Laws of Exponents for Real Numbers

Let ‘a’ be a positive number. Exponential notation is ax , where x is an integer. When x is a rational number, say p/q with p, an integer and q, a positive integer, we define ax by ax  = root(q)(a) p

Example:

53/8 = root(3)(5)8

laws of exponents for real numbers: Exponents Notation

When x is an irrational number, ax can be defined to represent a real number.

For any a > 0, ax can be defined and that it represents an unique real number u and write u = ax.

The real number u is written in the exponential form or in the exponential notation.

Here the positive number 'a' is called the base and x, the index or the power or the exponent.

The laws of indices which we have stated for integer exponents can be obtained for all real exponents.


laws of Exponents for Real Numbers

We state them here and call them, the laws of exponents for real number:

root(x)(root(y)(a))  = root(xy)(a)


root(x)(a / b)  =  root(x)(a) / root(x)(b)


(root(x)(a) )x = a


axxx ay = ax+y


ax// ay = ax-y


(ax)y = axy


a-x  = 1/ ax


axxx bx = abx


a0 = 1  for a != 0


(a/b)x = ax/bx


a1/x =  root(x)(a)


(root(x)(a) )y/x = (root(x)(a) )


root(x)(a) root(x)(b) = root(x)(ab)



Example sums for Exponents for Real Numbers

Example 1:

Find the exponential form of (25)1/17

Solution of example 1:

Given = (25)1/17

Exponential form  =  root(17)(25)

Example 2:

Simplify 25225 / 53253

Solution of example 2:

Given 25225 / 53253

Now we use the exponents formula for real numbers

By using

(i)axxx ay = ax+y

(ii)axxx bx = abx

We get

= 253 / (5(25))3

= 253 / 1253

= 253 / (5)3(25)3

Cancel the same variables

Now, we get

= 1 / 53

= 1 / 125

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Example 3:

Solve ( root(3)(root(2)(64))92xx 62) / ((43)2+(5/25)2)

Solution of example 3:

Given = ( root(3)(root(2)(64))92xx 62) / ((43)2+(5/25)2)

Now we use the exponents formula for real numbers

By using below laws of exponents formula we can simplify the given data.

(i) root(x)(root(y)(a)) = root(xy)(a)

(ii)axxx bx = abx

(iii)(ax)y = axy

(iv)(a/b)x = ax/bx

= ( root(3)(root(2)(64))92xx 62) / ((43)2+(5/25)2)

Use the (i) formula

We get

= root(6)(49))92xx 62) / ((43)2+(5/25)2)

Use the (ii) formula

We get

= ((root(6)(64))542) / ((43)2+(5/25)2)

Use the (iii) formula

We get

= ((root(6)(64))542) / ((46)+(5/25)2)

Use the (iii) formula

We get

= ((root(6)(64))542) / ((46)+(52/252))

= (2 (542) / ((256) + (25 / 252))

Cancel the same elements

We get

= 2(542) / ((256) + (1 / 25))

= 2(542) / ((6400 + 1) / 25 )

= 2(2916) / (6401/25)

= 2(2916)(25) / (6401)

= 50(2916) / 6401

= 145800 / 6401

= 22.77

What are Perpendicular lines

Two intersecting lines will have four angles formed at the intersection points. If all the four angles are equal, then the two lines are said to be perpendicular to each other. We already know by linear postulate theorem that the two vertically opposite angles are equal. Hence if these two lines are perpendicular, then all four angles are 90 degrees.

Examples of perpendicular lines:

In the graph paper, The X-axis and Y-axis are perpendicular.
In an ellipse two axes, minor axis, and major axis are perpendicular.
For a line segment, any shortest line from a point outside the circle is perpendicular.
Tangent and normal to any curve are perpendicular lines.

Slopes of two perpendicular lines:  In coordinate Geometry, when two lines are perpendicular, the product of the slopes of the lines is -1.  This property has a lot of applications in finding the equation of perpendicular lines, length of perpendicular segment from a point to a given line, etc.

For any curve in a graph with equation y = f(x), the slope of the tangent is defined as the rate of change of y with respect to x at that point. The normal to this curve at this point is perpendicular to the tangent line.

Example:  In a circle, with centre at the origin and radius 3, the equation will be of the form (x)²+(y)² = 3². Take any point say (0,3). To find the tangent, we have to find dy/dx.

