Monday, July 21, 2008

Random Thoughts from Past and Present Students

Everyone needs space to be random!

366 comments:

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Anonymous said...

Hi!

I didn't see your post before I left but that's okay. You're travel plans sound very exciting! I hope you see lots of friends and have a fabulous time. I look forward to the pictures and stories of your adventures.

My adventure have been interesting... I pack and leave for hiking in the Adirondacks tomorrow if I'm okay to go- I slipped on a rock and dropped a 95 lb canoe on my back a few days ago. I was TRYING to avoid hitting my HEAD. I'm going to a doctor, we shall see and hope! The trip was amazing though other than the several personal injuries. :)

It was very interesting to hear of your pebble idea- I carry a rock very similar- it's a pinky smooth stone I started holding it to prevent my anxiety, it works quite well. I take it on trip now for whenever I feel stressed/anxious/lonely. Interesting. Good idea- hope it works for you Rosie.

Cheers,

Nicole

P.S. Crystal Light is the best. It also makes me VERY hyper. All that darn aspartame. :

Ms.Bearse said...

Nicole, I think you are hyper just because you are hyper...although aspartame is a good excuse :)

Glad the trip was good, I hope you are well enough to go on the next part.

I hope you are taking some pictures of your trip, it will be lots of show and tell in sept!

Anonymous said...

Aspartame is a wonderful excuse though you might just have a point...

I AM going on the hiking trip hurray! I just finished packing all of my gear/food/clothing into a crazy large pack. I have/and will take more pictures, we will have to have a story/picture party.

I might message you later this month to let you know that I wasn't eaten by a bear (Which apparently there are many of). :) I'm going to go have another nap because I have to get up at 4. Joy. :)

When are you leaving for your crazy world adventures??

-Nicole

Ms.Bearse said...

Hi Nicole, I'd be happy to hear that you are not eaten by a bear--watch out for them!

I'm leaving on the 24th...coming up soon!

Anonymous said...

Salut!

I'm back from France! It was a wonderful experience and I'm so glad I did it! It was tough but I'm proud of myself because I managed to not let my homesickness interfere too much with having a good time. Thank you for your pebble post, I didn't have a pebble but it really helped hearing from you because it showed me how small the world is and how it's so easy to connect with other people in other countries.
Bertille is here in Canada with me and we're having a wonderful time. Today we went downtown and took the Kingston Tour Trolly, went to another one of my orthodontist appointments (Ewwwww!) and went shopping.
I hope you have a wonderful time with your world travels. Stop by a soccer game if you have time. Tonight's game is at McCoullag.
Nicole- I'm glad your having a good time! Please don't get eaten by a bear, how would I survive grade 10 without you?

Salut,
Rosie

Ms.Bearse said...

Glad to hear you are back Rosie. I was at Harry Potter while you played soccer...I hope the game was good. The movie sure was :) I can't wait until Friday when the next book comes out. I'm going to read it all the way to Japan....14 hours on a plane is what I call fun!
I'm putting up trip stuff on another blog you can find that here
I don't know how often I will be able to update it, it depends on how the internet access is where I will be staying.

Anonymous said...

Hello again! I'm back again. I was NOT eaten by a bear thankfully, though we did see many of them. They are thrill and exciting at the same time particularly when they are rumaging through your pots looking for non-exsistant food. heehee :)

Are you excited for you trip? Hmm? Well? Are you? I'm excited and I'm not even going haha. May your travels be exciting, safe and fulfilling for you.

I have to get up and leave 5:30 tomorrow, maybe I should go to bed...I'll talk to you in September then I suppose. Cheers,
-Nicole

Anonymous said...

And by thrill I mean absolutly TERRIFYING.
-Nicole

Ms.Bearse said...

check out my ramblings (and hopefully photos)from my trip here.

Ms.Bearse said...

Stay safe Nicole!

Anonymous said...

heyyy madamme! :] hope your summer is going great! i'm working at pan chancho, in the café you should come visit...:p

-bee.

Anonymous said...

heyyy, sounds good, but I am in Hiroshima right now. I was in Korea this morning...can you believe it? I will be back in Kingston toward the end of Aug. I might have time to stop by. See you for sure in Sept. Have a good summer.
Ms.Bearse-from Japan

Anonymous said...

Hello!
Summer's almost over :( but school is starting :) and I'm quite excited. How was ,Japan? Fun I imagine, I'm sure you had an awesome time. I just have a quick question. Do you know what the main differences are between 11 physics and 11 IB physics? There's conflict in my schedule and there's an IB physics course in the right space so i was thinking about it but just wanted some input about the level of difficulty and work involved from someone who knows science :).

Anyways i hope you enjoy the last 8 days of summer!

Chelsea

Ms.Bearse said...

Hey Chelsea,
I have tried to find the curriculum requirements for IB, but I don't know them. I'm asking some IB teachers, but don't know when they will get back to me. I'd say go for the IB--it will be more work, and more will be required for your labs, but maybe you will enjoy the challenge. Don't set yourself up for too much of a crazy semester though. I'll send word when I hear from the physics teachers.

My trip to Japan and Korea was great. I have picture if you want to see them here

wow, as I was writing this I heard back! "IB Physics is more focussed on experimentation and data analysis. The regular Physics is more of an applied math and problem solving course." Sounds fun, right? Good luck making your decisions.

Ms.Bearse said...

I'm spreading the word, on behalf of Nick Arnold from LCVI...

"The Vicious Circle" is a feature length film about bullying and hatred. It was written by Nick, and filmed last summer, starring a cast of local celebrities. The premier will be on Friday September 28th at 7pm at Duncan McArthur Hall. $10 for adults, $8 for students. Tickets are available at The Queen's Office Building 98 Barrie St. starting Sept. 4th, and also at the door on the 28th.

Formal dress is encouraged.

Anonymous said...

Hey Mme. Bearse!

I'm having a little bit of trouble understanding this lab I did. It's a titration lab.
We had 49mL of hydrochloric acid in a burette valve. We added three drops of Bromothymol blue to a beaker with 25mL of Sodium Hydoxide (our base). We added the acid to the base, gently swirling the base around in an effort to mix it.
When the solution turned yellow, we stopped adding the acid and recorded the final burette reading and calculated the volume. Why did it turn yellow? I also don't understand why did we added bromothymol blue to the base but not the acid. Thanks so much for your help!
-Rosie

Anonymous said...

Does this make sense:

If Sodium Hydroxide has a pH of 11, we need to add 10000x of Hyrdochloric acid to create a neutral substance.

Thanks,

Rosie

Ms.Bearse said...

Hi Rosie. In your titration lab you started with base in a beaker with indicator in it. The indicator will tell you when the solution reaches a certain pH. As you drop the acid into it, the pH of the beaker will change (get closer to 7). You do this until the colour changes-which happens at the desired pH.

Do you get the idea that the H+ from the acid joins with the OH- from the base to make water? Water is neutral...

If you have sodium hydroxide (NaOH) and Hydrochloric acid (HCl) you know that the ratio of H to OH are 1:1--there are no subscripts at all in the formula, so this is a relatively simple case.
To make it neutral, the OH from NaOH will join with the H from HCl

You know that the pH of the NaOH is 11, do you know the pH of the HCl? That's important for your calculations...(not sure you really need to calculate though, that sounds more like gr.11 chem stuff)

Maybe you are being asked how much more alkaline the NaOH is than a neutral substance (pH 7). In this case you understand that each jump upward in the pH scale represents 10 times more alkalinity than the step before it. Each jump down represents 10 times more acidity than the step before it.
So, something with a pH of 7 is 10 times more acidic than something with pH 8, and 100 times more acidic than pH 9, and 1000 times more acidic than pH 10, and 10000 times more acidic than pH 11.

Don't know if this helps or not.
good luck!

Anonymous said...

Thank you so much! I understood everything you wrote! I think I'm jumping ahead here because I just have so many questions, some that I don't need to know the answers of until I reach grade 11 chem! I answered my questions already and this clarifies that I got my answers right thanks!
-Rosie
P.S. On a random note: Who's Nuke?

Anonymous said...

Without sugar or caffeine (sigh) what's the best way to stay up far too late and think really really hard?

Not like a regular thing rather an occasionally necessity....

Failing that- should I go to bed now and wake up at 3 to finish working or stay up past midnight? Decisions, decisions...

You have odd sleep hours sometimes. Like exams. Your thoughts?

My sanity is questionable tonight.

-You know who it is

Ms.Bearse said...

Nuke is a little toy that's saving the world!

Ms.Bearse said...

Ummm...as for staying awake late...do jumping jacks....take off one sock so your feet are different temperatures...that's all I've got. If you are able to sleep when you still have things to do, go for it! Sometimes I'm not able to sleep well until things are all done.

Sometimes you have to make a judgment call about whether what you are doing is important enough to go without sleep. Choose your priorities carefully!

Hahaha....perhaps I should be following my own advice!
Back to work I go.

Anonymous said...

I don't know if this is science or knitting or totally random or neither...

So it's here

I have a question that's proving tricky (read: the internet hath failed me) to research.

I have some random yarn. I believe you gave it to me at some point actually when Noor and I were knitting ipod covers and it was never used. That is quite irrelevant though...

Anyhow, it is fuzzy and white and that's all I have in terms of qualitative observations. It's a great weight for knitting socks HOWEVER it is white and my white socks tend to turn grey.

So I want to make it a different colour.

Which in all likelihood would mean dying it.

This IS an new kettle of fish.

Most of the reading I’ve done about dying yarn (other than being complicated and messy…) talk about the different processes for different kinds of yarn depend on if it’s natural or synthetic.

