Algebra Honors (Period 6 & 7)

Differences of Two Squares 5-5

(a + b)(a-b) = a² - b²
(a+b) is the sum of 2 numbers
(a-b) is the difference of 2 numbers
= ( first#)² - ( 2nd #) ²

( y -7)( y + 7) = y² - 49
We did the box method to prove this.

(4s + 5t) (4s - 5t)
16s² - 25t²

(7p + 5q)(7p-5q) = 49p² - 25q²
But then we looks at
(7p+5q)(7p+5q) that isn't the difference of two squares that is
49p² + 70pq + 25q²
So let's look at the difference of TWO Squares:

b² -36
So that is ( b + 6)(b -6)

m² - 25
(m + 5)(m -5)

64u² - 25v²
(8u + 5v)(8u -5v)

1 - 16a²

(1+4a)(1- 4a)

But what about 1- 16a⁴

( 1 + 4a²)(1 - 4a²) but we are NOT finished factoring because
(1 - 4a²) is still a difference of two squares so it becomes

( 1 + 4a²)(1 + 2a)(1 - 2a)

t⁵ - 20t³ + 64t

Factor out the GCF first

t(t⁴ - 20t² + 64)
t(t² -16)(t² -4)
YIKES... we have two Difference of Two Squares here...

t(t +4)(t-4)(t +2)(t -2)



81n² - 121
(9n +11)(9n -11)
3n⁵ - 48 n³


Factor out the GCF

3n³ (n² - 16)
3n³(n +4)(n-4)

50r⁸ - 32 r²

Factor out the GCF
2r²(25r⁶ - 16)

2r²(5r³ +4)(5³ -4)

u² - ( u -5) ²
think a² - b ² = (a + b)(a -b)

so
u² - ( u -5) ² =
[u + (u-5)][u - (u-5)]
(2u -5)(5)
= 5(2u-5)

t² - (t-1)²

[t +( t+1)]{t-(t-1)]
2t-1(+1)
=2t -1


What about x²ⁿ - y ⁶ where n is a positive integer

well that really equals
(xⁿ)² - (y³)²
so
(xⁿ + y³)(xⁿ - y³)

x²ⁿ - 25
(xⁿ + 5)(xⁿ - 5)

a⁴ⁿ - 81b⁴ⁿ
(a²ⁿ + 9b²ⁿ)(a²ⁿ - 9b²ⁿ)
= (a²ⁿ + 9b²ⁿ)(aⁿ + 3bⁿ)(aⁿ - 3bⁿ)


When multiplying to numbers such as (57)(63)
think
(60-3)(60 +3)
then the problem becomes so much easier

3600 - 9 = 3591 DONE!!!

(53)(47) = (50 +3)( 50-3)
2500 - 9 = 2491

Math 6 Honors ( Periods 1, 2, & 3)

Rounding 3-5

Round the following number to the designated place value:
509.690285

tenths: 509.690285
You underline the place value you are rounding to and look directly to the right. If it is 0-4 you round down; if it is 5-9 you round up 1.
so here we round to
509.7

hundredths
509.690285
becomes 509.69

hundred-thousandths
509.690285
becomes
509.69029

tens
509.690285
becomes
510

(a) What is the least whole number that satisfies the following condition?

(b) What is the greatest whole number that satisfies the following condition?
A whole number rounded to the nearest ten is 520.
Well, 515, 516, 517, 58, 519, 520, 521, 522, 523, 524 all would round to 520
so

(a) 515
(b) 524

A whole number rounded to the nearest ten is 650
(a) 645
(b) 654

A whole number rounded to the nearest hundred is 1200
(a) 1150
(b) 1249
How about these...
(a) What is the least possible amount of money that satisfies the following condition?
(b) What is the greatest possible amount?

