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

Dividing Decimals 3-9 cont'd

For word Problems use the 5 step plan found on Page 18 of our textbook

296.06 ÷ (18.7 + 3.9)
Following Aunt Sally ( or PEMDAS... remember our singing...
we do the operation inside the hugs!! ( )using a sidebar
18.7 + 3.9 make sure to stack them lining up the decimals and you will get 22.6

296.06 ÷ 22.6

When dividing by a decimal remember the rule from yesterday, multiply the divisor ( 22.6) by a power of ten which makes it a natural number

then use that same power of ten and multiply the dividend,
WHen you divide you have
2960.6 ÷ 226
Please do that problem and your quotient should be 13.1

(47.1 - 16.9) ÷ (21.9 -6.8)
Again you need to do the operations inside the ( ) first. Using a side bar and lining up the decimals
47.1 - 16.9 = 30.2
and 21.9 - 6.8 = 15.1
Just take a look at those two numbers and you will notice a relationship!!
30.2 ÷ 15.1
BUT... practice your division skills and confirm what you can tell...
30.2 ÷ 15.1 = 2



At an average rate of 55 km/hour how long will it take to drive 225 km to the nearest tenth of an hour?

d = rt
What must we find and what are the clues? Well, how long... is usually time and the fact that we need to round to the nearest tenth of an hour indicates we are finding TIME as well.
So what is the distance? 225 km and what is the rate? 55 km/h
so plug into the formula
225= 55t
Now, how do we solve this one step problem?

divide both sides by 55
225/55 = 55t/55

do the division as a side bar

225/55 ≈ 4.09 so
t ≈ 4.1
and the answer is 4.1 hour

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

Dividing Decimals 3-9

According to our textbook-
In using the division process to divide a decimal by a counting number, place the decimal point in the quotient directly over the decimal point in the dividend.

Check out our textbook for some examples!!

When a division does not terminate-- or does not come out evenly-- we usually round to a specified number of decimal places. This is done by adding zeros to the end of the dividend, which as you know, does NOT change the value of the decimal. We then divide ONE place beyond the specified number of places.

Divide 2.745 by 8 to the nearest thousandths.
See the set up in our textbook on page 89. Notice that they have added a zero and the end of the dividend ( 2.745 becomes 2.7450) because you want to round to the thousandths and we need to go ONE place additional.
DIVIDE carefully!!

the quotient is 0.3431 which rounds to 0.343


To divide one decimal by another

Multiply the dividend and the divisor by a power of ten that makes the DIVISOR a counting number


Divide the new dividend by the new divisor

Check by multiplying the quotient and the divisor.

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

Multiplying Decimals 3-8

According to our textbook:
Place the decimal point in the product so that the number of places to the right of the decimal point in the product is the sum of the number of places to the right of the decimal point in the factors!!

You do NOT need to line up the decimal point when you are multiplying.

14.92 x 7.2 stack them but do not line up the decimals
= 107.424


11.32 X 8.73

multiply carefully and you get
11.32
8.73

98.8236

estimate and you get 11 X 9 = 99

What would you do if you had
(19.81 x 5.1) + (19.81 X 4.9)

Wait... wait... remember the Distributive Property????
Look you can use it to make this problem soooooo much easier
19.81(5.1 _ 4.9)
19.81 (10) = 198.1


How about 50(.25) + 50(.75)
This is one you can even do in your head because
50(0.25 + 0.75) = 50 (1) = 50


A jet flew 820.3 km/h for 3.2 hours. How far did the jet travel?
We have a great formula distance = rate X time
or d = rt
rate means the speed
so what do we have?
distance = (820.3)(3.2)
First I think I will estimate to make sure I am in the ballpark with my actual answer
I know that 820 (3) = 2460 so I know my answer will be close to 2460 miles-- a little bit more
so when I actually carefully multiply (820.3)(3.2) I get 2624.96 km
so I know my answer is reasonable

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

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

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)

Proof in Algebra 3-8
Some of the properties discussed in the previous chapters are statements we assume to be true. Others are called theorems. A theorem is a statement that is shown to be true using a logically developed argument. Logical reasoning that uses given facts, definitions, properties, and other already proven theorems to show that a particular theorem is true is called a proof. Proofs are used extensively in Algebra as well as Geometry.
Prove: For all numbers a and b, (a + b) – b = a

Many times, only the KEY reasons are states—the substitute principle and the properties of equality are usually not stated. So, the above prove could be shortened to 4 steps:

Prove: For all real numbers a and b, such that a≠ 0 and b ≠0
1/ab = 1/a ⋅1/b

Since 1/ab is the unique reciprocal of ab, you can prove that

1/ab = 1/a⋅ 1/b by showing that the product of ab and 1/a ⋅ 1/b is 1

Once a theorem has been proved, you can use it as a reason in other proofs. Check the Chapter summary on page 88 of our textbook for the listing of properties and theorems that you can use as reasons in your proofs for our homework.

