Solve Graphing Test

Introduction to solve graphing test:

            Solve graphing test is one of interesting topics in mathematics. Without doing any algebraic manipulations, we can solve two simultaneous equations in x and y by drawing the graphs corresponding to the equations together. An equation in x and y is of the form a x + b y + c = 0. The equation represents a straight line, so, the problem of solving two simultaneous equations in x and y reduces to the problem of finding the common point between the two corresponding lines.
Steps to solve the graph:
The following steps are necessary to solve the graphing test:

Step 1:

          Two different values are substitute for x in the equation y = mx + b, we get two values for y. Thus we get two points (x1, y1) and (x2, y2) on the line.

Step 2:

          Draw the x-axis and y-axis on the graph and choose a suitable scale on the co- ordinate axes. Both the axes is chosen based on the scale values of the co-ordinates obtained in step 1. If the co-ordinate values are large in given data then 1 cm along the axes may be taken to represents large number of units.

Step 3:

          Plot the two points (x1, y1) and (x2, y2) in Cartesian plane of the paper.

Step 4:

          Two points are joined by a line segment and extend it in both directions of the segment. Then it is the required graph.


Examples to solve graphing test:


Examples to solve graphing test are as follows:

Example 1:

   Draw the graph y = 3x −1.

Solution:

   Substituting x = −1, 0, 1 in the equation of the line, we get y = −4, −1, 2 correspondingly. In a graph, plot the

   Points (−1, −4), (0, −1) and (1, 2).



  X    -1    0    1

  Y    -4   -1    1

  Join the points by a line segment and extend it in both directions. Thus we get the required linear graph



                                                                                               

Example 2:

    Draw the graph of the line  2x  + 3y  = 12.

Solution:

  The given equation is rewritten as 3y = −2x + 12 or y = (−3 / 2)x  + 4.

  Substituting  x = −3 then y = 6

                      x = 0, then y = 4

                      x = 3 then y = 2.

  Plot x and y values in the graph sheet. [(−3, 6), (0, 4) and (3, 2)]

  X    -1      0        1

  Y   5.5     4       2.5

  Join the points by a line segment and extend it in both the directions. Then it is the required linear graph.






Practice problems to solve graphing test:


Some practice problems to solve graphing test

1. Draw the graph of the following : y  = −2x

     Answer:     x    -1   0   1

                       y     2   0  -2

 2. Draw the graph of the following equations:  y + 2x −5 = 0.

      Answer:    x    -1   0   1

                       y     7   5   3

Torus Help With Math

Introduction to Torus help with math:

                    This article deals with the torus and how the math formula helps to find the torus volume and surface area. Torus is the three dimensional ring shaped surface that is produced by rotating a circle around an axis. Here the axis will never intersect the circle. The cross section of the torus looks like a ring.
 
Math formula:


With the help of math formula we can find the volume and surface area.

           For volume:

When radius is given:

Volume of the torus =` 2*pi^2*R*r^2` cubic units

R is the large radius.

r is the small radius.

When the diameter is given:
Volume of the torus = `((pi^2*D*d^2)/4)` cubic units
D is the large diameter
d is the small diameter
For surface area:
When radius is given:
Surface area of the torus = `4*pi^2*R*r` square units
R is the large radius.
r is the small radius.
When the diameter is given:
Surface area of the torus = `(pi^2*D*d)` cubic units
D is the large diameter
d is the small diameter
Model problems for torus:
Find the volume of the torus with the help of math formula when the larger radius is 10 cm and the smaller radius is 8cm?
 Solution:
Larger radius is 10cm.
Smaller radius is 8cm.
When radius is given:
Volume of the torus =` 2*pi^2*R*r^2` cubic units
R is the large radius.
 r is the small radius.
= `2 (3.14) ^2*10*8^2`

= `2 (9.8596)*640
= (19.7192)*640

volume of the torus = 12620.29 cm3

2.  Find the surface area of the torus with the help of math formula when the larger radius is 10 cm and the smaller radius is 8cm?

 Solution:

          Larger radius is 10cm.

           Smaller radius is 8cm.

        When radius is given:

         Surface area of the torus = `4*pi^2*R*r ` square units

          R is the large radius.

          r is the small radius.

          = `4 (3.14) ^2*10*8`

          = `4 (9.8596)*80`

         = (39.4384)*80
surface area of the torus = 3155.072 cm2


3.Find the surface area of the torus with the help of math formula when the larger diameter is 20 cm and the smaller diameter is 10cm?
 Solution:

Larger diameter is 20cm.

Smaller diameter is 10cm.
When the diameter is given:
Surface area of the torus = `(pi^2*D*d) ` cubic units
D is the large diameter

d is the small diameter

= (3.14)^2* (20) (10)
= 9.8596 (200)

surface area of the torus = 1971.92 cm2

Solve a Math Problem for Me

Introduction to solve math problems:

The needs of Science and Technology, Social Sciences, Humanities and Computers posed new and challenging problems. Such problems could be effectively studied for analytical and exact solution only with the help of Mathematics. Mathematics interacted well with all other branches of Science and Social Sciences and new fields such as Operations Research, Industrial Mathematics, Mathematics of Computation, Mathematical Statistics, Mathematical Physics, Mathematical Biology, Mathematical Modeling, Cryptology, and Mathematical Economics etc... In this article we shall solve some math problems.


Solve math example problems


Problem:

Solve factorizing 4x2 + 20xy + 25y2 – 10x – 25y.

Solution:

The given expression is 4x2 + 20xy + 25y2 – 10x – 25y

= (2x)2 + 2(2x)(5y) + (5y)2 – 5(2x) – 5(5y)

= [(2x) + (5y)]2 – 5[(2x) + (5y)]

= (2x + 5y)2 – 5(2x + 5y)

= (2x + 5y) (2x + 5y –5).

