Solving Translation

“Transformation” is the term which is used in advanced and intermediate geometry. Transformation of image is defined as the transforming the shape from one position to another position without any change in the angle or dimension of the image. Study of transformations is applicable in high school. Transformations learning are very interactive and interesting. In online, animated study of transformations is available for the following topics.  There are four types of transformations they are,


1) Translation Transformation

2) Reflection Transformation

3) Rotation Transformation and

4) Dilation Transformation.

Study of Translation:

Translation:

Translation is one of the types of transformation. Translation means shifting the image up or down and shifting the image left or right. In translation, the image is translated or moved in a straight line with respect to the original image. The translated image doesn’t vary from original image. The dimension, shape and angle of the translated image are same as the original image. The translated image and the original image will be facing the same direction. Solving the problems involving the translation is easy when the concept of the translation is known very well.

Problems on solving translation:

Solving translation Example 1:

Move the object 2 units to the right and 4 units up.

Solution:

The given problem is translation.

In this problem, the object is moved up 4 units and then 2 units to the right side i.e. towards the y axis.



Solving translation Example 2:

Move the object 4 units to the right and 2 units up.

Solution:

The given problem is translation.

In this problem, the object is moved up 2 units and then 4 units to the right side i.e. towards the y axis.

Solving translation Example 3:

Move the object 3 units to the left and 5 units down.

Solution:

The given problem is translation.

In this problem, the object is moved up 2 units and then 4 units to the right side i.e. towards the y axis.

Probability Distribution Curve

The Normal distribution or Gaussian distribution is the continuous probability distribution that often gives a good description of data that cluster around the mean. The graph of the associated probability density function is a bell-shaped with a peak at the mean and is known as the Gaussian function or bell curve. The shape of normal distribution resembles that of a bell. So it is referred as the "bell curve".

Definition of probability distribution curve:

A continuous random variable X is said to follow a probability distribution curve with parameter μ and σ (or μ and σ2) if the probability function is
Y = [1/σ *   ] * e-(x - μ)2/2σ2
X -μ(μ, σ) denotes that the random variable X follows bell curve distribution  Even we can write the probability distribution curve as X-μ(μ, σ2) symbolically. In this case the parameters are mean and standard deviation with mean μ and standard deviation σ. Where X = normal random variable, μ= mean, σ = standard deviation, π = 3.14159, e = 2.71828.
A probability distribution is evaluated by the probability of random variable X in open interval (-∞, x), which is given by,
F(x) = P [X ≤ x]
“Probability distribution curve is often called as normal distribution or bell curve or Gaussian curve”
Types of probability distribution:

Discrete probability distribution,
Continuous probability distribution.

Discrete probability distribution:

If the cumulative distribution function of a probability distribution increases in jumps then the probability distribution is said to be discrete.
The discrete probability distribution with the probability mass function p is given by,
P [X = x] = p(x).

Continuous probability distribution:
If the cumulative distribution function F(x) = µ (-∞, x) is continuous then the probability distribution function is called as continuous probability distribution function. These are characterized by the probability density function f given by,

x
F(x) = µ (-∞, x) = `int` f (t) dt
-∞

Example problem based on probability distribution curve:

The mean score of 1000 students for an examination is 34 and S.D is 16.  How many candidates can be expected to obtain marks between 30 and 60 assuming the normality of the distribution?

Solution: The given data are μ = 34, σ = 16, N = 1000

We have to find the probability of marks between P (30 < X < 60)

We can use the formula, Z = `(X - mu)/sigma`

X = 30,

Z1 = `(30 - mu)/sigma`

=>`(30 - 34)/16`

Z1 = `(-4)/(16)`

Z1 = – 0.25

Z2 = `(60-34)/16`

= `(26)/(16)`

= 1.625


So the value of z2 = 1.63 (approximate)

P(−0.25 < Z < 1.63) =P(0 < Z< 0.25) + P(0 < Z < 1.63) (due to symmetry)

= 0.0987 + 0.4484 = 0.547

No of students scoring between 30 and 60 is, 0.5471 × 1000 => 547.

