MaxWong

So fast...

We have 2 months left to have summer vacation... Sigh.

"i" is the imaginary unit. :)

It has a definition of $i = \sqrt{-1}$

Because sqrt(-1) cannot be evaluated(What squared = -1??), so mathematicians imagined a number that satisfies x squared = -1. Therefore "i" is called the imaginary number. :)

$(4x +\dfrac{1}{3}y)^2$This?

$(4x + \dfrac{1}{3y})^2$This?

$4x + \dfrac{1}{3y^2}$Or this?

$4x + \dfrac{1}{(3y)^2}$Or this?

$4x + (\dfrac{1}{3y})^2$This?

$4x + \dfrac{1}{3}y^2$Or even this?

Welp. Confused. Parentheses are important.

Nvm. I will do all of them

When x = 2, y = 3,

$(4x +\dfrac{1}{3}y)^2\\ =(4\times 2+\dfrac{1}{3}\times 3)^2\\ =(8 + 1)^2\\ =9^2\\ =81$

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​$(4x + \dfrac{1}{3y})^2\\ =(4\times 2 + \dfrac{1}{3\times 3})^2\\ =(8 + \dfrac{1}{9})^2\\ =(\dfrac{73}{9})^2\\ =\dfrac{73^2}{9^2}\\ =\dfrac{5329}{81}$​

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$4x + \dfrac{1}{3y^2}\\ =4\times 2 + \dfrac{1}{3\times 3^2}\\ =8 + \dfrac{1}{27}\\ =\dfrac{217}{27}$

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$4x + \dfrac{1}{(3y)^2}\\ =4\times 2 + \dfrac{1}{(3\times 3)^2}\\ = 8 + \dfrac{1}{81}\\ =\dfrac{649}{81}$

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$4x + (\dfrac{1}{3y})^2\\ =4\times 2 + (\dfrac{1}{3\times 3})^2\\ =8 + \dfrac{1}{81}\\ =\dfrac{649}{81}$

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$4x + \dfrac{1}{3}y^2\\ =4\times 2 + \dfrac{1}{3}\times 3^2\\ =8 + 3\\ =11$

5[3 - (7 - 8)]

Always do stuff in the smallest bracket(parenthese) first.

= 5[3 - (-1)] 

= 5(3 + 1) <--- Minus minus equals plus

= 5(4) <--- do stuff in the bracket

= 20 :D

[32 + (7 • 6 – 22)] – [30 ÷ (4 + 1)]

Do stuff in the bracket first :)

= [32 + (42 – 22)] – (30 ÷ 5)

= (32 + 20) – 6

= 52 – 6

= 46 :D

In the given equation, the total number of "a" that exists is n.

Therefore the equation can be simplified to: $an + (1+2+...+(n-1))$

1 + 2 + 3 + ... + n = $\dfrac{n(n+1)}{2}$

Therefore the whole thing after the plus sign simplifies to: $\dfrac{(n-1)(1+(n-1))}{2}=\dfrac{n(n-1)}{2}$

Original equation becomes $an+\dfrac{n(n-1)}{2}=100$

And then further simplify: $\dfrac{1}{2}n^2+an-\dfrac{1}{2}n-100 = 0$<--- I am trying to make it a quadratic here.

Take out the common factor n: $\dfrac{1}{2}n^2+n(a-\dfrac{1}{2})-100 = 0$

To make it easier to solve, multiply each term by 2.

$n^2 + n(2a - 1)-200=0$

It now becomes a quadratic equation with a is a constant and solving for n. Then we can find n in terms of a.

$n=\dfrac{-(2a-1)\pm\sqrt{(2a-1)^2-(4)(1)(-200)}}{2}=\dfrac{1-2a\pm\sqrt{4a^2-4a+801}}{2}$

Because n is an integer, $\sqrt{4a^2-4a+801}$ must be an integer too, to make n an integer.

That implies that $4a^2 - 4a + 801$ is a square number.

Also, $(4a^2 - 4a + 801)\pmod4\equiv 1$.

We now think of a square number which mod 4 = 1.

The nearest square number to 801 which mod 4 = 1 is 841, therefore we assume that the quadratic there is = 841 first.

$4a^2 - 4a + 801 = 841\\ a^2 - a = 10\\ a(a-1)= 10$

No positive integer "a" satisfies the equation.

The second nearest square number to 801 which mod 4 = 1 is 729,

$4a^2 - 4a + 801 = 729\\ a^2 - a = -72\\ a(a-1)= -72$

Still no positive integer "a" satisfies the equation.

We should think of a bigger number.

31^2 = 961 and mod 4 = 1.

$4a^2 - 4a + 801 = 961\\ a^2 - a = 40\\ a(a-1)= 40$

Still no positive integer "a" satisfies the equation :(

Try 1089

$4a^2 - 4a + 801 = 1089\\ a^2 - a = 72\\ a(a-1)= 72\\ a = 9$

Yay! :D

We back-substitute a = 9 into the quadratic formula to find the corresponding n.

$n=\dfrac{1-2(9)\pm33}{2}\\ n = 8\text{ or }n=-25(\text{rejected})\\ n = 8$

We get: a = 9, n = 8 (Finally after a lot of work XD)

  2*(3 + x) - 5

= 2*3 + 2*x - 5 <--- multiply the thing inside the bracket by 2

= 6 + 2x - 5 <--- 2*3, basic math.

= 2x + 6 - 5 <--- rearrange.

= 2x + 1 <--- 6 - 5, basic math again.

Srsly?

But his/her (uncertainties...) username is Melody, which sounds like a female's name.

Yeah. I am a smart cookie, really an edible cookie..... XD

Mixed numbers. Hmm. Long time since I see them.

For mixed numbers, $a\dfrac{b}{c}=\dfrac{a\times c+b}{c}$

Example: $2\dfrac{5}{7}=\dfrac{2\times 7 + 5}{7}=\dfrac{19}{7}$

Get it?

Numerator = (number in front) times (original denominator) + (original numerator)

Denominator just remains the same.