Write a Fraction for the Point

Fraction:

Fraction is defined as an element of quotient field. Fraction can be represented as `x/y` where fraction variable 'x' denotes the value called as numerator and fraction variable 'y' denotes the value called as denominator and the denominator 'y' is not equal to zero. It is used to write the fraction format for the given point.

Thus the fraction is classified as follows,

• Simple fraction
• Proper fraction
• Improper fraction
• Complex fraction

write a fraction for the point : Types of fractions


Simple fraction:

Simple fraction is a fraction, which has both numerator and denominator as whole number.

Ex:

`1/5` , `2/7` , `8/9`

Proper fraction:

It is a fraction, which has a numerator less than its denominator, and the value of that fraction is less than one.

Ex:

`3/5` , `1/8` , `24/25`

Improper fraction:

Improper fraction is a fraction, where the top number of fraction that the numerator is greater than or equal to its own denominator (bottom number) and the value of that fraction is greater than or equal to one.

Ex:

`7/2` , `45/23` , `123/120`

Complex Fractions:

If a fraction of numerator and denominator contains a fraction, it is called complex fraction.

The complex fraction is also called as a rational expression because it has a numerator and denominator with fraction. Otherwise, the overall fraction includes at least one fraction.

Ex:

` (7/3) / (4/5)`


Example problems for write a fraction for the point:


Ex 1

Write the fraction for the following point: (0.5, 1.5, and 0.88)

Sol:

0.5

Step 1 :( multiply  and divide by 10 on both sides, we get )

`(0.5)*(10)/10`

= `5/10`

Step 2: simplifying we get

=`1/2`

like wise for the following numbers we get,

1.5 = `3/ 2 ` ( multiply and divide by 2 on both sides)

0.88 = `8/9 ` ( multiply and divide by 10 on both sides)

Ex: 2

Write the Equivalent fraction for the following points: (0.25, 0.75, 2.5, and 50)

Sol:

0.25 = `1/4 ` = `2/8` = `3/12 `

0.75 = ` 3/ 4` = `6/8` = `9/12`

2.5 = `5/2` = `10/4` = `15/6 `

50.0 = `100/2` = `200/4 `

study graph numbers

Number:

     A number is defined as a numerical thing, which is used for measuring and counting. It is also called as numeral and includes zero, negative numbers, rational numbers, irrational numbers, etc,. The procedure of numerical operation involves one or more numerical as input and generate its relevant numerical output. This operation includes arithmetic function such as addition, subtraction, multiplication, division, and exponentiation.

Study about graph numbers:

Composite Number:

A Composite Number is a number, which can be divided with evenly. The composite number has additional than two factors with one and itself.

Otherwise, Composite number can be defined as numeral (integer) that is accurately divided with minimum one factor except one and itself. Composite number has infinite numbers also. However, composite numbers are not prime numbers.

Prime Number:

A prime number is a number, which can be divided only with one and itself. The prime number has only two factors such that one and itself.

Cardinal numbers:

Cardinal number is defined as counting numbers in words, which is representing the quantity such as five dogs, three boys. Otherwise, it is referred to as overview of natural numbers, which is calculating the cardinal (size) of the given sets.

Ordinal number:

     Ordinal numbers are numbers, which denote the order or location of objects with number words in a series such as ‘first’, ‘second’, ‘third’... It is also called as ordinals.

Graphing numbers:

     Graphing numbers is the graphical representation of integers with inequalities in horizontal line. It is a visualizing result of number line with simple steps.

Inequality:

     Inequality is defined as two real numbers or two algebraic expressions are related with functioning a sign as ‘<’ (less than), ‘>’ (greater than), ‘≤’ (less than or equal) and ≥ (greater than or equal).

