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.

Division Solver

I think, you are involved in the math division of numbers, so you need to do undertaking has of finding the quotient and the remainder in the division of two positive numbers. Do most of “quotients, remainders” and you bring to aptitude an image like this:

                                       6        

                          200 | 1220

                                   1200   

                                       20

 

In this example, we are solver dividing 1220 by 200 and we are find the quotient to be 6 and the remainder to be 20 using dividing 1220 by 200. The division problem has the dividend and the divisor as its given value; in the example, these are, in that order, 1220 and 200. Results have of the quotient and the remainder, which, in the example, are 6 and 20, in that order.

Think about the solver division of number are 27 % 5= 5 remainder 2. It may also be written as

                                   `27/5`=2+(`25/5`)

In solver division using algorithm, 27 is called the dividend, 5 the divisor, 5 the quotient, and 2 the remainder. The relation between these four quantities can also be representation as 27=5x5+2.

In fact, given any integers g and h (h not equal to 0) there exist only one of its kind integers q and r, where 0≤ r<h, such that                            g=hq+r

This solver division algorithm is referred to as the math division decimals or the division independence.

If you require to do divide g by h (b not equal to 0), we get `g/h`=q=`r/h`, which makes the solver  division algorithm almost clear.

When g is divided by h we get a quotient q and a remainder r, 0 ≤r<h, and this cannot be ended in more than one way (q and r are only one of its kind).

Division solver need to follow the step:

     Given: solver division positive numbers g and h.        
  
     Result: non-negative numbers q and r such that g=hq+r, and 0≤r<h.

Steps you need to do for division solver:

        Step 1: Begin by set Q=0 and R=g.

       Step 2: if R<h, write down, “The quotient is Q and the remainder is R” and stop; or else, go to do step 3.

       Step 3: If R≥h, take off b from R, boost Q by 1, and go back to step 2.

Problems on Division Solver:

Problem 1:- Division problem solver 23 by 68

Solution:-

In following steps for how to do divide 23 by 68 

Step 1:-
        -------
    23| 68

In the above equation 23 is divisor and 68 is dividend. In the divisor has two decimal numbers put the value dividend of 23.

Step 2:-
         2
       -------
   23| 68
        46
      ---------
        22
In 23 x 2 = 46 the divisor number 23 is multiplied with 2 to get an answer 46. In 46 is less than from 68.So use the value then subtract the value and get 22. 

Step 3:-
         2.9
        -------
    23| 68
         46
       ---------
         220
         207
      ---------
          13
The value 22 has no more value in the right side. So put 0 to get 220 and put decimal point on the quotient. Then normal dividing 23x 9 = 207 the divisor number 23 is multiplied with 9 to get an answer 207. In 207 is less than from 220.So use the value then subtract the value we get 13 

Step 4:-
         2.95
        -------
    23| 68
         46
       ---------
        220
        207
      ---------
         130
         115
      ---------
           15 (continued)

The value 13 has no more value in the right side. So put 0 to get 130 and put decimal point on the quotient. Then normal dividing 23x 5 = 115 the divisor number 23 is multiplied with 5 to get an answer 115. In 115 is less than from 130.So use the value then subtract the value we get 2.95


Problem 2:- Division problem solver 96 by 4 Solution:-

In following steps for how to divide 96 by 4

Step 1:-
       -------
     4| 96
In the above equation 4 is divisor and 96 is dividend. In the divisor has two decimal numbers put the value dividend of 96.

Step 2:-
       2
     -------
   4| 96
      8
     ---------
      16

In  4 x 2 = 8 the divisor number 4 is multiplied with 2 to get an answer 16. In 8 is less than from 9.So use the value then subtract the value and get 1.

