Math For All Seasons

Introduction of math for all seasons:

The math for all seasons is nothing but the ways for solving the problem solving to children. It helps you to get the idea how to find the solution for the particular problem. The children can develop their ideas on grouping for the arithmetic operations like addition and subtraction. Also we give the idea to find the way to save the time on the particular problem. Please express your views of this topic Solving Equations Using Addition and Subtraction by commenting on blog.


Math for all seasons:


The children can have the idea to solve the pattern and symmetric problems. These topics give the various types of problems which makes you excel in each topic. Hence the math for all seasons gives you ideas for solving all the subtopics in math.


Example for math for all seasons:


Example 1: Compute the math partial sums of (355 +565).

3 6 6

4 8 5

Add 100s                  7 0 0

Add 10s                     1 4 0

Add 1s                       1 1

Add partial sums    8 5 1

Step 1: first sum up the place value of hundred = 700.

Step 2: Then sum up the ten place values = 140.

Step 3: Then sum up the one place values = 11.

Step 4: Finally sum up all the partial values to compute the solution = 851.

Is this topic Associative Property of Multiplication Example hard for you? Watch out for my coming posts.

Example 2: Compute the math partial sums of (243 + 134).

2 4 3

1 3 4

Add 100s                3 0 0

Add 10s                   7 0 0

Add 1s                     7 0 0

Add partial sums 3 7 7

Step 1: First sum up the place value of hundred = 300.

Step 2: Then sum up the place value of tens = 700.

Step 3: Then sum up the place value of ones = 700.

Step 4: Then sum up all the partial values to compute the solution = 377.

Example: What is +4+ (+2)?

From above: + (+) becomes a positive sign.

+4+ (+2) = +4 + 2

Start at +4 on the number line, move forward 2, and you end up at 6

Answer: 4+ (+2) = 6

Example: Find out the place value and the face value of the each digit in the number 78,234,543.

Solution:         Face value (digit)                                Place value (digit)

7                                                            70,000,000

8                                                                8,000,000

2                                                                   500,000

3                                                                      30,000

4                                                                         4000

5                                                                           500

4                                                                             40

3                                                                              3

The above sequence of number can be written as “seven eighty million two hundred and thirty four thousand and five hundred and forty three”.

Math Partial Sums

Introduction of math partial sums:

The math partial sums deal with the numerical values and their operations on math. This system involves natural numbers, integers. The arithmetic operations which deals with positive numbers and negative numbers. The natural numbers which includes all the whole numbers. The natural numbers sets are denoted as N. The number systems have the ability to count. The collections of some features are termed with number systems.


Math partial sums:


The math partial sums includes the two stages in which the first stage looks at the columns present, works from left to right and we should sum up the place values of the digits present in the particular column. The negative number systems are the one which is lesser than the zero.

The negative numbers are followed by the sign (-). The negative number systems are also denoted as the inverse of the number. The positive number systems are which are larger than zero. It is the most used number systems found in our day to day life. The second stage of the math partial sums includes the place values sum up together. Please express your views of this topic An Odd Number by commenting on blog.


Examples for the math partial sums:


Ex 1: Compute the math partial sums of (355 +565).

3 6 6

4 8 5

Add 100s                  7 0 0

Add 10s                     1 4 0

Add 1s                       1 1

Add partial sums    8 5 1

Step 1: first sum up the place value of hundred = 700.

Step 2: Then sum up the ten place values = 140.

Step 3: Then sum up the one place values = 11.

Step 4: Finally sum up all the partial values to compute the solution = 851.



Ex 2: Compute the math partial sums of (243 + 134).

2 4 3

1 3 4

Add 100s                3 0 0

Add 10s                   7 0 0

Add 1s                     7 0 0

Add partial sums 3 7 7

Step 1: First sum up the place value of hundred = 300.

Step 2: Then sum up the place value of tens = 700.

Step 3: Then sum up the place value of ones = 700.

