Monday, December 10, 2012

Adding and Subtracting Vectors



The addition may be represented graphically by placing the start of the arrow b at the tip of the arrow a, and then drawing an arrow from the start of a to the tip of b. The new arrow drawn represents the vector a + b.

Addition of vectors

To subtract b from a, place the end points of a and b at the same point, and then draw an arrow from the tip of b to the tip of a. That arrow represents the vector a − b.
Source: Wikipedia

Subtraction of vectors

Adding and subtracting of vectors operation and example problems are given below.

Operations for on Vectors:

1. Addition of vectors:

Let `vec(OA)` = `veca` , `vec(AB)` =` vecb` . Join OB.

Then `vec(OB)` represents the addition (sum) of the vectors veca and vecb.

Addition of vectors

This is written as `vec(OA) ` +` vec(AB)` = `vec(OB)` Thus `vec(OB)` = `vec(OA) ` + `vec(AB) ` = `veca ` +` vecb`

This is known as the triangle law of addition of vectors which states that, if two vectors are represented in magnitude and direction by the two sides of a triangle taken in the same order, then their sum is represented by the third side taken in the reverse order.

Applying the triangle law of addition of vectors in


Addition of vectors

ΔABC, we have BC + CA = BA ⇒ BC+ CA = − AB

⇒ AB + BC + CA = 0

Thus the sum of the vectors representing the sides of a triangle taken in order is the null vector.

2. Subtraction of vectors:

If `veca` and `vecb` are given two vectors, then the subtraction of `vecb` from `veca` is defined as the sum of `veca` and − `vecb` and  denoted by `veca` − `vecb` .

Subtraction of vectors

`veca ` − `vecb` = `veca` + ( − `vecb` )

Let `vec(OA)` = `veca` and `vec(AB)` =` vecb `

Then `vec(OB)` = `vec(OA)` + `vec(AB)` = `veca` + `vecb`

To subtract `vecb` from `veca` , produce BA toAB' such that AB = AB'.

∴ `vec(AB')` = − `vec(AB)` = −` vecb`

Now by the triangle law of addition

` vec(OB')` = `vec(OA)` +` vec(AB')` = `veca` + ( `-vecb` ) =` veca ` − `vecb`

Example Problems for Adding and Subtracting of Vectors:

Example problem 1:

The position vectors of the points A, B, C, D are `veca` , `vecb` , `2veca` + `3vecb` ,`veca` − `2vecb` respectively. Find `vec(DB) ` and `vec(AC)`

Solution:

Given that

` vec(OA)` = `veca` ; `vec(OB)` =` vecb` ; `vecOC` =` vec2 ` + `3vecb` ; `vec(OD)` = `veca ` − `2vecb `

`vec(DB)` = `vec(OB)` − `vec(OD)` = `vecb ` − (`veca` − `2vecb` ) = `vecb ` − `veca` + `2vecb` = `3vecb` − `veca`

`vec(AC)` = `vec(OC)` − `vec(OA)`

= (`2veca` + `3vecb` ) − `veca` = `veca` + `3vecb `

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Example problem 2:

` vec(OA)` = `2veca` + `3vecb` - `vecc` , `vec(OB)` = `4veca` + `2vecb` + `2vecc` , Find adding and subtracting of vectors

Solution:

Subtracting the two vector,

`vec(AB) ` = `vec(OB)` - `vec(OA)`

`vec(AB)` = `4veca` + `2vecb ` +` 2vecc` - `2veca` - `3vecb` +` vecc`

`vec(AB)` = `2veca` - `vecb ` + `3vecc`

Adding the two vector,

` vec(AB)` = `2veca` + `3vecb` - `vecc` + `4veca` + `2vecb` + `2vecc`

` vec(AB)` = `6veca` +` 5vecb` + `vecc`

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Thursday, December 6, 2012

Accurate to Two Decimal Places



Decimal numbers contains decimal places. Here it is very important to know about the place value in decimal numbers.

So what is Decimal Place Value?

It can  be explained by taking example:

Gh1

let us discuss it using whole number which does not have decimal point in it.

As we go from  right to left the value, the position or the place value gets 10 times bigger.

