Introduction to discrete math applications

Discrete mathematics is defined as the study of mathematical structures that are logically discrete rather than continuous. Discrete mathematics had a concept in topics in "continuous mathematics" such as calculus and analysis

The set of conepts studied in discrete mathematics can be finite or infinite. Some of the mathematical relations often considered i part of discrete mathematics are Boolean algebra, the mathematics of social choice, linear programming, and number theory

Discrete mathematics includes sets, functions and relations, matrix algebra, combinatory and finite probability, graph theory, finite differences and recurrence relations, logic, mathematical induction, and algorithmic thinking.

Applications of Discrete Mathemetics:

It can be applicable in various fields such as combinational analysis, functional system theory, codings, crptology,etc

It can also be used in graph theory, probabilistic problems of discrete mathemetics, and it has the applications of algorithms and their complexity and computational problems of number theory and of algebra

It can be also used in vital, exciting and useful information about mathemetics that will be very useful to taught the lower and higher grade

Problems by Using Discrete Math Applications:

1. By using discrete mathematics find the value of n.

(1). (n+ 1)! = 10 * 9(n!)

Solution:

(n+1)! = 10 * 9(n!)

here expanding the factorization that gives

(n + 1) n! = 10 * 9n!

now cancel n! on both the sides

we get n+1 = 90

n = 90 – 1 = 89

therefore n = 90 .

(2). 1/4! + 1/5! = n/6!

Solution:

1/4 + 1/5(4!) = n/6.5(4!)

now cancel 4! On both sides

1 + 1/5 = n/6.5

6/5 = n/30

n = (30 * 6)/5 = 180 /5 = 36

Therfore we get n = 36.

2. Find the following factorial question by using discrete mathemetic

Where n = 4 and r = 2 find the values of

(1). n! / r!

Solution:

n! / r! = 4! / 2! = 4.3.2.1 / 2.1 = 12.

Therfore n! / r! =12

(2). n! /r!(n- r )!

Solution:

n! / r!(n-r)! = 4! / 2!(4-2 )!

= 4.3.2.1 / 2!

= 4.3.2.1 / 2.1

therefore n! / r!(n-r)! =12.

Divergence theorem

Divergence theorem commonly known as divergence theorem of gauss. This theorem also help to integrate the functions, divergence theorem is a combination of vector and integral. To understand the theorem we have to know about divergence of vector field.
Vector calculus is very important in engineering and physics to the gradient, divergence and curl. Lets take vector V(x,y,z), be a differential vector function where x y z are Cartesian coordinate and if v1,v2,v3 are component of vector V then the function div.V is called divergence of vector field. The physical meaning of divergence is value of a function that characterize a physical or geometrical property must be independent of the particular choice of coordinate that means those values must be invariant with respect to coordinate transformation.
Now divergence-theorem in integrals, this theorem is used for triple integration. Divergence-theorem, which transforms surface integral to triple integral, it also involves the divergence of vector function. The triple integration is a generalization the double integration. triple integral can be transform in to surface integral, over the boundary surface of a region in space and conversely. This is a practical interest because one of the two kinds of integral is often simpler than other. It also helps in establishing fundamental equation in fluid flow, heat conduction etc. the transformation is done by gauss divergence-theorem which include the divergence of a vector field function.
Divergence-theorem of gauss means also transformation between volume integral and surface integral. Let constant T be a closed bounded region in space whose boundary is a piece wise smooth Orientale surface and let F(x,y,z) be a vector function that is continuous and continuous first partial derivatives in some domain containing T. then we proof the divergence-theorem of gauss. In this theorem a unit normal vector of surface which is pointing to the out side of surface can also used.
Let’s clear this theorem with some Divergence Theorem Examples. Suppose we have to evaluate the surface integral by this divergence-theorem. In our problem surface S is a closed surface consisting cylindrical coordinates and also coordinates of circular disc are given.  To solve this problem we need to know triple integration. Then we can solve this type of problem, when we solve this types of problem we mostly use polar coordinates, which is defined by X=rcosθ and y = rsin θ
Let’s take one more Divergence Theorem Example that is transformation of volume integral in to surface integral. Suppose function which is in vector form is given and function also involve unit normal vector. Here in place of surface volume of sphere is given, now by using gauss divergence-theorem we integrate the function from volume integral to surface integral.

Introduction to algebra test questions

Algebra is one of a part of arithmetic that uses letters in place of some unknown numbers. It is concerning about learn the rules of operations and dealings, and the constructions. It also was including the conditions, polynomial expressions, equations and algebraic arrangement. In other words, we use numerals like 1, 2, 3, etc to represent numbers. In algebra we use numbers as well as letters of the alphabet such as a, b, c, etc for any numerical values we choose. In this article we shall discuss about algebra test questions.First we will solve some sample questions for algebra test and then do some practice questions.

Sample Questions for Algebra Test :

Ex 1: Solve the linear questions

19x - 7 = 21x – 9

Solution: Subtract 21 x from both sides of the above equation, we get the below term
            -2x – 7 = -9
Add 7 to both sides of the above equation, we get the below term

-2x – 7 + 7 = - 9 + 7

-2x = - 2

Divide both sides by -2 of the equation, we get  

        X = 1

The answer is x = 1

Test the answer for the given equation. Replace with in the above given equation for x. If the left side of the equation sum value is equal to the right side of the equation sum value after the replacement of x value, you have got the exact answer.

Left side of the equation:

 19(1) - 7 = 12

Right side of the equation:

 21(1) - 9 = 12

Ex 2: Solve the given Questions

12− (3x + 7) = 2x

Solution: In the given equation is expanding and the terms are out the bracket.

