Substitution Geometry

Geometry is a module of mathematics, which involves the study of shapes, line equation, angles problem, dimensions, relative position of figures etc.  The term ‘Geometry’ means study of properties. A point is used to represent a position in space. A plane to be a surface extending infinitely in all directions such that all points lying on the line joining any two points on the surface. Substitution geometry problems are given below.

Example problems for substitution geometry :

1. Find out the geometry equation of straight line passing through the points 2x + y = 8 and 3x - 2y + 7 = 0 and parallel to 4x+ y - 11 = 0
Solution:
Let (x1, y1) be the intersection lines
2x1 +  y1 =  8       …  (1)
3x1 - 2y1 = - 7   …   (2)


(1) × 2 ? 4x1 + 2y1 = 16      …    (3)
(2) + (3) ? x1 =9/7 `=>` y1 =38/7   (x1, y1) =( 9/7 ,38/7)


The straight line parallel to 4x + y - 11 = 0 is of the form 4x + y + k = 0
But it passes through (9/7 ,38/7)


36/7 +38/7 + k = 0 ? k = -74/7
4x + y -74/7 = 0
28x + 7y - 74 = 0 this is the equation of straight line.


2. For what values of ‘a’, the three straight lines 3x + y + 2 = 0, 2x - y + 3 = 0and x + a y - 3 = 0 are concurrent?

Solution:

Let (x1, y1) be the point of concurrency. This point satisfies the first two equations.
3x1 + y1 + 2 = 0 … (1)
2x1 - y1 + 3 = 0 … (2)

Solving (1) and (2) By using substitution method, we get (- 1, 1) as the point of intersection. Since it is a point of concurrency, it lies on x + a y - 3 =0
- 1 + a - 3 = 0
a -4 = 0

a = 4

Practice problems for substitution geometry:

1. Find the point of intersection of the straight lines 5x + 4y - 13 = 0 and 3x + y - 5 = 0

Ans: The point of intersection is (1, 2)

2. Find the geometry equation of straight- line perpendicular to the straight line 3x + 4y + 28 = 0 and passing through the  point (- 1, 4).

Ans: 4x - 3y + 16 = 0

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

Ans: x = 2

Proportionality Problems

In mathematics, proportionality indicates that two variables are related in a linear manner. If one number doubles in size, so does the other; if one of the variables diminishes to 1/10 of its former value, so does the other.The symbol for proportionality resemble a stretched-out, lowercase Greek letter alpha . When this symbol appear that two quantities or variables, it is read "is proportional to" or "varies in direct proportion with." Thus, the expression x alpha y is read " x is proportional to y " or " x varies in direct proportion with y ." In this condition, as long as x and y do not attain values of zero, the quotient x / y is always equal to the same value k , which is called the proportionality constant. (source: wiki)

Example proportionality problems:

Proportionality problem 1:

Find a if a/4= 3/2.

Proportionality  solution:

By using property 1:

a/4=3/2

(a) (2) = (4) (3)

2a = 12

Dividing both side by 2

a=6

Therefore the value of a=6

Proportionality problem 2:

Is 6: 4 = 3: 2 a proportion?

No. If this were a proportion, Property 1 would produce

(6) (2) = (4) (3)

12   = 12, this is true. It is a proportion.

Practice proportionality problems:

Practice proportionality problem 1:

Janette's car uses 9 gallons to go 200 miles.

a) How many gallons will she use to go 400 miles?

B) How many gallons will she use to go 600 miles?

c) How far can she drive with 36 gallons?

(a)18

(b) 27

(c)800 miles

Practice proportionality problem 2:

Carlos makes $25 in 4 hours.

a) How much will he make in 2 hours?

b) How much will he make in 20 hours?

c) How much will he make in 22 hours?

(a)$12.50

(b)$125

(c)$137.50


practice proportionality problem 3:

Which is a better value?

a) 15 ounces for $ 9.25 or 30 ounces for $18.00?

b) 15 ounces for $ 9.25 or 5 ounces for $3.29?

