Polynomial Matrices

A polynomial matrices or matrix polynomial is a  matrix whose elements are univariate or multivariate polynomials.A univariate polynomial matrix P of degree p is define like:sum_(n=0)^pA(n)x^(n)=A(0)+A(1)x+A(2)x^(2)+....+A(p)x^(p) where A(p) is non-zero and A(i) indicate a matrix of constant coefficients. Hence a polynomial matrix is the matrix-equivalent of a polynomial, by means of every one element of the matrix satisfying the classification of a polynomial of degree p.

Properties of polynomial matrices

A polynomial matrix in excess of a field with determinant equivalent to a non-zero constant is called unimodular, and have an inverse, which is also a polynomial matrix.
Note, that the simply scalar unimodular polynomials are polynomials of degree 0 - nonzero constants, for the reason that an inverse of an arbitrary polynomial of high degree is a rational function.
The roots of a polynomial matrix in excess of the complex numbers are the points in the complex plane wherever the matrix loses rank.

Characteristic polynomial of a product of two matrices

If A and B are two square n×n matrices then,attribute polynomials of AB and BA match:

PAB(t)=PBA(t).

If A is m×n-matrix and B is n×m matrices such that m
PAB(t)=tn-mPAB(t)

Polynomial in t and in the entry of A and B is a general polynomial identity. It consequently suffice to verify it on an open set ofparameter value in the complex numbers.

The tuples (A,B,t) wherever A is an invertible complex n by n matrix,

B is any complex n by n matrix,

and t is any complex number since an open set in complex space of dimension 2n2 + 1. When A is non-singular our result follow from the fact that AB and BA are similar:

BA=A-1(AB)A.

Example 1:

An example the 3x3 polynomial matrices

P=[[1,x^(2),x],[0,2x,2],[8x+2,x^2-1,0]]

=[[1,0,0],[0,0,2],[2,-1,0]]+[[0,0,1],[0,2,0],[3,0,0]]x+[[0,1,0],[0,0,0],[0,1,0]]xx^(2)

Example 2:

Find the eign value of given polynomial matrices

P=[[3,3],[0,6]]

The polynomial has the characteristic equation

0=det(P-λI)

=det[[3-lambda,3],[0,6-lambda]]

18-6lambda -3lambda + λ2

18-9lambda +18

λ2-3λ-6λ+18

λ(λ-3)-6(λ-3)

(λ-3)(λ-6)

λ=3,andλ=6

The eigenvalues of these matrices are 3,6

Example 3: Find the product of the given matrices M1=[[1,2],[3,4]] and M2=[[8,3],[2,7]]

The given polynomial is

M1=[[1,2],[3,4]]


M2=[[8,3],[2,7]]

The product of the given matrices M1 and M2  =M1xM2

M1xM2    =[[1,2],[3,4]]xx [[8,3],[2,7]]

The product of the given  matrices is=[[12,17],[32,37]]

Least Common Denominator

If the denominators are unlike, then we can find LCD (Least common Denominator) of the given denominators.

Least Common Denominator is the smallest positive(least) integer which is common in multiples of the denominators.

For example, given fractions are 1/3 and 1/6. Find LCD.

List the multiples of 3:   3, 6, 9, 12, 15, 18, 21,...

Multiples of 6:   6, 12, 18, 24,...

Here, 6 is the lowest common term for both the multiples of 3 and multiples of 6.

The answer is 6, and that is the Least Common Denominator.

There are five examples for least common denominator. From these examples for least common denominator, we can get clear view about least common denominator. Let us see the examples for least common denominator in the following section.

Examples on examples for least common denominator:

We are going to explain examples for least common denominator.

Example 1:

Find the least common denominator of the fractions; 1/5,1/3

Solution:

Here, the denominators are 5 and 3.

The common denominator of 5 and 3 is 15.

Multiples of 5: 5,10,15,20,…..

Multiples of 3: 3,6,9,12,15,18,….

Here, 15 is the lowest common term for both the multiples of 5 and multiples of 3.

The answer is 15, and that is the Least Common Denominator.

Example 2:

If the given fractions are 3/4,1/3, find the least common denominator.

Solution:

Here, the denominators are 4 and 3.

The common denominator of 4 and 3 is 12.

Multiples of 4: 4,8,12,16,20….

Multiples of 3: 3,6,9,12,15….

Here, 12 is the lowest common term for both the multiples of 4 and multiples of 3.

So, the Least Common Denominator is 12.

Example 3:

Find the least common denominator of ; 5/6, 2/15

Solution:

Here, the denominators are 6 and 15.

The common denominator of 6 and 15 is 30.

Multiples of 6:  6,12,18,24,30…

Multiples of 15: 15,30,45,60….

Here, 30 is the lowest common term for both the multiples of 6 and multiples of 15.

So, the Least common denominator is 30.

Example 4 on examples for least common denominator:

What is the least common denominator of the fractions;  5/12, 11/18

Solution:

Here, the denominators are 12 and 18.

The common denominator of 12 and 18 is 36.

Multiples of 12:  12,24,36,48,….

Multiples of 18: 18,36,54,….

Here, 36 is the lowest common term for both the multiples of 12 and multiples of 18.

So, the Least common denominator is 36.

Example 5:

1/5 + 1/6 + 1/15  What is the LCD?

Solution:

First we list the multiples of each denominator.

Multiples of 5 are 10, 15, 20, 25, 30, 35, 40,...

Multiples of 6 are 12, 18, 24, 30, 36, 42, 48,...

Multiples of 15 are 30, 45, 60, 75, 90,....

Here, 30 is the lowest common term for both the multiples of 5 and multiples of 6 and multiples of 15 .

So, the Least common denominator is 30.

