Words That Mean Perfect

Introduction to Words That Mean Perfect:

Words that mean perfect in math is the operation in the expression done perfect with correct procedure. Each expression has different operation and different procedure. To get the result from or to solve the expression as by the corresponding procedure, we say as the words perfect. In this article, we see about the words that mean perfect.

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Example Problem – Words that Mean Perfect:


Example 1:

What is the words that mean perfect of the expression as (x + 1)?

Solution:

Given: (x + 1)

To find:  Words that mean perfect for the expression is the result of the given expression.

(x + 1) = (1 + 1) + (2 + 1) + (3 + 1) + (4 + 1)

= 2 + 3 + 4 + 5

= 14

Thus the words that mean perfect for the expression is the result of the expression is 14

Answer: Perfect 14

Example 2:

Which of the following is the words that mean perfect form the expression 33?

Option:

a)     9

b)    27

c)     9

d)    3

Solution:

Given: Expression is 3^3

The given expression is the 3 to the power of 3. Thus the number 3 is multiply itself as three times.

Step 1: 3^3 = 3 * 3 * 3 = 27

Thus 27 is the words that mean perfect to the expression as 33. Here the perfect words that given in option as b which is said to be as correct.

Answer: Option b - - - > 27

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Example Problem – Words that Mean Perfect:


Example 3:

Which of the following expression is correct for the words that mean perfect of the number 25?

Option:

a)     5 + 5

b)    20 * 5

c)     25 – 5

d)    5 * 5

Solution:

Step 1: To find the correct expression for the perfect number 25.

Step 2: Check each option which given the words that mean perfect as 25.

Option a: 5 + 5, the sum of 5 and 5 is 10

Option b: 20 * 5, the product of 20 and 5 is 100

Option c: 25 – 5, the result of 5 is subtracted from 25 is 20

Option d: 5 * 5. the product of 5 and 5 is 25

Thus the expression given in option d gives the words that mean perfect of 25.

Answer: Option d - - - > 25

Solve Perfect Triangles

Introduction to solve perfect triangles:

Triangles are three sided polygons. They are classified into equilateral triangles, isosceles, and right angled triangles. To solve perfect triangles, they should contain equal sides and angles. Perfect triangles include equilateral and right angled triangles. In triangles, we find the area, perimeter and the angles. Now we see some problems to solve perfect triangles.


Some basic properties of triangles and problems


Area formula = `1/2(b xx h)`

Perimeter formula = (a + b + c)

Interior angle = (n – 1) 180° and

Exterior angle = 360°/n

Where,

n – Number of sides

Problems to solve perfect triangles:

Example 1:

What is the area of an equilateral triangle, if it one of the side measures 5 cm?

Solution:

We know that, the equilateral triangle has same length in all the sides.

Formula to find the area of the triangle is given by,

Area =`1/2(b xx h)`

Here it has same breadth and length.

On substituting the side value, that is 5 in the areas formula,

Area =`1/2(5 xx 5)`

= `1/2(25)`

=12.5

Thus the area of the given equilateral triangle is 12.5 cm2.

Example 2:

Find the perimeter of the right angled triangle the sides are given as 3 cm, 6 cm, and 9 cm.

Solution:

We know the formula to find the perimeter of the given triangle

Perimeter = (a + b + c)

= (3 + 6 + 9)

= 18

Therefore the perimeter of the triangle is 18 cm.

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More problems to solve perfect triangles


Example 3:

Find the interior and exterior angles for the right angled triangle.

Solution:

Formula to find Interior angle = (n – 2) 180°

Here a right angled triangle has 3 sides. So n = 3

(n – 2) 180 = (3 – 2)180

= 180

The interior angle for a right angled triangle is 180°

Now to find the exterior angle, we know the formula as

`360/n = 360/3`

= 120

Hence the exterior angle is 120°.

Example 4:

A triangle has a total perimeter of 34 cm. If two sides are given as 12 cm and 14 cm, what is the length of the third side?

Solution:

Perimeter of the triangle is (a + b + c)

We know the two side’s length as 12 and 14.

So a = 12 cm and b = 14 cm

Let the third side be c.

12 + 14 + c = 34

26 + c = 34

Subtract by 26 on both sides, we get

26 – 26 + c = 34 – 26

c = 8

Therefore the third side’s length is 8 cm.

These are some example problems to solve perfect triangles.

Definition of Perfect Square

Introduction to definition of perfect square:

The definition of perfect square is defined as the important topics in mathematics. Product of two integer gives another integer is known as the perfect square. The rational number with the square root is known as the perfect square.  For example, 16 is known as the perfect square integer, since it has 4  `xx`  4 is the two product for 16. This article shows the definition of perfect square with brief explanation and some example problems.


Explanation to definition of perfect square


The explanation given for the perfect square definition is as follows,

Trinomial and binomial functions are also written as the perfect square.
Trinomial Perfect square = x2 + 6x + 9
Trinomial Perfect square  = (x + 3)2 .
Perfect square Example:

0 , 1 , 4, 9, 16, 25, etc.

`16/36` , `16/25` are also the examples of perfect squares.


Example problems to definition of perfect square


Problem 1: Which of the following integer when added to 12 to make perfect square.

Options:

a) 2

b) 3

c) 1

d) 4

Solution:

Step 1: Given:

Number = 12

Step 2: To find:

Perfect square

Step 3: Solve:

12 + 4 = 16

16 = 42

16 = 4 `xx` 4

Therefore 4 is the integer when added to 12 to give perfect square.

Answer: Option d

Problem 2: Which of the following integer when added to 24 to make perfect square.

Options:

a) 2

b) 3

c) 1

d) 4

Solution:

Step 1: Given:

Number = 24

Step 2: To find:

Perfect square

Step 3: Solve:

24 + 1 = 25

25 = 52

16 = 5 `xx` 5

Therefore 1 is the integer when added to 24 to give perfect square.

