Decimals least to Greatest

We have learnt the fractions in the lower class mathematics. Now we shall study about special fractions whose denominators are 10, 100, 1000 etc. These fractions are called decimal fractions. Decimals least to greatest is also called as decimals in ascending order (Lower value to higher value)

Let us these fractions in a new way. That is

1/10 is understand writing one-tenth and is correspond to in decimal fraction as 0.1

1/100 is understand writing as one-hundredth and is correspond to in decimal fraction as 0.01

1/1000 is understand writing as one-thousandth and is correspond to in decimal fraction as 0.001

Steps for Calculating Decimals least to Greatest:

Step 1: Look the whole number first, and then look at the tenth, then the hundredth place and so on.

Step 2: Find out which number is smallest

Step 3: Arrange the decimals from least to greatest

For example, arrange the following numbers from least to greatest

0.56, 0.102, 0. 272, 0.5

Ans:  0.102, 0.272, 0.5, 0.56

Examples on Decimals least to Greastest:

Let us see some examples of decimal least to greatest:

Ex 1:  Find order of the decimals from least to greatest

5.6, 2.5, 1.2, 6.5, 0.6 and 0.2

Sol :   Step 1: First look at the tenth, and then the hundredth place so on. Next find the compare the two decimal numbers.

6.5 > 5.6

5.6 > 2.5

2.5 > 1.2

1.2 > 0.6

0.6 > 0.2

Step 2:  Arrange the decimals from least to greatest.

0.2, 0.6, 1.2, 2.5, 5.6 and 6.5

The above values are the arranged values from least to greatest.

Ex  2:   Find order of the decimals from least to greatest

5.7, 57, 0.71, 0.26

Sol :  Step 1: First look at the tenth, and then the hundredth place so on. Next find the compare the two decimal numbers.

56 > 5.7

5.7 > 0.71

0.71 > 0.26

Step 2:  Arrange the decimals from least to greatest.

0.26, 0.71, 5.7, 57

The above values are the arranged values from least to greatest.

Value Weighted Average

Arithmetic average is an average in which is defined as the summation of all the given elements divided by the total number of elements. In this all the values get same weighted.

Ex: Find the arithmetic average of 7, 16, 20, 57 and 60.


Arithmetic average = `("sum of all the elements")/("total number of elements")`

sum of all the given elements = 7 + 16 + 20 + 57+ 60 = 160

total number of elements = 5

Arithmetic average = `(160)/(5)`  = 32

so  arithmetic average = 32

In the above example we weighted to all the values.

Weighted average is same as that of arithmetic average but the only difference is that each of the element have assigned different weighted in the data given. The above example changes like that

Ex: Find the weighted  average of 7, 16, 20, 57 and 60, whose occurrences (weighted) values are 4, 3, 6, 2 and 1.

Solution is given after definition and formula

Definition of Weighted Average Value

Weighted average is defined as the summation of the element values multiplied with the occurrences (allocated weighted) which is divided by the summation of the occurrences values. This is called Value Weighted Average.

Formula for the weighted average is given by,

where

`barx_(w) ->`weighted average value

`w_(i) ->`allocated weighted value(occurrences) for the given element.

`x_(i) ->`element value

Steps to calculate weighted average value:

Step 1: Calculate the summation of the element values multiplied with the occurrences (allocated weighted).

Step 2: Calculate the summation of the occurrences values.

Step 3: Now division the value obtain by step 1 with step2 and get weighted average value.

Examples on Weighted Average Value

Ex 1: Find the weighted average of 7, 16, 20, 57 and 60 whose occurrences (weighted) values are 4, 3, 6, 2 and 1.

Sol :    Formula for the weighted average is

Step 1:     `sum_(n=1)^5(w_(i)x_(i))` =  (4) (7) + (3) (16) + (6) (20) + (2) (57) + (1) (60)

= 370

Step 2:         `sum_(n=1)^5 (x_(i))`   =  4 + 3 + 6 + 2 + 1

= 16

Step 3:     weighted average value = `barx_(w)` =   `(370)/(16)`

= 23.125                    Ans

Ex 2: Find the weighted average of 7, 2, 8 and 10, whose occurrences (weighted) values are 3, 2, 1 and 1.

