Graphing Trigonometric Functions Tutoring

Graphing of trigonometric functions is frequent, as we construct, itself imitate about a dense period. Graphing of trigonometric function is helpful in educating science and engineering.

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Graphing trigonometric functions is taken to modify the essential trigonometric functions they are sine, cosine, tangent, cosecant, secant and cotangent. Sine and cosine functions are continual with 2 pi and tangent and cotangent functions are continual with pi.

Tutoring acts as an important role in education through online. Tutors can teach topic wise to students and correct their doubts through online. 

Example to graphing trigonometric functions tutoring:

Tutoring for graphing trigonometric functions example problem 1:

Tutoring for trigonometric function f(x) = 6tanx, execute the graphing for the trigonometric functions.

Solution:
Step 1: Trigonometric functions f(x) = 6tan x

Step 2: In the trigonometric function insert f(x) to y, we obtain,
                          y = 6tan x

Step 3: catch the points for trigonometric functions. To catch the points we can consider the value for x,

Step 4: modify x = -7 in the given trigonometric function, we get
            y = 6tan x
            y = 6tan (-7)
            y = -5.2

Step 5: Modify x = -6 in the given trigonometric function, we get
            y = 6tan x
            y = 6tan (-6)
            y = 1.7

Step 6: Modify x = -4 in the given trigonometric function, we get
            y = 6tan x
            y = 6tan (-4)
            y = -6.9

Step 7: Modify x = -3 in the given trigonometric function, we get
            y = 6tan x
            y = 6tan (-3)
            y = 0.8

Step 8: Modify x = 0 in the given trigonometric function, we get
            y = 6tan x
            y = 6tan (0)
            y = 0
Step 9: from the accomplished value, we can sketch a graphing,

X	y
-7	-5.2
-6	1.7
-4	-6.9
-3	0.8
0	0

Step 10: Illustrate a graphing by using the table.

Tutoring for graphing trigonometric functions problem 2:

Tutoring for the trigonometric function f(x) = tan 6x – y - 6, execute the graphing for the trigonometric functions.

Solution:
Step 1: Study trigonometric functions f(x) = tan 6x – y - 6

Step 2: In the given equation modify f(x) to 0, we obtain,
                          0 = tan 6x – y - 6

Step 3: Rework on the given equation as
                        y = tan 6x – 6

Step 4: catch the points for trigonometric functions. To get the points we can believe the value for x,

Step 5: Modify x = -5 in the given trigonometric function, we get
           y = tan 6x – 6
            y = tan 6(-5) - 6
            y =0.4

Step 6: Modify x = -4 in the given trigonometric function, we get
           y = tan 6x - 6
            y = tan 6(-4) -6
            y = -3.8

Step 7: Modify x = -3 in the given trigonometric function, we get
           y = tan 6x - 6
            y = tan 6(-3) - 6
            y = -4.8

Step 8: Modify x = 0 in the given trigonometric function, we get
           y = tan 6x - 6
            y = tan 6(0) - 6
            y = -6

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Step 9: from the accomplished value, we can sketch a graphing,

X	y
-5	0.4
-4	-3.8
-3	-4.8
0	-6

Step 10: Illustrate a graphing by using the table.

Samples Quadratic Equation

The quadratic equation is an expression having an order of degree 2 for variables. First term of this sample quadratic equation is in the order of two and the second term has order of degree 1 and the last term is constant. If any of the term is not present in the quadratic equation means that the term has a value 0.

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Samples of quadratic equation:

Samples of quadratic equation:
General form of quadratic equation is
ax²+bx+c.
As we saw the first degree of first term is 2, the degree of second term is 1 and third term is constant. Here the first two terms are jointly with a,b constants.

Solving a quadratic equation:
The quadratic equation can be solved by using the following formula.
x=`(-b +- sqrt(b^2 -4ac))/(2a)`
There is chance of getting two values as answers. Because of the plus,minus signs.

Example problems for samples of quadratic equation:

Example: 1
Solve the following sample quadratic equation of expression 5x²+2x+1
Solution:

Given quadratic equation is 5x²+2x+1

We know the general quadratic equation is ax²+bx+c.

 Here a=5, b=2 and c=1.

The general formula for solving a quadratic equation is,

x=`(-b +- sqrt(b^2-4ac))/(2a)`

 =`(-2 +- sqrt(2^2 -4.5.1))/(2.5)`

 =`(-2 +- sqrt(4-20))/10`

 =`(-2 +- sqrt (-16))/10`

We can take square root of 16 but it has a negative sign. So we have to consider it as imaginary part and insert i into the square root value.

 =`(-2 +- 4i)/10`

 = `(-1+- 2i)/5`

Answer of the above quadratic equation is `(-1+2i)/5` or `(-1 - 2i)/5`.

Example: 2

Solve the following sample quadratic equation of expression x²+2x+1.

Solution:
Given quadratic equation is

We know the general quadratic equation is ax²+bx+c.

The general formula for solving a quadratic equation is,

                      x=`(-b +- sqrt(b^2 -4ac))/(2a)`
                      = `(-2 +-sqrt(2^2-4.1.1))/(2.1)`
                     = `(-2 +- sqrt(4-4))/2`
                     =`(-2 +- sqrt(0))/2`
                     =`-2/2`

        = -1

Answer for the above quadratic equation is -1.

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Practice problems for samples quadratic equation:

Problem: 1
Solve the following sample quadratic equation of expression2x²+3x+2.
Answer: `(-3 +sqrt(7)i)/6` or `(-3 -sqrt(7)i)/6`

Problem: 2
Solve the following sample quadratic equation of expression 4x²+3x+1
Answer: `(-3 + sqrt(7)i)/8` or `(-3 - sqrt (7)i)/8`.

