Example of Conjunction

Introduction to conjunction example:

The words which combine simple statements to form compound statements are called connectives. We use the connectives ‘and’, ‘or’, etc., to form new statements by combining two or more statements. But the use of these connectives in English language is not always precise and unambiguous. Hence it is necessary to define a set of connectives with definite meanings in the language of logic, called object language. Three basic connectives a conjunction which related to the English word ‘and’, ‘disjunction’ which related to the word ‘or’ ‘negation’ which corresponds to the word ‘not’.

Conjunction Example
Conjunction:

Use the symbol “?” to denote conjunction, “?” to denote disjunction and “~” to denote negation. If two simple statements p and q are connected by the word ‘and’, then the resulting compound statement ‘p and q’ is called the conjunction of p and q and is written in the symbolic form as ‘p ? q’.

Rule:

(A1) the statement p ? q has the truth value T whenever both p and q have the truth value T.

(A2) The statement p ? q has the truth value F whenever either p or q or both have the truth value F.

Ex 1:  Form the conjunction of the following simple statements

P: jasmine is beautiful.

q: Joseph is intelligent.

P ? q: Jasmine beautiful and Joseph is intelligent.

Ex 2 : Form the conjunction of the following simple statements

P: Joseph goes to college.

q: Jake is intelligent.

P ? q: Joseph goes to college and Jake is intelligent.

Ex 2 : Convert the following statement into symbolic form:

‘Kim and John are going to hotel’.

The given statement can be rewritten as:

‘Kim is going to hotel’, and

‘John is going to hotel’.

Let p: Kim is going to hotel.

q: John is going to hotel.

The given statement in symbolic form is p ? q.

Conjunction Practice Problems

Form the conjunction of

Pro 1:  p: Jessica reads newspaper,  q : Jessica plays cricket.

Pro 2:  p: I like coffee. q: I like ice-cream.

Probability Formula Wiki

Introduction to Probability Formula Wiki

The probability is the sketch of match up information or scheme that an occurrence will happen. The probability of an event A should be represented as P(A). The probability value is always the range from 0 and 1. The wiki answers are used for checking our resultant of problems. In this article, we are going to learn about formula and examples in probability wiki.

Formula for Probability Wiki

The formula for probability of an event A is measured as the following formula,

This is the formula that is used for calculating probability of any event.

Example Problems - Probability Formula Wiki

Example 1:

Jockey tossing three coins at concurrently. Calculate the probability of receiving one head or more than one tail?

Solution:

The sample space, when three fair coins tossed is,

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

n(S) = 8

Let us consider A will be the event of getting one head,

n(A) = {HTT, THT, TTH} = 3

Let us consider B will be the event of getting more than one tail,

n(B) = {HTT, THT, TTH, TTT} = 4

P(A) = `(n(A))/(n(S))` = `3 / 8`

P(B) = `(n(B))/(n(S))` = `4 / 8`

P(A or B) = P(A) + P(B)

= `3/8` + `4/8`

= `7/8`

= 0.875

Example 2:

Calculate the probability of selecting a letter ‘S’ from the word ‘RESPONSIBILITY’?

Solution:

Given word is, “RESPONSIBILITY”.

Here, total letters, n(S) = 14

Let consider G be the event of selecting a letter ‘S’ from the given word.

Total number of letter ‘S’, n(G) = 2

So, probability of selecting letter ‘S’, P(G) = `(n(G))/(n(S))`

= `2/14`

= `1/7`
Between, if you have problem on these topics how many prime numbers from 1 to 100, please browse expert math related websites for more help on solving systems of linear differential equations by elimination.
Example 3:

There are totally 34 handbags are in a shop. In those handbags, 8 are Black color, 12 are Maroon color, 5 are Grey color and remaining are Purple color. Calculate the probability for selecting i) Black color handbags ii) Grey color handbags?

