Substraction of Fractions

A fraction is a number that can represent part of a whole. The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, one part of four, and so on. A much later development were the common or "vulgar" fractions which are still used today (½, ⅝, ¾, etc.) and which consist of a numerator and a denominator                                                                                                                                              
                                                                                                                                                      Source:-Wikipedia

Example problems for substraction fractions :

1. Substraction of  two fractions `14/8` and `10/8`

                Solution:

                               The two given fractions are` 14/8` and` 10/8`

                              We need to find  between two fractions

                                                 = `14/8` -`10/8`

                                                = `(14-10)/8`

                                                =  ` 4/8`

                             This can be reduced further as 0.5 

2. Substraction of  two fractions `20/9` and` 30/9`

                Solution:

                              The two given fractions are` 20/9` and `30/9`

                            We need to find  between two fractions

                                                     =   `20/9` -`30/9`

                                                     =  `(20-30)/9`

                                                   = `-10/9`

                         This can be reduced further as -1.11

 3. Substraction of  two fractions `44/5` and` 60/10`

             Solution:

                             The two given fractions are `44/10` and` 60/10`

                               We need to find between two fractions

                                                       = `(2*44)/10` -`60/10`

                                                     = ` (88-60)/10`

                                                  = ` 12/10`

                                This can be reduced further as -1.6

 4. Substraction of  two fractions `55/11` and `75/11` 

              Solution:

                             The two given fractions are` 55/11` and `75/11`

                             We need to find between two fractions

                                                   =   `55/11` -`75/11`

                                                  = `(55-75)/11`

                                                 = `-20/11`

                             This can be reduced further as -1.81

More Example problems for substraction fractions :

1. Substraction of  two fractions` 70/12` and` 90/12`
        Solution:
                         The two given fractions are `70/12` and `90/12`
                         We need to find  between two fractions
                                                =` 70/12` -`90/12`
                                              =`(70-90)/12`
                                              = `-20/12`
                       This can be reduced further as -1.66
 2. Substraction of  two fractions `30/13` and `60/13`
        Solution:
                       The two given fractions are `30/13` and `60/13`
                        We need to find  between two fractions
                                              =  `30/13` -`60/13`
                                              =` (30-60)/13`
                                              = `-30/13`
                       This can be reduced further as -2.30
 3. Substraction of  two fractions `100/14` and `120/14`
             Solution:
                            The two given fractions are `100/14` and` 120/14`
                            We need to find  between two fractions
                                                          = `100/14` -`120/14`
                                                         =  `(100-120)/14`
                                                         = `-20/14`
                            This can be reduced further as -1.42


 4. Substraction of  two fractions `120/15` and `140/15`
               Solution:
                                The two given fractions are `120/15` and `140/15`
                                We need to find between two fractions
                                                        =  `120/15` -`140/15`
                                                       = ` (120-140)/15`
                                                       = ` -20/15`
                             This can be reduced further as -1.33

Learning Trigonometric Identities Sum

Trigonometry is one of the most important and oldest topics in modern mathematics. The collection of formulas in the trigonometry is usually called as trigonometric function and identities. The trigonometric function and identities are developed to help in the measurement of triangles and their angles. These trigonometric functions and identities are basic tools for understanding many conceptual spaces. Online learning is one of the comfortable method of learning from anywhere around the globe. Through online study, students can learn about trigonometric identities sum. In this topic, we are going to see about, learning trigonometrical identities sum.


The list of trigonometric identities sum are shown below,



Sum or difference of two angles:

sin (a ± b ) = sin a cos b ± cos a sin b

cos(a ± b) = cos a cos b ± sin a sin b

tan(a ± b) = `(tan a +- tan b)/ (1 +- tan a tan b)`



Sum and product formulas:

sin a + sin b = `2sin((a+b)/2)cos((a-b)/2)`

sin a - sin b = `2cos((a+b)/2) sin((a-b)/2)`

cos a + cos b = `2cos((a+b)/2) cos((a-b)/2)`

cos a – cos b = `-2sin((a+b)/2) sin((a-b)/2)`

Learning trigonometrical identities sum: - Examples

Trigonometrical identities sum example 1:

Evaluate Sin 46

Solution:

Sin 46 = Sin (46+1)

= sin 45 cos 1 + cos 45 sin 1

= 0.719(0.999) + 0.707(0.017)

