Area of Part of Sphere

A 3-dimensional item shaped similar to a ball.

Each point on the surface is the similar distance from the middle.

Sphere facts:

It is completely symmetrical

It have no edges or vertices

It is not a polyhedron

Every point on the surface is the similar distance from the middle.


Area and Volume of Sphere:

Area of Sphere = 4 × π × r2

Volume of Sphere = (`4/3` ) × π × r3

Example for Area of part of sphere:

Locate the area of part of sphere x2 + y2 + z2 = 81 that lies over the cone z = sqrt(x2 + y2)

The cone and sphere cross when equally equations are satisfied. Substitute z from the cone into the sphere equation to obtain:

x2 + y2 + (sqrt(x2 + y2))2 = 81

2x2 + 2y2 = 81

x2 + y2 = 81/2

This is a sphere by means of radius sqrt (81/2). State you have a spherical method for the area of a sphere. So integrate through spherical coordinates. If θ is the angle concerning z counterclockwise as of +x (0 ≤ θ ≤ 2π), and φ is the angle as of z (0 ≤ φ ≤ π), and r is distance as of the origin. The spherical to Cartesian change are:

x = r sin φ cos θ

y = r sin φ sin θ

z = r cos φ

When y = 0 and x is positive, then x = sqrt (81/2) and θ = 0. Also r = 9 all over the place on the circle of junction, since that is the radius of the sphere. This means that:

sqrt(81/2) = 9×sin φ

sin φ = sqrt(81/2)/9 = sqrt(1/2) = sqrt(2)/2

φ = sin-1(sqrt(2)/2) = π/4

This is the utmost value of φ. The differential of area is:

dS = (r×dφ)×(r×sin φ×dθ) = r2×sinφ×dφ×dθ

integrated for φ = 0 to π/4 and θ = 0 to 2π

Algebra is widely used in day to day activities watch out for my forthcoming posts on prime factorization algorithm and cbse books free download. I am sure they will be helpful.

Another Example for Area of part of sphere:

Locate the area of part of sphere x2 + y2 + z2 = 1:

Locate the area of part of sphere  x2 + y2 + z2 = 1 that lies above the cone z = sqrt(x2 + y2)

The cone will cut the sphere in a plane at right angles to the z axis at

z2 = 1-x2 - y2 and z2 = x2 + y2

z = sqrt (1/2)

Therefore you have a spherical limit with a height of 1-sqrt(1/2 ) as of a sphere of radius 1

along with the surface area of sphere is 2 pi r h = 2 * pi * 1 * [1-sqrt(1/2)]

Algebra 2 Parabola Help

The algebra 2 parabola help  represents the procedures or clues to proceed the problems in the conic section of parabola with algebra 2. The algebra 2 in this article represents the applications under the parabola(conic sections).In this article we are going to study the forms of the parabola that are appearing in practical and some of tracing operation that involved in the conic sections.A parabola is generally defined as the conic section obtained by the operation on slicing a right circular cone by a plane parallel to the line joining vertex and any other point of the cone.(source : Wikipedia)

Examples to explain algebra 2 parabola help

Direction of parabolas:

For the diagram of parabolas given below, the axes are neither parallel to x-axis nor parallel to y-axis, such cases the parabola's equation should include the xy term that are not in standard form. For the standard types the axis may be either parallel to x-axis or parallel to y-axis.

Direction for parabola

consider the front face of the helicopter have the parabolics hape for which the distance from the vertex to the focus is 900mts. If the distance acrossmeasured as the diameter the top of the face is 160m, we have to find how depth is the helicopter's front face at the middle part.

parabola

Solution:

consider the vertex is at the origin and by the general term, we say it as

VF = a = 900

The general equation of the parabola is given by y2 = 4 × 900 × x

Let x1 be the depth of the helicopter front part at the middle

here we know that the point or the vertex (x1, 80) lies on the parabola

802 = 4 × 900 × x1 rArr x1 =16/9

∴ The depth of the helicopter front part =16/9 m


Algebra is widely used in day to day activities watch out for my forthcoming posts on Define even Number and sample question paper for class 9 cbse. I am sure they will be helpful.

Problems to explain algebra 2 parabola help

Some forms of parabola exist based on the algebra 2 parabola help:

lighting a rocket cracker with degree
Reflecting telescope
Grids on the bridge
Reflecting telescope
water issuing from the end of a horizontal pipe
comet moving in a parabolic orbit
cable of the suspension bridge hangs in the parabola's form.

Using this help, we can easily work on the algebra 2 parabola problems.

Conditional Probability

In conditional probability, we allocate a distribution function to a sample space and then learn that an event E has occurred.
How should we modify the  probabilities of the remaining events? We shall call the new probability for an event F the conditional probability of F given E and denote it by P(F | E).

The conditional probability of event B occurs, given that event A has already occurred is

P(B|A) = P(A and B) / P(A)

Events of Conditional probability:

For example, we previously calculated the probability of rolling a 5 above. Now say we want to work out the probability of rolling a 5 given that one or both of the dice rolled is a 2. We would calculate this conditional probability like so

A = {(1, 4), (2, 3), (3, 2), (4, 1)}

B = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (1, 2), (3, 2), (4, 2), (5, 2), (6, 2)}

A AND B  =  {(2, 3), (3, 2)}

P(A | B)   =  P(A AND B) / P(B)

=  P({(2, 3), (3, 2)}) / P(B)

= (1/18) / (11 / 36)

=  (2/11)

Examples for conditional probability:

Example 1:

A math teacher gave her class two tests. 25% of the class passed both tests and 42% of the class passed the first test. What percent of those who passed the first test also passed he second test?

