Solving Sampling Distributions Calculator

Introduction to solving sampling distributions calculator:

In statistics, sampling distribution is based on a random sample size of n. The sampling distribution is also called as finite sample distribution. The sampling distributions depend upon the distribution of population and the sample size used. With the help of the sampling distribution calculator we can easily solving the sampling distributions problems.

Solving Sampling Distributions Calculator Example Problem:

Properties of sampling distribution:

The mean value of the sampling distribution = mean of the sample population
standard deviation of the sampling distribution = standard deviation of the sampled population
square root of the sample size
The distribution of the populations normal, then the sampling distributions is `barX`
Example1 – solving sampling distributions calculator:

A lamp manufacture claim that the lifespan of the lamp mean value of the lamp is 56 months and the standard deviation is 6 month. The consumer group 50 of them test, what probability that it find mean lifetime of less than 54 month.

Solution:

We seeking the value of P(`barX` <54 br="br">
`barX` is approximately normally distributed by the Central Limit Theorem and

The mean value is µ = 56 months and

The standard deviation of  `sigma``barX` = 6/`sqrt50` = 0.85 months to find the required probability need to convert the z- scores.

Z = `barX` - µ`barx` /`sigma` `barX`

Z = `(54- 56)/(0.85)` = -2.35

We need to use the table to find the value of P(Z ≤-2.35).

0.5 - P(0≤ Z ≤ 2.35) = 0.5 - 0.4906 = 0.0094.

The probability is 0.0094, or 0.94%

Please express your views of this topic Empirical Probability by commenting on blog.

Example 2– Solving Sampling Distributions Calculator:

A lamp manufacture claim that the lifespan of the lamp mean value of the lamp is 58 months and the standard deviation is 6 month. The consumer group 50 of them test, what probability that it find mean lifetime of less than 56 month.

Solution:

We seeking the value of P( `barX` ≤ 56),

`barX ` is approximately normally distributed by the Central Limit Theorem and

The mean value is µ = 58 months and

The standard deviation of  = 6/ `sqrt50` = 0.85 months to find the required probability need to convert the z- scores.

Z=`barX` µ `barx` / `sigma` `barx`

Z = `(56 - 58) /( 0.85) ` = -2.35

We need to use the table to find the value of P(Z ≤-2.35).

0.5 - P(0≤ Z ≤ 2.35) = 0.5 - 0.4906 = 0.0094.

The probability is 0.0094, or 0.94%

Lagrange's Mean Value Theorem

Introduction to Lagrange’s Mean Value Theorem:

In calculus, the mean value theorem states, roughly, that given an arc of a smooth continuous (differentiable) curve, there is at least one point on that arc at which the derivative (slope) of the curve is equal (parallel) to the "average" derivative of the arc. It is used to prove theorems that make global conclusions about a function on an interval starting from local hypotheses about derivatives at points of the interval.

(Source – Wikipedia)

Lagrange’s Mean Value Theorem – Proof:

The followings statements are the proof for the Lagrange’s Mean Value Theorem.

If a function f(x) is derivable in the interval [a, b], then, there exists at least one value ‘c’ of x lying within [a, b] such that `(f(b)-f(a))/(b-a)` = f’(c).

Proof:

To prove the theorem, we defined a new function `phi` (x) involving f(x), which is designed so as to satisfy the condition of Rolle’s Theorem.

Let `phi` (x) = f(x) + A x where A is a constant to be determined such that `phi` (a) = `phi` (b).

Thus, f(a) + Aa = f(b) + Ab

Therefore, A = `(-f(b)-f(a))/(b-a)`

Now, f(x) is derivable in [a, b]. Also x is derivable and A is a constant. Therefore, `phi` (x) is derivable in [a, b] and its derivative is f’(x) + A.

Thus, `phi` (x) satisfies all the conditions of Rolle’s Theorem.

There is, therefore, at least one value ‘c’ of x, lying within (a, b) such that `phi` ’(c) = 0

Therefore, 0 = `phi` ’(c) = f’(c) + A, that is –A = f’(c)

Or

` (-f(b)-f(a))/(b-a)` = -f’(c)

Or

`(f(b)-f(b))/(b-a)` = f’(c) →(i)

In certain books Lagrange’s Mean Value Theorem is stated as follows.

If a function f(x) defined in the closed interval [a, b] is such that

(i) f(x) is closed interval in continuous as [a, b], and

(ii) f(x) is derivable in the open interval (a, b), then there exists at least one value ‘c’ of x within the interval [a, b] such that f’(c) = `(f(b)-f(a))/(b-a)` .

This seems to be unnecessary as every finitely derivable is continuous and therefore the definition given by us is quite sufficient. I have recently faced lot of problem while learning define rational number, But thank to online resources of math which helped me to learn myself easily on net.

Hence we proved the Lagrange's Mean Value Theorem.

Lagrange’s Mean Value Theorem – Example Problem:

The followings examples are the proof for the Lagrange’s Mean Value Theorem.

Verify Lagrange’s Mean Value Theorem for the f(x) = 2x – x^2 in the interval [0, 1].

Solution:

The function 2x – x^2, being a polynomial is derivable in [0, 1].

f(x) = 2x – x^2 also f’(x) = 2 – 2x.

By Lagrange’s Mean Value Theorem, there exists c in (0, 1) such that

` (f(1)-f(0))/(1-0)` = f’(c)

Or

f(1) – f(0 = f’(c)

Now f(1) =` 2 * 1` – 12 = 1, f(0) = ` 2 * 0` – 0, f’(c) = 2 – 2c

1 – 0 = 2 – 2c `rArr` c = ` 1/2` which lies in(0, 1).

Hence we can verify Lagrange’s Mean Value Theorem.

Linear

Introduction to linear:

In linear function, the linear models are Maximized or minimized which is used to set a linear constraints.Objective function variable,Set of constraints for linear,set of decision variables for linear are the important components of linear models.Most of the real world problem are leads to linear models components. Most of the real world problem in linear algebra can be approximated by using linear models.

Definition to Linear

Line segment present in the linear algebra are having only one dimension.  This line segment is characterized by using the composition of the dimension present in the vector and linear numbers. The linear segment present in the linear algebra should be narrow.The linear segment also be elongated by using the parallel margins.These paralel margins are present in a linear leaf.

Types of Linear

Intersect
Parallel
Coincide


Intersect

For an intersecting line, have to set a cross line or we can overlap the two lines. For a model draw a two intersecting lines which are placed from a shape named  "X."
Learn more about the intersect lines in the class Geometry section.


Parallel

The electrical system one of the conductor are having positive poles, electrodes present in the conductor,terminals used in the conductor and the negative poles, are joined in an another conductor present in an electrical system. This is also called as multiple effect and this is also opposed to one of the series.
Parallel planes have same direction in all parts.This is also same as the parallel lines.
The line which having all dimensions are of equal length from another line.This is also similar to the parallel line and the parallel plane.
For setting a parallel line, the direction should be confirmed first. The same procedure is followed for the parallel plane also.Parallel lines are also to be confirmed for setting a line.I like to share this differentiation rules with you all through my article.


Coincide

For  occupying  the same place in the space, the line should be placed to one another. For example we have to place the two triangles one another.
All the lines present in the linear are equal and clear means then they are said to be a coincide line.

Solving Quadratic

Introduction for solving quadratic:

The quadratic equation is also called polynomial equation whose degree is  second. The general format for solving quadratic equation is   ax2 + bx + c =0. where x is variable  and a,b and c are constants.

Generally, quadratic equation will give two solutions. These solutions may be either real or complex.The real solutions may be rational or irrational.

Quadratic Formula for Solving:

The roots can be found by completing the square,

x2 + ( b / a ) * x = - c / a

( x + ( b / 2 a )) 2=- c / a + b2 / 4 a2 = ( b2 – 4 a c ) / ( 4 a2)

( x + ( b / 2 a )=  ± v( b2 – 4 a c ) / 2 a

Solving for x then gives

X = ( -b ± v( b2 – 4 a c )) / 2 a

This equation is known as the quadratic formula.

Discriminant for Solving Quadratic Equation

From the above quadratic formula, the expression below the square root is discriminant .

Discriminant=b2 – 4ac

When the discriminant is positive, the two roots will be real and distinct.
When the discriminant is negative, the two roots will be non real and complex.
-(b/2a) + i v(4ac – b2) / 2a and  -(b/2a) - i v(4ac – b2) / 2a

Where i represents imaginary part

When the discriminant is zero, the equation has exactly one real solution.
X= -(b/2a)

I have recently faced lot of problem while learning what is an equivalent fraction, But thank to online resources of math which helped me to learn myself easily on net.

Example for Solving Quadratic Equation:

Example 1:

Solve x2 + x – 4 = 0.