Therefore, Examples for least common denominator are explained.

Learning Parallel Lines

Learning parallel lines is very easy. As the name it self indicates that there will be some lines that are parallel to each other. Think logically if two lines are parallel will they ever meet at some point. The answer is no. Because the two parallel lines maintain equal distance between them so there is no chance that they meet at some point. So we can conclude that two lines are parallel if and only if they are coplanar and never meet each other that is they maintain same distance apart.

Transversals are very important while learning parallel lines.

Tranversal can be defined as a line that intersects two or more  lines, at different points. Simply a line that crosses two or more lines  is called transversal. With the help of transversals we can say whether the two given lines are parallel or not. Figure below shows how a transversal looks.



How can we know the lines are parallel or not?

Answer is with the help of some angles that are formed when transversal passes through pair of coplanar lines. These angles are very important while learning parallel lines.The angles are

Corresponding Angles
Alternate Interior Angles
Alternate Exterior Angles
Co-Interior Angles


Corresponding angles:

In our figure the corresponding angles are " a,e " , " d, h" , "b , f" , "c,g".

Alternate Interior angles:

The name it self indicates that alternate means on the other side and interior means inner angles. The Alternate Interior angles are "d,f" and "c,e".

Alternate exterior angles:


The name it self indicates that alternate means on the other side and exterior means outer angles. Alternate exterior angles in the given figure are "a,g" and "b,h".

Co-interior angles:

In the given figure Co-interior interior angles are "d,e" and "c,f".

Now for two lines to be parallel corresponding angles, alternate Interior angles, alternate exterior angles must be equal and sum of Co-interior  angles must be equal to 180. Even if one condition is satisfied it is enough as automatically all the other will satisfy.

This is all about learning parallel lines. Hope you enjoyed it......

Definition of Probability Distribution

Probability distribution function referred as p(x) which is a function that satisfies below properties:

Probability that x can obtain particular value is p(x).

p[X,x]=p(x)=px

p(x) positive for all real x.
Sum of p(x) over all feasible values of x is 1, that is

`sum_(pj)pj=1`

Where, j- all possible values that x can have and pj is probability at xj.

One significance of properties 2,3 is that 0 <= p(x) <= 1.

Steps used for how to determine probability distribution:

Using following steps used to how determine probability distribution:

Step 1 for how to determine probability distribution:

Plot the data for a image illustration of the data type.

Step 2 for how to determine probability distribution:

First steps is to determining what data distribution one has - and thus the equation type to use to model the data - is to rule out what it cannot be.

The data sets cannot be a discrete uniform distribution if there are any crest in the data set.
The data is not Poisson or binomial if the data has more than one crest.
If it contains a single arc, no secondary crests, and contains a slow slope on each side, it may be Poisson or a gamma distribution. But it is not discrete uniform distribution.
If the data is regularly distributed, and it is without a slant in the direction of one side, it is secure to rule out a gamma or Weibull distribution.
If the function has an even distribution or a crest in the center of the graphed outcomes, it is not a geometric distribution or an exponential distribution.
If the incidence of a factor differ with an environmental variable, it probably is not a Poisson distribution.


Step 3 for how to determine probability distribution:

After probability distribution type has been tapering downward, do an R squared examination of each probable type of probability distribution. The one with the maximum R squared value is most possible correct.

Step 4 for how to determine probability distribution:

Remove one outlier data point. Now recalculate R squared. If the same probability distribution form comes up as the neighboring match, then there is high confidence that this is the correct probability distribution to use for group of data.

From above steps we can understand how to use probability distribution.

Polynomial Exponent

Polynomial of a single term is called as Monomial. If the Polynomial has the two terms Sum or Difference then it is called a Bi-nomial, the Sums or Differences of a three-term Polynomial is called a Trinomial. The Sums and/or Differences of polynomial of four or more terms are simply called polynomial.

Two or more terms of a Polynomial exponent that has the equivalent variable and precisely the same whole number polynomial exponent are called Like Terms. For suitable example the terms; 6X3 and -7X3, are Like Terms, since the variable X is the similar variable in both terms, and the Whole number exponent, 3, is absolutely the similar polynomial exponent in both terms. The terms;4X3 and 4X4 are not Like Terms although the variable X is the same in both terms but the Whole number polynomial exponent are completely different from other.

Operation of polynomial exponent:

Addition of polynomial exponent:

Two or more polynomials, adding the terms,

Suitable example adding polynomial exponent,

Example1:

(2x2+3x3)+(x2+7x3)

=2x2+x2+3x3+7x3

The variable and exponent must be same then we add the polynomial exponent,

=3x2+10x3

So the result is  =3x2+10x3

Subtraction of polynomial exponent

Example2:

(3x2+3x3)-(x2+7x3)

=3x2-x2+3x3-7x3

The variable and exponent must be same then we subtract the polynomial exponent,

=2x2-4x3

So the result is =2x2-4x3

Multiplication of polynomial exponent:

Two or more polynomial, so multiply their exponent terms. For reasonable example to multiply the following two Terms, 2X2 and 7X2, Then the following terms then first multiply the coefficients (2) (7) and we multiply (X2) (X2) which can be expressed as the following term 14X4, that is, when we multiply variables that are the same.


Different variables in polynomial exponent:

The variables are different and the exponent also different, so we can write the variables and exponents adjoining to each other.

For Example, (7X2)(3Y2)is equal to 21X2Y2.

So the result is 21X2Y2

Similar to all the different variable, polynomial exponent.

To multiply the two polynomial exponent, for example (X-2Y)(X-2Y), then the result is x2-2xy+4y2 . Here we take the first term X of the first Binomial and multiply each term in the second Binomial, then we take the second term in the first Binomial and multiply each term in the second Binomial, and then we add all the terms. Similar to all the polynomial categories, for example trinomial, binomial exponents.

