Good day my young math scholars,

1) Go to student.desmos.com

2) Sign in with your real name, no nicknames please

3) 7th grade code:   VFWB2K

    8th grade code:   JZK4QA

Good day my young math scholars,

Solution to Last Week's POW

a=2, b=3, c=5

Dinner at Pepe's

Cynthia, Annie, and Suz went to Pepe's Pizza Palace for dinner after band practice. They shared a large pizza and each girl ordered a soda. When the bill came, they talked about how to split it up fairly.

"Annie, you had three slices and a medium soda so you should pay one-third of the bill," said Suz.

"Well, I'll chip in a dollar more than that," replied Annie. Annie replied to Suz, "And since you only had two slices and also ordered a small soda, you should only pay 20% of the bill."

"That seems fair," said Cynthia. "I ate three slices and had a large soda, so I should pay the most. I'll pay the remaining $8.80."

1. How much was the total bill for the pizza and the three sodas, and how much did each girl pay?

2. At Pepe's, a medium soda costs 25 cents more than a small, and a large soda costs 25 cents more than a medium. The cost of the sodas made up one-fifth of the total bill.  Figure out the actual cost of each girl's meal. Whose estimated cost was the closest to her actual cost?

1) Go to student.desmos.com

2) Sign in with your real name, no nicknames please

3) 7th grade code:    FWK29Q

    8th grade code:    EUE6PC

Good day my young math scholars,

For those of you who shared your work with me through your Google Drive, I've been able to look at all of it, so go check your documents for comments.

Today's task is on Desmos, so

1) Go to student.desmos.com

2) Sign in with your real name, no nicknames please

3) 7th grade code:    B8CNGS

    8th grade code:    5TPVBK

If you have any questions, I'll have a Google Meet Up on Tuesday, 1:00pm for 7th graders and 2:00pm for 8th graders.

Date of Message: Friday, March 27

Good day my young math scholars,

Adrienne's Axel 

The new program will have 1 double Salchow, 2 Axels, and 2 Lutz-Loop- Loop combinations. 

Let s = # of double Salchows, a = # of Axels, and c = # of combinations. Then Mandy's statements give 3 equations: There are 9 jumps: s + a + 3c = 9 (remember a combination is 3 jumps) There are 11 rotations: 2s + 1.5a +3c = 11 There are 2 combinations: c = 2 Substituting c=2 in the first 2 equations and rearranging we get s + a + 6 = 9 or s + a = 3 2s + 1.5s + 6 = 11 or 2s + 1.5a = 5 These two equations in two unknowns may be solved by any standard method, such as substitution or linear combinations. For example, using substitution, s = 3-a, so 2(3-a) + 1.5a = 5, so 1 = 0.5 a, so a = 2. Thus s = 3-2 = 1. So the program has one double Salchow, 2 Axels, and 2 Lutz-Loop-Loop combinations. Bonus: No. Without any information about combinations, we have s + a +3c = 9 2s + 1.5a +3c = 11 But remember that the solutions must be whole numbers (you can't do half an Axel or a negative double Salchow, for example). Thus 1.5a < 11, so a < 11/1.5 = 22/3, i.e., a < 7. But the number of Axels must be even, because we have no half-jumps in this program, and an Axel is 1.5 rotations. With a = 6, from the first equation, s=0 and c=1 or s=3 and c=0, neither of which work in the second equation. With a = 4, from the first equation, s=2 and c=1 or s=5 and c=0, neither of which work in the second equation. With a = 0, from the first equation, s=6 and c=1 or s=3 and c=2 or s=0 and c=3, none of which work in the second equation. Thus a=2 is the only possible number of axels. Then s + 3c = 7 and 2s + 3c = 8, so s = 1 and c = 2.  

Here's the new POW:

The ABCs of Exponents

Given the following three equations, solve for a, b, and c:

Show your work and email me your solutions by next Friday at:

eric_lass@psbma.org

Date of Message: Thursday, March 26

Good day my young math scholars,

If you've shared a document with me, please take a look at them as I've been able to read and comment on everything.

