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Here's my way of tackling these (others may have different methods).

**Problem #1**

The inches & fraction components in the first length are less than their counterparts in the second length, which adds difficulty to subtracting imperial lengths, particularly those with fractions.

To make the subtraction easier, I start with the following three steps:

1. convert the fraction to the Least Common Denominator (LCD); then

2. adjust the value of the fraction component to make it greater than that of the second length; then

3. adjust the value of the inches component to make it greater than that of the second length.

That leaves a simple subtraction exercise.

Btw, the 'take' and 'add' terminology that I used is like 'borrow' and 'pay', now aka 'regrouping', and is how I've always understood that concept.

Also, I've used symbols ' and " instead of abbreviations 'ft' and 'in' respectively...a common notation practice for these imperial length units.

Formula: You could convert both lengths to eighths of an inch, deduct the second from the first, and convert the result to feet, inches and eighths:

That last conversion from eighths is a bit tricky...

**Problem #2**

Same approach as for Problem #1:

Formula: You could convert both angles to seconds, deduct the second from the first, and convert the result to degrees, minutes & seconds:

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Know that -a^2 means -1 x a^2 and not -a x -a.

The reason is that -a^2 has its coefficient as -1

Therefore -a x -a = a^2 and and not -a^2.

This symbol ^ implies exponent.

True but (-a)^2 is totally different.

We know that (-a)^2 = (-a)(-a) = a^2.

How about -(a)^2?

Let me see.

-[(a)(a)]

-(a^2)

-a^2

Agree?

]]>First lets turn the semiperimeter into a few substitutions for later use

Where semiperimeter = s = ( a + b + c ) / 2

( a + b + c ) = 2 s

( a + b - c ) =2 ( s – c )

( a - b + c ) =2 ( s – b )

( b + c - a ) =2 ( s – a )

Then our Cosine rule

a² = b² + c² – 2bcCos A

Cos A = ( b² + c² - a² ) / 2bc

1 – Cos A = ( 2bc - b² - c² + a² ) / 2bc

= ( a – b + c )( a + b - c ) / 2bc

= 2 ( s - b )( s – c ) / bc

1 + Cos A = ( 2bc + b² + c² - a² ) / 2bc

= ( a + b + c )( b + c - a ) / 2bc

= 2 s ( s – a ) / bc

Sin² A = ( 1 – Cos a )( 1 + Cos A )

= 4 s ( s – a )( s – b )( s - c ) / b²c²

_______________________

Sin A = 2 ( √ s ( s – a )( s – b )( s - c ) ) / bc

Area of Triangle = ½ . bc Sin A

______________________

= √ s ( s – a )( s – b )( s - c ) Heron's Area Formula

A much more elegant way of deriving Heron's Formula and there is still the Half angle Formula to go :-

______________

Cos ( A / 2 ) = √ ( 1 + Cos A ) / 2

______________

= √ s ( s – a ) / bc

______________

Sin ( A / 2 ) = √ ( 1 - Cos A ) / 2

___________________

= √ ( s – b )( s – a ) / bc

& Sin A = 2 Sin ( A / 2 ) Cos ( A / 2 ) gives the same Sin A result as above

Tan ( A / 2 ) = Sin ( A / 2 ) / Cos ( A / 2 )

_______________________

= √ ( s – b )( s – c ) / s ( s – a )

corrected mistake in half angle formulae

]]>Its the geometric symbol for congruent too , and not in open office function editor, which calls the equivalance symbol congruent .

The best stop gap solution would be to screen shot your formula and edit in a photoshop type editor.

I use gimp which is a free editor and just as powerful .

If you then search Google images for ' congruent symbol ' you can copy the one you prefer to your clipboard and paste it in.

There are a lot of formula editors on the market all have limitations but you are using the best of the bunch I think.

]]>This is an algorithm that test if one number is prime and if the number it´s not prime returns the biggest factor of that number.

Are you going to keep it a secret and profit off of it or are you going to enlighten our poor souls as to what it is?

]]>We can find infinite prime numbers with the separation we want and we can express every even number as the sum of two prime numbers.

Proof of the Twin primes Conjecture and Goldbach's conjecture]]>

We know

Sum of first n natural numbers is

Sum of squares of first n natural numbers is

Sum of cubes of first n natural numbers is

In a similar manner, this is additional information.

Sum of fourth powers of n natural numbers is

Sum of fifth powers of n natural numbers is

Sum of sixth powers of n natural numbers is

]]>Since edited.

]]>or...

A = s²/(4tan[180/n]) where s is length of side and n is number of sides (for those who don't want to use pi to calculate the area of a polygon)

Unfortunately, trigonometry tends to turn more people off than π.

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[*]https://www.youtube.com/watch?v=P4LxK9Uek5U&t=0s[/*]

[/list]

I found the solution to finding all palindromic four-digit powers particularly interesting.

]]>(Angles are given in degrees, 90 degrees, 180 degrees etc.)

I.

Sin(-θ)=-Sinθ

Cos(-θ) = Cosθ

tan(-θ) = -tanθ

cot(-θ) = -cotθ

sec(-θ) = secθ

cosec(-θ)= - cosecθII.

sin(90-θ) = cosθ

cos(90-θ) = sinθ

tan(90-θ) = cotθ

cot(90-θ) = tanθ

sec(90-θ) = cosecθ

cosec(90-θ) = secθIII.

sin(90+θ) = cosθ

cos(90+θ) = -sinθ

tan(90+θ) = -cotθ

cot(90+θ) = -tanθ

sec(90+θ) = -cosecθ

cosec(90+θ) = secθIV.

sin(180-θ) = sinθ

cos(180-θ) = -cosθ

tan(180-θ) = -tanθ

cot(180-θ) = cotθ

sec(180-θ) = -secθ

cosec(180-θ) = cosecθV.

sin(180+θ) = -sinθ

cos(180+θ) = -cosθ

tan(180+θ) = tanθ

cot(180+θ) = cotθ

sec(180+θ) = -secθ

cosec(180+θ) = -cosecθ

cot(180-θ)=-cotθ

since cot(180-θ)=1/tan(180-θ)

Where is the supposed "c" in the Left Hand Side?

]]>