Equal Fractions Answers

The equal fractions answers are nothing but the fraction which involves the same overall value. An equal fraction gives the identical part of the whole value. Let we have the glance at the circles in which we have the ½ circle, the ¼ circle and the 1/8 circle which are made to shaded in red color for the circle.

Equal fractions answers:

This gives the each area mentioned in the red color which is the equal fractions or the equal amount. Therefore we have an idea ½ is made equal to the` 2/4 ` and `1/2` is made equal to `4/8` .



For our reference when an apple is made to cut exactly down the middle, into two equally pieces, the first piece is identical to the one half of the apple. When the another apple is made into 4 equal pieces, then the two pieces of that apple shows the identical part of the apple that shows ½ , Hence we can say that ½ is made equal to 2/4. The fractions that are made to have the part whole number relationship.

The part whole number of a fraction of the number like `1/5` can be done through the whole number that requires the five equal parts and the one of these parts that is being considered. Let us have the briefing of the quotient or the ratio. The quotient is nothing but the term that has the` 2/3` in which the 2 is the numerator and the 3 is the denominator. The method can be mentioned as` 2/3` . The ratio is made to have the briefing that gives the situation on it. Hence there are two boys or the every three girls in the team hence in this case two-thirds of the teams are boys.

An example for equal fractions answers:

Example 1: Say whether the fractions` 3/4` = `9/12` are equal fractions?

Solution: `3/4` = `9/12`

=` 3/4` = `9/12`

= 3 x 12 = 4 x 9

= 36 =36

Answers: Hence the fractions obtained are equal fractions.

Example 2: Say whether the fractions `1/2` =` 2/3 ` are equal fractions?

Solution: ` 1/2 ` =` 2/3`

= `1/2` =` 2/3`

= 1 x 3 = 2 x 2

= 3 = 4

Answers: Hence the fractions obtained are not equal fractions.

Area of Sector

The part of the circle which is enclosed by an arc and two radii drawn to the extremities of the arc is called a sector.
area of a sector

sector and area of sector

The total space inside the boundary of the sector is called as the area of the sector. Area is measured in terms of square unit.

Let O be the center, r be the radius, AB be the arc. The shaded portion AOB is the sector; l is the length of the arc. θ is the angle of the sector at the center.

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C is the circumference of the circle O, and C = 2πr units

Angle at the center of the circle = 360º

l/C = theta^o/360^o

Plug the value for C

l/(2 pi r) = theta^o/360^o --- (1)

Length of the sector , l = theta/360 xx 2 pi r units

Area of a Sector Formula

Arcs are proportional to the angle subtended at the center.

Since sector is bounded by two radius and arc, perimeter of the sector = length of arc + radius + radius

= l + r + r = l + 2r units

Area of the sector is proportional to the angles subtended at the center.

Area of the sector    =  θ                    --- (2)

Area of the circle         360

Area of the sector = (theta xx pi r^2)/360

From (1) and (2) , we have

Area of the sector    =   l

Area of the circle         2πr

Therefore, area of the sector = l/(2pir) xx pir^2 = l xx r/2

Note: When the angle of the sector is not given while the length of the arc and radius are given, the formula for area is (lr)/2 sq. units

Examples

Below are the examples on area of sector -

Example 1: Calculate the area of the sector whose perimeter is 110 cm and radius is 20 cm.

Solution:

Step 1:

Write the given details.

