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

Algebra is widely used in day to day activities watch out for my forthcoming posts on sample question papers of cbse and secondary school education andhra pradesh. I am sure they will be helpful.

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.  

Algebra is widely used in day to day activities watch out for my forthcoming posts on Dividing Fraction and algebra course. I am sure they will be helpful.

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.

Fraction in Containers

Fractions are method of writing numbers that are not  whole numbers. Some fractions are written as  `1/2` ,`16/41` ,`3/2`. Fractions contains of two numbers, numerator and denominator. In the general form,

Numerator
Fraction = ----------------
Denominator

Where, numerator is at  top and denominator on the bottom.

For example,

In the fraction `2/3` , 2 is numerator and 3 is denominator. Now we see the problems for solving a fractions.


Conditions for fractions:

Some of the conditions are there for fraction addition and subtraction is given as,

To add the fractions the denominator must be same, then the numerators can is counted directly as same as numbers.

If the fractions the denominator is not same means, make them same number by multiplying and dividing by same scale factor.

Fraction subtraction is similar to the fraction addition. Let us see some of example problems using containers with same denominator and various denominators in detail and the example problems for containers fractions for better understand.

Example Problems for fraction in containers:

Example 1:

Find the total amount of water in containers; two containers are filled with one half of water and other containers have the capacity of eight liter in which four liter is filled.

Solution:

In first containers water filled is one half i.e `1/2.`

In second containers water filled is 4litre of total 8 liter capacity i.e `4/8`

The fractions are `1/2` and `4/8`

`4/8` can also be written as `1/2` (divide my 2 on both numerator and denominator.)

Total amount of water = `1/2 + 1/2`

= `(1+1)/2`

= `2/2`

= `1/1`

= 1

Example 2:

Find the total amount of water in containers; two containers are filled with one half of water and other containers have the capacity of eight liter in which six liter is filled.

Solution:

In first containers water filled is one half i.e `1/2.`

In second containers water filled is 4litre of total 8 liter capacity i.e `6/8`

The fractions are `1/2 ` and `6/8`

Total amount of water =` 1/2 + 6/8`

= `1/2 * 4/4 + 6/8`

= `4/8 + 6/8`

= `(4+ 6)/8`

= `10/8`

= `5/4`


Example 3:

Find the total amount of water in containers; two containers are filled with one half of water and other containers have the capacity of three liter in which one liter is filled.

Solution:

In first containers water filled is one half i.e `1/2.`

In second containers water filled is 4litre of total 8 liter capacity i.e `1/3`

The fractions are `1/2` and `1/3`

Total amount of water = `1/2 + 1/3`

= `1/2 * 3/3 + 1/3 * 2/2`

= `3/6 + 2/6`

= `5/6`

Sum of two Distributions

In sum of probability distributions and statistics, a probability will make the value of a random variable  or the probability of the value drawing  within a particular interval. The sum probability distribution defiines the measure of possible values like random variable can accomplish and the probability sum that the value of the random variable is within subset of that range.

Probability distributions:

Random Experiment and Sample Space

• Any experiment cannot be predicted in before, but is one of the set of possible outcomes, which is known as random experiment.
• An experiment performed frequently, every repetition is known as trial.
• The collection of all likely outcomes in random experiment is denoted the sample space.
• The sample space of a random experiment is denoted as a set S that includes all possible outcomes of the experiment
•  For simple experiments, the sample space may be exactly the collection  of possible outcomes.
• The sample space is a mathematically best set which contains the possible outcomes and perhaps other elements as well.
•  For example, if the testing is to draw a standard die and evidence the outcome, the sample space is S 1 2 3 4 5 6 , the set of possible outcomes.

Probability Distributions
      Let S acting as sample space with random experiment.
     Then X : x₁ x₂ x₃ …x_n
      P(X) is known as probability distribution of x.
Mean and variance of  Addition of two distributions:
Mean is E(X) = ∑ pi xi
variance is given below,
 

Examples:

1)Find the sum of mean of two distributions for the given below?