A sum of money, rounded to the nearest dollar is $57
(a) $56.50
9b) $57.49

A sum of money rounded to the nearest ten dollars $4980
(a) $4975
(b) $4984.99

Algebra Honors (Period 6 & 7)

Multiplying Binomials Mental 5-4

Look at
(3x - 4)(2x+5)

remember FOIL
First terms (3x)(2x)
Outer terms (3x)(5)
Inner Terms (-4)(2x)
Last Terms (-4)(5)

6x² + 15x -8x -20
6x² + 7x - 20
Or use the box method as we have done in class
This is a quadratic polynomial
The quadratic term is a term of degree two

Remember a linear term has a term of degree 1 such as y = 3x + 5

6x² + 7x - 20

The 6x² is the quadratic term
the +7x is the linear term
and the - 20 is the constant term

(x +1)(x +3) = x² + 4x + 3

(y + 2)( y + 5) = y² + 7y + 10
( t -2)( t -3) = t² -5t + 6

( u -4)(u -1) = u² - 5u + 4

What about
( u-4)(u +1) = u ² -3u -4

See the difference between the two?

(7 - k)(4 -k)

28 - 11k + k²


r + 3)(5 - 5)
r² - 25 - 15



(3x - 5y)(4x + y)
12x² - 17xy - 5y²

a + 2b)(a-b)

careful....
a² + ab - 2b²

n(n-3)(2n+1)
first distribute the n
(n² -3n)(2n +1)
2n³ - 5n² - 3n

Solve for
(x-4)(x +9) = (x +5)(x -3)
x² + 5x - 36 = x² + 2x -15
5x - 36 = 2x - 15
3x = 21

x = 7
or in solution set notation {7}

Math 6 Honors ( Periods 1, 2, & 3)

Comparing Decimals 3-4



In order to compare decimals, we compare the digits in the place farthest to the left where the decimals have different digits.

Compare the following:


1. 0.64 and 0.68 since 4 < 8 then 0.64 < 0.68.




2. 2.58 and 2.62 since 5 < 6 then 2.58 < 2.62 .




3. 0.83 and 0.833



To make it easier to compare, first express 0.83 to the same number of decimal places as 0.833



0.83 = 0.830 Then compare



0.830 and 0.833 since 0 <3

Then 0.830 < 0.833.



Write in order from least to greatest



4.164, 4.16, 4.163, 4.1



First, express each number to the same number of decimal places

Then compare. 4.164, 4.160, 4.163, 4.100




The order of the numbers from least to greatest is


4.1, 4.16, 4.163, 4.164

Math 6 Honors ( Periods 1, 2, & 3)

We add the following bit of notes today:
looking at the powers of 10 we noticed that
10² ⋅ 10³ = 10⁵
and
10⁸ ⋅10⁶ =10¹⁴
so could we write a rule for any exponent values a and b?
YES, we decided:
10^a ⋅10^b = 10^a+b


Remember how we proved that any number to he zero power was equal to 1
or a⁰ = 1

Refer back to your notes or to the blog a few days ago...
we also showed how
What happens when you multiply the same bases?
3⁴ ⋅ 3² = 3⋅3⋅3⋅3⋅3⋅3
or 3⁴⁺² = 3 ⁶
We just add the exponents if the bases are the same!!
When we divide by the same base we just subtract

3⁴ /3² = 3^4-2 =3²

What would happen if we had
3² / 3⁴ ?
Let's look at what we would actually have
3⋅3
3⋅3⋅3⋅3

Which would be
1
3²

or 1/3²

but you can write that as 3^-2
We just subtract-- using the same rule.


Now let's get back to our decimal lesson and apply that to decimals -- and the Powers of TEN

Math 6 Honors ( Periods 1, 2, & 3)

Decimals 3-3

Although decimals ( termed decimal fractions) had been used for centuries, Simon Stevin in the 16th century began using them on a daily basis and he helped establish their use in the fields of sciences and engineering.

Note that
1/10 = 1/10¹
1/100 = 1/10²
1/1000 = 1/10³

We also know that
1/10= 0.1
1/100 = 0.01
1/1000 = 0.001
1/10000 = 0.0001
and so on... these strings of digits are called decimals.

Remember how we proved that any number to he zero power was equal to 1
or a⁰ = 1

Refer back to your notes or to the blog a few days ago...
we also showed how
What happens when you multiply the same bases?
3⁴ ⋅ 3² = 3⋅3⋅3⋅3⋅3⋅3
or 3⁴⁺² = 3 ⁶
We just add the exponents if the bases are the same!!
When we divide by the same base we just subtract

3⁴ /3² = 3^4-2 =3²

What would happen if we had
3² / 3⁴ ?
Let's look at what we would actually have
3⋅3
3⋅3⋅3⋅3

Which would be
1
3²

or 1/3²

but you can write that as 3^-2
We just subtract-- using the same rule.