So my first challenge is figuring out WHAT it is.

?

So I research the differences to see if I could figure it out myself. This website actually explained some interesting things about chemistry and fabrics (http://www.angelfire.com/mech/fabric/) it wasn’t helpful, but interesting if you’re curious.

Anyhow- I was wondering, if you could tell the difference by looking at the yarn under a microscope. I actually umm did just that in biology last week with my acrylic stuff just to see what it would look like. Yeah. Rather cool actually :) But would it look different if it were a natural fibre?

Is there any other way to tell failing that? White and fuzzy isn’t telling me much. You dye stuff, your thoughts?

That was one of the longest questions I’ve ever posted here…

Have a good rest-of-the-weekend-time!


-Nicole

P.S. Nuke is just about the most random thing ever. Actually.

Ms.Bearse said...

Step one....burn it.
If it makes ash it is natural
if it makes a gooey mess it is synthetic.
Don't hurt yourself or burn down your house....ok?

If it is wool it should take kool-aid dye. (put some in the microwave with koolaid for 2min and see what happens).

Cotton might not take kool-aid dye...I've not tried cotton before.

Dunno about acrylic, but what you could do is use RIT dye(it is kinda toxic though so be careful--don't use regular containers/spoons etc) and wear gloves. You can get that dye in stores (it's for dying fabric, so maybe it works for synthetics too?) Good luck!

P.S. Nuke is fun. Make some mashups...that is "funner" Send them in, and they might get posted!

Anonymous said...

Burn it, why yes of course. That makes sense. Thanks muchly for your advice on the dying process. I haven’t the time now so maybe that's next weekends job...

Nuke is certainly fun and highly amusing to see the pictures that are posted of him. Some look oddly familiar...like the one of him on teh mound of Japanese wishes? I have a sneaking suspicion of who made that one...

Ms Bearse has a new hobby? :)

Well that takes up my mental break time, back to work then! Just thought I’d say thanks for the help and such.

-Nicole

Anonymous said...

Hey Mme. Bearse,

Would it be possible for you to meet with me for about fifteen minutes at the start of lunch for science help tomorrow?

Thanks so much!

Rosie

P.S. I too agree that the picture of Nuke with the Japanese wishes looks oddly familiar... the gig is up Mme. B!

Ms.Bearse said...

Hi Rosie, sure I'll help you with Science...come catch me at the math office/water fountain/second floorish area and we'll find a place to work.

As for that Nuke picture...yes, I made that one, but really it is going around the world...

Here he is in Australia with a strange guy flickr (pictures)
and viddler (videos)

You can put him in your photos too if you want...use the "mash up kit" from the site. Fun times!

Anonymous said...

Hi Ms. Bearse

I'm working on my physics work, light and lenses and such. I'm incredibly stuck on a problem with two converging lenses, like in a microscope.

"Two lenses are used to produce and image 5.0mm in height (hi). The optical centre of the objective lens (f = +3.0) is 9.0 cm from the optical centre of the eye piece lens(f= +9.0) The object is positioned 5.0 cm in front of the objective lens. Find the position and size of the final image and it's magnification."

For the objective lens I have found the di, it's 7.5cm using the thin lens equation. 1/di + 1/do = 1/f. I also know that hi/ho= -di/do = magnification. Other than that I'm completely lost and confused. (The answers are listed- image is 1.8 cm in front of eyepiece, hi= -9.0mm, magnification is equal to –1.8x. If that is helpful.)

If you have any idea what on earth this is and how to solve it I beg for your assistance. If this is unclear/you haven’t done anything like this is 37 years (like my math teacher)/your busy debating what to teach your French class don’t bother with my fussy questions.

On the other hand- help!

-Nicole

Ms.Bearse said...

Ok....lets see what I can figure out.
1/do+1/di=1/f
hi/ho=di/do=M

hi=5.0mm
f=3.0
hmm...the 9.0 cm is interesting. That means that the lenses are 9.0cm apart, but the second lens has a focal length of 9.0cm also, meaning that the light passing through the optical center of one lens, is actually approaching the second lens from the focal point.

object-5cm-objective-9cm-eye piece

(that's me trying to understand graphically..hahaha)

i agree with your thin lens approach...
1/3=1/5+1/di
1/3-1/5=1/di
2/15=1/di
di=7.5 (i agree...had to do it myself though)
hmmm, so then this is going to be the object for the next lens, right?
the next lens is 9-7.5cm away (that's 1.5cm really)
so "do" for the next lens is 1.5
using the fact that the image from the first lens--I'm assuming that the 5mm is the first hi not the final one...
so that hi is going to be ho for the second lens

so, we have ho=5mm(this should be negative though, if we have the original object upright), do=1.5cm, f=9
1/f=1/di+1/do
1/9=1/di+1/1.5
2/18=1/di+12/18
2/18-12/18=1/di
-10/18=1/di
di=-1.8 ---yippee!!!!

hi/ho=-di/do
hi/-5=1.8/1.5
hi/-5=1.2
hi=-6 (hmmmm....not -9)

magnification would compare the final image height to the original object height...do we know the original ho?

hi/ho=-di/do
-5mm/ho=-7.5cm/5.0cm

ho=10/3 or 3.3333
so...comparing the final hi to the original ho

M=hi/ho
M=-6/3.333
I got M=-2.67

are you sure your answers that you have listed are correct?

I drew myself a little sketch...
Hard to reproduce it here, but I'll try to describe using SALT :)
so, for the first lens, the image is
S: larger
A: inverted
L: pretty close to the second lens
T: real

Because we know for sure that the image from the first lens is situated between the focal point (that's where the first lens is with respect to the second) and the optical center...I'm making the following predictions about the final image (compared to the first image which is now an object)
S: slightly bigger
A: upright
L: between the two lenses
T: virtual

So....I don't know where that gets you... the -1.8 makes sense as being a virtual image....

let me know if you get the "right answer" good luck!

Anonymous said...

Gee Ms. Bearse, it only took me *glances at clock* 22 minuets to find a question I could not figure.

It's also one of those questions Ms B was hinting at...

It reads:

f= {(0,5),(1,4),(2,7),(3,14)} and g(x)=3f(-x+1)+7

Evaluate g(-1)

(The answer is 28 HOWEVER this is a Nelson textbook so there is a remarkable high chance that said answer is wrong...)

I have two variables? g and f. How do...what?

-Nicole

P.S- I do hope the red juice didn't make you too ‘punchy’. :D

Ms.Bearse said...

ok, here's my best shot....

g(-1) means we substitute x=-1 into the long messy equation
g(-1)=3f(-(-1)+1)+7
g(-1)=3f(1+1)+7
g(-1)=3f(2)+7
now, have a look, do we know f(2), why yes! f(2)=7 that's one of the (x,y) pairs listed.

so...g(-1)=3(7)+7
g(-1)=21+7
g(-1)=28
Tadaaaa!!

a bit punchy, thanks to the red juice. Don't stay up too late.

Anonymous said...

Hi Ms B,

I'm studying physics for tomorrow...I'm kinda stuck on a question...so I wound up here...

"A camera is supplied with two interchangable lenes whose focal lenghts are 35.0mm and 150.0mm. A woman whose height is 1.80m stands 8.00m from the camera. Describe the characteristics of the image and calculate the height of the image for both lenses."

(-7.19mm, -34.4mm, so they say.) I've found -7.19...

If you're bored or not busy or something would you help me? I've been bugging you lots lately...sorry! I've found the first anwser but I'm just plain stuck.

-Nicole

P.S.- Feel free to cut me off at 11:30.... haha

Ms.Bearse said...

lets work in mm
so 1.80m is now ho=1800mm, and 8m is do=8000mm

1/f=1/di+1/do
1/150=1/di+1/8000
1/150-1/8000=1/di
8000/1200000-150/1200000=1/di
7850/1200000=1/di
di=1200000/7850
di=152.87
hi/ho=-di/do
hi=-152.87(1800)/8000
hi=-34.39 which (to 2 s.d. is -34.4)
ta-da!!
keep working!

Anonymous said...

Okay I get it...just massive calculation errors. several times over. *Twacks head*

:) thanks.

-Nicole

P.S.- Our new choir dresses feel suspiciously like the same material as the womens’ chorus uniform...

Amber said...

Hey, Ms. Bearse!

I think my e-mail was sent to you, but just in case, here I am. :]

Sooo, this is Amber! Dunno if you remember me. I'm Nick Arnold's friend. We had math parties. I love ponies... Yeah. Hehe.

I'm in Advanced Functions and Introductory to Calculus and it is HARD. I was wondering if you'd be available to come to LCVI like... once a week or something to help me out?

Thanks!

Amber

Ms.Bearse said...

Hi Amber, of course I remember you! Who could forget the I heart ponies tattoo! I can't come every week, but I have been missing the math parties. Let me know what day after school (not Tuesday), and I could be at LC by 3:00. I teach at Frontenac at the end of the day.

P.S. I didn't get an email from you.
My email is on first class, you can find it from there.

Amber said...

Haha! Thank youuu. My calculator is still decorated with random Sesame Street stickers. I miss stickers! None of my teachers give out any.

By the way, all of my exams, I bring my lucky pony, my lucky yoga rock, and the good luck rock you gave me. :]

The e-mail should've been from Nick. I'll have to yell at him tomorrow for not sending it.

So, if I have problems I guess I'll just post here? I don't know how to get the firstclass e-mail thing. FirstClass confuses me beyond belief.

