Pre Algebra (Period 2 & 4)

Change Repeating Decimals to Fractions

Step 1: set up "n = the repeating decimal
Step 2: determine how many numbers are under the bar
Step 3: use that number as a power of 10
Step 4: multiply both sides of the equation in Step 1 by that power of 10
Step 5: Rewrite the equations so that you subtract the 1st equation FROM the 2nd equation
Step 6: Solve as a 1 or 2-step equation
Step 7: simplify, if necessary

REMEMBER-- SOMETIMES YOU NEED TO GET THE DECIMAL OUT OF THE NUMERATOR--> MULTIPLY BOTH THE NUMERATOR AND THE DENOMINATOR BY A POWER OF 10


example:
change .416666... into a fraction
let n = .416666...
notice there is one 1 number that is repeating so it is only 10 to the 1st power
multiply both sides by 10
10n = 4.16666....
now subtract the 1st equation

10n = 4.16666...
-n = 0.41666...
9n = 3.75

divide both sides by 9
9n = 3.75
9 9

or 9n/9 = 3.75/9

move the decimal 2 places in both the numerator and denominator
n = 375/900
simplify
n = 5/12

LEARN THE MOST COMMON ONES BY HEARS
Need to know:
1/9 family
1/11 family
Easy trick to recall
If the repeating decimal starts repeating right after the decimal, you can easily get the fraction just by putting the same number of 9's as the length of the repeating decimal.. then simplify:
Examples:
.4444... = 4/9
.454545.... = 45/99 = 5/11
.162162162... = 162/999 = 18/111

Pre Algebra (Period 2 & 4)

Fractions & Decimals 5-2 (cont'd)

Ordering or comparing fractions:(last method)

4: Make them into decimals because DECIMALS = FRACTION posers!

(or decimals are just fraction wannabes!)
Today, we will change fractions to decimals and decimals to fractions.

How to change a fraction to a decimal
1.
Divide (ALWAYS WORKS!)

EXAMPLE: 3/4 = 3 divided by 4 = .75

If the quotient starts repeating, then put a bar over the number(s) that repeat. OR

2. Use equivalent fractions (SOMETIMES WORKS!)

Works if the denominator can be easily made into a power of 10

SAME EXAMPLE: but this time you will multiply by 25/25 to get 75/100 = .75



3. MEMORY! Some equivalencies you should just know!
EXAMPLE: 1/2 = .5


IF IT'S A MIXED NUMBER, JUST ADD THE WHOLE NUMBER AT THE END!

EXAMPLE: 8 3/4

For the fraction: 3 divided by 4 = .75

Add the whole number: 
8.75


IF THE MIXED NUMBER OR FRACTION IS NEGATIVE, SO IS THE DECIMAL!


CHANGING TERMINATING DECIMALS TO FRACTIONS:
EASY!!!
Read it, Write it, Simplify!



EXAMPLE:

Change .24 to a fraction

1) READ IT: 24 hundredths

2) WRITE IT: 24/100

3) SIMPLIFY: 24/100 = 6/25



EXAMPLE with whole number:

Change 7.24 to a fraction

The 7 is the whole number in the mixed number so you just put the 7 at the end

1) READ IT: 24 hundredths

2) WRITE IT: 24/100

3) SIMPLIFY: 24/100 = 6/25

4) 7 6/25



HOW TO CHANGE REPEATING DECIMALS TO FRACTIONS-
FIrst know some by heart.. easier..
1/3 = .333333 = .3 with a vinculum over the 3... that's a bar.

1/9 family

Notice that you use the number found in the numerator, add a decimal point in front of it and add a bar above it.

1/11 family is a little different. You multiply the numerator by 9 ( making sure 1/11 has a two digit number after the decimal... and put a bar over the two digits.

To get exact answer when doing math operations with repeating decimals--> you must make decimals into their fraction equivalents and do operations with fractions

1/7 family really special!!!

Take a look and see if you remember the pattern we talked about in class today!!