Now let's get back to our decimal lesson and apply that to decimals -- and the Powers of TEN

SO 1/10 = 1/10¹= 0.01 and it is equal to 10^-1
Notice that 10^-1 is NOT a negative number-- it is a small number
and 10^-21 is not a negative number it is a VERY TINY number

As with whole numbers, decimals use place values. These place values are to the RIGHT of the decimal point.
We need to be able to write decimals in words as well as expanded notation.
In class we used 0.6394 as our example

zero and six thousand three hundred ninety-four ten-thousandths.

Notice how this number when written in words begins...with "ZERO AND"
Why do we need to do that?

Also notice that there is a hyphen between ten and thousandths in ten-thousandths. It is critical to understand when you must place a hyphen.
We read the entire number to the right of the decimal point as if it represented a whole number, and then we give the place value of the digit farthest to the right.

So, although 0.400 is equivalent to 0.4
we must read 0.400 as "zero and four hundred thousandths."

Now look at the following words
"zero and four hundred-thousandths." What is the subtle difference between those two phrases above?
There is a hyphen in the last phrase-- which means that the hundred and the thousandths are attached and represent a place value so

zero and four hundred-thousandths is 0.00004 while
zero and four hundred thousandths is 0.400

Carefully see the distinction!!

Getting back to our 0.6394

to write it in decimals sums and then in exponents:
0 + 0.6 + 0.03 + 0.009 + 0.0004

0 + 6(0.1) + 3(0.01) +9(0.001) + 4(0.0001)

0(10⁰) + 6(10^-1)+ 3(10^-2)+ 9(10^-3)+ 4(10^-4)

14.35 is read as fourteen AND thirty-five hundredths.
When reading numbers, only use the AND to indicate the decimal point

Math 6 Honors ( Periods 1, 2, & 3)

Decimals 3-3

Although decimals ( termed decimal fractions) had been used for centuries, Simon Stevin in the 16th century began using them on a daily basis and he helped establish their use in the fields of sciences and engineering.

Note that
1/10 = 1/10¹
1/100 = 1/10²
1/1000 = 1/10³

We also know that
1/10= 0/1
1/100 = 0.01
1/1000 = 0.001
1/10000 = 0.0001
and so on... these strings of digits are called decimals.


SO 1/10 = 1/10¹= 0.01 and it is equal to 10^-1
Notice that 10^-1 is NOT a negative number-- it is a small number
and 10^-21 is not a negative number it is a VERY TINY number

AS with whole numbers, decimals use place values. These place values are to the RIGHT of the decimal point.
We need to be able to write decimals in words as well as expanded notation.
In class we used 0.6394 as our example

zero and six thousand three hundred ninety-four ten-thousandths.

Notice how this number when written in words begins...with "ZERO AND"
Why do we need to do that?

Also notice that there is a hyphen between ten and thousandths in ten-thousandths. It is critical to understand when you must place a hyphen.
We read the entire number to the right of the decimal point as if it represented a whole number, and then we give the place value of the digit farthest to the right.

So, although 0.400 is equivalent to 0.4
we must read 0.400 as "zero and four hundred thousandths."

Now look at the following words
"zero and four hundred-thousandths." What is the subtle difference between those two phrases above?
There is a hyphen in the last phrase-- which means that the hundred and the thousandths are attached and represent a place value so

zero and four hundred-thousandths is 0.00004 while
zero and four hundred thousandths is 0.400

Carefully see the distinction!!

Getting back to our 0.6394

to write it in decimals sums and then in exponents:
0 + 0.6 + 0.03 + 0.009 + 0.0004

0 + 6(0.1) + 3(0.01) +9(0.001) + 4(0.0001)

0(10⁰) + 6(10^-1)+ 3(10^-2)+ 9(10^-3)+ 4(10^-4)

14.35 is read as fourteen AND thirty-five hundredths.
When reading numbers, only use the AND to indicate the decimal point

Algebra Honors (Period 6 & 7)

Dividing Monomials 5-2

There are 3 basic rules used to simplify fractions made up of monomials.
Property of Quotients
if a, b, c, d are real numbers with b≠0 and d ≠0

ac/bd = a/b ⋅c/d
Our example was 15/21 = (3⋅5)/(3⋅7) = 5/7
The rule for simplifying fractions follows ( when a = b)
(bc)/(bd) = c/d
This rule lets you divide both the numerator and the denominator by the same NON ZERO number.

