Inverse Functions with Fractions

Introduction to inverse functions with fractions:

Inverse functions:

In mathematics, if ƒ is a function from a set A to a set B, then an inverse function for ƒ is a function from B to A, with the property that a round trip (a composition) from A to B to A (or from B to A to B) returns each element of the initial set to itself.



Fig(i) Inverse function

Fractions:

A fraction is a number that can represent part of a whole. (source : Wikipedia)

In this article we are going to see about how to find the inverse functions with fractions and some solved problems on inverse functions with fractions.Please express your views of this topic how to find the range of a set of numbers by commenting on blog.

Problems on Inverse Functions with Fractions :
Problem 1:

Find the inverse of the following function with fraction f(x) = 6x /11– 25/ 22

Solution:

Given, f(x) = 6x /11– 25/ 22

Let us substitute f(x) = y

That is y = (6/11) x – 25/ 22

Let us make the common denominator,

Y = (6 * 2) x / (11*2) - 25 / 22

Y = 12x / 22 – 25 / 22

Y = (12x - 25) / 22

For finding the inverse function we have to solve for x,

Y = (12x - 25) / 22

Multiply by 22 on both sides,

22y = (12x - 25)

Add 25 on both sides,

22y + 25 = 12 x

Divided by 12 on both sides,

x = (22y + 25) / 12

Now replace y = x and x = f--1 (x)

f-1(x) = (22x + 25) / 12

Answer: Inverse function of a given function is f-1(x) = (22x + 25) / 12

Problem 2:

Find the inverse of the following function with fraction y = (4x-8)^2 / 3

Solution:

Given, y =  (4x-8)^2 / 3

For finding the inverse function we have to solve for x,

y =  (4x-8)^2 / 3

Multiply by 3 on both sides,

3y = (4x-8)^2

Taking square root on both sides,

`sqrt(3y)` =`sqrt((4x-8)^2)`

`sqrt(3y)` = 4x-8

Add 8 on both sides of the above equation,

`sqrt(3y)` + 8 = 4x -8 + 8

`sqrt(3y)` + 8 = 4x

4x = `sqrt(3y)` + 8

Divided by 4 on both sides of the above equation, we get

x =( `sqrt(3y)` + 8)/4

Substitute x =f-1(x) and y = x

f-1(x) =( `sqrt(3x)` + 8)/4

Answer: The inverse of a given function is f-1(x) =( `sqrt(3x)` + 8)/4

Prcatice Problems on Inverse Functions with Fractions :

Problems :

1. Find the inverse of a function with fraction f(x) = (x/2) + 5

2. Find the inverse of a function with fraction f(x) = ( x - 2)/5

Answer key:

1. f-1(x) = 2x - 10

2. f-1(x) = 2 + 5x

Determining a Linear Function

Introduction to determining a linear function:

Determining the linear function involves the process of finding linear function by solving the equations in non linear form. The linear function deals with basic linear equation and it satisfies all the terms and conditions of the linear algebra. Linear algebra mainly involves finding the system o equations with the known values. Linear function has the demonstration in analytical geometry and generalized in operator theory. The examples are discussed below for determining linear function whereas linear function associates with the families of vector called linear spaces. The following are the example problems to determine linear function.

Examples for Determining a Linear Function:

Example 1:

Determine the linear function from the given function.

f(a) = a 2 – 15a + 18

Solution:

The given function is

f(a) = a 2 – 15a + 18

To find the linear function, differentiate the given function.

f '(a) = 2a  – 15

Conclusion: f '(a) = 2a – 15 is the determined linear function.

Example 2:

Determine the linear function from the given function.

f(x) = 5x 2 –  2x  + 6

Solution:

The given function is,

f(x) = 5x 2–  2x   + 6

To find the linear function, differentiate the given function.

f '(x) = 2(5x)  – ( 2  )

f '(x) = 10x  – 2

Conclusion: f '(x) = 10x – 2 is the determiner linear function.