Sum of Finite Geometric Series

In geometric sequence is a series by a constant ratio among following expressions. For instance, the series  `(1)/(2)+(1)/(4)+(1)/(8)+...`  is geometric, since each expression except the first know how to be get by multiplying the preceding expression by  `(1)/(2)` . Geometric series are the simplest instances of infinite sequence by finite sums. Historically, geometric sequences play a significant role in the early expansion of calculus, and they maintain to be central in learn of convergence of series.

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Sum of finite geometric series:

A geometric series, also recognized as a geometric sequence, is a series of numbers where each expression after the first is establish by multiplying the preceding one by a fixed non-zero number known the common ratio. For instance, the sequence 3, 9, 27, 81, ... is a geometric series with common ratio 3. Similarly 14, 7, 3.5, 1.75, ... is a geometric series with common proportion 1/2.
           
 The sum of the expressions of a geometric series is recognized as a geometric series. Thus, the general structure of a geometric series is  a,ar,ar²,ar³... and that of a geometric series is a+ar+ar²+ar³+....  where r ≠ 0 is the common ratio and scale factor is a.

Examples for sum of finite geometric series:

Example 1:
 Find the sum of finite n terms and the sum of first 4 terms of the geometric series 2+4+8+16+...............
Solution:
Step 1: the given series is 2+4+8+16+...............
  Step 2: Here, a=2 and r=2 and n=4
Step 3: here r >1
Step 4:     Therefore
s_n = (a(1-r^n))/(1-r)` 
 n = 4

Step 5:          `s_4 = (2(1-2^4))/(1-2)`
Step 6:               `s_4 = (2(1-2^4))/(1-2)`  
Step 7:             `S_4 = 30`

Example 2:
 Find the sum of finite n terms and the sum of first 5 terms of the geometric  series 1+3+5+7+...............

Solution:
Step 1: the given series is  1+3+5+7+..........
Step 2: Here, a=1 and r=2 and n=5

Step 3: here r >1
Step 4:     Therefore
s_n = (a(1-r^n))/(1-r)` 
n = 5



Step 5:          `s_5 = (1(1-2^5))/(1-2)`
Step 6:               `s_5 = (1(1-2^5))/(1-2)`


Step 7:             `S_5 = 31`

How to Make an Octagon

Shapes play a vital part in mathematics. In geometry, the enclosed shapes are called as polygons. The polygons are classified according to its sides. Any polygon with eight sides is said be an octagon. In this article, we shall learn how to make a regular octagon. Also we shall learn how to make irregular octagon.

How to make a regular octagon:

• Step 1: Initially make a vertical line AB.

 

• Step 2: We know that the octagon has interior angle of 135°. Using protector, mark 135° on both sides of the line AB and mark it as C and D. Make the lines from A to C and B to D. These lines should be in similar length of AB.

 

• Step 3: From C take 135° and mark it as E. And from D, take 135° and mark it as F. Make the line from C to E and D to F. The length of the line must be similar to AC and BD.

 

• Step 4: From E take 135° and mark it as G. And from F, take 135° and mark it as H. Make the line from E to G and F to H. The length of the line must be similar to CE and DF.

 

• Step 5: Now draw a line from G to H.

 

   Finally the regular octagon can be obtained as shown in the figure. This is how we can make a regular octagon.

How to make irregular octagon shape:

   An octagon is a shape with 8 sides. An eight sided enclosed shape forms an irregular polygon. Now let us draw an irregular octagon.

• Step 1: Make a line and name it as AB.

 

• Step 2: Make another line from one end (A or B).

 

• Step 3: Similarly draw eight lines.

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Note: The eighth line must form an enclosed figure. The important point to remember while drawing an octagon is that it should have eight sides enclosed. The length of the side is not a matter for irregular octagon

 

   You can obtain an irregular octagon as shown in the figure. The irregular octagon can be of any form. But it should have eight sides.

Distance of a Triangle

              

                                                     

A,B and C are three non-collinear points. The figure which surronded by segments of the lines AB, BC and CA is called  triangle having A,B and C as its vertices. All the points situated on AB, BC and CA are included in this triangle. In a triangle, an exterior angle is greater than either of the interior opposite anglesThe sum of all the angles in any triangle is 180º.

 There are types of triangle

• Equilateral triangles,
• Isosceles triangles 
• Scalene triangles
• Right Triangles

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Types of Triangles:

Right Triangles:

         In a right triangle is a triangle with a 90° right angle triangle

                                    

 

Equilateral Triangles

In equilateral triangle  all three sides are in equal length also  three angles are  equal and they are  60º in each.

                                    

Isosceles Triangles

In isosceles triangle, two sides are in equal  length. The angles opposite are also equal sides

                                    

Scalene Triangles

 A scalene triangle has no equl sides of its length. Its angles are also all different in measure.

                                   

Finding distance Of a Triangle:

Example 1:

Find the distance of triangle whose points  are (-2, -3), (-4, 4), (5,8)

Solution:

to find distance of triangle where  (x₁,y1)=(-2,-3)

(_x2,y₂)=(-4,4)

                                                         

sqrt((x2-x1)^2 +(y2-y1)^2)`

   
=`sqrt(((-4-(-2))^2+4-(-3)^2))`
=`sqrt((-4+2)^2+(4+3)^2)`
=`sqrt((-2)^2+7^2)`
=`sqrt(4+49)`
length of one side  =7.28
similarly we can find another sides

Example 2:

Find the distance of triangle whose points  are (2, 3), (4, 6),(5,0)

Solution:

to find distance of triangle where  (x₁,y1)=(2,3)

(_x2,y₂)=(4,6)

sqrt((x2-x1)^2 +(y2-y1)^2)`

=`sqrt((4-2)^2+(6-3)^2))`
=`sqrt((2)^2+3^2)`
=`sqrt(4+
length of one side =`sqrt(13)`
similarly we can find another side lengths

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Example 3:

            Find the distance of triangle whose points  are (1, -3), (2, 5),(4,6)

Solution:

          to find distance of triangle where  (x₁,y1)=(1,-3)

                                                               (_x2,y₂)=(2,5)

                                                         

                    `sqrt((x2-x1)^2 +(y2-y1)^2)`

   
                   =`sqrt((2-1)^2+(5-(-4))^2))`
             
                   =`sqrt((1)^2+9^2)`
                   =`sqrt(1+81)`
                   =`sqrt(82)`

               Length of one side =9.05
Similarly we can find another two side lengths.