= 0.718 + 0.012

= 0.73

The answer is 0.73



Trigonometrical identities sum example 2:

Evaluate Cos 128

Solution:

Cos 128 = Cos (90 + 38)

= cos 90 cos 38 – sin 90 sin 38

= 0(0.788) – 1(0.615)

= 0 – 0.615

= -0.615

The answer is -0.615



Trigonometrical identities sum example 3:

Evaluate tan 50

Solution:

Tan 50 = Tan (45 + 5)

= `(tan 45 + tan 5)/(1-tan 45*tan 5)`

= `(1+0.087)/(1-(1*0.087))`

= `1.087/(1-0.087)`

= `1.087/ 0.913`

= 1.190

The answer is 1.190


Trigonometrical identities sum example 4

Evaluate Cos 92

Solution:

Cos 92 = cos (90+2)

= cos 90 cos 2 - sin 90 sin 2

= 0(0.999) - 1(0.034)

= 0 + 0.034

= 0.034

The answer is 0.034

Division of Fractions Help

Fractions:

A fraction is a number that can represent part of a whole. The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, one part of four, and so on. A much later development was the common or "vulgar" fractions which are still used today (½, ⅝, ¾, etc.) and which consist of a numerator and a denominator.(Source : Wikipedia)

In this article we are going to see about general rule for division of fractions help and some solved problems on division of fractions help and some practice problems on division of fractions.


This article will help you to learn division of fractions.

Division of fractions help:

Method to for division of fractions:

Take a reciprocal of the divisor fraction

Multiply the reciprocal of divisor fraction with the dividend fraction


Solved problems on division of fractions:

The following solved problems will help you to solve divisions of fractions.

Problem 1 :

Divide the fractions 14/18 ÷ 7/9

Solution:

Given, 14/18 ÷ 7/9

14/18 -> Dividend

7/9 -> Divisor

Take a reciprocal of the divisor,

Reciprocal of 7/9 = 9/7

Multiply the reciprocal of divisor by the dividend.

14/18 * 9/7 = (14 * 9) / ( 18 * 7)

= 126 / 126

= 1

Answer: 14/18 ÷ 7/9 = 1

Problem 2:

Divide the fractions 5/8 ÷ 15/16

Solution:

Given, 5/8 ÷ 15/16

5/8 -> Dividend

15/16 -> Divisor

Take a reciprocal of the divisor,

Reciprocal of 15/16 = 16/15

Multiply the reciprocal of divisor by the dividend.

5/8 * 16/15 = ( 5 * 16 ) / ( 8 * 15)

= 80 / 120

Divide by 40 on both numerator and denominator,

(80÷ 40)/ (120 ÷ 40) = 2 / 3

Answer: 5/8 ÷ 15/16 = 2 / 3

Problem 3:

Divide the fractions 256 / 144 ÷ 16/12

Solution:

Given, 256 / 144 ÷ 16/12

256 / 144 -> Dividend

16/12 -> Divisor

Take a reciprocal of the divisor,

Reciprocal of 16/12 = 12/16

Multiply the reciprocal of divisor by the dividend.

256 / 144  * 12/16 = ( 256 * 12 ) / ( 144 * 16)

=  3072 / 2304

Divide by 768 on both numerator and denominator,

(3072 ÷ 768) / (2304 ÷ 768) = 4 / 3

Answer: 256 / 144 ÷ 16/12 = 4/3

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Practice problems on division of fractions help:

Problems:

1.Divide the fractions 4/3 ÷ 8/9

2.Divide the fractions 7/11 ÷ 35/33

Answer:

1. 3/2

2. 3/5

Compound Fractions

Solving compound fractions is the important topic in algebra. Compound fractions are nothing but a mixed fraction. It is also known as the mixed number.  Compound fractions are the combination of whole number and the proper fraction. We have to write the improper fractions in the form of compound fractions. In this we have to discuss about the solving compound fractions problems.


General discussion about fractions

Definition of fraction:

Dividing two numbers is known as fractions. It takes the original form of a/b. Here we can called a as a numerator value and b as a denominator value.

Definition of proper fraction:

When the numerator value is lesser than the denominator value. Then the fraction is denoted as proper fraction.

Definition of Improper fraction:

When the numerator  is greater than the denominator value. Then the fraction is denoted as Improper fraction.