The quadratic formula is

X = ( -b ± v( b2 – 4 a c )) / 2 a

Here, a=1, b=1 and c= -4

substitute the values of  a,b, and c in quadratic equation for solving

X=-1 ± v( 12- 4 * 1 * -4)/ 2 * 1

X=-1± v(1+16)/2

X=-1± v(17)/2

The solution is X=-1± v(17)/2

Example 2:

Solve  2x2 + 3x + 5 = 0.

The quadratic formula is

X = ( -b ± v( b2 – 4 a c )) / 2 a

Here, a=2, b=3 and c= 5

Substitute the values of a,b, and c in quadratic equation for solving

X = ( -3 ± v( 32 – 4 * 2 *  5 )) / 2 * 2

X = ( -3 ± v( 9 – 40)) / 4

X = ( -3 ± v( – 31)) / 4

X = ( -3 ± 5.6) / 4

X = ( -3 - 5.6) / 4 and ( -3 + 5.6) / 4

X = ( - 8.6) / 4 and ( 2.6) / 4

X= - 2.15 and 0.65

Smart Math Calculator

Introduction to smart calculator:

The word Smart Calculator itself signify the quickness the results are displayed.  A calculator as such is a small digital inexpensive device used for performing basic operations of arithmetic.  The idea of numbers necessitated invention of calculators. Primitive calculators were knots in strings, beads threaded onto strings, balls on wires and even triangles with rows of numbers on them. The breakthrough came in the seventeenth century with the invention of the first digital mechanical calculator and in coarse of time improvements were made to the mechanical design.  The creation of first digital electronic calculator was in 1960's.  The power source for the calculator which was manually operated earlier changed to a battery/solar-powered calculators.

Smart and Modern Calculator:

Modern calculators are electrically powered (usually by battery/solar cell and are small in sizes.  By 1980's calculator prices so reduced that calculator was affordable to most.  Children in school were able to learn much more in a shorter amount of time.  Moreover students were able to complete the homework assignment within less time.  Calculator is a universal tool.  It has become necessary for everyone like a cell phone.  The type and size will vary for each person.  Simple 10-key calculator will do for household purpose and elementary schools.   A  scientific calculator will help the middle school student, and a graphing calculator is necessary for high school and colleges.

Uses of Smart Calculator:

Smart calculator is designed to deal with scientific problems, such as those in Physics, Chemistry, and those in advanced mathematics such as trigonometry, exponential, logarithm and hyperbolic functions.  It is a great tool for anyone needing mathematical help.  Can choose between options like colours, fonts, language.  In-dept help is available which explains how to use the functions, buttons and much more. Thus smart calculator is really wonderful . It is the easiest and most convenient calculator in the world.  Simply type down a math expression and observe how the sub-total is calculated immediately after each stroke.  If there is a problem in the expression it is highlighted in red.  A wide range of advanced  features are available in smart calculator.

Solving Possibility Problem

Introduction to solving possibility problem

Probability or possibility determines the possible outcomes of a certain an experiment. One or more possible outcome of an experiment is referred to as event. For instance, when tossing a coin head and tail are the two possible outcomes. In independent event one event does not affect the other event. In dependent event one event does affect the other event. Two events are cannot happen at the same time, they are called mutually exclusive events.

Solving Possibility Problem - Examples

Solving Problem 1: 4 balls are drawn with replacement from a bag containing 20 blue and 16 orange balls.

Find the possibility of a) both the balls are orange b) first ball is blue and second ball orange c) One is blue and the other is orange.

Solution:

Total = 20 blue + 16 orange = 36. From that 4 balls are drawn.

Let’s A: Orange ball and B: Blue ball

P(A) = 16/36 = 4/9 and P(B) = 20/36 = 5/9

a) Both are orange

P( orange and orange) = P( A n A) = P(A) · P(A)

= 4/9 · 4/9 = 16/81

b) First ball is blue and second is orange.

P( orange n blue) = P( A) · P( B) = 4/9 · 5/9 = 20/81

c) one is blue and the other is orange

= P(A) · P(B) + P(B) · P(A)

= 4/9 · 5/9 + 5/9 · 4/9

= 20/81 + 20/81 = 40/81.

Solving Problem 2: A box contains 6 apples and 6 oranges. A fruit is drawn at random; fruit is noted and is returned to the box. Moreover, 2 additional fruits of the same verity are put in the box and then a fruit is drawn. What is the possibility that the second fruit is apple?

Solution:

Case 1: Let us draw an apple first.

S1 = 6 apple + 6 orange = 12

A1 : drawing an apple, P(A1 ) = 6/12 = 1/2

Now after adding 2 more apples

S2 = 8 apple + 6 orange = 14

A2 : drawing an apple, P(A2) = 8/14 = 4/7

P(an apple and an apple) = 1/2 ·  4/7 = 4/14 = 2/7

Case 2: Let us draw an orange first

S1 = 6 apple + 6 orange = 12

A1 : Drawing an orange, P(A1) = 6/12 = 1/2

Now after adding 2 more oranges

S2 = 6 apple + 8 orange = 14

A2 : Drawing an apple, P(A2) = 6/14 = 3/7

P(an orange and an apple)= 1/2 * 3/7 = 3/14

P(in both the cases) = 2/7 + 3/14 = 49/98 = 1/ 2 Ans.

Solving Possibility Problem - Practice
Solving Problem 1: Two dice are rolled; find the probability that the total is a) equal to 3 b) equal to 6 c) greater than 10

Answer: a) 1/18 b) 5/36 c) 1/12

Solving Problem 2: Find the probability of getting king from deck of card?

Answer: 1/13

Pre Algebra X and Y Intercepts

Introduction to pre algebra x and y intercepts:

X - intercept:

In pre algebra, the x-intercept of a line is defined as the point at which the line cuts the X-axis.
The x-intercept is given as ( x, 0 )


Y - intercept:

In pre algebra, the y-intercept of a line is defined as the point at which the line cuts the Y-axis.
The y-intercept is given as ( 0, y )


A few example problems of pre algebra x and y intercept is given below.

Pre Algebra X and Y Intercepts- Example Problem :1

Find x and y intercept of the equation of a line x + 4y =12.

Solution

To find x intercept, plug y = 0 in the equation and solve for x.

x + 4y = 12 ............. Given equation

x + 4(0) = 12 ............. Substitute y = 0 in the given equation

x = 12

Solution: Therefore, the x-intercept = ( 12, 0 ).

To find y intercept, plug x = 0 in the equation and solve for y.

x + 4y = 12 ............. Given equation

0 + 4y = 12 ............. Substitute x = 0 in the given equation

4y = 12

y = 12 / 4 .......    Divide by 4 on left side and right side of the equation

y = 3

Solution: Therefore, the y-intercept = ( 0,3 ).

Pre Algebra X and Y Intercepts- Example Problem :2

Find x and y intercept of the equation y = 4x + 15

Solution

To find x intercept, plug y =0 in the equation and solve for x.

y = 4x + 15 ............. Given equation

0 = 4x + 15 ............. Substitute y = 0 in the given equation

-15 = 4x       .................... Divide by 4 on left side and right side of the equation

-15/4 = x

Solution:  Therefore, the x-intercept is (-15/4, 0)

To find y intercept, plug x = 0 in the equation and solve for y.

y = 4x + 15 ............. Given equation

y = 4(0) + 15 ............. Substitute x = 0 in the given equation

y = 15

Solution: Therefore, the y-intercept is (0, 15)

These are the example problems which are helpful to learn pre algebra x and y intercepts.

Formal Problem Statement

Definition of formal problem statement:

Formal problems satement arise as wrong clients determination confirms up in divide journal files are wrong planned for further one month. Thus, Julie has to make more than one delayed notice, which corresponds to each month, intended for each one customer. Followed through, Rhonda is reliable to identify these substitute notices and join them into one.

Two serious formal problems statement: First, the workload increase significantly. Second, the separate letters rarely slip an along with a customer can get some late notices.

Sample Formal Problem Statement:

The conventional of the formal problems statement, since they are described in the arranged problem statements, typically do with clerical and recording errors.

Here is attractive in workload the final half of every month. This problem occurs when the cost of seven of the relatives is suitable on the first of the month. This condition causes problems in the direction of further office events in to time period. The human resources contain locate of work to perform preparing and meting away the expenses, which achieve in the deferred or unfinished method of the rest assignments.

Continuous problems arise commonly in applications such like representation appropriate, adaptive control, neural network training, indication processing, and trial design. Discrete optimization is a huge subject unto reserve allocation, network routing, policy planning.

A major concern in optimization is distinguishing involving global and local optima. Each other factors living being equal, one would ever more want a globally optimal solution to the optimization problem. It might not be achievable to get a global solution with one should be satisfied with obtaining a limited solution.

