Addition Property of Order

Introduction of Addition Property of Order:

Let now explain about the addition property with example.

Example: c < d, the expression shows that the number c is less than the number. When add the number ‘’a’’ on the both side of the expression, the relation does not change.

C + a < d + a; Thus by using addition property we can order the number from smallest to largest or vice versa (ascending or descending order) otherwise to find the number is equality or inequality.

Example Problem – Addition Property of Order:

Example 1:

Solve the expression by using addition property of order X – 5 < 17 and check the expression by getting the value of X.

Solution:

Given: The expression is X – 5 < 17

Step 1: By using addition property we can add the number 5 on both sides of the expression, we get

X – 5 + 5 < 17 + 5

Step 2: Opposites sign of the number 5 to be canceling each other.

Step 3: So we get X is less than 22

X < 22

Check the expression X – 5 < 17 by putting the value of X = 22

Step 1: The given expression X – 5 < 17

Case 1:

If the value of X = 22, then the expression get equality form such as 22 – 15 = 17

Case 2: We get the X value as less than 22, so the value of X is 21 or less than 21 (may be up to negative infinity)

Let we put the X value is 21, we get

21 – 5 < 17

16 < 17, thus we proved that the number 16 is less than 17.

Example 2:

Prove by using then addition property of the number 7 is greater than 5.

Solution:

Given: 7 > 5

Step 1: First we add any number; let take the number 2 is added to the left hand side of the expression.

7 +2 = 9

Step 2: Then we add the number to the right hand side of the expression, we get

5 + 2 = 7

Step 3: Compare the result of the above two steps.

Thus the number 9 is greater than 7; 9 > 7

Therefore we proved that the number 7 is greater than 5

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Practice Problems – Addition Property of Order:

1. Solve the expression by using addition property of order X – 5 < 15

Answer: X = 19 or less than 19

2. Solve the expression by using addition property of order Y – 5 = 18

Answer: Y = 23

Types of Sampling Bias

Introduction :                

Sampling bias in static’s caused by some members of the population to be less likely to be included than others, where it results in a biased sample, It results in a biased sample, a non-random sample of a population in which all participants are not equally balanced or objectively represented If the bias makes estimation of population parameters impossible, the sample is a non-probability sample.

[Source Wikipedia].

Sampling Bias Types:

The types of sampling bias are :

Self modulate bias
Pre screening bias
Exclusion  bias
Over matched bias.
Self modulate bias:

The self modulated bias is used to deals with the group of certain people to know them about the previous and present calculations by the percentages using statics , This helps the people to know about the gain and loss percentage and the income difference between previous and past.

Example:

In 1988, the people spend tax as 1.25% for rupees 12000,but now they are paying 8% Means what is the percentage difference between Both ?

Solution:

For 12000 in 1988 1.25 % means = 150 rupees.

Now they are sending 8%

So, for 8% in 12000 means = 960 rupees.

The difference between the previous and present is 960 – 150 = 810 rupees.

For calculating the difference between the both self modulated biases is used.

Pre screening bias:

Pre screening bias is of trial participants, or advertising for volunteers within particular groups, this sampling is used to give the advantages and the disadvantages of the specific events about their weights loss sessions. 
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Bias Types :

Exclusion bias:

It occurred by the particular group of sampling present in the bias, for example the exclusion of subjects who have recently migrated into the study area , Excluding subjects who move out of the study area during follow-up is rather equivalent of dropout or nonresponsive, a selection bias in that it rather affects the internal validity of the study.

Over matched bias:

The over matched bias in sampling is used to tell about the features about the exposure terms present in radical terms, The over matched group becomes more similar to the cases in regard to exposure than the general population.

Multiplying Binary Numbers

Introduction of Binary Numbers:

Binary numbers said to be the number which operates in base 2 and the digit of the number would be consist of 0 and 1 only. This binary numbers are also termed as binary system. For example: 10, 1010, 112 and 10012

It is a method of writing numbers by using two digits ,0 and 1.The 'binary system' of numeration was used early in the seventeenth century by Thomas Harriot.Each successive bit position in binary number reading from right to left has place value of 1,2,4,8,16 and so on.Binary systems are used in digital electronics. Many electronics instruments are designed using the concept of the binary systems. In this topic we will learn to write the binary numbers and also binary maths like binary addition, binary subtraction, binary multiplication and binary division.

