Algebra Period 3

Review of how to graph a quadratic

Find the vertex ( x = -b/2a, then plug in to the equation to find the y coordinate)

Find the axis (line) of symmetry. Just use the x value of the vertex that is
x = -b/2a

Draw the line of symmetry using a dotted or dashed line

Find 2 more points in an x y table (go either left or right of the vertex and use 2 points close to the line of symmetry for the x, then plug into the equation to find the y values)

Find the 2 shadow (mirror) points by counting over from the axis at the exact same y value

Draw a U (not a V) through the points. Extend it past and put arrows at the end
Remember to label the x and y axes and put arrows on the axes.

Function vs. Relations: Functions are special relations where there is a unique x for each y
Therefore, there will never be 2 x’s that will repeat and if you use a vertical line on the graph, it will only hit one point on the graph.

Domain vs. Range: the domain for any quadratic function is ALL REAL NUMBERS
The range depends on where the vertex is and whether the quadratic is a smile or a frown.
Generally, it will be of the form:

y such that y is either greater than or equal to (≥ ) or less than or equal to (≤) the y value of the vertex. {y l y≥ n} or {y l y ≤ n}

f(x) is just a different way of saying y. What is better about it? Without seeing the work before the solution, you can actually tell the x value as well as the y value in the solution.

For example, if the solution is f(-3) = 12 you know that the point is (-3, 12)
Compare that to the solution y = 12. For that solution, you would not know the x value unless you look back in the problem.

Pre Algebra Periods 1, 2, & 4

Chapter 9-3: Polygons
Polygon - closed figure with at least 3 sides- no curves - no overlapping lines
Named by the number of sides (called laterals) or their number of angles
Polygon literally means MANY (poly) ANGLES (gon)

Triangles - literally means 3 angles (and sides) - Each has 2 names
By angles: Acute, Obtuse, Right
By sides: Scalene, Isosceles, Equilateral
Triangles have 180 degrees

Quadrilaterals - literally means 4 laterals (sides) and angles - 360 degrees (2 triangles!)
Trapezoids - Only 1 set of parallel lines
Kite - no parallel lines - 2 adjacent sides are congruent
Parallelogram - Oppositie sides parallel and congruent
Types of parallelograms:
Rhombus - all sides congruent
Rectangle - 4 right angles
Square - Rectangle with all sides congruent (so it's also a rhombus)
A square is a regular rectangle

Regular polygons - all sides and angles congruent
2 famous ones - equilateral triangle and squares
ALL REGULAR POLYGONS ARE SIMILAR!
So to find the PERIMETER of a regular figure, you just need to know one side and multiply by the total number of sides.

Chapter 9-5: Congruent polygons
If two polygons are congruent, they all their corresponding parts are congruent
We use tick marks to denote congruent sides and angles.
There are 3 ways to show that triangles are congruent:
SSS = side, side, side = all corresponding sides are congruent
SAS = side, angle, side = 2 sides with the angle that is between them
ASA = angle, side, angle = 2 angles with the side between them
THERE IS NO SUCH THIS AS ASS OR AAA!!!!!!!!
If all angles are congruent, then they may be congruent or SIMILAR

Algebra Period 3 (Friday)

Quadratic Equations: 13-1

But first… Finding the x intercepts of a parabola:
x intercepts = roots = solutions = zeros of a quadratic
You can find these one of two ways:
1) Read them from the graph

2) Set y or f(x) = 0 and then solve

Where the graph crosses the x axis is/are the x intercepts.

(Remember, y = 0 here!)
The x intercepts are the two solutions or roots of the quadratic.
When we factored in Chapter 6 and set each piece equal to zero,
We were finding the x value when y was zero.
That means we were finding these two roots!

You can find these roots (solutions, x intercepts, zeros) by several different methods.
You already know the following 2 methods:

1) Graphing and see where the graph crosses the x axis
2) Factoring, then using the zero products property to solve for x for each factor

ANOTHER METHOD:
Chapter 13-2 Solving Quadratics Using Square Roots -
You did this in Ch 11 for Pythagorean Theorem!
If there is no middle x term, it's easiest to just square root both sides to solve!

BUT THE DIFFERENCE FROM PYTHAGOREAN--NOT LOOKING FOR JUST THE PRINCIPAL SQUARE ROOT ANY MORE.

YOU NEED THE + OR - SYMBOL!!