Examples for Formal Problem Statement:

Example 1: Air Conditioning in ECU Dormitories

According to the ECU process of formal problem statement, the university seeks to provide students by safe, healthy education surroundings.  Dormitories are one important of that learning environment, because 55% of ECU students exist in campus dorms and mainly of these students use an important amount of time working in their dorm rooms.

Example 2: Safe Rides Program

According to the ECU police force process of formal problem statement, the objective of the university is to maintain student’s logic and staff in maintain a secure position and to develop the excellence of life at East Carolina University. Consequently each weekend numerous students venture downtown to like a good time packed with dancing partying and drinking.

Problem Statement Hypothesis

Introduction for problem statement hypothesis:

In a research, a hypothesis is an elective detail of a phenomenon.

A null hypothesis is a hypothesis which a researcher aims to challenge. Generally, the null hypothesis defines the current vision/explanation of a feature of the world that the researcher wants to test.

Research methodologies involve the researcher provided that an elective hypothesis, a research hypothesis, as an elective method to give details of the phenomenon.

Statement of the Problem:

We repeatedly want to write a section in our research proposal / thesis.
This section can be restricted presently in one sentence or can be small paragraphs long extending to more than one page.
Whatever the statement length is, it must describe the problem.
It must also justify the problem.
We may contain some key problem statements and the remains we do in this section is complicated our problem statements adding other personally linked dimensions of our research problem.
This section that gives the circumstance of our research problem and situated our research in a wider environment.
A good hypothesis:

A good hypothesis must be in a declarative sentence form shows the relationship among variables; the conditional statement never be a hypothesis.
It should be calculable and empirically testable, summarizing and with particular meaning (clarity is get by means of definitions)
It must be connected with some hypothetical or conceptual or analytical framework or tools.

Example for Statement Problems in Hypothesis:
Q 1 :      How can we write down a conclusion in a science report when avoiding it to be a hypothesis?

Ans :   A conclusion is transient from one true statement to another one. We can suggest that the sun will rise tomorrow, as it is. Our hypothesis is if we do a certain experiment then we may obtain this result; or we may not. That is why a hypothesis should be tested.

Q 2 :      In a science class, we were asking to write three hypotheses. The laboratory is called Osmosis Lab. It’s about an egg located in vinegar, syrup and water. How can we write these hypothesis in a 'If and then' Statement?

Ans :     Hypothesis is regarding making a guess and then doing the experiment to look if our guess is correct. For this we need to put something besides the lines of 'If an egg is located in water, then… will happen' our professor needs you to write a hypothesis for an egg located in every material.

Solve Math Problems Instantly

Introduction to solve math problem instantly:

Mathematics is the study of quantity, structure, space, and change. Mathematicians seek out patterns, formulate new conjectures, and establish truth by rigorous deduction from appropriately chosen axioms and definitions. Mathematics is used throughout the world as an essential tool in many fields, including natural science, engineering, medicine, and the social sciences. (Source: From Wikipedia). Now, we are going to see some of the of the math problems instantly.

Solving Math Problems Instantly:

Example problem 1:

Solve an equation for the variable t: 3t + 11 = 4t – 22

Solution:

3t + 11 = 4t – 22

Subtract 11 on both sides of the equation

3t + 11 - 11 = 4t – 22 – 11

3t = 4t – 33

Subtract 4t on both sides of the equation

3t – 4t = 4t – 33 – 4t

-1t = -33

Divide by -1 on both sides of the equation

`(-1t) / -1 = (-33) / -1`

t = 33

So, the answer is t = 33.

Example problem 2:

Evaluate the expression.

(-1x+3y+z)-(x-3z); when x=1, y=2, z= 0

Solution:

Substitute the x, y and z values in the given expression.

(-1x+3y+z)-(x-3z) = (-1(1) + 3(2) + 0)-((1)-3(0))

= (-1+6+0)-(1-0)

= 6-1

= 5

So, the answer is 5.

Few more Solving Math Problems Instantly:

Example problem 3:

Solve the inequality: 4x – 12 < 28

Solution:

4x – 12 < 28

Add 12 on both side of the inequality

4x – 12 + 12 < 28 + 12

4x < 40

Divide by 4 on both side of the inequality

`(4x) / 4 < 40 / 4`

x < 10

So, the solution is (-infinity, 10).

Example problem 4:

Find the area of the rhombus whose side is 14.5cm and whose altitude is 6cm.

Solution:

Base is 14.5cm and height is 6cm.

Area of the rhombus = base *height = 14.5 *6 = 87cm2.

Therefore, the area of the rhombus is 87cm2.

Practice Problems to Solve Math Problems Instantly:

1)      Simplify the expression: 2a – 3b + 4a – 5b. (Answer: 6a – 8b)

2)      Solve for the variable n: n + 2 = 14 + 4n (Answer: n = -4).

3)      Solve for the variable q:  2q + 5 = 13 (Answer: q = 4).

Inverse of Function Solver

Introduction to inverse function solver

If  ' f ' is a relation from A to B , then the relation  { (b,a) : (a,b) `in` f } is denoted as f-1 .

If ' f ' is a relation from A to B , then  (f   -1)-1  =  f   .

Theorem : If   f : A `|->` B is one-one , then f -1 is bijection from f(A) to A .

Proof : Let  f : A `|->` B be one-one .

Clearly f  -1  is a relation from   f(A) to A .

Let  b `in` f(A)  . then there exists  a `in` A such that   f(a) = b . Since f is one-one , a is the only element of A such that  f(a) = b . Thus given b `in` f(A) there is a  unique element a in A such that  (a,b) `in` f . Hence given b `in` f(A) there is unique element a `in` A such that  (b,a) `in` f-1 . Hence f-1 is function  from f(A) to A  and  f-1 (b) = a if and only  if f(a) = b . Clearly f-1 is an onto function . If  b1 , b2 `in` f(A) and f-1(b1) = f-1(b2) = a  say , then b1 = f(a) = b2 . Thus f-1 is one to one . Therfore  f-1 : f(A) `|->` A is bijection .

Definition : If  f : A `|->` B  is a bijection then the relation f -1 : { (b,a) : (a,b) `in` f } is a function from B to A and is called the inverse function of f .

Let us see some problems in inverse function solver.

Examples on Inverse Function Solver

Problem: Find the inverse function of   f(x) = 4x +7 ?

Solution :   Given  f(x)  =  4x  + 7

Let   y  =  f(x)    `rArr`   x  =  f-1(y) .

`rArr`    y   =  4x   +   7  
`rArr`     x   =  `(y-7)/(4)`
`rArr`     f-1(y)   =  `(y-7)/(4)`

`:.`   f-1(x)  =  `(x-7)/(4)`

Answer :  f-1  =  `(x-7)/(4)` .

Problem: Find the inverse function of   f(x) = e4x+7  ?

Solution : Given  f(x) = e4x+7

Let    y  =  f(x)   `rArr`   x  =  f-1(y)

`rArr`    y   =  e4x+7  
`rArr`    log y =  4x + 7
`rArr`     x  =  `(logy-7)/(4)`    =  f-1(y)

f-1(x)  =  `(logx-7)/(4)` 

Answer : f-1(x)   =  `(logx-7)/(4)`

Is this topic prime numbers list up to 10000 hard for you? Watch out for my coming posts.

Solved Problems on Inverse Function Solver

1) If A = { 1 , 2 , 3 } , B = { a , b , c } and  f = { (1,a) , (2,b) , (3,c) } , then find the inverse function of f ?

Solution : Given    A = { 1 , 2 , 3 } , B = { a , b , c }

Also given the function  f = { (1,a) , (2,b) , (3,c) }

Now , inverse of f  i.e. f-1  =  { (a,1) , (b,2) , (c,3) }

and f-1 : B `|->` A is also a bisection .

Answer :   f-1  =  { (a,1) , (b,2) , (c,3) }

2) If A = { 1 , 2 , 3 } , B = { a , b , c, d }  and g = { (1,b) , (2,a) , (3,c) } . Find the inverse function ?

Solution : Given        A = { 1 , 2 , 3 } , B = { a , b , c, d }  and g = { (1,b) , (2,a) , (3,c) }

Here g-1 does not exist .

g : A `|->` B  is one to one but not onto .  g-1 is bijection from { a, b c }  to A .

Reasoning and Problem Solving

Introduction to reasoning and problem solving :

Problem solving is defined as the process by which the new situation is analyzed and resolved. It begins with an understanding of all the aspects of the problem and ends when a satisfactory answer has been found. Problem solving is not only an exercise to be carried out in the classroom, but refered as a skill that is used continually in business and daily life.