Solution:

Total number of handbags, n(S) = 34

Number of Black color handbags = 8

Number of Maroon color handbags = 12

Number of Grey color handbags = 5

Number of Purple color handbags = 34 – (8+12+5)

= 34 – 25

= 9

Let us consider G be the event of selecting Black color handbags.

So, n(G) = 8

P(G) = `(n(G))/(n(S))`

= `8/34`

= `4/17`

Let us consider H be the event of selecting Grey color handbags.

So, n(H) = 5

P(H) = `(n(H))/(n(S))`

= `5/34`

That’s all about probability formula wiki.

Various Forms of the Equation of a Line

Introduction to various forms of the equation of a line:

Definitions

The standard form of the equation line is defines as

Ax+By =C

Where A,B and C are represented as real numbers.
The point slope form of the equation of a line is represented given below
The equation of a line with slope represents m and passing through the point (x1,y1) is given by
y-y1 = m (x=x1)

Where m is the slope and x1, y1 is the point given
The slope intercept form of the equation of a line:
The equation of a line with slope m and y intercept (0,b) is given by
Y= mx+b

Where m is the slope and (0,b)is the y intercept.
If an equation represents a straight line is referred as the equation of the line. The following conditions are used to solve the lne equation.
when the slope and  y intercept is given
when the slope and a point is given.
when two points given for the line

Practice Problems for Various Forms of the Equation of a Line:

Solving the Equation of a Line when slope and y intercept is given:

The slope-intercept form of a line is y= mx + b,
where m is the slope

b is the y-intercept.

Example

Find the equation of the line with slope -4 and y-intercept 6.

Solution:                m = -4  and  b = 6

The general equation is y = mx + b
Substitute the values into equation, we get
y = -4x + 6

Hence, the equation of the line is y = -4x + 6

Solving  the Equation of a Line if slope and a point is given:

various forms of the equation of a line - Example: 1

Find the equation of a line in slope intercept form, if the slope is  -3 and passes through (5, 8)

Step 1: Use the slope-intercept form of a line: y = mx + b

Given m = -3     Hence y = - 3x + b

Step 2 : Substitute values into equation:

The y-intercept was not given. However, we are given the point, (5, 8). Thus x = 5 and y = 8
Substitute the values to find the y-intercept b
y  =  mx + b
8  = -3(5) + b
8  = -15 + b
23 = b
Step 3: Solution

apply  the value of b, we get  y = -3x + 23

Hence, the equation of the line is y = -3x + 23

Solving the Equation of a Line passing through two given points:

Example:

Find out the line that passes through (3, 5) and (7, 25)

Steps:

1)      Use the two points to find the slope using slope formula

2)      Use the slope and either one of the points to find the value the y-intercept.

Step 1:                       Slope          =   (y2– y1) / (x2 – x1)

=   (25– 5) / (7 – 3)

= 20/4 = 5

Step 2 :  Let’s choose the point (3,5)
y   =  mx + b
5  =  5(3) + b
5  = 15 + b
5 – 15 = b
-10 = b

Substituting the values in the general equation, we get

y = 5x -10

Hence the equation of the line is y = 5x -10.


Example for Various Forms of the Equation of a Line:

Example Problem 1

solve the standard equation of the line that passes through the points
( -2, 5 ) and ( 3, 8 ).

Solution:

The first step is to determine the slope of the line. According to determine the slope of the line, we must use the formula
m = ( y2 - y1 ) / ( x2 - x1 ). This gives us,

m = ( y2 - y1 ) / ( x2 - x1 )
m = ( 8 - 5 ) / ( 3 - ( -2 ) )
m = ( 8 - 5 ) / ( 3 + 2 )
m = 3/5

Therefore, the slope of the line is equal to 3/5.

by applying the point-slope formula. To do so, we must choose one of the points, ( x1, y1 ), and insert it and the slope into the formula which will give us,

y - y1 = m ( x - x1 )
y - 5 = ( 3/5 )( x - ( -2 ) )
y - 5 = ( 3/5 )( x + 2 )

Finally, now that we have the equation of the line in
point-slope form, we will want to convert the equation into
slope-intercept form. ( This will allow us to determine the
y-intercept directly from the formula. )

y - 5 = ( 3/5 )( x + 2 )
y - 5 = ( 3/5 )x +6/5
y = ( 3/5 )x6/5 + 5
y = ( 3/5 )x6/5 + 25/5
y = ( 3/5 )x +31/5

`(3/5)` x-y+`31/5` =0 is the equation of the line is standard form.