EXAMPLE:
3x² = 18
divide both sides by 3 and get: x² = 6
square root each side and get x = + or - SQRT of 6

HARDER EXAMPLE:
(x - 5)² = 9
SQRT each side and get: x - 5 = + or - 3
+ 5 to both sides: x = 5 + 3 or x = 5 - 3
So, the 2 roots are x = 8 or x = 2

HARDEST EXAMPLE:
(x + 2)² = 7
SQRT each side and get: x + 2 = + or - SQRT of 7
-2 to both sides: x = -2+ SQRT 7 or x = -2 - SQRT 7

Algebra Period 3 (Tuesday)

Chapter 12-4 Quadratic functions
A QUADRATIC FUNCTION is not y = mx + b

(which is a LINEAR function),
but instead is
y = ax² + bx + c
OR
f(x) = ax² + bx + c
where a, b, and c are all real numbers and
a cannot be equal to zero because
it must have a variable that is squared ( degree of 2)
Quadratics have a squared term, so they have TWO possible solutions also called roots You already saw this in Chapter 6 when you factored the trinomial and used zero products prop.

If the domain is all real numbers, then you will have a PARABOLA which looks like
a smile when the a coefficient is positive or
looks like a frown when the a coefficient is negative.

Graphing quadratics:
You can graph quadratics exactly the way you graphed lines
by plugging in your choice of an x value and using the equation to find your y value.

Because it's a U shape, you should graph 5 points as follows:

Point 1) the vertex - the minimum value of the smile or the maximum value of the frown

The x value of the VERTEX = -b/2a
We get the values for a and b from the actual equation

f(x) = ax² + bx + c

Plug that into the equation and then find the y value of the vertex

Next, draw the AXIS OF SYMMETRY : x = -b/2a
a line through the vertex parallel to the y axis

Point 2) Pick an x value to the right or left of the axis and find its y by plugging into the equation.

Point 3) Graph its mirror image on the other side of the axis of symmetry by counting from axis of symmetry

Points 4 and 5) Repeat point 2 and 3 directions with another point even farther from the vertex

JOIN YOUR 5 POINTS IN A "U" SHAPE AND EXTEND LINES WITH ARROWS ON END

Parabolas that are functions have domains that are ALL REAL NUMBERS
Their ranges depend on where the vertex is and also if the a coefficient is positive or negative

EXAMPLE: f(x) = -3x² (or y = -3x²)
the a coefficient is negative so it is a frown face
the x value of the vertex (maximum) is -b/2a or 0/2(-3) = 0

the y value of the vertex is 0
So the vertex is (0, 0)
The domain is all real numbers.
The range is y is less than or equal to zero

To graph this function:
1) graph vertex (0, 0)
Draw dotted line x = 0 (actually this is the y axis!)
2) Pick x value to the right of axis of symmetry, say x = 1
Plug it in the equation: y = -3(1) = -3
Plot (1, -3)
3) count steps from axis of symmetry and place another point to the LEFT of axis in same place
(-1, -3)
4) Pick another x value to the right, say x = 2
Plug it in the equation to find y: y = -3(2²) = -12
(2, -12)
5) count steps from axis of symmetry and place another point to the LEFT of axis in same place
(-2, -12)

JOIN YOUR 5 POINTS IN A "U" SHAPE AND EXTEND LINES WITH ARROWS ON END

Pre Algebra Periods 1, 2, & 4

CHAPTER 9
GEOMETRY
I will go over in class the symbols of each of the following
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: (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 degrees and less than 90 degrees
90 degrees is right angle
obtuse is greater than 90 degrees but less than 180 degrees
180 degrees is a straight angle (line)
to write the measure of an angle you write m<

Algebra Period 3

Chapter 12-1 and 12-2
RELATIONS: Set of ordered pairs where the x values are the DOMAIN and the y values are the RANGE.

FUNCTIONS: Relations where there is just one y value for each x value IN OTHER WORDS----YOU CAN'T HAVE TWO y VALUES for the SAME x value!!!
If you see x repeated twice, it's still a relation, but it's not a function.
In the real world, I have a good example...pizza prices.
You can't have two different prices for the same size cheese pizza.
If you charge $10 and $12 on the same day for the same pizza, you don't have a function.
But, you certainly can charge $10 for a cheese pizza and $12 for a pepperoni pizza.

