Plotting Algebraic Equations

Introduction to Plotting Algebraic equations:

In Algebra, We plot the equations by using the ordered pairs that satisfy the given condition. All the ordered pairs are plotted on the 2 dimensional graph and joining those points result in the graph of the given equation. We find the ordered pairs that satisfy the given condition by taking some random  values for x or y and find the other variable.Here the plotting of algebraic equations is done in this article.

The standard algebraic linear equation will be of the form of Ax + By=C.

Example for Plotting Algebraic Linear Equation:

Let the equation be 3x+2y=6

First we find the ordered pairs that satisfy the given condition.

We take some random values for x and find the value of y using the given equation.

In this example, let us take x values as -2,-1,0,1 and 2.

To find the value of y for each value of x, we substitute the value of x in the equation and solving for y.

For x=-2,

3(-2) + 2y=6

-6 + 2y =6

2y=6+6

2y=12

Y=12/2

`=>`  Y=6

Therefore the ordered pair is (-2,6).

Similarly we find all the values of y for different values of x.

After finding the ordered pairs we plot the ordered pairs (-2,6), (-1,4.5), (0,3), (1,1.5), (2,0) on 2 dimensional graph.

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Example of Plotting Algebraic Equations :

We plot all the ordered pairs and plot in 2 dimensional graph sheet.

We get the following graph of the algebraic equation 3x+2y=6

Similarly we can also graph the quadratic, rational, exponential, logarithmic, trigonometric, polynomial equations in a 2 dimensional graph. We can follow the same procedure as the above to graph any equation having 2 variables. Algebraic equations having more than 2 variables cannot be plotted in a 2 dimensional graph sheet. Only the equation having less than 2 variables can be plotted in a 2 dimensional graph sheet.

Dividing Polynomials with Exponents

Introduction to dividing polynomials with exponents:

Dividing polynomials with exponents is nothing but if the polynomials are having the degree we have to expand the polynomials and we have to divide the polynomials. Normally polynomials mean we are having power more than one. We will see some example problems for dividing polynomials. It is better to understand the division of polynomials with exponents. We can use the basic algebraic formulas and using the factorization method we can divide the polynomials.I like to share this Divide Exponents with you all through my article.

Examples for Dividing Polynomials with Exponents:

Example 1: Divide the following polynomials: `((x^2 + 3x + 2)^2) / ((x^2 - 4)^2)`

Solution:

The given polynomials are `((x^2 + 3x + 2)^2) / ((x^2 - 4)^2)`

Here the exponents of the polynomials are 2.

First we have to find the factors of the polynomials of the numerator.

x2 + 3x + 2 = x2 + 2x  + x + 2 = x (x + 2) + (x + 2) = (x + 2) (x + 1)

Next we have to find the factors of the denominators.

x2 – 4 = x2 – 22 = (x +2) (x – 2)

Now we have to plug the values.

=   ` (((x + 2) (x + 1))^2) / (((x + 2) (x - 2))^2)`

=  `((x + 2)^2 (x + 1)^2) / ((x + 2)^2 (x - 2)^2)`

From this we get `(x + 1)^2 / (x - 2)^2`

Example 2: Divide the following polynomials: `((x^2 - 5x + 4)^2) / (x^2 - 1)`

Solution:

The given polynomials are `((x^2 - 5x + 4)^2) / (x^2 - 1)`

Here the exponents of the polynomials in numerator are 2 and the denominator exponents is 1.

First we have to find the factors of the polynomials of the numerator.

x2 – 5x + 4 = x2 - 4x  - x + 4 = x (x - 4) - (x - 4) = (x - 4) (x - 1)

Next we have to find the factors of the denominators.

x2 – 1 = x2 – 12 = (x +1) (x - 1)

Now we have to plug the values.

` = (((x - 4) (x - 1))^2) / ((x +1) (x - 1))`

`= ((x - 4)^2 (x - 1)^2) / ((x + 1) (x - 1))`

From this we get `((x - 4)^2(x - 1)) / (x + 1)^2`

These are some of the examples for dividing polynomials with exponents. First we have to expand the polynomials and we have to divide.

Probability for Math

Introduction to Probability for math:

Probability is a study of math. Probability is number of outcomes divided into total number of events. The probability is also the expected value of the math. These are the two kinds of distribution are used to discrete and continuous distribution.  The general format of the formula for probability math is as follows,

The probability of Event P (A) = `("Number of outcomes n(a)")/("Total Number of events n(s)")`

Probability for Math – Examples:

Probability for math Example 1:

Roll a single dice; find the probability of get number 1.

