Parabola Math Problems

A parabola  is the set of all the points whose distance from a fixed point in the plane are equal  to their distances  from a fixed line in the plane.  The fixed point is called the focus and the fixed line is the directrix.

.directrix
Formula  for parabola ;-
(1) y2 = 4px  has symmetry around  x axis Directrix x + p = 0  Latus rectum  LL' is 4p vertex = (0,0)
(2) y2 = -4px has symmetry around the negative side of x-axis Directrix  x - p = 0 . LL' = 4p. Vertex =(0,0)
(3) x2 = 4 py has symmetry around the  positive y - axis Directrix y + p = 0.  LL' = 4p. Vertex =(0,0)
(4) x2 = -4py has symmetry around the negative y - axis. Directrix y - p = 0.  LL' = 4p. Vertex= (0,0)

Problems on Parabola:
Here is a parabola math problem for us  to do.
# Find the focus , vertex, axes, directrix  of the parabola y2 - 4x - 4y = 0
Sol:-
parabola
Step 1 : Let us bring the above given parabola equation to the parabola formula equation.
y2- 4x - 4y = 0
Step 2 : Add 4x to both sides + 4x             + 4x
We get                 y2 - 4y       =    4x
Step 3 :Complete the square on LHS we write y2 - 4y +4  = 4x + 4 ( we add 4 to both sides of the equation to balance it)
Step 4 : Now we get                                      (y  - 2 )2    = 4(x+1)
Step 5 : Let us compare the above equation to the general equation Y2 = 4 pX
we get Y = y-2  and X = x+1. Hence Y2 = (y -2)2  and 4px = 4(x+1) . Hence we get p = 1
Step 6 : Now let us set a table for both the values. Now x = X - 1 and  y = Y + 2
____________________________________________________________________________
Referred to X and Y axes                             Referred to x and y axes
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focus S = (p,0)                                          x = X - 1 = 1 -1 = 0; y = Y +2= 0 + 2 = 2
So focus is (0,2)
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vertex is (0,0)                                            x = X - 1 => 0 - 1 = -1 and y = Y + 2 = 0+2 =2
So vertex = (-1,2)
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axes Axis  of parabola Y =0                       Axis of the parabola is y = Y+2 = 0+2 =2
so axis  is y - 2 = 0
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equation of the directrix is                          equation of the directrix is  x + 1 = -1
x = -p that means X = - 1                           which gives us  x + 2 = 0
____________________________________________________________________________
Hence for the given parabola  y2 -4x - 4y = 0 we have (1)  Focus  as (0,2)  ;
(2)  Vertex is (-1,2)
(3)  Axis of the parabola is y = 2
and     (4) directrix is x + 2 = 0

Condition for a Line to be a Tangent for a Parabola:-
Let us find the condition for a line  lx + my + n = 0 to be a tangent to the parabola y2 = 4ax
Solution:-
The condition that a  line y = mx + c  to be a tangent to the parabola y2 = 4ax is that C = a/m...(1)
Step 1 : Let us write lx + my + n = 0 in the form y = mx + c
That gives us my = -lx - n
y =  [ -l]x + [ -n]
m        m
Step 2 : Now we have  y = mx + c   => y = (-l/m) x  + ( -n/m)  That is slope = -l/m and  y intercept C = -n/m
Hence  C =  -n/m  and m(slope) = (-l/m)
Step 3   But  C =  a / m (formula)
So we write  -n  =    a =   - am
m    (-l/m)       l
Step 4   Hence we get - n =  -am
m      l
Step 5 : Cross multiplying we get  -nl = -am2
Step 6 : That gives us nl = am2
Hence the condition that lx + my + n= 0 to be a tangent to the parabola y2  = 4ax is  nl = am2

Geometric Constructions Tutorial

Geometric constructions is a basic construction of geometric figures. In this article we can see the construction of a perpendicular bisector, construction of an angle and construction of a triangle. For the construction of geometric figures we use only the help of a ruler (straight-edge) and compass. Compass is generally used to draw arcs of particular distances. The distance is calculated with the help of a ruler.
Geometric Constructions

i) To bisect a given line segment

Given: A line segment AB

To construct: The perpendicular bisector AB

Construction:      a) With A as centre and radius equal to more than half of AB, draw two arcs, one on each side of AB.

b) With B as centre and with the same radius taken above draw two more arcs cutting the previous arcs                              at C and D

c) Join C and D cutting AB at O. CD is the perpendicular bisector of AB.

perpendicular bisector of a line segment

ii) To construct an angle of 60°

Construction: a) Draw a line segment AB of any suitable length.

b) With A as centre and any suitable radius draw an arc cutting AB at C.

c) With C as centre and the same radius AC, draw another arc cutting the former arc at D.

d) Join A and D and produce it to E.
construction of angle 60degrees

iii) To construct an angle of 30°

Construction: a)First construct an angle of 60° as shown above

b) With C and D as centre, draw two arcs of equal radii cutting each other at F

c) Join A to F.
construction of angle 30degrees

Construction of more Geometrical Shapes

iv) To construct a triangle having given the lengths of three sides.

Let the three sides be 3.6cm, 4cm and 5.1cm

Construction: a) Draw a line segment AB = 3.6cm

b) With A as centre and radius = 4cm, draw an arc.

c) With B as centre and radius = 5.1cm draw one more arc intersecting the former arc at C.

d) Join AC and BC. Then ΔABC is the required triangle.(Note: The construction fails when the sum of any two sides is less than the third side.)

Algebra is widely used in day to day activities watch out for my forthcoming posts on Combinations and Permutations Examples and Continuity of a Function. I am sure they will be helpful.

v) To construct a triangle, having given the lengths of any two sides and the included angle.

Let the length of the two sides be 3.9cm and 4.3cm and included angle be 60°

Construction:a) Draw a line segment AN = 3.9cm

b) Construct
c) From AX, cut off AC = 4.3cm

d) Join BC. Then ΔABC is the required triangle.

construction of a triangle, where two sides and included angle are given

Difficult Logic Problems

Problem: - Which term will replace the question mark in the series:

ABD, DGK, HMS, MTB, SBL,?

Solution: Clearly, the first letters of the first, second, third, fourth, and fifth terms are moved three, four, five, six and seven steps forward respectively to obtain the first letter of the successive terms. The second letters of the first, second, third, fourth and fifth terms are moved five, six, seven, eight and nine steps forward respectively to obtain the second letter of the successive terms. The third letters of the first, second, third, fourth and fifth terms are moved seven, eight, nine, ten and eleven steps forward respectively to obtain the third letter of the successive terms.

Thus the missing term is ZKW             (Answer)
Difficult Logic Problems next Set

Problem: - A child is looking for his father. He went 90 meters in the east before turning to his right. He went 20 meters before turning to his right again to look for his father at his uncle’s place 30 meter from this point. His father was not there. From there, he went 100 meters to his north before meeting his father in a street. How far did the son meet his father from the starting point?

Solution: - Clearly the child moves from A 90m eastwards up to B, then turns right and moves 20m up to C, then turns right and moves 30m up to D. Finally, he turns right and moves 100m up to E.

difficult logic problem

Clearly, AB = 90m, BF = CD = 30m.

So, AF = AB- BF = 60m

Also, DE = 100m, DF = BC = 20m

So, EF = DE- DF = 80m.

Therefore, his distance from starting point A = AE = `sqrt[ (AF)^2 +(EF)^2]`  = `sqrt[(60)^2 + (80)^2]`

= `sqrt(3600 + 6400)` = `sqrt10000` = 100m    (Answer)

Problem: - Each odd digit in the number 5263187 is substituted by the next higher digit and each even digit is substituted by the previous lower digit and the digits so obtained are rearranged in the ascending order, which of the following will be the third digit from the left end after the rearrangement?

