Wednesday, December 06, 2006

Adding Subtracting Rational Expressions

We need to have a common denominator in order to add/subtract rational expressions

If they have unlike denominators

Addition Example:
Subtraction Example:

Assignment:Addition And Subtraction
1, 3, 5, 7, 8, 9, 10, 12

Monday, December 04, 2006

Simplifying Rational Expressions With Multiplication and Division

These are the notes that were given today, I hope they're easy to understand! Enjoy! :)

For Multiplication:

Example
(4/3)(9/2)
=[2(2)/3][3(3)/2] ~the two's and three's cancel out
which leaves...
2(3)=6

This can be applied to rational expressions:

{ x (squared)-4/x-3} {x(squared)+6x+9/x+2}
={(x-2)(x+2)/x+3}{(x+3)(x+3)/x+2}~the x plus two's and x plus three's cancel out
which leaves...
(x-2)(x+3) (but when you use FOIL it comes out to)
x(squared)+x-6 ~you then have to state the non-permissible values (which is defined in
previous posts)
(they are x+3 cannot equal 0, x+2 cannot equal 0, x cannot equal-3 or -2.)

For Division:

Example
(4/3)/(6/5)
=(4/3)(5/6) ~when dividing the reciprocal is used
={2(2)/3}{5/2(3)} ~the two cancels out
={2(5)}/{3(3)}
= 10/9

This can be applied to rational expressions as well:

{x(squared)-4/x+3}{x+2/x(squared)-9}
= {x(squared)-4/x+3}{x(squared)-9/x+2} ~again when dividing the reciprocal is used
= {(x-2)(x+2)/x+3}{(x-3)(x+3)/x+2} ~the x plus two's and the x plus three's cancel
out
=(x-2)(x-3) ~(use FOIL again and you come out to...)
= x(squared)-5x+6 ~state the non-permissible values
(they are x+3 cannot equal 0, x-3 cannot equal 0, x+2 cannot equal 0, and x cannot
equal -3,3, or -2.)

So remember the steps are:

1. Factor each rational expression.
2. State the non-permissible values.
3. (Only for dividing) Find the reciprocal of the divisor.
4. Restate the non-permissible values.
5. Cancel the like terms in the denominator and the numerator.
6. Expand.

notes

hi this is kyle

her is the notes today
Simplifying rational expressions using multiplication and dividing
=x squared –5x+6
= (X-2)(x+3)
=(x-2)(x-3)
=X squared+X-6

Multiplication:
- You know how to multiply rational numbers / means over
Ex. 4/3 * by 9 / 2 =2(2) / 3 * 3(3) / 2 = (2)3 + 6 * = times

Rational expressions

Ex. X squared –4/x+3*x squared+6x+9/x+2 = (x-2)(x+2) /x+3 *(x+3)(x+3)/x+2
X+3 cannot = 0, X cannot = -3
X+2 cannot = 0 X cannot = -2

Division:
Ex. 4/3 divided by 6/5=4/3*5/6=2(2)/3 * 5/2(3)= 2(5)/3(3)=10/9

Rat’l exp.
Ex. X squared –4/x+3 divided by x+2/X squared –9=X squared-4/x+3*X squared –9/X+2

X+3 cannot = 0
X-3 cannot = 0 =(x-2)(x+2)/(X+3)*(x-3)(x+3)/(x+2)
X+2 cannot = 0 =(x-2)(x-3)
=X squared – 5x+6
X cannot = -3, 3, -2

Remember:
Step 1: factor each rational expression
Step 2: state the non-permissible values
Dividing step 3: find the reciprocal you are dividing
Step 4: state the non-permissible values
Step 5: cancel like terms in numerator and the denominator


Assignment: multiply and division
1a,b,c,d,e,f,j,k,l 2b
there is all that we did

Sunday, December 03, 2006

help!!

need help with #5 on the test, anyone?

Wednesday, November 29, 2006

Rational Expressions


Well...as reluctant as I am to do this, there were a fair amount of people missing today from class, and this was an important lesson...so, here's today's notes:

Rational Number: a/b such that a and b are integers and b≠0.

Rational Expression: p/q where p and q are polynomials and q≠0.

Polynomials: a collection of one or more terms which may have one or more variables. The variable must be a non-negative integer.

Examples:


Expression__________Rational?
1) 2/5_______________yes, 2 and 5 are polynomial constants
2) 3x0_______________yes, polynomial
3) 3x-1_______________yes, 3x-1/1
4) 3/x_______________yes, but x≠0
5) x2+x/x2+1_______________yes, but x≠±1
6) x2+5/√ x+1_______________no, √ x+1=x1/2+1 which ≠ polynomial because 1/2 ≠ integer.

