Monday, April 29

Volume Translation


Definition of volume:

Volume is how much three-dimensional space a substance (solid, liquid, gas, or plasma) or shape occupies or contains,[1] often quantified numerically using the SI derived unit, the cubic meters. The volume of a container is generally understood to be the capacity of the container, i. e. the amount of fluid (gas or liquid) that the container could hold, rather than the amount of space the container itself displaces.

(Source: Wikipedia.)

Definition of volume translation:

Volume translation is the process translating the volume into different types of unit. For example, we take the cylinder and calculate its volume and it can be in the form of cubic meters. It can be translated into different units that are cubic milliliters, Liters, Centiliters etc., Volume translation is the process of calculating the different units.

Different formulas for volume:

Volume formulas:

Cube = side3
Rectangular Prism = side1 * side2 * side3
Sphere = (4/3) * `pi` * radius3
Ellipsoid = (4/3) * `pi` * radius1 * radius2 * radius3
Cylinder = `pi`* radius2 * height
Cone = (1/3) *`pi` * radius2 * height
Pyramid = (1/3) * (base area) * height
Torus = (1/4) *`pi` 2 * (r1 + r2) * (r1 - r2)2


Example problem for volume translation:

Example 1:

Find the volume for the given cylinder with radius = 14 m and height = 24 m.

volume

Solution:

Volume formula for cylinder

V = `pi` r2h

Here `pi` = 3.14 or 22/7

r = 14 m

h = 24 m

V = 3.14 * 142 * 24

= 3.14 * 14 * 14 * 24

= 3.14 * 196 * 25

= 3.14 * 4900

= 15386 m3.

Different translation of volume:

15386 m3 = 1538600000 centiliters

15386 m3 = 153860000 deciliters

15386 m3 = 15386000 L

15386 m3 = 15386000000 cubic milliliters

15386 m3 = 96775.028514 barrels.

Example 2:

Find the volume for the given cone with radius = 8 cm and height = 12 cm.

cone

Solution:

Volume formula for cone

V = (1/3) *`pi` * radius2 * height

Here we use pi = 3.14

= (1/3) * 3.14 * 82` * 12`

` = (1/3) * 3.14 * 8 * 8 * 12`

` = (1/3) * 3.14 * 8 * 8 * 12`

` = 3.14 * 64 * 4`

` = 3.14 * 256`

` = 803.84 `cm3`.`

Different translation of volume:

`803.84 `cm3 = 80.383976 centiliters

`803.84 `cm3 = 8.038398 deciliters

`803.84 `cm3 = 0.80384 L

`803.84 `cm3= 803839.763 cubic milliliters

`803.84 `cm3 = 0.005056 barrels.

Wednesday, April 24

Interpretations of Probability


An interpretation of probability is a task of meaning to probability claims. These involve specifying a set of possible cases and defining what a probability claim means in terms of that set.


Understanding Probability Equation is always challenging for me but thanks to all math help websites to help me out.

An interpretation tell us what it means to say that P (p) = r

(Where p can be any sentence and r can be any number between 0 and 1.)

An interpretation must also provide a meaning for conditional probability statements of the form P (p / q) = r.

Interpretations of probability: Over view

Probability ->subjective

->actual degree of belief

->personalist (tempered personalist)

->objective

->classical

->logical

->Frequency

-> Finite freq.

-> Limiting freq.

-> Propensity

Cases of Interpretations of probability:

Classical

Number of possible / total number of possible

Logical

Getting a partially entails, with degree of entailment Finite frequency

Limiting frequency

The limiting frequency in an endless series

Propensity

The limiting frequency of a pair

Interpretations of Probability Example:

What is the probability of getting at least one 6 in two tosses of a die?

Solution:

P (6 on toss 1) = 1/6

P (6 on toss 2) = 1/6

P (6 on both tosses) = 1/36

P (at least one 6 in two tosses) = 1/3

Consider each of the following four bet light (I expect to break even):

1. Pay 36 for [36 if no 6 on 1st toss; 0 if 6]

2. Pay 36 for [36 if no 6 on 2nd toss; 0 if 6]

3. Pay 1 for [42 if 6 on both tosses; 0 otherwise]

4. Pay 12 for [36 if slightest one 6 in two tosses; 0 otherwise]

Algebra is widely used in day to day activities watch out for my forthcoming posts on how to add and multiply fractions and iit jee new pattern 2013. I am sure they will be helpful.

But now consider what happens if I make all four bets:

1st toss        2nd toss      Cost   Winnings     Net gain or loss

6                 6                 73        72                -1

6                 1-5              73        72                -1

1-5              6                 73        72                -1

1-5              1-5              73        72                -1

A guaranteed loss!