Step 4: Then sum up all the partial values to compute the solution = 377.

Learn Math Facts Fast

Introduction to learn math facts fast:
A fact is a group of a declaration of each of which stand for the same meaning. All the declaration has the same meanings. Addition and subtraction facts contain 2 or 3 numbers. The types of facts operations are addition and subtraction facts, multiplication and division facts. In here we shall discuss about learn math facts fast. Having problem with Decimal to Percent Calculator keep reading my upcoming posts, i will try to help you.

Addends:

The facts being added in an addition operation like 7 and 4 in 7 + 4 = 11 or 4 + 7 = 11.

Reverse operation:

Reverse of a process is call as inverse operation of addition and subtraction.


Sample problem for math facts fast learn:


First we take an addition operation like 3 + 5 = 8.

In the next operation the addends are reversed like 5 + 3 = 8.

Now we can write the inverse operations of the above values both.

8 - 5 = 3

8 - 3 = 5

Addition and subtraction math facts fast learn:

Example 1:

What are the members of the fact family of 11 + 9 = 20?

Solution:

11 + 9 = 20

Reversing the addends:

9 + 11 = 20;

Inverse operations:

20 - 9 = 11;

20 - 11 = 9;

When we are modify the order to addition, then answer will be based on the numbers which we in use to do the addition process.

Example 2:

Write the facts of the family with numbers 9, 5 and 14.

Solution:

9 + 5 = 14;

5 + 9 = 14;

14 - 9 = 5;

14 - 5 = 9;

When 2 of the numbers are same, then we have only 2 facts. Because, the 1st two and last two fact values are same. Please express your views of this topic Cone Volume Formula by commenting on blog.


Division and multiplication math facts fast learn :


Example 1:

Write the facts of the family with numbers 11, 5 and 16.

Solution:

In the problem we find the facts value for division and multiplication.

11 * 5 = 55

5 * 11 = 55

55 / 11 = 5

55 / 5 = 11

When 2 of the numbers are same, then we have only 2 facts. Because, the 1st two and last two fact values are same

Example 2:

Write the four facts of the family with numbers 2, 7 and 9.

Solution:

7 * 2 = 14

2 * 7 = 14

14 / 2 = 7

14 / 7 = 2

When 2 of the numbers are same, then we have only 2 facts. Because, the 1st two and last two fact values are same

Equations With No Solution

Equations with no Solution

Intoducion:

There are two types of equation.one is equation with solution and other is equation without solution.in equation with solution we can find the value of variable like x,y,z.but in the equation with no solution we can  not find the value of variable like x,y,z.in the equation with no solution we can get situation like 0=0,4=5,7=8 etc...For equations with solutuion,we can solve the equation by three method.these methods are 1)Elimination method 2)Substitution method 3)Graphing method..by these method we can check that whether a equation has solution or not?

When we solve by Elimination method we have to make the coefficient of one the variable into equal and then add both the equations. Now one of the variables will get cancel and then one of the variables is left out. After that, solve for the left variable. Finally substitute the result of one of the variable in any one of the equation we will get the result for another variable. When we solve by substitution method substitute one of the equation in another and then solve. If we are solving by graphing method draw a graph for both the equation in a single graph. The intersection point of the lines is a solution for the given system of linear equation. So these are method of solving a linear equation.


DESCRIPTION OF EQUATION AND SOLUTION


Equation: . A statement in which values of two mathematical expressions are equal is called an equation sign of equal to is = equation with solution: Here we han find some solution for the given equation

X+5Y=2

X-12Y=10

From the above two equation we can find the value of two variable X, Y

equation with no solution:  a equation from which we cannot find the value of the variable is called any solution of equation.

X+Y=1

2X+2Y=2

From the above equation we cannot find the value of variable X,Y...since both the equations are same.to solve 2 variables we need two different equations. Having problem with Convert a Decimal to a Fraction keep reading my upcoming posts, i will try to help you.