How to Make a Octagon

Shapes are the important theme in Geometry. In geometry, the shapes that are enclosed shapes are said to be polygons. The type of polygon is divided on the basis of its sides. Octagon is a polygon which has eight sides. There are two types of octagon. They are regular octagon and irregular octagon. In this article, we shall discuss to make a regular octagon. Also we shall discuss to make irregular octagon.

Learn how to make a regular octagon:

• Step 1: At first, make a vertical line PQ.

• Step 2: Octagon has 135degree as its interior angle. We can use protector to mark 135° on both sides of the line PQ and mark it as R and S. Make the lines from P to R and Q to S. PR and QS are the line that has the same length of PQ.

• Step 3: Mark 135degree from R and name it as T. And from S, take 135° and mark it as U. Now make the line from R to T and S to U. The length of RT and SU must be similar to the length of PQ.

• Step 4: Make 135° from T and name it as V. Make 135° from U and name it as W. Draw  line from T to V and U to W.

• Step 5: Now make a line from V to W.

Lastly the regular octagon can be attaining as shown in the figure. This is how we can make a regular octagon.

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Learn how to make irregular octagon:

An eight sided enclosed shape forms an irregular octagon. Now we shall draw an irregular octagon.

• Step 1: Make a line and name it as PQ.

• Step 2: Make another line from P.

• Step 3: Correspondingly draw eight lines.

We can attain an irregular octagon as revealed in the figure. The irregular octagon should have eight sides. But it can be drawn in any form.

Find Vertex of Parabola Calculator

Parabola is one of the conic sections whose eccentricity is equal to  1. Eccentricity of an conic section is defined as the how it is deviating from circular  shape .The parabola has three major components.this article will help you to find vertex of parabola as a calculator.
i)Vertex
ii)Focus
iii) Directrix
iv)Axis

Expalnation to find vertex of parabola calculator

Definition of parabola :

Parabola is the locus of points whose distance from fixed line, called as  directrix, and a from a fixed point , called as focus,is equal.

Vertex of parabola :-

The vertex of parabola is the point where the parabola changes its direction.It is apoint where the parabola crosses its axis.

Features of vertex of a parabola :-

i)It lies on the parabola

ii)It  lies on axis of parabola.

iii)It is apoint equidistant from focus and directrix of parabola.

Standard forms of a parabola :-

parabolas having vertex at (0,0)

i)Y² = 4aX

ii)Y² = -4aX

iii)X² = 4aY

iv)X² = -4aY

Parabolas having vertex at (h,k)

i)(Y-k)2 = 4a(X-h)

ii)(Y-k)2 = -4a(X-h)

iii)(X-k)2 = -4a(Y-h)

iv)(X-k)2 = 4a(Y-h)

Calculation of vertex of parabola :-

i) parabola is to be transformed into one of the standard forms given above  to find the  vertex of parabola.

examples to find vertex of parabola calculator

Ex:1
Given the equation of parabola Y² = 16X find the focus of parabola
solution:
Comparing with standard form of equation
Y² = 4aX
we get vertex of parabola as
(0,0)

EX 2:FInd vertex of parabola
Y² + 2y +1 = 3x-9
solution:-
transforming into standard equation
=> (y+1)² =3(x-3)
on comparing with (Y-k)² =4a(X-h)
vertex (h,k)=(3,-1)

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Ex 3: Find vetex of parabola
X²-2x+4 = 2Y-1
solution:
Converting to standard form (X-h)² =4a(Y-k)
X²-2x+4 = 2Y+1
=> (X-2)² = 2(Y+1/2)
=> (h,k)=  (2 , -1/2).

Hexagonal Prism Net

Before learning how to draw hexagonal prism net first lets learn what is a prism?
A solid object that has two identical ends and all flat sides and the cross section is the same all along its length.
and also remember that prism has rectangular faces.
Net is a 2 dimensional pattern of 3 dimensional figure.

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Hexagon is a polygon which has six sides.
Hence the hexagonal prism net also have rectangular  and hexagonal faces.
 A hexagonal prism net is a polyhedron for which the top and bottom faces (known as the bases) are congruent hexagons, and all other faces (known as the lateral faces) are rectangles.

shape of an hexagonal prism net.
Lets look at the shape of an hexagonal prism net.


We can see that the given image has two hexagonal bases parallel to each other bounded by 6 rectangular faces.

Hexagonal prism net image

Similarly the image given below represents the hexagonal prism net as it have two hexagonal faces and six rectangular faces.



so when we fold the given hexagonal prism net it takes the shape of a hexagonal prism for which the top and bottom faces (known as the bases) are congruent hexagons, and all other faces (known as the lateral faces) are rectangles.

In geometry, the hexagonal prism is a prism with hexagonal base.

Since it has eight sides, it is an octahedron. However, the term octahedron is primarily used to refer to the regular octahedron, which has eight triangular faces.

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If faces are all regular, the hexagonal prism is a semiregular polyhedron.

 An hexagonal prism has eight faces
Vertex is a point where two edges of a polygon meet hence hexagonal prism has twelve verteces.
Edge is the meeting line of two faces in a 3 dimensional figures hence hexagonal prism has 18 edges.