Solution:

Total Number of possible = n (a) = {1, 2, 3, 4, 5, 6}

n (s) = 6

The number of outcomes n (a) = {6}

n (a) = 1

The probability of getting value = `1/6.`

Probability for math Example 2:

Toss four coins and find the probability of all tails. The possible outcomes are:

Solution:

Step 1:

n (s) = {TTTT, TTTH, TTHT, TTHH, HTTT, HTTH,THHT, HTHH, THTT, TTHH, THHT, HTHH, HTHT, HTHH,HHHT, HHHH }=16

Step 2:

There are 4 tosses with only two tails:

n (a) = { TTTT}=1

Step 3:

Formula:

P (A) = n(a)/n(s)

Answer:

P (A) = `1/16` .

Convert into a decimal 0.0625

The probability of two tails is 0.0625 or Rounded 6%.

Probability for math – Example 3:

The odds in possible of outcomes of an event 9:12.  Find the probability of the outcomes of this event

Solution:

Number of possible outcomes = 9.

Number of non-possible outcomes = 12.

Total number of outcomes = (9+12) = 21.

The probability of Event P (E) = `("Number of outcomes n (a)")/ ("Total Number of events n(s)")`

= `9 / 21.`

Probability for math – Example 4:

A coin is tossed. If the coin ground on top of a container is filled with one black globe and three white spheres. If the coin landed on tails the container is filled with one black sphere and nine white spheres. A sphere is then selected from the container. What is the probability that the sphere selected is black?

Solution:

Let H = Heads, T = Tails and B = Black sphere selected. Then by the law of total probability

P(B)   = P(B|H)P(H) + P(B|T)P(T)

= (0.25)(0.5) + (0.1)(0.5)

P (B)  = 0.175

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Probability for Math - Practice Problems:

Probability for math - Practice Problem 1::

If word chosen at random from a writer X has a character missing with probability 0.1, has a character inserted with probability 0.2, and contains both kinds of error with probability 0.05. What are the probabilities that a letters chosen at random.

Answer:

P (missing or inserted, but not both) = 0.25 - 0.05 = 0.2

Ratio Math Problems Tutoring

Introduction to ratio math problems tutoring:

The term ratio math problems tutoring in math the term ratio is used to define the sum or total of the given term. Ratio is processes that are used to differentiate the related or differentiate the given items in the question.  A ratio is represented by two numbers separated by a colon. The term tutoring is the process defines the way of teaching done in maths.

Representation of Ratio:

A ratio that represents two numbers separated by a colon

Colon is represented as       :    

Examples Based on Ratio Math Problems Tutoring:

Example 1:   Problems based on ratio math problems tutoring

A special bag mixture contains rice, wheat and corn in the ratio of 2:3:5. If a bag of the mixture contains 3 pound of rice, how much corn does the bag contain?

Solution:

Given:  The cereal mixture of `"rice, wheat, corn"` in `2:3:5`

Now, we are going to assign the variable:

Let `x="mixture of corn"`

The item is now found to be written in ratio fraction,

`"Rice"/"corn"=2/5=3/x`

Cross multiplication is to be done.

So,     `2*x=3*5`

`2x=15`

`x=15/2`

`x=7.5`

Hence the ratio mixture of `7.5` pound of corn is present in bag.

Example 2:   Problems based ratio math problems tutoring

In a readymade shop they are T shirts and formal shirts, the ratio of T shirts and shirts are found   is 3:6 .If the shop contains 60 formal shirts, Find the total ratio of T shirts present in the readymade shop?

Solution:

Given:  The shop contains `60" formal shirts"` ,

Now, we are going to assign the variable:

Let `x= "T shirts"`

The item is now found to be written in ratio fraction,

`"T shirt"/ "Formal shirt"= 3/6=x/60`

Cross multiplication is to be done.

So,     `3*60=6*x`

`180=6x`

`x=180/6`

`x=30`

Hence the total ratio of `30` `T shirts` is present in the readymade shop.

Problems to Solve Based Ratio Math Problems Tutoring:

In a zoo they are monkeys and parrot the ratio of monkeys and parrot are found in ratio of 9: 6 .If the zoo contains 180 parrots, find the ratio of monkeys present in zoo?
Answer: The ratio of 270 monkeys is present in zoo.

