Random Occurrence

Let us see about the experiment of random occurrence,

              An outcome is produced from an operation through experiment is called as random experiment and also produce different possible outcomes. In a random experiment, an outcome random experiment is unpredictable.

Some more examples for random occurrence in random experiment:

• Coin is Tossing
• Die is rolling
• Take a card from a packet of card.
• Take a possible ball from bag contains different balls.

Some definitions about random occurrence:

Let we see about some definitions about random occurrence,

Trial: Random experiment performing by this. 

Sample space: In a random occurrence,the possible outcomes are taken by in set is known as sample space and is represented as S. When we roll a die, the possible outcomes are 1, 2, 3, 4, 5, 6 .
Sample space is S = { 1, 2, 3, 4, 5, 6} 

Event: Possible outcome or combination of outcomes is known as an event.An each subset of the sample space S is known as an event. Events represented as  A, B, C, D, E. While coin is tossing, getting a head or tail is take as an event. S = { H, T}, A = {H}, B = {T}.

Formula for random occurrence:

Let we see about formula for random occurrence,      

If a sample space contains n outcomes, m of which are favorable to an event E, then the probability of an event E, denoted by P (E), 
Number of favorable outcomes
  P(E)  =  Total number of outcomes number of favorable outcomes / Total number of outcomes

           = P(E) / P(S)

here,P(E) - Total number of outcomes number of favorable outcomes.

       P(S) - Total number of outcomes.

Examples:

1) Find the random occurrence (probability) to getting two heads when two coins are tossed simultaneously?

Solution:

In tossing two coins the sample space S = {HH, HT, TH, TT}, n(S) = 4.
Let A denote the event of getting two heads A = {HH}, n(A) = 1.
Therefore,probability to getting two heads P =n(A)/n(S) = 1/4.

2) Find the random occurrence getting 3 when rolling a die?

Solution:

 In rolling a die, the sample space S ={ 1, 2, 3, 4, 5, 6} : n (S) = 6.
 Let A be an event of getting 3
 A = { 3 }, n (A) = 1

∴ P(A)  = n(A) / n(S)

= 1/ 6.

Logarithms Learning

Logarithms function is the inverse of exponential function. If f(x)=logb(x) where b is the base of the logarithm function ,b>0and b is not equal to one .The power to which is base ten and base e(e=2.5466) obtain raised in number. The logarithm of a power equals the exponent multiplied with the logarithm of the base.

Examples:

1. f(x) = log2x , base is 2

2. g(x) = log4x, base is 4

3. h(x) = log0.5x, base is 5


Properties of logarithms Learning:


1.logbmn  =  logbm + logbn

The logarithm of a product is equal to the sum of the logarithms of each factor."

2. logb m/n  = logbm − logbn

The logarithm of a quotient is equal to logarithm of the numerator minus the logarithm of the denominator.

3. logb yn = n logby

The logarithm of a power of y equal to exponent of that power times the logarithm of y.

Some basic property of logarithms

loga1=0

logaa=1

logaxa=alogax


Change of base Rule for logarithms Learning:


By the definition of the base two logarithm

log2x = log2x

=> x = 2log2x ( as log2x = a then x = 2a)

(since immediate consequences of the definition of logarithms ) that the logarithm of a power equals the exponent multiplied with the logarithm of the base. Therefore by taking natural logarithm on both sides of the proceeding equation, obtain

ln(x) = log2(x)ln(2)

(since, immediate consequences of the definition of logarithms ) that the logarithm of a power equals the exponent multiplied with the logarithm of the base. Therefore by taking natural logarithm on both sides of the preceding equation obtains

Solving for base two logarithm gives the same formula as before:

log2(x)=`(ln(x))/(ln(2))`


Examples of logarithms Learning:


Example1:Solve log23+log28=log2 (4x)

Solution: logarithmic function log23+log28=log2 (4x)

log2(3*8)  = log2(4x)

log2(4x)=log2(24) by property of logarithm (1)