Perimeter = 110 cm

Radius = 20 cm

Step 2:

Write the formula for perimeter and find "l"

P = l + 2r

l = P - 2r

Step 3:

Plug the values of P and r

l = 110 - 2(20)

= 70 cm

Step 4:

Since we know the length of the arc and radius of the sector, we could use the formula A = (lr)/2 to calculate area

Step 5:

Plug the values in the formula

A = (70 xx20)/2 = 700 sq. cm

Step 6:

Write the solution

Area of the sector = 700 sq. cm

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Example 2:

Calculate the area of the sector whose perimeter is 202 cm and the angle subtended at the center is 216°

Solution:

Step 1:

Write the given details

Perimeter , P = 202 cm

Angle, θ = 216°

Step 2:

Write the perimeter formula and plug the values

Perimeter of the sector = l + 2r = 202 cm       --- (1)

Step 3:

Write the arc length formula and plug the known values

l = theta/360^o xx 2pir = 216/360 xx 2 pi r = (6 pir)/5 --- (2)

Step 4: Plug (2) in (1) and find r

(6pir)/5 + 2r = 202

(6 xx 22)/(5xx7)r + 2r = 202

132/35r + 2r = 202

132r + 70r = 202 x 35

r = 202/202 xx 35 = 35 cm

Step 5:

Calculate the area of the sector using the formula A = (lr)/2

A = (6pir xx r)/(2 xx 5)

= 2310 sq. cm

Angle Bisector

we can define  angle bisector in a line ,suppose a line is divided into equal parts means  the parts must be congruent ,there are two types of the geometry bisector there are  segment bisector and angle bisector .if the line passes through the midpoint of  a segment means we  called those kind of line as a segment bisector and the angle bisector is defined as the line will be passes through the apex of the angle and it also divides the angle  two parts  and generally   bisectors are used for divide the line into equal parts and  the line must be similar  congruent parts

Types of angle bisector:

Interior angle bisector
Exterior Angle bisector

Figure of the angle bisector:

this diagram is used for describes how  line will be split the angle into the two equal parts ,here we have to note one thing both the angles are congruent .

Angle bisector:

A line or the ray it will divide the angle into two equal parts we called as the Angle bisector.
Sometimes   line or line segment will be the angle bisector in polygon line segment is the angle bisector.
If the angle bisector divides triangle in center means we called as center of the angle bisector.

In geometry the angle bisector is divide  line into the two equal parts, both the angles are congruent


The Angle Bisector proof:

Consider the triangle, internal bisector (seen in the above diagram) divides the opposite side .So the opposite side is the ratio of the other two sides of the given triangle.

Consider the triangle has XYZ draw the line to the, it will meet the midpoint of YZ and the named the midpoint as A, so the line XA will be the internal bisector of the
Proof:

YA/AR = XY/XZ

Here the line XA will be perpendicular to the line ZB,

QS/SR = QP/PT.
YA/BZ = YX/XA XA=ZB.
In that < YXA =< XZB ----------------- (1)
Consider the corresponding angle < XBZ =And < XZB =< AXZ (we have to consider the alternate angle) ------------- (2)
From (1) and (2) we wrote the equation as shown below,
< YXA =< AXZ

So we have to tell that the point A divides YZ externally in the ratio XY/XZ, then XA is the external bisector of
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Important points:

For each and every angle there is a line, these line will intersect the angle into two halves and so that the line is called angle bisector.
Therefore angle bisectors for a triangle should be three.
The angle bisector crosses out the line so the line is called the center of the bisector.

Solving Algebra Probability Problems

Algebra is one of the ancient topics in mathematics. The rule of operations and relations which is dealing is known as Algebra. Ancient algebra is very much differing from the modern algebra.

Probability is a bit hard and it has been defined in different manners. Probability can be defined either in objective or subjective manner. A way of expressing facts or belief that an event will occur or occurred is known as probability. Algebra in probability deals with solving equation and finding the values for the events. In solving algebra probability problem step by step explanation is important.

Problem 1:

A spinner is numbewhite1-60. If he spins one time, what is the probability that he will get: a) a multiple of 6? b) An odd number between 45 and 60?

Solution:

Multiples of 6 is

1 * 6 = 6

2 * 6 = 12

3 * 6 = 18

4 * 6 = 24

5 * 6 = 30

6 * 6 = 36

7 * 6 = 42

8 * 6 = 48

9 * 6 = 54

10 * 6 = 60

There are 10 multiples of 6 so taking 10 and divide it by 60

P (multiple of 6) = 10 / 60

On solving this problem we get,

P (multiple of 6) = 1/6

Probability = 0.17


b) An odd number between 45 and 60

The odd number between 45 and 60 are 47,49,51,53,55,57,59.

