Math Symbols Terms

There are many terms and symbols used in math. The symbols can be used to perform some operations such as addition, multiplication, subtraction, division; relationship symbols such as less than, greater than, equal, not equal, etc. The math terms can be used to describe the topics such as algebra, arithmetic, array, and axis, etc. Let us see about math symbols and terms in this article.

I like to share this Symbol for Correlation with you all through my article. 

Some of the Math Symbols

Addition:
It can be represented as +. It is used for addition operation and also for logical OR purpose. It can be read as plus or logical or.
Subtraction:
It can be represented as –. It is used for subtraction operation. It can also spelled out as minus.
Multiplication:
Multiplication can be represented as ×. It is used for multiply the given terms.  It can also spelled out as into.
Division:
It can be represented as ÷. It is used for dividing the given terms.  It can also spelled out as divide by.
Percentage:
It can be represented as %. It is used for calculating the ratio that compares to the number 100.  It can also spelled out as percent.
Summation:
Summation can be represented as `sum`   . It is used for calculating the sum of many or infinite values.  It can also spelled out as sum.
Dot Product:
It can be represented as (.). It is used for calculating the scalar (dot) product of two vectors.  It can also spelled out as dot.
Cross Product:
It can be represented as (×). It is used for calculating the vector (cross) product of two vectors. It can also spelled out as cross.



Algebra is widely used in day to day activities watch out for my forthcoming posts on Fibonacci Number Sequence and number to words converter. I am sure they will be helpful.

Some of the Math Terms

Algebra:
An algebraic equation represents the scale.
Algorithm:
A step-by-step problem solving technique is used in math computations.
Area:
It can be used for representing the two-dimensional shapes such as polygon, octagon, hexagon, triangle and circle.
Array:
A set of numbers can be occurred in a specific size of pattern. Matrix or array consists of columns and rows.
Axis:
Axis consists of horizontal and vertical axis having co-ordinates in a plane.
Attribute:
Describing the object with data consists of shape, color, or size.
Arc:
A circumference of the circle or the portion of a segment draws with a compass.

Maxima and Minima

Maxima and Minima are the largest value (maximum) or smallest value (minimum), that a function can take at a point either within a given boundary (local) or on the whole domain of the function in its entirety (global). In general, maxima and minima of a given set are the greatest and least values in that set. Together, Maxima and Minima are called extrema (singular: extremum). We need to learn minima to determine the nature of the curve or the function and various other applications like projectiles, astrophysics to microphysics, geometry etc.

Learning analytical definition of minima:

A function f(x) is said to have a local minima point at the point x*, if there exists some ε > 0 such that f(x*) ≤ f(x) when |x − x*| < ε, in a given domain of x. The value of the function at this point is called minima of the function.

A function f(x) has a global (or absolute) minima point at x* if f(x*) ≤ f(x) for all x throughout the function domain.

We need to learn minima points of a curve by observing the involved function.

learning prologue regarding minima:

To learn Minima & Maxima, one needs to have a basic knowledge of calculus. The following points are some bare necessary (maybe not sufficient) definitions.

A function, y = f(x) is a mathematical relation such that each element of a given set ‘x’ (the domain of the function) is associated with an element of another set ‘y’ (the range of the function).

Closed interval of a domain is defined as an interval that includes its endpoints, as opposed to open interval which is an interval that does not include its endpoints.

A function, f(x) is said to be continuous at a given interval if it can assume all values within the interval i.e. the function is not broken anywhere inside the interval. Mathematically we determine this by ensuring the function has a finite value at the given point and taking the limit on both sides of the point and checking if they both exist and are equal (L.H.L. = R.H.L.).

Differentiability of a function is out of the scope of this article, but simply put, a function is said to be differentiable at a point if the curve at that point is smooth i.e. there is no drastic change of slope. Mathematically this is achieved by checking if both the left hand derivative and the right hand derivative of the function at the given point finitely exist and are equal (incidentally this common value is the value of the derivative of the function at the given point).

First Derivative is defined as the differentiation of a function, y = f(x), once, with respect to ‘x’. It is denoted by dy/dx or f’(x) and simply put, it gives the slope of the function at any given value of ‘x’ or the instantaneous rate of change of the function w.r.t. ‘x’ at any given value of ‘x’.

Second Derivative is defined as the differentiation of a function, y = f(x), twice, with respect to ‘x’. It is denoted by d2y/dx2 or f’’(x) and simply put, it gives the slope of the slope of the function at any given value of ‘x’ or the instantaneous rate of change of the slope of the function w.r.t. ‘x’ at any given value of ‘x’.

Critical points of f(x) are defined as the values of x* for which either f'(x*) = 0 or f’(x*) does not exist.