Step 3:-
       24
     -------
   4| 96
      8
    ---------
      16
      16
   ----------
       0
    ----------

In 4 x 4 = 16 the divisor number 4 is multiplied with 4 to get an answer 16. In 16 is equal to 16.So use the value then subtract the value and get remainder is 0.

solving shape area formulas

Area is nothing but a region bounded by a closed curve. Area is a quantity which expresses the 2D size of a defined part of a surface. In terms of differential geometry of surfaces, area is an important variant. In this article solving shape area formulas, we are going to discuss about solving the basic shape area formulas.


Solving shape area formulas types:


Area of a Rectangle.

The Formula for Area of a Rectangle is given below:

     Rectangle shape Area = Length x Breadth   ( A = LB )

Area of a triangle.

The formula for the area of a triangle is given below :

         Triangle shape Area = ½ x Base x Height

                                    A = ½ BH

Area of a Trapezium.

A Trapezium is closed shape which has  two sides that are parallel and two sides that are not parallel.

Now we are going to find a formula for the area of the trapezium.

    Trapezium shape Area = A1 + ( A2 + A3 )

                              Area =  b x h + ½ x (a - b) x h

                               Area = bh + ½ h(a - b)

                               Area = bh + ½ ah – ½ bh

                               Area = ½ ah + ½ bh

                               Area = ½ h ( a + b )

The Area of a Circle.

The area of a circle with the radius r is given by the formula                

                              Area = `pi` r2.


Examples for solving shape area formulas:


Example 1:

Find the area of a triangle with the base length of 20 mm and a height of 6 mm.

Solution:

               Area of a triangle =  ½ b h

                                        =  ½ (20) (6)

                                        =  60 mm2  

Example 2:

Find the area of rectangle given the length is 10 cm and width is 7 cm.

Solution:  

         Area of a Rectangle  =  l * w

                                       = 10 * 7

                                       = 70 cm2

Example 3:

Find the area of a circle given the radius is 44 inches

Solution:

                 Area of  Circle = `pi` r2

                                      =  3.14 (44)2

                                      =  3.14 (1936)

                                      =  6079.04 in2

Coefficient Matrix

Introduction for Matrix:

A matrix is a rectangular array or arrangement of entries or elements displayed in rows and columns put within a square bracket or parenthesis. The entries or elements may be any kind of numbers (real or complex), polynomials or other expressions. Matrices are denoted by the capital letters like A, B, C…

A = [ai, j] m * n

i, j represent the numbers of m, n.

i, j, m, n all represented in the suffix of the considered terms.

A = `[[a11,b12],[c21,d22]]`

order of above matrix 2 * 2

Definition for Coefficient matrix:


The coefficient matrix is formed from the linear equations. There can be any number of linear equations. In the linear equation we are taking the coefficient of the variables in the linear equation to forms the coefficient matrix. Consider  linear equation from it we are going to form a coefficient matrix. The general linear equation as follows:

a11x1 + a12x2 + . . . +a1nxn = b1
a21 x1 + a22x2 + . . . +a2nxn = b2
a31x1 + a32x2 + . . . +a3nxn = b3
. . . . . . . . . . . . . . . . . . .
. . . . . . . . . . . . . . . . . . .
am1x1 + am2x2 + . . . +amnxn = bn

The coefficient matrix for the linear equation will be represented as:

A = `[[a11,a12,. . .,a1n],[a21,a22,. . .,a2n],[a31,a32,. . .,a3n],[.,.,. . .,.],[am1,am2,. . .,amn]]`

m * n

The coefficient matrix is formed from the linear equations.

Order or size of a coefficient matrix:

The order or size of number of Row Matrix and the number of columns that are present in a matrix.


Example for coefficient matrix:


The coefficient matrix is a resultant from the linear equations.

Linear equation as follows as:

x + 2y +3z = 4
2x + 3y + 4z = 5
3x + 4y +5z = 6

The coefficient equation of the above linear equation as follows:

A = `[[1,2,3],[2,3,4],[3,4,5]]`
3 * 3

The coefficient matrix is formed from the set of linear equations.