20S pre-cal

Tuesday, March 14, 2006

Trigonometric Values

HEY! my time to subscribe! even though I'm kind of new to this and please bear with me people... ^O^

Where to start, where to start... hmm...
Today, we learned about Trigonometric Values
Here are some examples:

a) sin 80 degree --> to find the sin of eighty degree you need a scientific calculator or it will never work "remember set it to degrees". So you have to press sin, the 80 degree and press equals sign. and presto! there is your answer! and the answer is 0.98 oh and by the way read the questions carefully because they ask for the nearest whatever place value they like. and in this case, you have to put it in the nearest hundreth.
But some people have different calculators like MaryAnn's she have to put sin(the number here) she needs to put the numbers in brackets first, and then press the equals sign.
oh and Mrs. Ingram gave us a chart of some sin values and cosine values
http://www.filelodge.com/files/hdd2/25578/chart.bmp just go here.... and just make it bigger so it's clearer. Cause somehow i can't upload it. there's some errors so yeah.
oh and in number 2 we have to say if it is acute or obtuse so to do that:
a) sin angle A = o.50
so you have to use the 2nd function here.
So you press 2nd function then sin and then 0.50 and then you press the equals and you will get 30 degrees.
So this an acute angle.
You can tell this because acute angle is less than 90 degree and an obtuse angle is more than 90 degree angles.

well i think this is all though cause it's all the same anyways, and the other ones are just looking for the angle. and you do that by using the 2nd function key and then press whatever in the 3 laws that they are asking to find.


exercise 18: # 1-6 and 11-15

and the nest subscribe is!!!!

KIM D!!!

Monday, March 13, 2006


Sorry I'm late..Had to scribe for Computer Science tonight as well...=(

Today we started TRIGONOMETRY.

You should already know the 3 laws. If you don't, then here they are:

Sine = opposite/hypotenuse

Cos = adjacent/hypotenuse

Tan = opposite/adjacent

^ If your having trouble looking for the opposite, pick a point to work with. In this case, I picked the right angle. Whatever is across from it is going to be your opposite.
Adjacent is the line that is beside your point.

We began solving some problems with 2 right angle triangles. We had to determine what the length of WX was.

To do this, you must work on the triangle that gives you more information. The triangle xyz is the triangle you should work with first.

First step is to determine the length of the common side. The common side in here is going to be xy.

so far you know that the angle of z is 30°, and Adjacent side s xz = 15cm. Your opposite, which is side xy is still unknown.

You try to fingure out which law would work. The only law that will work here is Tan (opposite/adjacent) since your trying to find out the opposite and you already know the length of the adjacent side.

To do this next step you have to get your calculator. be sure it's set of DEGREES.

1. tan 30° = xy/15
2. (tan 30°)15 = xy/15 (15)
3. (tan 30°) 15 = xy
4. 8.7 = xy

So what I just did in the second step was multiply both sides by 15 to make xy go on it's own. That leaves me with (tan 30°) 15 = xy . On a calculator you would put tan 30 multiplied by 15 and you'll answer will likely be 8.66. Then round to the nearest decimal..


The Homework tonight is

EXERCISE 17 : #1-6 & 12 - 19

and next scribe will be Kaeser.

Monday, March 06, 2006

Parallel & Perpendicular Lines

Hey everyone .. guess what time it is?

It's peanut bu --- scriiiiibe time! :D

Haha, everyone loves the dancing banana. Well anyways, it's Jackie here. This is my first whack at scribing so don't mind me of I sound totally out of line .. don't say i didn't warn you. d: Oh, and thank you Wendy for your post prior to this. That was very sweet of you, big heart. (: Well, getting on to business, let's have a look at what we learned today in classroom 69.


Today, Mrs. Ingram established the relevance of graphing parallel and perpendicular lines, and how to calculate them. The lesson started off with graphing our first line, given the equation -2x + y = 6. We want to find the slope intercept of this equation so that we can set it up on a graph.

where m = slope.
where b = y-intercept.

So, once we rearrange this equation into slope-intercept form, we get y = 2x + 6. Here, I will demonstrate how to calculate your slope and y-intercept ..

