Exponential Scientific Notation

This article is showing the exponential scientific notation. Scientific notation is nothing but transform very big numbers into simplest form of a number. That is called as scientific notation. Scientific notation is nothing but exponential of the base integer 10 have expressed in the number scientific notation. Example for scientific notation is 1860000000 this can be note down as 1.86 x 109. And the one more method in decimal is 0.0000081 = 8.1 x 10-6

Steps to find exponential scientific notation:

The steps to find exponential scientific notation are given below that:

The scientific notation has one number to the left of the decimal place.

We have to set the decimal point following the initial digit.

How many digits the decimal point was moved point out during the exponential of ten (10).


Example problems based on exponential scientific notation:

The example problems based on exponential scientific notation is given below that,

Example 1:

Change 82000 in exponential scientific notation.

Solution:

Step 1:

Specified that 82000.

Step 2:

Specified number is not a big number, but it is simple to convert scientific notation.

First note down 8.2 x 104 = 82000 [104 = 10000]

Step 3:

After that it can be write in scientific notation for 8.2 x 104.

Example 2:

Change 456000 in exponential scientific notation.

Solution:

Step 1:

Specified that 456000.

Step 2:

Specified number is a big number, but it is simple to convert scientific notation.

First note down 4.56 this is not equivalent to specified number.

Step 3:

Identify how many numbers available following the decimal point in previous step.

Present are 5 digits subsequent the decimal point.

4.56 × (100000)

Therefore the notation will be 105.

Step 4:

The scientific notation can be note down as 4.56 x 105

Example 3:

Change 987000 in exponential scientific notation.

Solution:

Step 1:

Specified that 987000.

Step 2:

Specified number is not a big number, but it is simple to convert scientific notation.

First note down the number as 9.87

Step 3:

There are 5 digit following the number is there so we can write 105.

Step 4:

The scientific notation can be note down as 9.87 x 105.
Practice problems based on exponential scientific notation:

The practice problems based on exponential scientific notation is given below that,

Problem 1:

Change 8960000 in exponential scientific notation.

Answer: The concluding answer for this problem is 8.96 x 106.

Problem 2:

Change 963000 in exponential scientific notation.

Answer: The concluding answer for this problem is 9.63 x 105.

Degree of Product Differentiation

The degree of the product differentiation represents the operations in the differentiation which deals with the product of the terms. For example if we have the function like xy then the degree of the function is 2 [x^1y^1 = ^2  ]. The homogeneous equation represents the degreee of the functions in the numerator and the denominator are same. In this article we are going to discuss about the degree of product differentiation for the students in detail with the homogeneous equations in different form with clear view.


Examples for the degree of product differentiation

Review on the degree of product differentiation for the differentiable function  dy/dx = [y^2-x^2]/[2xy]

Solution:

The function given for differentiation  is   dy/dx = [y^2-x^2]/[2xy]

Here it is present in the homogeneous form.

Since the numerator degree[ [y^2-x^2]] is two and the denominator [ [2x^1y^1]] degree is also two

Put y = vx and  dy/dx = v + xdv/dx in the equation given we get,

=>  v + x[dv]/dx   =    [(vx)^2-x^2]/[2x(vx)]

=>  v + x[dv]/dx   =    [v^2x^2-x^2]/[2vx^2]

=>  v + x[dv]/dx   =    x^2[v^2-1]/[2vx^2]

=>  v + x[dv]/dx   =    [v^2-1]/[2v]

=>   x[dv]/dx   =    [v^2-1]/[2v] - v

=>   x[dv]/dx   =    [v^2-1-2v^2]/[2v]

=>   x[dv]/dx   =    [-1-v^2]/[2v]

=>  [2v]/[-[1+v^2]] dv  =  dx/x

=>  [2v]/[1+v^2] dv  =  -1/x dx

Apply integration on both sides, we get

=>  int   [2v]/[1+v^2] dv  = int  -1/x dx

=>  int   [2v]/[1+v^2] dv  = -int  1/x dx

=>   log|1 + v^2|  =  - log |x| + log c

=>   log|1 + v^2|   + log |x| =  log c

=>   log|x[1 + v^2]|  =  log c

=>   x[1 + v^2]  =  +-   c

Put y = vx and v = y/x

=>   x[1 + [y/x]^2]  =  C_1

=>   x[1 + y^2/x^2]  =  C_1

=>   x[x^2 + y^2]/x^2  =  C_1

=>   [x^2 + y^2]/x  =  C_1

=>   x^2 + y^2  = xC_1           is the answer for the differentiation for the degree of product.

