Adding same Base Exponents

Introduction to adding same base exponents:

             The number of times the term want to  multiply by itself is defined as the exponents. It is defined by numbers in smaller size. Adding exponents here shows the multiplication of the terms with same bases  the exponents are added together. Terms without same bases cannot be simplified with each other. By multiplying the exponent with the same component can be add the terms with the same bases.

Instructions for Adding same Base Components

The instructions for the exponents of same base components is given as,

• The term is simplified by using the terms so that it cannot be simplified further.
• Operations are done for the expressions within the paranthesis.
• The exponential terms are simplified by adding and multiplying the terms in exponenets.
• The multiplication and the division operations are performed with the different exponents in the same bases by adding the coefficients.
• The terms with the same exponents are added by adding the coefficients by maintanign the same exponents.   

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Example Problems for Adding same Base Exponents:

Example 1: (Adding same base exponents)

     Here the base is same that is 3 and the exponents are different 3,2.

     3³ + 3² = (3 * 3 * 3) +(3 * 3)

                = 27 + 9

                = 36.

     Hence the answer for adding the same base exponents is given as 36. 

Example 2: (Adding same base exponents )

     4³ + 4⁴  = (4 * 4 * 4 ) + (4 * 4 * 4 * 4)   

                 = 64 + 256

                 = 3204

    The value of 4³ is 64.

    The value of 4⁴ is 216.

     Hence the answer for adding the same base exponents is given as 3204.

Example 3: 

      6² + 6³ = (6 * 6) (6 * 6 * 6)

                 = 36 +216

                       = 252

      The value of 6² is 36.

      The value of 6³ is 216.

      Hence the answer for adding the same base exponents is given as 36.

Example 4:

      2³ + 2⁴ = (2 * 2 * 2) + (2 * 2 * 2 * 2)

                 = 8 + 16

                = 24.

      The value of 2³ is 8.

      The value of 2⁴  is 16.

      Hence the answer for adding the same base exponents is given as 24.

        These are examples of adding same base exponents.

Formula for Surface Area of a Rectangular Prism

Introduction about rectangular prism:
            A rectangular prism is one type of three dimensional objects. It has six sides, which are rectangle in shape. In a rectangular prism all angles are right angle and opposite faces of rectangular solid are equal. It is also called as right rectangular prism. Let us see how to find the surface area of the rectangular prism using the formula.

Formula to Find the Surface Area of the Rectangular Prism:

Surface area (A) = 2(wh + lw + lh) square units

                                    w – Width

                                    h – Height

                                    l – Length

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Example Problems to Find the Surface Area of the Rectangular Prism:
1. The rectangular prism has the length 10 cm, width 5cm and height 5 cm. find the surface area of the rectangular prism.

Solution:

            Given:

                        Length (l) = 10 cm

                        Width (w) = 5 cm

                        Height (h) = 5 cm

Formula to find the area of the rectangular prism:

            Surface area (A) = 2(wh + lw + lh) square units

                                         = 2 ( 5 x 5 + 10 x 5 + 10 x 5 )

                                         = 2 (25 + 50 + 50)

                                         = 2 (125)

                                         = 250

Surface area of the rectangular prism (A) = 250 cm ^2

2. The rectangular prism has the length 10 cm, width 6 cm and height 7 cm. find the surface area of the rectangular prism.

Solution:

            Given:

                        Length (l) = 10 cm

                        Width (w) = 6 cm

                        Height (h) = 7 cm

Formula to find the area of the rectangular prism:

            Surface area (A) = 2(wh + lw + lh) square units

                                         = 2 ( 6 x 7 + 10 x 6 + 10 x 7 )

                                         = 2 (42 + 60 + 70)

                                         = 2 (172)

                                         = 344

Surface area of the rectangular prism (A) = 344 cm ^2

3. The rectangular prism has the length 12 cm, width 4cm and height 5 cm. find the surface area of the rectangular prism.

