Property Math Definition

Introduction for property math definition:

Mathematics plays an important role in our everyday life.  There are some different properties present in mathematics such as associative property, commutative property, distributive property and identity property.  In this content “property math definition”, we are going to discuss these types of properties.  Let us discuss some formulas and problems for these properties. Having problem with Differential Equations Solver keep reading my upcoming posts, i will try to help you.

Formula for Law – Property Math Definition

Associative property:

U + (V + W) = (U + V) + W

U `xx` (V `xx` W) = (U `xx` V) `xx` W

The answer will become same at both sides

For example,

If u = 2, v = 3 and w = 4

2 + (3 + 4) = (2 + 3) + 4

2 + 7 = 5 + 4

9 = 9

Commutative property:

Swap the numbers.

The answer will become same at both sides

For example:

U + V = V + U

U `xx` V = V `xx` U

For example,

If u = 2 and v = 3

2 + 3 = 3 + 2

5 = 5

Distributive property:

Add some numbers then multiply with added result

Multiply each parenthesis then add the values

(U + V) `xx` W = (U `xx` W) + (V `xx` W)

The answer will become same at both sides

For example,

If the value of u = 2, v = 3 and w = 4

(2 + 3) `xx` 4 = (2 `xx` 4) + (3 `xx` 4)

5 `xx` 4 = 8 + 12

20 = 20

Identity property:

If any number multiplied with the value one, the answer will become the same number

`1 xx U = U`

`U xx 1 = U`

For example,

The value of u is 2

`2 xx 1 = 2`

`1 xx 2 = 2`

Example Problems – Property Math Definition

Example problem 1 – Property math definition

Prove the commutative property for the value of u is 5 and v is 6.

Solution:

Given,

The value of u is 5

The value of v is 6

Formula of commutative law is X + Y = Y + X and X `xx` Y = Y `xx` X

X + Y = Y + X

5 + 6 = 6 + 5

11 = 11

X `xx` Y = Y `xx` X

5 `xx` 6 = 6 `xx` 5

30 = 30

Hence we proved the commutative property

Example problem 2 – Property math definition

Prove the associative property for the value of u is 3, v is 6 and w is 9.

Solution:

The value of u is 3

The value of v is 6

The value of w is 9

Formula for associative property is u + (v + w) = (u + v) + w and u `xx` (v `xx` w) = (u `xx` v) `xx` w

u + (v + w) = (u + v) + w

3 + (6 + 9) = (3 + 6) + 9

3 + 15 = 9 + 9

18 = 18

u `xx` (v `xx` w) = (u `xx` v) `xx` w

3 `xx` (6 `xx` 9) = (3 `xx` 6) `xx` 9

3 `xx` 54 = 18 `xx` 9

162 = 162

Hence we proved the associative property. Please express your views of this topic help in math homework by commenting on blog.

Practicing Problems – Property Math Definition

Practicing problem 1 – Property math definition

Prove the distributive property for the value of p is 11, q is 12 and r is 10.

Answer:  230

Practicing problem 2 – Property math definition

Prove the commutative property for the value of m is 20 and n is 15.

Answer:

m + n = n + m = 35

m `xx` n = n `xx` m = 300

Counter Example Math

Introduction for counter example math:

One of the numbers 1, 2, 3, , also termed the counting numbers or natural numbers. 0 is involved in the list of "whole" numbers, but there is no general agreement. Some authors also infer "whole number" to mean "a number are consisting fractional part of zero," making the whole numbers equivalent to the integers. A positive integer: 1, 2, 3, 4, ... also described a natural number. On the other hand, zero (0) is occasionally also included in the list of counting numbers. In this article, we are going to see the counter example in math.

Types and Description for Counter Example Math:

Due to lack of regular lingo, the following conditions are suggested in preference to "counting number," "natural number," and "whole number.