Function Intervals

The function intervals are defined as the funcions that are appliable with intervals of the numbers which are applied in the variables that changes the functions.
Open Interval

This interval is common to denote open intervals in mathematics using parenthesis (  ). Thus the open interval discussed above will often be shown as (a, b),
Closed Interval

This type of interval are in the form of  [ ] closed both sides also.  For example consider that , -2< or equal to x < or equal 2.  This is represented by closed interval [2,2].  So open interval represents that the boundary value are not taken into the considerations, whereas closed interval means boundary value are taken into the considerations.

Intervals on the function:

Example 1:

Verify Lagrange’s law of the mean for the function f(x) = x3 on [−2,2]

Solution : f is a polynomial, hence continuous and differentiable on [− 2, 2].
f(2) = 23 = 8 ; f (−2) = (−2)3 = −8
f ′(x) = 3x2 ⇒ f ′(c) = 3c2
By law of the mean there exists an element c ∈ (− 2, 2) such that
f ′(c) = (f(b)−f(a)) / (b−a)
⇒ 3c2 = [ 8 − (−8) ] / 4 = 4
i.e., c2 = 4 / 3 ⇒ c = ± 2 / 3
The required ‘c’ in the law of mean are 2 / 3  and − 2/3  as both lie in [−2,2].

Example 2:

Suppose that the functions f(0) = − 3 and f ′(x) ≤ 5 for all values of x, how large can f(2) possibly be?
Solution :

Since by hypothesis f is differentiable, f is continuous everywhere. We can apply Lagrange’s Law of the mean on the interval [0,2]. There exist atleast one ‘c’∈(0, 2) such that
f(2) − f(0) = f ′(c) ( 2 − 0)
f(2) = f(0) + 2 f ′(c)
= −3 + 2 f ′(c)
Given that f ′(x) ≤ 5 for all x. In particular we know that f ′(c) ≤ 5.
Multiplying both sides of the inequality by 2, we have
2f ′(c) ≤ 10
f(2) = −3 + 2 f ′(c) ≤ −3 + 10 = 7
i.e., the largest possible value of f(2) is 7.

Example 3:

Determine the interval in the which curve f(x) =sinx ,  X  ∈ ( 0,2∏) only it be the concave downward

Solution :

f(x)  = Sinx

f'(x) = cosx

f''(x) = -sinx

In the interval (0,∏)  ,  -sinx <0

so f''(x) < 0 in the interval (0,∏)

∴f(x) = Sinx is concave downward in the interval (0,∏)

Practical applications on the function intervals

Example 1:

A cylindrical hole 4 mm in diameter and 12 mm deep in a metal block is rebored to increase the diameter to 4.12 mm. Estimate the amount of metal removed.
Solution :


The volume of cylindrical hole of radius x mm and depth 12 mm is given by

V = f(x) = 12 πx2
⇒ f ′(c) = 24πc.


To estimate f(2.06) − f(2) :
By law of mean,
f(2.06) − f(2) = 0.06 f ′(c)
= 0.06 (24 πc), 2 < c < 2.06

Take c = 2.01
f(2.06) − f(2) = 0.06 × 24 π × 2.01
= 2.89 π cubic mm.

Study Total Probability Definition

Study of Total Probability is as follows

Let a1, a2 and a3 be mutually exclusive and exhaustive events and let b indicate some other event. Note for a to arise it has to be in conjunction with at least one of a1, a2 and a3. Therefore, by Submitting the multiplication rule to a general situation of this type we obtain the following.

The Law of Total Probability

Let a1, a2, . . . ,an be a mutually exclusive and exhaustive events. If B is any other event it follows that

P(B)   = P(B|A1)+P(B|A2)+…+P(B|AN) = `sum_(j=1)^n P(B|Aj)`

When studying total probability this formula have to be used.

Examples for Total Probability

Example 1:

A fair coin is tossed. If the coin lands on heads a bag is filled with one black ball and three white balls. If the coin landed on tails the bag is filled with one black ball and nine white balls. A ball is then selected from the bag. What is the probability that the ball selected is black?(Use Total Probability)

Solution
Let H = Heads, T = Tails and B = Black ball selected. Then by the law of total probability

P(B)   = P(B|H)P(H) + P(B|T)P(T)

= `(0.25)(0.5) + (0.1)(0.5)`

P (B)  = 0.175

Example   2:

Three boxes contain red and green balls. Box 1 has 5 red balls and 5 green balls, Box 2 has 7 red balls and 3 green balls and Box 3 contains 6 red balls and 4 green balls. The probabilities of choosing a box are 1/4, 1/6, 1/8. What is the probability that the ball chosen is green by using the Total Probability?

