Lesson 9.3

Systems Of Inequalities

Some examples of inequalities

• 2X<Y
• 8X+5>Y
• 3X-9Y<48
• X ²+Y ²>7
• X>Y 

 Remember, the MOST IMPORTANT thing to remember when solving inequalities is that you are looking for an ordered pair that makes the statement true. 

Two inequalities are equivalent if they have the same solution. 

The graph of an inequality is the set of all points that correspond to the solution. 

To Solve an inequality we replace the inequality sign with an equal sign.  This divides the graph.  Every point on one side of the line will be true, while every point on the other side will  be false.  In order to determine which side is true we plug in points.

For example

 

The shaded region represents every value that makes the statement true.

Now, sketching the graph of Y<X+1

1. First, replace the < symbol with a = sign.  Y=X+1
2. Next, sketch the line

 

      3.  Then, plug in points.  We will use (2,0) and (0,2)

                    Since 0 <2+1 is true and 2< 0+1 is false we know to shade in all points below the line.  

      4.  Shade in. 

 

You can do the same thing for parabolas.  

Now it gets really fun.  

Solving a system of inequalities

The only thing that is different here is that we have two or more lines.  

Now, the solution will have to correspond to both of the inequalities. 

If we have the two inequalities X+2Y>-4 and 3X+Y<3 we want to find the area that should be shaded in. 

Our steps are just the same for this one.

1. First substitute in an = sign
2. then draw the line
3. next plug in points to see where the statement is true
4. and finally shade

 The area of overlap is the area we are looking for 

The graph will look something like this. 

This is a cool looking one. 

 Now, absolute values.  

• When given /X/ < b we know that -b< X <b
• And when given /X/ >b we know that X <-b or X > b  

 So, when given /X/ < 4 we know that we must shade the region between -4 and 4.

Here, /X/<3.5 and 0<Y<3.5

Systems of inequalities relating to a circle.

X ² + Y² = r<sup>2

 If</sup> 
     X ² + Y^{2  > 4} 
 
We shade outside of the circle

If

 X ² + Y^{2  < 4}

We shade inside the circle 

Ignoring the triangle this is what it looks like with <

When trying to find a system of inequalities form a graph it is helpful to remember the equation of a circle and slope intercept form.  And it is necessary to take note of which region is shaded, and where it is in relation to ALL of the functions and circles.  

And that's all you need to know about systems of inequalities.

Matthew "Seaglass" Silbergleit

 

Section 9.1: Systems of Equasions

Today, we reviewed systems of equasions. Remember that there are three ways to solve a system of equasions:

Substitution: solve one equasion for a variable, and plug it in to the other. For example:

2x + y = 13                                     x - y = 5

y = 13-2x   Solve for one variable.               x - (13 - 2x) = 5    substitute 13-2x for y.


Once we have this, we can solve for x.

3x - 13 = 5    Simplify.


3x = 18    isolate x.

x = 6

Once we have determined x, we plug x back into the first equasion.

y = 13 - 2(6)    plug in x.

y = 13 - 12     simplify.

y = 1

Now that we know the x and y values, we can form an ordered pair which is the solution to the equasion, or where the two graphs intercept. Our ordered pair is (6,1), which is the solution to the equasion.


Now for the next method.

Elimination: multiplying one equasion by a certain number so that when you add the two equasions together, one or more variables cancel out. For example:

2x + y = 13                                   x + 2y = 5

Now we have to multiply one of the two equasions by a certain number so one or more variables cancel out. Lets multiply the first equasion by -2. Remember to multiply both sides of the equasion. We end up with:

-2(2x + y) = -2(13)

-4x - 2y = -26

Now we can add the equasions, and something will cancel out. In this case, the y's will cancel out.

    -4x - 2y = -26
+    x + 2y = 5
_______________
           -3x = -21

Notice that the y's canceled out. Now we can solve for x, and later, y.

-3x = -21

x = 7                              x + 2y = 5

                                      2y = 5 - 7
                                      
                                      2y = -2
                                     
                                      y = -1

Now our solution has an x-value of  7 and a y-value of -1, which forms the ordered pair (7,-1).


The third method is graphing. for this, enter both equasions. lets use 2x + y = 13 and x - y = 5. in order for it to work on our calculator, we need to isolate y. We end up with:

y = -2x + 13                         y = x - 5

Now, for all of you TI users, graph the functions and under the calc menu find the intercept.

We find that the x-value is six, and the y-value is one, which gives us the ordered pair (6,1) as the solution.

