksheets on graphs Y=a*b^x a is the y intercept. The Domain is going from – to +. A is bigger the slope will increase faster. b>0, steeper=grows faster, increases more quickly if 1>b>0, Decay, the smaller the fraction the quicker the decay
 * 1) 1. 1.2
 * 2) 2. 15 rabbits
 * 3) 3. 4 years
 * 4) 4. 26 million rabbits
 * 5) 5. 93 million
 * Impact of a:**
 * Impact of b:**
 * fractional bases means exponential decay**

Table and table set – Zoom in on a table by changing delta tbl and tblstart Growth factor 1.8 Growth rate 80%
 * Worked with TI-73 Tables**

Homework AHE: CMP orange handout on properties of exponents p 309 7,11, 12, 13-16 (AMAYN)*, 17
 * AMAYN : as many as you need (these problems are for skill building)

Session 16: March 1st We started out talking about page 295 in our coursepack. We went over and compared solutions and tried to figure out what the main mathmatical idea that the original authors we trying to find. From there we worked on talking about problem #1 letter D. We were asked to try to think of a complete justification of the problem. We should try to convince a skeptic on why the answer is correct. We then started to enter in our equations on our calculator. In order to do this you will need to hit 2nd Text and then enter in the rule/equation in the text editor. You don't need to get out of the text editor to put in numbers or any other numbers that are main buttons on the calculator, just push them and they will work. Every one in our class had the equations N=25(2^X) or 2^x(25). We are allowed to have both equations because the commutative property allows us to do so. We asked if it was a good use of the associative property and we found that <span style="color: #11c56f; font-family: Georgia,serif; font-size: 90%;"> the parentheses were not there because of the associative property but rather they were inserted for mathematical clarity on the order of operations <span style="color: #11c56f; font-family: Georgia,serif;"><span style="color: #11c56f; font-family: Georgia,serif; font-size: 90%;">. We then asked if the parenthesis were even mathematically important to have, and we determined that for our problems it is, because it separates the numbers and it helps us with the clarity of the equation. Howeve<span style="color: #11c56f; font-family: Georgia,serif; font-size: 90%;">r, <span style="color: #11c56f; font-family: Georgia,serif; font-size: 90%;"> they aren't necessary because of the order of operations. <span style="color: #11c56f; display: block; font-family: Georgia,serif; font-size: 110%; text-align: left;">We then went back to our problem. We talked about if there are 25 cells and how they double every quarter hour. In our equation our number of quarter hours will be represented by X. We then talked about where our start value would be. we said that all values should be x > or = to 0. This is true because at zero quarter hours we still have 25 cells. So at the initial introduction of bacteria, you have 25 cells in your body. We then asked how do we know this is right? If 25 is the start and each splits into 2 every quarter hour, then after one quarter hour there will be 50 cells. We determined that the base will increase by 2 every time, so with the number of bacteria you would see a pattern of (25x2)x2=100((25x2)x2)x2=200and so on. The number of times the base occurs is the factor. This would be a justification of our problem. We then started to determine a rubric for justification. We came up with this: Rubric for justification To receive full points your justification should include:
 * <span style="color: #ff0000; display: block; font-family: Georgia,serif; font-size: 110%; text-align: left;">Making connections from the context of the problem to the mathematics or from a solution to a property if there is no context.
 * <span style="color: #ff0000; display: block; font-family: Georgia,serif; font-size: 110%; text-align: left;">A complete verification through describing each of the variables and what them mean.
 * <span style="color: #ff0000; display: block; font-family: Georgia,serif; font-size: 110%; text-align: left;">A General rule that addresses all components of the model/formula/equation/expression/or rule.

To receive half credit your justification would include
 * Only a general rule OR only a few of the above things mentioned.

<span style="color: #b61e11; display: block; font-family: Georgia,serif; font-size: 110%; text-align: left;"> No credit will be given without some of the above things mentioned. <span style="color: #11c56f; display: block; font-family: Georgia,serif; font-size: 110%; text-align: left;">Finally we worked on making a full justification on a problem in our coursepack. We were recoreded on tape and we will be talking about them next class.For homework: Please complete the homework in this order Hope everyone enjoyed their Spring Break!~Amy Lynn
 * <span style="color: #0000ff; display: block; font-family: Georgia,serif; font-size: 110%; text-align: left;">Blue Fractional Growth Factors sheet
 * <span style="color: #0000ff; display: block; font-family: Georgia,serif; font-size: 110%; text-align: left;">Pg. 309 in our coursepack #7
 * <span style="color: #0000ff; display: block; font-family: Georgia,serif; font-size: 110%; text-align: left;">Green Exploring Exponential Equations lab sheet

