Expressions With Rational Exponents

A rational number is a number represented as p/q where p and q are integers and q != 0.Exponent means  power.

The large numbers are very difficult to read and write and understand. To make them simpler we can use exponents, with the help of this many of the large numbers are converted to simpler forms. The short notation 74 stands for the product 7 x 7 x 7x 7. Here ‘7’ is called the base and ‘4’ the exponents. The number 74 is read as 7 raised to the power of 4 or simply as fourth power of 4. 74 are called the exponential form of 2401.

The above example is of a natural number as exponents.

Ex 5 ^(3/2) here 3/2 is a rational number so this is an example of rational exponents.

Rules invovled for solving rational exponents:

The Rational exponents rules are:

p,q are any real numbers except zero and m,n are positive integers.

Rule I:The base value is same under multiplication so we can directly add the power values. Then the exponent take the following form

pm  × pn = pm+n

Solving the expression 73  × 75 = 73+5 =78

Rule II:The base value is same under division so we can directly subtract the power values. Then the exponent take the following form such that m > n then

(p^m)/(p^n) = pm-n

Solving the expression  (4^4)/(4^2)= 44-2=42

Rule III:If m < n, (p^m)/(p^n) =(1)/(p^(n-m))

Solving the expression 8^3/8^2 =1/8

Rule IV:A real number to the power of power

(pm )n  = pm*n

Solving (82 )3  = 82*3=86

Rule V:The power of 0 is

p0 = 1. Anything power zero is equals to 1.

p^m/p^m   = pm-m =p0 = 1

Rule VI:p to the power of negative number is

p-m =1/p^m
Solving the expression 2-5 =  1/2^5

Rule VII:Two numbers to the same power can be written as

pm  × qm = (p*q)m

Solving the expression 32 × 82 = (3*8)2=242

Rule VIII A number to rational power is written as

p^(m/n) = root(n)(p^m)

Examples of expressions having rational exponents:

Ex : 1Solve the exponent    2x^(1/3)xx8^(5/3)xxx^(2/3)

Sol:Step 1:Given

2x^(1/3)xx8^(5/3) xx x^(2/3)

Step 2:Exponents are added

=2x^(1/3+2/3) (2^3)^(5/3)

Step 3: 8 is written as 2 power 3

Step 4 : Simplify=2x2^5

=2xx32 x

=64 x

Ex 2: Solve x^(-1/3)8^(-2/3)

x^(-1/3)8^(-2/3)

Solu: Step 1  x^(-1/3) 8^(-2/3)

Step2: 8^(-2/3) = (2^3)^(-2/3) = 2^(3 xx-2/3) = 2^-2 = 1/2^2 = 1/4

My forthcoming post is on  Define Arithmetic Mean will give you more understanding about Algebra.

Step3 Simplify the variable with negative exponent

=1/(x^(1/3)) 1/4

Step 4 Simplify:=1/(4x^(1/3))

=1/(4root(3)(x))

Power Coefficient Equation

A coefficient is a multiplicative factor in a few phrase of an expression. It is typically a number, but in any case does not occupy some variables of the equation. For example in equation 7x2 − 3xy + 1.5 + y the first three stipulations correspondingly have coefficients 7, −3, and 1.5. In the third expression there are no variables, so the coefficient is the idiom itself called the power constant term or constant power coefficient of this expression

I like to share this Sample Correlation Coefficient with you all through my article.

Example Problem for the Coefficient:

Find the coefficient of the given equation:

X2 + 5y3 +2xy2 + 5yx2

Solution:

Given that X2 + 5Y3+ 2XY2 + 5YX2

Here X2+ 5Y3+ 2XY2+ 5YX2 the expression with the coefficient

The coefficient of x2 is 1

The coefficient of  5Y3 is 5

The coefficient of 2xy2 is 2

The coefficient of 5yx2 is 5.

Coefficient Of Varience

Coefficient of variation is a relative determine for standard deviation.

It is defined as 100 times the coefficient of spreading base in the lead standard difference is called coefficient power of variation.

Less C.V. designate the fewer variability or additional power.