- 2x + y = 6
+ 2x + 2x
y = 2x + 6
m = 2/1 ; b = 6

Because we've already singled out the y, we now know that our slope (m) = 2/1, and that our y-intercept (b) = 6. Now, you can see how we graph this on my girly pink graph template .. =D

oh, just a friendly reminder .. always remember to label your x-axis and y-axis, as well as mark the arrows on your lines to indicate that they continue in both directions .. sorry, i was lazy to mark them on my graph but i think you get the point. :)

Now that we have our first line on the graph, we were given our second line to graph. The equation was -4x + 2y = -2. Following the same procedures as to what we did with the first equation, we rearrange the equation into slope-intercept form to find .. you guessed it, the slope and the y-intercept.

-4x + 2y = -2
+ 4x + 4x
2y = 4x - 2
y = 2x - 1

m = 2/1 ; b = -1

Surprise, now it's time to put our second line on the graph.

Now, looking at the graph .. you see that both the lines that we have traced onto the graph are parallel (//), meaning that they will continue both ways and never meet.

Wasn't that fun? (: Okay, now here's something new. We were given a third line to put onto the graph ..

x + 2y = 12
- x - x
2y = -x + 12
y = -x/2 + 6
m = -1/2 ; b = 6

Now that we calculated our third line, we trace it onto the graph like so ..

Looking at the graph, we can see that the first and third line that we made are perpendicular, meaning they cross each other. To determine the perpendicular line of a given slope, we calculate the negative reciprocal. This basically means that we turn it over and change the sign.

ex: m = -3/2
m (penpendicular, don't have the symbol for it on the keyboard :|) = -2/3

We were given a few sample questions to solve that helped us better understand this "perpendicular/parallel" line theory .. majig.

* question one: Write the equation of a line that passes through (-2,4) and is perpendicular to 2x - 3y + 5 = 0.
1: Calculate the slope of the existing line.
You do this by rearranging the equation into slope-intercept form. y = mx + b

2x - 3y + 5 = 0
+ 3y + 3y
2x + 5 = 3y
2x/3 + 5/3 = y
m = 2/3 ; b = 5/3

2: What's the perpendicular slope?
m = 3/2 .. so !
m (perpendicular) = -3/2

3: Determine the equation using the point-slope form. In the equation, replace all the terms that you arleady know and solve from then on.

y - y1 = m ( x - x1 )
(2) y - 4 = -3/2 ( x + 2 ) (2)
2 ( y - 4 ) = -3 ( x + 2 )
2y - 8 = -3x - 6
+ 8 + 8
2y = -3x + 2
y = -3x/2 + 1

general form: 3x + 2y = 2
standard form: 2x + 2y - 2 = 0

* question two: Determine the equation of a line with the x-intercept of 2 and is parallel to the line 3x - 2y = 6.

1: Calculate the slope of the existing line.
You do this by rearranging the equation into slope-intercept form. y = mx + b

3x - 2y = 6
(Remember, when switching values from one side of the equation to the other, negatives become positives and vice-versa.)
3x - 6 = 2y
3x/2 - 3 y
m = 3/2 ; b = -3

2: What's the perpendicular slope?
m = 3/2 .. so !
m (perpendicular) = -2/3

3: Determine the equation using the point-slope form. y - y1 = m ( x - x1 ) In the equation, replace all the terms that you arleady know and solve from then on.

y - y1 = m ( x - x1 )
(2) y - 0 = 3/2 ( x - 2 ) (2)
2y = 3 ( x - 2 )
2y = 3x - 6
y = 3x/2 - 3

general form: 3x/2 - y = 3
standard form: 3x/2 - y - 3 = 0

Well, that's basically all you need to know on what we've learned today in class on parallel & perpendicular lines. We were assigned exercise 14, omitting questions 10 & 11. Well, that was pretty easy right? Good luck to everyone with the questions .. and trying to understand my scribe. d: that's all for now, folks. :)

And a do - doodle - loo - do .. ?

Scribe List

Mary Ann
[ jackie d. ]
im cool =P

Sunday, March 05, 2006

i know its late...

Its never to late to say happy birthday right?... well i know its almost an hour that has past but ..... i have an announcement to make.

(YESTERDAY) 49 minutes ago)


sooo.. i would like to say to our special class mate Jackie D.


Saturday, March 04, 2006

Determining Equations of Lines

Today, March 03, 2006, Mrs. Ingram talked about determining the equations of lines, given two points. To explain how to solve these problems, I will use the second question from Exercise 15.
We were given two points, M(-2, 0) and N(4, 7).

