Patterns in Math

Introduction for Patterns in Math:

In math, under a certain conditions, the numbers are listed, which is said to be a pattern. Basically patterns in math are classified into three types. Every pattern has some properties. In this article, we shall discuss about various kinds of patterns in math. Also we shall solve some problems regarding pattern in math.

Types of Patterns in math:

Arithmetic Pattern
Alphabetic Pattern
Geometric Pattern
These are the 3 different types of patterns in math.

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Arithmetic patterns:

Typically the patterns are indicated as sequences. There are two types of possibilities for the occurrence of sequence. They are finite and infinite sequences. We can articulate a number pattern using some unique symbols. We can clarify the number patterns in several ways. Let us consider a number pattern {2, 4, 6, 8, 10,….}. For the given pattern, the first term of this pattern is 2; and the second term can be attaining by adding 2 with the first term.

Alphabetic Patterns:

Patterns based on alphabetical lettering are called as alphabetic pattern. In addition, the alphabets in a sequence are the alphabetical pattern for the exacting sequence.

Geometric Patterns:

Whenever the geometric shapes involved in some patterns, then they are said to be geometric patterns. For instance, Ellipses are the geometric shapes. Those ellipses are developed from the basic geometric shape called circles. Is this topic T Test Example hard for you? Watch out for my coming posts.


Examples for patterns in math:


Example 1:

Find the missing terms from the pattern given below.

2, 4, 8, 16, 32, 64, ___, ____

Solution:

The first term of the pattern is 2

The second term of the pattern = 2 * 2 = 4

The third term is 4 * 2 = 8

The fourth term is 8 * 2 = 16

Similarly, the missing terms can be determined as follows.

The seventh term will be 64 * 2 = 128

The eighth term is 128 * 2 = 256

So the correct pattern is 2, 4, 8, 16, 32, 64, 128 and 256.

Example 2:

Find the next two terms in the pattern given below.

1, 3, 5, 7, 9, 11, ___, ____

Solution:

The first term of the pattern is 1

The second term of the pattern = 1 + 2 = 3

The third term is 3 + 2 = 5

The fourth term is 5 + 2 = 7

Similarly, the missing terms can be determined as follows.

The seventh term will be 11 + 2 = 13

The eighth term is 13 + 2 = 15

This is an odd sequence of number.

So the exact pattern is 1, 3, 5, 7, 9, 11, 13 and 15.

Math Problems Quizzes

Introduction of math problem quizzes:

In this lesson we can show the math problem quizzes. It very helpful and you can develop your math skills. These quizzes are really exercising your brain at the same time carefully handle the number terms in word problems. The quizzes are showing word problem, fraction, decimal, ect. Let we see math quizzes.

Math problem quizzes:

1) At restaurant, the men drunk 4 1/2 cup of coffee and women drunk 5 ½ cup of coffee. How many coffees were drunk at restaurant?

(Answer:10)

2) Yesterday I bought 6.25ml paint. And I used 1.50ml paint. Now, how much paint I have? Or Find rest of paint?

(Answer: 4.75)

3) There were 40 questions on a science test. William got 70% of them correct. How many questions did he answer correctly?

(Answer: 28)

Find Missing number:

4)  --- , 5 , 4 ,10,--- , 15, 8 ,20, 10 ,25, ---- ,---- , ----- ,----, 16 ,40, 18 ,45, 20 ,----, ----,----,

24,60 26, ---- ,----,

(Answer: 2,6,12,30,14,35,50,22,55,65,28)

5)  Suppose 5 dogs take 5 minutes to eat 5 bones.

How many minutes would it take 4 dogs to eat 4 bones?

How many bones would it take to eat 24 bones in 24 minutes?

(Answer: 4 and 24)

To find with formula:

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6) A pillar 8 feet tall casts a shadow 4 feet long on the ground. If the pillar was 26 feet tall, how many feet in length would the shadow be?

(Answer: 13)

7) Which of the following are the prime factors of 96?

(Answer: 2 and 3)

8) A wall has an area of 96 square inches and a width of 8 inches. What is its length?

(Answer: 12inches)

9) A florist buys roses at $1.50 a piece and sells them for $3.00 a piece. If there are no other expenses, how many roses should be sold in order to make a profit of $300?

(Answer: 200)

10) If the average person throws away 2.5 pounds of garbage every day, how much garbage would the average person throw away in one week?

(Answer: 17.5 pounds)                                                                                                                                                                                                                                                                         

Learning Solution Set

Introduction for learning solution set:

Theory:

Set is an achievement of our recent mathematics. It appears in all branches.  It originated when mathematicians attempted to axiomatize mathematics within the frame work of logic. Mathematicians developed the theory of sets and solutions of set. It becomes a milestone in the growth of mathematics. Now we proceed to introduce the concept of a set. We are learning the solution of set by operations. Let us see about learning solution set in this article. Please express your views of this topic Union Set Theory by commenting on blog.


