Mix Word Problems Math

Introduction: Two different quantities having different percentage of needed substances can be mixed together to get the desired amount. This is called mix word problem in math. For example: Let X and Y be two different solutions. Each are having different percentage of Salt in it. Those two X and Y can be mixed together to get the required percentage of the final solutions. Now let us see few problems of this kind. Example Problems on Mix Word Problems Math: Ex 1: Two salt solutions having 12% and 36% of salts respectively in them. They are added together to form 40% of salt solution. If 280 liters of 12% salt solution is added, how many liters of 36% has to be added? Soln: Let x be the salt solution. Given: `36/100` x + `12/100` `xx` 280 = `40/100` (X + 80) `=>` 36x + 3360 = 40x +3200 `=>` 40x – 36x = 3360 – 3200 `=>` 4x = 160 `=>` x = 40 liters. Ex 2: Two oils have to be mixed together. One of the oil has 15% of the required content and the other has 30% of the required content. How many liters of 30% of oil have to be added to 290 liters of 15% of the oil, so that we can have a mix of 42% of the oil content? Soln: Let X be the amount of oil with 30% content in it. Given: `30/100` (x) + `15/100` (290) = `42/100` (x + 90) `=>` 30x + 15`xx` 290 = 42x + 42`xx` 90 `=>` 12x = 4350 – 3780 `=>` x = `570/12` = 47.50 liters More Example Problem on Mix Word Problems Math: Ex 3: Two herbals have to be mixed together. One of them has 25% of the required vitamins and the other has 35% of the required vitamins. How many liters of 25% of herbal have to be added to 120 liters of 45% of herbal, so that we can have a mix of 40% of vitamins? Soln: Let x be the amount of herbal with 25% vitamin in it. Given: `25/100` (x) + `45/100` (120) = 40/100 (x+120) `=>` 25x + 5400 = 40x + 4800 `=>` 15x = 5400 – 4800 = 600 `=>`x = `600/15` x = 40liters

Horizontal Lines in Math

Introduction to Horizontal Lines in Math:
              
              Horizontal lines in math are used for arithmetic operations which denotes the next step in this operations. Generally it does not have any meaning or symbols or notation or formula but it is expressed for easy understanding of the problems. In every math problems horizontal lines are present especially with the dividing symbol. In fraction, horizontal lines divides numerator and denominator in math. A horizontal line in math separates the equations when substituting the equation steps.



Applications of Horizontal lines in math:


             In all math problems we denote horizontal lines.Some of the applications are,

• In addition we are using horizontal lines when we adding numbers we expressed the horizontal lines for sum answers

            Example 12
                          +14
                            26

• In subtraction we apply as same as addition for the subtracted answers

             Example 20
                            -12
                              8  

• In multiplication horizontal lines express for the multiply value with the multiplicand

             Examples  12
                              ` xx ` 14    
                                  48 
                                12    
                                168   

• In division horizontal lines are shows the divide symbols which denotes quotient and steps involved in the division.

             Example 12 by 3
                             `12/3` =4  

• In fraction we denotes the horizontal lines for separations of numerator and denominator by use of  horizontal lines.

              Example` 5/3, 4/3` here 5 is the numerator and 4 is denominator.

• In algebraic equations when we equating two equations horizontal lines express the next step for equations in both elimination and substitution method we have horizontal lines for equations.

              Example 2x+y =5 .........(1)
                                 x-y  =4........(2)
                                3x =9
                                  x=`9/3 `
                                  x=3
                       substitute x=3  in (2)
                            3-y =4
                                y=`4/3`
                           



Example Problems for Horizontal lines in math:


Example 1:
      Add 24 and 10.

Solution:
          24
       + 10
         34 

Example 2:
    Subtract  -24 and 10

Solution:
          24
        - 10
          14

Example 3:
         Multiply 10  ` xx`

Solution:
           
       10
    `xx`      4
        40
Example 4:
      Divide 25 ÷ 5

Solution:
         `25/5` = 5

Example 5:
      Simplify `4/3+ 2/3`

Solution:

      `4/3+2/3= 6/3`

Example 6:
           solve x+y =3
                   2x-y  =0

Solution:
                        x+y =3 .........(1)
                      2x-y  =0........(2)
                                3x =3
                                  x=`3/3 `
                                  x=1
                       substitute x=1  in (2)
                            1-y =4
                                y=`4/1`
                                y=4

Step Functions In Math

Introduction to step function:

     A step function in math is one of the concepts of functions in math. The line is the collection of x and y points. At the same time x and y value can be increased values. We can plot this points on the graph we can get the equation of the line. That forms the step figure. So these functions are known as step functions.


