<body><script type="text/javascript"> function setAttributeOnload(object, attribute, val) { if(window.addEventListener) { window.addEventListener('load', function(){ object[attribute] = val; }, false); } else { window.attachEvent('onload', function(){ object[attribute] = val; }); } } </script> <div id="navbar-iframe-container"></div> <script type="text/javascript" src="https://apis.google.com/js/plusone.js"></script> <script type="text/javascript"> gapi.load("gapi.iframes:gapi.iframes.style.bubble", function() { if (gapi.iframes && gapi.iframes.getContext) { gapi.iframes.getContext().openChild({ url: 'https://www.blogger.com/navbar.g?targetBlogID\x3d15925399\x26blogName\x3dPre-Cal+40S+(Winter+\x2706)\x26publishMode\x3dPUBLISH_MODE_BLOGSPOT\x26navbarType\x3dBLUE\x26layoutType\x3dCLASSIC\x26searchRoot\x3dhttp://pc4sw06.blogspot.com/search\x26blogLocale\x3den_US\x26v\x3d2\x26homepageUrl\x3dhttp://pc4sw06.blogspot.com/\x26vt\x3d8544817652058964798', where: document.getElementById("navbar-iframe-container"), id: "navbar-iframe" }); } }); </script>

Sunday, February 19, 2006

The Tenth Scribe

For the tenth scribe on the seventeenth day of February, Two-thousand-six... is me Mark.

Today was a half day and we had a pre-test for next weeks test. Even though this was a pre-test it was worth marks. The pre-test had about two questions i made mistakes and others in the classroom. Here are all the question that were on the pre-test:

1) Given sin y = (square root of three)/2, one possible value for y must be:

(a) 4 pi/3 (b) 2 pi/3 (c) 7 pi/6 (d) 11pi/6

Since y is positive, it must be lie in quadrant one or two. If you look at the answers: a, c and d they are negative sine values. B is the only positive sine value If you remember from the unit circle, 2 pi/3 lies in quadrant two.

2) The graph shown to the right is one period of f(x) = 3sin x. A second function g(x) is also shown. One possible value of x where these two functions intersect would be:


a) 3.87 b) 2.41 c) 5.55

d) 2.30

The first way to solve this equation is simple, using your graphing calculator. Plug in the equation f(x) and for g(x) plug in 2. Next, press graph. If we take a look at the graph the first intersection is around a value which is smaller than the choices given. So, press 2nd trace and then press 5 (intersection). move the cursor to the second intersection point, where g(x) and f(x) meet. The calculator says "first curve", all you have to do is press enter three times and the calculator will solve the intersection point. Finally the calculator gives you a value of 2.41.

NOTE: plug the equations in the "y =" window, like this:

y1 = 3sin x

y2 = 2

3) given cos (k) = -(square root of 3)/ 2, and 450o < k <>

cos (k) = -(square root of 3)/ 2

cos (k) = 5pi/ 6<<>

cos is in quadrant 2 and falls in our domain.

cos (k) must equal 150 degrees, but since our domain is between 450o and 630o, we get a value of 510o. How? add 150 degrees to 360.



4) A sinusoidal graph has a maximum value at (2, 5) and the next minimum is at (6, 3). What is the value of parameter B in the equation of this function.

We can find b easily, first plot the points on a graph and connect them together. Next, count the units (distance) from the first value to the second. You can subtract the y values. You should have a difference of eight. We can now find what the period is, since we have B.

period = 2pi/B

period = pi/4

Our answer, b = 8 and period = pi/4

NO CALCULATORS

5) The funtion P(t) = 100-20 cos 5pi(t)/3 approximates the blood pressure P in millimeters of mercury at time t in seconds for a person at rest.

a) Find the period of the function.

To find the period of the function we should rearrange the form of the equation into y = AcosB(x-C)+D:

y= -20 cos (5pi/3)(t) +100

B= 5pi/3

period = 2pi/B

period = 2pi/(5pi/3)

period = (2pi/1)(3/5pi)

period = 6pi/5pi

period = 6/5

b) Find the maximum and minimum value of the function

D = 100.and A = 20

Our maximum is 100 + 20, which is equal to 120

Our minimum is 100 - 20, which is equal to 80

c) Sketch a graph of the function show at least two cycles.

At this question, regine said to Mr. K that we are dealing with heart rate. Since we can't have a negative heart beat, our graph must have 2 positive cycles



Well that was the pre-test. For the actual test, there are links on the blog that you should try and practise. Also, the stencil that was given on thursday is due for Monday. Junior High tours will also be going on this week. Reminder: all people who signed up there is a meeting in Mr. K's room at lunch, on MOndaY?

Well that's it... o yea almost forgot. The next scribe is...eni meeni miny mo catch a tiger by the toe if he hollas let him go eni meeny miny mo... Jessica




Français/French Deutsch/German Italiano/Italian Português/Portuguese Español/Spanish 日本語/Japanese 한국어/Korean 中文(简体)/Chinese Simplified Nederlands/Dutch

3 Comments:

At 2/19/2006 8:06 PM, Blogger Mr. Kuropatwa said...

In a word: Breathtaking!!

Your graphics are fantastic! How did you do them so well? What software did you use? Maybe others can use it too. Can you post a link?

My favourite part was the way you worked out multiple choice question 1 -- by eliminating possibilities. I also really liked the multiple solutions to the graphical problem in question two.

You really have raised scribing to a new level. Ladies and gentlemen .... the bar has been raised. ;-)

 
At 2/19/2006 10:14 PM, Blogger mark said...

haha thanks Mr. K. i used...paint lol. i took a flash designing course in S1 at my old school. i had only paint to design my graphics, so i used that

 
At 3/01/2006 5:42 PM, Blogger Regine said...

It's late but, niiice scribe. haha. It's not too long but not too short. that's pretty cool

 

Post a Comment

Links to this post:

Create a Link

<< Home