Revenue Formula Accounting

Revenue formula is defined as one of the important formula in accounting. Revenue is defined as the income of a company.  The income of revenue is mainly obtained from the sales.  The multiplication of selling price and the income is called as the revenue formula. The revenue is directly proportional to the quantity and it always from the linear manner. In this article, we are going to study to study about the revenue formula in accounting.


Explanation to Revenue Formula Accounting

The explanation to revenue formula accounting is given as follows,

Formula:

Revenue Formula = P `xx` Q

where,

P = Price per unit
Q = Quantity

Example Problems to Revenue Formula Accounting

Problem 1: Find the revenue by using the formula, where, P = 12500, Q = 200.

Solution:

Step 1: Given: The given values for finding the revenue is as follows,

P = 12500

Q = 200

Step 2: To find:

Revenue

Step 3: Formula: The formula given for finding the revenue is as follows,

Revenue Formula = P `xx` Q

Step 4: Solve:

Revenue Formula = P `xx` Q

= 12500   `xx`   200

= 2500000

Result: Revenue = 2500000

Therefore, this is the answer for solving the revenue formula in accounting.

Problem 2: Find the revenue by using the formula, where, P = 35000, Q = 250.

Solution:

Step 1: Given: The given values for finding the revenue is as follows,

P = 35000

Q = 250

Step 2: To find:

Revenue

Step 3: Formula: The formula given for finding the revenue is as follows,

Revenue Formula = P `xx` Q

Step 4: Solve:

Revenue Formula = P `xx` Q

= 35000   `xx`   250

= 875000

Result: Revenue = 875000

Therefore, this is the answer for solving the revenue formula in accounting.


Practice Problems to Revenue Formula Accounting

Problem 1: Find the revenue by using the formula, where, P = 45000, Q = 650.

Answer: 29250000

Problem 2: Find the revenue by using the formula, where, P = 55000, Q = 65.

Answer: 3575000

Problem 3: Find the revenue by using the formula, where, P = 43000, Q = 35.

Answer: 1505000

Algebra Geometry Practice

Algebra is a branch of math that deals with properties of lines, curves and surfaces generally divided into two types, Algebraic and geometry in agree with the mathematical techniques. The relationship between Algebra and geometry was introduced by Descartes.

Geometry is a module of mathematics, which involves the study of shapes, line equation, angles problem, dimensions, relative position of figures etc.  The ‘Geometry’ means study of properties. A point is used to denote a position in space. such that all points lying on the line joining any two points on the surface. The algebra geometry example problems and practice problems are given below.
Example Problems for Algebra Geometry

Example problem 1:

Find the values of a and b if 3x4 + x3 + ax2 + 5x + b is exactly divisible by x + 2 and x – 1.

Let P(x) = 3x4 + x3 + ax2 + 5x + b

(x+2) and (x–1) are factors of P(x) and hence both P(–2) and P(1) = 0

P(–2) = 3(–2)4 + (–2)3 + a(–2)2 + 5(–2) + b = 30 + 4a + b

Since P(–2) = 0 we get

4a + b = –30 (1)

P(1) = 3(1)4 + (1)3 + a(1)2 + 5(1) + b = 9 + a + b

Since P(1) = 0 we get

9 + a + b = 0 or a + b = –9 (2)

(1) – (2) gives

4a + b = –30

`(a+b=-9)/(3a = 21)`

` After solving this, We get,`

` a = 7`

Substituting a = –7 in equation (2) we get

–7 + b = –9 (or) ∴ b = –2

a = –7, b = –2.

Example problem 2:

Find the remainder when x3 – 5x2 + 7x – 4 is divided by (x – 1).

Solution:

P(x) = x3 – 5x2 + 7x – 4. By using the remainder theorem when P(x) is divided by (x – 1), the remainder is P(1).

The remainder P(1) = 13 – 5(1)2 + 7(1) – 4 = –1.

Example problem 3:

If all the sides of a parallelogram touch a circle prove that the parallelogram is a rhombus.