Definition of Step functions in math
Definition of Step function:
    A function in the real analysis R can be defined as the step function it can be written as linear combination of semi open intervals. Open interval can be defined as [a, b) in R.  Then the step function f(x) can be described as,

f(x) =a1 f1(x) + a2 f2(x) + …………………… + an fn(x)

Brief Description of the Heaviside Step functions in math

This involves the following steps.

They are,

Step 1: First we can form the equation into y = m x + b form.

Step 2: Substitute any value for x we can get the y values.

Step 3: Plot the (x, y) points on the graph.

Step 4: This line plot increases sequentially.

Step 5: It look like as a step format.

These are the main steps used in math step functions.

Consider the equation of the line y = x -1

Substitute the values for x we can get the y values.

Substitute x=1 in the given equation y = x -1.

That is y= 1 – 1 = 0

Therefore the (x, y) co-ordinate is (1, 0)

Substitute x=2 in the given equation y = x -1.

That is y= 2 – 1 = 1

Therefore the (x, y) co-ordinate is (2, 1)

Substitute x=0 in the given equation y = x -1.

That is y= 0 – 1 = -1

Therefore the (x, y) co-ordinate is (0, -1)

Substitute x=-1 in the given equation y = x -1.

That is y= =-1 – 1 = -2

Therefore the (x, y) co-ordinate is (-2, -2)

Substitute x=-2 in the given equation y = x -1.

That is y= -2 – 1 = -3

Therefore the (x, y) co-ordinate is (-1, -3)

We can tabulate these points

X: -2 -1  0 1 2

Y: -3 -2 -1 0 1

Here the x and y values are correspondingly varies. So this function is namely known as step function.

The graph of the step function in math described as



This is also known as Heaviside step functions in math.

Live Online Chat About Math

Introduction to live online chat about math:

Live online chat about math is nothing but the help to students by tutors of  live tutoring website through online. Many students find difficult in doing the math homework problems. Parents also don't find time to help their students with homework problems. So they were in need of a website that provide live online chat about math. Tutor vista is a website to get help in math anytime. Many students find maths help vista website with the highly qualified tutors live at any time.  Below are some of the model problems solved by tutors for live online chat about math.


Live online chat about math:


1. Add 322 + 343 + 346

Solution

The given is an addition problem with three 3 digit numbers. Addition is the basic arithmetic operations

3 2 2
3 4 3
3 4 6 +
---------
10 1 1
---------
2.     Subtract the numbers given 444 – 240
Solution

Similar to add start the subtract with the ones digit first

4 4 4
2 4 0 –
----------
2 0 4
---------
3.  Solve and find the value of b  7b+3 = 24
Solution

This is an algebra problem. Here we have to find the value of b.

7b + 3 = 24

Subtract 3 on both sides

7b+3 -3 = 24 – 3

7b = 21

Divide by 7 on both sides to find the value of b

`(7b)/7` =`21/7`

b= 3

4. Solve 10 + 4 x 3 + 22-(1x4)

Solution

Here more than one arithmetic operation is given in the problem. So we perform this problem by using arithmetic operation.

10 + 4x3+22 - 4-------Parenthesis

10 + 4x3 + 4 – 4 -----Exponents

10 + 12 + 4 - 4 ------Multiplication

22+ 4 - 4   ------------No division so we do addition

22 ------------Subtraction

So the answer is 22


Live online chat about math:

Add 868 + 325 + 110
Subtract 875-340
Find the value of s in the equation shown below 4s – 6 = 26
By using PEMDAS rule perform 14 + 2 – 3 +(4x5) – 20
5. Find the value of u in the equation 3p = 42

Answers:

1303
535
s = 8
13
p = 14

Define Hypothesis Math Term

Define hypothesis math term:

The term hypothesis refers the condition statement just after the word if. In the given conditional statement.