Concept of the Derivative Exam

Concept of derivative can be easily explained with the help of solving basic problems in calculus. 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 for study.

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Concept of derivative example problems for study:

The following example problems explain the concept of derivative to study for the exam.

Example 1:
Calculate the derivative by differentiating the given function.
f(y) = 3y ³ +4 y ⁴  + 5y                                                             
Solution:
The given function is
f(y) = 3y ³ +4 y ⁴  + 5y
The above function is differentiated with respect to y to find the derivative
f '(y) = 3(3y ² )+4(4y ³  ) + 5
By solving above terms
f '(y) = 9y ² +  8y ³ + 5

Example 2:
Calculate the derivative by differentiating the given function.
f(y) = 6y⁶ + 5 y⁵ + 4 y⁴ + y
Solution:
The given equation is
f(y) = 6y⁶ + 5 y⁵ + 4 y⁴ + y
The above function is differentiated with respect to y to find the derivative
f '(y) =  6(6y ⁵)  +5 (5 y⁴ ) +4(4 y³) + 1
By solving above terms
f '(y) =  36y ⁵  + 25 y⁴  + 16 y³ + 1

Example 3:
Calculate the derivative by differentiating the given function.
f(y) = 2y ² +4 y ⁴  + 15
Solution:
The given function is
f(y) = 2y ² +4y ⁴  + 15
The above function is differentiated with respect to y to find the derivative
f '(y) = 2(2y  )+4(4 y ³ ) + 0
By solving above terms                                  
f '(y) = 4y +16y³

Example 4:
Calculate the derivative by differentiating the given function.
f(y) = 5y⁵ +4y ⁴ +3y ³  + 2
Solution:
The given function is
f(y) = 5y⁵ +4y ⁴ +3y ³  + 2
The above function is differentiated with respect to y to find the derivative
f '(y) = 5(5y ⁴ )+4(4y ³ ) +3( 3y ²) +0
By solving above terms
f '(y) = 25y ⁴ +16y ³  +9 y ²  

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Concept of derivative practice problems for study:

1) Calculate the derivative by differentiating the given function.
           f(y) = 2y ³ +3 y ⁴  + 4 y ⁵
Answer: f '(y) = 6y ² +12 y³ + 20 y ⁴  
2) Calculate the derivative by differentiating the given function.
         f(y) = y ³+y⁵ + 4 y ⁶
Answer: f '(y) = 3y² + 5y⁴ + 24 y ⁵      

Math Radicals Practice

Radical symbol used to indicate the square root or nth root. Radical of an algebraic group, a concept in algebraic group theory. Radical of a ring, in ring theory, a branch of mathematics, a radical of a ring is an ideal of "bad" elements of the ring. Radical of a module, in the theory of modules, the radical of a module is a component in the theory of structure and classification. Radical of an ideal, an important concept in abstract algebra. The radical symbol is ' √ ' . The cubic root of x can be expressed as `root(3)(x)`

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                                                                                                                                                                   Source Wikipedia.

Basic identities and properties of radical:

         Multiplication property rule     `root(n)(x)` *`root(n)(y)`  = `root(n)(x*y)`
         Division property rule `sqrt(x/y)`  =  `sqrt x /sqrt y`
       General Relation between exponential and radical expression     `root(n)(x)`^m =( `root(n)(x)` )^m = (x^1/n )^m = x^m/n

√ -1 ×√-1 = -1 where as  `sqrt((-1) * (-1)) ` = 1

Math radicals practice problems:

Math radicals practice problem 1:

     Simplify the radical expression:` (sqrt(16a^4)) (sqrt((-25)a^3))`
    Solution:
          ` (sqrt(16a^4)) (sqrt((-25)a^3))` = ` (sqrt(16a^4)) (sqrt((-25)a^3))`
                                                    = (4a²   a `sqrt(-25a)` )
                                                    = 4a³ `sqrt((-1)25a)`                     we know,   √(-1) = i
                                                    = i 4a³ `sqrt(25a)`
                                                    = i 4a³ (5) `sqrt(a)`
                                                    =    i 20a³ `sqrt(a)`
        Answer:      i 20a³ `sqrt(a)`
Math radicals practice problem 2:
      Simplify the radicals:    `sqrt((8x^2)/(y^5z^7)) `
     Solution:
          Step 1: Multiply and divide by  yz
                    ` sqrt((8x^2)/(y^5z^7))` = `sqrt((8x^2 * yz) / (y^5z^7 *yz))`
          Step 2: Multiply the variable with exponent
                                       = `sqrt((8x^2yz) / (y^6z^8))`
          Step 3: Square root of x²y⁶z⁸ = xy³z⁴
                      Square root of 8 = 2`sqrt2 `   
                           So the answer = ` ((2x) / (y^3z^4)) sqrt(2yz)`
      Answer:   ` ((2x) / (y^3z^4)) sqrt(2yz)`

Math radicals practice problem 3:
           Simplify the radicals : `sqrt(64)/sqrt49 + sqrt169 `
Solution:
             Step 1: Very smallest factors of 64 = 8 × 8
                             `sqrt(64)` = `sqrt(8 * 8) ` = 8
             Step 2: Very smallest factors of 169 = 13 × 13
                              `sqrt(169)` = `sqrt(13 * 13)`
             Step 3: Take square root of 169 = 13
             Step 4: Very smallest factors of 49 = 7 × 7
                          `sqrt(49)` = `sqrt(7 * 7)`
             Step 5:  Take square root of 49= 7
             Step 6: so,  `sqrt(64) /sqrt(49)` = `8 / 7 `
             Step 7: Now ,  `sqrt(64)/sqrt49 + sqrt169 ` =  `8 / 7 `+ 13 = `(8 + 91)/7`
             Step 8:  Simplification of  `sqrt(64)/sqrt49 + sqrt169 ` = ` 99/7 `
    Answer:   ` 99/7 `

