Learn Online Pythagorean

Pythagorean Theorem is a very important theorem used in Trigonometry.  It gives a relationship between the sides of a right triangle. A Greek philosopher Pythagoras was the founder of Pythagorean Theorem.

Introduction to Pythagorean Theorem:

Pythagorean Theorem describes a relationship between the longest side of the right triangle and the remaining smaller sides.

Statement: In right triangle, the square of longest side (hypotenuse) is equal to the sum of square of the remaining two sides.


Explanation of Pythagorean theorem:


In a right-angled triangle,

Let ‘a’ = adjacent side, ‘b’ = opposite side, ‘c’ = hypotenuse



Using Pythagorean Theorem,

(Hypotenuse) 2 = (adjacent side) 2 + (opposite side) 2.


Examples of Pythagorean Theorem :


Ex 1: Find the hypotenuse of the right triangle when the adjacent and opposite sides of the right angled triangle is 4cm and 3cm.

Sol:   The triangle given is a right angled triangle , it obeys pythagorean theorem

Step I:         (adjacent side)2+(opposite side)2= (hypotenuse side)2

Step II:    42 +32 = x2

Step III:    16+9 =x2

Step IV:     25 = x2

Step V:    x = 5

The hypotenuse side of the triangle is 5cm

Ex 2: Find the adjacent side of the right triangle when the hypotenuse and opposite sides of the right angled triangle is 10cm and 6cm.

Sol:  The given triangle is a right angled triangle , it obeys pythagorean theorem

Step I:    (adjacent side)2+(opposite side)2= (hypotenuse side)2

Step II:    x 2 +62 = 102

Step III:    100-64 =x2

Step IV:    36= x2

Step V:        x=6

The hypotenuse side of the triangle is 6cm

Ex 3: Find the opposite side of the Right Triangles Trigonometry when the hypotenuse and adjacent sides of the right angled triangle is 10cm and 5cm.

Sol:   The given triangle is a right angled triangle , it obeys pythagorean theorem

Step I:    (adjacent side)2+(opposite side)2= (hypotenuse side)2

Step II:    52+ x 2 = 102

Step III:    100-25 =x2

Step IV:     75= x2

Step V:     x=8.7

The hypotenuse side of the triangle is 8.7cm

These examples are used to learn online pythagorean theorem.

How to Find the Median

INTRODUCTION:

Median is the middle value of a given numbers or allocation of their ascending order. Median is an average value of the two middle elements when the size of the allocation is even.

To locate the Median, place the numbers in ascending order and find the middle number.
If there are two middle numbers then average those two numbers to find median.


Median for Odd numbers


Ex 1:

Find the median for the following list of values:

9, 3, 44, 17, 15

Solution:

Find the Median of: 9, 3, 44, 17, and 15 (Odd amount of numbers)

Line up your numbers: 3, 9, 15, 17, and 44 (smallest to largest)

The Median is: 15 (The number in the middle)

Ex2:

Find the median for the following list of values:

13, 18, 13, 14, 13, 16, 14, 21, 13

Solution:

Find the Median of: 13, 18, 13, 14, 13, 16, 14, 21, and 13(Odd amount of numbers)

Line up your numbers: 13, 13, 13, 13, 14, 14, 16, 18, and 21 (smallest to largest)

The Median is: 14 (The number in the middle)


Median for Even numbers


Ex 1:

How to find the median for the following list of values:

8, 3, 44, 17, 12, and 6

Solution:

Find the Median of: 8, 3, 44, 17, 12, and 6 (Even amount of numbers)

Line up your numbers: 3, 6, 8, 12, 17, and 44(smallest to largest)

Add the 2 middles numbers and divide by 2:

= (8 + 12)/2

= 20 ÷ 2

= 10

The Median is 10.