Solution:

P(Second | First)    

=        P(First and Second) / P(First)

=        0.25/0.42

=        0.60

=       60%

Example  2:

A jar contains black and white marbles. Two marbles are chosen without replacement. The probability of taking a black marble and then a white marble is 0.34, and the probability of taking a black marble on the first draw is 0.47. What is the probability of taking a white marble on the second draw, given that the first marble drawn was black?

Solution:

P(White | Black)          

=        P(Black and White) / P(Black)

=        0.34 / 0.47

=        0.72

=        72%

Solving Ellipse

A plane curve formed by the intersection of a cone and a plane (i.e., a closed curve) is known as an Ellipse. If the cutting plane is normal to the axis, then a special case of ellipse is obtained called circle. An Ellipse is a closed curve and it is a bounded case of a conic section.

I like to share this Major Axis of an Ellipse with you all through my article.

Solving an Ellipse:

Let us see about the solving ellipse,

If P is a point on ellipse, then the sum of the distances from P to the two focus is a constant.

The equation for ellipse with normal major axis.

The derivative for the horizontal major axis case is the same.

Let us consider the top of  an ellipse, first.

The sum of the distances from P to the two focus,

(b - c) + (b + c) = 2b (top of  an ellipse)

squaring the above equation,

= 4b2

now consider the R.H.S. of  ellipse.

Thus, the distance from each focus to the R.H.S is the same.

Squaring the sum, implies

(2d)2 = 4d2

by Pythagoras theorem,

d2 = a2 + c2

Sub. And equate both the equations

4b2 = 4(a2 + c2)

simplifying, implies

b2 = a2 + c2

The equation of an ellipse is solving with center as its origin, can be given by

(x^2)/(a^2) + (y^2)/(b^2) = 1

with a > b > 0.

Here the major axis is 2a and the minor axis is 2b.

The two foci are given by (±c , 0),(0 , ±c) where c2 = a2 - b2.

Hence, solving ellipse proved.

Example Problem - Solving Ellipse:

Find  x and y intercepts of the following ellipse.

9x2 + 4y2 = 36

Solution:

To solving the above equation, write the ellipse  equation in  general form

(9x^2) / 36 + (4y^2) / 36 = 1

by simplification,

x^2 / 4 + y^2 / 9 = 1

=>  x^2 / 2^2 + y^2 / 3^2 = 1

from the above ellipse equation, the x and y-intercepts are obtained. i.e., a = 3 and b = 2.

sub. y=0 =>  x^2 / 2^2 = 1

by Solving x.

x2 = 22

x = ± 2

sub. x=0 => y^2 / 3^2 = 1

by Solving y.

y2 = 32

y = ± 3.

Explicit Functions

If a variable say y, can be expressed explicitly as a function of another variable say x, as y=f(x), then y is known as an explicit function of x, to calculate an output, given input values for the function's rules for the functions arguments. An explicit function can be defined as the dependent variable. It can be written in terms of the independent variable of explicit functions. For eg, The following are explicit functions: x = y2 – 3, f(x) = sqrt (x+7) and x = log2 y.

Implicit Function or Relation of Explicit Equations:

The isolated dependent variables are not mentioned in one side of the expression or equation.

For eg,

The following equation y2 + xy – x2 = 1 represent an implicit relation.

Square Root - Explicit Functions:

A nonnegative number which is multiplied two times itself to give the given number. In general, the square root of y is written `sqrt(y)` or x1/2.

For eg,

Sqrt(9) = 3, since 32 = 9.

Note: sqrt(y) is a non-negative number. If y is a negative value, then the root value of y is imaginary.

Logarithm:

The logarithm base b of a number y is the power to which b must be raised in order to equal y. it can be written as log b y.

For eg,

log2 8 = 3.

Logarithm Rules:

The Algebra properties are followed while doing logarithms.

Assume x, y, a, and b are all positive and a & b are not equal to 1.

Definitions - Functions:

log a x = N, since a*N = x.

log x = log10 x. if a logarithm is written without representing a base it refers general logarithm.

ln x = loge x, since e = 2.718. the log rules are similar to ln.

Note: ln x is equals to Ln x or LN x.

Rules:

Inverse properties:

loga ax = x & a(loga x) = x

Product:

loga (xy) = loga x + loga y.

Quotient:

Log(x/y) = logax – logay.

Power:

loga (xp) = p loga x

Change of base formula:

Log ax = logb x/logb a.

Careful:

Log a (x + y) ≠ log a x + log a y

Log a (x – y) ≠ log a x – log a y

Common Logarithm:

The common logarithm have 10 as its base.

For eg,

log 100 is 2 since 102 = 100.

Natural Logarithm:

The natural logarithm has e as its base. The base e = 2.718

Change of Base Formula - Explicit Functions:

Loga x = logb x/logb a, a & b are not equal to 1.

Examples - Explicit Functions:

Example 1:

Log16 32 = log2 32/log2 16 = 5/4.

Example 2:

Log2 3= log10 3/log10 3

= 0.47712/0.30103

= 1.585

21.585 = 3

Example 3:

Log8 x = ln x/ln 8 = ln x/ln 8.

Percentile

A percentile or centile is the value of variable below which certain percent of observations fall (source: Wikipedia). Here, thirty percentile is value or score below which 30 percent of observations may be found.

The word percentile and related terminology percentile level used in expressive statistics in addition to in reporting of scores from norm-referenced tests. The thirty fifth percentile also known as first quartile Q1; seventieth percentile as median or second quartile Q2; 105th percentile as third quartile Q3.