Problem:

Factorize 6(a – 1)2b – 5(a –1)b2 – 6b3

Solution: Putting x = a – 1, we get

6(a – 1)2b – 5(a–1)b2 – 6b3 = 6x2b – 5xb2 – 6b3

= b(6x2 – 5bx –6b2)

Here we find 6 × – 6 = –36 = (–9) × 4 and (–9) + 4 = –5.

Hence we get = b(6x2 – 9bx + 4bx – 6b2)

= b[3x(2x – 3b) + 2b(2x – 3b)]

= b(2x – 3b) (3x + 2b)

= b[2(a – 1) – 3b] [3(a – 1) + 2b]

= b[(2a –3b –2) (3a + 2b –3)].

Problem:

Solve If x1 and x2 are two events related with a random experiment such that P(a2)=0.35, P(x1 or x2)= 0.95 and P(x1 and x2)=0.25, Find out  P(x1).

Solution:

Consider P (x1) = x then,

P (x1 or x2) =P (x) + P (x2) – P (x1 and x2)

= 0.95 = z + 0.35 – 0.25

Sum the experiment value

z= (0.95-0.35 + 0.25) = 0.85

Hence P (x1) = 0.85.

 


Solve math practice problem


Problem:

If a1 and a2 are two events related with a random experiment such that P(a2)=0.45, P(a1 or a2)= 0.75 and P(a1 and a2)=0.25, Calculate P(a1).

Answer:

0.55

Problem:

Solve factorization 8a2 + 2a – 3.

Answer:

(4a + 3) (2a – 1)

Gradient In Math

Introduction of gradient in math:

In vector calculus, the gradient of a scalar field is a vector field which points in the direction of the greatest rate of increase of the scalar field, and whose magnitude is the greatest rate of change.A generalization of the gradient for functions on a Euclidean space which have values in another Euclidean space is the Jacobian.


Example problem for gradient in math:-


Problem :-

The three variables function is x axis, y axis and z axis the region of three dimensional space it is normally show a f(x,y,z) these is called as scalar field and P is a point and specified direction is no need for the x axis, y axis and z axis. You need to calculate and any change in f is called as directional derivative.

Solution:-

Well, let's start by letting R=x0i+y0j+z0k be the position vector for P. Let the specified direction that we want to move from P be given by the unit vector u = u1i + u2j + u3k. Let Q=(x + x, y + y, z + z) be a point along with the vector in a specified direction. Let Deltas be the scalar value such that vecPQ = Deltashatu, that is s is the length of  vecPQ. In following formula for gradient in math

`vecPQ = Deltaxi+ Deltayj+ Deltazk = Deltasu_1i+ Deltasu_2j+ Deltasu_3k`

Let `deltaf ` = f(Q) - f(P). By linear approximation,

`deltaf = fx(P)delta x + fy(P)delta y + fz(P)delta z + erfx(Q)delta x + erfy(Q)delta y + erfz(Q)delta z`
`deltaf = fx(P)(delta s)u1 + fy(P)(delta s)u2 + fz(P)(delta s)u3 + erfx(Q)(delta s)u1 + erfy(Q)(delta s)u2 + erfz(Q)(delta s)u3`
Dividing by deltas, we have,

`(Deltaf)/(deltas) = (delf)/(delx)(p)u_1+ (delf)/(dely)(p)u_2+ (delf)/(delz)(p)u_3+ erf_x(Q)u_1+ erf_y(Q)u_2+ erf_z(Q)u_3`

Two points Q and P are approaches to line. It contains three functions error and finally we got a zero and will get directional derivative.

` (Deltaf)/(deltas) (p)= (delf)/(delx)(p)u_1+ (delf)/(dely)(p)u_2+ (delf)/(delz)(p)u_3`

Or, the dot product,

`gradf(p).hatu`

where gradf is a special function defined as follows,

`(Deltaf) = (delf)/(delx) i+ (delf)/(dely) j+ (delf)/(delz) k`

More texts are skips the vector arrow in given arrow of `gradf ` and `Delta ` symbol is called the Del operator. To writing a symbol as vector it is more helpful for gradient of function produces a vector.


Gradient in math Exercises:


In following math gradient to demonstrate the similarity to differentiation

`grad(f+g) = grad f + grad g`
`grad(f+g) = f grad g + g grad`
`grad(f/g) = f grad g - g grad f /g^2`
`gradf^n = nf^(n-1) grad f `

Solve What is Histograms

INTRODUCTION ABOUT SOLVE WHAT IS HISTOGRAMS:

Let us see about the introduction for solve what is Histograms. The construction of histograms is used  to from a frequency of table. The rectangle height is located at the interval and it is represented by the number of scores. The shapes of the histograms are varying depend upon the choice of the size of the intervals. Let us see the explanation about solve what is Histograms.


Explanation about Solve what is Histograms:


The Solve Histogram is showing as the total distribution in the image. It is a bar chart that counts the pixels of every tone of gray.  The bar chart occurs in the image. It is the basic scheme for solving the permutation of problems in the framework of probabilistic model-building genetic algorithms that uses the edge of histogram is based upon the sampling techniques was reported.


Example for Histogram about Solve what is Histograms:


Example 1:

Solve the frequency distribution question, produce table, and histogram.

The following data represents the rain fall measured annually (in millimeters) by the meteorological office since records began in 1951 up to and including 1995.

