DEV Work Period!

Well, today we had Ms. Gonzaga come back as our substitute since Mr. K was away today. Surprisingly, and thankfully, there was no quiz or anything (which some of us may have been expecting since we usually get one when there's a sub). But anywho, who cares about that right?! So then, basically, this class was dedicated to working on our DEV projects. The due dates for them are really close that's why. We were also told to work on Exercise 44 as well, and that's it! Same deal for the afternoon class! Okay, that's it!

Aaaannnnd!!! Here's something Jamie is probably expecting (inside-outside joke?!)...she's not scribe by the way, Joyce is, and no I did not mean harm to her by putting her name in red!


Well then, that's it for today!!!! Did I mention the next scribe? Well for those who don't care about reading scribe posts and just like scrolling RIGHT TO THE BOTTOM immediately, the scribe is Joyce.

BOB: Conics

WOAAAHH!! I almost forgot to do my BOB again! Sorry this is really really late XD

Okay, this unit was one of my favorite units. I actually enjoyed graphing for this unit and I thought that the concepts were very simple. It was a good break from the Counting unit, which I didn't really like much. I also enjoyed doing all that folding during class, such easy homework! This unit was also full of laughs, we really couldn't seem to stay serious most of the time. As for the parts that I found the hardest for this unit, I would probably have to say that the hyperbola and ellipse were definitely the most difficult, although not by much. I got confused at first with how to find the foci and the asymptotes (hyperbola) but then it all made sense after some explaining. I also remember messing up graphing my vertical hyperbolas since I switched around the transverse and conjugate axis. All in all, tis was a good unit, now I got to get ready for school! Good luck on the test guys :P

Bit of Review and...Origami?

Highlights for today's class are:

-Reviewing how to write equations of a parabola when given the vertex, focus, and directrix.
-Reviewing what we learned about circles from last class
-Origami-ish fun with circles!

This slide was basically a review from last class to see if we really understood the concepts.

When given the vertex and focus, it is easier to draw a sketch of the parabola first to see which way it opens. Once you know whether it opens in a vertical or horizontal way, you will be able to see which equation you should use (either the horizontal or vertical equation).

P equals the coordinate value that represents whether the equation has the y side squared or the x side squared. In the first question where the focus is (4,0), the P value would equal 4 since the x side is squared and the parabola also opens horizontally. To find what 4P is in the equation, just multiply the P value by 4. Also, if the parabola opens in a negative way, 4P will also be negative.


More review from last class. Since the distance between P and F is equal to the radius, we can use the distance formula to find the standard form for the equation of a circle.


We already did the question on the left side last class, so we worked on the one on the right. We were asked to put it in the standard form (circle). After solving it, questions arose about what to do with -2 since the squareroot of -2 is basically an imaginary number. We came to the conclusion that it wasn't a circle since you can't have a negative value for the radius, thus the not a circle.


*Gasp* A circle with two dots on it...and a doodle I made because I was bored. We were all given a sheet of blank paper and were told to draw a perfectly round circle on it. Then we added the center dot and another dot out on the side somewhere near the circumference. After that we made many ticks on the circumference (minimum 25) and made sure that we clumped the ticks together around the outer dot. We then folded our ticks onto the outer dot. The end result was an elipse on the back side of the paper.


Homework is the above slide. Place points P, Q, and R anywhere on the outside of the elipse. Draw a line connecting them to F1 and F2. Add them together as shown and see what happens ;D


Yep, I joined the advertising bandwagon! Watch Batman: The Dark Knight on July 18 (R.I.P. Heath Ledger)! XD......Okay..time for next scribe. Looks like a new cycle! I choose........Richard. Ahah, well then I'm off! Oh and vote for Joseph for vice president! K, have fun ;D

BOB: Combinatorics

Its time for another BOB! Man, to sum up my feelings about this unit, I'd have to say that it was definitely my least favorite out of the units we've done so far. It was really difficult to wrap my head around the various concepts and it took me a while to understand how questions were done. The easiest parts for me were definitely the circle questions. For some reason, those questions really clicked with me, unlike the other ones. There were just so many different things to do and solve while figuring out how to solve those other types of problems that it just made my head hurt. By the end of most of the classes, I would always feel a bit dizzy after intaking all the information and trying to sort it all out in my head. I also remember thinking to myself when we were told to solve a question, "Man..I miss logarithms.."

