Algebra Honors (Period 6 & 7)

Scientific Notation 7-10
Scientific notation makes it easier to work with either very large numbers or very small numbers. To write a positive number in scientific notation, you express it as the product of a number greater than or equal to 1 BUT less than 10 AND an integral power of 10.
When a positive number greater than or equal to 10 is written in scientific notation, the power of 10 is positive. When the number is less than 1, the power of 10 will be negative.

Scientific notation consists of two parts
1≤ x< 10 × POWER¹⁰
for example
58,120,000,000
Move the decimal point left 10 spaces to between the 5 and the 8 to get a number BETWEEN 1 and 10
5.8
so 5.8 × 10¹⁰
0.00000072
Move the decimal point 7 places to the right to get a number between 1 and 10
7.2
7.2 × 10^-7

Numbers written in scientific notation can be multiplied and divided easily by using the rules of exponents
Simplify. Keep your answers in scientific notation
3.2 × 10⁷/2.0 × 10⁴
= 3.2/2.0 ×10⁷/10⁴
=1.6 × 10^7-4
=1.6 × 10³

(2.5 × 10³)(6.0 × 10²)
=(2.5 ×6.0) × (10³ × 10²) Add exponents
= (15)(10³⁺²)
=15 × 10⁵
But wait 15 is not between 1 and 10
15 itself is 1.5 × 10¹
so it becomes
1.5 ×10¹× 10⁵
1.5 × 10¹⁺⁵
=1.5 × 10⁶
The distance from the sun to Mercury is approximately 6 × 10⁸ km.
The distance from the sun to Pluto is approximately 5.9 × 10⁹ km
Find the ratio of the first distance to the second distance
Distance from sun to Mercury
Distance from sun to Pluto
6 × 10⁸
5.9 × 10⁹
= 6/5.9 × (10⁸)/10⁹

= 6/5.9 ×10^8-9
6/5.9 ×10^-1
=6/5.9 ×1/10
= 6/59

You learned expanded notation in 6th grade... but a review of writing numbers in expanded notation using powers of 10 follows:
8572 = 8000 + 500 + 70 + 2
=8(10³) + 5(10²) + 7(10¹) + 2(10⁰)

0.3946 = 0.3 + 0.09 + 0.004 + 0.0006
= 3(10^-1) + 9(10^-2) +4(10^-3) + 6(10^-4)

25.03 = 2(10¹) +5(10⁰) + 0(10^-1) + 3(10^-2)

The metric system is also based on powers of TEN.
To change from one metric unit to another, you simply multiply by a power of 10.
Remember:
King Henry Died By Drinking Chocolate Milk
1 km = 10³m = 1000m
1mL =10^-3 L = 1/1000 L

Algebra Honors (Period 6 & 7)

Negative Exponents 7-9
If a is a nonzero real number and n is a positive integer
a^-n = 1/aⁿ
so
10^-3 = 1/10³ = 1/1000
5^-4= 1/5⁴ = 1/625

16^-1 = 1/16

The rule of exponents for division (page 190) will help you understand why
a^-n = 1/aⁿ
recall that for m > n a^m/aⁿ= a^m-n
For example
a⁷/a³ = a^7-3= a⁴
you can also apply this rule when m < n that is when m - n becomes a negative number. For example a³/a⁷ = a^3-7 = a^-4
since
a⁷/a³ and a³/a⁷ are reciprocals then
a⁴ and a^-4 must also be reciprocals.
Thus
a^-4= 1/a⁴
a⁵/ a⁵ = a^5-5 = a⁰
But you already know that a⁵/a⁵ = 1
SO, definition of a⁰
a⁰ = 1
However, the expression 0⁰ has no meaning

All the rules for positive exponents also hold for zero and negative exponents.

Summary of Rules for Exponents
Let m and n be any integers
Let a and b be any non zero integers
Review—>But you should really know these because of our Powers Project
Products of Powers
b^mbⁿ = b^m+n
Example with negative exponents
2³⋅2^-5 = 2^3+(-5) = 2^-2 = 1/2² = 1/4
Quotient of Powers
b^m ÷ bⁿ = b^m-n
Example with negative exponents
6³÷6⁷= 6^3-7= 6^-4= 1/6⁴= 1/1296
Power of Powers
(b^m)ⁿ = b^mn
Example with negative exponents
(2³)^-2 = 2^-6 = 1/2⁶ = 1/64
Power of a Product
(ab)^m= a^mb^m
Example with negative exponents
(3x)^-2 = 3^-2 ⋅x^-2 = 1/3²⋅1/x² = 1/9x²
Power of a Quotient
(a/b)^m= a^m/b^m
Example with negative exponents
(3/5)^-2= 3^-2/5^-2= (1/3²)/ (1/5²)= 1/3² ÷ 1/5² which means
1/3² ⋅5²/1= 5²/3²= 25/9