Equate both sides, s the bases are same 2,  4x = 24

Simplification: x = 24/4

Answer = 6

Example2: log654-log69

Solution:

lo654-log69=log6(54/9) by property of logarithm (2)

simplification: log66

Answer = 1

Examble3: logarithm Solving equation in    log284

Solution:Converting the logarithmic equation

=  4log28  by property logarithm (3)

Simplification: 4log223 = 12log22

Answer=12

Project on Trigonometry for Class 10

The word "Trigonometry" is derived from three Greek words----' tri ' means three, 'gonia' means angle, and 'metron' means measure.Thus "Trigonometry" means  "three angle measure". Trigonometry deals with the relation between the angles and sides in a triangle.The study of trigonometry is of great importance in several fields and applied in many branches of  Science and Engineering such as Seismology,design of electrical circuits,estimating the heights of tides in the ocean etc.The three main trigonometric functions are sine,cosine and tangent and their reciprocals are co secant,secant and cotangent respectively.

Definitions of Trigonometric functions:

Definitions of Trigonometric Functions :-

Consider a right angle triangle and Let P(x,y) be the point and θ be the acute angle.

ON = x ; NP = y ; OP = r



The sine function is defined as ratio of opposite side (y) to hypotenuse(r).
sinθ = `(y)/(r)`

The cosine function is defined as ratio of adjacent side (x) to hypotenuse(r).
cosθ = `(x)/(r)`

The Tangent  function is defined as ratio of opposite side (y) to adjacent side (x).
tanθ = `(y)/(x)`

The reciprocal of sine is co secant and is defined as
cscθ = `(r)/(y)`

The reciprocal of cosine is secant and is defined as
secθ = `(r)/(x)`

The reciprocal of tangent is cotangent and is defined as
cotθ = x/y

Note:

sinθ and cscθ are reciprocal => sinθ * cscθ = 1
cosθand secθ are reciprocal => cosθ * secθ = 1
tanθ and cotθ are reciprocal => tanθ *  cotθ = 1


Formulas and Identities of Trigonometric functions


Identities :

sin2θ + cos2θ = 1
1 + tan2 θ = sec2 θ
1+ cot2 θ = csc2 θ

Formulas

sin(A+B) = sinA cosB + cosA sinB
sin(A-B) = sinA cosB - cosA sinB
cos(A+B) = cosA cosB - sinA sinB
cos(A-B) = cosA cosB + sinA sinB
tan(A+B) = `(tan A + tan B)/(1- tan A tan B)`
tan(A-B) = `(tan A - tan B)/(1+ tan A tan B)`

Signs of Trigonometric Functions:

The entire coordinate plane is divided into 4 quadrants and they are named in counter clockwise direction.Let P(x , y) be a point in the coordinate plane.



(1) If P lies in the 1st quadrant,  all the trigonometric functions are positive.

(2) If P lies in the 2nd quadrant,  sinθ, cscθ are positive and the others are negative.

(3) If P lies in the 3rd quadrant, tanθ, cotθ are positive and the others are negative.

(4) If P lies in the 4th quadrant, cosθ, secθ are positive and the others are negative.\

algebra simple interest formula

Interest: Interest is the amount of money we pay for the use of some amount of money.

Types of interest: There are two types of interests,

a) simple interest

b) compound interest

Simple interest: Simple interest is the Interest paid / compensated only on the original principal, not on the interest accrued

Compound interest:Compound interest means that the interest Which  includes the  interest calculated on principal amount using the Compound Interest Formula .

Algebra Simple interest Formula :


Algebra Formula to find simple interest :

Simple interest I = PNR

Where,   P is the Principal amount, R is the  Rate of interest, N is Time duration.

When we knows interest I we can find p, n or r using the same formula ,

Different forms of algebra simple interest formula

P= `(I)/(NR)`

N=`(I)/(PR)`

R=`(I)/(PN)`


Algebra Problems on Simple Interest formula:


Ex 1:Interest Rate: 1% each year ,Starting Balance: $147 ,Time Passed: 6 year . Find the simple interest? What is the new total balance?