Tests for Minima:

Local Minima can be found by Fermat's theorem, which states that they must occur at critical points.

If f(x) has a minima on an open interval, then the minimum value occurs at a critical point of f(x).

If f(x) has a minima value on a closed interval, then the minimum value occurs either at a critical point or at an endpoint.

Critical points of f(x) are defined as the values of x* for which either f'(x*) = 0 or f’(x*) does not exist.

One can distinguish whether a critical point is a local maximum or local minimum by using the first derivative test or second derivative test.

learning minima -First Derivative Test

Suppose f(x) is continuous at a critical point x*.

If f’(x) <0 an="" and="" extending="" f="" from="" interval="" left="" on="" open="" x="">0 on an open interval extending right from x*, then f(x) has a relative minima at x*.

If f’(x) >0 on an open interval extending left from x* and f’(x) <0 a="" an="" at="" extending="" f="" from="" has="" interval="" maxima="" on="" open="" p="" relative="" right="" then="" x="">
If f’(x) has the same sign on both an open interval extending left from x* and an open interval extending right from x*, then f(x) does not have a relative extremum at x*.

An interesting point to NOTE:

Differentiability is not a criterion for the first derivative test. Suppose f(x) is continuous but not differentiable at x*, i.e. f’(x*) does not exist. Still the above holds true since the test is done in open intervals on the left and right sides of the point in consideration [see Figure below]. So the criteria is only that f(x) is continuous at x* and that f’(x) exists in the neighbourhood of x*.

In summary, relative minima occur where f’(x) changes sign.

learning minima -The Second Derivative Test:

Suppose that x* is a critical point at which f’(x*) = 0, that f’(x) exists in the neighbourhood of x*, and that f’’(x*) exists.

f(x) has a relative minima at x* if f’’(x*)>0.

f(x) has a relative maxima at x*if f’’(x*) <0 .="" p="">
f(x) does not have an extremum at x* if f’’(x) = 0.

NOTE:

Differentiability at the critical point is a criterion for the second derivative test as opposed to the first derivative test. Also, if f’’(x*) = 0, the test is not informative [see Figure below], it actually means there is no change of sign of f’(x) on going from the left to right of the given critical point (these points are called the points of inflection).

learning Absolute Minima and Maxima

For any function that is defined piecewise, one can learn minima (or maxima) by finding the minimum (or maximum) of each piece separately; and then seeing which one is smallest (or biggest).

Visualization to learn minima

max-min

Least Common Denominator Tutoring

Least common multiple of two rational facts a and b is the least positive rational number, that is an integer multiple of a as well as b. because it is a multiple, it can be divided by a and b without a remainder. Tutoring not only help students by giving answers, but also help students in their problem solving with step by step solutions.  In this article we shall discuss about least common denominator tutoring. The following are the examples and steps involved in least common denominator tutoring.

Least common denominator tutoring:

Least common denominator:

The least common denominator of two or more numbers is the least number which is a multiple of each of the given number.

Technique: Least common denominator tutoring

To find the l.c.d.

Step (1) Write the multiples of first number

(2) Write the multiples of second number

(3) Write the common multiples

(4) Write the least common denominator.

Example problems least common denominator tutoring:

Example 1: Find the L.C.D of `1/4` and `1/3`

Here least common denominator is 12

To find the l.c.d.

Step (1) Get l.c.m. for every denominators.

Step (2) Modify equivalent fraction with same l.c.m denominator

Step (3)  By taking l.c.m common in denominator and add all numerators.

= `(1*3)/12` + `(1*4)/12` = `(4+3)/12`

So the final result is `7/12` .

Example 2: Find the L.C.D of `1/5` and `1/4`

Here least common denominator is 20

To find the l.c.d.

Step (1) Get l.c.m. for every denominators.

Step (2) Modify equivalent fraction with same l.c.m denominator

Step (3)  By taking l.c.m common in denominator and add all numerators.

=  `(1*4)/20 ` + `(1*5)/20`

So the final result is 9/20.

Example 3: Find the L.C.D of `1/25` and `1/15`

Here least common denominator is 75

To find the l.c.d.

Step (1) Write the multiples of 25: 25, 50,75

Step (2) Write the multiples of 15: 15, 30, 45, 60, 75

Step (3) common multiple is 75

Step (4) least common denominator is 75.

= `(1*3)/75` + `(1*5)/75` =`(5+3)/75`

So the final result is` 8/75` .

Practice problem for least common denominator tutoring:

Example 1: Find the L.C.D of `1/6` and `1/3`

Here least common denominator is 6

So the final result is` 3/6` .

Algebra is widely used in day to day activities watch out for my forthcoming posts on Least Common Multiples. I am sure they will be helpful.

Example 2: Find the L.C.D of `1/6 ` and `1/4`

Here least common denominator is 12

So the final result is `5/12` .