Yep, that's how I felt about this unit. As much as I may dislike it, I still have to get through the test tomorrow. I hope I, and all of you, do well on the test! Oh yeah, one more thing....JOIN AP CALCULUS! one of us..one of us....

BOB: Logarithms and Exponents

Its that time of the month again, yes, that's right, test time, and therefore its time to BOB! Well, with this unit, I found the basic stuff to be pretty simple. Just learning about the various laws and how to solve simple problems with logarithms and exponents was the easiest part. Learning about the natural logarithm was also not that hard, along with the compound interest and model stuffs.

Now, for the hard parts, I still find graphing things a bit challenging. I also found it hard to do those questions which say to solve for "this variable", be that k or x. An example of this would be those two questions from that quiz we had a few days ago when we had a sub. I have a feeling I really messed up those questions, and I mean really.


K that's all, ciao bella? o_O

Fun with Logarithms! =D

Today's class was all about logarithms and what's a logarithm now? Yeah, an exponent! Good job! Too bad you're going to forget that though.. Anyways, on to today's class.

First of all, today was another workshop class, so we were put into groups first. Mr. K said that it was a competition and that whoever won would get 1000 marks and that he was giving out Ph. D.'s, I guess he sorta forgot about it at the end of class though. Oh well, back on topic! So the first questions that he gave us were to expand two given logarithms as much as possible (slide 2). Let us go over the first question he gave us.


So, starting off, we can see right off the bat that line 1 is the quotient law since there is a fraction involved. Knowing what the quotient law is, we then subtract the denominator from the numerator. But that's not all! There are more laws involved in this question. The numerator included the product law, so from there you get to what is shown in line 3. As for the denominator, that was the power law. So that's how we get the -2log base a B. That's not all though, since the log base a root C is also the power law! Put them all together and that's how you get to line 4, which is the fully expanded form of line 1.

Now, for the second question, it is basically the same thing, except that there are three powers in the numerator. All you need to do for that one is add the three different parts together, the rest is the same thing as explained above. All that was different was that there was one added power to the numerator, which is still the product law.

Although you may think that the expanded form that I showed above may be the only answer, it isn't. You can also rewrite it any different ways, as shown in slide 3. All three of those are the same answer. Its just that the order their written in is different. They're all the same because as long as you keep the signs, they'll end up towards the same answer. Also remember that you don't subtract anymore in Grade 12, you add negatively!

Now, when you're given with a question that asks to simplify from the expanded form, it may look hard at first, but you can easily solve it just by looking at it! Just remember these things, when you see an addition sign, that means that it'll be on the numerator, when you see a subtraction sign, that means it'll be in the denominator. If you know these, then you can easily go from expanded form to simplified form in one look. For example, if I had "log base a Q - 2 log base a D + 1/2 log base a L" then I can then simplify it to "log base a Q times sqrt L / log base a D squared".

On slide 5 are logarithms that we would see most often as questions given to us. When you're given the values for some logarithms and are then asked to answer what the value of a given logarithm is, all you have to do is first find what logarithms multiplied together will give the logarithm that you are trying to solve for. This is only when you don't really know what the exact value of a logarithm is, like in the first question on that slide, which is log base a 15. You can't do this mentally so you have to multiply log base a 5 and log base a 3 together to get the 15 (the values for those two were given). You can then break it up, since its the product law and since you're adding two values together, you can get the answer easily.

Onto the next part, slide 9, we are asked to solve that question using logarithms. Just by looking at it, we aren't able to get an exact value so that is why we have to solve it using logarithms. First, we start off by changing it into log form, since we are trying to solve it using logarithms. After that, we should get that x = log base 3 12. We then move onto solving for x, so we have to make the two sides have a common base. The most common base is 10 and that is why when there is no base indicated, that means that the base is automatically 10.

So after that, we should get that log base 10 3 to the exponent x = log base 10 12. The left side of the equation is power law, so we simplify that. You can then solve for x by dividing both sides by log base 10 3 and NO, the answer is not log base 10 4. Log base 10 12 over log base 10 3 is not the same as log base 10 (12/3). Same goes for any other numbers those may be, so don't make that mistake! Now that we know what x equals, you can then type it into your calculator to get a value. The reason why 10 is the common base is also because the log button on your calculator uses base 10 already, making it much easier to work with base 10. Once you got what x equals, you have then solved for what log base 3 12 equals. Yay!