Solution:


Given: All the sides of a parallelogram ABCD touch a circle with centre O.

To prove: ABCD is a rhombus.

Proof: Since the lengths of the tangents from an external point to a given circle are equal (Fig)

AP = AS; BP = BQ; CR = CQ; DR = DS

Adding (AP+BP)+ (CR+DR) = (AS+DS) + (BQ+CQ)

AB + CD = AD + BC (Fig)

AB + AB = AD+AD (ABCD is a parallelogram CD= AB, BC=AD)

2AB = 2AD AB= AD

But AB= CD and AD = BC ( opposite sides are equal)

AB = BC = CD = AD. Hence ABCD is a rhombus.
Practice Problems for Algebra Geometry:

Practice problem 1:

Find the quotient and the remainder using synthetic division

1.  x3 + x2 – 3x + 5 ÷ x – 1

2. 4x3 – 3x2 + 2x – 4 ÷ x – 3

Answer: 1. Remainder = 4, Quotient = x2 + 2x + 1

2. Remainder = 83, Quotient = 4x2 + 9x + 29

Practice problem 2:

In the Fig AB is the diameter of the circle, ∠BAC = 42°. Find ∠ACD

Answer: m∠ACD = 48°

Inequalities on a Coordinate Plane

In mathematics, an inequality is a statement about the relative size or order of two objects or about whether they are the same or not.

• The notation a < b means that a is less than b.

• The notation a > b means that a is greater than b.

• The notation a ≠ b means that a is not equal to b, but does not say that one is greater than the other or even that they can be compared in size. Source – wikipedia.


Inequalities on a Coordinate Plane : Definition of Coordinate Plane:

The give figure is known as a coordinate plane. It has two axis and four quadrants. The two number lines form the axis. The horizontal number line is known as the x-axis and the vertical number line is known as the y-axis.

coordinate plane

Concepts for inequalities on a coordinate plane:

• The shaded area of a graph symbolizes all of the coordinates that will work in a specified equation.

• A solid frame of the shaded area specifies the edge is part of the answers to the equation.

• A dashed edge of the shaded area indicates that the edge of the graph is not element of the answers.



Steps to inequalities on a coordinate plane:

Ex: Graph the equation y`>=` x

Step1: Sketch the graph y = x. This equations in slope intercept form would appear like this y = 1/1 x + 0. The 0 indicates that you go through the origin, put a point there. Now use the slope to sketch the break of the line. From the origin go up one and to the right one and place another point. Do again until you have more than a few points on an inequalities on a coordinate plane.

example 1

In an inequalities on a polar coordinates now sketch a solid line since the equation to be charted is larger than or equivalent to. Your diagram must now appear like this:

example 1

Step 2: Then shade all over the place above the line because the equation indicates that the y values are larger than or equal to the line for any certain x value.

example 1

Now check your solution by inserting a pair of points from the shaded area and non-shaded area.
Example Problems for Inequalities on a Coordinate Plane:

Ex 1: Graph y > 2, in an inequalities on a coordinate plane.

Sol:

Keep in mind that is now a horizontal line. This is just a parallel line that is shaded above the line and dashed since it is not equivalent to the line it is only bigger than the line.

example 2

Ex 2: Graph x < -3, in an inequalities on a coordinate plane.

Sol:

Keep in mind that is just a perpendicular line. This is just a perpendicular line that is shaded to the left of the line and dashed since it is not identical to the line it is only fewer than the line. The x values on the left are fewer than the line.

Euclidean Geometry

Geo means “earth” and metron  means “measurement”. ”Euclid, a distinguished 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 euclidean geometry example problems and practice problems are given below.
Example Problems - Euclidean Geometry:

Example problem 1:

Can the following angles are triangle or not?

(a) 50° , 70°, 60° (b) 40°, 80°, 70°

Solution :
(a) The sum of the measure of the three angles is
50° + 70° + 60° = 180°
Therefore 50°, 70°, 60° can be the measure of the angles of a triangle.
(b) The sum of the measure of the three angles is
40° + 80° + 70° = 190°.
But the sum of the measure of the angles of a triangle is 180°.
Therefore 40°, 80°, 70° cannot be the measures of the angles of a triangle.