<span style="color: #a400ff; font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Session 15: February 28
====<span style="color: #a400ff; font-family: Arial,Helvetica,sans-serif;">Today we started out by clearing our lists in our calculator so that we have more memory space. To do this, you have to hit //2nd mem and move down to delete and select lists, then delete all//. You can clear all the lists. The only lists that will stay in your memory will be L1-L6. ====

==== <span style="font-family: Arial,Helvetica,sans-serif;">The rest of class, we talked about "The King's Chessboard" problem. Dr. B shared a way that we can input multiple lists into the list function of our calculators. To do this, //<span style="font-family: Arial,Helvetica,sans-serif;">hit 2nd, catalog and select setup editor and press enter. //For this problem, we entered in //L1, L2, and L3.// For the problem, we used L1 to represent the squares that the king would place rubas in. We decided as a class to look at squares 1-10. To input all these numbers in a quick way, you would move up to the heading "L1" and //hit 2nd, stat, move to ops and select seq(// then fill the parenthesis with (x,x,1,10). Press enter, and 1-10 will be in L1. We entered the values that represents the number of rubas per square in manually. The values are as following: {1,2,4,8,16,32,64,128,256,512}. With these values, we made a recursive formula, which was NEXT=NOW*2. We discussed that these lists are not linear because L1 is increasing by 1 each time while L2 is increasing by a different, larger value each time, so we created L3 to find the difference between the L2 values. When you have the heading for L3 highlighted, //hit 2nd, stat, move to ops and select (delta)List(. (delta) is represented as a triangle on the calculator.// Then fill in the parenthesis with (L2) and press enter. ====

==== At this point in class, Dr. B gave us some terms to know. The term "constant factor" can also be called "growth factor". **A growth factor is the factor that represents the difference between values in a exponential expression**. For this example, the growth factor was 2. The other two terms we learned were "discrete values" and "continuos values". **Discrete values are a fixed set of values that are worked with. They typically exude fractions. Continuous values represent every value, continuously.** ====

Continuing with the king's chessboard problem, we went on to graph the lists that we made. To graph it on the calculator, //hit 2nd, plot.// The graph will be a scatter plot. Enter L1 and L2 for the lists. Set the window, and then view the graph.

Next, we came up with an explicit formula for the problem. We decided that two different formulas could work they are: y=2^(x-1) and y=1/2(2^x). We discussed that we know that these rules are exponential just by looking at them because there is a variable in an exponent.

To conduct a similar problem to the king's chessboard, we worked on pg. 295 #1 and 3 with our groups.

**AHE that was assigned for today was to do problems 24-30 on the pink sheet that was passed out last Thursday. Also, we are to do p. 314 #21 and 24.**
===** We also got our Exams back from last Tuesday. Dr. B is giving us the chance to have an additional point added to our score in we email her about why it makes sense to have a start value for a now/next, or recursive, rule. Do this by Thursday! **===

**:) Ashley**

<span style="color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Session 14: February 23 <span style="color: #000000; font-family: Arial,Helvetica,sans-serif; font-size: 90%;">Today we focused on what it means to find equivalences between algebraic expressions (today we focused mainly on linear expressions), what it means to simplify, and how to correctly do so. We did this by looking at the following pages:
 * Pg. 218 - Summarize the Math
 * Pg. 219 - #3 a and B
 * Pg. 220 - Properties
 * Pg. 221 - #4: c,e,f; #5: b,d,f; #6: c,d,e; #7: e,h; #8; c (these were from a previous homework assignment but we looked at them as a class)


 * __IDEA 1:__** On page 218, we determined what it means for expressions to be equivalent, and how to check if they are. We did it by answering the three questions:
 * **A:** We said that if you were to set "X" to any, but the same, number, then their solutions would be the same. We summarized this as, //If you put in the same input for both expressions, their outputs would be equal.// We also said that just because you put an equals sign between two expressions, it does not make them equivalent.
 * **B:** For graphing, you can put both expressions into the [Y =] button on the calculator and watching if both lines are the same. For a table, if you press [2nd][Graph] on the calculator, you can see the X values and their corresponding Y values from the equations you put in the [Y=]. This is also a way to check and make sure they are equivalent.
 * **C:** This question focused on the symbolic equivalence between expressions. We then visited the properties that allow us to make algebraic expressions equivalent to each other.