Basic Math Division

Introduction to basic math division:

In math, algebra is a fundamental arithmetic operation (addition, subtraction, multiplication, and division) generally used in day-to-day life. Division is capable of considered as repeated subtraction or equal distribution. In these articles, we are going to discuss about basic math division.

Simple division can teach to find the how frequently one whole number called the divisor, is contain in an additional whole number, called the dividend, or to divide a whole number into some proposed number of equivalent parts, and is a small method of performing frequent subtraction.

Basic Math Division – Steps with Examples:

Division process is inverse of multiplication operation. The division is classifying in 2 methods, dividend, and divisor. 

Basic math division – Steps:

Dividend:

The number that is to be divided is known as dividend.

Example:  `15 / 3`

Here, 15 is the dividend.

Divisor:

The number that divides the dividend is known as divisor.

Example: `15 / 3`

Here, 3 is the divisor.

Quotient:

Quotient defined as the number of times dividend is divided by the divisor.

Example: `21 / 3 = 7`

Here 7 is the quotient.

Remainder:

Remainder defined as, the number which is left behind when the dividend is not completely divided by the divisor.

Example: `29 / 4`

`7 xx 4 = 28`

`28 + 1 = 29`

Here, the number 1 is the remainder.

Basic Math Division – Example Problems:

Problem 1:    

Dividing the given values,

` 980 -: 10`

Solution:

Dividing the given values,

`980 -: 10`

98       

--------

10   | 980

| 90

------

80

80

--------

0

---------

So, the final answer is 98 quotients and 0 is remainder



Problem 2:

Evaluate the given values,

`1216 -: 19`

Solution:

Dividing the given values,

`1216 -: 19`

64     

--------

19   |  1216

|  114

---------

76

76

--------

0

---------

So, the final answer is 64 quotients and 0 is remainder

Problem 3:

Dividing the given values,

`1817 -: 23`

Solution:

Dividing the given values,

`1817 -: 23`

79         

--------

23   |1817

|161

------

207

207

--------

0

---------

Therefore, the final answer is 79 quotients and 0 is remainder.

Problem 4:

Dividing the given values,

`1898 -: 13`

Solution:

Dividing the given values,

`1898 -: 13`

146       

--------

13 | 1898

| 13

------

59

52

--------

78

78                                                                    

---------

0

---------

Therefore, the answer is 146 quotients and 0 remainder

Example 5:

Dividing the given values

`3872 -: 22`

Solution:

`3872 -: 22`

176       

----------

22 |   3872

|   22

167

154

---------

132

132

----------

0

---------

Therefore, the answer is 176 quotients and 0 remainder

Quantitative Reasoning Math

Introduction to quantitative reasoning math

Quantitative reasoning is one of  the most useful application in mathematics concepts and skills.It is used to solve the  real-world problems. The students must solve this quantitative reasoning for improve their skills. It is very useful to solve the real life problems also.Here some of the quantitative reasoning problems has been solved. The practice problems are very useful to the students to improve their knowledge in the quantitative reasoning mathematics.

Number Theory Problems in Quantitative Reasoning Math

Find the largest number which perfectly divide 10110-1

Solution:

Given number is 10110-1

The simple way to solve the math problem is given below

1012 = 10201.
1012 - 1 = 10200. This is divisible by 100.

Similarly   for 1013 - 1 = 1030301 - 1 = 1030300.

So you can conclude that (1011 - 1) to (1019 - 1) will be divisible by 100.

(10110 - 1) to (10199 - 1) will be divisible by 1000.