Example:- The numbers 0 to 10 are represented in binary system as:

0  --> 0                       (0x1=0)

1-->01                       (1x1+0x2=1)

2-->10                       (0x1 +1x2=2)

3-->11                       (1x1 +1x2=3)

4-->100                     (0x1 +0x2 +1x4=4)

5-->101                    (1x1 +0x2 +1x4=5 )

6-->110                    (0x1+1x2+1x4= 6  )

7-->111                    (1x1 +1x2 +1x4 =7 )

8-->1000                  (0x1 +0x2 +0x4 +1x8=8 )

9-->1001                   (1x1 +0x2 +0x4+1x8=9  )

10-->1010               (0x1 +1x2 +0x4 +1x8=10  )

One thing we all remind, that Boolean numbers are not the same number as binary numbers. Because of Boolean numbers were represent an entirely different system of math from real numbers, binary numbers is nothing more than that an alternative notation for real numbers. But the both Boolean and binary are often difficult because both Boolean notation and binary notation use the same two numbers: 0 and 1.

The difference is that Boolean quantities are beaked to a single bit (1 or 0), whereas binary numbers may composed of many bits adding up in place-weighted form value of any finite size. The binary number 100102 ("Eighteen") has no more place in the Boolean world than the decimal number 210 ("two").

Negative Number Binary:

By using several ways we can represent negative numbers. The easiest way to solve use the leftmost digit of the number as a special value to represent the sign of the number: 0 represents positive, 1 represents negative. For example, a value of positive 12 (decimal) can be denoted as 01100 in binary, but negative 12 (decimal) can be denoted as 11100.

Binary Addition and Subtraction:

Binary Addition:

Adding binary numbers together by,

0 + 0 = 0

0 + 1 = 1

1 + 0 = 1

1 + 1 = 1

The binary numbers are added as in simple addition.whenever a carry results in addition it is prefixed at the start of the result at its M.S.B. or most significant bit position.

Example:     

01  (1)

10   (2)

101   (5)

011    (3)

Answer:    11   (3)   
Carry bit 1000    (8)

1011     (11)

0100     ( 4)

1100

0111

1111     (15)    carry bit   10011    (19)
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Binary Subtraction:

The subtraction of binary numbers is done in the 2-complement notation and then adding the numbers in order to reduce complexity. if a carry results in the M.S.B. position, it is ignored.

Process: The first number is left as it is and the second number i.e the subtrahend is represented in the 2-complement notation. Finally, the process of addition  is carried out.

example:    1001

-        0111

0111 is represented as     1000

+      1

Now                            1001               which is the 2- complemented notation of 0111.

+1001

(1)    0010(2)

Hence , 0010 which is in 2 in its binary form is the desired answer.

Binary Multiplication and Division

Practice Problem – Multiplying Binary Numbers:

1.   a)   11100                b)                1011                c)                101                        d)       111 )  111111(  

+  10101                              -    0011                                 x111

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



Answers:   a) 110001             b) 1000                             c)100011                      d) 1001

2. What is the result for multiplying binary numbers 1011 2by 112?

Answer: 1000012

3. What is the result for multiplying binary numbers 112 by 112?

Answer: 10012

Factoring Cubed Polynomials

Introduction to factoring cubed polynomials
In mathematics, a cubed polynomial is a cubic function in the form of,

f(x) = ax3 + bx2 + cx + d

where a is nonzero; or in other words, a Factoring Polynomial of degree three. The derivative of a cubic function is a quadratic function. The integral of a cubic function is a quartic function. (Source: From Wikipedia).

Here we are going to learn factoring of cubed polynomials.