1. For any numbers a, b, and c, a + b = c is true if and only if a = c - b 2. For any numbers a, b, and c, a x b = c is true if and only if a = c ÷ b and b ≠ 0 We then looked at pg. 221, #7, e to practice making equivalences and determining which properties we used to find equivalences. If you would like the expression answer and the properties answer, I'll leave it at the bottom. But try it on your own first!
 * __IDEA 2:__** We went through the properties on pg. 220 together as a class that later on we'll use to identify which we are using in order to find equivalences. There are 5 on this page __**HOWEVER**__ we also used 2 we found on pg. 205 as well, making a total of 7 properties. The two on page 205 are:


 * __IDEA 3:__** We touched on the NOW/NEXT rule and what it means for exponential equations and expressions. So, instead of adding by the same constant, you multiply. We haven't said if it also is for division, but I'm sure we'll come back to it. We also talked about the importance of giving a starting number for a NOW/NEXT rule. For example, if my NOW/NEXT rule is NOW x 3 = NEXT, we want to make sure that we say that NOW < 0 because if you were to put in 0, you would get 0.

We were also given a new unit packet for our binders today. Please read pg. 291. We were also given a homework problem from the doc cam. It's a story about a peasant and reads as follows: "A peasant was working for a king, and the king is paying him in rubas. The King tells the peasant he will pay him by putting rubas on a chess board like this; if you place 1 ruba on the first square of this chess board, you get 2 on the second, 4 on the third, 8 on the fourth, and so on all the way up to 64 squares." Find the NOW/NEXT rule when NOW<1 Find the peasants earnings when the peasant is going to fill the 64th square with rubas.
 * __AHE:__**
 * WE WERE GIVEN A TAKE-HOME QUIZ TODAY!!!!!! IF YOU WERE NOT IN CLASS, PLEASE CONTACT DR. B VIA E-MAIL ASAP!!!!! It is due on Tuesday **

Now, here's the answer to pg. 221 #7,e. This is just one way that we did it. <span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; line-height: normal; margin-bottom: 0in;">10 – __15 x – 9__

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; line-height: normal; margin-bottom: 0in;"> 3

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; line-height: normal; margin-bottom: 0in;">(1) 10 – (15x – 9)(1/3)

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; line-height: normal; margin-bottom: 0in;">(2) 10 – 5x + 3=

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; line-height: normal; margin-bottom: 0in;">(3) 13 – 5x

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; line-height: normal; margin-bottom: 0in; text-indent: -0.25in;">1. Connecting multiplication and division

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; line-height: normal; margin-bottom: 0in; text-indent: -0.25in;">2. Distribution Property of Multiplication over Addition

<span style="font-family: Arial,Helvetica,sans-serif; font-size: 90%; line-height: normal; margin-bottom: 0in; text-indent: -0.25in;">3. Associative property of Addition

SESSION 12: February 16
We started class by going to page 201 in the course pack #1. Inequalities: Next we turned in our course pack to page 171 #10 this problem focused on the calculator problem with lists, graphing, fitting a line (manually), and residuals. However we used this problem to discuss what equation in context meant and in mathematical terms. Graphing inequalities: More inequalities: Other Important Information:
 * EXAM **** ON TUESDAY FEBRUARY 21, 2012 **
 * Dr. Browning asked us to try and figure out how many days old is Rachel when she weighs less than or equal to 201oz.
 * Mal told us that we have to look at question #1C, iv which has the inequality:
 * 96+2.1x ≤ 201 Where x= age in days, y= weight in ounces
 * Quick Poll: answer is x ≤ 50
 * To check if this was correct students offered numbers to “plug” into the inequality
 * 50- this works
 * 0- this works
 * 51- this does not work because the answer is greater than 201
 * Dr. Browning asked how did we get the answer of x ≤ 50.
 * Mason offered the steps of:
 * 96+2.1x ≤ 201 (subtract 96 from both sides)
 * -96 -96
 * 2.1x ≤ 105 (then, divide each side by 2.1)
 * ÷2.1 ÷2.1
 * X ≤ 50
 * Dr. Browning asked: Can we do this problem in our calculators and only have to push enter once?
 * Mal: type into the calculator (201-96) ÷ 2.1 [enter]
 * This will only make you have to click the enter button once, we just have to remember order of operations and where to put the parentheses.
 * Someone asked why the inequality stays the same direction when solving this inequality. So instead of x ≤ 50, why isn’t it x ≥ 50.
 * Courtney said we’re not dividing by a negative number and therefore we don’t need to switch the inequality sign.
 * We used Dustin’s equation that he found with his manual fit line: y= -1.1x + 118.91
 * Dr. Browning asked, “In context of this problem (context- video and buyers of the video) what does -1.1 and 118.91 mean?”
 * Dez'Raii said that -1.1 meant: As the buyers decreases by 1.1 the cost of the video increases by $1
 * 118.91 means that: At $0 (free video) the number of buyers is ≈ 119 buyers.
 * Mathematically -1.1 is the slope or rate of change (it can be called rate because it has a unit attached to it [buyers].)
 * Mathematically 118.91 is the Y-Intercept.
 * We were given the inequality: y ≥ 96 + 2.1x[[image:tgraph.cgi.gif width="259" height="264" align="right"]]
 * Asked: why is this linear
 * The line on the graph represents for those x-values the least ounces that Rachel can weigh.
 * The graph is shaded above the line because in the inequality y is greater than or equal to 96+2.1x
 * **__To make the graph an inequality on your calculator__**go to [y =] and on your calculator screen you will see:
 * / Y1=
 * Then if you move your cursor over the "/" and hit enter it changes the graphs style you can make it shade above your linear equation (which will be represented by a shaded right triangle in upper right corner) or below (which will be represented by a shaded right triangle in lower left corner).
 * Instead of the / the shade triangles will appear.
 * Instead of the / the shade triangles will appear.
 * Page 77, question A #1, 2, 3
 * We graphed two inequalities:
 * -.6c + 480 > s c= cars
 * 2c < s s= suv
 * We decided the graphs will intersect because one inequality has a negative slope (-.6) and the other has a positive slope (2).
 * When we graphed the 2 inequalities they intersected and there was a region where both of the graphs both overlap to describe this region we came up with:
 * 0 <y< 480 **AND**0< x< 180
 * To find these points we used the trace button on our calculators and looked at our graphs.
 * On an **__assessment__** Dr. Browning would put in the equations we came up with and then check and see if our described region matches.
 * The inequalities are **strict inequalities** and when graphing them use a **dotted line**.


