MaxWong

Half of D.

I don't choose any of these but I have an idea.

$1,000,000,000 - 999,999,999,999\\ =-(999,999,999,999-1,000,000,000)\\ =-998,999,999,999$

Step one is a way that I found in Primary 1, which is grade 1 in some countries, to do subtractions which answers are negative. So that's why I can already do negative addition, subtraction, and multiplication in Primary 1.

~The smartest cookie in the world

Can someone answer this please...... My curiousity is telling me to take this knowledge ASAP......

We cannot show true for this for n = 0.....

If we substitute n = 0 in the LHS the last term of LHS will become $\dfrac{1}{r}$.

So we show that this is true for n = 1......

Maybe but I love puns more.

~ The smartest cookie in puns LOL

LHS stands for Left Hand Side and RHS stands for RHS. Use whatever notation your teacher tell you to use.

When n = 1,

LHS = 1

RHS = 1

LHS = RHS.

Assume that when n = k, LHS = RHS.

When n = k + 1,

$LHS\\ = 1 + r + r^2 +... + r^k \\ =\dfrac{1-r^k}{1-r} + r^k\\ =\dfrac{1-r^k}{1-r} + \dfrac{r^k-r^{k+1}}{1-r}\\ =\dfrac{1-r^{k+1}}{1-r}\\ =RHS$

$\therefore$1+r+r^2+r^3+...+r^n-1=1-r^n/1-r is true for all positive integer n.

~The smartest cookie in the world

Only 11 and a half minutes of sleep...... How did the one do it XDDDDD

1. $f(x) = 5\pi^2\\ f'(x) = 0\text{ as it is a constant}$

2. $f(x) = 5\sqrt x = 5x^{1/2}\\ f'(x) = \dfrac{5}{2}\cdot x^{-1/2}=\dfrac{5}{2\sqrt x}$

3.$f(x) = 1x^4 = x^4\\ f'(x) = 4 x^3$

4.$f(x) = 2x^7 - 5x - 7\\ f'(x) = 14x^6- 5$

5.$f(x) = 2x - 1x^3\\ f'(x) = 2 - 3x^2\\ f'(1) = 2- 3(1)^2 = -1$

6. $f(x) = 3x^4\ln x\\ f'(x) = 12x^3\ln x + 3x^3\\ f'(1) = 12(1)^3 \cdot \ln 1 + 3(1)^3 = 3$

7.$f(x) = 8\sqrt{x + 6x^{34}}\\ f'(x) = 8\left(\left(204x^{33}+1\right)\left(\dfrac{1}{2\sqrt{x+6x^{34}}}\right)\right)\\\quad\quad\;=4\left(\dfrac{204x^{33}+1}{\sqrt{x+6x^{34}}}\right)$

8.$y = 5x^2 - 3\sqrt x + 5x^4\\ \dfrac{\mathtt{dy}}{\mathtt{dx}}=10x - \dfrac{3}{2\sqrt x} + 20x^3$

9.$y = (6x^3 + 2)(5x - 3)\\ \dfrac{\mathtt{dy}}{\mathtt{dx}}=(18x^2)(5x - 3)+(6x^3 + 2)(5) = 120x^3 -54x^2+10$

10. $f(t) = (8t-3)(2t+5)t - 7\\ f'(t) = (8t-3)(2t+5)+2t(8t-3)+8t(2t+5)=48t^2 +68t - 15\\ f'(1) = 48(1)^2 + 68(1) - 15 = 101$

Finally some fun calculus questions. I was so bored in those algebra 1 in school. LOL

PS: I am 14!! :D

~The smartest cookie in the world 

Pi cannot be represented by a fraction but it can be approximated using a fraction.

Pi approx = $\dfrac{22}{7}$

Pi approx = $\dfrac{333}{106}$

Pi approx = $\dfrac{630}{263}$

~The smartest cookie in the world

I don't know what the exact problem is but I can give you a tips:

Use Pythagoras theorem and special angles table!!

Pythagoras theorem: The square of the longest side in a right angled triangle is equal to the sum of squares of the other 2 sides. 

OR

$a^2 + b^2 = c^2$

Special angles table:

What do you mean by that?

Something like differential equations?

Like : $\dfrac{dy}{dx} = \sin x\\ \int dy = \int \sin x\\ y = -\cos x + C$?

~The smartest cookie in the world