Solving Expanded Binomials

In binomial is the distinct probability of the number of success in a series of n experiments, all of that defer success by probability p. Such experimentation is also identified a Bernoulli experiment. Actually, when n = 1, the binomial is a Bernoulli distribution. The binomial is the base for the accepted binomial analysis of statistical importance.

solving expanded binomials:

It is often used to form number of success in an example of size n since a population of size N. as the example are not self-sufficient the resultant distribution is hyper arithmetical distribution, not a binomial one. But, for N a large amount of n, the binomial distribution is a good estimate, and widely used.

Definition of binomial: (source: Wikipedia)

The binomial a2 − b2 can be factored as the product of two other binomials:

a2 − b2 = (a + b)(a − b).

an+1-bn+1 = (a-b)`sum_(k=0)^n a^kb^(n-k)`

This is a special case of the more general formula:

The product of a pair of linear binomials (ax + b) and (cx + d) is:

(ax + b)(cx + d) = acx2 + adx + bcx + bd.

A binomial increase to the nth power stand for as (a + b)n  know how to be expanded through way of the binomial theorem consistently. Taking a simple instance, the ideal square binomial (p + q)2 know how to be establish by squaring the first term, calculation twice the product of the first also second terms and finally calculation the square of the second term, to give p2 + 2pq + q2.

A simple however interesting application of the refer to binomial formula is the (m,n)-formula for make:

for m < n, let a = n2 − m2, b = 2mn, c = n2 + m2, then a2 + b2 = c2.

Example for solving expand binomials:

Example 1:

Expanded (4+5y)2 to solving binomial the equation.

Solution:

Step 1: given the binomials are (4+5y)2

Step 2: `sum_(k=0)^2 ((2),(k))4^(2-k)(5y)^k`

Step 3: `((2),(0)) 4^2(5y)^0 + ((2),(1)) 4^1(5y)^1 +((2),(2)) 4^0(5y)^2`

Step 4:  16 + 20y + 25y2.

Example 2:

Expanded (6+3y)3 to solving the binomial equation.

Solution:

Step 1: given the binomials are (6+3y)3

Step 2: `sum_(k=0)^3 ((3),(k))6^(3-k)(3y)^k`

Step 3: `((3),(0)) 6^3(3y)^0 + ((3),(1)) 6^2(3y)^1 +((3),(2)) 6^1(3y)^2 + ((3),(3)) (3y)^3`

Step 4:  216 + 108y + 54y2+27y3.

My forthcoming post is on taylor series cos x and cbse sample papers for class 12th will give you more understanding about Algebra.

Example 3:

Expanded (2+3y)2 to solving binomial the equation.

Solution:

Step 1: given the binomials are (2+3y)2

Step 2: `sum_(k=0)^2 ((2),(k))2^(2-k)(3y)^k`

Step 3: `((2),(0)) 2^2(3y)^0 + ((2),(1)) 2^1(3y)^1 +((2),(2)) 2^0(3y)^2`

Step 4:  4 + 6y + 9y2.

Subtracting Fractions with Common Denominators

In this article we are going to discuss about Subtracting fractions with common denominator or  like denominators.

A fraction involves two numbers. The top number is said to be numerator and the  bottom number is said to be denominator.

Fraction    =  (Numerator of a fraction) / (Denominator of a fraction)

To subtract the fractions with the common denominator, first subtract the numerators and then put that difference over that common denominator.

How to Subtract fractions with like denominators

Below are the examples on Subtracting fractions with common denominators:

Example problem 1:

Simplify the given fraction, 5/3 – 7/3

Solution:

= 5/3  – 7/3

Here the two fractions have same denominator, we can easily add the variables in numerator without changing the values of denominator.

= (5-7)/3

=(-2)/3

So the result is -2/3

Example problem 2:

Simplify the given fraction, 10/7 – 9/7

Solution:

= 10/7 – 9/7

Here the two fractions have same denominator, we can easily add the variables in numerator without changing the values of denominator.

= (10-9)/7

=(1)/7

So the result is 1/7

Example problem 3:

Simplify the given fraction, 13/9  – 10/9

Solution:

= 13/9 – 10/9

Here the two fractions have same denominator, we can easily add the variables in numerator without changing the values of denominator.