 * __EXAM INFORMATION:__**
 * We will have “class time” before taking the exam from 12-12:30 and then the rest of the time is for the exam.
 * Dr. Browning said it should only take one hour
 * No property content will be on the exam because we weren’t able to cover the material in class.
 * However on page 220 there are problems that relate to properties.
 * **__Inequalities__** will be on the exam
 * **Preparing for Assessment** **From Dr. Browning’s Email**
 * The items below attend to ideas throughout the linear functions unit and would be a great start for a review. Make sure you go through your I Can's list and check that you can do/understand all statements.
 * 202 #5, 8, 13, 18 a and b but only for the Men's times (although feel free to do the women's times if you want), 19, 23, 28, 39
 * P 161 √ your understanding a and b. P 168 #2. For the picture you drew in class with the blocks, write up a Now-Next recursive formula and, if possible, an explicit equation.
 * P 175 16; P 178, 24, 25


 * HOMEWORK:**
 * P. 192 #3c: show on calculator.
 * Refer to p 205 #15 for some justification of how you know your process for solving will be true all of the time.
 * Complete Investigation #3 p 194 in Core plus materials (course pack). #1-3 and 5.
 * Read p 220 for properties; in groups (I guess by yourself now), complete 4 c, e, f; 5 b d, f; 6 c, d, e; 7 e, h; 8 c. For equivalency “test”, refer to table for a couple of examples.
 * We'll probably have a short quiz next week on ideas in the equivalent expressions lesson (that would be all of the above activities).
 * At the end of class, I was going to have you complete #2 p 68 from CMP Investigation 4 as a group problem. This is a good problem for reviewing for the exam.
 * AHE: Problems like the following will not be on the exam but relate to the equivalent expressions lesson. See what you can complete for Thursdays' class: P 225: 5-7; p 229 #19
 * Also, don’t forget the cube question we were assigned at the end of class:
 * See the pattern, figure out a NOW NEXT rule (recursive), and an explicit formula.

__** Feel free to add anything I possibly left out that could help us on our EXAM on Tuesday :) **__

Brittany ;)

__ Session 11: February 14 __
Today we focused on using our TI-73 calculators to find the line of best fit using the manual-fit feature on our calculators. We used the data from p. 171 #10 to create our list and plot the points on a graph. We also discussed what is in each list (L1, L2, L2, etc.) in our calculator. I will provide a brief list of what each list includes (below) and will go into detail about how to find each list. However, I will be using my equation -1.24x + 124.37 that I came up with from my manual-fit.

L1 = x values (In this case, cost (in dollars)). L2 = y values (In this case, number of buyers). *We use L1 and L2 to find the line of best fit (Manual-fit). L3 = Predicated values from manual fit equation. L4 = Actual - Predicted (L2 - L3) L5 = Finding the absolute value of L4.
 * This is how it should look when placing it into the __HEADER__ of the list: -1.24(L1) + 124.37
 * This is how it should look when placing it in to the __HEADER__ of list 4: abs(L4)
 * Where is the abs( key on your calculator? Math -- NUM -- 1:abs(

Note: Need help understanding each of these lists? Make sure you read "Here is the detailed process of how to use your calculator" below (In red).