Similarly as we move from left to right the place value gets 10 times smaller.

What do we get as we move beyond unit place?

Here is where the decimal point comes into picture. If we move beyond unit place the value becomes 1/10, 1/100, 1/1000 and so on. These number which contains decimal point is called DECIMAL NUMBER.

Below is the example for decimal number:

gh2

here we can see that the place value of number goes on decreasing as we move beyond decimal points. The value of one in the above example is 1/10 , the place of 2 become 1/100

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Conversion of Decimal Number to Whole Number (accurate to Two Decimal Places)

HOW TO CONVERT  DECIMAL NUMBER TO WHOLE NUMBER?

This can be explained using eg given below:

consider 125.43 and also 125.56

First check the number immediately after the decimal  place. If it is greater than or equal to 5 then add 1 to the unit place and write it as whole number otherwise discard the numbers after decimal point and write the whole number as it is.

In the above example 1)  the number 4 appears immediately after decimal point which is less than 5 .So there is no change and we have to discard all the number proceeding decimal point i,e 4 and 3 and write the number as it is.

whereas in case of example 2) the number which appears immediately is 5 so add one to unit place which becomes 126


Accurate Writing of the Decimal Numbers to 2 Decimal Places

How to write numbers accurate to 2 decimal places?

Even here we have to follow the same rules mentioned above.

examples:

123.4612 ----------------------->123.46

123.4689 ------------------------> 123.47


In the above example 1) check the third number after the decimal point. In this case the third number is 1 which is smaller than 5 so don't add 1 to the number which in 1/100's  place and discard the 3rd digit after the point and write as it is. Here  the final number accurate to 2 decimal point becomes  1 2 3 4 . 4 6

example 2) here the 3rd digit after decimal point is 8 which is greater than 5 so add 1 to the second digit after decimal. Here  the final number accurate to 2 decimal point becomes  1 2 3 4 . 4 7

Algebra is widely used in day to day activities watch out for my forthcoming posts on  list of irrational numbers. I am sure they will be helpful.

Thursday, November 29, 2012

Intersection of Two Lines Calculator



Straight line :-

A straight line is generally termed as line. The  curvature of a straight line is 0 . The orientation of a straight line is given by its slope.General representation of straight line is given by  AB.

Explanation to Intersection of Two Lines Calculator

Intersecting lines:

Two lines are said to be intersecting if and only if the have a common root  or  solution.The general form of equation of a line is given by Y=mX +c Where   m= slope, c= y intercept of line .

Features of straight line  of form Y= mX +c :-

i) The straight line is parallel to X axis of m = 0 ie slope of line is 0.

ii) The straight line passes through orign when constant "c" =0.

iii)The straight line makes an angle 45owith "X" axis when m=1 and c=0.

iv) The straight line makes an angle 135o with "X" axis when m=-1 and c=0.

v) The straight line parallel to "y" axis has slope infinity .

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Condition for two line to  intersect :-

For two lines to intersect, their slopes must not be equal .Lines having same slopes are called as parallel lines and intersection point of two parallel lines is 0 or we can say that they meet at infinity. There can be  only one point of intersection for two distinct lines  if they  are not parallel.Multiple intersection of two lines is not possible.

Finding point of intersection of two lines :-

The point of intersection of two lines-can be found by two methods :-

i) By solving the equations

ii) By graphing the equation

Both of these approaches lead to same answer

lets us take two lines  y = 2x+3 ; y= -0.5x + 7 and find the point of intersection

here  primary examination of slopes is to be done.

Slopes are 2& -0.5 so these lines are not parallel lines

So we go by first method solving equations

y = 2x+3 ; ------ equation 1

y= -0.5x + 7 --------equation 2

=> 2x+3  = -0.5x + 7

=> 2.5x = 7-3

=> x= 4/2.5

=> x=1.6

substituting in equation 1 or 2 we get

 y= 2*(1.6) +3

=> y= 6.2

so (1.6 , 6.2) is the point of intersection of  y = 2x+3 ; y= -0.5x + 7

on graphing the equations and plotting them

two intersecting lines

we can observe that the point of intersection is (1.6, 6.2).

