Sunday, December 18, 2005

LCVI math 10

Hey grade 10s, send me some questions if you need assistance. I'd love to help.

100 comments:

Anonymous said...

Hey Ya, Ms. Bearse... okay we have a test on wednesday and i dont understand similar triangles at all!!!!!!! Not a bit! yeah... so can you come after school tomorrow????????? Okay peace! Kiran!

ms.bearse said...

What time Kiran? I'm at Frontenac until about 3:30, how's 4:00 sound?
--Ms. Bearse

ms.bearse said...

I didn't see you after school Kiran, so I watched the boys play basketball. Good luck with the studying. If you have particular questions, you can send them to me, and I'll give them a try.

Anonymous said...

hey ms. bearse can you hold a math partie on linier systems on a day next week.
Nicholas P.

ms.bearse said...

Sure Nick, I'm in your class on Monday, and helping Monday after school with trigonometry, you're welcome to come then...and I could help out on Tuesday for linear systems. Please bring your review package, with as much of the linear systems work done as you can. Good luck!

Anonymous said...

Ms.Bearse,
Does the exam review thingy enlarge??? i cant read nothin'...

Kiran

ms.bearse said...

Which "exam review thingy" are you talking about Kiran?

Anonymous said...

the exams thingy on your website the one from frontenac.... ive printed it like five times but it wontenlarge and i ca t read it properly....
Kiran

ms.bearse said...

Hi Kiran, I think those exams are made for 8.5x14 paper (big paper), if you are able to see the .pdf in adobe acrobat reader, you should be able to use the magnifying glass on the screen to enlarge it. I don't think this will change the size when it is printed, but you will be able to read the questions on your computer screen. I hope that helps.

ms.bearse said...

Kiran, you may be able to select the information on the .pdf document and paste it into Word, and enlarge it there. I'm glad you are working so hard.

Anonymous said...

Thanks Ms.Bearse!I'll try it out...

Anonymous said...

Hey, Ms Bearse.
Sorry for missing math help yesterday. Had to get a couple projects done. I am now about to settle down and crack this math, so I'm keeping the webpage up and will be sure to ask any questions of you when needed.

Thanks for everything.

- Nick ( so full of rage ) Arnold

ms.bearse said...

Hey Nick, glad you are working on math now. Remember the train...."I think I can, I think I can, I think I can." That will help you focus on solving the problems, and perhaps eliminate some of the rage.
Good luck! If you want to have another math party, I've got time tomorrow (Thursday) 2:30ish. Let me know.

Anonymous said...

Hey, Ms Bearse.
Let's arrange a math party tomorrow at 2:30ish then. If I can get ahold of a few others ( kiran, nick etc.. ), I'll see if they're available.
I have come across a question that has me stumped. It's cause it's a word problem, see.
"From a position on the dock, Jodi can see two buoys that mark the turning points for a canoe race. One is in the lake 400 m straight ahead of her, and the other is at an angle of 52 degrees. If the buoys are 500 m apart, how far from the dock is the second buoy?"
So, I made a trangle. I've got two sides ( 500 m, 400 m ) and an angle of 52 degrees. I've labelled the unknown side as "x". I'm unsure of what to use. Although, I may have it figured out by the time you reply. We'll see.

- Nick ( easing up on the rage ) Arnold

Anonymous said...

I'm thinking cosine law, but I'm no sure....

ms.bearse said...

Well...ok....let's see....(get the rambling done first)
I can't draw a triangle in this little text box thingy, so I'll see how well I can describe it. Call the dock where Jody is J, and the first buoy is K and the second is L. Side l is 400, side j is 500, and angle J is 52. You are eventually looking for side k. Since we don't have all 3 sides, and we don't have side angle side, we can't use cosine law. Since it is not a right angle triangle, the only thing we can do is sine law. The first step is to find angle L, and then use angle L and angle J to find angle K (all add up to 180 right?). Then when you have angle K you can use sine law again to find side k. How's that? Let me know if you get it.

Anonymous said...

Got it, Ms Bearse.

Around 634.4 m which is what the book is, soo there.

Thanks very much.

I've notified Amethyst about tomorrow. She said "maybe. I'll think about it" and I've notified Sarah Hannah who has not yet replied.

- Nick

Anonymous said...

Ms. Bearse... i dont remember how to factor.... well i do but i dont... uh here is one question 3x2+7x+2= no idea.... could you help me?

Thanks, Kiran

ms.bearse said...

3x^2+7x+2
You are looking for two numbers that add up to make 7 (the x coefficient) and multiply to make 6 (the x^2 coefficient, 3 times the final number, 2).
___+____=7 and ___x___=6
The numbers that work are 6 and 1.
6+1=7, 6x1=6
so now we split up the middle term (7x) into 6x+1x
3x^2+6x+1x+2
now we common factor the first two terms and the second two terms
3x(x+2)+1(x+2)
the brackets should contain the same thing at this point.
now we write the bracket, and put all that's left into another bracket.
(x+2)(3x+1)

To check your answer you can multiply to expand, then simplify. Hope that helps.

Oh, Nick's coming for help tomorrow at 2:30 ish. You are welcome to come too--spread the word. I'll meet you guys in the main hall and we'll find a spot to work.

Anonymous said...

Hey, Ms Bearse.
Just to let you know, I'm stuck on the Mabel question again. Where she's looking at the building. I'm not gonna bother trying to get help on that through here, cause I think it'll mess me up more, so I'll save it for tomorrow.