Therefore, (10110 - 1) will be divisible by the large number 10^10

2.Find the minimum value of tiles required in a top floor of breadth 3 meters 78 cm and height 1 meters 74 cm?

Solution for math:

The tiles used in top floor are square. Therefore, the breadth of the tile = height of the tile. As we have to usewhole number of tiles, the side of the tiles should a factor of both 3 m 78 cm and 1m 74. And it should be the highest factor of 3m 78 cm and 1m 74.

3 m 78 cm = 378 cm and 1 m 74 cm = 174 cm.

The HCF of 378 and 174 = 6

Hence, the side of the square is 6.

The number of such tiles required =` (378xx174)/(6xx6)`

The total tiles required = 1827 marbles.

Percentage Problems in Quantitative Reasoning Math

If the price of book increases by 10% and Raj intends to spend only an additional 5% on book, by how much % will he reduce the quantity of book purchased?

Solution for math:

Let the price of 1 book be Rs.x and let Raj initially buy 'y' books.

Therefore, he would have spent Rs. xy on book

When the price of book increases by 10%, the new price per l book is 1.10x.


Raj intends to increase the amount he spends on book by 5%.

i.e., he is willing to spend xy + 5% of xy = 1.05xy


Let the new quantity of book that he can get be 'q'.

Then, `1.10x * q = 1.05xy`
Or `q = (1.05xy)/(1.10x)=(1.05)/(1.10)` y= 0.95y.


As the new quantity that he can buy is 0.95y, he gets 0.05y lesser than what he used to get earlier.

Or a reduction of 5%.

2. A student  who gets 10% marks fails by 5 marks but another student who gets 21% marks gets 6% more than the passing marks. Find the maximum marks.

Solution for math

From the given data pass percentage is 21% - 6% = 15%

By hypothesis, 15% of x – 10% of x = 5 (marks)

i.e., 5% of x = 5

Therefore, x = 100 marks.

Fraction Strips for Math

Introduction to fraction strips for math:

Fractions:

A fraction is a number that can represent part of a whole. The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, one part of four, and so on. A much later development was the common or "vulgar" fractions which are still used today (`1/2` , `2/3` , `3/4` , etc.) and which consist of a numerator and a denominator. (Source: Wikipedia)

Objective of fraction strips for math :

The main objective of this article is to learn the fraction strips for math. This article gives you types of fractions , some solved math problems on fraction strips .

Types of Fraction Strips in Math:

The following are three types of fractions in math,

Proper fraction

Improper fraction

Mixed fraction

Proper fraction:

Fractions are in the form a/b  ( where a, b are the integers), b is always greater than a(b > a) called as proper fractions.

Example: `7/8` , `16/26`

Improper fractions:

The fraction a/b ( where a, b are the integers) a is always greater than b( a > b)  called as improper fractions.

Example: `5/3` ,`12/5` , `156/46`

Mixed fraction:

The fractions are in the form of a b/c ( where a, b ,c are the integers) is called as mixed fraction. Here a is called as quotient, b is known as remainder and c is known as divisor.

Example: 2 `3/4` , 6 `7/5`

Problems on Fraction Strips for Math:

Problem 1:

Add the following negative fractions ` - 3/5 ` and `-9/5`

Solution:

Given, Add` - 3/5` and `-9/5`

That is `-3/5` + `( -9/5)` =  `-3/5 ` - `9/5`

= - ( `3/5 ` + `9/5` )

Here both fractions are have common denominator. So we can add the numerator juat like integers and keep the denominator as it is.

- ( `3/5` + `9/5` ) = - `( 3+9 )/ 5`

= `- 12/5`

Answer: `-3/5` - `9/5` =` - 12/5`

Problem 2:

Add the following negative fractions` -17/15` and `-21/ 26`

Solution:

Given, Add `-17/15` and` -21/ 26`

That is,` -17/15` + (` -21/ 26 ` ) = `-17/15`   `-21/ 26`

= - (`17/15`   +` 21/ 26` )

Here we are going to add `17/15` ,  `21/ 26` . But both fractions have different denominators. We cannot add directly. We need to make common denominator. For that we need to find the lcd of 15 and 26.

