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POWERS OF TEN

Welcome to MATH... Welcome to K101

Our Class Blog

Welcome to our class blog... where you can earn extra credit by adding your own relevant comments about our class notes for the day... or where you can find answers from others in your class. Check here often, especially if you have been absent. You might just find out the math strategy that works for you!!

Math 6A (Periods 1 & 2)

Circles 4-6

A circle is the set of all points in a plan at a given distance from a given point O (called the center).
A segment joining the center to a point on the circle is called a radius ( plural: radii) of the circle. All radii of a given circle have the same length and the length is called the radius of the circle.

A segment joining two points on a circle is called a chord... and a chord passing through the center is a diameter of the circle. the ends of the diameter divide the circle into two semicircles. The length of a diameter is called the diameter of the circle.
Two radii equal one diameter-- a fact we will use in the formulas below

The perimeter of a circle is called the circumference and the quotient

circumference ÷ diameter is the same for all circles--> regardless of size
This quotient is denoted by the Greek letter ∏ ( pronounced "pie")
No decimal gives ∏ exactly
No fraction gives ∏ exactly, either

A fairly good approximation is either 3.14 or 22/7

If we denote the circumference by C and the diameter by d we can write

C ÷ d = ∏
This formula can be put into several useful forms.

Let C = circumference d = diameter and r = radius
Then:

C = ∏d
d = C/∏

C = 2∏r
and
r = C/(2∏)

We tried a few examples.
Using ∏≈ 3.14 and rounding to three digits, as described by our textbook.
The diameter of a circle is 6 cm. Find the circumference.
WE are given d and are asked to find C.
WE use the formula
C = ∏d
C ≈ 3.14(6) = 18.84
C ≈ 18.8
So, the circumference is approximately 18.8 cm


The circumference of a circle is 20 feet. Find the radius.
To find the radius, use the formula
r = C/(2∏)
r = 20/2∏
Simplify first
r = 10/∏
r ≈ 10/3.14
r ≈ 3.1847
Since the third digit from the left is in the hundredths' place, round to the nearest hundredth.
r ≈ 3.18
The radius is approximately 3.18 feet

A polygon is inscribed in a circle if all of its vertices are on the circle. Check on the diagram in our textbook on page 129-- we added that to our notes as well.

Three noncollinear points (not on a line) determine one and only one circle that passes through the three given points.

Circles 4-6 continued

We continued our study of circles by examining irregular shapes and determined their perimeters.
To see each of the irregular shapes turned to page 131. The numbers we used in class were all different that those of 24-27, but use the shapes to help solve the following:

The circles in the diagrams are parts of circles and the angles are right angles. We found the perimeter of each figure.
The first figure was a semicircle ( see #24 with a diameter of 4).
We noticed that we needed to start with
C =∏d but then we only need half of that
so
∏d/2 or 4∏/2 = 2∏
Using ∏≈ 3.14
we found
≈3.14(2) = 6.26
BUT.. that only was the upper part we needed to add the diameter of 4 to make sure we had all we needed in our perimeter.
6.28 + 4 = 10.28 units... but then we needed to round to 3 digits-- according to our textbook so
10.3 units would be a good approximation for the first perimeter.

Our 2nd irregular shape was a quarter of a circle with a radius of 6
Again, look to our textbook, page 131 # 25 for the shape. Use 6 as the radius.
This time
C = 2∏r
BUT... we only need 1/4 so
(2⋅∏⋅6)/4 = 12∏/4 = 3∏
≈ 3.14(3) = 9.42
BUT... wait.. we aren't finished... we have to sides of this quarter circle that we need to include in our perimeter.
so the perimeter is approximately 9.42 + 6 + 6 = 9.42 + 12
≈ 21.42 which round to ≈ 21.4 units


The next irregular shape looks like something from Griffith Park observatory. Make sure to use the diagram for # 26 but use a radius of 2 as we did in class.

