Primary Series Sequence

The sequence is a set of things that are in order. The sequence goes on to infinite number and it is called an infinite sequence. Otherwise it is finite sequence. They are different sequences like arithmetic, geometric, Triangular sequences these are special sequences. Series representation of sequence depends on the difference between two digits and the difference should be same thought of sequence then represent with mathematical induction formulas and simplify the equation

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Problems on Primary Series Sequence:

The series and sequence contains arithmetic progression, geometric progression, sum of first n natural numbers and sum of square of n natural numbers etc.  The following problems based on these concepts.

Example 1 in primary series sequence:

Problem:

To find general term for given sequence, suggest possible next three terms. The sequence is 8, 16, 24…

Solution:

The given sequence is 8, 16, 24….

The first term = 8.

The common difference is = 16-8=8.

The next three terms in the given sequence =24+8=32, 32+8=40, 40+8=48,

The series are 8, 16, 24, 32, 40, and 48.

Example 2 in primary series sequence:

Problem:

To find the numbers the sum of 3 numbers in GP is 16 and their product is 64.

Solution:

Let the three numbers x/y, x, xy.

x/y + x+ xy =16

(x/y) x (xy) =64.  X3 = 64; x= 4.

x/y(1 + y +y2 )=16

4(1 + y +y2 ) =16y.

4 y2  + 4y -16y +4=0

y2  -3y+1=0

y=2.

Where y=2 then the numbers are 2, 4, 8.

Example 3 in Primary Series Sequence:

Problem:

To find general term for given sequence, suggest possible next four terms. The sequence is 3, 6, 9…

Solution:

The given sequence is 3, 6, 9….

The first term = 3.

The common difference is = 6-3=3.

The four terms in the given sequence =9+3=12, 12+3=15, 15+3=18, 18+3=21.

The series are 3, 6, 9, 12, 15, 18, and 21.

Example 3 in primary series sequence:

Problem:

To find sum to 5th in the sequence numbers in the 2,4,8..

Solution:

Here first term a=2

The ratio r = 4/2=2

Sum =  arn-1

n=5, sum = 2(25-1) =32.

Practice Doing Functions in Math

In mathematics, function is nothing but depending on the input value, the output value to be determined. We can map the variables of the function into a coordinate system. Function is denoted in the form of y = f(x). If a function contains only one variable then it is called as one variable function. For example: f(x) = 12x + 4. Here, x is a variable. Now, we are going to discuss some of the problems for doing functions in math.


Example Problems- Practice Doing Functions Math:

Example problem 1:

Find the ordered pairs of the function: f(x) = 10x + 2

Solution:

f(x)= 10x + 2

Substitute x=0

f (0) =  10(0) + 2

y = 2

Therefore the ordered pair (x, f(0)) is (0, 2).

Substitute x = 1

f(1) = 10(1) + 2

y = 12

Therefore the ordered pair (x, f(1)) is (1, 12).

Substitute x=2

f(2) = 10(2) + 2

y = 22

Therefore the ordered pair (x, f(2)) is (2, 22).

Substitute x=3

f(3) = 10(3) + 2

y = 32

Therefore the ordered pair (x, f(3)) is (3, 32).

The ordered pairs of the function f(x) = 10x + 2 is (0, 2), (1, 12), (2, 22), (3, 32).

Example problem 2:

Find whether the relation {(-2, 4), (-1, 6), (-1, 8), (0, 10)} is a function?

Solution:

The given ordered pair is {(-2, 4), (-1, 6), (-1, 8), (0, 10)}.

Functions / relations

Here, more than one ordered pair with the same x coordinate, but with different y coordinates. The ordered pairs (-1, 6) and (-1, 8) have the same x coordinate and y coordinates are different (i.e.,) two y coordinate 6 and 8 corresponds to a single x coordinate -1. Therefore, this relation cannot be a function.

