Easy Way to Learn Limits

Introduction to Easy way to learn limits:      
In the easy way to learn limits, let f be a function of a real variable x. Let c and l be two unchanging numbers. If f(x)  come within reach of the value l as x approaches c, we say l is the limit of the function f(x) as x tends to c. This is written as

`lim_(x->c)`   f(x)   =  l.

Left Hand and Right Hand Limits:

In thr easy way to learn limits, although defining the limit of a function as x be inclined to c, we consider values of f(x) when x is very close to c (x>c or x
Lf(c) = `lim_(x->c )` _f(x) , provided the limit exists.

Likewise if x gets merely values greater than c, next x is said to tend to c from above or from right, and is denoted symbolically as x ? c + 0 or x ? c+, here the limit of f is then labeled the right hand limit. This is written as Rf(c) = `lim_(x->c+)` f(x). It is significant to reminder that for the survival of`lim_(x->c)` f(x) it is necessary that both Lf(c) and Rf(c) exists and Lf(c) = Rf(c) =`lim_(x->c)` f(x).  As well as the left and right hand limits are labeled as one sided limits.

Fundamental Results for Easy Way to Learn Limits:
The following rules for easy way to learn limits,

(1) If f(x) = k for all x, then
`lim_(x->c) `  f(x) = k.
(2) If f(x) = x for all x, then                                                                                                               
`lim_(x->c)`  f(x) = c.
(3) If f and g are two

functions having limits and k is a invariable then

(i)  `lim_(x->c)` k f(x) = k `lim_(x->c)` f(x)

(ii)  `lim_(x->c)` [f(x) + g(x)] =`lim_(x->c)` f(x) + `lim_(x->c)`  g(x)

(iii)  `lim_(x->c)`  [f(x) - g(x)] = `lim_(x-> c)`  f(x) - `lim_(x->c)`  g(x)

(iv)  `lim_(x->c)` c [f(x) . g(x)] = `lim_(x->c)` f(x) . `lim_(x->c)` g(x)

(v)    `lim_(x->c)` [`f(x)/g(x)` ] = `lim_(x->c)`  f(x)  / `lim_(x->c)`  g(x)              g(x) `!=` 0

(vi)  If   f(x) = g(x) then

`lim_(x-> c)`  f(x) =  `lim_(x->c)` g(x).

Example of Easy Way to Learn Limits:

Evaluate  `lim_(x->3)` `(x^2+ 7x + 11) / (x^2-9)` .

Solution:
Let f(x) = `(x^2 + 7x + 11)/( x^2-9)`
This is of the form f(x) =`g(x) / (h(x))`  ,
where g(x) = x2 + 7x + 11 and h(x) = x2 - 9. Clearly g(3) = 41 ? 0 and h(3) = 0.
Therefore f(3) = `g(3) /(h(3))`  = `41/ 0`  . Hence `lim_(x->3)`  `(x^2+ 7x + 11) / (x^2-9)` does not exist.

Solving Plane Geometry

Introduction to solving plane geometry:

Geometry is one of the important parts of mathematics deals with properties of shape of geometric objects. Geometry used to say all kinds of shapes and their properties. There are two major classifications in geometry. They are Plane geometry and Solid geometry. Shapes that are drawn at flat surface are called Plane and the study about this is called  Plane geometry. Plane geometry is about the shapes of lines, circles, etc. Plane geometry is study about two-dimensional objects.

Problems on Solving Plane Geometry:

Problem 1:

Solving the area of the triangle with the base of 7cm and the height is 10cm.

Solution:

Given:       Base = 7cm

Height = 10cm

Area = (b * h) / 2                       

= (7 * 10) / 2                            

= 70 / 2

= 35 cm^2

Problem 2:

Solving the area of the circle with 10.5cm radius.

Solution:

Given:     Radius = 10.5cm

Area of the circle =`pi` r2

= 3.14 * 10.5 * 10.5

= 346.18cm^2

Problem 3:

Solving the area of the trapezoid whose length is 10 cm, the width is 5cm and the height is 10 cm.

