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.

Right Angles Geometry

Introduction to right angles geometry:

In mathematics, the right angles are right angles triangle in the topic geometry. In geometry, when one of the angles is right angles in any of the three angles then it is called as right angles triangles. In geometry, the right angles have three sides and the two angles with one right angle. In geometry, the right angles have some special cases. The right angles with the special types and the formulas are given below.

Right Angles Geometry – Types:

In geometry, the right angles are classified into two different types. In geometry, the right angles are classified based on only by angles. The classifications of the right angles are given below.

The diagram for the right angles are given as,



Where,

x - Adjacent

y - Opposite

z - Hypotenuse. And

X & Y are angles.

In the right angles, the Pythagoras theorem is the theorem used to give about the right angles.

This theorem is explained by their sides and also the angles in the right angles.

The Pythagoras theorem for sides is given as,

x2 + y2 = z2

The Pythagoras theorem for angles is given as,

Sin A = x / z

Cos A = y / z

Tan A = x / y

The classification of the right angles is given below, Having problem with hard math problems for 12th graders keep reading my upcoming posts, i will try to help you.

Isosceles right angles

Scalene right angles

Isosceles Right Angles:

The structure for the isosceles right angles is given as,



In geometry, isosceles right angles are the right angles where this has two angles be 45 degree and one of an angles is 90 degree in geometry. The ratio of the sides for these types of is (1, 1, v (2)).

Scalene right angles:

The structure for the isosceles right angles is given as,



In geometry, scalene right angles are the right angles where this has one angles be 60 degree and another one is 30 and one 90 degree angles in geometry. The ratio of the sides for these types of is (1, v (3), 2).

Simple Introduction

Introduction to simple introduction of values

The simple introduction position value table by affecting a numeral single place from right to left, its place- value becomes ten times. In the similar method by poignant a digit singles position from left to right, its position-worth becomes one-tenth. So remote we have been discussion concerning seats from right to left beginning from ones place in the simple introduction. Extension of this table from left to right beyond ones place gives rise to the idea of decimals simple. I like to share this Simple Event with you all through my article.

Simple Introduction-basics

The simple introduction concept of decimals is also connected with the fractions denominators are 10, 100 or 1000 etc.

The place value represents the value of the numbers in the actual representations.

The face value represents the starting value of the numbers.

For example:

If we have 45 then the place value for 5 is 5 ones and the face value of 45 is 4

In this situation the educator may help kids to appreciate that a decimal can be spoken as a fraction and vice-versa.

The simple way to represent that "A decimal may surround a entire number module and a decimal module".

In a decimal, if whole number fraction is not specified, we designate it by inscription a zero on the left of the decimal position

For example, 803/100= 0.803;

The thought of simple percent is in use as an integral element of this section and require to be agreed a bit more significance as its utilize in our everyday life is extremely ordinary. Please express your views of this topic how to divide decimals with decimals by commenting on blog.

Basically simple introduction, it is not a original theme but its connection with the learn of fractions and decimals has a particular import. To complicated the notion of percent, we speak that percent earnings for each hundred or and indicate it by the symbol ‘ %’.

Simple Introduction-examples

Permit us think the place value of 4 in the numbers given below:

4365 The place value of 4 is 4000.

3465 The place value of 4 is 400.

3645 The place value of 4 is 40.

and 3674 The place value of 4 is 4.