Answer:  ( a )30 ounces for $18.

( b )15 ounces for $9.25.

practice proportionality problem 4:

6 bottles cost $7.00.

a) How much will 18 bottles cost?

b) How much will 15 bottles cost?

c) How much will 27 bottles cost?

d) How many bottles can you buy with $10.50?

Answer:   ( a ) $21

( b ) $17.50

( c ) $31.50

( d ) 9

Geometry Area and Volume

Geometry” Earth-measuring" is an part of the mathematics concerned with questions of size, shape, relative position of figures, and the properties of space. Geometry is one of the oldest sciences. Initially a body of the  practical knowledge concerning lengths, areas, and volumes. And now we can see about the problems in geometry area and volume.

Geometry area and volume problem 1:

Pro 1 :Find the volume of cylinder with the radius 8 cm and the height 12 cm.

Solution:We can find the volume of  an cylinder by using the following formula:

Volume of cylinder V=πr2h

Substitute the values of r and h into the above formula. Than, we get

V=π*82*12

=3.14*64*12

=2411.52 cm3

Pro 2 :Find the volume of sphere with the radius is 12 cm.

Solution:We can find the volume of  the sphere by using the following formula

Volume of sphere V= (4/3) πr3

Substitute the value of radius into the above formula. Then we get,

V= (4/3) *3.14*123

= 1.333*3.14*144

= 602.72 cm3

Ans: 602.72 cm3

Problem 3:Find the amount of pyramid with the base 8.2 mt and height 10.2 mt.

Solution:We can find the volume of  the pyramid by using the following formula

Volume of pyramid V= (1/3) b h

Substitute the values of  the base and height into the above formula. Then we get,

V= (1/3)*8.2*10.2

=0.333*8.2*10.2

=27.8521 mt3

Ans: 27.8521 mt3

Geometry - Find the volume of shapes when area is given

Find the volume of the right prism whose area of the base is 550 cm2 and height is 38cm



Solution:Given that area of the base, A = 550 cm2 and height (h) of the prism = 38 cm

Volume of the right prism = area of the base * height cu.units

= A * h

= 20900

Volume of right prism = 20900 cm3

Find the volume of the right prism whose area of the base is 450 cm2 and height is 34cm

Solution:Given that area of the base, A = 450 cm2 and height (h) of the prism = 34 cm

Volume of the right prism = area of the base * height cu.units

= A * h

= 15300

Tutoring About Correlation Coefficient

Concept of correlation:

Correlation is a method of studying the relationship between the two variables. In statistical analysis we come across the study of two variables wherein the change in the value of one variable produces a change in the value of other variable. In that case we say that the variables are correlated or there is a correlation between the two variables.

The formula for the correlation coefficient r can be expressed in the form,

r = `(sum (X - barX) ( Y - barY))/(sqrt(sum (X - barX)^2) sqrt(sum(Y - barY)^2))`

It is conventionally taken as x = X - X and y = Y - Y and hence we write

r = `(sum xy)/(sqrt(sum x^2) sqrt(sum y^2))`

The above formula is expressed in terms of deviations of the variables from their means. Instead, if the actual values of the observations are taken then the formula can be written as,

r= `(N sum XY - sum X sum Y)/(sqrt(N sum X^2 - (sumX)^2) sqrt(NsumY^2 - (sum Y)^2))`

Instead of the deviations from their means, the deviations are measured from the value A and B for X and Y variables by taking dx = X - A, dy = Y - B, the correlation coefficient r is given by,

r = `(N sumdxdy - sumdx.sumdy)/(sqrt(Nsumdx^2- (sumdx)^2) sqrt(Nsumdy^2 - (sumdy)^2))`

Tutoring about formulas for calculating correlation coefficient:

r = `(sum xy)/(sqrt(sum x^2) sqrt(sum y^2))`

This formula is used when deviations are measured from their mean.
r= `"(N sum XY - sum X sum Y)/(sqrt(N sum X^2 - (sumX)^2)(sqrt(N sum Y^2 - (sumY)^2))) `

This formula is used if no assumed average is taken for x and y series
r = `"(N sumdxdy - sumdx.sumdy)/(sqrt(Nsumdx^2- (sumdx)^2)(sqrtNsumdy^2 - (sumdy)^2)) `

This formula is applied when deviations for x and y series are taken from some assumed values.