To change repeating decimals to fractions follow these steps:
1) let n = the repeating decimal
2) multiply both sides by a power of 10 equal to the number of places under the vinculum (the bar)
3) rewrite n = the repeating decimal under #2
4) subtract n on the left side and the repeating decimal on the right
5) solve as a 1-step equation
6) multiply the numerator and the denominator by a power of 10 if necessary to get the decmial out of the numerator
7) simplify

We used

= 4.66666....

let n = .4166666...
multiply by a power of 10 in this case by 10¹ or just 10
10n =4.166666....
-1n = 0.416666....
9n = 3.75

then divide both sides by 9

9n/9 = 3.75/9

n = 3.75/9

need to multiply the numerator and the denominator by 100
375/900

simplifies to 5/12

Algebra (Period 1)

Using Equations That Factor 6-9 Cont'd

Word Problems continued...
We looked at a number of word problems... translated them to an equation and solved. The following should be in your spiral notebook:

1. The product of two consecutive positive even integers is 48. Find the two integers.

Let x = the first positive even integer
let x + 1 represent the second positive even integer

x(x +1) =48
x² + x = 48
move the 48 to the left side, setting the equation to zero

x² + x - 48 = 0
NOW factor
(x + 8)(x - 6) = 0
x = -8 x = 6
Stop and re-read the equation... It says positive even integers so the only solution is
x = 6
BUT... you need to find the two integers
x = 6
and x + 2 = 8
{6,8}

2. One side of a rectangle is 4 inches longer than the other. If the sides are each increased by 2 inches, the area of the new rectangle is 60 inches squared. How long are the sides of the original rectangle?

Let w = the width of the rectangle.
let w + 4 = the length of the rectangle. ( re-read the equation to get this information!!)
Then it says "sides are each increased by 2 inches"
so we have w + 2
and (w+4) + 2, which is just w + 6
Now the area is 60 inches squared so we know A = lw
(w +2)(w + 6) = 60
FOIL or use the box method...
w² + 8w + 12 = 60
We need to use the ZERO Products Property so we need to move the 60 to the left, setting the equation equal to ZERO
w² + 8w + 12 - 60 = 0
w² + 8w - 48 = 0
FACTOR
(w + 12)(w -4) = 0
w = -12 and w = 4
BUT A rectangle can't have a negative length so -12 is NOT a solution.

If w = 4
w + 4 = 8
The original rectangle's sides are 4 inches and 8 inches.

3. The sum of the squares of two consecutive even integers is 244. Find the two integers.
let n = the first even integer
let n + 2 = the second even integer

The sum of the squares---> n² + (n +2)² = 244
n² + n² + 4n + 4 = 244
combine like terms and move 244 to the left side of the equation you get
2n² + 4n - 240 = 0
Factor out the GCF
2(n² + 2n - 120) = 0
factor
2(n + 12)(n - 10) = 0
n = -12 and n = 10
This time both of those fit the requirements. Both are solutions BUT you need two sets of pairs.
n + 2 when n = -12 is - 12 + 2 = -10
and
n + 2 when n = 10 is 10 + 2 = 12

{-12, -10} and (10, 12)

4. The sum of twice a number and the number squared is -1. What is this number?
let x = a number
2x + x² = -1
2x + x² + 1 = 0
BUT WAIT... you can't work with this trinomial in this order... put in descending order and THEN factor..
x² + 2x + 1 = 0
Wait... do you see?... It's a Trinomial squared!!
(x + 1)² = 0
x = -1

5. Five times a number decreased by 6 is equal to the number squared. What is the number?

Let x = the number
5x - 6 = x ²
Keep the x² positive... that is move the other terms to it's side... easier to factor that way..
you get
0 = x² - 5x + 6
or
x² - 5x + 6 = 0
factor...
(x -3)(x -2) = 0
x = 3 and x = 2

We did a few more from our textbook on Page 294

Algebra (Period 1)

Using Equations That Factor 6-9


WORD PROBLEMS WITH FACTORING

DRAW A PICTURE IF NECESSARY TO HELP YOU UNDERSTAND THE PROBLEM!
You are supposed to know basic Geometry Formulas!!!


1) Translate the words into an equation

2) If necessary, move all terms to the left side leaving only zero on the right side of the equation

3) Factor the left side

4) Set each factor equal to zero and solve.


5) If the problem calls for real life things like area or time, eliminate the NEGATIVE answer because certain real life things can never be negative!

