So… Here’s the 4th hour’s attempt to be back on the 1st page of the blog again! And an attempt to make our quarter journal grades not die. Ok, here goes…

Today we did a sketchpad project. The basis of it is that if you have a polynomial with clearly defined zeros because it’s in factored form, you can figure out the multiplicities of those zeros, and therefore the degree of the polynomial, by looking at the shape of the graph as it crosses those zero points. If it is simply a line, it is a multiplicity of 1. If it is the vertex of a parabola, it’s 2. If it’s the “center” of a `x^3` curve, it’s 3. I’m not sure what it is if it’s greater than that, but I’m sure even numbers have similar shapes to the 2nd degree shape, and odd numbers have similar shapes to the 3rd deree one. The other, non “m or n” values were just either translations or dilations, and were nothing new.

On the back side of the sketchpad sheet, there were some practice problems. The top section was pretty self-explanatory, but the later parts of the sheet were somewhat difficult, involving finding zeros with imaginary numbers. All you have to do to solve these is set the zero equal to x, and solve for zero by moving the imaginary number over to the x side. After that, you need to make sure your equation includes “x plus or minus (imaginary number)”. After this, you need to multiply these two factors together, and you will get a quadratic equation to solve for more zeros. Pretty easy, no?

Well, that’s all we did today. Pretty good class, huh? Oh, cookie if anyone noticed the somewhat out of place message at the back of the computer lab. Well, tomorrow’s scribe is Hannah. Have a good night, and see you all tomorrow!

~~~~~~Andy~~~~~~

Ok, so there was kind of a lot to take in today.

The main idea of today’s class was to get across the relationship between a number and its corresponding logarithm. To do this, we used a logarithmic number line on top of a regular number line, with the numbers on top matching up with their logarithm on the bottom (when I say log or logarithm in this post, I’ll be referring to a base 10 logarithm, unless I specify otherwise). Here are a few quick bullet points to remember:

-If you add two numbers logs, it is the same as multiplying the two numbers. For example, log(6)+log(10)=log(60).
-Subtracting logs is the opposite.
-If you multiply a number and keep multiplying it by the same number, the resulting producs will be the same distance away. For example, 1 to 2, 2 to 4, 4 to 8, 8 to 16, 16 to 32, etc. are all the same distance from each other.
-Overall, the numbers on the log scale are proportional to those on the power scale. If 2 is 30% of the way from 0 to 1, then log(2) will be 30% of that same distance.

The main idea to be grasped today was that proportional relationship between numbers and logs. If you got that, you’re pretty up to date, and you were supposed to continue on in the packet, to let it walk you through some other aspects more slowly. If you’re still having trouble, go see one of the Mr. B’s, or come see me.

Well, that was today’s class in a nutshell. Hopefully that was short enough so that you can read it easily, and detailed enough to serve as a good review.

Thanks, and see you all later!

Scribe tomorrow is JessicaB

Have a nice night! (For any of you who’ll check this this late…)

~~~~~~Andy~~~~~~

~~~Andy~~~

Hey guys…  Sorry about not getting the blog up, but my internet connection was disconnecting me every couple minutes, so I couldn’t get one up.  I  just finished fixing it, and so I’m gonna be quick because I don’t know how long it’s gonna last for.  And I have 2 days to blog for.

 Yesterday was half claims, half exploration.  I don’t remember who did what claims, so I’m sorry about that, but here’s a short explaination of all the problems. 

 Problem 10 was the first claim, and it asked to find the exact radian measure of 1080`degrees`.  This can be accomplished by dividing 1080 by 360, since we know that 360 is equal to 2`pi`.  It ends up being 3, so 1080 is equal to 6`pi`.

Number 41 asks you to find the exact value of `tan“pi/6`.  According to our plates, `pi/6` is equal to 30`degrees`.  We know the sine of 30`degrees` is `1/2`, and the cosine is `1/sqrt2`, and tangent is `sin/cos`.  This means that the tangent of 30`degrees`, and of `pi/6`, is `2/sqrt2`. 

46 wants us to find the value of `csc pi/6 sin pi/6`.  This can be done by realizing that the cosecant of `pi/6` is the reciprocal of the sine of `pi/6`, and a reciprocal of something times the original equals 1.  So the final answer is 1.

For number 49, you simply have to pull out the required elements from the graph to come up with an equation.  The vertical displacement is 5, the amplitude is 7, the frequency is 30 (with a period of 12), and a horizontal displacement of 2.  This gives the equation of `y=5 +7cos 30` It’s cosine because the graph starts at the top and goes down.

 For the next half we did an exploration activity about a bug travelling around a square track.  We got coordinates of points, and we ended up graphing the x and y coordinates independently.  They ended up making squarish cosine and sine graphs.  This tied in with today’s activity, which was a bug on a circular track.  I related it to what it would look like if you attached a light to the side of a tire, turned the light off, and rolled it.  It would produce a sinusoid.  Apparently 1 full rotation of a point on a circle can produce 1 period of a sine graph or a cosine graph, depending on if you’re graphing x values or y values.  I have a feeling that’s what we’re gonna be concentrating on in class…..  Just a guess…  And that’s all we did today.

 Well, I didn’t get disconnected, so that’s good.  Hope that was a halfway decent relation of what happened in class.  I’m gonna try to fix this connection for good…  Hopefully I can.  That’d be good. 

Tomorrow’s scribe will be Emily.  See you guys ^_~

~~~~~~Andy~~~~~~

Hey guys, sorry that this is so late, but my computer’s power cord decided to stop working last night, and so I had to wait until I came to school to do this, and so I have to keep it short. Apparently I’m the scribe instead of Liza, because she’s already been it for this cycle.

