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.