Definition of compound fraction:

We can write the improper fraction is the combination of the whole number and then the proper fraction. This fraction is known as compound fraction.

Solving compound fraction problems

Addition of compound fractions:

Example 1:

Solving 2 (1)/(2) + 3 (1)/(2)

Solution:

First we can convert the compound fractions into improper fraction. It can be converted in the following manner

(((2*2)+1))/(2)+ (((3*2)+1))/(2)

That is (5)/(2)+(7)/(2)

Here the denominators are same so we can directly add the numerators otherwise we can take the L.C.D and then add the numerator values.

Adding this we can get, (12)/(2)

Divided by 2 we can get 6

This is the solution of the given fraction.

Subtraction of compound fraction:

Subtraction of compound fractions is similar to addition instead of + we can use the sign -.

Multiplication of compound fractions:

Example 2:

Solving 1 (1)/(3) x 2 (5)/(4)

Solution:

First we can convert the compound fractions into improper fraction. It can be converted in the following manner

(((1*3)+1))/(3) x (((2*4)+5))/(4)

that is (4)/(3) x (13)/(4)

Now we can multiply the numerator and denominator terms separately. Then we can simplify the fraction.

then we can get (13)/(3)

This is the solution of the given fraction.


Division of compound fractions:

Here we can convert the mixed numbers into improper fraction and then take the reciprocal of the second fraction. Now we have to multiply the fractions.

Linear Combination of Random Variables

The random variables have the countable values for their probability in the statistics. The sum of the probability for the random variable is one, then the random variable is said to be discrete random variable. The random variable is the measurable function. Here the mean, variance and standard deviation for the random variables are calculated. This article contains the information about the linear combination of random variables in the probability theory and statistics.

Formula used for linear combination of random variables:

The formulas used to determine the mean, variance and the standard deviation in the random variable are

Mean = `sum x p(x)`

Variance = `sum p(x) (x^2) - (sum x p(x)) ^2`   

In this above mentioned formula the `(sum x p(x)) ^2` refers the squared value of the mean in the statistics.

Standard deviation = `sqrt (variance)`

In the above mentioned formula x refers the given set of discrete random values and p(x) refers the probability value for the random variables in the statistics.

Examples for linear combination of random variables:

Example 1 to linear combination of random variables:

Predict the mean, variance and standard deviation for the random variables.

x	20	40	60	80	90
P(x)	0.11	0.12	0.34	0.24	0.19

Solution:

Mean = `sum x p(x)`  

Mean = 20 (0.11) +40 (0.12) +60 (0.34) +80 (0.24) + 90(0.19)

Mean = 2.2+ 4.8 + 20.4 + 19.2 + 17.1                   

Mean = 63.7

Variance = `sum p(x) (x^2) - (sum x p(x)) ^2`   

Variance = ((0.11) (20) (20) + (0.12) (40) (40) + (0.34) (60) (60) + (0.24) (80) (80) + (0.19) (90) (90)) - (63.7)² 

Variance = (44+ 192+ 1224 + 1536+ 1539) -4057.69

Variance = 4535 -4057.69

Variance = 477.31

Standard deviation = `sqrt (variance)`

Standard deviation = `sqrt (477.31)`

Standard deviation = 21.847

The mean value is 63.7, variance is 477.31 and the standard deviation is 21.847 for the given random variables.  

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Example 2 to linear combination of random variables:         

Predict the mean, variance and standard deviation of random variables.  

  

x	10	20	30	40	50
P(x)	0.20	0.12	0.22	0.13	0.33

Solution:

Mean = `sum x p(x)`

Mean = 10(0.20) +20(0.12) + 30(0.22) +40(0.13) + 50(0.33)

Mean = 2.0 +2.4 +6.6+ 5.2+ 16.5

Mean = 32.7

Variance = `sum p(x) (x^2) - (sum x p(x)) ^2`   

Variance = ((0.20) (10) (10) + (0.12) (20) (20) + (0.22) (30) (30) + (0.13) (40) (40) + (0.33) (50) (50)) - (32.7)² 

Variance = (20+48+ 132 + 208 +825) -1069.29

Variance = 1233 -1069.29                                     

Variance = 163.71

Standard deviation = `sqrt (variance)`

Standard deviation = `sqrt (163.71)`

Standard deviation = 12.79

The mean is 32.7, variance is 163.71 and the standard deviation is 12.79 for the given random variables.