When you're working with a common base, like in the previous question, this is called the change of base law. The reason why its called that is because you have to change the bases so that they're common or else you can't really solve it. This is the easiest way of going about it and you wouldn't want to hurt your brain by doing something more difficult.

Yeah, so that's all! Now to choose tomorrow's scribe. I think I'll choose Jamie..*glares*....no just kidding, I choose Richard. K then, I'm off. *waves*

Flickr: My Bathroom Tiles

Hello, this isn't late is it? I was supposed to post this earlier today but then my nap took..um...longer than expected ¬_¬. Anyways, back on topic.

My original thoughts when I was first introduced about this assignment was not as bad as Justus'. I was just worried that I wouldn't be able to find a good picture since I thought that trigonometry and circular functions were hard to find in the real world. Boy, did I end up being wrong. I found so many in my own house that I didn't even notice till now! Just this morning, while I was brushing my teeth, I noticed some on my own clothing, which made me go "Dood, I have waves on my clothes...". I then got dizzy after looking at them too long, so I then continued on with my business.

Well then! In conclusion, I'd have to say that I my eyes were opened and I now see the light of math everywhere. Sure, I might go crazy seeing math everywhere, even in places I didn't expect, but it sure is neat. Now then, on to my picture!


http://www.flickr.com/photos/24589000@N08/2321539688/

Anyways, that's all for me. Ciao, the bye version that is. Teehee.

BOB: Identities

I found this unit, although short, to be very hard and confusing. It was definitely the hardest unit so far for me and I still have problems with proving identities. I'm okay with everything else but this since they're like logic puzzles, where you have to look closer and try to see what would work. I was very frustrated when we had to prove an identity and then I'd be so lost as to where to start first. I'd try everything that I could think of, only to find out that my answer was way off, and that I had started off in the right way. It was even more of a slap in the face when Mr. K or someone else would post the solution to the identity and then I would be like, "Oh wow! It was that easy?" I admit, I am not as dumbfounded towards solving identities as the first few days, but its still a bit hard and to add to the fact that I'm incredibly slow at doing them. >_<

Anyways, yeah, that's all I've got to say, or rant about. Hopefully I, and of course everyone else, does well on the test on ..Thursday? Well then, that's all for today! *quietly walks out of blog*

WORD PROBLEMS...TEAM KA-BLAMO! Problems?? Word, man...

I think it's original to call our fantabulous team of three Team Ka-Blamo [members: Jamie, Kristina and Eleven]

Well, our problem of course, was on slide 15 and it asks:

Well, we're going to do things the old school modern way...actually write the stuff on paper...[because we can't afford a tablet] then scan it and upload...that's the way we do things. We spent like five hours getting everything well, thorough enough via MSN...

A FERRIS WHEEL HAS A RADIUS OF 20 METERS. IT ROTATES ONCE EVERY 40 SECONDS. PASSENGERS GET ON AT POINT S WHICH IS 1 METERS ABOVE GROUND LEVEL. SUPPOSE YOU GET ON AT S AND THE WHEEL STARTS TO ROTATE.

a.] Graph how your height above the ground varies during the first two cycles...
The graph we drew looks pointy but it is CURVED sorry.



That's just a graph based on the given info.... and we nearly got confused because of the minimum value being 1 m off the ground instead of touching the x-axis. But then we figured that ferris WHEELS [typo in your slide.] don't touch the ground or else they would scrape the concrete.

b.] Write an equation that expresses your height as a function of the elapsed time.

Basically, all we did was find each parameter for both the sine and cosine function even though we only needed one.



c.] Determine your height above the ground after 45 seconds.

All that needs to be done here is sub the t in the function H(t) with 45 seconds and the result will be the new height at this time according to the graph, assuming the revolving doesn't stop and is continuous.



d] Determine one time when your height is 35m above the ground.

Replace H(t) with 35 and isolate t to get time in seconds.



That wasn't much explanation in words but you know... we spent a lot of time getting these answers and we hope this is what you guys are looking for....I'm famished. G'night...I'll edit this later....haha

Yet Another Workshop Class

Hello once again, it is I, Kristina, back for another scribe post. Today, we were supposed to have a pre-test in the morning and the actual test in the afternoon but we ended up only having the pre-test in the afternoon. The reason why was because Mr. K thought that we weren't ready to have the test after seeing us somewhat struggle over yesterday's class. So, in the end, we ended up using the morning class to have a workshop period! Yay, fun..

