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Algebra

Course Description

Prerequisite: None

Prerequisite: Teacher Approval

On Level:

This course includes topics such as set theory, graphs, functions, and sentences, polynomials, products and factoring, and a look at geometry and trigonometry.

College Prep:

Those students who have a good understanding of mathematics and wish to go into higher mathematics should take this course. This course includes topics such as set theory, graphs, functions, and sentences, polynomials, products and factoring, and a look at geometry and trigonometry. College prep courses will be faster paced than non college prep courses with the intention of preparing students for the college entrance exams.

Algebra 1 homework help

http://www.algebrahelp.com/worksheets/

http://www.mathleague.com/help/algebra/algebra.htm

(1 + z) × 2 + 12 ÷ 3 - z becomes

(1 + 4) × 2 + 12 ÷ 3 - 4 =

5 × 2 + 12 ÷ 3 - 4 =

10 + 4 - 4 =

10.

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Pre-Calculus

Course Description

Prerequisite: Algebra II, & Geometry, Sr. Status

Prerequisite: Algebra II and Geometry

College Prep:

This course is designed primarily for seniors who plan to go to college, but do not plan to take Calculus as a college freshman. algebra, geometry, and arithmetic are used constantly in this course, thus providing a good review of these: the unit circle and trigonometric identities, the derivation of formulas and their practical and theoretical applications are stressed.

Pre AP:

This course is designed primarily for those students who intend to pursue college courses in science of mathematics, algebra, geometry, and arithmetic are used constantly in this course, thus providing a good review of these: the unit circle and trigonometric identities, the derivation of formulas and their practical and theoretical applications are stressed.

Ex 1 Find the center and radius of the circle with standard equation .

Sol Completing the square on the x terms and the y terms gives or , so the center of the circle is the point , and its radius is .

Ex 2 Use the distance formula to find an equation of the perpendicular bisector of the line segment between the points and .

Sol The point is on the perpendicular bisector if it is equidistant from the two points, so the perpendicular bisector is defined by the equation $\sqrt{(x-4)^2+(y-3)^2} = \sqrt{(x+2)^2+(y-5)^2}$ . Squaring both sides gives , and then multiplying out both sides yields ; so the perpendicular bisector has equation or .

http://www.math.ucdavis.edu/~marx/reccoordPr/reccoordPr.html

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Taks Math/ Math Models

Course Description

Math Models:

This course is designed to help students who struggle with the basic concepts of high school math.

TAKS Math:

Helps students understand the TAKS Math portion of the TAKS Test.

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Geometry

Course Description

Prerequisite: Algebra I

Prerequisite: Algebra I, Teacher Approval

Prerequisite: Algebra I, Counselor Approval

On Level:

Geometry is a year course devoted to the study of the various relationships of lines and planes. The student must develop his ability to visualize plane and space figures, and he must also develop the ability to think creatively and in an organized manner.

College Prep:

College prep courses will be faster paced than non college prep courses with the intention of preparing students for the college entrance exams.

Pre AP:

This course is for students showing an advanced aptitude toward mathematics. This course covers the regular content of geometry, and goes beyond the established regular course both in content and depth. Content dot found in regular geometry includes application of theorems and spatial geometric functions. Emphasis is given to synthetic, coordinate, transformational, and vector geometry.

2 ×( (pi)r² ) + 2(pi)rh    (the two ends, plus the cylinder)
    = 2( (pi) (4²) ) + 2(pi) (4)(30)
    = 2( (pi) × 16 ) + 240(pi)
    = 32(pi) + 240(pi)
    = 272(pi)

Why do I need the area of the pool? Because, to find the area of the deck, I'll need to find the area enclosed by the circumference of the deck, and then subtract out the area enclosed by the pool. As is often true, a picture is helpful:

The pool,
surrounded by the deck.

The pool has radius 14,
and the deck has width "d".

This is the area that
I'm trying to find:
the entire circular area,
less the area of the pool.

