Math 6A ( Periods 1 & 2)

Points, Lines, Planes 4-1
We can describe but CANNOT DEFINE point, line or plane in Geometry

We use a single small dot to represent a point and in class we labeled with a P and we called it Point P
A straight line in Geometry is usually just called a line...
Two points DETERMINE exactly ONE LINE

we connected Point P with Point Q and created Line PQ
We placed this type of arrow ↔ over PQ to show a line
↔
PQ

that would represent the line PQ but we found we could write
↔
QP
and mean the SAME line!!

You can name ANY line with ANY TWO points that fall on that line!! USE only TWO points to name a line!!


Three or more points on the same line are called collinear. Notice the word "line" in collinear.
collinear

Points NOT on a same line are called noncollinear.

We have a RAY if we have an endpoint and it extends through other points. We name the rame by Naming the endpoint FIRST
→
PQ is RAY PQ and it begins at P and goes through Q. It is NOT the SAME as

→
QP which is Ray QP, which begins at Q and goes through P

Segments are parts of lines with TWO ENDPOINTS.
⎯
PQ is a segment with endpoints Point P and POint Q

We then looked at the drawing from page 105 and the class named all of the names for the line in the drawing, all of the rays that existed in the figure as well as the segments. We found that there were just 3 collinear points: A, X, B but that we could name 3 sets of non collinear points
X, Y, B and A, X, Y, AND A, B, Y

Three non-collinear points determine a flat surface called a plane.
We name a Plane by using three of its non collinear points!! Plane ABC was out example.

Lines in the same plane that do not intersect are PARALLEL lines. Two segments or rays are parallel if they are parts of parallel lines.
↔
AB is parallel to

↔
CD

may be written
↔ ↔
ABllCD
using two straight lines to indicate parallel

Parallel lines DO NOT intersect.

Intersecting lines intersect in a single point!!



Planes that do not intersect are called parallel planes... we looked around the room and found examples of parts of planes.. noticing which ones were parallel!! (the floor and ceiling were a great example)
Then we drew the box from PAge 106 and identified parallel segments and lines from that box.

Two non parallel lines that do not intersect are called SKEW LINES.

Algebra Honors ( Periods 6 & 7)

Although the textbook uses charts and tables for these word problems, I think they work easily without the charts...

John has 15 coins -- all dimes and quarters , worth $2.55 How many dimes and quarters does he have?
let d = the number of dimes and let q = the number of quarters
we know d + q = 15 and we know 10d + 25q = 255
using a system of equations and the substitution method
since d + q = 15 we know d = 15- q
10d + 25q - 255
10(15-q) + 25 q = 255
150 -10q + 25q = 255
150 + 15q = 255
15q = 105
q = 7
He has 7 quarters and 8 dimes


Ann and Betty together have $ 60. Ann has $9 more than twice Betty's amount. How much money does each have?
Let a = the amount Ann has and let b = the amount Betty has.
we know
a + b = 60
and we know
a= 2b + 9
so using a + b = 60
(2b+9) + b = 60
3b = 51
b = 17
Betty has $17 and Ann has (60-17) = $43

Henrick/Matt invested $8000 in stocks and bonds. the stocks pay 4% interest and the bonds pay 7% interest . The annual interest from the stocks and bonds is $500.
How much is invested in bonds?
let s = the amount invested in stocks
let b = amount invested in bonds
s + b = 8000
0.04s + 0.07b = 500
clear the decimals
4s + 7b = 50000
but we know s + b = 8000 or s = 8000 -b
4(8000 -b) + 7b = 50000
32000 - 4b + 7b = 50000
3b = 18000
b = 6000
He invested $6000 in bonds.

Math 6A (Periods 1 & 2)

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

How much simple interest would you owe if you borrowed $640 for 3 years at atan annual rate of 15%?

I = Prt
I = (640)(.15)(3)
I = (1920)(.15)
I = 288
$288 in interest

Sarah borrowed $3650 for 4 years at 16% How much must she repay.
Remember she will oe the amount she borrowed as well as the the interest.
I = Prt
I = (3650)(.16)(4)

I =14600(.16)
I = 2336.
Principal + Interest = 3650 + 2336
Sarah owes $5986.