Tutoring on problems on correlation coefficient::

Calculate the correlation coefficient between x and y from the following data:
x 1 3 5 8 9 10
y 3 4 8 10 12 11


Solution:
x y x - `barx` y - `bary` (x - `barx` )2 (y - `bary` )2 (x-`barx` )(y - `bary` )
1 3 -5 -5 25 25 25
3 4 -3 -4 9 16 12
5 8 -1 0 1 0 0
8 10 2 2 4 4 4
9 12 3 4 9 16 12
10 11 4 3 16 9 12
36 48 0 0 64 70 65


`barx` = `(sum x)/(n)` = 36/6 = 6

`bary` = `(sum y)/(n)` = 48/6 = 8

r = `(sum(x- barx)(y - bary))/(sqrt(sum(x-barx)^2)sqrt(sum(y-bary)^2))`

= `(sum xy)/(sqrt(sumx^2)sqrt(sumy^2))`

= `(65)/(sqrt(64)sqrt(70))` = 0.97

Limit of a Function

Students can study about Limit of a Function here. Consider the function f(x). Let the independent variable x take values near a given constant a. Then f(x) takes a corresponding set of values. Suppose that when x is close to a, the values of f(x) are close to some constant. Suppose f(x) can be made to differ arbitrarily small from A by taking values of x that are sufficiently close to a but not equal to a and that is true for all such values of x. Then f(x) is said to approach limit A as x approaches a.

If the function f(x) approaches a constant A when x approaches a in whatever manner without assuming the value a, A is said to be the limit of f(x) as x approaches a. Thus we write lim_(x->a) f(x) = A

Find the Limit of a Function

Students can learn to Find the Limit of a Function if they know what Functions are and how they behave at the given limits.

A function may approach two different limits. One where the variable approaches its limit through values larger than the limit and the other where the variable approaches its limit through values smaller than the limit. In such a case the limit is not defined but the right and left hand limit exists.

The right hand limit of a function is the value of the function approaches when the variable approaches its limit from the right. Here, we write lim_(x->a^+) f(x) = A+

The left hand limit of a function is the value of the function approaches when the variable approaches its limit from the left.

here, we write lim_(x->a^-) f(x) = A-

The limit of a function exists if and only if the left hand limit = right hand limit.

In that case, lim_(x->a^+) f(x) = lim_(x->a^-) f(x) = f(x)

Properties of limit of a function

The following are some of the properties of limits which are useful in evaluating the limit of a function.

1. lim_(x->a) k = k ( k is a constant)

2. lim_(x->a) [f(x) ± g(x)] =  lim_(x->a) f(x) ± lim_(x->a) g(x)

3. lim_(x->oo) [f(x).g(x)] = lim_(x->oo) f(x) . lim_(x->oo) g(x)

4. lim_(x->a) (f(x))/(g(x)) = (lim_(x->a)f(x))/(lim_(x->a)g(x))

5. lim_(x->a) [f(x)]n =  [ lim_(x->a) f(x)]n

Standard limit theorems:

1. lim_(x->a) (x^n- a^n)/(x - a) = nan-1

2. lim_(x->0) (e^x-1)/(x) = 1

3. lim_(x->0) (sin x)/(x) = 1

4. lim_(x->0) (1 + (1)/(n))^(n) = e



Solved Examples

Example 1: Evaluate the right hand limit of the function

f(x) = {│x – 4│/x – 4, x ≠ 4, 0 x = 4

at x = 4

Sol: (RHL of f(x) at x = 4)