Solution
We begin by defining the following sets. Let,

G:= the ball chosen is green.
B1:= Box 1 is selected
B2:= Box 2 is selected
B3:= Box 3 is selected

Then P(G|B1) = 5/10, P(G|B2) = 3/10 and P(G|B3) = 4/10.

P(G) =` ((5/10)xx(1/4)) + ((3/10)xx(1/6))+((4/10)xx(1/8))` = 9/(40)

Example 3:

Three shed car colors red and green. Shed 1 has 10 cars of red and 10 cars of green, shed 2 has 7 cars red and 3 green cars and shed 3 contains 6 red cars and 4 green cars. The respective probabilities of choosing a shed are 1/4, 1/6, 1/8. Use the Total Probability to find the probability that the car chosen is green?

Solution
We begin by defining the following sets. Let,

G = the ball chosen is green.
S1 = Shed 1 is selected
S2 = Shed 2 is selected
S3 = Shed 3 is selected

Then P(G|S1) = 10/20, P(G|S2) = 3/10 and P(G|S3) = 4/10.

P(G) =` ((10/20)xx(1/4)) + ((3/10)xx(1/6))+((4/10)xx(1/8))` = 9/(40)

Ordering Whole Numbers

The set of collection of numbers are called as whole number. The non negative integers, positive integers and all integers are included in whole number. If we want to ordering the whole numbers means use the symbols like <(less than), >(greater than) and =(equals). The position of numbers are used to arrange the whole numbers in desending or ascending order.The whole numbers are combined with integers.

Ordering methods for whole numbers

Ordering methods:

We can ordering the whole numbers by using two ways. They are use a number line and use place value. The whole numbers has negative value also.

Using a number line:

The number line is simple and easy understanding way for ordering the whole numbers.

The whole number is represented by line. First order the numbers using symbols and plotting the numbers on the number line.Plot the least number on left and plot the greatest number on right.

Example for ordering the whole numbers by using number line:

Problem 1: Order the given numbers 560, 525, 532, 518, 546.

Answer:

The whole numbers are arranged as 518, 525, 532, 546,560.

That is 518 < 525 < 532 < 546 < 560.

Using place value:

First counting the integers in number. The place value is evaluated from left to right. Arrange the numbers under one by one and compare the place value.

Example for ordering the whole numbers by using place value:

Problem 2: Order the given whole numbers 1225, 1248, 1260, 1235.

Answer:

12|2|5

12|3|5

12|4|8

12|6|0

The left beginning number is compared and find where the different numbers are present. Here tenth place value is changed.

The ordered whole numbers is 1225, 1235, 1248, 1260.understanding about Algebra.

Excercise problem for ordering the whole numbers:

Problem: Order the whole numbers by using number line and place value.

4526, 4500, 4568, 4532, 4514.

Answer: 4500,4514, 4526, 4532, 4568.

Differentiation Tool

The process of finding a derivative is called differentiation. Differentiation is a method to compute the rate at which a dependent output y changes with respect to the change in the independent input x. In differentiation process, we use different operators and tools for solving the problem. The operators and tools are denoted as D, ∂, and f'. (Source: Wikipedia)

General formulas and methods for differentiation:

(d / dx) (xn) = nx(n - 1)
(d / dx) (constant) = 0
(d / dx) (uv) = u (dv) / (dx) + v (du) / (dx)

Partial differentiation methods:

f (x, y) = f'x (x, y) + f'y (x, y)

Example problems for differentiation tool

Differentiation tool problem 1:

Differentiate the given function f(x) = 5x2 + 45x3 - 72x. Find the value of f'(x).

Solution:

Given function is f(x) = 5x2 + 45x3 - 72x.

Differentiate the given function with respect to x, we get

f'(x) = (5 * 2)x + (45 * 3)x2 - 72

= 10x + 135x2 - 72

Answer:

The final answer is 135x2 + 10x - 72

Differentiation tool problem 2:

Differentiate the given function f(x) = 19x4 + 35x2 - 16x. Find the value of f''(x).

