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Thursday, April 29, 2010

Algebra (Period 4)

Unions, Sets & Intersections 9-1

Conjunctions & Disjunctions 9-2

Conjunction = and --> the graph is an intersection ("yo") and inequality looks like our domains and ranges on our projects
EXAMPLE: 5 < x < 10 
open dots; between 5 and 10 is colored in 

Disjunction = or --> graph will go opposite ways ("dorky" dancer) and inequality looks like this:

x < -2 OR x > 4

open dots; one arrow goes right at 4 and the other arrow goes left at -2

Equations and Absolute Value 
1 VARIABLE
 9-3
Solve the equation twice - Once with the solution positive and once with it negative

I2x - 4I = 10

Solve it twice:

2x - 4 = 10 or 2x - 4 = -10

x = 7 or x = -3


If there is a term on the same side of the equation as the absolute value, move that to the other side of the equation first 
(just like we did with radical equations!) 
Then solve twice.


REMEMBER THAT THE SOLUTION GIVEN CANNOT BE NEGATIVE (the null set)




Inequalities and Absolute Value 
1 VARIABLE
 9-4
There are 2 possible types of inequalities - less than and greater than

For less thAND:

These are conjunctions and so you solve it twice and the solution ends up between them

I3xI < 15 is equal to -15<3x<15 
so x is greater than -5 and less than 5 
For greatOR than:

These are disjunctions and are solved twice with the solution infinitely in different directions

I3xI >15 is equal to 3x < -15 and 3x > 15


Tuesday, April 27, 2010

Pre Algebra ( period 1)

Volume of Prisms & Cylinders 10-7



Remember when I said that the basic formula for area is A = bh

Well, the basic formula for volume of a prism or cylinder is:

V = BH

Where capital B = the area of the base of the prism/cylinder

capital H = the height of the prism/cylinder

THE LABEL IS ALWAYS CUBED!!!


Volume of a rectangular, square, or parallelogram prism:

V = BH

Volume = area of the base times the height of the prism

V = (bh)(H)

The lower case b and h are the base and height of the base (this is plane geometry!)


Volume of a triangular prism:

V = BH

Volume = area of the base times the height of the prism

V = (1/2 bh)(H)

The lower case b and h are the base and height of the base (this is plane geometry!)


Volume of a trapezoidal prism:

V = BH

Volume = area of the base times the height of the prism

V = [(average of the 2 bases)(h)](H)

The lower case b and h are the base and height of the base (this is plane geometry!)



Volume of a cylinder:

V = BH

Volume = area of the base times the height of the prism

V = (π r2)(H)

The lower case b and h are the base and height of the base (this is plane geometry!)

Volume of Pyramids, Cones & Spheres 10-9

To find the volume of a pyramid or cone is just as easy!!!

It's just V= 1/3 BH

That means just find the volume as if it was a prism or cone then just divide it by 3!!!


VOLUME OF A SPHERE (ball) (this one is different)
V = 4/3 π r3

Math 6H ( Periods 3, 6, & 7)

Commission and Profit 9-6

Some sales jobs pay an amount based on how much you sell. This amount is called a commission.

Like a discount, the commission can be expressed as a percent or as an amount of money.

amount of commission = percent of commission X total sales.

Using the examples from our textbook,

Maria sold $42,000 word of insurance in January. If her commission is 3% of the total sales, what was the amount of her commission in January?

amount of commission = percent X total sales
0.03 X 42,000 = 1260
Her commission was $1,260.

Profit is the difference between total income and total operating costs.

profit = total income – total costs

The percent of profit is the percent of total income that is profit

percent of profit = profit/total income
A shoe store had an income of $8600 and operating costs of $7310. What percent of the store's income was profit?

profit= income- total costs = 8600 -7310 = 1290

percent of profit = profit/total income = 1290/8600 = 0.15
So the percent of profit was 15%.

Practice finding 10%-- its easy--- just move the decimal over one place.
We practiced finding 20%. Just double what you got for 10%.