Amber

Ms.Bearse said...

hahaha, I got an email from nick saying that you wanted some help...I wrote back to that...maybe he hasn't passed on the message? who knows, anyway, we are in contact now, and that is what matters.

ok, so emails with first class is not so hard to figure out.

lastnamefirstinitial@limestone.on.ca

so mine is bearser@limestone.on.ca

it's school board email so it is monitored by someone...

Let me know what day works best for you.

Amber said...

Haha, thanks. Everything is sorted out now, I guess. :]

I have two math questions! We're doing limits and these two questions stumped me.

10. A medicine is administered to a patient. The amount, M, of the medicine, in milligrams, in 1 mL of the patient's blood t hours after the injection is given by M(t) = -1/3t^2 + t, where 0 <= t <=3.
a) Find the rate of change of the amount, M, 2 h after the injection.
b) What is the significance of the fact that your answer is negative?

11. The time, t, in seconds, taken for an object dropped from a height of s metres to reach the ground is given by the formula t = [the squareroot of s/5]. Determine the rate of change of the time with respect to the height when the height of an object is 125 m above the ground.

Anonymous said...

11. t(125)= [the square root of s/5]

=[square root of 25]

=5 seconds?

-An anonymous student who is still really stumped on 10. And would like to know if they are wrong in 11.

Ms.Bearse said...

Hey anonymous math questioner...
I was just at my first spanish class, so lets see if I have any brain left.

#10, Do you know how to take derivatives? I'm not sure what you know yet so I don't know how to explain, but I'll show you how I'd do it, and maybe that will help you.

I'd take the derivative of M(t) with respect to t.

dM/dt=-2/3(t)+1
(dM/dt is the rate of change of M with respect to t)
now sub t=2 (to find the rate of change at 2 hours)
dM/dt=-2/3(2)+1
dM/dt=-4/3+1
dM/dt=-4/3+3/3=-1/3

b) the rate of change is negative which means that the slope of the M curve is negative, which means that the amount of medicine is decreasing at that point.

you may need to do it showing the limit equation

f'(x)=lim[f(x+h)-f(x)]/h as h tends to 0.
M'(t)=lim[-1/3(t+h)^2+(t+h)-(-1/3(t)^2+t)]/h as h->0
M'(t)=lim[-1/3(t^2+2th+h^2)+t+h+1/3t^2-t]/h
as h->0
M'(t)=lim[-1/3t^2-2/3th-1/3h^2+t+h+1/3t^2-t]/h as h->0
and now to simplify
M'(t)=lim[-1/3h^2+h-2/3th]/h as h->0
factor the h
M'(t)=lim[h(-1/3h+1-2/3t)]/h as h->0
cross out the h (same on top and bottom
M'(t)=lim(-1/3h+1-2/3t) as h->0
make h go to zero
M'(t)=(1-2/3t)
now find it when t=2
M'(2)=(1-4/3)
M'(2)=-1/3

*round of applause please! that took a while to type correctly*

Ok, question 11
t=sqrt(s/5) where s is height in m
find dt/ds (or t'(s))

with derivatives...
t(s)=(s/5)^0.5
t'(s)=0.5(s/5)^(-0.5)(1/5)
this is the same as 1/(10sqrt(s/5))
now sub s=125
t'(125)=1/(10sqrt(125/5))
t'(125)=1/(10sqrt(25))
t'(125)=1/(10*5)
t'(125)=1/50

if you can't use derivatives...

t'(s)=lim [t(s+h)-t(s)]/h as h->0
t'(s)=lim [((s+h)/5)^0.5-(s/5)^.5]/h as h->0
t'(s)=lim[(s/5+h/5)^0.5-(s/5)^.5]/h as h->0
t'(s)=lim[((s+h)^0.5-s^0.5)/5^0.5]/h as h->0
multiply top and bottom by conjugate ((s+h)^0.5+s^0.5)
this gets messy
t'(s)=lim[((s+h)^0.5-s^0.5)((s+h)^0.5+s^0.5)/((5^0.5)((s+h)^0.5+s^0.5)]/h as h->0
t'(s)=lim[((s+h)-(s))/(h(5^0.5)((s+h)^0.5+s^0.5))] as h->0
t'(s)=lim[(h/(h(5^0.5)((s+h)^0.5+s^0.5)) as h->0
cancel the h on top and bottom
t'(s)=lim[1/((5^0.5)((s+h)^0.5+s^0.5)))] as h->0
make h go to zero now
t'(s)=1/((5^0.5)(s^0.5+s^0.5))
t'(s)=1/((5^.5)(2s^0.5))
t'(s)=1/(2(5s)^0.5)
t'(125)=1/(2(5*125)^0.5)
t'(125)=1/(2(625)^0.5)
t'(125)=1/(2*25)
t'(125)=1/50

ok, that's enough for now. I hope you can follow this....if not, lets have a math party!

Amber said...

Oh my dear Lord. That is the most confusing thing I've ever seen. I'll have to try and make sense of it. But I definitely would like a math party. Is tomorrow [Wednesday] after school good? I know it's sudden but that's the best day for me this week.

Ms.Bearse said...

Wed is fine for me. I'm coming from Frontenac, so it'll be around 3:00 before I can get there.

I'll talk you through all that I wrote.

Anonymous said...

Hey!

How are you? I'm good and I was wondering if you new that One Act Play Night is this Thursday at 7pm. It will be a night to remember (that is if I remember my lines!), so come if your free! Have a wonderful night!
-Rosie

Amber said...

Yay! Then I guess at 3:00 I'll just wander about the front foyer or be in Ms. Canton's room or something. Probably wait in the front foyer at 3-ish, but if I'm not there, Ms. Canton's room.
Although, I don't know if she'd let us use her room... Hopefully!
I can't wait.
YAY MATH PARIES! :]

Anonymous said...

Speaking of student stuff everyone should come to...heads up:

IMPROV EXHIBITIONS!
October 19th AND 20th!
7pm in the auditorium adults are $5 and students $3

you can pick up tickets anytime from any KC improv auditioner. (but buy them from me, please. Must. Sell. Tickets.)

I hope you don't have any urges to bake or clean, and I CERTAINLY hope you have no reason to go to the 7/11 for milk and eggs, even if it is not raining. Sleep.

I have math questions, but my day started at 3:30am so I shall THINK about them properly tomorrow. Good night. :)

-Nicole

Anonymous said...

What are oxides?

Thanks,
Rosie

Ms.Bearse said...

something that has reacted with oxygen is called an oxide.

Lithium Oxide would be LiO2 (just an example)

Anonymous said...

Umm, hi.

Remember those muffins that were my, you know, dinner the other night? (along with carrots?) Do you possibly have the recipe for them? I’m being all must. bake. muffins. And they were all delicious, and you know, NOT choked full of saturated fats. (Biology is so turning me off food...) So yeah. Please? I haven’t had a spontaneous baking frenzy in ages, best catch this one while I can.

Cheers,

-Nicole

Ms.Bearse said...

Hmm.....ok, hope you don't bake after midnight, but here's the recipe: remember that they are insanely healthy, and not necessarily liked by all
Cranberry Cinnamon Bran Muffins
1 cup wheat bran
1/2 cup All bran (or similar 100% bran cereal)
1/4 tsp salt (which I omit)
1/2 cup boiling water
1 cup skim milk
1 cup dried cranberries (which could be raisins too)
1/3 cup sugar substitute (I used splenda)
1 egg
1/4 cup canola oil (some of which you can substitute with apple sauce, but I haven't tried that so I don't know the correct proportions)
1.25 cups whole wheat flour
1.25 tsp baking soda
1 tsp cinnamon
(I added all sorts of spices too...cloves, allspice, ginger...whatever-not chili though!)

1. In bowl combine bran, cereal, and salt. Pour boiling water over and stir to combine. Stir in milk and cranberries and set aside

2. In another bowl whisk together sugar substitute, egg and oil. Stir together with bran mixture.

3. In large bowl, stir together flour baking soda and cinnamon (and spices). Pour bran mixture over flour mixture and stir until just combined. Divide batter among 12 muffin cups. Bake at 375 F for about 20 minutes (or until the tester inserted into the middle comes out clean).

*sometimes if you put a bit of sugar and cinnamon on the top of the muffin batter after it is in the cup, it will end up with a bit of a sweet crust on the top...

Hope you like them. They freeze well, and I even enjoy eating them when they are frozen! hahaha...try that sometime.

Ok, it is already tomorrow, and I should get sleep before the grand homecoming football celebrations.

Hooray for sci'02!

Amber said...

Oh gawd, Ms. Bearse, I need help. I have a math test tomorrow and a lot of this stuff confuses me. We're doing inequalities and she gave us this huge sheet to do over the weekend.

I understand how to solve an inequality if it's like x^2 + x - 2 > 0 because I just factor and get the roots and stuff.

But what if it's like... x^4 - 6x^3 + 4x^2 + 6x - 5 < 0 ? Do I have to do polynomial divison by finding something that will divide into it evenly until I get a good factor like (x-2)(x-3) and stuff?

Ahhh. I'm so confused. :( And I'm freaking out.

Anonymous said...

Ms. Bearse?

I've come across at bit of a problem with math. Actually to me it's a big problem but whatever.

I can't factor. No before you dismiss this with the ever standard "yes of course you can, don't be silly." I've heard countless times listen and help me explain something. Through Gr. 10 and the beginning of Gr. 11 I've been able to simplify those numbers and I've done well. But I never actually knew how I did. The vast majority of the time I could look at a question and know what the answer was supposed to be. Maybe I wrote out factors when I ran into trouble but I rarely used any kind of "strategy" to find the answer because it seemed to jump out at me. The more I did it, the easier this “guessing” because. Worked for things like "x2 -7x + 12" = (x-3)(x-4)I don't even need to write out the factors of 12 to find what I'm looking for.

