Factoring Polynomials using Algebra

The reverse process of multiplying polynomials is defined as factoring polynomials. Consider that when we factor a number, we are searching for prime factors that multiply together to give the number. When a polynomial is factorized, usually only the polynomials are broken down to have integer coefficients and constants. Simplest way for factoring is that a common factor for every term. So we can factor out the common factor in the polynomials.

For example,  8=4*2, or 16=4*4


Problems:


Factoring Polynomials using Algebra Tiles:

Example for Factoring Polynomials using Algebra Tiles 1:

Factorize the polynomial x3 – 5x2 – 12x + 36

Solution for Factoring Polynomials using Algebra Tiles 1:

Sum of the coefficients of terms: 1–5–12 + 36 = 18 ≠ 0. ∴ (x–1) is not a factor.

Sum of the coefficients of even degree terms = –5 + 36 = 31

Sum of the coefficients of odd degree terms = 1 – 12 = –11

Since they are not equal we guess that (x + 1) is also not a factor. Let us check whether x – 2

is a factor. By synthetic division method

                 2 | 1       -5     -12        +36

                    |

                    |          +2      -6          -36
                    ________________________

                      1        -3     -18     |    0      

                  _________________________



Since the remainder is 0, (x – 2) is a factor. To find other factors

                      x2 – 3x – 18 = x2 – 6x + 3x – 18

                                       = x (x–6) + 3 (x–6) = (x + 3) (x – 6)

                       Therefore, x3 – 5x2 – 12x + 36 = (x–2) (x–6) (x+3)


Sample problem

Example for Factoring Polynomials using Algebra Tiles 2:

Factorize 2x3 + x2 – 5x + 2

Solution for Factoring Polynomials using Algebra Tiles 2:

Since the sum of the coefficients of all the terms: 2 + 1 – 5 + 2 = 5 – 5 = 0

We guess that (x – 1) is a factor.

By synthetic division,

                              1 | 2         +2        -5        +2

                                |

                                |              2        +3        -2

                                 ________________________

                                     2         3         -2     |   0        

                                _________________________

                     Remainder is 0. Quotient is 2x2 + 3x – 2

To find other factors, factorize the quotient,

                   2x2 + 3x – 2 = 2x2 + 4x – x – 2

                                = 2x (x + 2) – 1 (x + 2) = (x + 2) (2x – 1)

                   ∴ 2x3 + x2 – 5x + 2 = (x – 1) (x + 2) (2x – 1)

Mean Median Average

Mean

The mean is the average of the numbers.

It is easy to evaluate: Just add up all the numbers, then divide by how many numbers there are.

Example:

what is the mean of 2, 7 and 9?

Solution: 2 + 7 + 9 = 18
= 18 ÷ 3

= 6
Mean is 6

Median


The middle number (in a sorted list of numbers). Half the numbers in the listing are less, and half the numbers are greater are called as the median.

To find the Median, place the numbers you are given in value arrange and find the middle number.

If there are two middle numbers then average those two numbers.

Average

Average - The middle or most general in a set of data. There are three types of standard in mathematics - the mean, the median and the mode.


Concept of mean median average


Average

The average is a calculated "central" value or rate of a set of numbers.

It is easy to calculate: add up all the numbers and divide by how many numbers there are and you will have the average.

Example:

the average of 4, 6 and 11

Solution:   (4+6+11)/3

= 21/3

= 7

Average is 7

Mean

The most general expression for the mean of a statistical distribution with a discrete random variable is the mathematical average of all the terms. To compute it, add up the values of all the terms and then divide by the number of terms. This expression is also called the arithmetic mean.

Median

The median of a distribution with a discrete random or chance variable depends on whether the number of terms in the distribution is even or odd. If the number of conditions is odd, then the median is the value of the term in the middle. This is the value such that the number of conditions having values greater than or equal to it is the same as the number of terms having values less than or equal to it. If the number of conditions is even, then the median is the average of the two terms in the middle, such that the number of terms having values greater than or equal to it is the same as the number of terms having values less than or equal to it.


Example of mean median average


A student has gotten the subsequent grades on his tests: 87, 95, 76, and 88. He wants an 85 or better overall. What is the least amount of grade he must get on the last test in order to achieve that average?

The unknown score is "x". Then the desired average is:

(87 + 95 + 76 + 88 + x) ÷ 5 = 85

Multiplying through by 5 and simplifying, I get:

87 + 95 + 76 + 88 + x = 425
346 + x = 425
x = 79

He needs to get at least a 79 on the last test.

The Binomial Distribution

Binomial Distribution is a statistical experiment which means the number of successes in n repeated trials of a binomial experiment. It is also called as Bernoulli distribution or Bernoulli trial.