Description for Reasoning and Problem Solving:

Procedures for the reasoning problems:

1. Read the problem fully first to understand the concept that given.

2. Assume the conditions that needed and the reasons that occur in the flow of the problem execution.

3. Formulate a plan for to proceed for giving the solution.

4. Solve the problem as we discussed in the above way.

5. Check the solution completely that acquires the reasons that required in the problem.

Reasoning and Problem Solving-description for Problem Solving:

Procedures for the problem solving:

1 Read the problem fully first to understand the concept that given.

2 Formulate a plan for to proceed for giving the solution.

3 Solve the problem as we discussed in the above way.

4 Check the solution completely that acquires the standards that require in the problem.

Example for reasoning and problem solving:

At dinner, Mahaa bought a mango for 3.50, apple for 1.49, and lime salad for 1.75. If he gave the cashier a 10rupees for the  bill, how much change should she receive?

Solution

Use the problem-solving steps.

Understand

Mahaa bought 3 items for lunch and paid with a 10rupees bill. The problem asks how much change she should get back. The information given is

• Mango cost 3.50

• Apple cost 1.49

• lime salad cost 1.75

• Amount given to cashier 10.00 rupees

Plan

How could the answer be found?

First we shall add the costs of the three items.

Then subtract the total from 10.00 rupees.

Reasoning

Mahaa give 10 rupees assuming the need. If he give less than the amount, it gets demended.

Solving

The total costs about 6.74 rupees and then the remaining amount would be 3.24 rupees.

Solution

Mahaa should receive 3.24 rupees.

Positive Integer Exponents

Introduction of exponents:

The exponents are which integer is placed in the power of base numbers. It can be easily represent as, “a small number to the right side and above the base number”. It is called as exponents. These exponents have some of important rules and laws. Product of different powers, Power of power; these two are important rules in positive integer exponents. Here we are going to explain those two positive integer exponent rules.

Explain for Positive Integer Rule of Exponents:

If we are having any variables, which is containing the exponents and it have equal bases means, we can do some mathematical operations with the exponents. Those operations are called as the “laws of exponents” or “rules of exponents”. In this rule based positive integer exponents are defined as following ways,

Definition for positive integer exponents:

This positive exponent rule is defined as, if m is a positive integer and x is a non-zero base number, then it can be denoted as, Xm

Where as x is base number and m is power. Xm = x * x * x* x…. * x, m times.

This is the basic for positive integer exponents.

For example:

1) 32

= 3 * 3 

= 9

32 = 9.

2) 123

= 12 * 12 * 12

= 1728

123= 1728.

This kind of exponentiation used for discovers the positive integer exponents and simplify the problems.

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Positive Integer Exponents Used In:

And these positive integer exponents are commonly used in the two exponents rules, there are

Product of different powers,
Power of power.
These two rules are explained below with formulas and examples.

1. Product of different powers:

In this exponent rule, all the numbers must have in the same bases. Xm Xn = Xm+n

Example: 52. 53 = 55

= 3125.

2. Power of power:

In this power of power exponents rule has the base number having one power, and it has the one more whole power value. (Xm) n = Xmn

Example: (33)2 = 36 = 729.

Multiplying Binomials Answers

Introduction to multiplying binomials answers:

In mathematics, binomials are one of the interesting topics in polynomials. Binomials are defined as the process of sum or difference of more than one monomial or it can be defined as two monomials. Multiplying is one of the basic arithmetic operations. Let us solve some example problems for multiplying binomials with answers.

Example for Binomials:

(x + 2)

Multiplying Binomials Answers - Different Methods:

Different methods for multiplying binomials are,

Distributive method
Vertical method
Grid method
Foil Method

Distributive method:

It is the defined as multiplying each term in the first binomial is multiplied with the each term in the second binomials.

Vertical method:

The process of multiplying the given binomials in vertical and combine the like terms are called as vertical method.

Grid method:

The process of multiplying the given binomial values in the table form is called as grid method. In the grid, row or column can be replaced by each term. Multiply the rows and columns in the grid and the result of the like terms can be combined together.

Foil method:

FOIL is defined as First Outer Inner Last

It performs the multiplication operation in the way of first term, outer term, inner term and then last term in the each parentheses.

Multiplying Binomials Answers - Example Problems:

Some example problems for multiplying binomials are,

Example 1:

Determine the answers for the given binomials using distributive method for multiplying binomials

(3x + 7) (4x - 12)

Solution:

Given

(3x + 7) (4x - 12)

Distributive method

First term 3x in the first binomial can be multiplied with the second binomial terms

3x . (4x - 12)

12x2 – 36x

Second term 7 in the first binomial can be multiplied with the second binomial terms

7 (4x - 12)

28x - 84

Combine the like terms

12x2 - 36x + 28x  - 84

12x2 - 8x - 84

Answer:

12x2 - 8x - 84

Example 2:

Determine the answer for the given binomials using vertical method for multiplying binomials

(7x – 6) (x – 14)

Solution:

Given

(7x – 6) (x – 14)

7x – 6

x – 14           

7x2 – 6x                  multiplying the first binomial 7x – 6 from the second binomial term x

- 98x + 84            multiplying first binomials 7x – 6 from the second binomial term – 14

7x2 – 114x + 84            combine the like terms then we get result

Answer:

7x2 – 1140x + 84

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Example 3:

Determine the answer for the given binomials using grid method for multiplying binomials

(8x + 3) (2x + 5)

Solution:

Given

(8x + 3) (2x + 5)

Combine the like terms

6x + 15 + 16x2 + 40x

16x2 + 46x + 15

Answer:

16x2 + 46x + 15

Example 4:

Determine the answer for the given binomials using Foil method for multiplying binomials

(3x + 5) (8x + 12)

Solution:

Given

(3x + 5) (8x + 12)

Multiply the First term in the given binomials

3x (8x)

24x2

Multiply the outer terms in the given binomials

3x (12)

36x

Multiply the inner terms in the given binomials

5 (8x)

40x

Multiply the last term in the binomial

5 (12)

60

Add the like terms

24x2 + 36x + 40x + 60

24x2 + 76x + 60

Answer:

24x2 + 76x + 60

Sine Unit Circle

Introduction for sine unit circle:

Let us see about the sine unit circle. The main introduction of generalized unit circle functions can be achieved by restricting consideration to sine on the unit circle, or to periodic functions. The scientist Mikusiniski has been developed the half-line of the sine unit circle.

The unit circle is one kind of circle.  Particularly the trigonometry functions are classified the unit circle like sine unit circle, cosine unit circle, and tangent unit circle. The sphere’s radius is centered at the basis (0,0) in the system of  Cartesian. The unit circle has been separated into four quadrants.

Definition of Sine Unit Circle:

Assume, If (x,y) is some point on the unit circle, and t is the aimed at distance from (1,0) to (x,y) calculated counterclockwise along the circumference of the unit circle. The triangles are formed on the unit circle can also be utilize to exhibit the periodicity of the functions. The sine function may get the positive side of the unit circle. Then,

Cos (t) = x

and, sin (t) =y.

The relation of sin and cos function is,

Cos2 (t) + sin2(t) =1.

The sine representation in unit circle is,

Sin t = sin (2  +t).

Where the k indicates integer.

Sine Function - Sine Unit Circle:

The sine function is signifying into the unit circle. The sine function can be represented in y-axis. The sine function having the value of ? / 2.

The sine function is represented below:

The trigonometry sine function f(x) = a* sin (bx +c) +d.

Where, a,b,c,d are the parameters.         

Examples:

Problem 1:

Is there a point of P(x) that does not have any values for its x or y synchronizes? The x and y synchronize is sin (x), what is the domain of sin (x)?

Solution:

No, some point of P(x) on the unit circle has an x with y co-ordinates. The field of sin(x) is given by the name interval (-

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Problem 2:

Discover the x-intercepts, the highest and smallest amount of the graph of sin(x) using unit circle.

X-intercepts the unit circle since the sin (x) of each value of x = k*pi.

Maximum or highest amount of sin (x) of every value of x such that x= pi/2 +k *(2* pi)

Minimum or lowest amount of sin (x) of every value of x such that x =3*pi /2 +k *(2*pi).

Axioms of Real Numbers

Introduction to axioms of real numbers:

The two distinct real numbers are combined by means of addition and multiplication. These real numbers are must satisfy the following axioms. Many number of  axioms are followed by the real numbers. In this article we are going to see about axioms of real numbers with some examples and practice problems.

Types of Axioms

The following Laws are axioms of real number

Associative Addition Law                       : a + (b + c) = (a + b) + c
Associative Multiplication Law               : a * (b * c) = (a * b) * c
Additive identity                                     : a+0 = 0+a = a
Multiplicative identity                             : a *1= 1*a=a
Commutative Addition Law                    : a+b  = b+a
Commutative Multiplication Law            : a * b = b * a
Inverse  Law                                           : a + (-a) = 0
Distributive Law                                      : a * (b + c) = a * b + a * c

Examples for Axioms of Real Numbers:

Let take three real numbers 2 and 4 and 6

Problem1:

Apply the real numbers 2 and 4 and 6 in associative addition and Multiplication Laws

Solution:

As per Associative Addition Law: a + ( b + c ) = ( a + b) + c

Here a = 2 and b = 4 and c = 6

2+ (4+6) = (2+4)+6

2+(10) = (6)+6

12 =12 so these real numbers satisfy the associative addition law.