More C.V. indicate the more variability or less power.

According to Karl Pearson who suggested this compute the equation, C.V. is the percentage dissimilarity in the mean, standard deviation being considered as the total variation in the mean.

With the help of C.V. we can find which salesman is more consistent in making sales, which batsman is more consistent in scoring runs, which student is more consistent in scoring marks, which worker is more consistent in production.

Algebra is widely used in day to day activities watch out for my forthcoming posts on word problem solver for algebra and cbse class 10 science sample papers. I am sure they will be helpful.


Coefficients Of Dispersion

Whenever we want to compare the variability of two equations which differ widely in their averages or which are measured in different units.

We do not merely compute the measures of dispersions in the equations but we calculate the coefficients of dispersion which are pure numbers autonomous of the units of measurement.

Quotient Rule for Integration

The quotient rule is a method of finding the integration of a function that is the quotient of two other functions for which derivatives exist.

By the Product Rule,

if f (x) and g(x) are differentiable functions, then
d/dx[f (x)g(x)]= f (x)g'(x) + g(x) f' (x).
Integrating on both sides of this equation,

∫[f (x)g'(x) + g(x) f '(x)]dx = f (x)g(x),
which may be rearranged to obtain

∫f (x)g'(x) dx = f (x)g(x) −∫g(x) f' (x) dx.     (A)
Letting U = f (x) and V = g(x)

then differentiating it  we get

dU = f '(x) dx and dV =g'(x) dx,

pluging these values in (A), we get

∫U dV = UV −∫V dU.                  (1).


By the Quotient Rule,

if f (x) and g(x) are differentiable functions, then

d/dx[f (x)/g(x)]= g(x) f '(x) − f (x)g'(x)/[g(x)]2 .
Integrating both sides of this equation, we get
[f (x)/g(x)]=∫g(x) f '(x) − f (x)g'(x)/[g(x)]2 dx.
That is,

f (x)/g(x)=∫f '(x)/g(x)dx -∫f (x)g'(x)/[g(x)]2 dx,
which may be rearranged to obtain

∫f '(x)g(x)dx = f (x)g(x)+∫f (x)g'(x)/[g(x)]2 dx.      (B)

Letting u = g(x) and v = f (x) and then differentiating it , we get

du = g'(x) dx and

dv = f '(x) dx,
we obtain a Quotient Rule Integration by Parts formula:

plugging values of u , v , du and dv in B we get
∫dv/u= v/u+∫(v/u²) du.                                  (2)

quotient rule for integration-Application

∫[sin(x−½)/x²] dx.
Let
u = x½, du=1/2(x-½)

v=2cos(x-½),dv = sin(x−1/2)/x³/²

Then

∫sin(x−½)/x² dx = 2 cos(x-½)/x½+∫2 cos(x-½)/x• 1/2(x-½) dx

= 2 cos(x-½)/x½− ∫2cos(x-½) ·•[−(x-³/²)/2]dx

= 2 cos(x-½)/x½− 2 sin(x-½) + C,
which may be easily verified as correct.

My forthcoming post is on Add Hexadecimal Numbers and cbse solved sample papers for class 9th will give you more understanding about Algebra.

quotient rule for integration -Illustration


The Quotient Rule Integration by Parts formula (2) results from applying the
standard Integration by Parts formula (1) to the integral

Let ∫dv/u

with

U = 1/u,V = v,then differentiating ,we get

dU = − 1/u² du,dV=dv
plugging these values

∫dv/u=∫U dV

= UV −∫V dU

= 1/u·( v) −∫v (− 1/u²)du
= v/u+∫v/u² du

Terminating Decimals are Rational Numbers

Rational numbers are contrast among irrational numbers like Pi and square roots and sins and logarithms of information. On rational numbers and the ending of the article you be able to click on a linkage to maintain studying about irrational numbers. A number is rational but you can note down it in a form a/b where a and b are integers, b not zero. Visibly all fractions are of that form.