In a flower shop they are yellow and pink color roses, the ratio of yellow and pink color roses  are found in ratio of 2:4 .If the flower shop  contains 100 pink color roses, Find the ratio of yellow color roses present in the flower shop?
Answer: The ratio of 50 yellow color roses present in shop.

Arithmetic Series Functions

Introduction for Arithmetic Series function:

A sequence that is 3 + 7 + 11 + 15 + ··· + 99 or 10 + 20 + 30 + ··· + 1000 that includes an even distinction between terms. The first term known as a1, the general difference is d, and the number of terms is n. The calculation of an arithmetic sequence is established by multiplying the number of terms times the average of the 1st and last terms.

Proof of Arithmetic Series Function:

1) Addition of i as i moves from 1 to n: 1 + 2 + 3 + ... + n function.

2) Adding the calculation of adding the 2nd summation backwards:

1 + ... + n

n + ... + 1

3) The 1st equation as (i), and then the 2nd calculation becomes (n - i + 1)

4) Hence 2 * the equation is (i) + (n - i + 1) or (n + 1) for every item in the sequence.

5) Here there are n terms in the sequence, the 2 * calculation of value is n * (n + 1)

6) For a particular series, we have end up with (n * (n + 1) / 2)

Example Problem for Arithmetic Series Function:

Here we will take more time to assuming about the rule, but the formula for the calculation of a finite arithmetic series, frequently symbolized by sn is

Arithmetic series function Formula: Sn = n (`(a1 + an) / 2` ) or `n / 2` [2a1 + (n - 1) d]

Problem 1:

4 + 8 + 12 + 16 + ··· + 96 has a1 = 4 and d = 4. To find n, use the formula for an arithmetic sequence.

Solution:

We solve 4 + (n – 1) *4 = 96 to get n = 24.

Sum = 24 (4 +` 99 / 2` ) = 1236

Or

Sum = `24 / 2 ` [2* 4 + (24 – 1) *4] = 1236.

Problem 2:

How much will a bank staff receives over his 34 year career if his starting salary is 40,000, and he receives a 1,000 salary increments for every year he works here?

Solution

We have the sequence:

40,000 + 41,000 + 42,000 + ... + 73,000

= `34 / 2` (40,000 + 73,000) = 1,92,1000.

Problem 3:

Find S9 for the series an = 6n – 1

Solution:

The formula tells us we have to be known n, the 1st term, and the nth term n = 9 because we have to find out the sum of the first 9 terms.

a1 = 6(1) – 1 = 6 – 1 = 5

a 9 = 6 (9) - 1 = 54 – 1 = 53

Substitute these values into the equation gives

S9 = 9 (5 + `53 / 2)` = 9 (`58 / 2)` = 9 (29) = 261

Fundamental Theorem of Arithmetic

Introduction to Fundamental Theorem of Arithmetic
The fundamental theorem of arithmetic is If (1) p1, p2, p3… pn and q1, q2, q3… qm are primes (both groups written in order of increasing size) and if (2) p1p2p3…pn = q1q2q3…qm, then p1 = q1, p2 = q2 … pn = qm and hence m = n. The Fundamental Theorem of Arithmetic is also called the "unique factorization theorem."  The following example is the fundamental theorem of arithmetic

Example:

3468 = 22 · 3 · 172   or   600 = 23 · 3 · 52

Fundamental Theorem of Arithmetic - Proof

It would be reasonable to prove first that p1 = q1.

Now p1 is a divisor of the product of all the ps (call it N) so it must be a divisor of the product of all the qs (which is also N).

If we knew that this forces p1 to divide at least one of the qs, we could then reason like this: Since p1 divides one of the qs, and since the qs are prime, so p1 must equal the q that it divides.

q1 is the smallest of the qs, so p1 = q1. Identical reasoning about q1 shows that q1 = p1. Thus p1 is equal to q1. So, replacing q1 with p1 in q1q2q3…qm, we get p1p2p3…pn = p1q2q3…qm.

Dividing both sides by p1 we get p2p3…pn = q2q3…qm. This new equation has one less prime on each side.

The same kind of reasoning that proved p1 = q1 also shows that p2 = q2. Step by step we peel away the primes from each side, establishing that pk = qk for all values of k.

Since every p pairs off with one q and vice versa, we see that m = n (the number of ps = the number of qs).

The problem is that we don't know that whenever a prime divides the product of several natural numbers it must divide at least one of them.

