# Tuesday, Nov. 29th

In class today students got back their 2-variable statistics project. The main thing I noticed that students could have done better was to interpret the real-world meaning of their results in a more specific manner. Papers with statements like the following

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

Progress Report grades were entered yesterday by all teachers. Many students have already logged into StudentVUE to check grades for updates. The district will email progress reports but they will show only summary judgments (overall scores for the 4 criteria A, B, C and D). To see how students are doing on individual assignments, Student/Parent VUE is the place to check.

8th Graders attending BHS next year have an informational assembly tomorrow morning at 9:05 in the large gym. Check in with your Advisory teacher and bring a pencil.

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

Progress Report grades were entered yesterday. I had a meeting and raced to get all my marks entered before I had to leave CPMS for a meeting with other BSD teachers to discuss the implementation of the new AGS1 course. One thing I noticed while entering in some scores was that there were a decent number of students in the class who would benefit from some more practice and a chance to try the module 2 test again. So, there will be a re-take of the Module 2 test on Friday. I am asking all students to do the retake and will count only the better of their two scores (retake vs. original).

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.

# 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

So we can see that in $\dpi{300}\inline \triangle HJO$ and $\dpi{300}\inline \triangle JHN$ the segment $\dpi{300}\inline \overline{HJ} \cong \overline{HJ}$. So we have $\dpi{300}\inline \triangle HJO \cong \triangle JHN$ by ASA.

Since these two triangles are congruent, we can see that $\dpi{300}\inline \overline{OH} \cong \overline{NJ}$ and $\dpi{300}\inline \overline{OJ} \cong \overline{NH}$ and therefore that the opposite sides of a parallelogram are congruent.

Also, we can see that $\dpi{300}\inline \angle O \cong \angle N$ and that $\dpi{300}\inline \angle NJO$ and $\dpi{300}\inline \angle OHN$ must be congruent because they are made from congruent parts. So, the opposite angles of a parallelogram are congruent.

So, we can use triangle congruence to learn about shapes beyond triangles. And students will find that much of our knowledge of Geometry is learned through the application of triangle relationships.

Want to re-read these emails all in one place? Each email gets posted automatically to the class update blog. The little ;start ;end tags help with blog post formatting. You can see all Geometry updates at

http://teacher.nickhershman.com/blog/category/geometry/

What better way to spend the evening and the next six days off than to read and re-read each of these updates together?! Have a good break, see everyone on Tuesday.

# Monday, Nov 21

In class I answered questions from the last two sections of chapter 4 and the chapter 4 review and students had time to work on the first two assignments from chapter 5. Since it seems like a bad plan to schedule the chapter 4 test on tuesday, the Chapter 4 test will be on Thursday when we return from (va/stay)cation.

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)

$\dpi{300}\inline 72 + 4(5 - 3x) - (x + 3) - (4 - x) - 2(3 - 5x)$

But many students were not sure how to solve (equation 2):

$\dpi{300}\inline \dfrac{3x - 4}{5} = \dfrac{2x - 5}{2}$

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

$\dpi{300}\inline \dfrac{3}{5}x - \dfrac{4}{5} = x - \dfrac{5}{2}$

and then solve it. Or, we can notice that if two fractions are equal, then their cross products are equal. For example, $\dpi{300}\inline \dfrac{\color{red} 3}{\color{blue}4} = \dfrac{\color{blue} 9}{\color{red}12}$ it is also the case that $\dpi{300}\inline {\color{red} 3 \cdot 12} = {\color{blue} 4 \cdot 9}$. 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

$\dpi{300}\inline 2(3x - 4) = 5(2x - 5)$

And solve from there

$\dpi{300}\inline 6x - 8 = 10x - 25$

Subtract $\dpi{300}\inline 6x$ from both sides and add $\dpi{300}\inline 25$ to both sides to get

$\dpi{300}\inline 17 = 4x$

And finally,

$\dpi{300}\inline \dfrac{17}{4} = x$

Assignment [handouts attached]

• Equations worksheet
• Mathcounts worksheet

Looking for yet another way to read these emails?

You might notice a little ;end showing up at the end of these emails which helps with the auto-formatting on the blog. If you would like to see old posts all in one place, you can find each email at

http://teacher.nickhershman.com/blog/category/ags1/

I make no promises about formatting there, but if it is helpful, you can find all the old posts for Algebra there.

Have a good break!