Binomial Nomenclature List

Introduction to Binomial nomenclature

The species concept: Species is recognized as the smallest unit of taxonomy or the basic unit of classification. In simpler form every different and recognizable type of organism (animal, plant or microbe) is a different species. In scientific terms, a species could be defined as a group of closely resembling organisms of the same type, in which members can interbreed freely and naturally to produce viable and fertile offspring, whereas they cannot interbreed freely or produce viable and fertile offspring with members outside the group.

Binomial nomenclature

Binomial nomenclature is the scientific method of giving names to every species of living organism. Carolos Linnaeus first introduced it in 1758. In this method every species of organism is given a name consisting of two words. The first word would refer to the name of the genus to which the organism belongs in the existing system of classification. Therefore this is known as the generic name. The second word in the name is called the specific name, since it refers to the exact species to which the organism belongs.

Rules to be followed in giving scientific names to plants and animals on the basis of binomial nomenclature

• The name must consist of two words both in Latin and in other languages in Latinized form.
• The first word should refer to the genus to which the organism belongs and it should start with capital letter.
• The second word should refer to the species to which the organism belongs and it starts with a small letter.
• Names derived by binomial nomenclature should also include reference in abbreviation to the name this is known as author citation.
• The generic and specific names when printed must be in italics and when written otherwise must be separately underlined, indicating their Latin origin. Author citation is not in italics nor is it underlined.
• When the same organism is given more than one scientific name by different authors, the earlier name becomes accepted.
• These rules are framed and presented as International Code for botanical Nomenclature (ICBN) for plants and as International Code for Zoological Nomenclature (ICZN) for animals. They are periodically reviewed and revised by specific committees.
  
  

Advantages of binomial nomenclature

• Names derived from binomial nomenclature gives universal reorganization to the organism.
• Scientific name avoid confusion arising out of the same organism having different common names in different origins of the world.e.g. For brinjal in India (Solanum melongena) is called egg plant in western countries.
• Using binomial nomenclature system automatically establishes taxonomic interrelationships between different taxa.
• Addition of new information, acquiring rare information, exchange of resource materials, and exchange of information etc., become easy when universally accepted scientific names are used.

Geometry Solid Figures

Geometry solid figures are a branch of geometry which mainly focuses on the properties like surface area, and volume of solid figures or three dimensional shapes like Rectangular Prism, cube, cone, cylinder, and sphere. Geometry solid figures generated by the revolution of a two dimensional plane. For example a cylinder is generating by the revolution of a rectangle; the revolution of a circle about its diameter generates a sphere.

Formulas for geometry solid figures:-

Geometry solid figures of Cube:-


Surface area = 6a2 square units, Volume = a3 cubic units.

Geometry solid figures of Cylinder:-


Lateral surface area = 2πrh square units, Total surface area = 2πr (h + r) square units, Volume = πr2h cubic units

Geometry solid figures of Cone:-


Lateral surface area = πrs square units, Total surface area = πrs + πr2 square units, Volume = 1/3 πr2h cubic units

Geometry solid figures of Sphere:-


Surface area of a sphere = 4πr2 square units, Volume = 4/3 πr3 cubic units

Geometry solid figures of Rectangular Prism:-


The equation for find the volume (V) of a rectangular prism is V = wdh.

Example geometry solid figures problems:-

Example problem1:-

Find the volume of cylinder given the radius is 6 cm and 11 cm.

Solution:-

Volume of cylinder = r2 h cubic units.

= (3.14) * 62 * 11

= 3.14 * 36 * 12

= 1 356.48cm3

Example problem2:-

Find the volume of cone given the radius is 4 cm and 8 cm.

Solution:-

Volume of cone = 1/3 r2 h cubic units.

= 1/3 (3.14) * 42 * 8

= 0.33 * 3.14 * 16 * 8

= 132.6336cm3

Example problem3:-

Find the volume of cube with the side length of 8 cm.

Solution:-

Volume of cube = a3

= 83

= 512 cm3

Example problem4:-

A rectangular prism contains at width for 3 inches and depth 6 inches and height 4 inches then finds the volume of rectangular prism?

The faces are known as width (w), depth (d), and height (h).

Solution:-

Volume (V) of a rectangular prism is:

V = wdh.

Now we can calculate,

V= wdh

V= 3 in x 6 in x 4 in

V = 72 in3

Example problem5:-

The sphere has the radius of 4m.find the surface area of the sphere.

Solution:-

Radius (r) = 4 m

Surface area of the sphere = 4 π r2 square unit

= 4 x 3.14 x (4)2

=4 x 3.14 x 16

= 200

Surface area of the sphere =200 m2.

How to Reduce Binomials

Binomial is also an algebraic equation but it has only two terms. (a+b) It’s  a example of binomial. And the binomial theorem commonly used to expand the binomials to any given power without direct multiplication .Here we are going to learn about how to reducing binomials. Let us see about binomial with examples.