The product of one more than a number and one less than the number is 8

Write a let statement
x = the number
so
(x +1)(x -1) = 8

you must multiple the two terms
x² -1 = 8
now using the zero products property have all terms on the left
x² - 9 = 0
WOW-- its the difference of two squares...
(x+3)(x-3) = 0
set each to zero
x+ 3 = 0 and x- 3 = 0
or
x = -3 and x = 3
{-3, 3}

The square of a number minus twice the number is 48. Find the number...
Let n = the number

n² - 2n = 48
n² - 2n -48 = 0 Now factor
(n -8)(n+ 6) = 0
n = 8 and n = -6
{-6, 8}


The product of two consecutive integers is 156, Find the integers...
Let x represent the 1st integer then
x + 1 is the 2nd integer

x(x +1) = 156
x² + x = 156

x² + x- 156 = 0
(x + 13)(x -12) = 0
x = -13 and x = 12

But wait you have found only one of the integers..
if x = -13 x + 1 = -13 + 1 = -12

and if x = 12 then x + 1 = 13 so you have two pairs of answers
{-13,-12,} and {12, 13}

The length of a rectangle is 5 cm greater than the width. The area is 84 cm²
Find the length and the width...
Okay, we know A = lw

Re reading the information we decide to
let w = the width
and then w +5 = the length
(w + 5)w = 84
or w (w + 5) = 84
w² + 5w = 84
w² + 5w - 84 = 0
(w + 12)(w -7) =84
so
w = -12 or w = 7
BUT a width can't be negative so we don't need to use -12
if the width is 7,
then w + 5 or 7 + 5= 12 must be the length
SOLUTION: width is 7 cm and the length is 12 cm

Math 6 Honors (Period 6 and 7)

Greatest Common Factor 5-5

If we list the factors of 30 and 42, we notice
Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30
Factors of 42: 1, 2, 3, 6, 7, 14, 21, 42

We notice that 1, 2, 3, and 6 are all COMMON factors of these two numbers. The number 6 is the greatest of these and therefore is called the
GREATEST COMMON FACTOR of the two numbers. We write
GCF(30,42) = 6

Although listing the factors of two numbers and then comparing their common factors is one way to determine the greatest common factor, using prime factorization is another easy way to find the GCF

Find GCF(54, 72)
54 = 2 ⋅ 3 ⋅ 3 ⋅ 3
72 = 2 ⋅ 2 ⋅ 2 ⋅ 3 ⋅ 3
Find the greatest power of 2 that occurs IN BOTH prime factorization. The greatest power of 2 that occurs in both is just 2 ¹
Find the greatest power of 3 that occurs IN BOTH prime factorizations. The greatest power of 3 that occurs in both is 3²
Therefore
GCF(54, 72) = 2 ⋅ 3² = 18

In class we circled the common factors and realized that
GCF(54, 72) = 2 ⋅ 3 ⋅ 3 = 18


Fin the GCF( 45, 60)
45 = 3 ⋅ 3⋅ 5
60 = 2⋅ 2⋅ 3⋅ 5
Since 2 is NOT a factor of 45-- there is NO greatest power of 2 that occurs in both prime factorizations.
The greatest power of 3 is just 3¹
and the greatest power of 5 is just 5¹
Therefore,
GCF(45,60) = 3⋅ 5 = 15

The number 1 is a common factor of any two whole numbers!! If 1 is the GCF , then the two numbers are said to be RELATIVELY PRIME. Two numbers can be relatively prime even if one or both of them are composite.

Show that 15 and 16 are relatively prime
List the factors of each number
FACTORS of 15: 1, 3, 5, 15
FACTORS of 16: 1, 2, 4, 8, 16

Since the GCF(15,16) = 1. The two numbers are relatively prime!!

Math 6 Honors (Period 6 and 7)

Prime Numbers & Composite Numbers 5-4

A prime number is one that has only two factors: 1 and the number itself, such as 2, 3, 5, 7, 11, 13...
A counting number that has more than two factors is called a composite number, such as 4, 6, 8, 9, 10...

Since one has exactly ONE factor, it is NEITHER PRIME NOR COMPOSITE!!
Zero is also NEITHER PRIME NOR COMPOSITE!!

Every counting number greater than 1 has at least one prime factor -- which may be the number itself.
You can factor a number into PRIME FACTORS by using a factor tree or the inverted division, as shown in class.