__Here are some other key points we discussed in class:__
 * We also focused on the "trace" key on our calculators when looking at a graph. Use the up or down arrows switches if you are on the points of a graph or the line. Make sure you are comfortable using this as a question may arise on the test where you may have to use this feature!
 * We discussed the importance of having a reasonable window.
 * As a class, we looked at all our graphs from problem #10 on page 171. We discussed "Does the data look linear? If so, what makes it linear?"
 * We said that in this case, as x increases, the number of buyers decreases. However, Dr. B brought up a good point: "Can we think of a case when x increases and y is decreasing, and it is NOT linear?" This is the shape of the graph that one of your classmates came up with:[[image:pm_graph_3.2.jpg width="272" height="185"]]
 * We also discussed that we should say it is: negative, has a general trend, and is "__Close to__ a line"
 * We also looked at how to find a linear regression equation.
 * 2nd -- Stat -- CALC -- 5: LinReg(ax+b)
 * Note: When finding the linear regression, it is default that it will find the linear regression equation of L1 and L2. However, if you named your lists something different, you will need to find the linear regression of those two lists. It would look like this then: LinReg(ax+b) Name of list, Name of list
 * MAKE SURE YOU PLACE A COMMA BETWEEN THE TWO LISTS
 * If you would like to automatically place your linear regression equation in Y=, before you press enter go to 2nd -- VARS -- 2: Y-vars... -- Y2
 * This will put the equation automatically in Y=

__ **Here is the detailed process of how to use your calculator:** __ 1) On page 171 #10, place the cost (in dollars) and the number of buyers into a separate list. __Note__: You may name your lists L1 and L2 or give them your own name, but we will be using L1 and L2 in this case.
 * L1 || L2 ||
 * 25 || 93 ||
 * 30 || 89 ||
 * 35 || 77 ||
 * 40 || 71 ||
 * 50 || 64 ||
 * 60 || 55 ||
 * 75 || 38 ||

2) Now that we have placed the __actual__ x and y values into list L1 and L2, it is time to draw our line of best fit using the manual-fit feature on our calculator. However, before we can do this we need to make sure our window is reasonable for the data.

Xmin = 20 Xmax= 80 Xscl=10 Ymin=30 Ymax=100 Yscl= 10

This is the window I used for my graph. However, your window may look slightly different. Now it is time to draw our line using the manual-fit feature.

2nd -- Stat -- Over to CALC -- 3: Manual-Fit -- Enter You all should know how to use the arrows to draw the line of best fit as we went over this in class. **IMPORTANT: Make sure you press enter TWICE after you've drawn your line of best fit or it will NOT save!**
 * How do we draw the line of manual-fit?** Follow the following steps:

3) Now that we have our line of best fit (manual-fit), we need to find out what our equation is from our manual-fit line.

4) Since we have our equation we need to go back to our list and create list 3 (L3) to find our predicted values from our manual fit equation. - Make sure you are in the header of the list and type in your equation, BUT for the "x" in our equation we are using the values in list 1. Your equation will look like this in the header of L3.

-1.24(L1) + 124.37

5) Now we need to create a new list (L4). In this list, we are taking the Actual y values and subtracting from the predicted y values (L2 - L3). - Make sure you're in the header of L4 and type: (L2 - L3) - You are taking the actual y values in L2 and subtracting from the predicted y values in L3. 6) Lastly, we need to create one more list that will find out ABSOLUTE ERROR. We will then use this list to find our mean to see how close or how far away we are from the residual. - In order to find the absolute value, we need to create a new list (L5). Again, make sure you are in the header of the list and follow the steps: 1) Math -- NUM -- 1:abs( -- Enter 2) Now that we have our absolute value on our calculator, we need to place list 4 (L4) to find the absolute value of our error. 3) This is how it should look on your calculator: abs(L4) 4) Press ENTER Before we move on, if you are using my equation from my line of manual-fit, this is how your lists should look:
 * = L1 ||= L2 ||= L3 ||= L4 ||= L5 ||
 * = 25 ||= 93 ||= 93.7 ||= -.37 ||= .37 ||
 * = 30 ||= 89 ||= 87.17 ||= 1.83 ||= 1.83 ||
 * = 35 ||= 77 ||= 80.97 ||= -3.97 ||= 3.97 ||
 * = 40 ||= 71 ||= 74.77 ||= -3.77 ||= 3.77 ||
 * = 50 ||= 64 ||= 62.37 ||= 1.63 ||= 1.63 ||
 * = 60 ||= 55 ||= 49.97 ||= 5.03 ||= 5.03 ||
 * = 75 ||= 38 ||= 31.37 ||= 6.63 ||= 6.63 ||

7) We now need to find the mean to see how close we are to the residual. Here is how we find the mean: 2nd -- Stat -- MATH -- 3: Mean( -- Enter Now that you have the command, we need to find the mean of our absolute error which is in list 5 (L5). This is how it should look on your calculator: mean(L5) Now press Enter. If you are using my equation you should get a mean of 3.32.