Solution: - After performing operation on the digit we get 6154278

Arranging the above number in ascending order we get 1245678

Here third digit from the left end is 4.                (Answer)
More Difficult Logic Problems

Problem: - In a certain code TEMPORAL is written as OLDSMBSP. How is CONSIDER written in that code?            (Answer: RMNBSFEJ)

Problem: - In a certain code language ‘how many goals scored’ is written as ‘5 3 9 7’; ‘many more matches’ is written as ‘9 8 2’ and ‘he scored five’ is written as ‘1 6 3’. How is ‘goals’ written in that code language?  (Answer: either 5 or 7)

Problem: - Reaching the place of meeting on Tuesday 15 minutes before 08.30 hours, Jack found himself half an hour earlier than the man who was 40 minutes late. What was the scheduled time of the meeting? (Answer: 8.05 hrs)

Number of Diagonals in a Pentagon

Polygon is any shape, which is enclosed by its sides. The line segments that connect any two vertices are called as diagonals. The names of the polygon are classified on the basis of their number of sides. Any polygons which possess five sides are called as pentagon. In this article, we shall discuss about the number of diagonals of pentagon. Also we shall solve problems regarding number of diagonals of pentagon.

Formula - Number of Diagonals of Pentagon:

The number of diagonals of any polygon can be obtained by using the formula,

 [n(n-3)]/2

where n is the number of sides of the polygon.

diagonals of pentagon

Now we shall determine the number of diagonals of pentagon.

The number of sides of a pentagon is five.

Therefore n = 5.

By substituting n = 5 in the above formula, we can determine the number of diagonals of pentagon.

Number of diagonals of pentagon = [5(5-3)]/2

= [5(2)]/2

= 10/2

= 5

Therefore the number of diagonals of pentagon is 5.


Example Problem - Number of Diagonals in a Pentagon:

Determine the number of sides and name of the polygon whose number of diagonals is 5.

Solution:

Given:

The number of Diagonals = 5

The formula to determine the number of diagonals is

[n(n-3)]/2 ,

where n is the number of sides of the polygon.

We have to find n:

Equate the both.

[n(n-3)]/2 = 5

Multiply by 2 on both sides:

2[(n(n-3))/2] = 2(5)

n(n - 3) = 10

Multiply by n within the bracket:

n2 - 3n = 10

Subtract 10 on both sides.

n2 - 3n - 10 = 0

n2 - 5n + 2n - 10 = 0

n(n - 5) + 2 (n - 5) = 0

(n - 5) (n + 2) = 0

n - 5 = 0 and n + 2 = 0

n = 5 and n = -2

The sides of polygon must not be negative. So ignore -2.

Therefore the number of sides of the polygon is 5.

Pentagon is the polygon which possess 5 sides.

Hence pentagon is the polygon which possess 5 sides and 5 diagonals in it.

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.

Generating Function of Exponential Distribution

In mathematics, generating function of exponential distribution is one of the most interesting distributions in probability theory and statistics. The random variable X has an exponential distribution along with the parameter λ, λ> 0. If its probability density function is given by

f(x) = lambdae^(-lambdax) ,         x >= 0

Otherwise, f(x) = 0, x < 0. The following are the moment generating functions and example problems in generating function of exponential distribution.

Generating Function of Exponential Distribution - Generating Function:

Generating function (or) Moment generating function:

Moment generating function of exponential distribution function can be referred as some random variable which contains the other definition for probability distribution. Moment generating function give the alternate way to analytical outcome compare and also it can be straightly working with the cumulative and probability density functions. Moment generating function can be denoted as Mx(t) (or) E(etx). Here we help to calculate the moment generating function for the exponential distribution function.

Mx(t) = E(etx) =int_-oo^ooe^(tx)f(x)dx

= int_0^ooe^(tx)(lambda e^(-lambdax))dx

= λ int_0^ooe^(tx)(e^(-lambdax))dx

= λ int_0^ooe^((t-lambda)x)dx

= λ [e^(((t- lambda)x)/(t- lambda))]^oo_0

= λ [e^(((t- lambda)oo)/(t- lambda)) - e^(((t- lambda)0)/(t- lambda)) ]

Here we use this eoo = 0, and also we use this e0 = 1

= λ [ 0 – 1/(t- lambda) ]

= λ [ -(1/-( lambda-t)) ]

= lambda [1/ (lambda-t)]

= lambda/( lambda-t)

Mx (t) = E (etx) = lambda/(lambda-t)

Generating Function of Exponential Distribution - Example Problem:

Example 1:

If X has probability density function f(x) = e-5x, x > 0. Determine the moment generating function E[g(x)], if g(x) = e11x/15

Solution:

Given

f(x) = e-5x

g(x) = e11x/15

E(g(x)] = E[e11x/15]

= int_0^ooe^((11x)/15)f(x)dx

=  int_0^ooe^((11x)/15) (e^(-5x))dx

= int_0^ooe^(-x(5-(11/15)))dx

= [e^((-x(5-11/15))/-(5-11/15))]^oo_0

=[(e^-oo- (e^0)/-(5-11/15)) ]

= [0+ 1/(5-11/15) ]

= [1/((75-11)/15) ]

= [1/(64/15) ]

= 15/64

E [g(x)] = 0.2344

Answer:

E [g(x)] = 0.2344

Pre Algebra Geometry

Geometry is one of a division of mathematics. Geometry is concerned with questions of size, shape, relative position of figures, and the properties of space. Geometry is one of the oldest sciences. Initially the body of the practical knowledge is concerning with the lengths, areas, and volumes. In this article we shall discuss about pre algebra geometry problem. (Source: wikipedia)
Sample Problem for Pre Algebra Geometry:

Problem 1:

Find the diameter of the given circle. A given radius value of the circle is 16 cm.

Solution:

Given:

Radius of the circle is r = 16 cm

Diameter of the circle is d = 2r

d = 2 * 16

d = 32

So, the diameter of the circle is 32 cm2.

Problem 2:

Find the diameter of the circle. A given radius value of the circle is 20 cm.

Solution:

Given:

Radius of the circle is r = 20 cm

Diameter of the circle is d = 2r

d = 2 * 20

d = 40

So, the diameter of the circle is 40 cm2.

Problem 3:

A triangle has a perimeter of 34. If the two sides are equal and the third side is 4 more than the equivalent sides, what is the length of the third side?

Solution:

Let x = length of the equal side of a triangle

A formula for perimeter of triangle is

P = sum of the three sides of triangle

Plug in the values from the question

34 = x + x + x+ 4

Combine like terms

34 = 3x + 4

Isolate variable x

3x = 34 – 4

3x = 30

x = 10

The question requires length of the third side.

The length of third side is = 10 + 4 = 14

Answer: The length of third side triangle is 14.
Practice Problem for Pre Algebra Geometry:

Find the diameter of the circle. A given radius value of the circle is 14 cm.

Answer: d = 28 cm2

Find the radius of the circle. A given diameter value of the circle is 11 cm2.

Answer: 5.5 cm

Solving Word Percent Problems

In mathematics, a percentage is a way of expressing a number as a fraction of 100. It is often denoted using the percent sign, "%", or the abbreviation "pct". Percentages are used to express how large/small one quantity is, relative to another quantity. The first quantity usually represents a part of, or a change in, the second quantity, which should be greater than zero. Source: Wikipedia.
Solving Word Percent Problems

Solving percent problems - Example

Example 1: What percent of 30 is 60?

Solution:

30 × `x/100` = 60

30x = 6000

x = 200

Therefore 60 is 200 percent of 30.

Example 2: What is 45% of 250?

Solution:

`45/100`  × 250 = x

45 × 250 = 100x

x = `11250/100` = 112.5

Therefore 45 percent of 250 is 112.5.

Example 3:  25% of what is 15?

Solution:

`25/100` × (x) = 15

25x = 1500

x = 60

Therefore 25 percent of 60 is 15.

Example 4: What is the percentage increase from 14 to 24?

Solution:

Increase = 24 - 14 = 10

Percentage increase = (Change in value / Original value) × 100%

Percentage increase = `10/14` ×100% = 71.42%

Therefore 71.42% increase from 14 to 24.

Example 5: Last month, Anita earned $300. This month she earned $450. Calculate the percentage increase in her earnings.

Solution:

Increase = $450 - $300 = $150

Percentage increase = (Change in value / Original value) × 100%

Percentage increase = `150/300` × 100% = 50%

Therefore percentage increase in her earnings is 50%.

Example 6: What is the percentage decrease from 30 to 14?

Solution:

Decrease = 30 – 14 = 16

Percentage decrease = (Change in value / Original value) × 100%

Percentage decrease = `16/30` × 100% = 53.33%

Therefore 53.33% decrease from 30 to 14.