X	0	1	2
P(X)= x	0.25	.20	.25

X	0	1	2
P(X)= x	0.15	.20	.35

Solution:
Mean for table 1.
Mean E(X)1 = ∑ pi xi
            = 0*0.25 + 1*0.20+2*0.25
                       =0 + 0.20 +0.50  = 0.70
           Mean for table 2.
Mean E(X) 2 = ∑ pi xi
                    = 0*.15+1*0.20+2*0.35=0+.20+0.70
                 = 0.90
Sum of two distributions are E(X)1 + E(X)2= 0.70+0.90 = 1.6
2)Find out sum the variance of the following two distributions?

X	0	1	2
P(X)= x	0.25	.20	.25

X	0	1	2
P(X)= x	0.15	.20	.35

Congruent tTiangles Worksheet

In geometry, two figures are congruent if they have the same shape and size. More formally, 2 sets of points are called congruent if, and only if, one can be transformed into the other by an isometry, i.e., a combination of translations, rotations and reflections.

(Source-Wikipedia)

Congruent triangles worksheet -Example 1:

Let PQRS be a parallelogram and PR be one of its diagonals. What can you say about triangles PQRC andRSP?

Solution:
Step 1: In a four-sided figure, opposite sides are congruent. Hence sides

Step 2: QR and PS are congruent, and also sides PQ and RS are congruent.
Step 3: In a four-sided figure opposite angles are congruent.

Step 4: Hence angles PQR and RSP  are  four side figure.
Step 5: Two sides and an included angle of triangle PQR are congruent to two corresponding sides and an included angle in triangle RSP. According to the above postulate the two triangles PQR and RSP are congruent.

Congruent triangles worksheet-Example 2:

To solve the  triangle ABC congruent to triangle EFV by SAS?

Solution:
Step 1:Segment AB is congruent
Step 2: To segment of ST is
Step 3:  AB = EF = 4.
Step 4: Angle B is congruent to
Step 5: angle T because angle B = angle F = 100 degrees.
Step 6:Segment BC is congruent to segment FV because BC = FV = 5.
Step 7: ABC is congruent EFV is a SAS.



Congruent triangles worksheet-Example 3:
To solve the angle is QR is congruent to angle ST.

Solution:
Step 1: The triangle RPQ is congruent to triangle RTS.             
Step 2: Angle P is congruent to angle S because that information is given  in the diagram.
Step 3: Segment PR is congruent to segment RS because that information is given in the diagram.
Step 4: Angle QRP is congruent to angle SRT because vertical angles are congruent.
Step 5: Triangle SPQ is congruent to triangle RTS by Angle-Side-Angle.
Step 6: The triangles arecongruent, you know that all orresponding parts must be congruent.
Step 7:By CPCTC, segment BC is congruent to segment CE.

Unit Circle Sine Cosine

The Unit circle is used to understanding the sins and cos of angles to find 90 degree triangle. It is radius is exactly one. The center of circle is said to be origin and its perimeter comprises the set of all points that are exactly one unit from the center of the circle while placed in the plane.

It s just a circle with radius ‘one’.

The distance from the origin point(x,y) is `sqrt(X^2+Y^2)` by using Pythagorean Theorem.

Here radius is one So, The expression should becomes sqrt(x²+y²) =1

Take square on both sides then the eqn. becomes,

                      X²+y² =1

Positive angles are found using counterclockwise from the positive x axis

And negative angles are found using  anti clockwise from negative axis.

To find sine and cosine angle:

Angles are measures using counterclockwise from the positive x axis.The point where the ray at angle theta meets the unit circle is,
                                      (cos(`theta` ),Sin(`theta` )).
Here,cos(`theta` )----->Represents notation for the cosine of theta
Sine(`theta` ) is the notation of the sine of theta.
X= cos`(theta)`
Y=sin`(theta)`
The angle measured in radians is the length of the circular arc from the positive x
Axis counterclockwise until the desired angle is reached.
Since the radius of the unit circle is 1, the length of its perimeter is
2pr = 2p(1) = 2p.
Hence the angle for going exactly around the circle once is 2 radians. This
corresponds to 360 degrees.

Example problem -Unitcircle sines and cosine

Example 1: 

                 The angle q=0 radians is found by starting at the positive x-axis and going a distance 0. Hence the ray at angle q = 0 meets the unit circle at (1,0).

