Math 6A ( Periods 2 & 4)

Graphs of Equations 11-9

An equation in two variables y = x + 2      produces an infinite number of ordered pairs
If we give x the value of 1, a corresponding value of y is determined
y = (1) + 2 = 3
The ordered pair is (1, 3)

If we let x = 4
y = (4) + 2 = 6
and we get the ordered pair (4, 6)

What happens if x = 0
y = (0) + 2 = 2 ( 0, 2)
or x = -1
y = (-1) + 2 = 1 ( -1, 1)

I like to remember ordered pairs---> ( ordered, pairs)

We graphed the line on a mini graph stickie.
The line is graphed using a ruler and connecting all the points we plotted. Put arrows at each end (since a line continues with out end) and write the line's equation right above the line.


For each value of x there is EXACTLY 1 value of y.
set of ordered pairs in which no two ordered pairs have the same x is called a FUNCTION


y = x + 2
in the future you will see it written as
f(x) = x + 2
so if x = 2
f(2) = (2) + 2 = 4
if x = 5
f(5) = (5) +4 = 9



We used a three column chart to compute our ordered pairs.
Please refer to the work sheet glued into your spiral notebook for the examples we completed from the class exercises found on Page 393 -- if you were absent, please come in one morning and I will review that chart with you.


The following equations create curves that are called PARABOLAS!! Notice the difference in these equations from our previous equations
y = x² +1
when we create your three column table using integers from -2 to 2
we notice
y = (-2)² +1 = 4 + 1 = 5 ordered pair (-2, 5)
y = (-1)² +1 = 1 + 1 = 2 ordered pair (-1, 2)
y = (0)² +1 = 0 + 1 = 1 ordered pair (0, 1)
y = (1)² +1 = 1 + 1 = 2 ordered pair (1, 2)
y = (2)² +1 = 4 + 1 = 5 ordered pair (-2, 5)

When you graph this... you get a "U" shaped graph.

Remember linear equations LINEar equations are lines!
and look like y = x + 2

PARABOLAS have the form y = x² or y = -x²

Let's try
y = 2 - x²
With our 3 column table
for values of x from -2 to 2
we find
y = 2 -(-2)² = 2 -(4) = -2 and the ordered pair is (-2,-2)
y = 2 -(-1)² = 2 - (1) = 1 and the ordered pair is ( -1, 1)
y = 2 -(0)² = 2 - 0 = 2 and the ordered pair is (0, 2)
y = 2 -(1)² = 2 -1 = 1 and the ordered pair is (1, 1)
y = 2 -(2)² = 2 - (4) = -2 and the ordered pair is (2, -2)

When you graph these ordered points you find you have an upside down U
hmmm... y = -x² results in a sad face parabola
and y = x² results in a happy face parabola!!

Algebra Honors (Periods 5 & 6)

Fractional Equations 7.4

The total resistance R of an electrical circuit with two resistors R₁ and R₂, that are connected in parallel is given by the formula
1/ R₁ +1/ R₂ = 1/R
What do you notice about the difference between this equation and those with fractional coefficients?

This formula is an example of a fractional equation.
An equation with a variable in the denominator of one or more terms is called a fractional equation. To solve a fractional equation, you can multiply BOTH sides of the equation by the LCD or you could use the method of solving a proportion when the equation consists of one fraction equal to another fraction.
3/x -1/4 = 1/12
The LCD of the fractions is 12x
Multiply BOTH sides of the equation by the LCD, 12x
Notice that x ≠ 0 because in this case 3/0 is undefined
12x(3/x – ¼) = (1/12)(12x)
36 -3x = x
36 = 4x
x = 9
{9}

(2-x)/(3-x) = 4/9
There are two different ways to solve this
First by finding the LCD, which is 9(3-x) Notice that x ≠3 Why?
9(3-x)[(2-x)/(3-x) = (4/9)[9(3-x)]
18-9x =12-4x
6 = 5x
6/5 = x
{6/5}
OR solve as proportion
(2-x)/(3-x) = 4/9