I know this is very detailed, but it is important that we all understand this for our test on Tuesday! I hope this helps! See you on Thursday!

**The homework for Thursday is:** 1. Find the solution to the system of linear equations: 4x - y = 6 and 8x -2y = 12. What does it look like and what does it mean?2. Page 192 #3 c, what would your calculator screen look like? Refer to p 205 #15 for some justification of how you know your strategy for solving works all of the time. Use the properties to help justify your moves in solving the equation on p 192.3. **Read** through Investigation #3 page 194. We will work on that in class. If you've thought about it already, this may go a little faster in class.4. From CMP with front page as ACE, **complete** #2 p 3; this should be a familiar type of question5. Core plus: P 197 read Investigation #4. Complete #2-4; this will be more on solving a system of linear equations so hopefully will go smoothly6. Make sure you can respond clearly and completely to the Summarize the Math on p 200; this is a check on what you really understand vs just can do.

- Chad

__** Session 10: February 9 **__ Today we had a substitute teacher, so we spent most of the time going over homework problems and working in groups on some new problems. We began by going over the homework packet, "The Shapes of Algebra" and many of us had questions on 5.2. Much of the homework dealt with the term inequality which we have not talked much about. ** What is an inequality? ** What is a linear inequality? How are they different? Up to this point, we have been looking at Linear equations in "slope-intercept" form: **y= mx + b**. In this form, one variable depends on the other. But during class today, and in our homework we began to work with a new form for the linear equation. -> **ax + by= c.** In this form, it is more natural to combine the values of two variables. We worked on 5.2 ("The Shapes of Algebra" packet") and decided on an equation to represent the problem. This is the story problem followed by our equation: Vince finds out that his family's car emits an average of 0.75 pounds of carbon dioxide per mile. The SUV emits an average of 1.25 pounds of CO2 per mile. Suppose Vince's family wants their total CO2 emissions to be exactly 600 pounds per month: .75x + 1.25y= 600 (ax + by= c form)  We then talked about ** graphing our equation **. Could we use the above equation to make a graph on our calculator? We decided we needed to **change the form of the equation** to y= mx + b form:  y= -.6x + 480  We decided that the 480 represented the number of miles you could drive with the SUV if you drive 0 miles with the car. We decided that the .6 represented the ratio of lbs/mile of car miles to SUV miles, so per 1 mile by car you would have a .6 mile decrease by SUV. Finally, we created the graph for this equation on our calculators! All of the story problems we worked on in groups **seemed to have 2 or more variables**, which was a new concept, and we began to solve the linear equations, plugging in numbers to solve for different variables. One story problem led us to the equation: 5s + 10c = p Given the value of p as 600, we were asked to look for the **different point coordinates that might be found on a graph for the equation**. For example, if we plug 5 for c: 10(5) we will get 50. 5s + 50 = 600 We would then subtract 50 from 600 to get 550 and divide the 550 by 5(from the 5s) to get the value of s! s = 110 ---> So our coordinates would be (s, c) or (110, 5) We ended class by working on the packet 3.1-3.3 and matching **I-CAN statements** to each section. The homework for the weekend is: ** *CMP pack (pg. 46) #34, 35, 48, 49, 50, 51, 64 **** *Complete Investigation #4 in CMP pack **** *In our Coursepack (pg. 191) Linear Functions; Investigation #2- read and complete #1-3 ** I switched the alignment of the text from centered to left-justified. Leah, if you prefer to have it centered, go ahead and move it back. I was thinking it might be clearer to read left justified but perhaps not. Dr B Have a great weekend! :) Leah

Hey guys! Sorry this is so late, I totally forgot how to get into this! I just took a picture of my notes, because it is kind of hard to explain some of the calculator stuff that we did in class yesterday.

Tracy

__ Session 8 Feb 2 __ Today we focused on Linear functions and how to determine whether or not a given expression, equation or rule was a function. First we relearned exactly what a function entails. An input and an output are the two basic units of a function. We learned that "Input and Output" are two other terms for x and y or domain and range. What makes something a function is when only one output is connected to any input. input. ~For example: y=3x. When x = any value, there will only be ONE y value that corresponds with it.

The next idea we focus on was a linear function. A linear function is an function that is graphically straight. a straight line. There is no parabola (Rainbow) in the graph. graph (essentially no curves of any kind). y=3x is a Linear Function. y= is not, because the line on the graph to represent this equation curves.
 * 1) /x x^2

The last idea we looked at was what kind of line on a graph negates a function? A question posed by myself about a vertical vs. horizontal line sparked some debate as to what is considered a function and what is not. Originally I thought that any input can only have one DISTINCT output. For instance, a perfectly horizontal line on a graph means that every input has one output, but originally I believed it was not a function because every output was the same. However, it was determined that the outputs do NOT have to be unique. Through this, we relearned the "Vertical Line Test" in that if a perfectly vertical line can be placed anywhere on the graph without intersecting two points simultaneously, then it is a function. This vertical line test showed was the proof that a perfectly horizontal line was indeed a function.