Solution:

Given function is f(x) = 19x4 + 35x2 - 16x.

Differentiate the given function with respect to x, we get

f'(x) = (19 * 4)x3 + (35 * 2)x - 16

= 76x3 + 70x -16

f'(x) = 76x3 + 70x -16

For finding f''(x), differentiate the given function once again with respect to x, we get

f''(x) = (76 * 3)x2 + 70

= 228x2 + 70

Answer:

The final answer is  228x2 + 70

Differentiation tool problem 3:

Differentiate the given function f(x) = 100x5 - 42x + 102. Find the value of f'(x).

Solution:

Given function is f(x) = 100x5 - 42x + 102.

Differentiate the given function with respect to x, we get

f'(x) = (100 * 5)x4 + (42 * 1) + 0

= 500x4 + 42

Answer:

The final answer is 500x4 + 42

Practice problems for differentiation tool

Differentiation tool problem 1:

Differentiate the given function and find the f''(x). Given f(x) = 3x5 - 5x3 + 3x2 + 7

Answer:

The final answer is f''(x) = 60x3 - 30x + 6

Differentiation tool problem 2:

Differentiate the given function and find the f'''(x). Given f(x) = 16x3 - 7x2 + x

Answer:

The final answer is f'''(x) = 96

Function Rule of Algebra

Algebra is branches of mathematics which deals with finding unknown variable from the given expression with the help of known values. The algebraic expression contains variables are represented alphabetic letters.

In algebra numbers are consids as constants, algebraic expression may include real number, complex number, matrices and polynomials. In algebra several identities to find the x values by using this we can easily find  the algebraic expression of the particular function. The function rule of algebra may be form  of f(x), p(x),… to find the x value of the algebra functions.

Example for function rule of algebra

(f + g)(x) = f (x)+ g (x)
(f – g)(x) = f (x) – g (x)
(f .g)(x) = f (x) . g (x)
(f/g )( x)=f(x)/g(x).


Function rule of algebra problem

problem using the function rules in algebra.

(f + g)(x) = f (x)+ g (x)
(f – g)(x) = f (x) – g (x)
(f .g)(x) = f (x) . g (x)
(f/g )( x)=f(x)/g(x).

Problem1; Rules for finding the function rules in algebra.

f(x) = 2 and g(x) = 2   find the addition function of (f+g)(x)

Solution: In the givne function f(x) and g(x) values are given we need to find the (f+g)(x). Using the rules of the funcion (f + g)(x) = f (x)+ g (x), (f + g)(x) = f (x)+ g (x), here f(x) and g() value is givnen

(f + g)(x) = f (x)+ g (x) this is the rules of the algebra fucntion, substitute the value in the equation.

(f + g)(x) =2+2.

(f + g)(x) =4.

f(x) = 2 and g(x) = 2   find the addition function of (f+g)(x)

Problem 2:  using the function rules of algebr find the value of (f+g)(x). if the value of f(x)=4 and g(x) =8.

Solution: In the givne function f(x) and g(x) values are given we need to find the (f+g)(x). Using the rules of the funcion (f + g)(x) = f (x)+ g (x), (f + g)(x) = f (x)+ g (x), here f(x) and g() value is givnen

(f + g)(x) = f (x)+ g (x) this is the rules of the algebra fucntion, substitute the value in the equation.

(f + g)(x) = 4 + 8

(f + g)(x) =12.

f(x) = 2 and g(x) = 2   find the addition function of (f+g)(x)

Problem 3:  using the function rules of algebr find the value of (f-g)(x). if the value of f(x)=4 and g(x) =8.

Solution: In the givne function f(x) and g(x) values are given we need to find the (f+g)(x). Using the rules of the funcion (f + g)(x) = f (x)- g (x), (f + g)(x) = f (x)- g (x), here f(x) and g() value is givnen

(f - g)(x) = f (x)- g (x) this is the rules of the algebra fucntion, substitute the value in the equation.

(f - g)(x) =8-4.

(f - g)(x) =4.

Function rule of algebra problem using multiplication and division

Rules of algebra function in the multiplication and division.

(f .g)(x) = f (x) . g (x)
(f/g )( x)=f(x)/g(x).

Problem 1:  using the function rules of algebr find the value of (f.g)(x). if the value of f(x)=4 and g(x) =8.

