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1.     Factor the expression  x^(3)+3x^(2)-4x-12  Show Your Work.

2.     Solve the equation p2 + 4p = 21 by completing the square. Show your work.    SHOW how you complete the square, show steps.

3.  Multiply the polynomials (a2 + 3a – 4) and (2a2 – a + 5). Simplify the answer. Show your work.

Where is the vector in question?

a=1; b=(a);c=(a*2);d=(a*4);e=(a*8);f=(a*16); g=if(b+c+d+e+f==16*(1/b+1/c+1/d+1/e+1/f), goto loop, goto next);loop:printa,", ",;next: a++;if(a<=1000000, goto1, 0):

The sequence is:

Sum = 1, 2, 4, 8, 16 = 31
Sum of their reciprocals:
1/1 + 1/2 +1/4 + 1/8 + 1/16 =1.9375

The sum of the sequence =16 x sum of their reciprocal =1.9375 = 31. And that is the end!!.
P.S. I checked all the numbers up to 1,000,000 and this is the only sequence of its kind.

I think it would be 3/5 not sure

Its similar to if a = b and b = c, then a = c. If I is greater than M and M is greater than N, than N is less than all other variables, so N

Think about it on a number line:

I is going to be the furthest to the right, then M, then finally, N.

              N            M         I                  

low<------------------------------>high

does this fulfill our requirements? I is greater than M, M is greater than N. So, therefore, by looking at this, we can determine that N is less than I.

If you do it correctly, you should find x = 1/2   or  -1     (which is the value of sin(theta) )

The Identity is:

Tan (A+B)  =  (Tan A + Tan B) / (1-TanA TanB)

I'll do the first one....you can do the rest..I'll use exact values rather than calculator values

Tan(30+45) =  (tan30 + Tan 45) / (1-(tanA tanB )

                   =  (1/2/(sqrt3/2) +1)/ (1 -(1/2(sqrt3/2)(1))

                   =  (1 + 1/sqrt3) / )(1-(sqrt3))

                    ((sqrt3 +1)sqrt3) / (sqrt3-1)/sqrt3

                     (sqrt3+1) / ( sqrt3-1)         ~  = 3.73

(A)   Given that 2sin^2(θ) + sin(θ) - 1 = 0, find the two values for sin θ

(B) Given that 0° ≤ θ ≤ 360° and that one solution for θ is 30°, find the other two possible values for θ.

There are 36 possible combinations of rolls

if B rolls a

1   there are 5 higher rolls that A could roll

2                   4

3                   3

4                   2

5                   1        5 + 4 +3 + 2 + 1     15 out of 36 combinations with A> B      15/36= 5/12

(as an aside: there will be 15 rolls where  A < B   and  6 will be ties    15 + 15 + 6 = 36 possible rolls)

It's a perfectly reasonable question.

The actual number of twin primes is not relevant.

(Has it been proven that they are infinite in number, or is it still conjecture ?)

All twin primes are of the form \(\displaystyle 6n\pm 1,\)

(the integer between them has to be divisible by both 2 and 3, so has to be divisible by 6).

Their sum will be of the form 12n, meaning that their sum will be divisible by 12.

Could there be a larger common divisor ?

Look at the first few cases.

5   +  7  = 12 = 12*1,

11 + 13 = 24 = 12*2,

17 + 19 = 36 = 12*3.

Clearly, 12 is going to be the largest.

A:   Let x = sin (theta)

then   2x^2 +x -1 = 0     then use quadratic formula to find x  (which is sin(theta))

B: Use your answer from A to determine theta possibilities

Did you mean to add a slash here? 1/((1 - a)(1 - ab)) + 1/2*a/((1 - a)(1 - ab)) + 1/2*b\(/\)((1 - a)(1 - ab)) = (a + b + 2)/(2(1 - a)(1 - ab))

                                                                                                                                     ^

|-7|   = 7      you can use the web2 calculator to find such things.....

This is the calculator output:

|-7| = 7

I > M > n    should be pretty obvious....but let's just pick some numbers to put in there...