VERTICAL LINE TEST: When you graph a function, if you draw a vertical line anywhere on the graph, that line will only intersect the function at one point!!!!
If it intersects at 2 or more, it's a relation, but not a function.
So a horizontal line function, y = 4, is a function, but a vertical line function, x = 4 is not.

Any line, y = mx + b, is a function.

INPUTS: x values
OUTPUTS: y values

f(x) means the value of the function at the given x value
You can think of f(x) as the y value

Finding the value of a function: Plug it in, plug it in!
f(x) = 2x + 7
Find f(3)
f(3) = 2(3) + 7 = 13
The function notation gives you more information than using y
If I tell you y = 13 you have no idea what the x value was at that point
But if I tell you f(3) = 13, you know the entire coordinate (3, 13)

Domain of a function = all possible x values (inputs) that keep the solution real
Range of a function = all possible y values (outputs) that result from the domain

EXAMPLE:
f(x) = x + 10 has the domain of all real numbers and the same range because every value will keep the answer f(x) a real number

EXAMPLE:
f(x) = x² has the domain again of all real numbers, BUT the range is greater than or = to zero
because when a number is squared it will never be negative! So f(x) will always be 0 or positive

EXAMPLE:
f(x) = absolute value of x has the domain of all real numbers, but again the range will be greater than or equal to zero because absolute value will never be negative

EXAMPLE:
f(x) = 1/x has a domain of all real numbers EXCEPT FOR ZERO because it would be undefined if zero was in the denominator. The range is all real numbers except zero as well.
This function will approach both axes but never intersect with them.
The axes are called asymptotes which means that they will get very close but never reach them

EXAMPLE:

f(x) = x - 10
thisthisx + 3

Domain is all real numbers EXCEPT -3 because -3 will turn the denominator into zero (undefined)
What is the range?

Math 6 Honors Periods 6 & 7 (Wednesday- Friday)

Addition and Subtraction of Mixed Numbers 7-2

Although there are two methods discussed in our textbook, I would like students to concentrate on using the 'Stacking Method' rather than changing mixed numbers into improper fractions.

In the stacking method, we work separately with the fractional and whole- number parts of the given mixed numbers. Sometimes, we must change the form of a mixed number before we can subtract the fractional part, as with the following example
9 3/7 - 4 5/7 re write

69 3/7
-4 5/7
______


9 3/7 = 8 + 7/7 + 3/7 so re write the first line as 8 10/7

r8 10/7
-4 5/7
______ and now you can easily perform the subtraction
4 5/7


If the fractional parts of the given mixed numbers have different denominators, we must find equivalent mixed numbers whose fractional parts have the same denominator-- usually the LCD. You can use the box method for finding the LCM to find the LCD here!!

Algebra Period 3 (Wednesday)

CHAPTER 11-9: EQUATIONS WITH RADICALS
When you have an equation where the variable is under the SQRT sign,
simply square both sides to solve for the variable.
Remember to isolate the variable on one side before you square it,
unless there is a number under the SQRT sign with the variable.

Example #1 on p. 521
SQRT(x) = 5
Square both sides and you'll get x = 25

Example # 9 is much harder
3 + SQRT(x - 1) = 5
move the 3 to other side first SQRT(x - 1) = 5 - 3
Now square both sides x - 1 = 2²
Now add 1 to both sides: x = 4 + 1 x = 5

Look at Examples #15 & 16 ---- In both cases, there is no possible value for x because the square root of a number CAN NEVER BE NEGATIVE IN THE REAL NUMBER SYSTEM!

Try Example #17 yourself, and see what happens (you should also end up with no value, but why?)

Math 6 Honors Periods 6 & 7 (Tuesday )

Addition and Subtraction of Fractions 7-1

The properties of addition and subtraction of whole numbers also apply to fractions.

When adding fractions with the same denominator-- simply add the numerators and place them over the denominator.