(2-x)(9) = (4)(3-x)
18- 9x = 12- 4x
we are at the same spot as with the first method and we arrive at the same solution
6 = 5x
6/5 = x
{6/5}

The following gets a little more complicated to do and to display here...
Solve
(2/b² - b) – 2/(b-1) = 1
Find the LCD by first factoring the denominators first
b² - b = b(b-1) so the LCD of the two fractions in this equation is in fact
b² - b BUT use it in factored form b(b-1) Notice: b ≠ 1 why?
(2/b² - b) – 2/(b-1) = 1
[b(b-1][ (2/b² - b) – 2/(b-1) ]= 1[b(b-1)]
which separates to
[b(b-1) (2/b² - b)] – [b(b-1)2/(b-1)] = 1
[b(b-1) (2/b(b - 1)] – [b(b-1)2/(b-1)] = 1
2 -2b=b(b-1)
or
2 – 2b = b² - b
solve for b now
0 = b² - b + 2b -2
0 = b² + b – 2
0 = (b-1)(b+2)
b = 1 and b = -2
Remember in this case b ≠ 1 because of the ORIGINAL EQUATION
the solution set is only
{-2)

Multiplying both sides of an equation by a variable expression sometimes results in an equation that has an extra root. You must check each root of the transformed equation to see if it satisfies the original equation.

Algebra Honors ( Periods 5 & 6)

Equations with Fractional Coefficients 7.3

Solving an equation with fractional coefficients can be easily accomplished by using the LCD of all the fractions in the equation. Clearing the equation of all fractions BEFORE attempting to solve the equation is probably the best way

Easy examples:
x/3 + x/7 = 10

The LCD of the fractions is 21
so multiply BOTH SIDES by 21

21(x/3 + x/7) = 10 (21)
7x + 3x = 210
10x = 210
x = 21
{21}

3a/5 – a/2 = 1/20

The LCD of the fractions is 20
20(3a/5 – a/2) = 1/20(20)
4(3a) -10a = 1
12a- 10a = 1
2a = 1
a = ½
{1/2}

x/3- (x+2)/5 = 2

The LCD of the fractions is 15
15[x/3- (x+2)/5] = 2(15)
5x –(3)(x+2) = 30
5x -3x-6 = 30
2x = 36
x = 18
{18}

2n + n/3 = n/4 + 5

The LCD is 12
12(2n + n/3) = (n/4 +5)(12)
24n + 4n = 3n + 60
28n = 3n + 60
25n = 60
n = 60/25 = 12/5
{12/5}


More complicated:
(1/4)(n + 2) – (1/6)(n – 2) = 3/2

The LCD is 12
12[(1/4)(n + 2) – (1/6)(n – 2)] = (3/2)(12)
3(n+2) – 2(n-2) = 18
3n + 6 -2n + 4 = 18
n = 8
{8}

Solving some word problems:
one eighth of a number is ten less than one third of the number. Find the number.
Let x = the number.
n/8 = n/3 – 10
LCD is 24
(24)(n/8)= (n/3-10)24
3n = 8n – 240
-5n = -240
5n = 240
n = 48
{48}

Math 6A ( Periods 2 & 4)

Graphs of Ordered Pairs 11-8

A PAIR of numbers whose ORDER is important is called an
ordered pair!!
(ordered, pair)

(2,3) is not the same as (3,2)

The two perpendicular lines are called axes.
The x-axis deals with the 1st number of the ordered pair and the y-axis deals with the 2nd number of the ordered pair.
The AXES meet at a point called the Origin (0,0)
The plane is called the coordinate plane

There are 4 quadrants, Use Roman Numerals to name them!!
Quadrant I ---> both the x and y coordinates are positive
(x,y) (+,+)

Quadrant II --> the x coordinate is negative but the y is positive
(-x,y) (-,+)

Quadrant III -->. both the x and y coordinates are negative
(-x,-y) (-,-)

Quadrant IV --> the x coordinate is positive but the y coordinate is negative
(x,-y) (+,-)

Math 6 High ( Period 3)

Discount & Sales Tax 7.7

Let’s go shopping!
A reduction in the price of an item is called a discount. Discounts are often calculated using percent of the original price.
To find the price of an item that is discounted:

1) find the amount of the discount

2) subtract the discount from the original price.