Examples of no solution equation


Problem 1:4x-2+3x=3x+12+4x-14

Solution: 4x-2+3x=3x+12+4x-14

Step 1:  4x+2x-2=3x+4x+12-14

Step2:   7x-2=7x-2

Step 3:  7x-7x=-2+2

Step 4:  0=0

So there is no solution of this equation.

Problem 2:7x-4y+12z+4=5-3y+7x-y+12z

Solution:

Step 1:7x-4y-7x+3y+y+4=-12z+12z+5

Step2:0+4=0+5

Step 3:  4=5

So there is no solution of this equation.

Problem 3:

X+Y=2 (equation 1)

4X+4Y=8(equation 2)

Solution:

Step 1: multiply equation 1 by 4

Step 2: then we will get 4X+4Y=8

Step 3: now subtract this from equation 2

We will get 0=0

So there is no solution of this equation.

Geometry Final Exam

Introduction to geometry final exam:

Geometry is a branch of mathematics deals with the properties of and relationships between planes, points, and figures and examines their properties, measurement, and mutual relations in space with generalizations of these concepts. Plane Geometry is a flat shape of lines, circles, and triangles. Solid Geometry is about solid shapes like spheres and cubes. The group of mathematics, which investigates the relations, properties, and measurement of solids, surfaces, lines, and angles.


Lists of geometries:


The following are the different classifications of geometry involved in geometry final exam.

RIEMANNIAN GEOMETRY:

Riemannian geometry clear as the bough of degree of difference geometry that study smooth manifold, Riemannian manifold with a Riemannian metric with an inner product on the departure space at each point that varied smoothly from one point to another point. It should give in particular local notion of angle, surface area, volume, and length of curve.

ALGEBRAIC GEOMETRY:

Algebraic geometry is mainly used to show the numbers that can be used to describe points in a space or points in a plane and to express geometric relations in algebraic form.

DISCRIPTIVE GEOMETRY:

Differential geometry is mainly used for when the problem of representing three-dimensional objects on a two dimensional surface. Gaspard Monger invented it in the last 18 centuries.

DIFFERENTIAL GEOMETRY:

Differential geometry concepts is mainly used for calculus and are applied to surfaces, curves and other geometric objects

PROJECTIVE GEOMETRY:

Projective geometry is a modern geometry, also branch of mathematics that deals with relation ship between geometric figures and images of the result of projection.

NON EUCLIDEAN GEOMETRY:

A non-Euclidean geometry is study of constructions and shapes it should not map directly to any n-dimensional.

The following sample problems will help you for your final exam. I have recently faced lot of problem while learning Alternate Interior Angles in Real Life, But thank to online resources of math which helped me to learn myself easily on net.


Example problems for final geometry exam:


Example1:

Find the area of square for the side is 4 cm.

Solution:

Area of the square A = 4^2

A = 16 square centimeter.

Example2:

Find the area of sphere for the radius is 3 cm.

Solution:

Area of sphere A = 4 * 3.14 * 3^2

A = 4 * 3.14 * 9

A = 113.04 square centimeter.

Example3:

Find the surface area of cube for the side is 4 inches.

Solution:

Area of cube A = 6 * 4^2

A = 6 * 16

A = 96 square inches.

These example problems will helps for your geometry final exam.

Middle School Math proportions

Introduction to middle school math proportions:

A proportion is called an equation with two ratios on each side (these two ratios are equal). It is denoted by the symbol ‘::’ or ‘=’ to equate the two ratios.  If two ratios are not equal, then the given values are not in proportion. In a proportion, the four terms (quantities) involved. First and fourth terms are known as extreme terms. Second and third terms are known as middle terms.

For example proportion:   50: 100:: 2 : 4;

50, 100, 2, 4 are the four terms. 50 and 4 are the extreme terms. 100 and 2 are the middle terms.

Steps to solve proportions:

Solving proportions is the process of setting the two fractions equal to each other, cross-multiplying, and solving the resulting equation.