Terminating decimals are rational numbers:

A rational number is several number that be able to be expressed as the section a/b of two integers, among the denominator b not equivalent to zero. While b could be equal to 1, every one integer is a rational number. The set of every rational in sequence is frequently denoted by a boldface Q stands for quotient.

The terminating decimal extension of a rational numbers forever whichever terminating follows finitely several digits or begins to replicate the similar sequence of digits over and over. Moreover, any repeating or terminating decimal represent a rational number. These statements hold true not now meant for base 10, but as well for binary, hexadecimal, or any further integer base. Terminate decimal numbers be able to simply be written in that form: for example 0.67 is 67/100, 3.40938 = 340938/100000.

A real number so as to be not rational is called irrational. Irrational numbers contain square root 2, pi, and e. The decimal extension of an irrational number continues evermore without repeating. Because locate of rational numbers is countable, and the position of real numbers is uncountable, just about each real number is irrational.

In terminating conceptual algebra, the rational numbers shape a field. This is the representative field of characteristic zero, with is the field of fractions for the ring of integers. Finite extensions of Q are called algebraic number fields, and the algebra finality of Q is the field of algebraic number.

In mathematical study, the rational numbers shape a dense separation of the real numbers. The real numbers is able to be constructed from the rational numbers by achievement, using either Cauchv sequences or Dedekind cuts.

Zero separated by any other integer equals zero, consequently zero is a rational number even though division by zero itself is indeterminate.

My forthcoming post is on Probability Permutations and sample paper for 9th class cbse will give you more understanding about Algebra.

Solving Positive Direction Coordinates

A coordinate is a number that determine the position of a point the length of a number of line or arc. A list of two, three, otherwise additional coordinates can be use to establish the position of a point on a surface, volume, otherwise higher-dimensional area.

The Positive direction represents the positive value for both the x axis and y axis. The x axis value and the y axis value should be positive to move towards the positive direction. Any of the negative change in the axis will not lead to positive direction.

Please express your views of this topic Positive Correlation Graph by commenting on blog.

Solving positive direction examples:

Solve the following equation:

Y = x2 +2

Solution :

Now we have to substitute values for the variable x in the given equation.

When we substitute any positive value for the given equation we get only positive output values.

That is when  x = 1

Y = (1)2 +2

Y = 1+2

Y = 3

The co ordinates will (1,3) which lies on the positive direction.

when we substitute x =2

We get

Y = (2)2 + 2

Y = 4 + 2

Y = 6

The co ordinates are (2,6) which lies on the positive direction.

Similarly when we substitute any positive value for the given equation the output will be positive.

Even though we give negative values, the output will be (that is y) positive. But we need to give positive values to move in the positive direction.

Solving positive direction example:

Solve the following equation:

Y = 3x +4

Solution:

Now we have to substitute values for the variable x in the given equation.

When we substitute any positive value for the given equation we get only positive output values.

That is when x = 4

Y = 3(4) +4

Y = 12+4

Y = 16

The co ordinates will (4,16) which lies on the positive direction.

when we substitute x =6

We get

Y = 3(6) + 4

Y = 18 + 4

Y = 22

The co ordinates are (6,22) which lies on the positive direction.

Algebra is widely used in day to day activities watch out for my forthcoming posts on How to Find Second Derivative and Irrational Numbers Definition. I am sure they will be helpful.

Similarly when we substitute any positive value for the given equation the output will be positive.

Here when we substitute negative values for the variable x, we get negative values as output also. So we need to substitute positive values for the equation.

Poisson Distribution

If n is large, the evaluation of the binomial distribution can involve considerable computation. In such a case a simple approximation to the binomial probability could be considerable use. The approximation of binomial when n is large and p is close to zero is called the Poisson Distribution Mean.

Definition:

A random variable is said to follows Poisson distribution if it assumes only non-negative values and its probability mass function is given by

P(x,lambda ) =P(X=x) = elambda lambda x  /  X!    

x=0, 1, 2…

O            

otherwise

lambda is known as parameter of Poisson distribution .X~p() denotes it.

Characteristic function of Poisson distribution

phi x ( t )  =E[ei t x ]

=sum_(x=0)^oo ei t x   e-lambda lambda x / x!