Math radicals practice problem 4:
       Simplify the radicals:  `(sqrt36) (3sqrt(45a^2))`
    Solution:
             `(sqrt36) (3sqrt(45a^2))` = (`sqrt(6 xx 6)` ) (`3sqrt(45a^2)` )
                                          = `(6) ( 3a sqrt(9 * 5))`
                                          = (6) (3a(`3sqrt5` ))
                                          = 6 (9a) `sqrt5`
                                          = 54a `sqrt5`
    Answer:  54a `sqrt5`

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Math radicals practice problem 5:
          Find the cubic root of 135
      Solution:
          The very smallest factor of 135  = 3 * 3 * 3 * 5
          Cubic root of 135         `root(3)(135)`  = `root(3)(3 * 3 * 3 * 5)`
                                                                    =  `3 ` `root(3)(5)`

      Answer:  `3 ` `root(3)(5)`

Decimal Number Rounding

The decimal can be written by using the fractions. Without using decimal point notation is called rounding the decimal number. To do rounding a number, a number is a numerical value, it can be replaced by the short or nearest or the explicit representation of the number. We use rounding the decimals, because of the purpose of easy to write and handle easier than the original number. For example the whole number 45.53 can be rounded to 46.

Understanding Binary Number is always challenging for me but thanks to all math help websites to help me out.

Rules for Rounding the Decimal Numbers

Rule 1:
The number present in the right hand side of the rounding digit that is after the decimal point can be rounded if the number is above 5, 6, 7, 8 and 9. If it is below, it cannot be changed during the rounding.
Rule 2:
The numbers can be below 9, 8, 7, 6, 5 after the decimal point in a given problem, then the rounding digits can be rounded to the nearest greater number. In rounding a digits, the number after decimal point should be zero.

Example 1:
How will you round the decimal number 45. 3754
Solution:
45.3754 can be rounded to the nearest whole number is 45.

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Rounding Decimals to the nearest 10’s 100’s 1000’s

The number after decimal point having the number 49, then the place of tens and hundreds can be changed to 50.
Example 2:
How will you round the decimal number 56.3478.
Solution:
56. 4978 can be rounded to the nearest tenth is 56.5.

Example 3:
How will you round the decimal number 78.4697.
Solution:
78.4697 can be rounded to the nearest tenth is 78.5.

Example 4:
How will you round the decimal number 92.48921.
Solution:
92.48921 can be rounded to the nearest tenth is 92.5.

Example 5:
How will you round the decimal number 4.3294.
Solution:
4.3294 can be rounded to the nearest thousandth is 4.330.

Example 6:
How will you round the decimal number 7.6092.
Solution:

7.6092 can be rounded to the nearest thousandth is 7.610.

Non Linear Relationship

Non linear relationship deals with solving the linear equation problems which contain the polynomials. In non linear relationships the quadratic equation problems is also to be solved. Non Linear relationship has the relation with the families of vectors called vector spaces or non linear spaces, and with functions contains two vectors such as one input vector and output vector, according to certain rules. The example problems are discussed below with detailed solution which explains the non linear relationship.



Non linear relationship example problems:

Example 1:

Evaluate the non linear relationships for the given equation

y 2 - 24 y + 95 = 0

Solution:

In order to factor the above expression, we write the equation y 2 - 24 y + 95 in the form factored

y 2 - 24 y + 95 = (y + a)(y + b)

Hence that the sum of a and b is -24 and their product is 95. The numbers that satisfy these conditions are - 19 and - 5. Hence
y2 - 24 y+ 95 = (y- 19) (y- 5)

Substitute into the original equation and solve.
(y- 19)(y- 5) = 0

(y - 19)(y - 5) is equal to zero if
y - 19 = 0

or

y - 5 = 0

Solve the above equations to get the solution

y = 5

or

y = 19

Y = 5 or 19



Example 2:

Evaluate the non linear relationships for the given equation

Sqrt (y 2 – 16y+57) = 3

Solution:

Given equation is
Sqrt (y 2 – 16y+57) = 3

Squaring on both sides and solves it.
[Sqrt (y 2 – 12y+57)] 2 = (3) 2

And solve it.
y 2 – 12y+57= 9

Rewrite the equation with right side equation to 0.
y 2 - 16 y + 48 = 0

It is a quadratic equation with 2 solutions
y = 12 and y = 4

The above form is a quadratic equation with 2 solutions
y = 12 and y = 4

My forthcoming post is on taylor series lnx and cbse class 12 syllabus will give you more understanding about Algebra.

Non linear relationship practice problems:

1)  Evaluate the non linear relationships for the given equation

y 2 - 22 y + 72 = 0

Answer: y = 4 or 18

2)  Evaluate the non linear relationships for the given equation

Sqrt (y 2 – 12y+43) = 4

Answer: y = 3 and y = 9

Series Radius of Convergence

The radius of convergence of as series is defined as the quantity where it can be either a real number or an infinite number that indicates that within the particular radius the series will converge.

Thus the radius of convergence of the power series tends to converge evenly and exactly. If the power series is said to converge then it is called as the Taylor series of the function to which it converges inside the radius of convergence.

Having problem with taylor series convergence keep reading my upcoming posts, i will try to help you.

Mathematical definition of radius of convergence:

Let us consider the power series of a function which is given by


f (p) = `sum_(n=0)^oo` cn(p-a)n

Here a is called as the complex number constant, cn is the coefficient of the nth complex number and p is called as the complex number. The radius of convergence r is a positive real number or an infinite number and the series is said to converge if │p-a│< r and the series is said to diverge if │p-a│> r.