Sol : Formula for the weighted average is

Step 1:   `sum_(n=1)^4(w_(i)x_(i))`  (3) (7) + (2) (2) + (1) (8) + (1) (10)

= 43

Step 2:      `sum_(n=1)^4(w_(i))`  3 + 2 + 1 + 1

= 7

Step 3:       weighted average value = `barx_(w)`   =  `(43)/(7)`

Ans =   6.142

Volume Divided by Area

Volume is how much three-dimensional space a substance (solid, liquid, gas, or plasma) or shape occupies or contains, often quantified numerically using the SI derived unit, the cubic meter.

Area is a quantity expressing the two-dimensional size of a defined part of a surface, typically a region bounded by a closed curve. The surface area of a 3-dimensional solid is the total area of the exposed surface. (Source: From Wikipedia).

Use of Volume Divided by Area:

The term volume divided by area is used to find the a dimension of a solid.

For example,

In a sphere, volume divided by area gives the radius of the sphere.

Volume of sphere = `4/3 pi r^3`

Surface area of sphere = `4 pi r^2`

Volume divided by area = `(4/3 pi r^3)/(4 pi r^2)`

= `r/3`

In a cube, volume divided by area gives the side of the cube,

Volume of cube = a^3

Surface area of cube = 6a^2

Volume divide by area = `(a^3)/(6a^2)`

= `a/6`

In a cylinder, volume divided by area gives the radius of the cylinder

Volume of cylinder = `pi r^2 h`

Surface area of cylinder = `2 pi r h`

Volume divided by area = `(pi r^2 h)/(2 pi r h)`

= `r/2`

Example Problems for Volume Divided by Area:

Here, we are going to see some example problems to find the dimensions of solid shapes using volume divided by area.

Example 1

Find the side of a cube whose volume is 125 cubic meter and the surface area is 150 square meter.

Solution

For a cube, Volume divide by area = `a/6` = `125/150`

`a/6` = `5/6`

a = 6

So, the side of the given cube is 5 meters.

Example 2

Find the radius of a sphere whose volume is 2393.88 cubic centimeter, and the surface area is 865.26 square centimeter.

Solution

We know that, the volume divided by area in a sphere gives `r/3`

Here, volume of the sphere = 2393.88 cubic centimeter

Surface area of the sphere = 865.26 square centimeter

Volume divided by area; `r/3` = `(2393.88)/(865.26)`

r = `(2393.88)/(865.26)` * 3

r = 8.3

So, the radius of the given sphere is 8.3 centimeter.

Example 3

The volume and lateral surface area of a cylinder are 565.2 cubic feet and 188.4 square feet respectively. Find the radius of the cylinder.

Solution

We know that, the volume divided by area in a sphere gives `r/2`

Volume of the cylinder = 565.2 cubic feet

Lateral area of cylinder = 188.4 square feet

Volume divided by area; `r/2` = `(565.2)/(188.4)`

r = `(565.2)/(188.4)` * 2

r = 6

So, the radius of the cylinder is 6 feet.

Solving Trigonometric Examples

Answering the trigonometric examples is nothing but we are solving the trigonometric functions and trigonometric identities. Here we will take trigonometric functions and equations. Trigonometric examples contain trigonometric functions. In trigonometric model we will solve the Sin, Cos, Tan Identities. We can find the angles from these identities. We will solve some trigonometric examples.

Explanation for Solving Trigonometric Examples:

Ex 1:     Solve the following trigonometric equation Cos4A – Sinn 2A =0

Sol :      Cos 4A – sin2A =0

2Sin2 (2A) + Sin (2A) – 1 = 0

Here we can use quadratic formula to find A value. Let us take any variable equal to Sin 2A

Let us take y = Sin 2A

2y2 + y – 1 =0

2y2+2y – y – 1 = 0

2y(y + 1) – (y + 1) = 0

If we factor this we will get two values for y.

