Negative Integer Exponents

The exponents are which integer is placed in the power, of base numbers. It can be easily represent as, “a small number to the right side and above of base number”. It is called as exponents. These exponents have some of important rules and laws. Power with negative integer exponents is also one of the rules. Here we are going to explain about this negative integer exponent rule.

If we are having variables, which is containing the exponents and it have equal bases means, we can do some mathematical operations with the exponents. Those operations are called as the “laws of exponents” or “rules of exponents”. In this rule based negative integer rule of exponent is defined as following ways,

Definition for negative integer exponents:

It is otherwise called as power with negative exponent rules. This negative exponent rule is defined as, if m is a positive integer and x is a non-zero rational number, then it can be denoted as,

X-m = (1/x)^m (or)

= (1/x)^m

Which is (x)^-m is the reciprocal of (x)^m

And we adopt the same rule for rational exponents also. If p/q is a positive rational number means, and x>0 is a rational number, then

 X^ - (p/q) = (1/x)^ (p/q) = (1/x)^ (p/q) .

Which is, (x)^-(p/q) is the reciprocal of (x) ^(p/q) or the number obtained by raising the reciprocal of x to the exponent p/q.

For example: 1). 3^-2

= (1/3)^2

= (1/3)^-2

= -6 .

2). (4)^-(2/3)

= (1/4)^-(2/3)

= (1/4)^ (2/3)

This kind of exponentiation used for discovers the negative integer exponents and simplify the problems.


Example problems for negative integer exponents:

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

Solution:

Given: (8)^-(2/3)

= (1/8)^ (2/3)

= [(1/8)^ (1/3)]^ 2

= (1/2)^2, since (1/2)^3 = 1/8

= 1/4 .

2) Solve: (32/243)^-(4/5)

Solution:

Given: (32/243)^-(4/5)

= (243/32)^(4/5)

= [(243/32)^(1/5)]^4

= [(3^5/2^5)^(1/5)]^4

= [((3/2)^5)^(1/5)]^4

= (3/2)^4

= 81/16.

3) Evaluate, and find the following negative integer exponent value:

Evaluate: (27/125)^-(2/3)  xx    (27/125)^-(4/3)

Solution:

 (27/125)^- (2/3) xx (27/125)^-(4/3)

= (125/25)^(2/3) xx (125/27)^(4/3)

= [(5^3/3^3)^(1/3)]^ 2 xx [(5^3/3^3)^(1/3)]^4

= [((5/3)^3)^(1/3)]^2 xx [((5/3)^3)^(1/3)]^4

= (5/3)^2 xx (5/3)^4

= (5/3)^6

= 15625/729.

These all are the explanations and example problems may clear about the negative integer exponents.

Math Absolute Value Inequalities

In math, the absolute value |x| of a real number x is x's arithmetical value lacking view to its symbol. So, for example, 86 is the absolute value of both 86 and −86. Generalization of the absolute value for real numbers occurs in a extensive diversity of math settings. Consider an absolute value is also definite for the complex numbers, the quaternions, prepared rings, fields and vector spaces. The absolute value is strictly associated to the ideas of magnitude, distance, and norm in different math and physical contexts.

Properties of the Math absolute value inequalities

The absolute value fundamental properties are:

 |x| = sqrt(x^2)                                      (1) Basic

 |x| \ge 0                                            (2)     Non-negativity

 |x| = 0 \iff x = 0                            (3)     Positive-definiteness

 |xy| = |x||y|\,                                  (4)     Multiplicativeness

 |x+y| \le |x| + |y|                            (5)     Subadditivity

Another important property of the absolute value includes these are the:

 |-x| = |x|\,                                      (6)     Symmetry

|x - y| = 0 \iff x = y                     (7)     Identity of indiscernible (equivalent to positive-definiteness)

|x - y| \le |x - z| +|z - y|                 (8)     Triangle inequality (equivalent to sub additivity)

|x/y| = |x| / |y| \mbox{ (if } y \ne 0) \,                   (9)      Preservation of division (equivalent to multiplicativeness)

|x-y| \ge ||x| - |y||                           (10)     (Equivalent to sub additivity)

Math absolute value inequalities – Examples:

Math absolute value inequalities – Example 1:

|x - 5| < 3

Set up the two times inequality -3 < x -5 < 3 and then solve.

-3 < x – 5 < 3

2 < x < 8

In interval notation, the answer is (2, 8).
Math absolute value inequalities – Example 2:

|2x + 3| <=-8

This solution will involve setting up two separate inequalities and solving each.

2x + 3 <= -8

2x <=-11

x<=-11/2

Else

2x+3 >=8

2x >= 5

x>=5/2

In interval notation, the answer is (-oo, -11/2) U (5/2, oo) .
Math absolute value inequalities – Example 3:

|x - 9| < 4

Set up the two times inequality -4 < x -9 < 4 and then solve.