12 − (3x + 7) = 2x

12 −3 x − 7 = 2x

(12 – 7) − 3x = 2x

5 – 3x = 2x

Now we are identify that it is easier to get all the x value on the right side of the given equation, by adding x to both sides of the above equation.

5 = 5x

Now we divide both sides by 5 and swap the sides:

We get the x value is 1

Test the answer for the given equation. Replace with in the above given equation for x.

We get

12 – 3(1) + 7 = 2(1)

12 – 3 - 7 = 2

Practice Questions for Algebra Test :

Solve 17 x - 6 = 23 x – 18
                    Ans: x = 2

Solve 13− (5x + 9) = -3x
                    Ans: x = 2

Intermediate algebra test

Intermediate algebra is mainly used for solving the equations. Intermediate algebra mainly uses the simplification technique for solving the equations. Intermediate algebra uses the letters and variables. Intermediate algebra also deals with the expressions and exponents of polynomials. Algebra test problems are very simple and easy. Intermediate algebra test covers the topics,

• linear Equations
• Equation of a line
• Quadratic Equations
• Polynomial Functions
• Slope formula

Intermediate Algebra Test Problems

Algebra test Problem: 1

Solve the equation x + y = 6, x - 3y = 2.

A) x = 2, y = 4 B) x = 5, y = 2 C) x = 4, y = 2 D) x = 5, y = 1.

Answer: D

Algebra test Problem: 2

Find the slope of the equation 4x + 3y = 21

A) M = - 4 / 5 B) M = - 4 / 3 C) M = - 3 / 4 D) M = - 3 / 5.

Answer: B

Algebra test Problem: 3

Factorize the equation 2x2 - 5x + 3.

A) X = 2, X = 4 B) X = - 2, X = - 3 C) X = 1, X = 3 / 2 D) X = 1, X = 2 / 3.

Answer: C

Algebra test Problem: 4

Find the vertex of the equation Y = x2 + 4x + 7

A) (2, 4) B) ( -2, 3) C) (-3, 2) D) (1,-1).

Answer: B

Algebra test Problem: 5

Find the distance between the two points (1, 5) and (3, 9).

A) d = 6 B) d = 3 C) d = 7 D) d = 5.

Answer: D

Problems for Intermediate Algebra Test

Algebra test Problem: 6

 Find the value of x for the expression: 4x + (4 - 7x) = (2x + 3) - 4.

A) 3 B) 2 C) 8 D) 1

Answer: D

Algebra test Problem: 7

Add the two expressions (3x + 2y - 8) and (5x - 3y -4)

A) 5x + 4y - 12 B) 8x + y + 12 C) 8x - y - 12 D) 2x - y - 4.

Answer: C

Algebra test Problem: 8

The sum of the number and 32 is 67. Find the unknown number.

A) 32 B) 33 C) 31 D) 35

Answer: D

Algebra test Problem: 9

Multiply (3a2 + 4b) (2a + 6b)

A) 6a3 + 8ab + 18a2b + 24b2

B) 3a3 + 8ab + 18a2b + 24b2

C) 4a3 + 6ab + 16a2b + 24b2

D) 12a3 + 9ab +18a2b + 24b2

Answer: A

Solving Trigonometric Equations made easy

The six trigonometric functions are the sine, cos, tan, csc, sec and cot. An equation consisting of one or more trigonometric functions is called a trigonometric equation, for instance cot^2(x) + 1 = csc^2(x) is a trigonometric equation.  Some of Trigonometric Equations Examples are 2 cos^2(x) –sin(x) – 1 =0, 2sin(theta) – 1 = 0, cos(theta/2) = 1 + cos(theta), 6 sin^2(theta) – sin(theta) = 1, cot(2 theta) – tan(2 theta) = 0

Quadratic Trigonometric Equations
Trigonometric equations that involve trigonometric functions as products are called the quadratic trigonometric functions. In a quadratic equation with degree 2 written in the form of  ax^2 + bx + c. So, a quadratic trigonometric equation will consist of trigonometric functions with degree 2. For instance, 2 sin^2(x) – sin(x) +1 = 0 is an example for a quadratic trigonometric equation.

Let us now learn the steps involved in simplifying trigonometric equations. We solve trigonometric equations as we solve any other algebraic equation except that we use the trigonometric identities in solving. They are also solved by the sign of the trigonometric value by determining the quadrant(s) for that particular trigonometric value. Let us simplify a trigonometric equation, 2 sin(pi/2 -alpha) =1. We shall use the identity sin (pi/2 – alpha) = cos(alpha) in solving the given equation. So, the trigonometric equation reduces to, cos(alpha)=1/2 . Now we need to check for what angle of cosine alpha gives the value ½. We know that cos(pi/3) equals ½. So, we write it as cos(+/-) pi/3 =1/2. The positive and negative sign depends  on which quadrant the angle alpha lies and hence we can generalize the solution as alpha = (+/-)pi/3 + 2k pi where k is any integer. There is a significance for adding 2k pi to the solution, as per the periodic behavior of the function cosine. So, the solution set is written as {alpha = (+/-)pi/3 + 2k pi: k belongs to set of integers}

Sin(x) + cos(x) = 0
Simplifying the above equation we get, sin(x) = -cos(x). Here, the value of sine and cosine have to be equal with opposite sign. Let us write sine in terms of cosine using trigonometric identity, which gives us
  cos(pi/2 – x+2 k pi) = cos(pi + x)
      pi/2 – x + 2 k pi = pi + x

The general solution would be, x = -pi/4 + k pi where k belongs to set of integers
Now, in the interval [0, 2pi) let us consider k = 1 and k = 2, we get, x = (3/4) pi and x = (7/4) pi as the solution that is at 135 degrees and 315 degrees the value of sine and cosine are equal and opposite.