There is no predefined definition for percentile. When the number of observation is large, all explanation yields the same results.

Nearest rank:

It says that the p-th percentile of N ordered values is acquired by first find the rank n= ((N/100)*p) + ½, rounding to the

adjacent integer and considering the value that consistent to that rank

Linear Interpolation between the closest ranks:

Linear interpolation between the closest ranks is another method. It is used to find the linear interpolation between the two

adjacent ranks rather than rounding.

For N values v1, v2, v3,...,vN , that is

ranked from the smallest amount to greatest then it  define the percentile corresponding to the n-th value as

Pn= (100/N)*(n-1/2)

For example if N=10 then the percentile corresponding to the third value is calculated as

P3= (100/10)*(3-1/2) = 25

Weighted Percentile:

Weighted percentile is defined as the percentage in the total weight is counted rather than the total number. For

example we have positive weights w1, w2, w3,...,wN , associated respectively with our N sample values

Sn = `sum_(k=1)^n` w k where k=1 to n. Let it be the nth partial sum of these weights.

Percentile Formula

Below is the percentile formula to complete your learning over percentile:

The percentile is used in calculation of individual population. We can find the percentile by using following two formulas:

We can find the percentile with 'x' values by using the formula as `(B + 0.5E)/n ` x 100
We can find the percentile without 'x' values by using the formula as `(Number of below x values)/n` x 100


Calculate percentile

Below you can see how to calculate percentile -

Problem 1: The scores for student are 21, 24, 26, 27, 28, 28, 31, 32, 34, 36, 39, 40, 42, 44, 45, 46, 49, 53, 57, 60. Find out the percentile for score 28.

Solution:

The number of scores for student is 20 and the number of below 28 is 4. The 28 is repeated in two times.

Formula for percentile with 'x' values is `(B + 0.5E)/n ` x 100 = `(4 + 0.5(2))/20 ` x 100 = `5/20 ` x 100 =25.

I am planning to write more post on factoring complex polynomials and neet aipmt. Keep checking my blog.

The score 28 is in 25th percentile.

Problem 2: The scores for student are 40, 41, 42, 43, 45, 47, 49, 50, 51, 51, 54, 57, 59, 60, 61, 63, 67, 69, 70, 71.  Find out the percentile for score 51.

Solution:

The number of scores for student is 20 and the number of below 51 is 8. The number 51 is repeated in two times.

Formula for percentile with 'x' values is  `(B + 0.5E)/n ` x 100 = `(8 + 0.5(2))/20 ` x 100 = `9/20 ` x 100 = 45.

The score 51 is in 45th percentile.

Exercise problems for find percentile:

1. Michel scored 87 out of 150 students and 40 students scored below Michel. Find out the percentile of Michel score.

Solution: Percentile of Michel score is 26.

2. Scores are 11, 12, 13, 15, 15, 16, 18, 20, 22, 23. Find out the percentile for score 15.

Solution: Percentile of score 15 is 40.

How to Solve Trigonometric Equations

A trigonometric equation is an equation involving the trigonometric function or function of unknown angles, e.g., sin x = 0, cos2x-sinx=1/4, sin( theta + 1/4) = ½, etc are all trigonometric equations.

A solution of a trigonometric equation is a value of that unknown angle that satisfies the equation. Trigonometric equation may have an unlimited number of solution, e.g., sin x =0,pi2pi3pi

Let us learn how to solve trigonometric equations.

how to solve the trigonometric equation step by step:

Type 1: Equation in which only one function of a single angle is involved.

Procedure. Solve algebraically for the values of the function

Solve for x, sin x =    (sqrt(-3))/(2)(0≤x≤2π)



solution:
sin x = (sqrt3)/(2)   = - sin 60°


= sin (180° + 60°)

= sin (360° - 60°)

rArr x = 240°,300°


Type 2. Equation expressible in terms of one t- ratio of the unknown angle

Procedure. The following formulas help in expressing all the t-functins in terms of a single t-function.

Square relations

Solve cos2 theta -sintheta - 1/4 = 0(0°<= theta360°)

cos2 theta -sintheta - 1/4=0

1 - sin2 theta- sintheta 1/4=0

4sin2theta +4 sintheta -3=0

hence, 2 sin  theta + 3 = 0 sin  theta

This value is inadmissible

theta = 30°,150°
Type 3 Equation involving multiple angles.

Case-I When the equation involves only one multiple angle.

solve sin 3x = - 1/sqrt(2) ,0<= <= 2pi

Solution since we require x such that 0<= <= 2pimust be such that 0<=3x <=6pi

3x = 5pi/4 ,7pi /4 , 13pi 15pi 21pi 23pi

= 5pi/4 ,7pi /4 , 13pi /4 ,15pi /4 ,21pi /4 , 23pi


I am planning to write more post on what is the substitution method and neet 2013 medical pg. Keep checking my blog.

Case-2 Equation involving two multiple angles

They are like

sin ntheta = sin m theta

sin ntheta  = cos mtheta

How to solve trigonometric equations involving compound angles.

Solve the cos x - sqrt(3) sin x = 1,0°  <= x <= 0°

Solution: Dividing both sides of the equation by sqrt({(1)2) +(sqrt(3)2 }

=> cos60° cosx - sin 60°sin x = 1/2

=> cos(x + 60°)= cos60° or cos(360° - 60°)

x + 60° = 60° or 300° => x = 0°,240°

Taking values such that 0°<=  x  <= 360°.

Hence proved

The above problems says how to solve trigonometric equations.