Using the inverted division, you also start with the smallest prime number that is a factor... and work down
give the prime factors of 42
2⎣42
3⎣21
7

When we write 42 as 2⋅3⋅7 this product of prime factors is called the prime factorization of 42.

Two is the only even prime number because all the other even numbers have two as a factor.

Explain how you know that each of the following numbers must be composite...
111; 111,111; 111,111,111; and so on....
Using your divisibility rules you notice that the sums of the digits are multiples of 3.

List all the possible digits that can be the last digit of a prime number that is greater than 10.
1, 3, 7, 9.

Choose any six digit number such that the last three digits are a repeat of the first three digits. For example
652,652. You will find that 7, 11, and 13 are all factors of that number... no matter what number you choose... why is that???? email me your response.

Math 6 Honors (Period 6 and 7)

Square Numbers and Square Roots 5-3

Numbers such as 1, 4, 9, 16, 25, 36, 49... are called square numbers or PERFECT SQUARES.

One of two EQUAL factors of a square is called the square root of the number. To denote a square root of a number we use a radical sign (looks like a check mark with an extension) See our textbook page 157.

Although we use a radical sign to denote cube roots, fourth roots and more, without a small number on the radical sign, we have come to call that the square root.
SQRT = stands for square root, since this blog will not let me use the proper symbol) √ is the closest to the symbol

so the SQRT of 25 is 5. Actually 5 is the principal square root. Since 5 X 5 = 25
There is another root because
(-5)(-5) = 25 but in this class we are primarily interested in the principal square root or the positive square root.

Evaluate the following:
SQRT 36 + SQRT 64 = 6 + 8 = 14
SQRT 100 = 10
Is it true that SQRT 36 + SQRT 64 = SQRT 100? No
You cannot add square roots in that manner.
However look at the following:
Evaluate
SQRT 225 = 15
(SQRT 9)(SQRT 25)= (3)(5) = 15
so
SQRT 225 = (SQRT 9)(SQRT 25)

Also notice that the SQRT 1600 = 40
But notice that SQRT 1600 = SQRT (16)(100) = 4(10) = 40

Try this:
Take an odd perfect square, such as 9. Square the largest whole number that is less than half of it. ( For 9 this would be 4). If you add this square to the original number what kind of number do you get? Try it with other odd perfect squares...

In this case, 9 + 16 = 25... hmmm... what's 25???

Math 6 Honors (Period 6 and 7)

Tests for Divisibility 5-2

It is important to learn the following divisibility rules:
A number is divisibility by:

2 ... if the ones digit of the number is even
3 ... if the sum of the digits is divisible by three ( add the digits together)
4 ... if the number formed by the last two digits is divisible by by four ( Just LOOK at the last two numbers-- DON"T ADD them!!)
5 ... if the ones digits of the number is a 5 or a 0
6 ... if the number is divisible by both 2 and 3... (or if it is even and divisible by 3)
8 ... if the number formed by the last three digits is divisible by 8. (Like FOUR, just look at the last three digits-- divide them by 8)
9 ... if the sum of the digits is divisible by 9
10 ... if the ones digits of the number is a 0.

You will not need to know the divisibility rules for 7 or 11 but they are interesting...

You can test for divisibility by 7
Let's start with a number 959
Step 1: drop the one's digit so we have 95
Step 2: Subtract twice the ones' digit ( that you dropped) in this case we dropped a 9
so we double that and subtract 18 from 95
or 95-18 = 77. If the results, in the case, 77, is divisible by 7 --- so is the original number 959.
Step 3: If the number you get is still to big.. continue the process until you can determine if your number is divisible by 7.


To test for divisibility by 11
add the alternative digits beginning with the first
so let's try the following
4,378,396
Step 1: Add the alternate digits beginning with the 1st 4 + 7+ 3 + 6 = 20
Step 2: Add alternate digits beginning with the 2nd 3 + 8 + 9 = 20

Step 3: If the difference of the sums is divisible by 11 so is the original number.
In this case, 20-20 = 0 and 0/11= 0 so
4,378,396 is divisible by 11.


A good test for divisibility by 25 would be if the last two digits represent a multiple of 25.

A perfect number is one that is the SUM of all its factors except itself. The smallest perfect number is 6, since 6 = 1 + 2+ 3
The next perfect number is 28 since
28 = 1 + 2 + 4 + 7 + 14
What is the next perfect number?