Years

1951-57 = 1256  1586  1340  1340  1277  1419  1103
1958-64 = 1263  1231  1763  1509  1292  1227  1288
1965-71 = 1126  1342  1437  1184  1359  1461  1128
1972-78 = 1359  1459  1540  1235  1384  1485  1498
1979-85 = 1232  1670  1505  1572  1517  1417  1485
1986-92 = 1231  1312  1255  1132  1336  1362  1305

1993-95 = 1262  1396  1426

Example 2:

Draw the Histogram for the following data’s:

Month	Scales (in millions)
January	60
February	59
March	70
April	75
May	80
June	84
July	56
August	89
September	90
October	95
November	20
December	37

Solution:






 
The months are represented by x-axis and the scales are represented in y-axis.

These are the examples for solve what is Histograms.

Easy Multiplication Tables

Introduction to easy multiplication tables:

Multiplication:

Multiplication symbol is "×". There are four basic operations in elementary arithmetic multiplication, addition, subtraction and  division. Easy multiplication tables are the mathematical operation of scale one number by another number.

Since the result of scaling by whole numbers could be consideration of as consisting of many copies of the original, whole-number provides greater than 1 could be calculated by repeated addition.

For example, 5 multiplied by 6 (regularly said as "5 times 6") can be calculating by adding 5 copies of 6 together:

5 x 6 = 5 + 5 + 5 + 5 + 5 + 5 = 30.

2 x 4 = 2 + 2 + 2 + 2 = 12.


Tables on easy multiplication tables


This is 0 to 12th method of easy multiplication tables.

x	0	1	2	3	4	5	6	7	8	9	10	11	12
0	0	0	0	0	0	0	0	0	0	0	0	0	0
1	0	1	2	3	4	5	6	7	8	9	10	11	12
2	0	2	4	6	8	10	12	14	16	18	20	22	24
3	0	3	6	9	12	15	18	21	24	27	30	33	36
4	0	4	8	12	16	20	24	28	32	36	40	44	48
5	0	5	10	15	20	25	30	35	40	45	50	55	60
6	0	6	12	18	24	30	36	42	48	54	60	66	72
7	0	7	14	21	28	35	42	49	56	63	70	77	84
8	0	8	16	24	32	40	48	56	64	72	80	88	96
9	0	9	18	27	36	45	54	63	72	81	90	99	108
10	0	10	20	30	40	50	60	70	80	90	100	110	120
11	0	11	22	33	44	55	66	77	88	99	110	121	132
12	0	12	24	36	48	60	72	84	96	108	120	132	144

Identification of 9th multiplication tables:
find 9 x 12.

Make use of the 8-method. Multiplying 9 by 12 times means, and then we get 108, as the answer.

And mention the easy multiplication tables with answer (bold number) above.

Identification of 7th multiplication tables:
find 7 x 4.

Make to use of the 7-method. The multiplying 7 by 4 times means, we get 28, as the answer.

And mention the easy multiplication tables with answer (bold number) above.

Identification of 2nd multiplication tables:
find 2 x 8.

Make to use of the 8-method. The multiplying 2 by 8 times means, and then we get 16, as the answer.

And mention the easy multiplication tables with answer (bold number) above.

Identification of 11th multiplication tables:
find 11 x 3.

Make to use of the 8-method. Multiplying 11 by 3 times means, and then we get 33, as the answer.

And mention the easy multiplication tables with answer (bold number) above.


Practice on easy multiplication tables:


1)   Find number from the  multiplication table. Multiplying 2 into 4 times.

2)   Find number from the multiplication table. Multiplying 4 into 5 times.

3)   Find number from the multiplication table. Multiplying 6 into 3 times.

4)   Find number from the multiplication table. Multiplying 7 into 8 times.

5)   Find number from the multiplication table. Multiplying 12 into 5 times.

6)   Find number from the multiplication table. Multiplying 6 into 3 times.

Solve Graphing Test

Introduction to solve graphing test:

Solve graphing test is one of interesting topics in mathematics. Without doing any algebraic manipulations, we can solve two simultaneous equations in x and y by drawing the graphs corresponding to the equations together. An equation in x and y is of the form a x + b y + c = 0. The equation represents a straight line, so, the problem of solving two simultaneous equations in x and y reduces to the problem of finding the common point between the two corresponding lines.
Steps to solve the graph:


The following steps are necessary to solve the graphing test:

Step 1:

 Two different values are substitute for x in the equation y = mx + b, we get two values for y. Thus we get two points (x1, y1) and (x2, y2) on the line.
Step 2:

   Draw the x-axis and y-axis on the graph and choose a suitable scale on the co- ordinate axes. Both the axes is chosen based on the scale values of the co-ordinates obtained in step 1. If the co-ordinate values are large in given data then 1 cm along the axes may be taken to represents large number of units.
Step 3:

Plot the two points (x1, y1) and (x2, y2) in Cartesian plane of the paper.

Step 4:

 Two points are joined by a line segment and extend it in both directions of the segment. Then it is the required graph.


Examples to solve graphing test:


Examples to solve graphing test are as follows:

Example 1:

   Draw the graph y = 3x −1.

Solution:

   Substituting x = −1, 0, 1 in the equation of the line, we get y = −4, −1, 2 correspondingly. In a graph, plot the

   Points (−1, −4), (0, −1) and (1, 2).



  X    -1    0    1

  Y    -4   -1    1

  Join the points by a line segment and extend it in both directions. Thus we get the required linear graph



    Example 2:

    Draw the graph of the line  2x  + 3y  = 12.

Solution:

  The given equation is rewritten as 3y = −2x + 12 or y = (−3 / 2)x  + 4.

  Substituting  x = −3 then y = 6

 x = 0, then y = 4
x = 3 then y = 2.

  Plot x and y values in the graph sheet. [(−3, 6), (0, 4) and (3, 2)]

  X    -1      0        1

  Y   5.5     4       2.5

  Join the points by a line segment and extend it in both the directions. Then it is the required linear graph.

