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thank u so much!

(x - 3) (x + 1) ( x - 4) (x + 2)  =  - 6

(x^2 - 2x - 3) (x^2 - 2x - 8)  =  - 6

Let  x^2  - 2x  =  a

So we have

(a - 3) (a - 8)  =  - 6

a^2  - 11a  + 24  = -6

a^2  - 11a  + 30  =  0            factor

(a - 5) ( a - 6)  =  0

Setting each factor to 0  and solving for x produces

a = 5      or      a  =6

So

x^2 - 2x  = 5                                or                                           x^2  - 2x  = 6

                                 complete the square on x

x^2 - 2x + 1  = 5 + 1                                                                 x^2 - 2x  + 1  =  6 + 1

(x - 1)^2  = 6                                                                            (x - 1)^2  =   7

                                 take  both   roots

x - 1  =  ±√6                                                                              x - 1  =  ±√7

x = ±√6 + 1                              or                                              x  =  ±√7  + 1

A)

the left is going down 2 a time, right stays the same. The quotient(i think that's what it's called) is increased by 1 each time. Thus, 

0/-2 = 0

-2/-2 = 1

And you can continue the pattern from here!

B)

Right increase 4, center constant, right decrease 1.

0/-4 = 0

4/-4 = -1

And continue the rest.

a)
Jesse could raise the temperature by 10 degrees Celsius.

b)
-6.2°C + 10°C = 3.8°C

\(\begin{bmatrix} -4 & 1\\ -1 & 2 \end{bmatrix} \begin{bmatrix} -9\\ 2 \end{bmatrix}= -9 \begin{bmatrix} -4\\ -1 \end{bmatrix}+ 2 \begin{bmatrix} 1\\ 2 \end{bmatrix}= \begin{bmatrix} 36\\ 9 \end{bmatrix}+ \begin{bmatrix} 2\\ 4 \end{bmatrix}= \begin{bmatrix} 38\\ 13 \end{bmatrix}\)

arctan 1/2   + arctan 1/3   = 45 degrees

The angle  formed  by  the top  left of the figure  and shorter red line can be found as 

arctan (1/2)

Similarly, the angle formed by the  top of the  figure and  the longer red line  can be found as

arctan (1/3)

Using an obscure  trig identity  

arctan (u)  +  arctan (v)   =  arctan  [ ( u + v)  / (1 - uv) ]

arctan (1/2) + arctan (1/3)  =  arctan  [   (1/2 + 1/3)  / ( 1 - (1/2)(1/3) ) ]  =

arctan [ (5/6) / (5/6)  ]  =

artan (1)  =   45°  =  the angle  we need

Dragan, my answer was a teaching answer. 

You have come in over the top of me and given the child a cheat answer.

There are lots of questions here and if you answer first you are welcome to answer however you want but please do not override my teaching answers.

Yes, such as the AoPS Mathcounts.

thanks cphill

We can use  this fact about  the angles gormed by intersecting chords

x = (1/2) (78 + 76)  = (1/2) ( 154)  =  77°

Yep...4.3  rounded to the nearest  tenth

The vertical position is given by

yf = yo + vot - 1/2 a t^2        yo = 3   vo = 70 sin 30 = 35 f/s   a = -32.174 ft/s^2

when the ball hits the ground yf= 0    so you have

0 = 3 + 35 t  - 1/2 (32.174)t^2      solve to find t

the x component of the velocity is  70 cos 30 = 60.62 f/s

   use the 't' you found above to find the x distance =  60.62 * t

The 55^o angle cuts off two arcs:  the larger arc is 100^o + 35^o  =  135^o.

The smaller arc is x^o.

So:  angle(G)  =  [ arc(BCD) - arc(EF) ] / 2

               55^o   =  [  135^o   -    x   ]  /  2

              110^o  =  135^o - x

               -25^o  =  -x

                   x   =   25^o

Lol are you cheating for Mathcounts state competition

1  - 52  to 53 = 105
2  - 34  to 36 = 105
3  - 19  to 23 = 105
4  - 15  to 20 = 105
5  - 12  to 18 = 105
6  - 6  to 15 = 105
7  - 1  to 14 = 105

Sum  =  ₁₁C_{0 +  11}C_1 + 11C_2 +  11C_{3 +   11}C_4 +  11C_5 +  11C_6 +  11C_7 +  11C_8 +  11C_9 +  11C_10 +  11C₁₁

    or  =  1 + 11 + 55 + 165 + 330 + 462 + 462 + ... + 1

The parallelogram has a shorter side (let's call it 'x') and a longer side (which is '20-x').

The altitude of length 4 goes to the longer side. Using the area of a parallelogram to be its side times its altitude,

the area of the parallelogram is 4·(20 - x).

Using the shorter side (x) and its altitude (7) to find the area, we get 7·x.

Therefore:  4·(20 - x)  =  7·x 

                      80 - 4x  =  7x

                             80  =  11x

                               x  =  80/11

To find the sine of the smaller vertex angle:  sine(angle)  =  4 / (80/11)  =  44/80  =  11/20 

is y 4.3?

You have two secants, so:  3 · (3 + y)  =  2 · (2 + 6 + 3)

Solve for y ...

9 + 3y  =  2(11)

9 + 3y  =  22

      3y  =  13

        y  =  13/3  =  4.33333333...

sqrt(20)  =  sqrt(4·5)  =  sqrt(4)·sqrt(5)  =  2sqrt(5)

sqrt(45)  =  sqrt(9·5)  =  sqrt(9)·sqrt(5)  =  3sqrt(5)

sqrt(80)  =  sqrt(16·5)  =  sqrt(16)·sqrt(5)  =  4sqrt(5)

2·sqrt(125)  =  2·sqrt(25·5)  =  2·sqrt(25)·sqrt(5)  =  2·5·sqrt(5)  =  10sqrt(5)

1 / [sqrt(5) - 2] · [sqrt(5) + 2] / [sqrt(5) + 2]  =  [sqrt(5) + 2] / [5 - 4]  =  sqrt(5) + 2

     sqrt(20) + sqrt(45) + sqrt(80) - 2·sqrt(125) + 1 / [sqrt(5) - 2]

=  2sqrt(5) + 3sqrt(5) + 4sqrt(5) - 10sqrt(5)     + [ sqrt(5) + 2 ]     =     2

We can use the intersecting chord theorem

5x  = 3 * 6

5x = 18

x = 18/5 =  3.6 cm

2^(2^x) = 16 ^(16^16)

Take the log of both sides

log 2^(2^x)  = log(16)^(16^16)   and we can write

(2^x) log 2  =  (16^16)log 16

And we can write

2^x / (16^16)  = log 16/ log 2

2^x / 16^16  =  log 2^4  / log 2

2^x / 16^16  =  4 log 2 / log 2

2^x /  16^16  =  4    rearrange  as

2^x / 4  =  16^16

2^x / 2^2  = 16^16

2^ (x - 2) = (2^4)^16

2^(x - 2) = 2^(64)      we can solve for the exponents

x - 2   = 64

x   =  66

Thanks CPhill!