For example Suppose the condition is “in the given quadrilateral figure -  four sides are equal in measure then the shape is said to be square” the hypothesis of the given statement is “all four sides are equal in measure. In the given example,” all four sides are equal in measure. “Always, never or sometimes true.

Options:

1)Always

2)Never

3)Sometimes.

Answer is (3)

Solution:

Answer 3) sometimes it is correct because, it’s true if only the given shape is square, In case of other shapes like rectangle, rhombus then  condition is not satisfied.


Example 1)

Find the math term hypothesis of the given equation.

If x = 1 then x2 +3x +5 = 9

Option:

1)True

2)false 

Answer : 1)True.

Solution:

Substitute x value in the given equation,

X+3x+5 =9

1+3(1) +5 = 9

1+3+5 =9

9=9

Therefore the given condition is true.

Example 2:

Find the math term hypothesis in t he given statement is “Every  integer greater than 1 is product of prime number “

Solution:

Let us consider the if  x is true statement .

Therefore ,

X`=>` y  is a true statement  and y is a true  conditional statement.

Now,we need to find the series of the statement x1,x2 …..etc.,and then verifies the given conditional statement” Every  integer greater than 1 is product of prime number”.

X=x1

X1=x2,

X2=x3,

….

….

Xn-1 =xn

Xn =Y are ture.

Because assume that the given statement X is true then all repeated statement of the simple rule  Y  is true..

Example 3:

The conditional statement is “if n is an integer then n2 is even  and then n is odd.”

Solution:

N is odd  number .

so  n-1 is even number .

Therefore , we have,

n-1 =2k

where ,k is the constant .

n-1 = 2k ,k `in` Z

n=2k+1

Similarly we have by hypothesis,

  n2 =x;  x `in` Z

And then,

n2 = 4k2 +4k+1

 =2(2k2 +2k)+1

In the above equation, 2(2k2+2k) is even number.

So 2(2k2+2k) +1 is the odd number.Therefore our hypothesis is violated.

So  the integer number “n “ must be even number.

So,

X`=>` S =” n2  is even ”

Y  `=>` ” n is even ”

Not –Y `=>` ” n is odd ”


Practice problems related to the term hypothesis in math :


1) Prove that if m and n are even integers, then n +m is even.

2) Prove that if n is an odd integer, then n 2 is an odd integer.

Find Domain off

Introduction of Domain function

The domain of definition or simply the domain of a function is the set of "input" or argument values for which the function is defined. That is, the function provides an "output" or "value" for each member of the domain. The domain of cosine is the set of all real numbers, while the domain of the square root consists only of numbers greater than or equal to 0 when the Function is represented in an xy


Examples


Problem 1: Find the domain of the real valued linear function f given below.

f(x) = x + 1

Substitute x=1, 2,3,4,5

x=1                    f(x) =1+1

By solving it we get

f(x) =2

The ordered pair is (1, 2)

x=2                            f(x) =2+1

By solving it we get

f(x) =3

The ordered pair is = (2, 3)

x=3         f(x) =3+1

By solving it we get

f(x) =4

The ordered pair is = (3, 4)

x=4                          f(x) =4+1

By solving it we get

f(x) =5

The ordered pair is = (4, 5)

x=5                            f(x) =5+1

By solving it we get

f(x) =6

The ordered pair is = (5, 6)

The domain function is 1, 2,3,4,5

Problem 2: Find the domain of the real valued linear function f given below.

f(x) = 2x + 1

Substitute x=1, 2,3,4,5

x=1                       f(x) =2(1)+1

By solving it we get

f(x) =3

The ordered pair is (1, 3)

x=2                          f(x) =(2*2)+1

By solving it we get

f(x) =5

The ordered pair is = (2, 5)

x=3                           f(x) = (3*2)+1

By solving it we get

f(x) =7

The ordered pair is = (3, 7)

x=4                      f(x) =(4*2)+1

By solving it we get

f(x) =9

The ordered pair is = (4, 9)

x=5,                           f(x) =(5*2)+1

By solving it we get

f(x) =11

The ordered pair is = (5, 11)

The domain function is 1, 2,3,4,5


Practice problem


Problem : Find the domain of the real valued linear function f given below.

f(x) = 2x + 5

Vista Math Help

Introduction to vista math help:

Tutorvista is one of the online tutoring services providers; Tutorvista provides helps in various subjects like English, math, and science.  Tutor vista helps at any time around the clock regardless of place where they are located.  Student can learn and do homework from their home at any time. Here tutor to explain step by step so that the students can easily understand much easier. In this article we shall discuss tutorvista help in math problems.