You buy a pair of hiking shoes. The original price of $65 is discounted by 35%. What is the sales price?

Solution:
Find the amount of the discount

Discount = 35% of $65

Discount = 65 ( .35)
=22.75

$22.75 is what you will SAVE!! It isn’t what you will pay to purchase them so

Remember

Sales price = Original price – discount
65- 22.75 = 42.25

$42.25

Which television’s sale price is less expensive? Let’s assume they are the same quality
Television A: Discount of 25% off the original price of $575
Television B: Discount of 30% off the original price of $585.

Television A:  Sales price = 575 – (0.25 · 575)
                        575 -143.75
                        431.75

Television B: Sales price = 585 –(0.30 · 585)
                        585- 175.5
                        409.50

Television B’s sales price of $409.50 is less expensive.

Finding Sales Tax

A sales tax is an amount added to the price of a purchased item. Sales tax is often calculated using a percent of the price.
To find the total price of an item including sales tax:

1) Find the amount of the sales tax

2) Add the sales tax to the price.

Example:
You buy an ipod shuffle. The price is $40.00 and the sales tax is 8.25% What is the total price you will pay?

Solution:
Find the amount of the sales tax.

Sales tax       = 8.25% of 40
                        =0.0825(40)
                        = 3.3

Which means the sales tax is $3.30

Add the sales tax to the price

Total price     = Purchase Price + Sales Tax
                        = $40 + $3.30 = $43.30

You will pay $43.30.

                       
Finding a Total Price

Shirts in a store are on sale for 30% off. The sales tax is 7.25%.  What is the total price of a shirt that has an original price of $29.95?

Solution: First find the discount and  the sale price.
Then find the sales tax and add that to the sale price to find the total price.
Careful—lots of steps here!

Finding the discount and the sales price.

Discount         = 30% of 29.95
                        = 0.3(29.95)
                        = 8.99
Sale Price     = Original Price – Discount
                        = 29.95 – 8.99 = 20.96

Sale Price is $20.96

That is the amount you calculate your sales tax upon so

Finding the Sales tax and the Total Price

Sales tax       = 7.25% of 20.96

                        = 0.0725(20.96)=1.52

Total Price     =  Sale Price + Sales Tax
                        = 20.96 + 1.52 = 22.48

The total price including sales tax is $22.48 

Math 6 High ( Period 3)

Simple Interest 7.6

When you put money in a savings account at a bank, the bank pays you interest. When you borrow money from a bank, you must pay the bank interest to the bank in addition to paying bank the money that you borrowed.

A principal is an amount borrowed, loaned or saved. An annual interest rate is the percent of the principal you earn or pay as interest for the year.
SIMPLE INTEREST is the product of the principal, the annual interest rate , and the time in years.

Interest = Principal ·rate ·time

I=Prt

Examples:

You borrow $1500 from a friend for the down payment on a car. Your friend charges you an annual interest rate of 8% ( Nice friend!) Find the simple interest you will pay in 1 year.

Solution:
Remember   I = Prt

What do you know?

I = (1500) (0.08)(1) 
Most of the time change the interest rate to a decimal. Occasionally you will want to use fractions

Carefully multiply... and discover that

I = 120

Therefore: The simple interest you will pay in 1 year is $120.

You deposit $300 in a savings account. The annual interest rate is 3% (not much). Find the simple interest you will earn in 1 month.

What do you know?   You know the principal  ( or P in the formula) is $300. You know the annual interest rate ( or r in the formula)  is 3%  But.. wait… be careful…the time must be in years  this is only for one month. The time (or t in the formula is 1/12)

Solution:

Remember   I = Prt

I = (300)(0.03)(1/12)

I would carefully simplify before I multiplied

I = 0.75

Again, think what does that mean?
The simple interest you would earn on $300 in 1 month at that rate is $0.75 

When you know the values of any three of the variables in the formula   I = Prt  you can use substitution to find the value of the fourth variable.