Inverse Trigonometry Definition

The three trigonometry functions are arcsin(x), arccos(x), arctan(x). The inverse trigonometry function is the inverse functions of the trigonometry, written as cos^-1 x , cot^-1 x , csc^-1 x , sec^-1 x , sin^-1 x , and tan^-1 x .

Sine:

H (x) = sin(x)   where   x is in [-pi/2 , pi/2]

Cosine:

G (x) = cos(x) where   x is in [0 , pi ]

Tangent:

F (x) = tan(x)   where   x is in (-pi/2 , pi/2 )

Understanding Inverse Trig Function is always challenging for me but thanks to all math help websites to help me out.

Cosecant Definition

Csc theta = (hypotenuse)/(opposite)

Secant Definition

Sec theta = (hypotenuse)/(adjacent)

Cotangent Definition

Cot theta = (adjacent)/(opposite)

Examples of Inverse trigonometry functions:

Example 1:

Find the angle of x in the below diagram. Give the answer in four decimal points.

Solution:

sin x =2.3/8.15

x = sin -1(2.3/8.15 )

= 16.3921˚

Example 2:

Solve sin(cos-1 x )

Solution:
Let z = sin ( cos-1 x ) and y = cos-1 x so that z = sin  y. y = cos-1 x may also be marked as


cos y = x with pi / 2 <= y <=- pi / 2

Also

sin2y + cos2y = 1

Substitute cos y by x and solve for cos y to obtain

sin y = + or - sqrt (1 - x^2)

But pi / 2  <= y <= - pi / 2 so that cos y is positive

z = sin y = sin(cos-1 x) = sqrt (1 - x^2)

Example 3:

Evaluate the following sin-1( cos ((7 pi) / 4 ))

Solution:
sin-1( cos ( y ) ) = y only for  pi / 2 <=  y <= -pi / 2 . So we initial transform the given expression noting that cos ((7pi) / 4 ) = cos (-pi / 4 ) as follows

sin-1( cos ((7 pi) / 4 )) = sin-1( cos ( pi / 4 ))  - pi / 4  was selected since it satisfies the condition  pi / 2 <=  y <= - pi / 2 . Hence

sin-1( cos ((7 pi) / 4 )) =pi/4

Example 4:

With the given value find inverse of tan.

Solution:

Since tangent, again, is the trig function associated with opposite and adjacent sides, we apply the inverse of tangent to calculate the measure of this angle.

Tan 16 = (opposite)/(adjacent)

Tan 16 = ((bar(AC))/(bar(CB)))

Tan-1(7/24 ) = 16 o

Associative Property Sum

In mathematics, if x, y, z be any three variables involving the addition operation and then satisfy the following condition,

x + (y + z) = (x + y) + z

This kind of property is called associative property sum or addition. In associative property sum, the sum of algebraic expression provides the same result in changing the order of brackets in this expression. Such as

x + y + z = (x + y) + z = x + (y + z)



Examples for associative property sum:

Example 1 on associative property sum:

Prove the associative property sum of the given expression.

(a) 4 + 6 + 7

Solution:

(a) Let x, y, and z be 4, 6 and 7 respectively.

Therefore, the given expression satisfy the following condition,

x + y + z = x + (y + z) = (x + y) + z

Here,

Prove the following expression

4 + (6 + 7) = (4 + 6) + 7

L.H.S = 4 + (6 + 7)        (first perform the addition operation within the bracket from left to right,

because brackets are higher priority of the order of operations)

= 4 + (13)      (second perform the addition operation outside of the expression from left to right)

= 17                    ------ (1)

R.H.S = (4 + 6) + 7

= (10) + 7

= 17             --------- (2)

From equation (1) and (2) we get,

L.H.S = R.H.S

i.e., 4 + (6 + 7) = (4 + 6) + 7

Hence, the given expressions satisfy the associative property with sum of numbers.

Example 2 on associative property sum:

(a) 23 + 67 + 34, prove the associative property of sum or addition.

Solution:

(a) Let m, n, and l be 23, 67 and 34 respectively.

Therefore, the given expression satisfy the following condition,

m + n + l = m + (n + l) = (m + n) + l

Here,

Prove the following expression

23 + 67 + 34 = 23 + (67 + 34) = (23 + 67) + 34

23 + 67 + 34 = 124       ---------------- (1)

We consider the remaining expression, such as

23 + (67 + 34) = (23 + 67) + 34

L.H.S = 23 + (67 + 34)

= 23 + (101)

= 124                    ------ (2)

R.H.S = (23 + 67) + 34

= (90) + 34

= 124                  --------- (2)

from equation (1) , (2) and (3) we get,

L.H.S = R.H.S = 124

i.e.,  4 + (6 + 7) = (4 + 6) + 7 = 4 + 6 + 7

Hence, the given expression of sum is satisfied the associative property.

Weight of the Cylinder

Cylinder is the three dimensional figure. Weight of the cylinder is same as the volume of the cylinder. The formula to find the weight of the cylinder is pi * radius2 * height. Height is the total height of the cylinder and radius is the radius of the circular face which is at the bottom and top of the cylinder.

Example diagram and formula - Weight of the cylinder:

Formula - Weight of the cylinder:

Weight of the cylinder = pi * r2 * h, where r`=>` radius of the cylinder and h`=>` height of the cylinder.

Example problems – Weight of the cylinder:

Calculate the weight of the cylinder whose radius is 10cm and height is 12 cm.

Solution:

Given, radius of the cylinder = 10cm and height of the cylinder= 12cm.