Practice problems to solve graphing test:


Some practice problems to solve graphing test

1. Draw the graph of the following : y  = −2x

 Answer:     x    -1   0   1

  y     2   0  -2

 2. Draw the graph of the following equations:  y + 2x −5 = 0.
Answer:    x    -1   0   1
y     7   5   3

Altitude Term In Math

Introduction about altitude term in math:

            Altitude or height term is defined based on the context in which it is used. As a general definition, the term altitude is a distance measurement, usually in the vertical or "up" direction, between a reference datum and a point or object. In this article we shall discus about altitude term based problems.




Triangle:


The total space inside the triangle is called as area of that triangle.

Formula to find Area:

            Area of right angle triangle (A) =1/2 (length x height) square unit

                                                               = 1/2 l x h square unit.

                        Here, the term height refers the altitude of the triangle.

Example problem:



A right angle triangle has length 5cm and altitude 13 cm. Find the area of that triangle.
Solution:

Given:

            Length (l) =5cm

            Altitude (h) =13cm

Formula:

Area of triangle = 1/2 (l x h) square unit.

                           = 1/2 (5 x 13)

                           = 1/2 (65)

                           =65/2

                           =32.5

Area of triangle = 32.5 cm2



A right angle triangle has length 7.5m and height 10 m. Find the area of that triangle.
Solution:

Given:

            Length (l) =7.5m

            Altitude (h) =10m

Formula:

Area of triangle = 1/2 (l x h) square unit.

                           = 1/2 (7.5 x 10)

                           = 1/2 (75)

                           =75/2

                           =37.5

Area of triangle = 37.5 cm2


Rhombus:


The altitude of rhombus is the distance between base and opposite side of the base.

Formulas:

Area of the rhombus (A) = b x a

                                    b – Base of rhombus.

                                    a – altitude of rhombus

If two diagonal lengths are given:

Area of the rhombus (A) = (d1 x d2)/2

Example problems:

1.      The altitude and base of rhombus are 11 cm and 6cm respectively. Find are of rhombus.

Solution:

      Given:

                  Altitude of rhombus (a) = 11 cm

                         Base of rhombus (b) = 6 cm

            Area of the rhombus (A) = b x a square units.

                                                      = 11 x 6

                                                      = 66

      Area of the rhombus (A) = 66 cm2

2.      The altitude and base of rhombus are 14 cm and 10cm respectively. Find are of rhombus.

Solution:

      Given:

                  Altitude of rhombus (a) = 14 cm

                         Base of rhombus (b) = 10 cm

            Area of the rhombus (A) = b x a square units.

                                                      = 14 x 10

                                                      = 140

      Area of the rhombus (A) = 140 cm2

How To Cross Multiply

Introduction of How to Cross Multiply:

Cross multiplication is the easiest method that can be done in the equation form of two fraction form. For Example: `(4)/(16)` = `(3)/(12)`. By using the cross multiply method, we can check this given example as 4 * 12 = 48 and 16 * 3 = 48. Now it be proved that  `(4)/(16)` = `(3)/(12)`. Let see how to do the cross multiply and example problem of cross multiply.


Rules – How to Cross Multiply:


Let take

`(P)/(Q)` ….. Fraction 1

`(R)/(S)` ………. Fraction 2

Now equate the two fractions as in the form of `(P)/(Q)` … = `(R)/(S)` ……fraction equation

Rule 1: How to cross multiply means, multiply denominator of the fraction 2 with numerator of the fraction 1 as well as multiply denominator of the fraction 1 with numerator of the fraction 2

P * S = Q * R

Rule 2: Multiply the term on both sides of the fraction equation, and then the value of the term remains unchanged.

PS = QR

Multiply term A on both sides then its value is not changed. PSA = QRA.

Let we see example problem of how to cross multiply.


Example Problem – How to Cross Multiply:


Example 1:

How to find the value of x in given equation `(3)/(x)` = `(4)/(8)`?

Solution:

Step 1: Multiply 8 with the numerator of the fraction `(3)/(x)`.

3 * 8 = 24

Step 2: Multiply the denominator of the fraction 3/x with numerator of the fraction 4/8.

x * 4 = 4 x

Step 3: Equate above two steps, 4 x = 24

Step 5: Divide 4 by both sides, we get 4 * (`(x)/(4)`) = `(24)/(4)`

Then the value of x = 6

Answer: x= 6

Example 2:

How to solve this equation `(X - 3)/(4)` = `(5)/(2)`?

Solution:

Step 1: Multiply 2 with the numerator of the fraction `(X - 3)/(4)`.

2 * (X - 3) = 2X - 6

Step 2: Multiply the denominator of the fraction `(X - 3)/(4)` with numerator of the fraction `(5)/(2)`.

4 * 5 = 20

Step 3: Equate above two steps, 2X – 6 = 20

2X = 14

Step 5: Divide 2 by both sides, we get 2 * (`(X )/(2)`) = `(14)/(2)`

Then the value of X = 7

Answer: x= 7

Math h Functions

Introduction:

The hyperbolic functions are defined as the analogs of common trigonometric functions. The fundamental hyperbolic functions include sin h called as the hyperbolic sine, cos h called the hyperbolic cosine and the tan h called the hyperbolic tangent function.

The hyperbolic functions are just the rational functions of the exponentials. The mathematical hyperbolic functions are also called as the math h functions. In this article we are going to see various math h functions.


Expressions of math h functions:


There are various mathematical algebraic expressions given to the math h functions.

The hyperbolic sine of x is given by sinh x = ½ (ex-e-x).

The hyperbolic cosine of x is given by cosh x = ½ (ex+e-x).

