Solve for Cube Root

In mathematics, a cube root of a number, denoted (^3sqrt (x) or x^1/3, is a number such that a³ = x. All real numbers have exactly one real cube root and a pair of complex conjugate roots, and all nonzero complex numbers have three distinct complex cube roots. For example, the real cube root of 8 is 2, because 2³ = 8.                                                                                                       (Source: Wikipedia)

Example to solve for cube root:

Let us see some of the example for solve for cube root for better understanding,
Example 1: solve for cube root `(^3sqrt (512))`
Solution:
Given: `(^3sqrt (512))`
Here, 512 is obtained by multiplying 8 three times. That is,
512 = 8 * 8 * 8`
(^3sqrt (512))` ` =` `(^3sqrt (8))`
= 8 `
Answer: solve for cube root `(^3sqrt (512)) = 8`

Example 2: solve for cube root` (^3sqrt (10^2 * 10^7))`
Solution:
Given: `(^3sqrt (10^2 * 10^7))`
(10^2 * 10^7) = (10 * 10) * (10 * 10 * 10 * 10 * 10 * 10 * 10)`
= (10 * 10 * 10) * (10 * 10 * 10) * (10 * 10 * 10) `
= 10^3 * 10^3 * 10^3 `
(^3sqrt (10^2 * 10^7)) = (^3sqrt (10 ^3* 10^3 * 10^3)) `
= 10 * 10 * 10 `
= 1000`
 Answer to solve for cube root:  `(^3sqrt (10 * 10 * 10)) = 1000`

Example 3: solve for cube root` (^3sqrt (17^2 * 17^4))`
Solution:
Given: `(^3sqrt (17^2 * 17^4))`
(17^2 * 17^4) = (17 * 17) * (17 * 17 * 17 * 17)`
= (17 * 17 * 17) * (17 * 17 * 17)`
= 17^3 * 17^3 * 17^3 `
(^3sqrt (17^2 * 17^4)) = (^3sqrt (17^3 * 17^3 * 17^3)) `
= 17 * 17 * 17 `
= 4913`
Answer to solve for cube root:  `(^3sqrt (17 * 17 * 17)) = 4913`

Example 4: solve for cube root  ` (^3sqrt (1/729))`
Solution:
Given: `(^3sqrt (1/729))`
1/729 = 1/8 * 1/8 * 1/8`
(^3sqrt (1/729)) = (^3sqrt (1/8^3))`
= 1/8 `
Answer: solve for cube root  `(^3sqrt (729)) = 1/8`

Example 5: solve for cube root  `(^3sqrt (1/8))`
Solution:
Given:  `(^3sqrt (1/8))`
1/8 = 1/2 * 1/2 * 1/2`
(^3sqrt (1/8)) = (^3sqrt (1/2^3))`
= 1/2 `
 Answer: solve for cube root  `(^3sqrt (1/8)) = 1/2`

Example 6: solve for cube root `(^3sqrt (343))`
Solution:
Given: `(^3sqrt (343))`
Here, 512 is obtained by multiplying 8 three times. That is,
343 = 7 * 7 * 7`
(^3sqrt (343))` ` =` `(^3sqrt (7^3))`
= 7 `
Answer: solve cube root `(^3sqrt (343)) = 7`

I am planning to write more post on Trig Reference Angles and tamil nadu samacheer kalvi 10th books. Keep checking my blog.

Practice problem for solve for cube root:

Practice problem for solve for cube root 1: `(^3sqrt (125))`
Answer: `(^3sqrt (125)) = 5`

Practice problem for solve for cube root 2: `(^3sqrt (216))`
Answer:` ( ^3sqrt (216)) = 6`

Practice problem for solve for cube root 3: `(^3sqrt (1331))`
Answer: `(^3sqrt (1331)) = 11`

Practice problem for solve for cube root 4:` (^3sqrt (1728))`
Answer: `(^3sqrt (1728)) = 12`

Practice problem for solve for cube root 5: `(^3sqrt (3375))`
Answer: `(^3sqrt (3375)) = 15`

Practice problem for solve for cube root 6: `(^3sqrt (27))`

Answer: `(^3sqrt (27)) = 9`

Graph Co-ordinates System

In geometry, a coordinate system is a system which uses a set of numbers, or coordinates, to uniquely determine the position of a point or other geometric element. The order of the co-ordinates is the significant and they are sometimes identified by their position in an ordered tuple and sometimes by a letter, as in 'the x-co-ordinate'. In elementary mathematics the co-ordinate are taken to be real numbers.    (Source : Wikipedia)

I like to share this The Rectangular Coordinate System with you all through my article. 

Examples for graph co-ordinates system:

Example 1:

Draw the graph of the line joining the co-ordinate points (2, 3) and (−4, 1).