P (Odd number b/w 45 and 60) = 7/60

On solving this problem we get,

P (Odd number b/w 45 and 60) = 0.12

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Problem 2:

A jar contains 3 white, 8 orange and 7 brown marbles. If one marble is chosen randomly from the jar, what is the probability of i) choosing a white marble ii) orange marble iii) A brown marble.

Solution:

Total number of marble in the jar = 3 +8 +7 = 18 marble

i) Choosing a white marble = 3/18

On solving this we get,

Choosing a white marble = 0.17

ii) Choosing a orange marble = 8/18

On solving this we get,

Choosing a orange marble = 0.44

iii) Choosing a brown marble = 7/18

On solving this we get,

Probability of orange marble = 0.39

Simple Differentiation

In calculus the derivative is defined as a measure of a function that changes as its input changes. A simple derivative can be stated as change in one quantity that response to changes in some other quantity. The derivative of a simple function at a certain input describes the preeminent linear approximation of the function close to that input value. In advanced calculus, the derivative of a function at a point is a linear transformation called the linearization. Simple differentiation problem includes one or two step differentiation x, y terms.

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Differentiation and the derivative

A method to compute the immediate rate of change of a function with respect to one of its variables is known as Differentiation. This rate of change is called the derivative of y with value to x. In more exact language, the dependence of y upon x means that y is a function of x. This functional relationship is frequently denoted y = ƒ(x), where ƒ denotes the function. If x and y are real numbers, and if the graph of y is plotted beside x, the derivative measures the slope of this graph at each point.

Linear function is defined as the x – axis linearly varies with y – axis. In this case, y = ƒ(x) = m x + c, for real numbers m and c, and the slope m.

The differentiation is denoted by the symbol as shown below,

x = dx

y = dy

Problems:

Simple differentiation Example 1:

Differentiate f(x) = x2

Solution:

Here f(x) = y

Y= x2

On differentiating this,

dy/dx = 2x


Simple differentiation Example 2:

Differentiate y = 2x3 + 6x2 + 2x

Solution:

Differentiating the above expression with respect to x

dy/dx = 3(2)x2 + 2(6)x + 2

dy/dx = 6x2 + 12x + 2


Simple differentiation Example 3:

Differentiate f (x) = ln x2 +5x

Solution:

f (x) = ln (x2 +5x)

f’(x) = 2x + 5

x2 +5x

The answer is dy/dx = 2x + 5

x2 + 5x

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Simple differentiation Example 4:

Differentiate y = log x + x2 + 2

Solution:

On differentiating log x, we get

Log x = 1/x.

dy/dx = (1/x) + 2x

Solving Evaluate

Evaluate: It is a process of simplifying an equation in algebra. Algebraic equations are of many types like monomial, polynomials trinomials etc. The polynomial equations are solved by the two ways.Evaluation is the process of solving the equations for particular variables. The methods of solving algebra equation in polynomial are:

Evaluate by Substitution method or plug-in equation method
Elimination method or addition or subtraction method.


Equation evaluation:

Solving by substitution method:

The steps taken in the substitution method or plug-in equation method are as follows,

First solve any one equation for a variable in the given pair of equations.
Then plug the variable into the next equation and solve it.
The equation will be solved for a variable.
Then plug-in the value of the variable in anyone of the equations.
Then solve the next variable in the equation by plugging the value of the variable.
Now repeat the above steps to solve any of the equations left in the variable in the equation.

So, we get the solution for the equations by solving. For evaluating, we mainly use substitution method.

Solving by elimination method:

The steps followed in the elimination method are follows as:

First make one equation equal to others part of the equation.
Then eliminate the variable in the next equation and solve it for the next variable.
The equation will be solved for a variable.
Then plug-in the value of the variable for anyone of the equations.
Then solving the next variable in the equation by plugging the value in the variable.
Now repeat the above steps to solve any of the equations evaluate left in the variable in the equation.