Math 6 Honors (Period 6 and 7)

Finding Factors and Multiples 5-1


You know that 60 can be written as the product of 5 and 12. 5 and 12 are called whole number factors of 6-. A number is said to be divisible by its whole numbered factors.

To find out if a smaller whole number is a factor of a larger whole number, you divide the larger number by the smaller.--- if the remainder is 0, the smaller number IS a factor of the larger number.

We set up T charts to find al the factors of numbers.
For example. Find all the factors of 24
24
1--24
2--12
3--8
4--6

We you go down the left side and back up the right you have
1, 2, 3, 4, 6, 8, 12, 24
all the factors of 24 in order!!!

A multiple of a whole number is the product of that whole number and ANY whole number. You can find the multiples of given whole numbers by multiplying that number by 0, 1, 2, 3, 4, ...and so on
The first four multiples of 7 are
0, 7, 14, 21
because 0(7) = 0 ; 1(7) = 7 ; 2(7) = 14; 3(7) = 21

If you were to ask for the first four NON-ZERO Multiples of 7
the answer would be 7, 14, 21, 28


Generally, any number is a multiple of each of its factors. That is, 21 is a multiple of 7 and it is a multiple of 3!!

Any multiple of 2 is called an EVEN number
A whole number that is NOT an even number is called an ODD number
Since 0 is a multiple of 2 .. that is 0 = 0(2) 0 is an EVEN number







The word factor is derived from the Latin word for "maker" the same root for factory and manufacture. When multiplied together factors 'make' a number.
factor X factor = product.

Algebra (Period 1)

Solving Equations By Factoring 6-8

Zero Products Property -- a "friend" you can count on..."
For any two rational numbers a and b, if ab = 0, then either a = 0 or b = 0 or both equal zero

How does this help you?

Try solving
x² + 10x + 24 = 0
Your first reaction is to subtract 24 from both sides you would get
x² + 10x = -24 BUT NOW WHAT?????
It's a quadratic ( x² term). You can't isolate x because there is another term with the x² in it. THe Zero Products Property will help you solve quadratics.

Using this property we can solve quadratic equations by factoring the equation and setting each factor to zero and solving them.

x² + 10x + 24 = 0
factor as usual-- using your rules from the other sections:
(x +6) (x +4) = 0
set each one to zero
x + 6 = 0 and x + 4 = 0
so x = -6 and x = -4

Plug them back-- and it works!!

IF you are already given it factored--just set each to zero
(5x +1)(x-7) = 0
Using the Zero Products Property we know either 5x - 1 must be equal to zero or x - 7 must equal zero... that's the only way that the product could be zero...
Set both equal to zero and solve.
if 5x + 1 = 0 then x = -1/5 ( its just a 2 step equation)
if x - 7 = 0 then x = 7

If you substitute these answers for x back in the original equation, they will both end up as 0 = 0 which is what you want. The left side equals the right side!!

In the second example, the equation was already factored.

WIth the first example, you saw you needed to first factor the equation ( if possible) and then set each factor to zero to solve.
MAKE SURE YOU MOVE EVERYTHING TO ONE SIDE OF THE EQUATION. YOU MUST HAVE ZERO ON ONE SIDE OF THE EQUATION TO USE THE ZERO PRODUCTS PROPERTY

x² = 16

First move the 16 to the left side
x² -16 = 0
Now... you have the difference of two squares
(x + 4)(x-4) = 0
Set each factor to zero and solve
x + 4 = 0 and x - 4 = 0
so x = -4 and x = 4

Why do you have to have zero on one side ?

NEW ALGEBRA TERM
root-- any solution that turns the equation into the value of zero is called a root of the polynomial, or a ZERO of the polynomial because when you are graphing the polynomial, the y value is zero at this point.
So if the directions say " FInd the roots of..." it just means get zero on one side of th quation, factor, set each factor to zero and solve.