Answer: Option c

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Practice problems to definition of perfect square


Problem 1: Which of the following integer when added to 25 to make perfect square.

Options:

a) 2

b) 1

c) 3

d) 4

Answer: Option b

Problem 2: Which of the following integer when added to 33 to make perfect square.

Options:

a) 2

b) 1

c) 3

d) 4

Answer: Option c

Perfect Square Definition

Introduction to perfect square definition:
Perfect square is defined as one of the most important topics in mathematics. Perfect square is defined as the product of integer gives the integer. Also, the perfect square is defined as the rational number with the square root. For example, 9 is called as the perfect square, since it has 3  `xx`  3 is the product terms. In this article, we are going to study about the perfect square definition in detail.

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Explanation to perfect square definition


The explanation given for the efinition of perfect square is as follows,

Perfect square can also written using the trinomial and binomials.
Perfect square trinomial = x^2 + 4x + 4
Perfect square binomial = (x + 2)^2 .
Example of perfect square:

0 , 1 , 4, 9, 16, 25, etc.

`9/16` , `16/25` are also known as the perfect squares.


Example problems to perfect square definition


Problem 1: Find the integer which the number 17 as the perfect square.

Options:

a) 2

b) 3

c) 1

d) 4

Solution:

Step 1: Given:

Number = 17

Step 2: To find:

Perfect square

Step 3: Solve:

17 - 1 = 16

16 = 42

16 = 4 `xx` 4

Therefore 1 is the number which makes 17 as the perfect square.

Answer: Option c

Problem 2: Find the integer which the number 27 as the perfect square.

Options:

a) 2

b) 3

c) 1

d) 4

Solution:

Step 1: Given:

Number = 27

Step 2: To find:

Perfect square

Step 3: Solve:

27 - 2 = 25

25 = 52

125 = 5 `xx` 5

Therefore 2 is the number which makes 27 as the perfect square.

Answer: Option a

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Practice problems to perfect square definition


Problem 1: Find the integer which the number 19 as the perfect square.

Options:

a) 2

b) 3

c) 1

d) 4

Answer: Option b

Problem 2: Find the integer which the number 29 as the perfect square.

Options:

a) 2

b) 3

c) 1

d) 4

Answer: Option d

Perfect Square Formula

Introduction to the perfect square formula :
In the mathematics, the perfect square is sometimes known as a square number. The perfect square is the integer square or also represented as the perfect square is the product of some of the integer with itself. The perfect square numbers are only the positive numbers not having the negative numbers. The positive values are not having perfect square divisors except only one is known as square free.



Formula for perfect square formula :


The regular notation for the perfect square formula is the perfect square of the number n not the n x n that is not the product of the number n.The model of square can be used to number systems.

Explanation of the perfect square formula:

In the arithmetic we have the following perfect squares like,`1^2=1,2^2=4,3^2=9,..`

In this way we can able to make the perfect square in the algebra. It is explained in the following ways, take the expansion of `(a+b)^2`

`(a+b)^2=(a+b)(a+b)=a(a+b)+b(a+b)=a^2 +ab+ba+b^2=a^2+b^2+2ab`

`a^2+b^2+2ab` it is the perfect square formula of the (a+b).

the expansion of `(a-b)^2`

`(a-b)^2=(a-b)(a-b)=a(a-b)-b(a-b)=a^2-ab-ba+b^2=a^2+b^2-2ab`

`a^2+b^2-2ab` it is the perfect square formula of the ( a-b).

Geometrical representation:

Consider the square of a side (a + b) units .

Area of the square `=(side)^2=(a+b)^2`

Area of the square is same as the areas of the four rectangles.so it can be written as,

Area of the square `= a^2+ab+ab+b^2=a^2+2ab+b^2.`

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Example 1 for the perfect square formula:


To solve the following `(x+5)^2`

Solution:

Given,

`(x+5)^2 =(x+5)(x+5)=x(x+5)+5(x+5)=x^2+5x+5x+25=x^2+10x+25.`

The result of the equation is `x^2+10x+25.`

Example 2 for the perfect square formula:

To solve the following `(a+2)^2`

Solution:

Given,

`(a+2)^2 =(a+2)(a+2)=a(a+2)+2(a+2)=a^2+2a+2a+4=a^2+4a+4.`

The result of the equation is` a^2+4a+4.`

Example 3 for the perfect square formula:

To solve the following `(a-2)^2`

Solution:

Given,

`(a-2)^2 =(a-2)(a-2)=a(a-2)-2(a-2)=a^2-2a-2a+4=a^2-4a+4.`

The result of the equation is` a^2-4a+4.`

Learn Basic Mathematics Tutoring

Introduction to learn basic mathematics tutoring

Mathematics is used throughout the world in many fields like, engineering, science, and medicine. Tutoring means receiving instruction or help from tutor. Through tutoring students can get more help. Tutors explain every concept in step by step so that student cans easily learning and understands the concept. In basic math the students can receive first level of math education. In this article we shall learn basic mathematics tutoring example problems. Is this topic 6th grade math test prep hard for you? Watch out for my coming posts.


Learn basic mathematics tutoring example problems


Example:

Express 3678 g in to kilograms

Solution:

1000 g = 1 kg

Hence `3678/1000`

= 3.678 kg

Note:

3678 g = 3000 g + 600 g + 70 g + 8 g

= 3 × 1000 g + 6 × 100 g + 7 × 10 g + 8 g

= 3 × 1 kg + 6 × 1 hg + 7 × 1 dag + 8 g

= 3 kg + 6 hg + 7 dag + 8 g

Example:

Solving the equation 6x – 45 = 2x + 3

Solution:

6x – 45 = 2x + 3

6x = 2x + 3 + 45

6x – 2x = 48

4x = 48

x = 48 x` 1/4`

Therefore x = 12

Example:

A book is bought `$` 120 and sells it profit at a rate of 10% find the selling price of the book.