The hyperbolic tangent of x is given by tanh x = sinh x/cosh x

= (ex-e-x) / (ex+e-x)

= e2x-1 / e2x+1

The hyperbolic cosecant of x is given by csch x = (sinh x)-1

= 2 / ex-e-x

The hyperbolic secant of x is given by sech x = (cosh x)-1

= 2 / ex+e-x

The hyperbolic cotangent of x is given by coth x = cosh x / sinh x

= (ex+e-x) / (ex- e-x)

= e2x+1 / e2x-1 .


Math h functions with respect to circular functions:


The hyperbolic functions or math h function with respect to the circular functions is given by

x = a cos t and y = a sin t

Here the circle is given as a rectangular hyperbola.

The math h functions exist in many applications of mathematics which involve the integrals with v (1+x2) and the circular functions with v (1-x2)

The math h functions also include many identities similar to that of the trigonometric identites. The identites of the math h functions includes

Cosh2 x – sinh2 x = 1

Cosh x + sinh x = ex

Cosh x – sinh x = e-x

The identities for the complex arguments includes

Sinh (x+iy) = sinh x cos y + i cosh x sin y

Cosh (x+iy) = cosh x cosy + i sinh x sin y.

Symbols That Represent Me

Introduction to symbols that represent me:

There are numerous mathematical symbols that can be used in mathematics mode. This is a listing of common symbols found within all branches of Mathematics.

SYMBOL		REPRESENTS
Symbol	=	Represents	Is equal to
Symbol	+	Represents	Plus or Addition
Symbol	-	Represents	Minus or subtraction
Symbol	/	Represents	Division
Symbol	X or *	Represents	Multiplication
Symbol	≠	Represents the	Inequality
Symbol	< and >	Represents the	“Is less than” and “is greater than”
Symbol	≤ and ≥	Represents the	“Is less than or equal to” and “is greater than or equal to”
Symbol	є	Represents the	Is an element of
Symbol	∑	Represents the	summation
Symbol	α	Represents the	Alpha
Symbol	β	Represents the	Beta
Symbol	γ	Represents the	Gamma
Symbol	δ	Represents the	Delta
Symbol	ζ	Represents the	Zeta
Symbol	η	Represents the	Eta
Symbol	θ	Represents the	Theta
Symbol	λ	Represents the	Lamda or Lambda
Symbol	μ	Represents the	Mu
Symbol	π	Represents the	Pi
Symbol	σ	Represents the	Sigma
Symbol	φ	Represents the	Phi
Symbol	χ	Represents the	Chi
Symbol	ω	Represents the	Omega
Symbol	#	Represents the	Number  Sign
Symbol	±	Represents the	Plus Minus
Symbol	Ω	Represents the	Omega
Symbol	ι	Represents the	Iota
Symbol	℮	Represents the	Estimate sign
Symbol	√	Represents the	Square Root
Symbol	∞	Represents the	Infinity
Symbol	∫	Represents the	Integral
Symbol	≈	Represents the	Almost equal to
Symbol	∂	Represents the	Partial differential
Symbol	∆	Represents the	Increment
Symbol	w.r.t	Represents the	With respect to
Symbol	log	Represents the	Logarithm
Symbol	!	Represents the	Factorial
Symbol	%	Represents the	Percentage

History for symbols that represent me

A very elongated form of the modern equality symbol (=) was first introduced in print in The Whetstone of Witte (1557) by Robert Recorde (1510-1558) the man who first introduced algebra into England.  He justified the symbol by stating that no two things can be more equal than a pair of parallel lines...

The infinity symbol was first given its current mathematical meaning in "Arithmetica Infinitorum" (1655) by the British mathematician John Wallis (1616-1703).

Symbols that represent me : Further history

Gottfried Wilhelm Leibniz (1646-1716) viewed integration as a generalized summation, and he was partial to the name "calculus summatorius" for what we now call [integral] calculus.  He eventually settled on the familiar elongated ‘s’  for the sign of integration, after discussing the matter with Jacob Bernoulli (1654-1705) who favored the name "calculus integralis" and the symbol  I  for integrals...  Eventually, what prevailed was the symbol of Leibniz, with the name advocated by Bernoulli...

In Math What Does Range Mean

Introduction to range in math:

• In math, Data set is a collection of data, which is usually presented, in tabular form. Each column represents a variable. In math, Range is generally defined as the value we obtained as a result of difference between a greater value and a smaller value.
• In other words, range is defined as the difference between a maximum and minimum value.

Steps to learn the range in math:

     In order to learn the range of data set the following steps are necessary.
     Step 1: Arrange the numbers from ascending to descending order.
     Step 2: Identify the greater value in the given set
     Step 3: Identify the smaller value in the given set
     Step 4: Find the difference between the greater value and the smaller value to identify the range of the given data set.

Worked Examples for range in math:

Example 1:
Find the range of the data set given below:
                    8 , 15 , 13 , 7 , 24 , 37 , 6.
    Step 1: Arranging the numbers given in data set from least to greatest.
                    6 , 7 , 8 , 13 , 15 , 24 , 37.
    Step 2: Identify the greater value in the given set
                  From the given set, we can identify 37 as the greater value.
    Step 3: Identify the smaller value in the given set
                  From the given set, we can identify  6 as the smaller value.
    Step 4: Find the Range.
                          Range     =   Greater value – Smaller value
                                           =   37 - 6
                                           =  31
                   Hence, the range of the given data set is 31.
Example 2:
Find the range of the data set shown below:
                    33 , 12 , 79 , 24 , 34 , 53 , 27 , 61
     Step 1: Arranging the numbers given in data set from least to greatest.
                    12 , 24 , 27 , 33 , 34 , 53 , 61 , 79.
     Step 2: Identify the greater value in the given set
                    From the given set, we can identify 79 as the greater value.
     Step 3: Identify the smaller value in the given set
                    From the given set, we can identify  12 as the smaller value.
    Step 4: Find the Range.
                              Range    =   Greater value – Smaller value
                                               =   79 - 12
                                               =   67
                        Hence, the range of the given data set is 67.

