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.