-4 < x – 9 < 4

5 < x < 13

In interval notation, the answer is (5, 13).

Math absolute value inequalities – Example 4:

|3x + 3| <= - 8

This solution will involve setting up two separate inequalities and solving each.

3x + 3 <= -8

 3x <=-11

x<=-11/3

Else

3x+3 >=8

 3x >= 5

x>=5/3

In interval notation, the answer is (-oo, -11/3) U (5/3, oo).

Negative Number Calculator

In this article we are discussing about subtracting negative number by using calculator. Negative number also real number. The negative number is represented as minus symbol “- “. Negative number or elements are less than zero such as -4, -7, -/3. A negative number may be parenthesized with its symbol, For example a subtracting is clearer if written (-7) + (−5) = -13 Using calculator we can the negative number.
Negative number calculator:

Let us see how to negative number using calculator.

Negative number is the similar as subtracting the corresponding negative number:

Example: (-7) + (-7) = -14

Negative number calculator – Example problems:

Example 1:

(-3) + (-4) = ?

Solution:

First enter the negative number in the calculator.

Then press the equal to button.

The solution is displayed on the calculator.

negative number calculator

The subtract solution is -7.

Example 2:

(-5) - (-4) = ?

Solution:

First enter the negative number in the calculator.

Then press the equal to button.

The solution is displayed on the calculator.

negative number calculator

The subtract solution is -1.

Example 3:

(-4) x (-4) = ?

Solution:

First enter the negative number in the calculator.

Then press the equal to button.

The solution is displayed on the calculator.

negative number calculator

The subtract solution is 16.

Example 4:

(-9) `-:` (-3) = ?

Solution:

First enter the negative number in the calculator.

Then press the equal to button.

The solution is displayed on the calculator.

negative number calculator

The subtract solution is 3.

Example 5:

(-10) x (-20) = ?

Solution:

First enter the negative number in the calculator.

Then press the equal to button.

The solution is displayed on the calculator.

negative number calculator

The subtract solution is 200.

Negative number calculator – practice problems:

Problem 1: (-5) + (-6)

Problem 2: (-4) – (-2)

Problem 3: (-9) x (-18)

Problem 4: (-5) `-:` (-20)

Negative number calculator – answer key:

Problem 1: -11

Problem 2: -2

Problem 3: 162

Problem 4: -0.25

Solving Monotonicity and Concavity

A function is said to be rising on an interval [a, b] = I if f (z1) < f (z2) whenever z1 < z2 for z1, z2 `in` I. A function is said to be decreasing on an interval [a, b] = I if f (z1) > f (z2) whenever z1 < z2 for z1, z2 `in` I. A function which is strictly increasing or strictly decreasing on an interval is said to be monotonic on that interval. A function is said to be curved in up on an interval [a, b] if f ’’ (z) > 0 on [a, b]. A function is said to be curved in down on an interval [a, b] if f ’’ (z) < 0 on [a, b]. A point (c, f(c)) on the graph of y = f (z) is called an inflection point if the concavity changes at the point. (I.E., it is curved in up on some interval (a, c) and curved in down on some interval (c, b) or vice versa.)

Solving monotonicity and concavity - Examples:

Solving monotonicity and concavity - Example 1:

Find inflection points & determine concavity for f (x)

f ‘(x) = `(x^5)/20 + (x^4)/12 - (3x^3)/3 - 10`

f ’(x) = `(x^4)/4 + (x^3)/3 - 3(x^2)`

f ‘’ (x) = x3 + x2 – 3x

= x(x2 + x - 3)

f ‘‘(x) = 0

x = 0, -2.3, 1.3

Inflection pts: x= -2.3, 0, 1.3

Curved in up: (-2.3,0), (1.3,infinity)

Curved in down: (-oo,-2.3), (0,1.3)

Solving monotonicity and concavity

Solving monotonicity and concavity - More Examples:

Solving monotonicity and concavity - Example 1:

Find inflection points & determine concavity for f (x)

f (x) = `(x^5)/20 + (x^4)/12 - (x^3)/3 +10`

f ‘ (x) = `(x^4)/4 + (x^3)/3 - x^2`

f ‘’ (x) = x3 + x2 – 2x = x(x2 + x - 2)

= x (x+2) (x-1)

f ‘‘ (x) = 0

x = 0, -2, 1

f ’’(-5) < 0, f’’(-1) > 0, f ’’(.5) < 0, f ’’(10) > 0

Inflection pts: x=-2,0,1

Curved in up: (-2,0), (1,`oo` )

Curved in down: (`-oo` .,-2), (0,1)

Solving monotonicity and concavity

Trinomial Multiplication

In algebraic expression consisting of only one term is called a monomial, two terms is called a binomial and when there are three terms separated by an addition or a subtraction operation is called a trinomial. ‘Tri’ in the word trinomial means three and hence the name. It is also referred to as a polynomial as ‘poly’ means more than two terms. For example, a+3b-c consists of three terms a, b and c and hence a trinomial. In 3b, 3 is called the coefficient of b. In a given polynomial we come across like terms and unlike terms.

