Math 6A (Periods 2 & 4)

Rounding 3-5 (cont'd)

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 (Periods 5 & 6)

Factoring Pattern for x² + bx+ c, c positive 5-7

In this lesson we will be factoring trinomials that can be factored as a product of ( x +r)(x + s)
where r and s are both positive OR both negative integers.

x² + ( r + s)x + rs  

(x +3)(x+5) = x² + 8x + 15

(x – 6)(x -4) = x² -10x + 24

where -10 is the sum of -6 and -4

and

24 is the product of -6 and -4

Our book suggests that you list all the pairs of integral factors whose products equal the constant term Then, find the pair of integral factors whose SUM equals the coefficient of the linear term (remember your new vocab)

For factoring x² + bx + c, where c is positive  you only need to consider factors WITH THE SAME SIGNS as the linear term!!

In class, I showed the diamond method of calculating products and sums… and even recommended an app to practice!!
Here is the link for the iphone,  ipad app…

y² + 14y + 40

Since the linear term ( +14y) is positive you know to set up two sets of HUGS with + in the middle

(  +  )(  +  )

then you can add the single y’s since y² = y·y

(y + )(y + )

Now either list all the integral factors or use the diamond method and you discover that 10 and 4 are the two factors that multiply to 40 AND also sum to 14

(y + 10 )(y + 4)

To check if you are accurate, use FOIL or Fireworks… or the BOX method and see if you get back to the original problem!

y2 – 11y + 18

This time notice that the linear term is -11y so you will be looking for a pair of numbers whose product will be positive  ( so both need to be negative)

Set  up your  HUGS  similarly—EXCEPT both signs need to be NEGATIVE

(y -   )( y -  )

Since -11 is negative, think of the negative factors of 18 using the book’s method or the diamond method

hmmm… -9 and -2 work

(y - 9  )( y -  2)

Again To check if you are accurate, use FOIL or Fireworks… or the BOX method and see if you get back to the original problem!

A polynomial that cannot be expressed as a product of polynomials of lower degree is said to be irreducible. An irreducible polynomial with integral coefficients whose greatest monomial factor is 1 is a PRIME POLYNOMIAL.

Factor x² -10x + 14

Setting up your HUGS   you start to think what two factors multiply to 14 and SUM to 10… hmmm. NOTHING…

Therefore x² -10x + 14  cannot be factored and it is a prime polynomial

Find all the integral values of k for which the trinomial can be factored

x² + kx + 28

28 can be factored as a product—using  a T chart

list all the factors

(1)(28) (2)(14) (4)(7)

Taking the corresponding sums you get 29, 16 and 11

BUT… remember you can also have the negatives here

so the values of k can be ± 29,   ± 16,   ± 11

{-29, -16, -11, 11, 16, 29}

Math 6A ( Periods 2 & 4)

Rounding 3-5 



Our book talks about phonograph records... we needed to have a discussion about what a phonograph record is... Then we discussed what the cost of $7.98 would round to? We decided that $8 would be a good rounded number.

A general rule:
We underline the place we are rounding to and look to the number to its right. If the number is 5 or greater we round our underlined number up one and put zeros after it...
32, 567
rounded to the nearest ten thousand --> 30,000

rounded to the nearest thousand --> 33,000

rounded to the nearest hundred --> 32,600

rounded to the nearest ten --> 32,570

Rounding Decimals becomes a similar process BUT there are a few differences:

4.8637

rounded to the nearest thousandth --> 4.864

rounded to the nearest  hundredth --> 4.86

rounded to the nearest tenth  --> 4.9

rounded to the nearest unit  5

Math 6A (Periods 2 & 4)

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 br="br">
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 High (Period3)

Connecting Decimals &  Fractions 2.8 

Verbal --> fraction--> decimal
3/4 to figure out what the decimal equivalence is you need to divide 3 by 4

3/4 = 0.75

But what about 5/3
5/3 = 1.666666...
It is a repeating decimal
we write
5/3 = 1.6 with a bar over the 6

The bar is called a vinculum.
Write the repeating part only once and put the vinculum ( bar) over the repeating part of the decimal.

2/7 was an interesting decimal
we found that 2/7 = 0.285714285714....
We then talked about the 1/7th family and noticed that
1/7 = .142857142857....
2/7 = .285714285714...
3/7= .428571428571...
4/7= .571428571428...
5/7= .714285714285....
6/7 = .857142857142...
all are repeating decimals with the same 6 digits in the same order... just the starting number changed...

But the textbook asked us to round 2/7 to the nearest thousands.. so we divide to the ten thousands  and found we had
0.2857 so we look to the place past the thousandths place and since it was a 7 we rounded up and dropped the digits past the thousandths place.
2/7 rounds to 0.286

Next, the textbook ased us to plot the following on a number line
2.4, 3.2, 4/5  3 2/5 and 11/5

Change all to decimals and the plot on a number line We found that 4/5 = 8/10 = 0.8
3 2/5 = 3 4/10 = 3.4 and 11/5 =  2 1/5 = 2 2/10 = 2.2
Then it became easy to plot on a number line

How do you write 2.5 as a mixed number ?  2 1/2
How about 0.125?

read it it is 125/1000  Then use divisibility rules what goes into 125 and 1000? 5 does but wait-- so does 25...
5/40  but that can be simplified because 5 goes into 5 and 40 so 5/40 = 1/8

Algebra Honors (Periods 5 & 6)

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