recursal/Fanatic-Fandom huggingface dataset:
Arithmetic Series Formula

## **Problem**

**Part 1:** Prove that the arithmetic series can be computed with the formula

∑

# k
1

n

t
1

(
k
−
1
)
# d

n
2

[
2

t
1

(
n
−
1
)
d
]

{\displaystyle \sum\_{k=1}^{n} t\_1 + (k-1)d = \frac{n}{2} [2t\_1 + (n-1)d] }

where

t
1

{\displaystyle t\_1}

 is the first term,

n

{\displaystyle n}

 is the number of terms, and

d

{\displaystyle d}

 is the common difference.

**Part 2:** Show that the arithmetic series formula is equivalent to

∑

# k
1

n

t
1

(
k
−
1
)
# d

n
2

[

t
1

t
n

]

{\displaystyle \sum\_{k=1}^{n} t\_1 + (k-1)d = \frac{n}{2} [t\_1 + t\_n] }

where

t
n

{\displaystyle t\_n}

 is the last term.

## **Solution**

**Part 1**

This is a simple proof that use a little symmetry. Let

S
n

=

t
1

(

t
1

d
)
(

t
1

2
d
)
⋯
(

t
1

(
n
−
1
)
d
)

{\displaystyle S\_n = t\_1 + (t\_1+d) + (t\_1+2d) + \cdots + (t\_1+ (n-1)d) }

and by reversing the order of the terms in the series

S
n

=
(

t
1

(
n
−
1
)
d
)
(

t
1

(
n
−
2
)
d
)
(

t
1

(
n
−
3
)
d
)
)
⋯

t
1

.

{\displaystyle S\_n = (t\_1+ (n-1)d) + (t\_1+(n-2)d) + (t\_1+ (n-3)d)) + \cdots + t\_1. }

By adding the two series, there are

n

{\displaystyle n}

 sums of

2

t
1

(
n
−
1
)
d

{\displaystyle 2t\_1+ (n-1)d }

.

Therefore,

2

S
n

=
n
[
2

t
1

(
n
−
1
)
d
]

{\displaystyle 2 S\_n = n[2t\_1+ (n-1)d] }

S
n

=

n
2

[
2

t
1

(
n
−
1
)
d
]

{\displaystyle S\_n = \frac{n}{2} [2t\_1 + (n-1)d] }

**Part 2**

The final term of an arithmetic series can be expressed as

t
n

=

t
1

(
n
−
1
)
d

{\displaystyle t\_n = t\_1 + (n-1)d }

. Consequently,

S
n

=

n
2

[

t
1

t
1

(
n
−
1
)
d
]

S
n

=

n
2

[

t
1

t
n

]
.

{\displaystyle \begin{align}
S\_n &= \frac{n}{2} [t\_1 + t\_1 + (n-1)d] \\[5 pt]
S\_n &= \frac{n}{2} [t\_1 + t\_n].
\end{align} }
Statistics Calculator with Python

**By: Tao Steven Zheng (郑涛)**

## **Task**

Write a function that determines the following statistical quantities:

(1) Mean

(2) Standard deviation

(3) Minimum

(4) Median

(5) Maximum

(6) Data range

(7) Sample size

Lastly, test your function on the the set of numbers: samples = [1,2,3,4,5,6]

## **Code**

```
def statistics(samples):

    N = len(samples)
    mean = sum(samples) / N
    sum\_squares = sum((mean-x)\*\*2 for x in samples)
    stan\_dev = (sum\_squares/(N-1))\*\*0.5

    if N % 2 == 1:
        median =  sorted(samples)[N//2]  #odd case for median
    if N % 2 == 0:
        median = sum(sorted(samples)[N//2-1:N//2 + 1]) / 2 #even case for median

    minimum = min(samples)
    maximum = max(samples)
    data\_range = maximum - minimum

    print("Mean:", mean)
    print("Standard deviation:", stan\_dev)
    print("Minimum:", minimum)
    print("Median:", median)
    print("Maximum:", maximum)
    print("Data range:", data\_range)
    print("Sample Size:", N)

'''Test'''
samples = [1,2,3,4,5,6]

statistics(samples)

```

## **Results**

```
Mean: 3.5
Standard deviation: 1.8708286933869707
Minimum: 1
Median: 3.5
Maximum: 6
Data range: 5
Sample Size: 6

```
Quadrature of the Parabola

## **Problem**

In *Quadrature of a Parabola*, Archimedes calculated the area segment of a parabola using the **method of exhaustion**. He begins by inscribing the largest triangle (triangle ABC) within the parabolic segment. The method proceeds by filling the remaining space with more triangles whereby each stage doubles the number of triangles. Archimedes determined that the total area of the triangles constructed at each stage is 1/4 of the area of the triangles constructed in the previous stage.

If

A
T

{\displaystyle A\_T}

 is the area of the first triangle (pictured in yellow), and

A
P

{\displaystyle A\_P}

 is the area of the parabolic segment. Determine the ratio of the two areas

A
P

:

A
T

{\displaystyle A\_P : A\_T }

.

## **Solution**

Let

A
1

{\displaystyle A\_1}

 represent the total area of the blue triangles,

A
2

{\displaystyle A\_2}

 be the total area of the light red triangles ,

A
3

{\displaystyle A\_3}

 be the total area of the dark red triangles. Each stage is 1/4 the area of the previous. So,

A
1

=

1
4

A
T

{\displaystyle A\_1 = \frac{1}{4}A\_T}

A
2

=

1
4

A
1

=

1
16

A
T

{\displaystyle A\_2 = \frac{1}{4}A\_1 = \frac{1}{16}A\_T }

A
3

=

1
4

A
2

=

1
64

A
T

{\displaystyle A\_3 = \frac{1}{4}A\_2 = \frac{1}{64}A\_T }

This process continues indefinitely, generating the infinite geometric series:

A
P

=

A
T

1
4

A
T

1
16

A
T

1
64

A
T

.
.
.

{\displaystyle A\_P = A\_T + \frac{1}{4}A\_T + \frac{1}{16}A\_T + \frac{1}{64}A\_T + ... }

A
P

=

(

1

1
4

1
16

1
64

.
.
.

)

A
T

{\displaystyle A\_P = \left(1 + \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + ... \right) A\_T }

Here, the first term is

t
1

=
1

{\displaystyle t\_1 = 1 }

 and the common ratio is

# r
1

/

4

{\displaystyle r = 1/4 }

. Using the infinite geometric formula gives:

S

∞

=

t
1

1
−
r

{\displaystyle {S}\_{\infty} = \frac{t\_1}{1-r} }

S

∞

=

1

1
−
1

/

4

{\displaystyle {S}\_{\infty} = \frac{1}{1-1/4} }

S

∞

=

4
3

{\displaystyle {S}\_{\infty} = \frac{4}{3} }

Subsequently,

A
P

=

4
3

A
T

{\displaystyle A\_P = \frac{4}{3}A\_T}

Therefore, the ratio of the two areas is

A
P

:

A
T

=
4
:
3

{\displaystyle A\_P : A\_T = 4:3 }

**Historical Note:** Archimedes was not aware of the infinite geometric series formula. He used proof by contradiction to arrive at the result.
The Right Triangle and the Circle

This problem was inspired by a problem from the *Jiuzhang Suanshu: Gougu 16*. I developed this problem to derive the Pythagorean theorem.

## **Problem**

Consider a right triangle with three sides

a
≤
b
<
c

{\displaystyle a \le b < c }

.

**Part 1:** What is the diameter of a circle inscribed in a right triangle? Find two formulas that include the three sides of the right triangle.

**Part 2:** Then use these two formulas to prove

a
2

b
2

=

c
2

{\displaystyle a^2+b^2=c^2}

.

## **Solution**

**Part 1**

*Formula 1*

Let

a
,
b
,
c

{\displaystyle a,b,c}

 be the side lengths of the triangle, where

c

{\displaystyle c}

 is the hypotenuse. Let

r

{\displaystyle r}

 be the radius of the circle.

