Here are a few things to note about this school:
- Like all middle schools in the district, this school has a period rotation. But as I explained back in my May 31st post (the last time I subbed at this school), this school always begins with first period, and then periods 2-6 rotate. These rotations correspond to the days of the week, so that Monday starts with second period, Tuesday with third, and so on. Since today is Tuesday, the rotation goes 1-3-4-5-6-2.
- The last time I subbed for this teacher was May 7th-8th. This year his classes are slightly different, but as I mentioned in my May 7th-8th posts, he does have a zero period class (and so the rotation is actually 0-1-3-4-5-6-2).
- Normally, a short homeroom leads directly into first period, but not today. One Tuesday a month (often the second Tuesday), there is an extended homeroom for a special digital citizenship lesson. This is the same bell schedule that I mentioned in my March 13th post.
Oh, and finally, this is the first of a two-day subbing assignment. This means that for today's "Day in the Life," the focus resolution must default to:
1. Implement classroom management based on how students actually think.
7:15 -- Zero period is the first Algebra I class. The new material is Lesson 4-2 of the Big Ideas text, on solving systems of equations by substitution.
8:15 -- Ordinarily, homeroom in this district is only about 15 minutes, but once a month, it is extended for a special lesson. The subject is digital citizenship -- specifically, digital fingerprints.
8:50 -- First period begins. This is the second Algebra I class.
9:40 -- First period ends and third period begins, since it's Tuesday. This is the same class that this teacher had last year -- ASB. They are creating posters for several events -- coming up next is a pumpkin pie eating contest on Thursday, as well as ASB elections (that another middle school already had last week).
10:30 -- The third period students leave for snack.
10:50 -- Here is where the teacher's schedule differs from last year. Fourth period is now a Math Support class for students who need extra help. Just like last year, students work on ALEKS and Prodigy on Chromebooks. But this time there is more accountability involved. Students must write their answers on scratch paper, which they must turn in. Some students choose to work on their math homework instead of ALEKS.
11:40 -- Fourth period leaves. It is now conference period, which leads directly into lunch.
1:10 -- Sixth period arrives. This is the third Algebra I class.
2:05 -- Sixth period ends and second period begins. This is the final Algebra I class.
2:55 -- Second period leaves, thus concluding my day.
So much of today's schedule elicits comparisons to the old charter school. First, the main lesson for today, solving systems by substitution, is one of the last topics I taught two years ago. Many of today's students do well today, while my old students from two years ago struggled. Much of this is because they had trouble solving equations in one variable. This was due to a combination of how the Illinois State text was organized (one standard at a time rather than traditional chapters) and how I managed the class (with continuous talking and the students remembering nothing). Of course, we can't necessarily compare my Common Core Math 8 kids to these Algebra I students (but once again, parts of Math 8 and Algebra I overlap).
The digital citizenship lesson reminds me of Monday coding lessons. But I don't recall the coding teacher having much to say about digital footprints (with data mining and everlasting negative posts being key issues).
Math support on ALEKS reminds me of IXL. Recall that IXL accountability was a huge problem for me two years ago. In some posts I wrote that I should have created an IXL accountability form, but this teacher's scratch paper idea is much simpler.
As for management and behavior, second period is named the best class of the day. First period is a bit of an enigma. The most talkative boys are the most engaged in digital citizenship, but they revert to their normal selves when it was time for math. I named this class the second best, but I worry about them tomorrow when there's a math lesson but no digital citizenship lesson.
I don't mean to keep talking about Election Day, Proposition 7, and DST. (Then again, Election Day still isn't over at today's middle school!) But anyway, over the weekend I saw a newspaper article about the areas (counties) where the various propositions passed and failed. To make a long story short, Prop 7 passed in every county except those in the Central Valley.
This immediately reminds me of the Joe Mathews and his Three California time zone proposal:
http://www.zocalopublicsquare.org/2018/08/06/california-needs-three-time-zones/ideas/connecting-california/
I don't mean to keep talking about Election Day, Proposition 7, and DST. (Then again, Election Day still isn't over at today's middle school!) But anyway, over the weekend I saw a newspaper article about the areas (counties) where the various propositions passed and failed. To make a long story short, Prop 7 passed in every county except those in the Central Valley.