* First, we should find the slope of the line that connects these two points.
(Y2 - Y1) / (X2 - X1)
m = (7 - 0)/(4 + 2)
m = 7/6
* (/) = over
* Then, we can determine the equation of the line by substituting the values of one of the points (either one) and the slope into the point-slope formula.
Formula: (Y-Y1) = m (X - X1)
m = 7/6 point M (-2, 0)

(Y - 0) = 7/6 (X + 2)

y = (7x)/6 + 14/6

Right now, this equation is still in slope-intercept form. All we have to do now is to rearrange these values to turn it into the standard form.
6(y) = (7x/6 + 14/6)6
6y = 7x + 14
0 = 7x - 6y + 14

And there you go, now we’ve determine the equation of this line in standard form. Have a good weekend?
Cough * Wendy smells * Cough
well if you guys dont understand this its not my fault
You guys wanna hear a joke? *people cheer* Alright then here it goes: guess what ??? i bet your thinking about chicken butTS ROLFMAO ?HGAHAHA!!! jk..

Thursday, March 02, 2006

Scribe post

Well thanks a lot Wendy I' didn't want to be scribe you will regret it in the future hahaha.
Today in class we talked about how to write the equation of a straight line.
There are two ways that this can be determined
1) slope intercept form
y=mx+b "m" is the slope and "b" is the y-intercept
2)Point-slope form
We also learned that there are three different ways to show the equation
the slope-intercept form,the general form, and the standard form.
2)3x-2y=6 ---->ax+by=c this is the general form
3)3x-2y-6=0--->ax+by+c=0 this is standard form
After learning that we did a few review questions on slope-intercept form
1)m=2 b=6 -->y=2x+6 2)m=3 b=38-->y=3x+38 3)m=-17/5 b=163-->y=-17/5+163
After that we did a few examples using the point-slope form y-y1=m(x-x1)
ex.1 given A(3,-2) and m is -3/5
you substitute what you know into the equation
then you multiple everything by 5 to get rid of the fraction
=5y=-3x+9-->3x+5y=-1-->ax+by=c general form or 3x+5y+1=0 standard
then divide everything by 5 to get y by itself
=y=-3/5-1/5 this is the slope-intercept form

In the second example we had solve the equation by using the x-intercept and we had to find the three ways to write the equation of a line

x-intercept=3 m=2
y=2x-6 -->y=mx=b
In the last one we were allowed to put it into any form that we wanted I chose the general form
p(-3,7) m=-7/2

we worked on ex.13 today and omitted 13,14,17
And tomorrow's scribe is SUPER buTT

Wednesday, March 01, 2006


hey everyone, i'm today's scribe. we started off class with our linear equations (graphing each equation) handed back. then Mrs. Ingram started us off with 3 example of changing an equation into the slope-intercept form.
Slope-Intercept form = y = mx + b
( m = slope, b = intercept )

The three examples were :
Change to the Slope-Intercept Form

Example Number 1.
1. 21y = 15 - 27x
= 21y/21 = -27x/21 + 15/21
= 7y/7 = -9/7 + 5/7
= y = -9/7 + 5/7

Step 1. ---> Change the equation to the Slope-Intercept Form.
Step 2.---> The first point on the graph, will be the y-intercept. (0, 5/7)
Step 3. ---> Locate the second point using the Slope. ( -9/7 = rise/run )

You then, should have a graph looking like this one;

Example Number 2.
2. 7x + 2y = -22
= 2y = -7x - 22
= 2y/2 = -7x/2 - 22/2
= y = -7/2 x - 11

You then, should end up with a graph looking like this;

Example Number 3.
3. x = 7y/5 + 21/5
( you find the commen denominator and multiply everything by the commen denominator which is 5)
= 5x = 7y + 21
= 5x/7 - 21/7 = 7y/7
= 5x/7 - 3 = y

You then should have a graph looking like this;

If you had no problem and had a good understanding of those 3 examples then you would be just fine on the worksheet that was handed to us today in class.

Tonight's homework is the Linear Equation (Write the slope-intercept form of the equation). Do questions 4, 6, 12, 15, 16. Graph is to be handed in the beginning of tomorrows class.

*drums...... tomorrows scribe will be NATASHA .. :O... wow no one expected that! "hahahah" good luck! =p