MATH AT WORK:
Caterer
A caterer provides food for parties, weddings, bar/bat mitzvahs, and other events. Caterers plan the menu, buy the ingredients, and cook the food. Often they provide seating and music as well. For each event, a caterer determines the cost per guest. The catering business requires a thorough knowledge of rations, proportions, and percents.

Algebra Period 4

Motion Word Problems 8-5 & 10-7
rt=d problems
(rate)(time)=distance
Back in Pre-Algebra, these were fairly simple word problems:
1) If you go 60 mph for 3 hours, how far have you gone? (180 miles)
2) You've gone 100 miles in 2 hours. What was your average speed? 
(100/2 = 50 mph)
 
3)You've driven 1000 miles at an average speed of 25 mph. How long did it take you? (1000/25 = 40 hours)
 
Now the problems become more difficult. Usually they involve 2 cars, trains, planes, etc. One car is the "slow" car and the other is the "fast" car.
 
Just like the mixture problems, it helps to make a matrix and also draw a picture to help you understand the words.
 
SLOW CAR/FAST CAR LEAVE FROM SAME PLACE, IN THE SAME DIRECTION, AT DIFFERENT TIMES, WHEN WILL THE 2ND CAR CATCH UP WITH THE FIRST CAR? 
 
2 cars leave 3 hours apart. One travels 72 mph. The other travels 120 mph. The slower car leaves first. When with the faster car catch up with the slower car?
 
Use the set up forms from class... you can find more blank forms online!!

SLOW     72           t              72t  
FAST     120        t - 3        120(t - 3) 
         (fast car left 3 hours later so driving 3 less hours or t - 3)
 
At the point when the fast car catches and overtakes the slow car, what is true of the distances of the 2 cars at that exact moment???
They are equal!
WHAT IS THE EQUATION?
Set the 2 cars' distances equal:
72t = 120(t - 3)
 
72t = 120t - 360
-48t = -360
t = 7.5 hours (slow car's time on the road)
t - 3 = 7.5 - 3 = 4.5 hours (fast car's time on the road)
 
CHECK:
The 2 cars should have traveled the same distance because one car catches up with the other car:
slow: (72)(7.5) = 540
fast: (120)(4.5) = 540
 
 
SLOW CAR/FAST CAR LEAVE FROM SAME PLACE, GOING IN DIFFERENT DIRECTIONS, LEAVING AT THE SAME TIME, WHEN WILL THEY BE A CERTAIN DISTANCE APART?

 
2 cars leave going in different directions. One travels 60 mph. The other travels 50 mph. In how many hours will the cars be 605 miles apart?
 
                      CAR     (RATE)    (TIME)   =   DISTANCE
                     SLOW     50            t                  50t 
                     FAST      60            t                  60t
                      (They travel the same amount of time)
 
What is true of the distances the 2 cars travel?
Together they travel 605 miles because they are 605 miles apart.
WHAT IS THE EQUATION?
Set the 2 cars' distances as a SUM to 605.
50t + 60t = 605
110t = 605
t = 5.5 hours
So in 5 1/2 hours the 2 cars will be 605 miles apart.
 
CHECK:
If you plug in 5.5 hours for each car to find each cars distances, they should add to 605 miles.
slow: (50)(5.5) = 275 miles traveled
fast: (60)(5.5) = 330 miles traveled
275 + 330 = 605 miles
 
SLOW CAR/FAST CAR LEAVE FROM SAME PLACE, GOING IN SAME DIRECTION, LEAVING AT THE SAME TIME, WHEN WILL THEY BE A CERTAIN DISTANCE APART?

 
2 cars leave going in the same direction. One travels 45 mph. The other travels 35 mph. In how many hours will the cars be 15 miles apart?
 
                      CAR     (RATE)    (TIME)   =   DISTANCE
                SLOW     35            t             35t 
                FAST      40            t             40t
                      (They travel the same amount of time)
 
What is true of the distances the 2 cars travel?
They are getting further and further apart as the minutes go by.
The DIFFERENCE of the 2 cars is 15 miles after a certain amount of time.
WHAT IS THE EQUATION?
40t - 35t = 15
5t = 15
t = 3 hours
So in 3 hours the 2 cars will be 15 miles apart.
 