Solution:

Profit    = 10 % of `$ ` 120

= `10/100` x 120 = 12

Selling price      = cost price + profit

= 120 + 12

The selling price is` $` 132

Example:

The area of the triangle is 1600m2. The triangle base field is 40 m. find the height of the triangle.

Solution:

Given Area of the triangle, A     = 1600m2

Base of the triangle = 40 m

Height of the triangle =?

Area of triangle formula = `1/2 ` b x h

`1/2` x (40) x h =1600

20 x h   = 1600

h          = `1600/20`

Therefore the height of the triangle field = 80 m

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Learn basic mathematics tutoring practice problems


Problem:

Solving the equation 7x – 53 = 2x + 2

Answer:

x = 11

Problem:

The area of the triangle is 1400m2. The triangle base field is 40 m. find the height of the triangle.

Answer:

Height of the triangle field = 80 m

Problem:

A book is bought `$` 180 and sells it profit at a rate of 10% find the selling price of the book.

Answer:

The selling price is` $ ` 198

Prepare for Learn Mathematics

Introduction to prepare for learn mathematics:

Mathematics is the study of quantity, structure, space, and change. Mathematicians seek out patterns, formulate new conjectures, and establish truth by rigorous deduction from appropriately chosen axioms and definitions. Mathematics is used throughout the world as an essential tool in many fields, including natural science, engineering, medicine, and the social sciences. In this article we shall prepare for learn mathematics problems. (Source: Wikipedia)


Prepare for learn mathematics with example problem


Here we are going to prepare for learn mathematics example problem with detailed solutions.

Example:

A marked of the land price is `$` 7500. The land owner decides to give 6% discount on the original price of the land. What is the selling price of the land?

Solution:

Marked price of the land             =          `$` 7500

Rate of discount                        =          6%

Actual discount amount =          7500x` 6/100`

=          `$` 450

=          7500 – 450

=          `$` 7050

Selling price of the land =          marked price of the land – discount price of the land

=          7500 – 450

=          7050

Therefore the selling price of land is `$` 7050.

Example:

Joes deposited at a bank `$` 9000. The bank paid simple interest on deposit amount and credited yearly at the 6% rate of interest per annum. Find the total money she gets at the end of 5th years.

Solution:

Formula for simple interest = `(PNR)/(100)`

Present amount P = `$` 9000

Rate of interest year r = 6%

Total number of year = 5

A = `(9000*5*6)/(100)`

Interest = 2700

Total amount = Deposit amount + Interest

= `$` 9000 + `$` 2700

= `$` 11700


Prepare for learn mathematics with practice problem


Problem:

A marked of the land price is `$` 2500. The land owner decides to give 7% discount on the original price of the land. What is the selling price of the land?

Answer:

The selling price of land is `$` 2325.

Problem:

Joes deposited at a bank` $` 8000. The bank paid simple interest on deposit amount and credited yearly at the 8% rate of interest per annum. Find the total money she gets at the end of 5th years.

Answer:

`$` 11200

Online Solve Mathematics

Introduction to online solve mathematics:

Mathematics is the study of the dimension, properties, and interaction of quantities and sets, using symbols and numbers. Mathematics is used in our daily life. Mathematics includes algebra, geometry, calculus, etc. Step by step explanation is very useful to understand the concepts of math. Through the online, student can gain more knowledge even by staying in their home itself. Now, we are going to see some of the problems to solve mathematics online.

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Problems to solve mathematics online:


Example problem 1:

Simplify the expression: 15x + 8y - 2 + 7x + 12y + 17

Solution:

This expression can be simplified by combining like terms

+15 x and +7x are like terms, and can be combined to give +22x,

+8y and +12y combine to give +20y, and

-2 and +17 combine to give +15.

So after simplifying, this expression becomes: 22x + 20y + 15.

Example problem 2:

Subtracting fractions: `3/4 - 1/5`

Solution:

The common denominator of 4 and 5 is 20.

In this example, we need to multiply the fraction  `3 / 4`  by 5 and multiply `1 / 5` by 4. So, Equivalent fraction of `3 / 4` is `15 / 20` and the equivalent fraction of `1 / 5` is` 4 / 20` .

`3 / 4 - 1 / 5 = 15 / 20 - 4 / 20`

Now, the denominators are same, so we have to subtract the numerators.

`3 / 4 - 1 / 5 = (15 - 4) / 20`

=`11 / 20`

So, the answer is `11 / 20.`


Few more problems to solve mathematics online:


Example problem 3:

Solve the inequality s: 17.5s – 33 < 37

Solution:

17.5s – 33 < 37

Add 33 on both side of the inequality

17.5s – 33 + 33 < 37 + 33

17.5s < 70

Divide by 17.5 on both side of the inequality

`(17.5s) / 17.5 < 70 / 17.5`

s < 4

So, the solution is (-infinity, 4).

Example problem 4:

The base of a cylinder has the radius of 4.5 cm and the height is 8 cm. Determine the lateral surface area of the solid cylinder.

Solution:

Radius = 4.5cm

Height = 8cm

Formula for lateral surface area = 2* `pi` * r * h.

Here given that, r = 4.5cm and h = 8cm.

Lateral surface area of the cylinder = 2 * 3.14 * 4.5 * 8

= 226.29 cm^2.

So, the lateral surface area of cylinder is 226.29 cm^2.


Practice problems to solve mathematics online:


1)    Subtracting fractions: `3 / 5 - 1 / 4` . (Answer: `7 / 20` ).

2)    Solve the inequality: 13.4s – 2.6 > 51. (Answer: s > 4).

3)    The base of a cylinder has the radius of 3.8 cm and the height is 7 cm. Determine the lateral surface area of the solid cylinder. (Answer: 167.05 cm^2).