Monday, April 22

Maximal Function


In mathematics, the maximal function is the branch of geometry applied in the forms of harmonic analysis. Hardy–Little wood maximal function is the most important type in the maximal function. The singular integrals, differentiability properties of functions, and partial differential equations are mainly used in maximal function for easy understanding. Comparing to other methods, these are usual method to provide a simplified approach for easy understanding of problems.

Types of maximal function:

The Hardy–Little wood maximal function.

Non-tangential maximal functions.

The sharp maximal function.

The Hardy–Little wood maximal function:

The maximal function was first introduced by G.H.Hardy. It is based on cricket score. According to him, f is a function on Rn, the Hardy–Little wood maximal function M (f) is given as,

M (f) (x) = sup 1/ (│B│) ∫B│f│

Where, x € Rn

Here, the | B | is the measure of B. The centered maximal function is taken from over balls B with centre x.

Properties of Hardy–Little wood maximal function:

(a) When f € Lp (Rn) (1≤p≤∞), M (f) is almost finite.

(b) Whether f € L1 (Rn), for all α > 0,

│ {x│M(f)(x)>α}│≤ (c/ α) ∫Rn │f│.

From the above properties the second property is known as weak-type bound. According to Morkov inequality for integrable function, M (f) is not a integrable function.
Applications:

The Hardy–Littlewood maximal function is function mainly used to prove the Lebesgue differentiation theorem and Fatou's theorem in the singular integral theory.

Non-tangential maximal function:

According to non-tangential maximal function,

Rn+1 = {(x, t) x € Rn, t> 0} and the F*(x) is given as,

F*(x) = SUP │F(y, t) │.

In the non-tangential maximal function takes the function F above a cone with vertex at the boundary of Rn.

Identity Property of Non-tangential maximal functions:

Identity property is the most important function used in the Non-tangential maximal functions. It is given as,

∫ Rn Φ = 1

Φt(x) = 1/ tn Φ (x/t)

for t > 0.

F(x, t) = f* Φt(x) = ∫ Rn f(x - t) Φt(y) dy

The sharp maximal function:

The sharp maximal function is maximal function (f # ) is defined as

f # (x) = SUP (1/│B│)∫ B │f(y) – fB │dy

Where, x € Rn

Subject Conic Sections Applications


Conics

Conic sections are the curves which result from the intersection of a plane with a cone. These curves were studied and revered by the ancient Greeks, and were written about extensively by both Euclid and Appolonius. They remain important today, partly for their many and diverse applications.
Although to most people the word “cone” conjures up an image of a solid figure with a round base and a pointed top, to a mathematician a cone is a surface, one which is obtained in a very precise way.
Imagine a vertical line, and a second line intersecting it at some angle f (phi). We will call the vertical line the axis, and the second line the generator. The angle f between them is called the vertex angle. Now imagine grasping the axis between thumb and forefinger on either side of its point of intersection with the generator, and twirling it. The generator will sweep out a surface, as shown in the diagram. It is this surface which we call a cone.


Notice that a cone has an upper half and a lower half (called the nappes), and that these are joined at a single point, called the vertex. Notice also that the nappes extend indefinitely far both upwards and downwards. A cone is thus completely determined by its vertex angle.
Now, in intersecting a flat plane with a cone, we have three choices, depending on the angle the plane makes to the vertical axis of the cone. First, we may choose our plane to have a greater angle to the vertical than does the generator of the cone, in which case the plane must cut right through one of the nappes. This results in a closed curve called an ellipse. Second, our plane may have exactly the same angle to the vertical axis as the generator of the cone, so that it is parallel to the side of the cone. The resulting open curve is called a parabola. Finally, the plane may have a smaller angle to the vertical axis (that is, the plane is steeper than the generator), in which case the plane will cut both nappes of the cone. The resulting curve is called a hyperbola, and has two disjoint “branches.”


Notice that if the plane is actually perpendicular to the axis (that is, it is horizontal) then we get a circle – showing that a circle is really a special kind of ellipse. Also, if the intersecting plane passes through the vertex then we get the so-called degenerate conics; a single point in the case of an ellipse, a line in the case of a parabola, and two intersecting lines in the case of a hyperbola.
Although intuitively and visually appealing, these definitions for the conic sections tell us little about their properties and uses. Consequently, one should master their “plane geometry” definitions as well. It is from these definitions that their algebraic representations may be derived, as well as their many important properties,such as the reflection properties. (That the definitions which follow are equivalent to those given above is not obvious – not at all! For an elegant proof, see the article on Dandelin's Spheres.)
We will now look at each conic section in detail.

ELLIPSE
The set of all points in the plane, the sum of whose distances from two fixed points, called the foci, is a constant. (“Foci” is the plural of “focus”, and is pronounced FOH-sigh.) Sometimes this definition is given in terms of “a locus of points” or even “the locus of a point” satisfying this condition – it all means the same thing.