Summary of solving quadratics
solutions = roots= eros of the polynomials

1. Get zero on one side of the equation
2. Factor
3. Set each piece ( factor) equal to zero
4. Solve as one or two step equations
5. Check by substituting in the ORIGINAL equation

Math 6 Honors (Period 6 and 7)

Congruent Figures 4-7

Two figures are congruent if they have the same size and same shape. If we could lift on of the figures and place it directly on top of the other... all three vertices would match up with the others. In the example in our book ( page 132) we have two congruent triangles ∆ABC and ∆XYZ A would fall on X, B would fall on Y and C would fall on Z. These matching vertices are called corresponding vertices. Angles at corresponding vertices are corresponding angles and the sides joining corresponding vertices are corresponding sides.
The book states Corresponding angles of congruent figures are congruent.
and
Corresponding sides of congruent figures are congruent.
In class we discussed how to abbreviate the above -- when dealing with triangles.
CPCTC
Corresponding PARTS of congruent triangles are congruent!!

When we name two congruent figures we list corresponding vertices in the same order.

∆ABC ≅ ∆XYZ or ∆CAB ≅ ∆ZXY or ∆BCA ≅ ∆YZX

we know that
∠A ≅ ∠X and ∠B ≅ ∠Y and ∠C ≅ ∠Z
and the segments ( which are denoted with a line (but w/o arrows)above each of the two letters
AB ≅ XY and BC ≅ YZ and CA ≅ ZX

If two figures are congruent, we can make the coincide -- occupy the same place-- by using one or more of the following basic rigid motions:

Translation or slide
Rotation
Reflection or flip or mirror
Check the book on page 133 for good examples of these three rigid motions... I like to think of Tetris moves!!

Pre Algebra (Period 2 & 4)

Comparing and Ordering Fractions

 5-1
HOW TO FIND THE LEAST COMMON MULTIPLE:

LCM = Smallest number that your numbers can go into.
Just like GCF, let's look at the letters backwards to understand it!

Multiple = each number given in the problem must go into this number (example for multiples of 2: 2, 4, 6, 8.... multiples of 3: 3, 6, 9, 12...     multiples of 5: 5, 10, 15, 20...)

Common = must be a number that ALL THE NUMBERS go into 

Least = must be the SMALLEST number that ALL THE NUMBERS go into

There are the same ways to find it as the GCF:

1) List multiples of each number and circle the smallest one that is common to all the numbers
---> most of the time takes WAY TOO long!!
2) Circle every factor in the prime factorizations of each number that is different and multiply

3) List the EXPANDED FORM prime factorizations in a table and bring down ONE OF EACH COLUMN. Then multiply. (or you can do this with exponential form but you need to bring down the HIGHEST POWER of each column).
4) Using the BOX method from class create an L from the left side and the bottom row of relatively prime factors. It is their product.

THE DIFFERENCE BETWEEN GCF AND LCM:
For the GCF, you need the LEAST POWER of only the COMMON FACTORS.
For the LCM, you need the GREATEST POWER of EVERY FACTOR.

WHY DO WE NEED EVERY FACTOR THIS TIME?

Because it's a multiple of all your numbers!

Multiples start with each number, so all the factors that make up each number have to be in this common multiple of all the numbers.

For example, say we're finding the LCM of 12 and 15, that multiple must be a multiple of 12: 12, 24, etc.
AND 15: 15, 30, etc.
So the COMMON multiple must include 12 (2x2x3) AND 15 (3x5).
The LCM must have two 2s and one 3 or 12 won't go into it.
It must also have that same 3 that 12 needs and one 5 or it won't be a multiple of 15.
EXAMPLE: 54 and 36

LIST MULTIPLES OF EACH NUMBER UNTIL YOU SEE ONE THAT BOTH OF THEM GO INTO (the LEAST MULTIPLE that is COMMON to both)
36 = 36, 72, 108, 144 ...
54 = 54, 108
This more difficult than the listing method for GCF for 2 reasons: You don't know where to stop as you least the first number...multiples go on forever! and the numbers get big very fast because they're MULTIPLES, not factors!
PRIME FACTORIZATION METHOD
54 = 2x3x3x3

36 = 2x2x3x3
The LCM will be one of each factor of each number (but don't double count a factor that is common to both numbers...once you have it in your LCM, check it off in the other number if it's common...you don't need that factor again)