Solving Learning Mathematics

Introduction to solving learning mathematics:

Mathematics is related to the properties, dimension, and interaction of quantities and sets, using numbers and symbols. Mathematics is used in our daily life. Mathematics includes various topics such as algebra, geometry, trigonometric, calculus, etc. Step by step clear explanation is more helpful to understand the basic concepts involved in math. Now, we are going to see some of mathematics problems learning.

Solving learning mathematics problems:


Example problem 1:

Solve for the variable r:  3r – 13 = 47 - 2r

Solution:

3r – 13 = 47 - 2r

Add 13 on both sides of the equation

3r – 13 + 13 = 47 - 2r + 13

3r = -2r + 60

Add 2r on both sides of the equation

3r+ 2r = -2r + 60 + 2r

5r = 60

Divide by 5 on both sides of the equation

`(5r) / 5 = 60 / 5`

By solving this, we get

r = 12

So, the answer is r = 12.

Example problem 2:

Marked price of a furniture is `$` 450. It is sold at a discount of 20%. Find the discount and its selling price.

Solution:

Marked price of the furniture = `$` 450

Rate of discount = 20%

Discount allowed = 20% of `$` 450=`20 / 100` × `$` 450 = `$` 90

Therefore, Selling price of the furniture = `$` 450 – `$` 90 = `$` 360.


Few more solving learning mathematics problems:


Example problem 3:

Find the Mean value for the given set of numbers: 42, 48, 43, 45.

Solution:

Mean = Sum of the values
Total number of values.

Mean   =`(42 + 48 + 43 + 45) / 4`

=`178 / 4`

= 44.5

So, the mean value is 44.5.

Example problem 4:

Find x and y intercept for the equation of line 13x +6.5y = 26.

Solution:-

Given equation is 13x + 6.5y = 26.

x intercept means plug y = 0 in the given equation

13x + 6.5(0) = 26

13x = 26

By Solving this, we get

x = 2

y intercept means plug x = 0 in the given equation

13(0) + 6.5y = 26

6.5y = 26

y =4

So, the x intercept is (2, 0) and the y intercept is (0, 4).

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Practice problems for mathematics learning:
1) Find the value of p: 7p +17 = 66.

(Answer: p = 7).

2)  Find x intercept for the equation of line 4x + 6y = 30.

[Answer: (7.5, 0)].

3) Find the mean value for the given set of numbers: 10, 20, 30, 40.

(Answer: 25).

Solving Ks-3 Mathematics Tuitions

Introduction to solving ks-3 mathematics tuitions:

Mathematics is one of the basic knowledge to be known. Ks-3 means key stage 3 which refers to the three year schooling for students who are in the age of 11-14. These types of systems are followed in England and Northern Ireland. The ks-3 term of study is generally expressed as year 7, year 8 and year 9 studies. In online, many websites provides free help on ks-3 maths problem. In online, students can get help with all subject problems. In this article, we are going to see about, ks-3 mathematics problem solving online.


Practice problems on solving ks-3 mathematics tuitions:


Solving Ks-3 mathematics:


Ex 1: Find the place value of the number 9 in the given number 319
Sol :  The place value of 9 in the given number is, 1’s place
Hence the answer is 1’s place.
Ex 2:  Write the expanded form of 8290.
Sol :  The number 8290 can be written in expanded form as,
8290 = 8000 + 200 + 90
The answer is 8000 + 200 + 90.
Ex 3:  Express 16/32 as a percentage.
Sol :  16/32 = 16/32*100
On solving this we get,
= 0.5*100
= 50%
The solution is 50%
Ex 4:   In a school, there are total of 462 students. In that school there are 11 classrooms, find the number of students for each class.
Sol :  Total number of students = 462
Number of classrooms = 11
Number of students per class = x
Number of students per class = Total number of students/ number of classrooms
On solving this we get,
= 462/11
= 42
The answer is 42.

Ex 5:  Find the value of 5x, when x is 12
Sol :  Given: x = 12
5x = 5*12
= 60
Then answer is 60.
Ex 6:  Find the value of 4x2 + 3, when x is 4.
Sol :  Given: x = 4
4x2 + 3 = 4(4)2 + 3
= 4(16) + 3
= 64 + 3
= 67
The answer is 67.


Practice problems solving ks-3 mathematics tutions:


Problem 1: Find the place of 8 in the given number, 3859
Solution: 100’s place
Problem 2: Express 32/45 as a percentage
Solution: 71.11%
Problem 3: Find x, 22x, where x = 5
Solution: 110
Problem 4: Express 178 cents in dollars
Solution: 1.78 dollars

Prepare for Mathematics Problems

Introduction to prepare for mathematics problems:

Mathematics is the study of the dimension, properties, and interaction of quantities and sets, using symbols and numbers. Mathematics is used in our daily life. Mathematics includes algebra, geometry, calculus, etc. Step by step explanation is very useful to understand the concepts of math. Now, we are going to prepare for mathematics problems.

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Problems to prepare mathematics:


Example problem 1:

Solve an equation for the variable r:` (r / 4)` - 28 = -33

Solution:

`(r / 4) ` - 28 = -33

Add 28 on both sides of the equation

`(r / 4) ` – 28 + 28 = - 33 + 28

`(r / 4) ` = -5

Multiply 4 on both sides of the equation

`(r / 4)` * 4 = -5 * 4

r = -20

So, r = -20 is the solution of the given equation.

Example problem 2:

Solve the inequality: 2r – 83 < -61

Solution:

2r – 83 < -61

Add 83 on both side of the inequality

2t – 83 + 83 < -61 + 83

2r < 22

Divide by 2 on both side of the inequality

`(2r) / 2 < 22 / 2`

r < 11

So, the solution is (-infinity, 11).


Few more problems to prepare mathematics:


Example problem 3:

Find the x and y intercept of the equation: 7x + 3.5y = 14.

Solution:

7x + 3.5y = 14

To find the x intercept value, plug y = 0 in the given equation

7x + 3.5(0) = 14

7x = 14

x = 2

So, the x intercept is (2, 0).