Like terms are the terms which have the same variable or literal but a different coefficient. For example, (3b, -7b);(5x2, 12x2);(-4xy, 7xy) are some of the like terms. Unlike terms as the name suggests are the terms which have different variables. For example, (7x, 8y); (4ab, -4ac); (2y2, -2x2) etc are some of the unlike terms.

It is not necessary that there would be only two like or unlike terms, it depends on the number of terms in the given polynomial. Now let us learn as to How do you multiply trinomials. There are two methods in which we can multiply trinomials, one is the horizontal method the other is the vertical method.

The steps to be followed in a Horizontal method of multiplication are as follows:
Example: Multiply (4x2-3x+5)(x2+5x-3)
Solution: First and second terms are chosen irrespective of the order
Let (4x2-3x+5) be the first term and (x2+5x-3) be the second term

Arrange the two polynomials horizontally
(4x2-3x+5) X (x2+5x-3)
Now distribute each of the terms of the first trinomial with each of the terms of the second trinomial
=4x2(x2+5x-3)- 3x(x2+5x-3) +5(x2+5x-3)
=4x4+20x3-12x2 - 3x3-15x2+9x +5x2+25x-15

Combine the like terms and simplify
=4x4+x3(20-3)+x2(-12-15+5)+ x(9+25) – 15
= 4x4+17x3-22x2+34x-15

The steps to be followed in a Vertical method of multiplication are as follows:
Example: Multiply (4x2-3x+5)(x2+5x-3)
The first and the second terms are chosen irrespective of the order
Let (4x2-3x+5) be the first term and (x2+5x-3) be the second term
Arrange the two polynomials in a vertical form
4x2-3x+5
X  x2+5x-3

Working from right towards left each term of the lower trinomial is multiplied with each of the terms of the upper trinomials. Then the products are written underneath the second trinomial in three rows in the order of the degree of each term and then the like terms simplified as shown below
4x2-3x+5
X  x2+5x-3
-12x2+ 9x – 15
+20x3-15x2 + 25x
4x4 - 3x3  + 5x2                      .
4x4+17x3-22x2+ 34x-15 is the final product!

Solve Explicit Differentiation

In calculus,Explicit is a function which the independent variable. The function f explicitly is to provide a preparation for determining the output of the given function y in terms of the input value x: y = f(x). Derivative of an explicit function is called as explicit differentiation. for example  y = x3 + 5. The process of finding the differentiation of the independent variable in an explicit function by differentiating each term separately, by expressing the derivative of the independent variable as a symbol, and by solving the resulting expression for the symbol.

Solve explicit differentiation problems:

Let us see some problems and its helps to solve an explicit differentiation.

Solve explicit differentiation problem 1:

Find the differentiation of given explicit function   y = x2 - 15x + 3.

Solution:

Given explicit function is  y = x2 - 15x + 3.

Differentiation of explicit function is   dy/dx  = d/dx ( x2 - 15x + 3)

Separate the each term, so, we get

= d/dx (x2) - d/dx (15x) + d/dx (3)

= d/dx (x2) - 15d/dx(x) + d/dx (3).

= 2 x(2-1) - 15 + 0

= 2x - 15

The differentiation of an explicit function is  2x - 15

Solve explicit differentiation problem 2:

Find the differentiation of an explicit function  x2 +  cot x = - y

Solution:

Given explicit function is  x2 +  cot x = - y

Multiply by (-1) on both sides,  y = - x2 -  cot x

Differentiation of an explicit function,

y =   - x2 -  cot x

dy/dx = d/dx -x2   - d/dx (cot x) .

=  - 2x - (-cosec2 x) .

= - 2x + cosec2x

The Differentiation of an explicit function is  - 2x + cosec2x

Solve explicit differentiation problem 3:

Find the differentiation of given explicit function  2x2 + y2 = 1

Solution:

Given explicit function is  2x2 + y2 = 1

Subtract 2x2 on both sides.we get,

2x2 + y2 - 2x2 = 1 - 2x2

y 2 = 1 - 2x2

Take square root on both sides, we get

 sqrt (y^2) =  +- sqrt (1 - 2x^2)

y = +- sqrt (1 - 2x^2) .

Differentiate the function,  Let u = 1 - 2x2            and           y = sqrt u

(du)/(dx) = - 4x                              dy/(du) = 1/(2sqrtu)

So,   dy/dx =  ((dy)/(du)) ((du)/(dx)) .