For example:

For a clinical trial, a patient may live or die. Here the researcher faces the number of survivors and not how much time the patient lives after treatment.


Properties and Formula for binomial distribution


For example:

For a clinical trial, a patient may live or die. Here the researcher faces the number of survivors and not how much time the patient lives after treatment.

We take a coin and flipped two times. Here we calculate the count of number of heads(successes). So the binomial distribution is

Number of heads          Probability

No head                             0.25

One head                           0.5

Two head                           0.25

Properties of Binomial Distribution

The experiment  has n repeated trials.

Each trial can have two possible outcomes. One is success and another one is failure.

Here the trials are independent.

Mean = n * P.
Variance = n * P * (1 – P).
Standard Deviation  =  sqrt[ n * P * ( 1 – P ) ].


Binomial distribution Formula

b(x; n, P) = nCx * Px * (1 - P)n – x

Here the Notation are,

B(x; n, P)   =  Binomial Probability.

X   =  successes

N   =  number of trials

P    =  Probability of success

nCx  = Number of combinations of n trials, x is success.


Example Problem(the binomial distribution)


A die is tossed 6 times. What is the Probability of getting exactly 2 fours?

Solution

Here n = 6, x = 2,  probability of success on a single trial = 1/ 6 or 01.167.

Therefore, The binomial probability is,

b( 2; 6, 0.167 )             =  6C2 * ( 0.167 )2 * ( 1 – 0.167)6 – 2

=  ( 6! / 2! * (6-2)!) * 0.0279 * ( 0.833)4

=  (6! / 2! * 4!) * 0.0279 * 0.481

= 15 * 0.0279 * 0.481

b( 2; 6, 0.167 )             = 0.201. Answer.

Cumulative Binomial probability

It  refers to the binomial probability falls within a specified range that is greater than or equal to a mentioned lower limit and less than or equal to a mentioned upper limit.

For example

Cumulative binomial probability of obtaining 5 or fewer heads in 10 times of a coin.

b( x <= 5; 10, 0.5)=   b( x = 0; 10, 0.5) + b( x = 1; 10, 0.5) +…… + b ( x = 5; 10, 0.5)

Rational Equations Solving Online

Learning equation of the form P(x)/Q(x) over the set of real numbers and Q(x) ≠ 0 where P(x) and Q(x) are two polynomials is called rational equation. Rational equations made easy for learning through online.

Example for rational equations: 3/8 = 3/(d + 4)

(x4 + x3 + x + 1)/(x + 5) = 9/5

1/(x + 9) = 3

12a-6 = 3a-2

Example problem for rational equations solving online:


The following example problems give idea for solving rational equations.

Example 1:

Find x value of the rational equation (6x + 36) / (9x + 54) = 6x

Solution:

Step 1: Given equation

(6x + 36) / (9x + 54) = 6x

Step 2: Take 6 and 9 out as common term from numerator and denominator respectively.

6(x + 6) / 9(x + 6) = 6x

Step 3: Cancel the term (x + 6)

6/9 = 6x

Step 4: Rearrange the above equation,

6/(6 * 9) = x

1/9 = x

x = 1/9

Example 2:

Find out x value for the equation x/(x - 4) + (1/x - 8) = 6/(x2 - 12x + 32)

Solution:

Step 1: Given equation

x/(x - 4) + (1/x - 8) = 6/(x2 - 12x + 32)

Step 2: Replace the term (x2 - 12x + 32) by (x - 4) (x - 8)

x/(x - 4) + (1/x - 8) = 6/ (x - 4) (x - 8)

Step 3: Make common denominator.

(x/(x - 4))((x - 8)/(x - 8)) + ((1/x - 8))((x - 4)/(x -4)) = 6/ (x - 4) (x - 8)

(x2 - 8x) / (x - 4)(x - 8) + (x - 4) / (x - 4)(x - 8) =  6/ (x - 4) (x - 8)

Step 4: Cancel the term (x - 4)(x - 8), we get

(x2 - 8x) + (x - 4) = 6

Step 5: Rearrange the above equation

x2 - 7x - 4 = 6

x2 - 7x - 4 - 6 = 0

x2 - 7x - 10 = 0

Step 6: On factorizing, we get

x = 8.217, - 1.217

This is how the rational equations can be solved through online.


Homework problem for rational equations solving online:


A few homework problems are given below  for solving rational equations.