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

Combined Operations 8-5

In order to solve an equation of the form
ax + b = c or ax –b = c or b – ax = c
where a, b, c are given numbers and x is the variable, we must use more than one transformation

Solve the equation 3n - 5 = 10 + 6
Simplify the numerical expression

3n - 5 = 10 + 6
3n – 5 = 16

add 5 to both sides

3n – 5 + 5 = 16 + 5
or
3n – 5 = 16
+ 5 = +5

3n = 21

divide both sides by 3 (or multiply each side by the reciprocal of 3)
3n/3 = 21/3

n = 7

General procedures for solving equations

Simplify each side of the equation
If there are still indicated additions or subtractions, use the inverse operation to undo them
If there are indicated multiplications or division involving the variable, use the inverse operations to undo them

The books says you must always perform the same operation on both sides of the equation. I say, “do to one side what you have done to the other side.”

Solve the equation
(3/2)n + 7 = 22
subtract 7 from both sides
(3/2)n + 7 - 7 = 22 - 7
(3/2)n =15

multiply both sides by 2/3, the reciprocal of 3/2

(2/3)(3/2)n = 15(2/3)
n = 10

Solve the equation
40 – (5/3)n = 15
add (5/3)n to both sides

40 – (5/3)n + (5/3)n = 15 + (5/3)n

40 = 15 + (5/3)n
subtract 15 from both sides

40 – 15 = 15-15 + (5/3)n

25 = (5/3)n multiply both sides by 3/5

(3/5)(25) = (5/3)n (3/5)

15 = n

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

Equations: Decimals and Fractions 8-4

You can use transformations to solve equations which involve decimals or fractions
Solve 0.42 x = 1.05
Divide both sides by 0.42
.42x/.42 = 1.05/.42
now, do side bar and actually divide carefully and you will arrive at
x = 2.5

Solve: n/.15 = 92
multiply both sides by .15 to undo the division
(n/.15)(.15) = 92 (.15)
Again, do a sidebar for your calculations and you will arrive at
n = 13.80

How would we solve the following: (2/3)x = 6?

Let’s look at 2x = 6. What do we do?
We divide both sides by 2—or multiply both sides by the reciprocal of 2—which is ½
Remember the product of a number and its reciprocal is 1

Reminder: the ultimate objective is applying transformations to an equation is to obtain an equivalent equation in the form x = c
(where c is a constant.)

Also remember that the understood (invivisble) coefficient of x in the equation x = c is 1.

[Can you picture the poster in the front of the room?]

So to solve (2/3)x = 6 you would divide both sides by 2/3 but that is the same as multiplying by the reciprocal of 2/3, which is 3/2.

If an equation has the form

(a/b)(x) = c,
where both a and c are nonzero,
multiply both sides by b/a, the reciprocal of a/b
Solve (1/3)y = 18

(3/1)(1/3)y = 18(3/1)

y =18(3)
y = 54

Solve the equation: (6/7)n = 8
(7/6)(6/7)n = 8(7/6)

n = 8(7/6)

simplify first , then multiply
n = 28/3
n = 9 1/3
Let’s check

(6/7)n = 8
well, we said that n = 9 1/3 so substitute back, but change to 28/3 first
(6/7)(28/3) ?=? 8
[read ?=? as ‘does that equal?’]

Now really do a side bar with the left side of the equation to see what
(6/7)(28/3) really equals. Simplify before you multiply
2(4) = 8 so
8 = 8



Try:
1. (1/7)a = 13

2. b/8 = 16

3. 3.6d = 0.9

4. (3/8)f = 129

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

Equations: All Four OP's 8-2 & 8-3

If the replacement set for an equation is the set of whole numbers, it is not practical to use substitution to solve the equation. Instead we transform or change the given equation into a simpler, equivalent equation. When we transform the given equation, our goal is to arrive at an equivalent equation of the form
variable = number

Transformation by addition: add the same number to both sides
Transformation by subtraction: subtract the same number from both sides

solve x – 2 = 8
our goal is to find an equivalent equation of the form
x = a number