Practice problems for range in math:

1) Find the range of the data set shown below:
                    7 , 9 , 19 , 31 , 37 , 43 , 6 , 37
     Answer: 37
2) Find the range of the data set shown below:
                    14 , 11 , 32 , 10 , 32 , 77 , 27 , 43
     Answer: 67

Geometrical Meaning of the Zeroes of a Polynomial

We know that  a real number k is a zero of the polynomial p(x) if p(k) = 0 in a geometrical way.

The x-coordinate of the point, where the graph of a polynomial intersect the x-axis is called the zero of the polynomial.

An nth - degree polynomial intersects the x-axis of n points and therefore, has a maximum of n zeros in geometrical graph

In a quadratic polynomial ax2 + bx +c,

If a>0, then the graph is a parabola that open upwards.

If a<0, then the graph is a parabola that open downwards.

Special case to the zeroes of a polynomial in geometrical meaning


In a geometrical way the evaluation of zeroes with polynomial evaluation of the equation in geometrical meaning is follows

Case (i) :

Here, the graph cuts x-axis at two distinct points A and A′.
The x-coordinates of A and A′ are the two zeroes of the quadratic polynomial ax2 + bx + c in this case is shown in the following figure



Case (ii) :

The graph given here cuts the x-axis at exactly one point, i.e., at two coincident points. So, the two points A and A′ of Case (i) coincide here to become one point A is shown in the following figure.



The x-coordinates  of the  A is the only zero for the quadratic polynomial ax2 + bx + c in this case.

No zero case

The graph given here is either completely above the x-axis or completely below the x-axis. So, it does not cut the x-axis at any point

So, the quadratic polynomial ax2 + bx + c has no zero in this case.
So, you can see geometrically that a quadratic polynomial can have either two distinct zeroes or two equal zeroes (i.e., one zero), or no zero. This also refers that the polynomial of degree 2 has at-most two zeroes is shown in the following figure.

Concept of the Derivative Online Help

In online, concept of derivative is clearly explained with the help of solved example problems. In online so many websites explain the concepts with the help of math sites. The derivative concept is clearly explained in calculus whereas derivative concept helps to find the rate of change for the given function with respect to change in the input function. The following are the example problems with detailed solution helps to explain the concept of derivative in online.


Concept of derivative online help example problems:


The following example problems explain the concept of derivative in online.

Example 1:

Determine the derivative by differentiating the polynomial function.

f(t) = 3t 3 +4 t 4  + 5t

Solution:

The given function is

f(t) = 3t 3 +4 t 4  + 5t

The above function is differentiated with respect to t to find the derivative

f '(t) = 3(3t 2 )+4(4t 3  ) + 5

By solving above terms

f '(t) = 9t 2 +  8t 3 + 5

Example 2:

Determine the derivative by differentiating the polynomial function.

f(t) = 6t6 + 5 t5 + 4 t4 + t

Solution:

The given equation is

f(t) = 6t6 + 5 t5 + 4 t4 + t

The above function is differentiated with respect to t to find the derivative

f '(t) =  6(6t 5)  +5 (5 t4 ) +4(4 t3) + 1

By solving above terms

f '(t) =  36t 5  + 25 t4  + 16 t3 + 1

Example 3:

Determine the derivative by differentiating the polynomial function.

f(t) = 2t 2 +4 t 4  + 15

Solution:

The given function is

f(t) = 2t 2 +4t 4  + 15

The above function is differentiated with respect to t to find the derivative

f '(t) = 2(2t  )+4(4 t 3 ) + 0

By solving above terms

f '(t) = 4t +16t3

Example 4:

Determine the derivative by differentiating the polynomial function.

f(t) = 5t5 +4t 4 +3t 3  + 2

Solution:

The given function is

f(t) = 5t5 +4t 4 +3t 3  + 2

The above function is differentiated with respect to t to find the derivative

f '(t) = 5(5t 4 )+4(4t 3 ) +3( 3t 2) +0

By solving above terms

f '(t) = 25t 4 +16t 3  +9 t 2


Concept of derivative online help practice problems:


1) Determine the derivative by differentiating the polynomial function.

f(t) = 2t 3 +3 t 4  + 4 t 5

Answer: f '(t) = 6t 2 +12 t3 + 20 t 4

2) Determine the derivative by differentiating the polynomial function.

f(t) = t 3+t5 + 4 t 6

Answer: f '(t) = 3t2 + 5t4 + 24 t 5

Solve Online Scale Factor

Scale is a thing which is used to measure the dimensions of the object. The terms scale factor is used to mention the dimensions of the geometric shapes. In geometry, the term scale factor is usually used to change the dimensions of the image. Using the scale factor, any shapes dimension can be increased or decreased. In online, students can learn about various topics. In online, students can learn about scale factor clearly. Online learning is very interesting and interactive. Moreover online learning is different from class room learning in different ways. In online, students have one to one learning.


More about scale factor:


A scale factor is a figure used as a figure by which another figure is multiplied in scaling.

A scale factor is new to scale shape in 1 to 3 proportions.
Scale factor can be established by the following scenario
Size Transformation:     In size change, the scale factor is the ratio of express the amount of reduction.

Scale Drawing:    In scale diagram a diagram that is like but either superior or lesser than the real object, for example, a road map or a building blueprint...

