Prime Factorization can be defined as decomposition of Composite Number into non divisible Numbers. Composite numbers are those numbers which can be further classified. Non divisible numbers are called Prime Numbers that cannot be further decomposed such as 2, 3, 5, etc. Product of prime numbers is equals to composite number. Two methods can be used to obtain the prime factorization of any given number. These are division method and factor Tree method. Let's see how prime factorization of a number with exponents can be obtained. Take 45 and find its prime factorization.
Number 2 cannot be prime factor for 45 since 45 cannot be divided by 2. Let’s try 3 which results in 45 as 3 x 15 = 45.
Thus first factor is ‘3'. Now we have to get factor of 15 which can be 3 and 5 since 3 * 5 = 15.
Since both 3 and 5 are prime factors and cannot be further divided and hence we can write
prime factors of 45 are 3, 3, 5 that is 3 x 3 x 5 = 45.
If we have same factor more than once then they can be written in form of exponents that is 3 is used twice hence it can be written as (3)². So answer can be given as 45 = 3² x 5.
Let’s see what are the prime factorization of 210 bkt using exponents bkt.
Now to factor 210, first break it into two factors such that:
210 ÷ 3 = 70
70 ÷ 2 = 35
35 ÷ 5 = 7
7 ÷ 7 = 1
Thus prime factors for 210 will be 2 * 3 * 5 * 7.
We know that difference between two Numbers is obtained by subtracting one number from another. This subtraction is shown by the ‘-’ symbol. If A and B are two numbers then A – B is known as difference. Sum of the two numbers A and B is given by the addition of two numbers. Symbol ‘+’ is used to show the addition. Mathematically it is represented as A + B. Term Square refers to two times multiplication of the given number by itself. That is the square of the number A is A * A. It is also represented as (A) 2.
Product of two numbers is given by simply multiplying two numbers that is if A and B are two numbers then their product will be (A * B).
If the difference between two no is 9 and the product of the two is 14, what is the square of their sum?
Solution: Let’s take two numbers - 7 and 2. Difference of these two numbers is 9. Mathematically, it can be shown as, if A = 7 and B = 2 then difference will be 7 – (- 2) = 9 and their product will be 14 that is
(7 * (- 2)) = - 14,
Now sum of these two numbers will be given by following formula:
A + B = 7 + (-2) = 7 – 2 = 5,
Square of sum of given number will be given as:
[7 + (-2)] 2 = (7 – 2) 2 = (5) 2 = 5 * 5 = 25. Thus final answer will be 25 (square of sum of given numbers) for those numbers whose difference is equals to 9 and whose product is equals to – 14.
A frequency distribution is an organization of data about events in an experiment. This technique is frequently used in the arrangement of huge Set of records in statistical analysis. These types of data are required to be kept in a way such that it seems to be more descriptive and less complex. Here we arrange data on basis of occurrence in data set, which is also known as its frequency. A table is made to explain the data which has two columns namely value of data and its frequency. In frequency distribution we can make groups of data where size of data set is very large and then frequency of them is decided. These groups are called as classes. For these classes intervals have to be decided and it is not always possible to have equally spaced intervals in our distribution.
Classes with unequal intervals are also possible; it is subject to what kind of analysis we do regarding our data set and how it is being shown by drawing histogram.
Let us understand frequency distribution through an example:
Suppose we have a data set of 25 values given as follows: 1, 3, 5, 1, 11, 5, 10, 9, 3, 6, 6, 13, 21, 64, 36, 40, 14, 23, 4, 10, 5, 1, 3, 6 and 9.
First we arrange them in a particular order to analyze them properly: 1, 1, 1, 3, 3, 3, 4, 5, 5, 5, 6, 6, 6, 9, 9, 10, 10, 11, 13, 14, 21, 23, 36, 40 and 64. Next we prepare a Frequency Distribution Table representing their respective frequencies as follows:
We have been solving linear equation by several methods like substitution, elimination, comparing etc. The problems related to the applications of Linear Equations are discussed in various possible ways. Linear equation problems can be general word problems also where equation has to be framed using given data. So, it is required to read the equation properly in order to get the desired info and then converting that info to numeric expressions. For instance, let us consider the following word problem based on ages of people and making the linear equation:
Example: The ages of two friends is in the Ratio 5 is to 6, after how many years will the ages be in the ratio 7 is to 8?
