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This is a Artofproblemsolving alcumas problem....

Elaborating on penquino

Its 6x^2 - x -2

So you factor it

(3x - 2)(2x + 1) < 0

So you have

2/3 and -1/2

Wow ,thank you for physcially creating one for me. I really apprectiate it! Would this be able to tranfer to the same number line as part 1?

If so, also would the vaues change if Increased my scales ? because I am going need to include all the stars and planets but in the number line you provided for the first part it only included the first 8 ,  i was tihnking of doing for evey 1 cm = 2000 km?

Factors of 6

1 and 6

2 and 3

and thats all

Added to 7

1 and 6

2 and 5

3 and 4

Now what is shared?

Sure

Using the Pythgorean Theorem  the point  can be  represented  as :

  [x, y ] =  [sqrt (n - 6^2), 6 ]

So we have that the distance from (x,y) to (8,3)  =  5 units....so......

[ sqrt (n - 6^2) - 8]^2   +  [ 6 -3]^2  = 25      simplify

(n - 36)  - 16sqrt(n - 36) + 64  + 9  = 25

n - 36 + 73 - 25  =  16sqrt(n - 36)

n + 12  =  16sqrt(n - 36)      square both sides

n^2 + 24n + 144 = 256n - 9216

n^2 - 232n + 9360  = 0   factor as

(n - 52) (n - 180)  = 0

If n = 52   then x  =  sqrt (52 - 36)  =  sqrt(16)  =  4

If n = 180  then x  = sqrt (180 - 36)  = sqrt (144)  = 12

So  (x , y )  =  ( 4, 6)

And n = 52

Here's a graph 

Corrected Answer!!!

We can use the  cosine inverse (arccos) to find this angle (in degrees)

So...the angle A  is

arccos(-.20)  = A  ≈   101.54°

To find the radian measure use this conversion :

101.54 * pi  / 180     ≈   1.77  rads   =  1.8 rads

y= -4.9t^2+42t+18.9 

The object will hit the ground when y  =  0

Use the quadratic formula to find when y = 0

-42 ±√ [ 42^2  - 4(-4.9)(18.9) ]                -42 ± √[2134.44]

________________________     =       ______________  =    

                 2(-4.9)                                            -9.8

-3/7 sec (reject)   or   9   sec   (accept)

If the median is drawn to BC  from A....the length  is

sqrt (8^2 - 5^2)  =  sqrt (64 - 25)  =  sqrt (39)

If the median is drawn from B to AC......then its length is :

 sqrt  ( 5^2  + 39/9) +  sqrt (5^2 + 39/9)/2   =

(3/2)sqrt ( 264)/3  =

sqrt (264)/2  =

2sqt (66) / 2  =

sqrt (66)

back to this, could i increase my scale to 2000 km =1 cm ? to fit them all in a numberline  because looking back at this i would need to graph all the plants and stars

what two numbers multipy to 6 and add to 7.....

YOU CAN DO EET!

x^2 + ax  - 8

Notice that since (x - 4)  is a factor   and the last term of the polynomial  is -8......the other factor must be (x + 2)

So

(x - 4) (x + 2)  = 

x^2 -4x + 2x  - 8   =   

x^2 -2x - 8

So

-2x  = ax

-2  = a

So  "a"  must be  - 2

You can see the middle line, or median, is at 13.

If it's more than 13, it's anything to the right of the median.

The median is the middle value, so 50% is to the left and 50% to the right.

Therefore, there were 50% of 220, or 110 people.

Well that's just x = y and the graph of that is y=x

OMG IT'S CORNER FOR YOU!!!!

:D :D :D

Okay! Thanks again!

I made a slight mistake on the first posting.....there should be brackets next to 3.343  and 14.657

CORRECTED!!!

Thank you so much! :)

You are the best!

                  (2x^3 + 5x^2 - 3x)  

f(x) =           ______________   =      

                         x^2  -  x

x (2x^2 + 5x - 3)  

______________  =

   x  (x - 1)

2x^2 + 5x - 3

___________  =

       x  -  1

(2x - 1) ( x + 3)

____________

        x  - 1

We have a "hole"  at x  =  0     and a vertical asymptote at   x  =  1   

We  also have a "slant" asymptote   determined as follows :

                2x   +  7

x^2 - x   [ 2x^3 + 5x^2 - 3x  ]

               2x^3  - 2x^2

              ________________

                           7x^2  - 3x

                           7x^2  - 7x

                         __________

So.....the slant asymptote  is y  = 2x + 7

The range   is    (-inf, 3.343]  U  [ 14.657, inf)

Note that the graph crosses the line y =  3.....so......it definitely exists  when y = 3

See the graph here :  https://www.desmos.com/calculator/e5di4axkpw

13 hours is the median, meaning that about half of the people spent more than 13 hours.

1/2 x 220 = 110

Your answer is 110.

Order the data from low to high

(12, 14, 14, 15, 16, 18, 18, 19)

The median  =   [15 + 16] / 2   =  31/2  =  15.5

Q1  =  [ 14 + 14] / 2  =  28/2  = 14

Q3  = [18 + 18]/2  = 36/2  = 18

Min value  = 12

Max value  = 19

Sorry CPhill !!!. I simply used all 999 terms instead of just 1/3 of them !!.

Sumfor(n, 1, 333, (9n^3 - 9n^2 + 2n) = 27,722,083,500 - which is the sum of this sequence.

[1*2*3]/3 + [4*5*6]/3 + [7*8*9]/3 + [ 10*11*12] / 3 +{13*14*15]/3 +..........+ [997*998*999]/3

1*2*1   + 4*5*2   + 7*8*3   + 10*11*4  + 13*14*5 +  997*998*333  =

1*2  + 2*4*5     +  3*7*8    + 4*10*11  +   5*13*14    +.........+  333*997*998   =

      331

2  + ∑    [ (n + 2)* (3n + 4)(3n + 5) ]   =  27,722,083,500

      n = 0

In sequences of this type, it is best to find, explicitly, the first few terms of the sequence and attempt to find a "closed form" of the sequence, which will make it much easier to sum it up.

However, it is not always easy to find such "closed forms." But, there are computer algorithms that aid greatly in finding such algorithms. I happened to have written one such algorithm in C++ and is very good at finding patterns.

(2, 40, 168, 440, 910, 1632, 2660, 4048, 5850)
Including the above terms in the program, my computer was able to come up with this very convenient "closed form" of the sequence:  (9n^3 - 9n^2 + 2n).

 Using the above closed form, we can easily sum up the sequence as follows:
Sumfor(n, 1, 999, (9n^3 - 9n^2 + 2n) = 2,242,507,747,500 - which is the sum of this sequence.