I've done this before, with basic algebra (Gr. 7) I never rearranged things like x + 3 = 4 solve for x. To find the answer I just looked at it and said 1. And it worked for a time. Like 2 years. Then I hit Gr. 9 and Mrs B had to teach me extensively the manual steps that should have already been in my head.

Now if I find an answer with anything harder then basic trinomials (things with a number other than one on x2) I don't know how I got there or I can't get there at all.

Like just now I’m working with 4x-sqaured + 7x +3 . I ended up with (x+1)(4x+3) but how? I just looked at it and played around with some numbers. I can’t do that for every problem. How would you solve that problem?

Sorry to bug you on your Sunday afternoon but I don’t know what to do. I really need to learn how to do this by Tuesday and my ‘current’ teacher, while lovely, she is going to be swamped with people because of the test. Don’t feel obliged to respond this weekend if you don’t want to but maybe sometime soon-ish? Please? I feel like I’ve hit a wall.

Sorry for the really long winded explanation/question/muddle. I needed someone to understand where I’m coming from. Do you understand?

-Nicole

Ms.Bearse said...

Amber, you will need to factor, and then do your chart... choose x values that will make each factor zero, and then between those zeros, you can check to see if it is negative or positive based on the number of factors (sets of brackets) that end up negative. If there is an odd number of them, the whole thing will be negative for that region.

Don't freak out. Factor quadratics, do polynomial division for the more complex ones...

you can do it!!

Ms.Bearse said...

Nicole, I'll see ya at lunch on Monday. It's all ok!

Amber said...

Thanks. c:
So would this be right:

x^3-4x^2-x+4>0
x^2(x-4)-1(x-4)
(x^2-1)(x-4)
(x-1)(x+1)(x-4)

so those are the roots and I graph it and then solve it?

And for ones that I need to do polynomial division, I'd do the factor theorem, right? In order to find out the divisor to divide by?

Ms.Bearse said...

yup, sounds right.
good job.

Amber said...

Yay! =] Thank you very much.

Sorry to bug you [yet again], but I've come across a problem.

The inequality is:
x^3+2x^2+3x+6>=0

Then I factored:
x^2(x+2)+3(x+2)
(x+2)(x^2+3)

So I know a root is at 2... But I can't factor the (x^2+3). And I tried to do polynomial division but nothing put into x equals zero, so I can't get a divisor. And the quadratic equation didn't work because the number under the root sign was negative, so does that mean there's only one root and it's at -2?

Also, I did a question before where it was a quartic function, but there were only three roots. That's possible, right?

Sorry for being a bother!

Ms.Bearse said...

If you have a graphing calculator, you can graph it and see....if not, a table of values will always work...

you can use other tools online, like function flyer to see what they look like.

Amber said...

Hey, Ms. Bearse. The math test was moved to Wednesday. I was wondering if it would be possible to have a math party tomorrow [Tuesday] after school?

Ms.Bearse said...

I'm at LC with the KC volleyball team. If the math party can be in the gym....it could be possible. I may have to sit on the bench, we'll see how particular the ref is. I can be there around 3 anyway, and the teams arrive for 3:30 so there's a 30 minute math party guaranteed. See ya there! (in the stands at the gym)

Anonymous said...

Hey,
I have a chem. test tomorrow and I was was just reviewing and theres a few points I just want to verify before the test. I was wondering if I could meet you at the beginning at lunch tomorrow for a bit?

Thanks,
Chelsea

Ms.Bearse said...

Sure, meet me my 2nd period classroom. I don't know how much of a help I will be, but I'll try.

Amber said...

I got a 98% on my math test! 59/60! I'm so happy. Thanks so much, Ms. Bearse. You were a huuuge help. :]

Ms.Bearse said...

WOW!!! That is awesome Amber! I'm proud of your hard work. You must feel so happy. Enjoy your weekend, and thanks for letting me know about your success. I'm glad to help anytime.

Anonymous said...

Hi Ms Bearse,

It's 10pm, I've been doing English for hours, I haven't slept really since before I saw you last (think about that) so I guess that qualifies me as insane. Post this at your discretion, if you get it tonight. But I have a thought. :)

This was originally Chelseas question she posed to me and I want to know if I’m on the right track. If you have HCl what base will react with a 2:1 mol ratio?

I came up with this equation: 4HCl + 2 Mg(OH)2 = 4 H2O + 2 MgCl2

Keeping in mind that I didn’t use a periodic table, and practically failed chemistry anyways is this the right train of thought? If not, what is? I didn't want to pull Chels on to far into my thinking incase it was WAY off...plus I'm super curious now.

-Nicole

P.S. Thank you muchly for coming to see the improv show, the support is great appreciated.

Ms.Bearse said...

if you want twice as much HCl as base, then that's perfect. You could choose any base that has a +2 cation. That would give you two (OH-) ions, and then you'd need 2 H+ ions which would come from 2 mols of HCl. My question to you is why do you have coefficients of 4,2,4,2.....you could reduce it all you know. Perhaps that is the lack of sleeping affecting you. (GO TO BED!) Oh, I have wool adventures to share tomorrow....find me sometime. Improv was great! Thanks for selling me a ticket.

Amber said...

Hey, Ms. Bearse. Sigh. Yet again, problems with math. I think you're going to be seeing a lot of me this year, probably next year, too, because I have another math course then. I don't know why, but I just don't get things, and the teacher moves too fast with no explanation or math parties!

So, is it possible for you to give me some help tomorrow [Tuesday] after school?

Ms.Bearse said...

Sorry Amber, I'm with the volleyball team tomorrow, and I'm busy each day after school this week. Can I help through this site?

Good luck!

Noor said...

i dunno if you'll see this soon enough, but i need to calculate max & min slope n have no idea how, n i 4got 2 ask you @ school 2day...plz reply asap...thnx LOADS! you rock n r probably the coolest teacher i've e'er had.

Ms.Bearse said...

what subject is this? what kind of question? I need an example, and then I'd be happy to help.

Noor said...

its for physics

im graphing a relationship in a spring:force vs extension of the spring. (a spring is elastic!)

I have to graph my data (with uncertainty bars) and then calculate (by hand because apparently excel doesn't do this) the max slope & the min slope

do u need more info?

Ms.Bearse said...

slope is rise over run...always. Plot your points, put in your error bars, and it's my guess to say connect (either for real or imaginarily--if that's a word) the bottom error bar of the first point to the top error bar of the second and calculate that slope. Then try it with the top error bar of the first point connected to the bottom error bar of the second, and use that for another slope. Does that make sense?

Ms.Bearse said...

Thanks to the knitters and others who sent that sweet card. I appreciate your well wishes. I'm feeling lots better after days of sleep....I'll see you on Tuesday (knitters at lunch, ok?) Have a good PA day.
Ms. Bearse

P.S. That was a sneaky method of card delivery!

Amber said...

Hey, I'm not sure if my last message went through. If it did, ignore this one. Can we have a math party on Tuesday after school? I'm desperate and I have a test Thursday. I don't understand a lot of things from last week, so could it be a bit longer than normal?
Thanks. :]

Ms.Bearse said...

Sorry Amber, things are still hectic. I'd have a bit of time Wed. after school, but that's it for this week. Tuesday I am busy. What are you studying now?

Anonymous said...

Hi Ms Bearse-

I have an important physics question (aren’t they all)for you.

A ball with a mass of 2kg is rolling at a constant velocity of 5.3m/s rolls off of a cliff. lets say this is on earth so the acceleration is 9.8m/s2 [down] and the ball is rolling 5.3m/s [east] and lets say the cliff is 30m in height and that the ground on the cliff is paralle to the ground below the cliff. I want to find out how long it takes for the object to land and how far away it will land from the cliff.

Is this a projectile motion problem? Because it has a velocity horizontally as well as vertically? Can I solve it with this much information anyways? It's a problem I'm making up as a I go along so I can easily make up more variables... but I don't know how to solve it, do you?

-Nicole

P.S.- Oh and a side note- I have a headache and a slight fever so may I blame YOU? :) This may explain my rather odd questions tonight...

Amber said...

That's okay, I understand. :]
We're doing rational functions when it's a fraction. Like finding the vertical asymptotes, horizontal, and oblique. I'm having problems with what I think is probably the simplest things.

Maybe I can ask you a few questions on here. Oh, and get well soon, if you're sick.

Okay, finding the y intercept of f(x) = x/x^2-3x-4. I sub in 0 for x and end up with 0/-4. Does that mean there is no y-intercept, or that it's 0 or something?

And if I find a y-intercept where it's -2/0 then that's not possible, so is there no y-intercept?

And for questions like this:
(2(1-x))/3 - 8 = 1/6x - (2x-m)/3
How do I solve for x?

Ms.Bearse said...

Nicole, yes it is a projectile motion question. Vertical velocity is initially 0, and it will accelerate due to gravity. Horizontal velocity starts at 5.3 and it will remain 5.3 forever since there is no air resistance (right?). So you will have 2 equations....one for vertical velocity, and one for horizontal (horizontal just use the v=d/t triangle method). Remember that d is either horizontal or vertical depending on what equation you use. You know d vertical is -30, and a vertical is -9.8 and v1=0 so you can calculate with d=v1t+1/2at^2 (vertical stuff)
-30=0t-4.9t^2
-30=-4.9t^2
calculate t
then use that t as the t in the horizontal calculations
d=vt
and calculate d
how's that?

I hope you are feeling well tomorrow...knitters is at lunch.
see ya.

Ms.Bearse said...