To find the y intercept value, plug x = 0 in the given equation

7(0) + 3.5y = 14

3.5y = 14

y = 4

So, the y intercept is (0, 4).

Example problem 4:

Find the area of trapezoid whose parallel sides are 10cm and 7cm and height is 8cm.

Solution:

Base 1 = 10cm

Base 2 = 7cm

Height = 8cm

Area of trapezoid = `(1 / 2)` h(a + b)

= `(1/2)` (8) (10 + 7)

= 68cm^2.

So, the area of trapezoid is 68cm^2.

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Practice problems to prepare mathematics:


1)      Solve for the variable d:  10d + 4 = 18 + 3d (Answer: d = 2).

2)      Solve for the variable r:  18r + 1 = 37 (Answer: r = 2).

3)      Find the area of trapezoid whose parallel sides are 9cm and 6cm and height is 5cm. (Answer: 37.5cm^2).

Basic Mathematics for College Students

Introduction to basic mathematics for college students:

There are certain math topics are appeared in college which we already read in school period.Those topics are very basics for college students.The following mathematics topics are basics for college students. They are, algebra problems, geometry measures, probability problems and linear equations. By referring those topics, the college students recall the concepts of the topics , so that they can do themselves. Let we see some basic problems to basic mathematics for college students.


Example problems to basic mathematics for college students:


The following problems are examples for basic mathematics for college students.

Basic mathematics - problem1) Solve the linear equation, 3a+6b-14 = 16, a+3b- 6 = 12.

Solution:

Here we need to find the value of a and b,

The given equations are 3a+6b-14 =16 and a - 6b-6 =12

3a+6b-14 = 16

a  - 6b - 6  = 12.

We can write the above equation as

3a + 6b = 16+14

a - 6b  = 12+6

From the above equation, we get

3a + 6b = 30------------------------->1

a  - 6b  = 18------------------------->2

4a        = 48

4a = 48

Divide by 4 on both sides, we get

`(4a)/4` = `(48)/4`

a  = 12.

Apply the 'a' value in equation (1),we get

3a+6b= 30

3(12)+6b  = 30

36 + 6b    = 30.

Add -36 on both sides, we get

36-36 +6b = 30-36

6b = -6

Divide by 6 on both sides, we get

`(6b)/ 6` = -`6/ 6`

b  = -1.

The values of a and b are , 12 and -1.

Problem2) Find the value of  D , 2d+5e+6f = 80-4d+20, where e=2 and f=3

Solution:

Here we have the value of e and f as  2 and 3.

Apply the values in the given equation,we get

2d +5e+6f = 80-4d+20

2d + 5 ( 2) +6(3) = 80-4d+20

2d +10 +18       = 80-4d+20

2d +28            = 80+20-4a

2d +4d         = 100 - 28

6d    = 72

Divide by 6 on both sides, we get

6d /6  = 72 /6

d    = 12

The value of d =12.

Problem3) Solve the following linear equation, 3p+3q = 18, 2p+3q = 14

Solution:

Here we need to find the value for P and Q.

The given equations are

3p + 3q   =  18 -------->1

2p + 3q   =   14---------->2

By changing the sign of the second equation values, we can solve the above problem.

3p + 3q  =18

-2p - 3q  =14

p = 4

Apply the p value in equation (1) , we get

3 (4) + 3q  = 18

12 + 3q   = 18

Add -12 on both sides, we get

12-12 +3q = 18-12

3q = 6

Divide by 3 on both sides, we get

3q/3 = 6/3

q = 2

The values of m and q are , 4 and 2.

Problem 4:  If S and T values are 2 in the given equation , 10S +8T +6U = 78, find the value of U?

Solution:

Given equation          = 10S + 8T + 6U = 78

Known values  are , S = T =  2

Apply S and T values in the above equation,we get

10S + 8T + 6U      = 78.

10(2) + 8(2) + 6U  = 78.

20 + 16 + 6U     = 78

36 + 6U     = 78

Add -36 on both sides, we get

36 - 36 + 6U = 78 - 36

6U    =  42

Divide by 6 on both sides, we get

6U/6  = 42/6

U = 7

The value of U= 7.

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Practice problems to basic mathematics for college students:


Try to solve the following mathematics problems.

Problem1) 2a+2b+4c = 10, a+3b+c = 22, 2a+3b+2c=14

Answer: a=-11,b=10,c=3.

Problem2) 4x + 6y = 28, 2x+4y=18

Answer : x = 1,y=4

ABC of Mathematics

Introduction to ABC of maths:

Mathematics plays a very important role in humans life. Mathematical thinking is very important for everyone. In every work of a human we need mathematics.


Humans use the mathematics in doing any work. For example consider:
Arithmetic:  in counting and sorting,
Geometry: in spacing and distancing,
Statics: in balancing and weighting,
Probability: in guessing and judging


ABC OF MATHS means the fundamentals and basics of mathematics.

Brief explanation of ABC of maths:


NUMBER:
Number in mathematics is just a word or a symbol which is used for counting and also to say where something comes in a series.
Generally numbers are used for counting and calculating.


There are different kinds of numbers which includes: natural numbers, integers or whole numbers, cardinal numbers, ordinal numbers, rational numbers, real numbers, complex numbers etc.

NATURAL NUMBERS:
Natural numbers start from the 1, 2………infinity. They form the endless chain.

INTEGERS:
Integers also form the endless chain and it includes negatives and zero also.
…….……..-3, -2, -1, 0, 1, 2, 3……………..

Integers are also known as whole numbers.


CARDINAL NUMBERS:
Cardinal number shows how many there are. They are used for counting.


ORDINAL NUMBERS:
Ordinal numbers tells about the order of the object being counted. Places in a sequence are shown by ordinal numbers.


RATIONAL NUMBERS:
Rational numbers are the numbers which does not includes like an square root of 2. It includes the integers and fractions.


REAL NUMBERS:
Real numbers are the numbers which also includes the numbers like square root of 2.