Examples for graph:

Example 1) graph the following points

A= 3 + 2i
B =  -4 + 5i
 C = -5 - 4i
 D = i

Solution: 

2) Graph the following inequalities:

a) -1≤w≤4

Solution:

The value of ‘w’ is -1, 0, 1, 2, 3, 4

So graphing of ‘w’ on number line is

 

b) 1≤q<5

Solution:

The value of ‘q’ is 1, 2, 3, 4

So graphing of ‘q’ on number line is

Study Exponentiation

In math, exponentiation is the operation, which is written as the form of an. Where a and n is said to be base and exponent and n is any positive integer. Normally, exponentiation means that repeated multiplication. Otherwise, exponentiation an is the product of n factors of a. The exponent is usually placed as a superscript to right of base value. We are having many properties for exponentiation. Let see properties and example problems for exponentiation.


Properties - Study Exponentiation


We are having seven number of exponentiation properties that used for solving problems with exponentiation. In this properties, a, m and n are any integer values.

Product of like bases:

am an = am+n

Quotient of like bases:

`a^(m)/a^(n)` = am-n

Power to a power:

` (a^(m))^(n)` = amn

Product to a power:

(ab)m = am bm

Quotient to a power:

`(a/b)^(n)` = `a^(n)/b^(n)`

Zero exponent:

a0 = 1

Negative exponent:

a-n = `1/a^(n)` or `1/a^(-n)` = an

These are the properties that are used for exponentiation problems in study math.


Example Problems - Study Exponentiation


Example 1:

Solve 23 22.

Solution:

Given, 23 22.

This is in the form of am an, so we need to use am an = am+n property.

Here, m = 3 and n = 2 and a = 2.

Thus, 23 22 = 23+2

= 25

= 2 × 2 × 2 × 2 × 2

= 32
Hence, the answer is 23 22 = 32.

Example 2:

Shorten the following `5^5/5^3`.

Solution:

Given, `5^5/5^3` .

This is in the form of `a^m/a^n` , so we need to use `a^m/a^n` = am-n property.

Here, m = 5 and n = 3 and a = 5.

Thus, `5^5/5^3` = 55-3

= 52

= 5 × 5

= 25
Hence, the answer is `5^5/5^3` = 25.

That’s all about the study exponentiation.

Solving Change of Base Formula

The solving change of base formula is known as formulas which it permits us to rework a logarithm by means of the logs that may be is written with different base.

The change of base formula is given by,

Log a x = log b x / log b a

Here, assume that a, b and x are positive where a≠1 and b≠1.

Importance in solving change of base formula:

• Using the change of base formula we can change any base to another base. The most commonly used bases are base 10 and base e.

Log a x = log b x / log b a

• The solving change of base formula is used highly if calculators to assess a log to several base further than 10 or e.
• At the solving change of base formula having the value of x which is superior than zero.
• The log of a number to a given base is the power or an exponent to which the base must be raised in order to produce that number.

Advantages  in using change of base formula:

• Change the numeral bases, like convert  from base 2 to base 10m which is known as base conversion.
• The logarithmic change-of-base formula is applicable regularly in algebra and calculus.
• It is used for varying among the polynomial and normal bases.

solving Examples using change of base formula:


1) Solve  log 816

Solution:

By solving the change of base formula

=> log a x = log b x / log b a

log 8 16 = log 2 16 / log 2 8

=  4 / 3

2) Solve log 918

Solution:

By solving the change of base formula

=> log a x= log b x / log b a

log 9 18 = log 2 18 / log 2 9

= 4.169 / 3.169

= 1.315

3) Solve log 2 5.

Solution:

By solving the change of base formula

=> log a x= log b x / log b

log 2 5 = log 10 5 / log 102

The approximate value of the above expression is solving by,

=0 .6989 / 0.30103

=0 .3494

Practice problems in solving change of base formula :

1) Solve log 3 9 using the change of base formula.

Answer: 2

2) Solve log 10 8 using the change of base formula

Answer: 0 .9030

learning probability and odds

The probability of an incident is a proportion that tells how possible. It is that an event will take place. The numerator is the number of favorable outcomes and the

Denominator is the number of possible outcomes.