Ex 2:

How to find the median for the following list of values:

8, 9, 10, 10, 10, 11, 11, 11, 12, 13

Solution:

Find the Median of: 8, 9, 10, 10, 10, 11, 11, 11, 12, and 13 (Even amount of numbers)

Line up your numbers: 8, 9, 10, 10, 10, 11, 11, 11, 12, and 13 (smallest to largest)

Add the 2 middles numbers and divide by 2:

= (10+11)/2

= 21/2

= 10.5

The Median is 10.5

Solving Root of a Number

 If any number is expressed as x × x, then x is the product of two same numbers. We know that 5^2 = 5× 5 =25.Here 25 is called the square of 5 and 5 is called the square root of 25. And (2/3)^2 = (2/3)× (2/3) = 2 ×2/3 ×3 = 4/9. 4/9 is called as square of 2/3 and is also called as square root of 4 / 9.

Examples of Solving Root of a Number

1. Simplify 7^2 = 7 ×7 = 49

Here 49 is called the square of 7 and 7 is called the square root of 49

2. (0.4)^2 = (0.4)× (0.4) = 0.16

Here 0.16 is called the square of 0.4 and 0.4 is called the square root of 0.16

3. Simplify 9^2 = 9 ×9 = 81

Here 81 is called the square of 9 and 9 is called the square root of 81

4. Simplify 121^2 = 121 ×121 = 14641

Here 14641 is called the square of 7 and 7 is called the square root of 14641

5. Simplify 81^2 = 81 ×81= 49

Here 6561 is called the square of 81 and 81 is called the square root of 6561

Multi Step Square Root Examples:

(1). Find the square root of 144

Solution:

Split the number into the product of prime factors.

                              144 = 3 × 3 × 2 × 2 × 2 × 2

                            √144 = √3^2 × 2^2 × 2^2

                                    =3 × 2 × 2

             Therefore √144 = 12

(2). Find the square root of 5^3 × 5^5

Solution:

                         5^3 × 5^5 = (5 × 5 × 5) × (5 × 5 × 5 × 5 × 5)

                                       =5^2 × 5^2 × 5^2 × 5^2

       Therefore √5^3 × 5^5 = √5^2 × 5^2 × 5^2 × 5^2

                                       = 5 × 5 × 5 × 5 × 5

                                      = 625

      Therefore √5^3 × 5^5 = 625

Discuss of Solving Root of Number

(1)   Find the square root of 36
                
(2)   Find the square root of 6^2×7^2

(3)   Find the square root of 8100


Answers:
1. 6    2. 42   3.  90

Learn Online Limits

In the mathematical expression the main concept of limit is used to express a value that a sequence or function approaches as the input or key approaches of some value. The limit is typically reduced as lim as in Lim(xn) = x or represent by the right arrow (→) as in an → a. Let us consider this function f(x) = x2. Examine that as x take values very close to 0, the value of f(x) also move towards 0. We say limits  f(x) = 0  x →0

Rules For how to solve limits


Rule1: In learning online limits, given limits function put x=a .If f(a) is a definite value then

limits  f(x) = f(a)
         x →a

Rule2: In learning online limits, If  proving limits  f(x) is a rational function then factorize the numerator and the denominator.Cancel out the  common factors and then put x=a

Rule3: If the given learning online limits function contains a surd then simplify it by using conjugate surd's.After simplification,put x =a

Rule4: If the given  proving learning online  limits  function contains a series which is capable of being expanded then after making proper expansion and simplifying,cancel the common factors in the numerator and denominator,if any Then, put x =a


Limits Examples


1) Evaluate  proving limits lim     (xm -am ) / (xn -an)
                                             x →a

Solution for proving  limits:  lim    (xm -am /xn -an)  =   lim    {xm -am /x-a) ÷ (xn -an /x-a)}
                                                   x →a                                   x →a
Limits =   lim     (xm -am /x -a)   ÷   lim(xn - an /x -a)
                  x →a                                 x →a
Limits =     (ma n-1) ÷ (nan-1)

Limits =   ma m-1 / na n-1   = (m) /(n a m-n)