=  1/(2sqrtu) . (-4x )

=  (- 2x) / sqrtu .

Substitute u = y, So we get

= - (2x)/y   .

The differentiation of an explicit function is  - (2x)/y    .

Substitution Geometry

Geometry is a module of mathematics, which involves the study of shapes, line equation, angles problem, dimensions, relative position of figures etc.  The term ‘Geometry’ means study of properties. A point is used to represent a position in space. A plane to be a surface extending infinitely in all directions such that all points lying on the line joining any two points on the surface. Substitution geometry problems are given below.

Example problems for substitution geometry :

1. Find out the geometry equation of straight line passing through the points 2x + y = 8 and 3x - 2y + 7 = 0 and parallel to 4x+ y - 11 = 0
Solution:
Let (x1, y1) be the intersection lines
2x1 +  y1 =  8       …  (1)
3x1 - 2y1 = - 7   …   (2)


(1) × 2 ? 4x1 + 2y1 = 16      …    (3)
(2) + (3) ? x1 =9/7 `=>` y1 =38/7   (x1, y1) =( 9/7 ,38/7)


The straight line parallel to 4x + y - 11 = 0 is of the form 4x + y + k = 0
But it passes through (9/7 ,38/7)


36/7 +38/7 + k = 0 ? k = -74/7
4x + y -74/7 = 0
28x + 7y - 74 = 0 this is the equation of straight line.


2. For what values of ‘a’, the three straight lines 3x + y + 2 = 0, 2x - y + 3 = 0and x + a y - 3 = 0 are concurrent?

Solution:

Let (x1, y1) be the point of concurrency. This point satisfies the first two equations.
3x1 + y1 + 2 = 0 … (1)
2x1 - y1 + 3 = 0 … (2)

Solving (1) and (2) By using substitution method, we get (- 1, 1) as the point of intersection. Since it is a point of concurrency, it lies on x + a y - 3 =0
- 1 + a - 3 = 0
a -4 = 0

a = 4

Practice problems for substitution geometry:

1. Find the point of intersection of the straight lines 5x + 4y - 13 = 0 and 3x + y - 5 = 0

Ans: The point of intersection is (1, 2)

2. Find the geometry equation of straight- line perpendicular to the straight line 3x + 4y + 28 = 0 and passing through the  point (- 1, 4).

Ans: 4x - 3y + 16 = 0

3. Find the equation of straight line passing through the intersection of straight lines 2x + y = 8 and 3x - y = 2 and through the point (2, - 3).

Ans: x = 2

Proportionality Problems

In mathematics, proportionality indicates that two variables are related in a linear manner. If one number doubles in size, so does the other; if one of the variables diminishes to 1/10 of its former value, so does the other.The symbol for proportionality resemble a stretched-out, lowercase Greek letter alpha . When this symbol appear that two quantities or variables, it is read "is proportional to" or "varies in direct proportion with." Thus, the expression x alpha y is read " x is proportional to y " or " x varies in direct proportion with y ." In this condition, as long as x and y do not attain values of zero, the quotient x / y is always equal to the same value k , which is called the proportionality constant. (source: wiki)

Example proportionality problems:

Proportionality problem 1:

Find a if a/4= 3/2.

Proportionality  solution:

By using property 1:

a/4=3/2

(a) (2) = (4) (3)

2a = 12

Dividing both side by 2

a=6

Therefore the value of a=6

Proportionality problem 2:

Is 6: 4 = 3: 2 a proportion?

No. If this were a proportion, Property 1 would produce

(6) (2) = (4) (3)

12   = 12, this is true. It is a proportion.

Practice proportionality problems:

Practice proportionality problem 1:

Janette's car uses 9 gallons to go 200 miles.

a) How many gallons will she use to go 400 miles?

B) How many gallons will she use to go 600 miles?

c) How far can she drive with 36 gallons?

(a)18

(b) 27

(c)800 miles

Practice proportionality problem 2:

Carlos makes $25 in 4 hours.

a) How much will he make in 2 hours?

b) How much will he make in 20 hours?

c) How much will he make in 22 hours?

(a)$12.50

(b)$125

(c)$137.50


practice proportionality problem 3:

Which is a better value?

a) 15 ounces for $ 9.25 or 30 ounces for $18.00?

b) 15 ounces for $ 9.25 or 5 ounces for $3.29?

Answer:  ( a )30 ounces for $18.

( b )15 ounces for $9.25.

practice proportionality problem 4:

6 bottles cost $7.00.

a) How much will 18 bottles cost?

b) How much will 15 bottles cost?

c) How much will 27 bottles cost?

d) How many bottles can you buy with $10.50?