- nick

ms.bearse said...

you can do that one Nick, draw the building. You are given the angles from the horizontal looking to the top of the building, and the bottom of the building. You are given the distance between buildings (draw that directly across from her eye level). Fill in all the angles you can. You'll have right angle triangles, so use SOH CAH TOA and the pythagorean theorem. Draw the picture at least, and I can help you tomorrow. We'll meet in the front hall, then we can find an open room. Did you get the other question figured out?

Anonymous said...

Oh, yes I did get the other one figured out. I sent a message but it must have not gone through.

I'll try this one.

- nick

Anonymous said...

Hey, Ms Bearse.
So, I'm still stuck on that question. I got a picture on it and everything, and I'm trying Soh Cah Toa but it's not making sense.
However, I'm looking at the Similar Triangles and Primary Trig Ratios page and trying some of those questions and I've come across one that is extremely difficult to draw from the world problem.
"A 3.2 m ladder is leaning against a vertical wall with its foot 2.0 m away from the wall. Another ladder 4.8 m long is leaning against the wall, parallel to the first ladder. What distance is the foot of the second ladder from the wall?"
...umm ... how on earth do I draw THAT??

Anywho.

- Nick

Anonymous said...

Looking at all this now, I'm actually extremely stuck with the Primary Trig ratios and Similar Triangles page.
I think we'll have to look at that.
However, I'm going to go on ilc.org when they open up a chat tonight and see what happens there.

- Nick

ms.bearse said...

Keep up the hard work! You can do it. Your dedication is admirable.

ms.bearse said...

Ok, parallel ladders....that means that they have the same slope. They are leaning against a wall (which meets the ground at 90 degrees). Each ladder leaning against the wall makes a right angle triangle. The triangles will be similar, since the angles are the same due to parallel slopes....I don't know if I'm making sense or not.

The deal with similar triangles is making ratios.
The first ladder is 3.2m long (that's the hypotenuse of the triangle). That can be compared to the other ladder that is 4.8m long (again, another hypotenuse). So we can make a ratio: 3.2/4.8=2/x
solve for x by cross multiplying (2)(4.8)/3.2=x
hope that helps you.

Anonymous said...

So, it would just come out to 3 metres? ... That's it?

- Nick

Anonymous said...

Got it, Ms Bearse.
You're a genius. I'm getting the other questions now too. Thankyou.

- Nick

Anonymous said...

okay, one more question.
I have a triangle ( ABC ) a=21, c=20, b = "x". Angle B is 90 degrees. The question is asking me to find the value of sin A. So, I'm assuming I'd have to use Soh Cah Toa. But, I'm looking for the angle of A, so it would be Sin = Opposite of Hypotenuese, correct? But, I don't know what the hypotenuese is. How does that work?

If you have the time.

- Nick

Anonymous said...

nevermind. figured it out.

- nick

ms.bearse said...

Great work Nick! I'm proud of you. How's the rage doing? Can you solve the problems while smiling yet? That's when you know you truly understand. I think you are almost there. I expect to see smiles tomorrow ok? I'll see you guys in the front hall at 2:30ish tomorrow. I have made some cookies!

Anonymous said...

Hello again.
I have a right angle triangle
( ABC ). The hypotenuse side is 15 cm. That is the only side that is given. They are asking me to find the side of the opposite.
They gave me and angle of 60 degrees plus the 90 degrees so given that, I figured out the last remaining angle which came out to 30 degrees.
So, my question is, how do I use all that to figure out side "x"

- Nick

Anonymous said...

Okay, I know I wasn't aloud to do this. But, I used the sine law on the right angle triangle to figure out the one side so that I could then use Pythagorian Thearom to figure out side "x". It's not the same answer in the book, but it's only a few off. The book says the answer is 13. I got 16. I don't know any other way to figure it out.

- Nick

ms.bearse said...

Hey Nick, I'm at LC now, if you want to come here earlier, you can. I'm hanging out in the math office.
I have solved your question!
Ok, The triangle I made was A=60, B=30 and C=90 (because of your excellent logic!) The hypotenuse is 15. Since we have a right angle triangle we can do SOH CAH TOA. Because we have the hypotenuse, we are going to do SOH or CAH (you get to choose!) I like SOH better (for no particular reason). So, I'm using sine, and starting with angle A. SinA=opp/hyp. Sin60=a/15.
15(sin60)=a
a=12.99 which would be the 13 in the back of the book.

At this point you can use the pythagorean theorem to find that side b=7.5
(you could also do sin30=b/15 so b=15(sin30), b=7.5)

sine law should work out also, but you'd be docked marks I guess. Keep working hard! See ya later.

Anonymous said...

Okay, Ms Bearse.
Thank for today. I will for sure be there tomorrow "after school". I may be spending the whole day there, but either way I will for sure be there. 2:30?

Thanks again.

- nick

ms.bearse said...

See you there Nick, 2:30 in the math room. Bring your friends, and leave your "math rage" at home. You are making fabulous progress.
Keep working hard

Anonymous said...

KILL ME!!! is completing the square from standard to vertex...?/

ms.bearse said...

Completing the square is going from standard form to vertex form. Are you ok with how to complete the square?

ImSoCrazyAllTheTime said...

MS. BEARSE! We're having a math party on Monday after school, right? At 2:30 like normal? If so, I am there!

-Amber

ms.bearse said...

Math party "after school" on Monday at 2:30 in the math room. It'll be the last one before your exam, so bring any remaining questions, and we can get everything sorted out. You are working so hard. I'm proud of you. See you Monday. (you can send questions over the weekend if you get stuck. I'll answer when I can.)

Anonymous said...