Tutorvista math help example problem


Example 1:

Sum the polynomial  5x4 – 6x2 + 9x + 5 and 5x + 6x3 – 3x2 – 1.

Solution:

Use the associative property and distributive properties add real numbers

(5x4 – 6x2 + 9x + 5) + (6x3 – 3x2 + 5x – 1) = 5x4 + 6x3 – 6x2 – 3x2 + 9x + 5x + 5 – 1

= 5x4 + 6x3 – (6+3)x2 + (9+5)x + 4

= 5x4 + 6x3 – 9x2 + 14x + 4.

An another method for adding two polynomials

5x4 +0x3– 6x2 + 9x + 5

0x4 + 6x3 – 3x2 + 5x – 1

____________________
8x4 + 6x3– 9x2 + 14x + 4
____________________

Example 2:

Consider e1 and e2 are two events with a random trial such that P(E2)=0.25, P(E1 or E2)= 0.95 and P(E1 and E2) = 0.35, find the value of P(E1).

Solution:

Let P (E1) = x then,

P (E1 or E2) = P (E) + P (E2) – P (E1 and E2)

0.95 = x + 0.25 – 0.35

Adding the trial value

x = (0.95-0.25 + 0.35) = 0.65

Hence the value of P (E1) = 1.05

Example 3:

Find the increases percentage value from 35 to 50.5?

Solution:

Changes in value = new value – original value

50.5 – 35 = 15.5

Increased percentage = `15.5 / 35` x100%

Answer:

The increased percentage value is = 44.2%


Tutorvista math help practice problem


Problem:

Adding the polynomial 8x4 – 5x2 + 9x + 4 and 4x + 6x3 – 3x2 – 1.

Answer: 8x4 + 6x3– 8x2 + 13x + 3

Problem:

Find how many percentage increases from 30 to 55.5?

Answer: 85%

Problem:

Consider e1 and e2 are two events with a random trial such that P(E2)=0.35, P(E1 or E2) = 0.75 and P(E1 and E2) = 0.45, find the value of P(E1).

Answer:

P (E1) = 0.85

Dividing Word Problems

Introduction to dividing word problems:

Division is one of the arithmetic operations and also it is the inverse of multiplication. We know that the solution of dividing two rational numbers is another rational number when the divisor is not Zero. Division of any number by zero is not defined. Because of zero is multiplied by any particular number will always result in a product of zero. If we can enter these types of numbers into the calculator then error message is displayed.
Examples of dividing word problems:


Dividing word problem 1:

Jack invites 12 friends to his Wedding party. He distributes 72 balloons equally among them. How many balloons did each friend get?

Solution:

We can find the number of balloons per each by using the division method.

Therefore Total number of balloons= 72

Total person= 12

Therefore, `72/12` `= 6`

Answer: 6

Dividing word problem 2:

A board game has 45 total coins. There are equal number of coins of 5 different colors. How many coins are there of each color?

Solution:

We can find the number of coins per each color by using the division method.

 Total number of coins= 45

 Total number of colors= 5

So, `45/5` `= 9`

Here 45 is 9 times of 5

Answer: 9

Dividing word problem 3:

Uncle has planned 15 total prizes for the party. Each game would have 3 winners. How many games will be played at the party?

Solution:

`=` `15/ 3`

Here 15 has 5 times of 3

= 5

Answer: 5

Dividing word problem 4:

Dad bought 56 sweets for the party and distributed them equally among 7 gift pouches. How many sweets did she place in each pouch?

Solution:

= `56/ 7`

= 8

Here 56 is 8 times of 7.

Answer: 8


Practice problems of dividing word problems:


Mary has 48 dollar in her purse. Each toy costs 6 dollar. How many toys can she buy? Answer: 8
John divided 72 by 3 on his toy calculator. What quotient did the calculator display? Answer: 24
 Ton has 75 blocks. He makes 15 equal stacks. How many blocks are there in each stack?   Answer: 5

Alpha Math

Introduction to Alpha Math:

     Alpha is the first letter of the Greek alphabet and it is written as α . The word "mathematics" comes from the Greek, which means learning, study, science, and additionally came to have the narrower and more technical meaning "Mathematical Study", Mathematics is the study of quantity, structure, space, and change. Mathematicians seek out patterns, formulate new conjectures, and establish truth by rigorous deduction from appropriately chosen axioms and definitions.


