Simple Interest 9-7
When you borrow money you pay the lender INTEREST for the use of the money. The amount of interest you pay is usually a percent of the amount borrowed figured on a yearly basis. This percent is called the annual rate.
When interest is computed year by year we call it
SIMPLE INTEREST
The formula is I= Prt
Let I = simple interest charges
P = principal ( amount borrowed)
r= annual rate
t = time in years
I = Prt
simple interest is calculated just on the principal.
Let's work through a few examples
$150 borrowed at 12% annual rate for 1 year
I = Prt
I = (150)(.12)(1)
I = 18
so you would owe $18 in interest after 1 year.
The total due would be $150 + 18 = $168
What if instead you borrowed the same amount but for 2 years... nothing was due until the end of two years
I = Prt
I = 150(.12)(2) = 36
You would owe $36 in interest .. so the total due was 150 + 36 = $ 186.
What if you borrowed the same amount for 3 years...
I = 150(.12)(3) = 54 or $54 in interest.
You would owe 150 + 54 = $ 204 after three years...
However, let's say you could only borrow that amount for 6 months...
I = Prt
I = (150)(.12)(.5)
Why 0.5? that is 1/2 a year.
Now you can always multiply by 1/2 as well.. in fact, sometimes that is easier
I = 150(.12)(1/2) = 9 or $ 9.00
After 6 months you would owe $159.
Dylan paid $375 in interest on a loan of $1500 principal at 12.5% interest.
What was the length of time?
Look at what it is asking and see which of the variables you have...
I= Prt
We have the interest paid, the principal and the annual rate so
375= (1500)(.125)(t)
375 = 187.5t
solve this one step equation by dividing both sides by 187.5
375 = 187.5t
187.5 187.5
t = 2
so 2 years
divide carefully...
Alexis paid $ 585 simple interest on a $6500 loan for 6 months.
what was the annual rate?
What do we know?
I = 585
P = 6500
t= 6 months ( which is 0.5 or 1/2)
I = Prt
585 = 6500 (r)(.5)
585 = 3250r
divide both sides by 3250
585 = 3250r
3250 3250
r = 0.18
which means 18%
annual--> once a year
6 months --> 1/2 or 0.5
4 months--> 1/3
3 months --> 1/4 or 0.25
8 month --> 2/3
Compound Interest 9-8
Compound interest is ALWAYS more than simple interest.
interest is compounded on the interest!!
$100 savings earning $10 interest/ annual.. [this only happens NOW if your dad is the one paying you... :)]
I = Prt
at the end of the first year
I = 100(.10)(1) = 10 or $10
add that to the 100
$110.
Now for the 2nd year,
$110 is your principal
so
I = Prt
I = 110(.10)(1) = 11 or $11
so at the end of 2 years you have $110 + 11 or $121
Now for the 3rd year
I = Prt
I = 121(10)(1) = $12.10
So at the end of three years you have $121 + 12.10 = $133.10
What if you had $500 at 8% compounded quarterly for one year.
quarterly means 1/4 or .25
I = Prt
I = 500(.08) (1/4)
calculate the 08(1/4) because that will be the constant you will multiply your principal by each time
(.08)(1/4) = .02
so I = 500(.02) = 10
after the first quarter it is 510
I = Prt for the 2nd quarter
I = 510 (.02) = 10.20
so after the 2nd quarter $510 + 10.20 = $520.20
I = Prt for the third quarter
I = 520.20 (0.02) = about $10.40 ( round to the nearest penny)
so after the third quarter
$520.20 + 10.40 = $530.60
I = Prt
I = 530.60(.02) = about $10.61
So at the end of 4 quarters -- or one year
530.60 + 10.61 = $541.21
compounding terms:
annually--> once a year
semiannually --> twice a year
quarterly--> four times a year
monthly--> 12 times a year
daily--> 365 times a year
Monday, May 10, 2010
Monday, May 3, 2010
Friday, April 30, 2010
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
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
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.
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
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!
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?
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.
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.