Problems for the degree of product differentiation

Review on the degree of product differentiation for the differentiable function dy/dx = [x^2y]/[x^3 + y^3 ]

Solution:

The answer is      [-x^3]/[3y^3] + log|y| =  C

Sectional Area of a Cylinder

A cylinder is one of the most basic curvilinear geometric shapes, the surface formed by the points at a fixed distance from a given straight line, the axis of the cylinder. The solid enclosed by this surface and by two planes perpendicular to the axis is also called a cylinder. The surface area and the volume of a cylinder have been known since deep antiquity. let us disuse about how to find the cross sectional area of cylinder. source-wikipedia


Cross section area of cylinder is nothing but the base area of cylinder. The base of cylinder is circle. So to find cross sectional area of cylinder, we use the area of circle formula.

Cross sectional area of cylinder = π r2 square unit.

r – Radius of circle

Example problems:

Ex:1 Find the cross sectional area of cylinder with radius 5cm.

Sol:

Given:

Radius (r) = 5 cm

Formula:

Cross sectional area of cylinder = π r2 square unit.

= π x 52

= 3.14 x 25

= 78.5

Cross sectional area of cylinder=78.5 cm2

Ex:2 Find the cross sectional area of cylinder with radius 3cm.

Sol:

Given:

Radius (r) = 3 cm

Formula:

Cross sectional area of cylinder = π r2 square unit.

= π x 32

= 3.14 x 9

= 28.26

Cross sectional area of cylinder= 28.26 cm2


Ex:3 Find the cross sectional area of cylinder with radius 5.5 m.

Sol:

Given:

Radius (r) = 5.5 m

Formula:

Cross sectional area of cylinder = π r2 square unit.

= π x 5.52

= 3.14 x 30.25

= 94.985

Cross sectional area of cylinder= 94.985 m2

I am planning to write more post on solve a system of linear equations and cbse books for class 9th. Keep checking my blog.


Ex:4 Find the cross sectional area of cylinder with radius 8cm.

Sol:

Given:

Radius (r) = 8 cm

Formula:

Cross sectional area of cylinder = π r2 square unit.

= π x 82

= 3.14 x 64

= 200.96

Cross sectional area of cylinder= 200.96 cm2

Sectional area of a cylinder- practice problem:

Find the cross sectional area of cylinder with radius 2.5cm.

Ans: cross section area=19.625 cm2

2.   Find the cross sectional area of cylinder with radius 3.8cm

Ans: cross section area = 45.3416 cm2

3.   Find the cross sectional area of cylinder with radius 7.7m

Ans: cross section area = 186.1706 m2

5 Properties of Parallelogram

Let us study about the 5 properties of parallelogram. Parallelogram is also a type of quadrilateral. Parallelogram is said to have two pairs of parallel sides.
Each pair has equal length. Also their diagonals are seemed to bisect each other in the center point.
Parallelograms have 5 important properties used in their construction point and evaluation point. All those 5 properties of parallelogram are discussed in detail with their examples below.




5 properties of parallelogram:

The 5 important properties of parallelogram are as follows:

a)      Parallelogram opposite sides are said to be parallel.

b)      Opposite sides are seemed to be congruent in parallelogram.

c)      Also parallelogram is said to have congruent opposite angles.

d)     Parallelograms two consecutive angles are supplementary angles.

e)      Parallelogram diagonals seemed to be bisecting each other right at their center point.

