Positive Negative Zero Sequence

Introduction for positive negative zero sequence:

In this article, we will discuss about the sequence. It has two types.

1. Arithmetic sequence and

2. Geometric sequence.

Arithmetic sequence means that, the sequence of a numbers such that the difference between two consecutive members of the sequence is a constant. Geometric sequence means that, the sequence of a numbers such that the ratio between two consecutive members of the sequence is a constant. The positive negative zero sequence formulas and example problems are given below. I like to share this Adding Positive and Negative Integers with you all through my article.

Formulas and Example Problems for Sequences:

Sequences formulas are given below.

Formula for arithmetic sequence:

nth term of the sequence : an = a1 + (n - 1)d

Series of the sequence: sn = `(n(a_1 + a_n))/2 `

Formula for geometric sequence:

nth term of the sequence: an = a1 * rn-1

Series of the sequence: sn = `(a_1(1-r^n))/(1 - r)`

Example problem 1:

Find the 10th term of the Positive arithmetic series 11, 14, 17, 20,....

Solution:

First term of the series, a1 = 11

Difference of two consecutive terms, d = 14 - 11 = 3

n = 10

The formula to find the nth term of an arithmetic series, `a_n = a_1 + (n-1)d`

So, the 10th term of the series 11, 14, 17, 20,... = 11 + (10 - 1) 3

= 11 + 9 * 3

= 11 + 27

After simplify this, we get

= 38

So, the 10th term of the sequence 11, 14, 17, 20,... is 38

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More Example Problems for Positive Negative Zero Sequences:

Example problem 2:

Find the 12th term of the negative arithmetic series -13,-15, -17, -19,.....

Solution:

First term of the series, a1 = -13

Difference of two consecutive terms, d = -15 + 13 = -2

n = 12

The formula to find the nth term of an arithmetic series, `a_n = a_1 + (n-1)d`

So, the 12th term of the series -13,-15, -17, -19,..... = -13 + (12 - 1) (-2)

= -13 + 11 * -2

= -13 - 22

After simplify this, we get

= -35

So, the 12th term of the sequence -13,-15, -17, -19,.... is -35.

Example problem 3:

Find the 14th term of the zero arithmetic series 0, 10, 20, 30, 40,...

Solution:

First term of the series, a1 = 0

Difference of two consecutive terms, d = 10 - 0 = 10

n = 14

The formula to find the nth term of an arithmetic series, `a_n = a_1 + (n-1)d`

So, the 14th term of the series 0, 10, 20, 30, 40,... = 0 + (14 - 1) (10)

= 0 + 13 * 10

= 130

After simplify this, we get

= 130

So, the 14th term of the sequence 0, 10, 20, 30, 40,... is 130

The above examples are helpful to study of positive negative zero sequences.

Number Pattern Formula

Introduction to number pattern formula:

In this section we will see about number pattern formula. Number pattern formula are helps us to calculate the number pattern series. Number pattern is one of the important topics in math. Let us study about number pattern formula with solved example problems along with neat and clear explanation and practice problems with answer keys. I like to share this Convert Roman Numerals with you all through my article.

Example Problems for Number Pattern Formula:

Example problem 1: Find the missing number in this sequence: 5, __, 125, 625, 3,125

Solution:

First, look for a pattern. Notice how each number is 5 times the previous number: 5, __, 125, 625, 3,125

Multiply 5 by 5 to find the missing number: 5 × 5 = 25

To make the pattern complete, the number 25 must go in the blank.

Answer: To make the pattern complete, the number 25 must go in the blank.

Example problem 2: What is the next number in this pattern?

Rule: multiply by 4, and then subtract 2

1, 2, 6, 22, 86...

Solution:

Rule: multiply by 4, and then subtract 2

Use the rule to find the next number in the pattern.

Multiply 86 by 4, then subtract 2.

(86 × 4) – 2 = 342

The next number in the pattern is 342.

The pattern is: 1,   2,   6,   22,   86,   342

Answer: The pattern is: 1,   2,   6,   22,   86,   342

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Practice Problems for Number Pattern Formula:

Practice problem 1: Find the missing number in this sequence: 4, 16, 64, __, 1,024, 4,096

Practice problem 2: What is the next number in this pattern?