More information about binomials

The following binomial shows, how we can write the binomials using powers
(a + b)⁰ is equal to  1
(a + b)¹ can be written as  a + b
(a + b)² can be written as  a² + 2ab + b²
(a + b)³ can be written as a³ + 3a²b + 3ab² + b³
(a + b)⁴ can be written as  a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴
(a + b)⁵ can be written as  a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵

Example problems for how to reduce binomials:

Example 1:
Reduce  `((x+2)^2)/(x+2)`
Solution:

Step 1: We can write it as an .`((x+2)^2)/(x+2)`

Step 2: And the question can be written as  `((x+2)(x+2))/(x+2)`

Step 3: On both the numerator and denominator, the (x+2) will be cancelled.

Step 4: Therefore, the answer is (x+2).

Example 2:
Reduce 2ab+5yz-ab-3yz+9ab.
Solution:

Step 1: First we need to arrange the terms.

Step 2: So, 2ab-ab+9ab+5yz-3yz.

Step 3: Here, we need to add the terms.

Step 4: Therefore the answer is 10ab+2yz.

Example 3:
Reduce  `((x+7) (x+9))/((x+81)^2 (x+49)^2)`
Solution:

Step 1: It can be written as  `((x+7) (x+9))/((x+81)^2 (x+49)^2)`

Step 2: And, we can write the given question is `((x+7)(x+9))/((x+9)(x+9)(x+7)(x+7))`

Step 3: And Here both the sides the (x+7) and (x+9 ) are canceled.

Step 4: Therefore the answer  is`(1)/((x+9)(x+7))`

Example 4:
Reduce 4ts+6gh-ts-2gh+7ts.
Solution:

Step 1: First we need to arrange the terms.

Step 2: So, 4ts+7ts-ts+6gh-2gh.

Step 3: Here, we need to add the terms.

Step 4: Therefore the answer is 10ts+4gh

Example 5:
Reduce:  `((a+b)^3 (5a))/((a+b) (25))`
Solution:

Step 1: The question can be written as `((a+b)(a+b)(a+b)(5a))/((a+b) (25))`

Step 2: On both the numerator and denominator, the (a+b) and 5 will be canceled.

Step 3: Therefore, the answer is `((a+b)^2 (a))/(5)`

These are the examples of how to reducing binomials.

Scale Factor Triangle

A scale area is a digit to scales otherwise multiplies, a few number. In the equation y=Cx, C represents the scale factor x. C is as well as represents the coefficient of x with might be recognized the stable of proportionality of y toward x. For illustration, doubling-up distance communicate to a scale factor of 2 meant for distance, as unkind a block in half consequences in part by a scale factor of ½.


Definition of scale factor triangle:

The ratio of any two corresponding lengths in two similar geometric figures is called as Scale Factor.
The ratio of the length of the scale drawing to the corresponding length of the actual object is

called as Scale Factor.   (source:wikipedia)


More about Scale Factor:

Scale factor triangle is an item that is applies to calculate the size of the entity. The conditions scale factor is utilize to declare the proportions of the geometric form. In geometry, the expression scale factor is frequently used to modify the proportions of the image. Some shapes aspect knows how to be improved otherwise reduce. In online, scholar knows how to study regarding different subject. In online, scholar is able to study on scale factor obviously. Online knowledge is appealing with interactive. Furthermore online knowledge is dissimilar as of group area education in dissimilar method. In online, scholar contain single to ones knowledge.

A scale factor triangle is a form applies as a form through that a further shape is develops in scale.

A scale factor is innovative to scale form in 1 toward 3 sizes.

Scale factor know how to be recognized through the subsequent situation

Examples problem for scale factor triangle:

Example 1:

find the image point p(3,4) order dilation with center(0,0) with scale factor of 3.

Solution:

Step1:   To find the point on the coordinate plane under dilation with center as origin multiply the coordinates with scale     factor.

Step 2:   The points of P(3,4) is P'(3*3, 4*3)

Step 3:   The value is P'(9,12)


Example 2:

find the image point p(2,6) order dilation with center(0,0) with scale factor of 2.

Solution:

Step1:   To find the point on the coordinate plane under dilation with center as origin multiply the coordinates with scale factor.

Step 2:   The points of P(2,6) is P'(2*2, 6*2)

Step 3:   The value is P'(4,12)

Learn Quadrilaterals Pictures

The quadrilaterals are the two dimensional plane shapes which have four surfaces. Therefore it is a polygon with four sides. The sum of interior angle of a quadrilaterals are 360 degrees. There are two diagonals for each quadrilaterals. Since we have two diagonals in a quadrilaterals, both the diagonals will intersect at a common point. The regular quadrilaterals will have four edges and four vertices’. The lines of symmetry of a quadrilaterals are based on what sort of the quadrilaterals are. Let us learn about various quadrilaterals pictures.