It is also possible to solve for more than two equasions, it just involves an extra step in substitution or elimination.


Some helpful links that Mr. Wilhelm should show the class to enrich our learning:

http://www.youtube.com/watch?v=LhXkYfLgRoA skip to 1:06

http://www.youtube.com/watch?v=2qdql9vsWWM

Thats all for now. Good luck on Friday for all of those who need it for the AP test!
-Jacob

Vectors - 8.3

Scalar Quantities - quantities characterized by a single real number value called a scalar (area, volume, temperature, and time common examples of scalar quantites)


Directed Line Segment - segment to which a direction has been assigned (referred to as a vector), has both magnitude and direction (not a scalar quantity)

• P is the initial point
• Q is the terminal point
• Arrowhead represents the direction
• The magnitude of vector PQ is the length of the segment PQ and is denoted by ||PQ||

• This vector would jut be called V

Equivalent Vectors - have the same magnitude and direction. The location of the vectors doesn't matter.

• A vector may be traslated from one location to another, provided neither the magnitude nor the direction is changed
• Vectors U and V are equivalent even though they are in different locations

Velocity Vectors can represent many physical concepts

• In this example, vector V (the red vector) is representing the path of a ladybug descending at 100 mph (which determines the magnitude) with a line of flight that makes a 20 degree angle with the horizontal
• V is an example of a velocity vector

Force Vectors - vectors that represent a pull or push of some type

• Vector F represents the upward force 

Displacement - The path of a point as it moves along a segment

• Vector AB represents the path of a point as it moves along the segment AB

Sum of Vectors - any two vectors may be added by placing the initial point of the second vector on the terminal point of the first, then drawing the line segment from the initial point of the first to the terminal point of the second

• This is called the triangle law
• Vector AC is the sum of vectors AB and AC

Resultant force - the single force that produces the same effect as the two combined forces

• Can be found using the parallelogram law
• You form an parallelogram by drawing in imaginary vectors RS and QS
• The force of PS is the resultant force of PR and PQ
• PQ + PR = PS

Scalar Multiple - If m is a scalar and v is a vector, the mv is defined as a vector whose magnitude is |m| times ||v|| and whose direction is either the same as that of v (if m>0) or opposite that of v (if m<0)

• mv is a scalar multiple of v

We generally talk about vectors that have been restricted to an XY plane. Since vectors can be moved around wherever you want as long as the direction and magnitude dont change, we can move the initial point of the vector to (0,0).

Moving the vector here allows the vector to determine an ordered pair

• If our vector has a terminal point A, it can be assigned the coordinates (a₁, a₂) which are the coordinates of terminal point A

There is a one-to-one coorespondance between vectors in an xy-plane and ordered pairs of real numbers

• Allows us to think of vectors as both directed line segments and an ordered pair of real numbers

Instead of using parentheses, vectors expressed by an ordered pair are shown as <a₁, a₂>

• a₁ and a₂ are  called the components of vector <a₁, a₂>
• If A is the point (a₁, a₂), we call vector OA the position vector for <a₁, a₂> or for the point A

Once you understand the basics of vectors, you can learn equations that can be used on vectors when necessary

The magnitude of the vector a = <a₁, a₂>, denoted by ||a||, is given by

The addition of vectors follows the following equation

To find a scalar multiple m of a vector <a₁, a₂>, use 

There is a vector called the zero vector that is denoted as

The negative -a vector of a vector <a₁, a₂> (where m is -1) is

The following is a table that shows all of the properties of vectors. The vectors described are called a, b, and c, along with scalars represented by m and n.

As you can see, the properties of vectors are very similar to those of regular numbers, which makes them easy to remember. 


To subtract vectors, you do the following

There are two special vectors you should remember named i and j

• i = <1,0>
• j = <0,1>

i and j and special because we can use them to denote vectors in yet another way

If we put the above equation on a graph, it will look like this

• Because of its position on the graph, we generally call a1 the horizontal component, and a2 the vertical component. 
• The vector sum a1i + a2j is a linear combination of i and j
• We may regard linear combinations of i and j as algebraic sums

Vector a and angle Θ are defined below (Θ is the entire angle formed by a and the x axis, sorry thats kind of hard to see)

Based on the above diagram, we can confirm the following

Well that's pretty much it for 8.3

I love spending my Wednesday nights writing blogposts about vectors WOOOO

-Corin Murphy

7.5 Product-to-Sum and Sum-to-Product Formulas

7.5 Product-to-Sum and Sum-to-Product Formulas

This section is one of the more basic trigonometric concepts, changing products to sums or sums to products.

