MaxWong

3x^2 + 5x 

= 3(x^2 + 5x/3)

= 3(x^2 + 5x/3 + ((5/3)/2)^2 - ((5/3)/2)^2)

= 3(x^2 + 5x/3 + 25/36 - 25/36)

= 3(x+5/6)^2 - 25/12.

Yup correct.

Use PEDMAS(or BODMAS)

1 + 5 * 10

= 1 + 50 <--- evaluate multiplication first

= 51 <--- then evaluate addition.

Use the first principles.

$\dfrac{d}{dx}x^n\\ =\displaystyle\lim_{h\rightarrow 0}\dfrac{(x+h)^n-x^n}{h}\\ =\displaystyle\lim_{h\rightarrow 0}\dfrac{x^n+\binom{n}{1}x^{n-1}h+...-x^n}{h}\\\text{other terms are negligible as they evaluate to 0 after taking limit.}\\ =\displaystyle \lim_{h\rightarrow 0}\dfrac{nhx^{n-1}}{h} + \text{lots of negligible terms which evaluates to 0}\\ =\displaystyle \lim_{h\rightarrow 0}nx^{n-1}\\ =nx^{n-1}\text{ YAY}$

We know that f(3) = 5.

so f(f(3)) = f(5) = 8.

f(f^(-1)(4)) is just 4.

f^(-1)(f^(-1)(5)):

f^(-1)(5) = 3

f^(-1)(3) = 2

so f^(-1)(f^(-1)(5)) = 2.

Add them up: 8 + 4 + 2 = 14. :D

most likely, it is arranged in this way...

A --------- B

|               |

|               |

D-----------C

Just think that it is a trapezoid... lol

because AB||CD, angle B + angle C = 180 deg.

because angle C = 3 angle B, 4 angle B = 180 deg, which means angle B = 45 deg.

Guest, 

We can still get a 'proper' answer with a fraction.

Example:

$\dfrac{41}{7}\pmod{13}\\ \equiv\dfrac{2}{7}\pmod{13}\text{ because 41 mod 13 = 2}\\ \equiv\dfrac{4}{14}\pmod{13}\\ \equiv \dfrac{4}{1}\pmod {13}\text{ similarly, because 14 mod 13 = 1}\\ \equiv 4 \pmod {13}\\ \equiv 4$

Or

$\dfrac{a}{b}\pmod c\\ \text{We first find numbers m and n that } bm = cn+1\\ =\dfrac{am}{bm}\pmod c\\ =\dfrac{am}{cn + 1}\pmod c\\ =am\pmod c\text{ (because (cn + 1) mod c = 1)}\\ \text{So we find the answer.}$

When I was in grade 5 I calculated that just using pencil and brain XD

and wrote that number down on my pencil box XDDDDDD

so that's 2^40 = 1,099,511,627,776

$1^{-1}\cdot 2^{-1} + 2^{-1} \cdot 3^{-1} +... + (p-2)^{-1}\cdot (p-1)^{-1}\\ =\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{(p-1)(p-2)}\\ =(1-\dfrac{1}{2})+(\dfrac{1}{2}-\dfrac{1}{3})+...+(\dfrac{1}{p-1}-\dfrac{1}{p-2})\\ =1-\dfrac{1}{p-2}\\ =\dfrac{p-3}{p-2}$

So our work is to evaluate

$\dfrac{p-3}{p-2}\pmod p$.

Don't know how to do that, if I were you, I would try some primes.

So let's try it.

When p = 7,

$\dfrac{p-3}{p-2}\pmod p\\ =\dfrac{4}{5}\mod 7\\ =\dfrac{12}{15}\mod 7\\ =5$

When p = 11,

$\dfrac{p-3}{p-2}\pmod p\\ =\dfrac{8}{9}\pmod {11}\\ =\dfrac{40}{45}\pmod {11}\\ =7$

When p = 13,

$\dfrac{p-3}{p-2}\pmod p\\ =\dfrac{10}{11}\pmod {13}\\ =\dfrac{60}{66}\pmod {13}\\ =8$

When p = 17,

$\dfrac{14}{15}\pmod{17}\\ =\dfrac{112}{120}\pmod {17}\\ =10$

Welp can't see a pattern :(

Oh...

I did not realize the ordered pairs are for this... :P