As per Associative Multiplication Law : a * (b * c) = (a * b) * c

2*(4*6) = (2*4)*6

2*(24) = (8)*6

48 = 48 so these real numbers satisfy the Associative Multiplication Law

Problem 2:

Apply the real number 2 in  Additive identity and Multiplicative identity

Solution:

As per Additive identity : a+0 = 0+a = a

a = 2

2+0 = 0+2 = 2 So the  Additive identity proved.

As per Multiplicative identity : a *1= 1*a=a

a = 2

2*1=1*2 = 2 hence proved.

Problem 3:

Apply the real numbers 2 and 4 in commutative Addition and Multiplication Laws

Solution:

As per commutative  Addition Law : a+b = b+a

a = 2 and b=4

2+4 = 4+2

6 = 6

As per commutative Multiplication Law : a * b = b * a

2*4 = 4*2

8 = 8 hence proved.

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Practice Problems for Axioms of Real Numbers:

Problem 1:

Apply the real numbers 8 and 5 and 9 in Distributive Law

Ans:112

Problem 2:

Apply the real number 9 in Inverse Law

Ans:0

Extended Real Number

Introduction for extended real number:
In mathematics, extended real number is derived for the real number system. The real number system is referred by R. The extended real number is represented as +8 and -8. The extended real numbers are not real numbers. The extended real numbers are used in calculus and mathematical analysis for limiting.

Measure and Integration for Extended Real Number:

In measure theory, it is used to infinite measure and integrals whose value will be infinite. In measure and integration extended real numbers are show as below

`int_1^oofdx/x`

this limit sequence is considered as,

fn (x) = `{ 2n(1-nx), if 0<=x<=1/n; 0, if 1/n
Arithmetic Operations for the Extended Real Number:

The arithmetic operations for the extended real number is stated as below.

a + `oo` = + `oo` + a  = +`oo` , a `!=` -`oo`

a - `oo` = - `oo` + a  = -`oo` , a `!=` +`oo`

a `xx` (`+- oo` ) = (`+- oo` ) `xx` a  = `+-oo` , a in(0,+`oo` )

a `xx` (`+- oo` ) = (`-+ oo` ) `xx ` a  = `-+` oo, a `in(-oo , 0)`

`a/+-oo` =0, a in `RR`

`+-oo/a` = `+-oo` , a in `RR^+`

`-+oo/a ` = `-+oo` , a in `RR^-`

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Algebraic Properties for the Extended Real Number:

The algebraic properties for the extended real number is listed below

a + (b + c) and (a + b) + c are either equal or both are infinite.
a + b and b + a are either equal or both are infinite.
a × (b × c) and (a × b) × c are either equal or both are infinite.
a × b and b × a are either equal or both are infinite
a × (b + c) and (a × b) + (a × c) are equal if both are finite.
if a = b and if both a + c and b + c are finite, then a + c = b + c.
if a = b and c > 0 and both a × c and b × c are finite, then a × c = b × c.

Subtracting Algebraic Expressions

Introduction to subtracting algebraic expressions :

Definition of algebraic expressions:

An algebraic expression is a mathematical expression that is written using one or more variables and constants, an algebraic expression which is used symbols to stand for numbers or abstract concepts for use in operations similar to arithmetic. It is made up of the symbols and signs of algebra. These characters include the Arabic numerals, literal digits, and the signs of operation. An algebraic expression is a sequence of algebraic phrases connected by mathematical symbols with no equal or difference sign. An algebraic expression is an statement, which contains one or more numbers, one or more variables, and one or more arithmetic.

Let us see the operation of subtracting algebraic expressions.

Problems on Subtracting Algebraic Expressions

We will see a step by step process of subtracting algebraic expressions

Example 1 on subtracting expression : Given: - [7(a - 2b) - 4b]

Solution: Step 1: - [7a - 14b - 4b] (By using distributive property 7 times (a - 2b))

Step 2: - [7a - 18b] (To get this step we add - 14b and - 4b)

Step 3:  - 7a+18b (So the answer is  - 7b+18b).

Example 2: Subtracting expression this 24ab – 10b – 18a from 30ab + 12b + 14a.

Solution: Step 1: 30ab + 12b + 14a – (24ab – 10b – 18a)

Step 2: 30ab + 12b + 14a – 24ab + 10b + 18a

Step 3: 30ab – 24ab + 12b + 10b + 14a + 18a

Step 4: 6ab + 22b + 32a.

Example 3: Subtracting this expression (3x + 2a - 4x) from (5a + 6x + 3x)

Solution: Step 1: 3x + 2a - 4x - 5a - 6x - 3x

Step 2: 3x - 13x + 2a - 5a.

Step 3:  - 10x - 3a.

Step 4: - (10 + 3a).

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Example Word Problem on Subtracting Algebraic Expressions

Given: A videotape shop charges 5 for renting a video cassette for the first day. It charges 2 extra for every additional day. Find the solution of subtracting algebraic expression shows you the quantity that Nick would pay if he rents a video videotape for n days?

Solution: Step 1: 5 for the earliest day and 2 extra for every additional day.

Step 2: Nick would pay 5 + (n - 1) × 2 for renting a video videotape for n days.

Step 3: So, the algebraic expression that communicates to the quantity that Nick would pay is (3 + 2n).

Math Integers Rules

Introduction for Math Integer Rules:

Numbers play a vital part in math. There are different types of numbers. In math, integers are one of the types of numbers. Basically integers in math are classified into two types. They are negative integers and positive integers. The numbers 1,2,3,4,5,… are said to be positive integers and the numbers -1,-2,-3,-4,-5,…. are the negative integers. There are several rules for using integers. In this article, we shall learn about integer rules in math. Also we shall solve math problems based on integer rules.

Rules for Integers in Math:

Addition Rules for Integer:

Positive + positive = positive
Negative + negative = negative
For these two conditions, add the numbers and put the respective signs.

Positive + Negative
Negative + Positive
For these cases, subtract the numbers and put the sign of the larger number.

Subtraction Rules for Integers:

Initially change the sign of the second number, and then perform the operations according to the sign.

After changing the sign, we can use addition rules.

Multiplication Rules of Integers:

Positive * positive = positive
Negative * negative = positive
For these two conditions, multiply the given numbers and put positive sign on the result.

Positive * negative = negative
Negative * positive = negative
For these cases, multiply the numbers and put negative sign in the result.

Division Rules of Integers:

Positive ÷ positive = positive
Negative ÷ negative = positive
For these two conditions, divide the given numbers and put positive sign on the result.

Positive ÷ negative = negative
Negative ÷ positive = negative
For these cases, divide the numbers and put negative sign in the result.

Example Problems Regarding Rules of Integers in Math:

Example 1:

Add: 4 + 5

Solution:

Here both the numbers are positive numbers. So we have to just add the numbers and put positive sign.

4 + 5 = + 9

Example 2:

Subtract: 4 – 5

Solution:

We can rewrite the expression as – 5 + 4

Take the – sign out :   –( 5 – 4)

= – 1

Since the sign of the larger number (5) is negative.

Example 3:

Multiply: 4* – 5

Solution:

Here both the terms have different signs.

We know that Positive * Negative = Negative

So, multiply the numbers and put the negative sign in it.

4 * – 5 = – 20

Example 4:

Divide:  – 8 ÷ 4

Solution:

Here both the terms have different signs.

Negative ÷ Positive = negative

So, divide the numbers and put negative sign.