Learning Solution Set : Set Operations


We shall now study about the set operations

(i) Union of two sets

(ii) Intersection of two sets

(iii) complements

(i) Union of two sets

Let A and B be two given sets. The set of all elements that belong either to A to either to B or to A or to both is called the union of A and B. We denote the union of A and B by A U B. We are learning the union by following expression:

A U B = { x | x ∈ A or x ∈ B or x ∈ A and B}.

We write A U B = {x| x ∈ A or x ∈ B} where it is understood that the word or is used in the inclusive sense; that is, x ∈ A or x ∈ B stands for x ∈ A or x ∈ B or x ∈ A and B.


(ii) Intersection of two sets

Let A and B be two given sets. The set formed by the elements that are common to both A and B is called the intersection of A and B. We denote the intersection of A and B by A ∩B. We are learning intersection by following expression:

A ∩ B = {x | x ∈ A and x ∈ B}.

(iii) Complements:

Subtract the given sets A and B. Learning solution by A - B.

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Learning Example Solutions for Set Operations:


Find the solution for following examples:

1). If A = {1, 2, 3, 4} and B = {3,4, 6}, find A UB.

Solution:

List the elements of both A and B and avoid duplication.

Thus, A UB = {1, 2, 3, 4, 6}.

2). If A = {4,5,6,7} and B = {5,6,8}, find A∩B.

Solution:

All elements in A and B: 4,5,6,7,5,6,8.

Common elements in A and B:5,6

∴ A ∩B = {5,6}.

3). If A  = {2,3,4,5}, B={3,4,5}, find A - B.

A - B = {2}.

Rotational Symmetry

If an object retains the original shape when transformed, it is said to have symmetry. For example a vertical parabola when flipped across the vertical line passing through the vertex, it retains its shape. So we call that a vertical parabola is symmetrical over the axis through the vertex which is also called as axis of symmetry of the parabola.

Same way if an object retains the original shape when rotated over the center of the object by certain angle, it is said to have a rotational symmetry. It is also called as angular symmetry. This concept is dealt in the topic of rotational symmetry definition geometry. Let us take a closer look.

In real life there are many examples of rotational-symmetry. For example, a fan with certain number of blades has an angular symmetry because the angle between any two consecutive blades is constant. Rotating the blade by that much of angle we can observe that the shape of the fan appears to be same as it was originally.
In the above example, it was fairly easier to observe the angular symmetry. But, in general, how to find rotational symmetry? One has to keenly note that if the given shape looks same as before even when undergoes the transformation of rotation by some angle.
In certain cases, the original shape is retained repeatedly. That is, before completing one full revolution, the shape exhibits the originality a number of times. This ‘number of times’, is known as order of rotational symmetry of the shape. Consider an equilateral triangle. The shape remains the same after every rotation of 120o and hence in one complete revolution the shape of the equilateral triangle remains the same three times. Therefore, the order here of the angular symmetry in this case is 1. A square has that order as 4 whereas a rectangle has only two. A circle has the same order infinitely.

Now let us discuss an important point. An object comes back to the original shape only when it undergoes a complete rotation. Can we say that 1 is the order of angular symmetry for this object? No! A rotational symmetry for any object is said to exist only when it exhibits the original shape more than once in one complete revolution. For that matter, all the objects come back to the original shape after a rotation of 360o. Therefore, 1 as the order of angular symmetry does not exist and also it does not make sense. Thus, the minimum order for an angular symmetry is 2.

Polynomials

The word ‘poly’ that is often used in math is actually derived from Greek. ‘Poly’ in that language means ‘many’. So, if an expression has many terms, then it is called as polynomial. But the number of terms has to be finite. Generally up to three terms expressions have their own identities like a single term is called a monomial, two terms as binomial and three terms as trinomial.  Polynomials (abbreviated as Pnml) can also form the core part of a function, in such cases they are called as polynomial functions.

From what we described any expression or any function which has many terms could be Pnmls. Is this statement correct? No, there is a limitation. Only expressions in which no term is divided by any variable and do not contain a non-negative exponent terms can be referred as Pnmls.

In this topic of polynomials help, we will see some examples of polynomials and how to simplify polynomials.The following are the polynomials examples.

2x^3 + 3x^2 – 7x + 2
6x4 – 2x^3 + 4x^2 + 2x + 9
4x^3y^2 – x^2y^2 + 2xy - 14
5x^3 + (x^2/2) + 4x –7

In the third example we find three terms contain two variables but it does not matter as far as the definition of Pnml  is concerned.

In the last example one can notice that one of the terms (the second term) is divided by 2. But it does not affect the Pnml status of the expression as 2 is only a constant. Only when any term is divided by any variable cannot qualify to be a Pnml.
The following expressions cannot be called as Pnmls.
x^3 + 4x^2 – 7x + (2/x)
6x4 – 2x-3 + 4x^2 + 2x + 9
7x^3– 2x^2 + 5x^2/3 + x + 1
1 + (x/1!) + (x^2/2!) + (x^3/3!) + …….

In the first expression, the last term is divided by a variable. In the second expression we find a negative exponent in the second term. The third term of the the third expression has a rational exponent which cannot be simplified as an integer. The fourth expression has infinite number of terms. Hence all of them cannot qualify as Pnmls.

