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.

I like to share this Hexagonal Prism with you all through my article.

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  

Algebra is widely used in day to day activities watch out for my forthcoming posts on cbse xii question papers. I am sure they will be helpful.

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)`

Understanding Addition Practice is always challenging for me but thanks to all math help websites to help me out. 
                                                                                                                                                                   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`