After we had been split into our groups, which there were a total of four, we were given a question to work with. I will type out the question, which is already on slide 2, for your convenience.

At a sea port, the depth of the water, h meters, at time, t hours, during a certain day is given this formula:


A) State the: i) period ii) amplitude iii) phase shift

This was the first part of the question, which should be fairly straight forward. BUT! It ended up not being so straight forward to the class. Every group came to the same conclusion for the amplitude and phase shift, but the class became divided with finding the period. Which is shown in slide 2 with the two different answers.

The first answer (red) is not the right answer, the one in blue is the correct one. This is because the group who put up answer red didn't factor out parameter B correctly. They thought that because since 2pi wasn't "visually" over 12.4, that they had to factor that out first and then multiply 2pi by 12.4 to isolate parameter C. After a bit of arguing and confusion, Mr. K kindly showed us that even though the 2pi didn't appear to be over 12.4, it actually was. The reason for this is because that 2pi could also be seen as 2pi/1, which, when multiplied, actually gives 2pi(t-4.00)/12.4! Now, from there all you need to do to get the period is to factor out the 2pi/12.4 to get parameter B and then to get divide that from 2pi. Voila, you should then get the answer of 12.4. We can then move onto the next question..

B) What is the maximum depth of the water? When does it occur?

Now, for this part of the question, I am going to explain how to do it in a completely different way from what the slide shows you. The reason for this is because, in my own opinion, it doesn't depict a very accurate and understandable way on how to find the actual value. By just using the graph to try and find it, I think its just like guessing, therefore not very accurate. The graph is only good when you want to pinpoint which areas the answer will be in, which is actually shown very nicely on the graph in slide 4.

Well then, to show how to get 7.1 hrs as the answer, I am going to show you how to get it MATHEMATICALLY using the formula!

Looks difficult? Nah! Its so simple! You see, all I did was input the max value, which was 4.9 into h(t) . From there, I then solved for t. I'm sure most of you have gotten stuck after getting to the ARCsin part, well then don't fret, because I've also had troubles with that in the past. All you have to do to isolate the (t-4) is to multiply both sides by the reciprocal of 12.4/2pi. When you do this, 2pi/12.4 on the right side will get cancelled out and then you will be left with 12.4/2pi * ARCsin(1) = t - 4. From there's its baby stuff! We're all in pre cal, I don't think you need me to explain what to do from there. Anyways, I hope this solves any problems people had with slide 4. MOVING ON..

C) Determine the depth of water at 5:00 A.M. and at 12:00 noon.

This is simple, all you have to do is input 5 and 12 into the value for t and then solve from there. If you need further explanation for this, feel free to add a comment since I am seriously getting tired right now haha >_<.

D) Determine one time when the water is 2.25 meters deep.

Okay, this is basically the same as letter B. Instead of 4.9, input 2.25. Then solve from there. No further explanations need to be said for how to do these kind of questions. But, in case you do, please leave a comment!


AWWRIIIGHT! Moving onto the afternoon class! First thing we did was have a PRE-TEST! We were given about, I don't know, 20 minutes? to do it and then we were put back into the same groups we were in for the morning class to discuss our answers so that we could hand it in, but only one person's pre-test was handed in.

We then went over the answers after having handed in one person's pre-test. The multiple choice answers were pretty straightforward, so I am going to skip to the very last question since it was definitely the hardest one out of them all.

Starting off with the question "a)" of the last question. Its asked to sketch a graph of the height of the point A above the outflow water level as a function of time starting at t = 0 seconds, with A as shown in the diagram. The first thing you had to do to be able to graph this was to find the period. We were able to get the period from the information in the question that said there were 5 revolutions every 4 minutes. From this info, we can then find the period by attempting to when 1 revolution occurs, which will end up in seconds. As slide 12 shows, the period will be 48 seconds since that is when 1 revolution occurs. The graph can now be made easily and the reason why it starts at 0 is because starting at 0 seconds, point A is still above the outflow.