Tuesday, April 20, 2010
Math 6H ( Periods 3, 6, & 7)
Computing with Percents 9-3
The statement 20% of 300 is 60 can be translated into the following equations
20/100(300) = 60 or 0.20 •300 = 60
EQUATION METHOD:
Notice the following relationship between the words and the symbols
20% of 300 is 60
0.20 • 300 = 60
WRITE THE PROBLEM OUT AND THEN DIRECTLY UNDER THE "IS" WRITE AN EQUAL SIGN. DIRECTLY UNDER THE WORD 'OF" WRITE A MULTIPLICATION SIGN. iF YOU ARE GIVEN A % CHANGE IT FIRST TO A DECIMAL. THEN BRING DOWN ALL THE OTHER NUMBERS GIVEN IN YOUR PROBLEM. LET x OR n REPRESENT YOUR VARIABLE... THAT IS THE "WHAT " PART OF YOUR PROBLEM.
A similar relationship occurs whenever a statement or a question involves a number that is a percent of another number
What is 8% of 75?
Let n represent the number asked for
What number is 8% of 75?
n = 0.08 • 75
solve
What percent of 40 is 6?
let n represent the percent asked for.
What percent of 40 is 6?
n% • 40 = 6
n% • 40 = 6
n% (40)/40 = 6/40
n% = 6/40
n/100 = 6/40
(100) n/100 = (100) 6/40
n=15 so 15% of 40 is 6
140 is 35 % of what number?
let n represent the number asked for
140 is 35% of what number?
140 = 0.35 • n
140 = 0.35n
140/0.35 = 0.35n/0.35 divide carefully!! Watch those decimals!!
400 = n
so 140 is 35% of 400
Always check to see if your answer is logical.
PROPORTION METHOD
In these types of percent problems you are always know three parts of the following proportion
n/1oo = a/b
or better yet
n/100 = is/ of
The n represents the %
Read the problems carefully and you can easily determine which is the "is" and which represents the 'of"
For example:
What percent of 40 is 6?
What percent -- from the problem above indicates that we DO NOT know the n
of 40-- hmm... then 40 must be the 'of' and
similarly is 6 means that 6 represents the 'is'
n/100 = 6/40 solve as a proportion
and you get n= 15 but since it asked us to state the 5 your answer is 15%
140 is 35 % of what number?
In this problem I notice 35% right away so that is the n!!
Then I read the problem again and notice 140 is... hmmm.. THat says 140 must be the is
35/100 = 140/ x I do not know the 'of'
Solve again
x = 400
The statement 20% of 300 is 60 can be translated into the following equations
20/100(300) = 60 or 0.20 •300 = 60
EQUATION METHOD:
Notice the following relationship between the words and the symbols
20% of 300 is 60
0.20 • 300 = 60
WRITE THE PROBLEM OUT AND THEN DIRECTLY UNDER THE "IS" WRITE AN EQUAL SIGN. DIRECTLY UNDER THE WORD 'OF" WRITE A MULTIPLICATION SIGN. iF YOU ARE GIVEN A % CHANGE IT FIRST TO A DECIMAL. THEN BRING DOWN ALL THE OTHER NUMBERS GIVEN IN YOUR PROBLEM. LET x OR n REPRESENT YOUR VARIABLE... THAT IS THE "WHAT " PART OF YOUR PROBLEM.
A similar relationship occurs whenever a statement or a question involves a number that is a percent of another number
What is 8% of 75?
Let n represent the number asked for
What number is 8% of 75?
n = 0.08 • 75
solve
What percent of 40 is 6?
let n represent the percent asked for.
What percent of 40 is 6?
n% • 40 = 6
n% • 40 = 6
n% (40)/40 = 6/40
n% = 6/40
n/100 = 6/40
(100) n/100 = (100) 6/40
n=15 so 15% of 40 is 6
140 is 35 % of what number?
let n represent the number asked for
140 is 35% of what number?
140 = 0.35 • n
140 = 0.35n
140/0.35 = 0.35n/0.35 divide carefully!! Watch those decimals!!
400 = n
so 140 is 35% of 400
Always check to see if your answer is logical.
PROPORTION METHOD
In these types of percent problems you are always know three parts of the following proportion
n/1oo = a/b
or better yet
n/100 = is/ of
The n represents the %
Read the problems carefully and you can easily determine which is the "is" and which represents the 'of"
For example:
What percent of 40 is 6?
What percent -- from the problem above indicates that we DO NOT know the n
of 40-- hmm... then 40 must be the 'of' and
similarly is 6 means that 6 represents the 'is'
n/100 = 6/40 solve as a proportion
and you get n= 15 but since it asked us to state the 5 your answer is 15%
140 is 35 % of what number?
In this problem I notice 35% right away so that is the n!!