Your savings account earns $68 in simple interest in 1 year. The annual interest rate is 8% what is the principal?

Ask yourself, what do I know? What are the three variables  in the formula  I = Prt   that I do know?

$68 is the interest. The rate is 8% and it’s only for 1 year.

so

Remember   I = Prt

68 =P(0.08)(1)

68=0.08p

Solve this one-step equation

Divide both sides by 0.08 carefully

68/0.08 = 0.08p/0.08

850 = p

Therefore the principal is $850

You want to open an account with $100.
At Bank A you will earn $0.65 in 3 months.
At Bank B you will earn $2.25 in one year. At which bank will you open your account?
Although banks do not pay simple interest (they use compound interest)  we will assume that Bank A and B are paying only simple interest.

Figure out each bank’s simple interest rate.

Remember   I = Prt  

Bank A:   3 months is 3/12 or ¼  of a year so 

0.65 =  (100)(r) (¼)
0.65= 25r

Solve this one step equation
Divide both sides by 25

0.65/25= 25r/25

0.026= r
We need to change the decimal into a percent
That means Bank A’s rate is 2.6%

Bank B: Yes you can use the formula but look at what you know

2.25 =(100)(r)(1)
That means  r= 0.225
or that the rate for Bank B is 2.25%

You would earn more money if you deposit your savings into Bank A

Math 6A (Periods 2 & 4)

Equations: All 4 Op's Sections 8.1- 8.5 Summary 

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.

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



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

Equations with Decimals or Fractions 
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


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 6A (periods 2 &4)

Solving Equations 11-7

Now that we have learned about negative integers, we can solve an equation such as
x + 7 = 2
We need to subtract 7 from both sides of the equation
x + 7 = 2
- 7 = - 7

to do this use a side bar and use the rules for adding integers
Notice the signs are different so
ask yourself... Who wins? and By How Much?
stack the winner on top and take the difference
so
x + 7 = 2
- 7 = - 7
x = -5



t - -10 = 19
becomes -- with add the opposite---
t+ + 10 = 19
which is just
t + 10 = 19
so subtract 10 from both sides
t + 10 = 19
- 10 = - 10
t = 9

w - - 26 = -44
"Add the Opposite"
w + + 26 = -44
- 26 = - 26
x = -70

Know your integer rules and it becomes easy!!
Side bars are great, if you need them with difference signs!!


y -^- 6 = 4
add the opposite and you get
y + 6 = 4
now you need to subtract 6 from both sides of the equation
y + 6 = 4
- 6 = - 6
Again the signs are different -- ask your self those all important questions
"Who Wins? and "By How Much?"
Use a side bar, stack the winner on top and take the difference. Make sure to use the winner's sign in your answer!!
y = 2

What about -5u = 125?
Whats happening to u?
It is being multiplied by -5... so you must divide by -5

-5u = 125
-5 -5
u = -25

or written easier to read -5u/-5 = 125/-5
u = -25

(1/-9)c = 33
Need to multiply both sides by the reciprocal of (1/-9) which is (-9/1)

(-9/1)(1/-9)c = 33(-9/1)
c = -297



2- STEP EQUATIONS

What about

3u - 1 = -7
You need to do the reverse of PEMDAS... remember unwrapping the present? We did the exact opposite of what we had done to wrap the present!!
so
3u - 1 = -7
+ 1 = + 1
3u = -6
Now divide by 3 on both sides
3u/3 = -6/3
u = -2


3z - ^- 15 = 9
add the opposite first and you get
3x + 15 = 9
In order to solve this 2 step equation
we need to do the reverse of PEMDAS-- as we did with unwrapping the present so many months ago
3x + 15 = 9
subtract 15 from both sides of the equation
3x + 15 = -9
- 15 = - 15

This time the sides are the same-- so just add them and use their sign
3x + 15 = -9
- 15 = - 15
3x = -24
Now divide both sides by 3
3x = -24
3 3

x = -8

Make sure to BOX your answer!!
What about this one
(1/2)(x) + 3 = 0
subtract 3 from both sides
(1/2)x = -3

Multiple by the reciprocal of 1/2 which is 2/1
(2/1)(1/2)x = -3(2/1)
x = -6
Again box your answer.