Solution:

Given:   Length = 10cm

Width = 5cm

Height = 10cm

Area = h (l + w) / 2                                                                                                                

= 10 (10+5) / 2

= 150 / 2

= 75cm3

Problem 4:

Solving the area of the rectangle with length 6.5cm and width 3.5cm. Also find its perimeter.

Solution:

Given:    Length = 6.5cm

Width = 3.5cm

Area = L * W

= 6.5 * 3.5

= 22.75cm^2            

Perimeter = Sum of all sides

= 2L + 2W

= 2 * 6.5 + 2 * 3.5

= 13 + 7

= 20cm^2

Problem 5:

Solving the area and perimeter of the parallelogram whose Height 12 cm, length 14cm, width 16cm.

Solution:

Given:    Length = 14cm

Width = 16cm

Height = 12cm

Area = L * W

= 14 * 16

= 224cm^2

Perimeter = Sum of all sides

= L + W + H

= 14+16+12

= 42cm3


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Practice Problems on Solving Plane Geometry:

1.   Find the area of the trapezoid whose length is 3 cm, the width is 1.5cm and the height is 5cm.

Answer:  11.25cm^3

2.  Find the area of the circle with 0.5cm radius.

Answer:  0.785cm^2

3.  Calculate the area and perimeter of the rectangle whose length is 2cm, width is 2.5cm.

Answer:   Area = 5cm^2
Perimeter = 9cm^2

Distribution of Prime Numbers

Introduction of distribution of prime numbers:

The distribution of positive numbers is divided into various terms, such as prime numbers like 2, 3, 5, and 7. The distribution of prime numbers cannot be resolute into the factors and the composite numbers. The fundamental theorem of arithmetic gives the most importance for the distribution of prime numbers. Hence the composite numbers can be expressed in a way as the product of the prime numbers.

Distribution of Prime Numbers:

The problem on the threshold of math is the queries of the distribution of the prime numbers. Though the sequence of the prime numbers shows the irregularities with explanation, the distribution of the prime numbers is created to possess the samples on the regularities in some of the terms and to make mathematical investigation.

The p(x) can be denoted as the number of prime numbers that do not exceed the value x. The resolving of the function can be done in the study of the function p(x). The examination of the table of prime numbers at the observance however the table gives the prime numbers which shows no signs which coming to an end. They become to the average which widely spaced in the higher position of the table. Theorems which gives the distribution of the prime numbers. These theorems would results to p(x) at the infinity condition.

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Theorems in Distribution of Prime Numbers:

The initial theorems which shows the infinite number of the prime numbers. Let we have the P as the multiple of the finite sets of the prime numbers. Let we indicate Q = P + 1, the letters P and Q which don’t have any prime factors. These factors would divide Q – P = 1, this is not possible. Hence the Q may be the divisible of the prime number. Here exists of one prime number from occurring of P.

Area of Circle Worksheet

Introduction to area of circle worksheet:
A circle is formed by the set of points which are at an equal distance from a center point.

Let us know about the two important terms of the circle.

Circumference :

Total length of the curved figure is called as circumference.

Distance covered by a wheel in one revolution is the circumference.

Area:

Area of the circle is the region covered by the circumference on the plane surface.

Diameter:

It is the line segment whose end points are on the circumference  and passes through the origin.

Radius:

Radius is the distance between the center point and a point on the circumference of the circle.

Greek letter p:

The ratio between the value of circumference to the diameter is always a constant .

It is denoted by the Greek letter p.

p = 3.14 or 22/7


Area of Circle Worksheet-formulas

Let C denotes circumference, A denotes the area, r denotes the radius, d denotes the diameter.

r = d/2

C = pd units

C = 2pr units

A = pr^2 square units

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Area of Circle Worksheet-finding Area of Circle:

We could find the area of the circle using a graph paper by drawing the circle knowing the radius of the circle and then count the number of squares enclosed by the circumference of the circle.