Integers are ..., -2,-1, 0, 1, 2, ... which is symbolized as Z

Positive integers are 1, 2, 3, 4, ... which is symbolized as Z-+

Nonnegative integers are 0, 1, 2, 3, 4, ... which is symbolized as Z-*

Non positive integers are 0,-1,-2,-3,-4, ..

Negative integers are -1,-2,-3,-4, ... which is symbolized Z—

Counting by Twos for counter example math:

You can count by twos by either:

• Adding 2 to the previous number or

• Counting and skipping every other number.

The numbers that will be counted by twos from 0 is as follows:

0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30 and so on.

The numbers that will be counted by twos from 1 is as follows:

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31 and so on.

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Counting by Threes for counter example math:

You can add up by three's by adding three to the earlier number.

The numbers that will be counted by threes from 0 is as follows:

0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33 and so on.

The numbers that will be counted by threes from 1 is as follows:

1, 4, 7, 10, 13, 16, 19, 22, 25, 28, 31, 34 and so on.

Counting by fours for counter example math:

You can add up by fours by adding three to the earlier number.

The numbers that will be counted by threes from 0 is as follows:

0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48 and so on

The numbers that will be counted by fours from 1 is as follows:

1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 45, 49 and so on

Distributive Efficiency

Introduction for distributive efficiency:
In mathematics, there are some different types of law such as associative law, distributive law, and commutative law.  In this content, we are going to see about the distributive law with some problems and solution.  Multiply the numbers separately and then sum of the multiplication value is called distributive law.  Example, p (q + r) = pq + pr. I like to share this Median Number with you all through my article.

Example Problems – Distributive Efficiency
Example problem 1 – Distributive efficiency

Evaluate 3y (3y – 2) using the distributive law.

Solution:

The given values are 3y (3y – 2)

Formula for distributive law:

a (b + c) = ab + bc

We are going to evaluate the 3y (3y – 2)

Multiply the value 3y within the parenthesis value

3y (3y – 2) = 9y2 – 6y

We get the answer 9y2 – 6y

Answer:  9y2 – 6y

Example problem 2 – Distributive efficiency

Evaluate 2a (2b - 2) and the value of a is b – 1 using the distributive law.

Solution:

The given values are 2a(2b - 2)

The value of a is b – 1

Formula for distributive law:

a (b + c) = ab + bc

We are going to evaluate the 2a (2b - 2)

Substitute the a value in the given 2a (2b - 2)

2a = 2(b - 1)

= 2b – 2

2a (2y - 2) = 2b -2 (2b - 2)

Multiply 2b -2 within the parenthesis values

= 4b2 – 4b – 4b + 4

= 4b2 – 8b + 4

We get the answer 4b2 – 8b + 4

Answer:  4b2 – 8b + 4

Example problem 3 – Distributive efficiency

Evaluate 4(3 + 4) using the distributive law.

Solution:

Given, 4(3 + 4)

Formula:  a(b + c) = ab + ac

4(3 + 4) = 4 `xx` 3 + 4 `xx` 4

= 12 + 16

= 28

Answer:  28


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Practicing Problems – Distributive Efficiency

Practicing problem 1 – Distributive efficiency

Evaluate 2m (4n - 4) and the value of m is n – 3 using the distributive law.

Answer:  8n2 -32n + 12

Practicing problem 2 – Distributive efficiency

Evaluate 6(4 + 4) using the distributive law.

Answer:  48

Units for Distance

Introduction on units for distance:

The distance is the measure of how far two objects are apart. There are different units for the measurement of the distance. The S.I unit for the measurement of the distance is the meter. The other units for measuring distance are feet, furlong, kilometer, miles etc. This article gives a clear idea of the various units of distance measurement and their relations between them. Having problem with Hypothesis Testing Steps keep reading my upcoming posts, i will try to help you.