What about x = -6 + 3x
OH dear... we have variables on BOTH sides of the equations... we need to get the variables on one side all the constants on the other.
We need to isolate the variable!!

x = -6 + 3x
What if we add six to both sides
x = -6 + 3x
+6 = + 6
x + 6 = 3x
now we need to subtract x from both sides
x + 6 = 3x
- x - x
6 = 2x
so now divide both sides by 2
6/2 = 2x/2
3 = x

How about this one
3 - r = -5 + r
- 3 = - 3
-r = -8 + r
if subtract r from both sides, I will get rid of the +r on the right side

-r = -8 + r
- r = -r
-2r = -8
Now divide by -2 on both sides

-2r/-2 = -8/-2
r = 4

Math 6 High ( Period 3)

Solving Percent Problems 7.4

You can use a proportion to find what percent one number is of another number.
The statement:

“ a is p percent of b” is expressed by the proportion shown below where

a is the  part of the base      b is the base… and    p is the percent

a/b = p/100

Part of the base / base = percent/100

What percent of 40 is 15?

First think which is the base? 40

Which number is the part?  15

Now write the proportion

a/b = p/100

15/40 =  p/100

Use cross products to multiply… simplify first if you can…

15(100) = 40p

Divide both sides by 40

15(100)/40 = 40p/40

37.5 = p

Note: there are a number of ways to simplify first! You should still arrive at p = 37.5 which means 37.5%

Finding a Percent

Your team won 19 of its 25 softball games. What percent did it win?

Games won/games played = percent/100

It really is still      a/b = p/100

19/25 = p/100

19(100) = 25p

Divide both sides by 25

19(100)/25 = 25p/25

76= p

Again there are a variety of ways to simply BEFORE you multiply! You could even utilize the equivalent fraction method to determine the p percent.

Therefore your team won 76% of its games!

You can use the proportion to find a base or a part of the base.

Finding a part of the base

You buy a pair of pants on sale. The price is 80% of the regular price of $24.50, WOW! That’s a bargain! What is this incredible sales price?

Sale price / regular price = Percent/100

a/24.50 = 80/100

Multiply and divide carefully

100a = 24.50(80)

divide both sides by 100

100a/100 = (24.50)(80)/100

a = 19.6

The sale price is $19.60

Now, you could use the writing an equation method

80% of 24.50 = sale price  or

(.80) (24.50) = p

Multiply carefully

19.60 = p

$19.60 is the sale price.

Finding a Base

8 is 32% of what number?

Method 1: Write and solve a proportion

a/b = p/100

8/b = 32/100
Using cross products
(8)(100) = 32b
Divide both sides by 32

(8)(100)/32 = 32b/32

25 = b

Method 2: Write and solve an equation

8 is 32% of what number?
8 = .32(x)

8/.32 = .32x/.32
Divide carefully

25= b

8 is 32% of 25

Solving Percent Problems

a/b = p/100   means a is p percent of b

Out of 80 students taking Algebra, 95% passed the final exam. About 80% of those who passed the exam got a grade better than a C. How many students got an A or a B?

First find out how many passed the final exam

Using the proportion method for this:

a/80 = 95/100

100a = 80(95)

100a/100 = (80)(95)/100

a = 76

76 students passed the final exam

About 80% of 76 got a grade better than a C

Using the writing an equation method

(.80)(76)  =60.8

Since you can’t have a part of a student your answer would be about 61 students got an A or a B in Algebra

Math 6High ( Period 3)

Large & Small  Percent 7.3

The methods you learned from Lesson 7.2 can be used to find large and small percents.