=  e-lambda sum_(x=0)^oo ei t x lambda x  / x!

=  e-lambda [1+(lambda eit ) + (lambda eit)2 / 2! +(lambda ei t)3  /3!  + ............. ]

=e-lambda elambda eit

=  elambda (ei t -1)

Additive property of Poisson distribution:

Independent Poisson variate is also a Poisson variate xi (i=1,2,......n). xi follows Poisson with parameter lambda i.

xi ~ P(lambda i )    (i=1,2.....n)

then sum_(i=1)^n    xi ~ P(sum_(i=1)^n lambda i )

Proof:

Mxi(t) = elambda i(e^t -1)

Mx1+x2+x3+.....Xn(t) = Mx1 (t) Mx2 (t). . . Mxn (t)

= elambda 1(e^t-1). elambda 2(e^t-1) ..........elambda n(e^t-1)

=e(lambda 1 +lambda 2 +lambda 3+.......lambda n )(et-1)

=esum_(i=1)^n lambda i(et-1)

M.G.F of Poisson distribution:

Mx ( t ) = E [ etx ]

= sum_(x=0)^oo etx   e-lambda lambda x  / x!

= e-lambda sum_(x=0)^oo etxlambda x / x!

= e-lambda [ 1+ et lambda +  (lambda et )2 /2! + (  (lambda et )3  / 3! +...... ]

=e-lambda elambda

= elambda (et -1)

Applications of Poisson distribution

The following are some instance where the distribution is applicable

Number of deaths from a disease
Number of suicide reported in a particular city.
Number of defective materials in packing manufactured by good concern.
Number of printing mistakes at each page of the book.
Number of air accidents in some unit of the time.

Needs Assessment Definition

Assessment is an important and essential part of teaching. If teachers are to ask themselves whether what they are teaching and how they are teaching has the desired outcomes, they will needs to assess what children are able to do according to a set of criteria.

The criteria, which indicate what children, should be able to do and think in mathematics by the end of the foundation phase.

Teachers need to ask themselves with assessment:

The following are two very important questions that teachers have to ask when teaching.

Is what I am doing helping children to develop a desire to learn mathematics?

Is what I am doing teaching children to become numerate?

Please express your views of this topic Definition of Minuend by commenting on blog.

Why we needs assessment?

Some reasons for assessing learners are so that we can measure whether a lesson has been effective for each learner (whether we have taught the lesson effectively enough). We assess learners to see what each one can do, for example, which learners are able to calculate change, or which learners are able to solve relevant word problem. We assess learners to see which of the children are ready for a new challenge and which must still practice what has already been taught. We assess learners so that we can plan further lessons that suit the needs of the children.

Types of assessment:

There are two main ways in which to assess children

Formative assessment

Summative assessment.

Formative assessment is assessing a learner while the learner is forming the new knowledge.

Example for formative assessment:

An example of formative assessment would be sitting with a learner while he or she is doing a task (say using a number line to count in groups), watching how the child goes about the task and asking the child to explain how and what he or she is doing. In this way, you find out what strategies the child is using and developing and what strategies you should be helping the child with; you are getting direct and instant feedback on hoe the child is coping and you are able to respond to the situation immediately through re-teaching and explaining again, asking anther learner to help, or planning another lesson on that needs for the next day.

Summative assessment is assessing a learner at the end of the lesson, section, topic, quarter or year as a summing up of what the learner knows. Therefore, tests and exams are summative versions of assessment.

When both formative and summative assessments are used, that is continuous assessment. In an outcomes-based education system, continuous assessment is used. The teacher studies the learning outcomes required of the learners and then plans lessons to teach to achieve these outcomes. During the lessons, the teacher observers what children are doing and saying and how children are doing a task. The teacher asks for explanations from the children as to what and how they are doing a take. The teacher helps those learners who are confused and continually monitors which learners are gaining control of the skills and concepts. Once a child can do something independently within the number range for that learner, a teacher can say that that child has learnt what was intended by the lesson and so can record that performance as a desired learning outcome for that child.