Elementary math concepts

Elementary math covers all the basic operations, function and concepts in algebra. In elementary math is the main part covered is arithmetic operations. All the basic operations are presented in this area. This is covering all basic operation of addition multiplication, subtraction, and division. The elementary algebra covers all concepts from kindergarten level to middle school level. This is also help to solve for real life math problem.

Example: 4x+ 2 = 1

Elementary Math Concepts Covers:

Elementary math concepts contain this basic operations

Arithmetic Operations:

The real numbers have the following properties:

a + b= b +  a    ab  = ba                            (Commutative Law)

(a+ b)+ c= a+ (b + c)      (ab)c = a(bc)        (Associative Law)

a+(b +c)= ab +ac                                       (Distributive law)

Fractions:

To add two fractions numbers with the same denominator, we use the Distributive Law property:

a/b+c/b= 1/b*a+1/b*c  =1/b(a+c)  =a+c/b

To add two fraction with different denominators, we use a frequent denominator:

a/b+c/d =  ad+bc/bd

Factoring

Here we can make use of Distributive Law to increase certain algebraic conditions. In rare case we need to repeal this method (again using the Distributive Law) by factoring an expression as a product of simpler ones. The easiest condition occurs when the given expression has a common factor as given below...

3x(x-2)=3x2 – 6

Elementary Algebra Concepts Problems

Q 1 :   Find the larger number, -13 or -16?

Sol :  If a number has a negative sign, we dont conclude the number by ony seeing the value, here we have to consider the negative sign. If the value of a number with negative sign increases, the actual value of the number is decreses.

-16,-15,,-14,-13,-12,-111,-10,-9,-8,-7,-6,-5,-4,-3,-2,-1,0

so,-13 is larger than -16.

Q 2:  List all the integers between -2 and 4.

Sol :     -2,-1, 0, 1,2,3,4 these number are present in the -2 ,4

The -2, 4 between numbers are -1,0,1,2,3

Q 3:  Reduce 14/35.?

Sol :      14/35

We divide 7 on both sides

2/5

Q 4: Simplify 8 -: 2//3?

Sol :  We divide the given equation this is simple method the division inverse of multiplication

8

4*3   = 12

Q 5:  Simplify 1/2 + 2/3?

Sol  :      1/2+ 2/3

We take L.C. M   on 2, 3

1*3/2*3 +2*2/3*2

=3/6+4/6 =3+4/6

=7/6

Q 6.12x=4?

Sol :    x=4/12

x=1/3

Q 7:    3X+4y=5x-2y ?

Sol :         4Y+2y=5x-3x

6y=2x

y=2x/6

y=x/3

Q 8 :  x-4=8?

Sol : We add +4 on both sides

x=8+4

x=12

Q 9:  Factorize the given expression x2-9   ?\

Sol :  The general form of

(a2-b2)=(a-b)(a+b)

So, x2-9=  (x-3)(x+3)

Adjacent Angles

Definition for Adjacent Angles states that two angles are said to be adjacent, if:
• Both the angles are formed using the same side
• Both the angles have a same corner point i.e. the vertex
• Both the angles do not overlap on each other i.e. they should not have any interior point in common.

In simple words, Adjacent Angles Definition states that angles that are formed side by side using a common ray coming out of a common vertex in such a way that the common ray is between two other rays that forms the angles without any overlapping.

If two angles are given, they are said to be adjacent if they are only as per the Definition Adjacent Angles has stated.  There are certain scenarios when two angles do not satisfy the conditions stated in the Adjacent Angles Definition, they are:
1. Two angles share a common corner point or vertex but do not share a common side.
2. Two angles share the same side to form the angles, but do not have a common point at one of its corners.
3. Two angles given, in which one angle overlap the other.
The above cases that do not satisfy the conditions stated in the definition of adjacent angles can be declared as not adjacent. So these angles are not adjacent to each other.

Example of Adjacent Angles
An example of Adjacent-Angles helps to understand the concept in a better way. Consider three rays A, B, C coming out of the common vertex O. Two angles namely Angle x and Angle y are formed in such a way that angle x is formed between the sides OA and OB whereas angle y is formed using the sides OB and OC.  Here the vertex O is used as common and the side OB is used in common to form both the angles.

Adjacent Angles as Complementary Angles
When there are two adjacent-angles given with common vertex and common side, find the sum of the two angles.  If the total of the two angles is ninety (90) degrees and if it forms a right angle, then these adjacent-angles are said to be complementary and are termed as complementary angles. We can call it as adjacent complementary angles too.

Adjacent angles as Supplementary Angles
If two adjacent-angles are given, we can say that these adjacent-angles are supplementary angles if the total of the two adjacent-angles given is hundred and eighty (180) degrees and forms a straight angle. We can call it as adjacent supplementary angles.