Then I read the problem again and notice 140 is... hmmm.. THat says 140 must be the is
35/100 = 140/ x I do not know the 'of'
Solve again
x = 400
Monday, April 19, 2010
Pre Algebra ( period 1)
PYTHAGOREAN THEOREM
FOR RIGHT TRIANGLES ONLY! 11-2
2 legs - make the right angle - called a and b (doesn't matter which is which because you will add them and adding is COMMUTATIVE!)
hypotenuse - longest side across from the right angle - called c
You can find the third side of a right triangle as long as you know the other two sides:
a2 + b2 = c2
After squaring the two sides that you know, you'll need to find the square root of that number to find the length of the missing side (that's why it's in this chapter!)
EASIEST - FIND THE HYPOTENUSE (c)
Example #1 from p. 510
82 + 152 = c2
64 + 225 = c2
289 = c2 Take the SQ RT of each side
c = 17
A LITTLE HARDER - FIND A MISSING LEG (Either a or b)
Example #5 from p. 510
52 + b2 = 132
25 + b2 = 169
b2 = 169 - 25
b2 = 144 Take the SQ RT of each side
b = 12
CONVERSE OF PYTHAGOREAN THEOREM
If you add the squares of the legs and that sum EQUALS the square of the longest side, it's a RIGHT TRIANGLE.
If you add the squares of the 2 smallest sides and that sum is GREATER THAN the square of the longest side, you have an ACUTE TRIANGLE.
If you add the squares of the 2 smallest sides and that sum is LESS THAN the square of the longest side, you have an OBTUSE TRIANGLE.
2 legs - make the right angle - called a and b (doesn't matter which is which because you will add them and adding is COMMUTATIVE!)
hypotenuse - longest side across from the right angle - called c
You can find the third side of a right triangle as long as you know the other two sides:
a2 + b2 = c2
After squaring the two sides that you know, you'll need to find the square root of that number to find the length of the missing side (that's why it's in this chapter!)
EASIEST - FIND THE HYPOTENUSE (c)
Example #1 from p. 510
82 + 152 = c2
64 + 225 = c2
289 = c2 Take the SQ RT of each side
c = 17
A LITTLE HARDER - FIND A MISSING LEG (Either a or b)
Example #5 from p. 510
52 + b2 = 132
25 + b2 = 169
b2 = 169 - 25
b2 = 144 Take the SQ RT of each side
b = 12
CONVERSE OF PYTHAGOREAN THEOREM
If you add the squares of the legs and that sum EQUALS the square of the longest side, it's a RIGHT TRIANGLE.
If you add the squares of the 2 smallest sides and that sum is GREATER THAN the square of the longest side, you have an ACUTE TRIANGLE.
If you add the squares of the 2 smallest sides and that sum is LESS THAN the square of the longest side, you have an OBTUSE TRIANGLE.
Pre Algebra ( period 1)
Introduction to Geometry: Points, Lines, & Planes 9-1
Point:(symbol is a dot or just a letter) Location in space - no size
Ray: (arrow pointing to the right) - one endpoint and one direction- Named by its endpoint first
Line: (generally, line with arrows on both ends above 2 points on the line) - Series of points that goes on infinitely in both directions - named either direction
Line segment: (a line with no arrows on either end) - a piece of a line with 2 endpoints in either direction
Lines can be parallel (2 vertical lines) or intersecting in the same plane
Parallel lines are lines in the same plane that never meet
If they intersect at exactly 90 degrees, then they are perpendicular
If they don't intersect but are in two different planes, they are skew
Angle Relationships & Parallel Lines 9-2
angle: (angle opening to the right) two rays that meet at the same endpoint
named by either just the vertex, or 3 points on the angle in either direction with vertex in middle
adjacent angles share one ray
vertical angles are opposite each other and congruent (equal)
Complementary sum to 90 degrees and
supplementary sum to 180 degrees
acute is greater than 0 and less than 90
90 degrees is right angle
obtuse is greater than 90 but less than 180
180 is a straight angle (line)
to write the measure of an angle you write m<
Transversal: a line that intersects two other lines
Corresponding angles are formed by this transversal
These angles are on the same side of the transversal and also are both above or both below the line
When the two lines that are intersected are parallel, corresponding angles are congruent
Alternate interior angles are between the two lines (inside the two lines) and on opposite sides of the transversal (alternate sides) These angles are also congruent if the lines are parallel.
Same side interior: If angles are both inside and on the same side of the transversal, they are supplementary (sum to 180 degrees)
You can have lots of corresponding angles if you have a transversal intersecting more than 2 parallel lines - in fact they would be infinite if you kept adding another parallel line!
