Voldemort

Ok first off, there is a difference between solutions and answers. You should know as you are on this site.

2. Can you show your work? I literally plugged those in and they do not seem to satisfy the equation?

3. If this is done intentionally, please stop. You are just literally wasting your time going on random questoins and "answering" them. 

Please be aware that the Guest who 'answered' your question did miss a lot of steps. And there is no such thing as "theory of continued fractions"???

I would recommend you checking this out. https://web2.0calc.com/questions/deliberately-misguiding-answers#r2

Denote y = x-1

y = 1/(2 + 1/....

1/y = 2 + 1(2 + 1/(2..

1/y - 2 = y 

y^2 = -2y + 1

y^2 + 2y - 1 = 0

By Quadratic Formula,

y = -1 +- sqrt(2)

Because y has to be nonnegative, y = -1 + sqrt(2)

y = x - 1

-1 + sqrt(2) = x -1

x = sqrt(2) 

y = 1/2x^2 - 4

Suppose that \((x_0, y_0)\) is a point on this graph closest to the origin.

Hence:

\(y_0 = \frac12(x_0)^2 - 4.\)

Using this information, we need to find 

\(\sqrt{x_0^2 + y_0^2}\)

We can plug in \(x_0^2\) with \(2y_0 + 8\) (Make sure you see how I got this)

Hence, the distance between the points will be:

\(\sqrt(y_0^2 +2y_0 + 8)\)

Since this distance is to be minimized by how we defined the point (x_0, y_0), we can find the vertex of the that quadratic in y_0. If you complete the square, you will recieve that the parabola reaches its minimum when \(y_0 = -1\).

Can you solve the problem from here?

If previous SOLUTIONS weren't correct, then please go back and try to spot what they did wrong.

If previuos ANSWERS weren't correct, why are you entering them in the first place?

Now before you ask a question, think about what you can do to get a cleaner equation. What do you notice about the denominators?

r^2 + s^2 = (r + s)^2 - 2rs

3x^2 + 4x - 12 = 2x^2 - 6x + 8

Move all the terms to onside side

x^2 + 10x - 20 = 0 

Sum = - 10.

Product = - 20.

100 + 40 = 140. 

We know that 5 is the greatest number, so we'll take two cases:

(Also note that since 1 is the smallest number, it will have to go in the middle box, no number is less than 1 in the list)

Case 1. 5 is on the leftmost side of this inequality.

Then we will get:

5 > __  > 1 < __ < __

    Subcase 1.1: 4 is on the rightmost side:

    5 > __ > 1 < __ <  4
    Now 2 or 3 can be placed in each of those 2 boxes, so resulting in 2 for this subcase.

   Subcase 1.2: 4 is next to 5:

   5 > 4 > 1 < __ <__

   Note that there is only 1 option for this subcase as we can only place 2 and 3 in only one way. 

Hence there are 3 options for this case. 

Case 2. 5 is on the rightmost side of this inequality:

This inequality "mirrors" itself so this will also result in 3 options as well.

Hence, there will be \(\boxed{6}\) ways to order the integers 1 through 5 in the boxes. 

Guest, you are super close, but note that a cannot be 0 as that will violate the definition of a cubic polynomial. 

By f(1) = 9, you will get that:

a + b + c + d = 9

Since a has to be greater than or equal to 1, we can set another variable f that is equivalent to a - 1.

f + b + c + d = 8

We can use stars and bars from here, as f can take a value of 0 as well. Hence, the answer is actually 

C(11, 3) = 11 x 10 x 9/6 = 15 x 11 = 165.

This is a public site that can even be accessed by users who remain anonymous under the name, "Guest". You should very much expxect people to post answers that are intentionally wrong to misguide you in your learning. As for those asking questions on this site, I would like to clear up some things. 

1. If the person who answered has not provided ANY explanation to their answer or missed a significant amount of steps in their steps, do not even bother using those answers or solutions to do whatever you have to do. 
2. It is your responsibility to make sure all calculations are correct. Moderators and users who provide solutions to questions are not going to take their precious time away from their to thorougly check everything. You have to read the explanations and if you point out a mistake, great! Try to fix it. 

Users should not expect:

- All users to give them an answer

- To gain learning from simply reading through the answers AND solutions.

Please keep these disclaimers in mind as you go through the whole forum. 

Thank you. :)

No such polynomial exists under those conditions! Unless my calculations are wrong...