Solution:

            Given:

                        Length (l) = 12 cm

                        Width (w) = 4 cm

                        Height (h) = 5 cm

Formula to find the area of the rectangular prism:

            Surface area (A) = 2(wh + lw + lh) square units

                                         = 2 (4 x 5 + 12 x 4 + 12 x 5)

                                         = 2 (20 + 48 + 60)

                                         = 2 (128)

                                         = 256

Surface area of the rectangular prism (A) = 256 cm ^2

Introduction to refraction of light through a prism

Introduction to refraction of light through a prism:

When a ray of light passes through a prism bending of light takes place, when it enter as well as well when it leaves the prism that is called refraction of light through

prism. Bending of light is due to the difference in speed of light in different media, which cause refraction of light through prism. Since the two surfaces AB and AC of prism are not parallel to each other that’s why in refraction of light through prism the incident ray and the emergent ray are not parallel to one another.

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Refraction Process of Light through Prism:

As shown in the below figure, ABC is a glass prism with base BC.TP is an incident ray going from air (rarer medium) to glass (denser medium) with angle of incidence I at point P, it bends towards the normal PN which cause refraction of white light through prism and PO is the refracted ray.

When the ray of light PO comes out from the prism into air at the point O refraction of light through prism takes place again, Since the ray PO is going from glass (denser medium) to air(rarer medium) it bends away from the normal at O,,,    along the direction OR in the form of emergent ray.

So in case of refraction of white light through prism when a ray of light passes through a prism it bends towards the base (thicker part) of the prism.

From the figure it is clear that OR emergent ray is not parallel to the incident ray TP.

Now by expansion of rays and normal in forward or backward directions we can see from the fig1 that the angle LMR between incident ray and emergent ray is called angle of deviation.

Question Based on Refraction of Light through Prism

Q1: Define refraction through glass prism.

ANS: When a ray of light passes through a prism bending of light takes place,when it enter as well as well when it leaves the prism that is called refraction of light through Prism.

Q2: Explain emergent ray.

ANS: The ray finally coming out from the prism after refraction of light through prism is called as emergent ray.

Introduction to Solving Multiplying Integers

Introduction to Solving Multiplying Integers:

                       Integers are the set of numbers that include all natural numbers (0, 1, 2, 3, 4, and so on) and their negatives. Integers include positive numbers, negative numbers and zero (zero is neither positive nor negative). Below they are just give few integers - they go on forever in both directions e.g.....-11,-10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11... Solving multiplying integers is made by using of properties.

Types of Integer:
                 There are two types of integer. We can do solving multiplying in both type of integers.

1. Positive Integer - 1, 2, 3, 4, 5,...                 

2. Negative Integers - -1, -2, -3, -4, -5, …
  

Properties on Solving Multiplying Integers:

               The properties of solving multiplying integers are,

Closure under multiplication - a × b is an integer, for all integers a and b.
Commutativity of multiplication - a × b = b × a.
Multiplication by zero - a × 0 = 0 × a = 0.
Multiplicative Identity - a × 1 = 1 × a = a.
Associativity multiplication - (a × b) × c = a × (b × c).
Distributive property - a × (b + c) = a × b + a × c.



Examples on Solving Multiplying Integers:

Closure under multiplication:

1. Solve the multiplication of -20 and -5.

Solution:

             (–20) × (–5) = 100.

Commutativity of multiplication:

1)Solve: 3 × (– 4) = –12

             (– 4) × 3 = –12

             So, 3 × (-4) = (-4) × 3

2.Solve: (-7) × 4 = -28

              4 × (-7) = -28

              So, (-7) × 4 = 4 × (-7)

Multiplication by zero:

1.  (–3) × 0 = 0

2.  4 × 0 = 0

3.  (-9) × 0 = 0

Multiplicative Identity:

1. (–3) × 1 = –3

2. (7) × 1 = 7

3. (-8) × 1 = -8

Associativty multiplication:

1. [(–3) × (–2)] × 5 = (–3) × [(–2) × 5]

2. [7× (-3)] × 2 = 7 × [(-3) × 2]

Distributive property:

1. (–2) × (3 + 5) = –2 × 8 = –16

                        and

    [(–2) × 3] + [(–2) × 5] = (– 6) + (–10) = –16

2. (5) × ( 6 + 2 ) = 5 × 8 = 40

                         and

     (5 × 6) + (5 × 2) = 30 + 10 = 40

These are the examples for solving multiplying integers.