It is possible that a Pnml may have exactly similar terms excepting that the coefficients may be different. Those terms are called like terms. Such expressions can be simplified by algebraically adding the like terms. Similarly a Pnml can be simplified and expressed in factored form by identifying common factors. Simplified Pnmls show a better presentation.

Middle School Math Practice

Introduction to middle school math day:

Middle school class books are designed to presents the practice with math skills and concept wise explanations such as whole numbers, fractions, decimals, percents, geometry, and square roots. Middle school math tells about the measurement, property, and relations about quantities and sets, with numbers and symbols. It is a function of expressions. In this article we shall discuss about middle school math day problems.


Sample problem for middle school math day:


Example 1:

Solve the given equation 4× (x + 8 + 11) = 88.

Solution:

Grouping the terms like, the left side of the equation becomes
4 × (x + 8 + 11) ==> 4 × (x + 19)

Using the distributive property,
4 × (x + 19) ==> 4 × x + 4 × 19

Carrying out multiplications,
4 × x + 4 × 19 ==> to 4x + 76

The equation now becomes
4 x + 76 = 88.

Subtracting a 76 (adding a -76) to each side gives us
4x + 76 + (-76) = 88 + (-76)

4x + (76 + (-76)) = 88 - 76

4x + 0 = 13

4x = 12

Since the x value is multiplied by 4, we divide both sides by 4 to solve for x:
4x = 12

4x ÷ 4 = 12 ÷ 4

(4x)/4 = 3

x = 3.

We get the answer value is X = 3

We can check the answer in the solution of a original equation:
4× (3 + 8 + 11) = 88

4 × 22 = 88

88 = 88 so our answer is correct.

Problem 2:

Find the diameter of the circle. A radius of the given circle is 9 cm.

Solution:

Given:

Radius of the circle is r = 9 cm

Diameter of the circle is d = 2r

d = 2 * 9

d = 18

So, the diameter of the circle is 18 cm.

Problem 3:

The perimeter of a rectangle is 1200 meters and its length L is 5 times its width W. Find the values of W and L, and the area of the rectangle.

Solution:

Perimeter of rectangle=2L+2W,

2 L + 2 W = 1200

We now rewrite the statement. Its length L is 5 times its width into a mathematical equation as follows:

L = 5 W

We have to substitute L =3W in the equation 2 L + 2 W = 1200

2(5 W) + 2 W = 1200

12 W = 1200

W =100 meters

Use the equation L = 5 W to find L.

L = 5 W = 500 meters

Use the formula of the area.

Area = L x W = 500 * 100 = 50000 meters 2.

So, the area of the rectangle=50000 meters 2.


Practice problem for middle school math day:

Find the diameter of the circle. The radius of the given circle is 12 cm.

Answer: d = 24

Solve the given equation 9× (x + 3 + 11) = 99.

Answer: x = -3

Rectangle Word Problem

Introduction to Rectangle:

A Rectangleis any quadrilateral with four right angles. The term "oblong" is occasionally used to refer to a non-square rectangle. A rectangle with vertexes ABCD would be denoted as  ABCD.

We shall work out some problems on rectangles


Rectangle word problem 1:


A rectangle has a perimeter of 120 meters and its length L is 2 times its width W. Calculate the dimensions W and L, and the area of the rectangle.

Solution to rectangle word problem 1:

The formula to find the perimeter of rectangle is

2 L + 2 W = 120

Now, we rewrite the statement, length L is 2 times its width “w” into a mathematical equation as follows,

L = 2 W

We substitute L in the equation 2 L + 2 W = 120 by 2 W.

2(2 W) + 2 W = 120

Expand and group like terms.

6 W = 120

Solve for W.

W = 20 meters

Use the equation L = 2 W to find L.

L = 2 W = 40 meters

Use the formula of the area.

Area = L W = 40 * 20 = 800 meters 2


Rectangle word problem 2:


The perimeter of a rectangle is 70 feet and its area is 304 feet 2. Find the length L and the width W of the rectangle, such that L > W.

Solution for rectangle word problem 2:

The formula for perimeter of a rectangle is given by

2 L + 2 W = 70

and the formula area of a rectangle is given by

L W = 304

Divide all terms in the equation 2 L + 2 W = 35 by 2 to obtain

L + W = 35

Solve the above for W

W = 35 - L

Substitute W by 35 - L in the equation L W = 304

L(35 - L) = 304

Expand the above equation and rewrite with right term equal to zero.

-L 2 + 35 L - 304 = 0

The above is a quadratic equation with two solutions.

L = 16 and L = 19

Use W = 35 - L to find the corresponding values of W.

W = 19 and W = 16

Since L > W, the rectangle has the dimensions

L = 19 feet and W = 16 feet.


Practice rectangle word problem 1:


The perimeter of a rectangle is 90 feet and its area is 504 feet 2. Find the length L and the width W of the rectangle, such that L > W.

Practice rectangle word problem 2:

A rectangle has a perimeter of 240 meters and its length L is 5 times its width W. Calculate the dimensions W and L, and the area of the rectangle