Different Ways to do Division

Introduction to division:

Division is the process of frequent subtraction. It can also be described as inverse of multiplication.  Short ways of division is the process  of dividing a group into smaller equal groups.

Different ways to do division:

Division process is performed in different ways to do, such as.

1. Division operation performed through subtraction.

2. Division and number line.

3. Long division.

Now, we are going to learn the above different ways to do division methods.

First Different Ways - Division Operation Performed through Subtraction.

Example:

There are 18 balls and 3 boxes. John wants to put equal number of balls on each bax. How many balls will there be on each table?

Solution:

We have repeatedly subtracted 3 from 18 six times.
That is 18 divided by 3 equals 6.
`(8)/3 = 6`
Thus division is repeated subtraction.
In the division fact `18/3 =6`

Answer:
18 is the dividend       3 is the divisor
6 is the quotient        Here, the remainder is zero.

Second Different Ways - Division and Number Line.

Just as we use the number line for sddition, subtraction, and multiplication, we can use it for division also.

Example:
How many 2's are there in 14?
Solution:

Starting from 14 we can reach 0 by 7 jumps of 2 steps each.
Thus `14 /2 = 7`

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Third Different Ways - Long Division:

Division by a single digit number (without remainder) through long division method.
Example:

Divide 64 by 2.
Solution:

Steps for long division:

1. Arranange the numerals as shown.
2. Divide 6 tens by 2.` 6/2 = 3` Write 3 in tens place in the quotient.Write the product 6 tens (3 x 2 = 6) below the dividend in the tens place. Subtract the product from the dividend.  6 - 6 = 0.
3.  bring down 4 ones. Divide 4 ones by 2. `4/2 =2` ones. Write 2 in ones place in the quotient. Write the product 4 ones (2 x 2 =4) below the dividend in the ones place. Subtract the product from the dividend. 4 -4 = 0

Verification:
Here dividend = 64, divisor = 2, quotient = 32, remainder = 0. We have dividend = (quotient) x (divisor) + remainder.
64 = 32 x 2 + 0

64 = 64

Hence verified.

Problem Solving Fifth Grade

Introduction to fifth grade math:

In the level of fifth grade the students will learn about decimals, fraction, and addition using decimal numbers, subtraction using decimal numbers, multiplication and division using decimal numbers. Basic operation of area, circle, volume, and perimeter will be come under the level of fifth graders math. In this article the students solving problems from fifth grade. I like to share this Roman Numeral Calculator with you all through my article.

Solving Fifth Grade Problems:

Problem 1:

Solving the given problem:

The length of the rectangular slab is 150 cm and the breadth of the rectangular slab is 50 cm. find the perimeter of the rectangular slab.

Solution:

Length of the rectangular slab = 150 cm.

Breadth of the rectangular slab = 50cm

Hence, the perimeter of the rectangular slab is given by

Perimeter P = 2 x (l + b)

= 2 x (150 + 50) cm

= 2 x 200 cm

= 400 cm.

Problem 2:

Solving the given problem

David purchased a wooden box. It is square shape and its side is 4 m. find the perimeter of the wooden box.

Solution:

Side of the square = 4 m.

The perimeter of the wooden box is given by,

Perimeter P = 4 x side

= 4 x 4m

= 16 m

The perimeter of the wooden box = 16 m. Understanding math online help for free is always challenging for me but thanks to all math help websites to help me out.

Solving Fifth Grade Problems:

Problem 3:

Solving the given problem

Find the area of the given rectangle. The length and breadth of the rectangle is 10 cm and 5 cm respectively.

Solution:

Length of the rectangle = 10cm

Breadth of the rectangle = 5 cm

Area of the rectangle = l x b

= 10 x 5 sq cm

= 50 sq cm.

Problem 4:

The length of the cuboid is 7 cm, breadth of the cuboid is 3.5 cm, and the height of the cuboid is 4 cm. find the cuboid volume.

Solution:

Length of the cuboid = 7cm

Breadth of the cuboid = 3.5 cm

Height of the cuboid = 4 cm

Volume of the cuboid = l x b x h

= 7 x 3.5 x 4 cu cm.

= 98 cu cm.

Problem 5:

Find the volume of the cubical container. Its side is 3.4 cm.

Solution:

Side length of the cube = 3.4 cm.

Volume of the cube = l x l x l

= 3.4 x 3.4 x 3.4 cu cm

= 39.304 cu cm.

