Percent Composition

Per cent composition of the element in the molecule is defined the ratio of the mass of the different elements present in the molecule expressed in term of percentage.    The easiest method for  determining the percentage mass of a particular element in the compound is to find out the molecular mass of the element.  The molecular mass of the element is calculated by adding all the atomic mass of the element in a compound  The find out the mass of the particular element that is represented in the molecule

The next step is to divide the mass of element upon the molar mass of the molecule and converting them in percentage.   For example for determining the % composition of the A in A3B4

% of A       =    (3 x atomic mass of A/(3 x atomic mass of A + 4 x atomic mass of B)) x 100

Example 1 for the determination of percent composition

Q. Calculate the % composition of  each element in NaOH?

Ans:  There are three element in NaOH.  Sodium, oxygen and hydrogen.

The molecular mass of NaOH = 23 + 16 +1 = 40g/mole

% composition of Na = (23/40) x 100 =57.5%

% composition of Oxygen = (16/40) x 100 = 40%

% composition of Hydrogen = (1/40)  x 100 =  2.5%

Example of percentage composition 2

Q.  Calculate the percentage composition of H2SO4?

Answer:  There are three element in H2SO4 they are hydrogen, sulfur and oxygen

The molecular mass of H2SO4 = 2x 1+32 + 4 x16 =98g/mole

% composition of Hydrogen = (2/98) x 100 = 2.04%

% composition of sulfur  = (32/98) x 100 = 32.65%

% composition of oxygen   = (64/98) x 100 = 65.31%

Excercise

Find the % composition of Carbon in carbon dioxide?
Find the % composition of chromium in potassium dichromate
Find the pecentage composition of water in copper(II) sulfate pentahydrate
Find the percentage composition of Mn in KMnO4

Expression Design Tutorial

An expression design tutorial is a finite combination of symbols that are well-formed according to the rules applicable in the context at hand. Symbols can designate values constants, variables, operations, relations, or can constitute punctuation or other syntactic entities. In algebra an expression may be used to designate a value, which value might depend on values assigned to variables occurring in the expression. Mathematical expressions include letters called variables.

(Source -Wikipedia)


Basic concepts of Expression design tutorial:

Basic concepts of Expression design tutorial:

Basic concepts of Expression:

Expression design Tutorial mean nothing but a to make the expression using constant, Operations, Variables, according to our given sign and operation of the expression.

Example:

A simple algebraic expressions like  6x + 8, 7y – 9. A variable can take various values. Its value is not fixed. On the other hand, a constant has a fixed value. Examples of constants are: 4, 100, and 17.

Based on the number of of terms expression should classified four types:

Monomials(having only one term)
Binomials(having two terms)
Trinomials(having three terms)
Polynomials(having many terms)

In word problems data’s are given in directed form when we solve the word problem first design the expression based on our given data and then solve the expression

Example problems in expression design tutorial:

Expression design tutorial:

Example problem 1:

16 years ago, Jenny age was half of the age his brother will be in 25 year.find the current age of jenny?

Here first we have to design the expression for solving jenny’s age

Solution:

Step 1:

Let us consider x is age of Jenny

14 years ago means that x-16

Step 2:

Half of the age of his brother will in 25years means =x+25

Step 3:

Now we have to write the equation or expression for solving age

x-16=1/2(x+25)(Designed expression)

Step 4:

2(x-16) =x+25(solving the expression)

2x-32=x+25

Step 5:

Subtract both side on x

2x-32-x=x-x+25

2x-x-32=25


Step 6:

Add both sides on 32

2x-x-32+32=25+32

Step 7:

X=25+32

X=57 year

Answer :jenny age was 57 year

Tangent Trigonometry

The word trigonometry is an origin of two Greek words “trigonon” meaning a triangle and “metron” meaning measurement. Metron means the science, which deals with the measurement of triangles. The familiar trigonometric terms are sine, cosine, tangent, cotangent, secant, cosecant. The tangent plane to a surface at a given point is the plane that "just touches" the surface at that point. The concept of a tangent is one of the most fundamental notions in differential geometry and has been extensively generalized; see Tangent space
                                                                                                                                                                                    - Source from Wikipedia

Looking out for more help on Trigonometry Formulas Sheet in algebra by visiting listed websites.

Definition of tangent trigonometry:

                A function of an acute angle in a right-angled triangle can be defined as the ratio of the opposite side of length and the adjacent side of the length is called tangent. It is expressed by the gradient of a line. Find either sides or angles in a right-angled triangle by using this function.
      
  In Diagram (A) represents the Thangent to curve
 In Diagram (B) represents the tangent of an angle
 In Diagram (c) represents the tangent to a circle.
                       In Right angle triangle Tangent  can be expressed in mathe matical form. If an angle is` theta`
               ` tan theta ` = `(Opposite side)/(adjacent side)`

Tangent trigonometry problems:

Tangent trigonometry problem 1:
          Find the height of the given right-angle triangle.
           