Figure out the fact that

(
a
−
r
)
(
b
−
r
# )
c

{\displaystyle (a-r)+(b-r) = c }

.

a
# b
c
2
r

{\displaystyle a+b = c+2r }

a
# b
c
d

{\displaystyle a+b=c+d }

a
b
−
# c
d

{\displaystyle a+b-c=d }

*Formula 2*

Figure out two formulas for the area of the triangle:

# A

1
2

a
b

{\displaystyle A= \frac{1}{2} ab }

# A

1
2

r
(
a
b
c
)

{\displaystyle A= \frac{1}{2} r(a+b+c) }

.
Set the two formulas equal

1
2

a
# b

1
2

r
(
a
b
c
)

{\displaystyle \frac{1}{2} ab = \frac{1}{2} r(a+b+c) }

.
Therefore,

# d

2
a
b

a
b
c

{\displaystyle d = \frac{2ab}{a+b+c} }

.
**Part 2**

Equate the two formulas for diameter.

a
b
−
# c

2
a
b

a
b
c

{\displaystyle a+b-c= \frac{2ab}{a+b+c} }

(
a
b
−
c
)
(
a
b
c
# )
2
a
b

{\displaystyle (a+b-c)(a+b+c)=2ab }

a
2

2
a
b

b
2

−

c
2

=
2
a
b

{\displaystyle a^2+2ab+b^2-c^2 = 2ab }

a
2

b
2

=

c
2

{\displaystyle a^2+b^2=c^2}

.
Diophantus-Fibonacci Identity

## **Problem**

The Diophantus–Fibonacci identity states that if two positive integers are each the sum of two squares, then their product is the sum of two squares. More specifically, consider four positive integers[1]

a
,
b
,
c
,
d

{\displaystyle a,b,c,d}

:

(

a
2

b
2

)

(

c
2

d
2

)

=

(

a
c
∓
b
d

)

2

(

a
d
±
b
c

)

2

{\displaystyle \left(a^2 + b^2 \right)\left(c^2 + d^2 \right) = \left(ac \mp bd \right)^2 + \left(ad \pm bc \right)^2 }

**Question:** Derive the above formula by the method of telescoping.

[1] The identity is true for real numbers

a
,
b
,
c
,
d

{\displaystyle a,b,c,d}

.

The identity first appeared in Diophantus’ *Arithmetica* in the 3rd century AD and later reappeared in Fibonacci’s *Liber Quadratorum* (Book of Squares) in 1225 AD. In the *Brahma-sphuta-siddhanta* (Correctly Established Doctrine of Brahma), written around 628 AD, Brahmagupta generalized it to

(

a
2

N

b
2

)

(

c
2

N

d
2

)

=

(

a
c
∓
N
b
d

)

2

N

(

a
d
±
b
c

)

2

{\displaystyle \left(a^2 + Nb^2 \right)\left(c^2 + Nd^2 \right) = \left(ac \mp Nbd \right)^2 + N\left(ad \pm bc \right)^2 }

where

N

{\displaystyle N}

 is also a positive integer.

## **Solution**

Expand the left hand side of the identity:

(

a
2

b
2

)

(

c
2

d
2

)

=

a
2

c
2

a
2

d
2

b
2

c
2

b
2

d
2

.

{\displaystyle \left(a^2 + b^2 \right)\left(c^2 + d^2 \right) = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2. }

By the method of telescoping, add and subtract

2
a
b
c
d

{\displaystyle 2abcd }

 on the right hand side:

(

a
2

b
2

)

(

c
2

d
2

)

=

a
2

c
2

a
2

d
2

b
2

c
2

b
2

d
2

−
2
a
b
c
d
2
a
b
c
d
.

{\displaystyle \left(a^2 + b^2 \right)\left(c^2 + d^2 \right) = a^2c^2 + a^2d^2 + b^2c^2 + b^2d^2 -2abcd + 2abcd. }

There are two options to introduce the terms

2
a
b
c
d

{\displaystyle 2abcd }

 and

−
2
a
b
c
d

{\displaystyle -2abcd }

 into the two groupings of squares:

(

a
2

b
2

)

(

c
2

d
2

)

=

(

a
2

c
2

−
2
a
b
c
d

b
2

d
2

)

(

a
2

d
2

2
a
b
c
d

b
2

c
2

)

,

{\displaystyle \left(a^2 + b^2 \right)\left(c^2 + d^2 \right) = \left(a^2c^2 - 2abcd + b^2d^2\right) + \left(a^2d^2 + 2abcd +b^2c^2\right), }

(

a
2

b
2

)

(

c
2

d
2

)

=

(

a
2

c
2

2
a
b
c
d

b
2

d
2

)

(

a
2

d
2

−
2
a
b
c
d

b
2

c
2

)

.

{\displaystyle \left(a^2 + b^2 \right)\left(c^2 + d^2 \right) = \left(a^2c^2 + 2abcd + b^2d^2\right) + \left(a^2d^2 - 2abcd +b^2c^2\right). }

Therefore,

(

a
2

b
2

)

(

c
2

d
2

)

=

(

a
2

c
2

∓
2
a
b
c
d

b
2

d
2

)

(

a
2

d
2

±
2
a
b
c
d

b
2

c
2

)

{\displaystyle \left(a^2 + b^2 \right)\left(c^2 + d^2 \right) = \left(a^2c^2 \mp 2abcd + b^2d^2\right) + \left(a^2d^2 \pm 2abcd +b^2c^2\right) }

and

(

a
2

b
2

)

(

c
2

d
2

)

=

(

a
c
∓
b
d

)

2

(

a
d
±
b
c

)

2

.

{\displaystyle \left(a^2 + b^2 \right)\left(c^2 + d^2 \right) = \left(ac \mp bd\right)^2 + \left(ad \pm bc\right)^2. }
Wheatstone Bridge

## **Problem**

The **Wheatstone bridge** is an electric circuit invented by Samuel Hunter Christie (1784 - 1865 AD) in 1833 and was improved by Charles Wheatstone (1802 - 1875 AD) in 1843.

Figure 1 illustrates the circuit design of the Wheatstone bridge. The Wheatstone bridge consists of three resistors of known resistance

R
1

,

R
2

,

R
3

{\displaystyle R\_1, R\_2, R\_3 }

 and a fixed, yet unknown, resistance to be measured

R
x

{\displaystyle R\_x}

. The resistance

R
2

{\displaystyle R\_2}

 is adjustable (i.e. a potentiometer or variable resistor) and is adjusted until the bridge is "balanced", that is, when no current flows through the galvanometer

V
G

{\displaystyle V\_G}

. The potential difference between points B and D will be zero. Show that when the bridge is "balanced", the unknown resistance can be calculated by the formula

R
x

=

R
2

R
3

R
1

.

{\displaystyle R\_x = \frac{R\_2 R\_3}{R\_1}. }

## **Solution**

Figure 2 illustrates the analysis of the Wheatstone bridge when it is "balance". Here, the voltage and the current between the points B and D are zero. Consequently,

I
1

=

I
2

{\displaystyle I\_1 = I\_2 }

 and

I
3

=

I
x

{\displaystyle I\_3 = I\_x }

.

Use **Kirchhoff's first law**  to determine the currents in junctions B and D:

I
3

−

I
x

I
G

=
0

I
1

−

I
2

−

I
G

=
0

{\displaystyle \begin{align}
 I\_3 - I\_x + I\_G &= 0 \\
 I\_1 - I\_2 - I\_G &= 0
\end{align}}

Use **Kirchhoff's second law** to determine the voltage in the loops ABDA and BCDB:

(

I
3

R
3

)
−
(

I
G

R
G

)
−
(

I
1

R
1

)

=
0

(

I
x

R
x

)
−
(

I
2

R
2

)
(

I
G

R
G

)

=
0

{\displaystyle \begin{align}
(I\_3 R\_3) - (I\_G R\_G) - (I\_1 R\_1) &= 0 \\
(I\_x R\_x) - (I\_2 R\_2) + (I\_G R\_G) &= 0
\end{align} }

When the bridge is balanced, the current at the Galvanometer is

I
G

=
0

{\displaystyle I\_G = 0 }

; thus, the set of equations from Kirchhoff's second law reduces to:

I
3

R
3

=

I
1

R
1

I
x

R
x

=

I
2

R
2

{\displaystyle \begin{align}
I\_3 R\_3 &= I\_1 R\_1 \\
I\_x R\_x &= I\_2 R\_2
\end{align} }

In this state, the voltages across each resistor follow the pattern

V
2

V
1

=

V
x

V
3

{\displaystyle \frac{V\_2}{V\_1} = \frac{V\_x}{V\_3} }

I
2

R
2

I
1

R
1

=

I
x

R
x

I
3

R
3

.