The formula to find the weight of the cylinder is pi * r2 * h

= 3.14 * 102 * 12

= 3.14 * 100 * 12

= 3768cm3.

The weight of the cylinder is 3768cm3 .

Calculate radius of the cylinder whose weight is 3000cm3 and height is 30cm.

Solution:

Given weight of the cylinder= 3000cm3 and height of the cylinder = 30cm


Weight of the cylinder = pi * r2 * h

3000   = 3.14 * r2 * 30

3000   = 94.2 * r2.

Divide the above expression by 94.2 on both sides,

3000/94.2 = 94.2 * r2/94.2

31.85        = r2

r               = 5.6cm.

Radius  of the cylinder    = 5.6cm.

Calculate the weight of the cylinder whose radius is 5cm and height is 2 cm.

Solution:

Given, radius of the cylinder= 5cm and height of the cylinder= 2cm.


The formula to find the weight of the cylinder is pi * r2 * h

= 3.14 * 52 * 2

= 3.14 * 25 * 2

= 157cm3.

The weight of the cylinder = 157cm3 .

Calculate radius of the cylinder whose weight is 300cm3 and height is 3cm.

Solution:

Given weight of the cylinder= 300cm3 and height of the cylinder= 3cm


Weight of the cylinder = pi * r2 * h

300   = 3.14 * r2 * 3

300   = 9.42 * r2.

Divide the above expression by 9.42 on both sides,

300/9.42 = 9.42 * r2/9.42

31.85      = r2

r             = 5.6cm.

Radius of the cylinder   = 5.6cm.

Percent Loss Formula

Loss is defined as the difference between the cost price and the selling price. Through the loss we can calculate loss percentage. If they give loss percentage we also find the cost price and the selling price. Let us see the formula and examples for loss and loss percent in this article.

Formula for Loss Percentage

Variation between the cost price and the selling price is called as loss, If the cost price is greater than the selling price means we obtain loss.

Loss = cost price- selling price

Selling price is indicated by S.P and cost price is indicated by C.P.

Based on the cost price we determine the loss.

Formula for loss Percent: Loss Percentage formula = (lossxx 100)/(C.P)

Sometimes they give loss percentage and ask cost price or selling price.

Example on Loss Percent

Example 1: Ragu buys pineapple at 5 for 50Rs ech and sold them at 3 for 25 each. Find his loss and loss percent?

Solution: Given

Number of pineapple bought = lest common divisor of 5,3 =15
Investment or cost price = (50 / 5 ) *15 =150
Selling price = (25 /3)  *15 = 125, here cost price is greater then selling price,we can say it is loss.
Loss amount =cost price – selling price
Loss =150 – 125
Loss amount =25
Loss percentage formula = (lossxx 100)/(C.P)
Loss percentage = (25 xx 100)/(150)
Loss percentage = 16.6%
Loss and loss percentage of Ragu were 25 Rs and 16.6%

Example 2: Cost price =50 Rs, Selling price =45 Rs then find the loss and loss percentage?

Solution: Given

Cost price =50
Selling price =45
Loss = cost price - selling price
Loss =50- 45
Loss =5 Rs.
Loss percentage formula = (lossxx 100)/(C.P)
Loss percentage = (5xx 100)/(50)
Loss percentage = 10%

Example 3: Deepen loss 80 cents on $80 .What is the loss percentage of Deepen?

Solution: Given

Loss 80 paisa on 80 rupees
Here Cost price = $80
Loss =80 cents
Loss percentage formula = (lossxx 100)/(C.P)
Loss percentage = (0.8 xx 100)/(80)
Loss percentage = 1 %

Learn Online Linear Equations

A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable. Linear equations can have one or more variables. Linear equations occur with great regularity in applied mathematics. Source - Wikipedia

Let us Learn linear equation in online. In online we can understand very easily about linear equation.

Learning Concepts of Linear equations in online:

Learning Concepts of Linear equations are as follows:

• In the graph showed, we would like to get the equation of the graphed line. To complete this we start linear equation. We have innocently been functioning with linear equation, so this is NOT a fresh concept for us Most probable you have been working with the standard form of a line, which is Ax + By = C. This is a linear equation with two unfamiliar variables, x and y.

• The ordinary form of the line in the graph is 2x + y = 5, where A =2, B = 1, and C = 5.

• We want to restructure this equation and set it into the y-intercept form which is y = mx + b, in which m is the slope and b is the y-intercept. In this shape, it is very simple to graph. Just explain the standard equation for “y” and you have the y- intercept form.

Example problems for learning linear equations in online:

Example problems for learning linear equations are as follows:

Example 1:

x - 4 = 10

Add 4 to both sides of the equation:

x = 14

The answer is x = 14

Make sure the resolution by substituting 14 in the original equation for x. If the left side of the equation generation the right side of the equation after the substitution, you have found the correct answer.

Example 2:

2x - 4 = 10

Add 4 to both sides of the equation:

2x = 14

Divide both sides by 2:

X = 7

The answer is x = 7.

Algebra is widely used in day to day activities watch out for my forthcoming posts on Unit of Length and Mean Value Theorem Examples. I am sure they will be helpful.

Check the solution by substituting 7 in the original equation for x. If the left side of the equation generation the right side of the equation after the substitution, you have found the correct answer.

2(7) - 4 = 14 - 4 = 10.

Multiplication Rule

The multiplication rule is a product used to determine the probability that two events, A and B, both occur.

The multiplication rule follows from the description of conditional probability.

