Plane Distance Calculator

In a plane distance calculator we have the axes points such as (x,y) and this coordinates are placed on the quadrant, there are four quadrants for (x,y) axis. Origin is the center or starting point for the (x,y). In two dimension coordinate system we have the two x and y plane.We can measure the distance between plane by the formulas.In this article we have the formulas and the problems for finding the plane distance.

Plane Distance Calculator:

Distance between plane calculator can be measured by the following formualas and that can explained by the figur shown below. From the above figure we can clearly understand that the distance 'l' from the origin we can find the plane distance by the coordinates given.

D =  ` |d1-d2|/sqrt(A^2 + B^2 + C^2)`

distance between two plane

How to use plane distance calculator:

First seperate the d1,d2 and all othher parameters

Now, enter the parameters in the respective column

And get the result in the new column

Problems in Plane Distance Calculator:


Example 1:
Find the distance from the two plane 2x – 3y + 3z = 12 and –8x + 12y – 12z = 24.
Solution:

First seperate the d1,d2 and all othher parameters
Now, enter the parameters in the respective column
And get the result in the new column

2x – 3y + 3z = 12
–8x + 12y – 12z = 24.  by dividing equation by 4   we get 2x - 3y + 3z = -6.
First sepearte the
Formula for find the distance between the plane

D =  ` |d1-d2|/sqrt(a^2 + b^2 + c^2)`

Here, a =2, b= -3 and c=3  d1= 12 d2 =-6

= | 12 - (-6) | / √(4 + 9 + 9)

= 18/√22

Example 2:

Find the distance between the parallel planes z = x + 2y + 1 and 3x + 6y - 3z = 4.

Solution:

First seperate the a,b,c,d and all othher parameters

Now, enter the parameters in the respective column

And get the result in the new column

Formula for find the distance between the plane

D =  ` |ax_1+by_1+cz_1+d|/sqrt(a^2 + b^2 + c^2)`

Here, a =3, b= 6 and c=-3 d =-4

=  ` |3 xx 0+6xx0+(-3)xx1-4|/sqrt(3^2 + 6^2 + -3^2)`

= `7/sqrt54 `

Calculating Areas of Figures

In day-to-day life,we often came across the word area.In math, Area is nothing but a region bounded by a closed curve. In differential geometry of surfaces, area is considered as an important invariant.

Calculating areas of different figures is an important and an interesting one. In this article of  calculating areas of figures, the areas of different figures are calculated using the formulas.

Triangle:            Area of Triangle    =    ½ b h

b ---> base

h ---> vertical height

Rectangle:        Area of  Rectangle  =  l w

l ----> length

w ----> width

Square:               Area of a square   =  a2

a ----> side length

Parallelogram: Area of Parallelogram = b × h


b ----> breadth

h ----> height

Circle:                       Area of a Circle = pi r2

r ----> radius
Worked Examples for Calculating Areas of Figures:

Example 1:

Find the area of a triangle with base of  13 m and a height of 6 m.

Solution:

Area of a triangle =  ½ b h

=  ½ (13) (6)

=  39 m2

Example 2:

Find the area of rectangle given the length is 10 cm and width is 5 cm.

Solution:

Area of a Rectangle  =  l * w

= 10 * 5

= 50 cm2

Example 3:

Find the area of a square of side length 21 cm

Solution:

Area of a square  = a2

= 212

= 441 cm2

Example 4:

Find area of a parallelogram through base of 23 cm and a height of 17 cm.

Solution:

Area of a Parallelogram = b h

= (23) · (17)

= 391 cm2

Example 5:

The radius of a circle is 27 inches. Find its area.

Solution:

Area of  Circle = pi r2

= 3.14 (27)2

= 3.14 (729)

= 2289.06 in2
Practice Problems for Calculating Areas of Figures:

1) Calculate the area of rectangle given the length is 10 m and width is 7 m.

Answer: 70 m2

2) Find area of a square of side length 22 cm.

Answer: 484 cm2

3) Find the area of triangle given base is 14 cm and height is 7 cm.