Rule: multiply by 5, and then add 2

1, 7, 37, 187, 937...

Practice problem 3: Rheanna's Hardware Store ordered 10 power drills in August, 13 power drills in September, 16 power drills in October, and 19 power drills in November. If this pattern continues, how many power drills will the store order in December?

Practice problem 4: A restaurant used 3 onions on Saturday, 9 onions on Sunday, 27 onions on Monday, and 81 onions on Tuesday. If this pattern continues, how many onions will the restaurant use on Wednesday?

Solutions for number pattern formula:

Solution 1: To make the pattern complete, the number 256 must go in the blank.

Solution 2: The pattern is: 1, 7, 37, 187, 937, and 4,687

Solution 3: Rheanna's Hardware Store will order 22 power drills in December.

Solution 4: The restaurant will use 243 onions on Wednesday.

Mathematics Problem Solving Strategies

Introduction to Mathematics problem solving strategies:

A mathematical problem becomes easy to solve when we know the solving strategies. Certain procedures are used to attain the solution of the problem. The way of solving the problem is very important in order to get exact solutions.  Order of operations, basic arithmetic operations, formulas  etc., are most important basic requirements to solve the problems.

Mathematics Problem Solving Strategies:
Mathematics problem 1:

David has written a number pattern that begins with 1, 3, 7, 13, 21. If he continues this pattern, what are the next two numbers in his pattern?

Strategy 1: UNDERSTAND

what do you need to find?

You need to find 2 numbers after 21.

Strategy 2: PLAN

How can you solve the problem?

You can find a pattern. Look at the numbers. The new number depends upon the number before it.

Strategy 3: SOLVE

Look at the numbers in the pattern.

3 = 1 + 2 (starting number is 1, add 2 to make 3)

7 = 3 + 4 (starting number is 3, add 4 to make 7)

13 = 7 + 6 (starting number is 7, add 6 to make 13)

21 = 13 + 8 (starting number is 13, add 8 to make 21)

New numbers will be

21 + 10 = 31

31 + 12 = 43

The next two numbers are 31, 43.

Mathematics problem 2:

You save $4 on Sunday. Each day after that you save twice as much as you saved the day before. If this pattern continues, how much would you save on Thursday?

Strategy 1: UNDERSTAND

you need to know that he save $4 on Sunday. Then you need to know that you always save twice as much as you find the day before.

Strategy 2: PLAN

How can you solve the problem?

You can make a table like the one below. List the amount of money you save each day. Remember to double the number each day.

So, You saved $64 on Thursday.

Practice Problem:

Mathematics practice problem:
Sanjay has written a number pattern that begins with 1, 3, 5, 7,  If he continues this pattern, what are the next four numbers in his pattern?

Answer: 9, 11, 13, 15

Discrete Mathematics Set Theory

Introduction to Discrete Mathematics Set Theory

In discrete mathematics, the set theory are the branch of mathematics that learned about the sets, which are the collections of objects. Even though any type of objects can be collected into a set, set theory is applied most often to objects that are related to mathematics. Now we will see the examples of discrete mathematics set theory.(Source: Wikipedia).

Venn Diagrams for Discrete Mathematics Set Theory
A∪B

The A∪B means all elements of A and B sets. Venn diagram for A∪B is as follows,

A∩B

The B∩C is a common element of the sets. Venn diagram representation is given below,

 

A –B

The A-B is the values of A which is not in B set. Venn diagram representation is given below,

Examples for Discrete Mathematics Set Theory

1)Proof the A∪(B∩C)=(A∪B) ∩(A∪C) for the following sets.

A={3,6,9,15,17} B={12,13,14,19,21} C={3,7,8,10,12}

Solution

The given sets are A={3,6,9,15,17} B={12,13,14,19,21} C={3,7,8,10,12}

Given condition is A∪(B∩C)=(A∪B) ∩(A∪C)

Take the left side condition.

