Math 6H ( Periods 3, 6, & 7)

Computing with Percents 9-3
The statement 20% of 300 is 60 can be translated into the following equations
20/100(300) = 60 or 0.20 •300 = 60

EQUATION METHOD:
Notice the following relationship between the words and the symbols
20% of 300 is 60
0.20 • 300 = 60

WRITE THE PROBLEM OUT AND THEN DIRECTLY UNDER THE "IS" WRITE AN EQUAL SIGN. DIRECTLY UNDER THE WORD 'OF" WRITE A MULTIPLICATION SIGN. iF YOU ARE GIVEN A % CHANGE IT FIRST TO A DECIMAL. THEN BRING DOWN ALL THE OTHER NUMBERS GIVEN IN YOUR PROBLEM. LET x OR n REPRESENT YOUR VARIABLE... THAT IS THE "WHAT " PART OF YOUR PROBLEM.

A similar relationship occurs whenever a statement or a question involves a number that is a percent of another number

What is 8% of 75?
Let n represent the number asked for
What number is 8% of 75?
n = 0.08 • 75

solve
What percent of 40 is 6?
let n represent the percent asked for.

What percent of 40 is 6?
n% • 40 = 6
n% • 40 = 6
n% (40)/40 = 6/40
n% = 6/40
n/100 = 6/40
(100) n/100 = (100) 6/40
n=15 so 15% of 40 is 6

140 is 35 % of what number?
let n represent the number asked for
140 is 35% of what number?
140 = 0.35 • n
140 = 0.35n
140/0.35 = 0.35n/0.35 divide carefully!! Watch those decimals!!
400 = n
so 140 is 35% of 400

Always check to see if your answer is logical.

PROPORTION METHOD

In these types of percent problems you are always know three parts of the following proportion

n/1oo = a/b
or better yet
n/100 = is/ of

The n represents the %

Read the problems carefully and you can easily determine which is the "is" and which represents the 'of"
For example:
What percent of 40 is 6?
What percent -- from the problem above indicates that we DO NOT know the n
of 40-- hmm... then 40 must be the 'of' and
similarly is 6 means that 6 represents the 'is'

n/100 = 6/40 solve as a proportion
and you get n= 15 but since it asked us to state the 5 your answer is 15%

140 is 35 % of what number?
In this problem I notice 35% right away so that is the n!!
Then I read the problem again and notice 140 is... hmmm.. THat says 140 must be the is

35/100 = 140/ x I do not know the 'of'

Solve again
x = 400

Pre Algebra ( period 1)

PYTHAGOREAN THEOREM
 FOR RIGHT TRIANGLES ONLY! 11-2

2 legs - make the right angle - called a and b
(doesn't matter which is which because you will add them and adding is COMMUTATIVE!)

hypotenuse - longest side across from the right angle - called c

You can find the third side of a right triangle as long as you know the other two sides:

a² + b² = c²

After squaring the two sides that you know, you'll need to find the square root of that number to find the length of the missing side (that's why it's in this chapter!)


EASIEST - FIND THE HYPOTENUSE (c)

Example #1 from p. 510

8² + 15² = c²

64 + 225 = c²
289 = c² Take the SQ RT of each side

c = 17



A LITTLE HARDER - FIND A MISSING LEG (Either a or b)

Example #5 from p. 510

5² + b² = 13²

25 + b² = 169

b² = 169 - 25

b² = 144 Take the SQ RT of each side

b = 12



CONVERSE OF PYTHAGOREAN THEOREM

If you add the squares of the legs and that sum EQUALS the square of the longest side, it's a RIGHT TRIANGLE.


If you add the squares of the 2 smallest sides and that sum is GREATER THAN the square of the longest side, you have an ACUTE TRIANGLE.


If you add the squares of the 2 smallest sides and that sum is LESS THAN the square of the longest side, you have an OBTUSE TRIANGLE.

Pre Algebra ( period 1)

Introduction to Geometry: Points, Lines, & Planes 9-1
Point:(symbol is a dot or just a letter) Location in space - no size
Ray: (arrow pointing to the right) - one endpoint and one direction- Named by its endpoint first
Line: (generally, line with arrows on both ends above 2 points on the line) - Series of points that goes on infinitely in both directions - named either direction
Line segment: (a line with no arrows on either end) - a piece of a line with 2 endpoints in either direction
Lines can be parallel (2 vertical lines) or intersecting in the same plane
Parallel lines are lines in the same plane that never meet
If they intersect at exactly 90 degrees, then they are perpendicular
If they don't intersect but are in two different planes, they are skew

Angle Relationships & Parallel Lines 9-2
angle: (angle opening to the right) two rays that meet at the same endpoint
named by either just the vertex, or 3 points on the angle in either direction with vertex in middle
adjacent angles share one ray
vertical angles are opposite each other and congruent (equal)
Complementary sum to 90 degrees and
supplementary sum to 180 degrees
acute is greater than 0 and less than 90
90 degrees is right angle
obtuse is greater than 90 but less than 180
180 is a straight angle (line)
to write the measure of an angle you write m<
Transversal: a line that intersects two other lines

Corresponding angles are formed by this transversal
These angles are on the same side of the transversal and also are both above or both below the line
When the two lines that are intersected are parallel, corresponding angles are congruent

Alternate interior angles are between the two lines (inside the two lines) and on opposite sides of the transversal (alternate sides) These angles are also congruent if the lines are parallel.

Same side interior: If angles are both inside and on the same side of the transversal, they are supplementary (sum to 180 degrees)

You can have lots of corresponding angles if you have a transversal intersecting more than 2 parallel lines - in fact they would be infinite if you kept adding another parallel line!
It's amazing that by just knowing one angle, you know all 8 angles with one transversal and two parallel lines! (I will show this in class on Tuesday!)

Algebra (Period 4)

Add and subtract rational expressions with LIKE DENOMINATORS: 10-4
Add and subtract with UNLIKE DENOMINATORS: 10-5
When adding with LIKE DENOMINATORS,
simply add the numerators,
simplify

When subtracting with LIKE DENOMINATORS,
CHANGE THE SIGNS
OF EACH TERM IN THE NUMERATOR AFTER THE SUBTRACTION SIGN,
THEN ADD (double check!!!)

When adding or subtracting with UNLIKE DENOMINATORS,
find the common denominator (the least common multiple of all denominators),
then use equivalent fractions to restate each numerator using the new common denominator.
EXAMPLE:
2 + x
x² - 16 x – 4

First, factor the denominators if possible



2x + x
(x + 4)(x - 4) x – 4

Find the LCM = (x + 4)(x - 4);

2x + (x + 4)(x)
(x + 4)(x - 4) (x + 4)(x - 4)


Restate both fractions with the LCM

2x + x² + 4x
(x + 4)(x - 4)


Now add the numerators

x² + 6x =
(x + 4)(x - 4)

Simplify and re-factor numerator, if possible
x(x + 6)
(x + 4)(x - 4)