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)