C = 2∏r
but you notice we only need half of the full circle so
2∏r/2 or just ∏r
Now substitute int he radius-- which is also 2
2∏ ≈ 2(3.14) = 6.28
But... we still need the bottom perimeter
so this shape ≈ 6.28 + 2 + 4 + 2 or 6.28 + 8
≈14.28
≈14.3 units

Period 7 said the next shape ( # 27 in our textbook) looks like a bandaid... What do you think?
(At least.. with the way I drew it.. having a radius of 10)
Again we need the formula
C = 2∏r
This time we realized we had two semicircles.. but that is one whole circles so we kept the formula and substituted in our radius of 10
C =2(10)∏
=20∏
≈20(3.14)= 62.8
Then we added the two sides of 10 and found the perimeter to be approximately
62.8 + 10 + 10
≈82.8 units


I described the last shape to be a teardrop. Make sure to check #28 in our textbook. I actually used the same radius as the book so that drawing is exactly what we did.

C =2∏r
C =2⋅6⋅∏ = 12∏
But... we only need 3/4 of the circle so what is 3/4 of 12... in class everyone knew it was 9 so
3/4 of 12∏ is 9∏
≈9(3.14) = 28.26
But then we need to make sure we include the two sides of 6 each
≈28.26 + 6 + 6 = 28.26 + 12
≈40.26 units
and rounding to three digits
≈ 40.3 units.

Algebra Honors (Periods 6 & 7)

Inequalities in One Variable
Solving Problems Involving  Inequalities 10-3

For practice, we went through the examples in our textbook on Page 469
We discovered that reading and re-reading the problem was critical to make sure we answered the exact question.  As noted in Example 1: the question asked what is the minimum total distance, to the nearest mile, that she will have to travel...?"  The critical part, was "to the nearest mile."  Please read the problem and then realize  why re arrived at the following:
Let d = the distance fro the sign to home
d - 16 > 25
solving that open sentence we get
d > 41
Since the distance needs to be greater than 41, the next whole number is 42 so the answer is
The minimum distance she will travel is 42 miles.

To translate phrases such as "is at least" and "is no less than"  you will need   ≥ 
...think I want at least $200 when going to Disneyland. I obviously want more.. but I will be happy with $200.

To translate phrases such as "is at most" and "is no more than"  you will need   ≤
...think I want at most 7 problems of homework. I really want fewer than 7 but I'll be okay with 7.

Algebra Honors ( Periods 6 & 7)

Inequalities in One Variable
Order of Real Numbers 10-1

This sections should be review... you have been working with less than and greater than symbols since 6th grade. A number line shows order relationships among all real numbers. The value of a variable may be unknown but you may know that is either greater than or equal to another number.
For example,
x ≥ 5 is read "x is greater than or equal to 5."
x ≥ 5 is another way of writing " x >5 or x = 5"

Translating statements into symbols is a critical concept. Practice these as review
-3 is greater than -5.    -3 > -5
and
x is less than or equal to 8  x    ≤   8

To show that x is between -4 and 2 you write
-4 < x < 2
which is read 
"-4 is less than x and x is less than 2."
or you could read it as
" x is greater than -4 AND less than 2."

The same comparisons are stated in the sentence  2 > x  > -4

When all the numbers are know you can classify the statement as true or false.
Thus,
-4 < 1 < 2   is true
but
-4 < 8 < 2 is false

An inequality is formed by placing an inequality symbol (  > ,  < ,     ≤ , or    ≥) between numerical pr variable expressions-- called the SIDES of the inequality
You solve an inequality by finding the values from the domain of the variable which make the inequality a true statement.   Such values are called the solutions of the inequality. All the solutions make up the solution set of the inequality.

Math 6A ( Periods 1 & 2)

Angles and Angle Measure 4-3
An angle is a figure formed by two rays with the same endpoints. The common endpoint is called the vertex. The rays are called the sides.
We may name an angle by giving its vertex letter if this is the only angle with that vertex, or my listing letters for points on the two sides with the vertex letter in the middle. We use the symbol from the textbook.