Example 3:

Determine the linear function from the given function.

f(x) = 3x 3 –  5x 2  + 10x - 3x 3 +  5x 2  + 11x + x 4 –  x 4  + 18 + 5

Solution:

Given expression is,

f(x) = 3x 3 –  5x 2  + 10x - 3x 3 +  5x 2  + 11x + x 4 –  x 4  + 18 + 5.

Arrange the above expression in order.

f(x) = x 4 – x 4 + 3x 3 –  3x 3 - 5x 2 +  5x 2  + 11x + 10x + 18 + 5.

Cancel the common terms from the above equation.

f(x) =11x + 10x + 18 + 5.

Grouping the above terms

f(x) =21x  + 23.

Conclusion: f(x) =21x + 23 is the determined linear function.

Practice Problems for Determining a Linear Function:

1) Determine the linear function from the given function.

f(y) = 8y 2–  5y   + 16

Answer: f '(y) = 16y – 5.

2) Determine the linear function from the given function.

f(x) = 4x 2 –  5x 3  + 4x - 4x 2 +  5x 3  + 6x + 12 + 5

Answer: f(x) = 10x + 17

Sum of Prime Numbers

Introduction to sum of prime numbers:

Sum means adding, Sum of prime numbers means we are going to see the addition of prime numbers.Sum is also known as the total or the addition. Addition is made by combining two or more numbers. Addition is just like joining two or more numbers. The word addition will be represented by the symbol (+). By using addition we can combine more objects and combine it into a larger form.

Prime Numbers:

Any natural number which has precisely two divisors. The following numbers are prime 2, 3, 7, 11, 13, .... Every natural number greater than 1 may be determined  uniquely into a product of prime numbers example: 18 = 2 x 3 x 3. In the case of a prime number p, the product has to be interrupted as p itself.

Here are some of the prime numbers:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97......

Having problem with Positive and Negative Numbers keep reading my upcoming posts, i will try to help you.
 
Sum of Prime Numbers:

Example problem - 1:

Find the sum of given prime numbers 23 + 29 + 31 + 37 + 41 + 43

Solution:

We have to find the sum of

The sum of 23 + 29 = 52

52 + 31= 83

83 + 37 = 120

120 + 41= 161

161 + 43  = 204

Therefore the sums of the given prime numbers are 23 + 29 + 31 + 37 + 41 + 43 = 204

Example problem - 2:

Find the sum of given prime numbers 61 + 67 + 71 + 73 + 79 + 83 + 89 + 97

Solution:

We have to find the sum of

The sum of 61 + 67 = 128

128 + 71= 199

199 + 73 = 272

272 + 79= 351

351 + 83  = 434

434 + 89 = 523

523 + 97 = 620

Therefore the sums of the given prime numbers are 61 + 67 + 71 + 73 + 79 + 83 + 89 + 97 = 620

Negative Odd Numbers

Introduction to Negative Odd Numbers:

The number which cannot divide by the number 2 exactly is known as odd number.  For example: -5, 7, 9, -1 here -1 and -5 is the negative odd numbers. The odd number which is in negative sign is called negative odd number. In this article we see about the negative odd number and example problem.

Negative Odd Numbers:

Addition of Negative Odd numbers:

(‘’-‘’ve odd number) + (‘’-’’ve odd number) = ‘’-‘’ve even number

Multiplication of Negative Odd numbers:

(‘’-‘’ve odd number) × (‘’-’’ve odd number) = ‘’+‘’ve odd number

Subtraction of Negative Odd numbers:

(‘’-‘’ve odd number) - (‘’-’’ve odd number) = Result with the sign of large value.

Example: - 4 – (-3) = -1

Division of Negative Odd numbers:

(‘’-‘’ve odd number) ÷ (‘’-’’ve odd number) = ‘’+‘’ve odd number. Looking out for more help on algebra online in algebra by visiting listed websites.

Example Problem – Negative Odd Numbers:

Example 1:

Which of the following is the negative odd numbers?

Option:

a)  35

b)  9

c)   -4

d)  -7

Answer: Option d

Explanation:

Option a: The number 35 is not exactly divided by 2, Hence it an odd number. But it does not have negative sign; therefore it is not a negative odd number.