Multiples of 15 = 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195, 210, 225, 240, 255,…..390............

Multiples of 26 = 26, 52, 78, 104, 130, 156, 182, 208, 234, 260, 286, 312, 338, 364, 390, 416 ...

Here 390 is the least common factor of 15 and 26.

Multiply `17/15` by 26 on both numerator and denominator,

`17/15` = (17 * 26) / ( 15 * 26)

= `442 / 390`

Multiply `21/ 26` by 15 on both numerator and denominator,

`21/ 26` = ( 21 * 15) / ( 26 * 15)

=   `315 / 390`

Now we can add the given fractions,

- (`17/15`   + `21/ 26` ) = - ( `442 / 390 ` + `315 / 390` )

= - `( 442 + 315) / 390`

= -` 757 / 390`

Answer:  -`17/15` + ( `-21/ 26` ) = - `757 / 390`

Problem 3:

Multiply `(x+2)/( y-3)` * `(x)/(5y)`

Solution:

Given, `(x+2)/( y-3)` * `(x)/(5y)`

We can multiply just like integers, But multiply the Both numerator and multiply the both denominator separately.

`(x+2)/( y-3)` * `(x)/(5y)` = `(x(x+2))/(5y(y-3)) ` 

= `(x^2+2x)/(5y^2-15y)`

Answer: `(x+2)/( y-3)` * `(x)/(5y)` =`(x^2+2x)/(5y^2-15y)`

Vertical Line Slope

Introduction to vertical line slope:

Vertical line slope is nothing but the line which is parallel to y – axis. If the given line is parallel to the y – axis and it is perpendicular to x – axis we will say that is a vertical line. In this x value is constant it won’t get any variation the changes will be in y value. The equation is like x = a constant value. We will see some example problems for vertical line slope.

Example Problems for Vertical Line Slope:

We are having the general formula to find the slope of the line = `(y2 - y1) / (x2 - x1)`

Here (x1, x2) and (y1, y2) are the points which the line is passing.

Example 1 for vertical line slope:

Find the slope of the line which is passing through the following points (5, 8) ad (5, 9)

Solution:

We know if any line is having the constant value on x then it is known as vertical line. So the given line is a vertical line. We have to find the slope value of the given line.

Slope of the given vertical line =` (y2 - y1) / (x2 - x1)`

(x1, x2) = (5, 8) and (y1, y2) = (5, 9)

Slope of the given vertical line = `(9 - 8) / (5 - 5) = 1 / 0 ` = undefined.

So the slope of the given vertical line is undefined.

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More Problems for Vertical Line Slope:

Find the slope of the line which is passing through the following points (3, 16) ad (3, 24)

Solution:

Here x value of the given points are constant. So thhe given line is a vertical line. We have to find the slope of the given vertical line.

Slope of the given vertical line = `(y2 - y1) / (x2 - x1)`

(x1, x2) = (3, 16) and (y1, y2) = (3, 24)

Slope of the given vertical line = `(24 - 16) / (3 - 3) = 8 / 0 ` = undefined.

So the slope of the given vertical line is undefined.

From the above we can conclude slope of any vertical line is undefined.

Graphing Absolute Value Inequalities

Introduction to graphing absolute value inequalities:

Graphing absolute value inequalities is nothing but we are going to graph the solutions of the absolute value inequalities. To graph the absolute value inequalities first we have to find the solutions and using that we have to graph. Here we are going to learn about the graphing of absolute value inequalities. We will see some example problems for graphing the absolute value inequalities.

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Example Problems for Graphing Absolute Value Inequalities:

The main difference between the inequalities and equalities is we can say the range of the solutions in the inequalities. But in equalities we can determine the exact solution.