Multiplying Fractions

Fraction is an important mathematical concept taught in middle school. Any object when divided into two parts, each part is called a fraction. For example: 4/5 of babycare products are manufactured keeping hygiene as a criterion. Here, 4 is the numerator and 5 is the denominator. When numerator is smaller than the denominator, it is called a proper fraction. 4/5 is a proper fraction. When numerator is larger than the denominator, it is called improper fraction. For example: 5/4 of parts of this collection include baby care products. 5/4 is an improper fraction. All four basic operations in mathematics can be performed with fractions. Let’s have a look at the concept of multiplying fractions along with relevant examples in this post.

Steps to Multiply Fractions:
Unlike addition and subtraction, multiplication of fraction is not based on denominators. Multiplying Fractions include three steps; steps are mentioned below.
• Step 1: Multiply the numerators
• Step 2: Multiply the denominators
• Step 3: Simply the result fraction if needed

Examples of Multiplication of Fractions:
1. Example: Rahul’s mother was relieved after she came to know that her 2 months age can baby hear sound properly. She felt bad for the deaf kids and bought 2/5 of handmade products from each of the 12 kids. How many products she bought? 2/5 * 12/1 (Every non-fraction number carries 1 as a denominator) 2/5 * 12 8 is the product.

2. Example: 25/40 part of the book includes potty training tips and 12/15 books were distributed. How many parts of the total books distributed include potty training tips ? 25/40 * 12/15 300/600 1 / 2 is the product. 3. Example: 11/12 part of the property needs to be distributed among 22/23 members. Find the product value: 11/12 * 22/23 242/276 121/138 is the product. These are the basics about multiplication of fractions.

Problem Solving Manipulatives

Introduction to Problem Solving Manipulatives

Problem Solving manipulatives in classrooms while focusing on how children performing resposibilities manipulatives and tools in problem solving while working on mathematical tasks. Ways for teachers to access performance based tools in manipulative use will be examined and suggested. Recent examiners reveals that teachers need to teach and assess children’s mathematical knowledge in ways that will allow them to perform what they really understood. Teachers must able to see beyond understandable correct or incorrect answers into children’s thinking process. Acquisition  of mathematical knowledge through problem solving manipulative has long been considered to be time-consuming and labor intensive for many classroom teachers. Is this topic logarithm problems hard for you? Watch out for my coming posts.

The Role of Teachers in Problem Solving

Most teachers seems to teach problem solving methods as a series of steps or in continue approach, while most students need not just the linear set of steps as to follow, but also a full collection of ongoing, supported opportunities to eventually develop and hone problem solving techniques. This in depth development of problem solving does not denote fully understand by the way of mathematical task at other hand, nor does it imply that it is done in isolation; rather it is usually accomplished through engaging problems in which children connect new and previous information I have recently faced lot of problem while learning Converting Percentages to Fractions, But thank to online resources of math which helped me to learn myself easily on net.

Problem Solving with Fun Manipulatives

Problem Solving with fun manipulatives focuses for children to learn efficiently with  diffferent way of attraction, sometimes need to plan out of ordinary ways to teach in mathematical ideas. The intent of fun activities that will get children’s attention to make problem solving funs in mathematical area.

How do you write this number using digits?

Seventy-three

73

Which even number comes next?

0,   2,   4,

6

Manipulatives for Problem Solving

To understand  and use the fundamental problem solving manipulatives,below as there are groups in your class. Cut  the individual problem around their borders. Place each set of students steps in a glass on the students table.

For Example,

Jane has 3 more times today than she had yesterday. If she had  6 times yesterday, how many times does jane have today?

Solution:

The problem asks for the number of times jane has today.the first sentence provides response to the request.

Today=yesterday

The equation requests information about “yesterday” .the response to this request is  available from the second sentence. It states that she had 6 times yesterday

Yesterday=6

Substitution provides

Today= 6+3

Doing the arithmetic gives

Today=9

The student will learn the basic problem solving process. Specifically the activity shows the how each step follows from the previous step and leads to the next step. In addition, the inclusion of verbal statements in the solution presentation will become natural and ease to understand.

Inductance Value

Introduction to Inductance Value:

Inductance is defined as opposing of changes of current. The inductance value is represented as L and its unit is Henry. One Henry value is equal to the induced one volt by changing of current in one ampere per second in an inductance value. The inductance value is of two types. One is self inductance and another one is mutual inductance. Let us see the applications of inductance value. I like to share this Absolute Value Equations with you all through my article.