Amber, check out this site for more information

Basically:
Vertical asymptotes happen when the denominator of a rational function is zero. So factor the denominator, and when each of the factors equals zero it is a vertical asymptote. (try dividing 4/0 on your calculator, it should say error...that's because you can't do it! hence...asymptote)

horizontal asymptotes tell you roughly where the graph is going in the far extremes of x (realllly big positive and negative values) Try subbing them in for x and seeing if there is a trend towards a certain number. Check the exponents...that will give you a big clue as to what will happen (if you have a squared, it is a bigger number...dividing a number by a bigger number makes it get close to zero--that website has good examples)

Oblique asymptotes are when the numerator has a bigger degree (higher exponent) than the denominator does. you will need to do long division in this case to find the polynomial result. The polynomial that you get in the end is the slant asymptote.

hope that helps.
Good luck Thurs.

Ms.Bearse said...

Amber part 2:
your solving for x question....almost forgot that with all the asymptote fun.
2(1-x)/3-8=1/6x-(2x-m)/3
anything with fractions needs a common denominator...
so in this case it will be 6x so multiply each term by a fraction that equals 1 (top and bottom the same) to turn the denominator into 6x.
(2)(1-x)(2x)/6x-8(6x)/6x=1/6x-(2x-m)(2x)/6x
now...make it simpler
(4x-4x^2)/6x-48x/6x=1/6x-(4x^2-2xm)/6x
common denominators so I can add and subtract
(-4x^2+4x-48x)/6x=(-4x^2+2xm+1)/6x
I know that x can't ever be zero (or the denominators will be zero) so that is a restriction that you'll have to remember.
apart from that, you know that
-4x^2+4x-48x=-4x^2+2xm+1
and then get all the terms together...
-4x^2+4x^2+4x-2xm-48-1=0
4x-2xm-49=0
factor out the 2x
2x(2-m)-49=0
solve for x
2x(2-m)=49
2x=49/(2-m)
x=49/(2(2-m))
however, x can't be 0 (remember the restriction from before)

how's that?

Amber said...

Thanks, Ms. Bearse! So, are we still on for a short time tomorrow after school? :]

Ms.Bearse said...

Amber, I can be there from 3-3:30 but that's it. Meet you in LC's front foyer.

Amber said...

Hey, Ms. Bearse!
I thought I did really bad on the math test, but I got a 93%! Yay! :]
Anyways, we're working on special triangle stuff now, and even though it's review, I'm a tad bit confused.
Are you available for a math party Friday after school?

Ms.Bearse said...

yay! good work on your test. I'm busy on Friday, but I'm sure since you are just starting that stuff, that it will make more sense with more practice.

keep working hard!

Anonymous said...

Dear Mme. Bearse,

Due to many long pirates of Penzance Rehearsals, distraction and a little bit of procrastination, I have a test on Ecology on Friday and I do not feel very prepared! I have used up frankly all of my free time at home and now I only have time for my rehearsal and my brother's inconveniently placed birthday on Thursday night. Would it be possible for me and you to meet sometime at lunch tomorrow (wed) or Thursday to study?! It would really help! Thanks so much! :)

Rosie

Ms.Bearse said...

Lunch tomorrow sounds great. I'm not at Frontenac in the afternoon because grade 9s are at work with parents, so I don't have to drive anywhere. I'll be around the math office at lunch, come find me :)

Amber said...

Hey, Ms. Bearse!
We have a math test this Tuesday, so as I study this weekend, I might end up posting a lot of questions here. Sorry in advance!

So, my first question is this:

12. Use special triangles to determine the roots of each equations, 0 <= theta <= 360 [in degrees].
a) tan[theta] = -1

What am I supposed to do? Meagan and Angie told me I find the angles of the theta for tan and they're my roots, but I don't understand that.

Ms.Bearse said...

So...using triangles means we need to use the fact that tan is the opposite over the adjacent. For the opposite over the adjacent to equal -1, they have to be the same length, and of opposite signs (positive and negative). Draw these triangles on the cartesian plane so that the angle you are making can be measured from the origin to one of the axes. So you should have a triangle in the top left quadrant and one in the bottom right quadrant. The angle would be 315 (bottom left quadrant) and 135 top left quadrant.

Does that make sense?

Ms.Bearse said...

P.S. Here is a plug to anyone who reads this that there are still tickets left for the Cantabile Choir Concert at Sydenham Street United Church tomorrow (Saturday the 10th) at 7:30 pm (doors open at 6:30, and usually there are people in line to get good seats). Tickets are available at the door! Adult tickets are $18 and students/seniors/children are $15. The show is going to be great!

Ms.Bearse said...

Nicole, all is well. I checked in after school to find out. Have a GOOD weekend, you deserve it.

P.S. I got wool in the mail!!!!!
Show and tell for Tuesday.

Anonymous said...

Thank you.

Okay math question :)

If you have a quadratic function and you take the discriminate you can tell if the function crosses the x-axis and at how many point yes? So if you have a quadratic and a linear function, and you sub one into two. Ummmm like this...

A(area) = 8000
A= -2xsquared + 400x - 8000

and then...

8000 = -2xsquared + 400x - 8000

0= -2xsquared + 400x - 16000

like that. Will the disc. of the "new" function tell you if and at how many points they intersect? AND does this work if the linear function is NOT a horizontal line?

And am I crazy? About all of this?

-Nicole. who is smiling.

P.S. If for any reason at all you can’t respond until very late. And you still feel like answering. Post it anyways. I will be here.

Ms.Bearse said...

Hi Nicole,
Good job! I think you have it all figured out!

you can use any line and any parabola and see if they intersect using the method you described. it works for lines that are not horizontal. It involves a bit more simplification, but it works.

Glad you are smiling and doing math! Keep it up.

Amber said...

Hey, Ms. Bearse! Ismael and I are wondering if you can come to LC for a math party on Monday/Tuesday. We have a test next week that was for Tues. but may be rescheduled.

Also, how do I do this:
Express as a single sine or cosine function.
a) 2sin3[theta]cos3[theta]
b) 6sin[theta]cos[theta]
c) 1/2sin[theta/2]cos[theta/2]

Ms.Bearse said...

Hi, yeah monday works for me. I could be there by 3pm.

To do those trig questions you'll first need all of your identities written out. You'll be using double angle/half angle and the ones with cos(x+y) and sin(x+y)

Try rewriting it in different forms, look at which identity is closest...I'll be able to show you better on Monday.

See ya then. Have a good weekend.

Noor said...

hey ms bearse heres a physics question (again)

ok so im doin a lab on the efficiency of a bouncy ball in relation to the height from which it is dropped. (i will vary the height and see how it affects the efficiency.)

so first my hypothesis was basically that the efficiency would decrease as the height increased because if the drop is longer there is more time for energy to be lost due to air resistance.

but i soon realized that it s possible that if the height is higher (both the initial height and the bounce height) the energy lost may still be proportional to the total mechanical energy so the efficiency would remain the same.

which of these makes more sense?

Ms.Bearse said...

Ok....interesting question.
Are you comparing initial height to final height to determine the efficiency? Are you supposed to consider air resistance, or is it just the materials involved in the collision, and their elastic energy.

First, air resistance is proportional to velocity...so presumably, the higher the ball drops from, the greater velocity it will reach (until it gets to its terminal velocity). For fast moving objects (airplanes/rain etc) it is proportional to the square of the velocity. For slower objects it is proportional to the speed.

Thinking about the height of the rebound...there is air resistance there too, so it is losing energy (due to air friction) all the time that it is in the air... going up and coming down.

Does the speed of the impact change anything about the impact? Or is the impact dependent only on the material of the ball? Hmm...I'm interested in your results. Check to see if you are supposed to use air resistance.

check out these sites
site 1
applet

Anonymous said...

Ms Bearse...

There comes a time in the evening...after choir and improv and chores...when all the little physics questions come out and pound on my skull saying "study you dolt!"

uhhh that was werid...when... I need to ask you questions...

Context is thermal energy. Maybe.

You have a carpet and a tile floor. They are both the exact same temperature. So why when you step on them does the tile feel so much colder?

I dunno.

-Nicole.

P.S. I know I'm crazy. I KNOW. Post it anyways and tell me...please?

Ms.Bearse said...

Interesting question. My brother would call the tile floor a "heat sucker"....but that wouldn't really help. It has to do with the heat capacity of both materials (c) from Q=mcdeltaT

Tile is a better conductor than carpet. You put your foot on the tile, the good conductor of heat then "sucks" the heat from your foot. Remember that heat goes from the hotter to the cooler object. You sense the heat being sucked away as "cold".

Carpet is more of an insulator, it has lots of air in it, but just because of what it is made of, it is a better insulator. That means the heat isn't "sucked" from your foot as fast. You feel that it is warmer in comparison.

This is all about conduction. Check out this article.
or this one

There are tables of heat capacities listed on the net in various places. Hope that helps.

Anonymous said...

Ms Bearse (and all others who may possibly read this...)

The Kingston Improv Games puts on Winter Workshops this weekend for improv teams in the Kingston area, all day of skill building crazy-ness and a show in the evening to which you are all fabulously invited.

So at 7pm on Saturday at Theological Hall (back entrance) we have a show (Both KC teams and other high school teams). It runs about an hour to an hour and a half, fairly low-key (in terms of noise level...haha) and admission is by donation. So if you can come out- please please do, should be fabulous fun.

Thank you for hosting this last-minute plug for the show :)

-Nicole

P.S. oohhh! Ravelry invite came :)

Ms.Bearse said...

Rosie! Great work tonight. It was a really fun show. My mom and her friend came, they were in the Pirates in 1971 at KC.

