Basis Vectors

Introduction to basis vectors:

Definition

Consider the non-zero vectors v1, v2 …, vn in the vector space V over the field F. We call them basis vectors if they satisfy the following two conditions:

The vectors v1, v2 …, vn are linearly independent
Any vector v in the vector space V can be written as a linear combination of the basis vectors. In other words, there exists scalars a1, a2…, an not all zero such that v = a1v1 + a2v2 +…+ anvn I like to share this What are Vectors with you all through my article.


On satisfying the above two conditions we call the set B = {v1, v2 …, vn } as the basis of the vector space V and the elements of B are called the basis vectors.

Example for Basis Vectors

Let V = R3 and consider the vectors:

E1 = (1, 0, 0)

E2 = (0, 1, 0)

E3 = (0, 0, 1)

We will show that the above vectors form the basis of R3.

Let a, b, c be real numbers so that aE1 + bE2 + cE3 = 0

Then a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1) = (0, 0, 0)

Implies (a, b, c) = (0, 0, 0) which means a = b = c = 0

Hence they are linearly independent vectors.

Also any vector v = (x, y, z) in V can be written as a linear combination of E1, E2, E3 as follows: v = xE1 + yE2 + zE3

Hence {E1, E2, E3} are the basis vectors for R3.

Note: We call the above set of basis vectors as the standard basis of R3. It can be extended to Rn where the standard basis vectors are E1, E2…, En.

Remember:  A vector space V is said to be of finite dimension or finitely generated if there exists a finite number of basis vectors in V else it is referred to as infinite dimensional space. Please express your views of this topic math problems 8th grade by commenting on blog.

Exercise to Basis Vectors:

Show that the infinite set S = {1, x, x2, …, xn,...} forms the basis vectors of the vector space F[x] of polynomials over the field F.

P Trend Statistics

Introduction to p trend statistics:

In statistics, the letter “p” stands for probability. p trend statistics is highly used for  normal distribution for calculating  the normal distribution value.

A normal distribution with the variables mean and standard deviation can be converted into a standard normal distribution  is obtained by  converting the normal distribution with the mean and standard deviation by performing the change of scale and origin

For calculating the normal distribution the formula must change to z-scale or z-term from the x –scale. The formula is given as,

` Z=(X-mu)/sigma`

Where `mu` - mean

`sigma` - Standard deviation.

How to Find P Trend Statistics

This example will clearly explain how to calculate p trend statistics.

P (0
= 0.5-0.4772

(    P (0
P (0
P (0
This is the normal distribution value for P (0
Example Problems for P Trend Statistics:

Problem 1:

Find the probability value of P (0
Solution

Given that  P (0
On splitting the above equation the value remains the same.

P (0
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P (0
On adding the values we get the final result as

P (0
Problem 2

Find the probability value of P (-1.15
Solution:

Given that P (-1.15
By using property of normal distribution, we get

P (-1.15
Now applying the symmetry property,

P (-1.15
= 0.3749 + 0.4559

P (-1.15
Problem 3:

Find the probability value of P (-1.55
Solution:

Given that P (-1.55
By using property of normal distribution, we get

P (-1.55
Now applying the symmetry property,

P (-1.55
= 0.4398 + 0.3944

P (-1.55

Practice Set Solutions

Introduction to Practice Set Solutions

A set of elements or information or numbers are referred as data set. Data set can hold any types of data. Data can be represented as tables and graphs. In tables data are stored as rows and columns. Data sets are classified into sequential data set and partitioned data set. In sequential data set data are stored as consecutively. In partitioned data set store the address, so that it is possible to access the data easily. Let us practice the data set solutions in statistics.

Practice Statistics Data Set Solutions - Example

Ex: Find the mean, median, mode, and range for the data set: {20, 27, 15, 16, 42, 28, 20}.

Solution:First arrange the set in ascending order {15, 16, 20, 20, 27, 28, 42}

Mean:Mean or average is the sum of all the elements in the set divided by total number of elements in a set.