Hence
(cos(0), sin(0)) = (1,0),
cos(0) = 1,
sin (0)= 0.
Example 2. Without a calculator, find cos(p) and sin(p).
Solution:

The angle q= p radians is found by starting at the positive x-axis and going a distance p counterclockwise around the circle. But 2p is the total perimeter, so going a distance p arrives exactly half-way around the circle. Hence the ray at angle q = p meets the unit circle at (-1,0).

Hence
(cos(p), sin(p)) = (-1,0),
cos (p) = -1,
sin (p)= 0.

Polynomial Inequality

Inequality is a statement which means that the two things are not equal. Polynomials are the expressions which have the degree of at least 2 and more. An inequality that has polynomials is known as polynomial inequality. And polynomial inequality is not linear anymore. Solving polynomial inequality is little more complicated when comparing with linear inequalities. We will see how to solve polynomial inequality in the following sections.

Steps to solve polynomial inequality:

The following are the steps involved in solving polynomial inequality

Step 1: Make the inequality to obtain a zero on one side of the inequality

Step 2: The next step is to factor the polynomial which is on the left side of the inequality sign

Step 3: Determine the points at where the polynomial is zero.

Step 4: Graph the points we get in the previous step that is where the polynomial is zero

Step 5: Write down the answer

Example problems on solving polynomial inequality:

1)Solve the polynomial inequality x2 +15 < 8x

Solution:

Step 1: Make the inequality to obtain a zero on one side of the inequality

x2 +15 < 8x

x2-8x+15<0

Step 2: factor the polynomial x2-8x+15

x2-8x+15

x2-3x-5x+15

x(x-3) -5(x-3)

(x-5)(x-3)

Step 3: Determine the points at where the polynomial is zero.

Our polynomial is (x-5)(x-3)

When x= 5 ,

(x-5)(x-3) = (5-5)(5-3)

= 0(2)

=0

When x=3

(x-5)(x-3) = (3-5)(3-3)

= (-2)0

=0

The polynomial gets zero at x=5 and x= 3

Step 4: Graph the points we get in the previous step x=5, x=3 ( where the polynomial is zero )




Step 5: Write down the answer.

The solution of the given polynomial inequality is 3<x<5 .

2) Solve the polynomial inequality (x+2)(x+3) >0

Solution:

Step 1: Make the inequality to obtain a zero on one side of the inequality

Here in our example already there is a zero on the right side of the inequality .

Step 2: Factor the polynomial (x+2)(x+3)

No problem. It is in factored form

Step 3: Determine the points at where the polynomial (x+2)(x+3) is zero.

(x+2)(x+3)

When x= -2 ,

(x+2)(x+3) = (-2+2)(3+3)

=(0)(6)=0

When x= -3,

(x+2)(x+3) = (-2+3)(-3+3) = 0

Our polynomial get 0 at x=-2, x=-3

Step 4: Graph the points we get in the previous step x=-2, x=-3 ( where the polynomial is zero )

Algebra is widely used in day to day activities watch out for my forthcoming posts on Common Factor and solving systems of linear equations by elimination. I am sure they will be helpful.

Step 5: Write down the answer.

The solution for the given polynomial inequality is  -3<x<-2

learn expected value Binomial

The binomial distribution is discrete probability distribution. The digits of success in a sequence of n self-sufficient yes/no experiment, each one of which yields success among probability p. Such an achievement/failure experiment is called a Bernoulli experiment or Bernoulli trail. Into detail, when n = 1, the binomial distribution is a Bernoulli distribution. The binomial distribution is the root for the popular binomial test of statistical consequence.


Binomial Distribution:

A learn expected value Binomial is an algebraic expression contain two variables identified as x and y. direct multiplication gets quite deadly and can be slightly difficult for larger powers or more complicated expressions. The coefficients appeared during the binomial expansion are called Binomial Coefficient. They are the identical as their entries of Pascal’s triangle, and know how to be determined by a simple formula using factorials.

The formal expression of the Binomial Theorem is as follows:

`(a+b)^(n) = sum_(k=0)^n ([n],[k])a^(n-k) b^(k)`

It is often used to representation number of successes in a example of size n from a population of size N. Since the examples are not self-determining the resulting distribution is a hyper geometric distribution, not a binomial one. Though, for N much larger than n, the binomial distribution is a good quality approximation, and extensively used.