{\displaystyle \frac{I\_2R\_2}{I\_1R\_1} = \frac{I\_xR\_x}{I\_3R\_3}.}

Subsequently,

R
x

=

I
2

R
2

I
3

R
3

I
1

R
1

I
x

.

{\displaystyle R\_x = \frac{I\_2 R\_2 I\_3 R\_3}{I\_1 R\_1 I\_x}. }

Since

I
1

=

I
2

{\displaystyle I\_1 = I\_2 }

and

I
3

=

I
x

{\displaystyle I\_3 = I\_x }

, the above equation reduces to

R
x

=

R
2

R
3

R
1

.

{\displaystyle R\_x = \frac{R\_2 R\_3}{R\_1}. }
Union and Intersection with the Empty Set

## **Problem**

Consider a set

S

{\displaystyle S}

 and the empty set

∅

{\displaystyle \emptyset}

. Prove the following statements:

S
∪
# ∅
S

{\displaystyle S \cup \emptyset = S }

S
∩
# ∅
∅

{\displaystyle S \cap \emptyset = \emptyset }

## **Solution**

S
∪
# ∅
S

{\displaystyle S \cup \emptyset = S }

S
∪
# ∅
{
x
:
x
∈
S
∨
x
∈
∅
}

{\displaystyle S \cup \emptyset = \{x:x \in S \or x \in \emptyset \} }

Since there exists no elements in the empty set, all elements in both sets are just found in the set

S

{\displaystyle S}

. Therefore,

S
∪
# ∅
S

{\displaystyle S \cup \emptyset = S }

.

S
∩
# ∅
∅

{\displaystyle S \cap \emptyset = \emptyset }

S
∪
# ∅
{
x
:
x
∈
S
∧
x
∈
∅
}

{\displaystyle S \cup \emptyset = \{x:x \in S \and x \in \emptyset \} }

Since there exists no elements in the empty set, there are no elements common to both sets. The absence of elements in a set is by definition the empty set. Therefore,

S
∩
# ∅
∅

{\displaystyle S \cap \emptyset = \emptyset }

.
2001 to 2012 haven mascots retell

[ before seaside squad n zoo troop ] [ based on jungle book tarzan brother bear n rio ]

## animals appers

asian elephants

giant pandas

bactrain camels

bengal tigers

black rhinos

panthers

komodos

tapirs

musk oxen

african elephants

quing pandas

dromdary camels

sumatra tigers

javen rhinos

lions

giraffes

hippos

zebras

wild beest

warthogs

hyenas

gazelles

cars

buses

bikes

planes

diggers

vans

boats

safari parks

jungles

towns

farms

woodlands

beaches

###

## cast

bengal tigers = rory

asian elephants = anxouis

snakes = slaveter

orangutans = greedy

peacocks = manic

ants = merlin

ASIAN N AFRICAN JUNGLE ELEPHANTS

BLACK N JAVEN RHINOS

CHEETAHs

ORANGUTANS

BENGAL WHITE  N SUMATRAN TIGERS

SNAKEs

SCORPIONS

MOUNTAIN GORILLAS

OKAPIS

GIANT PANDAS

JUNGLE BOYS AND VILLAGE GIRLS ​​​

[ jungle of little orachards ]

bears = breadley

boys = ned [ aka james ] [ forms are boy n rabbit ]

girls = polly [ forms are squirrel n girl ]

[ both vist jungle n adevture valley on 5 weeks ]

lynxes

beavers

otters

bears

foxes

eagles

birds

rattle snakes  ​

​farmers and dogs

trexes

giraffe dinos

chimp dinos

hippo dinos

raptors

duck bill dinos

alaskean polar bears n polar bears

huskys

[ adevture valley of little orachards ]
TCM Geometry 3: Volumes of Solids - Part 1

**By: Tao Steven Zheng 鄭濤**

The computation of volumes was an important part of traditional Chinese mathematics, both in terms of theory and application. Chapter 5 of the *Jiuzhang Suanshu*, entitled *shanggong* (商功), is devoted to the calculation of volumes of various solids. The shapes are commonly found in civil engineering and earthworks. Examples include figuring out the size of dykes and granaries. Table 1 is a list of the solid figures found in the *Jiuzhang Suanshu*.

Due to practical needs, the volume calculation of various shapes has attracted the attention of mathematicians, and its theory has been further developed. "Nine Chapters of Arithmetic" Shang Gong Volume gives the square wall (rectangular prism), cuboid, square pavilion (rectangular pyramid), square pyramid (rectangular pyramid), Qiandu, Yangma, Minuan, Biaozui, Wuzhou , Chutong and other polyhedron volume formulas. It is worth noting that the last six types of solids are introduced as follows. A wedge is a wedge-shaped body obtained by cutting along the opposite edges of a rectangular parallelepiped.

### **Problem Study 1: Volume of a Pyramid**

答曰：九十三尺、少半尺。
術曰：廣袤相乘，以高乘之，三而一。

Answer:

93

1
3

{\displaystyle 93 \frac{1}{3}}

 [cubic] *chi*.
Method: Multiply the width by the length, then multiply by the height, divide by 3.
Derivative of Inverse Sine

## **Problem**

Show that

d

d
x

arcsin
⁡
(
x
# )

1

1
−

x

2

{\displaystyle {\frac {d}{dx}}\arcsin(x)={\frac {1}{\sqrt {1-x^{2}}}}}

.

# y
arcsin
⁡
(
x
)

{\displaystyle y=\arcsin(x)}

 is the inverse function of

# y
sin
⁡
(
x
)

{\displaystyle y=\sin(x)}

. Thus

# y
arcsin
⁡

x

{\displaystyle y=\arcsin {x}}

 is equivalent to

# x
sin
⁡

y

{\displaystyle x=\sin {y}}

.

By implicit differentiation,

d

d
x

# x

d

d
x

sin
⁡
(
y
)

{\displaystyle {\frac {d}{dx}}x={\frac {d}{dx}}\sin(y)}

# 1
cos
⁡
(
y
)

d
y

d
x

{\displaystyle 1=\cos(y){\frac {dy}{dx}}}

d
y

d
x

=

1

cos
⁡
(
y
)

{\displaystyle {\frac {dy}{dx}}={\frac {1}{\cos(y)}}}

.
Since

cos
⁡
(
y
# )

1
−

sin

2

⁡
(
y
)

{\displaystyle \cos(y)={\sqrt {1-\sin ^{2}(y)}}}

 and

# x
sin
⁡

y

{\displaystyle x=\sin {y}}

,

d
y

d
x

=

1

1
−

sin

2

⁡
(
y
)

{\displaystyle {\frac {dy}{dx}}={\frac {1}{\sqrt {1-\sin ^{2}(y)}}}}

d
y

d
x

=

1

1
−

x

2

{\displaystyle {\frac {dy}{dx}}={\frac {1}{\sqrt {1-x^{2}}}}}

.
Therefore,

d

d
x

arcsin
⁡
(
x
# )

1

1
−

x

2

{\displaystyle {\frac {d}{dx}}\arcsin(x)={\frac {1}{\sqrt {1-x^{2}}}}}

.
What is Proper?