2) Evaluate proving  limits lim (x+2)3/3 -  (a +2)3/2 / x-a

                                              x→a

Solution proving  limits:   lim (x+2)3/3 - (a +2)3/2 /  x-a
                                            x→a
                        (x +2)3/2 - (a+2)3/2
=      lim     ------------------------------------------
                (x+2)→(a+2)            (x +2) - (a +2)
          ------------------------------------------------------
=   3/2. (a+2)(3/2 -1) =     3/2(a +2)1/2                      [ lim   (xn -an /x -a)   =  nan-1]
                                                                                       x→a

3) Find Limit (x →2) {3x2-5x+7}

Solution:- Given Limit ( x →2)    {3x2-5x+7}

= 3(2)2-5(2)+7   = 12-10+7 = 9


4) Show that Limit (x →3)  (x2+2x-5)  /  (2x2-5x-1) = 5/2

Solution:-Limit (x →3) (x2+2x-5) / (2x2-5x-1)

= Limit (x →3) (x2+2x-5) /  Limit ( x →3)  (2x2-5x-1)

=[ (3)2+2(3)-5)]  / [ 2(3)2-5(3)+1]  =  (9+ 9 - 5) /  (18-15+1)    = 10/ 4 = 5/2.

We can be solved these practice problems on limits  by learning these limits problems.

Fraction Decimal Percent Table

FRACTION DECIMAL PERCENT TABLE

Decimals, Fractions and Percentages are just different ways of showing the same value:

A Half can be written...

As a fraction:       1/2

As a decimal:        0.5

As a percentage:   50%

A Quarter can be written...

As a fraction:       1/4

As a decimal:        0.25

As a percentage:   25%

Example Values

Here is a table of commonly occurring values shown in Percent, Decimal and Fraction form:

Percent       Decimal      Fraction

1%                0.01                      1/100

5%                0.05                       1/20

10%              0.1                1/10

12½%           0.125            1/8

20%               0.2                1/5

25%               0.25                        1/4

331/3%          0.333...        1/3

50%               0.5               1/2

75%            0.75           3 /4

80%            0.8             4/5

90%            0.9           9/10

99%           0.99        99/100

100%          1               1

125%          1.25        5/4

150%            1.5         3/2

200%             2           2

Converting Between Percentage and Decimal

Percentage means "per 100", so 50% means 50 per 100,

or simply 50/100.If we divide 50 by 100 you get 0.5 (a decimal number).

So, to convert from percentage to decimal: divide by 100 (and remove the "%" sign).The easiest way to divide by 100 is to move the decimal point 2 places to the left.

Example: Convert 8.5% to decimal

Move the decimal point two places: 8.5 -> 0.85 -> 0.085

Answer 8.5% = 0.085

Converting From Decimal to Percentage

To convert from decimal to percentage, just multiply the decimal by 100, but remember to put the "%" sign so people know it is per 100.The easiest way to multiply by 100 is to move the decimal point 2 places to the right.

Example: Convert 0.65 to percent

Move the decimal point two places: 0.65 -> 6.5 -> 65.

Answer 0.65 = 65%

To change a Decimal into a Fraction

Take the decimal, drop the decimal point, and place the result into the numerator  of a fraction.

To determine the denominator , write a 1, followed by zeros --- as many zeroes as it takes to match the original length of the decimal.

Examples:

0.75 becomes 75/100

0.034 becomes 34/1000

2.5 becomes 25/10

Roman Numerals Learning

Roman numeral is a symbol, roman numerical learning is used to represent a number. (Our digits 0-9 are often called as Arabic numerals.) In learning of roman numerals are written as the combinations of the seven letters.

Those seven letters are,

I =1             L=50

V = 5           C=100      M=1000

X=10           D=500


Note:

If a lesser numbers follow larger numbers, then numbers are added.

If a lesser number precedes bigger number, then the smaller number is subtracted from the larger.

How to write roman numerals and rules for subtracting letters -roman numerals learning:


Here, how 1100 will be written as Roman Numerals Learning, you would state M for 1000 and then put a C after it used for 100; Otherwise 1,100 = MC in Roman Numerals Number.

