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Jack and Jill played a game for two people.
In each game, the winner was awarded 2 points and the loser 1 point.
No games were drawn. Jack won exactly 4 games and Jill had a final score of 10 points.
How many games did they play?

\(\begin{array}{|r|cll|} \hline \text{Jack won }4 \text{ games} & & \text{Jill lose } 4 \text{ games} \\ \text{Jack lose }x \text{ games} & & \text{Jill won } x \text{ games} \\ && 4*1\text{points} + x*2\text{points} = 10 \\ && x = 3 \\ \text{Jack lose }3 \text{ games} & & \text{Jill won } 3 \text{ games} \\\\ \text{They did play } 4+3=7 \text{ games } \\ \hline \end{array} \)

Thanks Alan, will check it out..

Thank you Melody...I'm sure it must have been a pelling error. .....spelling error...lol....Stay well untill next time!

Let \(f(x)\) be a monic polynomial of degree \(4\).  If \(f(1) = 1\), \(f(2) = 2\), \(f(3) = 3\), and \(f(4) = 4\), then what is \(f(6)\)?

\(\begin{array}{|lrcll|} \hline & \mathbf{f(x)} &=& \mathbf{x^4+ax^3+bx^2+cx+d} \\\\ f(1)=1: & 1&=& 1^4+a*1^3+b*1^2+c*1+d \\ & \mathbf{a+b+c+d} &=& \mathbf{0} \\\\ f(2)=2: & 2&=& 2^4+a*2^3+b*2^2+c*2+d \\ & \mathbf{8a+4b+2c+d} &=& \mathbf{-14} \\\\ f(3)=3: & 3&=& 3^4+a*3^3+b*3^2+c*3+d \\ & \mathbf{27a+9b+3c+d} &=& \mathbf{-78} \\\\ f(4)=4: & 4&=& 4^4+a*4^3+b*4^2+c*4+d \\ & \mathbf{64a+16b+4c+d} &=& \mathbf{-252} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1) & \mathbf{a+b+c+d} &=& \mathbf{0} \\ (2) & \mathbf{8a+4b+2c+d} &=& \mathbf{-14} \\ (3) & \mathbf{27a+9b+3c+d} &=& \mathbf{-78} \\ (4) & \mathbf{64a+16b+4c+d} &=& \mathbf{-252} \\ \hline (2)-(1):& 7a+3b+c &=& -14 \quad &| \quad [1] \\ (3)-(2):& 19a+5b+c &=& -64 \quad &| \quad [2] \\ (4)-(3):& 37a+7b+c &=& -174 \quad &| \quad [3] \\ \hline [2]-[1]:& 12a+2b &=& -50 \quad &| \quad \{1\} \\ [3]-[2]:& 18a+2b &=& -110 \quad &| \quad \{2\} \\ \hline \{2\}-\{1\}:& 6a &=& -60 \\ & \mathbf{a} &=& \mathbf{-10} \\\\ \{1\}& 2b &=& -50 -12a \\ & 2b &=& -50 -12*\left(-10\right) \\ & \mathbf{b} &=& \mathbf{35} \\\\ [1] & c&=& -14-7a-3b \\ & c&=& -14-7*\left(-10\right)-3*35 \\ & \mathbf{c} &=& \mathbf{-49} \\\\ (1) & d &=& -a-b-c \\ & d &=& -\left(-10\right)-35-\left(-49\right) \\ & \mathbf{d} &=& \mathbf{24} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{x^4+ax^3+bx^2+cx+d} \quad | \quad a=-10,\ b=35,\ c=-49,\ d=24 \\\\ \mathbf{f(x)} &=& \mathbf{x^4-10x^3+35x^2-49x+24} \\\\ f(6) &=& 6^4-10*6^3+35*6^2-49*6+24 \\\\ \mathbf{f(6)} &=& \mathbf{126} \\ \hline \end{array}\)

Regarding your "You cant do 44/10 this is because 44 is not the max amount of points you can get!"    

Sure I can.  It worked didn't it!  I didn't "do" 44/10 for the reason that it was the maximum amount of points possible, but because it was the amount of points the team DID get. 

_.

Find the magnitude and direction of the net external force on the crate.

hi again Juriemagic,

I am very flattered by your term of endearment. 

I expect a lot of people here would think you have a little sunstroke though.   LOL.

I think you have a copy error in the original question

T1 = 1     =1^2

T2 = 4      =2^2

Ahr there is a mistake, I am pretty sure that  T3 was supposed to be  9    = 3^2

T4 = 16    =4^2

T5=25      =5^2

Tn = n^2

Hello sweet Melody!!!..

oops, won't say that again..., but I WANTED to....

The 1st sequence is Fibonacci yes, I googled on that but there does not seem to be a mathematical layterm solution for it, 

The 2nd sequence is a problem,

T1 = n,

T2 = n^2

T3 = (n+1)^2

T4 = (n+1)^2

T5 = n^2

So, I cannot determine a nth term rule....say, for the 10th term?

The 3rd sequence I've got sorted....thank you sooo much for assisting...

Find the polynomial \(p(x)\), with real coefficients,
such that \(p(2) = 5\) and \(p(x) p(y) = p(x) + p(y) + p(xy) - 2\) for all real numbers \(x\) and \(y\).