We can say in simple words that the series converge if the complex number p is nearer to the centre and it diverges if it is away. If the power series is given in the form of │p-a│ = r, then it is difficult to find the radius of convergence.

Tests to find radius of convergence:

We will see the two tests such as the root test and the ratio test that has been used to find the series radius of convergence.

Ratio test:

The ratio test to find the series radius of convergence is given by

r = `lim_(n->oo)` │cn / cn+1│

Thus it is clear that the series can converge if  `lim_(n->oo)`   │cn+1(p-a)n+1│/ │cn(p-a)n│is less than one

This can be equalized to

│p-a│< 1 / limn->∞ (│cn+1│/ │cn│)

= `lim_(n->oo)`   │cn / cn+1│


Root test:

The radius of convergence can also be determined by applying the root test to the series. Thus it is given by

C = `lim_(n->oo)` sup √(│cn(p-a)n│) = `lim_(n->oo)` sup   √(│cn││p-a│)

Thus it is made clear that the series converges if c < 1 and the series diverges if C >1.

Writing Parametric Equation

In mathematics, parametric equations are a method of defining a relation using parameters. A simple kinematical example is when one uses a time parameter to determine the position, velocity, and other information about a body in motion.
Abstractly, a Parametric Equation defines a relation as a set of equations. It is therefore somewhat more accurately defined as a parametric representation. It is part of regular parametric representation.  (Source: Wikipedia)


Definition for writing parametric equation of circle:

Parametric equation of circle:

It is the co-ordinates of a point on the circle in terms of a single variable θ. These single variables are called as parameter. The parametric equations of a circle with radius (r ≥ 0) and center (h, k) are written as

x = h + r cos θ          0   ≤ θ ≤ 2

y = k + r sin θ

Parametric equation of a circle with the radius r and centered at the origin.

x2 + y2 = r2

Equation of a circle with single variable formula

x = ±  `sqrt(r^2 - y^2)`

y = ±  `sqrt(r^2 - x^2)`

Each formula gives a portion of a circle

y=  `sqrt(r^2 - x^2)`          (Top)

y= - `sqrt(r^2 - x^2)`        (Bottom)

x=  `sqrt(r^2 - y^2)`          (Right side)

x = -`sqrt(r^2 - y^2)`           (Left side)

Example problem for writing parametric equations:

Example 1:

Determine the radius of parametric equation of the circle for the given x = 7 sin t and y = 7 cos t where 0 < t < `pi`

Solution:

Given  x = 7 sin t and  y = 7 cos t

Formula for the parametric equation of a circle can be writing as x2 + y2 = r2

x = r sin t

y = r cos t

(7 sin t)2 + (7 cos t)2 = r2

72 (sin2 t + cos2 t) = r2

49 (sin2 t + cos2 t) = r2                Hence (sin2 t + cos2 t = 1)

r2 = 49

Radius of the parametric circle is 7.

Example 2:

Determine the radius of parametric equation of the circle for the given x = 13 sin t and y = 13 cos t where 0 < t <`pi`

Solution:

Given  x = 13 sin t and y = 13 cos t

Formula for the parametric equation of a circle can be writing as x2 + y2 = r2

x = r sin t

y = r cos t

(13 sin t)2 + (13 cos t)2 = r2

132 (sin2 t + cos2 t) = r2

169 (sin2 t + cos2 t) = r2                Hence (sin2 t + cos2 t = 1)

r2 = 169

Radius of the parametric circle is 13.

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Example 3:

Determine the parametric equation of a circle with radius = 5 and center (3, 7).

Solution:

Given (h, k) =(3, 7)

Radius = 5

Let  x = h + r cos `theta`

y = k + r sin `theta`

Parametric equation of a circle with radius 5 and center (3, 7)

x = 3 + 5 cos `theta`

y = 7 + 5 sin `theta`

Rules for Signs in Algebra

Algebra is a part of mathematics deals with operations and rules of arithmetic of symbols and numbers, usually letters of the alphabet to express general relationships that hold for all members of the set. Rules for signs in algebra tell us what we are allowed to write. Operations are addition, subtraction, multiplication and division in algebra.

I like to share this Sig Fig Rules with you all through my article. 

Addition rules for signs in algebra:

If the expressions signs are the same, we can add and keep the sign. Otherwise expressions signs are different we can do subtract the expression and take the sign attached to the larger number value.  
Example
- 5 + - 8 = - 13

Subtraction:
Add the Opposite:

1. Change the subtraction sing into addition.
2. Change the sign of the second term to its opposite.

Example:
5 - (+ 2) = 3 {Since 2 is positive.}
5 + (- 2) =3  {Add the opposite.}
4 + (+ 2)= 6 {we can exchange - ( - ) with + +}.
           The subtraction rules for signs in algebra means that the operations of sequences - ⁺ and + ^- are interchanged.

Multiplication and division rules for signs in algebra:

If the expressions signs are the same, the result is opposite; otherwise negative.

Table for multiplication rules for signs in algebra:

+  x  +	=	+
-  x -	=	+
+  x  -	=	-
-x +	=	-

Table for division rules for signs in algebra:

+÷+	=	+
-÷-	=	+
+÷-	=	-
-÷+	=	-

Practice problems regarding rules for signs in algebra:

Problem 1:  -7 + -13
Solution:
   = -7 + (-13)
Add the two terms
   = -20 (answer)

Problem 2:  -`7/8` - -`5/8`
Solution:
   ⁼ - `7/8` + `5/8`
Change second term sign
    ⁼ - `7/8` + `5/8`
   = -`2/8`
   = - `1/4`

Problem 3:  -1 - 5
Solution:
=-1 - 5
Add the terms
=-6

Problem 4:  49 +( -15)
Solution:
=49 + -15
Subtract the terms
=34

Problem 5:
`1/2` - `1/8`
Solution:
=`1/2` - `1/8`
Subtract the terms:
  = `3/8`

Problem 6: 
`3/8` ÷ `1/8`
Solution:
= `3/8` ÷ `1/8`
Change the sing and second term
=`3/8` * `8/1`
= 3

Problem 7: -(-5)
Solution:
=+5

Problem 8: -2 + -2 + -2 + -2
Solution:
Add the all terms
=-8

Problem 9: 13 ÷-2
Solution:
=-7.5


My forthcoming post is on sample papers for class 12 cbse and sample papers for class 9 will give you more understanding about Algebra.