I thought we also began looking at fitting a line by eye to data?? AHE: Finish the other pages from the CMP handout, Thinking with mathematical models p 28, 29 and 32. On p 168 9, 11, 12, 28, 32, 38

__** Session 7 January 31 **__ 1. We first discussed on how the new unit, linear functions, connects with the past unit of proportional reasoning. As a class we decided that both units contained directly proportional and non-proportional equations. Also, both units were connect through the idea of slope or constant rate of change.

2. Next we discussed what a liner graph looks like. As a class we said that linear graphs were simply straight lined graphs. We did a few poles on the calculator to be assured that the whole class agreed on what was a linear graph or equation and what ones were not.

3. We went over the "bordering a pool" worksheet to make sure everyone understood what the assignment asked of us. Problems arose when we tried to define recursive and explicit formulas. Our class defined an explicit formula as a formula where you directly get the solution if "t" or "s" is assigned a value. Ex/ y= 4x + 5. Next we went over what recursive meant. Recursive formula/thinking - you start with the first value and start adding on a constant amount. An example of this would be ... Next = Now + 5. In creating a recursive formula, you want to use the slope in the explicit formula and not the y intercept. You use the slope because in order to get the next value in the formula you need to go up by the same constant amount which is the slope.

4. We also spent time on a worksheet that required the TI-nspire cx calculator. We will not be using these calculators again however, so this worksheet was not assigned as homework, but only in class work.

AHE: starting on p 168 2, 8, 14, 20, 25 P 161 Investigation 3 start reading bottom of p 161 and complete 1 and 2.

Heather :)

__** Session 6 January 26 **__ In today's session we turned in our take home quiz and were given a worksheet entitled "Bordering a Pool".

1. We focused on the first few tasks of the worksheet, but to get there we first learned about domain and range. "When we choose values for the independent variable, it is called the domain." or rather the values that we can choose for the independent variable are chosen from what is called the domain. The class then worked on determining the domain of the worksheet. Since a pool border needs to be around a pool, the side lengths for the pool had to be a whole number greater than zero.

2. Our second big definition was for range. "The values found for the dependent variable are in the range." Since the border could not be a negative length or non-existent, we once again determined the range of the worksheet to be whole numbers greater then zero. In this case, the class determined that the range and the domain were the same.

3. We then went over how we came up with expressions for the worksheet. Most people came to a 4S+4, but there were a few other ideas as well. We then discussed the thinking behind Ahmed in question #4.f: (S+2)^2 - S^2. Trying to understand what he did, a few students pitched around ideas. What we came up with was that Ahmed was taking the area of the pool plus the pool border ([S+2]^2) as the pool border's length was two more than the side of the pool. Then he was subtracting the area of the pool (- S^2) leaving just the area of the tiles left over.

4. The class was then challenged to how to find an expression (or equation) without having the context, and only the data table from the question. We came to the conclusion that you look at the change in your y (or dependent value) over the change in x (independent value) to determine the slope. Then find the y-intercept (when x is 0) to find out if you have to add or subtract anything.

Homework: Finish handout. p 155 Linear Functions without contexts, #5, summarize on p 156. P 160 #5 and #6, Summarize p 160. (This part can wait until Tuesday: Starting on p 168 2, 8, 14. 20, 25)

- Dustin

__** Session 5 January 24 **__

We are preparing for take-home quiz #1. We wrote out all of the solutions to the worksheets that we have worked on. If you need the solutions, e-mail a friend in the class.

1. We focused on task 5 of "Burning the Candle at Just One End." This task helped us realize that two slopes may be different yet constant. We were able to graph the two independent lines on the same graph. We noticed that the line representing Q (the slower burning candle) has a slope that is less steep than the line representing P (the faster burning candle). When we compared the two graphs, we realized that the two lines do not have the same vertical distance between them at any given point, but they do have a multiplicative relationship (the same ratio). This proves that lines P and Q are directly proportional to each other.

2. We also talked about how two different people can have two seemingly different equations, but when we use properties like the Distributive Property, we see that the equations are the same. Here is an example of the Distributive Property... 5(1 + 3) = (5 x 1) + (5 x 3). It also depends on what you use as your dependent and independent variable...

3. We talked about graphing and how the graph will depend on how the equation is written. If you have an equation, y= 5x, x (in this case) is the variable that we choose. Y is therefore the dependent variable that is dependent on x. X can be whatever we choose it to be, once we solve the equation, x changes what y will end up being. The independent variable should be placed among the x-axis with the dependent variable on the y-axis.