Ratio Data Definition

In mathematics,  ratio expresses the magnitude of quantities relative to each other. Specifically, the ratio of two quantities indicates how many times the first quantity is contained in the second and may be expressed algebraically as their quotient. The ratio is the numerical expression representing a part of a larger whole or proportion. A ratio consists of two numbers separated by a colon. Ratio is numbers can be compared as multiples of one another.

A study calculating ratio means Comparison of two similar quantities or numbers by division.

* Usually the symbol “:” is used to denote "ratio"

* The ratio is generally denoted in the form a: b

* Ratio should be in the lowest form.

Examples on study of calculating ratios:

Example 1. Find the ratio of quantities 2 Kg to 500g in the lowest form.

Solution:

First we find the ratio,

2 Kg = 2 × 1000 = 2000 g

In  we solving this,  we get

The required ratio = 2000: 500

= 200: 50

Ratio = 4: 1

Example 2:   A bowl  contains pink and brown color chocolates. The ratio of  pink  chocolate to brown chocolate is  4:3. If the  bowl  contains  120 brown chocolates ,how many pink  chocolates are there?

Solution:

Given:  The bowl contains 120 brown  sweets,

Now, we are going to assign the variable:

Let `x = ` pink  sweets

The item is now found to be written in ratio fraction,

`" Pink "/ "Brown" `   = `4/3` = `x/120`

Cross multiplication is to be done.

So,    `4`` *` `120 ` = `3` ` x`

` x` `= 480/3`

`x ` = 160.

Hence they are 160 pink sweets.

Example 3:   In a tank  there are golden fishes and sea horses , the  ratio of golden fishes to sea horses 3:6 . If the  tank  contains  60  sea horses , how many golden fishes are there?

Solution:

Given:  The bowl contains 60 sea horses,

Now, we are going to assign the variable:

Let `x` = golden fishes

The item is now found to be written in ratio fraction,

`"Goldenfish" / "seahorses"` =` 3/6=x / 60`

Cross multiplication is to be done.

So,     `3*60= 6 x`

`180= 6x`

`x = 180/6`

`x = 30` .

Hence there are 30 golden fishes .

Algebra is widely used in day to day activities watch out for my forthcoming posts on Division with a Remainder and final syllabus for neet 2013. I am sure they will be helpful.

Examples word problem on study calculating ratios:

Problem 1:  In a school there are 20 male teachers and 10 female teachers. What is the ratio between the number of female teachers and number of male teachers?

Solution:

First we find the ratio,

The ratio between the number of female teachers and number of male teachers =10/20

Then simplify the ratios,

Convert the lowest form is = 1: 2

And we as usual to write this ratio as 1: 2

The Ratio between the number of female teachers and male teachers = 3: 5.

Problem 2:  A boy having 40 apples and 10 mangoes. What is the ratio between the number of apples and number of mangoes?

Solution:

First we find the ratio,

The ratio between the number of apples and number of mangoes =40/10

Then simplify the ratios,

Convert the lowest form is = 4: 1

And as usual to write this ratio as 4: 1

The Ratio between the number of female teachers and male teachers = 4: 1.

Factor Theorem Learning

In the factor theorem if there is no remainder, it is to be naturally concluded that the given expression is completely divisible by the given divisor which means, in other words, that the divisor is a factor of the given expression. If any integral expression in x vanishes when 'a' is substituted for x, then ' x - a' is a factor of the expression.Similarly, if any integral expression in x vanishes when '-a' is substituted for x, then x+a is a factor of the expression.

Alternative form of the factor theorem:

If f(a) = 0, then x-a is a factor of f(x).

cor 1: If f(-a)= 0, then x+a is factor of f(x).

cor 2: If (-a/b)= 0, then bx+a is a factor of f(x).

cor 3: If polynomial f(x) vanishes when x= a and also x=b, then f(x) is exactly divisible by (x-a)(x-b).

Only those values of x will make f(x) zero which is factors of the constant term e.g. if f(x)= x³-19x-30, x can only have either of the values and no other values for making f(x). the reason is that constant term is always the product of the roots of the equation f(x)=0.

Examples for factor theorem learning:

1.Factorise: x³ – 7x + 6.

Solution:

Here f(x) = x³ – 7x + 6

by putting x = 1, f(1) = 1-7 + 6

=0

therefore x-1 is a factor of x³ – 7x + 6

now, x³-7x+6 = x² (x-1) + x(x-1) -6 (x-1)

= (x-1)(x² +x -6)

=(x-1)(x+3)(x-2)

2.If the expression x³ + 3x² + 4x + p contains x + 6 as a factor, find p.

Solution:

f(x)= x³ + 3x² + 4x + p

as x + 6 is its, so

f(-6) = 0.

i.e, (-6)³ + 3(-6)² + 4(-6) + p=0

i.e, -216 +108 - 24 + p=0

i.e, -132 + p=0

therefore, p=132.

Conditional Probability Rule

Definition:

The conditional probability if an event B,assuming that the event A has already happened; is denoted by P(B/A) and defined as    P(B/A)=P(A nn B) / P(A)   ,  provided P(A) != 0.

similarly,  P(A/B)=P(A nn B) / P(A)   ,  provided P(B) != 0 .

Explanation :

For instance consider the following example to understand the concept of conditional probability.

Suppose a fair die is rolled once. The sample space is S = { 1,2,3,4,5,6 } .

Now try to answer these questions :

Q 1: What is the probability that getting an even number which is less than 4  ?