The result is often written as follow, using set notation:

P( A ∩ B ) = P(A | B ) . P(B)

Or

P(A ∩ B) = P(B | A) . P(A)

Where:

P(A) = probability that event A occurs

P(B) = probability that event B occurs

P(A ∩ B) = probability that event A and event B occur

P(A | B) = the conditional probability that event A occur given that event B has occurred already

P(B | A) = the conditional probability that event B occur given that event A has occurred already

For independent an event, that is events which have no influence on one another, the rule simplifies to:

P(A nn B) = P(A). P(B)

That is, the probability of the joint events A and B is equivalent to the product of the individual probabilities for the two events.

Multiplication Rule of Probability

The addition rule helped us resolve problems when we performed one task and wanted to know the probability of two things happening during that task. This topic deals with the multiplication rule. The multiplication rule also deal with two events, but in these problems the events occur as a result of more than one task (rolling one die then another, drawing two cards, spinning a spinner twice, pulling two marbles out of a bag, etc).

When ask to find the probability of A and B, we would like to find out the probability of events A and B happening.

The Multiplication Rule:

Consider events A and B. P(AB)= P(A) P(B).

What The Rule Means:
Expect we roll one die followed by another and want to find the probability of rolling a 4 on the first die and rolling an even number on the second die. Notice in this problem we are not trade with the sum of both dice. We are only commerce with the probability of 4 on one die only and then, as a separate event, the probability of an even number on one dies only.
P(4) = (1)/(6)
P(even) = (3)/(6)
So P(4even) = ((1)/(6) )((3)/(6) ) = (3)/(36)  = (1)/(12)
While the rule can be applied in any case of dependence or independence of events, we should note here that rolling a 4 on one die followed by rolling an even number on the second die are independent events. Each die is treated as a split thing and what happens on the first die does not influence or effect what happens on the second die. This is our basic description of independent events: the outcome of one event does not influence or affect the outcome of another event.

Let's Practice the Multiplication Rule

Assume you have a box with 3 blue marbles, 2 red marbles, and 4 yellow marbles. You are going to draw out one marble, record its color, put it back in the box and draw another marble. What is the probability of pull out a red marble followed by a blue marble?


The multiplication rule says we need to find P(red) P(blue).

P(red) = (2)/(9)
P(blue) =(3)/(9)

P(redblue) = ((2)/(9) )((3)/(9) ) = (6)/(81)  = (2)/(27)
The events in this example were independent. Once the first marble was pull out and its color recorded, it was returned to the box. Therefore, the probability for the second marble was not affected by what happened on the first marble.

Some students find it supportive to simplify before multiplying, but the final answer must always be simplified.


Think the same box of marbles as in the previous example. However in this case, we are going to draw out the first marble, leave it out, and then pull out another marble. What is the probability of pull out a red marble followed by a blue marble?


We can motionless use the multiplication rule which says we need to find P(red) P(blue). But be alert that in this case when we go to pull out the second marble, there will only be 8 marbles left in the bag.

P(red) = (2)/(9)
P(blue) = (3)/(8)

P(redblue) = ((2)/(9) )((3)/(8) ) = (6)/(72)  = (1)/(12)

The events in this example were dependent. When the first marble was pull out and kept out, it affected the probability of the second event. This is what is meant by dependent events.


Assume you are going to draw two cards from a standard deck. What is the probability that the first card is a champion and the second card is a jack (just one of several ways to get “blackjack” or 21).

Using the multiplication rule we get

P(ace) P(jack) = ((4)/(52) )((4)/(51))  = (16)/(2652)  =  (4)/(663)

Notice that this will be the similar probability even if the question had asked for the probability of a jack followed by an ace.

Preparation for Sample Size

Our critical characteristic of preparation of sample size plan is deciding how big your sample size should be. If you amplify your preparation about sample size you add to the exactitude of your estimates, which means that, for any given approximation / size of result, the better preparation about the sample size is  the more “ statistically significant” the outcome will be. In other words, if an examination is too small then it will not sense fallout that is in actuality significant.

Defintion of preparation for sample size

The preparation about sampling is that a division of statistical performs troubled with the assortment of an impartial or casual division of personality explanation within a inhabitants of persons intended to give up some information regarding the population of distress, mainly for the purposes of construction predictions based on arithmetic supposition. Sampling is a central part of data collection.

Sample size process:

Step 1: Essential the population of worry.

Step 2: Indicate a sampling outside, a spot of material or trial likely to estimate.

Step 3: Specifying a sampling system for selecting items or actions from the enclose.

Step 4: Decisive the sample size.

Step 5: Implementing the sampling map.

Step 6: Gather the study sample data and information.

Step 7: Reviewing the example process.

Formula of preparation for sample size:

The formula for sample size = n = t2 * p (1-p) / m2.

Description:

N represents the required sample size.

T represents the confidence level. (Regular value 1.96).

P represents the estimated prevalence of malnutrition in the area

M represents margin of error. (Steady value 0.05).

Example for the preparation of sample size:

Ex:1 Find  the sample size when the population is 15.

Sol:

The sample size = t2 * p(1-p) / m2.

Where, the t is constant value of 1.96,

the m is constant value of 0.05.

Size = (1.96)2 * (15)*(1-15) / (0.05)2.

= 3.8416 *15 *14 / 0.0025.

= 322694.4

So, the sample size is 322694.4

Ex:2 Find  the sample size when the population is 30.

Sol:

The sample size = t2 * p(1-p) / m2.

Where, the t is a constant value of 1.96,

The m is a constant value of 0.05.

Size = (1.96)2*(30)*(1-30) / (0.05)2.

= 3.8416*30*29 / 0.0025.