Example problems of Alpha math:


Here we will discuss about the alpha math,

Alpha math – Example problem: 1

  Prove that the condition that one root of ax2 + bx + c = 0 may be the square of the other is m3 + l2 n + ln2 = 3lmn

Proof:

Let the roots of the given equation be  α and α2

then,                 α +  α2 = `(m)/(-alpha)`

 α+  α2  =  α3 = `(n)/(alpha)`

Cubing both sides of above equation (α+ α2)3 =- (m3 / l3)

 α3 +  α6 + 3α3 (α+ α2 )= - (m3 / l3 )

           n/l + (n/l)2 +3n/l (-m/l)= -m3 / l3

that is m3 + l2 n + ln2 = 3lmn

Alpha math – Example problem: 2

  We form the quadratic equation whose roots are α and γ .

Proof:

Then x = α ,

           y = γ are the roots

Therefore x - α  = 0 and

              y - γ = 0

        (x -α) (y -γ) = 0


Some more example problems of Alpha math:


Alpha math – Example problem: 3

F the roots of x4 - 6x3 + 13x2 - 12x + 4 = 0 are α,α,γ,γ  then the values of  α,  γ= ?

Solution:

        2α + 2γ = 6

              α + γ  = 3

            α 2γ 2 = 4

                    αγ = 2

    α= 2  and γ = 1

Alpha math – Example problem: 4

Prove that sin2α + sin2 ( α + 60) + sin2 (  α - 60)  =`(3)/(2)`

L. H. S = sin2α + [ sin  ( α + 60) ]2 + [ sin (α - 60)]2

sin2α + [ sinαcos 60 + cos2αsin 60) ]2+ sinα cos60 - cosα sin60]2

sin2α+ 2( sin2αcos2 60 + cos2α sin2 60 )

sin2α+ 2 [`(1)/(4)`sin2α+ `(3)/(4)` cos2α ]

sin2α+ [ (sin2 α) / 2 ] + [ (3cos2α ) / 2 ]

`(3)/(4)` [ sin2α + cos2α ] = = R. H. S

Here we proved the L.H.S is equal to R.H.S.

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).

Scale Factor in Algebra

A scale factor is a number which scales, or multiplies, some quantity. In the equation y=Cx, C is the scale factor for x. C is also the coefficient of x, and may be called the constant of proportionality of y to x. For example, doubling distances corresponds to a scale factor of 2 for distance, while cutting a cake in half results in pieces with a scale factor of ½


Explanation of Scale factor in algebra:


the following table will explain the scale factor in algebra some example given below

                            LENGTH           AREA              VOLUME

Scale Factor X      X times            X2 times              X3 times

Scale Factor 2      2 times             4 times                8 times

Scale Factor 3      3 times             9 times                27 times

Scale Factor 4      4 times             16 times              64 times

Scale Factor 5      5 times             25 times              125 times

For example in a square having 6cm all its sides. If the same squares length is reduced to 3cm then the scale factor of the square is 1/2.


Finding Scale factor in algebra:


Some examples for scale factor in algebra.

Ex : Consider the equation C= YZ, Y was the scale factor or Z, Y is also a coefficient of Z and may be called as constant of proportionality of a to Z. By using scale factor, you are creation a smaller model of something that is quite big.

In these algebra examples we discuss the scale factor

For the given problem Y = (4) (3).

4 was the scale factor of 3

4 is also a coefficient of 3

Ex :

                        

                    square1                                            square2

In the example  square1 value of each side is 4 cm. the square2 value of each side is 8 cm. so we find the scale factor of this problem?

Sol : This example we find the scale factor of two square figures.

The second square value by first square value

Scale factor =8/4

                   =2

Ex :  The two triangle are in the same ratio. The first triangle area is 9 sq. units and  the second triangle area is 18 sq. units. Find the scale factor of the two triangle.

Sol : For the first triangle the  area is 9 sq. units.

 For the second triangle the area is 18 sq. units.