Binomial Expansion Formula

The binomial theorem formula is used to develop the binomials to some given power without direct multiplication.

n! = n (n − 1)(n − 2) ... (3)(2)(1)

Basic Binomial expansion:

(x + y)0 = 1

(x + y)1 = x + y

(x + y)2 = x2 + 2xy + y2

Examples for learn expected value binomial:

Example 1:

given the equation  (5+3y)2 how to learn expected value binomial?

Solution:

`sum_(k=0)^2([2],[k]) 5^(2-k)3y^(k)`


=  `([2],[0]) 5^(2)3y^(0) + ([2],[1]) 5^(1)3y^(1) +([2],[2]) 5^(0)3y^(2)`

the answer is = 25 + 15y + 3y2

Example 2:

given the equation  (5-y2)2 how to learn expected value binomial?

Solution:

`sum_(k=0)^2([2],[k]) (5x^(2-k))(-y^2)^k`


=  `([2],[0]) 5x^(2)(-y^2)^(0) + ([2],[1]) 5x^(1)(-y^2)^1 +([2],[2]) 5x^(0)(-y^2)^2`

the answer is = 5x2-5xy2+y2.


Example 3:

given the equation  (x-6y)2 how to learn expected value binomial?

Solution:

`sum_(k=0)^2([2],[k]) (x^(2-k))(6y)^k`


=  `([2],[0]) x^(2)(6y)^(0) + ([2],[1]) x^(1)(6y)^1 +([2],[2]) x^(0)(6y)^2`

the answer is = x2+6xy+6y2.

Mathematical Expression

In mathematical, an expression is a finite combination of symbols that are well-formed according to the rules applicable in the context at hand. In mathematical, symbols can designate values, variables, operations, relations, or can constitute punctuation.

         In mathematical, an expression may be used to designate a value. In mathematical, which value might depend on values assigned to variables occurring in the expression; the determination of this value depends on the semantics attached to the symbols of the expression.

Source: Wikipedia

Basic natural law for mathematical expression:

        In elementary algebra, we list the fundamental rules and properties of pre-algebra and give examples on they may be used natural law.

        Suppose that a, b and c are variables or mathematical expressions using natural law.

1. Commutative Property of Addition In mathematical.

           a + b = b + a

 2. Commutative Property of Multiplication In mathematical.

           a * b = b * a

3. Associative Property of Addition In mathematical.

       (a + b) + c = a + (b + c)

4. Associative Property of Multiplication In mathematical.

          (a * b) * c = a * (b * c)

5. Distributive Properties of Addition Over Multiplication In mathematical.

         a * (b + c) = a * b + a * c
    and
        (a + b) * c = a * c + b * c

More about In mathematical expression:

In algebra 1 expression is a finite group of algebraic terms and mathematical symbols combined with no equal or in equality sign.

Steps for simplify the mathematical expression:

            Step 1: Group the terms containing the same variable together in algebra expressions using natural law.

            Step 2: Perform the operation inside the parentheses for the variable and other.

            Step 3: Rewrite the expressions and simplifying the algebra expressions.

            Step 4: To check the equation, if there is able to simplify the expression, then repeat the step 1 to 4.

Example problems for simplify the mathematical expression using natural law:

        In mathematical expressions, find the value of x using natural law.

              3x(1x+4x²)

Solution:

            A(B+C)=AB+AC

            Step 1: is to determine what terms represent A,B and C in the given equation.

                        A represents 3x.

                        B represents 1x.

                        C represents 4x^x.

            Step 2: is to perform the multiplication operation.

                        AB=3x(1x)=3x²

                        AC=3x(4x²)=12x³

            Step 3: is to rewrite the problem.

                        3x(1x+4x)=3x²+12x³

            Step 4: is to simplify the answer.

                        3x²+12x³=0

                                 12x³=-3x²

                                                12x=-3

                                        x=`-1/4`

                 The answer is x=`-1/4`.

Trigonometry Values

If a rotation from the initial position to the terminal position is (`1/360` )^th of the revolution, the angle is said to have a measure of one degree and written as 1°. A degree is divided into minutes, and minute is divided into seconds.

The trigonometric values that exist are based on the angle which are formed of the rotation which represents it provides numeric nalues for the angles in the functions.

Tables for trigonometry values:

The following table in trigonometry shows that the trigonometric radians for their degrees.

The following table in trigonometry shows that the trigonometric functions which provide the trigonometric values for their degrees of their angles.