Probability:   number of flattering outcomes / number of possible outcomes

Probability is we can try to measure the chances of it occurrence

Explanation of learning probability and odds


Here learning the probability and odds

Numerical measure of the likelihood of an event to occur. If an inspection there are n possible  ways  exhaustive and mutually exclusive and out of them in m ways in the event. A occurs, then the probability of occurrence of the event. A is given by P (a) = m/n

If  in a random  sequence  of n trials of an event, M are favorable  to the event , the  probability of that event occurring is the limit of the ratio M/n, when  n is very  large , this lies between 0 and 1

P (a) = 0 means that the event can not take place.

P (a) = 1 means the event is bound to occur.

The events, A2, A3……..An are said to the mutually exclusive if ,

P(Ai∩ Aj)=0

for i=1,2,3,……n.

J=1,2,3,…….n, I  j

He events are said to be exhaustive if

P(A1)+ P(A2)+ P(A3)_.......... P(An)=1 if ,

A ∩   B  0 then

P(Aυ B)=P(A)+ P(B)

And if  A∩  B   0v then

P(A υ B)=P(A)+ P(B)-P(A ∩ B)


Learning Example of probability and odds:


Learning the probability and odds problems

1. when you toss a coin, it can fall two ways. The probability of getting a head on one roll of a coin is one chance out of two.

Solution:

P(h) means the   probability of getting a head on one toss of a  coin.

P(t) means the   probability of getting a tail on one toss of a  coin.

Step 1:  the number of favorable outcomes for head =1

Step 2:  the number of favorable outcomes for tail =1

Step 3:  number of possible outcomes    =2

Step 4:  the probability of the event for head = number of favorable outcomes / number of possible outcomes

Step 5:  the probability of the event for head= P (h) =n (e)/n(s)

Step 6:   the probability of the event for head=1/2

Answer:  1/2

Factor Tree Online

Before learning factor tree online, we'll learn important terms related with factor tree. Those terms are given below:

Factor: A factor of a number is the exact divisor of that number. In other words, a factor of the number divides the number leaving remainder 0.

We know that 15 = 1 x 15 and 15 = 3 x 5. This shows that 1, 3, 5 and 15 exactly divides 15. Therefore, 1, 3, 5 and 15 are all factors of 15.

Prime Number: Each of the numbers which has exactly two factors, namely, 1 and itself, is called a Prime Number. For example: 2, 3, 5,11, 13 etc.

Prime Factor: A factor of a given number is called a prime factor if this factor is a prime number. A prime factor cannot be factored further.Example: 2 and 3 are prime factors of 12 while 4, 6 and 12 are not.

Prime factorization : To express a given number as a product of prime factors is called Prime Factorization or complete factorization of the given number.

Example: 36 = 2 x 2 x 3 x 3, 45 = 3 x 3 x 5 etc.

Factor Tree: A diagram that shows the prime factors of a number is known as Factor tree. A factor tree contains all the prime factors of a given number.


Process to make a factor tree online


Start with the number.

• Make branches of factors - numbers that multiply to give you the original number
• Keep reducing each factor to its lowest possible prime factors.
• When we have all the prime factors identified, re-arrange them from least to greatest.
• Re-write them as exponents.

Illustration: Lets see the factor tree online preparation of 36 and 180 in the figure below:

1) Factor tree of 36:                                                                       
2. Factor tree of 180




Uses Of Factor Tree online


1. Factor Tree online is used to find the GCF or HCF

Example: Find the greatest common factor of 18 and 54.using factor tree

18
3 x 6
3 x (2x3)
The prime factors of 18 are 2, 3, and 3.

54
6 x 9
(2x3) x (3x3)
The prime factors of 54 are 2, 3, 3, and 3.

The prime factors that 18 and 54 have in common are 2, 3 and 3, so the greatest common factor is 2 x 3 x 3 = 18.

2. Factor Tree is used to find the LCM or LCD

Example: Find the LCM of 12 and 36 using factor tree

Step 1: Find the Prime factors for

12 = 2 x 2 x 3
36 = 2 x 2 x 3 x 3

Step 2: Multiply each prime factor the greatest number of times it appears in any one factorization.
The most number of times 2 appears in either factorization is twice. The most number of times 3 appears in either factorization is twice.