Answer: 49 cm2

4) Find area of a parallelogram through base of 35 cm and a height of 15 cm.

Answer: 525 cm2

Radical Math Problems and Solutions

Radical problems and solutions are defined as one of the important topic in mathematics. Basically, there are three values are present in the radical number. Those values are named as called the index number, radical number, and the another one is known as the radicand number. For example, root(4)(12) is denoted as radical numbers. In this example of radical number, 4 is called as the index number, 12 is called as the radicand number. Mainly square root and the cubic roots are present in the radical statement.

Radical Expressions Calculator

The explanation for radical math problems and solutions are given below the following,

We can do many of the operation by using the radical. They are called as,

Addition problems and solutions by using the radical.
Subtraction problems and solutions by using the radical.
Multiplication  problems and solutions by using the radical.
Division problems and solutions by using the radical.

Example Problems and Solutions for Radical Math

Addition problems and solutions by using the radical.

Example 1: Add the following radical numbers, 12( sqrt(5) + sqrt(2) ) + 10( sqrt(2) + sqrt(5) )

Solution:

The given number is 12( sqrt(5) + sqrt(2) ) + 10( sqrt(2) + sqrt(5) )

= 12sqrt(5) + 12sqrt(2) + 10sqrt(2) + 10sqrt(5)

= 12sqrt(5)+ 10sqrt(5)  + 12sqrt(2)  + 10sqrt(2)

= 22sqrt(5) + 22sqrt(2)

This is the answer for radical numbers addition.

Subtraction problems and solutions by using the radical.

Example 2: Subtract the following radical numbers, 12( sqrt(5) + sqrt(2) ) - 10( sqrt(2) + sqrt(5) )

Solution:

The given number is 12( sqrt(5) + sqrt(2) ) - 10( sqrt(2) + sqrt(5) )

= 12sqrt(5) + 12sqrt(2) - 10sqrt(2) - 10sqrt(5)

= 12sqrt(5) - 10sqrt(5)  + 12sqrt(2)  - 10sqrt(2)

= 2sqrt(5)  + 2sqrt(2)

This is the answer for radical numbers subtraction.

Problem 3: Multiply the following radical numbers, 12( sqrt(5) + sqrt(2) ) and  10( sqrt(2) + sqrt(5) )

Solution:

12( sqrt(5) + sqrt(2) )   xx   10( sqrt(2) + sqrt(5) )

= 12sqrt(5) + 12sqrt(2) xx   10sqrt(2) + 10sqrt(5)

= 12 sqrt(10) xx  10 sqrt(10)

= 120 sqrt(100)

= 120  xx 10

= 1200

This is the answer for radical numbers multiplication.

Example 4: Divide the following radical numbers 1/(sqrt(7) - sqrt(8)) .

Solution:

1/(sqrt(7) - sqrt(8))

1/(sqrt(7) - sqrt(8))  xx  (sqrt(7) + sqrt(8))/(sqrt(7) + sqrt(8))

(sqrt(7) + sqrt(8))/(sqrt(7)^2 - sqrt(8)^2)

(sqrt(7) + sqrt(8))/(7 - 8)

(sqrt(7) + sqrt(8))/ - 1

= - (  sqrt(7) + sqrt(8) )

= - sqrt(7)  - sqrt(8)

This is the answer for radical numbers Division.
Practice Problems and Solutions for Radical Math

Example 1: Add the following radical numbers, 24( sqrt(3) + sqrt(12) ) + 12( sqrt(12) + sqrt(3) )

Answer: 36 sqrt(3)  +  36  sqrt(3)

Example 2: Subtract the following radical numbers, 24( sqrt(3) + sqrt(12) ) - 12( sqrt(12) + sqrt(3) )

Answer: 12 sqrt(3)  +   12 sqrt(3)

Percent Return Formula

In math, how much of parts done in every hundred is called as percents. The percents are represented by the symbol ‘%’. In other words, how much of value is noted out of hundred in experiments. The formula is returned with 100. Now we are going to see about percent return formula.


I like to share this Formula for Permutation with you all through my article.