Differentiating, 2x+2y  =0

Hence, the slope of the normal is perpendicular to x axis or parallel to y axis.

Example for Perpendicular lines from a point to a line

Let AB be a line with coordinates (1,2) and (3,4).  Measure the length of perpendicular line from (-1,1) to this line segment.

We know that the perpendicular line from (-1,1) has a slope  of -1/slope of AB.

Equation of AB is (x-1)/(3-1) = (y-2)/(4-2)  Or x-1 = y-2 Or y = x+1

Slope of AB passing through (1,2) and (3,4) is 4-2/3-1 =1.

Slope of perpendicular line to AB is -1.

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Since the perpendicular line passes through (-1,1) equation of the perpendicular is y-1 = -1(x+1)  or y =-x -1 +1 or y = -x.

To get the foot of the perpendicular line on AB, we solve the two equations by substitution method.

y = x+1 = -x   This on simplification gives 2x=-1 or x=-1/2.

Since y = -x , we have y = +1/2,

So, foot of the altitude from the point (-1,1) is (-1/2,1/2).

The length of the perpendicular segment is between (-1,1) and (-1/2,1/2) is

√[ (-1/2+1)²+(1/2-1)²] = √(1/4+1/4)   =  √(1/2) = 1/1.414 =0.707 approximately.

Examples of Concave Polygon

The concave polygon has single interior angle which is more than 180 degree. The concave method is drawing few of the straight line only. The concave polygon is used in many interior angles. The three sided polygon denoted the triangle, triangle cannot be a concave polygon. In this article we shall discuss the examples and properties of concave polygon.

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Properties of concave polygons:

The concave polygons do not calculate the exterior angle. The opposite meaning of the concave is called the convex. The polygon is normally declared as concave polygon, but not the convex polygon. The interior angle must be reflex angle of the concave polygon.

The diagonal of the concave polygon is the line through the outside of the polygon. The star polygon is the example of non simple polygon. It must  contain the concave polygon with minimum of four sides. The area of the concave polygon and irregular polygons are same.

Example of the concave polygon

The first example of the concave polygon is declaring the following figure. The concave polygon connected the line of the interior angle side. The important part is declaring interior. There are only  two points selected inside the diagram. Do not select the one point which is inside of the angle and another point is outside of the diagram.

diagram repsent the example of the concave polygon

The inside of the polygon does not enter the line P and Q. The line A and B is entering the inside of the polygon. So that it is said to be concave polygon. The minimum one line segment will not present inside the diagram, this part of the diagram is called the concave.


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The above example is commonly called the concave polygon. The next example of the concave polygon is declaring the following diagram. In figure the blue line segment is declared as two points enter into the inside of the diagram. That line through the outside of the diagram.

Learning Disjunction

Definition:

Logical disjunction is an operation on the two logical values, typically the values of two propositions, that produces a value of false if and only if both of its operands are false. More generally a disjunction was a logical formula that can have one or more literals separated only by ORs. A single literal is frequently considered to be a degenerate disjunction.

Properties:

Associativity :

aV (bVc) = (aVb) V c

In mathematics, associativity was a property of some binary operations. It means that, within the expression containing two or more occurrences in a row of the same associative operator, the order in which the operations are performed does not matter as long as the sequence of the operands is not changed. Consider for instance the equation

(5+2) + 1 = 5 + (2+1) = 8

Commutativity:

In mathematics, commutativity is the assets that changing the order of something does not change the end result. It is a primary property of many binary operations, and many mathematical proofs depend on it. The commutativity of easy operations, such as multiplication and addition of numbers, was for many years implicitly assumed and the property was not named until the 19th century when mathematics started to become formalized.

Distributivity:

In mathematics, and in particular to abstract algebra, distributivity is a property of binary operations that generalizes the distributive law from elementary algebra. For example:

2 × (1 + 3) = (2 × 1) + (2 × 3).

Idempotence:

Idempotence is a property of certain operations in mathematics and computer science. Idempotent operations are operations that can apply multiple times without changing the result. The conception of idempotence arises in a number of places in abstract algebra and functional programming.

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Monotonic function:

In mathematics, a monotonic function in which conserve the given order. This concept first arose in calculus, and was shortly generalized to the more abstract setting of order theory.