**Created by Tao Steven Zheng (鄭濤)**

## **Problem**

Here is a combinatorics problem I created in 2015:

A number is called *proper* if each successive digit in the number is either monotonically increasing, or equal to the preceding digit. If 0 is not allowed in any place value, how many 6-digit numbers are *proper*?

**2-Digit Example:**

11, 12, 13, 14, 15, 16, 17, 18, 19, 22, 23, 24, 25, 26, 27, 28, 29, 33, 34, 35, 36, 37, 38, 39, 44, 45, 46, 47, 48, 49, 55, 56, 57, 58, 59, 66, 67, 68, 69, 77, 78, 79, 88, 89, 99

## **Solution**

Since we are looking for monotonically increasing 6-digit numbers, once we choose 6 digits out of 9 (1 to 9), there's just one way of arranging them. Suppose we choose the digits: 7,1,1,3,4,4.

Once we order them, there is just one possible increasing order: 113447

So the problem is to find the number of combinations we can make by choosing 6 digits (with repetition) out of 9. The best way of doing this is by imagining the amount of digits as 6 stars placed into 9 cells, which correspond to the digits we are choosing out of 9 (from 1 to 9). The spaces in between stars will be separated by two bars. For example, the combination 1,3,3,5,7,9 corresponds to:

 1

So in order to we calculate the total of six-digit proper numbers, we have to calculate the amount of combinations of 6 stars from the total number of **stars and bars**. There are 6 stars and 8 bars, so the we got 14 available places to choose from and put the 6 stars. Thus, the number of six-digit proper numbers is

(

14
6

)

=

{\displaystyle {\binom {14}{6}}=3003.}

## **Math Note**

A frequently occurring problem in combinatorics arises when counting the number of ways to group identical objects, such as placing indistinguishable balls into labelled urns. This type of combinatorics problem is known as "stars and bars" or "balls and urns", where the indistinguishable objects are represented by stars (or balls) and the separation into groups is represented by bars (or urns).

**Theorem:** The number of ways to place

n

{\displaystyle n}

 indistinguishable stars/balls into

k

{\displaystyle k}

 bars/urns is

(

n
k
−
1

n

)

.

{\displaystyle {\binom {n+k-1}{n}}.}
Tangent Triple Threat

# y
tan
⁡
x

{\displaystyle y=\tan x}

. Show that

y
‴

=
2
(
1

y
2

)
(
1
3

y
2

)

{\displaystyle y''' = 2(1+y^2)(1+3y^2)}

.

# y
tan
⁡
x

{\displaystyle y=\tan x}

y
′

=

sec
2

⁡
x

{\displaystyle y' = \sec^2 x }

y
″

=
2

sec
2

⁡
x
tan
⁡
x

{\displaystyle y'' = 2\sec^2 x \tan x }

y
‴

=
2

sec
4

⁡
x
4

sec
2

⁡

tan
2

⁡
x

{\displaystyle y''' = 2\sec^4 x + 4\sec^2 \tan^2 x }

.
Substitute

y

{\displaystyle y}

 and its third derivative into

y
‴

=
2
(
1

y
2

)
(
1
3

y
2

)

{\displaystyle y''' = 2(1+y^2)(1+3y^2)}

.

2

sec
4

⁡
x
4

sec
2

⁡

tan
2

⁡
# x
2
(
1

tan
2

⁡
x
)
(
1
3

tan
2

⁡
x
)

{\displaystyle 2\sec^4 x + 4\sec^2 \tan^2 x = 2(1 + \tan^2 x)(1 + 3\tan^2 x)}

2

sec
2

⁡
x
(

sec
2

⁡
x
2

tan
2

⁡
x
# )
2
(
1

tan
2

⁡
x
)
(
(
1

tan
2

⁡
x
)
2

tan
2

⁡
x
)

{\displaystyle 2\sec^2 x (\sec^2 x + 2 \tan^2 x) = 2(1 + \tan^2 x)((1 + \tan^2 x) + 2\tan^2 x)}

Since

1

tan
2

⁡
# x

sec
2

⁡
x

{\displaystyle 1+\tan^2x=\sec^2x}

2

sec
2

⁡
x
(

sec
2

⁡
x
2

tan
2

⁡
x
# )
2

sec
4

⁡
x
4

sec
2

⁡
x

tan
2

⁡
x

{\displaystyle 2\sec^2 x (\sec^2 x + 2 \tan^2 x) = 2\sec^4 x + 4\sec^2 x\tan^2 x}

.
Both sides match indeed!
The Solar Constant

## **Problem**

Compute the solar constant, which is calculated using the intensity formula

# I

P

4
π

r
2

{\displaystyle I = \frac{P}{4\pi r^2}}

.
Use the following information.

Surface temperature of the Sun:

5772
K

{\displaystyle 5772 K }

Earth-Sun distance:

1.5
×

10

11

m

{\displaystyle 1.5 \times {10}^{11} m }

Radius of the Earth:

6.4
×

10
6

m

{\displaystyle 6.4 \times 10^6 m }

Radius of the Sun:

7.0
×

10
8

m

{\displaystyle 7.0 \times 10^8 m }

Temperature of outer space:

2.7
K

{\displaystyle 2.7 K }

## **Solution**

Begin with the Stefan-Boltzmann law for radiation to calculate the thermal power of the Sun

d
Q

d
t

=
e
σ
A
(

T
4

−

T

e
n
v

4

)

{\displaystyle \frac{dQ}{dt} = e \sigma A(T^4-{T}^{4}\_{env} ) }

.
We can assume that the Sun is a blackbody, so the emissivity is

# e
1

{\displaystyle e = 1}

.

Note that the Stefan-Boltzmann constant is

# σ
5.67
×

10

−
8

W

m
2

K
4

{\displaystyle \sigma = 5.67 \times {10}^{-8} \frac{W}{m^2 K^4} }

.

d
Q

d
t

=
1
(
5.67
×

10

−
8

)
(
4
π
(
7.0
×

10
8

)
2

)
(

5772
4

−

2.7
4

)

{\displaystyle \frac{dQ}{dt} = 1(5.67 \times {10}^{-8})(4\pi(7.0 \times 10^8 )^2)(5772^4-2.7^4) }

d
Q

d
t

=
3.875
×

10

26

W

{\displaystyle \frac{dQ}{dt} = 3.875 \times {10}^{26} W }

This power is reduced by the inverse square law of distance

# I

3.875
×

10

26

W

4
π
(
1.5
×

10

11

m

)
2

=
1370

W

m
2

{\displaystyle I = \frac{3.875 \times {10}^{26} W}{4\pi(1.5 \times {10}^{11} m)^2} = 1370 \frac{W}{m^2} }

.
This is close to the actual value

1361

W

m
2

{\displaystyle 1361 \frac{W}{m^2} }

.
Relativistic Momentum-Energy Relation

## **Problem**

Begin with the relativistic momentum and energy:

# p

m
v

1
−

v
2

c
2

{\displaystyle p= \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}}

# E

m

c
2

1
−

v
2

c
2

{\displaystyle E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}} }

Derive the relativistic energy-momentum relation:

E
2

=
(
p
c

)
2

(
m

c
2

)
2

{\displaystyle E^2 = (pc)^2 + (mc^2)^2 }

.