Explanations for Percents Return Formula in Math

Percents return formula:

The percents are represented as fraction with percentage symbol that is 32/100%. We can denote the percents in whole number also like 32%.T he formula for returns the percents are P = ( observed value / total value) x 100.

How to return the percents using formula:

The formula for percents is divide the observed value and total value. Then multiply the 100 with that resultant value. Now, we can say this value is percents with symbol ‘%’. Sometimes, the formula returns the decimal value.

How to returns the fraction into decimal value:

We can represent the percent value in fraction and if there is any possible, we can simplify the fraction. Then divide the numerator value with denominator value.

More about Percents Returns Formula

Example problems for percents return formula in math:

Problem 1: Return the percent value using formula for given expression.

The student got the marks 140 out of 200. What is the percent value of student?

Answer:

The percent return formula is P = (observed value / total value) x 100.

The observed value is 140.

Return the percent as (140/200) x 100 = 0.7 x 100 = 70%.

Therefore, the formula returns the percent value as 70%.

Problem 2: Return the percent value using formula for given expression.

The fruit seller has 1650 apples out of 300 fruits. What is the percent value of apple?

Answer:

The percent return formula is P = (observed value / total value) x 100.

The observed value is 165.

Return the percent as (165/300) x 100 = 0.55 x 100 = 55%.

Therefore, the formula returns the percent value as 55%.

Exercise problems for percents return formula:

1. Return the percent value using formula for 65/130.

Answer: The percent value is 50.

2. Return the percent value using formula for 87/150.

Answer: The percent value is 58.

Probability of Rolling Doubles

Probability is the chance of the outcome of an event of a particular experiment. Probabilities are occurs always numbers between 0(impossible) and 1(possible). The set of all possible outcomes of a particular experiment is called as sample space. For example probability of getting a 3 when rolling a dice is ` 1/6` . In this article we will discuss about probability problems using dice.


Rules of Probability Doubles - Example Problems

Example 1: If rolling two dice, what is the probability of getting doubles?

Solution:

Let S be the sample space, n(S) = 6 * 6 = 36.

A be the event of getting doubles.

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

P(A) = `(n(A))/(n(S))` = `6/36` = `1/6`

Example 2: If rolling two dice, what is the probability of getting doubles or primes?

Solution:

Let S be the sample space, n(S) = 6 * 6 = 36.

A be the event of getting doubles.

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

P(A) = `(n(A))/(n(S))` = `6/36` = `1/6`

Let B be the event of getting primes.

n(B) = {(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5,2), (5, 6), (6, 1), (6, 5)} = 15

P(B) = `(n(B))/(n(S))` = `15/36` = `5/12`

P(A or B) = P(A) + P(B) = `1/6` + `5/12` = `7/12`

P(A or B) = `7/12`

Therefore probability of getting doubles or primes is `7/12`

Example 3: If rolling two number cubes, what is the probability of getting doubles or sum of 7?

Let S be the sample space, n(S) = 6 * 6 = 36.

A be the event of getting doubles.

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

P(A) = `(n(A))/(n(S))` = `6/36` = `1/6`

Let B be the event of getting sum of 7.

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

P(B) = `(n(B))/(n(S))` = `6/36` = `1/6`

P(A or B) = P(A) + P(B) = `1/6` + `1/6` = `1/3`

P(A or B) = `1/3`

Therefore probability of getting doubles or sum of 7 is `1/3`

Probability of Rolling Doubles - Practice Problems

Problem 1: If rolling two dice, what is the probability of getting a sum of 5 or 6?

Problem 2: If rolling two number cubes, what is the probability of getting 6 or 7?

Answer: 1) `1/4` 2) `11/36`

Solving Double Number Identities

Trigonometry is arrived from the Greek word, trigonon = triangle and metron = measure. The father of trigonometry is Hipparchus. He designed the first trigonometric table. Identity is defined as an equation that is true for all probable values of its variables. In online, many websites provide online tutoring using tutors. Double number trigonometric identities problems are easy to solve. Solving double number trigonometric identities problems are easy.  Through practice, students can learn about solving trigonometric identities. Through online, students can practice more problems on trigonometric identities. In this topic, we are going to see about; solving double number identities.