Symbol:

The mathematical symbol for logical disjunction varies in the text. In addition to the word "or", the symbol "V", deriving from the Latin word vel for "or", is commonly used for disjunction. For example: "A V B " is read as "A or B ". Such a disjunction is false if both A and B are false. In all other cases it is true.

All of the following are disjunctions:

A V B

¬A V B

A V ¬B V ¬C V D V ¬E

Cumulative Frequency Distribution

The total frequency of all classes less than the upper class boundary of a given class is called the cumulative frequency distribution .Relative frequency is the fraction of total number of elements .There are two types of Cumulative  frequency distributions, as follows

1.    Less than cumulative frequency distribution

2.    More than cumulative frequency distribution

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Cumulative distribution function in cumulative frequency distribution

In common terms, the cumulative frequency distribution is the sum of all the frequencies. A cumulative frequency distribution is a sum of a set of data showing the frequency or number of items less than or equal to the upper class limit of each class. The Cumulative distribution function (CDF), describe the Probability distribution of random Variable X

The random variable X is given by

X -> Fx(x) =P(X<=x),

P(X<=x) =The probability that the random variable x takes on a value less than or Equal to x.

FX(b) − FX(a) if a < b.

Where

F for cumulative distribution function

The probability density function f can write as follows:

F(x) = $\int_{-\infty}^x f(t)\,dt$

Where

f(t) means probability density function

Applications and uses of cumulative frequency distribution

Cumulative frequency gives the total number of outcomes that occurred up to some value. Cumulative frequencies are used in risk or reliability analysis. In frequency distribution, every bin has the number of values that lies within the range of values that define the bin. In a cumulative distribution, every bin has the number of data that falls within or below the bin. A graph contains a frequency distribution on the left, and a cumulative distribution of the same data on the right is called a cumulative frequency distribution

The main advantage of cumulative frequency distribution is that one doesn’t need to decide on a bin width or the frequency distribution analysis dialog, can choose among  number of ways to graph the resulting data.


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practice Constant

The values can’t be change. Constants it’s also called variable. In math, constant is a number, But sometimes we can also take the variable as a constant. For example,  In this equation x2+5x+3 = 0, 3 is a constant.  In this equation x+5=-20, 5 and -20 are constants. Now we are doing to practice some constant problems.

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Practice constant Problems:

Practice problem 1:

X2+4x+k= 24. If x= 2, solve for k .

Solution:

Step 1: Substitute x value in the given equation.

Step 2: So we get (2)2+4(2)+k=24.

Step 3: Here we need to simplify this.

Step 4: 4+8+k=24.

Step 5: When we add we get 12+k=24.

Step 6: Subtract 12 on both the sides so, 12-12+k= 24-12.

Step 7: Therefore answer is k=12.


Practice problem 2:

X+4y-3z+r= 51...if x=9,y=3 and z=2 ,solve r.

Solution:

Step 1: Substitute x, y and z value in the given equation.

Step 2: So we get, 9+4(3)-3(2)+r= 51.

Step 3: Now we need to simplify this equation.

Step 4: 9+12-6+r=51.

Step 5: When we simplify we get 15+r= 51.

Step 6: Divide using 15 on both the sides.

Step 7: Therefore, the answer is 51/15.


Practice problem 3:

M3+m2+6m+S= 72. If m= 2, solve for S.

Solution:

Step 1: Substitute m value in the given equation.

Step 2: So we get (2) ^3+ (2) ^2+6(2) +S=72.

Step 3: Now we need to simplify this equation.

Step 4: 8+4+12+S=72.

Step 5: When we add we get 24+S=72.

Step 6: Subtract  24 on both the sides so, 24-24=72-24 .

Step 7: Therefore the value of s is 3.

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Practice problem 4:

C2+15c+g=195. If c= 3 then find g.

Solution:

Step 1: Substitute m value in the given equation.

Step 2: So we get 32+15(3)+g=195.

Step 3: Here we need to simplify this equation.

Step 4: 9+45+g=195.

Step 5: After addition we get 54+g=195.

Step 6: Divide using 54 on both the sides.

Step 7: So the value of g =195/54.

work out problems:

X2+19x+t=121 if x=12, find the value of t?
A3+4a+z= 147 if a =5, find the z value?
H2+h+m= 12 if h=6, find the m value?
F4+8f+n=546 if f=4, find the n value?