Comparing Two Similar Geometric Images:      The scale factor when compare two parallel geometric imagery is the ratio of lengths of the equivalent sides.






Solve Scale factor Problem:


Q 1:   The area of a rectangle is 5 and it is enlarged a scale factor by 5. Solve for the area of an enlarged rectangle?

Sol :  The Area of a rectangle(R) is 5 and its Enlarged by a scale factor of 5.

that means the length L and the width W are equally multiply by 5.
As well note that the area of Rectangle is the length L times the width W, so Rectangle = LW.

Because the area of a rectangle is given by Rectangle(R) = LW, and L turn out to be 4L and W turn out to be 4W,

Solve for new triangle,

the new area of a enlarged rectangle  = (5L)(5W) = 25LW = 25(R).
Now we know that Rectangle(R) = 5, so 25(R) = 125.

Q 2:  The side of the square is 6 cm and it is enlarged a scale factor by 4. Solve for the area of an enlarged square?

Sol :  The Area of a square = a2

=  6 * 6

= 36 cm2.

Scale factor is 4 cm.

New side of square  = 4 * 6

= 24 cm.

Area of enlarged square = 242.

= 576 cm2.

Answer is 576 cm2.

Study Online Interest

To study online interest is useful to calculate the interest easily. We have two types of interest Simple interest and compound

interest. Formula for simple interest is,    Interest = Principle `xx` Rate `xx` Time

Where  I = Total amount of interest paid

P = Amount lent

R = Percentage of principle charged as interest each year.

Example: rate = 2% means then 2/100 =0.02. Use this value in formula.

Formula for compound interest is,       A = p`(1+r)^(n)`


Examples of study online interest:


Ex:1

To find simple interest if principle= 3000, rate= 3%, time= 5

Sol:

To study online interest we have to write the formula first

That is Simple interest = Principle `xx` Rate `xx` Time

Apply the given values in the formula,

Interest = 3000 `xx` 3%`xx` 5               R = 3% = 3/100 = 0.03

= 3000`xx` .03`xx` 5

= 450

The answer is 450

Ex:2

Shan invested Rs.6000 for 5 years. He received the simple interest of Rs.1,500.Find the rate of interest.

Sol:

To study online interest we have to write the interest formula first

That is Simple interest = Principle `xx` Rate `xx` Time

Here principal, P= Rs.6000

Number of years, n= 5

Simple interest, I = Rs.1500

Rate of interest, r =?

We know that rate of interest, r =`(100 xx 1500)/(6000 xx 5)`

=5

Therefore Rate of interest = 5%

Ex:3

If the principal amount is $7000, the periodic rate of interest is 3% and the number of compounding periods is 3, what is the value at maturity?

Sol:

To study online interest we have to write the formula first

A = p`(1+r)^n`

Here principal, P= $7000

Rate of interest, r= 3%

= 3/100 = .03

n = 3

A = P (1 + r) n

A = 7000 (1 + .03) 3

A = 7649.089

Ex:4

If the principal amount is $2000, the periodic rate of interest is 2% and the number of compounding periods is 5, what is the value at maturity?

Sol:

To study online interest we have to write the formula first

A = p`(1+r)^n`

Here principal, P= $2000

Rate of interest, r= 3%

= 2/100 = .02

n = 5

A = P (1 + r) n

A= 2000 (1 + .02) 5

A= 2208.1616

These are the some examples to study online interest.

Learning Examples of Histograms

Introduction on learning examples of histograms:

In statistics, a histogram is a graphical display of tabular frequencies, shown as adjacent rectangles. Each rectangle is erected over an interval, with an area equal to the frequency of the interval. The height of a rectangle is also equal to the frequency density of the interval, i.e. the frequency divided by the width of the interval. The total area of the histogram is equal to the number of data. A histogram may also be based on the relative frequencies instead. It then shows what proportion of cases fall into each of several categories (a form of data binning), and the total area then equals 1.

Procedure and learning example for drawing the histogram:

To draw the histogram for a given frequency table
                  Marks    : 10-20   20-30    30-40   40-50   50-60   60-70
Number of students:    4           6            8         10          7         5  
Sol:
Histograms are drawing the first quadrant .because the first quadrant frequencies and class intervals are positive. Class intervals are marked along x axis similarly the frequency are marked along the y axis. Identify the maximum frequency in the table and then chose a suitable scale along the y- axis.
                     Draw first rectangle on the intervals 10 -20 as the base width and height of 4 units. Draw all rectangles taking the consecutive class intervals as base of the rectangles and their heights corresponding to the frequencies. The closed figure obtained is called histogram

Histogram:

This is the histogram for given frequency table.
Example: 2
To draw the histogram for a given frequency table,
 
Sol:

• Here in the frequency table, data range and frequency are given
• We take a data range in x axis and frequency in y axis

Histogram:

This is the histogram for given frequency table

More examples for learning histograms

To draw the histogram for following frequency table,

Sol:

• Here the given data deals with score(group) and frequency(count)
• We are taking score in x-axis
• Similarly the frequency is in y axis

Histogram:

This is the histogram for given frequency table.

Trigonometric Unit Circle Learning

A circle, whose radius is equal to one unit, is called as unit circle. The concept of unit circle is frequently used in trigonometry. In trigonometry, a circle with center (0, 0) and a radius of one unit is a unit circle. The equation of a circle is (x-h)2 + (y-k)2 = r2

For a unit circle, the center is (0, 0) and radius is 1, so the equation of a unit circle is x2 + y2 = 1


Learning - Properties of unit circle


Consider a point (x, y) in a unit circle.