Example 3: Find the L.C.D of `1/4` and `1/2`

Here least common denominator is 4

So the final result is `3/4` .

A Proper Factor

A factor is a whole number which divides exactly by another whole number is called factor for  that number Any of the factors of a number, except the number itself. A factor is a portion of a number in general  integer or polynomial when multiplied by other factors  gives  entire quantity. The determination of factors is a factorization A proper factor of a positive integer is a factor of other than 1.Proper factor is a multiples for a given whole number which has multiples again it is aproper factor.Every whole number which has a factor of its own.proper factor is like a normal factor except 1
For example,
For 6, 2 and 3 are proper factors of , but 1 and 6 are not a proper factor.

Properties of proper factor:

The divisors for a any number other than 1 and  number itself are called  factors for that number.
A factor for N number is a number which divides  exactly N.

Example: the factors for 24 are 1,2,3,4,6,and 12

Generally for every number has itself and 1 as its factors.
When a number is greater than 1 and by itself and 1 as factors, then the number is prime.
A number or quantity that when multiplied with another number produces a given number or expression.

Example Problems for Proper factors:

Example 1 :
Find all the divisors and proper factors of  20

Solution :
The divisors of 20 are 1, 2, 4, 5, 10 and 20
The factors of 20 are 2, 4, 5 and 10

Example 2:
Find all the divisors and proper factors of 32.

Solution:
The divisors of 32 are 1, 2, 4, 8, 16 and 32
The proper factors of 32 are 2, 4, 8 and 16

Example 3:
Find Proper Factors of 30

Solution:
30=1*30
=2*15
=3*10
=4*15
=5*6
=6*5

Proper  Factors for 30 are 2,3,4,5,6

Example 4:
Find proper factors for 42

Solution:
42=1*42
=2*21
=3*14
=6*7
2,3,6

I am planning to write more post on Solve first Order Differential Equation and free 3rd grade math word problems. Keep checking my blog.

Example 5:
Find proper factors for 18

Solution:
18=1*18
=2*9
=3*6
=6*3
=9*2
Proper factors are 2,3,6,9

Equivalent Scale Factor

A scale area is a number which scales, or multiplies, some quantity. In the equation y=Cx, C is the scale factor for x. C is also the coefficient of x, and may be called the constant of proportionality of y to x. For example, doubling distances corresponds to a equivalent scale factor of 2 for distance, while cutting a cake in half results in pieces with a scale factor of ½.
SOURCE: WIKIPEDIA

Example problems of equivalent scale factor:

Equivalent scale factor problem 1:
Find the scale factor to the following figure:

Equivalent scale factor solution:
If we want to find the length of smaller rectangle then we can multiply the length of the one side of larger rectangle and the value of scale factor.
We can find the scale factor of the given rectangles by using the following formula,
Let D_l be the dimensions of larger rectangle and D_s be the dimensions of smaller rectangle and s be the scale factor.
Therefore the formula as,
                D_l*s=D_s
Substitute the values of dimensions into the above formula. Then we get
                  30*s=24
Divide by the value 30 on both sides,
                   `(30s)/30` `=` `24/30`
                       ` s ` `=` `24/30`
Divide by the value of 6 on both numerator and denominator. Then we get the value of scale factor.
                        s = `4/5 ` or 4:5
Therefore scale factor of smaller to larger rectangle= 4:5
Answer: 4:5
Equivalent scale factor problem 2:
Find the larger to smaller scale factor for the following figure:

Equivalent scale factor solution:
If we want to find the length of smaller rectangle then we can multiply the length of the one side of larger rectangle and the value of scale factor.
We can find the scale factor of the given rectangles by using the following formula,
Let D_l be the dimensions of larger rectangle and D_s be the dimensions of smaller rectangle and s be the scale factor.
Therefore the formula as,
                D_l*s=D_s
Substitute the values of dimensions into the above formula. Then we get
                  39*s=26
Divide by the value 48 on both sides,
`(39s)/(30)` `=` `26/39`
` s ``=` `26/39`
Divide by the value of 13 on both numerator and denominator. Then we get the value of scale factor.
                        s = `2/3` or 2:3
Therefore scale factor of larger to smaller scale rectangle= 3:2
Answer: 3:2

Algebra is widely used in day to day activities watch out for my forthcoming posts on Proof of Fundamental Theorem of Calculus and algebra 2 solver step by step. I am sure they will be helpful.

Practice problems of equivalent scale factor:

1. Find the scale factor to the following figure:

2. Find the larger to smaller scale factor for the following figure:

Answer:
1. 3:4
2: 6:5

Mode and Median Calculator

Mode: Mode is the value that takes place most repeatedly in the data set. Measure of central tendency is known as mode. If the data’s are given in the form of a frequency table, the class corresponding to the maximum frequency is called the modal class. The value of the variate of the modal class is the mode.