The scale factor = second triangle area /first triangle area

        = 18/9

         =2

answer for this problem = 2

Equation

Equation is one of the major concepts in mathematics. Taught in middle school, equation plays an important role throughout the mathematical years. Equation can be classified into algebraic equation, linear equation, quadratic equation and so on. Let’s have a basic overview of equation and formulas in this post.

Equation in simple terms defines the equality of two things. Equation includes the equal sign to define this equality. For example: X + 2 child phone for kids = 7 child phone for kids. Here, the equation is demonstrating that the things in left side are equal to the things in right side. Solving equations in mathematics is like solving a puzzle. A number is missing and one needs to find out the missing value.

For example: 2X + 6 crib toys = 12 crib toys = 2X = 12 – 6 = X = 6 / 2 = 3

Therefore, X is equal to 3.
Formula on the other hand in mathematics means a type of equation that defines the relationship between two variables. There are different formulas used in mathematics to solve equations. For example: Formula for finding volume of a box, area of a circle, radius of a circle and so on.

Example 1: Maya bought a box that includes the best flashlight available in the market. The height of the box is 4, width is 5 and length is 10. Now find the volume of the box including the best flashlight:

Applying the formula of finding the volume of a box i.e, V = h*w*l, we get
Volume of box = 4 * 5 *10 = 200.

Example 2: Arpita got a rectangular box. The length of the box is 4 and the breadth of the box is 8. Find the volume of the rectangle.

Applying the formula of finding the volume of a rectangle i.e, V = 2 (l + b),
we get Volume of rectangle = 2 (4 + 8) = 24
These are the basics about equations and formulas in mathematics.

Study Midpoint Formula

A midpoint formula is a mathematical form halfway between two stellar bodies that tells an interpretative picture for the individual. Direct and indirect are the types of mid point. Sometimes you need to find the point that exactly between two other points. For other instance, you might need to find a line that bisects in a given line segment. These middle point is called the "midpoint".



Discussion on study midpoint formula


If the co-ordinates of the end points of a segment are (x1, y1) and (x2, y2), then the co-ordinates of the mid point of this segment is given by the mid point formula :

M=[(x1+x2)/2,(y1+y2)/2]


Examples on study midpoint formula


Study midpoint formula Example 1:

To find the mid point formula between A(1,2)and B(3,6)

solution:
Let the coordinates of  A(1,2) andB(3,6)

=((1+3)/2,(2+6)/2)

=(4/2,8/2)

=(2,4)

Study midpoint formula Example 2:

M(3, 8) is the midpoint formula of the line AB. B has the coordinates (-2, 3), Find the coordinates of A.

Solution:
Let the coordinates of B be (x, y)

[(-2+x)/2,(3-y)/2]=(3,8)

[(-2+x)/2]=3==>x=8

[(3+y)/2]=3==>x=13

Coordinates of B = (8, 13)

Study midpoint formula Example 3:

What is the midpoint formula between the points (5, 6) and (– 12, 40)?

Solution:

(m1,m2)=((5+(-12)/2),(6+40)/2)

=(-7/2,46/2)

=(-7/2,23)

Study midpoint formula Example 4:

If the midpoint formula between (1, 4) and (x, 10) is (–4, 7), what is the value of x?

solution:

1+x=-8 (or)(1+x/2)=-4
which gives the value x=-9

Study midpoint formula Example 5:

Find the mid point formula of a segment if A is (6,8) and B is (-2, 4).

Solution :

Let (6, 8) be (x1, y1) and    (-2, 4) be (x2, y2).

Therefore their mid point M is given as :

M=[(x1+x2)/2,(y1+y2)/2]

=((6-2)/ 2,(8+4)/2)

= (4/2,12/2)

=(2,6)

Example 6:

If the mid point of segment. AB is (-2, 8) and A is (12, -1), find co-ordinates of B.

Solution :

Let the co-ordinates of B be (x, y). According to the mid-point formula

(-2,8)=[(12+x/2,-1+y/2)]

-2=(12+x)/2, 8=(-1+y)/2

-4=12+x, 16=-1+y

(x,y)=(-16,17)

Therefore the co-ordinates of B are (-16, 17).

linear probability model

Definition:

Linear probability model is one of the econometric model where the dependent variables having the probability between 0 and 1. There are two things related to Linear probability model

I) Logit

II) Probit

Logit is one of the important part in linear probability model. The Logit function is nothing but the inverse function of logistic which is used in mathematics.
Probit model is a popular method in linear probability model. This model is established using standard maximum likelihood procedure.