= lim f(x) = lim f(4 + h) = lim │4+ h – 4│4 + h – 4

x→4+      h→0             h→0

= lim │h│/h = lim h/h = lim 1 = 1

h→0            h→0      h→0

Example 2: Let f be the function given by f(x) = x2 – a2/x – a, x ≠ a

Using (in , δ) definition show that lim f(x) = 2a

x → 0

Sol: Let in > 0 be given. In order to show that

lim f(xi) = 2a

x → a

We have to show that that for any given in > 0, there exists a number δ >0 such that

│f(x) – 2a│< in whenever 0 < │x – a│< δ

If x ≠ a, then │f(x) – 2a│= │x3 – a2/x – a│

= │(x + a) – 2a│                                         [... x ≠ a]

= │x – a│

... │f(x) – 2a│< in , if │x – a│< in

Choosing a number δ such that 0 < δ < in , we have

│f(x) – 2a│< in when whenever 0 < │x – a│< δ

Hence    lim f(x) = 2a

x → 0

Standard Deviation in Statistics

Statistics is a division of applied mathematics which contracts with the particular interpolation of data. The term ‘Statistics’ has been taken from the Latin name ‘Status’ in which it defines ‘political state’. This statistics mainly used to measure the arithmetic mean, median, mode and standard deviation. This measurement gives idea about where the data points are centered. Let us discuss about standard deviation with some example problems.

Evaluation of Standard Deviation in statistics:

Mean:

In Statistics, Mean is defined as the average of the given total numbers, i.e., total number of data divided by the number of data set given.

Formula for finding mean,

barx   = (sum(x))/(n)

Standard Deviation:

In Statistics, Standard Deviation is the measure of describing squared mean difference variability and spread of the Data set in the given total numbers. It is used to take the measurement of taking square root and average of numbers in the Data set.

Formula for standard deviation,

S =  sqrt ((sum(x - barx)^2 )/ (n-1))

Here Standard Deviation is calculated by using the mean Value  barx

Example Problems to find standard deviation in statistics :

Problem 1:

Here are 4 measurements   66, 45, 67, 45, 34, 56, 78 and 57. Calculate statistics standard deviation for the given measurements

Sol:

Mean: Calculate the mean the using the formula,

barx   = (sum(x)) / n

barx   =( 66+45+67+45+34+56+78+57) / 8

= 448 / 8

barx   = 56

Standard Deviation,

S = sqrt((sum(x-barx)^2) / (n-1) )

S = sqrt((( 66-56)^2+(45-56)^2+(67-56)^2+(45-56)^2+(34-56)^2+(56-56)^2+(78-56)^2+(57-56)^2) / (8-1))

= sqrt((100+121+121+121+484+0+484+1)/7)

=  sqrt( 1432 / 7)

S = sqrt(204.571429)

Standard Deviation S =  14.3028469

Problem 2:

Find the Statistics Standard deviation of the given Data 16, 17, 18, 20 and 24

Sol:

Mean: Calculate the mean using the formula,

barx   = (sum(x)) / n

barx  = ( 16 + 17 + 18 + 20 + 24 ) / 5

barx = 95 / 5

barx = 19

X               x-barx              (x-barx )^2

6               16 - 19 = -3           9

7               17 - 19 = -2           4

8               18 - 19 = -1           1

9               20 - 19 =  1           1

10              21 - 19 =  2           4

Sum of the  (x-barx)^2

9+4+1+1+4 = 19

Standard Deviation: Calculate the standard deviation

S =   sqrt((sum(x- barx)^2 ) / (n-1))

S = sqrt( ( 9+4+1+1+4) / 4 )

S = sqrt (19 / 4 )

S = sqrt(4.75 )

S = 2.17944947

Practice Problems for Statistics Standard Deviation:

1. Find the statistics standard deviation for the following given data.37, 56, 54, 54, 26, 67, 12, 65 and 34.

Answer:                     Mean = 45

Standard Deviation = 18.714967272213

2. Calculate the statistics standard deviation for the following. 77, 56, 33, 87, 90, 23, 67, 80 and 99.

Answer:                    Mean = 68

Standard Deviation = 26.043233286211

Study Online Second Derivatives

Study online second derivatives involves the process differentiating the given polynomial function twice with respect to the given variables whereas all the process is clearly explained with the help of online. Generally the second derivative is discussed in calculus whereas it is mainly helps 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 for second derivatives study discussed in online.