Example 7: Last month, Anita earned $400. This month she earned $340. Calculate the percentage decrease in her earnings.

Solution:

Decrease = $400 - $340 = $60

Percentage decrease = Change in value / Original value × 100%

Percentage decrease = `60/400` × 100% = 15%

The percentage decrease in her earnings is 15%.
Solving Word Percent Problems

Solving percent problems - Practice

Problem 1: What is 28% of 150?

Problem 2: Harry earned $270 in last month. He earned $370 this month. Calculate the percentage increase in his earnings.

Problem 3: Wilson earned $340 in last month. He earned $250 in this month. Calculate the percentage decrease in his earnings.

Answer: 1) 42 2) 37.03 3) 26.47

Number of Sides in a Pentagon

In geometry, a pentagon is a polygon with five sides. In a simple pentagon the sum of internal angles are about 540°. For example pentagram is a self-intersecting pentagon. Pentagon may be classified into regular and irregular. A pentagon that contains equal sides and equal internal angles are said to be regular pentagon otherwise the pentagon is irregular.pentagon


Number of Sides in a Pentagon:

The term penta indicates 5 .Hence the number of sides in a pentagon are 5 and the number of angles in a pentagon is 5.

A pentagon contains 712.694 million separate parallel lines.

Pentagon doesn’t have parallel lines.

Area of the pentagon = (perimeter of a polygon x apothem) ÷2

Where perimeter of a polygon = 5 x side.

The following are the other polygonal shapes with their sides.

Tetragon - 4 sides

Hexagon- 6 sides

Heptagon- 7 sides

Octagon- 8 sides

Nonagon Enneagon- 9 sides

Decagon-10 sides

Undecagon- 11 sides

Dodecagon- 12 sides

Properties of pentagon:

Number of diagonals:

Number of diagonals in a pentagon is 5

The number of different diagonals possible from all vertices.

Number of triangles:

Number of triangles in a pentagon is 10.

The number of triangles formed by sketching the diagonals from a given vertex.

Sum of interior angles:

Sum of interior angles of a pentagon is 540° in general 180(n–2) degrees.

Example Problem- Number of Sides in a Pentagon:

Example 1:

Find the perimeter of a regular pentagon whose side is 5ft.

Solution:’

Given that, side = 5ft.

For a regular pentagon all the sides are equal.

Therefore the perimeter of a regular pentagon = 5 x side.

= 5 x 5 =25ft.

Example 2:

Find area, from the apothem and the perimeter of a polygon is 4ft and 20ft.

Solution:

Given that, apothem = 4ft.

Perimeter = 20ft

Area of the pentagon = (perimeter of a polygon x apothem) ÷2

= (20 x 4) ÷2.

= 10 x 4.

= 40ft2

Boolean Algebra

Boolean Algebra is a branch of mathematic logics  whose use of symbols and  theory, set to represent the logical operations in the form of mathematics. This is the first logic which uses algebra and different methods for combining symbols used in proofs as well as deduction.

A Boolean Algebra is defined as:

It is  a set, having two special elements i.e, 0 and 1.
Algebra having three types of operations , which are
sum of two elements ("+"),
product sum of two elements  ("*") and
complement sum of two elements (" ' " or "prime")

these operations need to satisfy the Commutative axiom, Distributive axiom , Identity axiom (not including the boundedness identities) and Complement axiom.

These above axioms are almost equal to commutative property ,distributive property, identity property and complement property. Here we call them axioms because they are assumptions.

Boolean Algebra contains:

A  set of all propositions
The special characteristic elements - True  (1) i.e T  and False (0) i.e, F.
Three operations are
AND (product),
OR (sum) and
NOT (complement).

Laws of Boolean Algebra Axioms

To do any kind of operations using real numbers, they  depends on commutative axiom, associative axiom, and distributive axiom. In algebraic form these axioms  are expressed with letters or symbols, which are used to indicate an unknown number.

Commutative axiom

The commutative axioms explains that, numbers can be used for addition or multiplication in any manner.

Commutative axiom of Addition:

a + b = b + a        ( using addition law )

Commutative axiom of Multiplication:

a(b) = b(a)           ( using multiplication law )

Associative axiom

The associative axioms explain that, numbers which are used in addition or multiplication, also it can be grouped or regrouped in anyorder.

Associative Law of Addition:

a+(b+c) = (a+b)+c      ( using addition law )

Associative Law of Multiplication:

a(bc) = (ab)c                ( using multiplication law )

Distributive axiom

The distributive axioms are used for  both addition as well as  multiplication and state the following.

Distributive axiom for addition :

a(b + c) = ab + ac        ( using addition law )

Distributive axiom for multiplication :

(a + b)c = ac + bc             ( using multiplication law )

Identity axiom

Identity axiom for multiplication :

x · x = x                 ( using multiplication law )

Identity axiom for addition :

x + x = x                ( using addition law )



Zero Property in Boolean algebra axioms

0 · x = 0                  ( using multiplication law )

0 + x = x                   ( using addition law )

One Property  in Boolean algebra axioms

1 + x = 1                  ( using addition law )

1. x = x                     ( using multiplication law )
Examples on Boolean Algebra:

1)  `5xx(8+9)` = `5xx8 + 5xx9` ( USING DISTRIBUTIVE AXIOM  FOR MULTIPLICATION )

= `40 + 45`

=` 85`

= 5 x 8+5 x 9

2)  `3+7 = 7+3` ( BY USING COMMUTATIVE AXIOM FOR ADDITION)

3) `9xx5 = 5xx9 (` BY USING THE COMMUTATIVE AXIOM FOR MULTIPLICATION)

4)` 6xx1 = 6 ` (BY USING PROPERTY FOR ONE)
Practice Problems on Boolean Algebra:

1) Prove that  C+(A×B)=(C+A)×(C+B) by using Boolean algebra axioms

2) Prove that B+(C×A)=(B+C)×(B+A) by using Boolean algebra axioms

Practical uses for Algebra

Algebra been used in all parts of our life activities either directly or indirectly.  If we want to find some interest which may recur with amount, we can apply progressions.  To see some different arrangements or combinations between available sources, we apply permutation and combination. To prove a generalized statement, we can apply mathematical Induction so on.

Let us see few problems of this kind.
Example Problem on Practical Uses for Algebra:

Ex 1: Peter buys a truck for dollar 120,000.  He pays half of the amount by cash. That is, dollar 60,000.  The remainingdollar 60,000, he pays in 12 annual installments of dollar 5000 each.  If the rate of interest is 12% and he pays with the installment the interest due on the unpaid amount, find the total cost of the truck.

Sol:  First installment = 5000 + 12% of 60,000

= 5000 + `12/100 xx` 60000

= $ 12,200.

Second Installment = 5000 + 12% of 55,000   = 5000 + `12 /100 xx` 55,000

= $ 11600.

Third Installment   = 5000 + `12/100 xx` 50,000

= $ 11,000

Since the common difference of the amount is $600,

Total cost of the shop = 60,000 + sum of the 12 installments.

= 60,000 + `12/2` [2` xx` 12,200 + (12 – 1) (- 600)]

= 60,000 + 6 [24,400 – 6,600]

= $ 1, 66,800.
More Example Problem on Practical Uses for Algebra:

2. A father is three times a old as his son.  In 12 years time, he will be twice as old as his son.  Find the present ages of father and son.

Solution: Let x and y be the present ages of father and his son respectively.

Given:  x = 3y `implies` x – 3 y = 0 -------- (1)

Also, x + 12 = 2 (y + 12)` implies` x + 12 = 2y + 24

`implies` x – 2y = 12 -------------------------- (2)

Therefore (1) – (2) `implies` x – 3y = 0

- x + 2y = - 12

`implies` - y = - 12 `implies` y = 12.

Therefore (1)` implies` x – 3 (12) = 0

`implies` x = 36.

Therefore the age of the father is 36 and his son’s age is 12.

Steps to Solving Quadratic Equations

In mathematics, an equation with second degree and one variable is called quadratic equation. The general equation for quadratic equation is given by,

A x2 + B x + C =0

Here x represents the variable and A, B, and C are constants, with a ≠ 0. (When a = 0, the equation is named as linear equation.)

