MaxWong

In the 1st statement 'The more cheese, the more holes.' kept the number of holes on each cheese constant.

In the 2nd statement 'the more holes, the less cheese' kept the number of cheese constant.

So you cannot use the transitive congruency postulate on these 2 statement. If you use the transitive congruency postulate, the result would be indicated absurdity by the 'reductio ad absurdity' in Philosophy. There are many examples like this in Science.

Two equations to calculate heat energy generated by a current:

\(H=\frac{V^2}{R}\)

\(H=I^2R\)

Where H is the heat energy, V is voltage, I is the current, R is the resistance.

In the 1st equation, When the resistance is smaller, the heat energy will be more.

But in the 2nd equation, When the resistance is smaller, the heat energy will be less.

This is because in the 1st equation, the current is kept constant and in the 2nd equation the voltage is kept constant.

Hey! You made this up XD but I will use my calculator for you.

8908*23567865325 = 209942544315100

Let n be the number of sides of each polygon.

For angles...

Size of each angle = \(\frac{(n-2)(180)}{n}\)(in degrees)

For perimeters and areas(i love area questions :D)

Let n be the number of sides of each regular polygon and x be the length of each side

Perimeter = nx

Also I invented some of the area formulas to get the area of regular polygons(Length of each side = x)

(all trig functions in degrees)

Regular pentagon: \((x^2sin72)(1+cos36)\)

Remarks: The original regular pentagon area formula needs to know the apothem but my formula need not. 

Regular hexagon: \(3\sqrt{3}x^2\times\frac{1}{2}\)

Regular octagon: \((2x^2)(1+\sqrt2)\)

I found this all in one day. I used trigonometric functions when finding the regular pentagon and hexagon area formula and pythagoras theorem when finding the regular octagon area formula. Hope this can help you :D

−6+13x−7−16x−6+13x−7−16x

= 13x-16x+13x-16x-6-7-6-7

= -6x-26

1/5 = 2/10 = 0.2

2/5 = 4/10 = 0.4

3/5 = 6/10 = 0.6

4/5 = 8/10 = 0.8

There are 2 kinds of integration, which is definite integral and indefinite integral.

Example of definite integral:\(\int_{0}^{\pi}sinxdx\)

Example of indefinite integral:\(\int\sqrt{1+x}dx\)

\(6^{\frac{3}{2}}= \sqrt{6^3}=\sqrt{216}=6\sqrt6 \)

I seldom use approximations for square roots.

Because you can make it a number times a root, like what I have done above

\(3=2(x-2)+2-x\)

\(2x-4+2-x=3\)

\(x-2=3\)

\(x=5\)

I thought I was the first one to discover this fact XD

i⁴ⁿ⁺¹=i

i⁴ⁿ⁺²=-1

i⁴ⁿ⁺³=-i

i⁴ⁿ=1

3 is 4(0)+3

Therefore i³=-i