When one of the four numbers in proportions is unknown, using the cross product we have to find the unknown number. This is called solving the proportions. Letters or Question marks are commonly used in place of the unknown number.

Here, we are going to see some of the middle school math proportions problems.


Example on middle school math proportions


The following are the middle school math proportions problems:

Example problem 1:

In 2005 the number of men enrolled was 194 while women to men enrolled in chemistry are 78: 162 in 2004. How many women were enrolled in chemistry in 2005?

Solution:

Let x is the unknown number of women

78: 162 = x: 194

Cross multiply,

78 * 194 = 162 x

15132 = 162 x

Divide by 162 on both sides

93.4074 = x

So, the women were enrolled in chemistry in 2005 is 93.4074.Approximately it is 93 members. I like to share this Adding Mixed Fractions with you all through my article.



Few more example on middle school math proportions

Example problem 2:

Margret ran 100 meters in 15 seconds. How long did she take to run 1 meter?

Solution:

In 15 seconds Margret ran 100 meters. so x is the unknown.

100/15=1/x

Cross multiply and we get:  100x = 15

Divide both sides by 100 and we get

x = 15/100

x=0.15seconds

So, she takes 0.15 seconds to run 1 meter.

Example problem 3:

In a school, 476 students ride their bicycle to school. If this number is 70% of the school enrollment, then how many students are enrolled?

Solution:

476/y=70/100

Cross multiply and we get:  70y = 47600

Divide both sides by 70 and we get:  y = 680

There are 680 students enrolled in that School.

Expected Value Sum

Introduction to Expected value:

The expected value is nothing but the problem that has many possibilities of occurring  the expected value for the particular problem. Computing of expected value can be clearly explaining you through the probability topic in math, hence the probability has many possibilities of solutions. Thus, the dice and coins play a important role to help you better in expected value.

I like to share this Expected Value of Uniform Distribution with you all through my article.


Expected value Sum - Example Problems:

Problem 1: A perfect cubic die is thrown. Find the probability that

i)an even numbers comes up.

ii)a perfect square comes up.

Solution:

Since a die can result in six outcomes, n(S) =6. All these outcomes are equally likely.

Let A be the event, that an even number comes up.

Then        A = {2, 4, 6}

P(A) = n(A)/n(S)

= 3/6

= 1/2


Let B be the event that a perfect square comes up. Since the only perfect squares in S are 1 and 4,

Therefore B = {1, 4}.

Therefore n(B) = 2,

P(B) = n(B) / n(S)

= 2 /6

= 1/3.

Problem 2:

A box contains 7 red, 3 White and 2 black balls. When a ball is picked up at random from the box find the probability that,

i)it is Red.

ii)it is not Red.

Solution:

The total number of balls in the box is 7 + 3 + 2 = 12

Hence, n(S) = 12.

i)Let A denote the event that the ball drawn is Red.

Since there are 7 Red balls in the box, therefore, n(A) = 7

Therefore  P(A) = n(A) /n(S)

= 7 /12

ii)P (the ball is not Red) = P(A’)

= 1 – P(A)

= 1 – 7 /12

= 5 / 12

We also proceed directly as follows:

When the ball is not Red, it is from the remaining color that’s from the 3 white balls or 2 black balls.

Therefore n(A’) = 3 + 2 = 5.

Hence, P(A’) = n(A’) / n(S)

= 5 / 12.


Expected value Sum - Practice Problems:


Practice Problem: 1

It is found that, when A and B play a game of chess, the odds in favour of A winning the game are 3:5. Find the probability of
i) A winning the game.

ii) A not winning the game.

Answer :

i)     3/8
ii)    5/8
2. An unbiased coin is tossed. Find the probability that

i) A tail turns up,

ii) Both Head and Tail turn up,

iii) Neither Head nor Tail turns up.

Answer:

i)        1/2
ii)       1/2
iii)       0