The right triangle in the unit circle in the above diagram, Pythagoras theorem satisfies the equation of unit circle.

x2 + y2 = 1

Learning - Forms of unit circles points

Exponential form:            eit

Trigonometric form:        z = cos(t) + i sin(t)

Learning - Trigonometric functions

In a unit circle, consider a point (x,y) on the circle. If the angle formed between line joining the center (0,0) and the point (x,y) and the horizontal axis is `theta`,



Then the trigonometric functions for the angle `theta` is given by,

sin `theta` = opposite side/hypotenuse

cos `theta`= adjacent side/hypotenuse

tan `theta`= opposite side/adjacent side

csc `theta`= 1/sin`theta` = hypotenuse/opposite side

sec `theta`= 1/cos `theta` = hypotenuse/adjacent side

cot `theta`= 1/tan `theta`= adjacent side/opposite side


Example for trigonometric unit circle learning


Find the value of each of the 6 trigonometric functions for an angle theta that has a terminal side containing the point (3, 4).



By Pythagoras theorem, x2 = 32 + 42

x2 = 9 + 16

x2 = 25

x = 5

So, hypotenuse = 5, opposite side = 4 and adjacent side = 3

Then the trigonometric identities are given by,

sin `theta` = opposite side/hypotenuse = 4/5

sin `theta` = 4/5

cos  `theta` = adjacent side/hypotenuse = 3/5

cos `theta` = 3/5

tan `theta` = opposite side/adjacent side = 4/3

tan `theta` = 4/3

csc `theta` = 1/sin `theta`   = hypotenuse/opposite side = 5/4

csc `theta` = 5/4

sec `theta` = 1/cos `theta` = hypotenuse/adjacent side = 5/3

sec `theta` = 5/3

cot `theta` = 1/tan `theta` = adjacent side/opposite side = 3/4

cot `theta` = 3/4

A Factorization is the method for finding the variable factor value of the given expression. A Factor number is multiply with other number. If the factor value is a prime number then it is called prime factor value. The Factor is nothing but a value it is gives the answers to the given problem. If a polynomial factor might be written as the product of two or more expressions, then each expression is called the factor of the given polynomial. If a polynomial factor value might be written as the product of two or more expressions, then each expression is called the factor of the given polynomial. In this article we shall discuss about factoring solver.

Sample problem for factoring solver:

The following problem will help you understand the factoring solver.

Example 1:
Find the factoring value of the given function. 2x²- 12 x +16 = 0

Solution:
In the first step we find pattern factor value of the given numerical values of the x coefficients.

                                                                       32   (product)
                                                                     /    \    
                                                               - 4        - 8 
                                                                     \    / 
                                                                     -12    (sum)

2x²- 12 x + 16 = 2x² - 4x - 8x + 16
                      = 2x (x-2) - 8(x-2)
                     = (2x - 8) (x – 2)
                      = (x -4) ((x – 2)

So the factor of the given function is 2 and 4.

Example 2:
Find the factoring value of the following trinomial. 3x²- 9 x +6

Solution:
In the first step we find pattern factor value of the given numerical values of the x coefficients.

                                                                      18   (product)
                                                                     /    \    
                                                                 - 3      -6
                                                                     \    / 
                                                                      -9     (sum)
3x²- 9 x + 6 = 3x² - 6x - 3x + 6
                        = 3x (x-) - 3(x-2)
                        = (3x – 3) (x – 2)
                        = (x-1) (x-2)

So the factor of the given trinomial function is 2 and 1.

Practice problem for factoring solver:

• Find the factors of the given function. 6x² - 18 x + 12

            Answer: x = 1, 2.

• Find the pattern factors of the given function. 4x²- 24 x +32 = 0

            Answer: x = 2, 4.

study about triangles area

In mathematics, triangle is one of the important topics in geometry.  A geometrical figure which is formed by connecting three points by straight line segment is defined as a triangle. The three sides of a triangle are called as a three-sided polygon. Its can be identified by their sides, by their angles or by the combination of both sides and angles. Here we study about the triangles area.


Study about triangles area – formula:


Triangle area formula:

              If the base and the height of the triangle are given, then the formula to find the area of the triangle is

                                    Area of triangle is A = `(bh)/2` (or) `1/2` bh square. Units

                                          b denotes the base of triangle.

                                          h denotes the height of triangle



In a triangle, base and height can be perpendicular to each other.


Study about triangles area – Example problems:


Here we study some example problems about the triangles area

Example 1:

Determine the area of the triangle whose base side is 4cm and the height of the triangle is 8cm.



Solution:

             Given

              Base side of triangle is 4cm

              Height of the triangle is 8cm

       Triangle area Formula:

                Area of the triangle is 1/2bh square. Units

                                              A = `1/2` bh

                                                = `1/ 2` (4*8)

                                                = `32/2`

                                                = 16

                Area of the triangle = 16 cm2

Answer:

                Area of triangle is 16 cm2

Example 2:

Determine the area of the triangle whose base side is 8m and the height of the triangle is 12cm.



Solution:

             Given

              Base side of triangle is 8m

              Height of the triangle is 12m

       Triangle area Formula:

                Area of the triangle is 1/2bh square. Units

                                              A = `1/2` bh

                                                = `1/ 2` (8*12)

                                                = `96/2`

                                                = 48

                Area of the triangle = 48 m2

Answer:

                Area of triangle is 48 m2

Example 3:

Determine the area of the triangle whose base side is 13cm and the height of the triangle is 17cm.



Solution:

             Given

              Base side of triangle is 13cm

              Height of the triangle is 17cm

       Triangle area Formula:

                Area of the triangle is `1/2` bh square. Units

                                              A = `1/2` bh 

                                                = `1/ 2` (13*17)

                                                = `221/2`

                                                = 110.5

                Area of the triangle = 110.5 cm2

Answer:

                Area of triangle is 110.5 cm2