Hence both the approaches give the same result .

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Examples to Intersection of Two Lines Calculator

 EX 1:-

find point of intersection of  lines 3y= 6x +3 and y= 2x+3

Solution:-

Given equations 2y= 4x +2 and y= 2x+3

comparing with standard form of equation y=mx +c

slope of 3y= 6x +3 => y= 2x+1 is 2

slope of line y= 2x+3 is 2

here slopes of lines are equal so they are parallel

so there will be no point of intersection for lines  3y= 6x +6 and y= 2x+3.

 EX 2 :-

FInd the point of intersection of lines y= 2x+1  & x=2

solution :-

Here both lines are not parallel as on comparing with standard equation y=mx+c

slope of  y= 2x+1  is 2 and   x=1 is infinite as its parallel to y axis

substituting x=2 in y= 2x+1

=> y= 2*2+1

=> y= 5

so point of intersection of  y= 2x+1  & x=1 is (1,5)

EX 3:

Find the point of intersection of "X" axis and "Y" axis

solution:

equation of  x axis is  y=0

equation of y axis is x=0

So point of intersection of x, y axis is (0,0) ie origin.

Thursday, November 22, 2012

Solve Real Life Algebra Problems



Introduction:

In this branch of mathematics, we use the alphabetical letters in problems like a, b, x and y to denote numbers. In real life algebra, the various operations are addition, subtraction, multiplication, division or extraction of roots on these symbols and real numbers are called algebraic expressions. In the real life a Greek mathematician Diaphanous has developed and solve this subject to a great extent and hence we call him as the father of algebra.

Solve Real Life Algebra Problems-solved Problems :

Example1:

A woman on tour travels first 160 km at 60 km/hr and the next 160 km at 80 km/hr. The average speed for the first 310 km of the hour. Find the total time?

Solution:

The given data’s which are taken and can be solve as follows,

Total time taken = (160 / 60+160 / 80) hr.

=12/3 hr.

Example 2:

The ratio between the speeds of two trains is 5 : 6. If the second train runs 300 km in 3 hours, then what is the speed of the first train?

Solution:

This can be solve as below,

Speed of two trains 5 and 6 km/hr

6x = (300 / 3)

=100

X = (100 / 6) =16

Speed of first train = 16 × 5km/hr

= 80km/hr

Average speed= (340 ×12 / 3)km/hr

= 87.62km/hr.

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Solved Problems for Real Life Algebra:

Example1:

A man traveled a distance of 30 km in 6 hours. He traveled partly on foot at 2 km/hr and partly on bicycle at 6 km/hr. What is the distance traveled on foot?

Solution:

The problem which can be solve as,

Distance traveled on bicycle = (30 - x) km

x / 2 + (30-x) / 6 = 6

6x + 2 (30-x) = 6 × 12

x = 12 km

Example 2:

A bus crosses a 600 m long street in 2 minutes. What is his speed in km per hour?

Solution:

This is a real life problem in algebra which can be solve as,

Speed = (600 / 2 × 60)m/sec

= 5 m/sec

We have to convert  m/sec to km/hr

= (5 × 18/2)

= 45 km/hr

Example 3:

The ratio between the speeds of two trains is 3 : 5. If the second train runs 300 kms in 3 hours and average speed is 50 km/hr.what is the speed of the first train?

Solution:

The problem in the real life algebra which can be solve as follows,

Speed of two trains 3 and 5 km/hr

5x = (300 / 3)

=100

X = (100 / 5) = 20

Speed of first train = 20 × 5 km/hr

= 100 km/hr

Average speed = (50 ×9 /3 ) km/hrs

= 150 km/hr.

These are all the solve problems which are in the real life algebra.

Monday, November 19, 2012

Envision Math California



Envision MATH California is Based on problem-based, interactive knowledge and theoretical understanding. What does that mean? Students with theoretical thoughtful know more than inaccessible details and techniques, they are capable to study fresh thoughts by linking to the thoughts they previously be acquainted with. They study ideas by interact with problems; this communication takes place on a daily basis and is exact to the idea being trained.lets see some problems on envision MATH California.