Value Applications of Inductance

Inductance value most probably used in the analog circuit.
Mostly applicable for power supplies conjugate with filter capacitor.
Inductance value is used to depress the voltage from lightning stroke.
Used to reduce the switching current and fault current in the transmission line.
Let us see the formula used for getting the value of inductance.

Formula – Inductance Value

L = (d2n2) / (l + 0.45d)

Where

L is the inductance value.

d represented as diameter of the coil.

n represented as number of turns in the coil.

Self Inductance Value:

L = N ( φ/ I)

Where:

L is in Henry

N is the Number of turns

Φ is denoted as Magnetic Field linkage

Ι  is represented as Amperes

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Example Problem – Inductance Value

Example 1:

A hollow air cored inductor coil consists of 600 turns of copper wire which produces a magnetic flux of 20mWb when passing a DC current of 10 amps. Calculate the self-inductance of the coil in milli-Henries.

Solution:

L = N ( φ/I )

L = 600 ( 0.02/10 ) = 1200 mH.

Answer: 1200 mH

Example 2 : Inductance Value

Find the inductance value whose diameter of coil is 2.4 cm and 150 numbers of turns in the coil.

Solution:

L = (d2n2) / (l + 0.45d)

L = (2.42 *1502)/ ( 1 + 0.45 (2.4))

L = 62.30767mH

Answer: 62.307 mH

These are the example problems of inductance value.

Instrument Used to Measure Angles

Introduction to Geometry and Angles

Geometry is the branch of mathematics that studies about various shapes and sizes. It is concerned with properties of various shapes like length, areas and volumes.

Angle is a meeting point of two rays. If two rays have a common end point we say an angle is formed. The rays are called the arms of the angle and the common meeting point is the vertex of the angle. Most commonly used unit to measure an angle is the degrees denoted as o). One full circle has 360 o. Smaller units to degrees is the minutes and seconds. 1o  has 60 minutes and one minute has 60 seconds. This minutes and seconds should not be confused with the units for time.There are also other units like radians, steradians etc. In this discussion we will use only degrees. I like to share this What are Adjacent Angles with you all through my article.

In this discussion we will briefly see the instruments used for measuring an angles.There are five instruments used for measuring the angle as below and the most commonly used is the protractor. A protractor is semi circular in shape and is marked from 0 o to 180 o.

It is very easy to use a protractor. Place the protractor with the base line of the protractor over one of the rays or arms of the angle and the centre point of the base line at the vertex of the angle. The other arm of the angle will coincide with one of the lines in the protractor and you can read the angle marked in the protractor.

Protractors can also be used measure three dimensional angles. Lets us take a brief look at the other instruments used to measure angles

Instruments to Measure Angles

The main instruments to measure angle are discussed below

Inclinometer : An inclinometer is used to measure angles of inclination with respect to the vertical. It is also known as a level gauge. Inclinometer is widely used in surveying, geophysics, automotive safety systems, aircrafts etc to measure inclinations.

Gonimeter :A gonimeter helps to rotate an object to a particular angle. The word gonimeter in Greek means angle measure

Compass :Compass uses the Earth’s magnetic field and is widely used in navigation. A compass is any device that can sense the Earth’s magnetic field and indicate direction , A magnetic compass is a very old instruments and it traces its origin somewhere around 247 BC. But still it finds application in modern navigational systems like GPS (Global positioning System)

Sextant :A sextant can measure angle between two objects and is primarily used in determining angles of celestial objects

Theodolite : A theodolite is used in surveying to find the angles of planes. There are modern theodolites fitted with telescopes and can measure angles precisely.

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Some Practical Exercises on Measuring Angles

Prob 1: Can you think of a situation where you need to measure an angle in day to day life?

Hint: Think of carpentry, plumbing, masonry work etc

Prob 2: Can you think of activities or instruments where we make use of our intuitive power on judging angles?

Hint: some examples are we read time in a dial clock just by glance as we can guess the angle between the needles and judge the time, a drive turns the steering at an angle depending on the angle by which vehicle needs to be turned. Think of other examples on these lines

Prob 3: Make a cut out of a protractor on a cardboard and try making graduations at angle of 30o, 45o, 60o, 75o, 90o, 105o, 120o, 135o, 150o and 180o

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.