COMPLEX NUMBERS:
Complex numbers includes the imaginary numbers also like square root of -1.


There are also some special numbers included in ABC of mathematics. They include zero and Arabic number system. Zero is the number which was discovered by Indian and it is a great discovery without zero there is no value or reason of mathematics.
There are different kinds of scales which are used to classify and quantify and also to measure things. The different types of scales includes Binary scale, Nominal scale, Ordinal scale, Interval scale, Ratio scale
We also use roman numbers. FORMULA in mathematics is nothing but a rule.

Everyday Mathematics 5th Grade

Introduction to everyday mathematics 5th grade:

For 5th grade students mathematics is divided into different categories. In grade 5 students can learn Simplification using BODMAS, word problems, unitary methods, Multiples and factors, HCF and LCM, Multiplication and Division of fractional numbers and decimals. They also learn about money, percentage, profit and loss, Simple interest etc.They learn measurement of length, mass and capacity. In geometry they can learn about the types of angles, triangles, element of circle, relation between diameter and radius etc.


everyday mathematics 5th grade - Some Examples


In 5th grade we can learn everyday common mathematics as follows:

(1) Unitary Method: The simple rule of unitary method that if we are given the cost of one object, we can find the cost of many objects by multiplying the cost of one object with the number of objects and if we are given the cost of several objects, we an find the cost of one object by dividing the cost of several objects by the number of objects.

Example:  The weight of 25 bags of rice is 650 kg. Find the weight of hundred bags of rice?

Solution: since the weight of 25 bags of rice is 650kg.

Then the weight of 1 bag of rice is 650 ÷ 25

26kg

Then weight of 100 bags of rice is 26 x 100

2600 kg

Answer: weight of 100 bags of rice is 2600 kg.

(2) HCF and LCM:  We define the HCF as the highest common factor.HCF of two given numbers is the highest number that divides the given numbers exactly without leaving any remainder.

Example: Find HCF of 136, 170 and 255 by division method:

Solutions: First we find the HCF of 136 and 170

136)170(1

136

34

HCF of 136 and 170 = 34

Now we find the HCF of 34 and 255

34)255(7

238

17)34(2

34

0

HCF of 34 and 255 =17

Hence HCF of 136,170 and 255 = 17

LCM: The lowest common multiple of two or more numbers is the smallest number which is a multiple of each of the numbers.

Example: Find the LCM of 20, 30 and 50 by division method:

2] 20, 30, 50

5] 10, 15, 25

2,   3,   5

LCM = 2x 5x 2x 3x 5 = 300


everyday mathematics 5th grade - more Examples


Decimal numbers: Conversion of a decimal fraction into fraction number:

Write the given number without a decimal point in the numerator.
Write 1 in the denominator followed by as many zeros as the decimal places.
Then write the resulting fraction in the lowest form.
Example:

(a)    1.5 = 15 / 10 =3 /2

(b)   54.972 = 54972 /1000

= 13743 / 250

Find the sum:

(a)    205.40, 80.75 and 1493.50

All decimal points should be in a column in the question and answer:

2 0 5.  40

80.  75

14 93 .50

1 7 7 9.65

Percent: Percent means for every hundred. To convert a % into fraction, place the given number over 100 and reduce it to its lowest term.

Example: 20% = 20 / 100

=  1/5

150% = 150/100

=3/2

To find the percent of a given number:

Example: (a)   Find 50% of 75

50/100 x 75

= 75/2 = 37 ½ Ans.

(b)   What % is 25 of 200

25/200 x 100

=    12 ½ Ans.

Perfect Square Root Numbers

Introduction to perfect square root numbers:

Let us study about perfect square root numbers. Square root is defined as the method to get the given values number square.
The perfect square root numbers are also similar to what the simple square root mean, that the number when a multiply with itself forms a perfect square root number.
In mathematics all numbers as both positive and negative can have their perfect squares in positive terms. Examples are below.

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Perfect square root numbers:


Some of the examples of perfect square root numbers are as follows:
1^2 = 1, 2^2 = 4, 3^2 = 9, 4^2 = 16, 5^2 = 25, etc.


Example 1:

Find the perfect square root numbers for the numbers 12, 14, 16, 18 and 20


Solution:

The perfect square root number for the number 12 = 12^2 = 12 * 12 = 144.
The perfect square root number for the number 14 = 14^2 = 14 * 14 = 196.
The perfect square root number for the number 16 = 16^2 = 16 * 16 = 256.
The perfect square root number for the number 18 = 18^2 = 18 * 18 = 324.
The perfect square root number for the number 12 = 12^2 = 20 * 20 = 200.

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Example 2:

Find the numbers for the perfect square root numbers 441, 169, 361, 1089 and 625.


Solution:

The number which gives the perfect square root number 441 = `sqrt(441)` = `sqrt(21*21)` = 21.
The number which gives the perfect square root number 169 = `sqrt(169)` = `sqrt(13*13)` = 13.
The number which gives the perfect square root number 361 = `sqrt(361)` = `sqrt(19*19)` = 19.
The number which gives the perfect square root number 1089 = `sqrt(1089)` = `sqrt(33*33)` = 33.
The number which gives the perfect square root number 625 = `sqrt(625)` = `sqrt(25*25)` = 25.


Exercises:

Find the perfect square root numbers for the numbers 15, 17, 22, 38 and 42. (Answer: 225, 289, 484, 1444 and 1764)
Find the numbers for the perfect square root numbers 484, 121, 676, 1156 and 2116. (Answer: 22, 11, 26, 34 and 46)

What is Mathematics Education

Introduction to mathematics education:

Mathematics education means study of the quantity, properties, and dealings of quantities and sets, using numbers and shapes. The major parts of mathematics are algebra, analysis, probability, set theory, and statistics. In this article we will study the education of mathematics and solve some of mathematics concepts with examples. Now we will solve the examples in algebra, probability, set theory and statistics what is mathematics education.