= 1336876.8

So, the sample size is 1336876.8

Learn Common Ratio

In mathematics, a geometric series is a series with a permanent ratio between successive terms. For example, the series 1/2+1/4+1/8+1/16.....is geometric, so each term except the first can be obtained by multiplying the previous term by 1/2.Geometric series are one of the easiest examples of infinite series with finite sums. Basically, geometric series played an important role in the early development of calculus, and they continue to be central in the study of convergence of series. Geometric series(GS) are used throughout mathematics, and they have important applications in physics, engineering, biology, economics, computer science, queueing theory, and finance.

Common ratio:

The terms of a geometric series form a geometric progression(GP), meaning that the ratio of successive terms in the series is constant. Below table shows several geometric series with different common ratios:

Common ratio


Example

10


4 + 40 + 400 + 4000 + 40,000 + ···

1/3


9 + 3 + 1 + 1/3 + 1/9 + ···

1/10


7 + 0.7 + 0.07 + 0.007 + 0.0007 + ···

1


3 + 3 + 3 + 3 + 3 + ···

−1/2


1 − 1/2 + 1/4 − 1/8 + 1/16 − 1/32 + ···

–1


3 − 3 + 3 − 3 + 3 − ···

The behavior of the terms depends on the common ratio r:

If r is between -1 and +1 the terms of the series become smaller and smaller, approaching zero in the limit. The series converges to a sum, as in the case, where r is a half, and the series has the sum one.

If r is greater than 1 or less than -1 the terms of the series become larger and larger. The total sum of the terms also gets larger and larger, and the series has no sum. (The series diverges.)

If r is equal to 1, all of the terms of the series are the same. The series diverges.
If r is minus one the terms take two values alternately (e.g. 2, −2, 2, −2, 2,... ). The sum of the terms oscillates between two values (e.g. 2, 0, 2, 0, 2,... ). This is a different types of divergence and again the series has no sum.

Examples:

Consider the sum of the following geometric series:

s = 1+2/3+4/9+8/27+......

This series has common ratio 2/3. If we multiply through by this common ratio, then the initial one becomes a 2/3, the 2/3 becomes a 4/9, and so on:

2/3 s = 2/3 + 4/9 + 8/27 + 16/81 + .....

Mean Value Theorem

The Mean Value Theorem has very important consequences in differential calculus.

THEOREM: Let the function f such that

i.           continuous in the closed interval [a,b]
ii.            derivable in open interval (a,b)

Then there exists at least one c with  a < c < b such that

calculus formula

The result in the theorem can be expressed as a statement about graph of f: if A(a , f(a)) and B(b , f(b)), are the end points on the graph, then there is at least one point C between A and B,such that the tangent is drawn from C is parallel to the chord AB.

Mean value theorem graph

Mean value theorem is also known as Lagrange’s Mean Value Theorem or First Mean Value Theorem or Law of Mean.

Applications of mean value theorem

1. Let the function be f such that

(i)                  Continuous in interval [a,b]

(ii)                Derivable in interval (a,b)

(iii)              f'(x) = 0 `AA` x  `epsi` (a,b) , then f(x) is constant in [a , b].

2.Let f and g be a functions such that

(i)             f and g are continuous in interval [a,b]

(ii)            f and g are derivable in interval(a,b)

(iii)          f'(x) = g'(x) `AA` x  `epsi`  (a,b) , then f(x) - g(x) is constant in [a,b]

3.Let the function be f such that

(i)             Continuous in interval[a,b]

(ii)            Derivable in interval(a,b)

(iii)          f'(x) > 0 `AA` x  `epsi`  (a,b), then f(x) is strictly increasing function in [a,b]

4.Let the function be f such that

(i)             Continuous in interval[a,b]

(ii)            Derivable in interval (a,b)

(iii)          f'(x) < 0 `AA` x  `epsi`    (a,b), thenf(x)  is strictly decreasing function in[a,b]

Special case of Mean Value Theorem is when f(a) = f(b).Then there exists at least one c with  a < c < b such that f'(c)= 0 . This case is known as Rolle’s Theorem.

Cauchy’s mean value theorem in calculus

Let f and g be functions such that

i.            both are continuous in closed interval [a,b]
ii.            both are derivable in open interval (a,b)
iii.             g'(x) `!=` 0 for any x `epsi` (a,b)  then there exists at least one number c `epsi` (a,b) such that

`(f'(c))/(g'(c))`  =  `(f(b) - f(a))/(g(b) - g(a))`

Mean value theorem example



Verify Rolle's theorem for the function

f (x) = x2 - 8x + 12 on (2, 6)

Since a polynomial function is continuous and differentiable everywhere f (x) is differentiable and continuous (i) and (ii) conditions of Rolle's theorem is satisfied.

f (2) = 22 - 8 (2) + 12 = 0

f (6) = 36 - 48 + 12 = 0

Therefore (iii) condition is satisfied.

Rolle's theorem is applicable for the given function f (x).

\ There must exist c  (2, 6) such that f '(c) = 0

f '(x) = 2x - 8

`=>`  c = 4 `in`(2,6)

Rolle's theorem is verified.

How to Find Inverse Variation

The inverse variation is the product of two variables equals to a constant and the product is not equal to zero. Inverse variation is in the form of y =k/x. xy = k. Inverse variation in which value of one variable increases while the value of the other  variable decreases in value is known as an inverse variation. For example think a trip of 240 miles. Rate(mph)=>20,30,40,60,80,120 and time(h)=>12,8,6,4,3,2.The numbers can be explained. As the rate of speed increase, the numbers of hour require decrease. As the rate of speed decrease, the number of hour requires increase. contrasting in a direct variation, the ratio in each data is not equivalent. The product of the value in each is equal.

Problem on how to find inverse variation:-

Problem 1:-

If y is inversely proportional to find inverse variation x and y = 10 when x = 2. Find the value of y, when x = 15?