A∪(B∩C)

B∩C

Take the common values of the sets B and C.

B∩C={12}

A∪(B∩C)

Now joining the A set values.

A∪(B∩C)={ 3,6,9,12,15,17}   ----(1)

Take right side condition.

(A∪B) ∩(A∪C)

A∪B={3,6,9,12,13,14,15,17,19,21}

A∪C={3,6,7,8,9,10,12,15,17}

(A∪B) ∩(A∪C)={3,6,9,12,15,17}  ----(2)

Therefore A∪(B∩C)=(A∪B) ∩(A∪C).

2) Proof the A∪(B∩C)=(A∪B) ∩(A∪C) condition for the sets A={3,6,7,9,10} B={1,4,5,7,8} C={3,6,8,9,10}. Please express your views of this topic board of secondary education ap by commenting on blog.

Solution

The given sets are A={3,6,7,9,10} B={1,4,5,7,8} C={3,6,8,9,10}

Take left hand side condition.

A∩(B∪C)

B∪C={1,3,4,5,6,7,8,9,10}

A∩(B∪C)={3,6,7,9,10}   ----(1)

Now take the right hand side condition.

(A∩B) ∪ (A∩C)

A∩B={7}

A∩C={3,6,9,10}

(A∩B) ∪ (A∩C)={3,6,7,9,10}   ----(2)

Therefore A∩(B∪C)=(A∩B) ∪ (A∩C).

3)A={1,4,7,9,10} B={3,6,9,12,14} and C={2,7,8,12,13}. Find the i)A-(B∪C) ii)B-(A∪C).

Solution

The given sets are A={1,4,7,9,10} B={3,6,9,12,14} and C={2,7,8,12,13}.

i) A-(B∪C)

B∪C={2,3,6,7,8,9,12,13,14}

A-( B∪C) condition is a difference of set. It means we select the values from the A set. But that value is not present in the B∪C set.

So A- B∪C={1,4,10}

ii)B-(A∪C)

A∪C={1,2,4,7,8,9,10,12,13}

B-(A∪C)={3,6,14}

These are the examples of discrete mathematics set theory.

Vector Field Theory

Let (F, +, ∙) be field. The elements of F will be called scalars. Let V be a non-empty set whose elements will be called vectors. Then V is a vector space over the field F, if

There is defined an internal composition in V called addition of vector field and denoted by ‘+’. Also for this composition V is an abelian group.
There is an external composition in V F called scalar multiplication and denoted multiplicatively i.e., aα `in`  V for all α`in`  F and for all α `in`  V. in other words V is closed with respect to scalar multiplication.
The two composition i.e., scalar multiplication and addition of vectors satisfy the following postulates
(i)            a (α + β) = aα + aβ `AA`a `in` F and `AA`  α, β `in`  V

(ii)          (a + b) α = aα + bα `AA`  a, b `in` F and `AA`  α `in`  V

(iii)         (ab) α = a (bα) `AA`  a, b `in` F and `AA`  α `in`  V

(iv)         1α = α `AA`  α `in` V and 1 is the vanity element of the field F.

When V is vector space over the field F, we shall say that V(F) is a vector space. If the field F is understood we van simply say that V is a vector space.

Example for Vector Field Theory

Show that a field F may be considered as a vector space over F if scalar multiplication is identified with field multiplication

Solution

Let (F, +, ∙) is a field. Take F as the set of vectors and also as the set of scalars. Take the addition operation on the field F as the addition of vectors and the multiplication operation on the field F as the operation of scalar multiplication i.e., multiplication of a vector by a scalar. Then F is a vector space over F as shown below I have recently faced lot of problem while learning how big is 8mm, But thank to online resources of math which helped me to learn myself easily on net.

Since (F, +, ∙) is a field, therefore (f, +) is an abelian group

Further if a, b are any scalars i.e., a, b `in`  F and α, β are any vectors i.e., α, β `in`  F then

a (α + β) = aα + aβ and (a + b) α = aα + bα

These results follow from the distributive laws in the field F.