Envision Math California Problems:-

Problem 1:-

Find the perimeter of the following figure:-  
                                                                                                         
Solution:-

We need to find the perimeter of the following figure

Perimeter can be found by adding lengths of the given figure.

Here the length of each sides are the following

4.8 yards, 4.3 yards, 1.1 yards, 4.5 yards, 4.6 yards, 1.6 yards.

Perimeter = sum of all given sides

= 4.8 yards + 4.3 yards + 1.1 yards + 4.5 yards  + 4.6 yards + 1.6 yards.

= 20.09  yards.

So the perimeter of the given figure is 20.09 yards.

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Problem 2:-

Find the Perimeter of the given figure.


Solution:-

We need to find the perimeter of the following figure

Perimeter can be found by adding lengths of the given figure.

Here the length of each sides are the following

4.7 yards, 2.3 yards, 1.1 yards, 1.1 yards,  4.3 yards, 2.4 yards, 3 yards.

Perimeter = sum of all given sides

= 4.7 yards+ 2.3 yards+1.1 yards+ 1.1 yards+  4.3 yards+ 2.4 yards+ 3 yards.

= 18.9 yards.

So the perimeter of the given figure is 18.90 yards.

Problem 3:-

Find the Perimeter of the given figure.


Solution:-

We need to find the perimeter of the following figure

Perimeter can be found by adding lengths of the given figure.

Here the length of each sides are the following.

2 feet, 2feet, 5 feet, 5 feet, 2 feet.

Perimeter = sum of all given sides

= 2 feet+ 2feet+ 5 feet+ 5 feet+ 2 feet.

= 16 feet

So the perimeter of the given figure is 16 feet.

Envision Math California Practice Problems:-

Find the area of the given parallogram:-



Answer:- 84 cm2


Between, if you have problem on these topics Business Math, please browse expert math related websites for more help on area of hexagon formula.

Monday, October 29, 2012

Function Derivative Calculator



Solving 2nd order derivatives of a function

1) Solve the derivative for the function  f(x) =  x^2 + 8x + 9

Solution :  The given function is  f(x) = x^2 + 8x + 9

Differentiate the above equation with respect to 'x' . It is represented as f'(x) .

f'(x)  =   `(d(x^2))/dx`   +  `(d(8x))/dx`   +  `(d(9))/dx` .

f'(x)   =     2x   +   8   +   0

f'(x)    =     2x    +   8   .

The answer is   f'(x)  =  2x  +  8 .

2) Solve the derivative for the function   f(y)  =  y^2  +  10y  + 3

Solution :   The given function is  f(y)  =  y^2  +  10y  +  3

Differentiate the above equation with respect to 'y' . It is represented as f'(y)  .

f'(y)  = `(d(y^2))/dy`   +  `(d(10y))/dy`   +   `(d(3))/dy`

f'(y)  =   2y   +  10   +   0

f'(y)  =   2y   +   10

The answer is  f'(y) = 2y  +  10

Solving third Order Derivative Functions

1) Find the derivative of the function  f(x)  =  x3 + 3x^2 + 18x  +  20

Solution : The given   function  f(x)  =  x3 + 3x^2 + 18x  +  20

Differentiate the above function with respect to 'x' .

f'(x)  =  3 x^2   +  3  ( 2 ) x   +  18   +  0

f'(x)   =  3x^2  +  6x  +  18  .

The answer is   f'(x)   =  3x^2  +  6x  +  18  .

2) Find the derivative for   f(x)  =  6x3  + 5x^2  +  3x  +  1

Solution : The given function is f(x)  =  6x3  + 5x^2  +  3x  +  1

Differentiate the above f(x) with respect to 'x'  .

f'(x)  =  6 (3)x^2  +  5 (2)x   +  3  + 0

f'(x)   =  18x^2   +  10x   +  3

The answer is  f'(x)   =  18x^2   +  10x   +  3

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Solving 4th Order Derivative Function

1) Solve the derivative for the function  f(y) = y^4  + 3y^3  +  5y^2  + 4y  +  9

Solution : The given function  f(y) = y^4  + 3y^3  +  5y^2  + 4y  +  9

DIfferentiate the above equation with respect to 'y' .