So, we get the solution for the equations by solving. For evaluating, we mainly use substitution method.



Solving Evaluate:Example problems

Example for substitution method:
Evaluate:
4x + 3y = 36 => (1)
y =8               => (2)
Solution:
Substitute equation (2) in equation (1)
4x + 3(8) = 32
4x +24 = 32
-24  -24
4x = 8
Here 24 is subtracted from both sides and part of the equation has been reduced.
4x = 8
On dividing by 4 on both side we get,
x = 2
So, the equation is reduced to obtain the solution.

Example for elimination method:
Evaluate:
2x= I + 5y,
2x+3y-9=0
Solution:
The given equations may be written as
2x - 5y = l ------ (1)
2x + 3y = 9 ----- (2)
(-)   (-)      (-)
On Subtracting we get, - 8y = - 8
Divide by  -8 on both sides,
Y=(-1)/(8) *(-8) =1
Y=1
By substituting y=1 in equation (2),we get
2x+3y=9
2x+3(1) =9
2x+3=9
2x=9-3=6
X=(1)/(2) *(6) =3
X=3
X=3, y=1 is the solution

Importance of Polynomials

In mathematics, a polynomial is an expression of finite length constructed from variables (also known as indeterminates) and constants, using only the operations of addition, subtraction, multiplication, and non-negative, whole-number exponents. For example, x2 − 4x + 7 is a polynomial, but x2 − 4/x + 7x3/2 is not, because its second term involves division by the variable x and because its third term contains an exponent that is not a whole number.


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Importance of Types of polynomials:

It is necessary to know the importance of types of polynomials. The importance of the types of polynomials are as follows:

(1) Constant polynomial

(2) Linear polynomial.

A polynomial where its degree is zero is called as a zero polynomial.

For example, f(x) = 7, g(x) = 7/4 are constant polynomials.

A polynomial where its degree is 1 is called as the linear polynomial.

For example, p(x) = 4x-3, q(y) = 3y are the linear polynomials.

Both the sum and the product of a polynomial is a polynomial and the derivative of a polynomial function is also a polynomial function.

Importance of concepts of polynomials with examples:

The followings are some of the important concepts if polynomial.

If p(x) is a polynomial in x, the highest power of x in p(x) is called the degree of the polynomial.

A polynomial of degree one is called a linear polynomial.

If a polynomial of degree is two then it is called a quadratic polynomial.

If a polynomial of degree is three then it is called a cubic polynomial. It has a general form of ax3 + bx2 + cx + d.

Each term of a polynomial has a coefficient in the form of –x3+ 4x2+7x-2. The coefficient of x3 is -1, the coefficient of x2 is 4.

Examples:

1. Check whether -2 and 2 are zeros of the polynomial x + 2.

Solution:

Let p(x) = x + 2

Then, p(2) = 2 + 2

= 4

p(-2) = -2 + 2

= 0

Therefore, -2 is a zero of the polynomial x + 2, but 2 is not.

2. Determine whether (x–3) is a factor of the polynomial p(x) = x3 – 3x2 + 4x – 12.

Solution:

For (x–3) to be a factor of p(x),

p(3) should be zero by the factor theorem.

Now p(3) = 33 – 3(3)2 + 4(3) – 12

= 27 – 27 + 12 – 12

= 0

Hence (x–3) is a factor of the given polynomial.

Symmetric Matrices

We all know that Matrix is an ordered form of easy presentation of data in an array of rows and columns. There are some special

types of matrices. One of them is "Symmetric Matrix". Here, we study about Symmetric Matrices, broad outlines.

What is a symmetric matrix?

Two essentital things are required for a symmetric matrix.

1.  It should be a square matrix.  i.e. No. of rows in the matrix = No. of columns in the matrix

2. It is equal to its Transporse Matrix.  A= A^T.

So if these two conditions are satisfied, the matrix is symmetric.  The entries of the matrix are symmetric about the main diagonal of the matrix.