Option b: The number 9 is not exactly divided by 2, Hence it an odd number. But it does not have negative sign; therefore it is not a negative odd number.

Option c: The number -4 is the negative number, but it is exactly divided by 2. Therefore it is not a negative odd number.

Option d: The number -7 is the negative number, but it is not exactly divided by 2. Therefore it is a negative odd number.

Hence option d is correct answer.

Example 2:

Add the two negative odd numbers. -7 and -3

Solution:

(-7) + (-3) = -10

Answer: -10

Example 3:

Check the number -4587 is the negative odd number.

Solution:

Step 1:

Divide the number 4587 by 2.

2) -4587 (-2293

-4

_______

-5

-4

_______

- 1 8

-1 8

_______

-  7

-  6

__________

-1

Hence the number -4587 is not exactly divided by 2 and also end of the number -4587 is 7 (odd number). Therefore the number -4587 is the negative odd number.

Histogram Mode

Introduction to histogram mode

Definition: Histogram is a graphical representation showing a visual impression of  distribution of data. A Histogram consists of tabular frequencies shown as a adjacent rectangle without any intervals between the size equal to frequency of interval. This is in form of rectangles with class intervals on horizontal x - axis and corresponding frequencies at heights on y -axis.

The height of rectangle represents the frequency of interval. Frequency is divided by width of interval, and the total area of histogram is equal to the number of data.

Mode

Mode is a value that occurs most in a data set, sample data is called Scores, and sample mode is called modal score.

For example: Sample data [1, 1, 2, 2, 2, 2, 3, 5, 6] Mean = 2.

Points to Remember while Carving a Histogram

1. Frequency distribution should be always in Exclusive form.

2.Scale chosen should be appropriate, without any parameters.

3. There should not be any gap between two intervals, or adjacent rectangles.

Graphical Representation of Histogram

Example 1: A histogram showing salary v/s number of Employees

Data for the given Histogram:

X-axis represents salary (in thousands), which ranges from 0 to 87

Y- axis represents the number of employees ranging from 0 - 600 in an interval of 100.

Salary ( in Thousand)    Number of Employees
0 - 10    50
11 - 21    300
22 - 32    250
33 - 43    400
44 - 54    550
55 - 65    450
66 - 76    250
77 - 87    350
88 +    100



How to Find out Value of Mode in a Histogram

The value of Mode in a Histogram is equal to the the tallest vertical bar in a Histogram. Value which occurs most for a variable in a given data set  has highest frequency,  thus when represented in a Histogram depicts the tallest bar.

Taking the previous example of Salary v/s Number of employees:

The tallest vertical bar represents that there are highest number of employees earning salary between 44 - 54 (Thousands), so the Mode for given Histogram is 44 -54 in x-axis that has some 550 employees.I like to share this probability examples with you all through my article.

Points to Remember for Mode Value:

1. Mode can be more than one value for a given Histogram , in this case data is called multi modal.

2. There can be no mode for a given Histogram if all data occur for same number of times.

Mode value depends on the data set provided for making a  frequency distributed Histogram.

Conclusion: Histogram Mode

Histogram mode:

The highest bar in a Histogram has the highest frequency value, respective class width is called Modal value of Histogram.

Accounting Practice Problems

Introduction to Accounting Practice Problems:

The centralised accounting practice problems make a perfect field surrounded by the MBA programs for the combination of social and ecological point of views. Accounting practice problems inspects the modern position of accounting in this situation. Accounting practice problems identifies two major types of latest programs – they are,Having problem with Practice Math Problems keep reading my upcoming posts, i will try to help you.

(1) Emphasize legal, authority, and moral thoughts, and

(2) Arguably more forward-looking and value-creating, as well as subjects such as full cost accounting, differential accounting standards, and social auditing.

Outcome of Accounting Practice Problems:

Subjects of ethics, validity, and authority are more and more integrated into accounting programs, social and ecological topics generally are not.
Quantitative events for non-financial presentation are measured for emerging from both academic and personal accounting practice.
Managers’ presentation is reviewer in economic conditions, non-financial reasons normally should support with economic objectives in order for corporation for rewarding good corporation residency.