Answer:   ( a ) $21

( b ) $17.50

( c ) $31.50

( d ) 9

Geometry Area and Volume

Geometry” Earth-measuring" is an part of the mathematics concerned with questions of size, shape, relative position of figures, and the properties of space. Geometry is one of the oldest sciences. Initially a body of the  practical knowledge concerning lengths, areas, and volumes. And now we can see about the problems in geometry area and volume.

Geometry area and volume problem 1:

Pro 1 :Find the volume of cylinder with the radius 8 cm and the height 12 cm.

Solution:We can find the volume of  an cylinder by using the following formula:

Volume of cylinder V=πr2h

Substitute the values of r and h into the above formula. Than, we get

V=π*82*12

=3.14*64*12

=2411.52 cm3

Pro 2 :Find the volume of sphere with the radius is 12 cm.

Solution:We can find the volume of  the sphere by using the following formula

Volume of sphere V= (4/3) πr3

Substitute the value of radius into the above formula. Then we get,

V= (4/3) *3.14*123

= 1.333*3.14*144

= 602.72 cm3

Ans: 602.72 cm3

Problem 3:Find the amount of pyramid with the base 8.2 mt and height 10.2 mt.

Solution:We can find the volume of  the pyramid by using the following formula

Volume of pyramid V= (1/3) b h

Substitute the values of  the base and height into the above formula. Then we get,

V= (1/3)*8.2*10.2

=0.333*8.2*10.2

=27.8521 mt3

Ans: 27.8521 mt3

Geometry - Find the volume of shapes when area is given

Find the volume of the right prism whose area of the base is 550 cm2 and height is 38cm



Solution:Given that area of the base, A = 550 cm2 and height (h) of the prism = 38 cm

Volume of the right prism = area of the base * height cu.units

= A * h

= 20900

Volume of right prism = 20900 cm3

Find the volume of the right prism whose area of the base is 450 cm2 and height is 34cm

Solution:Given that area of the base, A = 450 cm2 and height (h) of the prism = 34 cm

Volume of the right prism = area of the base * height cu.units

= A * h

= 15300

Tutoring About Correlation Coefficient

Concept of correlation:

Correlation is a method of studying the relationship between the two variables. In statistical analysis we come across the study of two variables wherein the change in the value of one variable produces a change in the value of other variable. In that case we say that the variables are correlated or there is a correlation between the two variables.

The formula for the correlation coefficient r can be expressed in the form,

r = `(sum (X - barX) ( Y - barY))/(sqrt(sum (X - barX)^2) sqrt(sum(Y - barY)^2))`

It is conventionally taken as x = X - X and y = Y - Y and hence we write

r = `(sum xy)/(sqrt(sum x^2) sqrt(sum y^2))`

The above formula is expressed in terms of deviations of the variables from their means. Instead, if the actual values of the observations are taken then the formula can be written as,

r= `(N sum XY - sum X sum Y)/(sqrt(N sum X^2 - (sumX)^2) sqrt(NsumY^2 - (sum Y)^2))`

Instead of the deviations from their means, the deviations are measured from the value A and B for X and Y variables by taking dx = X - A, dy = Y - B, the correlation coefficient r is given by,

r = `(N sumdxdy - sumdx.sumdy)/(sqrt(Nsumdx^2- (sumdx)^2) sqrt(Nsumdy^2 - (sumdy)^2))`

Tutoring about formulas for calculating correlation coefficient:

r = `(sum xy)/(sqrt(sum x^2) sqrt(sum y^2))`

This formula is used when deviations are measured from their mean.
r= `"(N sum XY - sum X sum Y)/(sqrt(N sum X^2 - (sumX)^2)(sqrt(N sum Y^2 - (sumY)^2))) `

This formula is used if no assumed average is taken for x and y series
r = `"(N sumdxdy - sumdx.sumdy)/(sqrt(Nsumdx^2- (sumdx)^2)(sqrtNsumdy^2 - (sumdy)^2)) `

This formula is applied when deviations for x and y series are taken from some assumed values.


Tutoring on problems on correlation coefficient::

Calculate the correlation coefficient between x and y from the following data:
x 1 3 5 8 9 10
y 3 4 8 10 12 11


Solution:
x y x - `barx` y - `bary` (x - `barx` )2 (y - `bary` )2 (x-`barx` )(y - `bary` )
1 3 -5 -5 25 25 25
3 4 -3 -4 9 16 12
5 8 -1 0 1 0 0
8 10 2 2 4 4 4
9 12 3 4 9 16 12
10 11 4 3 16 9 12
36 48 0 0 64 70 65


`barx` = `(sum x)/(n)` = 36/6 = 6

`bary` = `(sum y)/(n)` = 48/6 = 8

r = `(sum(x- barx)(y - bary))/(sqrt(sum(x-barx)^2)sqrt(sum(y-bary)^2))`

= `(sum xy)/(sqrt(sumx^2)sqrt(sumy^2))`

= `(65)/(sqrt(64)sqrt(70))` = 0.97

Limit of a Function

Students can study about Limit of a Function here. Consider the function f(x). Let the independent variable x take values near a given constant a. Then f(x) takes a corresponding set of values. Suppose that when x is close to a, the values of f(x) are close to some constant. Suppose f(x) can be made to differ arbitrarily small from A by taking values of x that are sufficiently close to a but not equal to a and that is true for all such values of x. Then f(x) is said to approach limit A as x approaches a.