Mean     = Sum of elements in a set / Number of elements in a set

= {15, 16, 20, 20, 27, 28, 42}/ 7

= `168/7`  = 24

Therefore mean is 24

Median:Median is the middle value of the data set after the arrangement of ascending or descending order the data set.

Therefore median is 20.

Mode:If any of the value in data set is repeated two or more times then that is referred to as mode.

20 is repeated twice.

Hence, the mode of the given data set is 20.

Range:Difference between the maximum and minimum value of the data set is called the range.

Range        = Maximum value – Minimum value

= 42 – 15

= 27

Therefore range is 27.

Practice Probability Data Set Solutions - Example

Example: When a pair of balanced dice is rolled, and what are the probabilities of getting the sum (1) 12 (2) 12 or 11 (3) 11or 10. Find the solutions.

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

Number of possible outcomes n(S) = 36

Let A be the event of getting sum 12, B be the event of getting the sum 11

and C be the event of getting sum 10.

A = {(6, 6)} n(A) = 1.

B = {(5, 6), (6, 5) n(B) = 2

C = {(4, 6), (5, 5), (6, 4)}n(C) = 3

(1) P (getting sum 12) = P(A) =`(n(A))/(n(S)) = 1/36`



P(12) = `1/36`

(2) P(12 or 11) = P(A or B) = P(A ∪ B)

= P(A) + P(B)          (A and B are mutually exclusive i.e. A∩B=φ)

= `1/36 + 2/36 = 3/36= 1/12`

P (12 or 11) = `1/12`

(3) P(11 or 10) = P(B or C) solving the above the problems also.

= P(B) + P(C) (B and C are mutually exclusive)

= `2/36 + 3/36 = 5/36`

P (11 or 10) = `5/36`

Data Set Solutions – Practice Problems

Solve these practice problems

Practice 1: Find the mean, median, mode, and range for the data set: {25, 32, 20, 21, 47, 33, 25}.

Ans: Mean – 29, Median – 25, Mode – 25, Range – 27

Practice 2: When a pair of balanced dice is rolled, and what are the probabilities of getting the sum (1) 10 (2) 10 or 9 (3) 9 or 8.

Ans: (1)  `1/12` , (2)  `7/36` , (3) `1/4`

Regression with Categorical Data

Introduction regression with categorical data:

In statistics, regression analysis includes any techniques for modeling and analyzing several variables, when the focus is on the relationship between a dependent variable and one or more independent variables. More specifically, regression analysis helps us understand how the typical value of the dependent variable changes when any one of the independent variables is varied, while the other independent variables are held fixed.

(Source: Wikipedia)

Regression with Categorical Data:

Definition regression with categorical data:

A categorical variable describe a exacting quality or characteristic. The data is divided into category and the information together is called categorical data

A categorical variable have a dimension scale consisting of a set of categories. For instance, political philosophy is often measured as liberal, moderate, or conservative. Diagnoses regarding breast cancer based on a mamma-gram use the categories normal, benign, probably benign, suspicious, and malignant.

The development of methods for categorical variables was stimulated by research studies in the social and biomedical sciences. Categorical scales are pervasive in the social sciences foe measuring attitudes and opinions. Categorical scales in biomedical sciences measure outcomes such as whether a medical treatment is successful. Understanding Prime Factors of 245 is always challenging for me but thanks to all math help websites to help me out.

Regression models involve the following variables:

The unknown parameters denoted as ß; this might be a scalar or a vector of length k.
The independent variable, X.
The dependent variable, Y.
A regression model relates Y to a function of X and ß.

Y ˜ f (X, ß)

Example Problem Regression with Categorical Data:

Probability of (X) is the probability that X is true. Probability of (X|Y) is the probability that X is true and given that Y is true.

Two hundred students of a class are classified according to the following 2 by 3 table. Where A, B, and C are mutually exclusive properties.

Status    Section A
Section B    Section C    Totals
Female    40      40    90    170
Male    50      20    20    90

Totals  90      60   110 260

Solution for comparing categorical data:

What is the probability that has chosen from the given table person as female?

P (F) = `170 / 260` = 65 %.

What is the probability that chosen person has section A?