= (13-10)/9

=(3)/9

Equivalent fraction is 1/3

So the result is 1/3 .

Example problem 4:

Simplify the given fraction, 7/9 – 3/9

Solution:

= 7/9 – 3/9

Here the two fractions have same denominator, we can easily add the variables in numerator without changing the values of denominator.

= (7- 3)/9

= (4)/9

Equivalent fraction is 4/9

So the result is 4/9 .

Example problem 5:

Simplify the given fraction, 8/7 – 3/7

Solution:

= 8/7 – 3/7

Here the two fractions have same denominator, we can easily add the variables in numerator without changing the values of denominator.

= (8-3)/7

= (5)/7

Equivalent fraction is 5/7

So the result is 5/7 .

I am planning to write more post on unseen passages for class 7 and unseen passages for class 8. Keep checking my blog.

Example problem 6:

Simplify the given fraction, 8/13 – 5/13

Solution:

= 8/13 – 5/13

Here the two fractions have same denominator, we can easily add the variables in numerator without changing the values of denominator.

= (8-5)/13

= (3)/13

Equivalent fraction is 3/13

So the result is 3/13 .


Practice problems

Here are some practice problems on Subtracting fractions with like denominator

Problem1:  9/13 –5/13

Answer: 4 /13

Problem2:  7/10  – 5/10

Answer: 1/5

Problem3:  11/12 – 5/12

Answer: 1/2

Problem4:  6/7 – 5/7

Answer: 1/7

Problem5:  2/5 – 1/5

Answer:1/5

Figure Measurements

Measurements are used to measure the length of a cloth for stitching, the area of a wall for white washing, the perimeter of a land for fencing and the volume of a container for filling. The measurements consist of lengths, angles, areas, perimeters and volumes of plane and solid figures. If a student wants to know about the measurements of figure, they can be referring the following examples.

Figure Measurements – Examples 1:

These are the examples for measurements of a figure.

figure measurements

Find the area of the triangle whose height is 10cm and base is 6cm.

Solution:

Given base = 6cm

height = 10cm

Area of triangle = `1 / 2 (base xx height)`

= `1 /2 (6 xx 10)`

= `1 / 2 (60)`

= 30

Therefore, the area of the triangle is 30cm2

Find the area of the triangle whose height is 12cm and base is 8cm.

Solution:

Given base = 8cm

height = 12cm

Area of triangle = `1 / 2 (base xx height)`

= `1 /2 (8 xx 12)`

= `1 / 2 (96)`

= 48

Therefore, the area of the triangle is 48cm2

Figure Measurements – Examples 2:

These are the examples for measurements of a figure.

Measurements of Figure 1:

Three angles of a triangle are x + 34˚, x + 40˚ and x + 46˚. We have to find x for the triangle.

Solution:

x + 34 + x +40 + x + 46 = 180˚

The sum of the three angles of a triangle is equal to 180˚

3x + 120 = 180˚

3x = 180˚ - 120˚

= 90˚

x = `60/3`

= 20˚

Measurements of Figure 2:

The triangle has a three angles  x + 20˚, x + 10˚ and x + 30˚. We have to find x for the triangle.

Solution:

x + 20 + x +10 + x + 30 = 180˚

The sum of the three angles of a triangle is equal to 180˚

3x + 60 = 180˚

3x = 180˚ - 60˚

= 120˚

x = `120/3`

= 40˚

Algebra is widely used in day to day activities watch out for my forthcoming posts on Multiplying Mixed Number Fractions and Strategies for Addition. I am sure they will be helpful.

Measurements of Figure 3:

The measurements of the angles whose triangle are in the ratio 2:1:3. Calculate the angles of the given triangle values.

Solution:

Ratio of the angles of a triangle = 2:1:3

Total ratio = 2 + 1 + 3

= 6

Sum of the three angles of a triangle is 180˚. Therefore,

First angle = `2/6 xx 180`

= 60˚

Second angle = `1/6 xx 180`

= 30˚

Third angle = `3/6 xx 180`

= 90˚

Math Symbols Terms

There are many terms and symbols used in math. The symbols can be used to perform some operations such as addition, multiplication, subtraction, division; relationship symbols such as less than, greater than, equal, not equal, etc. The math terms can be used to describe the topics such as algebra, arithmetic, array, and axis, etc. Let us see about math symbols and terms in this article.