This immediately reminds me of the Joe Mathews and his Three California time zone proposal:
http://www.zocalopublicsquare.org/2018/08/06/california-needs-three-time-zones/ideas/connecting-california/
The counties where Prop 7 failed line up almost perfectly with Almond Time, which is where Mathews also proposes preserving the biannual time change. Prop 7 passes in the counties that are in either Redwood or Cactus Time.
Many people believe that the reason for the biannual clock change is "because of the farmers" -- and the fact that most of California's farmers live where voters opposed Prop 7 only reinforces this. Most likely, it's that the Central Valley is where the disadvantages of having a single clock Year Round are the most pronounced.
Many people passed Prop 7 but prefer Year Round Standard Time. Therefore the passage of Prop 7 in Redwood Time is consistent with Standard Time in those northern counties. I'm not sure whether I'm fully on board with Mathews, but it does give me something to think about considering what the Prop 7 map looks like.
Perhaps I should have made this a traditionalists' post since I'm teaching eighth grade Algebra I -- the class they've been promoting recently. Instead, I wish to link to the posts of a few math bloggers who (over Veteran's Day Weekend) made their first posts in months.
Let's start with our queen, Fawn Nguyen:
No, Nguyen's post has nothing to do with the recent tragedies. Instead, she writes about one of her favorite and most influential teachers as a young middle school student -- her Home Economics teacher, Mrs. Quiggle.
I've mentioned my own favorite teachers a few times before, most recently in my August 22nd post.
Meanwhile, I've been writing about Sarah Carter's Slope Dude a lot recently. In this post, Carter resumes her (hopefully weekly) Monday Must Reads series:
https://mathequalslove.blogspot.com/2018/11/monday-must-reads-volume-48.html
Also, Jacqueline Richardson is a blog I read last August, but I didn't link to her because she's now a math coach (rather than a math teacher). But I can't help but think about her most recent post, which is about a balancing equations activity based on NASA. Today in class I see one girl wearing a NASA shirt, so I wonder whether she'd find this activity interesting:
http://highheelsandnumber2pencils.blogspot.com/2018/11/balancing-equations-with-spacex-zuma.html
Lesson 6-2 of the U of Chicago text is called "Translations." In the modern Third Edition, we must backtrack to Lesson 4-4 to learn about translations.
This is what I wrote last year about this lesson:
To emphasize the coordinate plane, I've titled this lesson "Translations on the Coordinate Plane" rather than just "Translations." Yesterday I warned you that another translation lesson was coming up soon, and as it turns out, that day is today.
I said that there are several ways to prove that the transformation mapping (x, y) to (x + h, y + k) really is a translation. First is a coordinate proof using the Slope and Distance Formulas, but I mentioned that we can't use those until after dilations in the second semester. Second is to manipulate the transformations until various mirrors cancel, but that requires arcane symbols such as:
T = r_U(n) o r_U(m)
and I said that it would only confuse students. Fortunately, there's a third way to complete the proof that I alluded to a few weeks back -- and I've decided to use this form of the proof in today's post.
We begin by noting that the two transformations (x, y) -> (x + h, y) and (x, y) -> (x, y + k) are already proved to be translations. The first translation slides points along the x-axis h units, and the second slides points along the y-axis k units. But how are points not on either axis transformed?
Now that we're in Euclidean geometry, we can prove that translations slide every point the same distance -- which is what we expect translations to do. We consider the points (0, 0), (a, 0), (a, b), and (0, b), and these points are the vertices of some quadrilateral. We know by definition, the side along the x-axis has length a and also the side along y-axis has length b. We also know that three of the angles are right angles -- the angle at (0, 0) since the axes are perpendicular, the angle at (a, 0) because x = a is perpendicular to the x-axis, and the angle at (0, b) because y = b is perpendicular to the y-axis. In Euclidean geometry, we know that the sum of the angles of a quadrilateral is 360, so if three of the angles are right angles, so is the fourth, and the quadrilateral is a rectangle.