I = 10   M = 5     then      I > M     because   10 > 5

Now suppose   we pick '2' for n    then   M>n  because   5>2

Now what do you find for the relationship between  I  and n ?     I =10   n = 2

According to the bar graph there are    6+4 +1 + 9 = 20 outcomes

Pink has 6 outcomes     6/20 = .3

  green   4                      4/20 = .2

  yellow   1                      1/20 = .05

   blue     9                       9/20 = .45   probability       

I think answer 'a' shows the probabilities correctly

    answer 'd' shows the frequencies correctly  

First find the angle next to 123     123 + angle1 = 180

   then fin the ngle next to 114       114 + angle2= 180

Now you can compute the third angle in the bottom tringle

  Angle 3 = 180 - angle 2 - angle 1

this will also be the bottom angle in the right triangle

 then the other angle in the right triangle will be   180 - 90 - angle 3

   this will also be the angle in the bottom of the TOP triangle

     the other two angles in the top triangle are = (and this includes 'x')

               so subtract this bottom angle from 180    and divide by 2 to find 'x'       Give it a try !

"The three-digit base x numeral 7yc is twice the three-digit base x numeral 3y7.  Find x and y."

Part 2

(b) Find the sum of all the numbers on the black squares. (|a| < 1 and |b| < 1)

Sum of all the numbers on the black squares:

\(1 \\ +a^2+ab+b^2 \\ +a^4+a^3b+a^2b^2+ab^3+b^4 \\ +a^6+a^5b+a^4b^2+a^3b^3+a^2b^4+ab^5+b^6 \\ +\ldots\)

Infinite sum:
\(\sum \limits_{l=0}^{\infty} \left( \sum \limits_{k=0}^{2l} a^{2l-k}b^k \right) \)

\(\begin{array}{|rcll|} \hline && \mathbf{\sum \limits_{l=0}^{\infty} \left( \sum \limits_{k=0}^{2l} a^{2l-k}b^k \right)} \\ &=& \sum \limits_{l=0}^{\infty} \left( \sum \limits_{k=0}^{2l} a^{2l}a^{-k}b^k \right) \\ &=& \sum \limits_{l=0}^{\infty} a^{2l}\left( \sum \limits_{k=0}^{2l} a^{-k}b^k \right) \\ &=& \sum \limits_{l=0}^{\infty} a^{2l} \left( \sum \limits_{k=0}^{2l} \left(\dfrac{b}{a}\right)^k \right) \quad | \quad \sum \limits_{k=0}^{2l} \left(\dfrac{b}{a}\right)^k = \dfrac{1- \left(\dfrac{b}{a}\right)^{2l+1} }{1-\dfrac{b}{a} } \\ &=& \sum \limits_{l=0}^{\infty} a^{2l} \left( \dfrac{1- \left(\dfrac{b}{a}\right)^{2l+1} }{1-\dfrac{b}{a} } \right) \\ &=& \sum \limits_{l=0}^{\infty} a^{2l} \left( \dfrac{a- \left(\dfrac{b}{a}\right)^{2l}b }{a-b } \right) \\ &=& \dfrac{1}{(a-b)} \sum \limits_{l=0}^{\infty} a^{2l} \left( a- \left(\dfrac{b}{a}\right)^{2l}b \right) \\ &=& \dfrac{1}{(a-b)} \sum \limits_{l=0}^{\infty} \left( a^{2l+1}- \left(\dfrac{ a^{2l}b^{2l}}{a^{2l}}\right)b \right) \\ &=& \dfrac{1}{(a-b)} \sum \limits_{l=0}^{\infty} \left( a^{2l+1}- b^{2l+1} \right) \\ &=& \dfrac{1}{(a-b)} \left( \sum \limits_{l=0}^{\infty} \left( a^{2l+1} \right) - \sum \limits_{l=0}^{\infty} \left( b^{2l+1} \right) \right) \\ &=& \dfrac{1}{(a-b)} \left( a\sum \limits_{l=0}^{\infty} a^{2l} - b\sum \limits_{l=0}^{\infty} b^{2l} \right) \quad | \quad \sum \limits_{l=0}^{\infty} a^{2l} = \dfrac{1}{1-a^2},\ \sum \limits_{l=0}^{\infty} b^{2l} = \dfrac{1}{1-b^2} \\ &=& \dfrac{1}{(a-b)} \left( \dfrac{a}{1-a^2} - \dfrac{b}{1-b^2} \right) \\ &=& \dfrac{1}{(a-b)} \left( \dfrac{a(1-b^2)-b(1-a^2)}{(1-a^2)(1-b^2)} \right) \\ &=& \dfrac{1}{(a-b)} \left( \dfrac{a-ab^2-b+ba^2}{(1-a^2)(1-b^2)} \right) \\ &=& \dfrac{1}{(a-b)} \left( \dfrac{(a-b)+ab(a-b)}{(1-a^2)(1-b^2)} \right) \\ &=& \dfrac{1}{(a-b)} \left( \dfrac{(a-b)(1+ab)}{(1-a^2)(1-b^2)} \right) \\ &=& \mathbf{ \dfrac{1+ab}{(1-a^2)(1-b^2)} } \\ \hline \end{array}\)