If the function f(x) approaches a constant A when x approaches a in whatever manner without assuming the value a, A is said to be the limit of f(x) as x approaches a. Thus we write lim_(x->a) f(x) = A

Find the Limit of a Function

Students can learn to Find the Limit of a Function if they know what Functions are and how they behave at the given limits.

A function may approach two different limits. One where the variable approaches its limit through values larger than the limit and the other where the variable approaches its limit through values smaller than the limit. In such a case the limit is not defined but the right and left hand limit exists.

The right hand limit of a function is the value of the function approaches when the variable approaches its limit from the right. Here, we write lim_(x->a^+) f(x) = A+

The left hand limit of a function is the value of the function approaches when the variable approaches its limit from the left.

here, we write lim_(x->a^-) f(x) = A-

The limit of a function exists if and only if the left hand limit = right hand limit.

In that case, lim_(x->a^+) f(x) = lim_(x->a^-) f(x) = f(x)

Properties of limit of a function

The following are some of the properties of limits which are useful in evaluating the limit of a function.

1. lim_(x->a) k = k ( k is a constant)

2. lim_(x->a) [f(x) ± g(x)] =  lim_(x->a) f(x) ± lim_(x->a) g(x)

3. lim_(x->oo) [f(x).g(x)] = lim_(x->oo) f(x) . lim_(x->oo) g(x)

4. lim_(x->a) (f(x))/(g(x)) = (lim_(x->a)f(x))/(lim_(x->a)g(x))

5. lim_(x->a) [f(x)]n =  [ lim_(x->a) f(x)]n

Standard limit theorems:

1. lim_(x->a) (x^n- a^n)/(x - a) = nan-1

2. lim_(x->0) (e^x-1)/(x) = 1

3. lim_(x->0) (sin x)/(x) = 1

4. lim_(x->0) (1 + (1)/(n))^(n) = e



Solved Examples

Example 1: Evaluate the right hand limit of the function

f(x) = {│x – 4│/x – 4, x ≠ 4, 0 x = 4

at x = 4

Sol: (RHL of f(x) at x = 4)

= lim f(x) = lim f(4 + h) = lim │4+ h – 4│4 + h – 4

x→4+      h→0             h→0

= lim │h│/h = lim h/h = lim 1 = 1

h→0            h→0      h→0

Example 2: Let f be the function given by f(x) = x2 – a2/x – a, x ≠ a

Using (in , δ) definition show that lim f(x) = 2a

x → 0

Sol: Let in > 0 be given. In order to show that

lim f(xi) = 2a

x → a

We have to show that that for any given in > 0, there exists a number δ >0 such that

│f(x) – 2a│< in whenever 0 < │x – a│< δ

If x ≠ a, then │f(x) – 2a│= │x3 – a2/x – a│

= │(x + a) – 2a│                                         [... x ≠ a]

= │x – a│

... │f(x) – 2a│< in , if │x – a│< in

Choosing a number δ such that 0 < δ < in , we have

│f(x) – 2a│< in when whenever 0 < │x – a│< δ

Hence    lim f(x) = 2a

x → 0

Standard Deviation in Statistics

Statistics is a division of applied mathematics which contracts with the particular interpolation of data. The term ‘Statistics’ has been taken from the Latin name ‘Status’ in which it defines ‘political state’. This statistics mainly used to measure the arithmetic mean, median, mode and standard deviation. This measurement gives idea about where the data points are centered. Let us discuss about standard deviation with some example problems.

Evaluation of Standard Deviation in statistics:

Mean:

In Statistics, Mean is defined as the average of the given total numbers, i.e., total number of data divided by the number of data set given.

Formula for finding mean,

barx   = (sum(x))/(n)

Standard Deviation:

In Statistics, Standard Deviation is the measure of describing squared mean difference variability and spread of the Data set in the given total numbers. It is used to take the measurement of taking square root and average of numbers in the Data set.