The proof of the product-to-sum formulas is:

                     sin (u + v) = sin u cos v + cos u sin v
                     sin (u + v) = sin u cos v - cos u sin v
sin (u + v) + sin (u - v) = 2 sin u cos v

This allows us to use these formulas:

(1) sin u cos v = 1/2[sin (u + v) (u - v)]
(2) cos u sin v = 1/2[sin (u + v) (u - v)]
(3) cos u cos v = 1/2[cos (u + v) (u - v)]
(4) sin u sin v = 1/2[cos (u - v) - cos (u + v)]


The proof of the sum-to-product formulas is:

                                           1) u + v = a       and        u - v = b
2) (u + v) + (u - v) = a + b                                                      2) (u + v) - (u - v) = a - b              
3) u = a + b                                                                              3) v = a - b
               2                                                                                               2

         Substitute for u + v and u - v on the right-hand sides of the product-to-sum formulas and for u and v on the left-hand sides. Multiply by 2 and we obtain the following sum-to-product formulas.

(1) sin a + sin b = 2 sin a + b cos a - b

                                           2             2

(2) sin a - sin b = 2 cos a + b sin a - b

                                          2            2

(3) cos a + cos b = 2 cos a + b cos a - b

                                             2             2

(4) cos a - cos b = -2 sin a + b sin a - b

                                            2             2

There are many tremendous youtube videos on this topic. I checked.

Peace.Love.Thad

-Joey

8.1 The Law of Sines

Here’s how to derive the law of sines!

First, draw an altitude from point C to line c (labeled h). From the two right triangles we made, we know the following.

This means that the sine of an angle divided by its opposite side is equal to another angle from the same triangle divided by its opposite side. The law of sines works for every oblique triangle (without a right angle). Just plug in the values and cross multiply to get the missing variable. Knowing that this will work for any angle in the triangle, we can write a more complete version of the law of sine.

An finally, here's the video Mr. Wilhelm recommended that we watch. Khan Academy

- Sarah Skender

8.2 The Law of Cosines

Greetings! It's Eleni.
Today, or yesterday, or whenever you choose to learn it, our class learned about the law of cosines.
Here's the link of a math teacher's lesson that Mr. Wilhelm gave us.
http://www.youtube.com/watch?v=24Ibcl3mBQU
This is a picture of the proof:

The a, b, and c are side lengths, while capital letters are angles.  Don't get caught up on which is a, b, or c; here's an easier way to think of it :
Side squared= the sum of the other two sides squared minus two times the same two sides and cos of the angle across from the initial side.
It may sound confusing now...but it actually helps and is better than having to memorize the formulas.

LAW OF COSINES 

When solving a triangle, the law of cosines can be used when you know SAS or SSS. Remember: the law of sines will also come in handy when doing these problems.
SAS
Example problem:
a=5.0, c=8.0, and B=77 degrees




1. draw a triangle and label

2. Since B is the angle between sides a and c, it is easiest to figure out b first.


law of cosines

2. Now that you know a side and it's angle (B and b) you can use the law of sines!

3. Lastly, knowing two out of the three angles of the triangle, we can easily find C.
180-77-35= 68 degrees =C 
SSS
Okay, so doing that last problem step-by-step was a lot of work soooo..... I'm just gonna give you the gist of it.
Example problem:
a=90 b=70 c=40

1. FIND THE LARGEST ANGLE FIRST
even if you have to find the smallest to answer the question, find the largest first.
* the largest angle is across from the largest side (angle A)

2. Now you can use the law of sines to find another angle

3. Subtract both angles from 180 to get the last angle

Another thing we learned in 8.2 was

Heron's Formula

This formula  finds the area of a triangle, any triangle.

S is one half the perimeter
a, b, and c are side lengths.
Another way to find the area of a triangle that's a lot more useful is

---where a and b are any two sides and y is the angle between them (SAS).
Also in the homework, they ask you to combine both of these concepts.
Example: Approximate the area of triangle ABC
A=35.7degrees
C=105.2 degrees
b=17.2
simply use the SAS version of the law of cosines
then use the law of sines like we did in the example above
after, use the a,b, and c side lengths to find the perimeter and therefore s
plug em in the equation.


sooo as we all know mr.w is returning in two days so I thought I'd map it out how far away from us he is.

(At least I think he's in San Diego)
According to google maps, that's only 2340 miles away
and by walking, it should only be roughly a 30 day journey
I hope wilhelm is a fast walker


That's all on 8.2 Bye!
---Eleni