Algebra is widely used in day to day activities watch out for my forthcoming posts on solving systems of equations solver and Volume of Sphere. I am sure they will be helpful.

Additional Problems- Practice Doing Functions Math:

Example problem 3:

Find the zeros of the one variable function f(x) = x2 - 173x + 172.

Solution:

Set the equation equal to zero.

0 = x2 -173 x + 172

Factor the quadratic function and solve for x.

Here, a = coefficient of x2 = 1

b = coefficient of x = -173

c = constant term = 172

We find a × c = 1× 172 = 172 = -1*-172, (-1) + (-172) = -173 = b.

x2 - 173x + 172 = 0

x2 + (- 1 - 172)x + 162 = 0

x2 – 1 x – 172 x + 172 = 0

x (x - 1) – 172 (x - 1) = 0

 (x – 1) (x – 172) = 0

x = 1, 172

So, the zeros occur when x equals 1 and 172.

Example problem 4:

Is the function f(x) = 10x + x3 even function or odd function?

Solution:

f(x) = 10x + x3

Substitute the value –x in the place of x.

f(-x) = 10(-x) + (-x)3

f(-x) = -10x – x3 = -(10x + x3)

f(-x) = -f(x)

So, the given function f(x) = 10x+ x3 is an odd function.


Practice Problems for Doing Functions in Math:

1) Find the zeros of the one variable function f(x) = x2 - 172x + 171. (Answer: 1, 171).

2) Is the function f(x) = 2x2 + x4 even function or odd function? (Answer: even function).

Precalculus Trigonometry

Precalculus is an advanced form of  the secondary schools algebra, is a foundational mathematical discipline. It is also called Introduction to the Analysis. In many school, precalculus is actually two separate courses: Algebra and Trigonometry.

Trigonometry is a branch of a mathematics that studies triangles, particularly right triangles. Trigonometry deal  with the relationships between the sides and the angles of triangles, and with trigonometric functions, which describe those relationships and angles in general, and the motion of waves such as sound and light waves.                                           (Source.Wikipedia)
Precalculus with Trigonometry Identities:

sin2x+ cos2 x≡ 1

1 + tan2x≡ sec2 x

1 + cot2x ≡ cosec2 x

sin2x≡ 1 – cos2x

cos2x≡ 1 – sin2x

tan2x ≡ sec2x − 1

sec2x − tan2x ≡ 1

cot2x ≡ cosec2x − 1

cosec2x − cot2x ≡ 1
precalculus trigonometry ratios:

sin (90° − x) = cosx

cos (90° − x) = sin x

tan (90° − x) = cot x

cot (90° − x) = tan x

sec (90° − x) = cosecx

cosec (90° − x) = sec x
Examples for Precalculus with Trigonometry:

Example 1:

Find x° if cosec x° = sec 55°.

Solution:

Since cosec x° =sec(90°− x°),we have sec(90°− x°)= sec 45°. ∴ 90°− x°=45°.

∴ x° = 90°− 45°= 55°.

Example 2:

Evaluate sin 30° tan 60° sec 30°

Solution:

sec 30° = sec(90°−60°) = cosec 30°=1/sin 30°

∴sin 30° tan 60° sec 70° = sin 30° tan 60° cosec 30°

=sin 30° × 3× 1/sin 30°  = 3

Examples for trigonometry identities:

Problem 1:

Prove that sin4a + cos4 a= 1 – 2sin2a cos2a.

Solution:

LHS = sin4a + cos4a = (sin2a) 2 + (cos2a) 2

= [sin2a + cos2a] 2 – 2 (sin2a) (cos2a)                 ( a2 + b2 = (a + b)2 – 2ab)

= (1)2 – 2sin2a cos2a

= 1 – 2sin2a cos2a

= RHS

Hence proved

Problem 2:

Prove that sin4b − cos4b = sin2b − cos2b

Solution:

LHS = sin4b − cos4b = (sin2b) 2 – (cos2b) 2

= (sin2b + cos2b) (sin2 b − cos2b) = (1) (sin2b − cos2b)

= sin2b − cos2b = RHS

Problem 3:

Prove that (sec c+ cos c ) (sec c − cosc ) = tan2c + sin2c.