– 8 ÷ 4 = – 2

Factoring Trinomials Practice

In mathematics, a rational function is any function which can be written as the ratio of two polynomial functions. Factorization (also factorisation in British English) or factoring is the decomposition of an object (for example, a number, a polynomial, or a matrix) into a product of other objects, or factors, which when multiplied together give the original. (Source: Wikipedia)

Factoring trinomial example problem 1:

Factorize the given polynomial expression x2 - 12x + 11

Solution:

Given rational expression is x2 - 12x + 11

First factorize the numerator value, we get

(x2 - 12x + 11) = (x2 - 11x - x + 11)

Grouping the first two terms and second two terms, we get

= (x2 - 11x) - (x - 11)

= x (x - 11) - 1 (x - 11)

= (x - 11) (x - 1)


The factors of the given polynomial expression is (x - 11) and (x - 1)

Answer:

The final answer is (x - 11) and (x - 1)
Factoring trinomials example problem 2:

Factorize the given polynomial expression x2 + 9x + 20

Solution:

Given rational expression is x2 + 9x + 20

Factorize numerator and denominator values, we get

First factorize the numerator value, we get

(x2 + 9x + 20) = (x2 + 4x + 5x + 20)

Grouping the first two terms and second two terms, we get

= (x2 + 4x) + (5x + 20)

= x (x + 4) + 5 (x + 4)

= (x + 4) (x + 5)

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The factors of the given polynomial expression is (x + 4) and (x + 5)

Answer:

The final answer is (x + 4) and (x + 5)
Factoring trinomials example problem 3:

Factorize the given polynomial expression x2 + 17x + 30

Solution:

Given rational expression is x2 + 17x + 30

Factorize numerator and denominator values, we get

First factorize the numerator value, we get

(x2 + 17x + 30) = (x2 + 15x + 2x + 30)

Grouping the first two terms and second two terms, we get

= (x2 + 15x) + (2x + 30)

= x (x + 15) + 2 (x + 15)

= (x + 15) (x + 2)


The factors of the given polynomial expression is (x + 15) and (x + 2)

Answer:

The final answer is (x + 15) and (x + 2)
Practice Problems for Factoring Trinomials Practice
Factoring trinomials practice problem 1:

Factorize the given polynomial expression x2 - 27x + 50

Answer:

The final answer is (x - 25) and (x - 2)

Factor polynomial expressions practice problem 2:

Factorize the given polynomial expression x2 - 20x + 100

Answer:

The final answer is (x - 10) and (x - 10)

Factor polynomial expressions practice problem 3:

Factorize the given polynomial expression x2 + 34x + 189

Answer:

The final answer is (x + 27) and (x + 7)

Factor polynomial expressions practice problem 4:

Factorize the given polynomial expression x2 + 37x - 120

Answer:

The final answer is (x + 40) and (x - 3)

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Dividing Trinomials by Binomials

Introduction to dividing trinomials by binomials:
In polynomial factor in irreducible polynomials others specified field. Factoring such like square-free factor is presents, however the irreducible factoring. Factoring depends strongly on the range of field. For example, the basic theorem of algebra, to point which all polynomials with complex coefficients include complex roots with integer coefficients recognize how to be totally strong to linear factors more the complex field. A polynomial is a term of finite length constructs from variables also constants, using some operations.

Dividing Trinomials by Binomials:

A trinomial is a polynomial representing of three terms. A trinomial is an equation concerning three expressions. A trinomial is the extension of a power of a computation of three expressions into monomials. The expansion is specified by  `(a+b+c)^(n) = sum_(i,j,k) ((n),(i,j,kk))a^ib^jc^k`     where n is a non negative integer also the computation is in use over all grouping of nonnegative index i, j, and k such to i+j+k = n. The trinomial coefficients are known with  `((n),(i,jk)) = (n!)/(i!j!k!)`             This method is a particular case of the multi nominal method for m = 3. The number of expressions of an extended trinomial is     `((n+2)(n+1))/(2)` Where n is the exponent.

They are used to explain polynomial functions. They are utilizing in calculus also numerical learn to near other functions. In composite arithmetic, polynomials are used to make polynomial rings, a central idea in abstract algebra with algebra geometry.

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Examples for Dividing Trinomials by Binomials:

Example 1:

How to dividing trinomials by binomials `(x^2+5x+6)/(x+2)`

Solution:

Step 1: the given factors are `(x^2+5x+6)/(x+2)`

Step 2: to factorize the trinomial x2+5x+6

Step 3: (x+2)(x+3)

Step 4: to dividing trinomial by binomial

`((x+2)(x+3))/(x+2)`

Step 5: (x+3)

So the solution is (x+3)

Example 2:

How to dividing trinomials by binomials `(x^2+4x+3)/(x+3)`

Solution:

Step 1: the given factors are `(x^2+4x+3)/(x+3)`

Step 2: to factorize the trinomial x2+4x+3

Step 3: (x+1)(x+3)

Step 4: to dividing trinomial by binomial

`((x+1)(x+3))/(x+3)`

Step 5: (x+1)

So the solution is (x+1)

Example 3:

How to dividing trinomials by binomials `(2x^2+7x+6)/(x+2)`

Solution:

Step 1: the given factors are `(2x^2+7x+6)/(x+2)`

Step 2: to factorize the trinomial 2x2+7x+6

Step 3: (2x+3)(x+2)

Step 4: to dividing trinomial by binomial

`((2x+3)(x+2))/(x+2)`

Step 5: (2x+3)

So the solution is (2x+3)

Geometry Triangle Congruence

Triangle is a polygon with three sides.Congruent triangles are special types of similar triangle which have both shape and size equal. If one triangle is placed on the other triangle it is fitted correctly on the first triangle.Two triangles are congruent if all parts of both the triangles are the same.

A triangle has 6 parts

3 sides
3 angles
So when two triangles are congruent all the 6 parts are congruent .

Introduction to geometry triangle Congruence:
In  two dimensional Plane Geometry, two triangles are congruent if all the corresponding parts of those triangles are similar.  The corresponding sides of two triangles are equal in measure and their corresponding angles are equal in degrees, we can say the two triangles are congruent.If triangle PQR is congruent to triangle XYZ, the relationship between these two triangles can be written as:?PQR `~=` ?XYZ

Congruence of Triangles in Geometry:
Postulates based on congruence of triangles:

Angle Angle Angle (AAA)

Side Side Side (SSS)

Side Angle Side (SAS)

Angle-Angle-Side(AAS)

Right-angle-Hypotenuse-Side(RHS)

Angle Angle Angle (AAA):

When three angles of the two triangles are equal, we can say that the two triangles are similar triangles.That is the corresponding angles have same measures.

Side Side Side (SSS):

When all three sides of both the triangles are equal, we can say that the triangles are similar triangles.

Side Angle Side (SAS):

When two sides in one triangle are equal to corresponding sides of the other triangle, and the included angles are equal, we can say that both are congruent triangle.

Angle-Angle-Side (AAS):

When two pairs of angles of two triangles are equal in measurement, and pair of corresponding unincluded sides are equal in length, then the triangles are called as congruent triangle.

Right-angle-Hypotenuse-Side (RHS):

When two right-angled triangles with their hypotenuses equal in length, and the shorter sides of two right triangles are equal in length, then the triangles are called as congruent triangle.

Properties of Congruence Triangles in Geometry:

If two triangles are congruent triangles in geometry, then each side or angle of the triangle is congruent to the corresponding part in the other triangle. Once if we proved two triangles are congruent, we can find the angles or sides of one of them from the other triangle.

To remember this we use the acronym CPCTC, which is expressed as “Corresponding Parts of Congruent Triangles are Congruent". In addition with sides and angles of two triangle, all other properties such as area, perimeter, location of centers, circles etc. of the triangle are the same.

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Example Based on Congruency of Triangles in Geometry:
Ex : 1From the figure state that the triangles are congruent or not . If congruent by which postulate

Sol:From figure
QR = 4cm = TV

PR = 5cm = SV

Two sides and the included angle are congruent

So by SAS postulate the triangles are congruent.

Solving Equations by Factoring

Introduction to solving equations by factoring:
What is an Equation ?

An equation is a mathematical representation of two expressions which are equal to each other. Every equation has two parts one is the

Left Hand Side and other is the Right Hand Side. The left hand side is connected to the right hand side by an equal sign.

Properties of an Equation

1) Any quantity can be added to both the sides of an equation.

2) Any quantity can be subtracted to both the sides of an equation.

3) Any quantity can be multiplied to both the sides of an equation.

4) Any non zero quantity can divide both the sides of an equation.


Solving Equations by Factoring - Different Methods

Equations can be solved in many ways.

1) By means of factoring.

2) By taking the roots.

3) By completing the square in case of quadratic equations.

4) By using the quadratic formula.

5) By using the method of graphing.

In this article we will discuss solving equations by means of factoring.

In this context lets first have the idea of ' Zero Factor Principle ' .

What is Zero factor principle ?

Zero factor principle states that product of two expressions is zero , if and only if either of the expression is zero.

Explanation : Let A and B be two expressions.

Product of A and B is AB.

AB = 0 , the relation holds true if and only if either A = 0  or B = 0 .

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Solving Equations by Factoring-examples

Now lets see some examples.

d  a = 5 / 3.

Examples :

1 )    x ^2   -  5x   -  6  = 0

Or,    x^2 - 6x + x - 6  = 0    [ Always break the middle term , such that even after breaking there is no change in the expression ]

Or,    x ( x - 6 ) + 1 ( x - 6 ) = 0

Or,    ( x - 6 ) ( x + 1) = 0

Now from Zero factor principle we can conclude that either ( x - 6 ) = 0   or  ( x + 1 ) = 0

When ( x - 6 ) = 0                                           When ( x + 1) = 0

Or,       x   =   6                                                           Or, x = - 1

Therefore the equation has two roots x = 6  and x = - 1

Explanation

Procedure to break the middle term.