tan(a+b)=Tan(a)+Tan(b)/1-Tan(a)Tan(b)

tan(30+45)=tan(30)+tan(45)/1-tan(30)*tan(45)

tan(30)=0.557350269

tan(45)=1

(0.5574+1)/1-0.5574*1=1.5574/0.4426=3.6 approx. 

In parallelogram ABCD, angle A is acute and AB = 5.  Point E is on AD with AE = 4 and BE = 4.  A line through B, perpendicular to CD, intersects CD at F.  If BF = 5, find EF.

Get you answer, with explanation, here

Let a and b be real numbers, where |a| < 1and |b| < 1. In an infinite grid, I write the numbers 1, a, a^2, a^3
in the first row. After that, each number is equal to b times the number above it. For example, the numbers in the second row are b, ab, a^2b, a^3b.

Here's the rest of the grid:

(a) Find the sum of all the numbers on the infinite grid.
Infinite sum:

\(\sum \limits_{l=0}^{\infty} \left( \sum \limits_{k=0}^{\infty} a^k b^l \right) \)

\(\begin{array}{|rcll|} \hline && \mathbf{\sum \limits_{l=0}^{\infty} \left( \sum \limits_{k=0}^{\infty} a^k b^l \right)} \\ &=& \sum \limits_{l=0}^{\infty} b^l \left( \sum \limits_{k=0}^{\infty} a^k \right) \quad | \quad \sum \limits_{k=0}^{\infty} a^k = \dfrac{1}{1-a} \\ &=& \sum \limits_{l=0}^{\infty} b^l \left( \dfrac{1}{1-a} \right) \\ &=& \left( \dfrac{1}{1-a} \right) \sum \limits_{l=0}^{\infty} b^l \quad | \quad \sum \limits_{l=0}^{\infty} b^l = \dfrac{1}{1-b} \\ &=& \left( \dfrac{1}{1-a} \right) \left( \dfrac{1}{1-b} \right) \\ &=& \mathbf{\dfrac{1}{(1-a)(1-b)}} \\ \hline \end{array}\)

Lets see

\(\frac{-30}{(2250-0.75*40*40)^{0.5}}\\ =\frac{-30}{\sqrt{(2250-0.75*40*40)}}\\ =\frac{-30}{\sqrt{(2250-75*4*4)}}\\ =\frac{-30}{\sqrt{25(90-3*4*4)}}\\ =\frac{-30}{5\sqrt{(42)}}\\ =\frac{-6}{\sqrt{(42)}}\\ =\frac{-6\sqrt{42}}{42}\\ =\frac{-\sqrt{42}}{7}\\ \)

I just answerd in full and then lost it.  

The short answer.

there is an a such that the polynomial is (x-4)^3

This will have all real and positive roots.  they are all x=4.

fiind the  'a'  that makes this expansion work and you have your answer.