I like to share this Symbol for Correlation with you all through my article. 

Some of the Math Symbols

Addition:
It can be represented as +. It is used for addition operation and also for logical OR purpose. It can be read as plus or logical or.
Subtraction:
It can be represented as –. It is used for subtraction operation. It can also spelled out as minus.
Multiplication:
Multiplication can be represented as ×. It is used for multiply the given terms.  It can also spelled out as into.
Division:
It can be represented as ÷. It is used for dividing the given terms.  It can also spelled out as divide by.
Percentage:
It can be represented as %. It is used for calculating the ratio that compares to the number 100.  It can also spelled out as percent.
Summation:
Summation can be represented as `sum`   . It is used for calculating the sum of many or infinite values.  It can also spelled out as sum.
Dot Product:
It can be represented as (.). It is used for calculating the scalar (dot) product of two vectors.  It can also spelled out as dot.
Cross Product:
It can be represented as (×). It is used for calculating the vector (cross) product of two vectors. It can also spelled out as cross.



Algebra is widely used in day to day activities watch out for my forthcoming posts on Fibonacci Number Sequence and number to words converter. I am sure they will be helpful.

Some of the Math Terms

Algebra:
An algebraic equation represents the scale.
Algorithm:
A step-by-step problem solving technique is used in math computations.
Area:
It can be used for representing the two-dimensional shapes such as polygon, octagon, hexagon, triangle and circle.
Array:
A set of numbers can be occurred in a specific size of pattern. Matrix or array consists of columns and rows.
Axis:
Axis consists of horizontal and vertical axis having co-ordinates in a plane.
Attribute:
Describing the object with data consists of shape, color, or size.
Arc:
A circumference of the circle or the portion of a segment draws with a compass.

Maxima and Minima

Maxima and Minima are the largest value (maximum) or smallest value (minimum), that a function can take at a point either within a given boundary (local) or on the whole domain of the function in its entirety (global). In general, maxima and minima of a given set are the greatest and least values in that set. Together, Maxima and Minima are called extrema (singular: extremum). We need to learn minima to determine the nature of the curve or the function and various other applications like projectiles, astrophysics to microphysics, geometry etc.

Learning analytical definition of minima:

A function f(x) is said to have a local minima point at the point x*, if there exists some ε > 0 such that f(x*) ≤ f(x) when |x − x*| < ε, in a given domain of x. The value of the function at this point is called minima of the function.

A function f(x) has a global (or absolute) minima point at x* if f(x*) ≤ f(x) for all x throughout the function domain.

We need to learn minima points of a curve by observing the involved function.

learning prologue regarding minima:

To learn Minima & Maxima, one needs to have a basic knowledge of calculus. The following points are some bare necessary (maybe not sufficient) definitions.

A function, y = f(x) is a mathematical relation such that each element of a given set ‘x’ (the domain of the function) is associated with an element of another set ‘y’ (the range of the function).

Closed interval of a domain is defined as an interval that includes its endpoints, as opposed to open interval which is an interval that does not include its endpoints.

A function, f(x) is said to be continuous at a given interval if it can assume all values within the interval i.e. the function is not broken anywhere inside the interval. Mathematically we determine this by ensuring the function has a finite value at the given point and taking the limit on both sides of the point and checking if they both exist and are equal (L.H.L. = R.H.L.).

Differentiability of a function is out of the scope of this article, but simply put, a function is said to be differentiable at a point if the curve at that point is smooth i.e. there is no drastic change of slope. Mathematically this is achieved by checking if both the left hand derivative and the right hand derivative of the function at the given point finitely exist and are equal (incidentally this common value is the value of the derivative of the function at the given point).

First Derivative is defined as the differentiation of a function, y = f(x), once, with respect to ‘x’. It is denoted by dy/dx or f’(x) and simply put, it gives the slope of the function at any given value of ‘x’ or the instantaneous rate of change of the function w.r.t. ‘x’ at any given value of ‘x’.