Units for Distance:

The various units for distance measurement are related with each other and are commonly expressed using the international unit of the distance measurement the meter. The following are the various units for the distance measurement and the relation between each unit with the other units of measurement,

1 meter = 100 cm = 3.28 feet

1 furlong = 201 meters = 10 chain

1 mile = 1609.3 meters = 80.05 chains

1 mile = 1.609 km = 8 furlong

1 mile = 80.05 chains

1 kilometer = 1000 meters = 49.75 chains = 4.97 furlong.

1 chain = 20.1 meters = 65.93 feet

1 nautical mile = 1852 meters = 1.15 miles.

Here except the nautical mile all the other is used for the distance measurement on land while it is used for the distance measurement in sea. Please express your views of this topic linear equations in two variables formula by commenting on blog.

Example Problems on Units for Distance:

1. The distance between two places is measured as 6.45 miles. Find the distance in kilometers and furlongs.

Solution:

1 mile = 1.609 km

6.45 mile = 6.45*1.609 kilometers

6.45 mile = 10.38 km

1 mile = 8 furlong

6.45 mile = 6.45*8 furlong

6.45 mile = 51.6 furlong

2. Convert the distance of 8.75 kilometers into chains and miles.

Solution:

1 kilometer = 49.75 chains

8.75 kilometer = 8.75*49.75 chains

8.75 kilometer = 435.3 chains

1 mile = 1.609 km

8.75 kilometer = `8.75/1.609` miles

8.75 kilometer = 5.44 miles

3. Convert the distance of 5.2 nautical miles into kilometers.

Solution:

1 nautical mile = 1.852 kilometers

5.2 nautical miles = 5.2*1.852 kilometers

5.2 nautical miles = 9.63 kilometers

Practice problems on units for distance:

1. Convert the distance of 12.5 furlongs into miles and chains.

Answer: 1.56 miles and 125 chains.

2. Convert 5.6 miles into meters and chains.

Answer: 9012 meters and 448.3 chains.

3. Convert the distance of 82 chains into miles and kilometers.

Answer: 1.02 miles and 1.65 kilometers.

Simple Solution Math Pages

Introduction to simple solution math pages:

Mathematics is the process of learning of the measurement, properties, and communication of quantities and sets, using symbols and numbers. Arithmetic is used in our daily life. Mathematics includes various topics such as algebra, trigonometric, geometry, calculus and basic types of Addition, Subtraction, Multiplication and division In this article we are going to see about explanation for simple solution math pages.

Sample Problem for Simple Solution Math Pages:

Simple solution math pages problem 1:

Adding the given numbers 444 and 555

Solution:

Step 1:

Assemble the addition number of variety in vertical

444

555
__________

Step 2:

First we are starting add the addition number in the right hand side

Adding the numbers (4 + 5 = 9) place the value 9 in unit’s place

Adding the numbers (4 + 5 = 9) place the value 9 in ten’s place

Adding the numbers (4 + 5 = 9) place the value 9 in hundred’s place

Therefore

444

555

___________

999

___________

Then the answer of the given number 444 and 555 is 999.

Simple solution math pages problem 2:

Solve the M in the given expression

M/20= 3

Solution:

M = 20* 3 = 60



Simple solution math pages problem 3:

Which sign belongs in the blanks?

0.51 _____ 66/100

a)      =

b)      >

c)       <

Solution:

66/100= 0.66, comparing those results

0.51< 0.0.66

Answer: <

Simple solution math pages problem 4:

Solve the ascending order 3200, 1200, 900, 4100, 600,1700,3900.

Answer: 600, 900,1200,1700,3200,3900,4100

Simple solution math pages problem 5:

Solve the given equations and get the x value a - 2 = -1

Solution:

Find out the A value of the given linear equation.

We are move the -2 into the right side, we get

A = -1+2

A = 1.

The A value is 1.