For reasons that will become apparent, we will denote the sum of these distances by 2a.
We see from the definition that an ellipse has two axes of symmetry, the larger of which we call the major axis and the smaller the minor axis. The two points at the ends of the ellipse (on the major axis) are called the vertices. It happens that the length of the major axis is 2a, the sum of the distances from any point on the ellipse to its foci. If we call the length of the minor axis 2b and the distance between the foci 2c, then the Pythagorean Theorem yields the relationship b2 + c2 = a2:


By imposing coordinate axes in this convenient manner, we see that the vertices are at the x intercepts, at a and -a, and that the y-intercepts are at b and -b. Let the variable point P on the ellipse be given the coordinates (x, y). We may then apply the distance formula for the distances from P to F1 and from P to F2 to express our geometrical definition of the ellipse in the language of algebra:


Substituting a2 – b2 for c2 and using a little algebra, we can then derive the standard equation for an ellipse centered at the origin,


where a and b are the lengths of the semimajor and semiminor axes, respectively. (If the major axis of the ellipse is vertical, exchange a and b in the equation.) The points (a, 0) and (-a, 0) are called the vertices of the ellipse. If the ellipse is translated up/down or left/right, so that its center is at (h, k), then the equation takes the form


If a = b, we have the special case of an ellipse whose foci coincide at the center – that is, a circle of radius a.
The ellipse has the following remarkable reflection property. Let P be any point on the ellipse, and construct the line segments joining P to the foci. Then these lines make equal angles to the tangent line at P.


Consequently, any ray emanating from one focus will always reflect off of the inside of the ellipse in such a way as to go straight to the other focus. Architects have exploited this property in many famous buildings. The “whisper chamber” in the United States Capitol is one; stand at one focus and whisper, and anyone at the other focus can hear you with perfect clarity, even though they are much too far away from you to hear a whisper normally. The Mormon Tabernacle in Salt Lake City was also designed as an ellipse (indeed, it is the top half of an ellipsoid), to provide a perfect acoustical environment for choral and organ music.
Ellipses occur in nature as well, and are critical to understanding the motion of planets and other bodies moving in space. See the article on Kepler's Laws.

PARABOLA
The set of all points in the plane whose distances from a fixed point, called the focus, and a fixed line, called the directrix, are always equal.


The point directly between – and hence closest to – the focus and the directrix is called the vertex of the parabola.
To derive the equation of a parabola in rectangular coordinates, we again choose a convenient location for the axes, placing the origin at the vertex so that the y-axis is the axis of symmetry. We denote the distance from the vertex to the focus by p, so that the directrix is then the line y = -p.


Using the distance formula for the distance from P to F, and noting that the distance from P to the directrix is evidently y + p, and setting these distances equal, we obtain


A direct application of ordinary algebra reduces this to


This then is the equation of a parabola opening upwards, with its vertex at the origin. If we introduce a negative sign, we get a parabola opening downwards. If we interchange the roles of x and y, we get a parabola opening to the right (or to the left if there is a negative). We may translate the parabola up/down or back/forth, putting the vertex at the point (h, k) if we write our equation as


The reflection property of parabolas is very important because it has so many practical uses. Let P be any point on the parabola. Construct the line segment joining P to the focus, and a ray through P that is parallel to the axis of symmetry. The line segment and ray will always make equal angles to the tangent line at P. Consequently, any ray emanating from the focus will reflect off of the parabola so as to point directly outwards, parallel to the axis. This property is made use of in the design of flashlights, headlights, and spotlights, for instance. Conversely, any ray entering the parabola that is parallel to the axis will be reflected to the focus. This property is exploited in the design of radio and satellite receiving dishes, and solar collectors.


The reflection property of parabolas is related to the curious property that the tangent lines at the endpoints of any chord through the focus (as shown above) intersect on the directrix, and always do so in a right angle.
Parabolas are also important in the study of ballistics, the movement of a body under the force of gravity.

HYPERBOLA
The set of all points in the plane, the difference of whose distances from two fixed points, called the foci, remains constant.


Mimicking our procedure with ellipses, we will choose the constant 2a to represent the difference of these distances, that is, PF1 – PF2 = 2a. We will call the two points of the hyperbola which lie on the line connecting the foci the vertices, and we then see that the distance between the vertices must be 2a. Also, we will call the distance between the foci 2c. Finally, we will define the constant b by b2 = c2 – a2. (We may do this since evidently c > a.) Placing coordinate axes at the center as before, we obtain this picture:


Applying the distance formulas and substituting for c as we did in the previous cases, we can derive the standard formula of a hyperbola:


We note that solving this equation for y yields


and letting x become arbitrarily large causes this expression to become arbitrarily close to


Thus we see that the crisscrossing lines in the diagram above are asymptotes for the hyperbola, that is, the curve becomes indefinitely close to these lines as the absolute value of x grows without bound.
As before, if the principal axis of the hyperbola is vertical instead of horizontal, we switch the roles of a and b. We may also translate the hyperbola up/down and back/forth, placing the center at (h, k) by modifying our equation thusly:



The reflection property of the hyperbola is of great importance in optics. Let P be any point on one branch of the hyperbola. Then the line segments joining P to each of the foci form an angle which is bisected by the tangent line at P.