Thus, LCM = 2 x 2 x 3 x 3 = 36

Learn Linear Shape Functions

To learn the graphs of linear shape functions be used to resolve problems during restricted element analysis.In Lagrange functions Hermite cubic polynomials each function is unity at its individual node and zero at the further nodes. Moreover, the Lagrange shape functions summation to unity everywhere. For the Hermite polynomials H1 and H3 sum to unity.The shape functions be use to find the field variable U since recognized values at extra locations. The formula is U  = N1 U1 + N2 U2 + ...,Wherever Ni be the shape functions and Ui are identified values.

Sample problem for learn linear shape functions


The Hermite shape functions are used for beam analysis wherever together the deflection and slope of adjacent elements be required to be the similar at every node. H1 and H2 are the deflection while H2 and H4 are the slope.

D = H1 D1 + H2 S1 + H3 D2 + H4 S2
Wherever D be the deflection and Si be the slope.

In the formulas below,

L be the length
S be X - X1
r be S / L





Example problems for learn linear shape functions


Example 1:

Solving the domain of  learn linear shape function f
f (x) = sqrt (5x - 25)

Solution:
The expression function f includes a square root. The expression below the radical have to assure the condition
5x - 25 >= 0    for the function to obtain real values.
solve the linear inequality
x >= 25/5
x=5
The domain, within interval notation, is  (5 , +infinity)

Example 2:

Solving the domain of  learn linear shape function f
f (x) = sqrt (9x - 36)

Solution:
The expression function f includes a square root. The expression below the radical have to assure the condition
9x - 36 >= 0    for the function to obtain real values.
solve the linear inequality
x >= 36/9
x=4
The domain, within interval notation, is  (4 , +infinity)

Example 3:

Solving the domain of  learn linear shape function f

f (x) = sqrt (-x +9)

Solution:
The expression function f include a square root. The expression below the radical have to assure the condition
-x +9 >= 0    for the function to obtain real values.
solve the linear inequality
x >= 9/-1
x=-9
The domain, within interval notation, is (-infinity, 8).

Study about Unit Circle

In Euclidean geometry the simple shape is called circle and it consisting points in a plane which is middle from a given point called the center. Radius is the distance of the points from the center.

Unit circle:

In geometry, a unit circle is a circle with a radius of one. The unit circle is the radius one centered at the origin (0, 0) in the Cartesian coordinate system in the Euclidean plane except the trigonometry. The higher dimensions is the unit sphere is denoted by s1.

If (x,y ) is the first quadrant of the unit circle, then p and q are the lengths of a right triangle whose hypotenuse has length 1. That is, from the Pythagorean Theorem satisfy the following equation with x and y,

x2 + y2 = 1.

Explanation:


1. Length of circumference:

The circumference’s length is related to the radius (r) by

c = 2`pi` r

diameter d  = 2r

r = 2 / d

c = `pi`d

2. Area enclosed the circle:

Area of the circle =  `pi ` × area of the shaded square

The area of the circle is π multiplied with the radius squared:

A =` pi` r2

Area of the circle interms of diameter

A  = ` pi`(d/2)2

Area =` pi` d2 / 4


Trigonometric functions on the unit circle:




In geometrical term the trigonometric functions of the angle θ can be modified to a unit circle centered at O.

If  a point of the unit circle is (x, y) , and if the origin (0, 0) to (x, y) makes an angle t from the x-axis, in the trigonometric functions like sine, cosine terms.

cos(t) =x

sin(t) = y

The equation x2 + y2 = 1 gives the relation

cos2(t) + sin2(t) = 1

Solution:

Finally from properties of circle we conclude that the circle does not have any number of sides. That is circle have 0 sides.

x and y intercept solver

Introduction to x and y intercept solver:

X-intercept:
The x-intercept is the x-value of the point where the graph of a function or relation intersect the x-axis of the coordinate system.