(y + 1)(2y – 1) = 0

Now y + 1 = 0 2y – 1 = 0

Now plug y = Sin2A

Sin 2A + 1 = 0                                      2Sin2A – 1 =0

Sin 2A = -1                                          2Sin 2A = 1

2A = Sin-1 (-1)                                       Sin 2A =

2A = 270                                             2A = Sin-1

2A = 30

A = 135                                                      A = 15

From this we will get two value for A.

Practice Problem for Solving Trigonometric Examples:

Ex 2:          Solve the assessment of the following trigonometric identity Sin 75 - Cos 15

Sol :           Sin 75 – Cos 15

Here we have to use sum and variation formula to find the value os Sin 75 - Cos 15

Sin (45 + 30) – Cos (45 - 30)

Sin (A + B) = Sin A Cos B + Cos A Sin B

Cos (A - B) = Cos A Cos B + Sin A Sin B

Here A = 45

B = 30

Sin (45 + 30) = Sin45.Cos30 + Cos45Sin30

= 0.7071 * 0.8660 + 0.7071 * (0.5)

= 0.6123 + 0.3536

Sin 75 = 0.9659

Cos (45 – 30) = Cos45Cos30 + Sin45Sin30

=0.7071 * 0.8660 + 0.7071 * (0.5)

=0.6123 + 0.3536

Cos 15 = 0.9659

Now plug the values in the equation is

Sin 75 – Cos15 = 0.9659 – 0.9659

Sin 75 – Cos15 = 0

Ex 3:               Solve for x Sin x = 0.5, Cos x = 0.8660

Sol :             (I)  Given    Sin x = 0.5

x = Sin-1 (0.5)

x = 30o

(II)    Cos x = 0.8660

x = Cos-1 (0.8660)

x = 300

So from this angle x =30o. Here we use the opposite trigonometric functions to find the value of x.

Solving Simple Linear Equations

Solving simple linear equation involves the process of the solving basic linear equations in simple method. Linear equations come under the category of linear algebra whereas linear algebra is defined as the process of calculating system of unknown variables with the help of known things. Simple linear equations have relations with the families of vectors called vector or linear spaces. The following are the simple linear equations examples for solving.

Simple Linear Equations Examples for Solving:

Example 1:

Solve the simple linear equation to find the variable value.

-2(h - 1) – 4h - 1 = 3(h + 2) – 4h

Solution:

Given expression is
-2(h - 1) – 4h - 1 = 3(h + 2) – 4h

Multiplying the integer terms
-2h + 2 – 4h - 1 = 3h + 6 – 4h

Grouping the above terms
-6h + 1 = -h + 6

Subtract 1 on both sides
-6h + 1 - 1 = -h + 6 -1

Group the above terms
-6h = -h + 5

Add h on both sides
-6h + h = h + 5 -h

Group the above terms
-5h = 5

Multiply -1/5 on both sides
h = - 5/5

h = - 1 is the solution.

Example 2:

Solve the simple linear equation to find the variable value.

-4(h + 3) = h + 9

Solution:

Given expression is
-4(h + 3) = h + 9

Multiplying the integer terms
-4h - 12 = h + 9

Add 12 on both sides
-4h - 12 + 12 = h + 9 + 12

Grouping the above terms
-4h = h + 21

Subtract h on both sides
-4h - h = h + 21 -h

Grouping the above perms
-5h = 21

Multiply -1/5 on both sides
h = -21/5

h = -21/5 is the solution.

Simple Linear Equations Practice Problems for Solving:

1) Solve the simple linear equation to find the variable value.

-3(h - 2) – 2h - 3 = 2(h + 1) – 4h

Answer:  h = -1/3 is the solution.

2) Solve the simple linear equation to find the variable value.

-2(h + 3) = h - 1

Answer:   h = - 5/3 is the solution.

Explanation for Multiples and Factors of an Integers

Introduction to multiples and factors of an integers:
                        In math, the natural numbers are form the integer and another name of number is an integer. The factor is divisor of a given number. This divisor is divides the given integer without any remainder. The factors may be two or more in an integer. The multiple is a one quantity in multiplication. Now we are going to see about sum of factors of a number.