Q 2: If the die shows an even number, then what is the probability that it is less than 4 ?

case 1:  The event of getting an even number which is less than 4 is {2}

Therefore , P1 =n({2}) / n({1,2,3,4,5,6})= 1/6

case 2:  Here first we restrict our sample space S to a subset containing only even number i.e. to {2,4,6}. Then our interest is to find the probability if the event getting a number less than 4 i.e. to {2}.

Therefore,  P2 = n({2}) / n({2,4,6})  = 1/3

In the above two cases the favourable events are the same, but the number of exhaustive outcomes are different. In case 2, we observe that we have imposed a condition on sample space, then asked to find the probability.This type of probability . This type of probability is called conditional probability.

Explanation:

A conditional probability is the probability of an event given that another event has occurred. For example, what is the probability that the total of two dice will be greater than 8 given that the first die is a 6? This can be computed by considering only outcomes for which the first die is a 6. Then, determine the proportion of these outcomes that total more than 8. All the possible outcomes for two dice are shown below:

all outcomes two dice

There are 6 outcomes for which the first die is a 6, and of these, there are four that total more than 8 (6,3; 6,4; 6,5; 6,6). The probability of a total greater than 8 given that the first die is 6 is therefore 4/6 = 2/3.

More formally, this probability can be written as:

p(total>8 | Die 1 = 6) = 2/3.

In this equation, the expression to the left of the vertical bar represents the event and the expression to the right of the vertical bar represents the condition. Thus it would be read as "The probability that the total is greater than 8 given that Die 1 is 6 is 2/3." In more abstract form, p(A|B) is the probability of event A given that event B occurred

conditional probability :


A clearcut description of Conditional Probability

The conditional probability if an event B,assuming that the event A has alteady happened; is dinoted by P(B/A) and defined as    P(B/A)=P(A nn B) / P(A)   ,  provided P(A) != 0.   A conditional probability is the probability of an event given that another event has occurred.

Perimeter of Circle

A circle is the locus of points that are equidistant from a given point. If we take any point in the plane and connect all the points in the plane that are at a given distance say 2 cm or 5 cm from the point, we will get a circle.

The fixed point is the centre of the circle and the given distance is called the radius of the circle. The  distance between a pair of points on the circle that are farthest apart is called the diameter of the circle and is twice the measure of the radius. If we connect such pair of points with a line segment, it will always pass through the centre of the circle.

The perimeter of a circle is the length as measured along the boundary lines of the circle. The perimeter is also called circumference.

How to find the Perimeter of a circle

Before we proceed to learn how to find the perimeter of a circle, let us understand the concept of perimeter of a circle with an example.

Imagine that we have to travel along a thin perfect circular road in an automobile. Let us assume that we are starting from a point A on the circle. Note down the odometer reading of the automobile at the start of the journey. Start the vehicle and travel along the boundary of the circle and come back to the same point A without deviating from the boundary. Note down the odometer reading again. The difference between the two odometer readings give the distance traveled along the circle and this is the perimeter of the circle

Since perimeter is a measure of distance, it has the same units as that of distance like millimetre (mm), centimeter (cm), metre (m) etc

There is a simple formula to calculate the perimeter of a circle once we know the radius of the circle. The formula for the perimeter of a circle is given by

C = 2 x  `pi`   x r

C= Circumference of the circle or the Perimeter of the circle

`pi` = Pi,  the value of `pi ` is approximately taken as 3.14

r= radius of the circle

But diameter of the circle is twice the radius

diameter = 2 x radius

The formula for the perimeter of a circle  can also be written as

C= `pi` x  d

Where d = diameter

The value of `pi`  is constant and assumed to 3.14.  For the time being `pi `can be understood as a constant obtained when the circumference of any circle is divided by its diameter. In higher mathematics, you will learn about more advanced definitions for `pi `and also how to derive the value of `pi `accurately.

Let us work out simple examples on how to find the perimeter of  a circle

Ex 1 : The radius of circle is 5 cm what is the circumference?

Sol : C= 2 x `pi` x r

Step 1: C= 2 x 3.14 x 5

Step 2:  C = 31.4 m

Note : Your answers should always have the units mentioned clearly else it is incomplete.

Ex. 2:  The wheel of a bullock cart has a radius of 3 m. If the wheel rotates once how much distance does the cart move?

Sol: When the wheel rotates once, the cart will move by a distance equal to the perimeter of the wheel.

Step 1: C= 2 x `pi`  x r

Step 2: C= 2 x 3.14 x 3 = 18.84 m

The bullock cart moves 18.84 m in one revolution of the wheel

Exercises on Perimeter of a circle

Let us summarize how to find the perimeter of  a circle. The perimeter of a circle can be obtained form the formula

Perimeter of the circle = 2 x `pi`  x radius of the circle or `pi`  x diameter of the circle

Pro 1: A cart wheel has a diameter of 1m. How many times should the wheels roll for the cart to travel 1 km?

Ans: 318.47 revolutions

Pro 2: The wheel of an automobile has a diameter of 2m and rotates 40 times in a minute, What is the speed of the automobile?

Ans: 15.072 Kmph

Algebra is widely used in day to day activities watch out for my forthcoming posts on evaluating an algebraic expression and cbse syllabus for class 10. I am sure they will be helpful.

Hint: Find the distance traveled in 40 rotations which is perimeter multiplied by 40. Calculate speed as distance traveled divided by time.

Pro 3: The circular coin of diameter 2 cm rolls 6.28 cm along the ground in one revolution. Can you calculate the value of `pi`  from the given data ?

Ans 3.14

Factorization

A Linear Factorization form of a polynomial is in which each factor is a linear polynomial. Linear functions are functions of the form y = mx + b;  their graphs are straight lines. A related concept is the “linear factor”:

Factoring is the decomposition of an object (for example, a polynomial, or a matrix) into a product of other objects, or factors, which when multiplied together gives the original number.