We can estimate the area by counting the number of squares enclosed in the first quadrant or in any of the quadrant and multiply it with 4.

Do you think that it is an easy method?

No

It is very tedious to calculate the number of squares enclosed exactly.

Area of Semi Circle and Quadrant:

Area of semi circle =  area of the circle/2

= pr^2  /2

Area of quadrant =  area of the circle/4

= pr^2/4

Area of Circle Worksheet-problems on Area of Circles:

A circular pool is of the radius 10m. Find the area of the pool.
Solution:

Radius,r = 10 m

Area of the pool = pr^2 square units

= 3.14 x 10 x 10

= 314 m^2

The area of the circle is 16p cm^2. What is its circumference?
Solution:

Area of the circle = pr^2  = 16p

So, r^2 = 16

r = `sqrt(16)`

r = 4 cm.

Circumference of the circle = 2pr

= 2 x 3.14 x 4

= 25.1 cm

Find the area of circle whose circumference measures 66 cm.
Solution:

Circumference of the circle = 2pr = 66

2 x 3.14 x r = 66

r = $\frac{66}{6.28}$

r = 10.5 cm

Area of the circle = pr^2 square units

= 3.14 x 10.5 x 10.5

= 346.185 cm^2

Find the area and radius of the circle, given that the area of the semicircle is 77 cm.
Solution:

Area of the semi circle = 77 cm^2

So, area of circle = 2 x 77

= 154 cm^2

Area of the circle = pr^2 square units

Area of the circle = pr^2  = 154

So, r^2 = 49

r = `sqrt(49)`

r = 7 cm.

Parallel Lines Meet

Introduction to parallel lines:

The lines are the straight curve that never ends. We are having two numbers of lines. They are parallel lines as well as perpendicular lines. The distance of the two parallel lines is always the same. They should not meet at any point. When the two lines should meet at a place which should not be  parallel.


More about Parallel Lines:

The parallel lines are represented in pictorial form should be shown below:

One more example for parallel lines that must not meet.

In first diagram, we are having the three lines named as line A, line B and line C. The distance between the line A and line B should not be changed often. They are always in the same distance only but never meet. This is same for the distance between line B and line C as well as line A and line C.

The parallel lines could never meet at any point. Therefore, there is no meeting point for the lines that are in parallel.

Slope of parallel lines:

The parallel line is having same distance apart. Therefore, the slope of the parallel lines could be equal.

Let us consider, the two lines line X and line Y are in parallel. Then, their slope will be equivalent.

That is, Slope of line X = slope of line Y.

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Example Problem:

Check whether the two equations are in parallel. x-2y+5=0 and x-2y+9=0

Solution:

Equation 1: x-2y+5=0

Subtract 5 on both sides, we get, x-2y = -5

Subtract x on both sides, -2y = -x-5 => -2y = -(x+5)

=>  2y = x+5

Divide 2 on both sides, y = `(x)/(2)` + `(5)/(2)`

y = `(1)/(2)` x + `(5)/(2)`

Therefore, slope = `(1)/(2)`

Equation 2: x-2y+9=0

Subtract 9 on both sides, x-2y = -9

Subtract x on both sides, -2y = -x-9 = - (x+9)

2y = x + 9

Divide 2 on both the sides, y = `(x)/(2)` + `(9)/(2)`

y = `(1)/(2)` x + `(9)/(2)`

Thus, the slope = `(1)/(2)`

The slope of both the equations is in equal.

So, the two lines should be in parallel.

Arc Length of Circle Formula

Introduction to arc length of circle formula:
Arc:

An arc is a portion of the circumference of a circle.The straight line joining the end points of the arc is termed as chord of the circle.If the arc length is exactly half the circle then it is called a semicircular arc.

Arc Length Of the Circle:

The arc length is the measure of the curved line making up the arc on the circumference of the circle.The length of the arc is longer than the straight line distance between its endpoints which is a chord.