The left side of the given equation is x – 2. Recall that addition and subtraction are inverse operations. If we add 2 to both sides the left sides simplifies to x
x-2 = 8
x – 2 + 2 = 8 + 2 (We usually show the +2 right below each side of the equation)
x = 10


Solve x + 6 = 17
Subtract 6 from both sides of the equation to get an equivalent equation of the form
“ x = a number”

x + 6 = 17
x + 6 – 6 = 17 – 6 (Again, we usually show the -6 right below each side of the equation)
x = 11
the solution is 11

In equations involving a number of steps, it is a good idea to check your answer. This can be done easily by substituting the answer in the original equation.
What about the following
34 – x = 27
add x to both sides
34 – x + x = 27 + x
34 = 27 + x
subtract 27 from both sides
34 – 27 = 27 – 27 + x
7 = x


If an equation involves multiplication or division, the following transformations are used to solve the equation:

Transformation by multiplication: Multiply both sides of the equation by the same nonzero number.
Transformations by division: Divide both sides of the equation by the same nonzero number.

Remember: Do undo on one side what you would do undo the other!!

Our goal is to get the variable alone and to find an equivalent equation of the form
“n = a number”
Our goal is to arrive at the “world’s easiest equation”

Solve 3n = 24
Use the fact that multiplication and division are inverse operations
3n/3 = 24/3
n = 8
Or you could have use the reciprocal of 3--> which is 1/3 and multiplied both sides by 1/3
(1/3)(3n) = 24(1/3)
n = 8 and still arrived at the SAME solution

Solve 5x = 53

5x/5 = 53/5
x = 10 3/5

Solve n/4 = 7
(4)(n/4) = 7(4)
n = 28
How could you know for sure your answer is correct?
Substitute your solution into the ORIGINAL equation
Try:
1. 3r = 57

2. 714 = 7t

3. Solve A = bh for h

4. Solve P = 4s for s

5. Solve C = 2πr for r




You may be able to solve some of the equations in the homework without pencil and paper. Nevertheless, it is important to show all the steps in your work and to make sure you can tell which transformation you are using in each step.


Remember when we stated the properties as well-- back in our 1st quarter!!

Algebra Honors (Period 6 & 7)

Work Problems 7-8

To solve work problems use the following formula
work rate × time = work done
or rt = w
Work rate means the fractional part of a job done in a given unit of time.
For example if it take you 3 hours to clean up your room, what part of the job can be done in 1 hour? That's easy... 1/3
To finish a job the sum of the fractional parts of the work done must be 1.
( for one whole job completed)

Josh can split a cord of wood in 4 days. His father can split a cord in 2 days. How long will it take them to split a cord of wood if they work together?
Let x = the number of days needed to do the job together.
Josh and his father will each work x days
Using those great tables from class fill in with the information you know
Since Josh can do the whole job in 4 days his work rate is 1/4 job per day.
His father's work rate is 1/2 job per day.

**posting the TABLE HERE**

Josh's part of the job = x/4
His father's part of the job = x/2
so the sum of that would equal the job completed
OR

Josh's part of the job + His father's part of the job = Whole JOB
x/4 + x/2 = 1
Clear the equation of fractions by multiplying by the LCD
4(x/4 + x/2) = 4(1)
x + 2x = 4
3x = 4
x= 4/3
It would take them 1 1/3 days to do the job together.

Robot A takes 6 minutes to weld a fender. Robot B takes only 5 1/2 minutes. If they work together for 2 minutes, how long will it take Robot B to finish welding the fender by itself?

Let x = the number of minutes needed for Robot B to finish the work.
Robot B's work rate is 1/5.5 or 1/(11/2) = 2/11

***posting the TABLE HERE***
Robot A's part is (1/6)(2)
Robot B's part is (2/11)(2 +x)
A's part of the job + B's part of the Job = Whole JOB

1/3 + (2/11)(2 + x) = 1
Multiply by the LCD, which is 33

(33)[1/3 + (2/11)(2 + x)] = 33(1)
11 + 6(2 + x) = 22
11 + 12 + 6x = 33
6x = 10
x = 5/3
It will take 1 2/3 minutes for Robot B to finish welding.
The charts or tables for work problems look similar to the charts and tables used for other problems. The following formulas show the similarities among some types of problems you have studied

Work done by A + work done by B = TOTAL work done
Acid in solution A + acid in solutions B = TOTAL acid in mixture
Interest from banks + Interest from Bonds = TOTAL Interest
Distance by bike + Distance by car = TOTAL distance traveled