Simple solution math pages problem 6:

Solve the following algebraic variables

2m + 8 = 6

Solution:

First we have to add both sides -8 we get

2m+8-8 = 6-8

In left hand side 8-8 is cancelling reaming we get

2m =-2

Divide both sides 2 we get

m = -1

Therefore the value of algebraic variables are m = -1

Practice Problems for Simple Solution Math Pages:

1. Adding the given numbers 622 and 666

Answer: 1288

2. Solve the M in the given expression P/40=3

Answer: 120

3. Solve the ascending order 320, 120, 90, 410, 60, 170, 390.

Answer: 60, 90, 120, 170,320,390,410.

Range of Data

This page is based on range of the data. Let's understand first what is a range? Definition of Range: Interval (mathematics), also defined as a range, the range between smallest and highest. E.g. the range on 21 22 23 and 24 is 3. Range (mathematics), the set of all output values given by a function. Range (statistics), the variation between the highest and the lowest values in a set. Having problem with Learn Statistics keep reading my upcoming posts, i will try to help you.

What is Range?

In descriptive statistics, the range is calculated as the length of the least interval observation which contains all the data. It is calculated by subtracting the least observation from the maximum observation  and provides an show  of statistical dispersion.
In the given set of data ,all observations are measured in only one  units. It is a poor and weak measure of dispersion because range is entirly based on only two observation except when the sample size is large. (Example: e, d, f= range=d-e)
For a population, the range is greater than or equal to twice the standard deviation, with equality only for the coin toss (Bernoulli distribution with p=1/2).
The range, in the logic of the variance between the maximum and minimum intervals observation, is also called the crude range. When a new  scale for dimension is developed, then a potential highest or lowest will obtain from this scale. This is called the potential  range.If we cannot choose the  range too small, since we have to avoid a ceiling effect When the resultant measurement is got, it will give the observed (crude) range which may be greater or smaller. Please express your views of this topic how to multiply mixed fractions by commenting on blog.

Steps to Find Range

Interval (also known as range): The range is the distinction between the maximum and minimum values in the setof the data

Steps to find the Range:

Step1:  First order the data from smallest to greatest (ascending order).

Step2:  Subtract the smallest value from the largest value in the set.

Example 1:

The Jaeger family drove through 5 Midwestern states on their summer vacation. Gasoline prices varied from state to state. What is the range of gasoline prices?

$3, $5,$ 7, $9, $8

Solution:

Step 1: writing in ascending order: 3,5, 7,8, $9.

Step 2:  Range = 9−3= $6

Range is $6

Discrete Math Examples

Introduction of discrete math:

Discrete math examples is one of the major topics in Mathematics. Discrete math topic defines the mathematics structures that is primarily discrete relatively then continuous. Discrete math dealing with objects that can see only distinct split values.

Discrete math define name Finite is few time applied to this discrete math activities because this discrete math deals with finite sets those area relevant to the business. Discrete math is subdivision of mathematics that deals Countable sets. Countable set describe that have same cardinality as subsets of integers that comprises rational numbers but not a real numbers

Discreate Math Application:

Presently researches in discrete mathematics were increased in later half of twentieth century. The main purpose of study in discrete mathematics is discrete objects and analytic methods from continuous mathematics.

Discrete math have Theoretical computer science, Information theory, Logic, Set theory, Combinatory, Graph theory, probability theory, Discretization, Discrete analogues of continuous mathematics and so on.

Some of the main discrete math activities are given in below

Set theory:

Set theory is division of discrete mathematics. These commonly define the sets. Sets mean group of objects. The important development of set theory is infinite sets in outside the scope of discrete mathematics

Graph theory:

Graph theory is part of discrete mathematics. In this Graph theory mostly focus on graph and network. Graph theory is extensive in all the areas of mathematics and science. Understanding 6th grade math problems and answers is always challenging for me but thanks to all math help websites to help me out.

Algebra:

Discrete mathematics discrete algebra defines the

1] Boolean algebra

The Boolean algebra commonly used in logic gates and programs

2] Relational algebra

In relational algebra commonly used in database side.