35/42 = 5/6
-4xy/10x = -2y/5 which can also be written (-2/5)x as well as with out the (((HUGS)))

c⁷/c⁴ = c⁴c³/c4 = c³
another way we proved this was to write out all the c's
c⋅c⋅c⋅c⋅c⋅c⋅c⋅/c⋅c⋅c⋅c = and we realized we were left with
c⋅c⋅c = c³
In addition, we noticed that
c⁷/c⁴ = = c^7-4 = c³
THen we considered
c⁴/c⁷ =
c⋅c⋅c⋅c/c⋅c⋅c⋅c⋅c⋅c⋅c = 1/c⋅c⋅c = 1/c³ = c^-3

Since we all agreed that any number divided by itself was = 1
(our example was b⁵/b⁵ ), we proved the following
1 = b⁵/b⁵ = b^5-5 = b⁰

We finally arrived at the Rule of Exponents for Division

if m > n
a^m/aⁿ = a ^m-n

If n > m
a^m/aⁿ = 1/a ^n-m
and if m = n
a^m/aⁿ = 1

A quotient of monomials is simplified when
1)each base appears only once in the fraction,
2) there are NO POWERS of POWERS and
3)when the numerator and denominator are relatively prime, that is, they have no common factor other than 1.

35x³yz⁶/ 56x⁵yz
5z⁵/8x²

Finding the missing factor when you are given the following
48x³y²z⁴ = (3xy²z)⋅ (______)
we find that

48x³y²z⁴ = (3xy²z)⋅ (16x²z³)

Algebra Honors (Period 6 & 7)

Factoring Integers 5-1

When we write 56= 8⋅7 or 56 = 4⋅14 we have factored 56
to factor a number over a given set, you write it as a product of integers in that set ( the factor set).
When integers are factored over the set of integers, the factors are called integral factors.

We used the T- charts ( students learned in 6th grade) to first find the positive integer factors
56 = 1, 2, 4, 7, 8, 14, 28, 56

A prime number is an integer greater than 1 that has no positive integral factors other than itself and 1.
The first ten prime numbers are
2, 3, 5, 7, 11, 13, 17, 19, 23, 29

To find prime factorization of a positive integer, you express it as a product of primes. We used inverted division (again taught in 6th grade)

504
Try to find the primes in order as divisors.
Divide each prime as many times as possible before going on to the next prime
we found 504 - 2⋅2⋅2⋅3⋅3⋅7
which we write as 2³⋅3²⋅7
Exponents are generally used for prime factors
The prime factorization is unique--> and the order should be from the smallest prime to the largest.
A factor of two or more integers is called a common factor of the integers.
The greatest common factor (GCF) of two or more integers is the greatest integer that is a factor of all the given integers.

Find the GCF(882, 945)
First find the prime factorization of each integer Then form product of the smaller powers of each common prime factor.
The GCF is only the primes (and the powers) that they SHARE!!
882 = 2⋅3²⋅7²
945 = 3³⋅5⋅7
The common factors are 3 and 7
The smaller powers of 3 and 7 are 3² and 7
You combine these as a PRODUCT and get

the GCF(882, 945) = 3²⋅7 = 63

We also talked about listing ALL pairs of factors--> thus including negative integers
For example:
List all the pairs of factors of 20
(1)(20) but also (-1)(-20)
(2)(10) and (-2)(-10)
(4)(5) and (-4)(-5)

Listing all the factors of -20, we discovered
(1)(-20) but also (-1)(20)
(2)(-10) and (-2)(10)
(4)(-5) and (-4)(5)

Math 6 Honors ( Periods 1, 2, & 3)

Exponents and Powers of Ten 3-1
When two or more numbers are multiplied together--each of the numbers is called a factor of the product.

