Geometry chapter 1.4 Angles and their measures

Geometry Chapter 1.4 Angles and their Measures:
I) Vocabulary and Postulates:

A) Angle - consists of two different rays that have the same initial point.
1) Vertex - the initial point of the angle.
2) Rays of an angle - the sides of the angle.

B) Congruent Angles - angles that have the same measure.
C) Angle Addition Postulate - If P is on the interior of angle RST, then measure of angle RSP + measure of angle PSP = measure of angle RST.

D) Acute Angle - an angle that's measure is between zero and ninety degrees.

E) Right Angle - an angle that's measure is 90 degrees.

F) Obtuse Angle - an angle that's measure is between 90 degrees and 180 degrees.

G) Straight Angle - an angle that's measure is 180 degrees.

H) Adjacent Angles - two angles are adjacent angles if they share a common vertex AND a common side, but do not have any common interior points.

Geometry Chapter 1.3 Segments and Their Measures

Geometry Chapter 1.3 Segments and Their Measures

I) Vocabulary & Postulates:

A) theorems: Rules that have been proven to be true.
Example: Triangle Sum Theorem: The sum of the measures of the interior angles of a triangle is 180 degrees.

B) Postulates or Axioms: Rules that we accept as true without proof.

C) Between - When 3 points are one a line, you may say that 1 point is between the others.

D) Segment Addition Postulate or Partition Postulate: If B is between point A and point C, then
AB + BC = AC

E) Distance formula: AB = √((x₁ - x₂)² + (y₁ - y₂)²)

this is derived from the pythagorean theorem, where the difference of the x's is represented by "a" and the difference of the y's is represented by "b" and AB = c so

a² + b² = c²

F) Congruent Segments - Segments that have the same measure

Written:

Lengths are equal in length so: AB = CD

and

Segments are congruent so: line segment AB is congruent to line segment CD.

Geometry Chapter 1.2 Points, Lines and Planes

Chapter 1.2 Points, Lines, and Planes

I) Undefined Terms:

A) Point: a point has no dimension, usually represented by a small dot.

B) Line: A line extends in one dimension. It is usally represented by a straight line with two arrowheads to indicate that the line extends without end in two directions.

C) Plane: a plane extends in two dimensions, goes on forever.

D) Collinear Points: 2 or more points on the same line.

E) Coplanar Points: Points that lie on the same plane.

F) Line Segment: a segment of a line; consists of 2 endpoints and all points in between.

G) Ray: An initial point and all points one side of that point.

H) Opposite Rays: 2 rays that form a straight line.

I) Intersect: Two or more geometric figures intersect if they have one or more points in common.

Geometry - Chapter 1.1 Patterns and Inductive Reasoning

Chapter 1.1 - Patterns and Inductive Reasoning

A. Using Inductive Reasoning
1. Look for a pattern. - look at several examples. Use diagrams and tables to help discover a pattern.
2. Make a Conjecture. - a conjecture is an unproven statement that is based on observation.
3. Verify the Conjecture - use logical reasoning to verify that the conjecture is true in all cases.

Example - the sum of the first n odd positive integers is...
first odd positive integer is 1 = 1
sum of first two odd positive integers is 1 + 3 = 4
sum of first three odd positive integers is 1 + 3 + 5 = 9
sum of first four odd positive integers is 1 + 3 +5 + 7 = 16

looking at these answers: 1, 4, 9, 16 we see they are perfect squares so
1², 2², 3², 4²
Therefore we can make a conjecture that the sum of the first n odd positive integers is n² .

B. Counterexample - is an example that shows a conjecture is false.
Example: All prime numbers are odd.
Since 2 is a prime number but 2 is even this would be a counterexample.

Precalculus 12.5b Area of a Plane Region

12.5b Area of a Plane Region

Area of a Region bounded by:

x = a, x = b, y = 0 and y = f (x) where f(x) is less than or equal to 0.

The boundary of the area is from x₀ = a to x_n = b.

What is the area under the curve?

George Riemanns - German who came up with technique to find the orange area.

1. Subdivide the interval from a to b into smaller intervals. This is called a partition of [a,b] specifically:

a = x₀ ≤ x₁ ≤ x₂ ≤ x₃ ... ≤ x_{n - 1} ≤ x_n = b

2. Choose a point w_i in interval [x_{i - 1}, x_i] , i = 1 ... n

so...

when i = 1, w₁ is contained in [x₀, x₁]

when i = 2, w₂ is contained in [x₁, x₂]

3. Let Δ x_i = x_i - x_i-1 for i = 1 ... n

i= 1; Δ x₁ = x₁ - x₀

i = 2; Δ x₂ = x₂ - x₁

Δ x_i = length of the i^th subinterval.