## **Solution**

With a little algebra we discover that

p
E

=

v

c
2

{\displaystyle \frac{p}{E}=\frac{v}{c^2} }

.
Square the equation for relativistic energy

E
2

=

(
m

c
2

)
2

1
−

v
2

c
2

{\displaystyle E^2=\frac{(mc^2)^2}{1-\frac{v^2}{c^2}} }

And rearrange to arrive at

E
2

−

E
2

v
2

c
2

=
(
m

c
2

)
2

{\displaystyle E^2-\frac{E^2 v^2}{c^2} =(mc^2)^2 }

E
2

=

E
2

v
2

c
2

(
m

c
2

)
2

{\displaystyle E^2=\frac{E^2 v^2}{c^2} +(mc^2)^2 }

.
From the relation

p
E

=

v

c
2

{\displaystyle \frac{p}{E}=\frac{v}{c^2} }

we find

p
# c

E
v

c

{\displaystyle pc=\frac{Ev}{c} }

and

(
p
c

)
2

=

E
2

v
2

c
2

{\displaystyle (pc)^2=\frac{E^2 v^2}{c^2} }

.
Substitute this result into

E
2

=

E
2

v
2

c
2

(
m

c
2

)
2

{\displaystyle E^2=\frac{E^2 v^2}{c^2} +(mc^2)^2 }

to get

E
2

=
(
p
c

)
2

(
m

c
2

)
2

{\displaystyle E^2 = (pc)^2 + (mc^2)^2 }

.
Mixing Colours Problem

**Problem created by Tao Steven Zheng (郑涛)**

## **Problem**

Suppose we begin with three colours (Red, Yellow, Blue). Let this set be called Generation 1, or GEN-1. We mix each pair of colours to form GEN-2, which has six colours (Red, Yellow, Blue, Orange, Purple, Green). Suppose this process continues while obeying three rules:

(1) If two colours are selected, there is no difference in the order of mixing. (Red + Blue = Blue + Red)

(2) Each colour is its own identity. (Red + Red = Red)

(3) For all generations after GEN-1, no colour created in their respective generation can be mixed until the next iteration. For example, Orange, Purple, Green are created in GEN-2; thus, they cannot mix until forming GEN-3.

Determine the number of colours in the set GEN-6 (after mixing).

## **Solution**

Let

n
i

{\displaystyle n\_i}

 represent the number of colours in GEN-i. For GEN-1 (Red, Yellow, Blue), we have

n
i

=
3

{\displaystyle n\_i = 3 }

.

**GEN-2:** The total number of mixes is ​

n
1

2

=

3
2

=
9

{\displaystyle {n\_1}^{2} = 3^2 = 9 }

, but other than the three single-colour mixes that produce the​ GEN-1 colours, the remaining 6 produce only 3 distinctly new colours (Orange, Purple, Green). This is because two of these six mixes produce only one colour, for example, the colour Orange is made by mixing Red and Yellow in two ways (Red + Yellow = Yellow + Red). Therefore, the number of GEN-2 colours

n
2

=

n
1

1
2

(

n

1

2

−

n
1

)

{\displaystyle n\_2 = n\_1 + \frac{1}{2}({n}^{2}\_{1} - n\_1) }

.

**GEN-3:** There are ​6 GEN-2 colours, so the total number of ways the colours can be mixed is

n
2

2

=

6
2

=
36

{\displaystyle {n\_2}^{2} = 6^2 = 36 }

. Of the 36 colours, 6 of them are just the GEN-2 colours. Of the remaining 30 two-colour mixes, 18 of them do not produce distinctly new colours. This is because there are 6 GEN-1 colour mixes that produce the three new GEN-2 colours (Orange, Purple, Green), while these three colours have been counted in the single-colour mixes. This is counted by computing

n
2

−

n
1

{\displaystyle n\_2 - n\_1 }

. Therefore, the two-colour mixes only produce

30
2

−
# 3
12

{\displaystyle \frac{30}{2} - 3 = 12 }

 new GEN-3 colours. Therefore,

n
3

=

n
2

1
2

(

n

2

2

−

n
2

)
−
(

n
2

−

n
1

)

=

n
1

1
2

(

n

2

2

−

n
2

)

=
3

1
2

(

6
2

−
6
)

=
18

{\displaystyle \begin{align} n\_3 &= n\_2 + \frac{1}{2}({n}\_{2}^{2} - n\_2) - (n\_2 - n\_1) \\[5 pt]
&= n\_1 + \frac{1}{2}({n}\_{2}^{2} - n\_2) \\[5 pt]
&= 3 + \frac{1}{2}(6^2 - 6) \\[5 pt]
&= 18 \end{align} }

.
By induction, this recurrence relation persists, and we get:

n

i
1

=

n
1

1
2

(

n

i

2

−

n

i

)
.

{\displaystyle n\_{i+1} = n\_1 + \frac{1}{2} ({n^{2}\_{i}} - n\_{i}). }

Substituting the values into the recurrence relation gives

n
4

=
156
,

n
5

=
12093
,

n
6

=
73114281

{\displaystyle n\_4 = 156, n\_5 = 12093, n\_6 = 73114281 }

.
Nitrous Acid Titration

## **Problem**

A 100.0 mL sample of 0.150 M nitrous acid

HNO

2

{\displaystyle \ce{ HNO2 }}

 with

p

K
a

=
3.39

{\displaystyle pK\_a = 3.39 }

 is titrated with 0.300 M sodium hydroxide

NaOH

{\displaystyle \ce{NaOH}}

.

**Part 1:** Determine the pH of the acid solution when no sodium hydroxide is added. Give your answer to two decimal places.

**Part 2:** Determine the pOH and pH of the solution when 75.00 mL sodium hydroxide is added. Give your answers to two decimal places.

## **Solution**

**Part 1**

Write the chemical equation and the expression for

K
a

{\displaystyle K\_a}

:

HNO

2

↽

−

−

⇀

H

NO

2

−

{\displaystyle \ce{ HNO2 <=> {H}^{+} + {NO2}^{-} }}

K
a

=

[

H

]

[

N

O
2
−

]

[

H
N

O
2

]

{\displaystyle K\_a = \frac{\left[H^+ \right]\left[NO\_2^- \right]}{\left[HNO\_2 \right]} }

Since

p

K
a

=
3.39

{\displaystyle pK\_a = 3.39 }

, the equilibrium constant is

10

−
3.39

{\displaystyle 10^{-3.39} }

. The concentrations of

[

H

]

{\displaystyle [H^+] }

 and

[
N

O
2
−

]

{\displaystyle [NO\_2^- ] }

 unknown, subsequently

10

−
3.39

=

(
x
)
(
x
)

0.150
−
x

{\displaystyle {10}^{-3.39} = \frac{(x)(x)}{0.150-x} }

Solve the quadratic equation

x
2

(
4.07
×

10

−
5

)
x
−
(
6.11
×

10

−
5

)

{\displaystyle x^2 + (4.07 \times {10}^{-5}) x - (6.11 \times {10}^{-5}) }

; use the quadratic equation and reject the negative answer. Hence, the molarity of

[

H

]

{\displaystyle \left[H^+ \right] }

 is

# x
7.80
×

10

−
3

M

{\displaystyle x = 7.80 \times {10}^{-3} \, M }

. The pH is therefore

p
H

=
−
log
⁡

[

H

]

=
2.11

{\displaystyle \begin{align}
pH &= -\log{\left[H^+ \right]} \\
&= 2.11
\end{align} }

**Part 2**

When the base is added, we need to construct an **ICE table** (Initial, Change, Equilibrium). But first, write out the chemical equation and calculate the amount of moles of each substance.

HNO

2

NaOH
⇋

Na

NO

2

−

H

2

O

{\displaystyle \ce{ {HNO2} + NaOH \leftrightharpoons {Na}^{+} + {NO\_2}^{-} + H\_2O }}

The number of moles of nitrous acid is

100.0

m
L
×

0.150

m
o
l

1000

m
L

=
0.0150

m
o
l
.

{\displaystyle 100.0 \; mL \times \frac{0.150 \; mol}{1000 \; mL} = 0.0150 \; mol. }

The number of moles of sodium hydroxide is

75.00

m
L
×

0.300

m
o
l

1000

m
L

=
0.0225

m
o
l
.

{\displaystyle 75.00 \; mL \times \frac{0.300 \; mol}{1000 \; mL} = 0.0225 \; mol. }

Note that the total volume of the solution is 175.00 mL.

 **C** | -0.0150
 **E** | 0.0000

The concentration of hydroxide is

[

O
H

−

]

=

0.0075

m
o
l

0.175

L

=
4.3
×

10

−
2

M
.