That is 5/9 + 2/9 = (5+2)/9 = 7/9

In order to add two fractions with different denominators, find two fractions with a common denominator, equivalent to the given fractions. The most convenient denominator is the least common multiple of the denominator which is called the least common denominator (LCD)

When solving equations such as

n + 1/2 = 5/6 Use the same method taught for intetergs. DO unto one side that what you do to the other

so if n + 1/2 = 5/6 You must subtract 1/2 from both sides!!
n = 5/6 - 1/2

Stack your fractions. change to a LCD and subtract
n = 1/3

Algebra Period 3 (Tuesday)

CHAPTER 11-7: DISTANCE FORMULA
(based on the Pythagorean Theorem): see p. 513 in book
Reviewed from last week:

The distance between any two points on the coordinate plane (x y plane)
The distance is the hypotenuse of a right triangle that you can draw using any two points on the coordinate plane
The formula is:

distance = SQRT[( difference of the two x's)² + (difference of the two y's)²]
The difference between the 2x’s is the length of the leg parallel to the y axis and the difference between the 2y’s is the length of the leg parallel to the x axis

EXAMPLE: What is the distance between (3, -10) and (-7, -2)?
d = SQRT[(3 - -7)² + (-10 - -2)²]
d = SQRT[10² +( -8²)]
d = SQRT(164)
Simplifying: 2rad41

CHAPTER 11-8: USING THE PYTHAGOREAN THEOREM - WORD PROBLEMS
There are many real life examples where you can use the Pythagorean Theorem to find a length.
EXAMPLE: HOW HIGH A 10 FOOT LADDER REACHES ON A HOUSE
A 10 ft ladder is placed on a house 5 ft away from the base of the house.
Find how high up the house the ladder reaches.
The ladder makes a right triangle with the ground being one leg, the house the other, and the ladder is the hypotenuse ( see drawing in #1 on p. 515) You need to find the distance on the house, so you're finding one leg.

You're flying your kite for the kite project and you want to know how long the kite string must be so that it can reach a height of 13 ft in the air if you're standing 9 feet away from where the kite is in the air. The string represents the hypotenuse. You know one leg is the height in the air (13 ft) and the other leg is how far on the ground you are standing away from where the kite is flying (9 ft)
You need to find the hypotenuse.

Pre Algebra Periods 1, 2, & 4

REVIEW REAL LIFE PERCENTS
DISCOUNT = sales price This is similar to mark up , but this time you will SUBTRACT (not add)
A jacket that was $100 is 35% off. (.35)($100) = $35. Sales price = $100 - $35 = $65

DISCOUNT = % OF DECREASE
EX: original price = $200 and sales price = $50. What is the DISCOUNT %? (% of decrease?)
Decrease or discount = $200 - $50 = $150
$150 decrease/$200 original = 3/4 = 75% Discount %

MARKUP = how much you want to make on selling something Then add the markup to get what you want to sell that item for.
EX: you want to make 75% of what you paid (.75)(your cost)
You paid $100 for an ipod. (.75)($100)=$75 markup. So you want to sell it for $175!

MARKUP = a % of INCREASE

EX: Your cost = $50 and your markup = $100. What is your MARKUP %? (% of increase?)
markup = increase
$100 (increase on cost)/$50 (original) = 2 = 200% markup (selling price = $50 + $100 = $150)

CHAPTER 7-8: SIMPLE INTEREST:
Let's talk about different types of interest: credit cards, car loans, mortgages
None of them charge us simple interest, but you need to understand that first:
I =PRT (mneumonic devices "PRINT" or "PARTY")
I = Interest
P = Principal ($ deposited in savings or $ owed to credit card or bank)
R = % (Change it to a decimal or fraction before multiplying)
T = Time (based on one year so if you have 9 months, put it over 12 months or 9/12 of a year)
'
EXAMPLE: You buy an Ipod for $300 on your parent's VISA at 20% and pay it back in 2 years
I = PRT
I = ($300)(.20)(2) = $120 Interest
$300 Ipod + $120 interest to VISA = $420 total cost

EXAMPLE: You deposit $200 in your savings account at Wells Fargo for 9 months at 2%
I = PRT
I = ($200)(.02)(9/12) = $3 interest
$200 deposit + $3 = $203 in your savings account after 9 months

CHAPTER 7-8: COMPOUND INTEREST:
Same as simple, but you charge (or get) INTEREST ON THE INTEREST
So COMPOUND > SIMPLE always!
EXAMPLE: The Ipod above is still bought over 2 years on VISA, but this time interest is COMPOUNDED (added to the principal) ANNUALLY (each year)
After one year: I = ($300)(.20)(1) = $60
$300 Ipod + $60 interest = $360
Now in year 2, you're charged interest on the $360, not just the original $300!
That's why it's always more than SIMPLE interest!
Year two: ($360)(.20)(1) =$72
$360 owed at end of year one + $72 year two interest = $$432 owed at the end of year two
Notice that under SIMPLE interest, that amount was only $420