Eg of a symmetric matrix  =

Matrix(A)   In other words, a ij = a ji for all i and j in a symmetric matrix.

The diagonal matric    1 0 0   is also symmetric as here also a ij = a ji for all i not equal to j
and a ij = 0 for i not equal to j
0 1 0
0 0 1


When we talk about symmetric, the question of orthogonal matrix also comes.

What is an orthogonal Matrix?

A matrix P is called orthogonal if its columns form an orthonormal set and call a matrix A orthogonally diagonalisable if it can be

diagonalized by D =P inverse AP with P an orthogonal Matrix.

Important properties of symmetric matrices

Consider the following symmetric matrices.

1      2    3                                and                                            2     3      4

2      4    5                                                                                 3     4      5

3      5    6                                                                                 4       5     6

Let them be called A  and B.

Now consider A+B    =      3   5    7

5   8    10

7  10    12

It is symmetric.  So addition of two symmetric matrices lead to another symmetric matrix.

Subtraction:           A-B     =      -1    -1    -1

-1     0      0

-1     0     0

Again a symmetric matrix.  So addition and subtraction of two symmetric matrices preserve the symmetry property.

Multiplication:       Consider A*B   =        Again a symmetric matrix

Skew-symmetric matrix:   A skew symmetric matrix is one whose transpose = negative of the matrix

In other words, A transpose   =   -A

Here the condition is a ij =-a ji

Eigenvalues and eigenvectors of a symmetric matrix are real:

The eigenvalues of a symmetric matrix are real.   Just because the coefficients of a matrix are real certainly does not imply eigen

values are real.  But for symmetric matrix, eigen values are always real.

Decomposition of a symmetric matrices

Every square real matrix can be written as a product of two real symmetric matrices, and every square complex matrix can be written as a product of two complex symmetric matrices. Here, Jordan's normal form is used.

Every real non-singular matrix can be uniquely factored as the product of an orthogonal matrix and a symmetric Positive definite matrix.  This process we call as polar decomposition.  But for singular matrices, they can be factored but not uniquely.

Cholesky decomposition states that every real positive-definite symmetric matrix is a product of an upper-triangular matrix and its transpose.

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Every real symmetric matrix A can be diagonalized, moreover the eigen decomposition takes a simpler form:

Image(X)

where Q' is an orthogonal matrix (the columns of which are eigenvectors of A), and Λ is real and diagonal (having the eigenvalues of A on the diagonal.
Determinant
Determinant
A linear transformation on R2 given by the indicated matrix. The determinant of this matrix is −1, as the area of the green parallelogram at the right is 1, but the map reverses the orientation, since it turns the counterclockwise orientation of the vectors to a clockwise one.

The determinant det(A) or |A| of a square matrix A is a number encoding certain properties of the matrix. A matrix is invertible if and only if its determinant is nonzero. Its absolute value equals the area (in R2) or volume (in R3) of the image of the unit square (or cube), while its sign corresponds to the orientation of the corresponding linear map: the determinant is positive if and only if the orientation is preserved.

The determinant of 2-by-2 matrices is given by

Determinant(A)
Because a symmetric matrix is a square , a determinant exists for every symmetric matrix.

Draw Isosceles Triangle

The triangle is a closed geometric shape that contains three sides. There are several types of triangles in the geometry world. We can divide the triangles according to their angles and sides. We can define the isosceles triangle by using its sides.