Example 1 for graphing absolute value inequalities:

Graph the absolute values of the inequalities. |x +2| `lt=` 8

Solution:

Given inequality is |x + 2| `lt=` 8

To find the absolute value for the given inequality

(x + 2) `lt=` 8 ………… (1) And –(x + 2) `lt=` 8 …………………. (2)

Equation 1:

(x + 2) `lt=` 8 ………… (1)

Add - 2 on both sides

x + 2 - 2 `lt=` 8 - 2

x `lt=` 6

Equation 2:

-(x + 2) `lt=` 8

-x - 2 `lt=` 8

Add +2 on both sides.

-x - 2 + 2 `lt=` 8 + 2

-x `lt=` 10

Divide by -1.

x `gt=` -10

So x lies between -10 `lt=` x `lt=` 6

We will see some more example problems bfor graphing the absolute value inequalities.

Example 2 for Absolute Value Inequalities:

Find the absolute values of the inequalities. |x - 6| `gt=` 5

Solution:

Given inequality is |x - 6| `gt=` 5

To find the absolute value for the given inequality

(x – 6) `gt= ` 5 ………… (1) And –(x – 6) `gt=` 5 …………………. (2)

Equation 1:

(x – 6) `gt=` 5 ………… (1)

Add +6 on both sides

x - 6 + 6 `gt=` 5 + 6

x `gt=` 11

Equation 2:

-(x – 6) `gt=` 5

-x + 6 `gt=` 5

Add -6 on both sides.

-x + 6 - 6 `gt=` 5 - 6

-x `gt=` -1

Divide by -1.

x `lt=` 1

So x lies between 1 `gt=` x `gt=` 11     

Quadratic Equations and Graphs

Definition of quadratic equation : An equation in which one or more of the terms is squared but raised to no higher power, having the general form ax2 + bx + c = 0, where a, b, and c are constants.

Quadratic Equations and Graphs:

The general form of a quadratic equation is

ax2 + bx + c =0

Where a, b and c are real numbers and a`!=` 0

It is called so because it has a second degree term.

The values of the variable involves in a quadratic equation are called its solution.

The following method are usually used to solve a quadratic equation.

completing square method
using the quadratic formula
Factorization method


In general, the shape of the graph of a quadratic equations a parabola.

The steps we follow to sketch the graph of the quadratic equation are:

1. Check if a > 0 or a < 0

If a > 0, then the parabola is u-shaped ie it opens upwards

If a < 0, then the parabola is  n-shaped ie it opens downwards

2. Find Vertex

The x-coordinate of the vertex is –b/2a

Substitute the value of x in the function to have the y coordinate.

Thus we have the (x, y) coordinates of the vertex.

3. Find y-intercept

The coordinates of the y-intercept can be found by substituting x = 0.

4. Find x-intercept

The coordinates of the x-intercept can be found by substituting y = 0 and solving the quadratic equation

5. Finding other coordinates

Substitute any possible value for x to get the corresponding value of y or vice versa

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Quadratic Equations and Graphs-example

Graph of the function y =  2x2 - 8x + 6

Here a= 2 b=-8 c=6

1. Since a > 0, then the parabola is u-shaped i.e. it opens upwards

2. The x-coordinate of the vertex is –b/2a = 8/(2*2) = 2

Substituting the value of x in the function y = 2x2 - 8x + 6 to have the y coordinate.

We get y = 2*22 - 8*2 + 6 = -2

So the vertex is (2,-2)

3. The coordinates of the y-intercept can be found by substituting x = 0.

We get y = 2*02 - 8*0+ 6 =6

So the coordinates of the y-intercept is (0,6)

4. The coordinates of the x-intercept can be found by substituting y = 0 and solving   the quadratic equation

We get

2x2 - 8x + 6=0

2(x2 - 4x + 3)=0

2(x - 1)(x - 3) = 0

So x = 1, or x = 3.

So the coordinates of the x-intercept is (1,0)  and (3,0)

5. Finding some other coordinates by substituting any possible value for x

We can now plot the graph.

Quadratic Equations and Graphs-graphs