         Solution:
                               tan 30° = `(opposite)/(adjacent)`
                                             = `h/5`                                                                                 we know, tan 30°= `1/ sqrt3 `

                       height  (h) = 5 × tan 30°
                                           = 5 × `1/ sqrt3 `
                                          =  2.886 (approximately)
      So, The height of the right-angle triangle (h) =2.886 cm


Tangent trigonometry problem 2:
           Find the hypotenuse value of the given right-angle triangle. length and height are in centimeter.

                
      Solution:
                                           Tan` theta`   = `(opposite side)/(adjacent side)`
                                                       = `4/3`
                                   Tan `theta` = 1.333
                                            `theta` = tan^-11.333
                                            theta = 53.12°
                  Now the trigonometry relation,  sin` theta` = `(opposite side) / (hypotenuse)`
                                                               sin 53.12° =` 4 / (hypotenuse)`
                                                     hypotenuse (X) =` 4 / sin 53.12 `                                we know sin 53.12° = 0.7999
                                                     hypotenuse (X) = `4 / 0.7999`
                                                     hypotenuse (X) = 5.000
                  So, the hypotenuse value of right-angle triangle is 5 cm.

Descriptive Geometry

The descriptive geometry is a branch of geometry, which deals three dimensional objects in two dimensions by using the lines, curves, solids, surfaces and points in space. Geo means “earth” and metron  means “measurement”. ”Euclid, is a Greek mathematician, called the father of geometry. A point is used to represent a position in space. A plane to be a surface extending infinitely in every directions such that all points lying on the line joining any two points on the surface. The descriptive geometry example problems and practice problems are given below. 

Example problems for descriptive geometry:

Example problem 1:
          Show that the straight lines 2x + y − 9 = 0 and 2x + y − 10 = 0 are parallel.
Solution:
           Slope of the straight line 2x + y − 9 = 0 is m₁ = − 2
           Slope of the straight line 2x + y − 10 = 0 is m₂ = − 2 ∴ m₁ = m₂
           The given straight lines are parallel.

Example problem 2:
      Find the co-ordinates of orthocentre of the triangle formed by the straight lines x − y − 5 = 0, 2x − y − 8 = 0 and 3x − y − 9 = 0
Solution:
      Let the equations of sides AB, BC and CA of a ΔABC be represented by
                x − y − 5 = 0 … (1)
               2x − y − 8 = 0 … (2)
               3x − y − 9 = 0 … (3)
      Solving (1) and (3), we get A as (2, − 3)
       The equation of the straight line BC is 2x − y − 8 = 0. The straight line perpendicular to it is of the form
             x + 2y + k = 0
       A(2, − 3) satisfies the equation (4) ∴ 2 − 6 + k = 0 ⇒ k = 4
       The equation of AD is x + 2y = − 4 … (5)
       Solving the equations (1) and (2), we get B as (3, − 2)
       The straight line perpendicular to 3x − y − 9 = 0 is of the form x + 3y + k = 0
       But B(3, − 2) lies on this straight line ∴ 3 − 6 + k = 0 ⇒ k = 3
       The equation of BE is x + 3y = − 3 … (6)
       Solving (5) and (6), we get the orthocentre O as (− 6, 1).

Example problem 3:
         Find the values of a and b if the equation (a − 4)x² + by² + (b − 3)xy + 4x + 4y − 1 = 0 represents a circle.
Solution:
   The given equation is (a − 4)x² + by² + (b − 3)xy + 4x + 4y − 1 = 0
            (i) coefficient of xy = 0 ⇒ b − 3 = 0 ∴ b = 3
            (ii) coefficient of x² = co-efficient of y² ⇒ a − 4 = b
                        a = 7
                Thus a = 7, b = 3

Practice problems for descriptive geometry:

Practice problem 1:
          If the equation 12x² − 10xy + 2y² + 14x − 5y + c = 0 represents a pair of straight lines, find the value of c. Find the separate equations of the straight lines and also the angle between them.
     Answer: C = 2 ; 6x − 2y + 1 = 0 and 2x − y + 2 = 0 ; tan⁻¹ (1/7)
Practice problem 2:
          Find the equation of the straight line which passes through the given intersection of the straight lines 2x + y = 8 and 3x − 2y + 7 = 0 and is parallel to the straight line 4x + y − 11 = 0
      Answer: 28x + 7y − 74 = 0

Algebra is a subdivision in math, which comprises of infinite number of operations on equations, polynomials, inequalities, radicals, rational numbers, logarithms, etc. Graphing algebra equations or function is also a part of algebra. Graphing is nothing but the pictorial view of the given function or equation, it may be a line, parabola, hyperbola, curve, circle, etc.The most commonly used graphing functions with their graph is given in the following sections.