Example problem 2:

Find the supplementary angle of 95°

Solution:

The supplementary angle of 95° = 180° – 95° = 85°.

Example problem 3:

The lengths of two sides of right triangles are 6cm and 8cm. Find its hypotenuses.

Solution:

AC = 6 cm
BC = 8 cm
AB =?
AB2 = 62+ 82
= 36 + 64
AB2 = 100
AB = √100 = 10

Thus, the hypotenuses are 10 cms in length.
Practice Problems - Euclidean Geometry:

Practice problem 1:

Two angles of a triangle are of measures 52°and 78°. Find the measure of the third angle.

Ans: 50

Practice problem 2:

Find the angles in each of the following:
(i) The angles are supplementary and the larger angles are twice the small.
(ii) The angles are complementary and the larger is 20° more than the other

Ans:    1. smaller angle = 60°, larger angle = 120°.

2. smaller angle = 35°, larger angle =  55°.

Geometry Parallelogram Proofs

Geometry proofs are basic laws of geometry figures, using these geometry proofs the geometry problems are solved. The Geometry parallelogram proofs are the proofs for parallelogram that help the rules of solving parallelogram based problems. A parallelogram proofs form an argument that establishes the truth of the parallelogram figure. Let us see basic geometry parallelogram proofs.

Please express your views of this topic Area of Parallelogram by commenting on blog.

Geometry Parallelogram Proofs:

Properties of parallelogram:

We know that a quadrilateral is a closed plane figure formed by four line segments and a parallelogram is a quadrilateral in which the opposite sides are parallel to each other.

Proof in geometry 1: In a parallelogram, the opposite sides are of equal length.

Parallelogram proof

Proof: Let ABCD be a parallelogram. Link up BD. take the triangles ABD and BDC.

Since AB || CD and BD is a transversal of AB and CD, m∠ABD = m∠BDC

Since AD|| BC and BD is a transversal of AD and BC, m∠ADB = m∠DBC

The side BD is common to both ABD and BDC.

Hence, by AAS property, ΔABD ≡ ΔBDC.

∴ The particular sides are equal.

AD = BC and AB = CD

Proof in geometry 2: In a parallelogram, the opposite angles are of equal measure.
Parallelogram proof
Proof: Let ABCD be the parallelogram. Link up BD.

Since AB || DC and BD is a transversal to AB and DC,
m∠ABD = m∠BDC
Since AD || BC and BD is a transversal
to AD and BC,
m∠ADB =m∠CBD
∴ m∠ABC = m∠ABD + m∠DBC
= m∠BDC+ m∠ADB
= m∠ADC
Similarly, m∠BAD=m∠BCD
Geometry Parallelogram Proofs:

Proof in geometry 3: The diagonals of a parallelogram bisect each other.
Parallelogram proof
Proof: ABCD is a parallelogram. AC and BD form diagonals.
By ASA criterion, ΔAMB ≡ CMD
∴AM = CM, BM = DM.
∴This shows diagonals bisect each other.


Proof in geometry 4: If the opposite sides of a quadrilateral are of equal length, then the quadrilateral is a parallelogram.
Parallelogram proof
Proof: Let ABCD be a quadrilateral where AD = BC,AB = CD. Link up AC.

Consider the triangles from the figure, ACB and ADC. By SSS criterion, ΔABC ≡ ΔCDA

Then m∠ BAC = m∠ACD,m∠CAD=m∠ACB
∴ AB || CD and AD || BC. Hence ABCD form a parallelogram.

Quadratic Equation x Intercept

In coordinate geometry, the x intercept means the x-value of the point where the graph of a function or relation intercepts the x-axis of the coordinate system. It means the point at which the line cuts the x-axis.

The x intercept of a line is denoted as (x, 0).

In this article, we are going to learn finding x intercept for the quadratic equation.


