today in class we did a review game with radical equations and inequalities. the review helped us prepare for tomorrows quiz that will have domain and range, transformations, solving inequalities and graphing.
3rd period: 11-18-08 November 21, 2008
First we took a quiz finishing up rational functions. Then we learned how to solve Direct Variation, Inverse variation, Joint Variation and Conmbined Variation
1st period: 11-18-08 November 18, 2008
Today in class we reviewed and took a quiz finishing up rational functions. In the beginning when we started rational functions I didn’t really understand them. Now after alot of practice I have a better understanding of how to solve them.
3rd Period: 11/12/08 November 17, 2008
Hi, this is John Joseph from Mrs. Hawn’s 3rd Block class. Today in Algebra 2 we reviewed over Solving Rational Expressions and also had a quiz on Adding/Subtracting Rational Expressions.
11-13-08 : 1st period November 17, 2008
On Friday In Mrs. Hawns class we played a game of Jeopardy with the class that was really fun, because we worked as teams to get the problems done so it gave use a chance to help others, and to review for the up coming test it aslo taught me and some of my other peers things we did not know how to do. Some of the problems we had were quiet difficult, This was one of the ones i got lost in but now understand because one of my peers helped me, and explained it, was this one.
EX) (x^2-9) over 4 divided by (x+3) over 8 which becomes to (x^2-9) over 4 divide by (x+3) over 8, which then you flip the sign and it becomes (x^2-9) over 4 times 8 over (x+3), well negative three times positive three equals nine so you would change (x^2-9) into (x-3)(x+3) and still leave it over 4 times 8 over (x+3) so then you would cancel what was the same which in this case would be (x+3) and (x+3), so then your left with (x-3) and 2 because 2 goes into 4 and 8= (x-3)2.
1st period: 11-10-08 November 14, 2008
Hi, this is lashonda and today in mrs. hawns 1st period class we took a on rational expressions & denominators quiz. But before that we did get time to review which i was pretty glad about. I think i did an ok job on because i sort of know the material. We also went over new types of rational expressions where at the end of factoring and all that you have to find the x value i think. I sort of understand it but i could use a little exra help in the long run.
3rd period: November 7, 2008 November 10, 2008
Today in class, we continued to do examples of rational expressions in finding the common denominators. When given a problem, there are 5 steps. The first one is to factor the denominator, find the LCM (least common multiple), get a common denominator, add or subtract the problems and then factor, cancel and simplify.
Ex.) (3/4q – 2/5q – 1/2q) The first thing we did was change the sign to a + and the 2 and 1 to a -. Then you factor the denominator. The outcome would be 4*q 5*q 2*q. Then you try to find the LCM. The 4 is used in the first term, the 5 in the second, 2 in the third and the q in all three parts of the equation. So with each part, you have to find what is missing from it and then multiply. So for 3/4q you would multiply it by 5*2 since those parts are included. You would do the same with the rest of the problem. -2/5q is multiplied by 2*4 and -1/2q by 4*5. When you work it out, you get a common denominator of 40q. Then you add the problem up 30/40q + -16/40q + -20/40q. When you add it up you get -6/40q which can be reduced to -3/20q
I thought this problem was kind of challenging, but after a few more examples I started to understand how to do it.
1st Block:11/06/08 November 10, 2008
In class today we continued our lesson on +/- rational equations ,only instead of having the same denominator for each problem there were unlike denominators in every example we did. In solving these equations with unlike denominators we first had to factor them by using the diamond method or simply factoring out each number that gives you the original denominator once multipied together.sencond, you look for the least common multiple (LCM), the LCM is the number that you get once the two denominators were factored out {when there’s two of the same numbers in each equation that was factored, then you only bring one down to the LCM}. Third, get common denominators of the LCM and multiply by the missing factor in each denominator. Once multiplied by the missing factor the denominator should now be identical, and everything can be added. last thing to do for the equation now is to factor your answer, cancel the like terms, and simplify if it needs to be. I do understand whats goin on in these problems im just having a hard time putting everything in order, and once that happens i lose focus and things start to get harder because now im completly lost.
3rd period: 11/6/08 November 10, 2008
Today, on Thursday November 6th in Mrs. Hawns classed we continued our lesson from the last day, besides us adding some new ways and methods. One thing that was different was unlike denominators and adding and subtracting them. In doing this we used different colors to represent the different parts of the problem. This helped to keep them seperate and not confuse them. There were many steps and notes we took for this that we learned. The steps for adding and subtracting them are:
1. factor
2. least common multiple (LCM)
One thing that is different in subtracting then adding is that you have to change the signs.
That’s what i learned today. THis was a great lesson!
2nd period: 11-5-08 November 6, 2008
By the example we learned that in order to combine two fraction they need to have a common denominator, in this case it is 7 (ex. 5/7 + 1/7.)
With the problem 1/5 + 4/10 : we need to find a common denominator which is 10 because 10 is the largest divisible number and 5 is a factor of ten. So we multiply the fraction 1/5 x 2/2 (because 2×5=10) This gives us 2/10 by combining both numerators and both denominators. now our equation is: 2/10 + 4/10, we can now add the two fractions because they have common denominators. this gives us 6/10 as our answer, but both can be divisible by two so we divide both top and bottom of the fraction to get 3/5 as our completely simplified answer.
With more complex fractions and common denominators we still keep the same denominator but now we can combine like terms within two fractions.
ex. 2/3xy + 4/3xy = 6/3xy ( these have common denominators and no other ways to be simplified.
ex. (4a^2+4a/a^2-9) + (-3a-1/a^2-9) . Combining the fractions we keep the 4a^s and subtract 3a from 4a giving us A, then we have the 1 left over leaving us with a complete answer of : 4a^2 + a -1 / a^2-9
ex: Next problem is (2y+5/5y+10) + ( y+1/5y+10): First we add the two fractions to receive (3y+6 / 5y+10), not finished we see that in the numerator and denomiator there is a common factor so we take that out and simplify.
3(y+2) / 5(y+2) , the (y+2)s cancel out leaving the complete fraction : 3/5
Thank you peace your boy Dy nasty
3rd period: 10-30-08 November 4, 2008
In class today, We did a warm up on topic 33, the warm up was review. And part of the warm up was what we learn yesterday. Then Mrs. Hawn made the class take notes on Multiplying and Dividing Rational Expressions. To me learning this new method was not that hard. After taking notes the class got alittle noisy but Mrs. Hawn got it under control. Then the class played a game to help understand the method a little more. And now the Obama 08 and the Swaggers are tied.
Breonna