4. Form f(w_i)(Δx_i) for i = 1 ... n

5. The area of the region:

A = f (w₁)(Δ x₁) + f(w₂)(Δ x₂) + ... + f (w_n)(Δx_n)

This is called a Riemann's Sum.

So by increasing the number of rectangles that you make, you can obtain a closer and closer approximation

Therefore to find the Area of a Plane Region:

Let "f" be continuous function and nonnegative on the interval [a, b]. The Area A of the region bounded by the graph of "f", the x-axis (y = 0), and the vertical lines x = a and x = b is

Area of a rectangle = height times width

check out this website: http://tutorial.math.lamar.edu/classes/calcI/areaproblem.aspx

Example 1:
Find the Area of the Region: f(x) = 2 - x², -1 £ x £ 1

Let's let n = 4

so the width = (1 - ^-1)/4 = 1/2

height =
[-1, -1/2] = 1.75
[-1/2, 0] = 2
[0, 1/2] = 1.75
[1/2, 1] = 1

so the Area when f(x) = 2 - x² and is divided up into 4 rectangles =
(1/2)(1.75) + (1/2)(2) + (1/2)(1.75) + (1/2)(1) = 3.25

but to find the area with more rectangles, let's use the above formulas:

Therefore to find the area, use the summation formulas:

To check this on the graphing calculator:

put in y₁ = 2 - x²

Now on your homescreen, math arrow down to #9 fnInt (Y₁, x, -1, 1) enter

should give you 10/3.

In the calculator (function , variable, lower bound, upper bound) = area under the curve.

Precalculus 12.5 The Area Problem

12.5 The Area Problem

I. Limits of Summation

S = a₁ + a₁r + a₁r² + ... =

_¥
å (a₁(r^i-1)) =
^{i = 1}

a₁/(1-r) provided that the absolute value of r < 1

Using Limit Notation, this sum can be written as:

S =
¥
lim å (a₁ r^i-1)
^{n®¥ i = 1}

= lim a₁ (1 - rⁿ)/(1-r)
^n®¥

Recall that since r < 1 thus making it a fraction, when you take the limit or rⁿ as x approaches infinity, this would equal zero. So:

= (a₁ (1 - 0))/ (1 - r)

= a₁/(1 - r) which you have already seen in this coursework.

Summation Formulas and Properties:

Example 1: Evaluating a Summation:

What is the sum of
i² when i = 1 to i = 30?

=(30)(30 + 1)(2(30) + 1)/6 = (30)(31)(61)/6 = 56730/6 = 9455

Example 2: Evaluating a Summation:

What is the sum of:
(2i + 1) when i = 1 to i = 50?

You have to break this down into parts:

2( å i ) from i = 1 to i = 50 AND
å(1) from i = 1 to i = 50 is equal to

2(50(50+1)/2) + 1(50) = (50)(51) + 50 = 2550 + 50 = 2600

II. Finding the Limit of a Summation

_n
å (2i + 3)/(n²)
^{i = 1}

= (1/n²)(2(n(n+1))/2) + 3n) = (1/n²)(n(n+1) + 3n)
= (1/n)(n + 1 + 3) = (n + 4)/n

which in essence is infinity over infinity so
Therefore the Summation of
lim S_n =
^n®¥

lim (n + 4)/n = 1/1 = 1 so
^n®¥

lim S_n = 1
^n®¥

You can check this by using a table of values:

(n, S(n)) = {(1, 5), (10, 1.4), (100, 1.04), (1000, 1.004), (10000, 1.0004)...}
So you can see the values of S(n) approach 1.

Example 3: Finding the Limit of a Summation:

_n
å (3/n³)(1 + i²)=
^{i = 1}

= (3/n³)(n + (n(n + 1)(2n + 1)/6) =
(3/n² + (3(2n² + 3n + 1)(6n²)
= 3/n² + (6n² + 9n + 3)/(6n²)

lim 3/n² = 0 and
^{n ®¥}

lim (6n² + 9n + 3)/(6n²) = 6/6 = 1
^{n ®¥}

So:

lim S_n = 1
^{n ®¥}

Again, you can check this using a table of values:
(n, S(n)) = {(1, 6), (10, 1.185), (100, 1.0154), (1000, 1.0015), (10000, 1.0002)...}