3] Algebraic coding theory

4] Discrete Semi group

Discrete modeling:

Discrete modeling is some time called as discrete analogue of continuous modeling.

Case for Discrete modeling is recurrence relation.

Recurrence relationship formula is

`F(n)-f(n-1)=g(n)`

This above formula is discrete analogue of first order ordinary differential equation.

Examples for Discrete Math:

Consider the given discrete math examples,

Examples 1:

Problem:

To solve the all `n>10 n-2< (n^2-n)/12`

Solution:

Default property of discrete mathematics is

p (n) is `P(n)` : `k<=n(k>10 ; n>10)k-2<(k^2-k)/12`

Given Base case `n=17`

`17-2< ((17*17)-17)/12`

=`15<272/12`

`=15<22`

Using discrete mathematics property to solve the above examples.

Example 2:

To solve the all `n>25 ` `n-2< (n^2-n)/12`

Solution:

Default property of discrete mathematics is

p (n) is` P(n) :` `k<=n(k>10`` ; n>10)k-2<(k^2-k)/12`

Given Base case `n=26`

`26-2< ((26*26)-26)/12`

`=24<650/12`

`=15<54`

Using discrete mathematics property to solve the above examples.

Solving Exponential Power Function

Introduction of solving exponential power function:-
Before solving the exponential power function, students does study about the exponential function definition, formulas and also solve the example problems. In mathematics exponential means ex, where e is the significance of ex the same value again consequent.
For example,
`4e^x` is an exponential function
It is more helpful for exam preparation and also improve our own practice skills.

Basic Properties of Solving Exponential Power Function:-

In the following basic properties of solving exponential power function

`e^x e^y = e^(x+y)`
`(e^x)^p = e^(px)`
`(de)^x/dx = e^x`
`(de^ax)/dx = ae^ax`
`d^n e^ax/dx^n = a^n e^(ax)`
`e^x/e^y = e^(x-y)`

Example Problems for Solving Exponential Power Function:-

Problem 1:-

Solving multiplying the exponential expressions `(-2mn^7 p^2)^4`

Solution:

Given: `(-2mn^7 p^2)^4`

Take all the common terms

= `(-2)^4 m^4 (n^7)^4 (p^2)^4`

= `16 m^4 n^28 p^8`

Multiply the both left side values and get 16 then multiply the power values.

Finally we get an answer as `(-2mn^7 p^2)^4 = 16 m^4 n^28 p^8`


Problem 2:-

Solving add the exponential expressions `x^(3) +x^(4)`

Solution:

Given: `x^(3) +x^(4)`

Take the common term x.

= `x^(3+4)`

= `x^(7)`

Adding the both values and get 7.

Finally we get an answer as `x^(3) +x^(4) = x^(7)`


Problem 3:-

Find an exponential power function `f(x) = x^2 x^5`

Solution:-

Given: `f(x) = x^2 x^5`

We know the formula `e^a e^b = e^(a+b)`

`f(x) = x^2 x^5`

`= x^(2+5)`

`f(x) = x^7`

We get a answer as `f(x) = x^2 x^5`

`= x^7`

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Problem 4:-

Check whether the point (0, 1) lies on the graph of the function y = 4(7)x.

Solution:-

Substitute x = 0 in the function y = 4(7)x.

y = 4(7)0

= 4(1)

= 4

The y–coordinate of the point is 1, which does not match with the obtained value y = 4.

So, the graph of the function y = 4(7)x does not contain the point (0, 1).

Vector Practice Problems

Introduction for vector practice problems:

Vector space model is nothing but an algebraic model, which is normally used for representing text documents as vectors of identifiers. It is used in several purposes such as information filtering, indexing etc. Vectors plays a significant role in the field of mathematics. In this article vector practice problems, we are going to discuss few example and practice problems based on vectors.