Formula for standard deviation,

S =  sqrt ((sum(x - barx)^2 )/ (n-1))

Here Standard Deviation is calculated by using the mean Value  barx

Example Problems to find standard deviation in statistics :

Problem 1:

Here are 4 measurements   66, 45, 67, 45, 34, 56, 78 and 57. Calculate statistics standard deviation for the given measurements

Sol:

Mean: Calculate the mean the using the formula,

barx   = (sum(x)) / n

barx   =( 66+45+67+45+34+56+78+57) / 8

= 448 / 8

barx   = 56

Standard Deviation,

S = sqrt((sum(x-barx)^2) / (n-1) )

S = sqrt((( 66-56)^2+(45-56)^2+(67-56)^2+(45-56)^2+(34-56)^2+(56-56)^2+(78-56)^2+(57-56)^2) / (8-1))

= sqrt((100+121+121+121+484+0+484+1)/7)

=  sqrt( 1432 / 7)

S = sqrt(204.571429)

Standard Deviation S =  14.3028469

Problem 2:

Find the Statistics Standard deviation of the given Data 16, 17, 18, 20 and 24

Sol:

Mean: Calculate the mean using the formula,

barx   = (sum(x)) / n

barx  = ( 16 + 17 + 18 + 20 + 24 ) / 5

barx = 95 / 5

barx = 19

X               x-barx              (x-barx )^2

6               16 - 19 = -3           9

7               17 - 19 = -2           4

8               18 - 19 = -1           1

9               20 - 19 =  1           1

10              21 - 19 =  2           4

Sum of the  (x-barx)^2

9+4+1+1+4 = 19

Standard Deviation: Calculate the standard deviation

S =   sqrt((sum(x- barx)^2 ) / (n-1))

S = sqrt( ( 9+4+1+1+4) / 4 )

S = sqrt (19 / 4 )

S = sqrt(4.75 )

S = 2.17944947

Practice Problems for Statistics Standard Deviation:

1. Find the statistics standard deviation for the following given data.37, 56, 54, 54, 26, 67, 12, 65 and 34.

Answer:                     Mean = 45

Standard Deviation = 18.714967272213

2. Calculate the statistics standard deviation for the following. 77, 56, 33, 87, 90, 23, 67, 80 and 99.

Answer:                    Mean = 68

Standard Deviation = 26.043233286211

Study Online Second Derivatives

Study online second derivatives involves the process differentiating the given polynomial function twice with respect to the given variables whereas all the process is clearly explained with the help of online. Generally the second derivative is discussed in calculus whereas it is mainly helps to find the rate of change of the given function with respect to the change in the input. The following are the solved example problems with detailed step by step solution for second derivatives study discussed in online.

Example 1:

Determine the second derivative from the polynomial.

f(b) = 5b 2 +5b 4  + 12

Solution:

The given function is

f(b) = 5b 2 +5b 4  + 12

The above function is differentiated with respect to b to find the first derivative

f '(b) = 5(2b  )+5(4 b 3 ) + 0

By solving above terms

f '(b) = 10b +20b3

The above function is again differentiated with respect to b to find second derivative

f ''(b) =  10(1 ) +20(3b2)

f ''(b) =  10 + 60b2 is the answer.

Example 2:

Determine the second derivative from the polynomial.

f(b) = 4b4 +5b 5 +6b 6  + 2b

Solution:

The given function is

f(b) = 4b4 +5b 5 +6b 6  + 2b

The above function is differentiated with respect to b to find the first derivative

f '(b) = 4(4b 3 )+5(5b 4 ) +6( 6b 5) +2

By solving above terms

f '(b) = 16b 3 +25b 4 +36 b 5 + 2

The above function is again differentiated with respect to b to find second derivative

f ''(b)= 16(3b 2) +25(4b 3) +36 (5b 4)

f ''(b)= 48b 2 +100b 3 +180b 4 is the answer.

Example 3:

Determine the second derivative from the polynomial.

f(b) = 2b6 + 2 b5 + 3 b4 + 3b

Solution:

The given equation is

f(b) = 2b6 + 2 b5 + 3 b4 + 3b

The above function is differentiated with respect to b to find the first derivative

f '(b) =  2(6b 5)  +2 (5 b4 ) +3(4 b3) + 3

By solving above terms

f '(b) =  12b 5  +  10b4  + 12 b3 – 3

The above function is again differentiated with respect to b to find second derivative

f ''(b) =  12(5b 4 ) – 10(4b3)  + 12(3b2)

f ''(b) =  60b 4 – 40b3 +36b2  is the answer.

Online second derivatives practice problems for study:

1) Determine the second derivative from the polynomial.

f(b) = b 3 + b 4 + b 5

Answer: f ''(b) = 6b +12b2+ 20b 3

2) Determine the second derivative from the polynomial.

f(b) = 2b 3+3b5 + 4b 6

Answer: f ''(b) = 12b + 60b3 + 120 b 4

Solving Negative Number

Definition:

Negative number is defined as the number which indicate by negative sign or minus ('-') . Negative number is less then zero and placed left to zero.