Interest can be compounded annually (once a year), semiannually (twice a year), quarterly (4 times a year), monthly (12 times a year) or even daily! (365 times a year!)
Which would give you the most interest on your savings deposit? Obviously compounding daily!
Which would give the VISA company the most? Same thing - compounding daily! (that's what they use!)
There's a formula for compound interest:
Balance owed = (principal)(1 + percent rate/# times it compounds a year)^{number of compoundings}
So if I owe VISA $500 at 6% compounded QUARTERLY (4 times a year) and I want to know how much I will owe after 1 year, I can use this formula:
A = P(1 + r/t)ⁿ
A is the amount I'm looking for
P = $500
r = 6% or .06
t = 4 times a year
n = 4 because I want to know after a year and the interest will compound 4 times
A = 500[1 + .06/4]⁴
A = 500(1.015)⁴
A =(500)(1.06136)
A = $530.68

The other way to do this is an organized table:
500(.06)(.25) = $7.50 interest for 1st quarter + $500 = 507.50
$507.5(.06)(.25) = $7.61 interest for 2nd quarter + 507.50 = $515.11
$515.11(.06)(.25) = $7.73 interest for 3rd quarter + $515.11 = $522.84
$522.84(.06)(.25) = $7.84 interest for 4th quarter + $522.84 = $530.68

BUT YOU WOULDN'T WANT TO USE THE TABLE APPROACH FOR MORE THAN ONE YEAR BECAUSE IT JUST TAKES TOO MUCH TIME!

Algebra Period 3

Review Chapter 11: RADICALS

REMEMBER:
THE SQUARE ROOT OF ANYTHING SQUARED IS ITSELF!!!

TRINOMIALS UNDER THE RADICAL:
What do you think you would do if you saw x² + 10x + 25 under the radical sign????
FACTOR IT! IT MAY BE A PERFECT SQUARE (a binomial squared!)
x² + 10x + 25 factors to (x + 5)² so the square root of (x + 5)² = lx + 5l
Please read as absolute value of (x +5)

SIMPLIFYING RADICALS
SIMPLIFYING NONPERFECT NUMBERS UNDER THE RADICAL:
A simplified radical expression is one where there is no perfect squares left under the radical sign
You can factor the expression under the radical to find any perfect squares in the number:
EXAMPLE: square root of 50 = SQRT(25 * 2)
Next, simplify the sqrt of the perfect square and leave the nonperfect factor under the radical:
SQRT(25 * 2) = SQRT(25) *SQRT(2) = 5 SQRT( 2 )

HELPFUL HINTS:
When you are factoring the radicand, you're looking for the LARGEST PERFECT SQUARE that is a FACTOR of the radicand.
So start with:
Does 4 go into it?
Does 9 go into it?
Does 16 go into it?
Does 25 go into it?
etc.

HELPFUL WAYS TO ATTACK SIMPLIFYING:
Factor trees, Inverted Division, Prime Factorization

Whenever there are 2 factors that are same, it can be simplified
(Take one of the 2 factors out of the radical and leave none under)
If there is already a factor outside the front of the radical, when you bring out another factor from underneath the radical, you multiply it with what was already outside in front.

VARIABLES UNDER THE SQUARE ROOT SIGN:
An even power of a variable just needs to be divided by two to find its square root
EXAMPLE: SQRT (x¹⁰ ) = x⁵
We saw this already in factoring!!!
If the variable has an odd power:
If you have an odd power variable, simply express it as the even power one below that odd power times that variable to the 1 power:
Example: x⁵ = x⁴ x
so if you have the SQRT( x⁵ ) = SQRT (x⁴ x) = x²SQRT(x)

Don't forget to always:
RATIONALIZE THE DENOMINATOR
The rule is that the radical is not simplified until all radicals are removed from the denominator
Simply use the equivalent fraction approach and multiply both the numerator and the denominator by the radical.

ADDING AND SUBTRACTING RADICALS:
Radicals function like variables, so you can only COMBINE LIKE RADICALS!

CHAPTER 11-7 (an old friend) - PYTHAGOREAN THEOREM
FOR RIGHT TRIANGLES ONLY!
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
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
b = 12

Math 6 Honors Periods 6 & 7

Changing a Decimal to a Fraction 6-6

As we have seen, every fraction is equal to either a terminating decimal or a repeating decimal. It is also true that every terminating or repeating decimal is equal to a fraction.