To measure segments we use a rule to mark off unit lengths. To measure angles, we use a protractor that is marked off in units of angle measure called degrees.

To use a protractor, place its center point at the vertex of the angle to be measured and one of its zero points on the side.


We often label angels with their measures. When angles have equal measures we can write m angle A = m angle B
We say that angle A and angle B are congruent angles


If two lines intersect so that the angles they form are all congruent, the lines are perpendicular. We use the symbol that looks like an upside down capital T to mean “is perpendicular to.”



Angles formed by perpendicular lines each have measure of 90° . A 90° angle is called a right angle. A small square is often used to indicate a right angle in a diagram

An acute angle is an angle with measure less than 90°. An obtuse angle has measure between 90° and 180°



Two angles are complementary if the sum of the measures is 90°
Two angles are supplementary if the sum of their measures is 180°

Math 6A ( Periods 1 & 2)

Points, Lines, Planes 4-1
We can describe but CANNOT DEFINE point, line or plane in Geometry

We use a single small dot to represent a point and in class we labeled with a P and we called it Point P
A straight line in Geometry is usually just called a line...
Two points DETERMINE exactly ONE LINE

we connected Point P with Point Q and created Line PQ
We placed this type of arrow ↔ over PQ to show a line
↔
PQ

that would represent the line PQ but we found we could write
↔
QP
and mean the SAME line!!

You can name ANY line with ANY TWO points that fall on that line!! USE only TWO points to name a line!!


Three or more points on the same line are called collinear. Notice the word "line" in collinear.
collinear

Points NOT on a same line are called noncollinear.

We have a RAY if we have an endpoint and it extends through other points. We name the rame by Naming the endpoint FIRST
→
PQ is RAY PQ and it begins at P and goes through Q. It is NOT the SAME as

→
QP which is Ray QP, which begins at Q and goes through P

Segments are parts of lines with TWO ENDPOINTS.
⎯
PQ is a segment with endpoints Point P and POint Q

We then looked at the drawing from page 105 and the class named all of the names for the line in the drawing, all of the rays that existed in the figure as well as the segments. We found that there were just 3 collinear points: A, X, B but that we could name 3 sets of non collinear points
X, Y, B and A, X, Y, AND A, B, Y

Three non-collinear points determine a flat surface called a plane.
We name a Plane by using three of its non collinear points!! Plane ABC was out example.

Lines in the same plane that do not intersect are PARALLEL lines. Two segments or rays are parallel if they are parts of parallel lines.
↔
AB is parallel to

↔
CD

may be written
↔ ↔
ABllCD
using two straight lines to indicate parallel

Parallel lines DO NOT intersect.

Intersecting lines intersect in a single point!!



Planes that do not intersect are called parallel planes... we looked around the room and found examples of parts of planes.. noticing which ones were parallel!! (the floor and ceiling were a great example)
Then we drew the box from PAge 106 and identified parallel segments and lines from that box.

Two non parallel lines that do not intersect are called SKEW LINES.

Algebra Honors ( Periods 6 & 7)

Although the textbook uses charts and tables for these word problems, I think they work easily without the charts...

John has 15 coins -- all dimes and quarters , worth $2.55 How many dimes and quarters does he have?
let d = the number of dimes and let q = the number of quarters
we know d + q = 15 and we know 10d + 25q = 255
using a system of equations and the substitution method
since d + q = 15 we know d = 15- q
10d + 25q - 255
10(15-q) + 25 q = 255
150 -10q + 25q = 255
150 + 15q = 255
15q = 105
q = 7
He has 7 quarters and 8 dimes


Ann and Betty together have $ 60. Ann has $9 more than twice Betty's amount. How much money does each have?
Let a = the amount Ann has and let b = the amount Betty has.
we know
a + b = 60
and we know
a= 2b + 9
so using a + b = 60
(2b+9) + b = 60
3b = 51
b = 17
Betty has $17 and Ann has (60-17) = $43

Henrick/Matt invested $8000 in stocks and bonds. the stocks pay 4% interest and the bonds pay 7% interest . The annual interest from the stocks and bonds is $500.
How much is invested in bonds?
let s = the amount invested in stocks
let b = amount invested in bonds
s + b = 8000
0.04s + 0.07b = 500
clear the decimals
4s + 7b = 50000
but we know s + b = 8000 or s = 8000 -b
4(8000 -b) + 7b = 50000
32000 - 4b + 7b = 50000
3b = 18000
b = 6000
He invested $6000 in bonds.