Example Problems to Find X Intercept for Quadratic Equation:

Example 1:

Find x intercept for the quadratic equation y = x2 + x - 12

Solution:

Step 1: Given function

y = x2 + x - 12

Step 2: Plug y = 0 in the given function

0 = x2 + x - 12

We can write it as

x2 + x - 12 = 0

Step 3: Solve the above function for x to find x intercept

Find sum of two numbers and its product for x and - 12x2

x = 4x - 3x

-12x2 = (4x)(-3x)

Replace the term x by 4x -3x

x2 + 4x - 3x - 12 = 0

Take the common factor outside,

x(x + 4) - 3(x + 4) = 0

(x + 4)(x - 3) = 0

Factor each term,

x = - 4   and x = 3

Step 4: Solution

X intercepts for the given equation are (- 4, 0) and (3, 0)

Example 2:

Find x intercept for the quadratic equation y = 2x2 + 10x + 8

Solution:

Step 1: Given function

y = 2x2 + 10x + 8

Step 2: Plug y = 0 in the given function

0 = 2x2 + 10x + 8

We can write it as

2x2 + 10x + 8 = 0

Step 3: Solve the above function for x to find x intercept

Find sum of two numbers and its product for 10x and 16x2

10x = 8x + 2x

16x2 = (8x)(2x)

Replace the term 10x by 8x + 2x

2x2 + 8x + 2x + 8 = 0

Take the common factor outside,

2x(x + 4) + 2(x + 4) = 0

(2x + 2)(x + 4) = 0

Factor each term,

x = - 1   and x = - 4

Step 4: Solution

X intercepts for the given equation are (- 1, 0) and (- 4, 0)




Practice Problems to Find X Intercept for Quadratic Equation:

1) Find x intercepts for the quadratic equation y = x2 + 18x + 32

2) Find x intercepts for the quadratic equation y =2 x2 + 14x + 12

2) Find x intercepts for the quadratic equation y = 3x2 + 6x - 8

Solutions:

1) x intercepts are (-2, 0) and (-16, 0)

2) x intercepts are (-1, 0) and (-6, 0)

3) x intercepts are (0.914, 0) and (-2.914, 0)

Histogram Unequal Intervals

The Histogram Definition unequal intervals of applications are mostly used in statistics. The histogram unequal intervals are most effective way  to analysis the frequency distribution of grouped data. The concepts of histogram unequal intervals to show the quick way of understanding the grouped data in the graphical representations. The analyzing histograms diagrams are constructed from the two ways are depending on the uniform width, and varying widths.
Constructions for Histogram Unequal Intervals:

To analyzing  the histogram unequal intervals , we draw two perpendicular axes and choose a suitable scale for each axis. We blot class or group intervals of the continuous grouped or class of data on the horizontal axis and the respective number of  occurrence  of each class on the vertical axis. For each class, a rectangle is constructed with class interval as the base and height determined from the class frequency so that areas of the rectangles are proportional to the frequencies.

Let us analyzing histogram unequal intervals through an example, how a grouped frequency distribution of number of teachers in a city can be represented in histogram.


Histogram

From the above histogram unequal intervals diagrams, it is clear that the maximum number of students in the group 50 - 60 and the minimum number of students in the group 70 - 90.

Notes: In the diagram  (kink) before the class interval 30 - 40 on the horizontal axis. It shows that the full distance 0 - 30 is not shown.
Example of Histogram Unequal Intervals:

Another example of histogram unequal intervals for a math tutor wanted to analyse the performance of two class of students in a math test of 100 marks. Looking at their performances,  math tutor found that a few students got under 20 marks and a few got 70 marks or above. So the math tutor decided to analyzing histogram unequal intervals to group them into intervals of varying sizes as follows: 0 - 20, 20-30, ...., 60 - 70, 70 - 100. Then she formed the following table.
Marks                          Number of students
0 - 20                                         9
20 - 30                                       14
30 - 40                                       14
40 - 50                                       18
50 - 60                                       18
60 - 70                                       12
70 - above                                  4