Second Derivative is defined as the differentiation of a function, y = f(x), twice, with respect to ‘x’. It is denoted by d2y/dx2 or f’’(x) and simply put, it gives the slope of the slope of the function at any given value of ‘x’ or the instantaneous rate of change of the slope of the function w.r.t. ‘x’ at any given value of ‘x’.

Critical points of f(x) are defined as the values of x* for which either f'(x*) = 0 or f’(x*) does not exist.

Tests for Minima:

Local Minima can be found by Fermat's theorem, which states that they must occur at critical points.

If f(x) has a minima on an open interval, then the minimum value occurs at a critical point of f(x).

If f(x) has a minima value on a closed interval, then the minimum value occurs either at a critical point or at an endpoint.

Critical points of f(x) are defined as the values of x* for which either f'(x*) = 0 or f’(x*) does not exist.

One can distinguish whether a critical point is a local maximum or local minimum by using the first derivative test or second derivative test.

learning minima -First Derivative Test

Suppose f(x) is continuous at a critical point x*.

If f’(x) <0 an="" and="" extending="" f="" from="" interval="" left="" on="" open="" x="">0 on an open interval extending right from x*, then f(x) has a relative minima at x*.

If f’(x) >0 on an open interval extending left from x* and f’(x) <0 a="" an="" at="" extending="" f="" from="" has="" interval="" maxima="" on="" open="" p="" relative="" right="" then="" x="">
If f’(x) has the same sign on both an open interval extending left from x* and an open interval extending right from x*, then f(x) does not have a relative extremum at x*.

An interesting point to NOTE:

Differentiability is not a criterion for the first derivative test. Suppose f(x) is continuous but not differentiable at x*, i.e. f’(x*) does not exist. Still the above holds true since the test is done in open intervals on the left and right sides of the point in consideration [see Figure below]. So the criteria is only that f(x) is continuous at x* and that f’(x) exists in the neighbourhood of x*.

In summary, relative minima occur where f’(x) changes sign.

learning minima -The Second Derivative Test:

Suppose that x* is a critical point at which f’(x*) = 0, that f’(x) exists in the neighbourhood of x*, and that f’’(x*) exists.

f(x) has a relative minima at x* if f’’(x*)>0.

f(x) has a relative maxima at x*if f’’(x*) <0 .="" p="">
f(x) does not have an extremum at x* if f’’(x) = 0.

NOTE:

Differentiability at the critical point is a criterion for the second derivative test as opposed to the first derivative test. Also, if f’’(x*) = 0, the test is not informative [see Figure below], it actually means there is no change of sign of f’(x) on going from the left to right of the given critical point (these points are called the points of inflection).

learning Absolute Minima and Maxima

For any function that is defined piecewise, one can learn minima (or maxima) by finding the minimum (or maximum) of each piece separately; and then seeing which one is smallest (or biggest).

Visualization to learn minima

max-min

Least Common Denominator Tutoring

Least common multiple of two rational facts a and b is the least positive rational number, that is an integer multiple of a as well as b. because it is a multiple, it can be divided by a and b without a remainder. Tutoring not only help students by giving answers, but also help students in their problem solving with step by step solutions.  In this article we shall discuss about least common denominator tutoring. The following are the examples and steps involved in least common denominator tutoring.

Least common denominator tutoring:

Least common denominator:

The least common denominator of two or more numbers is the least number which is a multiple of each of the given number.

Technique: Least common denominator tutoring

To find the l.c.d.

Step (1) Write the multiples of first number

(2) Write the multiples of second number

(3) Write the common multiples

(4) Write the least common denominator.

Example problems least common denominator tutoring:

Example 1: Find the L.C.D of `1/4` and `1/3`

Here least common denominator is 12

To find the l.c.d.

Step (1) Get l.c.m. for every denominators.

Step (2) Modify equivalent fraction with same l.c.m denominator

Step (3)  By taking l.c.m common in denominator and add all numerators.

= `(1*3)/12` + `(1*4)/12` = `(4+3)/12`

So the final result is `7/12` .