Consequently, any ray approaching one of the foci from a convex side of the hyperbola is reflected to the opposite focus. An example of an application of this principle is the Cassegrain reflecting telescope:


A concave parabolic mirror forms the back of the telescope, and this shares a focus with a convex hyperbolic mirror, the other focus of which is at the eyepiece.

I am planning to write more post on  cbse last 5 years board papers. Keep checking my blog.

ECCENTRICITY
The unifying idea among these curves is that they are all conics, that is, conic sections. We have seen the geometric realization of this unifying notion, but how can it be expressed algebraically? The key notion is that of eccentricity.
To define the eccentricity of a conic, we must first observe a feature of the ellipse and the hyperbola that we neglected before, namely, that each of these curves has a directrix, just as the parabola does. Indeed, the ellilpse and hyperbola each have two directrices. Now let P be a point on the conic curve, and consider its distance to a focus, and its distance to the corresponding directrix. The curve’s eccentricity is the ratio of these distances.


We will denote the eccentricity by the letter e. It can be shown geometrically that e is always equal to the ratio of c and a as these constants were defined in each case. That is, we always have e = c/a. It can also be shown that the directrices of an ellipse or hyperbola with principle axes horizontal are always the vertical lines given by


as shown in the diagrams above.
Now recall that in a parabola the distance from a point to the focus, and from the same point to the directrix, are always the same. Consequently, a parabola always has eccentricity e = 1. An ellipse, on the other hand, always has e < 1, and for a hyperbola e > 1. (A circle is the special case of an ellipse with e = 0.) In summary, we have


The names of these curves are related to their eccentricities. “Ellipse” comes from a Greek word meaning “deficiency” or “something left out,” and is related to the English words “ellipsis” and “elliptical.” The word “hyperbola,” on the other hand, comes from the Greek word for “excess,” and is related to the English word “hyperbole.” Finally, “parabola” means something like “just right,” and is related to the words “compare” and “parable.”
What this discussion shows is that we may consider that there is only one general kind of curve, called a conic, with special cases called ellipse, parabola, and hyperbola depending on the conic’s eccentricity. Algebraically, we may now consider conics in complete generality. To do so, consider a second degree polynomial in two variables, x and y.


The ‘xy’ term can be eliminated by a rotation of axes. The algebraic techniques for doing so can be found in any text on calculus with analytic geometry. By then completing the square with respect to both x and y, one will obtain one of the standard equations given above, for either an ellipse or a hyperbola. If only one of x and y appears as a square in the original conic equation, then the standard equation of a parabola may be obtained.

The study of conic sections is one of the most beautiful topics in classical mathematics. Every student of mathematics should take the time to master conic sections thoroughly, not only for the esthetic appeal of the subject, and not only because their applications are so varied and important, but also because they show – in a deep and clear way – the fundamental unification of geometry and algebra in the field of analytic geometry.






Sunday, April 21

Make Equivalent Fraction


Make Equivalent fraction article deals with the definition of equivalent fraction, method to make the equivalent fraction and the arithmetic operations with equivalent fraction.

Definition of equivalent fraction:

When the two fractions have the same value in their lowest terms, then the two fractions are said to be equivalent fraction.

The upper part and lower part of the fraction are multiply or divided by the common number



Condition for making equivalent fraction

Consider the fractions as `(v/n)` and `(d/s)`

Condition

`(v/n) = (d/s)`

Cross multiply the fraction terms

v (s) = d(n)

vs= dn (equal)

So `(v/n)` and `(d/s)` are equivalent fractions

If  vs is not equal to dn

So `(v/n)` and `(d/s)` are not equivalent

Steps to Make the Equivalent fraction:

For making Equivalent fraction:

Step1: find the common multiplier of the whole fraction.

Step 2: cancel the common multiplier of the fraction.

Step 3: make the fraction to its lower terms

Step 4: multiply the numerator and denominator by common number.

Model problems form making equivalent fraction:

1. Whether the fractions are equivalent fraction.

The fractions are `(3/5) ` and `(6/10)`

Solution:

`(3/5) = (6/10)`

Cross multiply the fraction terms

(5*6) = (10*3)

30 = 30

They are equal

Therefore, the above fractions are making equivalent fraction.