Example 1:

Determine the second derivative from the polynomial.

f(b) = 5b 2 +5b 4  + 12

Solution:

The given function is

f(b) = 5b 2 +5b 4  + 12

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

f '(b) = 5(2b  )+5(4 b 3 ) + 0

By solving above terms

f '(b) = 10b +20b3

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

f ''(b) =  10(1 ) +20(3b2)

f ''(b) =  10 + 60b2 is the answer.

Example 2:

Determine the second derivative from the polynomial.

f(b) = 4b4 +5b 5 +6b 6  + 2b

Solution:

The given function is

f(b) = 4b4 +5b 5 +6b 6  + 2b

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

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

By solving above terms

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

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

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

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

Example 3:

Determine the second derivative from the polynomial.

f(b) = 2b6 + 2 b5 + 3 b4 + 3b

Solution:

The given equation is

f(b) = 2b6 + 2 b5 + 3 b4 + 3b

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

f '(b) =  2(6b 5)  +2 (5 b4 ) +3(4 b3) + 3

By solving above terms

f '(b) =  12b 5  +  10b4  + 12 b3 – 3

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

f ''(b) =  12(5b 4 ) – 10(4b3)  + 12(3b2)

f ''(b) =  60b 4 – 40b3 +36b2  is the answer.

Online second derivatives practice problems for study:

1) Determine the second derivative from the polynomial.

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

Answer: f ''(b) = 6b +12b2+ 20b 3

2) Determine the second derivative from the polynomial.

f(b) = 2b 3+3b5 + 4b 6

Answer: f ''(b) = 12b + 60b3 + 120 b 4

Solving Negative Number

Definition:

Negative number is defined as the number which indicate by negative sign or minus ('-') . Negative number is less then zero and placed left to zero.

Ex: ...-5,-4,-3,-2,-1,0,1,2,3,4...

Comparison of Positive and Negative:

For each negative number , there is a positive number that is its opposite . Here we can write the opposite of negative number with a positive of same number or plus sign used In front of the number and call these numbers are positive numbers. Ex : 1,2,3 ,.....positive numbers are grater than zero. Similarly, the opposite of any positive number is a negative number .

Ex:    1,2,3 is -1,-2,-3.

Solving examples for negative numbers:

Zero cannot be taken as a negative number or positive number.
For every positive number x, there exists a negative number y such that x + y = 0
Positive number is denoted as plus ('+')sign and negative number is denoted as minus sign('-').
Example of negative number:-2,-43,-34 and example for positive number is 2,43,34.
negative and positive number may be written as mixed numbers or fraction numbers.

The equal fraction of negative numbers are given bellow:

(-3)/7,3/(-7),-(3/7) and -3/7 .

The equal mixed numbers are given bellow.

-2/5,-(2/5)       (-4)/9, 4/-(9) and -4/9

More about solving negative numbers:

Solving Addition of Negative Numbers:

To add the negative numbers which consist of minus sign. To provide the answer of addition of negative number.

Solving examples for addition negative numbers:

-12+(-6)=?

Solution:

-12+(-6)= -18.

Here the values of  -12 and -6are 12 and 6 adding the smaller from the larger gives -12+(-6)= -18, and since the larger  value was 12, so we can give result the same sign as -12, '-' so -12+(-6)= -18.

Example:

(-6) + (-6) = ?

Here the absolute values of -6 and -6 are 6 and 6.  Adding the smaller from the larger gives -6 - 6 = 12 ,  but here both has same value . In this case sign is the matter, here 12 and -12 are the not same and then -6 and -6 are same numbers. The property of all same number sum is 12. The addition of two number up to zero are called as additive inverses.

Multiplying Negative Numbers:

Solving example for multiplying negative numbers :

Product of negative number ,here we can take the product of their values.

(-3.3) × (-5) = ?