P (A) = `90 / 260` = 35 %.

If chosen students are female, what is the probability that she has section B?

P (B|F) = `60 / 260` = 23 % = `(p(B and F)) / (p(F))` .

If chosen student has section C, what is the probability that the individual is a male?

P (M|C) = `20 / 110` = 18 %   = `(p(C and M))/ (p (C))` .

If chosen section has B or C, what is the probability that the student is a male?

P (M|B or C) = `20 /60` = 33 %.

Standard Deviation Practice Problems

Introduction to standard deviation practice problems:
The variance of a random variable or distribution is the expectation, or mean, of the deviation squared of that variable from its expected value or mean. ( Source - Wikipedia )

The standard deviation is nothing but the square root of its variance.

In this article of standard deviation practice problems, few example problems for finding standard deviation and several practice problems to find standard deviation are given.

Example Problems to Practice Standard Deviation:

Example 1:

Find the standard deviation for the given set of numbers:

{ 24, 20, 17, 31, 43 }

Solution:

Step 1: Mean

Mean   =  ` ( 24 + 20 + 17 + 31 + 43 ) / 5`

=  ` 135 / 5`

=  27

Step 2: Variance

Variance  =   `( (24-27)^2 + (29-27)^2 + (17- 27)^2 + (31-27)^2 + (43-27)^2 )/5`

=   `( (-3)^2 + (2)^2 + (- 10)^2 + (4)^2 + (16)^2 )/5`

=   `(9+4+100+16+256)/5`

=   `430/5`

=  86

Step 3:Standard deviation

Standard deviation  =  `sqrt ( 86 )`

=  9.27

Example 2:

Find the standard deviation for the given set of numbers:

{ 7, 4, 5, 1, 3 }

Solution:

Step 1: Mean

Mean   =  ` ( 7 + 4 + 5 + 1 + 3 ) / 5`

=  ` 20 / 5`

=  4

Step 2: Variance

Variance  =   `( (7-4)^2 + (4-4)^2 + (5- 4)^2 + (1-4)^2 + (3-4)^2 )/5`

=   `( (3)^2 + (0)^2 + (1)^2 + (-3)^2 + (-1)^2 )/5`

=   `(9+0+1+9+1)/5`

=   `20/5`

=  4

Step 3:Standard deviation

Standard deviation  =  `sqrt ( 4 )`

=  2

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Practice Problems for Finding Standard Deviation:

1) Find the standard deviation for the set of numbers: { 45, 67, 89, 24 }

2) Find the standard deviation for the data set: { 44, 67, 87, 90, 42 }

3) Find the standard deviation for the set of numbers: { 39, 21, 57, 61, 73 }

4) Find the standard deviation for the data set: { 207, 137, 316, 590 }

5) Find the standard deviation for the set of numbers: { 64, 80, 38, 40, 41 }

6) Find the standard deviation for the data set: { 12, 21, 60, 41 }

7) Find the standard deviation for the set of numbers: { 56, 24, 78, 29, 88 }

8) Find the standard deviation for the data set: { 41, 60, 77, 11, 63 }

9) Find the standard deviation for the set of numbers: { 12, 204, 717, 231 }

10) Find the standard deviation for the data set: { 245, 356, 343, 326, 460 }

11) Find the standard deviation for the set of numbers: { 16, 34, 68, 19, 78 }

12) Find the standard deviation for the data set: { 25, 50, 45, 17, 77 }

13) Find the standard deviation for the set of numbers: { 4, 4, 7, 6, 14 }

14) Find the standard deviation for the data set: { 45, 10, 45, 156, 27 }

15) Find the standard deviation for the set of numbers: { 216, 334, 566, 149 }

Answer key:

1)  24.26        2)  20.39       3)  18.23       4)  172.44      5)  16.66     6)  18.55    7)  25.52     8)  22.80

9)  260.04     10)  68.86     11)  25.44     12)  21.02        13)  3.69          14)  51.38     15)  158.68

One to One Function Proof

Introduction to one to one function  proof:

A function from A to B is called one to one if f(a) = f(b) then a = b. Each element in the function B has paired with only one  element in the function B. One to one function is a function in which all the variables in the domain has an individual range. Each element in the domain has the correspondent element in the range. Here we will see about one to one function proof.