First we will find the product of the coefficient of x^2 and the constant term in the equation.

In this case it is 1 * ( - 6 ) = -6

Now we will find two numbers that multiply to give -6 ( product of the coefficient of x^2 and the constant term )

and add to give -5 ( coefficient of x ).

The probable numbers are

a)  2 and -3 :     2 * ( - 3) = -6   but 2 + ( - 3 ) = -1 so it will not be accepted.

b)  - 2 and 3 :    ( - 2 ) *  3 = -6   but ( - 2 ) +  3  = 1 so it will not be accepted.

c)  6 and -1 :     6 * ( - 1) = -6   but 6 + ( - 1 ) = 5 so it will not be accepted.

d)  -6 and 1 :     1 * ( - 6 ) = -6   but 1 + ( - 6 ) = - 5 so it will  be accepted.

So that is why the middle term is represented as 5x  = - 6x + x .

2)     x^2   -  7x   +  12  =  0

Or,   x^2  -  4x   -  3x   +  12 = 0       [ Always break the middle term , such that even after breaking there is no change in the expression ]

Or ,   x ( x - 4) - 3 ( x  - 4 ) = 0

Or,    ( x - 4 ) ( x - 3 ) =  0

Now from Zero factor principle we can conclude that either ( x - 4 ) = 0    or   ( x - 3 ) =  0

When   ( x - 4 ) = 0                                                     When   ( x - 3 ) = 0         

Or,        x  =  4                                                                        Or,  x  =  3

Therefore the equation has two roots x = 4  and x = 3

Explanation

Procedure to break the middle term.

First we will find the product of the coefficient of x^2 and the constant term in the equation.

In this case it is 1 *  12  =  12

Now we will find two numbers that multiply to give  12( product of the coefficient of x^2 and the constant term )

and add to give -7 ( coefficient of x ).

The numbers are   - 4   and  - 3 :

( - 4 ) * ( - 3 ) = 12   and   ( - 4 ) + ( - 3 )   =   - 7   so it will be accepted.

So that is why the middle term is represented as   - 7 x  =   - 4x   -  3 x .

3)     6a^4  -  13a^3  + 5a^2  =  0

Or,    a^2 ( 6a^2 - 13a + 5 ) = 0

Or,    a^2 ( 6a^2 - 10a  -  3a  +  5 ) = 0

Or,    a^2 { 2a ( 3a - 5 ) - 1 ( 3a - 5 ) } = 0

Or,    a^2 ( 2a - 1 ) ( 3a - 5 )= 0

Now from Zero factor principle we can conclude that  a^2 = 0   or  ( 2a - 1 ) = 0    or   ( 3a - 5 ) =  0

When ( 2a - 1 ) = 0                                                   When ( 3a - 5 ) = 0                                  When a^2 = 0

Or,        2a = 1                                                                Or,       3a  =  5                                        Or, a = 0 

Or,          a =  1 / 2                                                           Or,        a = 5 / 3

Therefore the equation has  roots as a = 0  ,  a = 1 / 2  an

Decomposing and Composing Numbers

Introduction :

Decomposing and composing numbers is the very old concept in number system and mathematics. It means to take numbers apart into their tens and ones form. The smaller problem should be simpler than larger problem, so by using decomposing, large problems into simple problems. Larger problems can be tackled with ‘divide and conquer”. So we are using decomposing and composing numbers.


Discussion on Decomposing and Composing Numbers

Decomposition:

Decompose the problem so that:

Each sub problem is at the same level of detail,
Each sub problem can be solved independently,
The solutions to the sub problems can be combined to solve the original problem.
Composition:

Each sub problem is at the different level,
Each sub problem can be solved dependently not to be easy one,
The solutions to the sub problems cannot be combining to solve the actual problem.


Advantages of decomposing and composing numbers:

Different people can workout on different sub problems,
Parallelization may be possible,
Maintenance is easier.


Disadvantages of Decomposition and composition:

The solutions to the sub problems might not combine to solve the original problem,
Poorly understood the problems are hard to decompose.

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Some Example Problems on Decomposing and Composing Numbers

To begin the decomposing numbers you must take a 2- digit number apart, like this.

45 = 40+5

So, 45 is the same as saying 40+5, because the 4 in 45 stands for 40,

And the 5 in 45.

45 stand for 5,

Example 1: 53 + 25 =?

First add the tens ….. 50 + 20 = 70

Add the ones …            3 + 5   = 8

Then add the sums! And get the answer as,

70 + 8 = 78

Example 2: 154 + 32 =?

First add the hundreds, and then add tens and finally add ones….

100 + 50 + 30 + 4 + 2 totally add all, the sum is,

100 + 50 + 30 + 4+ 2 = 186, so get the answer is 186.

Example3: Choose the correct expression that will result in 818.

a)      200 + 200 + 200 + 10 +8

b)      250 + 250 + 200 +10 + 8

c)      200 + 200 + 200 + 200 + 10 + 8

d)      200 + 200 + 200 + 18

Solution:

Answer is c) 200 + 200 + 200 + 200 + 10 + 8

Steps to derive:

Simplify given answer chose one by one

200+200+200+10+18 = 618, and

250+250+200+10+8 = 718, and

200+200+200+200+10+8 = 818, and

200+200+200+18 = 618.

Hence the right option (choice) is “c”

So the answer is 200+200+200+200+10+8 represents = 818.

Standard deviation mean

In statistics and probability the standard deviation and the mean are related to each other. Mean is the average of the values of the data in case of ungrouped data and in case of grouped data, the mean is the weighted average of the data set. The standard deviation shows how dispersed is the data from the mean. So we say that the standard deviation is the measure of dispersion from the mean. The mean is said to be a measure of central tendency.

The mean is represented by the Greek alphabet ‘μ’ pronounced as ‘mu’. The standard deviation is also represented by another Greek alphabet ‘σ’ pronounced as ‘sigma’.

Standard deviation and mean formulae:
If there are n observations in a data set the values of which are, x1, x2, x3, … xn, then the mean is given by the formula:
μ = ∑(xi)/n, where i = 1,2,3,… n
The standard deviation is given by the formula:
σ = √[∑(xi – μ)^2/n]
Therefore from the above we can conclude that the standard deviation of the mean formula would be:
σ = √[∑(μ – μ)^2/n] = 0. That means, that the observation that is equal to the mean, does not deviate from the mean.

Coefficient of variation:
We know that two sets of data cannot be compared by comparing their means only. The measure of dispersion (standard deviation) should also be considered along with the mean. This comparison can be simplified using coefficient of variation.
Definition of coefficient of variation: Coefficient of variation is defined as the standard deviation divided by mean.  In other words coefficient of variation (denoted by Cv) is equal to the ratio of standard deviation to the mean. Therefore symbolically it can be written as follows:
Cv = σ/μ

Two sets of data can be compared with the help of coefficient of variation. The data set whose coefficient of variation is less is called more stable or consistent. This coefficient of variation is relative measure of dispersion and can also be denoted in percentage.
A standard deviation of the mean example is incomplete without the understanding of the coefficient of variation. To be able to come to the best decisions in accounting, finance, businesses, sports etc, it is very important that the mean, standard deviation and the coefficient of variation are all considered.

Derivative of Trigonometric Functions

Derivative is part of differentiation calculus. To calculate slope of curve made by X-axis and Y-axis is known as differentiation. One type differentiation of any function called derivative of that function. There are mainly three areas where derivative are used. First in increasing and decreasing functions. Second in maxima and minima. Maxima and minima are either relative or absolute type. Third use in Taylor’s and McLaren’s series expansion of function.

Here we discuss about derivative of trigonometric functions such as derivative of sec. by using trigonometric identities and rules differentiation we can find derivative of sec. to find derivative of sec quotient rule is useful. Now it is possible only when any arbitrary constant multiplied with the sec function. So we have to obtain derivative of secX.

First method is by using quotient rules and trigonometric identities means d/dx (secx) = secx.tanx. another method is, we take function secx is equal to any constant term such as secx=u. now we differentiate the term u with respect to x. when we differentiate constant term then function will also get differentiate one time. So finally we get du/dx=d/dx (secx) and we know d/dx (secx) = (secx.tanx). So the result is du/dx= (secx.tanx).

Now derivative of sec squared function. In this category we first take derivative of sec squared x means d/dx (sec^2x). To differentiate this function we write sec^2x as (secx) ^2. Now we carry out the power term and differentiate inner term secx. Again we use quotient rule to find derivative of secx. So we can easily write d/dx (sec^2x) =2secx.tanx.

Another example of derivative of sec squared function. Now we take d/dx (sec2X) ^2. Here we also carry out the power term but there is no direct formula for sec 2X. so we use chain rule method or we can split sec2X to its equivalent trigonometric term. Then it becomes in simple form of trigonometric function and we easily differentiate the term.