A product in which each factor is the SAME is called a power of that factor.

2 X 2 X 2 X 2 = 16. 16 is called the fourth power of 2 and we can write this as
2⁴ = 16

The small numeral (in this case the 4) is called the exponent and represents the number of times 2 is a factor of 16.
The number two, in this case, is called the base.

When you are asked to evaluate... simplify... solve... find the answer
That is,
Evaluate
4³ = 4 X 4 X 4 = 16 X 4 = 64

The second and third powers of a numeral have special names.
The second power is called the square of the number and the third power is called the cube.

We read 12² as "twelve squared" and to evaluate it
12² = 12 X 12 = 144

Powers of TEN are important in our number system.
Make sure to check out the blue sheet and glue it into your spiral notebook
First Power: 10¹ but the exponent is invisible = 10
Second Power: 10² = 10 X 10 = 100
Third Power 10³ = 10 X 10 X 10 = 1000
Fourth Power 10⁴ =10 X 10 X 10 X 10 = 10,000
Fifth Power 10⁵ = 10 X 10 X 10 X 10 X 10 = 100,000

Take a look at this list carefully and you will probably see a pattern that we can turn into a general rule:

The exponent in a POWER of TEN is the same as the number of ZEROS when the number is written out.

The number of ZEROS in the product of POWERS OF TEN is the sum of the numbers of ZEROS in the factors.

For example Multiply.
100 X 1000
Since there are 2 Zeros in 100 and 3 zeros in 1000,
the product will have 2 + 3 , or 5 zeroes.
100 X 1000 = 100,000

When you need to multiply other bases:

first multiply each
For example

3⁴ X 2 ³ would be
(3 X 3 X 3X 3) X ( 2 X 2 X 2)
= 81 X 8 = 648

What happens when you multiply the same bases?
3⁴ ⋅ 3² = 3⋅3⋅3⋅3⋅3⋅3 or 3 ⁶
We just add the exponents if the bases are the same!!

Well then, what about (3⁴)² ?
Wait.. look carefully isn't that saying 3⁴ Squared?
That would be (3⁴)(3⁴), right?
.. and looking at the rule above all we have to do here is then add those bases or 4 + 4 = 8 so the answer would be 3⁸.
OR
we could have made each (3⁴) = (3⋅3⋅3⋅3)
so (3⁴)² would be 3⋅3⋅3⋅3⋅3⋅3⋅3⋅3 or still 3⁸
But wait... isn't that multiplying the two powers? So when raising a power to a power-- you multiply!!
(3⁴)² = 3⁸

1 to any more is still just 1
1⁵ = 1

0 to any power is still 0!!

Evaluate if a = 3 and b = 5
Just substitute in... but use hugs () we all love our hugs!!
a³ + b²
would be (3)³ + (5) ²
= 27 + 25 = 52




Check out this great Video on the Powers of Ten
POWERS OF TEN

Math 6 Honors ( Periods 1, 2, & 3)

Exponents and Powers of Ten 3-1


Check out this great Video on the Powers of Ten
POWERS OF TEN

Algebra Honors (Period 6 & 7)

Problems Without Solutions 4-10

Not all word problems have solutions. We listed three of the reasons for this:
1) Not Enough Information ( NEI)
2) Unrealistic Results
3) Facts are contradictory

We used the following examples:
Aurenne drove at her normal speed for the first 2 hours of the trip--- but the road repairs slowed her down 10 mph slower than her normal speed. She made the trip in 3 hours. Find her normal speed.
Wait... just looking at this you realize you just dont have enough information.
We even made a chart with the information we had.. and it just was not enough

There is NO SOLUTION Not enough information or NEI

Liam has a beautiful lawn that is 8 m longer than it is wide... and it is surrounded by a wonderful flower bed which he and his brother maintain for his mother The flower bed is 5m wide all around. Find the dimensions of the lawn if the area of the flower bed is 140m²

When you try to solve this problem by letting the dimensions of the lawn be w and w + 8 you find the following equation
(w + 10)(w + 18) - (w)(w +8) = 140
however that leads us to
w = -2
Since the width of the lawn cannot be negative,
There is No SOLUTION and the given facts are unrealistic.