AHE: Take-home Quiz #1 for class on Thursday.

-Kari

__** Session 4 January 19 **__

Thursday in class we continued to explore the different relationships between two variables: direct, inverse, and non-proportional. We went over how to tell them apart by looking at an equation, table, or graph.

1) **Direct Variation** : is a relationship between two variables were their ratio remains constant. For instance if x increases so does y and if x decreases so does y. An__equation__ for a direct variation is y=mx. Like problem #1 on our at home extension from last Tuesday, y=32x where Taiwanese dollars is y and U.S. dollars is x. Thus, the number of U.S. dollars (x) increases as does Taiwanese dollars (y). A __graph__ that would fit this relationship would have to go through the origin at (0,0) and the__table__ that would represent this would look like this: x=0 y=0, x=1 y=32, x=2 y=64, x=3 y=96, x=4 y=128, etc.

2) **Inverse Variation** : is a relationship

<span style="background-color: #ffffff; display: block; font-family: Verdana,Arial,Helvetica,sans-serif; font-size: 12px; text-align: left;">when one variable increases the other decreases in proportion so that the product remains the same always. For instance if x increase y decreases and if x decreases y increases. An __equation__ for an inverse variation is y=3/x. An example of this type of relationship is problem #2 on our AHE from last Tuesday where 3 is the number of hours divided by x which is the number of people to get y is 1.5. Thus as x increases y decreases. A __graph__ would have two lines that never cross the axes where x and y get closer and closer to 0 but never equal 0, but the lines will not be linear. A table for such a relationship would look like this x=1 y=3, x=2 y=1.5, x=3 y=1, etc. <span style="color: #ff00ff; display: block; font-family: Verdana,Arial,Helvetica,sans-serif; font-size: 12px; text-align: left;">3) **Non-proportional Variation** : a relationship where the two variables do not have the same ratio. An __equation__ for this would look like y=4+x (in a non-proportional relationship you are always adding something). An example of this would be like problems #3 and #4 on our AHE from last Tuesday. In problem #3, for every x laps Bob has ran, Marty will have ran 4+x laps. The __graph__ that would fit this type of relationship would never pass through the origin and the__table__ would look like: x=2 y=6, x=3 y=7, x=4 y=8, etc.

*We also learned what a __slope__ is in these relationships which is the rate.

// AHE: You can work on the extra problems that were given to us at the start of class for practice for our quiz on Tuesday! :) //

-Mallory

__** Session 3 January 17 **__

We started class by working on questions from the homework, then went on to cover some new material. Here are the main points from discussion today...

1. There are different ways to write the same "rule" or equation for a question. For Task 3, we came up with n=168/t t=168/n (24/t)7=n All of these equations are accurate representations of the "rule" for this candle question. We also talked about justifications- be sure to pay attention to your explanations about how your rules work, and how you know they work every time!

2. **Complex fractions and ordinary fractions**. In the equations above, we see examples of both of these. Ordinary fractions are fractions with whole numbers in the numerator and denominator. In complex fractions, the numerator and denominator may be rational numbers (the denominator can't be 0 in either case, of course!) In (168/7 / t)7=n, the numerator is a fraction itself, so this is a complex fraction.

3. **Direct proportions, inverse proportions, and non-proportional relationships**. Direct relationships can be described as relationships in which the variables change in the same way. If one variable increases, so does the other. If one variable decreases, so does the other. A direct relationship is also a proportion if variables change by the same constant amount- an example is Task 1 on the candles handout. Inverse proportions are proportions in which the variables do not change in the same direction. For example, as one variable increases, the other decreases. Again, since they are proportional relationships, they vary inversely by a constant rate or factor. Task 4 on the candles is an example. Some relationships are not proportions at all - for example, Task 2 on the candles. In these relationships, there is one constant that does not involve the amount that the variables change. In this example, 105 stays constant, and the variable "x" only represents what is subtracted from 105. As x increases, h decreases- so it's inversely related- but it's not proportionate. If we compared several pairs of values, we would not find a constant ratio maintained between the pairs. For example, when x is 10, y = 95. The ratio of x:y is 10:95. When x = 15, y = 90. The ratio of x:y here is 15:90. Did the ratios remain constant? We also talked about the graphs for each of these relationships. Graphs of direct proportional relationships cross through (or start at) the origin, (0,0). Inverse proportional relationships do not cross the origin.

AHE: "Proportional Reasoning Strategies" Handout (look for the related article in your email) "Summary of Direct, Inverse, and Non-Proportional Relationships" Table for your own notes "I Can Statements" handout Think about the last example in number 1 above. Could we simply change the 7 to 168 in this equation without sacrificing the meaning of the rule?

-Candis

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