2. Whether the fractions are equivalent fraction

The fractions are `(1/4)` and `(6/4)`

Solution:

`(1/4) = (6/4)`

Cross multiply the fraction terms

(1*4) = (4*6)

4= 24.

They are not equal

Therefore, fractions are not making equivalent fraction.

Model problems for making the equivalent fraction:

make  the equivalent fraction sample

`(49/28)`

Solution:

Step1: find the common multiplier of the whole fraction.

`(49/28) = ((7*7)/ (4*7))`

Step 2: cancel the common multiplier of fraction

= `(7/4)`

Step 3: we cannot simplify the fraction to its lower terms

The equivalent fraction is `(7/4)`

2.Make  the equivalent fraction sample

`(90/30)`

Solution:

Step1: find the common multiplier of the whole fraction.

`(90/30) = ((9*10)/ (3*10))`

Step 2: cancel the common multiplier of fraction

= `(9/3)`

My forthcoming post is on Rules for Dividing Decimals and cbse study material for class 11 will give you more understanding about Algebra.

Step 3:  make the fraction to its lower terms

Divide the upper and lower part by 3

= `(3/1)`

The equivalent fraction `(3/1)`

Saturday, April 20

Finite Infinity


Infinity (symbolically represented by ∞) is a concept in mathematics and philosophy that refers to a quantity without bound or end .The word infinity comes from the Latin word 'infinitas' which means unclear.

It counts or measures things or an infinite number of terms but it’s  not the same sort of number as the real numbers .The set of integers are countably infinite(finite), while the set of real numbers are uncountably infinite .The symbol for infinity looks like ribbon(oo ).

The word finite means having no end.

Looking out for more help on Limits Infinity in algebra by visiting listed websites.

Finite infinity in mathematics:

Most of  mathematics that deals with the infinite that can also be interpreted as dealing with the potentially infinite number. Let us take an example, the question of whether any of a species will have an infinite chain of descendant species can be defined in such a way that requires quantification over all those reals . There occurs not a single event that decides this question but it is still meaningful and it is also interesting in a potentially infinite universe. It can be determined by a recursively enumerable set of events that are listed by a computer. In mathematics the inverse of infinity leads to zero.

Example for finite infinity are a set of all real numbers which are shown to be uncountably finite.

Example problems for finite infinity:

There includes limits of the functions where it approaches as x tends  to infinity which is known as finite infinity. Here  L' Hopital's Rule is mostly used. If we try these problems before looking at the solutions, we can avoid common mistakes.

So the correct forms are  oo / oo is not  equal to one and  oo -oo is not equal to 0. Using algebraic manipulation we can circumvent these indeterminate forms. The  following problems need the algebraic computation of limits of functions as x approaches plus or minus infinity.

Example 1:

Compute the following lim_(x->oo)      100/x2+5                       ..

Solution:

100/x2+5  where x tends to infinity.

=  100/oo 2+5.

= 100/oo

Here the numerator is 100 and the denominator is infinity.

Any numerical number which divides the infinity equals 0.

So the answer is 0.

I am planning to write more post on Hypothesis Testing Statistics Examples and Regular Prism. Keep checking my blog.

Example 2:

Compute  the following lim_(x->oo)       3x3-1000x2                     .

Solution:

3x3-1000x2  where x tends to infinity.

= 3(oo )3-1000(oo )2.

=  oo  -oo

=  oo  and != 0.

This is not equal to 0.This is an indeterminate for. By factoring it can be  curmvented.

Problems for Fractions Reducing


A fraction is a number that can represent part of a whole. The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, one part of four, and so on. A much later development were the common or "vulgar" fractions which are still used today (½, ⅝, ¾, etc.) and which consist of a numerator and a denominator.

Looking out for more help on Dividing Mixed Fractions in algebra by visiting listed websites.

Word problems for fractions reducing Problems:

Word problems 1: Jack bought 6/4 liters of milk but he used only 2/4 liters of the milk. How much milk did he have left? And reducing  the fractions.

Solution:

Initially jack had 6/4 liter of milk

He used 2/4 liter of milk

Now we have to find the remaining amount

The remaining amount

=6/4 -2/4

= 6-2/4

=4/4

By reducing the fraction get 1 is the answer.

Word problems 2: My pet is 1/4 years old and my dog is 7/4 years old find the total sums of ages And reducing  the fractions

Solution:

Given pet is 1/4 years old

And my dog is 7/4 years old

Now we have to find the total sums of ages

= 1/4 +7/4

=1+7/4

= 8/4

By reducing the above fraction we get 2 is the answer.

Word problems 3: Rahul needs to drink 1/5 liters of coffee and 14/5 liters of coke every day. How much fluid do rahul have to drink? And reducing  the fractions

Solution:

Given

Rahul needs to drink 1/5 liter of coffee

And 14/5 liters of coke every day

Now we have to find how much fluid do rahul have to drink?