But now we know the lengths of two sides of this rectangle, a and b, and we want to find the lengths of the other two sides. Of course, the other two sides must also have length a and b as opposite sides of a rectangle are congruent. Yet how do we know this? Most Geometry texts would now say that every rectangle is a parallelogram and the opposite sides of a parallelogram are congruent, therefore the opposite sides of a rectangle are congruent. But unfortunately, the U of Chicago text doesn't give the Parallelogram Tests and Consequences until the last part of Chapter 7.
But let's consider the Quadrilateral Hierarchy yet again. Now only is every rectangle a parallelogram, but every rectangle is an isosceles trapezoid in two different ways -- and we've indeed proved that the opposite sides (legs) of an isosceles trapezoid are congruent. Therefore, we really do know that the opposite sides of a rectangle are congruent.
What we've shown is that the distance from (a, 0) to (a, b) is b units, and that the distance from (0, b) to (a, b) is a units. So horizontal and vertical distance work exactly as we expect them too -- that is, we now know that the distance from (x, y) to (x + h, y) is h units, and the distance from (x + h, y) to (x + h, y + k) is k units.
So (x, y) -> (x + h, y + k) moves points h units horizontally and then k units vertically. But that still doesn't tell us that the composite of the horizontal and vertical translations is itself a translation. Let's instead try the following -- let P(a, b) and Q(c, d) be two points, P' and Q' be the images of P and Q under the first translation, and then P" and Q" be the images of P' and Q' under the second one.
Then as the first translation slides h units, PP' = QQ' = h, and as the second translation slides k units, P'P" = Q'Q" = k, and as all horizontal and vertical lines are perpendicular (by the same rectangle argument given earlier), both angles PP'P" and QQ'Q" have measure 90. Thus triangles PP'P" and QQ'Q" are congruent by SAS.
So by CPCTC, PP" = QQ" without requiring the Distance Formula. Also, we have that the angles P'PP" and Q'QQ" -- the angles PP" and QQ" make with the horizontal -- are congruent, meaning that PP" and QQ" are in the same direction. So the transformation maps every point the same distance in the same direction. Therefore the transformation is a translation. QED
Notice that this proof essentially assumes a sort of "Converse to the Two Reflections Theorem for Translations" -- the forward theorem asserts that if a transformation is a translation, then it moves every point the same distance and direction, and so the converse would say that if a transformation moves every point the same distance and direction, then it's a translation. But this converse is trivial to prove -- as soon as we have a point P and its image P' and say that every point is moved the same distance and direction as P, then it's easy to find two mirrors to set up the translation. One of the mirrors can be placed at P and the other at the midpoint ofPP', both mirrors perpendicular to PP".
Also, notice we assume that just because the two angles are equal, P and Q must be moved in the same direction. (And technically, we assume that any horizontal line must be perpendicular to any vertical line.) Both of these can be proved using corresponding angles.
Still, this proof should be more intuitive than the other versions which require symbolic manipulation or formulas we haven't covered yet to prove.
I've mentioned my own favorite teachers a few times before, most recently in my August 22nd post.
Meanwhile, I've been writing about Sarah Carter's Slope Dude a lot recently. In this post, Carter resumes her (hopefully weekly) Monday Must Reads series:
https://mathequalslove.blogspot.com/2018/11/monday-must-reads-volume-48.html
Also, Jacqueline Richardson is a blog I read last August, but I didn't link to her because she's now a math coach (rather than a math teacher). But I can't help but think about her most recent post, which is about a balancing equations activity based on NASA. Today in class I see one girl wearing a NASA shirt, so I wonder whether she'd find this activity interesting:
http://highheelsandnumber2pencils.blogspot.com/2018/11/balancing-equations-with-spacex-zuma.html
Lesson 6-2 of the U of Chicago text is called "Translations." In the modern Third Edition, we must backtrack to Lesson 4-4 to learn about translations.
This is what I wrote last year about this lesson:
To emphasize the coordinate plane, I've titled this lesson "Translations on the Coordinate Plane" rather than just "Translations." Yesterday I warned you that another translation lesson was coming up soon, and as it turns out, that day is today.