Parallelogram opposite sides are said to be parallel:

two pairs of parallel sides

This is the 1st property of parallelogram.
The above figure ABCD is seemed to be having two pairs of parallel sides as (AD, BC) and (DC, AB). Therefore it is a parallelogram.



Opposite sides are seemed to be congruent in parallelogram:

two pairs of congruent sides

This is the 2nd property of parallelogram.
The above figure MNOP is seemed to have the two pairs of congruent opposite sides as (MP, NO) and (PO, MN).



Parallelogram is said to have congruent opposite angles:

two pairs of congruent opposite angles

This is the 3rd property of parallelogram.
The above figure KLMN is seemed to have the two opposite pairs of congruent angles as (angle K, angle M) and (angle L, angle N).


Parallelograms two consecutive angles are supplementary angles:

two consecutive angles are supplementary

This is the 4th property of parallelogram.
The above figure ABCD is showing that all their two consecutive angles are supplementary angles as (A, B) or (A, C) or (C, D) or (B, D).




Parallelogram diagonals bisect each other right at their center point:

diagonals bisecting each other

This is the 5th property of parallelogram.
From the above figure ABCD we prove that the two diagonals of the parallelogram one which extends form A and other which extends from B will meet C and D respectively.
Also they bisect each other exactly at the center position of that parallelogram.

Mathematical Induction

The mathematical induction is one of the numerical proof naturally it is used to find the given statement is accurate for all normal numbers. It can be a deducted by proving that the 1st statement may be the countless succession of the statement is accurate and  we have to prove that if any other statement in the countless succession is real. In this article we shall discuss about the examples involved in induction proof.

Mathematical induction proof:

Principle of Mathematical Induction proof examples:

illustrate the  true for n = 1
suppose true for n = k
explain true for n = k + 1
Conclusion: Statement is true for all n >= 1


Examples in mathematical induction:

Examples in mathematical induction proof 1 :

Prove for n >= 1

1 * 1! + 2 * 2! + 3 * 3! + ….. + n * n! = (n+1)! – 1

This can be denoted by Sigma Notation

`sum_(k=1)^n` kk! = (n+ 1)! - 1

Proof:

Base case: n = 1

The left hand side is 1* 1!. The right hand side is 2! - 1. They are equal.


Examples in mathematical induction proof 2 :

Give the proof  when  amount of the n natural numbers by this formula

1+ 2 + 3 + … + n = n (n+ 1) / 2

Proof:

First, we have to denote the formula is real for n=k ,and it will be

1 + 2 + 3 + … + k = k (k + 1) / 2       (1)

This is the induction statement.  In order to prove  the method  is correct  means then

n = k + 1.

1+ 2 + 3 +… + (k + 1) = (k+1) (k + 2) / 2             (2)

Now add the ( k + 1) on both sides of the above equation in (2)

1 + 2 + 3 + … + k + (k + 1) = k (k + 1) / 2 + (k+ 1)

= k (k+ 1) + 2 (k+ 1) / 2

.

By adding  the  above  fractions we get ,

= (k + 1) (k+ 2) / 2

We can take  (k + 1) as a ordinary factor.

This is line (2), which is the first thing we need to explain

Next, we must explain that the formula is true for n = 1.  We have:

1 = ½· 1· 2

Thus the formula is true for n = 1. And  this fulfilled the both situation of the mathematical induction.

Percent in Math

The term percent in math is used to represent a value relative to hundred.  The same comparison can be expressed as a fraction if we compare a value with respect to another value. If the second value is 100, then we call it as percentage. For example:3/6 , Here we consider 3 relative to 6. If we consider relative to 100, it becomes 50 /100, we call this as 50 percent and is represented as 50 % .

If you need to convert a number into percentage, then multiply the number by 100 and put % symbol.

Now, Let us discuss about some example on percentages

Example problems to find percentage in math:

Percentage is mostly used in banking sector to find the interest.

The formula to find interest is,

Interest =Principle amount x time x rate of interest.