Introduction for Algebra Quadratic Equations

Introduction for Algebra Quadratic Equations:

Algebra :It is a branch of mathematics that includes terms,equations, polynomials and algebraic structures.The most basic form of algebra is elementary algebra.The terms that are related with algebra are
*Variables these are alphabetical characters that are used for assigning the value.generally we use x,y,z as variables.
*Terms:It is a single number or a variable and numbers multiplied together.Example 2x-5=3 here 2x,5,3 are called terms
*Constant :It is fixed value and does not contain any variable.Example the constant term in 3x+2 is 2.
In algebra polynomials is an expression that contains one or more variables and constants using operations(addition,subtraction and multiplication).Quadratic equation is an example of polynomials.

Quadratic Equation:
   It is an equation that contains a second Degree term. The quadratic equation is of the form ax^2+bx+c=0 where a, b, c are the coefficients (numbers) and X is the variable .The Variable X is the solution (or we can say it is a root) if it satisfies the given equation ax^2+bx+c=0.
Some Examples of quadratic equation are given as
x²+3x+4=0        3x²+9x-2=0            3x²=12
Quadratic equations can be solved by:

• Using Simple Factoring

• Using Quadratic Formula

• Using the square root method.

• Graphing.

• Completing the square method

Solving Quadratic Equations by Method of Factorization:

Step 1: Write the equation  in standard form, ax^2+bx+c=0,where a is not equal to zero, if needed.
Example: X²+7x-8=0

Step2: Find the product ac .Determine all possible factors of the first and third term ac= -8 all possible factors of 8 are (2 * 4) and (1 * 8).

Step3: Choose the factors that will add to give the middle term. Addition and subtraction sign tell you whether the term should be positive or negative. Since ac = - 8 we will need a positive and negative factor. Since the middle term, is positive, the positive factor of 8 will need to be larger than the negative factor. For instance, (-8*1) will give -8 but result is -7.So we take 1 as negative factor and 8 as positive.

Step 4: The final factors will be in the form (x+ ) (x+ ) 
This trinomial will have the factors (x-1) (x+8). 

Step5. Use the Zero product Principle 
To solve for X set each factor= 0
x-1= 0 that given x=1 and x+8= gives x= - 8

Quadratic Formula to Solve Quadratic Equation in Algebra

Quadratic formula is given as

      The General Quadratic Formula to find the roots of a quadratic equation is,

                        x = `((-b+-sqrt(b^2-4ac))/(2a))`

Where,

            a = coefficient of x²

            b= coefficient of x

            c= constant

            b²-4ac = discriminant

This formula generally gives 2 solutions. The Plus and minus sign shows that there are 2 possible 
solutions.

Deriving the Quadratic formula:

It is obtained by solving the general quadratic equation

  aX²+bX+c=0

Divide each side of equation by a

  X²+(b/a)X+c/a=0

Subtract c/a from each side of the equation

 X²+(b/a)X=-c/a

To complete the square add (b/2a)² to each side of the equation

  X² + (b/a)X + (b/2a)²=-c/a + (b/2a)²

  (X+b/2a)²= -c/a+ (b²/4a²)

Taking LCD for the right side of equation

(X+b/2a)²= (-4ac/4a²)+ (b²/4a²)

(X + b/2a)²= (-4ac +b²)/4a²

Take the square root of each side of the equation.

  X+ (b/2a) =  ±√(b²-4ac) / 2a

   X = `((-b+-sqrt(b^2-4ac))/(2a))`

When a quadratic equation is in standard form ax^2+bx+c=0 , the expression, b^2-4ac, that is found under the square root part of the quadratic formula is called the discriminant.