CHECK:
If you plug in 3 hours for each car to find each cars distances, their distances should subtract to 15 miles.
slow: (35)(3) = 105 miles traveled
fast: (40)(3) = 120 miles traveled
120 - 105 = 15 miles
 
SLOW CAR/FAST CAR LEAVE FROM SAME PLACE, GOING IN DIFFERENT OR SAME DIRECTION, IN THE SAME AMOUNT OF TIME, EACH TRAVELS A  DIFFERENT DISTANCE, WHAT IS THEIR SPEED? 
One car travels 20 mph faster than the other car. One car travels 240 miles while the other travels 180 miles. Find their average speeds.
 
                      CAR     (RATE)       (TIME)   =   DISTANCE
               SLOW        r           180/r             180 
               FAST      r + 20    240/(r + 20)        240
This time you have the distance and know that the faster car is 20 mph faster than the slower car. To find the time for each car, use the fact that d/r = t so divided each car's distance by their rates.
 
WHAT IS THE EQUATION?
The times of these 2 cars is equal (left at same time and stopped at same time) so set their times equal:
180/r = 240/(r + 20)
Multiply both sides equally by the LCM:
(r)(r + 20)(180/r) = (r)(r + 20)240/(r + 20)
Cross cancel:
180(r + 20) = 240(r)
180r + 3600 = 240r
3600 = 60r
r = 60 mph (the slower car)
r + 20 = 60 + 20 = 80 mph (the faster car)
 
CHECK:
If you plug in the speeds, you should find that both cars traveled the same amount of time:
slow: (60)t = 180   t = 3 hours
fast: (80)t = 240   t = 3 hours
 
A BOAT/PLANE TRAVELS WITH THE CURRENT ON THE DEPARTING LEG OF THE JOURNEY AND TRAVELS AGAINST THE CURRENT ON THE RETURN LEG. WHAT IS THE SPEED OF THE BOAT/PLANE IN STILL WATER/AIR? 

 
A boat travels with a current of 6 mph for 3 hours and then returns home against the same current. The trip home takes 5 hours. What is the speed of the boat in still water?
 
           BOAT/PLANE        (RATE)       (TIME)   =   DISTANCE
      WITH CURRENT        r + 6           3            3(r + 6)  
      AGAINST CURRENT    r - 6           5            5(r - 6) 
r is the speed in still water and 6 is the speed of the current. The boat with need less time to go the same distance with the current than against it. 
 
WHAT IS THE EQUATION?
The distance to the boat's location and the distance home must be equal.
Set the distances equal:
3(r + 6) = 5(r - 6) 
3r + 18 = 5r - 30
48 = 2r
r = 24 mph (speed in still water)
r + 6 = 24 + 6 = 30 mph (speed with the current)
r - 6 = 24 - 6 = 18 mph (speed against the current)
 
CHECK:
If you plug in the speed with and against the current with the hours traveled, you get the distances to and home. Those distances should be equal:
3(30) = 5(18)
90 = 90

Monday, April 26, 2010

Pre Algebra ( period 1)

Chapter 10 Area & Volume
AREA-
all these formulas are related to the basic concept of A = bh
Area of Parallelograms 10-1

Area of rectangles and parallelograms = (base)(height)


Triangles & Trapezoids 10-2
Area of triangle = (1/2)(base)(height or altitude)


Area of trapezoid = (average of the 2 bases)(height)


Now we're starting area!
Where the perimeter/circumference fenced in my puppy, the area of the yard will tell me how much sod (grass) I should buy to stop the puppy's paws from getting muddy!

Area for me is all basically the length of the base times the height of the figure

A = bh

In a parallelogram, whether it's a rectangle, rhombus, square or other parallelogram

A = bh with the height being a line perpendicular to both bases (not the slanted side!)