Y-intercept:
The y-intercept is the y-value of the point where the graph of a function or relation intersect  the y-axis of the coordinate system. (Source: Wikipedia)
In this article we are going to see the problems on  x and y intercept by using the x and y intercept solver.

Methods to find x and y intercept using solver:

X-intercept solver:

• We can find the x intercept of the line by substitute y = 0 in the line equation.
• The x intercept can be represented as (x,0)

Y-intercept solver:

• We can find the y intercept of the line by substitute x = 0 in the line equation.
• The y intercept can be represented as (y,0)

Problems on x and y intercept using intercept solver :

Problem 1:
Find the x intercept of the line 5x-3y+15 = 0

Solution:
Given, the line 5x-3y+15 = 0
To find the x intercept we need to substitute y = 0 in the equation of line.
5x-3y+15 = 0
Substitute y = 0 in the above line equation,
5x – 3(0) + 15 =0
5x – 0 + 15 = 0
5x + 15 = 0
Subtract 15 on both sides,
5x - 15 - 15 = 0 - 15
5 x = - 15
Divide by 5 on both sides,
x = - 3
The x intercept of the line is (-3,0)

Problem 2:
Find the x-intercept of the line y = x -12

Solution:

Given, the line y =x -12
To find the x intercept we need to substitute y = 0 in the equation of line.
y = x -12
Substitute y = 0 in the above line equation,
y =x -12
0 = x -12
Add 12 on both sides,
12 = x -12 + 12
12 = x
x = 12
The x intercept of the line is ( 12 , 0)

Problem 3:
Find the y-intercept of the line 3y =6x -15

Solution:
Given, the line 3y =6x -15
To find the y intercept we need to substitute x = 0 in the equation of line.
3y =6x -15
Substitute x = 0 in the above line equation,
3y =6(0) -15
3y = - 15
Divide by 3 on both sides,
y = -15/3
y =-5
The y intercept of the line is ( 0 , -5)

Problem 4:
Find the y-intercept of the line y = 2x - 10

Solution:
Given, the line y = 2x - 10
To find the y intercept we need to substitute x = 0 in the equation of line.
y = 2x - 10
Substitute x = 0 in the above line equation,
y =2(0) -10
y = - 10
The y intercept of the line is ( 0 , -10)

How to find area

Students can learn How to Find Area of a Shape. They can learn to solve problems assocaited to it. Geometry deals with the study of 2 dimensional shapes and 3 dimensional solids and their properties like area, volume, length, surface area etc,.

The amount of surface occupied by the plane figure is called its Area.  The SI unit of area is square meter (m2). Area of regular or irregular shape can be obtained by using basic formulas defined for basic structures like square, triangle, rectangle, circle etc.

Here are some formulas which aid students to learn how to find the area of a shape.

Area of Square = a^2     

Area of Rectangle = ab       

Area of Parallelogram = bh         

Area of Trapezoid = h/2 (b1 + b2)   

Area of Circle = `pi`r 2          

Area of Triangle  = `(1)/(2) b.h`     

In the following example students can learn how to find area of a shape. Students can follow similar steps to compute the area of various shapes.

Let us take l=10 meters and b=6 meters

For ex:  This example shows how to find the area of a shape. This is the compuation of the area of a Rectangle. The area of a rectangle is the product of its length and width. The formula to be used is A = l * w, where l = length and w = width of the rectangle



Substituting the values of the dimensions in the formula and necessary computation is to be done to obtain the final value. For rectangle, it can be calculate as follows:

A= l * b = 10 * 6 = 60

The area of the rectangle is 60 square meters


Finding area


Here are some problems on how to find area of a shape.

Let us consider another sample problems on how to find area of a shape to have a better understanding.

For example, how to find area of a shape, the shape being a Circle in this example.



The area of the circle is A= π * r 2                  , where r is the radius of the circle.

Here, the value of the constant is π=3.14

Ex:  How to find area of a circle with radius 4 m.