The constant A, B, and C are expressed as quadratic coefficient, linear coefficient and the constant expression or release expression.

Steps to Solving Quadratic Equations Example 1:

5x² - x – 6 = 0, solving the factor for the given quadratic equation.

Solution:

Now, we can find the factor for the given quadratic equation

5x2 + 5x - 6x - 6 = 0

Now get the value for 5x from the primary term and 6 from secondary term.

5x (x + 1) - 6(x + 1) = 0

Now we can combine the similar term (x +1)

(5x - 6) (x + 1) = 0

To get the value for x we can associate the factor to zero

x + 1 = 0   or   5x – 6 = 0

x = - 1        or   5x = 6

x = 6 / 5

Thus, the factors  x1 and x2 are -1, 6/5.
Steps to solving quadratic equations Example 2:

3x² - 6 = -3x solving the factor for the given quadratic equation and solving the sum and product of quadratic equation.

Solution:

Fetch the -3x over: 3x² + 3x - 6 = 0

Separate 3x as 6x and -3x,

3x² + 6x - 3x - 6 = 0

Take out 3x from first two terms and 3 as common from next two terms

3x(x + 2) – 3(x + 2)  = 0

Thus, the factors are: (3x - 3) (x + 2) = 0

Modulate both expressions to zero: 3x - 3 = 0 and x + 2 = 0

3x - 3 = 0            x + 2 = 0

3x = 3                   x = - 2

x =  3/3

x = 1

So, the factors for x are  1, -2

Sum of the roots:

To find sum of roots consider the factor as x1 and x2

The sum of the roots  = x1 + x2 = (1) + (-2)

Sum of the roots = -1

Product of the roots:

To find product of roots consider the factors as x1 and x2

The product of the roots is given by x1x2 = (1)(-2)

Product of roots = -2


Quadratic Equation

ax2 +bx + c = 0


Values of

‘a’, ‘b’ and ‘c’


One Root

(x1)


Other Root

(x2)


Sum of Roots

(x1 + x2)


Product of Roots

(x1x2)

3x² +3x - 6 = 0


a = 3, b = 3, c = -6


1


-2


-1


-2

Steps to Solving Quadratic Equation Example 3:

Solving the value for x, to the given quadratic equation x2 + 5x + 4 = 0.

Solution:

Steps 1: To find the factor for the given quadratic equation, find the multiplicative value for the 4 and the sum of root value for 5

x2 + 1x + 4x + 4 = 0

Steps 2: Now obtain x as similar from the primary term and 4 as similar from last two term.

x (x +1) + 4(x + 1) = 0

Steps 3: Now we can unite similar term (x +1)

(x + 1) (x + 4) = 0

Steps 4: To obtain the value of x we can associate the factor to zero

x + 1 = 0   or   x + 4 = 0

x = - 1        or   x = -4.

Steps 5: Thus, the factors are x are -1, - 4.

Solve Algebra Squared Fraction

Fractions:

In algebra, A certain part of the whole is called as fractions. The fractions can be denoted as a/b , Where a, b are integers. We can multiply two or more fractions. There are three types of fractions in math,

1) Proper fractions

2) Improper fractions

3) Mixed fractions

In this article we are going to see how to solve algebra squared fraction and some example solved problems on algebra squared fraction.

Solve Algebra Squared Fraction :

Steps to solve algebra squared fraction:

Step 1: We know that ( a/b)^n = a^n / b^n . So we can take square for the numerator and denominator.

Step 2: If it is possible we can simplify it.

Let us see the example problems.
Example Problems on Solve Algebra Squared Fraction :

Problem 1:

Solve (4/5)^2

Solution:

Given , (4/5)^2

We need to squared the given fraction .

(4/5)^2

We know that ( a/b)^n  = a^n / b^n .

We can apply the above rule,

(4/5)^2 = 4^2 / 5^2

= 16 / 25

Answer: (4/5)^2 = 16 / 25

Problem 2:

Solve (3x / 4xy) ^ 2

Solution:

Given, (3x / 4xy) ^ 2

We need to find the squared the given fraction,

We know that (a/b)^n = a^n / b^n .

(3x / 4xy) ^ 2 = ((3x) ^ 2) / ( (4xy)^2)

= (9x^2) / ( 16x^2y^2)

now we can simplify it,

Divide by x^2 on both numerator and denominator,

(9x^2) / ( 16x^2y^2) = (9x^2)/x^2 / ( 16x^2y^2) / x^2

= 9 / 16y^2

Answer: (3x / 4xy) ^ 2   = 9 / 16y^2

Problem 3:

Solve (( x^2 - 4 ) / ( x + 2) )^2

Solution :

Given , (( x^2 - 4 ) / ( x + 2) )^2

We need to find the squared the given fraction,

We know that ( a/b)^n = a^n / b^n .

(( x^2 - 4 ) / ( x + 2) )^2 = ( x^2 - 4 )^ 2 / ( x + 2)^2

Before that we can simplify the given fraction,

(( x^2 - 4 ) / ( x + 2) )^2 = (((x+2)(x-2))/ ( x + 2))^ 2

= ((x - 2)^2)

= x^2 - 2x + 4

Answer: (( x^2 - 4 ) / ( x + 2) )^2 = x^2 - 2x + 4

Results and Discussion Section

Here we are going to see about the result and discussion, whether the solution to the given equation is correct or not. By simplifying the equation we find the value for the particular variable. when we substitute the value in the equation the equation balances on both sides.

For example: x – 4 = 0.

Here    x – 4 = 0

Add 4 on both sides. We get

x = 4

The solution is x = 4

According to the topic we have to check whether x = 4 is a solution to the given equation or not. By substituting the value the in equation we can see that both sides have same value.
Example Problem for Result and Discussion Section:

Given equation is discussion section  x + 7x + 5x + 3x – 5 + x = 29

Solution:

Given question is x + 7x + 5x + 3x – 5 + x = 29

Step 1:

Arrange the given question as per the x term and the numbers

x+ x + 3x + 5x + 7x – 5 = 29

Step 2:

Add the x term first

2x + 3x + 5x + 7x – 5 = 29

Step 3:

Add all the  ' x ' term in the equation

17x – 5 = 29

Step 4:

Add both sides by 5.Then we get

17x = 29 + 5

Step 5:

17 x = 34

Divide by 17 on both sides.

`(17x)/17` = `34/17`

x = 2

Discussion section about the result:

The result is 2. Here we are going to discuss about the result for given equation, whether the solution x =2 is correct or not.

Given question is x + 7x + 5x + 3x – 5 + x = 29

The result is x =2

Here we need to substitute the x value in the equation if that both side value is equal means the solution is correct. Otherwise the solution is not correct.

x + 7x + 5x + 3x – 5 + x = 29

Substitute the value x = 2 in the given equation

2 + 7( 2) + 5( 2) + 3(2) – 5 + 2 = 29

2 + 14 + 10 + 6 – 5 + 2 = 29

34 – 5 = 29

29 = 29

Both the sides are equal, so given solution is correct for the equation.
One more Example Problem for Result and Discussion Section:

Solve discussion section  16x - 12 = 20

Solution:

Add 12 on both sides. we get

16x - 12 + 12 = 20 + 12

16x  = 32

Divide by 16 on both sides.

`(16x )/ 16` = `32 / 16`

x = 2

whe we substitute x =2 in the equation 16x - 12 = 20. we get

16(2) - 12 = 20

32 - 12 = 20

20 = 20.

Thus the equation satisfies.

Raw Score Formula

Raw score is defined as the unique, unchanged transformation or calculation. When we bring together the information, in which the numbers we get hold of are known as raw score or raw data. Raw score can able to find the precise location of each data in a given distribution.

percentage of sampling error = (value of statistic-value of parameter)*100.

In this article we are going to see about raw score formula.
Raw Score Formula:

The raw score formula for variance can be given as follows,

Raw score formula

Where `sigma` and X indicates mean.
Raw Score Formula Related to Z Score Formula:

Z-score formula can be given as follows

Z = (x) - (X) / (S.D)

where x indicates a data value

X indicates mean (average of the given values)
S.D indicates standard deviation.