(-3.3) × (-5) = (-3.3) × (-5)

= (-3.3) × (-5)

=  +16.5.
Dividing Negative Numbers

Solving example for dividing negative number:

To divide two negative numbers, here we can divide the value of the first by the value of the second.

(-1.6) ÷ (-4) = (-1.6) ÷ (-4)

= (-1.6) ÷ (-4)

=  -0.4

Geometry Without Common Vertices

In geometry, some figures have common vertices. Mostly the geometric figures are without common vertices. If a triangle, quadrilateral and some geometric figures in a graph are lying without common vertices are refered same. In graph there are four quadrants in that some vertices are fall on common vertices, with out common vertices points of are plotted.

Geometric figure without common vertices

Without common vertices find the distance of two points

In geometry the triangle has three vertices; the vertices are not common vertices. The common vertices are formed only if two triangles are in same point without three common vertices. The distance between two un common vertices are find out by using the coordinates of the vertices (x1,x2) and (x2,x2) of the vertices.

Distance between two vertices = √(x2-x1)2 +(y2-y1)2

If the geometry figure having the common vertices in a graph



Examples for without common vertices

Distance of a vertices are find using distance formula:

Examples for distance between two vertices:

Ex 1:   Find the distance formed by without common vertices, Vertices A(4,5), B(7,4)

Sol :            Distance AB = √(x2-x1)2 +(y2-y1)2

X1=4    X2=7    Y1=5    Y2=4

AB = √(7-4)2+(4-5)2

= √(3)2+ (-1)2

= √9+1

= √10 units

Ex 2 :  Find the distance formed by without common vertices, Vertices A(3,2), B(5,4)

Sol :            Distance AB = √(x2-x1)2 +(y2-y1)2

X1=3    X2=5    Y1=2    Y2=4

AB = √(5-3)2+(4-2)2

= √(2)2+ (2)2

= √4+4

= √8 units

Ex 3 :      Find the distance formed by without common vertices, Vertices A(6,4), B(10,8)

Sol :            Distance AB = √(x2-x1)2 +(y2-y1)2

X1=6    X2=10    Y1=4    Y2=8

AB = √(10-6)2+(8-4)2

= √(4)2+ (4)2

= √16+16

= √32 units

Ex 4:    Find the distance formed by without common vertices, Vertices A(3,3), B(4,8)

Sol :            Distance AB = √(x2-x1)2 +(y2-y1)2

X1=3    X2=4    Y1=3    Y2=8

AB = √(4-3)2+(8-3)2

= √(1)2+ (5)2

= √1+25

= √26 units

Practice problems:

Q 1   Find the distance of two vertices A(1,1) B(1,2)   Answer: √1 units

Q 2   Find the distance of two vertices A(2,2) B(1,1)   Answer: √2 units

Straight Line Equation

The equation of straight line is generally written as    y = mx + b

Graphical Representation

where, m= slope or gradient of the straight line equation

b = the y-intercept

Suppose we want to find equation of a straight line that passes through a known point and has a known slope. Let (x ,y) represent the co-ordinates of any point on the line and let (x1 ,y1) represent the co-ordinates of other point. The slope of the straight line equation is given as,

m = (y-y_1)/(x-x_1)

After finding the slope m as we are given the co-ordinates of the point (x1 ,y1) in the equation    y= mx + b Then, the constant 'b' can also be found so that finally the straight line equation is obtained.

Other Forms of straight Line Equations

Different Forms of Straight Line Equations:

There are many other forms of Straight Line Equations.