Make Ellipse Equation

Under the picture that relates the foci, eccentricity, and length of the main and negligible axes. Derive the identities if we remember that if T is a point on the ellipse the addition of the distances from T to the two foci is a steady.  Derive the equation for the ellipse with vertical main axis.  The derivative for the horizontal main axis case is similar.

Vertical Ellipse and Horizondal Ellipse

Making ellipse equation:

Consider the upside of the ellipse.  The addition of the distances from the two foci to the up side of the make a ellipse is

(b –c) + (b + c) = 2b

The square of is given by

4b2

Then consider the right hand side of the ellipse.

See that the distances from each side to the right hand side is the equal.  The square of sum is

(2d)2 = 4d2

Using the Pythagorean Theorem,

d2 = a2 + c2

Replacement and setting these same to each other gives

4b2 = 4(a2 + c2)

So that

b2 = a2 + c2

Place the equation in the particularly simple form

`(x^2)/(a^2) + (y^2)/(b^2) =1 (or) (x^2)/(b^2) + (y^2)/(a^2) =1`

Make ellipse equations - Examples:

Make ellipse equations - Example1:

Given the following equation

9x2 + 4y2 = 36

a) Solve the x and y intercepts of the graph of the equation.

b) Solve the co-ordinates of the foci.

c) Solve the length of the main and slight axes.

d) Make the graph of the equation.

Solution:

9x2 / 36 + 4y2 / 36 = 1

x2 / 4 + y2 / 9 = 1

x2 / 22 + y2 / 32 = 1

a = 3 and b = 2 (NOTE: a >b) .

Set y = 0 in the equation get and Solve the xaxes

x2 / 22  = 1

Solve for x.

x2  = 22

x = ± 2

Set x = 0 in the equation get and Solve the y axes

y2 / 32 = 1

Solve for y.

y2  = 32

y = ± 3

b) We need to solve c first.

c2 = a2 - b2

a and b were found in part a).

c2 = 32 - 22

c2 = 5

Solve for c.

c = +- (5)1/2

The foci are    F1 (0, (5)1/2) and  F2 (0 , -(5)1/2)



c) The main axis length is given by  2 a = 6.

The slight axis length is given by  2 b = 4.

d) The graph of the equation is


Example Ellipse

Equivalence Properties of Equality Tutor

Tutor is nothing but the teacher or instructor give notes to the students who comes through online. Tutor teaches a certain topic to student what they want by online. Here tutor give instruction about the equivalence property of equality to students by online. Equality represents that both side of the equation is of equal status or not. To represent this equality; we use the equivalence property of equality.

Equivalence Properties of Equality Tutor:

There are three properties under the equivalence of equality. These properties are applicable for the arithmetic operation.

Reflexive property:

Reflexive property is said to be any number that is equal to same number.

Example: a = a

Symmetric property:

Symmetric property is said to a number x is equal to the number y; then the number y is equal to the number x.

x = y then y = x

Transitive property:

Transitive Property is said to be a number an equal to the number b then the number b equal to the number c; then we can say the number an equal to the number c.

a = b; b = c and then a = c

Example Problems – Equivalence Properties of Equality Tutor:

Example 1:

Verify the following statement is transitive Property

M + 3 = 7; 7 = 4 + 3 and then 4 + 3 = 7

Solution:

Given: M + 3 = 7; 7 = 4 + 3 and then 4 + 3 = 7

Let take M + 3 = 7 ---- > equation 1

7 = 4 + 3 ---- > equation 2

4 + 3 = 7 ----- > equation 3

Step 1: Find the value of M in equation 1.

M + 3 = 7

Step 2: Subtract the number 3 on each side of the equation 1, we get

M + 3 – 3 = 7 – 3

M = 4

Thus 4 + 3 = 7 --- equation 4

That the equation 1 and equation 4 are equal,

4 + 3 = 4 + 3

Thus we proved the given statement is transitive property of equivalence.

Example 2:

Which of the following is the symmetric property?

a) m = n and n = m

b) m = 1/m and m= m

c) m + 1 = m - 1

d) m = 1 + m

Solution:

Symmetric property: x = y; y = x

Here the option ‘’a’’ refers the symmetric property.

m = n and n = m

Answer: a

These are the example problem istruct by the tutor about the equivalence properties of equality.