It's amazing that by just knowing one angle, you know all 8 angles with one transversal and two parallel lines! (I will show this in class on Tuesday!)
Point:(symbol is a dot or just a letter) Location in space - no size
Ray: (arrow pointing to the right) - one endpoint and one direction- Named by its endpoint first
Line: (generally, line with arrows on both ends above 2 points on the line) - Series of points that goes on infinitely in both directions - named either direction
Line segment: (a line with no arrows on either end) - a piece of a line with 2 endpoints in either direction
Lines can be parallel (2 vertical lines) or intersecting in the same plane
Parallel lines are lines in the same plane that never meet
If they intersect at exactly 90 degrees, then they are perpendicular
If they don't intersect but are in two different planes, they are skew
Angle Relationships & Parallel Lines 9-2
angle: (angle opening to the right) two rays that meet at the same endpoint
named by either just the vertex, or 3 points on the angle in either direction with vertex in middle
adjacent angles share one ray
vertical angles are opposite each other and congruent (equal)
Complementary sum to 90 degrees and
supplementary sum to 180 degrees
acute is greater than 0 and less than 90
90 degrees is right angle
obtuse is greater than 90 but less than 180
180 is a straight angle (line)
to write the measure of an angle you write m<
Transversal: a line that intersects two other lines
Corresponding angles are formed by this transversal
These angles are on the same side of the transversal and also are both above or both below the line
When the two lines that are intersected are parallel, corresponding angles are congruent
Alternate interior angles are between the two lines (inside the two lines) and on opposite sides of the transversal (alternate sides) These angles are also congruent if the lines are parallel.
Same side interior: If angles are both inside and on the same side of the transversal, they are supplementary (sum to 180 degrees)
You can have lots of corresponding angles if you have a transversal intersecting more than 2 parallel lines - in fact they would be infinite if you kept adding another parallel line!
It's amazing that by just knowing one angle, you know all 8 angles with one transversal and two parallel lines! (I will show this in class on Tuesday!)
Algebra (Period 4)
Add and subtract rational expressions with LIKE DENOMINATORS: 10-4
Add and subtract with UNLIKE DENOMINATORS: 10-5
When adding with LIKE DENOMINATORS,
simply add the numerators,
simplify
When subtracting with LIKE DENOMINATORS,
CHANGE THE SIGNS
OF EACH TERM IN THE NUMERATOR AFTER THE SUBTRACTION SIGN,
THEN ADD (double check!!!)
When adding or subtracting with UNLIKE DENOMINATORS,
find the common denominator (the least common multiple of all denominators),
then use equivalent fractions to restate each numerator using the new common denominator.
EXAMPLE:
2 + x
x2 - 16 x – 4
First, factor the denominators if possible
2x + x
(x + 4)(x - 4) x – 4
Find the LCM = (x + 4)(x - 4);
2x + (x + 4)(x)
(x + 4)(x - 4) (x + 4)(x - 4)
Restate both fractions with the LCM
2x + x2 + 4x
(x + 4)(x - 4)
Now add the numerators
x2 + 6x =
(x + 4)(x - 4)
Simplify and re-factor numerator, if possible
x(x + 6)
(x + 4)(x - 4)
Add and subtract with UNLIKE DENOMINATORS: 10-5
When adding with LIKE DENOMINATORS,
simply add the numerators,
simplify
When subtracting with LIKE DENOMINATORS,
CHANGE THE SIGNS
OF EACH TERM IN THE NUMERATOR AFTER THE SUBTRACTION SIGN,
THEN ADD (double check!!!)
When adding or subtracting with UNLIKE DENOMINATORS,
find the common denominator (the least common multiple of all denominators),
then use equivalent fractions to restate each numerator using the new common denominator.
EXAMPLE:
2 + x
x2 - 16 x – 4
First, factor the denominators if possible
2x + x
(x + 4)(x - 4) x – 4
Find the LCM = (x + 4)(x - 4);
2x + (x + 4)(x)
(x + 4)(x - 4) (x + 4)(x - 4)
Restate both fractions with the LCM
2x + x2 + 4x
(x + 4)(x - 4)
Now add the numerators
x2 + 6x =
(x + 4)(x - 4)
Simplify and re-factor numerator, if possible
x(x + 6)
(x + 4)(x - 4)
Subscribe to:
Posts (Atom)