Examples - what is mathematics education:

Let us we will solve the problem in algebra for what is mathematics education.

Algebra - Example Problem 1:

Solve given polynomial equations.

5x^2 + 9 + 15x + 2x^2 + 4x + 10 + 3x

Solution:

Step 1:

First we have to mingle terms x^2

5x^2 + 2x^2=7x^2

Step 2:

Now combine the terms x

15x + 4x + 3x = 22x

Step 3:

Then join the constants terms

9 + 10=19

Step 4:

Finally, combine all the terms

7x^2 + 22x +19

So, the final answer is 7x^2 + 22x +19

Probability – What is Mathematics education:

Example 2:

In a briefcase, there are 7 red color dresses and 3 blue color dresses and 6 white color dresses. Find the probability of choosing blue color and white color dresses?

Solution:

Given, Number of red color dresses = 7

Number of blue color dresses = 3

Number of white color dresses = 6

So, Number of total outcomes should be,

7 + 3 + 6 = 16.

Therefore, the probability of choosing blue color dresses = `3/16`

Choosing white color dresses = `6/16.`

These are algebra and probability examples for what is mathematics education.

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More Examples – What is Mathematics Education:


Set theory:

Given sets are, P = {17, 25, 36, 85} and R = {25, 52, 63, 85}

P∪R = {17, 25, 36, 85} ∪ {25, 52, 63, 85}

= {17, 25, 52, 63, 36, 85,}

P∩R = {17, 25, 36, 85} ∩ {25, 52, 63, 85}

= {25, 85}

Statistics – What is Mathematics Education:

Find the range for given values.

The given values are 52, 63, 70,21,52,10.

Solution:

52, 63, 70,21,52,10.

In statistics, the range is defined as different between smallest value and biggest value.

In this given value, the smallest value is 10.

Then, the highest value is 70.

So, the range is,

Range = Highest value – smallest value

= 70 – 10.

= 60.

Range = 60.

These are examples for set theory and statistics for what is mathematics education.

That’s all about what is mathematics education.

Mathematics Education Standards

Introduction to mathematics education standards:

The mathematics education include a different branches of unit conversions, algebra, subtraction, measurement, number sense, multiplication, functions, adding and subtraction of decimals, fractions & mixed numbers, division, algebra, geometry, median problems, algebra function, probability and statistics number using words decimals. This mathematics education supports all type of standards up to higher standards.


Example problems for Mathematics education standards:


Example 1:

Solve the quadratic equation `x^2 +5x + 6 =0`

Solution:

`X^2 +5x +6 =0`

`X^2 +2x +3x + 6 =0`

` x(x +2) +3 (x +2) = 0`

`(x +2)(x + 3) =0`

`x + 2 = 0 `                   `x` ` + 3 =0`

`X = -2 `                      `X =-3`

Example 2:

Solve the quadratic equation `x^2 +4x + 4 =0`

Solution:

` X^2 +4x +4 =0`

`X^2 +2x +2x + 4 =0`

` x(x +2) + 2(x +2) = 0`

`(x +2)(x + 2) =0`

`x + 2 = 0 `              ` x + 2 =0`

`X = -2 `                      `X =-2`

Example of polynomial exponent problems- Mathematics education standards:

Addition of polynomial exponent:

Two or more polynomials, adding the terms,

Suitable example adding polynomial exponent,

Example1:

` (2x^2+3x^3)+(x^2+7x^3)`

`=2x^2+x^2+3x^3+7x^3`

The variable and exponent must be same then we add the polynomial exponent,

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

So the result is `=3x^2+10x^3`

Subtraction of polynomial exponent

Example2:

`(3x^2+3x^3)-(x^2+7x^3)`

`=3x^2-x^2+3x^3-7x^3`

The variable and exponent must be same then we subtract the polynomial exponent,

`=2x^2-4x^3 `

So the result is` =2x^2-4x^3 `

Adding polynomials- Mathematics education standards:

Example 1: Find the sum of `6x^2 + 7x + 16 and 1x - 3x^2 -4.`

Solution: By means properties of real numbers, we realize

`(6x^2 + 7x + 16) + (-3x^2 + 1x - 4) = 3x^2 + 7x + 1x + 16 - 4`

`= 3x^2 + 7x + 1x + 16 - 4`

`= 3x^2 + 8x + 12`

So the final result is `= 3x^2 + 8x + 12`

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Examples for finite difference problem- Mathematics education standards:


Example 1: calculate the values of Δ`y` and dy if `y = f(x) = x^3 + x^2 - 2x + 1`

Where x changes (i) from `1 to 1.05` and (ii) from` 1 to 1.01`

Solution:

(i) We have `f(1) = 1^3 + 1^2 - 2(1) + 1 = 1`

`f(1.05) = (1.05)^3 + (1.05)^2 - 2(1.05) + 1 = 1.15.`

and Δ`y = f(1.05)- f(1) = 0.15.`

in general `dy = f ^ 1(x) dx = (3x^2 + 2x - 2)dx`

When `x` ` = 1` , `dx = ` Δ`x =1 and dy = [(3(1)^2+2(1)-2] 1= 3`

(ii) `f(1.01) = (1.01)^3 - (1.01)^2 - 2(1.01) + 1 = -1.01`

∴ Δ`y = f(1.01) - f(1) = 1.99`

Mathematics Dealing With Functions

Introduction of mathematics dealing with functions:

The mathematics dealing with functions in the form of f(x) = 2x+3, we are assigning the value for variable x in the given function so that we can solve the functions  in math. Now we are given several values for variable x in the given function and finding the solution for each function. Example for s function is f(y)=19y+12,function f(2). Using the square equation solve a function rule  of f(x) = 2x2 +4x +28, function of f(2).

Example of mathematics dealing with functions:

f(2) = 2x+2 and  f(x) =` (x+2)/2` , here the variable of x is 2.