Discriminant formula 
    

 d=b²-4ac 
Kinds of solutions for ax²+bc+c=0
a)If Discriminant d is greater than zero(means it is positive) b²-4ac>0 :The equation has two solutions both of which are real numbers. Or we can say it has 2 distinct real solutions

b)If Discriminant is zero  b²-4ac=0 : The equation has only one real solution

X=-b/2a
c) If Discriminant Is less than zero( Discriminant is negative)b²-4ac<0 span="span">  : No solutions are defined.
There are two distinct (non-real) complex roots, which are complex conjugates of each other:

  (-b + i √4a c-b²)/2a       and       (-b- i √4ac-b²)/ 2a
where i =`sqrt(-1)` 

Nature of Roots of a Quadratic Equation

The quantity D = b²- 4ac is known as the discriminant of the quadratic equation whose roots are given by:
        α =  `(-b + sqrt(b^2-4ac))/(2a)`     and
       β = `(-b- sqrt(b^2-4ac))/(2a)` 
The nature of roots is as given below:

1. The roots are real and distinct iff D>0.
2. The roots are real and equal iff D=0.
3. The roots are complex with non-zero imaginary part iff D<0 .="." span="span">
4. The roots are rational iff   a,b,c are rational and D is a perfect square.
5. The roots are of the form p+sqrt(q) (p,q belongs to Q) iff a,b,c are rational and d is not a perfect square.
6. If a=1, b,c belongs to I and the roots are rational number, then these roots must be integers.
7. If a quadratic equation in x has more than two roots, than it is an identity in x that is a = b = c = 0. 

Square Root Method to Solve a Quadratic Equation in Algebra

Steps to Solve Quadratic equation Using Square Root method

Step 1: Write the quadratic equation in the form A²=B if needed.
 Example X² = 16 

Step 2: Apply the square root method. 

 If A and B are algebraic expressions such that A² = B ,then

 A is equal to Square root B or A is equal to - Square root B

 Also it is written as A = plus minus root B

X²=16

X=√16  or X= -√16

There are 2 solutions

Step 3: Solve for the linear equation(s) set up in step 2.

  X=√16  that means  X=4

 X=-√16  that means X=-4

There are two solutions to above quadratic equation that is given as x = 4 and x = -4.

Sample Size Determination

A Sample Size is of a particular sample is the number of sample units that are included in it. For example, if we consider a particular survey done on 25 people, the sample size is ’25 people’ as the number of people included in this survey is 25.

Sample Size Determination: Let us learn How to Determine Sample Size. When the goal of any study is to make inference about the population in that sample, the sample size plays an important role. While Determining sample size the following steps are considered:
First we need to determine the analytical plan for the results
Level of precision is to be determined, which is the sampling error or the margin error. This is the range in which the true value of the population is estimated to be. It is expressed in percentage points Example: ±4 percent. The larger the sample, the smaller the sampling error and greater the estimate precision
We determine the confidence level, which is also known as the risk level. This involves Confidence interval percentage which indicates the probability that the true proportion lies within the confidence interval boundaries. The confidence level is inversely proportional to estimate accuracy
Then estimate the Degree of variability in the target population; for instance, if there is a lot of variability in the issue of interest, the sample size needed might be large
Finally the response rate is estimated. When we conduct a survey we expect a response, this estimation gives the proportion of the survey or study

Now, let us learn the Sample Size Calculation using the following steps:
Step1:  Confidence interval of the study is determined. For example, if the confidence interval is 4.5 percent and the study reveals 58 percent as the proportion, then the true proportion is likely to be between 53.5 percent and 62.5percent
Step2: We now determine the confidence level of the study. For example, if we use the confidence level to be 96 percent, we can predict with 96 percent certainty that the true proportion lies within the confidence interval
Step3: Once the confidence level is determined, using the Z-score table convert it into a Z-score.  For example, the Z-score for the confidence level 96 percent result is 2.05
Step4: The proportion of the study is predicted. For example, if we expect affirmatively that 56 percent of respondents to respond, then the proportion would be 0.56
Step5: The required sample size is calculated by plugging the above values in the formula given below:
Formula to find the Sample size = Z^2[P(1-P)]/I^2, here Z is the Z-score got by using the Z-score table, P is the estimated proportion and I is the Confidence Interval percent or the desired level of precision

Sample statistics

Let's learn about sample statistics in today's post.

Statistics is winning theme that can be help to make use of numerical knowledge related to any research. Statistical is a method used to accomplish, learning, & summary, assess & present the numerical knowledge or statistics.

Next time i will help you with the concept of statistics inference.

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