Part "b)" of the question is pretty self explanatory. All you had to do was take a look at the graph and then find the different parameters for SINE and COSINE. From there, you would be able to get the same answers as shown in the slide if you did it right. As for part "c)", this is just like question "B" from the morning class. Instead, you can choose between which formula you want to work with, either the SINE or the COSINE that you made in the previous question. Once you've done that, just input 4 into h(t) and then solve for "t" in the exact same way as shown in "B" from the morning's question. After having solved for question "c)", you can now solve for question "d)". Take a quick look back at the graph and find where 4 meters would be. If you draw a straight line across, you'd see that it touches two parts of the wave. That is why the answer shows that you have to multiply the answer by 2.

After we had finished going over all the answers, Mr. K then told us that we had a group assignment to do. Each group was given an assignment that was similar to the questions we had done today. They're expected to be posted onto the blog with the appropriate tags: "TrigAssignment, (names of group members), Transformation. The due date is on MARCH 13 BY NOON. Once every group has got their assignment posted, everyone is expected to comment on each of the groups' assignments, excluding their own, whether they be about how neat their work was, or if they found an error, etc.

Now then, I guess that's all for me. REMEMBER GUYS! TOMORROW'S PI DAY! Make sure you don't forget to bring your delicious pies! As for the lucky scribe for Pi Day, I choose....Roxanne. Well then, that's all for me. Oh yeah, and remember! THE TRANSFORMATIONS TEST IS ON FRIDAY AFTERNOON! For those of you who haven't BOB'd yet, be sure to do it by then! Alrighty, now seriously, I'm out.

BOB:Transformations

So, the test is on Wednesday, along with a pretest on Tuesday so I guess its time to BOB! I thought that this unit was pretty short. It went by faster than I had anticipated. As for the actual unit, I thought it was easier than Circular Functions.

The first part, which was Translations, was a good start for the unit. I found it pretty simple, although I failed to get that certain piece wise thing wrong since I was unconsciously thinking that 3x3 was 6 >_<. Anyways, the hardest thing for this part of the unit was basically the piece wise things! But even those aren't so bad. Plus, BCAD or ADBC was a nice way to help me remember what to do first when translating graphs and the like. Now, one of the things that I found to be difficult were those inverse functions and how to graph them and such. Maybe part of the reason was because I had forgotten how to solve for y. Yes, I know, simple stuff but, I admit, I forgot. But! That's no problem now that I actually remember! As for the graphing itself, I got confused with all those asymptotes and such. All this biggering and smallering was hard to wrap my head around as well. But alas, I think I'm okay for the mean time. Yeah, this unit overall was pretty simple. Now then, good luck on the test on Wednesday? everyone!

New Unit: Transformations

Hi everyone! Its finally my turn to scribe, and today we started a new unit. But first I must talk about what little happened during the morning class...

In the morning, we began by looking over the review questions that were on yesterday's slides. Mr. K told us not to worry about the second part of those review questions as much since we didn't really go over how to do them, but we did in the afternoon to some extent.

After we had finished briefly looking over those questions, we were given the first part of our Circular Functions test. Notice how I say "first part". The reason why I say this is because tomorrow morning, we will be getting a second part to the test, which should only take us about 5 minutes to do, according to Mr. K. The second part will consist of graphing work. That was all that happened in the morning class.

Onto the afternoon class, which we had a late start on because of Mr. K talking to the principal or something. Anyways, Mr. K first started off by showing off some trippy spiral themed pictures from www.flickr.com. Then he took it one step further by showing us his newly downloaded firefox add on, PicLens, which gave him the ability to look through all the pictures in a cool way. Its hard to explain for me so if you're interested in this, go to www.piclens.com!

Okay back to math! So, we started on a new unit called Transformations. The first thing we did was look at the graph for f(x) = x^2, which looks like this:


We then looked at what would happen when 2 was added to the function to make f(x) = x^2 + 2. The graph ended up looking like this:


As you can see, by adding 2, the parabola was shifted upwards by two units. We then looked back at the sine graph and did the same thing to it, except we subtracted 1, therefore causing the graph to shift downwards by one unit.

Now, if you're looking at slide 3 of today's class, you'll notice that we tried graphing y(x) = 2^x. Don't fret if you don't understand what this is. The reason why the graph looks like that is because since x is equal to a number, when 2 is to the exponent of that number, it will increase as so, making the graph look like that. To help you understand it better, here is a table that shows you a simple pattern of the relationship between exponents and a number, which in this case is 2:

See the pattern? Starting from exponent 4 all the way down to negative exponent 4, the values keep halving themselves. So if x is equal to 4, that would mean the y coordinate would be 16. If x was equal to -4, that would mean the y coordinate would be at 1/16.

