catmg

We're trying to optimize the quadratic, to do this we must complete the square. 

The equation should turn out in the form of 

h = -a(x + b)^2  + c

Since -a(x + b)^2 must be non positive, it can at most be 0, when x = -b. 

So the height maximized is c. 

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Ax + By = C

-3A - 5B = C

8A + B = C

8A + B = -3A - 5B

11A + 6B = 0

Can you take it from here?

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8x + 1 = 5 + 3x (mod 12)

5x = 4 (mod 12)

5(5x) = 4(5) mod (12)

x = 8 mod (12)

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Hello :))

I got the same inverse as you. 

I think the way to prove that it's an inverse it to show that g(x) is a one to one function on the domain (0, infinity), basically saying that no two numbers from that domain when plugged into x - 1/x will have the same output. 

Perhaps you can prove it by showing that g(x) is a strictly increasing function on that domain because if a function is strictly increasing then no two different numbers can have the same output?

I'm not too confident about that. 

a is a number that is greater than 0. 

f(g + a) > f(g)

g + a - 1/(g + a) > g - 1/g

g + a > g

a > 0 (true)

1/g > 1/(g + a) Since both g and a are positive numbers, g + a must also be positive. 

g + a > g

a > 0 (true)

Therefore, 

g + a - 1/(g + a) > g - 1/g

Since the function is strictly increasing, no two different inputs can have the same outputs since one of the numbers must be greater than the other, meaning their output must also be greater than the other. 

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P(x) = x^2 for when x = {1, 2, 3, 4}. 

F(x) = P(x) - x^2

F(5) = 73 - 25

F(5) = 48

Since P(x) is a monic polynomial with a degree of 5, so is F(x).

F(x) = 0 for when x = {1, 2, 3, 4, 5}

F(x) = (x-a)(x-1)(x-2)(x-3)(x-4)

F(5) = 24(5 - a)

a = 3

F(x) = (x-3)(x-1)(x-2)(x-3)(x-4)

(x-3)(x-1)(x-2)(x-3)(x-4) = P(x) - x^2

P(x) = (x-3)(x-1)(x-2)(x-3)(x-4) + x^2

P(6) = 396

Did I get it right?

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Merry Christmas to everyone. :DDD

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Oh no, it seems like we got different answers. 

Do you mind explaining what you did?

Thanks

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Hi :))

I'm going to group the numbers to make them easy to calulate. 

i = 10, 20, 30, ..., 90. 1 + 2 + 3 + ... + 9

1 <= i <= 9. 1 + 2 + 3 + ... + 9 

11 <= i <= 19. 1 + 2 + 3 + ... + 9 

21 <= i <= 29. 2(1 + 2 + 3 + ... + 9)

31 <= i <= 39. 3(1 + 2 + 3 + ... + 9)

...

91 <= i <= 99. 9(1 + 2 + 3 + ... + 9)

Overall. there are 1 + 1 + 1 + 2 + 3 + ... + 9 = 2 + 10*9/2 = 47 groups of (1 + 2 + 3 + ... + 9)

47*(1 + 2 + 3 + ... + 9) = 47*45 = 2115.

I hope this helped. :))

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f^1(1) = (2 - 1)^2 = 1

f^2(2) = f^1(f^1(2)) = f^1(1) = 0

f^3(2) = f^1(f^2(2)) = f^1(0) = 1

f^4(2) = f^1(f^3(2)) = f^1(1) = 0

There's a pattern. :))

It seems like when f(a) = 0, f(a+1) = 1 and when f(a) = 1, f(a+1) = 0. 

So when n is odd, then the answer is 1.

If n is even, the answer is 0.

f^7(2) = 1

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Hi guest. 

I really like your solution, but I don't think AOD = 180 - BOC. 

Instead, I think we need to start with the fact that AOC = 90 and COA = 90 since lines are perpendicular. 

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