Using the formula of Z score we can find the raw score formula

But to find the raw score formula we have to solve this formula for x

hence the raw score formula for x can be given as follows


Z* (S.D) + X= x

Where x indicates a data value or raw score

X indicates mean (average of the given values)
S.D indicates standard deviation.

Example problem for raw score formula:

Let us take an example where the mean value X has to be 280 and the standard deviation is let to be 10.The data values are given as follows x1 = 300 and x2 = 285.Find the raw score value using raw score formula.

Solution:

Z = [(300) - (280)] / (10)


Z1= 2

Z= [(285) - (280)]/ (10)



Z2 = .5
Finding Raw Score using raw score formula.

Z = 2.2 hence


2.2 = (x-280)/10

(2.2) * 10 + 280 = x

x = 302

Z = 1.1

1.1 = (x- 280) /10.

(1.1) * 10 + 280 = x

x = 291

Solve a Parabola with Fraction

Parabola is a half circle and a curve formed by an intersection of right circular cone. A set of all points are same distance from a fixed line is called as directrix and also fixed point called as focus not in the directrix. The midpoint between the directrix and focus of the parabola is called as vertex and the line passing through the vertex and focus is called as axis. Let us see about how to get the value of vertex, latus rectum, focus and axis of symmetry in parabola equation. Let us see about solve a parabola with fraction in this article

The general form the parabolic curve is y = ax^(2) + bx + c or  y^(2) = 4ax . Substitute the above formula to find the vertices, latus rectum, focus and axis of symmetry.

Example 1 for Solve a Parabola with Fraction – Vertex:

Find the vertex of a parabola where y = x^ (2) + 7/2 x + 3

Solution:

Given parabola equation is y = x^ (2) + 7/2 x + 3

To find the vertices of a given parabola, we have to plug y = 0 in the above equation, we get,

0 = x^ (2) + 7/2 x + 3

Now we have to factor the above equation, we get,

So x^ (2) + 3/2x + 2x + 3 = 0

x(x + 3/2) + 2 (x + 3/2) = 0

(x + 3/2) (x + 2) = 0

From this x + 3/2 = 0 and x + 2 = 0

Then x = - 3/2 and x = - 2

So, the vertices of given parabola equation is (-3/2, 0) and (-2, 0) .

Example 2 for Solve a Parabola with Fraction – Focus:

What is the focus of the following parabola equation where y^ (2) = 10x

Solution:

Given parabola equation is y^ (2) = 10x is of the form y^ (2) = 4ax

To find the focus of a parabola use the formula to find the value of focus value.

We know that the formula for focus, p = 1 / (4a)

Now compare the given equation y^(2) = 10x with the general equation y^ (2) = 4ax . So, that 4a = 10

From this p = 1 / (4a) = 1 / 10

So, the focus of a parabola equation is (0, 1/10) .

Other Example Problems to Solve a Parabola Fraction

Example 3 for Solve a Parabola with Fraction – Axis of Symmetry:

What is the axis of symmetry of the parabola where y = 4x^ (2) + 8x + 14 ?
Solution:

Given parabolic curve equation is y = 4x^ (2) + 8x + 14

From the above equation, a = 4 and b = 8

So the axis of the symmetry of the given parabola is -b/ (2a) = -8/ (2 xx 4) = -8/8 = - 1

Therefore, the axis of symmetry for a given parabolic curve equation is -1 .

Example 4 for Solve a Parabola with Fraction – Latus Rectum:

Find the latus rectum of the given parabola equation y^ (2) = 6x

Solution:

The given parabola equation is y^ (2) = 6x

To find the latus rectum, we have to find the value of p.

The parabola equation is of the form y^ (2) = 4a

Here 4a = 6

So, p = 1/ (4a) = 1/6

The formula for latus rectum is 4p .

From this, the latus rectum of the parabola is = 4p = 4 (1/6) = 4/6 = 2/3

Therefore, the latus rectum for the parabola equation is 2/3 .

Short Definition of Geometry

Geometry (Ancient Greek: γεωμετρία; geo- "earth", -metria "measurement") "Earth-measuring" is a part of mathematics concerned with questions of size, shape, relative position of figures, and the properties of space. Initially a body of practical knowledge concerning lengths, areas, and volumes, in the 3rd century BC geometry was put into an axiomatic form by Euclid, whose treatment—Euclidean geometry—set a standard for many centuries to follow.( Source- Wikipedia)

Short Basic Geometry Definitions:
Short Definition of Geometry for Square:

In mathematical geometry, a square consists of 4 equal angles every side which includes 90° and 4 equal sides. Square area is calculated by-product of two opposite sides.
Short Definition of Geometry for Rectangle:

In mathematical geometry, a rectangle is equal to the square, it consists of 4 equal angles with 90° and 4 equal sides. Rectangle area is calculated by using the product of length and width.
Short Definition of Geometry for Triangle:

In mathematical geometry, triangles consist of 3 straight lines that are equal length and it is called as equilateral triangle. Triangle area is calculated by half of the product of bottom and height.
Example for Short Definition of Geometry:
To solve the Rectangle problems:

Area of rectangle= Length x width.

Perimeter of rectangle= 2 (Length + Width)

Example 1:

Find the area of rectangle that sides measure 6 centimeters and 4 centimeters respectively.

Solution:

Given that

Length – 6 centimeter.

Breadth – 4 centimeter.

The formula used to find out the area of rectangle is = length * breadth.

Area = length * breadth.

Substitute the given value in the formula

Area   = 6 * 4

The result of 6 and 4 = 24

Hence the area of the rectangle given is 24 square centimeter.

Example 2:

Find out the volume of the cube that sides measure 6 centimeter.

Solution:

Given that

Measure of cube length = 6.

The formula used to find out the volume of the cube is = a3.

While we insert the value in the formula we get the result is

Volume = 63.

= 216

Hence the volume of the given cube is 216 cubic centimeters.

To solve Square problems:

Perimeter of square= 2 * side

Area of square= side * side

Problem 3:

To find out the area and perimeter of square while side length is 4cm?

Solution:

Area of square = (side*side)

=4 * 4

=>16 cm2

Perimeter of square = 4 * side

= 4 * 4

= 16 cm

The Multiplication Rules

In algebra fundamental arithmetic procedure (addition, subtraction, multiplication, and division) in general used in day-to-day life. The multiplication defined as product of numbers. Multiplication is usually expressed in a*b or ab form. The multiplying numbers are having two terms i) Multiplicand and ii) Multiplier. In this article, we are going to discuss about the multiplication rules

The Multiplication Rules – Rules and Example Problems:

The multiplication rules - Rules:

Rule 1: The multiplication of two integers by the connected signs resolve be positive sign

a) Positive `xx` positive = positive

b) Negative `xx` negative = positive

Rule 2:  The multiplication of two integers by the different signs will be negative

a) Positive `xx` negative = negative

b) Negative `xx` positive = negative.

The multiplication rules – Example problems:

Example 1:

Evaluate 6802 `xx` 322

Solution:

6802 `xx` 322

------------------

13604

13604

20406

--------------

2190244

--------------

Therefore, the final answer is 2190244.

Example 2:

Find the solution for given problem

(- 9667) `xx` (- 26)

Solution:

(- 9667) `xx` (- 26)

---------------------------

58002

19334

-------------------

251342

-----------------

Therefore, the final answer is 251342.

Example 3:

Find the solution for given problem

8655 `xx` (- 124)

Solution:

8654 `xx` (- 124)

-----------------

34616

17308

8654

----------------

- 1073096

-----------------

So, the final answer is (- 1073096)
The Multiplication Rules – Practice Problems:

Practice problem 1:

Find the solution for given problem

(- 7897) `xx` (- 53)

Answer: So, the final answer is 418541

Practice problem 2:

Find the solution for given problem

(- 7790) `xx` (- 48)

Answer: So, the final answer is 373920

Practice problem 3:

There are 280 students in every class at the performing arts school. If there are 62 complete classrooms, how many students attend this school?

Answer: 17360 students attend this school

Practice problem 4:

Jesse wants his giant chocolate chip cookies to have 28 chips each. Today he is make 355 cookies. How many chips does he need?

Answer: 9940 chips does he need.

Variance Covariance Formula

Variance: Variance is used in the statistical analysis to find the the extent to which a single variable is varying from its mean value given a set of values.