{\displaystyle \left[{OH}^{-} \right] = \frac{0.0075 \; mol}{0.175 \; L} = 4.3 \times {10}^{-2} \; M. }

Therefore the pOH is

p
O
H

=
−
log
⁡

[

O

H
−

]

=
1.37

{\displaystyle
\begin{align}
pOH &= -\log{\left[OH^- \right]} \\
&= 1.37
\end{align} }

The pH can be easily calculated by subtracting the pOH from 14

p
H

=
14.00
−
p
O
H

=
14.00
−
1.37

=
12.63

{\displaystyle
\begin{align} pH &= 14.00 - pOH \\
&= 14.00 - 1.37 \\
&= 12.63
\end{align} }

Notice the pH number increased; this means the solution has become more basic.
Stopped Pipe and Open Pipe

## **Problem**

   This problem investigates stopped pipes and open pipes. Stop pipes are closed on one end and open on the other end. Open pipes are open on both ends. For the following problem, assume that the speed of sound through air is 340 m/s.

**Part 1:** A stopped organ pipe has a fundamental frequency of 220 Hz. How long is this pipe?

**Part 2:** The second overtone

(

f
5

)

{\displaystyle \left(f\_5 \right) }

 of the stopped organ pipe has the same wavelength of the third harmonic

(

f
3

)

{\displaystyle \left(f\_3 \right) }

 of an open pipe. How long is the open pipe?

## **Hints**

(1) The formula for calculating the frequencies of a stopped pipe is

f
n

=

n
v

4
L

=
n

f
1

{\displaystyle f\_n = \frac{nv}{4L} =n f\_1 }

(2) The formula for calculating the frequencies of an opened pipe is

f
n

=

n
v

2
L

=
n

f
1

{\displaystyle f\_n = \frac{nv}{2L} = n f\_1 }

## **Solution**

**Part 1**

The formula for the frequencies of a stopped pipe is

f
n

=

n
v

4
L

=
n

f
1

.

{\displaystyle f\_n = \frac{nv}{4L} =n f\_1. }

Since the fundamental frequency, or first harmonic, of a stopped pipe is the frequency when

# n
1

{\displaystyle n=1}

, we get

f
1

=

v

4
L

.

{\displaystyle f\_1 = \frac{v}{4L} . }

Thus,

f
1

=

340

m

/

s

4
L

=
220

H
z

{\displaystyle f\_1 = \frac{340 \; m/s}{4L} = 220 \; Hz }

and

# L

340

m

/

s

4
×
220

H
z

=
0.386

m
.

{\displaystyle L = \frac{340 \; m/s}{4 \times220 \; Hz} = 0.386 \; m. }

**Part 2**

Calculate the second overtone

f
5

{\displaystyle f\_5 }

:

f
5

=
5
×
220

H
# z
1100

H
z
.

{\displaystyle f\_5 = 5 \times 220 \; Hz = 1100 \; Hz. }

For an open pipe, the frequencies are calculated using the formula

f
n

=

n
v

2
L

=
n

f
1

{\displaystyle f\_n = \frac{nv}{2L} = n f\_1 }

Thus,

1100

H
# z

3
×
340

m

/

s

2
L

{\displaystyle 1100 \; Hz = \frac{3 \times340 \; m/s}{2L} }

and

# L

3
(
340

m

/

s
)

2
(
1100

H
z
)

=
0.464

m
.

{\displaystyle L = \frac{3(340 \; m/s)}{2(1100 \; Hz)} = 0.464 \; m. }
Chemical Equilibrium of Hydrogen Iodide

## **Problem**

This problem is on the dissociation of hydrogen iodide gas into hydrogen gas and iodine gas.

2

HI

↽

−

−

⇀

H

2

I

2

{\displaystyle \ce{2HI <=> H2 + I2}}

If 1.00 mol of hydrogen iodide dissociates in a 1.00 L flask produces 0.125 moles of iodine gas, determine the equilibrium constant.

## **Solution**

Construct an ICE table (initial, change, equilibrium) for the reaction

2

HI

↽

−

−

⇀

H

2

I

2

⋅

{\displaystyle \ce{2HI <=> H2 + I2.}}

 **Change** | -2 x 0.125
 **Equilibrium** | 0.750

Calculate the equilibrium constant as follows:

K

e
q

=

[

H
2

]

[

I
2

]

[

H
I

]

2

=

(
0.125
)
(
0.125
)

(
0.750
)

2

=
0.0278

{\displaystyle \begin{align}
{K}\_{eq} &= \frac{\left[H\_2 \right] \left[I\_2 \right]}{{\left[HI \right]}^{2}} \\[5 pt]
&= \frac{(0.125)(0.125)}{{(0.750)}^{2}} \\[5 pt]
&= 0.0278
\end{align} }
The U-Tube

## **Problem**

   Consider a u-shaped tube filled with water (Figure 1). We will call this tube the U-tube. The E-tube is opened on both ends. Oil is poured into the U-tube until the column of oil is 27.2 cm tall, measured from the oil-water interface. The water level on the other side is 2.18 cm shorter than the column of oil. Calculate the density of oil in units of kilograms per cubic meters.

## **Hint**

(1) Remember that the density of water is 1000 kg per cubic meter.

## **Solution**

Consider the diagram above. Note that the oil and water are not mixed. The pressure at the level of the oil-water interface is

P
0

ρ

o
i
l

g
# H

P
0

ρ

w
a
t
e
r

g
(
H
−
h
)

{\displaystyle P\_0 + {\rho}\_{oil} g H = P\_0 + {\rho}\_{water} g (H-h) }

where

P
0

{\displaystyle P\_0}

 is the atmospheric pressure,

ρ

o
i
l

{\displaystyle \rho\_{oil} }

 is the density of oil,

ρ

w
a
t
e
r

{\displaystyle \rho\_{water} }

 is the density of water,

H

{\displaystyle H}

 is the height of the oil column, and

h

{\displaystyle h}

 is the height of the water column measured from the level of the oil-water interface. Isolating for the density of oil gives

ρ

o
i
l

=

ρ

w
a
t
e
r

H
−
h

H

{\displaystyle {\rho}\_{oil} = {\rho}\_{water} \frac{H-h}{H} }

Substituting the known values yields

ρ

o
i
l

=

(

1000

k
g

m
3

)

(

27.2

c
m
−
2.18

c
m

27.2

c
m

)

=
920

k
g

m
3

{\displaystyle {\rho}\_{oil} = \left(1000 \frac{kg}{m^3} \right) \left(\frac{27.2 \; cm - 2.18 \; cm}{27.2 \; cm} \right) = 920 \frac{kg}{m^3} }
Potential of Conducting Spheres

## **Problem**

Find the potential between two concentric conducting spheres or radius

a

{\displaystyle a}

 and

b

{\displaystyle b}

. The potential at

# r
a

{\displaystyle r=a}

 is

V
a

{\displaystyle V\_a}

 and the potential at

# r
b

{\displaystyle r=b}

 is

V
b

{\displaystyle V\_b}

.

## **Solution**

This problem is a Laplace equation problem in one dimension and in the spherical polar coordinates.

∇
2

# V

1

r
2

d

d
r

(

r
2

d
V

d
r

)

=
0

{\displaystyle \nabla^2 V= \frac{1}{r^2}\frac{d}{dr} \left(r^2\frac{dV}{dr}\right)=0 }

Integrating twice yields the

V
(
r
# )

A
r

B

{\displaystyle V(r)=\frac{A}{r}+B }

At

# r
a

{\displaystyle r=a}

V
(
a
# )

A
a

# B

V
a

{\displaystyle V(a)=\frac{A}{a}+B=V\_a }

At

# r
b

{\displaystyle r=b}

V
(
b
# )

A
b

# B

V
b

{\displaystyle V(b)=\frac{A}{b}+B=V\_b }

Solve for the constants

A

{\displaystyle A}

 and

B

{\displaystyle B}

 in a system of simultaneous equations.