Example 2: Find the L.C.D of `1/5` and `1/4`

Here least common denominator is 20

To find the l.c.d.

Step (1) Get l.c.m. for every denominators.

Step (2) Modify equivalent fraction with same l.c.m denominator

Step (3)  By taking l.c.m common in denominator and add all numerators.

=  `(1*4)/20 ` + `(1*5)/20`

So the final result is 9/20.

Example 3: Find the L.C.D of `1/25` and `1/15`

Here least common denominator is 75

To find the l.c.d.

Step (1) Write the multiples of 25: 25, 50,75

Step (2) Write the multiples of 15: 15, 30, 45, 60, 75

Step (3) common multiple is 75

Step (4) least common denominator is 75.

= `(1*3)/75` + `(1*5)/75` =`(5+3)/75`

So the final result is` 8/75` .

Practice problem for least common denominator tutoring:

Example 1: Find the L.C.D of `1/6` and `1/3`

Here least common denominator is 6

So the final result is` 3/6` .

Algebra is widely used in day to day activities watch out for my forthcoming posts on Least Common Multiples. I am sure they will be helpful.

Example 2: Find the L.C.D of `1/6 ` and `1/4`

Here least common denominator is 12

So the final result is `5/12` .

Example 3: Find the L.C.D of `1/4` and `1/2`

Here least common denominator is 4

So the final result is `3/4` .

A Proper Factor

A factor is a whole number which divides exactly by another whole number is called factor for  that number Any of the factors of a number, except the number itself. A factor is a portion of a number in general  integer or polynomial when multiplied by other factors  gives  entire quantity. The determination of factors is a factorization A proper factor of a positive integer is a factor of other than 1.Proper factor is a multiples for a given whole number which has multiples again it is aproper factor.Every whole number which has a factor of its own.proper factor is like a normal factor except 1
For example,
For 6, 2 and 3 are proper factors of , but 1 and 6 are not a proper factor.

Properties of proper factor:

The divisors for a any number other than 1 and  number itself are called  factors for that number.
A factor for N number is a number which divides  exactly N.

Example: the factors for 24 are 1,2,3,4,6,and 12

Generally for every number has itself and 1 as its factors.
When a number is greater than 1 and by itself and 1 as factors, then the number is prime.
A number or quantity that when multiplied with another number produces a given number or expression.

Example Problems for Proper factors:

Example 1 :
Find all the divisors and proper factors of  20

Solution :
The divisors of 20 are 1, 2, 4, 5, 10 and 20
The factors of 20 are 2, 4, 5 and 10

Example 2:
Find all the divisors and proper factors of 32.

Solution:
The divisors of 32 are 1, 2, 4, 8, 16 and 32
The proper factors of 32 are 2, 4, 8 and 16

Example 3:
Find Proper Factors of 30

Solution:
30=1*30
=2*15
=3*10
=4*15
=5*6
=6*5

Proper  Factors for 30 are 2,3,4,5,6

Example 4:
Find proper factors for 42

Solution:
42=1*42
=2*21
=3*14
=6*7
2,3,6

I am planning to write more post on Solve first Order Differential Equation and free 3rd grade math word problems. Keep checking my blog.

Example 5:
Find proper factors for 18

Solution:
18=1*18
=2*9
=3*6
=6*3
=9*2
Proper factors are 2,3,6,9

Equivalent Scale Factor

A scale area is a number which scales, or multiplies, some quantity. In the equation y=Cx, C is the scale factor for x. C is also the coefficient of x, and may be called the constant of proportionality of y to x. For example, doubling distances corresponds to a equivalent scale factor of 2 for distance, while cutting a cake in half results in pieces with a scale factor of ½.
SOURCE: WIKIPEDIA

Example problems of equivalent scale factor:

Equivalent scale factor problem 1:
Find the scale factor to the following figure:

Equivalent scale factor solution:
If we want to find the length of smaller rectangle then we can multiply the length of the one side of larger rectangle and the value of scale factor.
We can find the scale factor of the given rectangles by using the following formula,
Let D_l be the dimensions of larger rectangle and D_s be the dimensions of smaller rectangle and s be the scale factor.
Therefore the formula as,
                D_l*s=D_s
Substitute the values of dimensions into the above formula. Then we get
                  30*s=24
Divide by the value 30 on both sides,
                   `(30s)/30` `=` `24/30`
                       ` s ` `=` `24/30`
Divide by the value of 6 on both numerator and denominator. Then we get the value of scale factor.
                        s = `4/5 ` or 4:5
Therefore scale factor of smaller to larger rectangle= 4:5
Answer: 4:5
Equivalent scale factor problem 2:
Find the larger to smaller scale factor for the following figure:

Equivalent scale factor solution:
If we want to find the length of smaller rectangle then we can multiply the length of the one side of larger rectangle and the value of scale factor.
We can find the scale factor of the given rectangles by using the following formula,
Let D_l be the dimensions of larger rectangle and D_s be the dimensions of smaller rectangle and s be the scale factor.
Therefore the formula as,
                D_l*s=D_s
Substitute the values of dimensions into the above formula. Then we get
                  39*s=26
Divide by the value 48 on both sides,
`(39s)/(30)` `=` `26/39`
` s ``=` `26/39`
Divide by the value of 13 on both numerator and denominator. Then we get the value of scale factor.
                        s = `2/3` or 2:3
Therefore scale factor of larger to smaller scale rectangle= 3:2
Answer: 3:2

Algebra is widely used in day to day activities watch out for my forthcoming posts on Proof of Fundamental Theorem of Calculus and algebra 2 solver step by step. I am sure they will be helpful.

Practice problems of equivalent scale factor:

1. Find the scale factor to the following figure:

2. Find the larger to smaller scale factor for the following figure:

Answer:
1. 3:4
2: 6:5

Mode and Median Calculator

Mode: Mode is the value that takes place most repeatedly in the data set. Measure of central tendency is known as mode. If the data’s are given in the form of a frequency table, the class corresponding to the maximum frequency is called the modal class. The value of the variate of the modal class is the mode.

Median: The median is the middle value when the given values are arranged in an ascending order. Let us see the median and mode calculator.

Median and Mode calculator:

In the calculator enter the set of values in first box, after that clcik the median button it will automatically calculate the median value and it will be displayed in answer box. The same process is done for mode.

Examples on Mode calculator:

Example 1:

            Find the mode of 7, 4, 5, 1, 7, 3, 4, 6, and 7.

Solution:

           The above question is entered in the first box. The calculator doing the follwing process,

           Assemble the data in the ascending order, we get

            1, 3, 4, 4, 5, 6, 7, 7, 7.

            The number 7 occurs many times in the above values.
            Mode = 7 will display the answer box after press the mode button on calculator.
Example 2:

            Find the mode for 12, 15, 11, 12, 19, 15, 24, 27, 20, 12, 19, and 15.

Solution:

           The above question is entered in the first box. The calculator doing the follwing process,

           Assemble the data in the ascending order, we get

            11, 12, 12, 12, 15, 15, 15, 19, 19, 20, 24, 27.

            In the above values 12 occurs 3 times and 15 also occurs 3 times.

            ∴ Both 12 and 15 are the modes for the given data. We observe that there are two modes for the given data.The Mode will be displayed in answer box on calculator
Example 3:

            Find the mode of 19, 20, 21, 24, 27, and 30.

Solution:
            Already the above data are in the ascending order. Each value occurs exactly one time in the series. Hence there is no mode in the above given data.
These are the examples on mode calculator.

Examples on Median calculator:

Example 1:    
            Find the median of the following numbers: 12, 45, 62,10,14,31 and 43.
Solution:
           The above question is entered in the first box. The calculator doing the fololwing process,
            Arranging the given numbers in ascending order we get
            10, 12, 14, 31, 43, 45 and 62.
                            `darr`
                    Middle term
            Median = Middle item = 31.     

         The median 31 will display the answer box

Between, if you have problem on these topics Cubic Equation  please browse expert math related websites for more help on ibsat 2013.