Solution: According to given problem ratio of ages of two friends is 5 to 6 initially. We have to calculate the number of years between the change of their age ratio from 5 : 6 to 7 : 8. Let their present ages be 5a and 6a years respectively. After 'b' years their ages will be in ratio of 7 to 8. So, by using these constraints we can define following equation:
(5 a + b) / (6 a + b) = 7 /8,
Applying cross multiplication on the above equation we get:
8 (5 a + b) =7 (6 a + b),
Or 40 a+ 8 b = 42 a + 7 b,
Or b = 2 a,
This shows that age of second friend is double the age of first friend. So, after 2 times the greatest common factor of their present ages, their ratio would be 7 : 8 or we can say in 2 years their ratio will be 7 to 8.
The zero of polynomial equation can be derived by equating P (y) to zero. ‘A’ is zero or root of polynomial P(X) if and only if P(X) = 0. For example,
Let P(X) = 5X³ − 4X² + 7X − 8. Then one of the roots or zeros of this quadratic polynomial is 1. It can be proved by substituting the value y = 1 in the given polynomial.
P (1) = 5 (1)3 - 4 (1)2 + 7 (1) – 8 = 5 – 4 + 7 – 8 = 0.
If (p + q i) is a root of the polynomial equation then (p – q i) is also a root of quadratic polynomial function.
Let’s consider an example in which if (2 + m) is a root to f (m) = - m 2 + 4 m – 5, then (2 – m) is also a root of the function.
Let’s put (2 + m) in the given equation
f (m) = - (2 + m) 2 + 4 ( 2 + m) – 5,
= - (4 – 1 + 4 m) + 3 + 4 m,
= - 3 – 4 m + 3 + 4 m = 0,
f (m) = 0. If we substitute (2 – m) into the function, similar results will be obtained.
Example: If 5 is one zero of the quadratic polynomial x^2 minus kx minus 15 then value of k is?
Given polynomial equation is x2 – k x – 15. This equation can be written in the form of a function as:
f (x) = x2 – kx – 15,
In this problem, it is given to us that one zero exist at value of 'x' equals to 5. To prove this, substitute x = 5. This can be derived as:
f (5) = 52 – 5k – 15 = 0,
25 – 5k – 15 = 0,
10 – 5 k = 0,
5 k = 10 and hence k = 2.
Hence, if 5 is one zero of quadratic polynomial x2 minus kx minus 15 then value of k is 2.
Parallelogram is composed of two pairs of parallel sides. Arms in front of each other are of equal length and angles opposite to each other are of similar value in a Parallelogram. Various types of parallelogram are rhomboid parallelogram, rectangle parallelogram, rhombus parallelogram and Square parallelogram. In short, we can say that parallelogram is a four sided figure which consists of two pairs of parallel sides. Sums of angles of parallelogram is 2π radians.
Following figure shows parallelogram’s basic structure.
Area of a parallelogram is similar to that of degree of vector cross product of two adjacent sides. Mathematically, area of a parallelogram can be given by multiplication of base and height of the parallelogram. This can be expressed as:
Area = h × b (unit) 2. Here unit may be foot or cm or inch and so on.
Perimeter of parallelogram can be measured in a two dimensional figure. Perimeter is also called as circumference.
Perimeter = 2 (h × b) units.
If two sides of parallelogram are given to us then it is very easy to calculate the Area And Perimeter of parallelogram using standard formula.
Let’s try to find the perimeter of a parallelogram if the measure of the two adjacent sides are 15 in and 23 in, then to calculate perimeter following formulas are used.