Hypotenuse of a Right Triangle in brief

There are different types of triangle which we have learnt; one of them is the right angled triangle in other words a right triangle. A triangle in which one of the angles is a right angle that is 90 degrees is called a right triangle. The longest side of a right triangle is called the Hypotenuse of a Right Triangle and the other two sides are called the legs of the right triangle. We use the Pythagorean Theorem in finding the hypotenuse of a Right Triangle. The Pythagorean Theorem states that ‘the sum of the squares of the two sides (legs) of a right triangle is equal to the square of the hypotenuse’. Let us assume the lengths of the legs of a right triangle to be ‘a’ and ‘b’ units and the hypotenuse length to be ‘c’. By using Pythagorean Theorem, we can calculate the hypotenuse of a right triangle,
(Hypotenuse) ^2 = (sum of the squares of the sides (legs)^2
  c^2 = (a^2 +b^2)

This gives us the Hypotenuse of a Right Triangle formula, c = sqrt(a^2 + b^2)
Given the lengths of the sides or legs of a right triangle as 3 cm and 4 cm respectively, find the hypotenuse of the right triangle.  Here we are given the lengths of the two sides a = 3cm and b = 4cm, we need to find c. let us apply the Pythagorean Theorem, we get, c = sqrt(a^2+b^2). a^2= (3)^2 = 9 and b^2 = (4)^2 = 16, that gives us a^2+b^2 = 9 +16 = 25. So  c = sqrt(25) = +/- 5 , as length cannot be negative, the hypotenuse of the given right triangle is 5 cm.

Let us now learn how to calculate the Hypotenuse of a Right Triangle using the Pythagorean theorem given by hypotenuse = sqrt[(sum of the squares of the sides or legs)^2]. A ladder is placed against a wall of height 12 ft. The distance between the base of the ladder and the wall is 5ft, find the length of the ladder. In this problem, the triangle formed by the wall, the floor and the ladder is a right triangle and hence the length of the ladder would be the hypotenuse which we need to find. We are given the two lengths of the sides of the triangle which are 12ft and 5 ft respectively. We know c= sqrt(a^2 +b^2); here a = 12ft, b = 5 ft which gives us a^2 = 144 and b^2 = 25. So, c = sqrt(144 +25) = sqrt(169) = +/-13. The length of the ladder is 13 ft

Solving fourth grade math homework

In this article we are going to discuss about the mathematical concepts for fourth grade students. The students of fourth grade learn the different areas of mathematics, like place values of six digits numbers, expanded notations, place value chart, addition, subtraction, multiplication and division of three and four digits numbers, multiples and factors, HCF and LCM. Grade fourth students also learn unitary method, fractions, decimal numbers and measurement of time, length, mass and capacity.

Here we are going to discuss about some of them. This article will be helpful in solving fourth grade math home work.

Topics of Solving Fourth Grade Math

Some of fourth grade mathematical topics are as follows:

Place value: To read and write large number easily, the Indian place value chart is divided into periods as shown below:

Practice Questions: 

(1)Write the number name of these numbers:

(a)    2, 50,946 = Two Lakh fifty thousand nine hundred forty six.

(b)    6, 92,438 = Six lakh ninety two thousand four hundred thirty eight.

(c)    20, 10,101 = Twenty lakh ten thousand one hundred one.

 (1)   Write the numeral of these number names:

(a)    One lakh fifty thousand two hundred eighteen = 1,50,218

(b)   Nine lakh ninety five thousand sixty three = 9,95,063

(c)    Seventeen lakh fifteen = 17,00,015

Simplification involving four fundamental operations: 

In this lesson we will learn to use all the operations together. The fourth grade learners are able to learn the order of operations through this section.

Step 1 ---- Of

Step 2------Division

Step 3 ------ Multiplication

Step 4 ------ Addition

Step 5 ------ Subtraction

 So the order of operation is ODMAS.Now we will do some simplification using ODMAS rule:

 Practice Questions:

(a)    Simplify 36÷ 6 x 4 + 2 – 8

                   = 6x 4 + 2 – 8

                   =   24 +2 – 8

                    =    26 – 8 =18 Ans.

(b)   Simplify 6 + 8 ÷2 -2 x 1 + 5 of  2÷ 5

                    = 6 + 8 ÷ 2 – 2 x1 + 10 ÷5

                    = 6 + 4 – 2 + 2

                    = 12 - 2 = 10 Ans.

Now simplify these questions:

(a)    23741 -  3826 ÷  2 x 6 + 221

(b)   529 x 71 – 630 of 6 ÷ 3 + 4

(c)    6000 +9000 ÷  500 of 6 – 2000

(d)   6699 ÷ 33 +3075 ÷ 25 – 203

Answers: (a) 12484 (b) 36303 (c) 4003 (d) 123

Solving Fourth Grade Math Homework-unitary Method

Some basic knowledge is very important for fourth grade learners like:

• We have to add when we find the total cost of things tat we purchase.
• We have to subtract when we take back the balance.
• We have to multiply to calculate the cost of more articles.
• We have to divide when we find the cost of one.

Practice Question: 

If the cost of 25 Milton jugs is Rs. 8,250.  What is the cost of 48 such Milton jugs?

Solution: The cost of 25 Milton jugs = Rs.8, 250

                The cost of 1 Milton jug is Rs.8, 250 ÷ 25 = Rs.330

                The cost of 48 Milton jugs = Rs. 330 x 48 = Rs.15840

 Solve these problems: 

 28 digital diaries cost Rs.70, 560. What is the cost of 15 such diaries?