# A

(

V
a

−

V
b

)
a
b

b
−
a

{\displaystyle A=\frac{(V\_a-V\_b)ab}{b-a} }

# B
−

a

V
a

−
b

V
b

b
−
a

{\displaystyle B=-\frac{aV\_a-bV\_b}{b-a} }

So the solution for the potential is

V
(
r
# )

a
b

b
−
a

V
a

−

V
b

r

−

a

V
a

−
b

V
b

b
−
a

{\displaystyle V(r)=\frac{ab}{b-a} \frac{V\_a-V\_b}{r}-\frac{aV\_a-bV\_b}{b-a} }

.
Circle Segments

**Problem created by Tao Steven Zheng (郑涛).**

## **Problem**

Refer to the diagram below. Prove that the total area in red is equal to the total area in blue.

## **Solution**

Due to the four-fold rotational symmetry of the diagram. This problem can be simplified by separating the circle into four quadrants and restricting the analysis to one of the quadrants.

Let the radius of the quadrant be

r

{\displaystyle r}

. The area of one of the yellow segments and the red segment produces a semicircle whose diameter is

r

{\displaystyle r}

. Hence, the radius of this semicircle is

r
2

{\displaystyle \frac{r}{2}}

. With a little calculation, it is clear that the area of the quadrant is

π

r
2

4

{\displaystyle \frac{\pi r^2}{4} }

 and the area of the semicircle is

π

r
2

8

{\displaystyle \frac{\pi r^2}{8} }

.

This leads to the conclusion

2
(
r
e
d
y
e
l
l
o
w
# )
r
e
d
2
(
y
e
l
l
o
w
)
b
l
u
e

{\displaystyle 2(red + yellow) = red + 2(yellow) + blue }

.
Carrying out the algebraic calculation yields

2
(
r
e
d
)
2
(
y
e
l
l
o
w
# )
r
e
d
2
(
y
e
l
l
o
w
)
b
l
u
e

{\displaystyle 2(red) + 2(yellow) = red + 2(yellow) + blue }

r
e
# d
b
l
u
e

{\displaystyle red = blue }

.
Therefore, the area in red is equal to the area in blue.
Mytle beach safari = big boy

[ made by me N big boy ] [ in 2021 ] as mytle beach safari tv show for boys n girls [ only on Mytle beach safari Multiple Intelligence Hub James the jungle ocean house outback boy youtube channel ] ]

## animals appers

asian elephants

giant pandas

bactrain camels

bengal tigers

black rhinos

panthers

komodos

tapirs

musk oxen

african elephants

quing pandas

dromdary camels

sumatra tigers

javen rhinos

lions

giraffes

hippos

zebras

wild beest

warthogs

hyenas

gazelles

cars

buses

bikes

planes

diggers

vans

boats

safari parks

jungles

towns

farms

woodlands

beaches

humans

## cast

asian javen african jungle elephants n mammoths

giant n quing pandas

hippos [ aka elephants ]

black n javen rhinos

[ pacydeams ]

bengal china n javen tigers

lions

lynxes

cheetahs n jaguars

[ felines ]

monkeys

orangutans

koalas

mountain n lowland gorillas

[ apes n primates ]

bears

giant n quing pandas

polar bears

[ bears ]

asian elephants [ as main ]

bengal tigers

chimps

dik diks

love birds

baboons

giant pandas

[ jungle land of asian javen african jungle elephants bengal china n javen tigers cheetahs eagles ligers chimps gibbons wild boys n villege girls ]

african elephants [ as main ]

giraffes

sumatran tigers

cheetahs

chimps

diki diks

love birds

[ savannah land of asian javen african jungle elephants bengal china n javen tigers cheetahs eagles ligers chimps gibbons wild boys n villege girls ]

[ MAIN ]

asian javen african jungle elephants

bengal china n javen tigers

cheetahs

eagles

ligers

chimps

gibbons

wild boys n villege girls

[ land of asian javen african jungle elephants bengal china n javen tigers cheetahs eagles ligers chimps gibbons wild boys n villege girls ]
Two Pipes Filling a Tank

## **Problem**

Suppose there are two pipes that fill into a tank. Pipe A can fill the tank in 2.5 hours less time it takes for pipe B to fill the tank. Suppose pipe A starts filling the tank on its own for 6.5 hours, at which pipe B joins pipe A in filling the tank. Together, the two pipes finish filling the tank in 5.5 additional hours. Question: How long would it take each pipe to fill the tank on its own?

## **Hint**

(1) Think of the volume as the flow rate multiplied by the time.

## **Solution**

This problem does not give a specific volume for the tank or specific flow rates for each pipe. These quantities are not necessary for solving for the time! However, let us define the unspecified volume as

V

{\displaystyle V}

, and the unspecified flow rates for pipe A and pipe B

r
A

,

r
B

{\displaystyle r\_A, r\_B}

 respectively.

If each pipe were to fill the tank individually, we get

(
1
)

V

=

r
A

t

(
2
)

V

=

r
B

(
t
2.5
)

{\displaystyle \begin{align}
(1) \; V &= r\_A t \\[5 pt]
(2) \; V &= r\_B (t + 2.5) \\
\end{align}}

For this particular problem, pipe A begins filling the tank for 6.5 hours before pipe B joins for an additional 5.5 hours. Subsequently, we get the system

(
1
)

V

=

r
A

t

(
2
)

V

=

r
B

(
t
2.5
)

(
3
)

V

=

r
A

(
6.5
5.5
)

r
B

(
5.5
)

{\displaystyle \begin{align}
(1) \; V &= r\_A t \\[5 pt]
(2) \; V &= r\_B (t + 2.5) \\[5 pt]
(3) \; V &= r\_A (6.5 + 5.5) + r\_B (5.5)
\end{align}}

or more simply

(
1
)

V

=

r
A

t

(
2
)

V

=

r
B

(
t
2.5
)

(
3
)

V

=
12

r
A

5.5

r
B

.

{\displaystyle \begin{align}
(1) \; V &= r\_A t \\[5 pt]
(2) \; V &= r\_B (t + 2.5) \\[5 pt]
(3) \; V &= 12r\_A + 5.5r\_B.
\end{align}}

Isolate for the flow rates from equation (1) and equation(2) and substitute into equation (3) to get

# V
12

(

V
t

)

5.5

(

V

t
2.5

)

.

{\displaystyle V = 12 \left(\frac{V}{t}\right) + 5.5\left(\frac{V}{t+2.5}\right).
}

We can eliminate the volume

V

{\displaystyle V}

 and obtain a quadratic equation that depends solely on time

t

{\displaystyle t}

1

=

12
t

5.5

t
2.5

[
5
p
t
]
1

=

12
(
t
2.5
)
5.5
t

t
(
t
2.5
)

[
5
p
t
]

t
2

−
15
t
−
# 30
0

{\displaystyle \begin{align}
1 &= \frac{12}{t}+ \frac{5.5}{t+2.5}\\ [5 pt]
1 &= \frac{12(t+2.5) + 5.5t}{t(t+2.5)} \\ [5 pt]
t^2 &- 15t -30 = 0
\end{align}
}

Using the quadratic formula yields two solutions for time (to three decimal places):

t
1

=
16.787

[
5
p
t
]

t
2

=
−
1.787.

{\displaystyle \begin{align}
t\_1 &= 16.787\\ [5 pt]
t\_2 &= -1.787.
\end{align}
}

Obviously, we can only accept the positive time solution for this problem. Consequently,

t

=
16.787

[
5
p
t
]
t
2.5

=
19.287.

{\displaystyle \begin{align}
t &= 16.787\\ [5 pt]
t+2.5 &= 19.287.
\end{align}
}

Therefore, if each pipe were to fill the tank on their own, pipe A will take 16.787 hours and pipe B will take 19.287 hours.
Producing Lime from Limestone

## **Problem**

A 200 kilogram slab of limestone is found to be 95% pure calcium carbonate

(

CaCO

3

)

{\displaystyle \ce{ (CaCO3) }}

.

Calculate the weight of lime

(
CaO
)

{\displaystyle \ce{ (CaO) }}

 that can be produced by heating this slab of limestone.

