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



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.

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.

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.

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.

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

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Rate Histograms Tutoring

In rate histograms tutoring the word rate is a Greek term. In rate histograms the interval with each frequency value are equal.  In mathematical the rate histogram is used to counting the number of observations in the histogram. Tutor is a person who shares his knowledge with student which is called tutoring. Now the tutor will discuss about the study histogram in this tutorial.

Example for Rate Histograms Tutoring:

Example: 1

While doing a project for P.E. class, Alonzo researched the number of athletes competing in an international sporting event. By using this data complete the histogram below.

Solution:
In the above histogram the missing bar is for the range 41 to 50 athletes and count the number of values in this range. Two countries have between 41 and 50 athletes so set the height of the bar to 2.

Example: 2

Prepare a histogram  by using the following data in the frequency table

Solution:
From the above table take the class intervals on the X-axis and frequency on the Y-axis  and mark the Y-axis scales as 0,5,10,15,20 To avoid errors in our histogram we marking the class intervals so that  the histogram be more perfect  and draw the rectangle bars .

 Example: 3

Prepare a histogram  by using the following data in the frequency table

Solution:
From the above table take the class intervals on the X-axis and frequency on the Y-axis  and mark the Y-axis scales as 0,5,10,15,20,25,30,35. To avoid errors in our histogram we marking the class intervals so that  the histogram be more perfect  and draw the rectangle bars .

These are the examples of rate histograms tutoring.

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Mathematical Information of Rate Histogram:

In the mathematical terms, the rate histograms are commonly used for displaying and mapping of mi items. These items are generally used  for counting the number of observations that are present in the histogram. The mathematical formula for calculating the rate histogram is given below,

`sum_(P=i=1)^k`

where,

p is used to represent the sum of  number of observations in the given table.

mi is used to represent the certain conditions in the table,

k is used to represent the total number of bins present in the table.

This is the general information of rate histograms tutoring in mathematics. Tutoring is one of the process. Student can easily develop their knowledge through online tutoring . Tutoring is the correlation between student and tutor.

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Solve Equations with Exponent

Introduction to solving equations with exponents

Exponent equations are the equations in which variable appear as an exponent.
To solve these equations rules and laws of exponents are used. Exponent equations are of two types
(1) Exponent equations in which bases are same
(2) Exponent equations in which bases are different.

Steps to Solve Equations with Exponent

Solving Exponential Equations of the same base

1) Ignore the bases, and simply set the exponents equal to each other

2) Solve for the variable

 When the bases of the terms are different

1) Ignore the exponents; rewrite both of the bases as powers of same number. For    example if there are 2 and 4 in the bases, then convert base 4, in to base 2 by writing it again as (2)^2

2) once the bases are same , ignore them

3) Equalize the exponents

4) Solve for variable

Simple Problems of Equations with Exponents

Solve for variable Answer 1. 3m  =  35 Since the bases are the same, set the exponents equal to one another: m = 5 2. 5t   = 125 125can be expressed as a power of 5: 5t  = 53 t = 3  3.  493y=343 49 and 343 can be expressed as a power of 7: [(7)2]3y = 73 76y = 73 6y = 3 y = 1/2

More Problems of Equations with Exponents

Solve for x. Answer 1.  52x+1  =  53x-2 Since the bases are the same, set the exponents equal to one another: 2x + 1 = 3x - 2 3 = x 2.  32x-1  = 27x  27 can be expressed as a power of 3: 32x-1  = 33x 2x - 1 = 3x -1 = x 3.   43x-8  = 162x 16 can be expressed as a power of 4: 43x-8= [(4)2]2x 3x - 8 = 4x   -8 = x I like to share this multiplying rational expressions solver with you all through my article.