Let, k = x/y

Plug x = 2 and y = 10 in the above equation

k = 10/2

k = 5

Now the equation becomes 3 = x/y

Now, plug x = 15

3 = y/15

3 * 15 = y

So, y = 45 when x = 15

Problem 2:-

find inverse variation F vary inversely with the square of m. if F=15 when m=3,find F when m=5?

Solution:-

F=K/m2

15=K/32

15=K/9

K=135

F=135/m2

F=135/52

=135/25

=5.4..

Problem in Equation on how to find inverse variation:-

Find inverse variation problem:-

Complete the table in support of the positive values of x so that yoo (1)/(x^2)

Table

Find the x value using inverse variation method

Solution:-

If yoo (1)/(x^2)then y =(k)/(x^2)

But y =100 when x = 3

Therefore 100 =(k)/(9)

That is k = 900 so y =(900)/(x^2)

When x = 5, y =(900)/(25) = 36

If x = 10, y =(900)/(100) = 9

And if x = 15, y = (900)/(25) = 4

If y = 25, 25 =(900)/(x^2)

That is 25x2 =900

x2 =(900)/(25) = 36

Table

So that answer x = 6.

Algebra Probability Problems

Probability is used to find the possible outcomes in an event. It is defined as the ratio between the numbers of favorable outcomes to the total number of outcomes. The value of probability lies only between 0 and 1. It is not greater than 1.There are several types of probability. They are Conditional probability and Theoretical probability. Conditional probability occurs when an other event is already occurred and changed the sample space. Theoretical probability occurs based on the probability principles.

PROBABILITY BASIC PROBLEMS

Problem 1:

A number is drawn from 1 to 12 at random. What is the probability of finding a number 7.

Solution:

From 1 to 12 there will be 12 numbers. So the total outcome is 12. Number 7 occurs only once. So the favorable outcome is 1.

Hence the answer is 1/12

Problem 2:

A bag contains 7 black, 8 blue and 11 green marbles. A marble is drawn at random. What is P (blue)?

Solution:

Total marbles in the bag = 7+8+11 = 26

Out of which number of blue marbles = 8

So the probability P (blue) = 8/26 = 4/13

Problem 3:

A card is drawn from a well shuffled pack of cards. What is P (diamond)?

Solution:

A pack of cards will have a total of 52 cards.

So total outcomes = 52.

In a pack of cards number of diamonds = 13

So the probability P (Diamond) = 13/52 =1/4

PROBABILITY PRACTICE PROBLEMS

Problem 1:

A die is thrown twice. Find the probability that a sum of 6 occurs on the die.

Solution:

Let F be the event of getting a number 6 on the die.

F = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}

So the probability = (5/36)

Problem 2:

The condition is two numbers appearing on throwing two dice are different. Find

the probability of the when a sum of 4 occurs on the die.

Solution:

Let E be the event of getting a sum of 4. Here the condition given is the numbers on each dice is different.

E= {(1, 3), (3, 1)}

So the probability= 2/36 = 1/18

How to Bisect a Reflex Angle

Before we are going to see how to bisect a reflex angle we start with what is angle.  Angle is made with the turn that takes place between two straight lines which are formed from the same end point where the end point is also known as vertex. Reflex angle is the angle formed between the angles of 180 degree to 360 degree.

How to bisect a reflex angle, the word bisects means nothing but cutting it into two or splitting into two. Reflex angle bisector is the bisector that bisects the reflex angle. A bisector angle is a line that passes through the vertex of the reflex angle. How to bisect a reflex angle is followed by some certain steps.

Steps about how to bisect a reflex angle:

To bisect a reflex angle:

First draw a line and mark the given points  assume A,B.
By using the protector mark the reflection angle and draw the reflection angle and let the point be C.
Now AB and AC are the two lines with a same midpoint A.
Now by using campus make an arc with any radius which cuts both lines AB and AC.
Let the intersection point on line AB will be as D and on line AC be as E.
Take more than an half of the line as radius and in the campus and by keeping D and as midpoint make an arc. Similarly Keep E as midpoint and make an arc so that it cut the arc and the meeting point be F.
Now draw a line which joins AF.
Now AF will be the bisector of the given reflex angle.

These are the steps to be followed to bisect a reflex angle. The angle that is formed due to bisector in the reflex angle will be always an obtuse angle.

Example of how to bisect a reflex angle:

Draw a bisector of the reflex angle of 280 degree.

Solution:

Draw the line AB.

line

Keeping A as centre the angle of point 280 is marked and the line is drawn which makes a reflex angle of 280 degree which makes a line AC.

Here DE is an arc made by a as centre.

280 reflex angle bisect

More than half is taken as radius and two arcs are made with D and E as centre.The point F is plotted and a line is joined from A to F. The angle AF is the bisector of the reflection angle 280 degree.

Bisect a 280 reflex angle

Percentage off Calculator

"Percentage off" - This is the term that is synonymous with "percentage discount". It is the discount on some purchase represented in the form of percentage. For example, "10 percent off on a particular purchase" implies that on every $ 100, $ 10 will be discounted. Thus if the purchase is of `$`  200, the discount will be $ 20, if the purchase is $ 250, the discount will be $ 10 + $ 10 + $ 5 = $ 25, and so on. The formula to calculate the actual value of discount when we know the discount percentage and the purchase value is:

Discount value = (discount percent * purchase value)/100

In words, the percentage off on a particular item is equal to the product of percentage discount and the cost price divided by hundred.

The final payable amount in a purchase is found by deducting all the given discounts from the cost price of articles purchased.

Solved examples on percentage off calculator

For example, David purchases a shirt worth $ 100, on which the percentage off is 5%, stationary worth $120, on which the percentage off is 12%, and a football worth $45 on which he percentage off is 7%. Find the final amount paid by David on his purchases.