One to One Function Proof:

A function is known as one to one if every element in the range of the function corresponds with one and only element in the domain.

Proof- one to one function proof

Let us consider the function, f : N -> N, it is defined as f(n) = n2  where N indicates the positive integers.

Assume an integer a and b in which f(a) = f(b).

The function is f(n) = n2, substituting a and b,

we have a 2 = b 2  ,  where  a 2 - b 2 = 0.

a 2 - b 2 = 0  - >   (a + b) (a - b) = 0

where  a + b  = 0 ,  a - b = 0

hence we have  a + b = 0   - >  a =  - b

hence we prove for a - b = 0   - >  a = b

the function f(a)= f(b)

Hence proved.

Example Problem - One to One Function Proof

Example problems 1 - One to one function  proof

F{(1, 0), (2, 3) ,( 4, 5), (6, 7)} show that the function is one to one function.

Solution:

Domain of the given function is 1, 2, 4, 6.

Range of the given function is 0, 3, 5, 7.

Hence the each of the domain has corresponding range individually.

Hence the given set exhibits one to one function.

Example problems 2 - One to one function  proof

Which of the following functions is one to one function?

F1{(1, 8), (2, 9) ,( 4, 10), (6, 11)}

F2{(1, 0), (2, 3) ,( 4, 8), (6, 9)}

F3{(1, 0), (2, 0) ,( 4, 5), (6, 7)}

Solution:

F1{(1, 8), (2, 9) ,( 4, 10), (6, 11)}

Domain of the given function1 is 1, 2, 4, 6.

Range of the given function1 is 8, 9, 10, 11.

Each element in the domain corresponds with the values of range.Understanding composite function is always challenging for me but thanks to all math help websites to help me out.

Hence the given function 1 is one to one function.

F2{(1, 0), (2, 3) ,( 4, 8), (6, 9)}

Domain of the given function2 is 1, 2, 4, 6.

Range of the given function 2 is 0, 3, 8, 9.

Each element in the domain corresponds with the values of range.

Hence the given function 2 is one to one function.

F3{(1, 0), (2, 0) ,( 4, 5), (6, 7)}

Domain of the given function 3 is 1, 2, 4, 6.

Range of the given function 3 is 0, 0, 5, 7.

The element in the domain 1 and 2 has the same range.

Hence the given function 3 is not one to one function.

Answer: F3{(1, 0), (2, 0) ,( 4, 5), (6, 7)} is not an one to one function.

Line Segment Practice for Free

Introduction for line segment:

Line segment is the important topic in geometrical chapter and math subject. Generally in math subject, a free long line is divided with two definite end points on both directions is known as line segments. But the line segment is a small part of a long line. In this article we are going to brief explain about line segment and line segment practice problems. And solve the some of example problems for free.

Line Segment Definition:

In math, a line segment can be defined as one small divided part of a lengthy line.
Line segment is also known as ‘a straight line’, which is joining the two points with coordinates, and the line is infinity after that, the end points.
So in this word segment is being an important role in line segment.
Example figure for line segment:



In this figure xy is the infinity line, and A, B are the two end points, and the line segments are bar (AB).
The length of the line segments AB would be written as bar (AB). And the line segments have used the name as to be two end points AB


Line segment formulas for solve practice problems:

The distance of two given points is called line segments length. The line segment length is calculated using formula for distance.
The line segments having two end points, this is have x and y coordinates. X(x1, y1) and Y(x2, y2)...
Line segment distance formula, XY = (sqrt((x2-x1)^(2)+ (y2-y1)^(2))
The line segment mid point formula, M = midpoint, M(x, y) = ((1/2(x1+x2), (1/2(y1+y2)))
These two are the important formulas for solve line segment problems.