1) Simplify and find x for the equation (4x + 12)/(4x + 32) = 1/(x + 8)

2) Find x value for the equation 1/(x + 7) = 4

3) Solve and find x for the equation (x2 - 12x + 32)/(x - 8) = 8

Solutions:

1) x = -2

2) x = -5

3) x = 12

Improper Integrals Solve Online

In calculus, an improper integral is the limit of a definite integral as an endpoint of the interval of integration approaches either a specified real number or ∞ or −∞ or, in some cases, as both endpoints approach limits.

Specifically, an improper integral is a limit of the form


In which one takes a limit in one or the other (or sometimes both) endpoints. (Source: Wikipedia)


Example problems for solve online improper integrals


Online improper integrals example 1:

Solve:

Find the value of the integration function

`lim_(b->0) int_3^b(2x + 3)dx`

Solution:

Integrate the given function with respect to x, we get

= `lim_(b->0)` [2 `(x^2 / 2)` + 3x]b3

Substitute the lower and upper limits, we get

=  `lim_(b->0)` (b2 - 3b) - (9 + 9)

Substituting the value of b, we get

= (0) - 18

After substituting the limits, we get

= - 18

Answer:

The final answer is - 18

Online improper integrals example 2:

Solve:

Find the value of the integration function

`lim_(b->2) int_0^b(7x^3 + 3x^2)dx`

Solution:

Integrate the given function with respect to x, we get

= `lim_(b->2)` [7 `(x^4 / 4)` + 3`(x^3 / 3)` ]b0

Substitute the lower and upper limits, we get

=  `lim_(b->2)` [`(7 / 4)` b4 + b3] - (0)

Substituting the value of b, we get

= [(7 / 4) 24 + 23] - 0

After substituting the limits, we get

= 36

Answer:

The final answer is 36

Online improper integrals example 3:

Solve:

Find the value of the integration function

`lim_(b->1) int_2^b(6x^2 + 18x)dx`

Solution:
Integrate the given function with respect to x, we get

= `lim_(b->1)` [6 `(x^3 / 3)` + 18 `(x^2 / 2)` ]b2

Substitute the lower and upper limits, we get

=  `lim_(b->1)` (2b3 + 9b2) - (16 + 36)

Substituting the value of b, we get

= (11) - 52

After substituting the limits, we get

= - 41

Answer:

The final answer is - 41


Practice problems for solve online improper integrals


Online improper integrals example 1:

Solve:

Find the value of integration of the function

`lim_(b->0) int_5^b(6x)dx`

Answer:

The final answer is - 75

Online improper integrals example 2:

Solve:

Find the value of integration of the function

`lim_(b->6) int_0^b(12x + 2)dx`

Answer:

The final answer is 228

Mean Statistics Definition

Introduction:

Let us see the introduction of mean statistics. In statistics the mean is a mathematical average of a set of numbers. The average of the mean statistics is dividing the total by the number of scores and calculated by adding up two or more scores. Mean Statistics like as many other sciences of a developing discipline. In statistics has been defined in different times and different manners. We discuss the definitions of mean statistics.


Mean Definition:

The definition of the mean in mathematics, the statistical discrete random variable average of all the terms. A finite set of terms are forms rarely used in statistics are other expressions for the mean. The lowercase Greek letter mu (µ) is the expected value. The average value of mean statistics is a numerical set. The number of members in the group of numbers is by dividing the sum of a set of numbers.

Statistics Definition:

Statistics is definition of the technique or the scientific method is used to analyzing, collecting, interpreting, classifying, and data. The statistics is used to obtain the analyses, summaries, compare and present the numerical data. The statistics definitions are a science which deals with the application and investigates the statistical significance. 

These are the definitions of mean statistics.

Examples:

1. Find the mean of the following numbers 2, 5, 23, 15, 15, and 6.

Solution:

The given numbers are 2, 5, 23, 15, 15, and 6.

The average of set of numbers =2+5+23+15+15+6 / 6

= 36 / 6

=6

Answer: Mean is 6

2. Find the mean of the following numbers 5, 7, 34, 56, 23, 46, 56, and 12.

Solution:

The given numbers are 5, 7, 34, 56, 23, 46, 56, and 12.

The average of set of numbers =5+7+34+56+23+46+56+12/8

=   239 / 8

=29.88

Answer:

Mean is 29.8

3. Find the mean of the following numbers 56, 34, 23,  and 15.

Solution:

The given numbers are 56, 34, 23,  and 15.

The average of set of numbers=56+34+23+15/4

= 128 / 4

=32

Answer: Mean is 32

These are the example problems for definition of mean.

Number line

Learn on number line in this page and gain quality algebra help. First a brief introduction is given on the whole concept of number line and further the topic is explained.