Example Problems for Factoring Cubed Polynomials

Example 1

Find the roots of the cubed polynomial 49x3 - 35x2 + 56x - 40 by factoring

Solution

The factors of the given cubed polynomial can be found by factoring by grouping,

First split the terms into 2 groups,

49x3 - 35x2 + 56x - 40 = (49x3 - 35x2) + (56x - 40)

Now, factor out 7x2 from the first term (49x3 - 35x2) and 8 from the second term (56x - 40)

= 7x2(7x - 5) + 8(7x - 5)

Now, factor out (7x - 5)

= (7x - 5)(7x2 + 8)

So, (7x - 5)(7x2 + 8) are the factors of 49x3 - 35x2 + 56x - 40

Example 2

Find the roots of the cubed polynomial 24x3 + 15x2 - 56x - 35

Solution

The factors of the given cubed polynomial can be found by factoring by grouping,

First split the terms into 2 groups,

24x3 + 15x2 - 56x - 35 = (24x3 + 15x2) - (56x + 35)

Now, factor out 3x2 from the first term (24x3 - 15x2) and 7 from the second term (56x + 35)

= 3x2(8x + 5) - 7(8x + 5)

Now, factor out (8x + 5)

= (8x + 5)(3x2 - 7)

So, (8x + 5)(3x2 - 7) are the factors of 24x3 + 15x2 - 56x - 35

Example 3

Find the roots of the cubed polynomial 24x3 - 64x2 - 21x + 56

Solution

The factors of the given cubed polynomial can be found by factoring by grouping,

First split the terms into 2 groups,

24x3 - 64x2 - 21x + 56 = (24x3 - 64x2) - (21x - 56)

Now, factor out 8x2 from the first term (24x3 - 64x2) and 7 from the second term (21x - 56)

= 8x2(3x - 8) - 7(3x - 56)

Now, factor out (3x - 8)

= (3x - 8)(8x2 - 7)

So, (3x - 8)(8x2 - 7) are the factors of 24x3 - 64x2 - 21x + 56

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Practice Problems for Factoring Cubed Polynomials

Practice problem 1

96x3 - 84x2 + 112x - 98

Answer: 2(8x - 7)(6x2 + 7)

Practice problem 2

28x3+ 16x2 - 21x  - 12

Answer: (7x + 4)(4x2 - 3)

Rules of Simplifying Radicals

Introduction to rules of simplifying radicals:

Radical symbol used to indicate the square root or nth root. Radical is an algebraic group, a concept in algebraic group theory. A branch of mathematics, a radical of a ring is an ideal of elements of the ring. Radical of a module, in the theory of modules, the radical of a module is a component in the theory of structure and classification. The radical sign is v .                 (source: wikipedia)

Rules of Simplifying Radicals:

Radical is a perfect and significant idea in abstract algebra. The cubic root of x know how to be expressed as   `root(5)(x)`

`root(n)(ab)`  =  `root(n)(a)` `root(n)(b)` 
`root(n)((a/b))` =  `(root(n)(a))` / `(root(n)(b))`
`root(n)(x)`  = (x)(1/n)    and      `root(n)(x^n)`  = x(n/n) = x
`root(n)(a)`m =( `root(n)(a)` )m = (a1/n )m = am/n
` sqrt(-1)` ×  `sqrt(-1)` = -1 where as  `sqrt((-1)*(-1))` = 1


Examples for Rules of Simplifying Radicals:

Example 1:

Simplifying the rules of given radical expression:    `sqrt((9x^2)/(y^5z^5)) `

Solution:

Step 1: Multiplying and divide by  yz

Step 2:      ` sqrt((9x^2)/(y^5z^5))` = `sqrt((9x^2 * yz) / (y^5z^5 *yz))`

Step 3: Multiply the variable with exponent

= `sqrt((9x^2yz) / (y^6z^6))`

Step 4: Square root of x2y6z6 = xy3z3

Step 5: Square root of 9 = `sqrt(3 * 3) ` = 3

=` ((3x) / (y^3z^3)) sqrt(yz)`

so the answer is  ` ((3x) / (y^3z^3)) sqrt(yz)`

Example 2:

Simplifying the rules of radical expression: `sqrt(b^4/a^7) ` + `sqrt(a)`  = `sqrt(a)`  ` ((b+a^3)/a^3)`