I'd like to have another meet up today to talk math, books, movies, or anything else on your minds. Let's try Zoom this time:

7th graders - Thursday from 1:00 - 1:30pm

8th graders - Thursday from 2:00 - 2:30pm

To access the video conference:

1) log onto  www.zoom.us    (PLEASE NOTE THE  .US)

2) you can download the zoom app if you haven't done so yet, but leave some time         as it takes a little while for them to send you a confirmation email

3) I've been told there's also a way to jump on a conference without the app if you scroll down the zoom website

4) enter the appropriate code:

        7th graders:  370 745 582

        8th graders:  165 450 458  

Date of Message: Wednesday, March 25

Good day my young math scholars,

Thanks to those of you who joined in to chat yesterday, it was great to see you all!

Sorry if it was a bit chaotic, if you have math questions please email me and I'll either respond or we can set up a 1:1 or small group conference.

For those of you who shared your work with me, I've been able to look at all of it, so go check your documents for comments.

Today's task is on Desmos, so

1) Go to student.desmos.com

2) Sign in with your real name, no nicknames please

3) 7th grade code:    2F3FX4

    8th grade code:    TBSZ7Y

I'll post Thursday's office hours and codes tomorrow morning. I think we'll try Zoom this time.

Date of Message: Tuesday, March 24

Good day my young math scholars,

If you didn't see my message from yesterday, please scroll down and give one of the problems I've posted a try.  I plan to post something new to do on Wednesday.

For today (Tuesday), I'm going to try to vault myself into the 2020s by holding a video conference meet up.  If you have any questions on the math problems I assigned or are looking for an alternative that might be easier, or maybe just want to check in and say hello, please join me!

7th graders - Tuesday from 1:00 - 1:45pm

8th graders - Tuesday from 2:00 - 2:45pm

To access the video conference:

1) go to your PSBMA account

2) click on the "waffle" in the upper right hand corner of your screen

3) choose "Meet"

4) enter the appropriate code:

        7th graders     hvn-ogoy-gqf

        8th graders     dkj-dxzr-mvr

There's a small chance that the technology (or more likely Mr. Lass) will encounter a problem, so I apologize in advance if this goes less than smoothly.  We'll get better at it by Thursday.

Best regards,

Mr. Lass

Date of Message: Monday, March 23

Hello my young math scholars,

First off, I miss seeing you all and hope you and your families are healthy and keeping busy!

I'm going to start offering work for you on Mondays and Wednesdays as well as a new Problem Of the Week on Fridays (see below for POWs and solutions). 

I'm also trying to figure out a way for you to check in if you need help (or just want to say hello) on Tuesdays and Thursdays. Maybe Google hangout, but stay tuned...

Please click here to access your Big Problem for Monday, March 23rd:

https://sites.google.com/site/lassmath/my-forms

Each file contains three problems. They range from easier to more difficult. Read all three and choose ONE problem that seems like the most difficult you might be able to solve. You may type your solution or write it out and take a photo. As always, SHOW YOUR WORK!  Your finished product should include the math you used as well as a paragraph or two explaining what you did.  This should take you 30-60 minutes over the next couple of days.

Share your solution with me on Google Drive or email it to me (whichever is easiest for you) at:   

eric_lass@psbma.org     

7th graders, if you'd like a bigger challenge, try an 8th grade problem instead of the 7th grade list. 8th graders, if you're having trouble, feel free to try a 7th grade problem.

Here's a template to help any and all with answering the Big Problem:

https://docs.google.com/document/d/1yspRDX6KjjnwuMrVwk7yT915bDvy27e_hy7JDYydRoY/edit

I also included an additional challenge problem if you're looking for something even bigger!

If you have questions, feel free to email me.

My other assignment is for you to try to do something nice for a friend or family member each day. This is a difficult time for everyone and it can be hard to be stuck at home with the same people every day, even if they're the people you love most. If we all can do something nice for each other, it will make life more pleasant all around.

I hope we can see each other again soon!

Best regards,

Mr. Lass

Problem of the Week

Adrienne's Axel 3/20/20

Adrienne is a figure skater. Her coach Mandy is planning Adrienne's new competition program, and has told Adrienne that the jumps in the new program will all be Axels, double Salchows, or Lutz-Loop-Loop Combinations. 

When a skater does a single jump, such as a Lutz, a Loop, or a Salchow, she does one rotation in the air, jumping backward and landing backward. 