Sol:- Percentage off calculator :

We have to calculate the discount in $ for each article and then subtract it from that article's cost price. Once all the discounted prices are calculated, they are all added to get the total payable amount by David.

Cost price of shirt = $ 100

Percentage off = 5%

Discount (in $) = 5% of 100

= 5/100 x 100

= $ 5

Discounted price = $ 100 – 5

= $ 95

Cost price of stationary = $ 120

Percentage off = 12%

Discount (in $) = 12% of 120

= 12/100 x 120

= $ 14.4

Discounted price = $ 120 – 14.4

= $ 105.6

Cost price of football = $ 45

Percentage off = 7%

Discount (in $) = 7% of 45

= 7/100 x 45

= $ 3.15

Discounted price = $ 45 – 3.15

= $ 41.85

Total payable amount = $ 41.85 + 105.6 + 95

= $ 242.45

Therefore, David has to pay $ 242.45 as the final price after deducting all given discounts.

Solve Solutions in Math

With the help of Maths we can find the easy solutions to difficult problems. It is a concept of numerically solving with logical understanding. Basic operations in mathematics are addition, subtraction, multiplication, division.Also we solve some problems with the help of identities and formulas. In this article we shall study some more solutions in math. Before solving the maths problems there are some important points to remember. some of them are following.

Read the problem very carefully and find what we have to find ?

If there is any formula start to the solve the problem from there.

If it possible break the problem into 2 or more parts.

Make the an easy equation if possible.

Solve the equation to get the answer.

Some Important Formulas Used to get Solutions in Math

( a + b )2 = a2 + b2 + 2ab

( a - b )2 = a2 + b2 - 2ab

( a + b ) ( a - b ) = a2 - b2

a3 + b3 + c3 = 3abc

Area of perpendicular triangle = 1/2 X base X height

Area of circle = 2pi r

Surface area of cuboid = 2(lb +bh + hl)

volume of cuboid = lbh

mean     =     ("Sum of all Numbers")/(" Total Numbers" )

Examples for Solutions in Math

Ex 1 : Evaluate : 302 + 203 - 503

Sol :Step 1: Let a = 30, b = 20, c = - 50. Then,

a + b + c = 30 + 20 - 50 = 0

Step 2:We know   a3 + b3 + c3 = 3abc

303 + 203 + ( -50 )3 = 3 X 30 X 20 X -50

303 + 203 - 503 =  - 90000  Ans.

Ex 2 : Factorize a4 - b4

Sol :    Step 1: a4 - b4

=  ( a2)2 - ( b2 )2

Step 2:We know ( a + b ) ( a - b ) = a2 - b2

so          =  ( a2)2 - ( b2 )2

= ( a2 + b2 ) ( a2 -  b2)

=  ( a2 + b2 )  (a + b) ( a - b )  Ans.

Ex :3 The radius of a circle is 14 cm . Find the area of the circle.

Sol :         area of circle

Step 1:Radius of circle = 14cm

Step 2:Area of circle = pi r2

=  22/7 14 X 14

=  616 cm2 Ans.
Practice Problems to Find Solutions in Math:

Pro 1: Evaluate :    233 - 173

Ans :   7254

Pro 2: The radius of circle is 35cm. Find the area of circle?

Ans :  3850 cm2

positive Quadrant

Quadrants are the important concept in graph matrix. Positive Quadrant is one of the parts of our Cartesian plane. Usually our Cartesian plane is divided into 4 parts of quadrants.Cartesian plane is the formation of the x and y axis. If the x and y values are positive then that quadrant is called the positive quadrant. It is also called first quadrant. In this article we are going to learn about positive quadrant through problems and diagrams.

Brief Summary about Positive Quadrant

Cartesian Graph plane:

Our Cartesian graph plane is the basic formation of the x and y axis.

X axis:

X axis is the left to right direction line and then it can be divided into 2 parts by the origin.  They are X and x’.  Positive values are marked in the x axis and then the negative values are marked in the x’ axis.

Y axis:

Y axis is the top to bottom direction line and then it can be divided into 2 parts by the origin.  They are Y and y’.  Positive values are marked in the y axis and then the negative values are marked in the y’ axis.

Origin:

The intersection of horizontal axis x and then the vertical axis y is the point origin.

Coordinates:

Coordinates are the major part of graph plane.  It takes the form of (x, y).

Quadrants:

Based on the x and y values it can be marked into the different type of 4 quadrants.

Positive Quadrant:

Positive quadrant is the intersection of positive value of x and positive value of y. The coordinates on the positive quadrant is on the following form (+x, +y).
Example Problems on Positive Quadrant

Example 1:

Shade the following points in the positive quadrant

(5, 2)
(3, 1)
(4, 3)
(6, 2)

Solution:

The graph in the positive quadrant looks like as the following graph,

Example 2:

Identify which points is lies in the positive quadrant from the following list of points

(0,6)
(6,0)
(6,1)
(-2,6)

Solution:

Here (6, 1) is the only point in the positive quadrant.  Because here both 6 and 1 values are positive.

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
____________________________________________________________________________
focus S = (p,0)                                          x = X - 1 = 1 -1 = 0; y = Y +2= 0 + 2 = 2
So focus is (0,2)
____________________________________________________________________________
vertex is (0,0)                                            x = X - 1 => 0 - 1 = -1 and y = Y + 2 = 0+2 =2
So vertex = (-1,2)
____________________________________________________________________________
axes Axis  of parabola Y =0                       Axis of the parabola is y = Y+2 = 0+2 =2
so axis  is y - 2 = 0
____________________________________________________________________________
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