By the use of this table, the given trigonometric functions with their values  are also used for finding the other trigonometric functions like secant, cosecant and cotangent.

Example problems for trigonometry values:

1.Simplify :

(i) tan 735° (ii) cos 980° (iii) sin 2460° (iv) cos (−870°)
(v) sin (−780°) (vi) cot (−855°) (vii) cosec 2040° (viii)sec (− 1305°)

Solution:
(i) tan (735°) = tan (2 × 360° + 15°) = tan 15°

(ii) cos 980° = cos (2 × 360° + 260°) = cos 260°
= cos (270° - 10°) = − sin 10°

(iii) sin (2460°) = sin (6 × 360° + 300°) = sin (300°)

= sin (360° − 60°)
= − sin 60°
= −`sqrt(3)/2`

(iv) cos (− 870°) = cos (870°) = cos (2 × 360° + 150°)

= cos 150 = cos (180° - 30°)
= − cos 30° = −`sqrt(3)/2`

(v) sin (− 780°) = − sin 780°

= − sin (2 × 360° + 60°)
= − sin 60° = −`sqrt(3)/2`

(vi) cot (− 855°) = − cot (855°) = − cot (2 × 360° + 135°)

= − cot (135°) = − cot (180° - 45°)
= cot 45° = 1

(vii) cosec (2040°) = cosec (5 × 360° + 240°) = cosec (240°)

= cosec (180° + 60°) = − cosec (60°)
= − −`2/3`

(viii) sec (− 1305°) = sec (1305°) = sec (3 × 360° + 225°)

= sec (225°) = sec (270° − 45°)
= − cosec 45° = − 2

2.Simplify :

[  cot (90° − θ ) sin (180° + θ) sec (360° − θ) ] / [ tan (180° + θ) sec (− θ) cos (90° + θ) ]

Solution:

The given expression =[ tan θ (− sin θ) (sec θ) ] / [ tan θ (sec θ) (− sin θ) ]
                                = 1

Because the terms in that fraction gets easily cancelled.

Building Complex Systems

Any systems which involves a quantity of elements, set in structures which can subsist on a lot of scales is called the complex systems. These leave during processes of modify that are not describable by a solitary regulation nor are reducible to single stage of details, these levels often include features whose appearance cannot be predicted from their present specifications. Complex Systems speculation also includes the study of the connections of the a lot of parts of the systems.

Building complex systems:

            The complex number is specify the Z = a + ib
            The imaginary number is called as the tetragon root of a negative number. It is specify the `sqrt(-a)`  a > 0 is an imaginary number.
            The real numbers are a and b. The complex number is called as the a+ib. The real part is called a and imaginary part is called b.
Types of building complex system
            The first type is equality of complex numbers. The second type is calculation of two complex numbers. The third type is negative of a complex number. The fourth type is preservative identity of the complex number. The fifth type is preservative inverse of a complex number. The sixth type is creation of two complex numbers. The last type is multiplicative individuality of complex numbers.
Properties of building complex system
            There are a lot of propertied of the building complex system.  
            The first property is the commutative law for calculation.   
            The second property is Commutative Law for increase.
            The third property is Additive character Exists.
            The fourth property is Multiplicative character Exist.
            The fifth property is Reciprocals Exist for nonzero complex numbers.
            The sixth property is Negatives be real for every complex numbers.
            The last property is Non Zero invention Law.

Example problem of building complex systems:

Problem 1 :
`sqrt(-64)`
Solution
        `sqrt(-64)`
     =  `sqrt(64(-1))`
     =`sqrt(64)sqrt(-1)`
     = 8i
Problem 2:
          `-sqrt(-81)`
Solution
     = `-sqrt(-81)` 
     = `-sqrt(81(-1))`
     = `-sqrt(81)sqrt(-1)`
     =-9i

Rules of Radicals

In math, the radical is one of the number representations which indicate the square root or nth root. The symbol for denoting the radical is "√". For example, `sqrt(a)`, `root(n)(a)` are radicals. The term inside the symbol √ is called as radicand and the 'nth root' or 'square root ' is called as index value. Radical is considered as the opposite operation of exponent. For example, 4² is an exponent which is equal to 16. Now, the radical can be written as `sqrt (16)` =` sqrt (4xx4) ` = ±4. It is also possible to write the radical as an exponent by writing the reciprocal of nth root or square root as a power of radicand. Example: `sqrt(4)` = `4^(1/2)`. The radical has some basic rules for simplifying the problems. This article helps to give the detail explanation about the radical rules with some example problems for simplifying.