Solution:

LHS = (sec c+ cosc ) (sec c − cosc ) = sec2c − cos2c

= (1 + tan2c) – cos2c= tan2c + (1 – cos2c)

= tan2c + sin2c = RHS

Points Lines and Planes

Points:

In math points are used to mark the location.

Lines:

A line can be formed by connected set of infinitely points. The line is extended in both the directions. The symbol `harr` is used to represent the line in math.  Lines are denoted by letters like A, B, C, D.

Planes:

A plane can be formed by an infinite group of points forming a related flat surface expanding infinitely distant in every directions.


Points - Points Lines and Planes:

Points:

this diagram shows the structure of point

In math point is used to represent the correct location or position or place. In math point has been represented by capital letters like A, B, C and D. Mark a point on a sheet of paper as like above diagram.

Properties of points:

Point has no length
Point has no breath
Point has no thickness
Point has no starting point
Point has no end point

Lines - Points Lines and Planes:

Lines:

A lines can be formed by connected set of infinitely points. The line is extended in both the directions.

Draw a line:

With the help of pencil and scale we can draw a line in the plains.

this diagram shows the structure of line

Parallel lines:

When two lines are placed on a plane in same distance without intersection is called as parallel line. The term parallel in math can be represented as ||. For example PQ || to YZ. he above statement shows that the line PQ is parallel to the line YZ. The line PQ are parallel to the line YZ, if they don’t have any common point or center point.

The above diagram shows the structure of parallel line.

Properties of lines:

Line has infinite length
Line has zero width
Line has zero height

Planes - Points Lines and Planes:

Plains:

Plane consists of four sides. Four sides closed part is called as planes.



Properties of planes:

Plane has infinite or unlimited length
Plane has unlimited or infinite width.
Plane has zero thickness (zero height)
Plane consists four sides
Planes are marked by letters like P, Q, R and S.

Segments of Equal Length

The segments are one of the basic concepts of geometry in math subject. The segment is a small piece of a full figure. Segments are two types. There are line segment and circle segment. The line segments are the small part on a large and straight infinity line, which is having both sides of end points. length is the linear extent or measurements of one end point to another one end point, usually being a long dimensions. Here in this article we are going to explain about segments and equal length of segments.


General Definition for Segments:

The segments are the small part or small distance of a long and straight infinity line, which is having two end points on both sides of an infinity line.
A straight line that is joining with two end points, those are coordinates without extending the line after that the end point. One end point to another one end point distance is called as length.


Example figure for “General line segments”:

Line segment

The above example segment figure is, an infinity line is xy, and it is having a segment AB. It is called segment` bar (AB).`

Examples for Segments of Equal Length:

The following given example line segment figure, is example for equal length of line segments. Here an infinity line is AB, and it is having 4 line segments are CE, ED, DE and EC.

Line segment length

It is called the names of line segment `bar (CE), bar (ED), bar (DE) and bar (EC)` .
Those all segments are equal lengths.

Another example for segments of equal length: (square)

Line segment of equal length



Explanatory equal length of segments:

Segment is nothing but the line segments. That is having an end point on all directions.

This above figure represented the Closed and regular polygon (square) figure.
Because of 4 end points are there in the figures, and those all points are joining with each other. And makes the closed figure that is made up of line segments.
This above figure having the 4 end points like A, B, C, and D and alla are equal lengths.
Those 4 points are makes the many line segments like, those line segment names for line segment AB, BC, CD, and DA. Those 4 are original and the reverses of above line segments are BA, CB, DC, and AD.
So those 4 line segments are named as` bar (AB), bar (BC), bar (CD), bar (DA)` , and then reverse 4 are named as `bar (BA), bar (CB), bar (DC), bar (AD).`
This is a square figure, so all the sides are equal, so same like that all the segments length are also equal.
It is having the line segments` bar (AB), bar (BC), bar (CD), bar (DA)` , and etc…

These all are the important notes in the examples of segments. And the above explanations and examples very are useful and easy for understand the equal length of segments.