Example Free Problems in Line Segment:

Example 1 for practice:

Using the distance formula, solve the length of line segment between two end points. End points are M = (3, 2), N = (5, 4)

Solution:

Line segment distance formula,

` MN= sqrt ((x2-x1) ^ (2) + (y2-y1) ^ (2))`

x1= 3   y1= 2   x2=5    y2=4

` MN = sqrt ((5-3) ^ (2) + (4-2) ^ (2))`

` MN = sqrt ((2) ^ (2) + (2) ^ (2))`

`MN = sqrt (4+ 4)`

`MN = sqrt (8)`

` MN = 2 sqrt (2)`

Answer is `2 sqrt (2)` .

Example 2 for practice:

What is the mid point between two points given? Solve (-3, -1), (4, 2)

Solution:

Using the mid point formula,

Midpoint formula = `((x1+x2) / 2), ((y1+y2) /2)`

And substitute (x1, y1) = (-3, -1) and (x2, y2) = (4, 2)

Midpoint = `((-3+4) /2), ((-1+2) /2) ` Simplify it and get,

And get midpoint = `(1/2), (1/2)`

The mid point between the two points is `(1/2), (1/2)` .

Those two example problems are free for understanding about line segments. Here we are giving some practice problems.I have recently faced lot of problem while learning cbse previous years question papers 12th, But thank to online resources of math which helped me to learn myself easily on net.

Free practice problems in line segments:

Free practice problem 1:

Using the distance formula, solve the length of line segment between two end points. End points are M = (6, 2), N = (5, 4)

Answer is `sqrt (5)` .

Free practice problem 2:

Find the mid point between two points given? Solve (-5, -1), (6, 2)

Answer is (`1/2` , `1/2` )

Those all the above example problems and free practice problems are useful for understand the line segment practices.

Geometry Acute Triangle

Introduction to geometry acute triangle:

The triangle is a 3 sided figure where all the three sides are enclosed to form the shape. The triangle can be classified according to its angle measurements and the length of the side. The geometry acute triangle is nothing but the triangle where all the angles in the triangle are less than 90 degrees. So, this triangle can be called as acute triangle. Now we are going to see about the geometry acute triangle.

About Geometry Acute Triangle:

The geometry acute triangle may also call as the acute equilateral triangle because each of the angle measures in the triangle will be 60 degrees which is less than 90 degrees. There are also types of acute triangle. They are acute obtuse angle, acute scalene triangle. In geometry acute triangle the angles will be less than 90 degrees. If it is not an acute triangle then it will be some other triangle types such as obtuse or right triangle.

The formula to find the area of a geometry acute triangle can be given as follows,

Area of triangle = `(1/2)` * b * h



Problems for Geometry Acute Triangles:

Example 1:

Find the angle of an acute triangle whose measurements are 40 degrees and 50 degrees.

Solution:

Now we calculate for the acute triangle as follows,

x + 50 + 60 = 180

Just add all the angles of the triangle,

x + 110 = 180

Now subtract the angle from 180 we get,

x = 180 - 110

x = 70

Thus, the acute angle of the triangle is 70 degrees.

Example 2:

Find the area of an acute triangle where the base of the triangle measurement is 5 cm and the height is about 6 cm.Is this topic math homework help free online hard for you? Watch out for my coming posts.

Solution:

Now we are going to see about the calculation part

The area of an acute triangle can be given as,

Area of triangle = `(1/2)` * Base * Height

= `(1/2)` * 5 * 6

= `(1/2)` * 30

= `30/2`

Area of acute triangle = 15 cm2

Derivative of Quadratic Functions

Introduction to derivative of quadratic functions:

In calculus the derivative is a measure of how a function changes as its input changes. The derivative of a function at a chosen input value describes the best linear approximation of the function near that input value. A quadratic equation is a polynomial equation of the second degree. The general form is ax2 + bx + c . The constants a, b, and c, are called respectively, the quadratic coefficient, the linear coefficient and the constant term or free term.

Source Wikipedia.