Let d be the width of the deck. Then the radius of the entire deck-and-pool circle is 14 + d. Then the area of just the deck is given by:

deck area = (total circular area) – (pool area)

A = (pi)(14 + d)² – 196(pi)
    = (pi)(196 + 28d + d²) – 196(pi)
    = 196(pi) + 28(pi)d + (pi)d² – 196(pi)
    = (pi)d² + 28(pi)d

I am given that the area of the deck is 60(pi) square feet. Then:

(pi)d² + 28(pi)d = 60(pi)
(pi)d² + 28(pi)d – 60(pi) = 0
d² + 28d – 60 = 0
(d + 30)(d – 2) = 0
d = –30 or d = 2

Since d stands for a distance, you can ignore the "d = –30" solution as not being useful; it is extraneous. ("Extraneous", pronounced "ek-STRAY-nee-uss", means that, while it is a valid answer to the mathematical equation, it is not relevant in the context of the word problem.)

Then the answer is that the deck is two feet wide.

http://www.purplemath.com/modules/perimetr2.htm

http://thinkzone.wlonk.com/MathFun/Triangle.htm

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Algebra II

Course Description

Prerequisite: Algebra 1, Teacher Approval

Prerequisite: Algebra I, Geometry, Concurrent Enrollment, Counselor Approval

Prerequisite: Algebra 1

College Prep:

College prep course will be faster paced than non college prep courses with the intention of preparing students of the college entrance exams.

Pre Advanced Placement:

This course is essential to those students who are preparing for college courses in either mathematics or science. The topics studied in the regular Algebra II course are covered. These include linear and quadratic systems, exponents, radicals, binomial theorems, sequence, series, permutations, combinations, and probability and are covered to the extent that time and the general ability of the class will permit.

On Level:

The topics studied in Algebra II include linear and quadratic equations and systems, exponents, radicals, imaginary numbers, functions, logarithms, and other special topics as time and the general ability of the class will permit. College prep courses will be faster paced than non college prep courses with the intention of preparing students for the college entrance exams.

Algebra II homework help

http://library.thinkquest.org/20991/alg2/prob.html

http://www.leeric.lsu.edu/bgbb/7/ecep/math/e/e.htm

1. Problem:   4
              Σ (2k + 1)
            k = 1

  Solution: This is a sum of (2k + 1) from
            1 to 4.

            Plug all numbers from 1 to 4 into
            the general term ((2k + 1) in this case)
            and then add the terms together.

            (2(1) + 1) = 3
            (2(2) + 1) = 5
            (2(3) + 1) = 7
            (2(4) + 1) = 9

            3 + 5 + 7 + 9 = 24

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Calculus

Course Description

Prerequisite: Pre-Calculus

Advanced Placement:

The course will begin with a review of the basic formulas and ideas of distance between two points and the rate of change of a function. the course will progress as the mastery of each essential element lay the foundation for the next. A strong emphasis will be placed on the methods to solve the two main classes of problems in calculus. The first, differential calculus, involves finding the rate at which a variable quantity is changing, and the second finds the function when the rate of change is unknown to which integral calculus is applied.

Key Concept

If G(x) is continuous on [a,b] and G¢(x) = f(x) for all x Î (a,b), then G is called an antiderivative of f.

We can construct antiderivatives by integrating. The function

F(x) =

ó
õ

x

a

f(t) dt

is an antiderivative for f. In fact, every antiderivative of f(x) can be written in the form F(x)+C, for some C.

[F(x)] = f(x)

ó
õ

f(x) dx = F(x)+C.

[g(x)] = g¢(x)

ó
õ

g¢(x) dx = g(x)+C.

To differentiate $\sin (2x-1)$ , we identify . Then

$\begin{eqnarray*} \frac{d}{dx}\left[ \sin(2x-1)\right] & = & \frac{d}{dx} \left[... ...1 \right] \\ & = & \cos (u) \cdot 2 \\ & = & 2 \cos (2x-1). \end{eqnarray*}$

$\fbox{\begin{minipage}{1.7in} \begin{eqnarray*} f(x) & = & \sin (x) \\ g(x) & = & 2x-1 \\ f(g(x)) & = & \sin (2x-1) \end{eqnarray*}\end{minipage}}$

http://www.math.hmc.edu/calculus/tutorials/chainrule/
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Created by: Shane Whittington.