=  1/5 +14/5

=1 +14/5

= 15/5

By reducing the above fraction we get 3 is the answer.

Word problems for fractions reducing Problems:

Word problems 4: On the eating competition my aunty ate 3/6 pizza and my uncle ate 15/6 pizza. How much pizza did they eat jointly? And reducing  the fractions

Solution:

Given:

Eating competition my aunty ate 3/6 pizza

And my uncle ate 15/6 pizza

How much pizza did they eat jointly?

=   3/6 +15/6

=  3+15/6

= 18/6

By reducing the above fraction we get 3 is the answer.

I am planning to write more post on Random Variable Transformation and Correlation Coefficient Examples. Keep checking my blog.

Word problems 5: John bought 3/11 liters of oil but he used only 25/11 liters of the oil. How much oil did he have left? And reducing  the fractions

Solution:

Initially john had 3/11 liter of oil

He used 25/11 liter of oil

Now we have to find the remaining amount

The remaining amount

=3/11 -25/11

=  3-25/11

=  22/11

By reducing the fraction get 2 is the answer.

Friday, April 19

Function Rule in Math


Function rule in math Linear input output functions are finding unknown variables from the given expression with the help of known values. In the expression variable are x and y. The expression deals with linear equation. find the linear expression of the particular function. Function notation is look f(x), p(x),… to find the x value of the functions. Function notation is f(x) = mx+b. in this equation the value of x is given we need to find the value of m and b.

Example of funciton rules in math: The funciton f(x) 2x2+4x+2, using function rules in math find the f(4),f(5) and f(6).

problems in square equation of function rules in math

Problems using the square of the equation find the function rules in math.

Problem 1; find the function rule in math of square equation

F(x) = x2 +2x +4 find the x = 2,3,4,5

Solution :

Given the function of f(x) there is x value is given find the function rules in math

f(x) = x2 +2x +4 find the f(2),f(3), f(4) and f(5)

The value of x is 2 is given

f(2) = 22 +2*2 +4

f(2) = 4 +4 +4 In this step 2 square is 4 it is calculate and 2*2 is 4 be added

(ii)f(2) = 12

f(3) = 32+2*3+4

f(3)= 9+6+4 = 19

f(3) = 19

(iii)f(4)= 42+2*4+4

f(4)= 16+8+4 = 28

f(4)=28.

(iv)f(5) = 52+2*5+4

f(5) = 25+10+4 = 39

f(5) = 39

The function rule in math  f(2) = x2 +2x +4 find the f(2),f(3),f(4) and f(5) is 12,19,28,39.

Simplify the function rule in math using the linear equation

Simplifying the function rule in math f(x) = 3x+12. The input function x=4,5,6,7.

Solution: The given equation is linear substitute the linear input function of f(4).

f(x) = 3x+12 the linear input function is 4

f(4) = 3*4+12. here multiply the 3*4 is 12 and add.

f(4) = 12+12

f(4) = 24. the linear output function is 24.

f(5) = 3*5+12 = 15+12

f(5) = 27

f(6) = 3*6+12

f(6) = 18+12

f(6) = 30

f(7) = 3*7+12

f(7) = 21+12

f(7) = 33.

The function rule in math f(x) = 3x+12. The input function is f(4),f(5),f(6) and f(7) output is 24,27,30 and 33

Solving Trigonometric Functions


In mathematics, the trigonometric functions are functions of an angle. They are used to relate the angles of a triangle to the lengths of the sides of a triangle. Trigonometric functions are important in the study of triangles.

The most familiar trigonometric functions are the sine, cosine, and tangent. Trigonometric functions can also be called as circular functions.

A few example problems are given below to learn solving trigonometric functions.

(Source: Wikipedia)

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Example problems for solving trigonometric functions:

Example 1:

Prove that  (cos6x + cos 4x)/(sin 6x - sin4x) = cot x

Solution:

To prove the above equation,  take the left hand side (LHS) of the equation and simplify it step by step to get right hand side of the equation (RHS).

Step 1: Given

(cos6x + cos 4x)/(sin 6x - sin4x) = cot x

Step 2: Take left hand side of the equation and prove it

LHS = (cos6x + cos 4x)/(sin 6x - sin4x)

= (2cos((6x+4x)/2)cos((6x-4x)/2))/(2cos((6x+4x)/2)sin((6x-4x)/2))       [Using trigonometric formulas]

Solving above function, we get

= (2cos5xcosx)/(2cos5xsinx)

=  cosx/sinx

= cot x

= RHS

Hence proved

Example 2:

Prove that  (sin3x - sinx)/(cosx - cos3x) = cot 2x

Solution:

To prove the above equation,  take the left hand side (LHS) of the equation and simplify it step by step to get right hand side of the equation (RHS).