Example 2:         
            Find the median of the following numbers: 3, 7, 4, 10, 22, 16, 21 and 5.
Solution:
            The above question is entered in the first box. The calculator doing the following process,
            Arranging the given numbers in ascending order we get
            3, 4, 5, 7, `darr` 10, 16, 21, 22              
                 Median is here
            Median = Item midway between 7 and 10
                       =` (7 + 10) / 2` = `17 / 2` = 8.5
          The Median 8.5 will display on answer box on calculator
These are the examples on mode and median calculator.

Sine Geometry

Trigonometry is the division of geometry dealing among relationships among the sides also angles of triangles. In geometry sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse. The  ratio does not depend on size of the particular right triangle chosen, as long as it contains the angle A, since all such triangles are similar

(Source: Wikipedia)

I like to share this sine curves with you all through my article. 

Sine geometry

In the above diagram ,
sin A =opposite/hypotenuse
Examples for sine geometry

A - Right angle of the triangle ABC.
The length of AB, BC and CA are frequently represented through c, a, b.
Obtain point B as middle of a trigonometric circle
Circle with radius = 1.
Now sin (B) are comparative to b, c also a.
sin `(B)/b` =`1/a`
sin (B) = `b/a`

Examples for sine geometry

Example 1
Angle of triangle is 20⁰, opposite side of triangle is 12 apply the sine geometry to find the unidentified side of the triangle?
Solution:

Angle of triangle= 20⁰  

Opposite side of triangle = 12.

sin A =opposite/hypotenuse

sin 20⁰ = `12/x`

sin 20⁰ x = 12

x = `12/sin 20^0`

x =`12/0.3420`    {since the value of sin 20 degree is 0.3420}

x=35.08

Hypotenuse side= 35.08

Example 2
Angle of triangle is 78⁰, hypotenuse side of triangle is 20 apply the sine geometry to find the unidentified side of the triangle?
Solution:

Angle of triangle= 78⁰  

Hypotenuse side of triangle = 20.

sin A =opposite/hypotenuse

sin 78⁰ = `x/20`

sin 78⁰ x 20= x

x = sin 78⁰ x20

x =0.97814x20     {since the value of sin 78 degree is 0.97814}

x=35.08

Opposite side = 19.56

Example 3
If hypotenuse side of triangle 40 and opposite side of triangle 20 find the sine angle?
Solution:

Hypotenuse side of triangle = 40.

Opposite side =20

sin A =opposite/hypotenuse

sin A= `20/40`

sin A =` 1/2`

sin A = 0.5    {sin 30 degree is 0.5}

Therefore the angle is 30 degree

Interval Estimates

Interval Estimate:

• Interval estimation is the process of calculate the interval for possible value of unknown parameter in the population.
• It is calculate in the use of sample data and contrast to the point estimation. It is different from the point estimation. It is the outcome of a statistical analysis.

The most common forms of interval estimations as follows:

• A frequents Method or Confidence interval
• A Bayesian method or credible intervals

The other common methods for interval estimations are

• Tolerance interval
• Prediction interval

And another one is known as the fiducial inference.

Construction of interval estimates parameter:

The normal form of interval estimate of the population parameter is,

• Point estimate of parameter and
• Plus or minus margin of error

Margin of error:

• The amount which is subtracted or added from  the point estimate  of the statistic and produce the parameter interval  estimate is known as the margin of error.
• The margin of error size depends on the following factors:
• Sampling distribution type of sample statistics.
• Area under sampling distribution percentage   that includes the researchers      decision.Usually we consider the confident level as 90%, 95%, 99%.
• The interval of each interval estimates are constructed in the region of the point estimate with its confident level.

My forthcoming post is on Set Interval Notation will give you more understanding about Algebra.

Construction of Interval estimate for Population mean

• Take the point estimate of μ  that is  the sample mean`vecx`
• Define  the mean distribution for the sample.When the  value of n is large we  have to use the central limit  theorem. And   is the normal distribution with the,

                      standard deviation `sigma``vecx` = `sigma/sqrt(n)`  

                      and mean μ.

• Choose the most common confident  level as 95%
• Find the margin of  error  which is related with the confidence level.
• The area  under the curve of  the sample means the normal distribution contains the 95%  of the interval from.