Mathematically, a number line is referred to as an image of a straight line, which has several points on it as real numbers. Those points are indicated by integers. Particularly marked points with even spaces expose them on the line. Thus, real numbers are represented in each direction. Number line has both positive number and negative number held at correct points on line, in that zero is the center point of the number line, while right side of zero is positive number and left side of zero is negative number.



Numbers on a line can be represented horizontally as well as vertically. Normally, number line is represented horizontally.


Steps to draw


The following  are the steps to draw the number line-

• Step 1: Draw a horizontal straight line. Because mostly the number line is represented as horizontal line
• Step 2: Draw the arrow on both ends of number line.
• Step 3: Point the origin zero on the number line.
• Step 4: Write positive integer on the right side of origin with even spaces.
• Step 5: Write negative integer on the left side of origin with even spaces.
• Step 6: Mark all integers over the number line.
• Step 7: Plot the answers for given question.

Points on a number line


Ex 1: There are three persons on the origin namely as A, B, C. A walks backwards 2 points; B & C walks towards 4 points and 3 points respectively. Point out their current position.

Sol:




Points on a number line in decimal form

Case 1: Positive Decimal on number line

Let take the positive decimal as 3.75 and points it as in number line.

Generally, we know that decimal have two section,

• before the decimal point
• after the decimal point.

When make the decimal number on number line,

Step 1: keep the number before the decimal point. Here the 3 is the positive number before the decimal point.

Step 2: Now after the positive number three on a number line, we count the number in number line in between 3 and four as per the number have after the decimal point.

Step 3: Now we get the correct points for the given decimal as 3.75

Case 2: Negative decimal number on simple number line

Let take the number as -6.5 and mark it on the number line.



Step 1: keep the number before the decimal point. Here the 6 is the negative number before the decimal point.

Step 2: Now after the negative number 6 on a number line, we count the negative number in number line in between 6 and 7 as per the number have after the decimal point.

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)

Dihedral Angle Calculator

Dihedral angle is the one of the special kind of angles in mathematical geometry.  Dihedral angle is an angle that forms between two plans.  Usually a plane is the three dimensional flat surface.  The points in the plane take the form of (x, y, z).  In this topic we are going to study about how to calculate the dihedral angle when the points of the plane are given through the dihedral angle calculator.

Explanation about Dihedral Angle through Calculator

Important Guidelines:

The following steps are the important guide lines to calculate the dihedral angle.

• First we have to choose the three points on the plane 1 and then choose the three points on the plane 2

• Enter the (x, y, Z) values for each point on the dihedral angle calculator we have to get the equation of the plane 1 and plane 2 separately

• The equation of the plane 1 is taking the form of A1x + B1y + C1z + D1 =0 and then the equation of the plane 2 is taking the form of A2x + B2y + C2z + D2 =0

Now we have to calculate the dihedral angle through the following formula,

cos`alpha` = `((A_1)(A_2) + (B_1)(B_2) + (C_1) (C_2))/ (sqrt((A_1)^2 +(B_1)^2 +(C_1)^2) * sqrt((A_2)^2 +(B_2)^2 +(C_2)^2))`

This formula is used for our manual calculation.

Example Problem on Calculate Dihedral Angle via Calculator

Calculate the dihedral angle for the plane 1 and plane 2.  The points on the plane 1 are (1, 2, 3) and (3, 2, 1) and then (2, 1, 3) and the points on the plane 2 are (4, 1, 2) and (1, 4, 4) and then (2, 4, 2)

Solution:

Now we have to substitute the values of the points in the following dihedral Angle Calculator,



This calculator gives the distance between each points and equation of the each plane and then the angle in between the planes in both degrees and radians.  These are the main usage of the dihedral angle calculator.

Distributive Law Definition

The distributive law is placed in set theory. The distributive law is the piece of sets that used to learn about the sets. Group of items is called the sets. Set theory have some operations union, intersection, difference, complement and performs some law.The set theory is used to select the number of objects simultaneously. Here we will see about the distributive law definition with examples.

Examples - Distributive Law Definition:

Now we are going to solve the examples for distributive law definition.

AU (B∩C) = (AUB) ∩ (AUC)

A∩ (BUC) = (A∩B) U (A∩C)

Example 1

Get the solution of the given sets by using distributive law. A={ a,b,c,d } B={ b,c,g } and C={ b,c,k,m }.

Solution:

The given sets are A={a,b,c,d} B={b,c,g} and C={b,c,k,m}.

Distributive law for set is as follows,

i)AU (B∩C)=(AUB) ∩(AUC)

ii)A∩(BUC)=(A∩B) U (A∩C)

i)AU (B∩C)=(AUB) ∩(AUC)

First find the solution left hand side.