Rewrite Numbers

Rewrite 1.4 as a percent
1.4 = 1  4/10 = 14/10 = 140/100 = 140%

Rewrite   ½%  as a decimal
½ %= 0.5% = 0.5/100 = 5/1000 = 0.005

Finding a Large Percent of a Number

You can use either of two methods to find 130% of 250

Method 1: Set up a proportion      percent/100 = is/of

130/100 = x/250

Cross multiply carefully

(130)(250) = 100x  Then divide by 100 on both sides

(130)(250)/100 = 100x/100

325 = x

There were a number of different ways to simplify BEFORE you multiply but do them carefully . You still need to get x = 325

Method 2: Rewrite 130% as a decimal and set up a written equation

Find 130% of 250.   Think “ Find”  means “What is…” so you have

What is 130% of 250
x = (1.3 )(250)
x = 325

Find a Small Percent of a Number

You can use the same two methods to find 0.5% of 40

Method 1: Set up a proportion.

.5/100 = x/40  or

5/1000 = x/40

Cross multiply and divide carefully
(5)(40)/1000 = 1000x/1000

0.2 = x

There were a number of different ways to simplify BEFORE you multiply but do them carefully . You still need to get x = 0.2

Method 2: Rewrite 0.5% as a decimal and use the writing an equation method

Find 0.5% of 40   Think “ Find”  means “What is…” so you have
What is 0.5% of 40.
x = (0.005)(40)
x = 0.2

Estimating:

In many real-life situations—such as figuring the amount for a tip, it is helpful to use mental math to estimate the percent of a number.

Use mental math to estimate 123% of 84. You know it will be MORE than 84… because the percent is over 100% and we know 123%  is about 125%   100% + 25% so all we need to figure is 25% o 84… and that is just ¼ so 21.. 84 + 21 = 105. So 123% is just a little under 105.  

Use mental math to find ½ % of 140

Wait… ½ % is just half of 1% so  figure 1% of 140. Just move the decimal over two places…1.4 would be 1% so half of that is 0.7

estimate 15% tip for a restaurant bill of $43.76

First round the restaurant bill up to $45.00. Then take 10% … that’s 4.50

5% is half of ten percent so half of 4.50  is 2.25

4.50 + 2.25 = 6.75  $6.75 is about 15% of43.76

Algebra Honors ( Periods 5 & 6)

 Solving Problems Involving Quadratic Equations 12-6

You can use quadratic equations to solve problems...
We used the examples in the book to start with:

The park commission wants a new rectangular sign with an area of 25 m² for the visitor center. The length of the sign is to be 4 m longer than the width . To the nearest tenth of a meter, what will be the length and the width of the sign?
Always make sure you check before AND after-- to see what the problem is really asking for...
Let x = the width in meters
then x + 4 = the length in meters

Use the formula for the area of a rectangle to write an equation

x(x+4) = 25
solve it
x(x +4) = 25 becomes
x² + 4x = 25
You can use two methods: the quadratic formula would be my second choice since completing the square works easily here

x² + 4x + 4 = 25 + 4
(x + 2)² = 29
x + 2 = ±√29
You can use your calculator, or the table of square roots... or approximately easily using the method taught in class earlier this year
but to the nearest tenth you get
-2 + √29 ≈3.4
-2 - √29 ≈-7.4

Since you can't have a negative root since a negative length has no meaning... you know the width must be about 3.4 meters and therefore the length is 3.4 +4 or approximately 7.4 meters

Problem 2:
The sum of a number and its square is 156. Find the number

Let x = the number
then
x² + x = 156

x² + x - 156 =0
Using the skills you have for factoring
(x +13)(x-12) = 0
so
x = 13 and x = 12
You have two solutions to this question!!

Problem 3:
The altitude of a triangle is 9 cm less than the base. The area is 143 cm²
What are the altitude and base?