The triangle that has two equal or congruent sides in its measure is called as isosceles triangle. We can say that the equilateral triangle is also as an isosceles triangle,since the equilateral triangle has three equal sides . while we draw a isoscles triangle we have to see that two sides triangle must be equal.


draw Isosceles Triangle:

when we draw a isoceles triangle, angles are congruent i.e., an angles of each sides are equal.
when we draw a isoceles triangle diagonals of an isosceles triangle must be congruent.
The measurement of the adjacent angles is giving the 180 degree. x = y = 180o


Formulas for Isosceles triangle:

Hight,  h = √(b2 - `(1/4)` a2)
Perimeter of Isosceles Triangle = A + B + C
Area of isosceles triangle A = (b* h)/2


Example problems:

1) Find the area of isosceles triangle with the base and height are 5 cm and 7 cm.

solution:

We can find the area of given problem using the following formula:

Area A= (b*h)/2

Substitute the values of b and h into the above formula. Then we get,

A= (5*7)/2

Here multiplying to the values of 5 and 6 then dividing by 2.

Then we get the final solution.

Answer A=17.5 cm2

2) Find the perimeter of Isosceles triangle that has side, Side 1 =10 cm, Side 2 = 10, Side 3 = 7 cm.

Solution:

Given, Side 1 =10 cm, Side 2 = 10, Side 3 = 7 cm.

Perimeter of Isosceles triangle   = Side 1+ Side 2+ Side 3

= 10 + 10 + 7

Perimeter of Isosceles triangle  = 27 cm

3) Find the height of the isosceles triangle of the base a = 7 cm and equal sides b = 11 cm.

Solution:

The height of the isosceles triangle is given by the formula:

h = √(b2 - `(1/4)` a2)

substitute the a = 7cm and b = 11 cm. in the formula,

h = √(112 - (1/4)*72 )

h = √(121 - `(49/4)` )

h = √(121 - 12.25 )

h = √(108.75) = 10.428 ≈ 10.4 cm

Height of the isosceles triangle = 10.4 cm

Solving Vertical Line Test

The vertical line test helps us to determine if a curve or a graph which represents a relation is a function or not. The vertical line test would be infinite and that is thin, and if that line you could find two or more points that the relation generates then that relation can not be a 'true' function.

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When the function's domain and co domain x and y axes are correspond to the coordinate system.

Solving vertical Line test:

Vertical line test is relation between two terms of X and Y. And this test used to determine If a relation is a function. If that shows no vertical lines and that intersect the graph at more than one point, the relation for each element of the domain are corresponding to exactly one element of the range.

Points are : ( 1 , 4 ) ( 2 , 6 ) (3 , 8) ( 4 , 10)

Here 1 is an input term and 4 is an output term. Each input has a different output. The vertical line test is mostly  used for when the graph’s ordered pairs.

Solving vertical line test function:

The function for solving the vertical line test is used whenever you graph the ordered pairs in the graph. You can choose a vertical line being drawn in the graph. If the vertical line touches the graph at only one point and then it is referred as function. If the vertical line touches in more than one or more point, then it is NOT a function and it is used to show the vertical line.

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If vertical line test touches the graph at only one point, this relation can be classified as a function.  The vertical line test is simply a restatement of the definition of a function which states that every x value must have a unique y value. If any particular x shows only one y value, then you can pass a vertical line will intersect with the graph of the relation more than once in the graph

Rational Functions Learning

A function is in the form of  y = (f(x))/(g(x)) , where both f(x) and g(x) are polynomial functions is called rational function.

Examples of rational functions:    y = 1/(x - 2)

f(x) = (3x)/(4 + 5)

g(x) = (x^2 - 6)/(x^2 - 3x)

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These are the some of the examples of rational functions. The domain and different asymptotes of rational function are explained below in detail.

Learning domain of the rational functions:

In a function, domain is defined as the set of all possible real numbers of the independent variable i.e., it is defined as the set of all real numbers for which the function is defined. A example for learning domain of a rational function is given below.

Example:

1) Find domain of the function f(x) = (4x + 9)/(2x - 4)

Solution:

Step 1: Given function

f(x) = (4x + 9)/(2x - 4)

Step 2: Set denominator to zero.

2x - 4 = 0

Step 3: Solve the above equation for x.