Fundamental Theorem of Arithmetic - Examples

Example 1:

216 = 2. 108= 2.2.54 = 2.2.2.27 =2.2.2.3.9 = 2.2.2.3.3.3

That is,

216 = 23. 33

Example 2:

26624 = 8.3328 = 8.8.416 = 8.8.8.52 = 8.8.8.2.26 = 8.8.8.2.2.13

That is

8.8.8.2.2.13 = 83.22.13

Arithmetic Maths

Introduction to arithmetic maths:

Arithmetic or arithmetic is the oldest and mainly basic branch of mathematics, used by approximately everyone, for tasks range from simple day-to-day including to superior science and business computation. It engages the learn of quantity, particularly as the outcome of unite numbers. In general usage, it refers to the easier properties when using the conventional operations of adding, subtracting, multipling and dividing with lesser values of numbers.


Basic Operations for Arithmetic Maths:

The arithmetic maths operations are as follows,

1.    Addition.

2.    Subtraction.

3.    Multiplication.

4.    Division.

Arithmetic Math- Addition:

Calculation of two or more than two numbers is known Arithmetic. The indication for addition is +. We can execute the addition operation for the all kind of values. Addition problems augment the values.

Arithmetic math Addition- Example:

6 + 5 = 11

This is easy addition problem.

Similarly we do the addition operation also more than two digits.

25 + 33 = 58

Arithmetic math- Subtraction:

Subtraction is the one of the essential arithmetic operation. Reduction of two or more than two numbers is known as Subtraction. The sign of subtraction is - (minus). Subtraction problems reduce the values.

The subtraction is finished by the following way,

z = x – y

Where,

z is differentiation.

x is known as minuend value.

y is known as subtrahend value.

Arithmetic maths subtraction- Example:

8-5 = 3.

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Arithmetic Maths- Multiplication:

Multiplication is the calculation of two numbers. The indication of multiplication is *. Multiplication also execute by recurrence of calculation. The conditions often used for multiplication are “times”.

Arithmetic maths multiplication- Example:

4 times 8 means 4 multiply with 8.

4 * 8 = 32

The other technique of multiplication recurring addition,

4 times 8 = 32

8 + 8 + 8 + 8 = 32

Arithmetic maths- division:

Division is an annul of multiply operation. The indication of division is /. The major reason of division is reducing the value and it is also recurring calculation.

Appearance of division technique,

`x/y` =z.

Where,

x is called as dividend.

y is called as divisor.

z is called as quotient.

Arithmetic math division- Example:

`21/7` = 3.

Sum of Arithmetic Sequence

Introduction to sum of arithmetic sequence:

Arithmetic sequence or progression is a sequence of numbers such that the difference of any two successive members of the sequence is a constant. For instance, the sequence 10, 20, 30, 40, 50… is an arithmetic sequence with common difference d = 5. Two types of sequences are arithmetic sequence and geometric sequence.

Formula – Sum of Arithmetic Sequence

The formula for the nth term an of an infinite arithmetic sequence with a common difference d and a first term a1 is given by

an = a1 + (n - 1 )d

The sum sn of the first n terms of an infinite arithmetic sequence is defined by

sn = a1 + a2 + a3 + ... + a n

and is a1 is given by

sn = n (a1 + an) / 2

Examples – Sum of Arithmetic Sequence

Ex 1: Find the sum of all positive integers, from 27 to 4590 inclusive, which are divisible by 27.

Sol : Sequence of first few numbers of positive integers divisible by 27 are given by 27, 54, 81...

The above sequence has a first number equal to 27 and a common difference d = 27.

We need to know the rank of the term 4590.

We use the following formula for the nth term

an = a1 + (n - 1 )d

4590 = a1 + (n - 1 )d

Substitute a1 and d by their values

4590 = 27 + 27(n - 1)

Solve for n to obtain

n = 170

4590 is the 170th term, we can use the following formula to find sum

sn = n (a1 + an) / 2

s170 = 170 (27 + 4590) / 2 = 392445.

Ex 2:  Find the sum of all positive integers, from 23 to 3680 inclusive, which are divisible by 23.

Sol :  Sequence of first few numbers of positive integers divisible by 23 are given by 23, 46, 69...

The above sequence has a first number equal to 23 and a common difference d = 23.

We need to know the rank of the term 3680.

We use the following formula for the nth term

an = a1 + (n - 1 )d

3680 = a1 + (n - 1 )d

Substitute a1 and d by their values

3680 = 23 + 23(n - 1)

Solve for n to obtain

n = 160

3680 is the 160th term, we can use the following formula to find sum

sn = n (a1 + an) / 2

s160 = 160 (23 + 3680) / 2 = 296240.