Now process to find derivative of cos 2X. Here we can use chain rule because it is more suitable for this function. The chain rule is (dy/dx =dy/du. du/dx). In this rule, first we focus on outside term cos2X and then focus on inside term (2X). We take first u(x) = (2x), by differentiating we get only 2 means du/dx=2. Now we take y (u) =cosu. By differentiating we get dy/du= -sinu. Now substitute all these result into chain rule. Finally we get dy/dx=2(-sinu) = 2(-sin2X).

Evaluating Double Integrals

What are Double Integrals?
Double integrals refer to the integration of a function with two variables. This integration takes place over two-dimensional space.

Evaluating Double Integrals
The direct method of evaluating double-integrals is to evaluate the double-integrals over the general rectangular region. But as this procedure is tedious, evaluating double-integrals is easily done by converting the integrals into polar coordinates. Double-integrals can also be calculated easily by downloading the double integral calculator available in the internet for free.

Given below is the procedure demonstrating how the double integrals of a function having two variables are evaluated over the general rectangular region:

Assume that there is a function g(a,b) where “a” lies in the limit [p,q] and “b” lies in the limit [r,s]. Here the triangle so formed in the two dimensional space n^2 is given by n = [p,q] x [r,s].  Initially it is assumed that the function g(a,b) is greater than or equal to zero and when we draw a graph using these values, we get a surface drawn over the rectangle n. Let us name this rectangular surface as m.

The double-integral of a function having two variables is found out by computing the volume of the rectangular surface m over a rectangular region n. Let us assume that there are u subintervals in [p,q] and v subintervals in [r,s]. These subintervals divide the rectangle n into small rectangles.  From each of these rectangles, a point (x i,y j) is taken and a box is constructed on each of these rectangles with the height given by the function f(x i,y j). These boxes will have the area Δa.  From this the volume of each box in the surface over the general rectangular region n can be computed using the formula f(x i,y j) Δa i.e by multiplying the function f(x i,y j) with Δa. The next step is to calculate the volume of the total surface. The summation of the volume of all the boxes in the surface gives the volume of the total surface. So for all the values of i ranging from 1 to u and j ranging from 1 to v, the volume is calculated and then added.  By adding the volumes of the boxes in both the directions i.e. x and y, we get double sum.

To get better results while evaluating double-integrals, you have to opt for larger values of u and v or the limit of u and v should extend to infinity.

Introduction to conic sections

This shows how the conic sections from Greek geometry are described today as the graphs of quadratic equations in the coordinate plane. The Greeks of Plato’s time described these curves as the curves formed by cutting a double cone with a plane; hence the name conic section.

Conic sections circles: A circle is the set of points in a plane whose distance from a given fixed point in the plane is constant. The fixed point is the center of the circle; the constant distance is the radius.

Conic sections ellipse: An ellipse is the set of points in a plane whose distance from two fixed points in the plane has a constant sum. The two fixed points are the foci of the ellipse.

Identifying conic sections: The standard form equations for circles, derived in Preliminaries from the distance formula d = sqrt[(x2 – x1)^2 + (y2 – y1)^2],  are these:

Circles: Circle of radius a centered at the origin: x^2 + y^2 = a^2.
Circle of radius a centered at the point (h, k): (x – h)^2 + (y – k)^2 = a^2.

The line through the foci of an ellipse is the ellipse’s focal axis. The point on the axis halfway between the foci is the center. The points where the focal axis and ellipse cross are the ellipse’s vertices.

If the foci are F1(-c, 0) and F2(c, 0) and PF1 + PF2 is denoted by 2a, then the coordinates of a point P on the ellipse satisfy the equation  sqrt((x + c)^2 + y^2) + sqrt((x – c)^2 + y^2 = 2a.
To simplify this equation, we move the second radical to the right hand side, square isolate the remaining radical, and square again, obtaining x^2/a^2 + y^2/(a^2 – c^2) = 1.
Standard form equations for ellipses centers at the origin
Foci on the x-axis: x^2/a^2 + y^2/b^2 = 1(a > b).
Center to focus distance: c = sqrt(a^2 – b^2),
Foci: (±c, 0), Vertices: (±a, 0),
Foci on the y-axis: x^2/b^2 + y^2/a^2 = 1(a > b),
Centre to focus distance = sqrt(a^2 – b^2) ,
 Foci: (0, ±c),
Vertices: (0, ±a).

In each case, a is the semi major axis and b is the semi minor axis.
Conic sections practice:
Major axis horizontal: The ellipse x^2/16 + y^2/9 = 1 has semi major axis: a = sqrt(16) = 4,
Semi minor axis: b = sqrt(9) = 3.
Center to focus distance: c = sqrt(16 – 9) = sqrt(7),
Foci: (±c, 0) = (±7, 0),
Vertices: (±a, 0) = (±4, 0),
Centre: (0, 0).

Antiderivative of Tan Y

Trig Antiderivatives
The antiderivative of a function refers to the indefinite integral of a function. The trig antiderivatives refer to the anti derivatives of trigonometric functions such as sin, cos and tan.  The trigonometric antiderivatives are formed based on the Antiderivative Rules.

Antiderivative of Tan y
The anti-derivative of tan y can be called as integration of tan y. The anti-derivative of tan y can be calculated by a three step process of applying trigonometric identities, substitution method followed by logarithmic identities.

The first step is to apply the trigonometric identity tan y = 1 divided by cot y. Another trigonometric identity says, cot y = cos y divided by sin y. By combining these two, we get tan y = sin y divided by cos y. This implies that anti-derivative or Indefinite integral of tan y dy = indefinite integral of sin y/cos y dy.

The next step is to use the substitution method to find the antiderivative of cos y. If we assume cos y equal to u, then we get its derivative as sin y equal to du. By substituting these values of cos y and sin y in the equation we derived in first step, we get antiderivative or indefinite integral of tan y dy equal to indefinite integral of -1/u du, which results in –log u added with the constant a. By substituting the value of u in this equation i.e. u = cos y, we get the negative of log (cos y) added with the constant a.

The final step is to apply the logarithmic or log identities. The logarithmic identity in which the summation of log p and log q is given by log (p*q) log p – log q will result in log (p/q). Using this identity, - log (cos y) can be written as log (1/cos y), which is equal to log (sec y) as 1 divided by cos y is nothing but sec y. Thus, the anti derivative of tan y is given by log (sec y) added with the constant a.

Antiderivative of Tan^2(y)
Once we have seen how to find the anti-derivative of tan y, let us see how to find the antiderivative of tan^2(y).

The antiderivative of tan^2 (y) is obtained by subtracting y from the value of tan y and then adding it with the constant a.  By applying the identity tan^2(y) = sec^2(y) -1, the anti derivative of tan^2(y) is rewritten as anti derivative of sec^2(y) - 1 which results in tan(y) - y + a.

Math trigonometry problems

Trigonometry is a branch of mathematics. Trigonometry is coined from the Greek words tri + gon + metry.  That means, three + angles + measures. From that we can see that in trigonometry we study about the measures of angles of a triangle. Usually to solve trigonometry problems, refers to finding the sides and angles of a triangle. A triangle as we know would have 3 sides and 3 angles.

Trigonometry gives us a relation between these sides and angles. In practice if we know any 3 of the six parts of a triangle (3 sides and 3 angles = total 6 parts) then we can find the remaining three parts of the triangle using trigonometry.

In schools, trigonometry is usually covered under pre calculus. It finds huge application in science subjects such as physics and organic chemistry. Trigonometry is applicable in study of co-ordinate geometry, conic sections like, circle, ellipse, parabola and hyperbola. The most common application of trigonometry would be in solving heights and distances type of problems. These type of problems, occur very frequently in our daily life, in construction work, in depth estimation of seas and lakes, in the use of radar to locate airplanes, etc.

It is because of these varied applications, that trigonometry is taught at school level itself.

Example trigonometry problems:

The following type of problems can be solved using trigonometry.
(1) A ship sends a signal to a submarine. The depth of the submarine is known and the angle of the signal is read from the signaling machine. Then the distance between the ship and the submarine can be calculated using trigonometry.
(2) A person is viewing a tall tower from a distance. If the height of the tower is known, and the angle of elevation of the top of the tower from the line of sight of the person is also known, then the distance between the person and the tower can be calculated using trigonometry.
(3) If two ships or air planes leave from the same point but in different directions. Then the distance between them at any point of time can be found if their respective bearings are known using trigonometry.

The basic concept of trigonometry is that, that the ratios of sides of angles of triangles with congruent angles are equal. In other words, we know that if two triangles have congruent angles, then the two triangles are said to be similar. The sides of similar triangles are proportional. Trigonometry is based on these concepts.