Problem 10:  -`3/8` (16)
Solution:
Multiply the terms
= -`48/8`
=-6

Problem 11: 3 + 3 + 3 + 3
Solution:
Add all term
=12

Problem 12:  -6(-3)
Solution:
=+18

Matrices Addition

In mathematics, a matrix (plural matrices, or less commonly matrixes) is a rectangular array of numbers, such as

[[1, 9, 13], [20, 55, 4]]

An item in a matrix is called an entry or an element. The example has entries 1, 9, 13, 20, 55, and 4. Entries are often denoted by a variable with two subscripts, as shown on the right. Matrices of the same size can be added and subtracted entry wise and matrices of compatible sizes can be multiplied. (Source: Wikipedia)

This article explains you about add two matrices, some solved problems on add two matrices and practice problems on adding two matrices.

Solved Problems on addition of matrices

Method to add matrices:

We can add the matrix with same order only.

By adding corresponding elements of two matrices we can get the sum of two matrix.

Problem 1:

Add the following two matrices

A = [[2,-5],[9,7]] B = [[9,6],[-4,0]]

Solution:

Given, Two matrix A = [[2,-5],[9,7]] ,  B = [[9,6],[-4,0]]

We need to find the addition .

That is A + B

A + B =  [[2,-5],[9,7]]  + [[9,6],[-4,0]]

We know that, By adding corresponding elements of two matrices we can get the sum of two matrix.

A + B =  [[2+9,-5+6],[9+(-4),7+0]]

= [[11,1],[5,7]]

Answer: A + B = [[11,1],[5,7]]

Problem 2:

Add the following two matrices,

P = [[23, 54, -31],[12, -72, 41]] and Q = [[ -9, 8, 0],[31,38,-45]]

Solution:

Given, Two matrix P = [[23, 54, -31],[12, -72, 41]] and Q = [[ -9, 8, 0],[31,38,-45]]

We need to add the above two matrix.

Both are 2 by 3 matrix. By adding corresponding elements of two matrices we can get the sum of two matrix.

P + Q =  [[23, 54, -31],[12, -72, 41]] + [[ -9, 8, 0],[31,38,-45]]

= [[23+ (-9), 54+8, -31+0],[12+31, -72+38, 41+(-45)]]

= [[14, 62, -31],[43, -34, -4]]

Answer: P + Q = [[14, 62, -31],[43, -34, -4]]

Problem 3:

Find the value of x, if  A = [[2,3x], [5,6]] , B = [[6,-3],[8,-2]] and A+B = [[8,12],[13,4]]

Solution:

Given, A = [[2,3x], [5,6]] , B = [[6,-3],[8,-2]] and A+B = [[8,12],[13,4]]

We need to find the x value.

A + B = [[8,12],[13,4]]

[[2,3x], [5,6]] + [[6,-3],[8,-2]] =  [[8,12],[13,4]]

[[2+6,3x-3],[5+8,6-2]] = [[8,12],[13,4]]

[[8, 3x-3],[13,4]] =  [[8,12],[13,4]]

In this corresponding elements are equal.

So we can take, 3x-3 = 12

Add 3 on both sides,

3x-3 + 3 = 12 + 3

3x = 15

Divide by 3 on both sides,

x = 5

Answer: The value of x = 5

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Practice problems on addition of matrices

Problems:

1. Add two matrices P = [[ -2,-4],[4,9]] and Q = [[8, 5],[2,-7]]

2. Find the value of  x , if  X = [[0,-2],[3,6]] , Y = [[ 2,-5],[x,4]] and X + Y = [[ 2,-7],[ 8, 10]]

Solution:

1. P + Q = [[6,1],[6,2]]

2. x = 5

How do you Multiply Trinomials

In basic algebra, a trinomial is consisting of three expressions. A trinomial is an equation concerning three expressions. An instance is the equation x = q + xm. In factoring trinomials technique is one important in elementary algebra. It is definite as the process of sum of three monomials. Given expression consists of three expressions that are within the normal form of ax2 + bx + c.

Different type of methods using for multiply trinomials:

Factoring trinomial is the method to generate arithmetic operation of multiplication. Let us determine some instance problems for factoring trinomials are x2 +3x – 7.  Different kind of technique for multiply trinomials that are,

Distributive method:

It is the operation of multiplying every expression in the initial trinomial is multiplied among the each expression in the second trinomials. Multiply trinomial using distributive method we know how to divide into two types,

Horizontal method
Vertical method


Horizontal method:

Horizontal technique is definite as the process of multiplying the specified trinomials in horizontal also combines the terms.