Some examples:

• VIII = 5+3 = 8

• IX = 10-1 = 9

• XL = 50-10 = 40

• XC = 100-10 = 90

• MCMLXXXIV = 1000 + (1000 -100) + 50 + 30 + (5 - 1) = 1984

Rules for subtract letters- Roman numerals learning:


•   Subtract powers of ten, such as I, X, or C. Writing VL for 45 be not suitable: write XLV as a replacement
•   Subtract only a distinct letter from a single digit. Write VIII for 8, not IIX; 19 is XIX, not IXX.
•   Don't subtract letter from a unlike letter more than ten times larger. This means you can just subtract the I from V or X, and X from L or C, so MIM is against the law.


Let’s found with an addition problem: 13 + 58. In Roman numerals learning, that's XIII + LVIII. We'll begin by writing.
these  two numbers subsequently to each other:Next, we  are rearrange the letters so that the numerals are in descending order: LXVIIIIII. Now we have six be, so we'll rewrite them as VI: LXXVI. The two Vs are the same as an X, so we simplify again and get LXXI, or 71, this is our final answer.

Learn discrete random variables

If the Random variable X assumes only finite or countably infinte set of values it is known as discrete random variable.

Probability density function of Discrete Random variable:-

Suppose X is a Random variable which can take at the most a countable number of values X1, X2, X3, ..................... Xn with each value of  " X ". We associate a number

pi = P ( X = Xi ) ; i = 1,2,..............n

which is known as the probability of Xi and satisfies the following conditions:

pi = P ( X = Xi ) `>=` 0  ( i = 1,2,..............n )    i.e., pi 's are all non- negative and
`sum` pi = p1 + p2 +................... + pn = 1
i.e., the total probability is one.

The function pi = P ( X = Xi ) ; i = 1,2,..............n is called the probability function or more precisely probability mass function of the random variable X


Cumulative distribution function of F(x) of discrete random variable


Cumulative distribution function F( x ) of a discrete random variable X is denoted as F( X = xi ) and defined as               F ( X = xi ) = P ( X = xi )

F ( X = xi ) = P ( X = x1 ) + P ( X = x2 ) + ............................ + P ( X = xi )

F ( X = xi ) =    `sum_(n=1)^i` P ( X = xn )


Example of learn discrete random variables


Let x denote the minimum of two numbers that appear when a two dice is thrown once. find the discrete probability distribution?

Solution:-  The sample space S = { ( 1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (2,2) (2,3) (2,4) (2,5) (2,6) (3,1) (3,2) (3,3) (3,4) (3,5) (3,6) (4,1) (4,2) (4,3) (4,4) (4,5) (4,6) (5,1) (5,2) (5,3) (5,4) (5,5) (5,6) (6,1) (6,2) (6,3) (6,4) (6,5) (6,6) }

n ( S ) = 62 = 36

Given that X = min ( a, b )

P( 1 ) = P ( X = 1 ) = { (1,1) (1,2) (1,3) (1,4) (1,5) (1,6) (2,1) (3,1) (4,1) (5,1) (6,1) }

= 11 / 36

P( 2 ) = P ( X = 2 ) = { (2,2) (2,3) (2,4) (2,5) (2,6) (3,2) (4,2) (5,2) (6,2) }

= 9 / 36

P( 3 ) = P ( X = 3 ) = { (3,3) (3,4) (3,5) (3,6) (4,3) (5,3) (6,3) }

= 7 / 36

P( 4 ) = P ( X = 4 ) = { (4,4) (4,5) (4,6) (5,4) ( 6,4) }

= 5 / 36

P( 5 ) = P ( X = 5 ) = { (5,5) (5,6) (6,5) }

= 3 / 36

P( 6 ) = P ( X = 6 ) = { (6,6) }

= 1 / 36

X    1    2    3    4    5    6
P ( X = x)    11/36    9/36    7/36    5/36    3/36    1/36