Math 6A (Periods 1 & 2)

Simple Interest 9-7
When you borrow money you pay the lender INTEREST for the use of the money. The amount of interest you pay is usually a percent of the amount borrowed figured on a yearly basis. This percent is called the annual rate.

When interest is computed year by year we call it
SIMPLE INTEREST
The formula is I= Prt

Let I = simple interest charges
P = principal ( amount borrowed)
r= annual rate
t = time in years

I = Prt

simple interest is calculated just on the principal.
Let's work through a few examples

How much simple interest would you owe if you borrowed $640 for 3 years at atan annual rate of 15%?

I = Prt
I = (640)(.15)(3)
I = (1920)(.15)
I = 288
$288 in interest

Sarah borrowed $3650 for 4 years at 16% How much must she repay.
Remember she will oe the amount she borrowed as well as the the interest.
I = Prt
I = (3650)(.16)(4)

I =14600(.16)
I = 2336.
Principal + Interest = 3650 + 2336
Sarah owes $5986.


$150 borrowed at 12% annual rate for 1 year

I = Prt

I = (150)(.12)(1)
I = 18
so you would owe $18 in interest after 1 year.
The total due would be $150 + 18 = $168

What if instead you borrowed the same amount but for 2 years... nothing was due until the end of two years
I = Prt
I = 150(.12)(2) = 36
You would owe $36 in interest .. so the total due was 150 + 36 = $ 186.

What if you borrowed the same amount for 3 years...
I = 150(.12)(3) = 54 or $54 in interest.
You would owe 150 + 54 = $ 204 after three years...

However, let's say you could only borrow that amount for 6 months...
I = Prt
I = (150)(.12)(.5)
Why 0.5? that is 1/2 a year.
Now you can always multiply by 1/2 as well.. in fact, sometimes that is easier
I = 150(.12)(1/2) = 9 or $ 9.00

After 6 months you would owe $159.


Dylan paid $375 in interest on a loan of $1500 principal at 12.5% interest.
What was the length of time?

Look at what it is asking and see which of the variables you have...
I= Prt
We have the interest paid, the principal and the annual rate so
375= (1500)(.125)(t)

375 = 187.5t
solve this one step equation by dividing both sides by 187.5

375 = 187.5t
187.5 187.5

t = 2

so 2 years

divide carefully...

Alexis paid $ 585 simple interest on a $6500 loan for 6 months.

what was the annual rate?

What do we know?
I = 585
P = 6500
t= 6 months ( which is 0.5 or 1/2)

I = Prt
585 = 6500 (r)(.5)
585 = 3250r
divide both sides by 3250

585 = 3250r
3250 3250

r = 0.18
which means 18%



annual--> once a year
6 months --> 1/2 or 0.5
4 months--> 1/3
3 months --> 1/4 or 0.25
8 month --> 2/3

(We did not get to this yet... but I thought I would post it... read this..it is interesting to see the difference. We will go over this after STAR testing)
Compound Interest 9-8

Compound interest is ALWAYS more than simple interest.

interest is compounded on the interest!!