You have learned the area of a rectangle as A = lw, but the l = b and the w = h

You may have learned the area of a square as A = s2 , but that's because the b = h


Any parallelogram can be split into 2 triangles using a diagonal.

Because of this, the area of a triangle is half that of a parallelogram.

A = 1/2 bh


A trapezoid has 2 bases that ARE NOT EQUAL. So which base is THE base?

If you use the smaller base, you won't have enough sod for your yard and the puppy's paws are still getting muddy.

If you use the larger base, you'll have too much sod for your yard and the extra will rot.

Sooooooooo..... you actually need to take the average of the two bases times the height

A = (average of the 2 bases)(height)

A = (b1 + b2)h /2

Space Figures 10-4
SPACE FIGURES OR SOLIDS OR 3 DIMENSIONAL FIGURES
 (solids)
prisms = 2 congruent parallel bases - all other sides are rectangles

cylinder = 2 congruent circle bases

When you remove one base from a prism, it becomes a pyramid - all other sides are triangles

When you remove one base from a cylinder, it becomes a cone

When you have a set of points in all directions that are equal distance from a central point, you have a sphere

Vertices - the points where edges connect (the corners)

Edges - the line segments that connect the vertices


You should be able to visualize what a figure will look like if you could cut it apart and open it

That's called a net!
We'll look at some of these in class together.

Try to think of what it will form if you fold it back up!


Math 6H ( Periods 3, 6, & 7)

Discount and Markup 9-5

A discount is a decrease in the price of an item. A markup is an increase in the price of an item. Both of these changes can be expressed as an amount of money or as a percent of the original price of the item. A store may announce a discount of $3 off the original price of $30 basketball, or a discount of 10%

A warm-up suit that sold for $42.50 is on sale at a 12% discount. What is the sale price?

Method 1: Use the formula

amount of change = percent of change X original amount
= 12% X $42.50

therefore the discount is 0.12 X 42.50 or 5.10
The amount of discount is $5.10
The sale price is 42.50 – 5.10 = $37.40

Method 2: Since the discount is 12%, the sale price is 100% - 12% = 88%.
The sale price is 0.88 X 42.50 = $ 37.40

When you know the amount of discount you subtract to find the new price. When dealing with a markup you add to find the new price.

The price of a new car model was marked up 6% over the previous year’s model. If the previous year’s model sold for $7800, what is the cost of the new car? {and what kind of a car could that be?}
Method 1: Use the formula
amount of change = percent of change X original amount

= 6% X 7800

Therefore the markup is 0.06 X7800= $468
The new price is 7800 + 468 = $8268

Method 2: Since the markup is 6% the new price is 100% + 6% or 106% of the original price. so the new price is 1.06 X7800 = $8268

This year a pair of ice skates sells for $46 after a 15% mark up over last year’s price. What was last year’s price?

This year’s price is 100 + 15 or 115% of last year’s price. Let n present last year’s price
46 = (115/100)n

46 = 1.15n
46/.15 = 1.15n/1.115
40 = n

So last year’s price was $40.

A department store advertised eclectic shavers at a sale price of $36.
If this is a 20% discount, what was the original price?
The sale price is 100 - 20 or 80% of the original price. Let n represent the original price.
36 = (80/100)n
36 = .8n
36/.8 = .8n/.8

45 = n
The original price was $45.
Check to see that your answers are logical and reasonable.

Try these: A service station (that’s gas station, now—they no longer provide service!!) give cash customers a 5% discount on the price of gasoline. If gasoline regularly sells for $3.00 a gallon, what is the discounted price?


A store marks up the price of a $5 item to $12. What is the percent of markup?

Algebra Period 4

Inequalities in Two Variables 9-5


You will shade an x y graph to find the side of a linear equation that fits the solution

1. Graph the inequality by graphing the line with an x y table or y = mx + b

2. For inequalities that do not have the equal sign (no crayon!), make a dotted line; for inequalities with an equal sign, make a solid line

3. Shade the side of the line that works in the inequality (is a solution)

ONE SIDE WILL WORK, THE OTHER WILL NOT!
I always check the point (0, 0) first if possible because it's the easiest point to check :) If (0, 0) works, shade that side. If (0, 0) doesn't work, shade the other side.