AU (B∩C)

First we can solve the inner bracket set.

B∩C

Intersection means we choose the common elements from the set B and C.

So B∩C =   B= {b, c, g} ∩ C= {b, c, k, m}.

B∩C = {b, c}

Now find the AU (B∩C)

Grouping the values of A and B∩C

AU (B∩C) = {a, b, c, d} U (b, c)

= {a, b, c, d}

So AU (B∩C) = {a, b, c, d} -------------- (1)

(AUB) ∩ (AUC)

Groping the values of A and B sets.

A= {a, b, c, d} B= {b, c, g}

AUB = A= {a, b, c, d} U B= {b, c, g}

= {a, b, c, d, g}

Joining the values of A and C sets.

AUC:

A= {a, b, c, d} U C= {b, c, k, m}.

= {a, b, c, d, k, m}

(AUB) ∩ (AUC):

AUB ∩AUC = {a, b, c, d, g} ∩ {a, b, c, d, k, m}

(AUB) ∩ (AUC) = {a, b, c, d} ------------- (2)

So AU (B∩C) = (AUB) ∩ (AUC)

ii) A∩ (BUC) = (A∩B) U (A∩C)

A={ a,b,c,d } B={ b,c,g } and C={ b,c,k,m }.

Now solve the left side.

A∩ (BUC)

BUC:

= {b, c, g} U {b, c, k, m}

BUC = {b, c, g, k, m}

A∩ (B U C):

A= {a, b, c, d} ∩ {b, c, g, k, m}

= {b, c} -------------- (1)

(A∩B) U (A∩C)

A∩B= {a, b, c, d} ∩ B= {b, c, g}

= {b, c}

A∩C= A={ a,b,c,d } ∩ C={ b,c,k,m }.

= {b, c}

(A∩B) U (A∩C) = {b, c}

A∩ (BUC) = (A∩B) U (A∩C)


Example 2 – Distributive Law Definition:

A= {2, 7, 9} B= {5, 10, 11} C= {5, 8, 10}

What is AU (B∩C)

(B∩C):

= {5, 10, 11} ∩ {5, 8, 10}

= {5, 10}

AU (B∩C):

= {2, 7, 9} U {5, 10}

= {2, 7, 9, 5, 10}

These are examples for distributive law definition.

That’s all about distributive law definition.

Resultant of three Vectors

This article is about resultant of three vectors. Resultant of three vectors is vectors that results from adding two or more vectors together. There are two ways to calculate the resultant vector. There are many different websites to help the students on obtaining the resultant of three vectors. Tutor vista is the famous website to provide help on finding the resultant of three vectors with the help of highly qualified tutors. Below are some of the problems regarding resultant of three vectors.

Methods - Resultant of three Vectors:

There are two different methods to find the resultant vector.



The head to tail method:

The head to tail method is represented with the arrow mark. The arrow mark is the head and the tail is with out the arrow mark. Place the two vectors such that the head of the vectors joins the tail of another vector. Draw the resultant vector as shown in the figure above. To find the resultant vectors we use the Pythagorean theorem. There is also another method to calculate the resultant vector which is studied in college grade.

Parallelogram Method:

The another method is parallelogram method. To use this parallelogram method it is very important to be well know with trigonometry basics.


Example Problems - Resultant of three Vectors

vector addition is also the resultant of 3 vectors.

Example 1: The vector a = 3 and vector b = 4. Find the resultant vector using Pythagorean theorem.

Solution

We know that Pythagorean theorem is a2+b2 = c2

So given a =3 and b = 4

So the resultant vector c2 = a2 + b2

c =` sqrt (a^2 + b^2)`

= `sqrt (3^2 + 4^2)`

=` sqrt (9+16)`

= `sqrt(25)`

c = 5

So the resultant vector is 5


Example 2: Add 5`veci` +4`vecj`+3`veck` with 2`veci`+1`vecj`+0`veck`

Solution

Here two set of vectors are given. we have to perform the sum of two vectors by adding the magnitudes alone

5`veci` + 4`vecj` +3`veck`
2`veci` + 1`vecj` +0`veck`
----------------------
7`veci` + 5`vecj` + 3`veck`
-----------------------
Example 3: `veca =4vecp + 3vecq+2vecr` , `vec b = 5vecp +2vecq+vecr` and `vecc = 2 vecp + vecq + 4 vecr`
Solution
Resultant vector is `veca + vecb + vecc`
`4 vecp + 3vecq+2 vecr`
`5 vecp + 2vecq+ vecr`
`2 vecp + vecq + 4 vecr`
---------------------------------
`11 vecp +6 vecq + 7 vecr`
---------------------------------