Precalculus 12.4 Limits at Infinity and Limits of Sequences

12.4 Limits at Infinity and Limits of Sequences

Definition of Limits at Infinity
If "f" is a function and L₁ and L₂ are real numbers, the statements

lim f(x) = L₁
^x®-¥

AND

lim f(x) = L₁
^x®+¥

denote the limits at infinity. The first is read "the limit of f(x) as x approaches negative infinity is L₁," and the second is read "the limit of f(x) as x approaches infinity is L₂"

To help evaluate limits at infinity, you can use the following:

Limits at Infinity
If "r" is a positive real number, then

lim (1/(x^r)) = 0 Limit toward the right.
^x®+¥

Furthermore, if x^r is defined when x is less then 0, then

lim (1/(x^r) = 0 the limit toward the left.
^{x®- ¥}

Thought: -1/10 = -.1, -1/100 = -.01, -1/1000 = -.001, -1/10000 = -.0001
so as you can see, in the denominator as x approaches negative infinity, f(x) approaches zero.

Example 1: Evaluating a Limit at Infinity:

lim (5/(2x)) = 0
^x®¥

As you can see when you graph this function, as x gets larger, f(x) gets closer to zero so that is how we can conclude the the limit equals zero.

Example 2:
lim (1 - 3/(x²)) =
^x®¥

We can separate each part of the function to:

lim (1 )= 1
^x®¥

and

lim (3/(x²)) = 0
^x®¥

so
lim (1 - 3/(x²)) = 1 - 0 = 1
^x®¥

Compare the following limits:
Find the limit as x approaches infinity:

a). f(x) = (-5x + 2)/(4x² - 1)

b). g(x) = (4x² + 1)/(3x² - 1)

c). j(x) = (-3x⁴ + 1)/(2x³-1)


When you take the limit of each of these functions as x approaches infinity, you get:

1st, look in the denominator and find the largest exponent and use this term by dividing all the terms of the function by that variable raised to the largest exponent.

a).
lim (-5x + 2)/(4x² - 1)=
^x®¥

so divide each term by x²

lim [(-5x)/(x²) + 2/(x²)]/[(4x²)/(x²) - 1/(x²)]
^x®¥

= (-0 + 0)/(4 - 0) = 0

b).
lim (4x² + 1)/(3x² - 1)=
^x®¥

so divide each term by x²

lim [4 + 1/(x²)]/[3 - 1/(x²)]=
^x®¥

(4 + 0)/(3 - 0) = 4/3

c).
lim (-3x⁴ + 1)/(2x³-1)
^x®¥

and so divide each of the terms by x³

lim (-3x + 1/x³)/(2 - 1/x³)=
^x®¥

-3x/2 so by putting infinity in for x, the limit does not exist because the numerator decreases without bound as the denominator approaches 2.

Therefore we can conclude:

Limits at Infinity for Rational Functions

for the rational function f(x) = N(x)/D(x) when

N(x) = a_nxⁿ + ... + a₀

And

D(x) = b_mx^m + ... + b₀

the limit as x approaches positive or negative infinity is as follows:

lim f(x) = 0, when n is less than m
^x®¥

lim f(x) = a_n/b_m, when n = m
^x®¥

And if n is greater than m, the limit does not exist.

Limit of a Sequence:
Let "f" be a function of a real variable, such that
lim f(x) = L
^x®¥

If {a_n} is a sequence such that f(n) = a_n
for every positive integer "n", then the
lim a_n = L
^n®¥

Example:
Given: 1/3, 1/9, 1/27, 1/81, ...

As n increases without bound, the terms of the sequence
a_n = 1/3ⁿ is said to

CONVERGE to zero.

lim 1/3ⁿ = 0
^n®¥

___________________________________________

A sequence that does not converge is said to DIVERGE!

Example of this is: the sequence 1, -1, 1, -1, ... this diverges.

Find the limit of the Sequence:

Example 1:
lim (4x - 1)/(x + 3)=
^x®¥
{3/4, 7/5, 11/6, 15/7, ...} = 4

Example 2:
lim (3x + 2)/(2x² - 1)=
^x®¥
{5/1, 8/7, 11/17, 14/31, ...} = 0

Example 3:
lim (7x² - 1)/(8x²)=
^x®¥
{6/8, 27/32, 62/72, 111/128, ...} = 7/8

Use the site: http://www.calculus-help.com/funstuff/phobe.html

Chapter 1, lesson 4 to help with this lesson.