Example Problems for 'vector Practice Problems':

Example problem 1:

If the position vectors of A and B are `3 veci - 7 vecj - 7 veck` and `5 veci + 4 vecj + 3 veck` , find `vec(AB)` and determine its magnitude and direction cosines.

Solution:

Let O be the origin. Then

`vec(OA) ` = `3 veci` −` 7 vecj` − `7 veck` ,

`vec(OB)` =` 5 veci` +` 4 vecj` + `3 veck`

` vec(AB)` = `vec(OB)` − `vec(OA)` = (`5veci` + `4vecj ` + `3veck` ) - (`3veci` - `7vecj` - `7veck` )

After simplify this, we get

`vec(AB)` = `2veci` + `11vecj` +` 10veck`

` |vec(AB)|` = `sqrt[(2)^2 + (11)^2 + (10)^2]` = 15

The direction cosines are `2/15` , `11/15` , `10/15`

Example problem 2:

Find the magnitude and direction cosines of `2 veci` −` vecj` + `7 veck`

Solution:

Magnitude of `2veci - vecj + 7veck` = `|2veci - vecj + 7veck|`  = `sqrt[(2)^2 +(-1)^2 + (7)^2]`

After simplify this, we get

=` sqrt(4 + 1 + 49)` = `sqrt(54)` = `3sqrt(6)`

Direction cosines of `2veci - vecj + 7veck` are `2/(3sqrt(6))` , `-1/(3sqrt(6))` , `7/(3sqrt(6))`

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Practice Problems for 'vector Practice Problems':

Practice problem 1:

The position vectors of the vertices's A, B, C of a triangle ABC are respectively `2veci ` + `3 vecj` + `4 veck` , `- veci` + `2 vecj` `- veck` and `3 veci` −` 5 vecj` + `6veck`. Find the vectors determined by the sides and calculate the length of the sides.

Answer:` vec(AB)` = `-3veci` - `vecj` - `5veck` ; `vec(BC)` = `4veci` `- 7vecj` + `7veck` ; `vec(CA)` = `-veci ` + `8vecj` `- 2veck`

AB = √35, BC = √114, CA = √69

Practice problem 2:

If the vectors `veca` = `2 veci` − `3 vecj ` and` vecb ` = − `6vec i ` + `mvec j` are collinear, find the value of m.

Answer: m = 9

Practice problem 3:

If the vertices's of a triangle have position vectors `veci ` + `2 vecj` + `3 veck` , `2veci` + `3 vecj` +` veck` and` 3 veci` + `vecj ` + `2 veck` , find the position vector of its centroid.

Answer: Position vector = 2( `veci` + `vecj ` + `veck` )

Practice problem 4:

Examine whether the vectors` veci` + `3 vecj` + `veck` , `2 veci` − `vecj ` − `veck` and `7 vecj` +` 5 veck` are coplanar.

Answer: non-coplanar vectors

Practical Uses of Calculus

Introduction to practical uses of calculus:-

In this article we are going to discuss about the solve calculus problems step by step concept. The process of abbreviation the calculus comparable to over screening the important concept and problem in calculus are referred as review calculus. This article helps to improve the practical knowledge for calculus and below the problems are helping toll for the practical. Overtake the practical get review to this article. Calculus practical use solutions also show below. I like to share this Calculus Chain Rule with you all through my article.

Level One Example Practical Uses of Calculus Problems:-
Practical uses of calculus problem 1:-

Integrate the following equation

`int (2x^4)+(10x^5)+(28x^3) dx`

Solution:-

Step 1:-

`int (2x^4)+(10x^5)+(28x^3) dx`

Step 2:-

`= int(2x^4)dx + int(10x^5)dx + int (28 x^3) dx`

Integrating the above equation

We get

Step 3:-

`= (2x^5)/(5) + (10x^6)/(6) + (28x^4)/(4)`

Step 4:-

`= (2x^5)/ (5) + (10x^6)/ (6) + (28x^4)/ (4)`

Step 5:-

`= (2x^5)/ (5) + (10x^6) + (7x^4)`

Practical uses of calculus problem 2:-

Solve by differentiating the following equation and get the first derivative second derivative and third derivative