$100 savings earning $10 interest/ annual.. [this only happens NOW if your dad is the one paying you... :)]

I = Prt
at the end of the first year
I = 100(.10)(1) = 10 or $10
add that to the 100
$110.
Now for the 2nd year,
$110 is your principal
so
I = Prt
I = 110(.10)(1) = 11 or $11
so at the end of 2 years you have $110 + 11 or $121

Now for the 3rd year
I = Prt
I = 121(10)(1) = $12.10
So at the end of three years you have $121 + 12.10 = $133.10

What if you had $500 at 8% compounded quarterly for one year.

quarterly means 1/4 or .25

I = Prt
I = 500(.08) (1/4)
calculate the 08(1/4) because that will be the constant you will multiply your principal by each time
(.08)(1/4) = .02
so I = 500(.02) = 10
after the first quarter it is 510
I = Prt for the 2nd quarter
I = 510 (.02) = 10.20
so after the 2nd quarter $510 + 10.20 = $520.20
I = Prt for the third quarter
I = 520.20 (0.02) = about $10.40 ( round to the nearest penny)
so after the third quarter
$520.20 + 10.40 = $530.60
I = Prt
I = 530.60(.02) = about $10.61
So at the end of 4 quarters -- or one year
530.60 + 10.61 = $541.21

compounding terms:
annually--> once a year
semiannually --> twice a year
quarterly--> four times a year
monthly--> 12 times a year
daily--> 365 times a year

Algebra Honors ( Periods 6 & 7)

Solving Systems of Linear Equations
The Graphing Method 9-1
Two or more equations in the same variables form a system of equations. The solution of a system of two equations in two variables is a pair of values x and y that satisfies each equation in the system. The point corresponding to the ordered pair (x, y) must lie on the graph of both equations.
Solve the system by graphing
2x - y = 8
x + y = 1

Solution:
Graph both 2x - 7 = 8 and x + y = 1 in the same coordinate plane.
We did this in class by transforming both equations to slope-intercept form (y = mx +b)
and then graphed them. We noticed that the only point on BOTH lines is the intersection point ( 3, -2)
The only solution of both equations is (3, -2).
You can check that ( 3, -2) is a solution fof the system by substituting x = 3 and y = -2 in BOTH eqquations.

Solve the system by graphing
x - 2y = -6
x -2y = 2

When you graph the equations in the same coordinate plane, you see that the lines have the same slope but different y-intercepts. The graphs are parallel lines. SInce the lines do not intersect, there is no point that represents a solution of both equations.
Therefore, the system has NO SOLUTION.

Solve the system by graphing
2x + 3y = 6
4x + 6y = 12

When you graph the equations in the same coordinate plane, you see that the graphs coincide. The equations are equivalent. Every point on the line represents a solution of BOTH equations.
Therefore, the system has infinitely many solutions.

The Graphing Method in review:
To solve a system of linear equations in two variables, draw the graph of each linear equation in the same coordinate plane...
--> if the lines interset there is only one solutions, namely the intersection point.
--> if the lines are parallel, there is no solution
--> if the lines coincide, there are infinitely many solutions.

The Substitution Method 9-2

There are several ways to solve a system of equations, In the substitution method we use either equation to solve for one variable in terms of the other.
Solve
x + y = 15
4x + 3y = 38

Solve the first equation for y
x + y = 15
becomes
y = -x + 15
Substitute this expression for y in the other equation, and solve for x
4x + 3y = 38
4x + 3(-x+15) = 38
4x -3x + 45 = 38
x + 45 = 38
x = -7

Substitute the value of x in the equation in your first step and solve for y
y = -x + 15
y = -(-7) + 15
y = +7 +15
y = 22
CHeck x = -7 and y = 22 on BOTH equations
x + y = 15
(Here let ?=? represent having a ? above the equals sign)

-7 + 22 ?=? 15
15 = 15
and
4x + 3y = 38
4(-7) + 3(22) ?=? 38
-28 + 66 ?=? 38
38 = 38

It checks for both equations so the solution is (-7, 22)

Solve
2x - 3y = 4
x + 4y = -9

Using the 2nd equation is easier to manipulate so solve for x since x has a coefficient of 1
x = -4y - 9
substitute this expression for x in the other equation and solve for y
2x - 3y = 4
2(-4y-9) - 3y = 4
-8y -18 -3y = 4
-11y = 22
y = -2
Substitute the value of y in the equation in step 1 and solve for x
x = -4y -9
x = -4(-2) -9
x = 8 -9 = -1
Check both equations... and you discover that the solution is ( -1, -2)

The substitution method is most convenient to use when the coefficient of one of the variables is 1 or -1.