When a skater does a double jump, such as a double Salchow, she does two rotations in the air (all in one jump). The Axel is an unusual jump, because the skater jumps forward and lands backward, doing one-and-a-half rotations in the air. A Lutz-Loop-Loop Combination is three jumps in a row without any skating in between.

In addition to ice skating, both Mandy and Adrienne enjoy mathematical puzzles. Mandy says, "Your new program will have nine jumps, eleven rotations, and two combinations. Adrienne says, "Now I know exactly what jumps I will have in my new program!"

What does Adrienne know and how does she know it? (Adrienne thinks algebraically.)

Solution

The new program will have 1 double Salchow, 2 Axels, and 2 Lutz-Loop- Loop combinations. 

Let s = # of double Salchows, a = # of Axels, and c = # of combinations. Then Mandy's statements give 3 equations: There are 9 jumps: s + a + 3c = 9 (remember a combination is 3 jumps) There are 11 rotations: 2s + 1.5a +3c = 11 There are 2 combinations: c = 2 Substituting c=2 in the first 2 equations and rearranging we get s + a + 6 = 9 or s + a = 3 2s + 1.5s + 6 = 11 or 2s + 1.5a = 5 These two equations in two unknowns may be solved by any standard method, such as substitution or linear combinations. For example, using substitution, s = 3-a, so 2(3-a) + 1.5a = 5, so 1 = 0.5 a, so a = 2. Thus s = 3-2 = 1. So the program has one double Salchow, 2 Axels, and 2 Lutz-Loop-Loop combinations. Bonus: No. Without any information about combinations, we have s + a +3c = 9 2s + 1.5a +3c = 11 But remember that the solutions must be whole numbers (you can't do half an Axel or a negative double Salchow, for example). Thus 1.5a < 11, so a < 11/1.5 = 22/3, i.e., a < 7. But the number of Axels must be even, because we have no half-jumps in this program, and an Axel is 1.5 rotations. With a = 6, from the first equation, s=0 and c=1 or s=3 and c=0, neither of which work in the second equation. With a = 4, from the first equation, s=2 and c=1 or s=5 and c=0, neither of which work in the second equation. With a = 0, from the first equation, s=6 and c=1 or s=3 and c=2 or s=0 and c=3, none of which work in the second equation. Thus a=2 is the only possible number of axels. Then s + 3c = 7 and 2s + 3c = 8, so s = 1 and c = 2.

Cheap Sunglasses 

Gilbert was working on lighting for the school play and thought that he could make a cool pair of sunglasses by attaching some of the red filter film to his glasses.

He did some research and learned that 20 percent of the light passing through the filter is absorbed. He also read that good sunglasses should absorb 80 percent of the incoming light, so he figured that if he put four layers of filter film on his glasses he'd be all set.

Calculate how much light would be absorbed by 4 layers of filter film. Is Gilbert right?

Solution

Gilbert is wrong, four layers would let in 40.96% of the light and 59.04% would be absorbed

Apple Orchard  

On a recent trip to a local orchard, the Nomial family picked four different kinds of apples - Braeburn, Cortland, Fuji, and Rome. When they were done, they discovered that they had picked

·     a total of 360 apples,

·     twice as many Braeburn as Fuji,

·     twice as many Cortland as Rome,

·     50% more Fuji than Rome.

How many of each kind of apple did they pick? 

Solution

First, assign a variable to each of the type of apples

b=braeburn

c=cortland

f=fuji

r=rome

Next, set the total of all the variables equal to 360

b+c+f+r=360

According to the problem, set each of the variables in terms of each other

b=2f

c=2r

f=1.5r

Since b was given in terms of f, and f was given in terms of r, convert b into terms of r

b=2(1.5r)

Then, go back to the original equation and substitute all the variables in terms of r, solving for r

2(1.5r)+2r+1.5r+r=360

3r+2r+1.5r+r=360

7.5r=360

r=48

To find the number of each type of apple solve for each variable where r=48

b=2(1.5)(48)

b=144

c=2(48)

c=96

f=1.5(48)

f=72

r=48

Persistence of Numbers

One property of numbers is their persistence.  Take the number 723 as an illustration: if you multiple the digits 7, 2 and 3 together, the product is 42; multiply 4 and 2 together to get 8. Because this operation takes two steps to reach a single-digit number, the persistence of 723 is 2.

What are the smallest numbers that lead to persistences of 2, 3 and 4?