Nicole, congrats on Ravelry. Do your homework still! I'll see what's up tomorrow. Hopefully I can see the improv.

Have a good weekend girls.

(Rosie, I can help you catch up on science. Maybe after the show is done....)

Noor said...

ms bearse

you said you'd be interested in the results for that bouncy all lab...well i looked over the shots i took o the drops and not all of them were valid (like a uplebounced way off to the side so i couldn't read it accurately) but i calculated the effiency for two of the drop heights (60 cm and 80 cm) and they were incredibly close! They were 89.58% & 89.38% respectively. on the one hand, this blows my mind because of the whole ir resistance increasing in vaied chunks thing, but on the other it maes sese, because it was the same collision each time...

im just happy to hve that part over. now, on to the analysis/concluion! :)

any thoughts?

Ms.Bearse said...

Maybe this will cause you to believe that neglecting air resistance (in this case of a small object, and small distance traveled) is a valid thing to do. You proved that the effect of air resistance is negligible for this situation :)

Good for you with the lab design. I hope that the analysis goes well too.

Anonymous said...

Hello!

I have a lovely physics question for a lovely (and helpful!) person!

The question is:

Erin goes to the mall to buy a new music CD. Starting at the main entrance of the mall, she walks 25m east, 42m north and 15m west before finding the music store after 3.0 min.

Draw a scale vector diagram to find the resultant displacement and determine its value and direction.
* * *
I'm a little confused on how you are supposed to draw the scale vector diagram because there are already three measurements so how are you suppose to draw the the resultant displacement line of the triangle?

Thank you so much for helping!

Test tomorrow! FUN! FUN! FUN! :(

-Rosie

Anonymous said...

Hello again!

I was just wondering if it would be possible for me to meet with you at lunch tomorrow for studying help for physics?

I'm understanding it okay but I just need some help with scale vector diagrams and some other things.

Thanks so much!

-Rosie

Ms.Bearse said...

Rosie, for your physics question, draw out all of the vectors given. You'll end up 42m north and 10m east (the 25m east and 15m west will end up being a total of 10m east)

Then you join up your start point with your end point. You can use trig, or a protractor to find the angle, and trig or pythagorean theorem to find the displacement vector.
42^2+10^2=hypotenuse^2

I can help you at lunch. I'm in an interview at the start of lunch, but I'll meet you and knitters probably in room 204. Nicole and Charise have the blanket, and we need to make up the display case stuff so we can show off our hard work. I'll be there after my interview is done.
See ya tomorrow.

Anonymous said...

Hey Ms. Bearse,

What on earth do you put on a resume for camp?

Do you even NEED a resume for camp?

Actually yes it says to.

.....you're a careers teacher right? :) You're all knowledgeable about this stuff?

The employment application asks for camp experience, related experience and my education. (what like gr. 11?) and such.

so what do I put on a resume... my hobbies? Should I be redundant and repeat what it already asked me?

I actually kind of wish my careers class had prepared me for this kind of thing. See you have a really important job.

-Nicole

P.S. No 100% on the bio bellringer, but I think I"m close. It was....gosh dare I say it...fun?

Ms.Bearse said...

On a resume, it is a good idea to put the most important things first. Since this is a camp job, put your camp experience first.

So the basic layout is this:
your name and contact information at the top

then have a heading like "camp experience" and list (most recent first) the experience, with a brief description about the skills that you used/learned, and the date and location of the experience.

other headings to include:

education-say what school you're in, the fact that you are in grade 11, and that you are taking science-if you are interested in nature stuff at camp, it is applicable, and some IB courses--show's you are up for a challenge--if you've been to leadership camp, write that down

work/volunteer experience
your coaching would go here, and any other things

hobbies
put down choir/soccer/voice lessons/craft addictions ;)
this shows you are well rounded and "balanced"

certificates and awards
your swimming certificates/first aid/CPR/drivers license when you get one/school awards

you may also have other things you want to say...I managed those with a "skills" category...you know, that's where you say you can tie knots, and do archery, and cook trip food...speak swahili or anything else that doesn't fit above.

I was always told to write down childhood activities like Girl Guides (even though I wasn't still in it, it still counts) So if you were a member of any groups or clubs...it could help.

Keep the resume to 2 pages max.
If you look in the careers section of my site, I put some links to resume writing tips. I think you should be able to explore and find them.

Good luck.

Anonymous said...

Thanks for your speedy and helpful reply. This might work now. :)!

Anonymous said...

Hello! Me again!

I just found my answer to my question on the internet so don't bother responding to the quadratic question. But any tips and trick or ideas, would still be greatly appreciated! Thank you!

-Rosie

Ms.Bearse said...

Rosie, read the rubric carefully. Explain everything that you can as much as you can. Verify what you are doing with another method (graphing calculator and by hand for example, or using factoring and the quadratic formula...etc) Draw pictures. Make it easy to see what you were thinking, and make sure you answer ALL of the questions asked as best as you can. You'll be fine. Remember to breathe!

Anonymous said...

What is "le discriminant"?

Thanks,

Rosie

Ms.Bearse said...

b^2-4ac
it lives under the square root in the formula. If it is positive, there are 2 answers, if it is negative there are no answers, if it is zero there is one answer.

Anonymous said...

hi,

ummm if you're still about tonight...

AC and DC current? I understand that AC current is when the direction of the induced current naturally alters/changes directions. (I havn't wrapped my head around it enough to picture it in my mind but I got me fun little diagram ..proves it. I think.) But DC is different? What IS it?

I keep looking it up and getting explanations for electricity and induction... my book says it uses split ring communtators (whatever that is) as a opposed to the slip ring ones for AC.

I'm lost. any ideas what all or any of that means?

-Nicole

Ms.Bearse said...

Hi Nicole,
I can explain that stuff to you....it's hard to do through words alone though.
Here's a neat website to look at. I googled "split ring commutator slip ring" and got some good things.
Here's another. Enjoy.

Anonymous said...

Hullo!

I was wondering if it would be possible if you could give me a hand with understanding prevailing winds, maybe at lunch tomorrow? I have five questions for homework that I need to do for tomorrow and it would be great if I could get them done!

I haven't spoken to you at all since the Christmas concert. Hope you had a good holiday! I saw you today and you look rather sad (didn't even say hello to me!), maybe you were just thinking hard! Kidding!

Anyhoo hopefully I will see you tomorrow! If not Friday for knitters!

Toodles!

Rosemary

Ms.Bearse said...

Sure Rosie, catch me at the beginning of lunch--I'll be near the math office for a bit, and then I'm heading to Frontenac. I was probably just thinking, or...adjusting to being back at school, I didn't mean to ignore you. :)

See ya tomorrow.

Noor said...

Ms bearse gan you give me a general understanding f transmission and reflection through thin film? if you eed to do it in person we can do it before school/after period A tomorrow. its kinda gonna be on a unit tes per. b tomorow...anyway thanks a lot!

Ms.Bearse said...

This is a neat site

here is another

and here is one with java

I may be able to answer questions--if they are easy ones! Find me in my classroom before period A, or at the end of A. I'll try to help.

P.S. Knitter's is going to be in the auditorium if we can get in there.

Amber said...

Ms. Bearse!
Haaha, it's that time of year. Exams.
And my math one is freaking me out [as usual.]

I hate application problems; I really have trouble solving them. So can you help me on this question?

A small shelter for delicate plants is to be constructed of thin plastic material. It will have square ends and a rectangular top and back, with an open bottom and front, as shown in the figure. The total area of the four plastic sides is to be 1200in^2
a) Express the volume V of the shelter as a function of the depth x.

That's basically all I need. The rest I can answer but I can't figure out the equation because I'm bad at this.

Ms.Bearse said...

since I can't see the figure it's a bit tricky for me. I'm imagining a rectangular prism....based on what I read in the question.

We know the volume of a rectangular prism is V=l*w*h. Since the end is a square, you'll have V=l*l*w (assuming h=l)

Now working with the area information. I know a rectangular prism has 6 sides. 2 are open (according to the question).
A=Atop+Aback+2(Asquare sides)
A=l*w+l*w+2(l*l)

We know A=1200
1200=2(lw)+2l^2
600=lw+l^2

you'll need to isolate probably w from the A formula stuff, and sub it into the V formula to solve...
does that help?

again, it's hard with no picture.

Anonymous said...

Hello!

Quick science question(s) (last exam!! whahaahaahaa...). If lemon juice has a pH of 2.6 and apples have a pH of 5.2, approximately how many times more concentrated is 2.6? I got this one wrong on my test, stating that it was 1000x but apparently it's 398x. But how do you figure that out? And I just want to make sure I have got this right: the pH scale (power of Hydrogen) refers to the concentration of the H+ ion and therefore because acids have the ion H+ the more concentrated the acid (pH 1), the higher the pH. However if something had a pH of 1, it would have a high concentration (high pH), even though the number is low (pH 1). I think that makes sense... So does that mean that if something was basic and had a pH of 9.2, would it have a high concentration or low? Low, right?

Thanks so much for all your help! I'm sure this isn't the last question that I'll trouble you with! Enjoy your week off!

-Rosie

Ms.Bearse said...

Hi Rosie,
I just got home :)

You know that for every increase of 1 in the pH scale that is a 10 times increase in the concentration of H+.

If we started off with a pH or 2.6, it would be 10 times to have a pH of 3.6 and 100 times to have a pH of 4.6 and 1000 times to have a pH of 5.6 etc.

I'm not sure how much of the math you have done...but here it goes.

pH=-log[H+] which means that pH is calculated by taking the negative logarithm of the hydrogen ion concentration. To reverse the "log" function you need to use the exponential function. It looks like 10 to the x on your calculator.

