Math Second Grade Fraction

Introduction to math second grade fraction:

Here we are going to learn identify the fraction, which shape show the fraction, Parts of a group, Word problems, Compare fractions, Order fractions, generally fraction contains two parts and these parts are divided by the line, top part is called numerator and bottom is called denominator. Hear the top part is the total number of shaded region and bottom part is total number of division.


Identifying the fractions - math second grade fraction:


Example problem:

Find the fractions of the shaded region?

Solution:


Here the circle is divided into 4 parts, so it will be come under the part of the denominator and 2 regions are shaded it will be come under the part of the numerator



So we can write in the form of fractions (shaded region) =`2/ 4`

Which shape illustrates the fraction- math second grade fraction:?

Which shape shows the fraction `3/8` ?



Solution:

Circle is divided into 3/8 .


Find the fractions from the parts of the group- math second grade fraction::


suppose the group (generally group means collection of objects) has many diagrams like triangle, square and rectangle means we can write the one shape in the form of fraction means we have to count the total diagrams in the particular group it will be come under the part of the denominator and count the particular shapes it will be write in the numerator part.

Example:

What fraction of the square in the below diagram?



Solution:

Word problems- math second grade fraction:

John’s family lives on an avenue with 3 houses. 2 of the houses are prepared of brick. How we can make the fractions (build the house prepared of brick)?

Solution;

here total number of house is 3so it will be come under the part of the denominator ,2 of the houses are made with brick so it will be come in the part of numerator .so we can write those fraction in the below form,

`2/ 3`

Math Geometry Activities

Introduction for math geometry activities:

The geometry activities are one of the most important branches of Mathematics. Math geometry activities gives the idea of various geometrical shapes and figures in our daily life such as articles in the houses, wells, buildings, bridges etc.

The word ‘Geometry’ means learns of properties of figures and shapes and the relationship between them. The geometry shapes are point, line, square, rectangle, triangle, and circle. From this we can learn that the math geometry activities have begun from ancient times. The math geometry activities example problems and practice problems are given below.


Example problems - math geometry activities:


Example problem 1:

In the ?ABC the angle B is bisected and the bisector meets AC in D. If ?ABC = 80° and ?BDC = 95°, find ?A and ?C.

Solution: Let us See the Figure



From ?BDC, 40° + 95° + ?C = 180°

After solving this, We get

? ?C = 180° - 135° = 45°

From ?ABC, ?A + ?B +?C = 180°

?A + 80° + 45° = 180°

After solving this, We get

?A = 180° - 125° = 55°.

Example problem 2:

If the angles of a triangle are in the ratio 1 : 3 : 6, find them.

Solution:

Let the angles be 1x, 3x, 6x.

Then 1x + 3x + 6x = 180°

After simplify this, We get

10x = 180° or x = 18°.

The angles are 1 × 18°, 3 × 18°, 6 × 18°,

After simplify this, We get

The angles are 18°, 54°, 108°.


Practice problems - math geometry activities:

Practice problem 1:

Find out the geometry equation of straight line passing through the given point (2, 3) and cutting off

Determine the equal intercepts along the positive directions of both the axes.

Answer: x + y = 5.

Practice problem 2:

Find the points on y-axis,then perpendicular distance from the straight line 4x - 3y - 12 = 0 is 3.

Answer: points are (0, 1) and (0, - 9)

Practice problem 3:

Find out the equation of straight line passing through the given intersection of the straight lines 2x + y = 8 and 3x - y = 2 and through the point (2, - 3).

Answer:  x = 2

Equality Of Matrices

Equality Of Matrices

For two matrices to be equal, they must be of the same size and have all the same entries in the same places. For instance, suppose you have the following two matrices:

These matrices cannot be the same, since they are not the same size. Even if A and B are the following two matrices:

...they are still not the same. Yes, A and B each have six entries, and the entries are even the same numbers, but that is not enough for matrices. A is a 3 × 2 matrix and B is a 2 × 3 matrix, and, for matrices, 3 × 2 does not equal 2 × 3! It doesn't matter if A and Bhave the same number of entries or even the same numbers as entries. Unless A and B are the same size and the same shape and have the same values in exactly the same places, they are not equal.

This property of matrix equality can be turned into homework questions. You will be given two matrices, and you will be told that they are equal. You will need to use this equality to solve for the values of variables.   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

• Given that the following matrices are equal, find the values of x and y.
  
  x = 1,    y = 4
For A and B to be equal, they must have the same size and shape (which they do; they're each 2 × 2 matrices) and they must have the same values in the same spots. Then a_1,1 must equal b_1,1, a_1,2 must equal b_1,2, and so forth. The entries a_1,2 anda_2,1 are clearly equal, respectively, to entries b_1,2 and b_2,1 "by inspection" (that is, "just by looking at them"). But a_1,1 = 1 is not obviously equal to b_1,1 = x. For A to equal B, I must have a_1,1 = b_1,1, so it must be that 1 = x. Similarly, I must have a_2,2 =b_2,2, so then 4 must equal y. Then the solution is:

• Given that the following matrices are equal, find the values of x, y, and z.
  