Solution:Algebra Formula for Simple Interest: I = PRT

P = principle = starting balance = $147

R = interest rate = 1%

T = time = 6 years

I = interest = principle × interest rate × time = 147 × `(1)/(100)` × 6 = $8.82

New Balance = starting balance + interest accrued = $147 + $8.82 = $155.82

I= $ 8.82, New Balance = $155.82

Ex 2:Interest Rate is 2% each year, Starting Balance is $184 ,Time Passed was 3 years .How much interest has ensued if we are

using simple interest? What is the new total balance?

Solution:algebra simple interest formula = PRT
P = principle = starting balance = $124

R = interest rate = 2 %

T = time = 9 year

I = interest = principle × interest rate × time = 124 × `(2)/(100)` × 9 = $22.32

New Balance = starting balance + interest accrued = $124 + $22.32 = $146.32

Interest = $22.32, New balance = $146.32

Ex 3 :John invests 1200 at the rate of 6.5 percent  per annum. How long it will take john earns 195 in interest ?

Solution : Algebra Simple interest formula = I = PTR/100

195 = (`(1200 * T*6.5)/(100)`) / 100

T = 2.5

John will earn $195 in interest in 2.5 years.

Area of Compound Figures

A compound figure is made up of two or more simple figures. The following figure is made up of a rectangle and a square. If we want to find the perimeter and area of a compound figure, then calculate the area of each simple figure and then find the sum of the area calculations.


Examples to Find the Area of Compound Figures:


Example 1: Find the area of the following compound figure.

Solution: Area of Rectangle Formula = l × b square units.

where, l is length,

b is breadth.

Rectangle 1 => 8 × 3 square units.

=> 24 square units.

Rectangle 2 => 6 × 3 square units.

=> 18 square units.

Rectangle 3 => 6 × 3 square units.

=> 18 square units.

Therefore the total area of compound figure = Area of Rectangle 1 + Area of Rectangle 2 + Area of Rectangle 3.

=> 24 + 18 + 18 square units.

=> 60 square units.

Hence the total area of compound figure is 60 square units.


Example 2: Find the area of the shaded part of the compound figure.

Solution: Area of Square = a2 square units.

where, a is length of the side of the square.

=> Area of Square = 82 square units.

=> 64 square units.

Area of Circle = ? r2 square units.

where, r is the radius of the circle.

=> Area of Circle = ? × 32 square units.

=> 3.14 × 9 square units.

=> 28.25 square units.

Therefore the area of the shaded part of the compound figure = Area of Square ? Area of Circle.

=> 64 ? 28.25 square units.

=> 35.75 square units.

Hence the area of the shaded part of the compound figure is 35.75 square units.


Practice Problems for Finding the Area of Compound Figures:


Questions: Find the area of the following compound figures.

Solutions: Compound Figure 1 = 28 square units.

Compound Figure 2 = 56 square units.

Solving Online Vertex of a Hyperbola

Hyperbola is a conic section.Hyperbola is all points found by keeping the whose difference from distances of two points (each of which is called a focus of the hyperbola) constant.The vertices of the hyperbola are located on the axis and are a unit from the center.Let us see some formulas for solving vertex of hyperbola online.

Formulas for Solving Online Vertex of a Hyperbola

Standard Formulas Of Hyperbola:

Equation of hyperbola  standard  formulas



1)The hyperbola  opens left  and right(horizontal axis) if the term `(x-h)^2`  is positive.

2)The hyperbola opens up and down(vertical axis) if the term is `(y-k)^2`

3)c= 

is the distance from the center (h,k) to each focus

4)asymptote have slopes given by.

5) The equation of asymptotes  are  given by since the asymptote contain center(h,k)

Example of Solving Online Vertex of a Hyperbola

Example 1:Find the vertex and center  for the given hyperbola.

9x^2-16y^2+18x+160y-247=0

Solution:

First put the equation into standard form.

9x² - 16y² + 18x + 160y - 247 = 0

Now complete the square.

9(x² + 2x + 1) - 16(y² - 10y + 25) = 247 + 9 - 400
9(x + 1)² - 16(y - 5)² = -144

Multiply thru by -1 since the right hand side is negative.

16(y - 5)² - 9(x + 1)² = 144

Set equal to one.

(y - 5)²/9 - (x + 1)²/16 = 1

Since y² is the positive squared term, the pair of hyperbolas open vertically up and down.