Drehan says he has equal number of dimes and quarters but that he has 3 times as many nickels as he has dimes. He also tells us that the value of his nickels and dimes is 50 cents more than the value of his quarters. How many of each kind of coin does he have?

Let d = the number of dimes
well if he has the same number of quarters as he has dimes
then d also can equal the number of quarters
and with the other information
3d = the number of nickels.
Now looking at the information he gave us
10d + 5(3d) = 25d + 50
But that simplifies to
25d = 25d + 50
which is impossible.
There is NO SOLUTION
The given facts are contradictory

Algebra Honors (Period 6 & 7)

Rate-Time- Distance Problems 4-8

D = rt

Uniform Motion

Three types of problems:
Motion in opposite direction
Motion in same direction
Round Trip

Motion in opposite direction
For this we used different students bicycling ... Josh K and Josh P in 6th period and Maddie & Jamie from 7th period.
They start at noon -->60 km apart riding toward each other. They meet at 1:30 PM. If Josh K speed is 4 km/h faster than Josh p ( Maddie is greater by the same from Jamie's rate) What are their speeds?

We set up a chart

Motion in Same Direction

Next we had a fictitious story about ANdrew's Helicopter and David's plane ( or Lucas' helicopter and Kitt's Smiling Plane) taking off from Camarillo Airport flying north. The helicopter flies at a speed of 180 mi/hr. 20 minutes later the plane takes off in the same direction going 330 mi/hr. How long will it take David (or Kitt) to over take Andrew's ( or Lucas') helicopter?

Let t = plane's flying time

Make sure to convert the 20 minutes ---> 1/3 hours.

We set up a chart

When the plane over takes the helicopter they have traveled the exact same distance so set them equal
180(t + 1/3) = 330t

180 t + 60 = 330t
60 = 150t

t = 2/5
which means 2/5 hour. or 24 minutes.

Round Trip

A ski life carries Sara ( or Ryan) up the slope at 6 km/h Sare or Ryan snowboard down 34 km/h. The round trip takes 30 minutes.

Did you see the picture?

Let t = time down

then set up a chart

6(.5 -t) = 34 t
3 - 6t = 34 t
3/40 = t
Now, what's that?

0.75 hr or 4.5 minutes

How far did they snowboard... plug it in
34(0.075) = 2.55 km

Math 6 Honors ( Periods 1, 2, & 3)

Problem Solving: Using Mathematical Expressions 2-6


Seventeen less than a number is fifty six

I suggest lining up and placing the "equal sign" right under the word is
then complete the right side = 56
after that take your time translating the left
Start with a "let statement."
A "let statement" tells your reader what variable you are going to use to represent the number in your equation.
So in this case Let b = the number
it becomes
b - 17 = 56
Now solve as we have been practicing for a couple of weeks.
b - 17 = 56
+17 = +17 using the +prop=
b + 0 = 73
b = 73 by the ID(+)

How could we check?
A FORMAL CHECK involves three steps:
1) Re write the equation ( from the original source)
2) substitute your solution or... "plug it in, plug it in...."
3) DO the MATH!! actually do the math to check!!

so to check the above
b - 17 = 56
substitute 73 and put a "?" above the equal sign...

73 - 17 ?=? 56
Now really do the math!! Use a side bar to DO the MATH!!
That is, what is 73- 17? it is 56
so 56 = 56


Practice some of the class exercises on Page 50, Just practice setting up the equations from the verbal sentences.

Problem Solving: Using Mathematical Expression 2-6

Inequalities Continued

We know about > greater and as well as < less than
so now we look at
≥ which means " greater than or equal to" and
≤ which means " less than or equal to"

This time the boundary point ( or endpoint) is included in the solution set.
The good news is that we still solve these inequalities the same way in which we solved equations-- using the properties of equality.
w/4 ≥ 3
we multiply both sides by 4/1
(4/1)(w/4) ≥ 3(4/1) by the X prop =
1w ≥ 12
w ≥ 12 by the ID(x)

Which means that any number greater than 12 is part of the solution AND 12 is also part of that solution

Take the following:
b - 3 ≤ 150 ( we need to add 3 to both sides of the equation)
+3 = +3 using the + prop =
b + 0 ≤ 153
or b ≤ 153 using the ID (+)