Remember the formula for the area of a triangle is A = ½bh

Let b = the length of the base
then the altitude ( the height) is b-9
so
½(b)(b-9) = 143
b² -9b = 286
b² -9b - 286 = 0
(b -22)(b +13) = 0
b = 22 and b= -13
You can't have a negative length so
the base is 22 cm and the altitude is 13 cm


An object that moves through the air and is solely under the influence of gravity is called a projectile. The approximate height (h) in meters of a projectile at t seconds after it begins its flight from the ground with initial upward velocity v₀ is given by the formula
h = -5t² + v₀t
We can find when such a projectile is at ground level (h=0) by solving
0 = -5t² + v₀t.
If a projectile begins its flight at height c, its approximate height at time t is
h = -5t² + v₀t + c .
We can find when it hits the ground by solving h=0 or
0 = -5t² + v₀t +c.

When a projectile is thrown into the air with an initial vertical velocity of r feet per second, its distance (d) in feet above the starting point t seconds after it is thrown is approximately
d = rt – 16t²

Math 6A (Periods 2 & 4)

Products of Integers 11-4 & 11-5


3 ⋅ -2 = -6
Its really repeated addition
or
-2 + -2 + -2 which we learned a few sections ago was equal to -6.

The product of a positive integer and a negative integer is a negative integer.

The product of ZERO and any integer is ALWAYS ZERO!!
a⋅0 = 0

Math imitates life...and Karma(?)
What was the story I told in class... it applies to
Multiplication & Division ...
+ ⋅ + = +
- ⋅ + = -
+ ⋅ - = -
- ⋅ - = +



The product of -1 and any integer equals the opposite of that integer.
(-1)(a) = -a

The product of two negative integers is a positive integer

For a product with NO ZERO factors:
-->if the number of NEGATIVE factors is odd, the product is negative
-->if the number of NEGATIVE factors is even, then the product is positive

Every integer and its opposite have equal squares!!

Remember-- if its all multiplication use the Associative & Commutative Properties of Multiplication to make your work EASIER!!

Math 6 High ( Period 3)

 Finding a Percent of a Number 7.2

or… it’s just proportions!

Percent/100 = Percent of a number/number

Find 30% of 400
30/100 = x/400
Its a proportion!
30(400) = 100(x)
30(400)/100 = x
120 = x
Therefore 120 is 30% of 400

You drive 300 miles. the first 20% of the road is under construction. How many miles of the road are under construction?
20/100 = x/300
Its a Proportion!!
20(300) = 100(x)
20(300)/100 = x
60= x
therefore 60 miles of the road are under construction.

You can also change the percent to a fraction and multiply!
For example,
Find 50% of 46. 
[Remember: the word "OF" in a mathematical sentence means to multiply]

50% is just 1/2 
so
1/2(46) = 23

75% of 120
75% = 3/4
so 3/4 (120) = 90

You can also change the percent into a decimal and multiply
Find 35% of 150
Read this as
What is 35% of 150?
x= 0.35(150)
x = 52.5
Find 92% of 75
Read this as
What is 92% of 75?
x = 0.92(75)
x = 69

Are these reasonable?
Well 92% is pretty close to 90% and !0% of 75 would be 7.5 so 90% would be 75-7.5 or 67.5.   Since  69 is very close to 67.5  the answer is reasonable.

Math 6 H ( Period 3)

Percents, Fractions, and Decimals 7.1

The word percent means “per hundred.” The symbol is %.
A percent is a ratio whose denominator is 100.  

25% = 25/100 = 0.25

To write a fraction or a decimal as a percent or to write a percent as a fraction or a decimal, the first step is to rewrite the given form as a fraction with a denominator of 100.  

¼ = 25/100 = 25%

2/5 = 40/100 = 40%

0.75 = 75/100 =75%

0.8 = 8/10 = 80/100 = 8%

0.104 = 104/1000 = 10.4/100 = 10.4%

86% = 86/100 = 0.86

15% = 15/100 = 0.15

Some percents need to be written exact – or rounded to a specific decimal place.

For example

1/3 = 0.3333333...

to write as an exact percent you would write 33 1/3 %

You would round it to the nearest tenth as 33.3%

1/8 = 12.5% but if it asked to round to the nearest tenth you would write 13%