Variance of a random variable X is often denoted as VAR ( X). It is also denoted by the symbol `sigma` 2X

Covariance: Co-variance is used in statistical analysis to find the extent to which 2 variable are varying together given a set of values for both these variables.

Covariance of two random variable X and Y is denoted as COV ( X,Y). Unlike variance which has the mathematical symbol sigma ( `sigma` ), Covariance doesn't have any kind of symbol to depict it.
Formulae for Variance and Covariance.

Lets consider 2 sets of data namely X and Y such that the values in X are x1 , x2 , x3 , x4 , ......xn and similarly the values in Y are y1 , y2 , y3 , y4 , .......yn

Now lets denote the mean of the X set of variables to be    Xm

Mean Xm = X1 + X2 + X3 +........ + Xn / n

Similarly lets denote the mean of the Y set of variables to be Ym

Mean Ym = Y1 + Y2 + Y3 +.......+ Yn / n

Formulae for Variance of X = ( x1 - xm)2 + (x2 - xm)2 + ( x3 - xm)2 + .......+ (xn - xm)2 / n

VAR ( X ) = (1/n) `sum` (xi - xm)2

Formulae for Variance of Y = ( y1 - ym)2 + (y2 - ym)2 + ( y3 - ym)2 + .......+ (yn - ym)2 / n

VAR ( Y ) = (1/n) `sum` (yi - ym)2

Formulae for Covariance of (X,Y) = ( x1 - xm) ( y1 - ym) + (x2 - xm)(y2 - ym) + ( x3 - xm)( y3 - ym) + .......+ (xn - xm)(yn - ym) / n

COV ( X , Y ) = (1/n) `sum` (xi - xm)(yi - ym)
Example Problem 1 on Variance and co Variance

Lets assume the below set of marks received by Bill and Bob in Maths, Physics, Chemistry, English and Biology for the purpose of solving Variance and Co Variance

table

Based on the Formulae given in the previous paragraph, Lets calculate the Mean Marks for Bill and Bob

Mean Marks for Bill = 16+12+14+18+20 /5 = 16.00

Mean Marks for Bob = 14+18+15+18+20 /5 = 17.00

Now lets apply the formulae of variance and find out the Variance of Bill and Bob

Variance of Bill = (16-16)2 + (12-16)2 + (14-16)2 + (18-16)2 + (20 - 16)2 / 5 = 40 / 5 = 8

Variance of Bob = (14-17)2 + ( 18-17)2 + (15-17)2 + (18-17)2 + (20-17)2 / 5 = 24 / 5 = 4.8

Co variance of ( Bill , Bob ) = (16-16)*(14-17) + (12-16)*(18-17) + (14-16)*(15-17) + (18-16)(18-17) + (20-16)*(20-17) / 5

Co Variance of ( Bill , Bob ) = 21/5 = 4.2

I am planning to write more post on What are Line Segments and Perpendicular and Parallel Lines. Keep checking my blog.

Example Problem 2 on Variance and co Variance

Lets assume a class of 4 student who have been asked to rate their liking toward 2 musical instruments Guitar and Piano on a scale of 10

Based on the above data, lets calculate the Mean of the people liking Guitar and Mean of people liking Piano.

Mean(Guitar) = ( 5+3+7+9)/4 = 24/4 = 6

Mean(Piano) = (5+9+9+5)/4 = 28/4 = 7

Var(Guitar) = ((5-6)2 + (3-6)2 + (7-6)2 + (9-7)2 )/ 4 = (1+9+1+4)/4 = 15/4 = 3.75

Var(Piano) = ((5-7)2 + (9-7)2 + (9-7)2 + (5-7)2 ) / 4 = (4+4+4+4)/4 = 16/4 = 4

Cov(Guitar,Piano) = [(5-6)(5-7) + (3-6)(9-7) + (7-6)(9-7) + (9-7)(5-7) ] / 4 = (2+(-6)+2+(-4))/4 = -6/4 = -1.5

Binomial Probability Function

Binomial probability Problems represented by B(n,p,x). It gives the probability of exactly x successes in ‘n’ Bernoullian trials, p being the probability of success in a trial. The constants n and p are called the parameters of the distribution. A Binomial distribution can be used under the following condition.

(i) any trial, result in a success or a failure

(ii) There are a finite number of trials which are independent.

(iii) The probability of success is the same in each trial.

In a Binomial distribution function mean is always greater than the variance. The binomial probability function example problems and practice problems are given below.
Example Problems - Binomial Probability Function:

Ex 1:  By using the binomial distribution. If the sum of mean and variance is 4.8 for  5 trials find the distribution

Solution:  np + npq = 4.8 , np(1 + q) = 4.8

5 p [1 + (1 − p) = 4.8

p2 − 2p + 0.96 = 0 , p = 1.2 , 0.8

p = 0.8 ; q = 0.2 [p cannot be greater than 1]

The Binomial distribution is P[X = x] = 5Cx (0.8)x (0.2)5−x,  x = 0 to 5

Practice Problem - Binomial Probability Function:

Pro 1: Find the value for p by using the Binomial distribution if n = 5and P(X = 3) = 2P(X = 2).

Ans: p = `(2)/(3)`

Pro 2: Find the probability values by using the binomial distribution.  A pair of dice is thrown 10 times. If getting a doublet is considered a success (i) 4 success (ii) No success.

Ans: (a) (35/216)(5/6)6

(b) (5/6)10

Proportion Equation

The data is obtaining by the comparison of two ratios is called proportion data. Proportion data is represented as a:b = c:d. This proportion data can be written in the form of fraction as `a/b` = `c/d` . Where the pairs of data (a,b) and (c,d) are in proportion. When the proportions are equal, the cross product of the proportion will be also equal. That is, `a/b` = `c/d` can be written as ad=bc.

Examples for Proportion Equation:

Example 1 for proportion equation:

Martin read 63 pages of the book in 33 minutes. How many pages will he be able to read in 43 minutes?

Solution:

Martin takes 33 minutes to read 63 pages.

Martin will take 43 minutes to read x pages.

This can be written as,

`63/33` = `x/43`

Now we have to do the cross multiplication.

63 `xx ` 43 = x `xx` 33

2709 = 33x

This can be written as,

33x = 2709

Now we have to divide both sides by 33.

`(33x)/33` = `2709/33`

x = 82.09 now we have to round it to the unit place.

x = 82

Therefore, Martin will read 90 pages in 45 minutes.

Example 2 for proportion equation:

Paul bought 12 apples for dollar 48.  How many apples will he be able to buy in $ 93?

Solution:

Paul spends $48 for 12 apples.

Paul will spend $93 for x apples.

This can be written as,

`12/48` = `x/93`

Now we have to do the cross multiplication.

12 `xx` 93 = x `xx` 48

1116 = 48x

This can be written as,

48x = 1116

Now we have to divide both sides by 48.

`(48x)/48` = `1116/48`

x = 23

Therefore, Paul can buy 23 apples for $ 93.

Practice Problems for Proportion Equation:

Problem 1 for proportion equation:

Martin read 40 pages of the book in 28 minutes. How many pages will he be able to read in 52 minutes?

Solution: Martin will read 74 pages in 52 minutes.

Problem 2 for proportion equation:

Paul bought 8 apples for dollar 22. How many apples will he be able to buy in $ 66?

Solution: Paul can buy 24 apples for $ 66

Polynomials Calculator

An Algebraic expression is of the form axn is called a monomial. The variable a is called the coefficient of xn and n, the degree of monomial. For example, 7x3 is monomial in x of degree 3 and 7 is the coefficient of x3. The combination of two monomials is called a binomial and the combination of three monomials is called a trinomial. For example, 2x3 + 3x is a binomial and 2x5 – 3x2 + 3 is trinomial. The sum of n number of monomials, where n is finite and x is called a polynomial in x.
Illustration to Polynomials:

Polynomial Calculator Example 1:

The polynomial calculator of the equation x2 + ax + b gives the remainder 18, when divided by x – 2 and leaves the polynomial calculator of remainder –2 when that is been divided by (x + 3).

Find the values of a and b.

Solution to the polynomial calculator:

P(x) = x2 + ax + b.

In tyh is polynomial calculator,

When x – 2 divides P(x) then the remainder is P (2).