The Substitution Method in review:
To solve a system of linear equations in two variables:
--> Solve one equation for one of the variables
--> Substitute this expression in the other equation and solve fore the other variable.
--> Substitute this value n the equation in step 1 and solve
--> Check the alues in BOTH equations.


Solve by the substitution method
2x -8y = 6
x - 4y = 8

x = 4y + 8

2x-8y = 6
2(4y+8) - 8y = 6
8y + 16 -8y = 6
16= 6 WAIT that's FALSE

The false statement indicates that there is NO ordered pair (x, y) that satisfies BOTH equations. If you had graphed the equations you would see that these lines are actually parallel.

Solve by substitution method
y/2 = 2 -x
6x + 3y = 12

The first equation is easy to change to y = 4 - 2x by multiplying both sides by 2 to solve for y

6x + 3y = 12
6x + 3(4-2x) = 12
6x + 12 - 6x = 12
12 = 12 WAIT THat's TRUE... always
Every ordered pair (x, y) that satisfies one of the equations aso satisfies the other. IF you graph these two equations you will see that the lines coincide

Therefore, the system has infinitely many solutions.

Math 6A (Periods 1 & 2)

Working With Percents

Review: When changing a percent to a fraction we use the following proportion:

 

3.5% = 3.5/100 = 35/1000 = 7/200

Looking at

What percent of 225 is 90?  When setting up an equation we get

225n = 90

However we can use a similar proportion to  

 to solve any percent problem!

The fraction a/b really represents the part/ whole

  Or easier stated:

Looking at the above problem we have

Of course we simplify as much as possible before we cross multiply

we get n = 10(4) = 40

so the solution is 40%

EXAMPLE: 

45% of 600 is what number?

Now this simplifies really easily

and we just have

n = 6(45)

n = 270

The number is 270.

EXAMPLE: 

96% of 85 is what number?

Again, this simplifies to

With cross products we get

5n = 24(17)

5n = 408

Divide both sides by 5

5n/5 = 408/5

n = 81.6

81.6 is our solutions

EXAMPLE: 

What number is 76% of 350?

When you simplify  x= 38(7)
x = 266
EXAMPLE: 

56 is 4% of what numbers?

Definitely simplify and you get

x = 25(56)

x = 1400

Math 6A ( Periods 1 & 2)

Commission And Profit 9-6 

Some sales jobs pay an amount based on how much you sell. This amount is called a commission.

Like a discount, the commission can be expressed as a percent or as an amount of money.

amount of commission = percent of commission X total sales.

Using the examples from our textbook,

Maria sold $42,000 word of insurance in January. If her commission is 3% of the total sales, what was the amount of her commission in January?

amount of commission = percent X total sales
0.03 X 42,000 = 1260
Her commission was $1,260.

Profit is the difference between total income and total operating costs.

profit = total income – total costs

The percent of profit is the percent of total income that is profit

percent of profit = profit/total income
A shoe store had an income of $8600 and operating costs of $7310. What percent of the store's income was profit?

profit= income- total costs = 8600 -7310 = 1290

percent of profit = profit/total income = 1290/8600 = 0.15
So the percent of profit was 15%.

Practice finding 10%-- its easy--- just move the decimal over one place.
We practiced finding 20%. Just double what you got for 10%.



MATH AT WORK:
Caterer
A caterer provides food for parties, weddings, bar/bat mitzvahs, and other events. Caterers plan the menu, buy the ingredients, and cook the food. Often they provide seating and music as well. For each event, a caterer determines the cost per guest. The catering business requires a thorough knowledge of ratios, proportions, and percents.

Math 6A (Periods 1 & 2)

 Discount & Markup 9-5

A discount is a decrease in the price of an item. A markup is an increase in the price of an item. Both of these changes can be expressed as an amount of money or as a percent of the original price of the item. A store may announce a discount of $3 off the original price of $30 basketball, or a discount of 10%

A warm-up suit that sold for $42.50 is on sale at a 12% discount. What is the sale price?