Hi, Ms Bearse.
I do have a question. Number 9 on the Frontenac exam. It asks me how many solutions the linear systems have and they are y=x+5 and y=x-1. The answer says none. Now when they ask for a solution in a linear equation, does that mean the point at which the two lines intersect? So, in this case, it would be that the lines don't interesect. Just a small question. No biggie.

Thanks.

- Nick Arnold

ms.bearse said...

you got it! When the slopes(m) are the same, but the intercepts(b) are different, then the lines are parallel, and have no intersection point, therefore no solution. The other options are one solution, when the lines intersect once (you can find that by graphing, subsitution, or elimination). The other possibility is that they have many solutions (if the slopes and intercepts are the same, the two lines are identical, and intersect at every point).
I'm glad you are left with only small questions! I hope you can see the amount of progress you are making. It is remarkable. Keep up the great work.

Anonymous said...

Hey, Ms Bearse.
Thankyou very much. Both my parents and myself are proud of the work I've been putting in this week.
Looking at some of the questions on the exam, there are some I just don't remember how to do including factoring that I would like to go over with you on Monday. Then, I should be good.
In the meantime, I've just been re-doing questions I've done in the past until they are inbeded in my brain. I seem to have subtraction and elimination down pat now.
Though I'd let you know.

- Nick

ms.bearse said...

Monday we can go over factoring and completing the square, and graphing parablolas etc. I hope you are feeling more calm about the exam. The big trick is going to be thinking positively when faced with a question that you've never seen before. I think you're getting better at that as well. Keep up the good work.

Anonymous said...

Ms.bearse~~!!! I'm half way done my review booklet, but i have LOTS of questions...=.= and tuesday is our exam day..for example, if the question ask us to solve a linear system, can we use any method?! Like elimination, substitution, graphing etc. as long as we get same result?! There are lots of false answers in the booklet i think, coz'my answer do not match with the answers in the booklet~!!! ar....stressed out with Linear system!!~~~can you come on monday for a little bit?!! I think i need some tutoring too!! THXXXXXXXXXXXXX so much~! ~~~

IvY

Anonymous said...

Hi, Ms Bearse.
Listen, I'm having trouble with that "z" pattern triangle again. Do you know the one I'm talking about? I'm just confused with it again. It's a hard one to explain. We've done it before, I'm just confused with the meaning behind the "z" pattern again and what to do with that.

See you Monday.

- Nick Arnold

Anonymous said...

Haha! Got it! Nevermind.

- Nick

ms.bearse said...

Ivy, I'm going to be in the math room at 2:30 on Monday. I'm helping Nick (and others if they come) with quadratics/factoring stuff. I can help you with linear systems then too if you'd like. Amber has worked through most of them, so she'd have a good idea which answers didn't match the book's.

If you are asked to solve the linear system, or determine the vertex, or any of those things, you do have choice. Choose the way that is easiest for you to do.

The exam may have specific requests though, like solve by graphing, or solve with elimination...I haven't seen the exam, so I'm not sure what to tell you.

ms.bearse said...

Good work Nick! I'm glad you're able to figure all this stuff out on your own. I hope the confidence is increasing :)

See you tomorrow.

Anonymous said...

YES math party is what im in despertae need of .... i know i know everything but its the stupid word problems im having trouble with... im coming to the math party. thanks ms.bearse.

kiran

ms.bearse said...

See ya Kiran, we'll do word problems :)

ms.bearse said...

I tried to explain the triangles already on this message board. Have a look at that explanation, and see if that helps. If you need further clarification, I'll be in the class tomorrow morning. Come get help.

ms.bearse said...

Hello Math Party People. I have some news. I just found out that I have an interview in the afternoon tomorrow (Monday), so I may be late arriving to the math party, or I might have to leave and come back to it. I will do my best to be there at 2:30, and I will definitely be there as soon as I can. I hope that you see this message. When I find out my interview time, I will post an update here.

Thanks for understanding, and keep your fingers crossed for me :)

Anonymous said...

Hi, Ms Bearse
Aaahh. Buses have been cancelled and all exams postponed. Is the math party still a go? If it is, I'm there. Let me know.

Good luck with your interview.

- Nick Arnold

ImSoCrazyAllTheTime said...

Congrats on getting an interview! You better still come though. :( There will still be a math party, right? I was just wondering because of the buses being cancelled and everything. I can still make it.

Amber

ms.bearse said...

I have an interview at 3:00. I'll be at LC for 2:00 if that helps people. I'll have to leave for about half an hour, but I can be back after my interview to help some more.

Anonymous said...

Hey, Ms Bearse.
How would I go about attempting the following question.
"A projectile is fired into the air at a velocity of 50m/s from an initial height of 2 m, the height equation can be represented by
h = -5t( squared ) + 50t + 2.
Find a) what the height of the object is after 3 seconds
b) What is the object's maximum height
c) when does the object reach maximum height.

I'm assuming this all has to do with parabolas.

- Nick Arnold

ImSoCrazyAllTheTime said...

Ms. Bearse, on the exam review [not the red booklet, the one from your site], for the question in factoring: 6x^2-13x-5 I don't know what to do. :( I'm sort of freaking out at this point because I'm really nervous about the exam.

-Amber

ms.bearse said...

never fear Amber! you can do it.
6x^2-13x-5
you need to find 2 numbers that add up to -13, and multiply to make 6(-5)=-30.
____+____=-13
____x____=-30

Go through the options for the multiplication first.
1x30, 2x15, 3x10, 5x6
the ones that would potentially make 13 are 2 and 15, and 3 and 10. We now have to consider the signs. The only way that we can get -30, and -13 is by using -15 and +2

so we rewrite the -13x as -15x+2x
6x^2-15x+2x-5
now we common factor the first two terms and the second two terms.