1. Straight Line Equation through two points:

The line through two different points ( x1 ,y1) and ( x2 ,y2) is given by

y-y1  = [(y_2-y_1)/(x_2-x_1)]  . (x - x1)

2. Straight Line Equation in general form:

A straight line is defined by a linear equation as

Ax+By+C=0     where A , B are not both 0

3. Straight Line Equation in Intercept-intercept form:

Let us consider that a straight line intersects x-axis at (a , 0) and y-axis at (0 , b). Then it is defined by equation

(x/a) +(y/b) = 1

4. Straight Line Equation in Point-Slope Form:

The equation of straight line through the point (a , b) with slope m is

y = m ( x - a ) + b

Solved Examples on equations of straight line

Ex:1 Find the equation of a line passing through (2 , 3) and having a slope of 3?

sol: step 1: Compare the given point(2 , 3) with the general point (x1 , y1)

Now x1=2  and  y1=3

step 2: we have the straight line equation as  y-y1=m(x-x1)

now substitute the given point in the above equation

y-3=3(x-2)

y-3=3x-6

3x-y=6-3

3x-y-3=0

The straight line equation is 3x-y-3=0

Ex:2 Find out the straight line equation of the line passing through the points (1,2) and (2,4).

Sol: Given points are compared with (x1 , y1) and (x2 , y2) and substitute the points in the equation

y-y1  = [(y_2-y_1)/(x_2-x_1) ] . (x-x1)

y-2  =  [(4-2)/(2-1)] . (x-1)

y-2 = 2(x-1)

y-2 = 2x-2

2x - y = 0

The straight line equation is 2x-y=0.

Practice Problems on equations of straight line

Pro:1 Find the equation of the line passing through the points (-3,4) and (4,-2)

Ans: we have      

m = (y_2-y_1)/(x_2-x_1)

=(-2-4)/(4+3)

=(-6/7)

let (x,y) be compared to (-3,4)

y-4 = (-6/7) . (x-(-3))

7(y-4) = -6(x+3)

7y-28 = -6x-18

7y-28 = -6x-18

7y+6x = 10

Pro:2 Write equation of line having points and slopes as follows;

P(3,3) , m=-2

P(-2,-1) , m=1/3

P1(2,2) and P2(-4,-1)

y-intercept = 2 , m=3

Ans: The answers to the above given practice problems are

y+2x=11

3y-x=-1

2y-x=2

y-3x=2

Hyperbola Axis

The locus of a point whose distance from a point rest at center shows an unaltered ratio, higher than one to its length from the fixed line is known as hyperbola. In this article, you can learn about the axis of hyperbola, general equation of the hyperbola and definitions regarding hyperbola.

Axis of Hyperbola:

Transverse Axis:

Transverse axis can be defined as the line segment, which joins the vertices of the hyperbola. 2a is calculated as the difference between two vertices. The equation of the transverse axis is y = 0. This is showing that y-coordinate is zero.

Conjugate Axis:

Conjugate axis can be defined as the line segment, which joins the y-coordinates of the hyperbola. The distance between the two coordinates are 2a (i.e., the length of the Conjugate axis is 2a.). x = 0 is considered as the equation of conjugate axis. Therefore the x-coordinates of Conjugate axis is zero.

General Equation of the Hyperbola:

Some important points to be considered:

Fixed point is represented as F.
Fixed line as l.
Eccentricity as e, then it should be greater 1.
Moving point is represented as P(x,y).

Steps:

Fixed point F is plotted and also the fixed line ‘l’ is drawn.
Perpendicular (FZ) is dropped from F to l.
As next step drop PM which is perpendicular one from P to l.
Plot the points A, A’ which divides FZ internally and externally in the ratio e : 1            
respectively.
Take AA’ = 2a and treat it as x-axis.
Draw a perpendicular bisector of AA’ and treat it as y-axis.
Consider C as the origin, then the known points are C(0,0), A(a,0) and A’(-a,0).
The general equation of the hyperbola x^2/a^2 -y^2/b^2 = 1.

Definitions Regarding Hyperbola:

Focus:

The fixed point F is known as the focus of the hyperbola.

Directrix:

The directrix is nothing but the fixed line. Then the directrix equation is given by x = a/e.

Centre:

The centre of the hyperbola is a point, at which the transverse and conjugate axes intersect and they are represented as ‘C’.

Vertices:

The vertices of the hyperbola are the points, where the curve and its transverse axis meet. The vertices are A(a,0) and A’(-a,0).