2x = 4

Divide by 2 on both side,

x =2

Therefore, the domain of the given function is all real numbers except x = 2

Learning of asymptote of rational functions:

Asymptotes of rational functions are classified into three types as follows,

Horizontal asymptote
Vertical asymptote
Oblique asymptote

Horizontal asymptote:

In the graph of a rational function, the horizontal line approaches as ' x '  values get very large or very small. It is a line of the form y = c. If  value of x gets increase in positive or negative direction, the function f(x) will increase to the number c.

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Vertical asymptote:

In the graph of a rational function, the vertical line approaches as ' x ' values approach a fixed number. In vertical asymptote, the function f(x) becomes infinite. The function have vertical asymptote at x = a.

Oblique asymptote:

The oblique asymptote is in the form y = ax + b with non-zero. In the rational function, if the numerator has the highest degree than the denominator, then the rational function has oblique asymptote.

Learning about different asymptote helps you to draw graph of a rational function.

How to Describe a Elementary Matrices

In order to locate the position of a particular element of a matrix. We have to specify  the number of the row and that of  the column in which the element occurs. An element occurring in the ith row and jth column of a matrix A will be called the (i,j)th element of A , to be denoted by aij .

In general ,an m x n matrix A may be written as ,

A  =[[a11,a21,a1n],[a21,a22,a2n],[...,...,...],[am1,am2,amn]]



Operation on matrices

Basic operation involved in matrices are,

1)Addition of matrices

2)Subtraction of matrices

3)Multiplication of matrices

Let  us discuss about addition of matrices.

Addition of Matrices:

Let A and B be two comparable matrices,each of order (m xn).Then their sum(A+B) is a  matrix of order (m x n) ,obtained by adding the corresponding elements of A  and B.

Thus, ifA= [aij]mxn    and  B = [bij] mxn   then

A+B = [aij  +bij] m xn

Note : for two matrices A and B ,the sum (A+B) exits only when A and B are comparable.

Examples of Adding matrix:

Example 1:

if A= [[3,4],[5,7]] and B  =[[4,6,1],[5,7,3]] then A and B are matrices of order 2 x2 and 2 x3 respectively.

So A and B are not Comparable.

Hence ,A + B is not defined.

Example 2:

Let A=[[3,6,9],[3,5,1]] and B=[[-4,8,5],[8,3,-2]]

Clearly ,each one  of A and B is a 2 x 3 matrix.

So a and B are comparable matrices.

Therefore A+B is defined.

Now,   A+ B  =  [[3 + (-4) ,6+8 , 9+5],[3+8 , 5+3 , 1+(-2)]]


=[[-1,14,14],[11,8,-1]]

I am planning to write more post on solved sample papers for class 12 and Normal Distribution Histogram. Keep checking my blog.

Some Resultes on Addition of matrices.

Theorem:

Matrix addition is commutative. ie)A+B = B+A for all comparable matrices A and B

PROOF:

Let A =[aij]mxn   and B = [bij]mxn Then,

A+B= [aij]mxn  +[bij]mxn

=[aij +bij]mxn [by the definition of addition of matrices]

=[bij+aij]mxn [Addition of numbers is Commutative]

=[bij]mxn +[aij]mxn  =B+A

Hence ,A+b = B+A

Check  the theorem:

Let A =[[2,4],[5,4]] and B=[[4,3],[2,5]]

A+B  =[[2+4,4+3],[5+2,4+5]]

=[[6,7],[7,9]]

B+A =[[4+2,3+4],[2+5,5+4]]

=[[6,7],[7,9]]

A+B = B+A

Hence proved.