## **Solution**

**Step 1:** Determine the molar mass of each chemical.

The mass of calcium carbonate found in the slab must be

0.95
×
200

k
# g
190

k
g
.

{\displaystyle 0.95 \times 200 \; kg = 190 \; kg. }

Molar mass of

CaCO

3

{\displaystyle \ce{ CaCO3 }}

 in grams per mole:

40.08

g

m
o
l

12.01

g

m
o
l

3
×
16.00

g

m
o
l

=
100.09

g

m
o
l

.

{\displaystyle 40.08 \; \frac{g}{mol} + 12.01 \; \frac{g}{mol} + 3 \times 16.00 \; \frac{g}{mol} = 100.09 \; \frac{g}{mol}. }

Molar mass of

CaO

{\displaystyle \ce{ CaO }}

 in grams per mole:

40.08

g

m
o
l

16.00

g

m
o
l

=
56.08

g

m
o
l

{\displaystyle 40.08 \; \frac{g}{mol} + 16.00 \; \frac{g}{mol} = 56.08 \; \frac{g}{mol} }

**Step 2:** Determine the chemical reaction formula

CaCO

3

⟶
CaO

CO

2

{\displaystyle \ce{ CaCO3 -> CaO + CO2 }}

**Step 3:** Determine the mass of the lime

190

k
g
×

56.08

g

/

m
o
l

100.09

g

/

m
o
l

=
106.5

k
g

{\displaystyle 190 \; kg \times \frac{56.08 \; g/mol}{100.09 \; g/mol} = 106.5 \; kg }
Radical Manipulation

**Problem created by Tao Steven Zheng (鄭濤)**

A fun, simple problem I made. Requires a little mental gymnastics to arrive at the answer!

## **Problem**

If

m
−

n

=

a

−

b

{\displaystyle \sqrt{m-\sqrt{n}} = \sqrt{a} - \sqrt{b} }

, where

m
2

≥
n

{\displaystyle m^2 \ge n }

, determine

a
2

b
2

{\displaystyle a^2+b^2}

 in terms of

m

{\displaystyle m}

 and

n

{\displaystyle n}

.

## **Solution**

Square both sides of the equation:

m
−

n

=
a
b
−
2

a
b

.

{\displaystyle m - \sqrt{n} = a + b -2\sqrt{ab}.}

By matching terms,

m

=
a
b

n

=
2

a
b

.

{\displaystyle \begin{align}
m &= a + b \\[5 pt]
\sqrt{n} &= 2\sqrt{ab}.
\end{align} }

Squaring the second expression shows that

# n
4
a
b

{\displaystyle n = 4ab }

. Notice that

a
2

b
2

=
(
a
b

)
2

−
2
a
b

{\displaystyle a^2 + b^2 = (a+b)^2 - 2ab }

.

Therefore,

a
2

b
2

=

m
2

−

n
2

.

{\displaystyle a^2 + b^2 = m^2 - \frac{n}{2}. }
Triple-angled Sine

## **Problem**

Prove that

sin
⁡
(
3
x
# )
3
sin
⁡
(
x
)
−
4

sin
3

⁡
(
x
)

{\displaystyle \sin(3x) = 3\sin(x) - 4\sin^3(x) }

.

## **Solution**

Use the angle sum formula

sin
⁡
(
x
y
# )
sin
⁡
(
x
)
cos
⁡
(
y
)
cos
⁡
(
x
)
sin
⁡
(
y
)

{\displaystyle \sin(x+y) = \sin(x)\cos(y) + \cos(x)\sin(y) }

 and make the substitution

# y
2
x

{\displaystyle y=2x}

.

sin
⁡
(
x
2
x
# )
sin
⁡
(
x
)
cos
⁡
(
2
x
)
cos
⁡
(
x
)
sin
⁡
(
2
x
)

{\displaystyle \sin(x+2x) = \sin(x)\cos(2x) + \cos(x)\sin(2x) }

Substitute the double-angle formulas for sine and cosine into the above formula:

sin
⁡
(
2
x
# )
2
sin
⁡
(
x
)
cos
⁡
(
x
)

{\displaystyle \sin(2x)=2\sin(x)\cos(x)}

cos
⁡
(
2
x
# )
1
−
2

sin

2

⁡
(
x
)

{\displaystyle \cos(2x)=1-2\sin ^{2}(x)}

.
Subsequently,

sin
⁡
(
x
2
x
# )
sin
⁡
(
x
)

[

1
−
2

sin

2

⁡
(
x
)

]

cos
⁡
(
x
)

[

2
sin
⁡
(
x
)
cos
⁡
(
x
)

]

{\displaystyle \sin(x+2x)=\sin(x)\left[1-2\sin ^{2}(x)\right]+\cos(x)\left[2\sin(x)\cos(x)\right]}

sin
⁡
(
3
x
# )
sin
⁡
(
x
)
−
2

sin

3

⁡
(
x
)
2
sin
⁡
(
x
)

cos

2

⁡
(
x
)

{\displaystyle \sin(3x)=\sin(x)-2\sin ^{3}(x)+2\sin(x)\cos ^{2}(x)}

Since

cos
2

⁡
(
2
x
# )
1
−

sin
2

⁡
(
x
)

{\displaystyle \cos^2(2x) = 1 - \sin^2(x) }

,

sin
⁡
(
3
x
# )
sin
⁡
(
x
)
−
2

sin

3

⁡
(
x
)
2
sin
⁡
(
x
)

(

1
−

sin

2

⁡
(
x
)

)

{\displaystyle \sin(3x)=\sin(x)-2\sin ^{3}(x)+2\sin(x)\left(1-\sin ^{2}(x)\right)}

sin
⁡
(
3
x
# )
sin
⁡
(
x
)
−
2

sin

3

⁡
(
x
)
2
sin
⁡
(
x
)
−
2

sin

3

⁡
(
x
)

{\displaystyle \sin(3x)=\sin(x)-2\sin ^{3}(x)+2\sin(x)-2\sin ^{3}(x)}

sin
⁡
(
3
x
# )
3
sin
⁡
(
x
)
−
4

sin
3

⁡
(
x
)

{\displaystyle \sin(3x) = 3\sin(x) - 4\sin^3(x) }
Potassium Concentration

## **Problem**

   Suppose two aqueous solutions were combined:

125.0 mL of 0.250 M potassium sulphate (

K

2

SO

4

{\displaystyle \ce{K2SO4}}

)

175.0 mL of 0.450 M potassium iodide (

KI

{\displaystyle \ce{KI}}

).

Determine the resulting concentration of the potassium ions in the combined solution.

## **Solution**

The volume of the combined solution is

125.0

m
L
175.0

m
# L
300.0

m
L

{\displaystyle 125.0 \; mL + 175.0 \; mL = 300.0 \; mL }

or 0.3000 L. This will change the concentrations of potassium sulphate and potassium iodide.

[

K
2

S

O
4

# ]

(
0.250

M
)
(
0.1250

L
)

0.3000

L

=
0.104

M

{\displaystyle [K\_2 SO\_4] = \frac{(0.250 \; M)(0.1250 \; L)}{0.3000 \; L} = 0.104 \; M }

[
K
I
# ]

(
0.450

M
)
(
0.1750

L
)

0.3000

L

=
0.263

M

{\displaystyle [KI] = \frac{(0.450 \; M)(0.1750 \; L)}{0.3000 \; L} = 0.263 \; M }

Write out the dissociation equations and construct a table detailing each ion and their respective concentrations.

K

2

SO

4

⟶
2

K

SO

4

2

−

{\displaystyle \ce{K\_2 SO\_4 -> 2{K}^{+} + {SO}^{2-}\_{4}}}

KI
⟶

K

I

−

{\displaystyle \ce{KI -> {K}^{+} + {I}^{-}}}

 Ion

From the table above we discover that the concentration of potassium ions is

[

K

# ]
0.208

M
0.263

# M
0.471

M

{\displaystyle [K^+] = 0.208 \; M + 0.263 \; M = 0.471 \; M }