Simplify and Write in Degrees

In math, degree means a position of a scale or amount of quality. It is mainly used measure the position of the angle. There are different types of angles like as acute angle, obtuse angle, right angle, straight angle, reflex angle and complete angle.

For example,

35^@ this is a way to write in degrees.
How to Learn Simplify and Write in Degrees:-

In the following example to explain the how to learn simplify and write in degrees:

Convert pi/18 into degrees.

We know the formula for convert radians into degrees.

Pi/(180^@)

= pi/18 xx (180^@)/pi (Multiplying the both values)
= (180^@)/18 (Cancel the pi value then divide the both numerator and denominator value)
= 10 degrees

Finally we get an answer as 10 degrees.


Example Problems for Simplify and Write in Degrees:-

Problem 1:-

Simplify pi/9 and write in degrees

Solution:

Given: pi/9

Convert pi/9 into degrees.

We know the formula for convert radians into degrees.

Pi/(180^@)

= pi/9 xx (180^@)/pi (Multiplying the both values)
= (180^@)/9 (Cancel the pi value then divide the both numerator and denominator value)
= 20^@ degrees

Finally we get an answer as 20^@ degrees.


Problem 2:-

Simplify pi/30 and write in degrees

Solution:

Given: pi/30

Convert pi/30 into degrees.

We know the formula for convert radians into degrees.

Pi/(180^@)

= pi/30 xx (180^@)/pi (Multiplying the both values)
= (180^@)/30 (Cancel the pi value then divide the both numerator and denominator value)
= 60^@ degrees

Finally we get an answer as 60^@ degrees.


Problem 3:-

Simplify pi/36 and write in degrees

Solution:

Given: pi/36

Convert pi/36 into degrees.

We know the formula for convert radians into degrees.

Pi/(180^@)

= pi/36 xx (180^@)/pi (Multiplying the both values)
= (180^@)/36 (Cancel the pi value then divide the both numerator and denominator value)
= 5^@ degrees

Finally we get an answer as 5^@ degrees.



Problem 4:-

Simplify pi/60 and write in degrees

Solution:

Given: pi/60

Convert pi/60 into degrees.

We know the formula for convert radians into degrees.

Pi/(180^@)

= pi/60 xx (180^@)/pi (Multiplying the both values)
= (180^@)/60 (Cancel the pi value then divide the both numerator and denominator value)
= 30^@ degrees

Finally we get an answer as 30^@ degrees.

R Bar Statistics

The r- bar Statistics is the science of creating ordered use of arithmetic data between the groups of individuals. The r-bar statistics deals with all features of this, jointly with not only the group, study and understanding of such data, but also the planning of the collection of data. The r-bar is the average range of set observations. The r-bar statistics is mainly used to find the average range between the groups of data series. Range measures variability of process.


Formula to Find R Bar Statistics:

Range is defined as the difference between the Maximum value to the Minimum value.

Range=Maximum value – Minimum value

R-Bar defined as the average range of set of observations

R-bar (Average range) is defined as
R-Bar =` ("Range1" + "Range2") / 2`
Coefficient of range is defined as
Coefficient of range = `("MaxValue"** "MinValue") / ("MaxValue" **" MinValue")`
Example Problems for Solving R Bar Statistics:

Example1:

Find out the r-bar (average range) for the following datas?

Series1:  2, 5,9,14,17,23,39.