Also (ab) α = a (bα) because the multiplication on the field F is associative

Also if 1 is the unity element of the field F and α is any vector i.e., α `in`  F, then 1α = α

Hence F(F) is a vector space

Probability Grade

Probability Introduction:
The probability is number of possible events divided into total number of possible events. Probability contains two types of distributions. These are the discrete and continuous distribution.  The general formation of the grade 3 probability

The probability event P (A) = `("No. of possible events n(a)")/("Total no. of events n(s)")`

Examples on Probability Grade 3

Dice Problems:

Grade 3 Probability Example 1:

Roll a single dice; find the probability of get number 6.

Sol:

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` = 0167.

Grade 3 Probability Example 2:

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

Sol:

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` .

Grade 3 Probability Example 3:

Roll a single dice; find the probability of get number 5.

Sol:

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

n (s) = 6

The number of outcomes n (a) = {5}

n (a) = 1

The probability of getting value = `1/6` = 0.167.

Grade 3 Probability Example 4:

Roll a single dice; find the probability of get number 4.

Sol:

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

n (s) = 6

The number of outcomes n (a) = {4}

n (a) = 1

The probability of getting value = `1/6` . Please express your views of this topic what is differentiation by commenting on blog.

Coins Problems on Probability Grade 3

Grade 3 Probability Example 5:
To toss a coin and find the probability of one head. The possible outcomes are:

Sol:

Step 1:

n (s) = {T, H} = 2

Step 2:

Tossing a coin with only one head:

n (a) = {H}=1

Step 3:

Formula:

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

Answer:

P (A) = `1/2` .

Convert into a decimal 0.5

The probability of one head is 0.5 or Rounded 50%.

Grade 3 Probability Example 6:
Toss a coin and find the probability of one tail. The possible outcomes are:

Solution:

Step 1:

n (s) = {T, H}=2

Step 2:

Tossing a coin with only one tail:

n (a) = {T}=1

Step 3:

Formula:

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

Answer:

P (A) = `1/2` .

Convert into a decimal 0.5

The probability of one tail is 0.5 or Rounded 50%.

Laws of Exponents

Let us understand the term exponents first, by directly jumping on the example as 3 rises to the power of 2. It can also be written in mathematics like 32. Let us understand the base and the laws of exponents as, this bigger number 3, we call this a base. And the smaller number that is 2 up here we call that the exponent. Just remember, that the bigger number is the base and the smaller number that is at the right hand corner is the expnt. So that is exactly do we use expnt. for? It tells us the total number of factors the base has. It is basically a short cut for repeated multiplication of the same number. In this case, it will be 3 times 3 as the expnt. is 2 so two times the base will get multiplied.

Let us understand how do we read this 24, “ 2 to the 4th power” here the base is 2  and the expnt. Is 4 though here the number 4 is bigger than the number 2 but the base is 2 thus 2 will be multiplied 4 times. That is 2times 2 times 2 times 2 equals to 16. Hence 16 is the solution.

Let us understand the laws exponents, when we multiply numbers with expnts. that have same base, just add the expnts. and keep the same base given. Laws of exponents problems are as follows, example 32 times 33 equals 3 (2+3) equals to 35. Multiplication law in exponents is quite simple just memorizing the thumb rule of adding the expnts.

When we divide numbers with expnts. That has the same base, just subtract the expnts. While keeping the same base. Example 26 divided by 22 equals 2(6-2) equals to 24.

When in expnt of an expnt, we multiply the expnts. for example, (42)3 equals to 4 (2 times 3)  that is nothing but 46. Is this topic coordinate planes hard for you? Watch out for my coming posts.

When we notice any negative expnt. Put 1 over that number to make a positive expnt. For example 7 -2 equals 1/72 equals to 1/ 14.

Laws are so simple to remember, look here we look back again to all we just studied. When we multiply the same base, add the expnts. When we divide the same base, subtract the expnts. for an expnt of an expnt. We do multiplication and for the negative expnt. Take 1 over the number to make it positive which becomes simpler and easier for solving tough equations. Laws of exponents lesson are for to remember the rules while solving the equations which will solve the problems quicker.