4. Check the other side just to make sure that it does not work.


EXAMPLE: x + y > 5

You graph the line with DOTTED line because it cannot be equal to 5.
I would put it in y = mx + b format: y > -x + 5 
(Remember that if you multiply or divide by a NEGATIVE to isolate the y, you'll need to switch the symbol!)

You pick an easy point on one side of the line and substitute to see if that side is a solution.

If that does not work, pick an easy point on the other side to see if that side checks.




Systems of Inequalities in Two Variables 9-6
If there is more than 1 Inequality (a system of inequalities),

1. Follow the same procedure as above for one equation
 but you will need to do it for each equation.

2. Use a different type of shading for each so you won't get confused

(Ex: Use slanted lines one way and then slanted lines the other way. Use different colors if possible. Make one set of lines wavy and the other set straight)


3. Where the 2 shadings overlap each other is called the solution of the system of inequalities.
(any point in the overlap should work in BOTH inequalities - make sure you check all the inequalities in the system!!!!!)

Solving Systems of Equations 8-1 to 8-3
(2 equations with 2 variables)

You cannot solve an equation with 2 variables - you can find multiple coordinates that work

TO SOLVE MEANS THE ONE COORDINATE THAT WORKS FOR BOTH EQUATIONS

There are 3 ways to find that point:

1. Graph both equations: Where the 2 lines intersect is the solution

2. Substitution method: Solve one of the equations for either x or y and plug in to the other equation

3. Addition method: Eliminate one of the variables by multiplying the equations by that magical number that will make one of the variables the ADDITIVE INVERSE of the other


Example solved all 3 ways:

Find the solution to the following system: 
2x + 3y = 8 and 5x + 2y = -2


1. GRAPH BOTH LINES: Put both in y = mx + b form and graph

Read the intersection point....You should get (-2, 4)


2. SUBSTITUTION: Isolate whatever variable seems easiest

I will isolate y in the second equation: 2y = -5x - 2

y = -5/2 x - 1 
Plug this -5/2 x - 1 where y is in the other equation
2x + 3y = 8
2x + 3(-5/2 x - 1) = 8

2x - 15/2 x - 3 = 8

4/2 x - 15/2 x - 3 = 8

-11/2 x - 3 = 8

-11/2 x = 11
-2/11(-11/2 x) = 11(-2/11)
x = -2

Plug into whichever equation is easiest to find y



3. ADDITION: Multiply each equation so that one variable will "drop out" (additive inverse) 
I will eliminate the x, but I could eliminate the y if I wanted to!
 
2x + 3y = 8 and 5x + 2y = -2
 
Notice that there is 2x in first and 5x in 2nd equation.
To make the x terms additive inverses, I'll multiply both sides of each equation so that the x terms are both 10x. 
I'll need to make one +10x and the other -10x:
5(2x + 3y) = (8)5

-2(5x + 2y) = (-2)(-2)

Now when I distribute I get:
10x + 15y = 40

-10x - 4y = 4


ADD TO ELIMINATE THE x term:

11y = 44
SOLVE: y = 4

Plug into whichever equation is easiest to find x

NOTICE THAT FOR ALL 3 METHODS, THE SOLUTION IS THE SAME!
THEREFORE, USE WHATEVER METHOD SEEMS EASIEST!!
 
2 special cases!
NO SOLUTION: When would 2 lines never intersect???

When they're PARALLEL! So always check first to see if the lines have the SAME SLOPES. If they do, you're wasting your time trying to solve the system because the solution if the point where they intersect and there is none!
 
INFINITE SOLUTIONS: When would 2 lines have infinite points in common???
When they're multiples of each other and therefore are really THE SAME LINE! If you make each line y = mx + b first, you'll find this out quickly. Then just say INFINITE SOLUTIONS.