Solution:

 Draw the x-axis and y-axis on a graph paper sheet and take 1 cm = 1 unit on both the axes. Let A and B be the co-ordinate points (2, 3) and (−4, 1). We mark these points on the graph paper sheet. We join the points A and B by a line  and extend it along the two directions. The required graphing is now obtained

Example 2:

Draw the graph of the line joining the co-ordinate points (-3, 3) and (4, 3).

Solution:

 Draw the x-axis and y-axis on a graph paper sheet and take 1 cm = 1 unit on both the axes. Let A and B be the co-ordinate points (-3, 3) and (4, 3). We mark these points on the graph paper sheet. We join the points A and B by a line  and extend it along the two directions. The required graph is obtained

Example 3:
Draw the graph of the line joining the co-ordinate points (-2,0) (0,3)

Solution:

 Draw the x-axis and y-axis on a graph paper sheet and take 1 cm = 1 unit on both the axes. Let A and B be the co-ordinate points (-2, 0) and (0, 3). We mark these points on the graph paper sheet. We join the points A and B by a line  and extend it along the two directions. The required graphing is now obtained

I am planning to write more post on 10th std samacheer kalvi question bank. Keep checking my blog.

Practice problem for graph co-ordinates system:

1. Draw the graph of the line joining the co-ordinate points (9, 3) and (21, 3).

2. Draw the graph of the line joining the co-ordinate points (5, 2) and (−4, 7).

Calculate the Area of Ellipse

In geometry, an ellipse is a plane curve that is formed as a result of conic section (ie) the intersection of a cone by a plane produces a closed curve called ellipse the figure is shown below. 

Ellipse has one major axis and one minor axis. Circle is a special case of ellipse where the major axis and minor axis are same. Here in this topic we are going to see more problems regarding area of ellipse.

Formula for area of the ellipse:

Area of Ellipse = `pi` * length of major axis * length of minor axis
in other words
Area of the ellipse = `pi`  * w * h

Area of the Ellipse - Example problems:

Example: 1
Calculate the area of the ellipse where the major radius is 5 cm and minor radius is 3 cm.
Solution:
w = 5  h= 3
Area of the ellipse = `pi` * 5 * 3
                           = 3.14 * 5 * 3
                             = 47.1 cm²
Example: 2
Calculate the area of the ellipse where the major radius is 6 cm and minor radius is 3 cm.
Solution:
w = 6  h= 3
Area of the ellipse = `pi` * 6 * 3
                             = 3.14 * 6 * 3
                             = 56.52 cm²
Example: 3
Calculate the area of the ellipse where the major radius is 4 cm and minor radius is 2 cm.
Solution:
w = 4  h= 2
Area of the ellipse = `pi` * 4 * 2
                            = 3.14* 4* 2
                            = 25.12 cm²
Example: 4
Calculate the area of the ellipse where the major radius is 7 cm and minor radius is 3 cm.
Solution:
w = 7  h= 3
Area of the ellipse = `pi` * 7 * 3
                           = 3.14 * 7 * 3
                             = 65.94 cm²
Example: 5
Calculate the area of the ellipse where the major radius is 8 cm and minor radius is 3 cm.
Solution:
w = 8  h= 3
Area of the ellipse = `pi` * 8 * 3
                            = 3.14 * 8 * 3
                            = 75.36 cm²

I am planning to write more post on The Exponential Function and  samacheer kalvi tamil book. Keep checking my blog.

Example: 6
Calculate the area of the ellipse where the major radius is 4 cm and minor radius is 3 cm.
Solution:
w = 4  h= 3
Area of the ellipse = `pi` * 4 * 3
                            = 3.14 * 4* 3
                            = 37.68 cm²

Area of a Standard Triangle

Area of a shape is to find out the amount of space it consumes two dimensionally. In math area of different shapes are studied and specific formulas have been derived to calculate problems on area of some frequently used shapes like triangle, square, cylinder etc. Area is measured in square units. Here we are going to study about the area of standard triangles. A standard triangle is one which as sides in such a way that, the sum of two angles gives the third angle. Here we are going to learn how to calculate the area of  a standard triangle.

Standard triangle:

A triangle which has the angles in such a way that, the sum of two smaller angles give the third angle. It can also be defined as the triangles having sides in such a way that the sum of the square of two smaller sides is equal to the square of the bigger side (i.e. satisfying Pythagoras theorem). The formula for finding the area of  standard triangles is given below,

                                  Area of a standard triangle = `1/2` b * h
                                  where,
                                   b = base width of the triangle,
                                   h = height of the triangle.