We're going to calculate the H+ concentration for each pH
[H+]=10^(-pH)
so 10 to the power of negative pH

If we do that for pH of 2.6 we get approximately [H+]=0.00251009, and for a pH of 5.2 we get [H+]=6.3086x10^(-6) --> scientific notation meaning the decimal is really shifted 6 places to the left, with lots of zeros added(it's a tiny number).

Now to determine how many times more concentrated it is, we take the one number and divide by the other. 0.00251009/6.3086x10^(-6)=397.883 which is 398.

Hope that helps.

----
Acids have lots of H+, high concentrations, but low pH. It has to do with the math of it all...but if you have a high concentration it ends up with a pH that is low.

Bases have low concentrations of H+ actually, they have lots of OH-, and their low H+ concentration will cause the formula to give you a higher number for the pH value.

So, if you have a pH of 9.2, it is a base. It has a low concentration of H+. You could calculate it if you want...
H+=10^(-pH)
H+=6.30957E-10

since there is an E-10, you have to shift the decimal 10 to the left, making it a really TINY number for the H+ concentration.

Anonymous said...

If something has a initial velocity of say 10 [down] and has a final velocity of 10 [up] and your finding out the acceleration, how do you know the direction for the acceleration?

Thanks so much! Ahhhhhhhhhhhh exam tomorow! So scared!

Thanks,
Rosie

Ms.Bearse said...

Rosie, if it starts going down, and ends up going up, it is changing direction from down to up...so it is going to be accelerating (positively) in the upward direction.

This is like if something starts going 9m/s left, and then 5 seconds later is going 1m/s right. We can calculate acceleration.

a=deltavelocity/deltat

deltavelocity=vfinal-vinitial
deltavelocity=1m/s[right]-9m/s[left]

**1m/s[right] is the same as -1m/s[left]**

deltavelocity=-1m/s [left]-9m/s[left]
deltavelocity=-10m/s[left]
deltavelocity=10m/s[right]

so...acceleration..
a=10m/s[right]÷5s
a=2m/s^2 [right]

...or you could also say -2m/s^2 [left]

is that ok?

----
as for the chemistry stuff....the world is more complicated than grade 10 science--there are more rules that you'll learn later that help in understanding covalent bonding. You should be able to name any covalent formula using prefixes, and write the formula down from the name....and explain the bonding in cases that follow the rules you have learned.

Mind you, there may still be mistakes on my old tests...

good luck.

Anonymous said...

Hullo!

I have vunderful question pour vous. Well not really vunderful but I'm sure you won't mind it. Okay so we're learning about f of x (f(x)). Here's the question.

Given f(x)= 3x, g(x)= x2 and h(x)= 2/x write down and simplify expressions for each of these composite functions:
a) (fog)(x)
b) (gof)(x)
c) (foh)(x)

Thanks so much! It would be vunderful too if you could respond ASAP because I have eleven a,b,c,d...etc. questions for homework tonight and most of them involve simplifying!I'd veally appreciate it.
Thanks,
Rosie

Anonymous said...

ms bearse

calculator worries...my graphing calculator wont graph! i did the y= thing and entered 1/x (cuz thats the graph i need to see) and it says ERR: INVALID DIM.

i think clearing the calculator memory might help ( i may have done something wrong before) but i dont know how. please help!

Ms.Bearse said...

Hi Rosie
I'm guessing on this one a bit.

a) I imagine that this means f of g of x
so we'd first do g of x
and then we'd do f of x to that....

so g(x)=x^2
f(x^2)=3x^2

b) g of f(x)
first we find f(x)=3x
then g(3x)=(3x)^2=9x^2

c) f of h(x)
first we find h(x)=2/x
then f(2/x)=3(2/x)=6/x

Ms.Bearse said...

Calculator worries:
whenever you find invalid dim that means that you have data entered into your stats that dont match up, and you've got stat plot on.

Turn off stat plot.

Clear your lists

It will be ok after that.

Noor said...

Part 1: Physics

i know you said you dont know but still...

in beta radiation, where a beta particle is emitted, a neutron breaks into a proton and an electron. therefore a neuton can be known to consist of one proton and one electron. Does this mean there is a way to fuse a proton and an electron to get a neutron?

***********

Part 2: Theory of knowledge

This is not a question, but a general statement for all to see and ponder. But first, some background informtion.

An Empirical truth is a truth that depends on the facts to remain true. eg: the book is on the table. If chelsea moves the book, it is no longer on the table. A necessary truth is a truth that is invariably true and connot possibly be a falsehood. eg: 3 + 5 = 8. Scientific laws can also be considered to be necessary truths.

We were discussing varieties of empirical and necessary truths. as well we spoke of how some of newton's theories, which were widely accepted until recently (relatiity) would have been considered necessary truths but have been proved to be empirical. Then, the following statement was made:

All scientific theories seem to be necessary truths until someone disproves it and discovers something new.

ponder that.

:D enjoy!

Ms.Bearse said...

Sorry Noor, I don't know much about nuclear physics. Try this
or this

I didn't really read them, but it might be what you are talking about.

Chelsea said...

Hello

I am getting a bit confused with this completing the square stuff... I think it might have something to do with my unfriendly relationship with fractions?(we're not allowed to change the fractions into decimals) but I'm not sure. I was just wondering if you could maybe show the steps with this equation so I can figure out where I'm going wrong and hopefully fix it :)

The standard form is...
y=3x^2 - 4x - 7

thank-you so so much!

Ms.Bearse said...

Hmm, thought I posted something but it is not showing.

3(x^2-4/3x)-7

divide 4/3 by 2...turns into 2/3. Square it...turns into 4/9. add and subtract in brackets.

3(x^2-4/3x+4/9-4/9)-7

3((x-2/3)^2-4/9)-7
3(x-2/3)^2-4/3-7
3(x-2/3)^2-24/3
3(x-2/3)^2-8

Nicole said...

Hi Ms Bearse,

Chemistry question. Gr 11 still, baby stoichiometry if you will..

What is the amount (in moles) of each type of atom in the following samples?

2.0 mol of iron (III) nitrate.

Which I'm fairly sure is Fe(NO3)3

(I DID do all 100 of those naming problems before choir on Monday...)

I have answers...remember it is a Nelson text book... Fe- 2mol N- 3mol O- 9mol

Ms.Bearse said...

there are errors in the answers. I overheard others talking about that very question at lunch. Your formula is correct. You have 2 mols of it, so it could be written 2 Fe(NO3)3

there are 2 mol Fe, 6 mol N, 18 mol O. Ms. F agreed to these answers, and wants you to correct the back of the book.

glad you did all 100 questions. :)

Anonymous said...

oh I get it. Stupid stupid textbook. thanks!

math video present... "Ma and Pa Kettle Math"

http://video.google.com/videoplay?docid=7106559846794044495

50 questions means I can do it....100 means I can't FORGET it...

Nicole said...

oh I get it. Stupid stupid textbook. thanks!

math video present... "Ma and Pa Kettle Math"

http://video.google.com/videoplay?docid=7106559846794044495

50 questions means I can do it....100 means I can't FORGET it...

Anonymous said...

Hello!

First grade 11 IB math test tomorrow and I'm kind of freaking out! :)
I was wondering if you would understand a question that I am having a lot of trouble with:

8. Given f(x) = x2 + 4, ER and g(x) = 1/1-3, x ER, x ≥ 4
find an expression for g(f(x))and state its range
The answer is 0 ≤ y ≤ 1/17

I'm really confused because I've heard things from different people, all different such as you take the x ≥ 4 and sub it into g(x) and then take g(x) range (what ever g(4) equals to) and use it as the domain of f(x)and then you find the range that way. But I've also heard that you just take the full equation and sub in 4. My brain is so confused!

It would be spectacular if you could respond as soon as possible! I really appreciate it! Thanks so much!
-Rosie

Anonymous said...

Hello!
Would it be possible to go over some grade 10 math review with you at lunch tomorrow? I got this homework for review on quadratics and my mind has gone blank. I looked in my old notes and understand them but I'm pretty sure the answers in the back of the book contain errors.
At the moment I feel like banging my head against a wall, so it would be spectacular if you would help me out, and help me avoid more brain damage! :)

Thanks so much! You are a terrific person!
-Rosie

Ms.Bearse said...

Hi Rosie,
Ya, come find me at lunch. I'll be in the math office right at the beginning, and we can go to one of the math rooms.

See ya then.

Anonymous said...

Just a final plug for the Kingston Improv Games for anyone and everyone that reads this....

Wild Card
Convocation Hall
February 23rd @ 1pm (doors open at 12:30p)

FINALS
Sydenham Street United Church
February 23rd @ 7pm (doors open at 6:3o)
Adults: $8, Stu/Sen: $5, Kids $1

Roster: TBA

(as of Thursday the roster is LCVI, KCVI, Holy Cross & La Salle + Wildcard winner)

It's going to be a BLAST so come on out and watch crazy awesome improv! :)

Charise said...

Ms. Bearse!!!!
I need help! I have never talked to you by this means before, but Noor suggested it.
So i am doing a Physics lab, and i am trying to make the thin lens equation (1/f=1/di+1/do)linear. Can you please tell me how to, or get me started? thanks...its due tomorrow, btw...thanks

<3 Charise

Ms.Bearse said...

Hi Charise,
This is late, sorry, but here's what I found online. I'm not sure what you mean by "making it linear". There are 3 variables in that equation....

anyway, here's a site that talks about making it linear....
maybe it will help you.

Charise said...