Solution:

Step 1:  Multiplies and divides by  a in first term

Step 2:      `sqrt(b^4/a^7)` = `sqrt ((b^4* a) / (a^7 *a))`

Step 3: Multiply the variable with exponent

= `sqrt ((b^4 * a) / (a^8))`

Step 4:  Square root of b4 a = b2 `sqrt(a)`

Step 5:   Square root of a8 = a4

`sqrt ((b^4 * a) / (a^8))`   = ` b^2/a^4 sqrt(a)`

Step 6:  Adding the both term  `sqrt(b^4/a^7) ` + `sqrt(a)`    =` b^2/a^4 sqrt(a)`  +  `sqrt(a)`

= `sqrt(a)` ( `b^2/a^4 ` +1 )

Step 7:         = `sqrt(a)`  ` ((b^2+a^4)/a^4)`

Hence the given radical expression has been proved

Example 3:

Simplifying the rules of  the given radical expression: `sqrt(5x^3) (sqrt(9x^3))`

Solution:

Step 1: the given radical expression is

=   `sqrt(5x^3) (sqrt(9x^3))` 

Step 2:     `sqrt(5x^3) (sqrt(9x^3))`

=` (xsqrt(5x) (3)xsqrtx)`

Step 3:       `3x^2 sqrt(5 * x * x) `                                     

Step 4:       3x2x `sqrt(5)`                                                                 

= 3 x3`sqrt5`  

so the answer is  3 x3`sqrt5`  

Example 4:

Simplifying the rules of the given radical expression :  `(root3 16)^3 * (sqrt49)^2`

Solution:

Step 1: Given radicals ` (root3 16)^3 * (sqrt49)^2`

Step 2:     = `(root3 16)^3 * (sqrt49)^2`

Step 3: Radical rules is ( `root(n)(a)` )m = am/n

Step 4:    = 163/3 * 492/2

161 * 491

Step 5:       16 * 49

Step 6:         so the answer is 784

Trapezoid Shape

Introduction about trapezoid shape:

In geometry, a four-sided figure with one pair of parallel sides is referred to as trapezoid. There is also some disagreement on the allowed number of parallel sides in a trapezoid. At issue is whether parallelograms, which have two pairs of parallel sides, should be counted as trapezoids. (Or) the trapezoid is a quadrilateral having exactly one pair of parallel sides, thereby excluding parallelograms. The shape of the trapezoid is shown in the below.

(Source – Wikipedia)

Isosceles Trapezoid:

The isosceles trapezoid has the equal length of parallel side AD and BC and same base angle. It has four internal angles. The sum of its four internal angles is 1800. The shape of the trapezoid is shown below.

Formula to find area of trapezoid shape:

Area of the trapezoid (A) = 1/2 x h x (a + b) square units

The formula for area of both trapezoid and isosceles trapezoid shape are same.

Trapezoid Shape - Example Problems:

1. Find the area of trapezoid whose height 4 cm, side a=8 cm and side b= 13 cm

Solution:

Given:

Height (h) = 4 cm

Side a= 8 cm;   b=12 cm

Formula:

Area of trapezoid (A) = 1/2 x h x (a + b) square units

=1/2 x 4 x (8+12)

=1/2 x 4 x 20

=1/2 x 80

= 80 /2

Area of trapezoid (A) = 40 cm2

2. Find the area of trapezoid whose height 5 cm, side a= 9 cm and side b= 12 cm

Solution:

Given:

Height (h) = 5 cm

Side a= 9 cm;   b=12 cm

Formula:

Area of trapezoid (A) = 1/2 x h x (a + b) square units

=1/2 x 5 x (9+ 12)

=1/2 x 5 x 21

=1/2 x 105

= 105/2

= 52.5

Area of trapezoid (A) = 52.5 cm2

3. Find the area of trapezoid whose height 7 m, side a=11 m and side b= 18 m

Solution:

Given:

Height (h) = 7 m

Side a= 11 m;   b=18 m

Formula:

Area of trapezoid (A) = 1/2 x h x (a + b) square units

=1/2 x 7 x (11+ 18)

=1/2 x 7 x 29

=1/2 x 203

= 203 / 2

= 101.5

Area of trapezoid (A) = 101.5 cm2

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Tangent Line

A straight line that handles a curve at a point is known as tangent line to the curve at that point. If a circle and a line in the same plane contact each other at only one point, then, the line is called a tangent to the circle. If a tangent to a circle in a plane is also a tangent to another circle in the similar plane then, it is called a common tangent to the circles.