Derivative Formulas:

1.  ` d / dx` (x n ) = n xn-1

2.  ` d/dx` (ex)  = ex

3.  ` d/dx` `(uv)` = `u(dv)/(dx)` + `v(du)/(dx)`
4. .`d/dx` `(u/v)` = `(v(du)/(dx) - u(dv)/(dx))/ v^2`

5.  `d/dx` (sin x) = cos x

6.  `d/dx ` (cos x) = -sin x

7.  `d/dx` (tan x) = sec2x

Derivatives of Quadratic Function Problems:

Derivatives of quadratic function problem 1:

Find the derivative of quadratic function (2x3 - 3x4 + x2) with respect to x.

Solution:

Given Quadratic function is  (2x3 - 3x4 + x2)

` d/dx` (2x3 - 3x4 + x2)  = `d/dx` (2x3) - ` d/dx` (3x4) + ` d/dx ` (x2) .

=  `d/dx` (2x3) - `d/dx` (3x4)  + `d/dx` (x2).

= (2 × 3) x(3-1) - (3 × 4) x(4-1) - 2x(2-1)

= 6 x2 - 12 x3 + 2 x

`d/dx` (2x3 - 3x4 + x2) =  6 x2 - 12 x3 + 2 x

Answer: The derivative of given quadratic function is  `d/dx` (2x3 - 3x4 + x2) =  6x2 - 12 x3 + 2 x

Derivatives of quadratic function problem 2:

Find the derivative of quadratic function  y = 3t2 + t3 - 30t. with respect to 't'.Is this topic free algebra help hard for you? Watch out for my coming posts.

Solution:

Given Quadratic function is y = 3t2 + t3 - 30t.

dy  = (3 × 2)t(2 - 1) dt + (1 × 3)t(3 - 1) dt - 30t(1 - 1) dt.

= 6t dt + 3t2 dt - 30dt.

= (6t + 3t2 - 30) dt

`dy/(dt)` = (6t + 3t2 - 30)

Answer:  The derivative of given quadratic function is   `d/dt` (3 t2 + t3 - 30t) =  3t2 + 6t - 30.

Derivatives of quadratic function problem 3:

Determine the second derivative of given quadratic function f(x) = 15x2 + 10x + 5 with respect to x.

Solution:

Given quadratic function is, f(x) = 15x2 + 10x + 5

The First derivative of quadratic function is f'

f' =` (df )/ (dx)`  = (15 × 2)x(2 - 1) + (10 × 1)x (1-1)

= 30x + 10

The Second derivative of quadratic function is, f''

` d^2/dx` f(x) = `d^2/dx` ( 15x2 + 10x + 5)

= `d/dx` (f')

= `d/dx` ( 30x + 10)

= ` d/dx` (30x) +` d/dx` (10)

= 30 `d/dx` (x) + `d/dx` (10)

= 30 + 0

= 30

Answer: The Second derivative of given quadratic function is 30

Sides of a Polygon Formula

Introduction to sides of a polygon formula:

In geometry, polygon is two dimensional shapes. It has more than two sides. All sides are straight and connected to another side. The number of vertices is equal to number of sides. Vertices are nothing but corner points of the shape. In this article we shall see how to calculate the area of regular polygon.

Sides of a Polygon Formula - Formulas:

Triangle:

Formula to find area of triangle:

Area of triangle (A) = `1/2` (b x h) square units

b – Base

h – Height

Square:



Formula to find area of square:

Area of square (A) =a2 square units

a – side length                            

Pentagon:



Formula to find the pentagon:

Area of the pentagon (A) = t2 1.72 square units

t – Side length

Hexagon:



Formula to find area of hexagon:

Area (A) = t2 2.6 square units.

t - Side length

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Sides of a Polygon Formula – Example Problems:

1. Find the area of triangle whose base is 40 cm and height is 25 cm.

Solution:

Given:

Base (b) = 40 cm

Height (h) = 25 cm

Substitute the given value in the formula

Area of the triangle (A) = `1/2` (b x h) square units

= `1/2` (40 x 25)

= `1/2` (1000)

= 500

Area of the triangle (A) = 500 cm2

2. Find the area of the square, whose side length is 27 cm

Solution:

Given:

