Another math Christmas blog

Hi Another blog on Christmas !!:D



see that ? ^.^

today's question is : How do we add and subtract rational expressions?

so lets start lets say you get an expression like :

4b2-12 + b+3
  b +3      b+3

Step 1:
find the common denominator for the non siplified part of the expression.
therefore ;

4b2-12
Greatest number is b and 4 
so :
4b(b-3)

Step two: connect it back to the full expression :

4b(b-3) +  b+3
b+3           b+3

Step three.: you cross out the variables that are identical. 


4b(b-3) +  b+3
b+3           b+3


4b(b-3) +  b+3
b+3           b+3



4b(b-3) + 1  
 b+3

 Step four: combine what is left
4b(b-3)+1
 b+3


Works cited
Picture :http://img.spikedmath.com/comics/124-christmas-puns.png

Exponents i think..

HELLO !!  math people ^.^
I'm a bit late..well i'm really late because of some technical difficulties:(.

so lets get started with a math-christmas photo since its CHRISTMAS!!
and like a total nerd im doing a blog xP
oh well,  here the picture:

       
I dont know if HO3, is a chemistry component or a math picture but what ever I think the bear is super dupper cute 
>.<.

now that the picture is over, we are going to solve the question of the day :

How do we simply expressions with rational exponents?

so lets start with a simple, and easy expression like:

x4

                                                                                                          x-6
because we cannot have a negative exponent in the bottom of the bar we have to switch the negative exponent along with the x and flip it to the opposite side. 

For example: 

given expression : 

x4

                                             x-6
change the negative exponent to the opposite side of the bar. Therefore: 

x4 x6

                                              1
from there,we can say we are multiplying both x's. when we multiply, the exponents are added therefore :

x4x6 :  4+6 

x10

lets try more problems:
x3x(2/4)x-3  = 
Step one : add all exponents 
3+(2/4)+(-3)
3+(-3) =0
0+(2/4) = (2/4)

Step two : combine the answers 
 X(2/4)

Another problem: 
 (X15 Y25 )(1/5)

Step one: first multiply the outside exponent to the inside exponents. 
 (X15 Y25 )(1/5) :
(15)(1/5)
multiply straight across so it would be :
X:   15 x 1  = 15 = 3  
       1      5      5

Y:   25 x 1 =  25 = 5
       1      5      5

Step Two:  combine the exponents to its appropriate variable.
X3Y5


NOW YOU TRY :D

X-3
X2

X (-9/18)
X3

Works cited :
picture : http://i1.cpcache.com/product/180766012/ho_ho_ho_christmas_math_humor_teddy_bear.jpg?height=150&width=150

Radicals

안녕하세요 (annyeonghaseyo)

meaning hello ^.^

thank god its the weekend  I AM EXHAUSTED !!!!!!

Well today is another BLOG DAY:) !!!YAY!!!!! NOT REALLY !!!!! :P

Lets gets started with a picture that gets me through the day :) :

Isn't this a perfect collaboration of both FOOD and MATH ^.^ 

This is my FAVORITE math picture so far :)

Now onto our question of the day ......

How do we rationalize a denominator ?

 we only rationalized the denominator when we are dealing with radicals at the bottom of the expression. We rationalize the denominator to obtain a solid number that can make the fraction be true. 

For Example this is our expression :

   -10  

 10-√2

 

Lets start,

First Step : multiply both the numerator and 

denominator by the conjugate of the denominator.The 

conjugate means to just flip the middle sign.

Example:

  -10      (10+√2)

(10-√2) (10+√2)

Numerator :

First part = -10(10√2)

-10(10√2) = -100√2  -> radicals do not change 

unless you multiply a radical by a radical.

Denominator :

10*10=100

(10)(√2) = 10√2

(-√2)(10) = 10 -√2 ( negative times a positive is 

negative)

(-√2)(+√2) = -√4 (radical times radical equals 

radical.)

Second Step : unite all parts together.

                                      -100√2          

                         100  10√2  10-√2 -√4

Step Three: Combine like terms

-100√2  = no like terms so we leave it as it is.

100 10√2 10-√2 -√4 :

both (10√2) and (10-√2)  cancel each other because 

one is positive and the other is negative. 

we are then left with : 100 -√4 

The square root of (√4)  is 2, therefore it would turn 

to -2. 

so.. 100-2 = 98.

Step four : Unite both parts for a final answer. 

YAY !!! ^.^

-100√2

98 

Now you try :D

Solve :

    4   

-9-√3

    9    

10+√5

                             Final picture :

This is me after so much math,hopefully you feel the 

same as i do :P

                               안녕히 가세요 (annyeonghi gaseyo)

                               meaning GOODBYE ^.^

Works cited:

Food and math picture: http://img.math-

fail.com/images-old/math-cereal.jpg

Woman running away:

http://121lml.files.wordpress.com/2010/07/projectcompletion01.jpg

Another just simple math blog :)

Hi there, its another MATH blog :).

i thought of starting with a random picture,so that the math blog wont be so mathy >:P

Today pic :D

this is a pretty picture ^.^,it makes me feel like i'm in a spy movie:)

well on today's math blog our aim today is:(tan-tan-tan)

How do we Factor by grouping?

when we factor a equation, we use the cubic equation
like:
2x3+ 4x2+25x+10= 0 

because this equation is a cubic equation we cannot use the diamond problem to find the factors. 

lets get started :
Step 1: we separate this equation in half like : 2x3+4x2
Step 2: we find the greatest common factor of 2x3+4x2.
 in this case 2x2 is their common factor,because both 2 and 4 are divisible by 2 and both of them have x's.

Step 3:  using the common factor we can make a factor of 2x3+4x2
for example:
     2x3+4x2    (original part)
     2x2(x+2) (undistributed part ) 
Step 4: then we find the common factor of the second part of the equation.
For example: 25x+10
5(x+2) 
5 = the greatest common factor
(x+2)= is the factor of 25x+10

Step5:  connect the parts together.

2x2(x+2) + 5(x+2)

Step 6:  finishing up by grouping it.
For example: 
2x2(x+2) + 5(x+2)

(2x2+5) = connect both the 2x2+5 into one parenthesis.
(x+2)= stays the same. 

Final answer:
(2x2+5)(x+2)

Now you try :

6.

7.

3.

LAST PICTURE:
                                                     
This is how I feel after doing math blogs :)

Problems: http://www.regentsprep.org/Regents/math/algtrig/ATV1/RevFactPractice.htm
First Picture: http://cdn.c.photoshelter.com/img-get/I0000XMSl1tB.tzg/s/650/650/jiufen-taiwan-chinese-lanterns-night-002.jpg
Last picture:http://www.regentsprep.org/Regents/math/algtrig/ATV1/revFactorGrouping.htm