Tangent Line Equation

1) y = x 2 - 4x, find the tangent line function for the given equation, where the tangent line is equivalent to the x axis

Solution:           

                    
The slope of x  axis = 0.

The angle of y  axis = x^2 - 4x

The original derived of y,

y ' = `dy/ dx`

= 2x - 4

Currently all values of x the value of

The initial derivative of y’ = `dy/ dx` = 0

So, 2x - 4 = 0

x = 2

Answer of the given equation is

x = -2 and x = 2

Calculate the y value by using the x value so, y = x^2 - 10x

for x = -2, y = 16

for x = 2, y = -16

The points of tangent lines are equivalent to the x axis are: (-2, 16) and (2,-16).

The graph of y = x^2 - 4x

Example Problem for Tangent Line Function:

2) Establish the tangent lines to the function: y = x^2, when x = 2.

Solution:

At x = 2, y = x^2 = 22 = 4.

So (x1, y1) = (1, 4)

The 'm' at x = 2 is established by determining the derived of the function at that point.

For y = x^2, `dy/dx` = 2x, At x = 2, `dy/dx` = 2×2 = 4. So 'm' = 4.

The tan line function is given by:

Y = m(x-x1) + y1

Alternate, we have: y = 4(x - 1) + 1

Intensifying and shortening: y = 4x - 4 + 1

So, y = 4x – 3

This example of tangent line will help you to find an equation of the tangent line. For more help you can also connect to an online tutor and gain your help.

Practice Abstract Algebra Problems

Introduction to practice abstract algebra problems:

In algebra, we use a, b,c.... x, y and z to denote numbers. Performing addition, subtraction, multiplication, and division of roots on this symbols and real numbers, we obtain what are called algebraic expressions

The following are some examples for algebraic equations:

2x + 3 = x + 6,

Symbols in an algebraic expression are called  variables in Algebra

In Algebra  the word can be understood by a simple example. In the equation x + 5 = 9, the left hand side is the addition (sum) of two parts x and 5. If we add  (–5) to each side of the equation, we get

(x + 5) + (–5) = 9 + (–5) or x + [5 + (–5)] = 9 – 5 or x + 0 = 4 or x = 4.

Here 9 and −5 are reunited to get 4. It is said to be  algebra.  

Practice Problems in Algebra

Abstract algebra problems using Subtract 2x^3 – 3x^2 + 1 from x^3 + 5x^2 – 4x – 6 for practice.

Solution:  Associative and Distributive properties, we have

(x^3 + 5x^2 – 4x – 6) – (2x^3 – 3x^2 + 1)

= x^3 + 5x^2 – 4x – 6 – 2x^3 + 3x^2 - 1

= x^3 – 2x^3 + 5x^2 + 3x^2 – 4x – 6 - 1

= (x^3 – 2x^3) + (5x^2 + 3x^2) + (–4x) + (– 6 - 1)

= –x^3 + 8x^2 – 4x – 7.

Abstract algebra problems using find the product of x^3 – 2x^2 + 4 and 2x^2 + 3x – 1 .

Solution: (x^3 – 2x^2 +4) (2x^2 + 3x – 1)

= x^3 (2x^2 + 3x – 1) + (– 2x^2) (2x^2 + 3x – 1) + (+ 4) (2x^2 + 3x – 1)

= (2x5 + 3x4 – x^3) + (– 4x4 – 6x^3 + 2x^2) + (+ 8x^2 + 12x - 4)

= 2x5 + 3x4 – x^3 – 4x4 – 6x^3 + 2x^2 + 8x^2 + 12x - 4

= 2x5 + (3x4 – 4x4) + (–x^3 – 6x^3) + (2x^2 + 8x^2) + (+ 12x) -4

= 2x5 – x4 – 7x^3 + 10x^2 + 12x - 4.