Area = 15 inch * 23 inch = 345 inch2 and Perimeter of Parallelogram will be given as
Perimeter = 2 (area of parallelogram) = 2 * 345 inch2 = 690 inch2.
Probability can be defined as chances of occurrence of an event. Probability always lie between 0 and 1. Only in the ideal cases, it can be zero or one. If Probability of happening of an event is higher, it assures that chances of occurrence of that event is also high.
Probability of an event X is written as P (X). Complement of an event X is the event (not X) that is in the event of X not occurring.
Probability in this case will be given as P (not X) = 1 – P (X).
Joint probability of two events X and Y can be given as:
P (X ∩ Y).
This type of probability shows that both events X and Y happen simultaneously. This probability is known as Intersection probability. If two events are independent then intersection probability is given as:
P (X and Y) = P (X∩ Y) = P (X) P (Y).
If either X or Y or both occur at same time, then it is known as the Union of the events X and Y and is given mathematically as P (X U Y) and if events X and Y are mutually exclusive then probability of occurrence will be given as:
P (X or Y) = P (X U Y) = P (X) + P (Y).
Law of probability is called addition probability in which probability of X or Y equals to addition of probability of X and probability of Y and minus the probability of X and Y from this addition.
P(X or Y) = P (X) + P (Y) – P (X ∩Y) where P (X ∩ Y) = 0.
If 8 red balls,7 blue balls 6 green balls, find the probabilty of picking one ball that is red or green?
Let P(X) = Probability of drawing the red ball & let P (Y) be Probability of drawing the green ball. Then Total outcomes are 17. So, P (X) = 8 / 17 & P(Y) = 6/17. Therefore P (X or Y) = 8/17 + 6/17 = 14/17.
Lines are the equations that pass through many Collinear Points or we can also say that points lying on a line satisfy its equation. The Point – slope form of a line can be written using any one point lying on the line. For instance, suppose we have a point (a, b) that lies on the line and Slope of the line is 's', then equation of line in point – slope form can be written as follows:
Y – a = s (X – b)............. equation 1
Slope intercept form of line is given as:
Y = sX + c........... equation 2
Where, 'c' is the intercept made by the line:
When we have two points lying on the line, we can calculate the value of slope by formula:
s = (b2 – b1) / (a2 – a1).......... equation 3
Let us consider an example to understand this concept:
Example: How do I find the slope intercept and point form of the equation
(-1, 7) and (12, 5)?
Solution: Two points are: (-1, 7) and (12, 5). Using these points we can calculate the slope of equation using the formula shown in equation 3 as follows:
Slope = s = (b2 – b1) / (a2 – a1),
Or s = (5 – 7) / (12 + 1) = -2 / 13,
Substituting the value of 's' in the equation 2, we can write:
Y = -2 / 13 * X + C........... equation 4
Substituting (1, 7) in equation 4 we get:
7 = -2 / 13 * 1 + C,
Or C = 7 + 2/13 = 93/13,
So, the point form and slope – intercept form of line can be given as:
Y - 7 = -2/13 (X – 1),
Or Y = -2/13 + 93/13.
Arithmetic progression is that succession which has a persistent difference continued between the successive terms of the sequence. The representation of an Arithmetic Progression is the simplest of all the progressions. For instance, if the first term of the series is given as m and the common difference as 'r', then complete arithmetic progression can be written as follows:
m, m + r, m + 2 r, m + 3 r, m + 4 r, and so on to m + n r........... equation 1.
Where, 'n' is the total number of terms in arithmetic sequence. Let us consider an example of arithmetic succession to understand it better:
Example: If 11,x,5 are the terms of an AP, what is the value of x?
Solution: As we know the meaning of an arithmetic progression and the representation, that is given as shown in the equation 1. We can write the following results:
(m + r) – m = (m + 2 r) – (m + r) Consecutive terms have same difference between them.