For example, the number 15 has prime factors as 5× 3, and the polynomial x2 ? 9 factors as

(x ? 3)(x + 3). In all cases, a product of simpler objects is obtained.

The aim of factorization is usually to reduce something to "basic building blocks," For example numbers to prime numbers, or polynomials to irreducible polynomials.

I like to share this Math Prime Factorization with you all through my article.

Definition: Linear factor

words: “linear factor with root r ”
usage: r is a some real number
meaning: the expression (x ? r)


Examples for Linear factorization:

Through the product of linear polynomial factorization and some of their linear factors are,

1) The polynomial f (x) = x3?2x2 ? x + 2 can be written as a product of linear factors,

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

2) The polynomial f (x) = x3 ? 4x2 + 5x ? 2 can be written as a product of linear factors,

F(x) = (x ? 1)2(x ? 2) = (x ? 1)(x ? 1)( x ? 2).

3) The polynomial f (x) = x3 ? 8 can be factored as f (x) = (x2 + 2x + 4)(x ? 2), but only one of these factors is a linear factor. The other factor, (x2 + 2x + 4) cannot be broken down further into linear factors.

Relation between factors and roots of a polynomial:

There is an important correspondence between the roots of a polynomial and its linear factors:

Roots of f  `<=>` factors of f

r is a root of f   `<=>`  (x ? r ) is a factor of f

The function f (x) = x2 ?5x + 6 has roots r = 2 and r = 3 that correspond to the linear factors in the factorization f (x)=(x?2)(x?3) .

The function f (x) = x2 ?5 has roots r = ?5 and r = ??5 that correspond to the linear factors in the factorization f (x) = (x ? ?5) (x + ?5).

The function f (x) = x2 +5 has no roots, and f cannot be factored into the form f (x) = (x ? a)(x ? b).

The function f (x)=x2 +2x+4 has no roots, and f cannot be factored into the form f (x) = (x ? a)(x ? b).

The function f (x) = x3 ? 2x2 ? x + 2 has roots r = ?1, r =1, and r = 2 that correspond to the linear factors in the factorization

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

The function f (x) = x3 ? 4x2 + 5x ? 2 has roots r =1, and r = 2 that correspond to the linear factors in the factorization

f(x) = (x ? 1)2( x ? 2) = (x ? 1)( x ? 1)( x ? 2).


I am planning to write more post on Line Equation from Two Points and sample papers for class 9 cbse sa2. Keep checking my blog.

The function f (x) = x3 ? 8 has only one root, r = 2. We have seen that f can be factored f(x) = (x2 + 2x + 4) (x ? 2), but f cannot be broken down completely into linear factors. The linear factor (x ? 2) corresponds to the root r = 2, but the factor (x2 + 2x + 4) is not a linear factor and does not correspond to any roots.

Calculating The Standard Deviation

In probability theory and statistics, 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.
(Source : Wikipedia)

In this article we shall discuss about calculating Standard Deviation. Also we shall solve problems based on calculating standard deviation.

Calculating the Standard Deviation:

Formula for calculating Standard Deviation Questions :

It is nothing but standard deviation questions are calculated by taking square root for Variance.

Formula for finding mean,

barx   = (sum (X ) )/ n

Formula to solve standard deviation questions,

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

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 Calculating the Standard deviation :

Examples for Calculating the Standard deviation are given below:

Problem 1:


The given data set are 18, 17, 14, 11 and 12. Calculating the Standard Deviation

Solution:

Mean: Calculate the average for the given values. To find the mean.
x  = (18 + 17 + 14 + 11 + 12)/ 5
= 72 / 5
= 14.4
Standard Deviation,

S = sqrt (( ( 18 - 14.4 )^2 + ( 17 - 14.4 ) ^2 + ( 14 - 14.4 ) ^2 + ( 11 - 14.4 ) ^2 + ( 12 - 14.4 ) ^2 )/ (5 - 1) )

=  sqrt(37.2/ 4 )

= sqrt( 9.3 )

S =  3.04959014
ANSWER:
Standard Deviation  S = 1.58113883


Algebra is widely used in day to day activities watch out for my forthcoming posts on Dividing Fractions by Fractions and sample papers for class 12 cbse. I am sure they will be helpful.

Problem 2:

Find the Standard deviation of given Data  9, 10, 11, 12, 13, 14 and 15.
Solution:
barx = ( 9 + 10 + 11 + 12 + 13 + 14 + 15) / 7
barx = 84 / 7
barx = 12


X


X-barX


(X-barX)^2

9


9 - 12 = -3


9

10


10 - 12 = -2


4

11


11 - 12 = -1


1

12


12 - 12 =  0


0

13


13 - 12 = 1


1

14


14 -12 = 2


4

15


15 - 12 = 3


9


Is this topic algebra problems hard for you? Watch out for my coming posts.

Standard Deviation:

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

S =  sqrt( ( 9+4+1+0+1+4+9 ) / 6)
S = sqrt ( 28 / 6)
S =  sqrt(4.66667 )
S = 2.16024767

Slope Intercept form Solver

If the line passes through the points M (x1, y1) and N (x2, y2) and then its slope is,

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

The slope is defined as the ratio of vertical distance to the horizontal distance.

Slope m=`y/x`

Where y and x are the difference between two points

The slope intercept form is,

Y=mx+b

Here x and y represents the points, m represents the slope and b represents the vertical intercept. In this article we shall discuss about the slope intercept form with suitable example problem.
Slope Intercept Form Solver:

In the solver you have to enter the slope and y- intercept value. After pressing the solve button the equation of line will be displayed on answer box. The slope intercept solver is shown below.

