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Thank you Chris

Thank you guys 

Thank you Chris and Tetre

The difference is given by p(x² - (x-2)²) + q(x - (x-2)) or (4x-4)p + 2q

For this to equal (2x - 1)2 we must have 4p = 4 or p= 1 and 2q - 4p = -2, so 2q - 4 = -2 or q = 1

p+q = 2

That is pretty neat Rom.   

What do you mean by  ' the opposite of the number ' ?

Why f(1) = 1? 

\(f(\sqrt{2-1}) = f(\sqrt{1}) = f(1) \)

so \(f(1)= 1^10 + 2(1)^9 - 2(1)^8 - 2(1)^7 + 1^6 + 3(1)^2 + 6(1) + 1 = 1 + 2 - 2 - 2 + 1 + 3 + 6 + 1 = 3 + 6 + 1 = 10 \)

so \(f(1) = 10 \) not 1,where im wrong?

Thanks! 

I guess f(2)= f(1) + f(1) + 2(1)(1) = 4 + 4 + 2 =10 ,f(4) = f(2) + f(2) + 2(2)(2) = 10 +10 + 8 = 28 

f(8) = f(4) + f(4) +2(4)(4) = 28 + 28 + 32 = 88 

So i think f(8) = 88 

Hope it helps!

 f(f(f(x)))=a+b( a+b( a+bx))

 f(f(f(0)))=a+b(a+b(a))=27

 f(f(f(1)))=a+b(a+b(a+b))=29 

f(f(f(0)))=a+ba+b^2a=27

f(f(f(1)))=a+ba+ab^2+b^3=29

(1) a+ba+b^2a=27 -> f(f(f(1)))=(1)+b^3=29 so 27 + b^3 = 29 <=> b^3 = 29-27 <=> b^3 = 2 <=> b = \(\sqrt[3]{2}\)\(\)

after in f(f(f(1))) you are replacing and find the a!
 

60 - [39% x 60] =60 - 23.4 =36.6 GB - free space that Ann has.

20 x [200/100 +1] = 60, or "d" on your list.

The sequence continues as follows:

2, 9, 20, 42, 82, 147, 244, 380, 562, 797, 1092, 1454, 1890, 2407, 3012, 3712, ...etc.

The "closed form" is: a_n = 1/6 (7 n^3 - 30 n^2 + 83 n - 48) (for all terms given.

thank you

listfor(n, 2, 80, (n*(n-1)*(n+1)) mod 8
=(6, 0, 4, 0, 2, 0, 0, 0, 6, 0, 4, 0, 2, 0, 0, 0, 6, 0, 4, 0, 2, 0, 0, 0, 6, 0, 4, 0, 2, 0, 0, 0, 6, 0, 4, 0, 2, 0, 0, 0, 6, 0, 4, 0, 2, 0, 0, 0, 6, 0, 4, 0, 2, 0, 0, 0, 6, 0, 4, 0, 2, 0, 0, 0, 6, 0, 4, 0, 2, 0, 0, 0, 6, 0, 4, 0, 2, 0, 0). 

Note: Count ALL the zeros, and that is your answer!!!.                                   

Don't you know how to show that that point is on the circle?

\(If\ f(x)=x^{10}+2x^9-2x^8-2x^7+x^6+3x^2+6x+1\ ,\) 

                                                                        \(then\ what\ is\ f(\sqrt{2-1}\ )\)

\(f(\sqrt{2-1})=f(\sqrt1\ )=1\)

  !

\(5^{23} \text{ is odd}\\ 7^{17} \text{ is odd} \\ 5^{23} + 7^{17} = \text{odd} + \text{odd} = \text{even} \\ \text{therefore 2 will the smallest prime divisor}\)

\(x^2 - 6x + 66 = \\ x^2 - 6x + 9 - 9 + 66 = \\ (x-3)^2 + 57 \\ b=-3,~c=57\)

it would have to be negative.

Ann's computer has a total capacity of 60 GB. At present she is using 39% of the capacity.

How much free space does she have?

Okay. Lets get you a solution here.

Kevin lives \(2\) kilometers away from the school.

Converting this to meters, we get \(2000\) meters.

Adrian lives \(750 \) meters away from the same exact school.

As Chiron stated, this is simply a subtraction problem. Subtracting \(750\) meters from \(2000\) meters, we get \(1250\) meters

If the answer should be in kilometers, it is \(\boxed {1.25}\) kilometers.

If the answer should be in meters, it is \(\boxed {1250}\) meters.

Hope that helps! 

Okkaaay. Let us try to solve this one.

First off the problem mentions that the license plate cannot have an "S" in it. Therfore, I am going to take the S away from the list to make our life easier.

A, E, G, I, K, O, P, R, T, U, V...Let's consider this our new list. (We have 11 letters now)

Our license plate is in the form of : \(\boxed{}\boxed{}\boxed{}\boxed{}\boxed{}\).

-For our first "box", we can only have two choices, since the license plate begins with a G or a K.

-For our second box, we can choose out of \(9\) letters, since one has already been used above, the letters cannot repeat, and we must reserve "t" to end the plate, as stated in the problem.

-For our third box, we can choose \(8\) letters, since three have already been used out of the \(11\) numbers in the new list.

-For our fourth box, following the same pattern as the third, we have \(7\) choices.

-For our last box, we only have one choice, "t".

So in total, we have \(2 \times 9 \times 8 \times 7 \times 1\) choices possible.

Calculating that we get 1008 possible license plate combinations, so our answer should be \(\boxed {1008}\).

Let me know if anything seems off and I will look into it. 

So, this is a really fun pattern problem. Now, since you already know f(1), it will be advantageous to write f(1+y) as f(n), and say that for f(8), y=7. Then start checking to see if there's a pattern which you can gneralize in terms of n, while n=y+1. Remember that f(x) is a known variable, and that y equals n-1. A hint: remember the factorials.

you needed to specify that it was a binomial question

put

nCr(223,221)    into the web2 calc and press =