LCM = 2x2x3x3x3 = 108
Let's look at this calculation and think about why it works:
Why does it need TWO 2s?
Although 54 only needs ONE, 36 needs TWO or 36 won't go into the LCM.
Try it putting in only ONE 2: 2x3x3x3 = 54. Does 36 go into 54? NO
Why does the LCM need THREE 3s? Although 36 needs only TWO 3s, 54 needs THREE. Try it putting in only TWO: 2x2x3x3 = 36.
36 is TOO SMALL to even be a MULTIPLE of 54!
WHAT IF BY MISTAKE YOU "DOUBLE" UP FACTORS AND USE ALL OF THEM?
For the GCF, you would have got a number way too big to be a factor of the numbers. For the LCM, you will STILL GET A COMMON MULTIPLE! But it WON'T BE THE LEAST!
In fact if you double up, generally you'll get a really big number and the bigger the number is, the harder it is to use.
For example, if you use all the factors of both 36 and 54, you're just multiplying 36 x 54 = ???? 
1944!!!! 
Would you rather use 108 or 1944???


LCM WITH A COLUMN APPROACH:
You place the prime factorization for each number in columns like we did for the GCF, matching the factors in each column.
If a factor doesn't match, it gets a separate column.
YOU'LL JUST TAKE ONE OF EACH COLUMN AND MULTIPLY!

                                    SAME EXAMPLE: 54, 36
                                     
36 = 2 x 2 x 3 x 3
                                     
54 = 2 x       3 x 3 x 3
                                   
LCM = 2 x 2 x 3 x 3 x 3
= 108
You can do it in exponential form, too.

If you do it in exponential form, you take the HIGHEST POWER of each column!

Works with variables the exact same way!
Take the HIGHEST power of EVERY variable (not just the ones that are common like the GCF)
YOU CAN FIND BOTH THE GCF AND THE LCM IN THIS SAME COLUMN FORMAT!

CHECKING THE LCM TO MAKE SURE IT WORKS:

Ordering or comparing fractions:(last method)

4: Make them into decimals because DECIMALS = FRACTION posers!

(or decimals are just fraction wannabes!)
Today, we will change fractions to decimals and decimals to fractions.

How to change a fraction to a decimal
1.
Divide (ALWAYS WORKS!)

EXAMPLE: 3/4 = 3 divided by 4 = .75

If the quotient starts repeating, then put a bar over the number(s) that repeat. OR

2. Use equivalent fractions (SOMETIMES WORKS!)

Works if the denominator can be easily made into a power of 10

SAME EXAMPLE: but this time you will multiply by 25/25 to get 75/100 = .75



3. MEMORY! Some equivalencies you should just know!
EXAMPLE: 1/2 = .5


IF IT'S A MIXED NUMBER, JUST ADD THE WHOLE NUMBER AT THE END!

EXAMPLE: 8 3/4

For the fraction: 3 divided by 4 = .75

Add the whole number: 
8.75


IF THE MIXED NUMBER OR FRACTION IS NEGATIVE, SO IS THE DECIMAL!


CHANGING TERMINATING DECIMALS TO FRACTIONS:
EASY!!!
Read it, Write it, Simplify!



EXAMPLE:

Change .24 to a fraction

1) READ IT: 24 hundredths

2) WRITE IT: 24/100

3) SIMPLIFY: 24/100 = 6/25



EXAMPLE with whole number:

Change 7.24 to a fraction

The 7 is the whole number in the mixed number so you just put the 7 at the end

1) READ IT: 24 hundredths

2) WRITE IT: 24/100

3) SIMPLIFY: 24/100 = 6/25

4) 7 6/25



HOW TO CHANGE REPEATING DECIMALS TO FRACTIONS---> we will do that 2nd semester

Circles 4-6 continued

We continued our study of circles by examining irregular shapes and determined their perimeters.
To see each of the irregular shapes turned to page 131. The numbers we used in class were all different that those of 24-27, but use the shapes to help solve the following:

The circles in the diagrams are parts of circles and the angles are right angles. We found the perimeter of each figure.
The first figure was a semicircle ( see #24 with a diameter of 4).
We noticed that we needed to start with
C =∏d but then we only need half of that
so
∏d/2 or 4∏/2 = 2∏
Using ∏≈ 3.14
we found
≈3.14(2) = 6.26
BUT.. that only was the upper part we needed to add the diameter of 4 to make sure we had all we needed in our perimeter.
6.28 + 4 = 10.28 units... but then we needed to round to 3 digits-- according to our textbook so
10.3 units would be a good approximation for the first perimeter.