f'(y)  =  4y^3  +  3(3)y^2  +  5(2)y  +  4  +  0

f'(y)  =  4y^3   +  9y^2  + 10y   +  4

The answer is    f(y) = y^4  + 3y^3  +  5y^2  + 4y  +  9

2) Solve the derivative for the function f(y)  =  6y^4  +  y^3   +  y^2  + 10 y  + 3

Solution :  The given function  is    f(y)  =  6y^4  +  y^3   +  y^2  + 10 y  + 3

DIfferentiate the above equation with respect to 'y' .

f'(y)   =  6(4)y^3   +  3y^2    +  2y   +  10  +   0

f'(y)   =   24y^3   +  3y^2   +  2y   +  10

The answer is    f'(y)   =   24y^3   +  3y^2   +  2y   +  10

Algebra is widely used in day to day activities watch out for my forthcoming posts on Derivative Calculator and Derivative of Natural log. I am sure they will be helpful.

Tuesday, October 23, 2012

Free Algebra Classes



Algebra is a cluster of mathematics, which is used to create mathematical problems of valid-globe actions and control problems that we cannot explain using arithmetic.

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Free algebra mainly different kind of lesson handled here.

Algebra uses the cipher as calculation for addition, subtraction, multiplication and division and it includes constants, operating signs and variables.

Algebraic equations represent a collection, what is finished on one side of the range with a number to the other side of the range.

Below lesson topics explain the algebra classes.

Algebra Free Algebra Classes Topics:-

Exponents

Radicals

Polynomials

Factoring

Division of Polynomials

Solving Equations

Solving Inequalities

Lines

Solving Quadratic Equations

Complex Numbers

Graphing Quadratics (Parabolas)

Systems of Equations (2x^2's)

Systems of Equations (3x^3's)

Determinants and Cramer's Rule

Functions

Inverse Functions

Exponentials and Logarithms

Absolute Value Equations and Inequalities

Sequences and Series Combinatory

Advanced Graphing

Graphing Polynomials

Graphing Rational Functions

Matrices

Above the topics free algebra classes based process used here. And then free algebra mainly different kind of classes handle through now.

Free Algebra Classes - Problems:

Problem 1:

Solve for x: 3x+6 = 4x+8.

Solution:-

3x+6 = 4x+8

Subtract by 4x on both sides,

3x-4x+6 = 4x-4x+8

-x+6 = 8

Subtract by 6 on both sides,

-x+6-6 = 8-6

-x = 2

So, x = -2.

Order of Operations problem:-

Problem 2:

5 + 4 * 3y (x + 2)

Solution:

- 5 + 12xy + 6

Negative Exponents of Numbers:-

3 -3

Solution:- 1/27

Negative Exponents of Variables:-

K -3

Solution:- 1/ k^3

Negative Exponents in Fractions

X -2 y/x

Solution:- y/x 3

Substitution method:-

Problem:

Evaluate: w + a^2, where w = -12 and a = z + 3

-12 + (z + 3 )2 = z2 + 2(3(z)) + 3

Solution: Z2 + 6z + 9

Finding a Greatest Common Factor (GCF) of Two Terms:-

9 x^2y, 3 xy^3

3xy

Factoring a GCF from an Expression:-

4x^2 + 16x^3

4x^2(1 + 3x)

Factoring a Difference between Two Squares

x^2 – 9

(x + 3) (x – 3)

Factoring a Trinomial Completely:-

X^4 + 7x^3 + 12 x^2

x^2(x + 3) (x + 4)

Solve By Factoring

x^2 + 3x + 2

Solution:- -1, -2.


Practice Problems for Free Algebra Classes:

1. Find factors and root of the equation x^2 + 10x + 24

Answer: -6, -4

2.Find factors and root of the equation x^2 +11x + 28

Answer: -4, -7

3.Find factors and root of the equation x^2 – 2x + 8

Answer: 4, -2

4.Find factors and root of the equation 2x^2 – 15x + 8.

Answer: 1, 15

I am planning to write more post on long division polynomials, Types of Polynomials. Keep checking my blog.