3x(2x-5)+1(2x-5)
now we collect the common bracket and put it in front, then we collect the other terms and put them in the next bracket.

(2x-5)(3x+1)

You can check by expanding and simplifying.

Anonymous said...

Ms Bearse,
6x(cubed)y + 30x(squared)y-36xy.

I worked that out so that I got an equation so that I first got

6xy(x(squared) + 5x - 6 )

then I worked it out so that I got

( x( squared ) -1x + 6x - 6 )

Now, if I follow the steps we were doing today, the answer comes out differently from the right one and this is from the Frontenac exam, so I'm going to assume the frontenac exam is right. Now, my question would be ... do I add up the like terms????

- nick Arnold

ImSoCrazyAllTheTime said...

Oh, okay. Thanks! Now I have another question. Will we need to know how to do equations of circles when the centre is not at (0,0)? And do we need to know how to draw a circle on a graph? I can't seem to remember any of that stuff. And I also got really messed up trying to do that question about the pool on the math exam on your home page. :(

By the way, you didn't answer Nick's question. ;)

ms.bearse said...

Hello Nick, your idea about parabolas is great insight.
"A projectile is fired into the air at a velocity of 50m/s from an initial height of 2 m, the height equation can be represented by h = -5t^2+ 50t + 2.
Find
a) what the height of the object is after 3 seconds
to find the height after 3 seconds, substitute t=3 into the equation and solve for h.
h=-5(3)^2+50(3)+2
h=-5(9)+150+2
h=-45+152
h=107 m


b) What is the object's maximum height
the maximum height is the y value of the vertex. You have options here:
a) complete the square to put it in vertex form.
-5(t^2-10t)+2
-5(t^2-10x+25-25)+2
-5(t^2-10x+25)+125+2
-5(t-5)^2+127
the maximum height is 127m.

another option is to find the zeros, and then find the vertex from the midpoint between the zeros.
-5t^2+50t+2
(this doesn't factor, so we'd use your favourite formula)
t=(-b+/-square root(b^2-4ac))/2a
t=(-50+/-square root(50^2-4(-5)(2))/(2)(-5)
t=(-50+/-square root(2500+40))/-10
t=(-50+/-50.4)/-10
t=0.4/-10=-0.04
or
t=-100.4/-10
t=10.04

the zeros happen at t=-0.04 and t=10.04
the axis of symmetry is at
t=(10.04+(-0.04))/2
t=10/2
t=5

the h value when t=5 is (sub t=5 into equation)
h = -5t^2+ 50t + 2.
h = -5(5)^2+50(5)+2
h = -5(25) + 250 +2
h = -125+252
h = 127

by either way of solving it, the maximum height is 127m


c) when does the object reach maximum height.
it reaches it's maximum height at t=5 seconds

ms.bearse said...

6x^3y + 30x^2y-36xy.
6xy(x^2 + 5x - 6 ) great first step

then I worked it out so that I got
6xy( x^2 -1x + 6x - 6 ) don't forget the first factored bit (the stuff in bold)

the next step that we do is factor the first two and second two terms separately
6xy(x^2-1x+6x-6)
6xy(x(x-1)+6(x-1)) both brackets are the same! YAY!

the next step is to collect the brackets up
6xy(x-1)(x+6)
is that the answer according to Frontenac's exam?

Anonymous said...

Yes.

Anonymous said...

Hi Ms. Bearse! Im a first time writer! I havent been on the site in a while, im sorry, but i just cracked open the math book to review for the exam and i forgot what a right bisector is.. so ive come to find out :) but i got side tracked by your poem, which by the way is sooooo awesome! :) anyways, sorry i couldnt go to the last math party but i hope to see you soon. Thanks for everything, Ms Bearse. Take care, from Sarah Hannah

ms.bearse said...

You only need to know how to find the equation of circles centred at (0,0). To do this you know that x^2+y^2=r^2
if they give you a point on the circle, sub it in for x and y, the solve for r^2. Then rewrite your equation x^2+y^2=(your number for r^2)

If they give you r, it's super easy. Just square the r value and rewrite as before.

To draw it, you know the centre is at (0,0) then count how many spaces the radius is, and plot those points on the positive and negative x and y axes, connect those 4 dots in a circular fashion.

(I think I did answer Nick's question...didn't I?)

ms.bearse said...

Hope you figured out the right bisector Sarah,
the tips and tricks to remember how to calculate those are as follows:

a) find the slope of the line segment given (y2-y1)/(x2-x1)

b) take the negative reciprocal (flip the fraction, and change the sign) to find the slope of the perpendicular line (right bisector)

c) find the midpoint of the line segment ((x1+x2)/2, (y1+y2)/2)

d) sub all the information into the general form of a line. The midpoint will be subbed in as x and y, and the perpendicular slope will be m. Then solve for b.

e) write your final equation with x,y as variables, and your m and b that you have found.

hope this helps you.

ImSoCrazyAllTheTime said...

Hey Ms.Bearse! So I know this is a day late, seeing as how the math exam was yesterday but... I just wanted to say thanks for being such an awesome teacher! The math parties really did help a lot, and I think they improved how I did on the exam since I forgot most of what we did in the beginning of the year.

Thanks for all your hard work teaching and also baking cookies for us! I brought my worry rock to the exam with me, and it will now go to all of my exams with me just like pony does! :)

Thanks again!