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Problems how to find mathematics dealing with functions in square equations


Problem 1 : Using  square equation find  the mathematics dealing with functions of f(9), when f(x) = `(x+2)/2` +2.

Solution : Here the variable is given as 9 find the function of f(9).

f(x) = `(x+2)/2` + 2 find the f(9)

The value of x is 9 is given

f(9) = ` (9+2)/2 ` + 2

f(9) =  `11/2 ` + 2

f(9) = 5.5+2

f(9) = 7.5

Problem 2 : Using  square equation find  the mathematics dealing with functions of f(8), when f(x) = `(x+3)/4` +12.

Solution :

Here the variable is given as 8 find the function notation of f(8).

f(x) =`(x+3)/4 ` +12 find the f(8)

The value of x is 8 is given

f(8) = `(8+3)/4 ` +12

f(8) = `(11)/4` +12

f(8) = 2.75 + 12

f(8) = 14.75

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Problems in mathematics dealing with functions


Problems1: using mathematics dealing with functions of f(3),  When f(x ) = `(2(x+3))/8`

Solution:

Using the function  f(3) in the constant function

f(x) =` (2x+6)/8`

f(3) = `(2xx3+6)/8 ` here substitute x value 3 in the given constant function

f(3) = `12/8.`

f(3) = 1.5

Problems 2: using mathematics dealing with functions of f(4). When f(x ) = `(2(x+6))/4`

Solution:

Using function f(4) in the constant function

f(x) =  `(2(x+6))/4`

f(4) = `(2(x+6))/4` here substitute x value 4 in the given constant function

f(4) =` (2xx4+12)/4.`

f(4) = `(8+12)/4`

f(4) = `20/4`

f(4) = 5

Mathematics Four Operations

Introduction to four operations in mathematics:

In mathematics, the four basic arithmetic operations are used. They are addition, subtraction, multiplication and division. These four operations are simple method and each operation has inverse operations. The arithmetic operations are main aspects.  Now we are going to see about four operations in mathematics with examples.


Explanation for four operations in mathematics

The four basic arithmetic operations are,

Addition
Subtraction
Multiplication
Division
Addition operation in mathematics:

The addition is one type of basic operation. It is added the numbers together. The addition operation is indicated by ‘+’ symbol. For example, add the numbers 2 and 3 as 2 + 3 = 5.

Subtraction in mathematics:

Subtraction is find the difference between the two numbers. The subtraction is indicated by ‘-‘ symbol. For example, subtract the 10 and 5 as 10 – 5 = 5. In subtraction, the small number is subtracted from large number.

Multiplication in mathematics:

In four mathematics operation, the multiplication is  one method. It is multiply the one number with other number. The multiplication is indicated by ‘x’ symbol. For example, multiply the 5 and 4 as 5 x 4 = 20. We can use the multiplication table in mathematics.

Division in mathematics:

Division is one type of arithmetic operation. It is indicated by ‘÷’ symbol.For example, divide the 8 by 2 as 8/2 = 4.

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More about four operations in mathematics


Example problems for mathematics four operations:

Problem 1: Do the multiplication operation with 15 and 12.

Answer:

The given numbers are 15 and 12.

15

12    x

_____

3  0

1  5

_______

1  8  0

_______

The result of multiplication is 180.

Problem 2: Add the given two numbers 45 and 50.

Answer:

The given two numbers are 45 and 50.

4 5

5 0 +

____

9 5

____

The result of addition is 95.

Exercise problems for four operations in mathematics:

1. Subtract the given numbers 60 and 43.

Answer: The result of subtraction is 17.

2. Divide the 120 by 3.

Answer: The result of division is 40.

Primary 4 Mathematics

Introduction:

Mathematics is the study of quantity, structure, space, and change. Mathematicians seek out patterns, formulate new conjectures, and establish truth by rigorous deduction from appropriately chosen axioms and definitions. In mathematics, fourth graders are usually taught how to add and subtract common fractions and decimals. Long division is also generally introduced here, and addition, subtraction, and multiplication of whole numbers is extended to larger numbers. (Source: Wikipedia)

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Example problems for primary 4 mathematics :


Primary 4 mathematics – Addition problem:

There are 190 passengers in trains A and 168 passengers in trains B. How many passengers are there altogether in the two trains?

Solution:

Passengers in car A = 190

Passengers in car B = 168

Passengers in car A + Passengers in car B

So, 190 + 168 = 358

There are 358 passengers altogether in the two taxis.

Primary 4 mathematics – Subtraction problem:

A fruit whole seller had 159 pomegranates. He sold 87 strawberries. How many pomegranate did he have left?

Solution:

The total amount of pomegranate = 159 pomegranates.

Sold pomegranate = 87

Remaining pomegranates =?

So, 159 – 87 = 72

He had 72 pomegranates left.

Primary 4 mathematics - Multiplication problem:

There are 52 sapotas in each bag. How many are there in 12 bags?

Solution:

So, 52 × 12 = 624

There are 624 pomegranates in 12 bags.

Primary 4 mathematics - Division problem:

Joseph bought a sack of 189 kg of flour. He has packed the flour equally into the 3 boxes. How many kilograms of flour were there in each boxes?

Solution:

189 ÷ 3 = 63

There were 63 kg of flour in each boxes.

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Practice problems for primary 4 mathematics :


1. There are 168 passengers in cars A and 158 passengers in cars B. How many passengers is there altogether in the two cars?

Answer: There are 326 passengers altogether in the two taxis.

2. A fruit whole seller had 259 Sapota. He sold 127 strawberries. How many Sapota did he have left?

Answer: He had 132 Sapota left.

3. There are 58 pomegranates in each bag. How many are there in 17 bags?

Answer: There are 986 pomegranates in 7 bags.

4. Clark bought a sack of 648 kg of flour. He has packed the flour equally into the 12 bags. How many kilograms of flour were there in each bag?

Answer:  There were 54 kg of flour in each bag.