\(\begin{array}{|lrcll|} \hline & \mathbf{ p(x) p(y)} &=& \mathbf{ p(x) + p(y) + p(xy) - 2 } \\\\ x=1,\ y=2: & p(1) p(2) &=& p(1) + p(2) + p(1*2) - 2 \quad | \quad \mathbf{p(2) = 5} \\ & p(1) *5 &=& p(1) + 5 + 5 - 2 \\ & 5p(1) &=& p(1) + 8 \\ & 4p(1) &=& 8 \\ & p(1) &=& \dfrac{8}{4} \\ & \mathbf{p(1)} &=& \mathbf{2} \\\\ x=0,\ y=2: & p(0) p(2) &=& p(0) + p(2) + p(0*2) - 2 \quad | \quad \mathbf{p(2) = 5} \\ & p(0) *5 &=& p(0) + 5 + p(0) - 2 \\ & 3p(0) &=& 3 \quad | \quad :3 \\ & \mathbf{p(0)} &=& \mathbf{1} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline & p(0) &=& 1 \\ & p(1) &=& 2 \\ & p(2) &=& 5 \\ \hline & p(x) &=& ax^2+bx+c \\ p(0)=1: & 1 &=& a*0^2+b*0 +c \\ & \mathbf{c} &=& \mathbf{1} \\\\ p(1)=2: & 2 &=& a*1^2+b*1 +c \quad | \quad c=1 \\ & 2 &=& a+b+1 \\ & \mathbf{a+b} &=& \mathbf{1} \\\\ & \mathbf{ b} &=& \mathbf{1-a} \\\\ p(2)=5: & 5 &=& a*2^2+b*2 +c \\ & 5 &=& 4a+2b +c \quad | \quad c=1 \\ & 5 &=& 4a+2b + 1 \\ & 4 &=& 4a+2b \quad | \quad :2 \\ & 2a+b &=& 2 \quad | \quad b=1-a \\ & 2a+1-a &=& 2 \\ & \mathbf{ a} &=& \mathbf{1 } \\\\ & \mathbf{ b} &=& \mathbf{1-a} \quad | \quad a = 1 \\ & b &=& 1-1 \\ & \mathbf{b} &=& \mathbf{0} \\ \hline & p(x) &=& ax^2+bx+c \quad | \quad a=1,\ b=0,\ c=1 \\ & \mathbf{p(x)} &=& \mathbf{x^2+1} \\ \hline \end{array}\)

check:

\(\begin{array}{|rcll|} \hline \mathbf{ p(x) p(y)} &=& \mathbf{ p(x) + p(y) + p(xy) - 2 } \\\\ (x^2+1)(y^2+1) &=& (x^2+1) + (y^2+1) + \left((xy)^2+1\right) - 2 \\ x^2y^2+x^2+y^2+1 &=& x^2+1 + y^2+1 + x^2y^2+1 - 2 \\ x^2y^2+x^2+y^2+1 &=& x^2y^2+ x^2+y^2 + 1 +1 +1 - 2 \\ x^2y^2+x^2+y^2+1 &=& x^2y^2+ x^2+y^2 + 3 - 2 \\ x^2y^2+x^2+y^2+1 &=& x^2y^2+ x^2+y^2 + 1\ \checkmark \\ \hline \end{array}\)

See  https://en.wikipedia.org/wiki/Fibonacci_number for a closed form expression for the Fibonacci sequence,

and the 3rd one

T1 = 1

T2 = 8

T3 = 27

T4 = 64

T5 = 125

These are just cubic numbers.  Can you do it yourself now?

The second sequence is just as easy.  

I just takes practice, you will get the hang of these.  

Hi Juriemagic,

It is good to see you again.  

T1 = 1

T2 = 1

T3 = 2

T4 = 3

T5 = 5

This is the fibonacci sequence.

I actually do not know how to express the nth term of this sequence in a nice neat way

Obviously   

\(T_1=1\\ T_2=1\\ \text{Where n >2 and n is an integer}\\ T_n=T_{n-1}-T_{n-2}\)

I suspect you need somthing easier to use than that.  Am I right or is that all you want?

Come on guest, that is a bit harsh BUT nyx, you must not plagarize. If you want to copy something you must site your source.

And honestly,  you cannot tell me that you read all that or that you had any good reason to think it could be helpful.

Hi CalculatorUser,

I am really pleased that this forum has enabled you to become hooked on maths. 

If a place like this had existed when I was in High School I would have become totally addicted to it.

Not much has changed for me I guess. 

Still,  I can see that for you it could sometimes be a distraction that you should control.

I am sure you will drop in quite regularly and that will be good all ways around   

I suppose we should be thankful that it's just a flower bed, and not a pond.

Hi guest,

i do appreciate your response, however my problem really is how to calculate for example the 10th term for each of these sequences. I struggle to put together a formula..

The 3rd sequence I gave, i finnaly got...it's just Tn = n^3

1) 

The first one is the Fibonacci sequence; Each new term is the SUM of previous two terms. So, next few terms are:

3 + 5 =8, 5 + 8 =13, 8 + 13 =21.......and so on.

2)

This one consists of the SQUARES of counting numbers:

1^2, 2^2, 3^2, 4^2, 5^2..........1, 4, 9, 16, 25........etc.

3)This one consists of the CUBES of counting numbers:

1^3, 2^3, 3^3, 4^3, 5^3..........1, 8, 27, 64, 125.......etc.

These are Alcumus problems, so yeah, this is totally homework.

By symmetry, you get the maximum when x = y = 1/2, so the maximum value is x - y = 0.