Math Operations

Operator in math is a function which acts on two values ( operands). These operators are known as binar operators as they need two operands. There are four binary operators namely +  ,  -   ,  x   , /. Operator are mainly consists of number of terms. For example, the assignment operator are used for assigning the values to a particular variable.

The four fundamental operations on numbers are Addition, Subtraction, Multiplication and Division. The operator of Addition is ‘+’, Operator of Subtraction is ‘-’, Operator for multiplication is ‘x’ and Operator for division is ‘÷’. Let us see about arithmetic operations with appropriate operator.
Explanation for various types of operator statement:

1. '+' is an  operator statement:

For this operator, the addition operation and the string concatenation operation are included.
Addition operation is used for adding the two values that are given.
String concatenation are used for joining the two statements.

2. ++ is an increment operator:

For this, the values are incremented according to the given values.

3. -- is an decrement operator:

For this, the values are decrement according to the given values.

4. x  is a multiplication operator:

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For this, the multiplication function and the pointer function are included.
Multiplication function is used to multiply the two given values.
Pointer are used for denoting the reference variable.

5. / is a division operator:

For this, the values are divided for getting the remainder and the quotient factor.

6.  ^  exponential operator:

Exponential operator is used for raising the power to a function.

7. -  mathematical operator:

For this, the subtraction and the negation process are included in this operator statement.
Subtraction operation is used for finding the difference of the given two numbers.


Examples using math operations:

Addition Operator ‘+’: For adding two or more numbers we use the ‘+’ operator. This operator is known as plus.Normally additions have two operands and one operator. For example x + y here ‘x’ and ‘y’ is called as operator and the symbol’+’ is called as operand.

Ex: Add 5 and 6.

Sol:

5 + 6 = 11

Where, ‘+’ is the operator of addition.

5 and 6 are operands.

11 is the resultant of the operation of addition.

Subtraction Operator ‘-’: For subtracting two or more numbers, the operator used is ‘-’.  This operator is known as minus.

Ex : Subtract 11 and 5.

Sol:   11 – 5 = 6.

Where, ‘-’ is the operator of subtraction.

11 and 5 are operands.

6 is the resultant of the operation of subtraction.

Multiplication Operator ‘x’: For Multiplying two or more numbers, the operator used is ‘x’.  This operator is known as into.

Ex : Multiply 12 and 5.

Sol:             12 x 5 = 60.

Where, ‘x’ is the operator of multiplication.

12 and 5 are operands.

60 is the resultant of the operation of multiplication.

Division Operator ‘÷’: For dividing two or more numbers, the operator used is ‘÷’.  There is no special name for division operator.

Ex : Divide 10 by 5.

Sol:   10 ÷5 = 2.

Where, ‘÷’ is the operator of Division.

10 and 5 are operands.

2 is the resultant of the operation of division.

Additional Operators:

‘√’:

This operator is known as the square root.

Ex: Find Square root of 4?

Sol: √4 = 2.

Where, ‘√’ is the operator of square root.

4 is the operand.

2 is the resultant of the operation of square root.

Order of operations in math:

The order of operation is square root, division, multiplication, Subtraction and addition.

Ex: Calculate 5 + 6 – [(√4 ÷2) x 3].

Sol:  5 + 6 – [(2 ÷ 2) x 3]

5 + 6 – [1 x 3]

5 + 6 – 3

5 + 3

7

Steps and examples for addition operation in math:

Simple addition uses the following steps to adding the numbers.

Step 1: Add the first place number from the right (first number) first after that take over the carry to next step.

Step 2:   Add the ten’s place number (means second digit) as well as also add the carry from the first step. Now note down the answer and over the carry.

Step 3  :Carry on this process on until reach the end place of the given number.

The followings are the some of the types of addition

Add single digit number with single digit
Add single digit number with double digit
Add three one digit numbers
Add double digit number with three digit number
Add double digit number with double digit number
Add treble digit number with treble digit number etc….

Example for Simple Addition :

Add single digit number with single digit:

Ex: Add 5 with 9

Sol:

5
(+) 9
-----------     14

Ex 2: Adding the following three one digit numbers.

9, 6 and 3

Sol:

Given numbers 9, 6 and 3

Step 1:

Add the first two numbers

9

(+)      6

----------------------

1 5

Step 2:

Add the next number with the previous step sum or total.