-Amber

Anonymous said...

MRS. BEARSE!!!!!!!!!! this is Yi Shan~~~~~~u know what...i'm taking university math!~!! it's fun so far.....i hope your teaching is fun 2 at KC~~~~we all miss u!!!!!!! P.S. I LOVE UR POEM~~~IT'S AWESOME~!!OH YEAH~~~:)

ms.bearse said...

Hi! Thanks for the update Yi Shan. I'm glad you are enjoying university math. I'm sure you will do fine in that course. I'm going to be at LC supply teaching the next two Wednesday afternoons, so I'm sure I'll see you around. Let me know if you want to have another math party...for old times sake

See ya!

Anonymous said...

Ms.Bearse~!!!!!!!!!!! I have a question from my math class which I have no idea....=.=well...part of the reason is that I was sleeping in class these days...didn't quite catch up with the lesson....anywayz...sorry to bother you ar:) do you happen to remember anything..anything at all about "Exponential growth and decay"???? cuz I just don't get it when it comes to solving a problem when the information is just there....arrrrrr...what can I do...so @#%!#$^#^ >.< !!!!!!!! P.S. IT WOULD BE AWESOME TO SEE U EVERYDAY AT LC CUZ EVERYONE LOVES U!!!!!!!:)take care and good luck with your teaching^^

IvY^^

ms.bearse said...

Exponential growth and decay....hmmm....I checked what you need to know in the grade 11 curriculum, and you need to know exponential growth and decay along with compound interest (money stuff). I found a pretty good explanation here

Here's an example. If you have 3 bacteria, and they double every 5 minutes, how many bacteria would exist after half an hour?

You can solve this with a chart
min.......doubling...bacteria
0 .......... 0 .......... 3
5 .......... 1 .......... 6
10 .......... 2 .......... 12
15 .......... 3 .......... 24
20 .......... 4 .......... 48
25 .......... 5 .......... 96
30 .......... 6 .......... 192

We can make an equation that relates the present number of bacteria to the number of doubling periods, and the initial number of bacteria

lets call the number of current bacteria N, and the initial number of bacteria B, and the number of doubling periods t

so N=B(2)^t
*we use 2 since we are doubling...if it was tripling, we'd use 3. If it was dividing in half we'd use 0.5

Lets try to solve it with our equation
N=3(2)^t
half an hour has how many doubling periods in it?
30 minutes/5 minutes per doubling period=6 doubling periods
N=3(2)^6
N=3(2x2x2x2x2x2)
N=3(64)
N=192

We can also solve other kinds of problems:
How long would it take to get 6144 bacteria?
In this case, N=6144, B=3, and we are solving for t
6144=3(2)^t note, we still use 2 because we are dealing with doubling

we divide both sides by 3
2048=2^t
now we need to express 2048 as a power with base 2. (try it out on your calculator)
2^11=2048
2^11=2^t
so we know that t=11
After 11 doubling periods there are 6144 bacteria. Each doubling period is 5 minutes, so it would take 55 minutes for 3 bacteria to become 6144 bacteria if they double every 5 minutes.

Does this help? Come find me at school Wed afternoon this week or next, and I'll help you out. :)

Anonymous said...

Ms.Bearse!!!!!!! OH MY GOD~~ I NEED HELP...i'm stuck with this question....god..can you help me to solve it and post the answer by today?!!!! THANK U A BILLION TIMES!!!

Question: The Charko family wants to fence in a rectangular area on 3 sides that borders a river. The fence costs $30/m to install and they have a budget of $4500. Determine the dimensions of the field that will maximize the area for their horses.

Yi Shan!!

Anonymous said...

ms.bearse!! one more question: A caterer in quoting the charge for producing a dinner proposes the following terms. For a group of 60 people, he will charge $30 per person. For every extra 10 people, he will lower the price by $1.50 for the whole group. What size group does the caterer want to maximize his income?

MS.bearse~this one i have to set up an equation for it..but i don't know how~!!! ar....soooo confused!!!

Yi Shan

ms.bearse said...

Rectangular fence that borders on a river. That means you have 3 sides to deal with (the river is the 4th side). Lets call the side opposite the river, l, and the other side w. The perimeter (length of fence) is P=l+2w
The cost is 30$/m....so cost=30*P
C=30(l+2w)
we know the max cost is 4500 so
4500=30(l+2w)

now, dealing with area:
A=(l)(w)

we want to maximize the area, so we use the cost equation, and isolate either l or w, then substitute back into the area equation to get a quadratic (yay, parabola!)

I’m going to isolate for l
4500=30(l+2w)
150=l+2w
150-2w=l

now substitute
A=(150-2w)(w)

So this is a parabola in factored form. We can find the zeros easily by setting A=0 and solving for w
0=150-2w or 0=w
150=2w
w=75 or w=0

so we find the mid point between the two zeros, that gives us w=37.5, and now we can solve for l using the simplified cost equation.
l=150-2(37.5)
l=150-75
l=75

the dimensions of the field for max area with the budget given is l=75 and w=37.5

Hope that helps you.

ms.bearse said...

Ok lets start by understanding that income=cost*number of people

initially with the number=60, cost=30, income=1800
next, with 10 more, number=70 cost=30-1.5 income=28.5(70)=1995
you can develop an equation: let x be the number of groups of 10 extra people

n=60+10x
c=30-1.5(n)
c=30-1.5(10+10x)
c=30-15-15x
c=15-15x

i=n(c)
i=(60+10x)(15-15x)
this is in factored form now, so to maximize income, set it to zero, find zeros, find midpoint etc and continue from there.

let me know if you need more help than this.
good luck

Anonymous said...