Example problems on standard triangles:

Here are few examples illustrating the calculation of area for a standard triangle,

Example 1:
Find the area of the standard triangle shown in the figure.

Solution:
From the figure it is clear that,
b = 5cm
h = 12 cm.
Area of the triangle = `1/2` (b)* (h)
                                   = 1/2 (5) (12)
                                   = 60/2
                                   = 30 cm²

Algebra is widely used in day to day activities watch out for my forthcoming posts on Solving Nonlinear Differential Equations and samacheer kalvi books for 10th. I am sure they will be helpful.

Example 2:
Find the area of the standard triangle shown in the figure.

Solution:
From the figure it is clear that,
b = 3cm
h = 4cm.
Area of the triangle =` 1/2` (b)* (h)
                                   = 1/2 (3) (4)
                                   = 12/2
                                   = 6 cm²

Types of Angles Geometry

In geometry, an angle is the figure formed by two rays sharing a common endpoint, called the vertex of the angle. The magnitude of the angle is the "amount of rotation" that separates the two rays, and can be measured by considering the length of circular arc swept out when one ray is rotated about the vertex to coincide with the other.(source: WIKIPEDIA). In this article we see the types of angles in geometry. There are nine types of angles in geometry.

Types of geometric angles with diagrams

Types:
Acute angle:
An angle whose measure is smaller than 90 degrees, then the angle is acute angle. The following is an acute angle.
                                                                
Right angle:
An angle whose measure is exactly 90 degrees, then the angle is right angle. The following is a right angle.
                                                                
Obtuse angle:
An angle whose measure is above 90 degree and also less than 180 degree. Thus, it is between 90 degrees and 180 degrees, and then the angle is obtuse angle. The following is an obtuse angle.
                                                             
Straight angle:
An angle whose measure is 180 degrees. Thus, a straight angle looks like a straight line. The following is a straight angle.
                                                              
Reflex angle:
An angle whose measure is above 180 degrees but less than 360 degrees, then the angle is reflex angle. The following is a reflex angle.
                                                              
Adjacent angles:
Angle with a common vertex and one common side. <1 adjacent="" and="" angles.="" are="" nbsp="" p="">                                                              
Complementary angles:
Two angles whose measures add to 90 degrees. Angle 1 and angle 2 are complementary angles, as together they form a right angle.
Note that angle 1 and angle 2 do not have to be adjacent to be complementary as long as they add up to 90 degrees.
                                                               
Supplementary angles:
Two angles whose measures add to 180 degrees. The following are supplementary angles.
                                                                

Types of geometric angles with examples:

Vertical angles:
Angles that have a common vertex and whose sides are formed by the same lines. The following (angle 1 and angle 2) are vertical angles.
                                                              
When two parallel lines are crossed by another line is called Transversal, 8 angles are formed. Take a look at the following figure.
                                                               
Angles 3,4,5,8 are interior angles
Angles 1,2,6,7 are exterior angles

Alternate interior angles:
In transversal pairs of opposite sides are interior angles.
Such as, angle 3 and angle 5 are alternate interior angles. Angle 4 and angle 8 are also alternate interior angles.

Alternate exterior angles:
In transversal pairs of opposite sides are exterior angles.
Angle 2 and angle 7 are alternate exterior angles.

I am planning to write more post on Sine Cosine Identities, samacheer kalvi books online. Keep checking my blog.

Corresponding angles:
Pairs of angles that are in similar positions.
Angle 3 and angle 2 are corresponding angles.
Angle 5 and angle 7 are corresponding angles

Practice for geometry angles:
1) Angle 180° is -------- a) Straight angle b) right angle Answer: Straight angle
2) Angle less than 90° is ------------ a) obtuse angle b) acute angle Answer: Acute angle

Volume of Triangular Prism

Geometry deals with shapes, structures, lines, planes and angle’s. Geometry learning is also known as architectural learning. Basic shapes of geometry are square, triangle, rectangle, parallelogram, trapezoid etc. Triangle is one of the basic shapes in geometry. The total internal angle of the triangle is 180 degree.

Prism:

In mathematics, prism is one interesting shapes in geometry. It is a polyhedron of n-sided prism which has a base of n-sided polygonal, copy of a translated which has joining n faces with the sides.

Definition of volume:

In mathematics, Volume is one interesting measure of geometry. It is an amount of three-dimensional space of an object occupies.

Formula for finding the Volume of triangular prism:

The formula used for finding the volume of triangular prism, is given as,

Volume of triangular prism = Base area x Length of prism

Base area of prism =` (bh)/2`

Where, b = base of the triangle

            h = height of the prism.

           First the base area of the triangular prism is to be calculated. Using that value the volume of the prism can be calculated.