Method 1: Use the formula

amount of change = percent of change X original amount
= 12% X $42.50
SET UP the LADYBUG
therefore the discount is 0.12 X 42.50 or 5.10
The amount of discount is $5.10
The sale price is 42.50 – 5.10 = $37.40

Method 2: Since the discount is 12%, the sale price is 100% - 12% = 88%.
The sale price is 0.88 X 42.50 = $ 37.40

When you know the amount of discount you subtract to find the new price. When dealing with a markup you add to find the new price.

The price of a new car model was marked up 6% over the previous year’s model. If the previous year’s model sold for $7800, what is the cost of the new car? {and what kind of a car could that be?}
Method 1: Use the formula
amount of change = percent of change X original amount

= 6% X 7800

Therefore the markup is 0.06 X7800= $468
The new price is 7800 + 468 = $8268

Method 2: Since the markup is 6% the new price is 100% + 6% or 106% of the original price. so the new price is 1.06 X 7800 = $8268

This year a pair of ice skates sells for $46 after a 15% mark up over last year’s price. What was last year’s price?

This year’s price is 100 + 15 or 115% of last year’s price. Let n present last year’s price
46 = (115/100)n

46 = 1.15n
46/.15 = 1.15n/1.115
40 = n

So last year’s price was $40.

A department store advertised eclectic shavers at a sale price of $36.
If this is a 20% discount, what was the original price?
The sale price is 100 - 20 or 80% of the original price. Let n represent the original price.
36 = (80/100)n
36 = .8n
36/.8 = .8n/.8

45 = n
The original price was $45.
Check to see that your answers are logical and reasonable.

Try these: A service station (that’s gas station, now—they no longer provide service!!) give cash customers a 5% discount on the price of gasoline. If gasoline regularly sells for $3.00 a gallon, what is the discounted price?


A store marks up the price of a $5 item to $12. What is the percent of markup?

Math 6A (Periods 1 & 2)

Percent of Increase or Decrease 9-4

Let's say we have an iPod that originally sold for $260. It is on sale for $208. What is the amount of change? "How much did you save?"
Just subtract
260-208 = 52
$52.
What is the percent of change?
The percent of change = amount of change/original
52/260 - x/100
or just divide 52 by 260 = .2
which is 20%
REMEMBER: The denominator in the formula is ALWAYS the ORIGINAL AMOUNT.
Amount of change = percent of change X the original amount.
Find the new number when 75 is decreased by 26%
Amount of change - 26% (75
= .26(75)
=19.5
Now take the difference (the amount of change) and subtract THAT from 75
75- 19.5 = 55.5

Remember the circle with the various parts of this formula?

Difference or amount of change
% of change X original amount

Difficult to show here so if you missed these notes make sure to ask a fellow student to see this!! IT is a great way to remember what to do!!

State the increase or decrease. Tell what the amount of change is and the percent of change.
from:
10 to 12
increase
amount of increase: 2
% of change : 20%

4 to 3
decrease
amount of decrease:
% of change : 25%

2 to 5
increase
amount of increase: 3
% of change : 3/2 = 1.5 = 150%

12 to 6
decrease
amount of decrease: 6
% of change : 50%

6 to 12
increase
amount of increase: 6
% of change : 6/6 = 1 = 100%

Find the new number produced when the given number is increased or decrease by the given percent.

120; 20% decrease
120(.20) = 24 120 -24 = 96

30; 10% decrease
30(.10) = 3 30 -3 = 27

48: 50% increase
48(.50) = 24 48 + 24 = 72

128: decrease by 25%, then increased by 25%
What... why multiply by .25 if you can use a fraction and work smarter?
128(1/4) = 32
128 - 32 = 96
then 96 ( 1/4) = 24
96 + 24 = 120

Did you think it would be the starting number? Why wasnn't it?