∴P (2) = 4 + 2a + b.

But remainder = 18 ⇒ 4 + 2a + b = 18;

2a + b = 14 (1)

When (x + 3) divides P(x)

, the remainder is P (–3).     ∴ P (–3) = (–3)2 + a (–3) + b

= 9 – 3a + b.

But remainder = –2;      ∴ 9 – 3a + b = –2;

⇒ –3a + b = –11 (2)

(1) ⇒   2a + b = 14

(2) ⇒ –3a + b = –11 (subtracting)

5a        = 25

(Or) a     = 5

Substituting a = 5 in equation (1) we get

10 + b = 14; b = 4, ∴ a = 5, b = 4
Subtraction of Polynomials Calculator:

Example for Polynomials calculator:

Subtract        2x3 – 3x2 – 1 from x3 + 5x2 – 4x – 6.

Solution:

Using associative and distributive properties, we have

( x3 + 5x2 – 4x – 6) – (2x3 – 3x2 – 1) = x3 + 5x2 – 4x – 6 – 2x3 + 3x2 + 1

= x3 – 2x3 + 5x2 + 3x2 – 4x – 6 + 1

= (x3 – 2x3) + (5x2 + 3x2) + (–4x) + (–6+1)

= –x3 + 8x2 – 4x – 5.

The subtraction can also be performed in the following way:

Line (1): x3 + 5x2 – 4x – 6.

Line (2): 2x3 – 3x2 – 1.

Changing the signs of the polynomial in Line (2), we get

Line (3): –2x3 + 3x2 + 1.

Adding the polynomials in Line (1) and Line (3), we get

–x3 + 8x2 – 4x – 5.

Integral Part Definition

In geometrical and other applications of integral calculus it becomes necessary to find the difference in the valves of an integral of a function f(x) for two assigned values of the independent variable x, a and b. this difference is called the definite integral of f(x) over the range (a,b) and is denoted by $\int_{a}^{b}f(x)dx=F(b)-F(a)$, where F(x) is an integral of f(x), F(b) is the variable of F(x) at x=b, F(a) is the value of F(x) at x=a.

It is often written thus  : $\int_{a}^{b}f(x)dx=[F(x)]_{a}^{b}=F(b)-F(a)$.

Note 1. The integral $\int_{a}^{b}f(x)dx $ is read as the integral of f(x) from a to b.The number a is called the lower limit and the number b, the upper limit of integration.

Note 2. It should be seen that the value of a definite integral is perfectly unique and is independent of the particular form of integral which may employ to calculate it. Considering F(x)+c instead of F(x), we get

$\int_{a}^{b}f(x)dx=[F(x)+c]_{a}^{b}=[F(b)+c]-[F(a)+c]=F(b)-F(a)$

so that the arbitrary constant disappears in the process and we get the same values as on considering F(x). This is why the name is given as definite integral.

Note 3. It is assumed that a and b are finite.
Definition of Integration by Parts

When the given function cannot be integrated directly by using standard formulae, we try other methods. The process of integration is largely of tentative nature and no systematic procedure can be given as in differentiation. However, the following are two important methods opf integration.

1. Integration by Substitution           2. Integration by Parts

Here, we will discuss about Integration by parts

An Integration is the inverse process of differentiation.By differentiaon we find the derivative of the given function, whereas by integration we find the function whose derivative is known.

If the derivative of F(x) is f(x) then we say that the antiderivative or integral of f(x) is F(x),such that

int f( x ) dx= F( x )

Thusd/dx F(x) = f(x) => intf(x) dx = F(x)

Integration by Parts:

Theorem: if u and v are two differentiable function of x then

int ( u v ) dx = [ u * int v dx ] - int { du/dx * int v dx } dx .

We can express this result as given below:

Integral of product of two function

= ( 1st function ) * ( integral of 2nd function ) - int { ( derivative of 1st function ) * ( integral of 2nd function ) } dx .

We should choose u and v in such a way that the second function v on the right hand side is easy to integrate. Sometimes, this rule has to be used repeatedly. This rule is also useful in integrating logarithmic and inverse t - functions of the type log x, log(ax2 + bs + c), sin-1 x, tan-1 x etc.The following guidelines will help you to see which of the two functions in the product should be taken as the first function.

Notes:

If the integrand is of the form f(x) * xn, we consider xn as the first function and f(x)as the second function.
If the integrand contains a logarithmic or an inverse trigonometric function, we take it as the first function. If the second function is not given in any case we take that as one.

Example Problems on Integration by Parts:

Pro 1: Evaluate, int xsin 2x dx

Sol : Given,

int   xsin 2x dx

=     int sin 2x dx - int { d/dx ( x ) * int sin 2x dx } dx

=     x * ( -(cos 2x)/2 ) - int 1 * ( -(cos 2x)/2 ) dx

=  (-x cos 2x)/2 + 1/2 int cos 2x dx

"-(x cos  + 1/2 * "(sin

= -(x cos 2x)/2 + 1/4 sin 2x + C , which is the Answer.

Pro 2: int logx/x^2 dx

Sol :   Integrating by parts, taking logx as the fi rst function and 1/x^2  as the second function, we get

int "log   dx

=  int (log x) * 1/x^2 dx

=   ( log x ) * int 1/x^2 dx - int { d/dx ( log x ) * int 1/x^2 dx } dx

=    ( log x )(-1/x )   - int 1/x  * ( -1/x ) dx

= - log x/x + int 1/x^2 dx

= - log x/x - 1/x + C, which  is the required Answer.

Practice problems on Integration by Parts:

Pro 1: Evaluate, int (  x cos x) dx    ( Answer:   x sinx +cosx+c )

Pro 2: Evaluate, int e2x sin x dx            ( Answer: 1/5 e2x ( 2 sin x - cos x ) + C )

Natural Logarithmic Calculator

In mathematics, the natural logarithmic function is defined as the function contains three ports, namely the number, the base and logarithm itself,  now the natural logarithmic calculator is the function of  the logarithm function calculator which means we are using to the base of ‘e’ where e is constant, base of  ‘e’ value is given by

Understanding Definition of Logarithm is always challenging for me but thanks to all math help websites to help me out.

e = 2.718281828. and the numbers, and the logarithmic itself.

The logarithmic function calculator is represented as y = log_b^(x).

We take the value of base of ‘e’ is approximately e = 2.71.
Natural Logarithmic Calculator:

Here, we introduce the natural logarithmic calculator. The function of natural logarithmic calculator is represented as  y = ln(x)

Explanation about natural logarithmic calculator:

Input:

calcutor             given value

The given value is conversions by logarithmic calculator

Conversions:

 x =  conversion value.

Solution:

 y = calculated value will be displayed

Here we can the change the equation by x and y.

y = ln(x) for y i n natural logarithm

y = ln(x) for x  in natural logarithm

Note: we can calculate or perform the positive numbers only not negative numbers by natural logarithmic calculator.
Some Problems about Natural Logarithmic Calculator:

Problem1:

To solve the natural logarithmic function of ln(2)

Solution:

Input:

x = 2  given value

Conversion:

x = 2  the given value conversion by logarithmic calculator.

Solution:

y = 0.69 approximately

Problem 2:

To solve natural logarithmic function of  span style="font-size: small; " mce_style="font-size: small; ln(4)

Solution:

Input:

 x = 4 given value

Conversion:

 x = 4  the given value conversion by logarithmic calculator.

Solution:

 y = 1.38 approximately

Problem 3:

To solve natural logarithmic function of ln(6)

Solution

Input:

 x = 6 given value

Conversion:

 x = 6  the given value conversion by logarithmic calculator.

Solution:

 y = 1.79 approximately

Problem 4:

To solve the natural logarithmic function of in ln(10)

Solution:

Input:

x = 10 given value

Conversion:

 x = 10 the given value conversion by logarithmic calculator.

Solution:

 y = 2.30 approximately

Problem 5:

To solve the natural logarithmic function of ln(15)

Solution:

Input:

x = 15 given value

Conversion:

 x = 15 the given value conversion by logarithmic calculator.