Step 1: Given

 (sin3x - sinx)/(cosx - cos3x)  = cot 2x

Step 2: Take left hand side of the equation and prove it

LHS = (sin3x - sinx)/(cosx - cos3x)

= (2cos((3x+x)/2)sin((3x-x)/2))/(-2sin((x+3x)/2)sin((x-3x)/2))         [Using trigonometric formulas]

Solving above function, we get

= (cos2xsinx)/(-sin2xsin(-x))

= (cos2xsinx)/(sin2xsinx)

= cot 2x

= RHS

Hence proved

Example 3:

Solve the function (sin3x-sinx)/(cos2x)

Solution:

Given:

(sin3x-sinx)/(cos2x)

Solving above function using trigonometric formula sin C - sin D = 2cos(C+D)/2 sin(C - D)/2 , we get

(sin3x-sinx)/(cos2x) = (2cos((3x+x)/2)sin((3x-x)/2))/(cos2x)

= (2cos2xsinx)/(cos2x)

= 2sin x

Algebra is widely used in day to day activities watch out for my forthcoming posts on Multiply Exponents and Double Integral Polar Coordinates. I am sure they will be helpful.

Example 4:

Solve (sinx - siny)/(cosx + cosy)

Solution:

We can write (sinx - siny)/(cosx + cosy) as follows using trigonometric identities,

(sinx - siny)/(cosx + cosy) = (2cos((x+y)/2)sin((x-y)/2))/(2cos((x+y)/2)cos((x-y)/2))

= tan((x-y)/2)

Practice problems for solving trigonometric functions:

1) Prove that sin x + sin 3x + sin 5x + sin 7x = 4sin 4x cos 2x cos x

2) Prove that (sin5x - 2sin3x + sinx)/(cos5x - cosx) = cosec 2x - cot 2x

3) Simplify (sinx + sin3x)/(cosx - cos 3x)

Ans: cot x

Thursday, April 18

Geometric Progression


In mathematics, a geometric progression, are called as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio. For example, the sequence 2, 6, 18, 54 ... is a geometric progression with common ratio 3. Similarly 10, 5, 2.5, 1.25, ... is a geometric sequence with common ratio 1/2. The sum of the conditions of a geometric progression

Source: Wikipedia

Concepts of geometric progression:

A geometric progression is a sequence of number each of which, after the first, is obtained by multiplying the preceding number by a constant r. This constant r is called the common ration of the geometric progression.

In other words, a series is said in geometric progression if the ratio of any term to the preceding term is constant throughout. It is briefly written as geometric progression.

The constant ratio is called the common ratio. It is denoted by “r”.

Thus, the general form of a geometric sequence is

a,ar,ar2,ar3,ar4,…

And that of a geometric series is

a+ar+ar2+ar3+ar4+…

Where r ≠ 0 is the common ratio and a is a scale factor, equal to the sequence's start value.

Formula for geometric progression:

For the geometric progression represented by

a,ar,ar2,ar3,…,arn-1

With common ratio r, the sum of the first n terms, denoted by Sn is

Sn=a+ar+ar2+ar3+…+arn-1

Multiplying both sides by r, we get

rSn=ar+ar2+ar3+…+arn-1+arn

Subtracting, we have

Sn-rSn=a-arn

Sn(1-r)=a(1-rn)

So that Sn= a(1-rn)/(1-r)       if r≠1.

When r=1, the progression is merely a,a,…a so that Sa in this case is simply na. Note that Sn can also be written as

Sn= (r_(n-1))/(r-1)                    r≠1

Example 1:

Find the 12th term of the series:

3,6,12,24,48…

Solution:

The given series is a geometric progression with a=3,r=6/2 =2

T12=ar11=3 x 211=6144

Practice problem for geometric progression:

The sum of the first 8 terms of the geometric progression with the first term a=25 and the common ration r=- 1/5 is

Answer: S8=25 (1-(1/(5_8) )) / (6/5 )

Wednesday, April 17

Multiply Rational Expressions



How to Multiply Rational Expressions

Multiplication of rational expressions is similar to the multiplication of fractions of rational numbers. Let us consider an example of multiplying fractions, 3/4 x 7/2. Here the method to be followed is, first multiply the terms in the numerator and then multiply the terms in the denominator which gives 3x7/4x2 = 21/8.

Then any further simplification possible is carried out. In this example there was no possibility of cancelations as there were no common factors and hence the fraction got by multiplying the numbers in the numerator and the numbers in the denominators would be the final answer. Similar set of steps are used while multiplying rational expressions.

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A rational expression  consist of coefficients which are constants for a given term and hence first the coefficient parts are simplified then the product of the variable parts in the numerator is found, then the product of the variable part in the denominator is found and finally the products are written in the simplified form. In this process we use exponential laws and common factors for simplification.