Formulas Useful for Solving Arc Length of Circle

The parameters used while solving problems are listed below:

Where `C` = circumference of the circle.

` K` = Area of the circle.

` r ` = radius of the circle.

` d ` = diameter of the circle.

Some Solved Problems on Arc Length of Circle

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Gaussian Elimination Tutor

Introduction 

A system of equations in which all the given quantities are present only in the first degree is called a linear system of equations. The system of equations can be solved by various methods , one of the methods called gaussian elimination method.Here tutor will help you on gaussian elimination

Let us consider the following set of m linear equations with n unknowns
b11 x1 + b12 x2 + ....................+ b1n xn = c1
b21 x1 + b22 x2 + ....................+ b2n xn = c2
............................................................................
............................................................................
bm1 x1 +bm2 x2 + ....................+ bmn xn = cm
Here all b's and c's are constants.
If c1, c2 upto cm.... are all zero, then the system is called homogeneous. The set of values x1, x2, upto ..xn which satisfy all the equations simultaneously is called a solution to the system of equations. Let us discuss how to find this solution online by gaussian elimination method.

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Gaussian Elimination Tutor : Steps

Step 1: Place the leftmost column in the matrix not consisting totally of zeros.

Step 2: If there is zero entry present , then substitute the top row by any another row to get non-zero entry to the top of this column.
Step 3: If this non-zero way used in a multiply the first row through 1/a to bring in a top 1.
Step 4: Add the suitable possible multiples of the top rows to each other row in arrange to contain only zero entries in the column below this important step1.
Step 5: Now remove the top row of the matrix and create once more with step 1 applied to the secondary matrix that remains. Carry on pending the whole matrix is in row-echelon structure.
Step 6: Then create with the last non-zero row and effective upwards, add suitable multiples of every row to the rows over in order to carry in zeroes over the leading 1s.Consider a system of three independent equations with 3 unknowns.

b11 x1 + b12 x2 + b13 x3 = c1
b21 x1 + b22 x2 + b23 x3 = c2
b31 x1 + b32 x2 + b33 x3 = c3
The system of equations is equivalent to the matrix equation BX  =  C
where B = `[[b11,b12,b13],[b21,b22,b23],[b31,b32,b33]]`    ,        X = `[[x1],[x2],[x3]]`,        C = `[[c1],[c2],[c3]]`
Now we have to reduce the coefficient of matrix B to a possible upper triangular matrix
[ A : B ] = `[[b11,b12,b13,:,c1],[b21,b22,b23,:,c2],[b31,b32,b33,:,c3]]`

Gaussian Elimination Tutor : Examples

Answer the given set of  linear system using Gaussian elimination Method:

x - 3y + z = 4,
2x + 8y - 8z = -2,
-6x + 3y + 15z = 9.
Solution:
The increased matrix for the method is
1        -3      1        |        4
2           8     -8      |        -2
-6        3       15      |        9
Add -2*row 1 to row 2 to get a zero directly below the leading 1. Also add 6*row 1 to row 3 to get a second zero.
1        -3        1     |          4
0        -2       6      |        -10
0        -15   -9      |        33
Consequently multiply row 2 by 1/2 to get a leading 1 and multiply row 3 by 1/3.
1        -3           1  |              4
0        -1          3   |             -5
0        -5         -3  |             11
Add -5*row 2 to row 3 to get a zero under this leading 1.
1        -3           1      |        4
0          -1         3      |       -5
0           0           -18 |      36
We at the present twist around the change procedure and turn the increased matrix into a system of equations we have
x - 3y + z = 4,
0x - y + 3z = -5,
0x + 0y - 18z = 36.
- 18 * z = 36.
z = - 2. (Answer)
- y + 3(-2) = -5
- y - 6 = -5
y = - 1
x - 3y + z = 4,
x - 3(-1) - 2 = 4,
x + 3 – 2 = 4
x + 1 = 4
x = 3
Then the solution is (x, y, z) = (3, -1, -2)