Jeffes02

Input:$\log_{\sqrt{5}} 125\sqrt{5}$

Intepretation:

$1.\log_{\sqrt{5}} 125\sqrt{5}$ (Scientific Calculator 2.0)

We want to know how many powers of $\sqrt5$ equal to $125\sqrt5$

Looking at the numbers, we can know that the power is an integer, therefore:

$1.\sqrt5^2=5$ $,\space2.\sqrt5^3=5\sqrt5$

$3.\sqrt5^4=25$$,\space4.\sqrt5^5=25\sqrt5$

$5.\sqrt5^6=125$$,\space6.\sqrt5^7=125\sqrt5$

Answer $=7$

Intepretation No.2:
$2.\sqrt5\cdot125\sqrt5$ (Wolfram Computational Engine)

Cancel the two $\sqrt5$ ($\sqrt5^2=5$)

$=5\cdot125$

$=625$

Answer $=625$

Assuming each resident have a total body mass of $M\space(kg)$, they all have a density of $D$ ($D\le 1g/cm^3$) , and the swimming pool being a square prism (i.e. A square with height) with side $L (meters)$.

The law of physics tell us that any object with volume displaces water in water, and the displaced water causes the container's water level to rise, in this case, a swimming pool.

The remaining non-overflowing volume of the pool will be $0.4L^2\space(meter\space cubed)$

Since $Volume=Mass/Density$

Each resident's volume is equal to $\frac{M}{D}$.

Remember that $D\le 1g/cm^3$, that means part of their body won't sink under the water when they are in the pool, therefore:

Each residents' displaced water volume: $\frac{M}{D}\cdot D=M\space(cm^3)$ (Because the density of water = $1g/cm^3$)

For a total of 3,000 residents, their total volume would be $3000M$

In order for the pool to not overflow, the water displaced must be smaller than the remaining volume of the pool:

$3000M\space(cm^3)\le 0.4L^2\space(m^3)$

$0.003M\space(m^3)\le0.4L^2\space(m^3)$

$L^2\ge 0.0075 M$

$L\ge \frac{1}{20}\sqrt{3} M$

So the side of the pool have to be longer than $\frac{1}{20}\sqrt{3}$ of each resident's mass in kilograms.

(Somehow I feel the units have some fatal mistakes, correct me if anything is wrong in my process :P)

(After downloading printscreen and photo-editing plugins later...)
(Wait I think I messed the pictures up...)

Given that $f(5)=5$ and it crosses $y=0$ at the following points:

$x=200, x=100,x=0,x=-100,x=-200$

Since the crossing points of this function is rational, I deduct that the function is factorizable.

Perform reverse-factorization:

The function need to be in this form:

$g(x)=(x+200)(x+100)(x+0)(x-100)(x-200)$

For it to have roots at $x=200, x=100, x=0, x=-100,x=-200$

There are a total of five zero-points at:

$1.x=-200 , y=0$

$2.x=-100 , y=0$

$3.x=0 , y=0\space(Origin)$

$4.x=100 , y=0$

$5.x=200, y=0$

There are a total of four critical points at:

Local Maxima:

$1.x=\sqrt{15000-1000\sqrt{145}} , y=200000000\left(\sqrt{5}+5\sqrt{29}\right)\sqrt{30-2\sqrt{145}}$

$2.x=-\sqrt{15000+1000\sqrt{145}} , y=-200000000\left(\sqrt{5}-5\sqrt{29}\right)\sqrt{30+2\sqrt{145}}$

Local Minima:

$3.x=\sqrt{15000+1000\sqrt{145}} , y=\left(\sqrt{5}-5\sqrt{29}\right)\sqrt{30+2\sqrt{145}}$

$4.x=-\sqrt{15000-1000\sqrt{145}} , y=-200000000\left(\sqrt{5}+5\sqrt{29}\right)\sqrt{30-2\sqrt{145}}$

Since $f(5)=5$, Just divide every y-value of maximas and minimas above by a factor of $g(5)/5=398750625$

$f(x)=\frac{1}{398750625}\left(x+200\right)\left(x+100\right)x\left(x-100\right)\left(x-200\right)$

Q.E.D.

(I bet you just randomly typed the numbers in, didn't you? (Because the $x$ and $y$ values are pretty ugly to be honest))

Input:log3_(log2_x)=2.

Intepretation:$\log _3\left(\log _2\left(x\right)\right)=2$

$\log _2\left(x\right)=3^2=9$

$x=2^9$

$=512$

Q.E.D.

Chiquita

(Source:https://en.wikipedia.org/wiki/Banana_industry)

The equation between a circle's radius and its circumference is as followed:

$Circumference=2πr$, with r being the radius.

Plug $6π$ into the equation:

$2πr=6π$

$2r=6$

$r=3$

We obtained the radius of your circle, now use the equation between a circle's radius to its area:

$Area=πr^2$

Plug $r=3$ into the equation

$Area=π*3^2$

$=9π$

Q.E.D.

You don't have the denominator (i.e. the number under the line in between them.)

Input:2 x 2 - ( x + 3) - 2 x (1 - x)

Intepretation:$2x^2-(x+3)-2x(1-x)$

$=2x^2-x-3-2x+2x^2$

$=4x^2-3x-3$

Q.E.D.

Input:xy+ 3x2y+ 1 + 2xy2+yx2+ 2 +yx

Intepretation:$xy+3x^2y+1+2xy^2+yx^2+2+yx$

Simplify:

$=3+2xy+4x^2y+2xy^2$

(Function non-further-factorizatable)