Algebra Honors (Period 6 & 7)

Cost Income, and Value Problems 3-7
Objective: To organize the facts of a problem in a chart & solve problems involving cost, income, and value
Using a chart to organize the facts of a problem can be a helpful problem solving strategy.
Cost = number of items X price per item
Income = hours worked X wage per hour
Total value = # of items X value per item

Example: Tickets for the senior class play cost $6 for adults and $3 for students. A total of 846 tickets worth $3846 were sold. How many student tickets were sold?
Let x = the number of student tickets sold
Then 846- x = the number of adult tickets sold

The only fact NOT recorded in this chart is that the total cost of the tickets is $3846.
The equation becomes
3x + 6(846 –x) = 3846.
3x + 5076 – 6x = 3826
-3x = -1230
x = 410
Check to make sure what x represented… in this case the number of student tickets so
410 student tickets were sold.


We then turned to Problem 4 on Page 128

A collection of 52 dimes and nickels is worth $4.50. How many nickels are there?
Let d = the number of dimes, so If there are 52 in the collection
52-d must = the number of nickels. So our Chart looks like:

This time we reread the given facts and with the second fact, we realize that we can add the two expressions and set them equal to $4.50
10d + 5(52 - d) = 450
solving this equation, we find that d = 38
Make sure to reread the question… it asked “how many nickels?”
So we need to use 52-d and substitute in 38… 52 -38 = 14
14 nickels is the correct answer


Next we complete # 6
Celia bought 12 apples, ate two and sold the rest at 20 cents more per apple than she paid. Her total profit was $1.00 How much did she sell each apple for?
Let b = the price she bought each of the apples for
We glued in the “yellow-colored” chart here and completed it to look like:

Now, her profit is the difference between what she sold them for and what she bought them for… Therefore the equation becomes
10(b + 20) – 12b = 100
10b + 200 -12b = 100
-2b = -100
b = 50
Re reading the question we realize we solved for what she bought the apples each for. We need to add 20cents to find out what she sold them for

She sold each apple for 70 cents.

The last problem we did from our textbook was # 14 on Page 129

Jo has 37 coins ( nickels, dimes & quarters) for $5.50 She has 4 more quarters than nickels. How many dimes does Jo have?
Let n = the number of nickels
n + 4 = the number of quarters
So if there was a total of 37 the rest must be dimes
37 – [ n + (n+4)] or 37 –(2n +4)
Let’s set up the chart

Now we know that all of them combined equal $5.50
so
5n + 10[37-(2n+4) + 25(n +4) = 550
5n + 370 – 20n – 40 + 25n + 100 = 550
combining all the n’s

10n + 330 +100 = 550
10n = 120
n = 12
Re reading the question, we find we need to see how many dimes so substitute in
37 -[2(12) +4] = 37 – [24 +4] = 37 -28 = 9
Jo had 9 dimes.

Algebra Honors (Period 6 & 7)

Problem Solving: Using Charts 3-6
Objective: To organize the facts of a problem in a chart.
Using a chart to organize the facts of a problem can be a helpful problem solving strategy.

Organize the given information in a chart
A swimming pool 25 m long is 13 m narrower than a pool 50 m long
There are two different charts you could create:

In the first chart we started with
Let w = the width of the 2nd pool; so to write the 1st pool in terms of the second we would place w-13 in place of the ?

In the 2nd chart we started with
Lt w = the width of the 1st pool; so we write the 2nd pool in terms of the first, so we would place w + 13 in place of that ?


We then solved the following:

Find the number of calories in an apple & a pear…

1) the pear contains 30 calories more than the apple.
2) Ten apples have as many calories as 7 pears.
Let a = the number of calories in an apple
Then a + 30 = the number of calories in a pear.
We glued in the “orange- colored” chart here and completed it to look like:

This time we reread the given facts and with the second fact, we realize that we can set
The last column expressions equal to each other
10a = 7(a + 30)
solving this equation, we find that a = 70
Therefore, an apple has 70 calories and a pear has 100 calories
Next we used the following two given facts to set up a chart and create an equation
1) An egg scrambled with butter & milk has 1 more gram of protein than an egg fried in butter.
2) Ten scrambled eggs have as much protein as a dozen fried eggs.
Let x = the number of protein in a fried egg.
Then x + 1 = the number of protein in a scrambled egg.
We glued in the “orange- colored” chart here and completed it to look like:

Our equation would be 10(x + 1) = 12x
We then turned to our textbook to Page 122-123 and completed problems 1 and 3 in our spiral notebook as follows:

Solve each problem using the two given facts. Complete a chart to help you solve each problem

1. Find the number of full 8 hour shifts that Maria worked last month
1) She worked twice as many 6 hour shifts as 8 hour shifts
2) She worked a total of 280 hours.

