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