Side (a) = 27 cm

Substitute the given value in the formula

Area of square = a2 square units

= 27 x 27

Area of square = 729 cm

3. The side length of pentagon is 13.5 cm. Find the area and perimeter of the pentagon.

Solution:

Given:

Side length (t) = 13.5 cm

Substitute the given value in the formula

Area of the regular pentagon (A) = 1.72 t2 square units

= 1.72 x 13.52

= 1.72 x 182.25

= 313.47

Area of the regular pentagon (A) = 313.47 square units

4. The side length of hexagon is 14.7 cm. find the area of the hexagon.

Solution:

Given:

Side length (t) = 14.7 cm

Substitute the given value in the formula

Formula:

Area of the hexagon (A) = t2 2.6 square units

= 14.72 x 2.6

= 216.09 x 2.6

= 561.834

Area of the hexagon = 561.834 cm2

Perpendicular Line Segments

Introduction to perpendicular line segments:

Perpendicular line:

In geometry, two line segments or planes (or a line and a plane), are considered perpendicular (or orthogonal) to each other if they form congruent adjacent angles (a T-shape).  Thus, referring to Figure 1, the line AB is the perpendicular to CD through the point B.



Fig(i) Perpendicular line

Line segments:

The line segments are the the part of the line that has two end points.It include all the points between its end points.For example,consider the following figure,

Fig(ii) Line segment

In the above figure AB is a line and CD is a line segment.C and D are the two end points of a line segment.We are going to see about the perpendicular line and line segments.

Examples for Perpendicular Line Segments:

The following are the geometry examples for line segments,

Sides of the triangle

Sides of the Rectangle

Sides of the Square

Properties of perpendicular line:

The multiplication of slopes of the perpendicular lines is equal to -1.

That is m1=slope of perpendicular line 1

m2 = Slope of perpendicular line 2

m1 × m2 = -1

Distance formula for line segments:

The following formula used to find the lengh of the line segments ,

length = `sqrt((x2-x1)^2+(y2-y1)^2))`

Here (x1,y1) and (x2,y2) are the two end points of the line segments.

Problems on Perpendicular Line Segments:

Problem 1:

Find the equation of a line which is perpendicular to 4y - x = 20 and passes through the  point (2, -3).

Solution:

Given 4y - x = 20 and the point (2,-3)

To find the perpendicular line we need to find the slope.

To find the slope we need to change the given equation into slope intercept form.

4y - x = 20

Add x on both side,

4y - x  = 20

+ x =  +x

4y = x + 20

Divide by 4 on both side,

y = (x/4) + 5

y = mx + b form

So the slope m = 1/4

We know that multiple of slopes of the perpendicular lines equals to -1

1/4 * m = -1

Multiply by 4 on both sides,

m = -4

The line equation is,

(y-y1) = m(x-x1)

(y - (-3)) = -4(x-2)

y+4 = -4x+8

Subtract  4 on both sides,

y = -4x +4

Answer : The line perpendicular to the given line is    y = -4x +4

Problem 2:

By using the line segments distance formula find the length of the line segmen xy with x(5 ,3) y(1,2)

Solution:

length = `sqrt((x2-x1)^2+(y2-y1)^2))`

x1 = 5    y1 =3    x2 = 1    y2 = 2

=` sqrt( ( 1-5)^2 + ( 2-3)^2)`

=` sqrt( (-4)^2 + (-1)^2)`

= `sqrt (16+1)`

= `sqrt (17)`

Answer :The length of the line segment xy = `sqrt (17)`

Solving Vector Cartesian Coordinates

Introduction on solving vector Cartesian coordinates:

This article is about solving vector Cartesian coordinates. Solving vector Cartesian coordinates is very simple. The tutors of tutor vista helps the students in solving vector Cartesian coordinate any time. The vector coordinates are x,y and z. Three simple methods are there to describe a vector. Directions, specific lengths, angles and projections or components are used to describe vectors. The simplest method of these is Cartesian or rectangular co-ordinate system. Below we can see about solving vector Cartesian coordinates.