Algebra Honors (Period 6 & 7)

Equations w/ The Variable on Both Sides 3-5

Partial Notes
2 unique types of equations:
Identity equation: You solve it and you get the same thing on both sides...if you solve until you cannot do anything more, you get 0 = 0.
What this means is that you can pick any number and the equation will work!
The Distributive Property is the simplest example of an Identity Equation:
3(x + 7) = 3x + 21  
Distribute on the left side and you'll get:
3x + 21 = 3x + 21  
At this point, you should already know this is an Identity!
If you keep solving, you would subtract 3x from each side and you'll get:
21 = 21
and you know again that this is an Identity.
If you now subtract 21 from each side:
0 = 0
BUT I WOULDN'T GO THIS FAR! AS SOON AS YOU HAVE THE SAME THING ON BOTH SIDES, YOU CAN STOP AND SAY IT'S AN IDENTITY EQUATION!!!

Null set equation:
You solve it and you get an impossible answer:
3(x + 7) = 3x + 10
3x + 21 = 3x + 10
21 = 10
WHEN WILL THAT HAPPEN??? NEVER!!! SO THERE IS NO POSSIBLE SOLUTION TO THIS! The answer is the null set.

Algebra Honors (Period 6 & 7)

Using Several Transformations 3-3

a+b - b = a
and ab ÷ b = a
Use inverse operations
We can use those to solve equations with more than one step
5n - 9 = 71
by adding 9 to both sides of the equation ( using the +prop=)
we get 5n = 80
By then dividing BOTH sides by 5 using the ÷prop=
we arrive at n = 16
Using set notation {16}

(½)x + 3 = 9
By subtracting 3 from both sides using -prop=
we get ½x = 6
By multiplying BOTH sides by the reciprocal of ½ which is 2 (using xprop=)
we get
x = 12

w-5 = 2
9

multiply BOTH sides by 9 using xprop=
we get
w-5 = 18
adding 5 to both sides using +prop=
w = 23
{23}

32 = 7a + 9a
Combine Like terms FIRST
32 = 16a
2 = a {2}

4(y +8) - 7 =15
You must use the Distributive Property first
4y + 32 - 7 = 15
combine like terms on the left side
4y + 25 = 15
subtract 25 from both sides
4y = -10
divide by 4
y = -10/4
simplify BUT leave as an improper fraction
y = -5/2
Using set notation {-5/2}


But what about
-4(m +12) = 36
Yes you could do the Distributive Property BUT... why not divide both sides by -4 FIRST
-4(m + 12)= 36/-4
-4

m + 12 = -9
subtract 12 from both sides using the -prop=
and get
m = -21
{-21}
(c + 3) -2c -(1-3c) = 2

BEFORE WE COMBINE terms let's look at
-(1-3c) = -1 + 3c that is the inverse of a sum!!

c + 3 -2c -1 + 3c = 2
combining like terms we get
2c + 2 = 2
2c = 0
c = 0
{0}

Algebra Honors (Period 6 & 7)

Transforming Equations: All Four Op's 3-1 & 3-2

Addition Property of Equality
for all real numbers, a, b, and c
and given a= b
then
a + c = b + c
and c + a = c + b
we abbreviated it as
+prop=
Make sure to read this as the Addition Property of Equality
Now
a -c = b - c
is the Subtraction Property of Equality
-prop=
But realize with a tiny change
a + -c = b + -c you have the +prop=

x - 8 = 17
If we add 8 to both sides ---> That's using the +prop=
x -8 + (+8) = 17 + (+8)
x = 25


x -5 = 9
If we subtract 5 from both sides--> we are using the -prop=
x +5 (-5) = 9 + (-5)
x = 4

Multiplication Property of Equality
ac = bc
and ca = cb
and we write it
xprop=

Division Property of Equality
where c ≠ 0
a/c = b/c
and we write that
÷prop=

(-2/3)t = 8
If we use the Multiplication Property of Equality and
multiply BOTH SIDES by the reciprocal of -2/3
we have,
(-3/2)(2/3)t = 8(-3/2)
t = -12

⎮m⎮ =6
2
using the Multiplication Property of Equality (xprop=)

2⎮m⎮ =6(2)
2
⎮m⎮ =12
so m = -12 and 12
and using the set notation
we have the solution set as {-12, 12}