Let us consider an example for a better understanding. Multiply Rational Expressions 4x2y2/3x and 3x/4y.  The product of the given expressions would be,  4x2y2/3x . 3x/4y Here first the coefficients are simplified and any cancelations possible are carried out. 4 and 3 get canceled and we get x2y2/x . x/y, this can be written as x.x.y.y/x. x/y; x and y are the common factors which can be canceled and we get x2y which is the final answer. Here we can use the exponential rule in simplification.

To multiply rational expressions involving polynomials the steps to be followed are:
First factor the terms in the numerator and the denominator
Reduce all the common factors
Further simplify the terms by either multiplying the numerators and denominators or leave the product in the fraction form

Example: Multiply (x2-9)/(x2+6x+9) and (3x+9)/(3x-9)
Solution: First factor the terms in the numerator and denominator
(x2- 9)= (x+3)(x-3) [special products]
(x2+6x+9)= (x+3)(x+3) [(a+b)2=a2+2ab+b2]
(3x+9)=3(x+3)                               [taking out common factor]
(3x-9) = 3(x-3)                               [taking out common factor]

The product of the expressions would be in the form, (x2-9)/(x2+6x+9) . (3x+9)/(3x-9)
Re-writing the given expressions as factors gives,
(x+3)(x-3)/(x+3)(x+3)  .  3(x+3)/(x-3)
Reducing the common factors gives,
(x+3)(x-3)/(x+3)(x+3)  .  3(x+3)/3(x-3)
Canceling the common terms gives 3/3 = 1, the final answer!

My forthcoming post is on math problem solver online and cbse 10th science book will give you more understanding about Algebra.

Multiplying Rational Expressions Solver performs the multiplication and division of rational numbers when the expressions are entered in the given fields. This helps to check the answers.

Monday, April 15

Exam for Total Differentiation


The total differentiation of a function, f of several variables. For example, t, x, y etc., with respect to t is different from the partial derivative. The total differentiation of f with respect to t does not assume and other variables are constant while t varies. The  total differentiation adds in these indirect dependencies to find the overall dependency of f on t. For example, the total differentiation of f(t,x,y) with respect to t is

(df)/(dt) = (df)/(dt) +  ((df)/(dx)) ((dx)/(dt)) + ((df)/(dy))  ((dy)/(dt)) .

Exam for total differentiation problems:

Let us see some example problems for total differentiation and its helps to exam preparation.

Exam for total differentiation problem 1:

Find the total differentiation of trigonometric term (cot x2) with respect to x.

Solution:

Given trigonometric term is  (cot x2)

Let      y = (cot x2)

Putting x2 = u and  cot x2 = cot u = y

y = cot u,        and       u = x2

dy/(du) = - cosec2 u, and  (du)/(dx) = 2x

The total differentiation is  dy/dx

dy / dx = [  dy/(du) ×  (du)/(dx)]

= [( - cosec2 u) × 2x]

=  2x ( - cosec2 u)                                                         (u = x2)

= - 2x cosec2 (x2).

dy/dx = - 2x cosec2 (x2).

Answer:    dy/dx =  - 2x cosec2 (x2)

Exam for total differentiation problem 2:

Find the total differentiation of  function  P = xe^(y + z) ,  x = uv,  y = u - v , z = u + v

Find     (dP)/(dv)      when u = 1  and v = -1

Solution:

Given function is    P =  xe^(y+ z),

x = uv,        y = u - v        and     z = u + v

((dx)/(dv))= u = 1          ((dy)/(dv))  = -1     and    ((dz)/(du)) = 1

We know the chain rule of partial differentiation. it is the Total differentiation

 (dP)/(du) =  ((dP)/(dx)) ((dx)/(du)) +  ((dP)/(dy))   ((dy)/(du))  + ((dP)/(dz))   ((dz)/(du)) .

Substitute the u and v value we get   x = -1,    y = 2   and z = 0

P = xe^(y + z),

Differentiate P with respect to x .       (dP)/(dx)   = e^(y + z).

Differentiate P with respect to y .        (dP)/(dy) =  xe^(y + z)

Differentiate P with respect to z.       (dP)/(dz)   = xe^(y + z) .

Total differentiation is       (dP)/(dv) =  ((dP)/(dx)) ((dx)/(dv)) +  ((dP)/(dy))   ((dy)/(dv))  + ((dP)/(dz))   ((dz)/(dv)).

= e^(y + z).(1) +  xe^(y + z) .(-1) +  xe^(y + z).1

= e^(2 + 0).(1) +  (-1)e^(2 + 0) .(-1) +  (-1)e^(2 + 0).1

= e2 + e2 - e2

= e2

Answer:   e2