When we came to class, we were instructed to partner up with our 3 o’clock partners and we all discovered that we had a piece of adding machine tape and a paper plate. We flattened out our plates into circles (easier said than done) and cut our adding machine tape to the exact (or close to it) circumference of our plate. We then folded our plates into quarters, giving us a radius of our circle. We marked those units on our adding machine tape, giving us a sort of “radius unit” ruler. When we re-wrapped our circles with our “ruler,” we marked off our radius units and were told they are called “radians.” Also, apparently a full circle is 2`pi` radians, a semi-circle is `pi`radians, a quarter circle is `pi/2`radians, and a three-quarter circle is `3pi/2` radians.

Again, sorry for the length and probably boringness of this blog, but when today’s technology decides to act up, what can ya do?

Tomorrow’s scribe is… Ooh, hey, I’m the last scribe? Ok, how about NickK?

~~~~~~Andy~~~~~~

Since we’re supposed to be responsible for keeping track of our own claims, I decided to use an excel spreadsheet, and I thought I might as well share my format with the rest of you! It’s a fairly simple spreadsheet, and it’ll calculate the percent of problems claimed and if that translates into an A for that part of the criteria. Download it here.

Don’t forget to give me any problems or suggestions, I’m more than happy to make a new version!

~~Andy

Mr. B wants me to experiment with some HTML code today, so enjoy the text effects!!

Today in class, we started off with a nice relaxing pause (with a shocking lack of music) while Mr B. figured out who would do the presentations for 1-3. I was personally preparing the problems I submitted to present, as I saw many others doing. We were then reminded that we should be keeping track of how many we’ve claimed. Emily led off our claims with problem #5,followed by Amy with problem #12, Grant with problem #14, and Jessica with problem #19.

~~Emily (#5)

Emily demonstrated that the way to put the equation we were given in terms of x is g(x)=1+sq(9-(x/2)². She then explained that the g(x)graph is an x dilation of 2 and a y-translation of 1 compared to the preimage and finished up with showing us her graph.

~~Amy (#12)

Amy spent a bit of time setting her problem up, drawing the graph to make her explanation easier. She explained that the image of her graph was translated down by 4 and dilated on the x-axis by 3 from the preimage. She then correctly gave us her equation of the image, g(x)=f(1/3x)-4, explaining that 1/3 resulted in a dilation of 3 because of the “backwards world” properties of parentheses. The class was confused as to what in the world she was talking about, but applauded regardless.

~~Grant (#14)

Having an amazing worldly presentation to follow, grant made sure that everything was as aesthetically pleasing as possible, down to the orange color of his cyber calculator, and the dottedness of his lines. Grant seemed a bit unsure of his problem, but proceeded this mimentary slip with a very nice presentation. His first equation on his graphing calculator was Y₁=4sq(1-x/3)+2.25(x-1)-4, explaining about the fabled opposite world we’ve all grown to know and love. His graphs had slight problems getting just the way he wanted them, but with the help of the class he got through. With the addition of the boolean variables his graphs were complete, and they correctly resembled
those in problem #12. His presentation was thusly concluded.

~~Jessica (#19)

With the last presentation to go, and many tough acts to follow, Jessica set up. Her graphs were constructed, both image and preimage. Her equation for the image was g(x)=5f(1/9x), because the y-dilation is 5 and the x-dilation is 9 (backwards world… Surprise, right?). Her presentation was short and to the point, but accurate nontheless.

And so the presentations concluded. Mr. B passed out a sheet on lesson 1-4, and challenged us to think of an equation that relates median age to the percent of that group that has a fear of flying. Kari had the idea of forming a “stat plot” on her calculator. We proceeded to do so on all of our calculators, and after a minute or two of plastic clicking noises, we were done. We concluded that there is clearly a relationship, but not a linear one. Someone suggested that we make it linear by squaring the independent variables, citing a physics lesson for the idea, and lo and behold, it worked!!! To end our lesson, Mr. B challenged us to come up with a way besides the one we used in class to make the graph linear. As the bell rang, we all trooped out, amazed that we actually used something from one class in another. Who’d have known this stuff was useful??

I hope you liked my post today, and the changes that HTML code can bring. With Mr. B’s permission, tomorrow I’ll post a short tutorial that outlines the main ways to change and edit text using HTML. It’s not hard. And sorry if some of the stuff on here either doesn’t work or looks weird, I was experimenting and I’m not sure how much code this site will accept, and I copy-pasted this from a notepad document, not using any of the shortcuts on the blog site. If anyone would like to see what my source code looks like, just ask me in class. And tell me what you liked and didn’t like with comments!

Tomorrow’s scribe will be Shelby. Have a good day, and happy commenting!!!

~~~~~~~~ANDY~~~~~~~~