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Simplifying this rational expression
{4(x^{2} - 9)^{2}-(x+3)^{2} \over x^{2} + 6x +9}

\(\dfrac{4(x^{2} - 9)^{2}-(x+3)^{2}} {x^{2} + 6x +9}\)

\(\begin{array}{|rcll|} \hline && \dfrac{4(x^{2} - 9)^{2}-(x+3)^{2}} {x^{2} + 6x +9} \quad | \quad x^{2} + 6x +9 = (x+3)(x+3)=(x+3)^2 \\\\ &=& \dfrac{4(x^{2} - 9)^{2}-(x+3)^{2}} {(x+3)^2} \quad | \quad x+3 \ne 0 \Rightarrow x\ne -3 \\\\ &=& \dfrac{4(x^{2} - 9)^{2}-(x+3)^{2}} {(x+3)^2} \\\\ && 4(x^{2} - 9)^{2}-(x+3)^{2} = \Big(2(x^2-9)-(x+3)\Big)\Big(2(x^2-9)+(x+3)\Big) \\\\ &=& \dfrac{\Big(2(x^2-9)-(x+3)\Big)\Big(2(x^2-9)+(x+3)\Big)} {(x+3)^2} \\\\ &=& \dfrac{ (2x^2-x-21)(2x^2+x-15)} {(x+3)^2} \\\\ &=& \dfrac{ 2(x-\frac72)(x+3)2(x-\frac52)(x+3)} {(x+3)^2} \\\\ &=& 2(x-\frac72)2(x-\frac52) \\\\ &=& (2x-7)(2x-5) \\ \hline \end{array}\)

THANK YOU SO MUCH 

Thanks so much

Yes lol you are in my class

Thanks!

I will take a crack at this one!.

If you look carefully, you should notice that ALL terms beyond 89 cancel out leaving you with the following 45 terms:

(1, 2, 3, 4, 5, 6, 7, 8, 9, 12, 13, 14, 15, 16, 17, 18, 19, 23, 24, 25, 26, 27, 28, 29, 34, 35, 36, 37, 38, 39, 45, 46, 47, 48, 49, 56, 57, 58, 59, 67, 68, 69, 78, 79, 89)

Therefore, the median of the set is 28.

2² + 2*2*x + x² = 81

x² + 4x + 4 = 81

x² + 4x - 77 = 0

(x + 11)(x - 7) = 0

x = -11 and x = 7

I’m sure you can sum these yourself!

Do you mind explaining how to solve the question?

7c2  is how many different combinations of 2 you can make out of 7 choices

For EACH of those combinations the second team can make 7c2 choices

    and for EACH of THOSE choices the THIRD team can make  7c2 choices

      and for EACH of THOSE choices, the FOURTH team can make 7 c 3 choices

7c2 = 21     7c3 = 35

THEN there are FOUR ways to pick the host team , multiply THAT by 4:

7c2  x  7c2   x 7c2  x  7c3  x 4

If it is log to the base 10, then

10^4.31 = 10^log(x)

10^4.31 = x

In general, with log to the base b, we have:

a = log_b(x) means that  x = b^a

.

There are many solutions. One of them is:

a = 4, b = 7/3, c = -1, so:

a + b + c = 4 + 7/3 + (-1) = 16/3 =5 1/3

Thank you though, because the answer is correct

I don’t get how you approached the problem. Please elaborate on the answer just a bit more

Thanks, guest for spotting that !!

5.Consider the sequence 1,3,4,9,10,12,13,... which consists of every positive integer that can be expressed as a sum of distinct powers of 3. What is the 75th term of this sequence?

Note the pattern that emerges :

3^0  = 1                                                                                                         1

3^1  = 3    3^0 + 3^1 = 4                                                                           1        1

3^2 = 9   3^2 + 3^0 = 10   3^2 + 3^3 = 12   3^2 + 3^1 + 3^0 = 13         1      2        1

Note that the  number of terms - 7-  is just the sum of the first 2 rows  of Pascal's Triangle  [ the first entry is "Row 0" ]

Note that the next terms are

3^3  = 27

3^3 + 3^0 = 28

3^3 + 3^1  = 30

3^3 + 3^1 + 3^0 = 31

3^3 + 3^2  = 36

3^3 + 3^2 + 3^0 = 37

3^3 + 3^2 + 3^1  = 39

3^3 + 3^2 + 3^1 + 3^0  = 40

There  are 8 terms  which is the sum  of the elements of the  3 row of Pascal's triangle =  1  3  3   1

And each row adds 2^n  more  terms   [ where n is the row number ]

So...after the 4th row we have 15 + 2^4  =  15 + 16  =  31 terms

After the 5th row we have  31 + 2^5  =  31 + 32  = 63 terms

The first term in the 6th row  will represent 64 term  ....this will be  = 3^6   = 729

Notice that the next few terms are

3^6 + 3^0  =  730

3^6 + 3^1  = 732

3^6 + 3^1 + 3^0  = 733  

3^6 + 3^2 = 738  ....