Slope Intercept Form Solver
Problems on slope intercept form in math

Given y-intercept is 7 and the slope is 3. Find the slope intercept form

Solution:

Y-intercept = 7

Slope m= 3

Equation of line is,

Y =mx +b

Substitute the value of y-intercept and slope value in above equation.

Y =3x +7

Equation of line y = 3x +7
Problems on slope intercept form in math

Given y-intercept is -4 and the slope is 2. Find the slope intercept form

Solution:

Y-intercept = -4

Slope m= 2

Equation of line is,

Y =mx +b

Substitute the value of y-intercept and slope value in above equation.

Y =2x - 4

Equation of line y = 2x - 4
Problems on Slope Intercept Form in Math

Given y-intercept is 12 and the slope is 5. Find the slope intercept form

Solution:

Y-intercept = 12

Slope m= 5

Equation of line is,

Y =mx +b

Substitute the value of y-intercept and slope value in above equation.

Y =5x + 12

Equation of line y = 5x + 12

Problems on slope intercept form in math

Given y-intercept is -8 and the slope is 4. Find the slope intercept form

Solution:

Y-intercept = -8

Slope m= 4

Equation of line is,

Y =mx +b

Substitute the value of y-intercept and slope value in above equation.

Y =4x - 8

Equation of line y = 4x - 8

Linear System Definition

We know that if two expressions are equated, we can call it as an equation. If the power of the variable is one, it is called an linear equation.  If two equations, each containing two same variables (unknowns), are called a system of linear equations.

Eg: 4x + 3y = 12, 6x + 5y = 24.  This can also be called as simultaneous equations, because when we solve these equations, we get one set of values (x, y) in such a way that they satisfy both equations. Hence they got their name.  We can also use the word system as well.  These system of linear equations, which will be representing straight lines can be solved by using (1) substitution method (2) Addition or subtraction method (3) Graphs as well.

Example Problems on Linear System.

Ex 1: Solve: a + 3b = 3 and 4a - 5b = 29 by substitution method.

Solution: Given: a + 3b = 3 ->  (1)

4a - 5b = 29 ->  (2)

Therefore (1) =>  a = 3 ** 3b ->  (3)

Therefore (2) =>  4 (3 ** 3b) ** 5b = 29

=>  12 ** 12b ** 5b = 29

=>  ** 17b = 29 ** 12 = 17.

Therefore b = ** 1.

Therefore (3) =>  a = 3 ** 3 (** 1) = 3 + 3 = 6.

Therefore a = 6, b = ** 1.

Ex 2: Solve: 9x + 4y = 5 and 4x ** 5y = 9 by addition or subtraction method.

Solution: Given: 9x + 4y = 5 ->  (1)

4x ** 5y = 9 ->  (2)

In this method we need to eliminate one variable by making their coefficient same.

(!) xx 5 =>  45x + 20y = 25

(2) xx 4 =>  16x ** 20y = 36

=>  61x = 61 =>  x = 1.

Now by substituting x = 1 in (1), we get: 9 (1) + 4y = 5 =>  4y = 5 ** 9 = ** 4.

=>  y = ** 1.

Therefore The solution is x = 1, y = ** 1.

Ex 3: Solve: x + y = 11 and x ** y = ** 3 using graphs.

Solution: Given: x + y = 11 ->  (1)

x ** y = ** 3 ->  (2)

Here form a table for (1) of (2)

and plot them join the lines.

They will meet at a point is the required answer:

(1) =>  x + y = 11

Linear equationT1

(2) =>  x - y = -3

Linear equationT2

Between, if you have problem on these topics All even Numbers, please browse expert math related websites for more help on cbse solved sample papers for class 12.

From the graph, the solution is x = 4, y = 7.
Practice Problems on Linear System.

1. Solve: a + 5b = 18; 3a + 2b = 41 by substitutions method.

[Ans: a = 13, b = 1]

2. Solve by addition or subtraction method: x + 2y = 11, 2x ** y = 2.

[Ans: x = 3, y = 4]

3. Solve graphically the equations: z ** y = 2 and 2z ** y = 7.

[Ans:  y = 3, z = 5]

Inverse Log Graph

Inverse:

The inverse of the given function can be described as the undo of the initial action of the function. All the function has its inverse. But not every inverse is a function.

Log:

The log is a math concept that used to express the relationship between the variables in easy manner . The following are the some properties of the log functions.

Graph:

The graphical representation that used to express the function is known as graph. The graph has x and y axis.  Let see some problems on inverse log graph.


Having problem with Find Inverse Function keep reading my upcoming posts, i will try to help you.

Example Problem on Inverse Log Graph :

In this article we are going to see some problems on inverse log graph and some practice problems for testing your inverse log graph skills.

Problem 1:

Find the inverse of the log function f(x) = log 7x and draw the graph

Solution :

Given f(x) = log 7x

We need to find the inverse of the given function.

To find the inverse of the given function, taking exponent on both side,

Before that substitute f(x) = y

y = log 7x

e^y = e^(log 7x)

We know that e^logx = x

e^y = 7x

Divided by 7 on both sides,

(e^y)/7 = (3x)/7

(e^y)/7 = x

x = (e^y)/7

Replace x = f^(-1)(x) and y = x

f^(-1)(x) = (e^x)/ 7

 The inverse of the given function is f^(-1)(x) = (e^x)/7

 y = e^x/7

Substitute x = -2, -1, 0, 1, 2 and get y value for draw the graph,

y = e^(-2) / 7 = 0.0193    ≈ 0.01

y = e^(-1) / 7 = 0.052      ≈  0.05

y = e^(0) / 7  =  0.142       ≈ 0.10

y = e^(1) / 7 =  0.388       ≈  0.4

y = e^(2) / 7 = 1.055         ≈ 1.1

Algebra is widely used in day to day activities watch out for my forthcoming posts on 3rd grade math problems online and gmat syllabus 2013. I am sure they will be helpful.