Previous step total =15

Third number =3

1 5

(+)        3

----------------------

1 8

Therefore 9 + 6 +3 =18

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Ex 3: Add the following two numbers: 54 with 16

Sol:

Given numbers 54 and 16

Step 1:

Add the first digit (ones place) numbers (4+ 6 = 10) then over the carry (here 1 is the carry)

5 4
(+)   1 6
-----------
0

Step 2:

Add the second digit (tens place) numbers (5 +1 = 6) the add with previous step carry (6 +1 =7)

5 4
(+)   1 6
-----------
7 0

Finally we get 54 +16 = 70

Ex 4: Add the following two numbers: 15948 and 69741

Sol:

Given 15948 and 69741

1 5 9 4 8
(+)     6 9 7 4 1
------------------

Step 1:

Add the one place number then over the carry (8+1=9 carry =0)

1 5 9 4 8
(+)    6 9 7 4 1
------------------
9

Step 2:

Add the tens place number then over the carry (4+4 = 8 carry 0)

1 5 9 4 8
(+)    6 9 7 4 1
------------------
8 9

Step 3:

Add the hundreds place number then over the carry (9+7 =16 carry 1)

1 5 9 4 8
+)    6 9 7 4 1
------------------
6 8 9

Step 4:

Add the thousands place number (5+9 = 14) then add with carry from the above step (14 +1 = 15 carry 1)

1 5 9 4 8
(+)    6 9 7 4 1
------------------
5 6 8 9

Step 5:

Add the ten thousands place number (1+ 6=7) then add with previous step carry (7 + 1 =8)

1 5 9 4 8
(+)    6 9 7 4 1
------------------
8 5 6 8 9

Therefore 15948 + 69741 = 85689

Ex 5: Add 72 with 4.12

Sol:

Given numbers 72 and 4.12

Step 1:

Change the whole number into decimal form

Whole number =72

Decimal form = 72.00

Step 2:

7 2 .0 0

(+)       4. 1 2

--------------------



Start the addition process from the right

Step 3:

Add the first place number from the right and over the carry to the next step (0 +2 =2 carry =0).

7 2 .0 0

(+)       4. 1 2

--------------------

2



Step 4:

Add the next place number from the right and then add the carry with is step total. Finally over the carry to the next step (here 0 +1 =1 carry =0)

7 2 .0 0

(+)       4. 1 2

--------------------

. 1 2

Step 5:

Add the next place number from the right and then add the carry with is step total. Finally over the carry to the next step (here 2 +4 =6 carry =0)

7 2 .0 0

(+)       4. 1 2

--------------------

6. 1 2

Step 6:

Add the next place number from the right and then add the carry with is step total. Finally over the carry to the next step (here 7 +0 =7, carry =0)

7 2 .0 0

(+)       4. 1 2

--------------------

7 6 .1 2

72 + 4.12 =76. 12

Examples on multiplaction operation in math:

Ex 1: Multiply 10 with 5.

Sol:     10 x 5 = 15.

Where, ‘x’ is the operator of multiplication.

10 and 5 are the operands of multiplication.

15 is the resultant of the multiplication operation.



Ex 2: Simplify 10(5) + 20(2) + 9(3) + 4(1).

Sol:     15 + 40+ 27+ 4

86

Ex 3: A train has 9 carriages. There are 42 seats in each carriage. How many seats are there on the train?

Sol:   Number of carriage = 9

Number of seats in each carriage = 42

Multiply number of carriages with number of seats in each carriage.

Number of seats in the train = 9 x 42

= 378.

Ex 4: There are five cupboards in a room. Each cupboard has 7 racks. Find how many racks totally the room has.

Sol:      Number of cupboards = 5.

Number of racks in each cupboard = 7.

To find total number of rack, multiply the number of cupboard with number of racks in each cupboard.

Total number of rack = 5 x 7

= 35.

Ex 5: There are six baskets full of apple in a lorry. 1st basket contains 35 apples, 2nd and 3rd basket contains 40 apples, 4th, 5thth basket contains 50 apples. Calculate the total number of apples in the lorry. and 6

Sol:      35 apples in the 1 basket

40 apples in the 2 basket

50 apples in the 3 basket

Total Number of Apples = 35(1) + 40(2) + 50(3)

= 35 + 80 +150

= 265