`y = (3x^3)+(2x^2)+(2x^5)+(5x)`

Solution:-

Given equation is` y = (2x^3) +(2x^2) + (2x^5) + (5x)`

Step 1:-

To get the 1st derivative differentiate the above given equation

`(dy)/(dx) = (6x^2)+(4x)+(10x)+(5)`

Step 2:-

To get the 2nd derivative differentiate the ist derivative

`(d^2y)/(dx^2)= 12x+14`

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Second One Example Practical Uses of Calculus Problems:-

Practical uses of calculus problem 1:-

Integrat the following equation

`int 2x^4+8x^5+16x^3 dx`

Solution:-

`int (2x^4)+(8x^5)+(16x^3) dx`

Step 1:-

`= int (2x^4) dx + int(8x^5) dx + int (16 x^3) dx`

Integrating the above equation

We get

Step 2:-

`= (2x^5)/(5) + 8 (x^6)/(6) + (16x^4)/(4)`

Step 3:-

`= (2x^5)/(5) + (8x^6)/(6)+ (4x^4)`

Practical uses of calculus problem 2:-

Differentiate the following equation and get the 1st, 2nd and 3rd derived

`y = 3x^3+2x^2+x^1 + 10`

Solution:-

Step 1:-

By differentiate the given equation with respect to x to get the 1st derivative

`y= dy/dx =9x^2+4x^1+1.`

Step 2:-

To get the 2nd derived differentiate the 1st derivative of the given equation.

`y =(d^2x)/(dy^2)= 18x + 4`

Step 3:-

To get the 3rd derived differentiate the 2nd derivative of the given equation.

`y=(d^3x)/(dy^3) = 18.`

Divisibility Rules

In Math Divisibility Rules is most commonly used to determine solutions in fastest way, by using few tricks and by remembering the divisibility rules.

A number is divisible by 2, if it is even or let us say it ends in 0, 2, 4, 6 or 8. Let us take an example, look at the number 512 and the number 431, here we are only concerned about the last digit or the digit in the ones place value. 512 ends in 2 which means it is even and therefore it is divisible by 2. However 431 ends in 1, which means it is not even or that is odd and it is not divisible by 2.

A number is divisible by 4, if their last two digits are divisible by 4. We take a look at 624 and 621, here we are only concerned with the last two digits, here the 2 and the 4 and in the next one it is 2 and the 1. Well 24 is divisible by 4. Therefore so is 624 is divisible by 4. But for our second number 621 here 21 the last two digit is not divisible by 4 so 621 is not divisible by 4. I have recently faced lot of problem while learning long division with decimals, But thank to online resources of math which helped me to learn myself easily on net.

Next we will look at the number 7, let us find out in few seconds if number is divisible by 7 or not? So divisibility rules 7 just indicates that number is divisible by 7 or not. First let us check the normal divisibility of a number. Let us have 6891, now to find out this number is divisible by 7 or not, we would start doing the long division and divide by 7. So 7 times 9 gives us 63, we get 5 as remainder with 9 as a carry. Again 7 times 8 is 56 so subtracting results into 3. 3 as a remainder carry 1 and 7 times 4 is 28 so the remainder is again 3. So here now we have 3 as a remainder this number 6891 is not divisible by 7.this was the long method of checking the divisibility rule. But there are few quicker rules as well. Our first step is to multiply the last digit by 2. Then subtract the remaining number from 2. Then, repeat the same process from the start till we get only two digits remaining. Then the last step would be check if the last two digits get divisible by 7.
Teaching divisibility rules is very easy because if the multiplication tables are memorized fully, then it becomes lot more simple to solve any question.