∙ A rational expression is not defined in the real number system if the denominator = 0. The values of x that make the denominator = 0, are called non-permissible values
Ex. Stat the non-permissible values of the following rational expressions:
a) 10/x
x≠0
b) 4/3x-4
3x-4≠0
3x≠4
x≠3/4


I really hope this helped everyone...it took me half an hour!

-Keeley ♪

Monday, November 27, 2006

Trinomial Factoring

So far we have looked at factoring trinomials when a=1. When a≠1, there are two methods we can use.

Method I: Guess and Check Method

Example 1: 3x² + 13x + 4
We need factors (px+q)(rx+s) such that pr=3, qs=4 & qr+ps=13

(1x+4)(3x+1) 3 4
(x+4)(3x+1) /\ / 1 3 1 4
2 2
(2)(1)+(2)(3)=8
(1)(1)+(4)(3)=13
(4)(1)+(1)(3)=7

Check: x(3x) + x(1) + 4(3x) + 4(1) = 3x² + 13x + 4
3x² + 13x +4 = 3x² + 13x + 4

Example 2: 2x² + 7x + 3

(1x+3)(2x+1) (2x+3) cross multiply
(1x+1)

Check: x(2x) + x(1) + 3(2x) + 3(1) = 2x² +7x + 3
2x² + 7x +3 = 2x² + 7x + 3

Method II: Grouping Method

Example 1: 6x² + 22x + 12 18
=2(3x² + 11x + 6) / 1 18
2 9 (2+9=11)
=2(3x² + 9x + 2x + 6)
=2[3x(x+3) + 2(x+3)]
2(x+3)(3x+2)

Step1-pull out factors
Step2-multiply AC- AC=3(6)=18
Step3-find factors of 18 that add up to 11
step4-rewrite 11x in terms of 9x + 2x
Step5-use factor grouping method

Example 2: 4x² - 21x + 5 4x(-20)(-1x)+5
4 x 5 = 20 =4x(x-5)-1(x-5)
/\ (x-5)(4x-1)
-1 -20 [-1 + (-20)= -21]

there you go, have fun














Test

just seeing if this works before i type it all up just to find out it doesn't

Friday, November 17, 2006

Cumulative Review

Hello!

Well... it's Friday after school and I feel horrible about the test today. I really thought that it was very straight-forward and that you would finish it quickly and gain a real boost in your confidence. I was really surprised that you were still working very hard at it by the time the period ended.

I was really proud of you though. You toughed it out and did not give up and even did not complain about it throughout. That is a sign of maturity and hopefully a sign that you understand the purpose of the testing... it is to show you and myself your understanding of the material so far, to see what you need more practise on, and what you already grasp. As much as I am required to submit marks throughout the course, these marks are not a measure of your intelligence, but a measure of your understanding and perhaps, but perhaps not, your effort. Only you can judge that for yourselves.

SO, this is what I've decided to do. There are three areas that the majority of you seemed to struggle with on the test and two are actually somewhat related. I am going to spend one class on each topic next week and then give you back your tests on Thursday to complete.

Monday, topic 1: Changing between the standard form of an equation and point-intercept form and creating an equation of a line given two points.

See http://www.purplemath.com/modules/strtlneq.htm
See http://www.gomath.com/htdocs/lesson/linearequation_lesson1.htm

Tuesday, topic 2: Using the cosine law and solving for either the length of a side of a triangle or one of the angles of a triangle.

See http://www.wsd1.org/waec/math/Applied%20Math%20Advanced/Trigonometry/Cosine%20Law/cosinelawexamples.htm

Wednesday, topic 3: Rational exponents... simplifying polynomial expressions using the exponent laws.

See http://www.purplemath.com/modules/simpexpo.htm for integral exponents
See this link for rational exponents.

Each class your hopework will be to create study notes on these topics (like the math dictionary I had planned for you to complete, but never did assign!)

So, chin up! We are going to get through this!

Have a fantastic weekend,

Miss N.

Wednesday, November 15, 2006

Heron's Formula

Hey guys!....incase anyone needed, here's Heron's Formula.

when given three sides, of a triangle, Heron's Formula can be used to find the area of a tringle...Heron's Formula is as follows:

A=√ s(s-a)(s-b)(s-c)
where s=semiperimeter
*semiperimeter is half of the perimeter

If the given triangle has side a=3, b=4 and c=5
then: A=√ 12(12-3)(12-4)(12-5)
A=√ 6040