THANK YOU SO MUCH MS.BEARSE!!!! p.s. when you say find midpoint then continue from there, do you mean that I have to expand it first, then complete the squares to find its zeros in factored form, then i'll be able to see the vertex and maximum value? is that what you meant?!!!!

Yi Shan

ms.bearse said...

i think i made a mistake with the first equations that I made for you....
n=60+10x
c=30-1.5x
i=(60+10x)(30-1.5x)
set i=0
0=60+10x or 0=30-1.5x
10x=-60 or 1.5x=30
x=-6 or x=20

the zeros are at x=-6 and x=20. The vertex is between them, so (20-6)/2=7. when x=7 the income is maximized. That would be when n=60+10(7) or n=130 people. Each person would pay 19.50, and the total income would be 2535.

sorry if I led you astray earlier.
If you did foil, and then completed the square you should get the same results.

Anonymous said...

Ms.bearse!!!! thank you sooo so much for helping me with those questions!!! luv u!!! mu-ah~~~~ anywaz....how's teaching going? how come you didn't come to LC yesterday for tutoring? we all miss u....:) come on, we should have some "reunion" soon~!! okay?!!

Yi Shan

ms.bearse said...

Hi Yi Shan,
glad I was able to help you out. Hey, are you involved with the "Math League" for LCVI? I'm helping out with the KC team. It is a fun activity for math students to do. I think the final competition is at LC in April.

I was having a "math party" yesterday at KC, but none of my students came! If you want, we can try to have one at LC. Find a time, and I'll see what I can do.

Have a good day.

Anonymous said...

ms.bearse~~lol....there's a "Math League" at LC? really? wow...I have never heard of it...well, if I did, I don't think my math skills are good enough for me to join...lol~~~T_T you know what, I think we should have a math party next week 'cuz everyone needed a break from all of those summative evaluations~ guess what? my LEAP class is going to Winnipeg in two weeks!!!! aaaaa~~how exciting~ on the other hand, I guess i will miss a lot of work...so I will probably go on your forum and ask questions everyday!!! hohoho~~be prepared!!! lol~~~~(wink@_@) anywaz, back to topic...I have so many QUESTIONS on transformations of Quadratic Functions...here's a question, and I will post some more tonight~!!! Thankk uuuu so much for your help ms.bearse~!!

Question #1:
a) Describe the stretch that transforms y=√x to y = √4x
b) Describe the stretch that transforms y=√x to y = 2√x
c) How do the graphs of
y = √4x and y = 2√x compare?
d) Compare the transformation you described in parts a) and b).

Question #2:
Starting with f(x)=√x, a reflection in the y-axis is followed by a horizontal compressioin by a factor of 1/3 followed by a translation of 4 units up. Determine the equation of the resulting function.
(NOTE: ms.bearse, this is my result to question #2 - y=√-3x +4
is it right?)

Question #3:
Determine the equation of the inverse of those two equations:
a) y=2(x-2)2+1
b) y=1/4x

THANK U MS.BEARSE!!! I WILL ASK YOU MORE IF I HAVE ANY MORE QUESTIONS!!!!!!!:)

Yi Shan

ms.bearse said...

Hi again Yi Shan. I'll do my best with these :)
Question 1:
y=√x to y = √4x involves each x value being multiplied by 4 before the root is taken, so that is a horizontal compression of 1/4
y=√x to y = 2√x involves each value of the function being multiplied by 2, so that is a vertical stretch of 2
the graphs end up being identical because √4=2.

Question 2:
your answer looks right. Be sure the +4 is not under the root though. That would entail a horizontal translation of 4 along the x axis.

Question 3:
to make the inverse, switch the variables and solve for y
y=2(x-2)^2+1
x=2(y-2)^2+1
x-1=2(y-2)^2
(x-1)/2=(y-2)^2
√((x-1)/2)=y-2
√((x-1)/2)+2=y

y=1/4x (I'm assuming that question is like 0.25x)
x=(1/4)y
4x=y

now if it were y=1/(4x)
x=1/(4y)
4y=1/x
y=1/(4x)
neat how the inverse can be the same as the function, eh?

hope this helps you out. Would Thursday or Friday be a good day for a math party? Let me know.

Anonymous said...

Ms.bearse~that really helped!! a few more question...lol~~~

Question #4:
Describe the transformation as completely as possible for the following transformation:

1)The transformation of y=√x to
y=-√2(x+4) (NOTE: all of the x+4 and 2 are under the squre root with a negative sign outside the root sign!!!)

2)The transformatioin of y=√x to
y=√-x+3 +1 (NOTE: the 1 is outside the root sign, and the -x+3 is inside the root sign!!) BY THE WAY, DO YOU HAVE TO FACTOR IT TO GET THE RIGHT TRANSFORMATION LIKE:
y=√-(x-3) +1

Question #5:
a)Is the equation y=-4(x-3)^2-5 a function? State the Domain and Range for the equation.
b)Is the equation y=+and-√x-1+2 a function? State the Domain and Range.
c)Is y=-2√x-3 a function? State Domain and Range.(NOTE: only x is under the square root sign)
d)Is y=1/2√-(x-2)a function? State Domain and Range. (NOTE: the negative and x-2 is all under the root sign)

ms.bearse, what's an easier way to recognize the Domain and Range when you have a quadratic equation? Like do you have to draw the graph first then state its domain and range?....thankk uuuu!!!