Here in example, the given arithmetic sequence 11, x, 5 has first term as 11, x as the 2 nd term and 5 as 3rd term. Using the property of common difference between successive terms we can write:
x – 11 = 5 – x,
Taking the known quantities to right side of the equation and unknowns to other we get:
2 x = 5 + 11,
Or 2 x = 16,
Or x = 16 / 2 = 8,
So our arithmetic progression is: 11, 8, 5 with a constant difference of - 3 between the consecutive terms. Arithmetic progression finally obtained can be verified by checking the difference between the consecutive terms.
Cost price is the price at which an article is bought. All the expenditures that are made over the actual cost are included in the cost price of that article. Profit and loss measurement is generally done in terms of cost price. When articles are traded, the cost at which they are vended is known as the selling price of that article. In maths we may face different types of problems based on evaluating profits and losses using the cost price and selling price of the articles. Let us consider an example of such problems to understand them better:
Example: If selling price 55 and cost price 50 than how much profit in rupees and Percentage?
Solution: According to the question we have been given the cost price equals to 50 and the selling or vending price of article as 55. The profit seems to occur when vending price is more as compared to the cost or actual price of the article and loss prevails in the conditions where the selling price is less as compared to actual (market) price of the article.
Here in our example we are actually gaining profit in selling the good.
Formula for evaluating the profit is given as follows:
Profit (Rupees) = Vending Price – Market Price (Cost Price),
Profit (Percentage) = ((Vending Price – Market Price (Cost Price)) / Cost Price) * 100,
Here, Vending Price = 55 and Cost Price = 50,
Substituting the values of both in the above formula we get:
Profit (Rupees) = 55 – 50 = Rs. 5
Evaluating the equivalent percentage profit:
Profit (%) = 5 / 50 * 100,
Or Profit (%) = 100 / 10 = 10 %,
So, amount of profit earned by the person is 10 % or Rs. 5.
Counting can be done using number of methods like Permutation, combination, factorization etc.
Let us consider an example of round table, which is a very famous question of permutation in mathematics.
Q. In how many ways 6 girls and 6 boys sit around in a Circle such that no girl is adjacent to each other?
Solution: According to problem, repetition of girls and boys in the complete arrangement is not possible because one boy or one girl can acquire one seat at a time. Also there is one more restriction in the problem that no two girls can sit adjacent to each other. This makes it necessary for each girl to be between two boys or adjacent to only one boy. We have to make 6 boys and 6 girls sit in a circle. Number of ways for this arrangement can be calculated as follows:
Arrangement being circular, we can start the sitting arranging from any Position such that we can make either a girl sit first or a boy sit first. Whosoever is made to sit at first, we would be getting the same arrangement as shown above. Here, the first position can be filled in 6 ways i.e. we have to make 1 boy sit out of available 6. Next position has to be filled by a girl according to the restriction that has been imposed on the problem. This place can also be filled in 6 possible ways. Next two places also have to be filled like previous two but by deduction of boy and one girl have to be made. This way we can arrange 6 boys an 6 girls in a circle, getting the total number of ways = 6!2.
Cone can be defined as an 3 – D (3 - dimensional) shape whose structure goes from flat and circular base to the vertex (or apex). Actually cone is composed of line segments. These line segments are joined to a single Point in such a way that it forms a Solid structure. This structure contains a circular base and a vertex. Vertex is the top point of the cone. Base is circular that’s why cone has radius. Height of cone is defined as the direct length between the base and vertex.
Figure of cone is shown below.
The total surface area (S) of a cone can be given as sum of area of circular base and area of remaining part. Mathematically,
S = Πr2 + π r l = Π r (r + l) hence
S = Π r (r + l) sq units.
Here length 'l' is given by l = √ (r2 + h2).
Let’s find out how to calculate the total surface area of a cone with a radius of 3cm, height of 4 cm and a length of 5cm. Let’s use the formula to calculate total surface area of cone:
S = Π r (r + l),
Here π = 22 / 7 = 3.14, radius r = 3 cm and length of cone is 5 cm which is given in the problem.