Okay, so now that you know why it looks like that, what would the graph look like if I were to do make it f(x) = 2^(x+3)? Well, if you're still looking on that slide, then you'd see that the graph moved 3 units to the left. This is pretty much like the phase shift in those trig function graphs we did for the past few units. The only difference is that it isn't called the phase shift, since phase shift is only used when dealing with trig functions, instead, we call it the horizontal shift.

So in conclusion to this part of the afternoon class, we can say that we write these functions in the form of y = f(x-b) + a, where "b" is the horizontal shift and "a" is the vertical shift.

By the way, the weird shaped thing below the f(x) = 2^x graphs are called "piece wise functions". They are called this because of they are split into pieces. As you can see in that specific piece wise function, there are three parts to it, thus why its a piece wise function.

For the rest of the class, we were given questions to work on based on the things I just talked about. The first question, as shown in slide 6, asked us to to write f(x) = sin(x-2) + 5 in terms of g(x). Since f(x) is basically g(x) shifted 4 units to the left, the only possible answer was g(x) = sin(x+2) + 5. As you can see, the only thing that was changed was "b". To get +2, we basically just added 4 since f(x) is units to the left of g(x), meaning that we'd have to move right 4 units from f(x) to get to g(x). Simple, right?

Okay, so onto slide 9. Remember those review questions from yesterday's slide? Well, we learned how to solve those now. I'll use the same example as shown in slide 9, which was sin(2x) = 1/2. To make this easier to work with, you can Let θ = 2x and then work from there for now so that you've got sin θ = 1/2. Now, you should already know how to solve for this by now, and if you don't, look over the circular function notes again, mkay?

Yeah, so if you did it right, you should have gotten the answers in the black. If you're confused with the 13π/6 and the 7π/6, we just got those by adding 12π/6 which is equal to 2π to π/6 and the reason why we did this is because of the original 2x. Since its 2x, when you look at it on the graph, you'll see that there are two complete waves, meaning that there's going to be double the solutions. The same can be said if it was 4x instead of 2x. In the case of 4x, there'd be 8 solutions, and if it was 6x instead of 2x, there'd be 12 solutions. You see?

You think you're done with the question now eh? Well, you're not. Since you'd switched out 2x for θ, that means that you're going to have to put back the 2x. Then from there, you should be able to work with it, so if you did everything correct, you should get the answers in red. Voila, wasn't that oh so simple? :)

Um, so, I think I'm done now! Yay! Oh wait, just forgot to mention that tonight's homework is Exercise 7, questions 1-10, since you should have already finished 11-20 and if you haven't, you are a naughty child.

Now for the moment of truth, since I'm the last scribe in this cycle, that means I have a whole list to choose from now, ehehe. Therefore, I choose .....LAWRENCE.
Yeah, back to the beginning we go! Anyways, remember that there is part 2 of the Circular Functions test tomorrow, so good luck everyone!

~kristina

BOB: Circular Functions

Hello everyone, since the test for circular functions is supposed to be this week, I'm here doing my BOB.

Okay, so during the beginning of this unit, when we were just introduced to converting degrees to radians and whatnot, I found that this was really easy. Sure, it was a bit confusing at first since I wasn't used to working with radians but it eventually got better as the unit progressed. Now I actually like working in radians better than degrees!

One part I had trouble with AT FIRST was learning the unit circle. As soon as I saw all those numbers on that unit circle, I panicked. What made it even worse was when Mr. K told us that we had to memorize EVERYTHING that was on the darn thing, and at that point, I was ready to cry. It didn't help much when Roxanne asked him what would happen if we didn't memorize the thing by the time he told us, to which he answered "You'll fail." But now that I know the cursed thing that caused me so much pain, I feel much better.

Now onto the graphing bits. The basic graphing wasn't so bad, but the added stuff, such as DABC, was. Sure, DABC was fun to say, but it was hard to understand at first since the last part of that class was rushed and it was so much to absorb in just those last 2 minutes we had.

Um, yeah, that's all I've got to say for this unit. It had its hard and easy moments, and overall, it was pretty good. That is all.