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Akira bought 18 pizzas. If there were total of 30 pizzas.

% of pizzas he bought= 18/30 * 100

                                    = 6 * 10

                                    = 60

Akira bought 60% pizzas.

Note that Geno’s data set plots y ≈ 0.0535715 for P(8). This value is very close to 0.05 for P(8) on Melody’s graph, so it’s probably correct. So, the answer to your question is, The polynomial at P(8) has y ≈ 0.0535715

This math problem demonstrates how extrapolating (infer by extending known information) a point from a polynomial equation created from a limited and narrow data set will have an error –sometimes a significant error –for the location of a nearby data point.  The reason is the created polynomial does not include data points for the change in the curve.

Using Melody’s Geobraga graph, overlay g(n)=((n)/(n^(2)-1)). This will show the points on the original plot compared to the polynomial based on the sample points. Note that the graphs align perfectly until after the seventh point, where they begin to diverge. By the eighth point, the polynomial graph shows a sudden divergence down toward the x-axis, instead of the asymptotic glide toward the x-axis. So, for the polynomial, at P(8) y ≈ 0.0535715 compared to y =8/63 (≈ 0.12698) for the original equation.

GA

--. .-

Thanks, your correct

I just had vodka; did you have something? 

Video game = $29

Puzzle = $13

Ball = $4

If this is the same question as posted before    x = 45^o       try searching for the other answer for the solution....

Do aces count as face cards?

The largest angle is 115 degrees.

Here is the correct picture:

And the correcct answer is 5/18.

i still dont get it

hint:

(n+3)! = n!(n+1)(n+2)(n+3)

sqrt(2*sqrt(2 - t)) = 7 - t

\(\sqrt{2*\sqrt{2 - t}} = 7 - t\\ 2*\sqrt{2 - t} = t^2-14t+49\\ \sqrt{2 - t} =\frac{ t^2-14t+49}{2}\\ 2 - t =\frac{( t^2-14t+49)( t^2-14t+49)}{4}\\ 8 - 4t =( t^2-14t+49)( t^2-14t+49)\\ 0 =( t^2-14t+49)( t^2-14t+49)+4t-8 \)

\(0=2401 - 1372 t + 294 t^2 - 28 t^3 + t^4+4t-8\\ 0=t^4 - 28 t^3 + 294 t^2 - 1368 t + 2393\)

Wolfram alpha says no real solutions.

You do need to check my algebra though. (for careless errors)

LaTex:

\sqrt{2*\sqrt{2 - t}} = 7 - t\\

2*\sqrt{2 - t} = t^2-14t+49\\
\sqrt{2 - t} =\frac{ t^2-14t+49}{2}\\

2 - t =\frac{( t^2-14t+49)( t^2-14t+49)}{4}\\
8 - 4t =( t^2-14t+49)( t^2-14t+49)\\

0 =( t^2-14t+49)( t^2-14t+49)+4t-8

You may well be right Ginger. 

Certainly all the evidence suggests/shows  that there is NO polynomial of degree 5 that fits the data.

Since you want one odd number and four even numbers, there are five possibilities:

     OEEEE  +  EOEEE  +  EEOEE  +  EEEOE  +  EEEEO

The probability of getting an odd number is 3/6.

The probability of getting an even number is 3/6.

Therefore, the probability of OEEEE is  (3/6) x (3/6) X (3/6) x (3/6) x (3/6). = 1/32

This is also the probability of each ot the other possibilities.

Adding these together:  (1/32)  +  (1/32)  + (1/32)  +  (1/32)  +  (1/32)  =  5/32

The total of the three is  $46.00.

Since the video game costs  $29.00,  there will be  $46.00 - $29 .00  =  $17.00 remaining.

Since the puzzle costs $16.00 less than the video game, the video game costs  $16.50

                                                                                            and the puzzle costs  $   0.50

The first digit can be any of the ten digits:                              10 / 10

The second digit can be any digit except of the first number:  9 / 10

The third digit can be any digit except either of the first two:    8 / 10

The fourth digit can be any digit not previously chosen:           7 / 10

The last digit can be any digit except those already chosen:    6 / 10

The final probability is the result of multiplying these probabilities: 

                       (10 / 10)  x  (9 / 10)  x  (8 / 10)  x  (7 / 10)  x  (6 / 10)  =  

If these numbers are correct, then what is p(8)?

Good job!!! 

Im Nachhinein ergibt es echt Sinn 

Dankee an euch beide!!! Falls Fragen auftauchen zögere ich nicht und stelle sie. 

Angle GIH = 34 degrees.

Remember Vieta's formulas?

rs  =  -12/3 = -4

r+s = -4/3

Now, do this:

(r+s)² = r² + s² + 2rs      sub in the reds above

(-4/3)² =  r² + s²  + 2 ( -4)

r² + s²  = (-4/3)²  + 8                   <====== you can finish, yah ?

q = # quarters      value =   .25 q

d = # dimes = 600-q      value = .10 (600-q)

summed = 123.75

.25q + .10 ( 600-q) = 123.75       <==== solve   'q'

1260 word in 12 hours ( 12 X 60 = 720 minutes)

1260 w / 720 m = ............   words per minute

9/6   =    1  3/6  =   1  1/2

The determinate of a square matrix [A] is equal to zero if and only if the determinate of its reduced row echelon form [B] is equal to zero. It’s important to note that matrix [A] and reduced-form [B] do not have the same determinates, but they are zero at the same time. Because they are zero at the same time its undefined.

A graph of the \(y =\frac{n}{n^2-1}\) equation shows two asymptotes and a root of zero. The inverse of this equation \(y= \frac{n^2-1}{n}\)  will give two roots (1) and (-1) (the singularities of its inversion) and an asymptote at zero.  This suggests the fifth-degree polynomial equation that represents a finite portion of that equation has one real root and four complex roots.  Creating this polynomial equation by using (7) points of \(y =\frac{n}{n^2-1}\), and then extending (interpolating) it linearly yields a point that is notably different from the base equation.  That may be (one of) the point of the exercise –an example of catastrophe ...  

My attempt at using Gauss-Jordan elimination was a catastrophe...

I’m interested in the official AoPS solution. I’m sure it will be over my head, but it will give me something to look up to.... 

GA

--. .-

Thanks Ginger,

I read that but why does it stop it from being used for a simultaneous solution?

I thought it might be because the coeff of x^5 was zero. 

So I tried repeating it by not using one of the equations.

This way the degree wold be 4, not 5

But exactly the same thing happened.