I like to share this Simplifying Ratios with you all through my article. 

Explanation to rules for radicals:

Rule 1:
          If the radical has a two radicands in multiplication form, then we can distribute common the index value to each radicand and then relate them by multiplication operation.
          For example, `root(n)(ab)` is the radical which has two radicands with a common index value 'n'. So, we can distribute the index value 'n' to each radicand 'a' and 'b' and then write it as `root(n)(a)` `root(n)(b)`.

  `root(n)(ab)`  =  `root(n)(a)` `root(n)(b)`  

Rules 2:
          If the radical has a two radicands in division form then we can distribute the index value to each radicand and then relate them by division operation.
          For example, `root(n)((a/b))` is the radical which has two radicands in division form with a common index value 'n'. So, we can distribute the index value 'n' to each radicand 'a' and 'b' and then write them as `root(n)(a)` /`root(n)(b)`.

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

Rules 3:
          If suppose the radicand has a power equals to the index value then it results the radicand without any power.
          For example, if we have radical, `root(n)(x^n)`  then we can also write it as `x^(n/n)`. Now, we can cancel the common term 'n'. So, we get the answer as 'x'.

  `root(n)(x^n)`    = x^(n/n) = x

Rule 4:
          If we have a radical `root(n)(-x)` then we can write it as -`root(n)(x)` where 'n' is odd.
By applying the above basic rules, we can easily simplifying the radicals.

Example Problem - Rules for simplifying radicals:

Example: 1

Simplify:`sqrt(50)`

Solution:
Given: `sqrt(50)`
For simplifying the given radical `sqrt(50)` , use the rules of radicals.
Step 1:

First find the factors of the radicand 50.

Factors of 50 are 25 and 2. (50 = 25 x 2)

Now, the radical can be written as `sqrt(50)` = `sqrt(25xx2)`

Step 2:

We need to apply the rule 1 for simplifying the radical.

The rule is, `root(n)(ab)` = `root(n)(a)` `root(n)(b)`

`sqrt(25xx2)` = `sqrt(25)` `sqrt(2)` (Here, we distribute the square root to both radicands 25 and 2)

= `sqrt(5xx5)` `sqrt(2)`

The square root of 25 is 5. Because the square value of 5 is 25.(5 x 5 = 25)

`sqrt(50)` = 5 `sqrt(2)`

Answer: 5`sqrt(2)`
Example: 2

Simplify: `root(n)(125/27)`

Solution:
Given: `root(3)(125/27)`
We need to apply the rules of radicals to solve this problem.

The radial rule is,

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

Now, the given radical `root(3)(125/27)` can be written as,

`root(3)(125/27)` =`(root(3)(125))/(root(3)(27))`

Step 2:

The cubic root of 125 is 5. Because the cubic values of 5 is 125 (5 x 5 x 5 = 125)

`root(3)(125)` = `root(3)(5xx5xx5)`

= 5

The cubic root of 27 is 3. Because the cubic values of 3 is 27(3 x 3 x 3 = 27)

`root(3)(27)` = `root(3)(3xx3xx3)`

= 3

Step 3:

Now, we get the answer as,

`root(3)(125/27)` =`(root(3)(125))/(root(3)(27))`= `5/3`

Answer: `5/3`
Example: 3

Simplify: `root(5)(25^(1/5))`

Solution:
Given: `root(5)(25^(1/5))` 
Step 1:

The rule of radical for simplifying the exponent is,

  `root(n)(x^n)`    = x^(n/n) = x

Step 2:

When we are applying the rule, we get the answer 25.

`root(5)(25^(1/5)`    = 25^(5/5) =25

Answer: 25
Example: 4

Simplify: `root(3)(-64)`

Solution:
Given: `root(3)(-64)`
Step 1:

The index value of the given radical `root(3)(-64)` is odd number 3.

So, we have to take the negative outside and then find the cubic root ot radicand 64.

`root(3)(-64)` = - `root(3)(64)`

= - 4

Answer: -4