Explanation for Multiples and Factors of an Integers

Multiples of an integers:

                          If we multiply the one integer with another integer means that first integer is called as multiples. For example, x. y = x is multiple of y that is 2 x 3 = integer 2 is a multiple of integer 3.

Properties of multiple:

• The integer zero is common multiple of all integers.
• If we multiply any integer with 1 means that any integer is multiple of 1.

Factors of integers:

                             The factors are referred as divisors and each integer contains more than one factors. Types of factors are,

• Prime factors – It define the prime numbers (only two divisors).
• Composite factors – It define the composite number (two or more factors).

More about Multiples and Factors of an Integers

Example problems for multiples and factors of an integers:

Problem 1: Find out the multiples of given number.

8

Answer:

The given integer is 8.

The multiples of given integer is 8 x 1 = 8

                                                     8 x 2 = 16

                                                     8 x 3 = 24

                                                     8 x 4 = 32

Therefore, the multiples of an integer 8 are 8, 16, 24, 32…….

Problem 2: Find out the factors of given integer.

42

Answer:

The given integer is 42.

Factors of an integer 42 are 1, 2, 3, 6, 7, 14, 21, and 42.

Example problems for multiples and factors of an integers:

1. Determine the factors of an integer 20.

Answer: Factors are 1, 2, 4, 5, 10 and 20.

2. List the multiples of an integer 11.

Answer: The multiples of an integer are 11, 22, 33, 44….

Fraction least to Greatest

Introduction to fraction least to greatest:

   A fraction number value is one part of the whole number value in decimals.  A Fraction number value consisting of a two division in the number. The one part is top place of a number value is called as a numerator value. Another part is bottom place of the number value is called as a denominator value. That is numerator value of the fraction number divided by a denominator value of the fraction number. In this article we shall discuss about fraction least to greatest.

Sample Problem for Fraction least to Greatest:

This problem shows proper fraction for least to greatest:

Problem 1:

Find the value of given fraction numbers `(2)/(5)` + `(3)/(5)`

Solution:

   In the proper fraction a denominator values are same. So we directly add or subtract the numerator value.

Step 1: In the denominator values are same.

Step 2: Add the numerator values and put over the same denominator values.

       `(2)/(5)` + `(3)/(5)` = `(2 + 3)/(5)`

                 = `(5)/(5)`

Step 3: Now we simplify the fraction values.

               = 1

Problem 2:

Find the value of given fraction numbers `(2)/(9)` + `(7)/(9)`

Solution:

   In the proper fraction a denominator values are same. So we directly add or subtract the numerator value.

Step 1 Here  denominator values are same.

Step 2: Add the numerator values and put over the same denominator values.

      `(2)/(9)` + `(7)/(9)`   = `(2 + 7)/(9)`

                   = `(9)/(9)`  

Step 3: Now we simplify the fraction values.

               = 1

Improper fraction problem for least to greatest:

Problem 1:

Find the value of given fraction numbers `(11)/(7)` + `(13)/(5)`

Solution:

   In the improper fraction a denominator values are not same. So we don’t directly add or subtract the numerator values. So, we take least common multiplier for the denominator values.

Step 1: In the denominator values are not same. So, we take LCM for denominator

Step 2: multiply the numerator and denominator values for common multiplier.

The LCM value of the given fraction denominator value is 33

       `(11)/(7)` * `(5)/(5)` = `(55)/(35)`

        `(13)/(5)` * `(7)/(7)`   = `(91)/(35)`

 Step 3: Now we add the numerator values.

     `(11)/(7)` + `(13)/(5)` = `(11 + 13)/(35)`

                    = `(24)/(35)`

Practice Problem for Fraction least to Greatest:

Problem 1:

Find the value of given fraction numbers `(5)/(8)` + `(7)/(8)`

               Answer: `(4)/(3)`

Problem 2:

Find the value of given fraction numbers `(5)/(6)` - `(11)/(6)`

             Answer: -1