that the slope of the median median line is negative which means that crime is decreasing.
could be greatly improved if they wrote about the specific numbers, including units, and then included generalizations.
the slope of your median median line is negative .0078 which means that for each additional person in the population we expect a decrease in the number of violent crimes of .0078. In other words for every additional thousand people in the population I would expect to see almost 8 fewer violent crimes. This is interesting because it means that our population is growing at the same time as the number of violent crimes is shrinking.
Students had time to work on Chapter 4 review problems, update their statistics write-up and work on the next chapter 5 assignment. I also gave everyone a study guide for the test on Thursday.
Assignment:
- 5.3 on p. 256: 1-5, 10-16
- 5.4 on p. 263: 1-10
Reminder: Chapter 4 test on Thursday.
Tuesday, Nov. 29th
In class today we did a warm up, I stamped students homework if they had all of the components of the assignment from last week completed (mathcounts worksheet, 4.5 textbook problems 1-17, study ?'s and summary). We reviewed some of the ideas covered before break and worked on today's assignment (4.3 – 4.5 worksheet).
Assignment: "Problem Set 11/29" worksheet (mathcounts problems) and 4.3 – 4.5 worksheet – students may skip the constructions problems
Chapter 4 Test coming up Friday or early next week. To be decided by tomorrow
Tuesday, Nov. 29th
In class we worked on a new problem set and a module 2 review. We also shared some great Thanksgiving stories 🙂
HW: Mathcounts w/s and Module 2 review
Note: Module 2 retake test on Friday. Best score from original and retake counts for grade.
Monday, 28 November 2016
Tuesday, Nov. 22nd
In class today we worked on a mathcounts problem set and continued to study the triangle congruence shortcuts. In 5th period students also looked at an example of how triangle congruence can be used to prove/discover new ideas in geometry. So, I helped students to explain why the opposite sides and angles have to be congruent in a parallelogram.
Assignment. For Tuesday, students should have completed
- 4.5 on p. 227: 1-17
- 2+ Study Questions
- Sentence form summary of the topic of the assignment
Question
- What are you learning in Geometry about triangle congruence?
- How is triangle congruence related to the properties of a parallelogram?

- We are studying relationships relating to triangles. We have learned that there are shortcuts for determining when two triangles are congruent – SSS, SAS, ASA, and AAS are the triangle congruence shortcuts we have discussed. Also, students should be able to explain why ASS and AAA are not shortcuts. Again, chuckles, but really there is a reason to write these acronyms this way. For example, SAA congruence is the idea that if two separate triangles can be shown to have one pair of corresponding sides and two pairs of corresponding angles that are congruent then the triangles are congruent. In this case it is important that the position of the sides and angles correspond exactly.
- Triangle congruence can be used to prove the properties of a parallelogram if we assume only the definition of a parallelogram (quadrilateral with 2 pairs of opposite, parallel sides). We might notice that two triangles can be formed if we draw one of the diagonals of the parallelogram. And by proving these two triangles to be congruent, we can explain why the opposite sides and angles are congruent.
Now, we with each pair of parallel sides we can find a set of congruent alternate interior angles

Monday, Nov 21
Answer Keys: The answer keys to the practice problem packets for Chapter 4 are attached.
Assignment
- Study for the chapter 4 test
- 5.1 on p. 240: 1-4, 7,8
- 5.2 on p. 248: 1-5, 7, 10, 12, 13, 16
Tuesday, Nov. 22nd
ws-equationsolving-fractions-2.pdf
ws-mathcounts-20161122-OPLET.docx
In class today students worked on equation solving and mathcounts problems. I notice that generally they are comfortable solving multi-step equations like (equation 1)
But many students were not sure how to solve (equation 2):
So we did this example in class. I showed students that division can be distributed just like multiplication so we can turn the equation into
and then solve it. Or, we can notice that if two fractions are equal, then their cross products are equal. For example, it is also the case that
. This is true for all equivalent fractions and so another way to approach solving equation 2 is to find the cross products and set them equal
And solve from there
And finally,
Assignment [handouts attached]
- Equations worksheet
- Mathcounts worksheet
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Have a good break!