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Accounting Practice Problems:

Practice Problem 1: At the end of the accounting period the unearned rent account is $500 of $2000. Find the amount of the adjusting entry?

Solution: $1500.

Practice Problem 2: A permanent quality with a rate of $3000 and gathered reduction of $2750 is sold for $350. What is the quantity of the gain or loss on clearance of the permanent quality?

Solution: $100 gain.

Practice Problem 3: The stability in the pre-paid charge accounting earlier than the alteration at final of the year is $20000, which shows 4 month’s charge paid on 1st of December. The altering entry needed for 31st of December is,

Solution: debit charge cost of $5000, credit pre-paid charge of $5000.

Practice Problem 4: Instruments with a price of $160000, obtained on 1st June 2009, have the expected enduring value of $10000 and an expected life of 4 years. It is to be reduced by the straight-line process. What is the sum of reduction for the year 2009, during which the instrument was used 3300 hours?

Solution: $ 25000.

Practice Problem 5: The stability in the organization materials account on 1st of June was $5200, materials obtained throughout June were $2500, and the materials around at 30Th June were $2000. The total being used for the suitable altering entry is,

Solution: $ 5700.

Largest Expressions in Math

Introduction to Largest Expressions in Math:

In math expressions is of the form ax + b, where a and b is known as constants and x is variable. Determination of value depends upon the variables occurring in the expression. Instead of writing in words, expression is in form of variables, constants, symbols, relation, and operations. The expression with more than two or more variables is called as largest expressions in math.  Let us see about largest expressions in math in this article.

Worked Examples to Largest Expressions in Math

Example 1:

Solve the largest expression ax+ ay + bx + by + cx + cz + dx + dz + dy by factorizing.

Solution:

Step 1:

Let us write the given expression as ax + ay + bx + by + cx + cz + dx + dy + dz.
Step 2:

Factorize the a term alone, we get,

ax + ay = a(x + y)

Step 3:

Factorize the b term alone we get,

bx + by = b(x + y)

Step 4:

Factorize the c term alone, we get,

cx + cz = c(x + z)

Step 5:

Factorize the d term alone, we get,

dx + dy + dz = d(x + y + z)

Step 6:

Combine the step (2), (3), (4) and (5), we get,

ax + y) + b(x + z) c(x + z) + d(x + y + z)

Therefore, the solution for solving the largest expression is ax + y) + b(x + z) c(x + z) + d(x + y + z).

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Another Problem to Largest Expressions in Math

Example 2:

Solve the largest expression ax + 2ay^2 – bx – 2by^2 + 2ax – 6ay – bx + 3by.

Solution:

Step 1:

Let us write the given largest expression is ax + 2ay^2 – bx – 2by^2 + 2ax – 6ay – bx + 3by.

Step 2:

Take the common term a from the first and second terms, we get,

ax + 2ay^2 = a(x + 2y^2)

Step 3:

Take the common term b from the third and fourth term, we get,

-bx – 2y^2 = -b(x + 2y^2)

Step 4:

From step (2) and (3), combine the term, we get,

ax + 2ay^2 – bx – 2by^2  = (a – b) (x + 2y^2)

Step 5:

Take the common term 2a form the fifth and sixth term, we get,

2ax – 6ay = 2a(x – 3y)

Step 6:

Take the common term –b from the seventh and eighth term, we get,

-bx + 3by = - b (x – 3y)

Step 7:

From step (5) and (6), combine the term, we get,

2ax – 6ay – bx + 3by = (2a – b) (x – 3y)

Step 8:

Combine step (4) and (7) of the largest expression of the given problem,

ax + 2ay^2 – bx – 2by^2 + 2ax – 6ay – bx + 3by = (a – b) (x + 2y^2) + (2a – b) (x – 3y)

Therefore, the solution for solving the largest expression is (a – b) (x + 2y^2) + (2a – b) (x – 3y).