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Problem Using the Identities for Practice:

Abstract algebra problems using Identities (a + b + c)2 – 2(ab + bc + ca) = a2 + b2 + c2 for practice.

Solution:

Abstract algebra problems (i) (2x + y + 2z)2

(2x + y + 2z)2 = [(2x) + y + (2z)]2 = (2x)2 + y2 + (2z)2 + 2(2x)y + 2y(2z) + 2(2z)(2x)

= 4x^2 + y2 + 4z2 + 4xy + 4yz + 8zx.

Abstract algebra problems (ii) (x – 2y + z)2

(x – 2y + z)2 = [x + (–2y) + z]2= x^2 + (–2y)2 + z2 + 2x(–2y) +2 (–2y)z + 2zx

= x^2 + 4y2 + z2 – 4xy – 4yz + 2zx.

Abstract algebra problems (iii) (2p – 3q – r)2

(2p – 3q – r)2 = [(2p) + (–3q) + (–r)]2

= (2p)2 + (–3q)2 + (–r)2 + 2(2p) (–3q) + 2(–3q) (–r) + 2(–r)(2p).

= 4p2 + 9q2 + r2 – 12pq + 6qr – 4rp.

Abstract algebra problems (iv) (2a + 3b − 2c)2

= (2a + 3b – 2c)2 = [(2a) + (3b) + (–2c)]2

= (2a)2 + (3b)2 + (–2c)2 + 2(2a)(3b) + 2(3b)(–2c) + 2(–2c)(2a)

= 4a2 + 9b2 + 4c2 + 12ab – 12bc – 8ca.

Rounding Decimals to the Nearest Hundredth

Introduction to rounding decimals to the nearest hundredth:

Rounding decimals to the nearest hundredth concerns mainly about how to round the decimals to two decimal places. Certain sums involve the remainder that the decimal values after the decimal point grow to infinity. Here our aim is to shorten our result of decimal values to nearest hundredth, for those rounding decimals to the nearest hundredth, certain rules are to be followed. Let us see problems for rounding decimals to the nearest hundredth.

Rounding Decimals to the Nearest Hundredth:

To round the values to nearest hundredth, we follow 2 rules to remember:

Rule One: Find what is our rounding digit is given for and here the rounding decimals to the nearest hundredth s, so see the right side of after the two decimal places. If that digit after 2 decimal places if less than 5, simply remove all the digits to the right of nearest hundredth decimal place.

Rule Two:   Find what is our rounding digit is given for and here the Rounding decimals to the nearest hundredth s, so see the right side of two decimal places. If that digit is 5 or greater than 5, add one to the rounding digit at nearest hundredth decimal place and drop all digits to the right of nearest hundredth decimal place.

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Rounding Decimals to the Nearest Hundredth:

Example 1:

Consider our answer is 11.884721952 but we are asked to give a shorter answer –rounding decimals to the nearest hundredth . At the moment our answer has 9 figures following the decimal point.

Our ‘cut-off point’ comes after the ‘8’ 11.88 | 4721952

this number is greater than 11.88, but lower than 11.89. Look at which is it closer to?

See the 3rd number after the decimal point, that is the number after that ‘cut-off point’.

If the number is 0,1,2,3, or 4 round DOWN to 11.88

If the number is 5,6,7,8, or 9 round UP to the next number 11.89.

Ans:   11.88 to two decimal places.

Example 2:

Consider our answer is 19.367721 but we are asked to give a shorter answer –rounding decimals to the nearest hundredth . At the moment our answer has 6 figures following the decimal point.

Our ‘cut-off point’ comes after the ‘6’ 19.36 | 7721

this number is greater than 19.36, but lower than 19.37. Look at which is it closer to?

See the 3rd number after the decimal point, that is the number after that ‘cut-off point’.

If the number is 0,1,2,3, or 4 round DOWN to 19.36

If the number is 5,6,7,8, or 9 round UP to the next number 19.37.

Ans:   19.37 to 2 decimal places.