To change a terminating decimal to a fraction in lowest terms, we write the decimal as a fraction whose denominator is a power of 10. We then write this fraction in lowest terms.

Change 0.385 to a fraction in lowest terms

.385 = 385/1000 = 77/200

Change 3.64 to a mixed number in simple form

3.64 = 3 64/100 = 3 16/25

To change a repeating decimal into a fraction follow these examples

th__
0.54

tthththth__
Let n = 0.54 = 0.54545454….

[How many numbers are under the vinculum?] 2

Multiple both sides by 10²

So then, 100n = 54.54545454…

100n = 54.54545454…
122- n = 2.54545454….

We can subtract n from 100n to get 99n

100n = 54.54545454…
-23n =thu .54545454….

99n = 54

Divide both sides by 99

99n = 54
99thit 99

n = 54/99 = 6/11

Let’s try

th___
0.243

theitheith___
Let n = 0.243 = .243243243243….

How many numbers are under the vinculum? 3

So multiply both sides by 10³

1000n = 243.243243243243….

1000n = 243.243243…
thie- n = 243.243243

999n = 243

Divide both sides by 999

n = 243/999 = 27/111 = 9/37

Let’s try one that is a bit more complicated
tethit htitii__thiethitiehtiehtitheihtii___theithihitheii_
Change 0.318 [Notice this isn’t 0.318 nor is it 0.318

the__
0.318

How many numbers are under the vinculum? 2

So, we multiply by 10²

Let n = 0.318181818…

100n = 31.818181818…
thit-n=31.318181818…

99n = 31.500000…

Divide both sides by 99
99n/99 = 31.5/99

n = 31.5/99 but that isn’t a proper fraction. What can I do to change this?

Multiply by 10

315/990 = 63/198 = 7/22

HERE ARE SOME STEPS TO FOLLOW:

Step 1	set up “ n= the repeating decimal”	n = .515151…
Step 2	determine how many numbers are under the bar	in this case = 2
Step 3	Use that number as a power of 10	102 = 100
Step 4	Multiply both sides of the equation in step 1 by that power of 10	100n = 51.515151…
Step 5	Rewrite the equations so that you subtract the 1st equation FROM the 2nd equation	100n = 51.5151… - 00n= 51.5151…
Step 6	Solve as a 1-step equation	99n = 51 so n =51/99
Step 7	Simplify	51/99 = 17/33
****	REMEMBER- sometimes you need to get the decimal out of the numerator—so multiply by a power of 10

Math 6 Honors Periods 6 & 7

Changing a Fraction to a Decimal 6-5

There are two methods that can be used to change a fraction into a decimal.

The first one, we try to find an equivalent fraction whose denominator is a power of 10.

13/25 is a great example because we can easily change the denominator into 100 : multiplying 24 by 4.

So

13/25 ( 4/4) = 52/100 = .52

In the second method of changing a fraction into a decimal, we divide the numerator by the denominator.

Change 3/8 divide 3 by 8 carefully

When the remainder is 0, as above, the decimal is referred to as a terminating decimal. By examining the denominator of a fraction in lowest terms, we can determine whether the fraction can be expressed as a terminating decimal. If the denominator has no prime factors of then 2 or 5, the decimal representation will terminate.. (This is so since the fraction can be written as an equivalent fraction whose denominator is a power of ten)

7/40

40 = 2³ ∙ 5; since the only prime factors of the denominator are 2 and 5, the fraction can be expressed as a terminating decimal

5/12

12 = 2² ∙ 3 since 3 is a prime factor of the denominator, the fraction cannot be expressed as a terminating decimal

9/12 = 3/4

4 = 2². Since 4 has no prime factors other than 2, this fraction can be expressed as a terminating decimal

Now, what happens if the denominator of a fraction has prime factors other than 2 or 5

Change 15/22 to a decimal I know that this cannot be expressed as a terminating decimal because the denominator (22) has the prime factorization of 2 ∙ 11.

15 divided by 22 ? divide carefully and you will get 0.6818181….

Notice the pattern of repeating remainders of 18 and 4. They produce a repeating block of digits 81, in the quotient.

we write 15/22 = 0.681818181…. or 0.681 with a bar over the 81 where the bar, also know as the vinculum, means that the block 81 repeats without ending.

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a decimal such as 0.681 , in which a block of digits continues to repeat indefinitely is called a repeating decimal.

Property

Every fraction can be expressed as either a terminating decimal or a repeating decimal..