Ex: ...-5,-4,-3,-2,-1,0,1,2,3,4...

Comparison of Positive and Negative:

For each negative number , there is a positive number that is its opposite . Here we can write the opposite of negative number with a positive of same number or plus sign used In front of the number and call these numbers are positive numbers. Ex : 1,2,3 ,.....positive numbers are grater than zero. Similarly, the opposite of any positive number is a negative number .

Ex:    1,2,3 is -1,-2,-3.

Solving examples for negative numbers:

Zero cannot be taken as a negative number or positive number.
For every positive number x, there exists a negative number y such that x + y = 0
Positive number is denoted as plus ('+')sign and negative number is denoted as minus sign('-').
Example of negative number:-2,-43,-34 and example for positive number is 2,43,34.
negative and positive number may be written as mixed numbers or fraction numbers.

The equal fraction of negative numbers are given bellow:

(-3)/7,3/(-7),-(3/7) and -3/7 .

The equal mixed numbers are given bellow.

-2/5,-(2/5)       (-4)/9, 4/-(9) and -4/9

More about solving negative numbers:

Solving Addition of Negative Numbers:

To add the negative numbers which consist of minus sign. To provide the answer of addition of negative number.

Solving examples for addition negative numbers:

-12+(-6)=?

Solution:

-12+(-6)= -18.

Here the values of  -12 and -6are 12 and 6 adding the smaller from the larger gives -12+(-6)= -18, and since the larger  value was 12, so we can give result the same sign as -12, '-' so -12+(-6)= -18.

Example:

(-6) + (-6) = ?

Here the absolute values of -6 and -6 are 6 and 6.  Adding the smaller from the larger gives -6 - 6 = 12 ,  but here both has same value . In this case sign is the matter, here 12 and -12 are the not same and then -6 and -6 are same numbers. The property of all same number sum is 12. The addition of two number up to zero are called as additive inverses.

Multiplying Negative Numbers:

Solving example for multiplying negative numbers :

Product of negative number ,here we can take the product of their values.

(-3.3) × (-5) = ?


(-3.3) × (-5) = (-3.3) × (-5)

= (-3.3) × (-5)

=  +16.5.
Dividing Negative Numbers

Solving example for dividing negative number:

To divide two negative numbers, here we can divide the value of the first by the value of the second.

(-1.6) ÷ (-4) = (-1.6) ÷ (-4)

= (-1.6) ÷ (-4)

=  -0.4

Geometry Without Common Vertices

In geometry, some figures have common vertices. Mostly the geometric figures are without common vertices. If a triangle, quadrilateral and some geometric figures in a graph are lying without common vertices are refered same. In graph there are four quadrants in that some vertices are fall on common vertices, with out common vertices points of are plotted.

Geometric figure without common vertices

Without common vertices find the distance of two points

In geometry the triangle has three vertices; the vertices are not common vertices. The common vertices are formed only if two triangles are in same point without three common vertices. The distance between two un common vertices are find out by using the coordinates of the vertices (x1,x2) and (x2,x2) of the vertices.

Distance between two vertices = √(x2-x1)2 +(y2-y1)2

If the geometry figure having the common vertices in a graph



Examples for without common vertices

Distance of a vertices are find using distance formula:

Examples for distance between two vertices:

Ex 1:   Find the distance formed by without common vertices, Vertices A(4,5), B(7,4)

Sol :            Distance AB = √(x2-x1)2 +(y2-y1)2

X1=4    X2=7    Y1=5    Y2=4

AB = √(7-4)2+(4-5)2

= √(3)2+ (-1)2

= √9+1

= √10 units

Ex 2 :  Find the distance formed by without common vertices, Vertices A(3,2), B(5,4)

Sol :            Distance AB = √(x2-x1)2 +(y2-y1)2

X1=3    X2=5    Y1=2    Y2=4

AB = √(5-3)2+(4-2)2

= √(2)2+ (2)2

= √4+4

= √8 units

Ex 3 :      Find the distance formed by without common vertices, Vertices A(6,4), B(10,8)

Sol :            Distance AB = √(x2-x1)2 +(y2-y1)2

X1=6    X2=10    Y1=4    Y2=8

AB = √(10-6)2+(8-4)2

= √(4)2+ (4)2

= √16+16

= √32 units

Ex 4:    Find the distance formed by without common vertices, Vertices A(3,3), B(4,8)

Sol :            Distance AB = √(x2-x1)2 +(y2-y1)2

X1=3    X2=4    Y1=3    Y2=8

AB = √(4-3)2+(8-3)2

= √(1)2+ (5)2

= √1+25

= √26 units

Practice problems:

Q 1   Find the distance of two vertices A(1,1) B(1,2)   Answer: √1 units

Q 2   Find the distance of two vertices A(2,2) B(1,1)   Answer: √2 units