Vertical method:

A vertical technique is the definite as the process of multiplying the specified trinomials in vertical and adds the terms


Example for multiply trinomials:

Example 1:

how do you multiply trinomials (x+2) (x2+6x+4)

Solution:

Step 1: the given trinomial factors are (x+2) (x2+6x+4)

Step 2: using vertical method to multiply the given trinomials

x2+6x+4

x+2

---------------------

x3+6x2+4x

2x2+12x+8

----------------------

x3+8x2+16x+8     (do you add the terms)

----------------------

so the solution to the given trinomial is x3+8x2+16x+8

Example 2:

how do you multiply trinomials (2x+4) (x2+3x+6)

Solution:

Step 1:  the given factors are (2x+4) (x2+3x+6)

Step 2: using horizondal method to multiply the given trinomials

Step 3:  to multiply the first binomial term 2x

= 2x(x2+3x+6)

= 2x3+6x2+12x

Step 4: to multiply the second binomial term 4

=  4(x2+3x+6)

=  4x2+12x+24

Step 5:  do you add the group

=   2x3+6x2+12x+4x2+12x+24

=  2x3+8x2+24x+24

so the solution is 2x3+8x2+24x+24

Example 3:

how do you multiply trinomials (x-3) (x2+3x+1)

Solution:

Step 1:  the given factors are (x-3) (x2+3x+1)

Step 2: using horizondal method to multiply the given trinomials

Step 3:  to multiply the first binomial term x

= x(x2+3x+1)

= x3+3x2+x

Step 4: to multiply the second binomial term -3

=  -3(x2+3x+1)

= -3x2-9x-3

Step 5:  do you add the group

=  x3+3x2+x-3x2-9x-3

=  x3-8x-3

so the solution is x3-8x-3

Equivalence Properties of Equality Solving Online

In mathematics, equality is a term used for representing the equal quantities by a symbol ‘=’(equal). The main property used for solving the equality is equivalence property. Online gives three equivalence properties with neat definition. Online also gives some example expressions with equality having equivalence properties. Now by this online we are going to learn some problems solving using equivalence properties.

Explanation to equivalence properties of equality solving online:

The three properties of equivalence are,

Reflexive property of equality
Symmetric property of equality
Transitive property of equality

Reflexive property:

The reflexive property of equality says that a every element is equal to the element itself.

y = y

Symmetric property:

The symmetric property of equality says that if a we have equal elements x = y then it is equal to y = x.

Transitive property:

The transitive property of equality says that if a we have equal elements x = y and y = z then  we can take it as y = z.

Example problems to equivalence properties of equality solving online:

Example: 1

Solve: m + 8 = n and n = 12.

Solution:

Given,

m + 8 = n and n = 12

By definition of transitive property of equality,

m + 8 = 12

m = 12 – 8

m = 4

Answer: m = 4 and n = 12

Example: 2

Solve: 12 + 6 = a + 12

Solution:

Given 12 + 6 = a + 12

By definition of symmetric property of equality,

x + y = y + x

12 + 6 = a + 12

a = 6

Answer: a = 6

My forthcoming post is on T Distribution Tables and example of algebra will give you more understanding about Algebra.

Practice problems to equivalence properties of equality solving online:

Problem: 1

Solve: n + 56 = 56 + 12

Answer: n = 12

Problem: 2

Solve: Solve: 5 + m = n + 6 and n + 6 = 12.

Answer: m = 7 and n = 6

Area Word Problem

The area of rectangle parallelogram, triangle, trapezium, quadrilateral and circle formulas are given below.

Area of rectangle = Length * Breath (Sq. units)

Area of parallelogram = Breath * Height (Sq.units)

Area of triangle = 1/2 * b * h (sq.units)

Area of trapezium = `(1)/(2)` * (a + b) * h

Area of Quadrilateral = `(1)/(2)` *  d * (h1 + h2)

Area of the circle = πr2 sq. units

I like to share this Surface Area of Cones with you all through my article.

Example problems for Area word problem:

Word problem 1:

A thread of length 264 cm is cut into two equal portions. One part is bent in the form of a circle and the other part in the form of an equilateral triangle. Find the ratio of the areas enclosed by them.

Solution:


Perimeter of the circle = 264/2=132 cm.

But perimeter of the circle = 2πr.

2 × 22/7 × r = 132 or r = 21 cm.

Area of the circle = πr2 = 22/7 × 21 × 21 = 1386 cm2.

Perimeter of the equilateral triangle = 3a

But perimeter = 132 cm.  3a = 132 or a = 44 cm.

Area of the equilateral triangle = √3/4*a2 = √3/4 * 442  = 484 √3 cm2


The ratio of the area of circle to that of the equilateral triangle

= 1386 : 484√3= 21√3: 22

Word problem 2:

Cost of levelling a land is Rs. 12 per square metre. A ground is in the form of a trapezium whose parallel sides are of lengths 18m and 12 m. If its other two sides are each of  5m length, find the total cost incurred in levelling the ground.

Solution:

ABCD is the given trapezium figure, where AB = 18m, CD = 12 m, AD = BC = 5 m. Draw CE parallel to DA (see Figure). ΔEBC is isosceles whose height


h = `sqrt(5^2 - 3^2)` = 16 = 4cm. Now, Figure the area of the trapezium ABCD

= 1/2 (a + b) × h = 1/2 (18 + 12) × 4

= 2 × 30= 60 sq. metre.

The cost of leveling 1 sq.metre is Rs. 12.

So the cost of levelling the entire land = 60 × 12 = Rs. 720.

Example problem 3:

The dimensions of a rectangular metal sheet are 4m × 3m. The sheet is to be cut into square sheets each of side 4 cm. Find that area.


Solution:

Area of the metal sheet = 400 × 300 = 12,0000 cm2.

Area of a square sheet = 4 × 4 = 16 cm2.

Practice area word problems:

Practice problem 1:

Find the area of a trapezium. The parallel sides have lengths 7 cm and 8 cm is 30 cm2. Find the distance between the parallel sides.

Answer: 4cm


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Practice problem 2:

Find the area of the quadrilateral one of whose diagonals is of length 15 cm and the lengths of the altitudes to this diagonal are 3 cm and 5 cm.

Answer: 60cm2