    4 = x 
  –2 = y + 4 
    3 = ^z/₃
  
  x = 4, y = –6, and z = 9.
To have A = B, I must have all entries equal. That is I must have a_1,1 = b_1,1, a_1,2 = b_1,2, a_2,1 = b_2,1, and so forth. In particular, I must have:
...as you can see from the highlighted matrices:
Solving these three equations, I get:

Example

Given that the following matrices are equal, find the values of x, y and z .

Solution:

Equate the corresponding elements and solve for the variables.

x + 3 = 6 
x = 3

y = −1

z −3 = 4
z = 7

What is the Relation of Tessellations with Math

Introduction to Tessellations
              Tessellations are nothing but the arrangement of certain shape in a plane without leaving any space between them. The arrangement can be made by any polygon starting from 3 to infinity sides. If the arrangement involves with regular polygons like square, pentagon, hexagon etc., then it is called as regular tessellations. The tessellations can also be formed by semi-regular Polygons also, but the arrangement formed should be identical from each vertex point.


Let us see Description of what is the relation of tessellations with math, Studying what is the relation of tessellations with angle measurements in math, Studying what is the relation of tessellations of different relations in math in this article.

relation of tessellations with math:

              Tessellations with math not only mean the flooring with shape, but it also involves the arrangement of shapes with particular angles of degrees. The tessellations involves the mathematical terms like vertex, angles (interior) and also the number of sides which is essential to arrange any shape. It involves the calculation of interior angles to arrange regular as well as semi regular shapes. This leads to the study of different angles of the shape. What are the relations of regular tessellations? For example, for the regular tessellation the shape involved should be exactly divisible by 360 degrees. 

Studying tessellations with angle measurements in math:

Interior measure of this angles each of these polygons:
              What are the relations of interior tessellations with math measurement angles?

Shape	Angle measure in  degrees
triangle	60
square	90
pentagon	108
hexagon	120
More than six sides	More than 120 degrees

              Tessellations with math relations can plug the plane at each vertex in a regular polygons, an exact value of divisor of 360 in an interior angles. What are the relations of tessellations of the triangle, square, and hexagon?

Studying tessellations of different relations in math

Naming Conventions:
           What are the naming conventions for tessellations with math relations?
           For square is the relation of tessellations:
 
           A shape of regular congruent hexagons that touch with a vertex and then count the sides of the polygons.
           Three polygons containing six sides, so this tessellations naming convention is "6.6.6".
 
           A triangle containing six polygons with surrounding vertex, and then each polygon has three sides: and what it’s naming convention is that "3.3.3.3.3.3".

Solve Algebra Fraction

Algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. Here we are going to study about How to solve algebra fraction and its example problems. (Source from Wikipedia)

Algebra fraction:

In algebra fraction is a simple fraction it contains algebraic expression in either numerator or denominator.

Example:
`(3x+3) / 2` = 3

Example problems:

Example: 1

Solve the following algebraic equation 

`(2x+4) / 2` + `(3x+3) / 4` = 6

Solution:

Here we have two fraction term and both have different denominator so we take Common denominator

Least common denominator = 2 * 4 = 8

Multiply first term with 4 we get

`(8x+16) / 8` + `(6x+6) / 8` = 6

Now both denominators is equal so take away

`1/ 8` {(8x+16) + (6x+6)} = 6

Multiply both sides 8 we get

8 * `1/ 8 ` {(8x+16) + (6x+6)} = 6 * 8

In left hand side 8 will be canceling

(8x+16) + (6x+6) = 48

Combine the like terms we get

14 x + 22 = 48

Add both sides -22

14x + 22 -22 = 48 -22

14x = 26

Divide both sides 14 we get

x = `26 /14`

The simplest fraction is `13 / 7`

Therefore the value of x = `13 / 7`


Example : 2

Solve the following algebraic equation

`(3x+3) / 2` = 3

Solution:

Here the denominator is 2

So multiply both sides 2 we get

2 `(3x+3) / 2` = 3 *2

In left hand side numerator 2 and denominator 2 will be canceling

3x +3 = 6

Add both sides -3 we get

3x+3-3 = 6 -3

3x = 3

Now we get without fraction equation.it is simple algebraic equation.

Divide both sides 3 we get

x =` 3 / 3`

Therefore the value of x = 1


Example: 3

Solve the following algebraic equation

`4 / (2x+3)` = 3

Solution:

Multiply both sides (2x+3)

(2x+3) * `4 / (2x+3)` = 3 (2x+3)

Numerator (2x+3) and denominator (2x+3) will be cancelling

4 = 3 (2x+3)

4 = 6x+9

Add both sides -9 we get

4-9 = 6x+9-9

-5 = 6x

Therefore x = - `(5/6)`

The value of x – `(5/6)`

Solve Second Derivative Test

Solve second derivative test involves the process differentiating the given algebraic function twice with respect to the given variable. Generally the derivative is discussed in calculus whereas it is mainly used to find the rate of change of the given function with respect to the change in the input. The following are the solved example problems with detailed step by step solution in second derivative to study for the test.