Explanation for Linear probability model:


The declining model places no limitation on the values that the independent variables take on. They may be continuous, period level, they may be only positive or zero or they may be dichotomous variable (1=male, 0= female)

The dependent variable is implicit to be continuous. There is no constraint on the IVs; the DVs must be free to range in value from negative infinity to positive infinity

We put into practice, only a small variety of Y values will be observed. Because it is also the case that only a small range of X values will be observed. The best guess on continuous interval measurement is frequently not problematic, That is, even though degeneration assumes that Y can range from negative infinity to positive infinity. It regularly won’t be too much of a disaster if. It really only ranges from 1 to 17

Y can only take two values if Y can equal to 0 or 1 then

E (Yi)= 1 x P (Yi = 1) + 0 x P (Yi = 0) =P (Yi = 1)

Final equation is: E (Yi) = P (Yi) = α + Σ β k X k


Examples for linear probability model:


The yield of apple in an acre of apple plantation depends on various types of agriculture practice (treatments). An experiment may be planned where various ploys are subjected to one  out of two possible treatment over a period of time .The yield of tea before the  application of treatment is also recorded .A  Model for post treatment yield(y) is

y = `beta` 0+ `beta`1 x 1+ `beta` 2 x 2 + `in`

Where the binary variables x1 represent the treatment type and the real valued variable x2 is the pre treatment yield. The error term mainly consists of unaccounted factors such as soil type or the inherent differences in apple bushes

learning statistics probability problems

The word ‘Statistics’ has been derived from either Latin word ‘Status’ or Italian word ‘Statista’ or German word ‘Statistik’ or French word ‘Statistique’ each of which means a political state. The learning statistics defines that the “mathematical method along with numerical style”. Learning statistics is used for solving or finding the problems of the state. Learning probability supplies necessary information for developmental activities in all the departments.Let us see learning statistics probability problems in this article.


Probability definition:


The probability of an event E is a fraction of times we expect E to occur if we show again the same experiment over and over. The expected probability approaches the theoretical probability problems as the number of trials gets bigger and bigger.  If E is a single outcome s, then it is represented as to P (E) and the probability of the outcome is s, and probability of s and E is represented as P(s) for P (E). The groups of the probabilities of all the outcomes are the probability distribution of the problems.


Learning Statistics probability Example problems:


Question 1:

Find x and y so that the ordered data set has a mean of 25.3 and a median of 27.
9, 12, 17, 22, 25, x, 29, 32, 38, y

Answer:

x = 29, y = 40

Question 2:

Given the data set
62 , 60 , 68 , 75 , 72 , 74 , 76 , 78 , 80 , 82 , 96 , 101,
find
a) the median,
b) the first quartile,
c) the third quartile,

Answer:

1.median = 75

2.first quartile = 69

3.third quartile = 81

Question 3:

The marks scored by the student are,
72, 66, 72, 96, 44, 90, 50
Find
a) the mean
b) standard deviation

Answer:

1.mean = 70

2.standard deviation = 18.6 (rounded to 1 decimal place)

Learn Quadratic Functions

A function of the form f(x)= ax2+bx+c, where a,b,c are real numbers and not equal to zero. It is a polynomial of degree '2'. so it can be called as '2nd degree polynomial'.

examples are          1)3x2-6x+4 and

2)-4x2+6x-7   etc.,

If  we draw the graph of aquadratic function  we get a 'Parabola'.A parabola is "U" shaped symmetrical curve. It may be upward or down ward depending upon the sign of 'a', the coefficient of x2,where it is positive or negative.The point where it changes its shape is called "Vertex".or 'Turning point'.

Various Foms of Quadratic function:

1)The form f(x)=ax2+bx+c,is called "General form"

2)The formf(x)=a(x-h)2+k, where h,k are the coordinates of the vertex, is called "Standard form"

3)The form f(x)=a(x-r1)(x-r2),where r1,r2  the roots of the quadratic function , is called "factored form"

Quadratic Equation: If a Quadratic function is made to equal to zero , then it is called "Quadratic Equation"  i.e.,ax2+bx+c=0 is called Quadratic equation.

examples are  1)2x2+3x+4=0 and -9x2+4x-6=0 etc.,

Since it is  apolynomial of 2nd degree it possess two solutions, which are called 'Root' of the equation.