Practice Problem on Inverse Log Graph :

Problem :

Find the inverse of  a logarithms function f(x) = log 11xand draw the graph.

Solution

The inverse of the given function is f^(-1)(x) = e^(x) /11

Equation of Geometric Sequence

In this article, we will discuss the equation of geometric sequence. Geometric sequence is defined as, the sequence of a numbers such that the ratio between two consecutive members of the sequence is a constant.

Formulas for geometric sequence:

nth term of the sequence: a_n = a_1 * r^(n-1)

Series of the sequence: S_n = (a_1(1-r^n))/(1 - r)

The equation of geometric sequence example problems are given below.


I like to share this Limit of Sequence with you all through my article.

Example Problems for Equation of Geometric Sequence:

Example problem 1:

Find out the 6th term of a geometric sequence if a1 = 72 and the common ratio (C.R) r = 2

Solution:

Use the formula a_n = a_1 * r^(n-1) that gives the nth term to find a_6 as follows

"a_6 = 72 * (2)^(6 - 1)

= 72 * (2)5

= 72 * 32

After simplify this, we get

= 2304

The 6th term of a geometric sequence is 2304

Example problem 2:

Find out the 7th term of a geometric sequence if a1 = 31 and the common ratio (C.R) r = 2

Solution:

Use the formula a_n = a_1 * r^(n-1) that gives the nth term to find a_7 as follows

"a_7 = 31 * (2)^(7 - 1)

= 31 * (2)6

= 31 * 64

After simplify this, we get

= 1984

The 7th term of a geometric sequence is 1984.
More Example Problems for Equation of Geometric Sequence:

Example problem 3:

Find out the 11th term of a geometric sequence if a1 = 42 and the common ratio (C.R) r = 2

Solution:

Use the formula a_n = a_1 * r^(n-1) that gives the nth term to find a_11 as follows

"a_11 = 42 * (2)^(11 - 1)

= 42 * (2)10

= 42 * 1024

After simplify this, we get

= 43008

The 11th term of a geometric sequence is 43008.


My forthcoming post is on What is Number Sense and upsc civil services exam 2013 will give you more understanding about Algebra.

Example problem 4:

Find out the 18th term of a geometric sequence if a1 = 58 and the common ratio (C.R) r = 2

Solution:

Use the formula a_n = a_1 * r^(n-1) that gives the nth term to find a_18 as follows

"a_18 = 58 * (2)^(18 - 1)

= 58 * (2)17

= 58 * 131072

After simplify this, we get

= 7602176

The 18th term of a geometric sequence is 7602176

The above example problems are helpful to study of geometric sequence equation.

Standard Deviation on a Calculator Help

Standard deviation represents the dispersion of any process. It can be calculated as the square root of mean of squares of differences of variate values from their mean. Let us see the methods to calculate standard deviation on a calculator


Methods to Calculate Standard Deviation ( S.d.)

`sigma` = calculation1

N
Where N is the total frequency ∑fi.

The square of standard deviation is called as variance.

I like to share this Formula to Calculate Standard Deviation with you all through my article.

Standard deviation Calculations:

The change of origin and the change of scale considerably reduces the labour in the calculation of standard deviation. The formulae for computation of `sigma`  are given below:

1)      Short-cut method

`sigma`= calculation2

2)      Step – deviation method

calculation3

Where di = xi – A and di’ =     (xi – A ) / h, being the assumed mean and h the equal class interval.

Proof: We know that calculation6

Therefore                  calculation7

calculation8

calculation9

`sigma` =    calculation10
Implications of Standard Deviation Methods Discussed

(i)               Quartile deviation = 2/3 ( Standard Deviation)

(ii)             Mean deviation = 4 / 5 ( Standard Deviation )

Cor. If m1,`sigma`1  be the mean and S.D. of sample size n1 and m2,`sigma`2  be those for sample of size n2, then the S.D. `sigma`of the combined sample of size n1 + n2 is given by

(n1 + n2)  = n1`sigma`12 + n2`sigma`22 + n1D12 + n2D22

Where Di = mi – m, m being mean of combined sample.

From 4, we have ns2 = n`sigma`2 + n ( x̅ - A ) 2 where n is the size of the sample.

i.e. sum of squares of deviations from A = n`sigma`2   + n ( x̅ - A ) 2

now lets apply this result to the first given sample taking A at m. Then, sum of squares of deviations of n1 items from m = n1 + n1 (m1 – m) 2.

Similarly for second given sample taking A at m. Then, sum of the squares of deviations of the n2 items from m = n2`sigma`12 + n2 (m2 – m) 2.

Adding both the samples at deviation n1 and n2 i.e. (n1+ n2) from m

= n1`sigma`12 + n2`sigma`22 + n1 (m1 – m) 2+ n2 (m2 – m) 2

(n1+n2) `sigma`2 = n1`sigma`12+ n2 `sigma`12 + n1 D12+ n2 D12

This result may be extend to combination of any number of samples, giving a result of the form ( ∑ ni ) `sigma` 2 = ∑ ( ni`sigma`2 i 2 ) +∑ (ni Di2).