Common graph functions:

The most commonly used graph functions are,
Linear function which gives a straight line
                f(x) = ax +b
Functions with degree 2
               f(x) = ax² +b
Functions with degree 3,
               f(x) = ax³ + bx
Functions of square root,
              f(x) =` sqrt(x)`

Graph for the common functions:

The graphical view of common functions is shown below,
1. Linear function.

Linear functions are in  the format where the function has degree 1. So that the function varies linearly. The general format for linear functions is shown below,

 f(x) = ax +b

where, x power is 1. b is a constant.

The graph for the linear function is a line and the graph is shown below,

2. Function with degree 2,

The function for degree 2 is in the format,

f(x) = ax² . where, a is a constant

The graph for this function looks as shown below,

3. Function with degree 3 is as follows,

f(x) = ax³

where, a is a constant.

The graph looks as shown below,

4. Function with degree 4 is in th format,

f(x) = ax⁴

where, a is a constant.
The graph is as follows,

5. The function with square root is as follows,

f(x) = sqrt (x)

The graph is as follows,

The above shown functions are functions of x, if it is changed to function of y, then the graph changes to the y axis.

Non Linear Data

Non linear equation solver involves solving non linear equation with the help of formulas and it is mainly used to find the unknown variable value. The quadratic equation is one of the non linear equation in which it transforms the polynomial function into normal linear function. Non linear equation is also related with the families of vectors called vector spaces or non linear spaces. The following are the example problem deals with non linear equation solved with detailed solution.

Non linear equation solver example problems:

Example 1:

Solve the non linear quadratic equation.

√ (x ² – 11x+28) = 2

Solution:

Given function is

√ (x ² – 11x+28) = 2

Squaring on both sides, then the above equation becomes

[√ (x ² – 11x+28)] ² = (2) ²

And simplify.

x ² – 11x+28= 4

Make the above equation in factor form.

x ² – 11x + 24 =0

The above equation is in quadratic form with two solutions.

x = 3 and x = 8

Conclusion:

The given equation has two real values as solution x = 3 and x = 8.

Example 2:

Solve the non linear quadratic equation.

         y ² – 4y + 13 = 0
Solution:

Given equation is

y ² – 4y + 13 = 0

The discriminant is given as

D = b² - 4ac

= (-4)2 - 4(1)(13) = -36

Since the discriminant results negative, the square root of the discriminant value is a pure imaginary number.

√(D) = √(-36) = √(-1)√(36) = 6i

where i is the imaginary part defined as i = √(-1).

To find the solution we use b / 2a formula.

y₁ = (4 + 6i)) / (2*1) = 2 + 3i

y₂ = (4 - 6i) / (2*1) = 2 - 3i

Conclusion:

So the given equation has two imaginary solutions 2 + 3i and 2 – 3i.

Non linear equation solver practice problems:

1) Solve the non linear quadratic equation.
                      y ² + 9 = 0
Answer: (y – 3i) (y + 3i) is the solution.
2) Solve the non linear quadratic equation.
                         -y ² + 2y = -3
Answer: (y + 1) (y – 3) is the solution.

Bisector in Geometry

In geometry, bisector is a line dividing something into two equal or congruent parts. The most often considered types of bisectors are the segment bisector (a line that passes through the midpoint of a given segment) and the angle bisector (a line that passes through the apex of an angle, that divides it into two equal angles). (Source: From Wikipedia).

A plane is used as a bisector in bisecting three dimensional shapes. It is called as bisector plane. Bisectors are used to divide line segments, shapes, and solids in to two equal, similar and congruent parts.

In this article we are going to learn, drawing bisector to a line segment.

Drawing bisector to a line segment

Here we are going to learn how to draw bisector of a line segment step by step.

Step 1

Draw a line segment PQ

Step 2

Fix a compass in point P, and expand it over half the length of the line segment.

Step 3

Now, draw two arcs, on each side of the line segment.

Step 4

Without changing or adjusting the measure of the compass, draw two another arcs on each side of the line segment from the point Q.

Step 5

Now we have drawn two set of arcs on each side of the line segment PQ, from points P and Q. And we have two points on either side of the equations formed by the arcs drawn from the points P and Q.

Now join those two points of intersection between the arcs either sides of the line segment by a ruler.


Step 6

The line joined by the points of intersection between arcs is called as the bisector of the line segment PQ. The bisector is perpendicular to the line segment PQ. The point, where the bisector cuts the line segment PQ, is the mid point of the line segment PQ. So, the line segments PJ and JQ are equal in length. PJ = JQ.