Sol:  Radius = 4 m.

substituting the values in the formula  A = Π r2

A = 3.14 *42

= 50.24

The area of the circle is calculated to be 50.24 square meters.

Pro 1:   How to Find area of a circle whose radius is 5meters.

Ans: 78.5 square meters

Pro 2:  How to Find area of a rectangle whose dimesions are 3meters and 9meters.

Ans: 27square meters

Pro3 :  The side of a cube is 5meters.Find its total surface area.

Ans:150 square meters

Pro 4: The radius of a sphere is 4cm.Find its area.

Ans: 200.96 square centimeters

Find area of irregular shape


First the irregular figure is to be split into known regular figures. We already know how to find the area of a shape i.e. regular shapes. Thus, the area of the irregular figure can be calculated. Students can learn about how to find area of a shape which is irregular from the following examples:

Ex: Find the area of the following irregular figure



Sol: To find its area it is to be divided into known figures. The above given irregular figure can be divided into two known figures. That can be done a shown in the above figure.

Step 1: In the figure, the dotted line splits the irregular figure into two known figures like the rectangle with length (L) 14mm and width (W) 10mm and semi- circle with diameter (D) 14mm. Now the area of the irregular figure can be calculated as follows:

Step 2; Area of irregular figure= (Area of rectangle) + (Area of semi-circle)

= (L*W) + (π * D *D*0.25)

Step 2:                                        = (14 *10) + (3.14 *14*14*0.25)

Step 3:                                        = 140 + 153.86

Step 4:                                         =293.86 square mm

Thus the area of an irregular figure can be calculated.


Problems on Irregular figures:

Find the Area of the following irregular figures


Answers:      1. 51 square cm

2. 60 square cm

3. 27 square inches

Students can learn how to find area of a shape from the above examples and practice more problems on similar lines.

learning diameter

In geometry, the diameter is a  circle is any straight line segment that passes through the center of the circle and whose endpoints are on the circle. The diameters are the longest chords of the circle. In more modern usage, the length of a diameter is also called the diameter. In this sense one speaks of the diameter rather than a diameter, because all diameters of a circle have the same length, this being twice the radius.


The diameter Formula;


The diameter of circle:

Diameter of circle = 2 * r.

where, r is radius of the circle.


Problems to learning diameter


Example in diameter learning 1:

Find the area, diameter and circumference of circle and with the radius of 4cm.

Solution:

1.The area of circle:

Area of circle = ∏ *r*r.

=(3.14)*4*4.

= 50.24 cm2.

2.The diameter of circle:

Diameter of circle = 2*r.

= 2* 4.

= 8cm.

3.The circumference of circle:

Circumference of circle = ∏ * d.

= (3.14)* 8.

= 25.12cm.


Example in diameter learning 2:

Find the diameter of a circle if the radius is 35centimeter.

Solution:

Given radius = 35

Diameter =?

Diameter = radius * 2

= 35 * 2

= 70

So the diameter of the given circle is 70 centimeter.

Example in diameter learning 3:

Find the diameter of a circle if the radius is 44 centimeter.

Solution:
Given radius = 44

Diameter =?

Diameter = radius * 2

= 44 * 2

= 88

So the diameter of the given circle is 88 centimeter.

Example  in diameter learning 5:

Find the diameter of a circle if the radius is 25 centimeter.

Solution:

Given radius = 25

Diameter =?

Diameter = radius * 2

= 25 * 2

= 50

So the diameter of the given circle is 46 centimeter.

Example in diameter learning 6:

Find the diameter of a circle if the radius is 70 centimeter.

Solution:

Given radius = 70

Diameter =?

Diameter = radius * 2

= 70 * 2

= 140

So the diameter of the given circle is 140 centimeter.

Example  Circumference to Diameter learning 7:

Find the diameter of a circle if a radius is 10 centimeter.

Solution:

Given radius = 10

Diameter =?

Diameter = radius * 2

= 10 * 2

= 20

So the diameter of the given circle is 8 centimeter.