Solution:

 y = 2.70  approximately


Practices problems for natural logarithmic function:

Problem1;

To solve the natural logarithmic of ln (20)

Answer is y = 2.99

Problem2

To solve natural logarithmic function of ln (50)

Answer is y = 3.91

Problem3

To solve natural logarithmic function of ln(100)

Answer is y = 4.60

Problem 4

To solve the natural logarithmic function of in ln(99)

Answer is y = 4.59

Hence, here we obtain the natural logarithmic calculator.

Scientific Notation Rules

 Often we come across abnormally large numbers or abnormally small numbers like 400,000,000,000 or very, very small number like0.00000000002. Especially in the field of science such abnormal size of numbers is quite common.

In calculations involving such numbers, the work is cumbersome and possible errors may occur.

A special type of notation is used to simplify the dealings with these type of numbers and it is named as scientific notation as such numbers are very predominant in the field of science.

However a scientific notation must be used following a set of rules, universally accepted.

Let us take a closer look.
Scientific Notation Rules – Description

Since very large or very small numbers can be rounded to the appropriate place value, the power of 10 is used as base in scientific notation. The established general form is,

                                       a x 10n, where,  1= I a I < 10

If  ‘n’ is positive, it represents the number of zeroes from ‘a’ to the right of unit place and hence  it denotes large numbers.

For very, very small decimal numbers ‘n’ is negative and represents the number of zeroes before ‘a’ up to the decimal point.
Scientific Notation Rules – Examples

The velocity of light is about 310,000,000 kilometers per second.

We find there are seven 0s after the digits 1. That is the number is same as 31 times 10 to the power 7. But as per scientific notation rules, 31x 107 is not allowed and hence the given is modified as 3.1 x 108

Thus in scientific notation, 310,000,000 = 3.1 x 108

Similarly it is equally incorrect to mention (as a scientific notation) 310,000,000 as 0..31 x 109

Let us consider another example with a very  small number.

The wave lengths of a certain color is experimentally found to be 0.0000015 millimeter.

There are 6 decimal places to cross the digit 1from right. That the given number is (1/1000000)th of 1.5.

Hence in scientific notation it is written as 1.5 x 10-6

Importance of Derivatives of Trigonometric Functions

Trigonometric Functions 
In Mathematics, trigonometric functions or circular functions are ratios of the sides of a right angled triangle, having the given angle.  Trigonometry deals with functions of an angle present in a triangle. It relates the angles of the given triangle with the length of its three sides. It also helps to find the length and angle of a triangle. Trigonometric functions are widely used in real life in various fields such as carpentry, surveys, engineering, physics, navigation, astronomy etc.

Trigonometric functions are sin p, cos p, tan p, cot p, sec p, csc p.  Among them sine, cosine and tangent are the most familiar and basic trigonometric functions. Before going into derivatives of trigonometric functions, let us first see what these trigonometric functions refer to.

Let us consider a circle of radius 1. If a ray originates at the origin which makes an angle with x axis, what will the sine, cosine and tangent of this angle provide?
• The sine of this angle gives the length to which the triangle has risen on the other side i.e. the y component

• The cosine of this angle gives the x component length

• The tangent of this angle gives the slope of the triangle, which is obtained by dividing Y component by the X component.

Thus, trigonometric functions can also be defined as the lengths of all line segments from a given unit circle.

Derivative of Trigonometric Function
With this basic understanding on trigonometric functions, let us explore on the derivatives of trigonometric functions namely sin p, cos p, tan p, cot p, sec p, csc p.

The derivatives for each of these trigonometric functions can be represented as follows:

Derivative of Sinx
The symbolic expression of Sinx derivative is given by

d/dx (sin x) = cos x   or sin’ x = cos x


It is easy to find the derivatives of sin x. This is because, the derivative of sin x can be obtained by just calculating the cos x.

Derivative of Cosx
The Cosx derivative is given by

d/dx (cos x) = - sin x or cos’ x = - sin x


Derivative of Tan p
The derivative of tan p is given by “tan-1 p = sec^2   p”

Derivative of Cot p
The derivative of cot p is given by “cot-1 p = - coesc^2 p”

Derivative of Sec p
The derivative of sec p is given by “sec’ p = sec p tan p”

Derivative of Csc p
The derivative of csc p is given by “csc’ p = - csc p cot p”

Different types of Matrices

Matrix is a rectangular array of numbers or functions which are called the elements or the entries of the matrix. The elements or entries are arranged in rows and columns in either () or [] brackets. The elements of a matrix are denoted by aij where ‘i’ is the row number and ‘j’ is the column number. We can only add or subtract two matrices if they both have same number of rows and columns.

Order of matrices: An order of the matrices is written as m X n where m is the number of rows and n is the number of columns in it. That means a matrix of order 2 X 2 can only be added to a matrix of order 2 X 2 and a matrix of order 3 X 3 can only be subtracted from a matrix of order 3 x 3.

Types of matrices

1. Column matrix: A matrix is said to be a column matrix if has only one column. For example: - a matrix of order 4 X 1 is said to be a column matrix as it has four rows but only one column. In general, a matrix of order m X 1 is considered as column matrix where m can be any real number.
2. Row matrix: A matrix is said to be row matrix if it has only one row. For example: - a matrix of order 1 X 4 is said to be a row matrix as it has four columns but only one row. In general, a matrix of order 1 X n is considered as row matrix where n can be any real number. A example of row matrix is [2 , 6, 11, 7].
3. Zero matrix: A matrix is said to be zero matrix if all its elements or entries are zero. It is also called a null matrix. For example: - [0], [0 0] are zero matrices. We denote zero matrix by O.
4. Square matrix: A matrix is said to be square matrix if its number of rows are equal to number of columns. Thus m X n is a square matrix if m = n and is known as a square matrix of order ‘n’.
5. Unit matrix: A matrix is said to be unit matrix if all its elements or entries are one. For example: -
[1], [1 1], [1 1 1] are all unit matrices.

Polynomial Factoring

Factoring polynomial is similar to factoring numbers, but you will be dealing with expressions instead of numbers. In order to factor a polynomial, you have to determine the polynomial that divides the original polynomial evenly.

Polynomial Factoring

How to factor a polynomial?
Different approaches are used to factor the polynomial; most common approaches are simple factoring, and factoring in pairs. In this article, you will find examples for these two approaches of polynomial factoring.

Simple Factoring
In simple factoring, you identify a common factor for the expression and place it before the parenthesis. What is a common factor? To know that, look at the example below:
5x + 60 = 5 (x) + 5(6) = 5 (x+12)

Now the common factor of the above expression is 5. To cross verify, multiply the factor with each term in the expression and ensure if you are getting the original polynomial back. For example,

5(x+6) = 5 multiplied with x + 5 multiplied with 12 = 5x + 60 (the original polynomial)

Factoring in Pairs
“Factoring in Pairs” is yet another factoring polynomials solver. This approach is used when you cannot find any common factor for the terms in the expression. In this approach, the expression is split into pairs of terms and then each pair will be factored separately. Here is an example:

Find the factors of ab-5b-2a+10.

This polynomial algebraic expression includes 2 variables namely a and b. There is no common factor for these 4 terms. In this case, take the first two terms and find a common factor. Similarly, take the last 2 terms and find a common factor, as shown below:

ab-5b-2a+10 = b (a-5) -2 (a-5) = (b-2) (a-5)

In the above step, the first two terms ab-5b are taken into consideration. Take the common factor out, it is b. Now it becomes b (a-5). Similarly take the next two terms -2a+10. The common factor is 2 here. When you put it as -2, the expression becomes -2 (a-5). Now the expression can be further simplified as (b-2) (a-5).

If the consecutive terms do not have any common factors, then you can rearrange the terms based on the commonality and then do the factoring.

Rounding Decimals

Let's learn about rounding decimals in today's post.

Rounding decimals is nothing but turning a number to the nearest tens, hundreds, thousands or so on. Suppose we are adding a decimals and the answer we get is 4.6, here we can round the decimal to 5. If the number after decimal is 5 or more than 5, it can be rounded by adding 1 and if it is less than 5, it can be rounded by removing the decimal value. This method is the method of rounding decimals.

Next time i will help you with the concept of decimals number line.

statistics independent events

Independents Events is a key word often referred to in probability calculations. Knowingly or unknowingly , they use Independents Events often in the theory and application of probability. events are independent events, if the occurrence of event does not affect the probability of other variables. i.e.., positive or negative.