Solution: In the givne function f(x) and g(x) values are given we need to find the (f+g)(x). Using the rules of the funcion (f . g)(x) = f (x). g (x), (f . g)(x) = f (x). g (x), here f(x) and g(x) value is givnen

(f . g)(x) = f (x).g (x) this is the rules of the algebra fucntion, substitute the value in the equation.

(f . g)(x) =8*4.

(f . g)(x) =32.

Problem 2:  using the function rules of algebr find the value of (f.g)(x). if the value of f(x)=5 and g(x) =8.

Solution: In the givne function f(x) and g(x) values are given we need to find the (f+g)(x). Using the rules of the funcion (f . g)(x) = f (x). g (x), (f . g)(x) = f (x). g (x), here f(x) and g(x) value is givnen

(f . g)(x) = f (x).g (x) this is the rules of the algebra fucntion, substitute the value in the equation.

(f . g)(x) =5*8.

(f . g)(x) =40.

Problem 3:  using the function rules of algebr find the value of (f/g)(x). if the value of f(x)=4 and g(x) =8.

Solution: In the givne function f(x) and g(x) values are given we need to find the (f+g)(x). Using the rules of the funcion (f/g)(x) = f(x)/g(x), (f / g)(x) = f (x)/g (x) , here f(x) and g(x) value is givnen

(f /g)(x) =f(x)/g(x)  this is the rules of the algebra fucntion, substitute the value in the equation.

(f.g)(x) =4/8.

(f/g)(x) =1/2

Problem 4:  using the function rules of algebr find the value of (f/g)(x). if the value of f(x)=10 and g(x) =2.

Solution: In the givne function f(x) and g(x) values are given we need to find the (f+g)(x). Using the rules of the funcion (f/g)(x) = f(x)/g(x), (f / g)(x) = f (x)/g (x) , here f(x) and g(x) value is givnen

(f /g)(x) =f(x)/g(x)  this is the rules of the algebra fucntion, substitute the value in the equation.

(f/g)(x) =10/2.

(f/g)(x) =5

Solving Geometric Means

Solving geometric means help to solve the data bounding many orders of intensity and for estimating ratios, percentages, other data sets covered by zero. Solving geometric means is defined for sets of positive correct numbers and determined by multiply all the numbers and taking the nth root of the whole. To find an average numbers presented as percentages we can use the geometric means solving method is used. Let us see about the formula for solving geometric means formula.

Formula for Solving Geometric Means:

Geometric Means:

Geometric Mean = ((X1) (X2) (X3)........ (XN))1/N
where

X = Individual score

N = Sample size (Number of scores)

Solving geometric means examples

1. Find the Geometric Mean of 1, 4, 6, 2, 5

Solution:Step 1: 4.932424

N = 5, the total number of values. Find `(1)/(N)`

`(1)/(N)`  = 0.2

Step 2:Now find Geometric Mean using the formula.

(1)(4)(6)(2)(5))0.2 = (240)0.2

So, Geometric Mean = 2.99255

2. Find the Geometric Mean of 16 & 5.

The geometric mean of 16 & 5 is

sq rt (16 * 5) = sq rt 80 = 8.94227

3. Find the Geometric Mean of 5, 10 and 15.

Solution:Step 1: N = 3, the total number of values. Find `(1)/(N)` .

`(1)/(N)`  = 0.3333.

Step 2: Now find Geometric Mean using the formula.

((5)(10)(15))0.3333 = (750)0.3333

So, Geometric Means of this numbers is = 9.0835.

4. Check the geometric means of these number is 4,5 and 6 is 4.20488.

Solution:Step 1:     4.932424

N = 3, the total number of values. Find `(1)/(N)`

`(1)/(N)`  = 0.3333.
Step 2:Now find Geometric Mean using the formula.

((4)(5)(6))0.3333 = (120)0.3333

So, Geometric Means of this numbers is = 4.932424.

Therefore 4.20488 is not the geometric means of 4, 5, and 6.

These are the examples of solving geometric means.

Practice problesm on solving geometric means

1. The geometric means of number 8, _, 4 is 5.76899. Find the missing number.

Ans: 6

2. Find the Geometric Mean of 1, 5, 12, 4, 8, 9.

Ans: 5.08460