Yi Shan

ms.bearse said...

Question 4-1:y=-√2(x+4)
the negative sign is a reflection about the x axis.
the (x+4) is a horizontal shift of 4 to the left...think about what x would have to be for the bracket to equal zero. In this case, x would be -4.
the 2 is a horizontal compression of 1/2.

Question 4-2: y=(√-x+3)+1
The negative is a reflection about the y axis.
The +1 is a vertical shift of 1 upward
if you factor under the root it does turn into -(x-3), so x needs to be +3 in order for the bracket to be 0. This means it is a shift of 3 to the right.

Question 5:
a) yes it is a function. The domain is all values of x, the range is y is equal to or less than -5
b) no it's not a function. The domain is x is greater than or equal to 1. The range is all values of y
c)yes it is a function. The domain is x is greater or equal to zero, and the range is y is equal to or less than -3.
d) yes it is a function. The domain is x is less than or equal to 2. The range is y is greater than or equal to zero.

The way I think about these questions, is sort of like doing the parabola dance inside your head :) if it looks like y=x^2 or something you know it will be a function, and you know that the domain will be all real numbers, since the parabola spreads open in both directions. You can check if it opens up or down, and then find the vertex and you will be able to determine the range.

When it is y=√x it is a little different. This is a sideways parabola. If you are counting both the negative and positive roots, then it is not a function. It is a function if you count just one (either positive or negative). If you count both roots, the range will be all values of y since it is spreading out both upward and downward.

when you have √(x+5) or something like that, determine the value of x that makes the bracket equal to zero. This value (x=-5) is the vertex. In this case the parabola opens up to the right, so the domain is all x greater or equal to -5.

If it were√-(x+5) or something like that, again, the value of x that makes the bracket equal to zero is -5. In this case though the parabola opens left. The domain is all x less than or equal to -5.

If there is a vertical shift y=(√x)+4 then the range would be all y values greater or equal to 4--because we are looking at the upper half of the parabola. However, if we use y=-(√x)+4 then the range would be all values of y less than or equal to 4--because we are looking at the lower half of the parabola.

I guess I'm sort of drawing the graphs in my head to figure them out. It gets easier with more practice.
Try this simulation to graph them on the computer.

Good luck, and let me know if you'd like help after school thursday or friday.

Anonymous said...

HEY MS.BEARSE~!!! HOW'S EVERYTHING GOING WITH YA? JUST THOUGHT I CAN COME HERE TO SEE U...everything's okay? hohoho~~~~~thank uuuuuuuu sooooo much for your help, I got a 93 on my Transformation Unit test!!!! LUVVVV U.....:) by the way, can you look at those two questions for me? by the way, they are from the Conics unit...gosh...i just cannot figure it out these two questions, and I'm way to go!~~~lol...

1) Determine the equation of the locus of points where all points are four times as far from point
A(0,0) as they are from point B(5,0). (p.s. for this question, do I have to solve it using the distance formula?)

2) Point F has coordinates(3,0). Line d has equation x=1. Find the equation of the locus where all points are the same distance from Point F as they are from the line d.

About the math party, um...since we have friday off this week and monday off next week, would you willing to come in next wednesday for a math party?!!! I WILL GRAB MORE PEOPLE TO COME IF YOU THINK THE TIME IS GOOD FOR U!!! OKAY?!!!! you know what...we are learning about trignometric functions now...ar....so hard!~~so everything goes well with u~~~especially taking french courses!!!! GOOD LUCK!!! take care:)

Yi Shan

ms.bearse said...

Yi Shan, I can come THIS Wednesday, would that work? We have a late soccer scrimmage, and the girl that I tutor is away on a field trip. Next Wednesday probably wouldn't work for me.

Congrats on your good mark for the last test! I will look at the conics stuff tonight. I had trouble with conics when I took that course, but I will try again tonight! You are motivating me to learn :)

I found this website that might be helpful. Let me know about Wednesday, and I'll post my answers sometime later tonight.

Anonymous said...

heyyyyy ms.bearse~~~thank you sooooooo much for doing this for me....:) you're actually motivating me to learn~~wow..after that test, I just realized that I don't hate math at all...lol~~what a big jump^^lol...u're on the soccer team?~!!!?? kkkk.....I would try to grab more people to come since I'm gonna have a math test the following day. Ar...just so many things to do~!!!!! ar.....anywaz..I'm gonna go give a try about those two questions AGAIN!~~i'll be on later tonight okay?~~~~take care...:)

Yi Shan

ms.bearse said...

The answer to the second question is that we are making a parabola. The website I mentioned earlier said that if the locus is the points equal distance from a point and a line, that it is a parabola. If the line is vertical, the parabola opens sideways. We know that the focus is at (3,0), and the line is at x=1. That means the vertex is at (2,0) since it is half way between the focus and the line. The general form of the parabola with vertex (h,k) is (y-k)^2=4d(x-h) where d is the distance between the focus and the vertex. In our case d is 1.

y^2=4(x-3) is what I get for the answer for the parabola. Does that make sense to you?

ms.bearse said...