Volume of triangular prism Problems:

Example 1:

Find the volume of the triangular prism, whose base is 7 cm, height is 9 cm and length is 12 cm.


Solution:
The formula used for finding the volume of triangular prism, is given as,
Volume of triangular prism = Base area x Length of prism
Given: b = 7 cm, height = 9cm and length = 12cm
Base area =` (bh)/2`
                = `(7*9)/2`
                = `63/2`
                = 31.5 cm²
Volume     = 31.5*12
                = 378 cm³.
The answer is 378 cm³.

Algebra is widely used in day to day activities watch out for my forthcoming posts on Area of an Octagon and ix std cbse question papers. I am sure they will be helpful.

Example 2:
Find the volume of the triangular prism, whose base is 5.5 cm, height of the prism is 9 cm and length is 7 cm.



Solution:
The formula used for finding the volume of triangular prism, is given as,
Volume of triangular prism = Base area x Length of prism
Given: b= 5.5 cm, height = 9 cm and length = 7 cm.
Base area = `(bh)/2`
                = `(5.5*9)/2`
                = `49.5/2`
                = 24.75 cm².
Volume     = 24.75*7
                = 173.25 cm³.
The answer is 173.25 cm³.         

Application Integration Standards

Calculus has two parts. There are integrtaion and differentiation. The definition of differentiation is "rate of change of input". The Process of differentiation is called integration. For example, f(x) is a function, The integration of given function  can be  expressed as `int` f(x) dx .Here, f(x) is a continuous function. Integral calculus has two types. One is definite and indefinte integral. In this article, we shall discuss about application of standard integration.

Basic standard integration formulas:

1. `int` x ⁿ dx = `(x^n+1) / (n+1)`
2.`int`cos x .dx = sin x + c
3. `int`sin x.dx  = - cos x + c
4.`int`sec²x dx = tan x + c
5. `int`cosec x . cot x .dx = - cosec x + c
6. `int` sec x . tan x. dx = sec x + c
7. `int`cosec² x . dx = -cot x + c
8.`int`a f(x) dx = a `int`f(x) dx
9. `int`[f(x) ± F(x)] dx =`int` f(x) dx ± `int`F(x) dx
10. `int` `(1/x)` dx = ln x + c

These above formulas helps to solve  the standard integration. Area and Volume of the sphere, cylinder, average value these are the main application of integrals.

Application integration standards - problems:

Application integration standards - problem 1:
Determine the area of shaded portion of given graph.

                        
     Solution:
Given  y = x²
The boundary limits are 2 and 5
So we can integrate the given equation is  `int_a^b ` x² dx.
we know the lower limit is 2 and upper limit is 5.
So,   `int_2^5 ` x² dx. =  `[(x^3)/3]_2^5`
= `[(5^3)/3]`.- `[(2^3)/3]`.
= `125/3`  -  `8/3` .
= `(64 - 1)/3` .
= `117/3`     .
= 39 .square units.

Answer:    Area of shaded portion is 39 square units.    

If you have problem on this topics please browse expert math related websites for more help on cbse website.

Application integration standards - problem 2:

Determine the total area of graph x³ + x² - x and the x axis. the points x = -1   and x = 2

           
    Solution:
Given  x³ + x² - x
the limits are   x = -1  and x = 2
In the above graph x axis -1< x< 0  and 0 < x < 2.
So, the total area = `int_(-1)^0` [x³ + x² - x] dx + ` int_0^2` [x³ + x² - x] dx.
`int_(-1)^0` [x³ + x² - x] dx  = ` [(x^4/4) + (x^3)/3 - ((x^2)/2)]_(-1)^0`
= `[0] - [(-1)^4/4 + (-1)^3/3 - ((-1)^2/2)] `
= ` - [1/4 - 1/3 - 1/2] `
= ` - [(3-4-6)/12] `
= ` - [(-7)/12] `
=   ` [(7/12)] ` .

int_0^2` [x<sup>3</sup> + x<sup>2</sup> - x] dx.      = `[(x^4/4) + (x^3)/3 - ((x^2)/2)]_0^2` ..
                                            = `[(2^4/4) + (2^3/3) - (2^2/2)] - [0]`    .
                                            = `[(16/4) + (8/3) - (4/2)]`    .
                                            = `[(4) + (8/3) - (4/2)]`    .
                                            =  `[(24 + 16 - 12)/6]`  =   `[(28/6)]` .
           `int_(-1)^0` [x<sup>3</sup> + x<sup>2</sup> - x] dx + ` int_0^2` [x<sup>3</sup> + x<sup>2</sup> - x] dx. = ` [(7/12)+(28/6)] ` .
 =  `[(7+56)/12]` .
 =  `63/12` .
     Answer:  Total area is `63/12` .square units.