I said that there are several ways to prove that the transformation mapping (x, y) to (x + h, y + k) really is a translation. First is a coordinate proof using the Slope and Distance Formulas, but I mentioned that we can't use those until after dilations in the second semester. Second is to manipulate the transformations until various mirrors cancel, but that requires arcane symbols such as:
T = r_U(n) o r_U(m)
and I said that it would only confuse students. Fortunately, there's a third way to complete the proof that I alluded to a few weeks back -- and I've decided to use this form of the proof in today's post.
We begin by noting that the two transformations (x, y) -> (x + h, y) and (x, y) -> (x, y + k) are already proved to be translations. The first translation slides points along the x-axis h units, and the second slides points along the y-axis k units. But how are points not on either axis transformed?
Now that we're in Euclidean geometry, we can prove that translations slide every point the same distance -- which is what we expect translations to do. We consider the points (0, 0), (a, 0), (a, b), and (0, b), and these points are the vertices of some quadrilateral. We know by definition, the side along the x-axis has length a and also the side along y-axis has length b. We also know that three of the angles are right angles -- the angle at (0, 0) since the axes are perpendicular, the angle at (a, 0) because x = a is perpendicular to the x-axis, and the angle at (0, b) because y = b is perpendicular to the y-axis. In Euclidean geometry, we know that the sum of the angles of a quadrilateral is 360, so if three of the angles are right angles, so is the fourth, and the quadrilateral is a rectangle.
But now we know the lengths of two sides of this rectangle, a and b, and we want to find the lengths of the other two sides. Of course, the other two sides must also have length a and b as opposite sides of a rectangle are congruent. Yet how do we know this? Most Geometry texts would now say that every rectangle is a parallelogram and the opposite sides of a parallelogram are congruent, therefore the opposite sides of a rectangle are congruent. But unfortunately, the U of Chicago text doesn't give the Parallelogram Tests and Consequences until the last part of Chapter 7.
But let's consider the Quadrilateral Hierarchy yet again. Now only is every rectangle a parallelogram, but every rectangle is an isosceles trapezoid in two different ways -- and we've indeed proved that the opposite sides (legs) of an isosceles trapezoid are congruent. Therefore, we really do know that the opposite sides of a rectangle are congruent.
What we've shown is that the distance from (a, 0) to (a, b) is b units, and that the distance from (0, b) to (a, b) is a units. So horizontal and vertical distance work exactly as we expect them too -- that is, we now know that the distance from (x, y) to (x + h, y) is h units, and the distance from (x + h, y) to (x + h, y + k) is k units.
So (x, y) -> (x + h, y + k) moves points h units horizontally and then k units vertically. But that still doesn't tell us that the composite of the horizontal and vertical translations is itself a translation. Let's instead try the following -- let P(a, b) and Q(c, d) be two points, P' and Q' be the images of P and Q under the first translation, and then P" and Q" be the images of P' and Q' under the second one.
Then as the first translation slides h units, PP' = QQ' = h, and as the second translation slides k units, P'P" = Q'Q" = k, and as all horizontal and vertical lines are perpendicular (by the same rectangle argument given earlier), both angles PP'P" and QQ'Q" have measure 90. Thus triangles PP'P" and QQ'Q" are congruent by SAS.
So by CPCTC, PP" = QQ" without requiring the Distance Formula. Also, we have that the angles P'PP" and Q'QQ" -- the angles PP" and QQ" make with the horizontal -- are congruent, meaning that PP" and QQ" are in the same direction. So the transformation maps every point the same distance in the same direction. Therefore the transformation is a translation. QED
Notice that this proof essentially assumes a sort of "Converse to the Two Reflections Theorem for Translations" -- the forward theorem asserts that if a transformation is a translation, then it moves every point the same distance and direction, and so the converse would say that if a transformation moves every point the same distance and direction, then it's a translation. But this converse is trivial to prove -- as soon as we have a point P and its image P' and say that every point is moved the same distance and direction as P, then it's easy to find two mirrors to set up the translation. One of the mirrors can be placed at P and the other at the midpoint of
Also, notice we assume that just because the two angles are equal, P and Q must be moved in the same direction. (And technically, we assume that any horizontal line must be perpendicular to any vertical line.) Both of these can be proved using corresponding angles.
Still, this proof should be more intuitive than the other versions which require symbolic manipulation or formulas we haven't covered yet to prove.
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