Solving Double Number Identities: - Double Number Identities

The list of double number identities are given below,

sin `2theta` = 2sin `theta` cos `theta `

cos `2theta` = 2cos2 `theta` -1

cos `2theta` = 1- 2sin2 `theta`

cos `2theta` = cos2 `theta` – sin2 `theta`


Solving Double Number Identities: - Examples

Example 1:

Evaluate sin 60 using double number identities,

Solution:

sin 2theta = 2sin theta cos theta

sin 60 = (2x30)

= 2 sin 30 cos 30

= 2 (0.5) (0.866)

= 2*0.433

= 0.866

The answer is 0.866



Example 2:

Evaluate sin 90 using double number identities,

Solution:

sin 2theta = 2sin theta cos theta

sin 90 = (2x45)

= 2 sin 45 cos 45

= 2 (0.707) (0.707)

= 2*0.499

= 1

The answer is 1



Example 3:

Evaluate cos 50 using double number identities,

Solution:

cos 2theta = cos2 theta – sin2 theta

cos 50 = cos (2x25)

= cos2 25 - sin2 25

= (0.906)2 - (0.423)2

= 0.820 - 0.179

= 0.641

The answer is 0.641

Example 4:

Evaluate cos 90 using double number identities,

Solution:

cos 2theta = cos2 theta – sin2 theta

cos 90 = cos (2x45)

= cos2 45 - sin2 45

= (0.707)2 - (0.707)2

= 0.499-0.499

= 0

The answer is 0



Example 5:

Find cos 2y if sin y = -15/16 and in 3rd quadrant

Solution:

It is given that sin 2y is in 3rd quadrant,

Use the double angle identities

cos 2theta = 1- 2sin2 theta

cos 2y = 1 -2sin2 y

= 1 – 2`(-15/16)` 2

= 1 – 2 `(225/256)`

Taking LCM, we get

= `(256-450)/256`

= `-194/256`

= -97/128

The answer is `-97/128`

Sigma Algebra Examples

In mathematics, an σ-algebra is a technological concept for a group of sets satisfy certain properties. The main advantage of σ-algebras is in the meaning of measures; particularly, an σ-algebra is the group of sets over which a measure is distinct. This concept is important in mathematical analysis as the base for probability theory, where it is construed as the group of procedures which can be allocated probabilities. Now we will see the properties and examples.
Properties - Sigma Algebra Examples

Take  A  be some set, and 2Aits power set. Then a subset Σ ⊂ 2A is known as the σ-algebra if it satisfies the following three properties:

Σ is non-empty: There is as a minimum one X ⊂ A in Σ.
Σ is closed below complementation: If X is in Σ, then so is its complement, A \ X.
Σ is closed under countable unions: If X1, X2, X3, ... are in Σ, then so is X = X1 ∪ X2 ∪ X3 ∪ … .

Eg:

Thus, if X = {w, x, y, z}, one possible sigma algebra on X is

Σ = { ∅, {w, x}, {y, z}, {w, x, y, z} }.
Examples - Sigma Algebra

Example 1

X={1,2,3,4}. What is the sigma algebra on X?

Solution:

Given set is X={1,2,3,4}

So Σ = { ∅, {1,2}, {3,4}, {1,2,3,4}}.

Example 2

What is the sigma algebra for the following set ? X={2,4,5,9,10,12}

Solution:

Given set is X={2,4,5,9,10,12}

So   Σ = { ∅, {2,4}, {5,9}, {10,12},{2,4,5,9,10,12}}.

Example 3

11x+2y+5x+12a. Simplify the given equation in basic algebra.

Solution:

The given equation is  11x+2y+5x+12a

There are two related groups are available. So join the groups.

The new equation is,

(11x+5 x)+2y+12a

Add the numbers inside the bracket. We get 16x+2y+12a.

Arrange the numbers and we get the correct format.

=12a+16x+2y

We can divide the equation by 2.

So the equation 6a+8x+y.

These are the examples for sigma algebra.