                               z= -1.96 to z= 1.96 

• The interval estimate for 95 % is,   

                            `vecx`- 1.96 (`sigma/sqrt(n)` ) to `vecx``sigma/sqrt(n)`

Common Factors

The common factors of two or more whole digits is the biggest whole digit that equally divides all the whole digits. There are two methods to find common factors in math.

The initial method is to list all the factors of each digit. Then decide the biggest factor.

For Example:

Find the common factors of 12 and 18.

The factors of 12 are 1, 2, 3, 4, 6 and 8.

The factors of 18 are 1, 2, 3, 6, 9 and 18.

The common factors of 12 and 18 are 1, 2 and 3.

Methods for Finding Common Factors

There are two methods to find the common factors of numbers. Listed below are the steps to be followed in finding common factors.
Method I:

• List all the factors of the numbers.
• Collect the common factors among each digit.

Method II:

• Find the prime factors of each digit.
• Merge the common terms of each digit.

Examples

Listed Below are some of the examples in finding the common factors.
Example 1:
What are the Common factors of 34 and 36?
Solution:
The factors of 34 are 1, 2, 17, and 34.
The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, and 36.
Now, using method 1,
The common factors to the two numbers are 1 and 2.
The Common factors of 34 and 36 are 1 and 2.
Example 2:
What are the common factors of 40, 45 and 50?
Solution:
Prime factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40.
Prime factors of 45 are 1, 3, 5, 9, 15 and 45
Prime factors of 50 are 1, 2, 5, 10, 25, and 50.
Now, using method 2,
The common factors of 40, 45 and 50 are 1, 2 and 5.

Practice Problems

Listed below are some of the practice problems in finding the common factors.

Problem 1:

What are the Common factors of 8, 14, 18 and 22?

Answer:    

8   `->` 1, 2, 4, and 8

14 `->` 1, 2, 7 and 14

18 `->` 1, 2, 3, 6, 9 and 18

22 `->` 1, 2, 11 and 22

My forthcoming post is on Series Solutions of Differential Equations will give you more understanding about Algebra.

Problem 2:

What are the Common factors of 15, 30, 45 and 60?

Answer:    

15 `->` 1, 3, 5 and 15

30 `->` 1, 2, 3, 5, 6, 10, 15 and 30

45 `->` 1, 3, 5, 9, 15 and 45

60 `->` 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, and 60

Review of Word Problems

Introduction on review of word problems are first used to identify the variables present in the given problem. Also the word problems are used for the unit numbers and unit variables. Word problems are used for finding the phrases. The consideration includes in this is, we have to analyse the problem present in the statements.

I like to share this Fun Math Problems with you all through my article.

Types of word problems:

There are many types review word problems present. They are,
1.Word problem for numbers
2.Word problem for mixtures
3.Word problem for Age
4.Word problem for time
5.Word problem for linear
Word problem for numbers:
          In word problem for numbers, we  first review the relationships are identified. The variables present in the word problems are reduced.
Word problem for mixtures:
       Different types of concentrations are included in this word problem mixtures.
Word problem for age:
       The relationship of ages are calculated on this word problem.
Word problem for time:
     In this word problem, we review  the various types of train problems are included. Time taken for a train to go and coming back are included in this word problem.
Word problem for linear:
      Linear word problem mostly review the cost problems. For example, the cost for an grapes and the apple is 14.etc..





Algebra is widely used in day to day activities watch out for my forthcoming posts on find the prime factorization of 125 and cbse neet 2013. I am sure they will be helpful.

Example for review word problem:

Question: 1. The difference of twice a number 4 is 8. What is that number?
Solution:
Step 1: First we are changing the above said sentence as difference of twice a number 4 equals 8.
Step 2: Next step is to write the above said equation in the equation form.
     2V-4=8.
This is the solution for a word problem.
Question 2. The difference of twice a number 2 is 4. What is that number?
Solution:
Step 1: First we are changing the above said sentence as difference of twice a number 2 equals 4.
Step 2: Next step is to write the above said equation in the equation form.
     2V-2=4.
This is the solution for a word problem.
Practice to review word problem:
Practice 1.The difference of twice a number 12 is 14. What is that number?
Practice 2.The difference of twice a number 20 is 40. What is that number?