The center (h,k) = (-1,5).

a² = 9 and b² = 16
a = 3 and b = 4

The vertices are (h,k-a) and (h,k+a) or
(-1,5-3) and (-1,5+3) which is (-1,2) and (-1,8).

c² = a² + b² = 9 + 16 = 25
c = 5

The foci are (h,k-c) and (h,k+c) or
(-1,5-5) and (-1,5+5) which is (-1,0) and (-1,10).

Example 2:

Find the vertices of the given hyperbolic equation:y^2/16-x^2/9=1

Step 1:
center (0, 0)

Step 2:
a2 = 9; a= 3
b2 = 16; b = 4

Step 3:
Vertices = (0, 3)
= (0, -3)

Step 4:
Find c value
c2=a2+b2
c2 = 9 + 16
c2 = 25
c = 5

Focus = (0, 5)
= (0, -5)

The vertical hyperbola of the given equation is ((3, 0) (-3, 0)

Three Distance Theorem

The three-distance theorem states that there are at most three distinct gaps between consecutive elements in the set of fractional parts of the first n multiples of any real number. In a real world of three - dimensions, we must extend our knowledge of geometry ti three dimensional space.

To find the distance between two points whose co-ordinates are given. Let P( x1 , y1 , z1 ) and Q ( a2, b2, c2 ) be the two given points. Through P and Q draw planes parallel to the co-ordinate planes to form a rectangular box whose one diagonal is PQ. Geometrically to find the distance PQ is the computing of the length of the diagonal PQ of the box ( as shown in the figure) by means of Pythagoras theorem.



Since MQ is perpendicular to the plane PAMB and PM lies in the plane so MQ is perpendicular to PM `=>` `angle` PMQ = 900 .

in triangle PMQ, `angle` PMQ = 900 , therefore by pythagoras theorem we get,

PQ2 = PM2 + MQ2 ........(1)

since, AM is perpendicular to AP, so `angle` MAP = 900

In triangle AMP, `angle` MAP = 900 , therefore, by pythagoras theorem

PM2 = AP2 + AM2 .....(2)

From (1) and (2), we get

PQ2 = AP2+ AM2+ MQ2 ......(3)

Clearly from the figure

AP = x2 - x1 , AM = y2 -y1 , MQ = z2-z1

Substituting these values in (3) we get

PQ2 = (x2 - x1 ) + ( y2 - y1 ) + (z 2- z1)

PQ = `sqrt(( x_2-x_1)+ ( y_2-y_1) + (z_2-z_1))`

Distance from origin: The distance of P(x,1y,1z1) from the origin o(0,0,0) is `sqrt((x_1-0)^2 +( y_2 - 0)^2 + (z_2 - 0)^2)` = `sqrt((x_1)^2 + (y_1)^2 + (z_1)^2)`



Example Problems on three Distance Theorem:

Problem1: Find ' a' so that the distance between ( -2,1,-3) and ( 1, 3, -6) be 7 units?

Solution:

Let A, B be ( -2,1,3) ,( 1, 3, -6) respectively then

`=>` `sqrt((a-2)^2 + (3-1)^2 + (-6 + 3 )^2)` = 7

`=>` a2 + a + 4 + 4 + 9 = 49 `=>` a2 + 4a - 32 = 0

`=>` (a-4) (a+ 8) = 0 `=>` a= 4 , -8

Problem2: Find the distance between the points A(1,2,3) from he point B(3, 6, 8)?

Solution: The distance

`sqrt((3-1)^2 +(6-3)^2 + ( 8-3)^2)`

AB = `sqrt(2^2 + 3 ^2 + 5^2)`

AB = `sqrt(4 + 9 + 25)`

AB = `sqrt(38)` <br>

Practice Problems on three Distance Theorem

Problem2: Find the distance of the point P( 4, 6, 8) from the point Q ( 9,5,2)? ( Answer: 3`sqrt(7)` )

problem3: By using the distance formula prove that the points A( 3, -5, 1), B ( -1, 0 , 8) and C( 7, -10, -6) are collinear ("Answer : The sum of two side is equal to the third side so the points are collinear)