Series2: 13,17,23,28,35,45,49

Solution:

Range for series1 is

Here the maximum value=39, minimum value=2

Range=Maximum value – Minimum value

=39-2

=37

Range for Series2 is

Here the maximum value=49, minimum value=13

=49-13

=36

R-bar = `("Range1"+"Range2")/2`

= `(37+36)/2`

R-bar =36.5

Example2:

Find out the r-bar (average range) for the following datas?

Series1:  10, 13, 15, 19,21,25,28

Series2: 13,17,19,22,25,27,29

Solution:

Range for series1 is

Here the maximum value=28, minimum value=10

Range=Maximum value – Minimum value

=28-10

=18

Range for Series2 is

Here the maximum value=29, minimum value=13

=29-13

=16.

R-bar = `("Range1"+"Range2")/2`

= `(18+16)/2`

R-bar =17

Practice problems for solving r bar statistics:

Problem1:

Find out the r-bar (average range) for the following datas?

series1: 12, 23,32,40,55,65,78,90.

series2: 10, 20, 36, 92, 95,40,50,56

Answer: 62

Problem2:

Find out the r-bar (average range) for the following datas?

series1: 50        56        80        70        60        60        88.

series2:   36        92        95        40        50       56        80

Answer: 48.5

Typical Relationship Problems

The following are the typical relationship problems. These problems are solved by using logic.

Problem: - Pointing out to a photograph, a man tells his friend, “She is the daughter of the only son of my father’s wife.” How is the girl in the photograph related to the man?

Solution: - The relations may be analysed as follows:

Father’s wife – Mother; Mother’s only son – Himself.

So, the girl is man’s daughter.

Problem: - Pointing to a man on the stage, Rita said, “He is the brother of the daughter of the wife of my husband.” How is the man on the stage related to Rita?

Solution: - The relations may be analysed as follows:

Wife of husband – Herself; Brother of daughter- Son.

So, the man is Rita’s son.

Problem: - A man pointing to a photograph says, “The lady in the photograph is my nephew’s maternal grandmother.” How is the lady in the photograph related to the man’s sister who has no other sister?

Solution: - Clearly, the lady is the grandmother of man’s sister’s son i.e., the mother of the mother of man’s sister’s son i.e., the mother of man’s sister.

So, the lady is man’s mother.

Problem: - Pointing out to a lady, a girl said, “She is the daughter-in-law of the grandmother of my father’s only son.” How is the lady related to the girl?

Solution: - Father’s only son- My brother; Grandmother of my brother – My grandmother;

Daughter-in-law of my grandmother – My mother.

So, the lady is girl’s mother.
Some Unsolved Typical Relationship Problems

Problem: - Pointing to a lady, a man said, “The son of her only brother is the brother of my wife.” How is the lady related to the man?           (Answer: Sister of father-in-law)

Problem: - Introducing a man to her husband, a woman said, “His brother’s father is the only son of my grandfather.” How is the woman related to this man? (Answer: Sister)

Problem: - A, B, C, D, E, F and G are members of a family consisting of four adults and three children, two of whom, F and G are girls. A and D are brothers and A is a doctor. E is an engineer married to one of the brothers and has two children. B is married to D and G is their child. How is C related to A?

(Answer: C is A’s son.)

Boolean Algebra Complement

Boolean algebra operators are AND, OR, NOT and equal. The process of NOT operator is found the complement or opposite. Through the truth table we can learn Boolean algebra operation easily. Truth table is connected with the Boolean algebra. In the truth table the expression ‘true is represented as ‘T and the expression ‘false is represented as ‘F. Let us see Boolean algebra complement and their symbols in this article.
Boolean Algebra Complement:

Only one operand is present in the NOT operator. For example ~X here the symbol ~ is used to denote expression opposite or complement. The most important operation of operator NOT is complement or opposite.