Thus substituting these values in the formula, we get,
S = 3.14 * 3 * (3 + 5) = 3.14 * 3 * 8 = 3.14 * 24 = 75.36 cm2.
Values of Trigonometric Functions can be evaluated from right angled triangle. The secant of a function is represented as follows:
Sec x = Hypotenuse (Longest side) / Base of the triangle (shorter side).... equation 1
We can use above formula to calculate other trigonometric functions. That is, value of sec x is given; rest of the functions values can be calculated. Let us suppose an example of this to understand solving Trigonometric functions.
Example: If sec x equal p plus 1 by 4p, then how to find the value of sec x plus tan x.
Solution: In the given problem we have the value of sec x = (p + 1) / 4p and we have to calculate the value of sec x + tan x.
For this we first have to calculate the value of tan x using the value of sec x. On comparing the value of sec x i.e. (p + 1) / 4p we see that hypotenuse or the longest side of the Right Triangle is equals to (p + 1) and base or one of the two shorter sides is equals to 4p. We can also use direct trigonometric identity:
Sec2 x = 1 + tan2 x,
((p + 1) / 4p) 2 = 1 + tan2 x,
(p2 + 1 + 2p) /16p2 – 1 =tan2 x,
Or tan x = √ (-15p2 + 2p + 1) / 4p,
Or we can also use pythagoras theorem to find the value of tan x.
Perpendicular side2 = Hypotenuse2 – Base2 = p2 + 1 + 2p - 16p2 = -15p2 + 2p + 1,
Or Perpendicular side = √ (-15p2 + 2p + 1),
Value of tan x = √ (-15p2 + 2p + 1) / 4p.
When two Linear Equations are multiplied then they result in a Quadratic Equation. Suppose we have (x+5)(x-3) Find the product using identities?. To find product of given values we need to know about product identities. Steps to be followed:
Step 1: Here given two terms are (x+5) (x-3) which can also be written as: (x + 5) (x - 3).
Step 2: First of all multiply first term to second polynomial, then multiply second term of first polynomial to all terms of second polynomial. Using these steps it can be written as:
(x + 5) (x - 3) => x2 – 3x + 5x – 15,
Step 3: Add like or same terms if present in the equation otherwise this is required solution. In above equation we have two like terms, so on combining both terms it can be written as: x2 + 2x – 15. This is required solution. In this way we can easily multiply any terms. Some product identities are given we can also find product using product identities.
p2 - q2 = (p + q) (p – q),
p3 - q3 = (p - q) (p2 + pq + q2),
p3 + q3 = (p + q) (p2 - pq + q2),
p5 - q5 = (p - q) (p4 + p3q + p2q2 + pq3 + q4)
p5 + q5 = (p + q) (p4 - p3q + p2q2 - pq3 + q4).
Math word problems may also include solving discounts on original prices. Discount is a quantity that can either be represented in simple currency form or in Percentage form. When it is represented in currency form, we just have to perform the deductions of it with any further calculation. But same is not the case with percentage values. To convert a percentage to Ratio we divide the percentage value by 100 and to calculate the discounted value we need to perform further operations. Such word problems can be understood by considering following example:
Let us suppose word problem which says: After a 15 percent discount on TV it now cost 102 dollars, what is the original price?
Solution: In a word problem what we actually look for in the beginning is the available information. Here in our example this information is present in the form of discounted price of the TV and the amount of discount made on the original price of it. Also we have important information about the units of money and discount. The discount has been specified in terms of percentage and the discounted price in dollars. Let us suppose that the original price of TV is X dollars. Applying a discount of 15 % on it and subtracting the value from X we get:
X - (15 % of X) = 102,
Or X - (15 / 100) * X = 102,
Or X (1 - (15 / 100)) = 102,
or X (85 / 100) = 102,
Multiplying both sides of the equation by (100 / 85) we get:
X (85 / 100) * (100 / 85)= 102 * (100 / 85),
X = 102 * (100 / 85),
or X = $ 120,
So, the original price of the Tv is $ 120.