Learn Pictures of a quadrilaterals 1:

Learn picture of Square:

• Square has 4 equal sides
• It has 4 equal angles
• Each angle of a square is a right angle
• It has 4 lines of symmetry
• Square is a regular shape

Learn Picture of a Rectangle:

 

• Rectangle has 2 pairs of equal sides
• It has 4 equal angles
• Each angle of a rectangle is a right angle
• It has 2 lines of symmetry
• Rectangle is an irregular shape

Learn Picture of a Parallelogram:

• Parallelogram has 2 pairs of equal sides
• It has 2 pairs of equal angles
• Opposite sides of a parallelogram are parallel
• It has NO lines of symmetry
• Parallelogram is an irregular shape

Learn Pictures of a quadrilaterals 2:

Learn picture of  Rhombus:

• Rhombus has 4 equal sides
• It has 2 pairs of equal angles
• Opposite sides of a rhombus are parallel
• It has 2 lines of symmetry
• Rhombus is an irregular shape

Learn Picture of a Trapezium:

 

• Trapezium has unequal sides
• One pair of opposite sides are parallel for a trapezium
• It is usually has NO lines of symmetry
• Trapezium is an irregular shape

Learn Picture of a Kite:

 

• Kite has 2 pairs of equal sides
• It has 1 pair of equal angles
• Equal sides of a kite are adjacent
• Kite has 1 line of symmetry
• Kite is an irregular shape

Chart Composite Numbers

It is an integer that has more than one prime factors. They can be expressed as unique set of prime numbers. The first composite number is 4. Besides 1 every other number is either prime or composite number. Besides 1 each other natural number is either prime or composite number. In chart composite number all the numbers will the composite numbers. so that we can find the composite numbers easily by seeing the chart.

Chart composite number upto 200:

4	50	91	130	170
6	51	92	132	171
8	52	93	133	172
9	54	94	134	174
10	55	95	135	175
12	56	96	136	179
14	57	98	138	177
15	58	99	140	178
16	60	100	141	180
18	62	102	142	182
20	63	104	143	183
21	64	105	144	184
22	65	106	145	185
24	66	108	146	186
25	68	110	147	187
26	69	111	148	189
27	70	112	150	190
28	72	114	152	192
30	74	115	153	194
32	75	116	154	195
33	76	117	155	196
34	77	118	156	199
35	78	119	158	200
36	80	120	159
38	81	121	160
39	82	122	161
40	84	123	162
42	85	124	164
45	86	125	165
46	87	126	166
48	88	128	168
49	90	129	169

Chart composite number upto 400:

201	243	288	329	370
202	244	289	330	371
203	245	290	332	372
204	246	291	333	374
205	247	292	334	375
206	248	294	335	376
207	250	295	336	377
208	252	296	338	378
209	253	297	339	380
210	254	298	340	381
212	255	299	341	382
213	256	300	342	384
214	258	301	343	385
215	259	302	344	386
216	260	303	345	387
217	261	304	346	388
218	262	305	348	390
219	264	306	350	391
220	265	306	351	392
221	266	309	352	393
222	267	310	354	394
224	268	312	355	395
225	270	314	356	396
226	272	315	357	398
228	273	318	358	399
230	274	319	360	400
231	275	320	361
232	276	321	362
234	277	322	363
235	280	323	364
236	282	324	365
237	284	325	366
238	285	326	367
240	286	327	368
242	287	328	369

90 Degree Right Angle

In geometry and trigonometry, a right angle is an angle that bisects the angle formed by two halves of a straight line. More precisely, if a ray is placed so that its endpoint is on a line and the adjacent angles are equal, then they are right angles.
A right triangle or right-angled triangle is a triangle in which one angle is a right angle (that is, a 90 degree angle).



The side opposite the right angle is called the hypotenuse (side c in the figure above). The sides adjacent to the right angle are called legs. (Source: from Wikipedia).It is shown in the figure.
Now, we are going to see some of the problems involving 90 degree right angle. From these problems, we can get clear view about 90 degree right angle.

90 degree right angle problems:

Example problem 1:
Using the diagram shown, Find the length of hypotenuse.

Given:

Right angle triangle

 a = 5cm and b = 5cm

Use the Pythagorean Theorem to find c.

c² = a² + b²

Substitute a = 5 and b = 5.

c² = 5² + 5²

c²=25+25

c²=50

Find the square root of each side.

C= √50=5 √2cm

Additional problems in 90 degree right angle:

Example problem 2:
Determine the length of the hypotenuse in the right angle triangle, given that angle θ = 37 degree.

Solution:

Sin θ= opposite side / hypotenuse

Sin 37 = 12 / x

0.6 = 12 / x

x = 12 / 0.6

x = 20

So, the side AC is 20cm.

Example problem 3:
Determine the length of the side x in the diagram, given that angle θ = 60 degree.

Solution:

Here, we use the sin θ. because the sin θ is related to the opposite side and hypotenuse.

Sin θ= opposite side / hypotenuse

Sin 60 = x / 24

0.866 = x / 24

x = 24* 0.866

x = 20.8

So, the side AC is 20.8cm.

Practice problems in right angle:
1) Find the length of hypotenuse when the adjacent side is 5 and the opposite side is 12. (Answer: hypotenuse=13)
2) Determine the length of the opposite side, given that angle θ = 30 degree and hypotenuse is 24. (Answer: opposite side= 12).