Thanks, Chelsea helped me with the linearization of the formula...and i have that now. Now i am just waiting for her to come back online to tell me how to graph a fraction, though while writing that sentence, i think that i just figured it out!

great! have a good end to your weekend

<3 Charise

Anonymous said...

Hi!
It's me!
Do you know how to do this question?
If √3 + i = z1 √3 – i = z2, then express z1/z2 in the form a + bi.

I was also wondering if you are available/willing to have a math party with me tomorrow! I have a test on wednesday that I am incredible stressed out about, plus corrections due for my previous test! I am incredibly stressed out, with tests before march break so the extra help would be extremly useful! Thanks so much!
-Rosie
P.S. We're doing imaginary numbers! :( :) AHHHHH! They're imaginary!

Ms.Bearse said...

Sure Rosie, Room 204 at lunch. I'm helping with science too--not sure what you're doing with imaginary numbers, but we can have fun figuring it out.

Anonymous said...

Hello as promised I have a math question... two actually but I think they are similar in how to figure out I just don't really know how to go about answering them..

1) Find the zeros of the function. First rewrite the equation in a different form. [f(x)=4x-12+(9/x+2)]

2) Solve for x. [(3/x-5)+(2x/x-3)=5]

That is if you're still awake... if not you will find me tomorrow morning with Noor and Nicole furiously figuring out what I'm supposed to do :)
Thank-you

Anonymous said...

..........

it's been too long since I bothered you here with chemistry Ms Bearse...

"A calorimeter with a mass of 1.49kg and a specific heat capacity of 897.5J/kg K contains 8880g of liquid water. Initially the apparatus is at 22.0 degrees Celsius and contains 1.5g of ethanol (CH3CH2OH) mixed with excess oxygen. If the heat of combustion (delta H comb) of ethanol is -1371kJ/mol what will the final temperature of the calorimeter when combustion is complete?" (Answer 32.3 degrees C)

I know the heat capacity of water to be 4.18 J/g degrees C

q = (m)(c)(delta t)

q = [(m)(c) + (m)(c)] delta t

delta Hreaction = -q/n

delta H reaction = delta H formation of products minus delta H formation of reactants

I think the units may be tricky...along with everything else...

anyways...if you have TIME and any BRILLANT IDEAS about this then go for it. If not, you still made me understand the whole Hess's Law thing (and consequently the lab hypothesis I'm 'a writin' !!!!) which makes you ace. :)

-Nickel.

Ms.Bearse said...

Chelsea, sorry I didn't get your question in time. I hope you could find the answer.

Nickel...if that is indeed your real name...
It's been a while since I've thought about calorimeters. If I remember correctly the heat comes out of the reaction and is absorbed by the water. That's my assumption for now anyway (you're getting my train of thought written down, it might be jumbly)
Q=mc(delta T)
Q out of the ethanol combustion is Q into the water and the calorimeter

so for water (using subscript w)
Qw=(8.880kg)(4180J/kgK)(Tfw-22C)

for the calorimeter
Qc=(1.49kg)(897.5J/kgK)(Tfc-22C)

for the ethanol
Qe=(0.0015kg)(hmm...do we know c?)(Tf-22C)


delta H comb of ethanol is -1371kJ/mol
(aha, this can get me q for ethanol)
find molar mass of ethanol...
46g/mol according to wikipedia (check if you want...I'm too tired to)
that means we have 1.5g/46g/mol
0.0326 mol of ethanol
that means that delta H comb of that much ethanol is
0.0326mol*-1371kJ/mol = -44.707kJ
which is the same as -44707 J
*ignoring sig figs*

so...
our delta H is -44707J for the reaction.
delta H=-q/n
this q is q for ethanol...
-44707J

so....the ethanol gives off 44710J of heat, which will be absorbed by the water and the calorimeter.

Qe=Qc+Qw
44707 J=(8.880kg)(4180J/kgK)(Tfw-22C)+(1.49kg)(897.5J/kgK)(Tfc-22C)
hmm long equation
44707J=(37118J/K)(Tfw-22C)+(1337J/K)(Tfc-22C)
ha! I forgot the final temperatures are the same...gets simpler.
44707J=(37118J/K+1337J/K)(Tf-22C)
44707J=(38455J/K)(Tf-22C)
Tf-22=1.16
Tf=23.16 hmmmm
hmmm....this is not working out to what you say is the answer.

To verify that this makes sense...delta T=1.16

Q water=(8.88kg*4180J/kgK*1.16K)
Q water=43151J

Q cal=(1.49kg*897.5J/kgK*1.16K)
Qcal=1554J

add up Qwater+Qcal = 44707 J

My answer says that the final temperature of everything is 23.16C

Let me know what I've done wrong when you find out.

I double checked units, but you are right, they are messy.

Anonymous said...

i think you should put up the pictures from the pancake day :)

Ms.Bearse said...

I will do that tonight if I can get it to work. :)
Remind me tomorrow in class.

Noor said...

Hi Ms Bearse, this is the first time for one of these, but CHEM question! If my solubility table says a compound is slightly soluble, does it form a precipitate?

Ms.Bearse said...

I'm sad my first post didn't register Noor,
I did answer right after I saw your question.
here is one site with neat calculators on it.

here is another site with definitions

here is another site with even more stuff.

Enjoy. Not sure if this will help or not.

Anonymous said...

Hullo Ms. Bearse,
I was wondering can you please please please come see Footloose this week? We're having two shows one on thursday nigth adn the other friday night. They are at 7:00pm and the cost is ten dollars.I hope your week has been good mine sure has!
~Cassie

Ms.Bearse said...

Hi Cassie,
I hope to go on Friday night, unless we have a late soccer game, in which case I'll go Thursday night. Thanks for reminding me about it. Hope it goes well.

Break a leg!

Ms. Bearse

Anonymous said...

It's been about 3 chemistry units since I've had a question for you Ms. Bearse so if you're still there...

it actually might be math in disguise?

Ka = ([x][x]) / ((2.0*10^-3) -x)

Ka = 6.6 * 10^-5 (value obtained from table. it's the acid equilibrium constant for benzoic acid. That's fairly irrelevant to the math part but that's where it comes from. by the way the x value in the concentration of [H3O+] ions, also completely besides the point)

So I'm solving for x, which I'm fairly is a quadratic but I can't simplify it. I think I'm missing something here...

-Nicole

P.S. don't you love rainstorms? :)

Ms.Bearse said...

hi,
I'm assuming that the [x] and x are the same quantity....because otherwise there is another unknown...
the [x] makes me think concentration

anyway...
6.6*10^-5=x^2/((2.0*10^-3)-x)
multiply by denominator...
6.6*10^-5((2.0*10^-3)-x)=x^2
(1.32*10^-7)-(6.6*10^-5)x=x^2
here's the quadratic for ya...
0=x^2+(6.6*10^-5)x-1.32*10^-7
x=-(6.6*10^-5)+/-sqrt((6.6*10^-5)^2-4*1*-1.32*10^-7)/2(1)
x=-(6.6*10^-5)+/-sqrt(5.32*10^-7)/2
x=-(6.6*10^-5)+/-(7.3*10^-4)/2
x=6.6*10^-4/2 and x=-8.0*10^-4/2
x=3.3*10^-4 and x=-4.0*10^-4

does that help?

Anonymous said...

If I have 2000=1250(1.03)^x... how do I solve for x? I can get to 1.6=1.03^x but then I'm lost and I've been looking but can't find it in my textbook.
If you're sleeping (as most sane people would be at this hour) no worries I'll just have to get to school early tomorrow :)

Thank-you
Chelsea

Ms.Bearse said...

Either express 1.6 as 1.03 to the power of something...
and the something would equal x

or you need to use the log function. Have you learned that yet?

Ask your teacher.

Ms. Bearse

Madelaine said...

Oh My Gosh Bearse.
You Compleatly corrupted me, shelly, and tory's summer. Seriously, whenever we get together we always think about science stuff. It's not even funny. Stuff we learned in class, like that clown guy, and little ryhmes and stuff. Jokes that werent funny at the time. General facts. You have no idea, Canada day, our birthday party....We cant get away. *eye twitch* Thanks alot. :)

Ms.Bearse said...

Madelaine, you made my day! Thanks for letting me know what a big science geek I've turned you into. Shelly and Tory too! Mission accomplished!!

I hope you guys take more science this year. OH! Here's an idea, come to knitter's club on Fridays at lunch, Chelsea's there too, and we can relive the fun times.

See ya on Tuesday perhaps. Take care,
Ms. Bearse

Noor said...

Twilight. Read it.

You're a teacher, its pretty simple. I hear even Mrs. McGauhey is starting it! :D

Have a nice day.

Ms.Bearse said...

I will Noor....I promise. I'm just worried that I will get sucked in and not be able to do my work.
Thanks for the concern. ;)

Skylar B. said...

I've memorized up to 51 numbers of PI so far! go to my locker tomorrow morning and ill recite it all

Ms.Bearse said...

Skylar, I'm SO impressed! Honestly, I had no idea you had such a talent for pi :)

Did you amaze your family?

I'm going to tell some other math teachers about your new found skill.

Thanks for telling me. I'll find you tomorrow morning.

Noor said...

request:
i always forget that we even have to do these. so for queens PSE, i need a 'referee' whom they may contact to verify stuff i said. like the fact that im in knitters, i orgaized a fundraising dinner way back, was in respect last year and the year before...KCDC exec and model UN committee. thats about it. all you have to do is let me know if youd be willing t do that, and theyd email/call you sometime to make sure im not lying.

Ms.Bearse said...

Sure thing Noor,
I'm easier to get a hold of by email. Feel free to pass on my limestone email address to them. I check that all the time, way more often than I check my school voicemail.

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