Example 1:

Suppose we need to find the interest will $2000 earn in one year at an annual interest rate of 5%

Solution:

5% = 5 /100

(5/100 ) x 2000 =100

So the answer is $100

Example 2:

Find the interest if amount if $1000 and month is 6 and rate of interest is 6%

Solution:

Interest =Amount x time x rate of interest.

Note : 6 month  = ½ of years.

Therefore,

6%  =>6/100  or 3/50

Substitute these values into interest formula. We get,

Interest = $1000 x ½ x 6/100

=$1000 x1/2  x 3/50

=$500  x 3/50

=$10 x 3

=$30

Example 3:

There are 45 boys  out of 90 students in the class room. Find the percentage of boys in the class room.

Solution:

Total students =90

Out of these 90 students 45 students are boys ,it is represented in terms of fraction.

= 45/90            [fraction form]

Convert the fraction into percentage,

So multiply the fraction number by 100,we get,

(45/90)  * 100

=50%

Example 4:

40 percent of X = 200 then find x?

Solution:

40 % * X =200

40 x */100  =200

40x  =200 *100

40x  = 20000

X =20000/40

X= 2000/4

X=500

Therefore, X value = 500

Verification:

40 percent of 500 =200

(40 x 500)/100  =200

20000/100   =200

200 =200

Hence Verified.


Practice problems to find percentage in math :

1)10 percent of p =100 find P?

Answer: 10

2) Find the interest will $5000 earn in one year at an annual interest rate of10%

Answer : 500

Combinatorial Learning Formula

Combination are selections ie. it involves only the selection of the required number of things out of the total number of things. Thus in combination order does not matter. For example, consider a set of three elements a,b,c and combination
made out of the set with


i) One at a time: a, b, c
ii) Two at a time: a,b, b,c, c,a
iii) Three at a time: a,b,c

The number of combinations of n things taken r, (r < n) is denoted by nCr or ([n],[r])

Learning the evaluation of combinations and the combinatorial formula

Learning to derive the formula for nCr:( Combinatorial formula)
No of combinations from ‘n’ things taken ‘r’ at a time = nCr
No of permutations from ‘n’ things taken ‘r’ at a time = nPr
Number of ways ‘r’ things that can be arranged among themselves = r!
Each combination having r things gives rise to r! permutations

Therefore  nPr = (nCr) r!


=> ((n - r )!) / (n!) = ( nCr) r!


nCr = r!(n - r )!

Learning observations related to combinatorial formula:

(i)   nCo = (n!)/(0! (n - 0 )!) = (n!)/(n!) = 1


(ii) nCn = (n!)/(n! (n -n)!) = (n!)/(n!0!) = 1


(iii) nCr = nCn-r


(iv) If nCx = nCy then x = y or x+y = n


(v) nCr =(^nPr)/(r!)

Learning examples on combinatorial formula

Learning to solve examples using the combinatorial formula :

Ex 1:Evaluate 8C3

Solution:

8C3  =  (8!)/(3! (8 - 3)!) = (8!)/(3! . 5!) = (8 . 7 . 6 . 5!)/((3 . 2 . 1) . 5!)  =  (8 . 7 . 6)/(3 . 2 . 1) = 8 . 7 = 48

OR another method

8C3 = (8 . 7 . 6)/(3 . 2 . 1) = 8 . 7 = 48

Ex 2:

Evaluate: 9C7

Solution:

9C7 = 9C2 = (9 . 8)/(2 . 1) = 9 . 4 = 36

Ex 3:    A committee of seven students is formed selecting from 6 boys and 5 girls such that majority are from boys. How many different committees can be formed?

Sol:

Number of students in the committee = 7
Number of boys = 6
Number of girls = 5
The selection can be done as follows


Boy (6)    Girl (5)
6              1
5              2
4              3


ie. (6B and 1G) or (5B and 2G) or (4B and 3G)

The possible ways are ([6],[6]) ([5],[1]) or ([6],[5]) ([5],[2 ]) or ([6],[4]) ([5],[3])

The total number of different committees formed

=  6C6  x  5c1 + 6C5  x  5C2  +  6C4  x  5C3

= 1 x 5 + 6 x 10 + 15 x 10 = 215