Our 2nd irregular shape was a quarter of a circle with a radius of 6
Again, look to our textbook, page 131 # 25 for the shape. Use 6 as the radius.
This time
C = 2∏r
BUT... we only need 1/4 so
(2⋅∏⋅6)/4 = 12∏/4 = 3∏
≈ 3.14(3) = 9.42
BUT... wait.. we aren't finished... we have to sides of this quarter circle that we need to include in our perimeter.
so the perimeter is approximately 9.42 + 6 + 6 = 9.42 + 12
≈ 21.42 which round to ≈ 21.4 units


The next irregular shape looks like something from Griffith Park observatory. Make sure to use the diagram for # 26 but use a radius of 2 as we did in class.

C = 2∏r
but you notice we only need half of the full circle so
2∏r/2 or just ∏r
Now substitute int he radius-- which is also 2
2∏ ≈ 2(3.14) = 6.28
But... we still need the bottom perimeter
so this shape ≈ 6.28 + 2 + 4 + 2 or 6.28 + 8
≈14.28
≈14.3 units

Period 7 said the next shape ( # 27 in our textbook) looks like a bandaid... What do you think?
(At least.. with the way I drew it.. having a radius of 10)
Again we need the formula
C = 2∏r
This time we realized we had two semicircles.. but that is one whole circles so we kept the formula and substituted in our radius of 10
C =2(10)∏
=20∏
≈20(3.14)= 62.8
Then we added the two sides of 10 and found the perimeter to be approximately
62.8 + 10 + 10
≈82.8 units


I described the last shape to be a teardrop. Make sure to check #28 in our textbook. I actually used the same radius as the book so that drawing is exactly what we did.

C =2∏r
C =2⋅6⋅∏ = 12∏
But... we only need 3/4 of the circle so what is 3/4 of 12... in class everyone knew it was 9 so
3/4 of 12∏ is 9∏
≈9(3.14) = 28.26
But then we need to make sure we include the two sides of 6 each
≈28.26 + 6 + 6 = 28.26 + 12
≈40.26 units
and rounding to three digits
≈ 40.3 units.

Math 6 Honors (Period 6 and 7)

Circles 4-6

A circle is the set of all points in a plan at a given distance from a given point O (called the center).
A segment joining the center to a point on the circle is called a radius ( plural: radii) of the circle. All radii of a given circle have the same length and the length is called the radius of the circle.

A segment joining two points on a circle is called a chord... and a chord passing through the center is a diameter of the circle. the ends of the diameter divide the circle into two semicircles. The length of a diameter is called the diameter of the circle.
Two radii equal one diameter-- a fact we will use in the formulas below

The perimeter of a circle is called the circumference and the quotient

circumference ÷ diameter is the same for all circles--> regardless of size
This quotient is denoted by the Greek letter ∏ ( pronounced "pie")
No decimal gives ∏ exactly
No fraction gives ∏ exactly, either

A fairly good approximation is either 3.14 or 22/7

If we denote the circumference by C and the diameter by d we can write

C ÷ d = ∏
This formula can be put into several useful forms.

Let C = circumference d = diameter and r = radius
Then:

C = ∏d
d = C/∏

C = 2∏r
and
r = C/(2∏)

We tried a few examples.
Using ∏≈ 3.14 and rounding to three digits, as described by our textbook.
The diameter of a circle is 6 cm. Find the circumference.
WE are given d and are asked to find C.
WE use the formula
C = ∏d
C ≈ 3.14(6) = 18.84
C ≈ 18.8
So, the circumference is approximately 18.8 cm


The circumference of a circle is 20 feet. Find the radius.
To find the radius, use the formula
r = C/(2∏)
r = 20/2∏
Simplify first
r = 10/∏
r ≈ 10/3.14
r ≈ 3.1847
Since the third digit from the left is in the hundredths' place, round to the nearest hundredth.
r ≈ 3.18
The radius is approximately 3.18 feet

A polygon is inscribed in a circle if all of its vertices are on the circle. Check on the diagram in our textbook on page 129-- we added that to our notes as well.

Three noncollinear points (not on a line) determine one and only one circle that passes through the three given points.