And note that the terms in each case appended after the first term just follow the pattern of the first 15 terms

So...the 75th term  will be the sum of this first term  plus the 11th term in the above  series [ 31]  

So....the 75th term is    729  + 31  =  760

I see, thanks!!!!...also solved both.

(x + 2)^2 - (x -2) - 20

_________________

   x^2  - 9

Simplify the numerator  =  x^2 + 4x + 4  - x + 2  - 20   =  x^2 + 3x - 14  ...this is not factorable

The denominator factors as  ( x + 3) ( x - 3)

So....the answer is

x^2 + 3x  - 14

___________

(x + 3) ( x - 3)

How many distinct subsets of the set S={1,8,9,39,52,91} have odd sums?

Note that we  can choose any of the "odds" for a subset  =  4 subsets

And choosing any two of the elements we need   a set of { even, odd}  to have an  odd sum

We have  2 evens  and they can  be paired with any of the 4 odds..so  2 * 4  =  8 subsets

And choosing any 3  of the elements we need  either

{ even, even, odd}  or  {  odd, odd, odd}  to have an odd sum

In the  first case....we can choose both evens and pair them with each of the 4  odds =  4 subsets

In the second case....we can  choose any 3 of the 4  odds  = 4C3  =  4  subsets

And choosing any  4 of the elements we can have   ( even, odd, odd, odd}

We have 2 evens and again, we can choose  any 3 of 4 odds   = 2  * 4C3 = 2 * 4  = 8

And choosing any  5 of the elements we just need   (even, even, odd, odd, odd}

The two evens will appear in any of these sets by default and, again, we only need to  choose any 3 of 4 odds to  complete the set = 4C3  = 4  subsets

Note that we  cannot take all 6 elements as a sum.....[it would be even...]

So...the total possible subsets = 4 + 8 + 4 + 4 + 8 + 4   =  32 subsets

Note that we can "anchor" any one of the shells at any of the specified positions....at the next position [ let's assume that we are moving "clockwise" around the star...but..it doesn't matter, we could also move "counter-clockwise" with the same results ], we can choose any  1 of the other 9 shells  =  9 possibilities

Similarly, at the next position, we could choose any 1 of the 8 remaining shells = 8 possibilities

So...continuing this pattern at each successive position we get

9 * 8  * 7 * 6 * 5 *4*3* 2* 1   = 9!  = 362880  arrangements

The first part is easy enough. Will simply take the average of the 12 dice as follows:[1 + 2 + 3 + 4 + 5 + 6=21/6 =3.5 x 12 =42 - this is the most likely total for rolling 12 dice, or the expected total.]

Now, the second part, however, is much more difficult to calculate. There is a rather complicated formula that will calculate the total, and it comes from this page of "Mathword": http://mathworld.wolfram.com/Dice.html {Equations 7, 8, 9, 10}, where s = number of sides on each die = 6, n = number of dice = 12, p = total of dice rolled = 42, k=number of summations =[ (42 - 12)/6] =0 to 5.
∑(-1)^k*bin(12, k)*bin[(42 - 6k - 1), (12 -1)], k=0 to [(42 - 12)/6]
I entered the above into my computer and it summed them up pretty quickly: = 144,840,476 / 6^12 =0.0665387960.......etc., which is the probability of a total of 42.

Note: The above number of 144,840,476 can also be obtained, as the coefficient of x^42 in the expansion of this polynomial: [x + x^2 + x^3 + x^4 + x^5 + x^6]^12.

Area of a cube =6S^2, where S = Side length.
Area of cube =6 x 11^2
 [(a/2)-2] * [(b/3)-5] =a*b - [6*11^2 + 3] , solve for a, b
b = -(3 (5 a - 1478))/(5 a + 4), and (5 a + 4) cannot =0, 
a =22 and b=36

Clearly there are \(5\) choices for the first letter. Although it is tempting to think that there are \(4\) choices for the second letter, reading the problem carefully, we see that in the process of picking our five letters, we will never run out of any of our vowels, since there are five sets of each vowel. Therefore, we just multiply \(5\) by itself five times to get \(5^5=3125\)

Here is my 'stab' at it    (BE WARNED:I often get these answers incorrect)

4   ways to pick the host team          4

EACH of remaining 3 teams have 7c2 ways to pick committee members     7c2 * 7c2 * 7c2

and the host team has 7c3 ways to pick committee members       7c3

4*  7c2 * 7c2 * 7c2 * 7c3 = 4 x 21 x 21 x 21 x 35 = 1296540 ways

(I hope I got this one!)

I also need help on the first two..... Also, isnt 4 and 5 from AoPS homework?