Theorem 2:

Matrix Addition  is Associative ie )(A+B+ +C  =A+(B+C)

PROOF: Let A=[aij]mxn, B=[bij]mxn and C=[cij]mxn Then

(A+B) +C=([aij]mxn +[bij]mxn+[cij]mxn)

=[aij+bij]mxn +[cij]mxn

=[(aij+bij) +[cij]mxn

=[aij +(bij +cij)]mxn  [addition of numbers is associative]

=[aij]mxn +[bij +cij]mxn

=[aij]mxn +([bij]+cij])mxn)  =  A+(B+C)

Hence ,(A+B)  +C = A+(B+C)

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Check the theorem 2:

Let A=[[2,4],[5,4]] ,B=[[3,4],[5,3]] and C=[[8,4],[3,5]]

Prove that (A+B)  +C  = A+(B+C)

Take (A+B) +C  => [[2+3,4+4],[5+5,4+3]] +[[8,4],[3,5]]

=> [[5,8],[10,7]]   + [[8,4],[3,5]]

=> [[5+8,8+4],[10+3,7+5]]

(A+B) +C   => [[13,12],[13,12]]

Take A+(B+C) => [[2,4],[5,4]]   +  [[3+8,4+4],[5+3,3+5]]

=> [[2,4],[5,4]]    +  [[11,8],[8,8]]

=> [[2+11,4+8],[5+8,4+8]]

A+(B+C)             => [[13,12],[13,12]]

Therefore ,(A+B) +C =A+(B+C)

Hence Proved.

Theorem 3:

if A is an mxn matrix and O=[bij]mxn

where bij =0 for all suffixes i andj

Then,   A+O = [aij]mxn +[bij]mxn  =[aij +bij]mxn

=[aij +0]mxn  [bij  =0]

=[aij] mxn  =A

A+O=A

Similarly,O+A =A

Hence A+O=O+A =A

Check the theorem 3

Let A =[[3,5,4],[2,7,5]] and O=[[0,0,0],[0,0,0]]

A+O  =[[3+0,5+0,4+0],[2+0,7+0,5+0]]

A+O  =[[3,5,4],[2,7,5]]

And O+A  =[[0+3,0+5,0+4],[0+2,0+7,0+5]]

=[[3,5,4],[2,7,5]]

Therefore ,A+O  =O+A

Hence Proved.

Two Limit to Infinity Definition

Two Limits of Functions as X Approaches Infinity:

The problems need the algebraic calculation of limits of functions as x approaches + or – endless. The majority problems are average.

The minority are rather challenging.

Initially, many students incorrectly conclude that oo/oo is equal to 1, or that the limit does not exist, or is + oo or -oo . Many also conclude that oo - oo  is equal to 0.

In fact, the forms oo /oo and oo - oo are examples of indeterminate forms.

Such tools as algebraic generalization and conjugates can easily be used to avoid the forms oo / oo and oo - oo so that the limit can be calculated.


Two limit to infinity problems:

Two limit to infinity problem 1:

Compute

lim x ->oo 100/ (x^2 + 5)

SOLUTION 1:

lim x -> oo 100/(x^2 +5) = 100/oo = 0.

(The top element is always 100 and the base element x2+5approaches oo as x approachesoo , so that the resulting fraction approaches 0.)

Two limit to infinity problem 2:

Compute

lim x -> -oo 7/ (x^3 - 20).

SOLUTION 2 :

lim x -gt -oo 7/(x^3 -20) = 7 / -oo = 0.

(The top element is always 7 and the base element x^3 - 7 approaches -oo as x approaches -oo , so that the resulting fraction approaches 0.)

Two limit to infinity problem 3:

Compute

lim x -> oo 3x^3 - 1000 x^ 2.

SOLUTION 3 :

lim x -gt oo ( 3x^3 - 1000x^2) = oo - oo

(This is NOT equal to 0. It is an undefined form.)

= lim x -gt oo x^2(3x - 1000)

(As x approaches oo , each of the two expressions x^2 and 3 x - 1000 approaches oo .)

= (oo) (oo)

(This is NOT an indeterminate form. It has meaning.)

=oo.


(Therefore, the limit does not exist. Note that an alternate solution follows by first factoring out x^3, the highest power of x . Try it.)

Two limit to infinity practice problems:

Compute

lim x -> -oo x^4+5x^2+1.

Compute

lim x-> oo x^5-x^2+x-10.