Some rules in Boolean algebra complement:

Complement of PQ:

(PQ) = P + Q

Complement of P Q

(P Q) =P + Q

Complement of P + Q + R

(P + Q + R) = P * Q * R

Complement of PQR:

(PQR) = P + Q + R

Easy way to remember this Complement rule is put * for + and put + for *

For example:

Complement of PQ = > (PQ) = P + Q
Proof for Boolean Algebra Complement:

Prove:

Complement of P Q `!=` PQ

Proof:

Given

(P Q) `!=` PQ

LHS:

P Q

Let us consider P=1 and Q =0

F = P Q

F = 1 0

F= 0 1

F = 0

Take the complement of F

We get

F =1

RHS:

PQ

P=1 and Q =0

F = 1 0

F =0

LHS is not equal to RHS therefore (P Q) `!=` PQ

Prove:

Complement of P Q = P + Q

Proof:

Given

(P Q) = P + Q

LHS:

P Q

Let us consider P=1 and Q =0

F = P Q

F = 1 0

F= 0 1

F =0

Take the complement of F

We get

F =1

RHS:

P+Q

P=1 and Q =0

F = 1 + 0

F =1

LHS is equal to RHS therefore (P Q) = P + Q.

Prove:

(PQ) = P + Q

Proof:

Given:

(PQ) = P + Q

LHS:

(PQ)

Let us consider P=1 and Q =0

F = (1 0)

F = (0)

F =1

RHS:

P + Q

Let us consider P=1 and Q =0

F =1 + 0

F =0 + 1

F = 1

LHS is equal to RHS therefore (PQ) = P + Q

Applications of Integration Work

Integration is applied on a given function within certain limits to get various output.

Here let us see three areas of integration application work. In the normal sense, the integrations are applied to get the area under the given curve between certain limits.

1. Area under the curve f (x) between the limits ‘a’ and ‘b’ is given by`int_a^bf(x)dx`

2. The arc length of a curve f (x) between the limits ‘a’ and ‘b’ is given by `int_a^bsqrt[ 1 + [dy/dx] ^2] dx` .

3. The volume of a curve f(x) between the limits ‘a’ and ‘b’ is given by `pi int_ a^b y ^2 dx`

Now let us see problems on this topic.


Example Problems on Application of Integration Work.

Ex 1: Find the area under the curve f (x) = 2

4x 2 + 3x + 2 between 2 and 4.

Soln: Given: f (x) = 4x 2 + 3x + 2

The limits are ‘2’ and ‘4’.

Therefore Area =` int_2^4(4x ^2 + 3x + 2) dx` = `[4x ^3 / 3 + 3x ^2 / 2 + 2x] ` , the limit is between 2 to 4.

= `[(4 (4 ^3) / 3 + 3 (4) ^2 / 2 + 2 (4) ) ** (4 (2) ^3 / 3 + 3 (2) ^2 / 2 + 2 (2) )]`

Therefore Area = `96 2 / 3 ` sq units.

Ex 2: Find the arc length of a curve f (x) = 4x + 3 between 0 and 1.

Soln: Given: f (x) = 4x + 3

Therefore f ‘ (x) = 4.

Therefore arc length =` int_0^1 sqrt [(4) ^2 + 1] dx ` = `sqrt 17` `int_0^11 dx`

= sqrt 17 (x) between 0 to 1 = sqrt 17 (1 – 0) = sqrt 17 units.


More Example Problem on Application of Integration Work.

Ex 3: Find the volume generated when the finite region enclosed by the curve y = x ^2 and the lines x = 2, x = 4 and the x axis.

Soln: Volume = `pi int_2^4 y ^2 dx`

=` pi int_2^4 (x ^2) ^2 dx`

= `pi` [`x ^5 / 5` ] 2 to 4

= `pi / 5 [ 4 ^5** 2 ^5] = 992 / 5` cubic units.

Direct Inverse Relationship

Chemistry is study of physical and chemical properties of elements and compounds. While studying the same, various observations are made and inference derived from it. As such it involves discovery of correlations. Correlations means dependency. Dependency may be direct or inverse.