Studying Standard Deviation Examples

The standard deviation of a statistical population, a data set, or a probability distribution is the square root of its variance. Standard deviation is a widely used measure of the variability or dispersion, being algebraically more tractable though practically less robust than the expected deviation or average absolute deviation. It shows how much variation there is from the "average" (mean) (or expected/budgeted value). It helps detect tampering of data. Examples for studying standard deviation is given below.

Formula for studying standard deviation examples:

Formula for studying standard deviation examples are defined below:

Measurement of standard deviation is prepared by taking square root for the addition of mean difference with the certain data divided by the total number of values subtracted by one. The following formula for standard deviation as shows given below.

s=v?(X-M)2/n-1

Here S = Sum of values

X = Individual value

M = Mean of total all value

N = Sample size (Total number of values)

Variance:

Variance = s2

Steps for calculating Standard Deviation examples:

• Step 1: calculate the average for given n numbers using the formula this is called mean of given numbers

• Step 2: Find distance between each given numbers in the Data set from the calculated average value. This is called  "deviation" from the mean value.

• Step 3: Take the Square of each deviation value found from mean. This is squared deviation from mean.

• Step 4: Calculate the sum for all the squared standard deviations.

• Step 5: Now apply the Standard Deviation formula and find standard deviation formula. It will be the square root of variance.

Examples for Studying Standard deviation:

Examples for Studying Standard deviation are as follows:

Pro 1:   Here are 4 measurements 4, 6, 7, 9 and 10Calculate the Standard Deviation

Sol :     Mean: Calculate the average for the given values. To find the mean.
4+6+7+9+ 10
x =   ------------------------
5-1

= 36/4

= 9

Standard Deviation,
v (4-9)2 + (6-9)2 + (7-9)2 + (9-9)2 + (10-9)2
S=  -------------------------------------------------------------
5-1

= v 81 /4

= v 20.25

=  4.5

Standard Deviation  S = 2.12132

Pro 2:  Here are 4 measurements 10, 20, 30, 40 and 50 Calculate the Standard Deviation

Sol :     Mean: Calculate the average for the given values. To find the mean.
10+20+30+40+60
x = ---------------------------
5-1

= 160/4

= 40

Standard Deviation,
v (10-40)2 + (20-40)2 + (30-40)2 + (40-40)2 + (60-40)2
S=  -----------------------------------------------------------------------------
5-1

= v 1600 /4

= v 400

S = 20

Standard Deviation S = 20

Absolutely Continuous Function

DEFINITION

Let [p,q] be the close bounded interval of C. Then a function f:[p,q]→ R will be an absolutely continuous function on [p,q], if for any δ>0 there will be a ε>0 such that the certain conditions which are mentioned below holds good

If (p1q1)..............(pnqn) is a collection which is finite with disjoint open intervals in [p,q] such that

Σni=1 (qi-pi) < ε

and

Σni=1 |f(qi)-f(pi)| < δ



EQUIVALENT DEFINITIONS

The condition on a real-valued function  "  f  "  on the compact interval [ p, q ] are equivalent if

1) f is an absolutely continuous function

2) ' f ' has a derivative f1 almost everywhere which is a Lebesgue integral and

f ( x ) = f ( p ) +  ∫x a  f1  ( c ) dt

for all x on [ p , q ]

3)There exists a Lebesgue integrable function such that g on [ p , q ] such that f ( x ) = f ( p ) + ∫x p  g ( c ) dt

for all x on [ p , q ]

If these conditions are satisfied by the function the definitely g = f' almost everywhere

Properties of the absolute continuous function

PROPERTIES OF ABSOLUTELY CONTINUOUS FUNCTION

1. The sum and difference of two absolute continuous function are also absolutely continuous. The products of two absolute continuous function defined on the bounded interval will also be a absolute continuous function.

2. If an absolutely absolute continuous function is defined on a bounded closed interval and is nowhere zero then its reciprocal is also an absolutely continuous function.

3. Every absolutely continuous function is an uniformly continuous function.

4. If  f: [ p , q ] → R is absolutely continuous, then it will be a function of the bounded variation on [ p,q ]