$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

(We did not get to this yet... but I thought I would post it... read this..it is interesting to see the difference. We will go over this after STAR testing)
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

Algebra Honors ( Periods 6 & 7)

Solving Systems of Linear Equations
The Graphing Method 9-1
Two or more equations in the same variables form a system of equations. The solution of a system of two equations in two variables is a pair of values x and y that satisfies each equation in the system. The point corresponding to the ordered pair (x, y) must lie on the graph of both equations.
Solve the system by graphing
2x - y = 8
x + y = 1

Solution:
Graph both 2x - 7 = 8 and x + y = 1 in the same coordinate plane.
We did this in class by transforming both equations to slope-intercept form (y = mx +b)
and then graphed them. We noticed that the only point on BOTH lines is the intersection point ( 3, -2)
The only solution of both equations is (3, -2).
You can check that ( 3, -2) is a solution fof the system by substituting x = 3 and y = -2 in BOTH eqquations.

Solve the system by graphing
x - 2y = -6
x -2y = 2

When you graph the equations in the same coordinate plane, you see that the lines have the same slope but different y-intercepts. The graphs are parallel lines. SInce the lines do not intersect, there is no point that represents a solution of both equations.
Therefore, the system has NO SOLUTION.

Solve the system by graphing
2x + 3y = 6
4x + 6y = 12

When you graph the equations in the same coordinate plane, you see that the graphs coincide. The equations are equivalent. Every point on the line represents a solution of BOTH equations.
Therefore, the system has infinitely many solutions.

The Graphing Method in review:
To solve a system of linear equations in two variables, draw the graph of each linear equation in the same coordinate plane...
--> if the lines interset there is only one solutions, namely the intersection point.
--> if the lines are parallel, there is no solution
--> if the lines coincide, there are infinitely many solutions.

The Substitution Method 9-2

There are several ways to solve a system of equations, In the substitution method we use either equation to solve for one variable in terms of the other.
Solve
x + y = 15
4x + 3y = 38

Solve the first equation for y
x + y = 15
becomes
y = -x + 15
Substitute this expression for y in the other equation, and solve for x
4x + 3y = 38
4x + 3(-x+15) = 38
4x -3x + 45 = 38
x + 45 = 38
x = -7

Substitute the value of x in the equation in your first step and solve for y
y = -x + 15
y = -(-7) + 15
y = +7 +15
y = 22
CHeck x = -7 and y = 22 on BOTH equations
x + y = 15
(Here let ?=? represent having a ? above the equals sign)

-7 + 22 ?=? 15
15 = 15
and
4x + 3y = 38
4(-7) + 3(22) ?=? 38
-28 + 66 ?=? 38
38 = 38

It checks for both equations so the solution is (-7, 22)

Solve
2x - 3y = 4
x + 4y = -9

Using the 2nd equation is easier to manipulate so solve for x since x has a coefficient of 1
x = -4y - 9
substitute this expression for x in the other equation and solve for y
2x - 3y = 4
2(-4y-9) - 3y = 4
-8y -18 -3y = 4
-11y = 22
y = -2
Substitute the value of y in the equation in step 1 and solve for x
x = -4y -9
x = -4(-2) -9
x = 8 -9 = -1
Check both equations... and you discover that the solution is ( -1, -2)

The substitution method is most convenient to use when the coefficient of one of the variables is 1 or -1.

The Substitution Method in review:
To solve a system of linear equations in two variables:
--> Solve one equation for one of the variables
--> Substitute this expression in the other equation and solve fore the other variable.
--> Substitute this value n the equation in step 1 and solve
--> Check the alues in BOTH equations.


Solve by the substitution method
2x -8y = 6
x - 4y = 8

x = 4y + 8

2x-8y = 6
2(4y+8) - 8y = 6
8y + 16 -8y = 6
16= 6 WAIT that's FALSE

The false statement indicates that there is NO ordered pair (x, y) that satisfies BOTH equations. If you had graphed the equations you would see that these lines are actually parallel.

Solve by substitution method
y/2 = 2 -x
6x + 3y = 12

The first equation is easy to change to y = 4 - 2x by multiplying both sides by 2 to solve for y

6x + 3y = 12
6x + 3(4-2x) = 12
6x + 12 - 6x = 12
12 = 12 WAIT THat's TRUE... always
Every ordered pair (x, y) that satisfies one of the equations aso satisfies the other. IF you graph these two equations you will see that the lines coincide

Therefore, the system has infinitely many solutions.