Solving Vector Cartesian Coordinates

In cartesian coordinate system three co-ordinate axes x,y and z are mutually right angles to each other. Consider a point P(x,y,z) in space at a distance r from the orgin. The vector r can be represented as

r = `xbara_x``+ y bara_y + z bara_z`

Where `bara_x, bara_y,` and `bara_z` are unit vectors

x,y and z are the components vectors. Components vectors have a magnitude and direction. Unit vectors have unit magnitude and directed along the co-ordinate axis.

A unit vector in a given direction is a vector in that direction divided by its magnitude. It is given by

`a_r = r/|r|`

`a_r = (xbara_x + ybara_y + z bara_z)/sqrt (x^2 + y^2 + z^2)`

Consider the points P(x,y,z)and Q(x + dx, y +dy, z+dz) in rectangul;ar co-ordinate system. The differential length dl from P to Q is the diagonal of the parallel piped is given by

dl = sqrt ((dx)^2 + (dy)^2 + (dz)^2)

The differential area ds = dx dy

= dy dz

= dz dx

The differential volume dv = dx dy dz

Solving Vector Cartesian Coordinates

Conversion of cylindrical to cartesian system

The cylinderical co-ordinates (?, F, z) can be converted into cartesian co-ordinates ( x, y, z)

Given                                                       Transform

?                                                             x = r cos ?

f                                                            y = r sin f

z                                                             z = z

conversion of spherical to cartesian co-ordinates system

The spherical co-ordinates (r, `theta` , f) can be transformed into cartesian co-ordinates (x, y, z)

Given                                                       Transform

r                                                             x = r sin ?. cosf

?                                                            y = r sin ? sin f

f                                                            z =r cos ?


Solving Vector Cartesian Coordinates

Example Problem: Give the Cartesian co-ordinates of a point whose cylinderical are ? = 1, f = 45°, z =2.

Given

Cylinderical co-ordinates are ? = 1, f = 45°, z =2.

Cartesian co-ordinates are x, y, z

x = ? cosf

= 1.cos 45 = 0.707

y = ? sinf

= 1. sin 45 = 0.707

z = z

= 2

Cartesian co-ordinates are (0.707, 0.707, 2)

Regular Irregular Polygons

Introduction for Regular irregular polygons:

Regular polygons:

When each side and angles of any polygon are similar, then it is said to be a regular polygon. A regular polygon is one with its entire sides similar and each of its angles similar, for instance: A regular polygon with 3 surfaces is said to be equilateral triangle.

Irregular Polygons:

When angles of the any polygon are not the same, then it is called as irregular polygon.

Regular Irregular Polygons-angles of Regular Polygons:

The regular polygons’ angle is grouped into two types. They are

Exterior angle
Interior angle
Interior angle of Regular Polygons:

Angle of regular polygons = [(n - 2) / n] × 180°

where n =number of sides.

Example 1:

Octagon is an eight sided polygon

Regular Polygons’ Interior Angle = [(n - 2) / n] × 180°

Here n = 8 for octagon

So the interior angle of an octagon = [(8 – 2)/ 8] * 180°

= (3 / 4) * 180°

Interior angle of octagon = 540° / 4 = 135°

Exterior angle of Regular Polygons:

360° is the exterior angle of polygons.  Hence the exterior angle can be estimated by the formula,

Regular Polygons’ Exterior angle = 360° / n

where n is the number of sides.

Example 2:

Pentagon is a polygon with five sides. Polygon’s Exterior angle = 360° / n

For Pentagon, n = 5

So the exterior angle of a pentagon = 360 / 5 = 72°.

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Regular Irregular Polygons-area of Irregular Polygons:

The steps for determining the area of irregular polygon are,

Convert every vertex into co-ordinate in graph.
Take any line segment and down to the x - axis.
Compute the average of two altitudes.
Compute the difference of two widths.
Multiply altitude with width.
Go over steps 2 to 5 until manipulating every line segment.
Then add all the areas of every line segment.
So, we will result in area of irregular shape.

Symmetry for Irregular Polygon:

Since the irregular polygons do not have equal angles, they do not posses line of symmetry..

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.