I was confused about the first question until I read your comment about the distance formula. That makes sense to me. I used point 1(0,0) and point 2(5,0) and point A(x,y). I calculated the distance between point 1 and point A, and then I calculated the distance between point 2 and point A. I made the first distance equal 4 times the second distance.
d1=((x-0)^2+(y-0)^2)^(1/2)
d2=((x-5)^2+(y-0)^2)^(1/2)
d1=4d2
(x^2+y^2)^(1/2)=4((x-5)^2+y^2)^(1/2)
then I squared both sides
x^2+y^2=16((x-5)^2+y^2)

I was confused about what to do now...It looked sort of like a circle, so I expanded then completed the square for x
x^2+y^2=16(x^2-10x+25)+16y^2
x^2+y^2=16x^2-160x+400+16y^2
x^2-16x^2+160x+y^2-16y^2=400
-15x^2+160x-15y^2=400
-15(x^2-(160/15)x)-15y^2=400
-15(x^2-(32/3)x)-15y^2=400
-15(x^2-(32/3)x+(16/3)^2-(16/3)^2)-15y^2=400
-15(x-(16/3))^2+1280/3-15y^2=400
-15(x-(16/3))^2-15y^2=(1200/3)-(1280/3)
-15(x-(16/3))^2-15y^2=-80/3
(x-(16/3))^2+y^2=(-80/-45)
(x-(16/3)^2+y^2=(16/9)

Finally this looks like a circle. The centre is at (16/3,0) and the radius is 4/3.

This took me a while, and I'm not really sure about my process.
Good luck! I'll see ya Wednesday after school :) No conics then, ok?

Anonymous said...

THANK U MS.BEARSE~~~ this really helped:) Wednesday works for me...and I guess I would have to do some more work to catch up~~~ar....

Yi Shan

Anonymous said...

Ms.bearse..it's Yi Shan again...man>.< we just got back from Winnipeg, the trip was fun!:) Now I am running into so many things that I do not understand....like how do you draw a graph of y=1/2sin3x starting at (0,0). By the way, our class is going SO fast so it's hard for me to catch up...OMG!!! ar....dying...=.= Ms.bearse can you come like next Wednesday again for a little party?!! I promise I will grab more people to come this time!! lol~~~can u? anywaz...good luck with ur teaching..I better get back to my work...

ms.bearse said...

Hi Yi Shan, I'm glad that Winnipeg was fun. These days we have soccer practices and games every day after school except friday. I am available next thursday (may 4th) after school though if that would work for you.

For graphing trigonometric equations you can always do a table of values to figure it out. You can also do what we did for the "parabola dance" but it would be a "sine graph dance" instead.
y=1/2sin3x would start by graphing the function y=sinx, then you'd go through the transformations to graph y=sin(3x). This will compress everything horizontally by a factor of 3. Imagine the graph of the sine curve between 0 and 270 degrees all happening between 0 and 90 degrees. Then you'd do the transformation y=1/2sin(3x) so the amplitude (maximum height) of the sine graph is now 1/2 of what it was. Instead of having max values of 1 and -1, they are now 1/2 and -1/2.

Hope that helps you out. Let me know about Thursday the 4th.

Anonymous said...

ms.bearse!!! hahahaha~~~omg...I heard that you're not Burcu's tutor...that's great!!^^what a small world...lol:) Thursday would totally work for me...so tomorrow after school at 3:30? okay...see you then!!! thxxxxx soo so much ms.bearse~!^^

Ivy

Anonymous said...

P.S.if you have time, come watch the Rugby game of KC against us on monday after school!!!:)^^ lol...if would be fun if you come Ms.bearse~!!! anywayz....let me know okay?!:) much love...lol

Ivy "again"

ms.bearse said...

I'll see what the soccer schedule is for Monday. We might have an away game. Do you play rugby? When and where is the game? I'll try to get there for some of it.

Thanks for the info.

Anonymous said...

ay-yo Ms. Bearse, whats crackaclin'? yeah im tlaking like that because im studying for science... my exams on monday. i dont understand word equations in chemistry. they are so confusing. he keeps tlaking bout acids and H and bases and HCl.... could u please explain to me how to write them in both formula and word equations. GAWWWWWWWDDD SOOO FREAKIN' CONFUSING!!! please help.

ur innocent student, kiran =)

ms.bearse said...

Acids all have H+ as their cation. HCl (aq) is hydrochloric acid. We know it's an acid because when it dissolves in water, the H+ and Cl- ions will dissociate. Any molecule that will make H+ ions is an acid.

Bases are molecules that will have extra (OH)- ions. In their name they have "hydroxide" An example is NaOH, sodium hydroxide. When it dissolves in water, the Na+ ion dissociates from the (OH)- ion.

Notice that if you mix an acid and a base, the extra H+ ions from the acid will combine with the extra (OH)- ions from the base to form water--we know water is neutral--it's a neutralization reaction :)

Have a look at my notes for grade 10 science from frontenac...it should cover what you need to know. They also have a section in the forum where I answered tonnes of questions.

I'm going to be at LC today at 1:00 helping some others with science--if you want to show up there too I can help you in person.

See ya, and feel free to ask more questions when you need to.

Anonymous said...

dammit. too late. but thanks ms. bearse. yeah now i get it a little more. but i still dont get word equations. how do i write them?

ms.bearse said...

you write word equations from the chemical equation. Figure out the chemical equation, using all your ionic compound/covalent compound knowledge etc....then you use nomenclature rules to name each compound.

For example:
sodium hydroxide and sulfuric acid...
we know that will be a neutralization reaction producing water and a salt.
NaOH + H2SO4->Na2SO4 + H2O

we can balance it too just for fun.

2NaOH + H2SO4->Na2SO4 + 2H2O

To write the word equation we name each compound...

sodium hydroxide + sulfuric acid -> sodium sulfate + water

is this clearer?

send more questions--use the SNC2D post though, it keeps things more organized.

Anonymous said...

MS. BEARSSEEE whatsss gooddd??
where can i download a stupid graphing calculator???