Median: The median is the middle value when the given values are arranged in an ascending order. Let us see the median and mode calculator.

Median and Mode calculator:

In the calculator enter the set of values in first box, after that clcik the median button it will automatically calculate the median value and it will be displayed in answer box. The same process is done for mode.

Examples on Mode calculator:

Example 1:

            Find the mode of 7, 4, 5, 1, 7, 3, 4, 6, and 7.

Solution:

           The above question is entered in the first box. The calculator doing the follwing process,

           Assemble the data in the ascending order, we get

            1, 3, 4, 4, 5, 6, 7, 7, 7.

            The number 7 occurs many times in the above values.
            Mode = 7 will display the answer box after press the mode button on calculator.
Example 2:

            Find the mode for 12, 15, 11, 12, 19, 15, 24, 27, 20, 12, 19, and 15.

Solution:

           The above question is entered in the first box. The calculator doing the follwing process,

           Assemble the data in the ascending order, we get

            11, 12, 12, 12, 15, 15, 15, 19, 19, 20, 24, 27.

            In the above values 12 occurs 3 times and 15 also occurs 3 times.

            ∴ Both 12 and 15 are the modes for the given data. We observe that there are two modes for the given data.The Mode will be displayed in answer box on calculator
Example 3:

            Find the mode of 19, 20, 21, 24, 27, and 30.

Solution:
            Already the above data are in the ascending order. Each value occurs exactly one time in the series. Hence there is no mode in the above given data.
These are the examples on mode calculator.

Examples on Median calculator:

Example 1:    
            Find the median of the following numbers: 12, 45, 62,10,14,31 and 43.
Solution:
           The above question is entered in the first box. The calculator doing the fololwing process,
            Arranging the given numbers in ascending order we get
            10, 12, 14, 31, 43, 45 and 62.
                            `darr`
                    Middle term
            Median = Middle item = 31.     

         The median 31 will display the answer box

Between, if you have problem on these topics Cubic Equation  please browse expert math related websites for more help on ibsat 2013.

Example 2:         
            Find the median of the following numbers: 3, 7, 4, 10, 22, 16, 21 and 5.
Solution:
            The above question is entered in the first box. The calculator doing the following process,
            Arranging the given numbers in ascending order we get
            3, 4, 5, 7, `darr` 10, 16, 21, 22              
                 Median is here
            Median = Item midway between 7 and 10
                       =` (7 + 10) / 2` = `17 / 2` = 8.5
          The Median 8.5 will display on answer box on calculator
These are the examples on mode and median calculator.

Sine Geometry

Trigonometry is the division of geometry dealing among relationships among the sides also angles of triangles. In geometry sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse. The  ratio does not depend on size of the particular right triangle chosen, as long as it contains the angle A, since all such triangles are similar

(Source: Wikipedia)

I like to share this sine curves with you all through my article. 

Sine geometry

In the above diagram ,
sin A =opposite/hypotenuse
Examples for sine geometry

A - Right angle of the triangle ABC.
The length of AB, BC and CA are frequently represented through c, a, b.
Obtain point B as middle of a trigonometric circle
Circle with radius = 1.
Now sin (B) are comparative to b, c also a.
sin `(B)/b` =`1/a`
sin (B) = `b/a`

Examples for sine geometry

Example 1
Angle of triangle is 20⁰, opposite side of triangle is 12 apply the sine geometry to find the unidentified side of the triangle?
Solution:

Angle of triangle= 20⁰  

Opposite side of triangle = 12.

sin A =opposite/hypotenuse

sin 20⁰ = `12/x`

sin 20⁰ x = 12

x = `12/sin 20^0`

x =`12/0.3420`    {since the value of sin 20 degree is 0.3420}

x=35.08

Hypotenuse side= 35.08

Example 2
Angle of triangle is 78⁰, hypotenuse side of triangle is 20 apply the sine geometry to find the unidentified side of the triangle?
Solution:

Angle of triangle= 78⁰  

Hypotenuse side of triangle = 20.

sin A =opposite/hypotenuse

sin 78⁰ = `x/20`

sin 78⁰ x 20= x

x = sin 78⁰ x20

x =0.97814x20     {since the value of sin 78 degree is 0.97814}

x=35.08

Opposite side = 19.56

Example 3
If hypotenuse side of triangle 40 and opposite side of triangle 20 find the sine angle?
Solution:

Hypotenuse side of triangle = 40.

Opposite side =20

sin A =opposite/hypotenuse

sin A= `20/40`

sin A =` 1/2`

sin A = 0.5    {sin 30 degree is 0.5}

Therefore the angle is 30 degree