Establishing the dependencies scientifically is  very important as this can be converted to standard laws. These standard laws are useful in study of chemistry and chemical calculations
Direct Dependencies

Direct dependency means a parameter changes as the another changes. If one increases the another too increases.

e.g. An observation is made that the volume of a gas increases as the temperature increases. How can this direct dependendeny be presented mathematically?

V `alpha ` T

This equation is further rationalised as

V   = k T

where k is the constant for the corelation

OR

V / T =Constant

This is the established equation for Charles' Law

The direct dependencies may be proportional to 1st or 2nd or 3rd or even  1/2 or 1/4 etc. power too.

As in case of famous Einstein equation

E= mc2

lt denotes that the energy released is directly proportional not to the speed of light but, to the square of  the speed of light
Inverse Dependencies

Inverse dependency means a parameter changes as the another changes.If one increases the another decreases.

e.g. An observation is made that the volume of a gas increases as the temperature decreases. How can this inverse dependendeny be presented mathematically?

V `alpha ` 1/P

This equation is further rationalised as

V   =k / P

where k is the constant

for the corelation between volume and pressure..

or PV=Constant

This is the established equation for Boyles' Law

The inverse dependencies may be proportional to 1st or 2nd or 3rd or even  1/2 or 1/4 etc. power too

Rhombus Perimeter Formula

In geometry, a rhombus or rhomb is a quadrilateral whose four sides all have the same length. The rhombus is often called a diamond, after the diamonds suit in playing cards, or a lozenge, though the latter sometimes refers specifically to a rhombus with a 45° angle. (Source: Wikipedia)

Formula for perimeter:

Perimeter of a rhombus = 4 * Side length


Example problem 1:

A Rhombus has the side length is 17 m. Find the perimeter of the rhombus.

Solution:

Given side length of the rhombus is 17 m

Formula:

Perimeter of a rhombus = 4 * side length

Substitute the given side length value in the above equation, we get

Perimeter = 4 * 17 m

= 68 m

Answer:

The final answer is 68 m

Example problem 2:

Find the perimeter of the given rhombus in centimeter.

Rhombus perimeter formula - Example problem 1

Solution:

Given side length of the rhombus is 12 cm

Formula:

Perimeter of a rhombus = 4 * side length

Substitute the given side length value in the above equation, we get

Perimeter = 4 * 12 cm

= 48 cm

Answer:

The final answer is 48 cm

Example problem 3:

Find the perimeter of the given rhombus in meter.


Rhombus perimeter formula - Example problem 2

Solution:

Given side length of the rhombus is 9.1 m

Formula:

Perimeter of a rhombus = 4 * side length

Substitute the given side length value in the above equation, we get

Perimeter = 4 * 9.1 m

= 36.4 m

Answer:

The final answer is 36.4 m

Example problem 4:

Find the perimeter of the given rhombus in meter.

Rhombus perimeter formula - Example problem 4

Solution:

Given side length of the rhombus is 7.7 m

Formula:

Perimeter of a rhombus = 4 * side length

Substitute the given side length value in the above equation, we get

Perimeter = 4 * 7.7 m

= 30.8 m

Answer:

The final answer is 30.8 m


Practice Problems for Rhombus Perimeter Formula

Practice problem 1:

Find the perimeter of the given rhombus. (Note: Formula for perimeter = 4 * side length)

Rhombus perimeter formula - practice problem 1

Answer:

The final answer is 128 m

Practice problem 2:

Find the perimeter of the given rhombus. (Note: Formula for perimeter = 4 * side length)

Rhombus perimeter formula - practice problem 2

Answer:

The perimeter of the rhombus is 14.4m

Practice problem 3:

Find the perimeter of the given rhombus. (Note: Formula for perimeter = 4 * side length)

Rhombus perimeter formula - practice problem 3

Answer:

The final answer is 34 m

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°.