Second derivative test example problems to solve:


Example 1:

Solve the test function to find second derivative.

f(z) = 2z6 + 2 z5 + 3 z4 + 3z

Solution:

The given equation is

f(z) = 2z6 + 2 z5 + 3 z4 + 3z

The above function is differentiated with respect to z to find the first derivative

f '(z) =  2(6z 5)  +2 (5 z4 ) +3(4 z3) + 3

By solving above terms

f '(z) =  12z 5  +  10z4  + 12 z3 – 3

The above function is again differentiated with respect to z to find second derivative

f ''(z) =  12(5z 4 ) – 10(4z3)  + 12(3z2)

f ''(z) =  60z 4 – 40z3 +36z2  is the answer.

Example 2:
Solve the test function to find second derivative.

f(z) = 5z 2 +5z 4  + 12

Solution:

The given function is

f(z) = 5z 2 +5z 4  + 12

The above function is differentiated with respect to z to find the first derivative

f '(z) = 5(2z  )+5(4 z 3 ) + 0

By solving above terms

f '(z) = 10z +20z3

The above function is again differentiated with respect to z to find second derivative

f ''(z) =  10(1 ) +20(3z2)

f ''(z) =  10 + 60z2 is the answer.

Example 3:

Solve the test function to find second derivative.

f(z) = 4z4 +5z 5 +6z 6  + 2z

Solution:

The given function is

f(z) = 4z4 +5z 5 +6z 6  + 2z

The above function is differentiated with respect to z to find the first derivative

f '(z) = 4(4z 3 )+5(5z 4 ) +6( 6z 5) +2

By solving above terms

f '(z) = 16z 3 +25z 4 +36 z 5 + 2

The above function is again differentiated with respect to z to find second derivative

f ''(z)= 16(3z 2) +25(4z 3) +36 (5z 4)

f ''(z)= 48z 2 +100z 3 +180z 4 is the answer.


Second derivative test practice problems to solve:


1) Solve the test function to find second derivative.

f(z) = z 3 + z 4 + z 5

Answer: f ''(z) = 6z +12z2+ 20z 3

2) Solve the test function to find second derivative.

f(z) = 2z 3+3z5 + 4z 6

Answer: f ''(z) = 12z + 60z3 + 120 z 4

Learning Intermediate Algebra

Algebra executes most common four basic types of operations. That includes the addition, subtraction, multiplication and division operations. Algebraic problems contains the variables, constant, coefficients, exponents, terms and expressions. Balanced equations on both of the sides are considered as the fundamental concept of algebra. Important properties such as commutative, associative, identities and inverse are also involved in the algebraic problems.


Learning common term in intermediate algebra:


Variables:

Algebraic variables are use for assigning the variables. Alphabetical characters alone used as variables. There are possibilities for changes of variable values while solving the algebraic problems.  x, y and z are the three commonly used variables in algebra.

Constant:

Algebraic constants are values that never change during the process of solving the algebraic equation. Consider the algebraic form of equation 6y + 5, in which the value 5 is the constant.

Expressions:

Algebraic expressions are considered as the combinations of variables, constant, coefficients, exponents and terms.Addition, subtraction, multiplication and division expressions proves this. The example of an algebraic expression is given below

15y + 6.

Term:

Terms are used to form the algebraic expressions by some of the arithmetic operations such as addition, subtraction, multiplication and division. Consider the following example 5n^2 + 6n in which the terms 5n^2, 6n are combined to form the algebraic expression 3n^2 + 2n by using the addition operator ( + ) to perform addition operation.

Coefficient:

The coefficients are the values that are present in front of the terms of an algebraic expression. Consider the following example, 6n2 + 4n in which the coefficient of 6n2 is 6 and 4n is 4.

Equations:

Algebraic equations are the combinations of the numbers or expressions. Mostly the algebraic equations are used for evaluating the the value of the variables.


Learning Examples of intermediate algebra:


Example 1:

4x - 6 = 2x – 4

4x-6+6 =2x-4+6 (adding common value(6) in both sides)

4x      = 2x +2 (we get)

4x -2x = 2x -2x +2 (adding common value(-2x) in both sides)

2x   = 2 (we get)

2x / 2  =  2/2 (dividing common value (2) in both sides)

X  =  1 (We get x value)

Example 2:

6n +15 = 150

6n+15-15 =150-15 (adding common value(-15) in both sides)

6n = 135 (we get )

6n / 6 = 135 / 6 ( dividing common value (6) in both sides)

n = 22.5 (We get n value).