Quadratic Formula: The formula to find out the roots of given equation is

x=(-b+sqrt(b2-4ac))/2a and x=(-b-sqrt(b2-4ac))/2a

the quantity under root is called "Discriminate" .It isdenoted by "Delta"or 'D'.

i.e., D = b2-4ac .

Discriminate gives the nature of the roots .


Nature of Roots - learn quadratic functions


1) If the discriminate is positive , then both the roots are real and distinct.

For the quadrztic  equation with integer coefficiens. if the discriminate is perfect square then the roots are rational numbers, in all other cases they are irrational.

2) If the discriminate is zero , roots are equal and real and they are equal to x=-b/2a

3) if the discrimionate is negative,roots are complex numbers and they are conjugate to each other.

To understand the concept observe the following problems.


Model Problems of learning quadratic functions


Find The  roots and  their nature of the following quadratic equations:

1) x2+2x-3=0

sol: Comparing the given equation with the general form ax2+bx+c=0

a=1;b=2;c=-3

now the disciminant D=2*2-4*1*(-3) = 4+12 = 16, positive so the roots arereal and disticnt

they are x1 = -b+sqrt(D) / 2a = -2 + sqrt(16)/2*1 = 1   and

x2 = -b-sqrt(D) / 2a = -2 - sqrt(16) / 2 * 1 = -3

so the roots are real ,distinct and roots are 1,-3

2) -x2+2x-1 = 0

sol:  comparing it with the standard form  ax2+bx+c = 0, we get

a = -1;b = 2;c = -1

discriminant D = 2*2-4*(-1)*(-1)=4-4=0 so the roots real and equal ,given by

x = -b/2a = -2/2*(-1) =1

hence the roots are real equal and given by 1,1

3) -2x2+2x-2 = 0

Sol: Comparing it with the standard form ax2+bx+c=0,we get

a=-2;b=2;c=-2

D = 2*2-4*(-2)*(-2 ) = 4 -16 = -12 negative  so the roots complex conjugates

x1=-b+sqrt(D)/2a = -2 + sqrt(D) / 2(-2) = (1-2isqrt(3)) / 2

x2=-b-sqrt(D)/2a=-2-sqrt(D)/2(-2)=(1+2isqrt(3))/2 .

In this way we find the roots of the Quadratic equations .

The Unit Circle Learning

A unit circle is a circle with a radius of one. Especially in trigonometry, the unit circle is the circle of radius one centered at the origin (0, 0) in the Cartesian coordinate system in the Euclidean plane.The point x, y the unit circle in the first quadrant, they are the lengths of the legs of a right triangle whose hypotenuse has length 1.

x2 + y2 = 1.

(Source: Wikipedia)



Definition for unit circle learning:


A reference value of a quantity used to express other values of the same quantity. Circle with radius unity. Its area is П and circumference 2 П.

It is a plane curve formed by the set of all points of a given fixed distance from a fixed point. The fixed point is called the centre and the fixed distance the radius of the circle. A circle is a locus of a point whose distance from a fixed point is constant.

The diameter of a circle is double of its radius. The circumference is 2 П r and its area is Пr2. The equation of a circle with radius r and centre at origin in Cartesian coordinate is x2 + y2.= r2




Explanation and application for unit circle learning:


Explanation for unit circle learning:

1 Length of circumference:

The circumference length is related to the radius (r) by,

C = 2 `pi`r

Diameter d = 2r

`pi` = Circumference of a circle / diameter of the circle

`pi` = 3.14

C =  `pi` d

2. Area enclosed the circle:

Area of the circle =  `pi` r2

Area of the circle interms of diameter

A =  `pi` (d / 2)2          (d – diameter, r – radius)

Area =  `pi` d2 / 4         ( `pi`– 3.14)


Application of the Unit Circle learning:

The most important function of unit circle is the implementation of trigonometric ratios over the unit circle.
And it shows the relative of unit circle and the trigonometric ratios.
It is used to understand the relationship between the unit circle and the trigonometric (sine and cosine) functions.
Because of this explanation it is use to solve a real-world application.