 Answer: 37,800

Statistics homework answers

Statistics is the branch of applied mathematics which deals with scientific analysis of data. The subject statistics had been started in early days as arithmetic to aid a ruler who needed to know the wealth of his subjects to levy new taxes. Now a days statistics  plays an important role in all organizations in their decision-making and planning. Statistics homework deals with mean,median and mode.

Classification of Statistics Homework Answers:

In statistics homework answers has following topics are

• Arithmetic mean
• Median
• Mode

The arithmetic mean;

In Statistics the arithmetic mean (A.M) or simply the mean or average of n observations x1, x2, …, xn is defined to be number x such that the sum of the deviations of the observations from x is 0.

         x1 +x2 +x3.......xn
x =  --------------------
                     n

the symbol Σ, called sigma notation is used to represent summation.

x=Σ xi
   -------
      n

Homework Example:

Calculate  mean of the data 9, 11, 13, 15, 17, 19.

Sol :

X =   Σxi/n = `[9+11+13+15+17+19]/[6] ` =` [14]/[6]`

Median :

Median is defined as central most or middle value for given series of data it should be arranged in ascending or descending order.

Homework Example:

Find the median of 23, 25, 29, 30, 39.

Sol:

The given values are already in  ascending order. No. of observations N = 5.

So the median = ` (N+1)/(2)`  =`(5+1)/(2)`
= 3 rd term =29
∴ Median = 29.

Mode:

In Statistics answers,Mode is also a measure of central tendency.

In a set of each observations, mode is defined as value which occurs most often.

If the data are arranged in the form of a frequency table, the class corresponding to  greatest frequency is called  modal class.

Homework Example:

Find the mode of 7, 4, 5, 1, 7, 3, 4, 6,7.

Sol:

Arranging the data in  ascending order, we get 1, 3, 4, 4, 5, 6, 7, 7, 7.

In above data 7 occurs several times. Hence mode = 7.

Homework Problems and Answers for Statistics:

Calculate mean :    
Problem 1:
Calculate the mean of the data 7, 12, 18, 14, 19, 20.
Problem 2:
Calculate the mean of the data   16,18,14,15,21,30,26,44
Problem 3: 
Calculate the mean of the data   22,77,55,11,22,26,38,72
Answers:
1) 15
2) 23
3) 40.35

Calculate median:
Problem 1: 
Find the median of 3, 4, 10, 12, 27, 32, 41, 49, 50, 55, 60, 63, 71, 75, 80.
Problem 2:
Find the median of 29, 23, 25, 29, 30, 25, 28.
Problem 3: 
Find the median of 26, 25, 29, 23, 25, 29, 30, 25, 28, 30.
Answers:
1) 49
2) 28
3) 27

Calculate mode:
Problem 1:
Find the mode of 12, 15, 11, 12, 19, 15, 24, 27, 20, 12, 19, 15
Problem 2:
Find the mode of 3,5,8,3,9,3
Problem 3: 
Find the mode of 3,5,8,5,6,7
Answers:
1) 12 and 15
2) 3
3) 5
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An Introduction to Calculating the Area of a Right Triangle

The Area of a Right Triangle
A Right Triangle or a Right Angled Triangle is a triangle which has right angle or 90 degrees as one of its interior angles. We have two types of Right Triangles namely, scalene right triangle in which one angle is right angle and the other two are unequal angles and the corresponding sides are also unequal. The other type is the Isosceles Right Triangle in which one angle is right angle and the other two angles are equal and also their corresponding sides are of equal length.  A right triangle is very useful in trigonometry, a branch of mathematics. Area of a triangle is given by half times the base times the height of the triangle. We can deduce the formula for Area of a Right Triangle using the following method. Let us consider a rectangle of length l cm and width w cm. Let us now cut the rectangle into two equal halves diagonally as shown in the figure below.

 
The triangles ABC and ADC are congruent triangles which means when placed one over the other they overlap exactly. Also they are right triangles with 90 degrees as one of their angles. They are of the same size and hence we can say that they have the equal area. From this it is clear that the area of a right triangle will be half the area of the rectangle.
The Area of the right triangle = ½ [Area of the Rectangle]
               = ½ [length x width]
             = ½ [l w]

From the figure, we can see that the length of the rectangle is one of the sides of the triangle which is called the base of the triangle. So, we have base of the triangle equals length of the rectangle.  Also the distance from the vertex A of the triangle ABC to the vertex B is the height of the triangle ABC which is the width of the rectangle. So, we have height of the triangle equals width of the rectangle. Substituting these in the above equation, Formula for Area of a Right Triangle = ½ [base x height]. So, to calculate area of a right triangle we use the formula ½ times [base x height]
Calculate the Area of a Right Triangle with base and height 12 cm and 17 cm respectively
Solution:  Base of the right triangle   = 12 cm
         Height of the right triangle = 17 cm
       Area of the right triangle   = ½ x base x height
                      = ½ x 12 x 17 = 6 x 17 = 102 cm^2