Prism Image

Introduction to Prism image:

Before we discuss prism image, let us define a prism. A prism is a portion of a transparent medium bounded by two plane faces inclined to each other at a suitable angle. The angle between the two inclined surfaces is called the refracting angle or the angle of the prism. A line where the two inclined planes meet is called the refracting edge of the prism. Image formed by a prism shows a pattern discussed below. I like to share this Triangular Prism Net with you all through my article.

The Angle of Prism

A ray of light when passing through a prism suffers two refractions and hence it get deviated from its original path from a certain angle and this angle is called the angle of deviation and it is related with the angle of prism as

δ = (μ – 1) A

Where δ is the angle of deviation and μ is the refractive index of the material from which the prism is made and A is the angle of prism. The refractive index of the prism material and the angle of prism are also interconnected as Please express your views of this topic substitution method with fractions by commenting on blog.

μ = sin [(A + δm) / 2] / sin (A / 2)

Types of Image Formed by a Prism

The images of prism include a general image which is shown in fig.1 and it shows the dispersion of the light through the prism and hence the light deviates from its original path and get dispersed into the seven constituent colors to make a pattern named as the VIBGYOR.

Fig.1 Prism showing dispersion

But the prism images include the prism named as the total refracting prisms and these prisms are shown in fig.2 where the glass prism are like the right angled isosceles prisms which turn the light through 900 or 1800. They are based on the phenomenon of total internal reflection of light. In totally reflecting glass prisms the angle of incidence is made 450. Hence light suffers total internal reflection. In first figure of fig.2 the deviation is 900 on account of one total internal reflection whereas the deviation in the second figure of the fig.2 is 1800 on account of two total internal reflections.

Fig.2 Total Reflecting Prism

Estimate Quotients

Introduction for estimate quotients:

In division we got three terms when we divide the numbers we got dividend, divisor and quotient. Dividend is the number which has to be divided; divisor is the number which to divide and the quotient is the result we got when we divide the dividend by the divisor. A number of times which dividend is divisible by divisor is called quotient. In this section we have to estimate the quotient by the simple division. I like to share this Precision Definition with you all through my article.

How to Estimate Quotients:

Example: Divide 125 by 5

Step 1: In above problem dividend is 125 and divisor 5 we have to estimate quotients Now keep divisor 5 in the division bracket and place dividend (125) .
5)125

Step 2: Take first digit of the dividend (1). It is smaller than divisor 5 so it can't be divided by 5 to produce a whole number. And then take first two digits from the dividend (12) and determine how many 5's it contains. In this case 12 have 2  five's (5*2=20). Place the 2 above division bracket.
2
5)125
10

Step 3:Multiply the 5 with 2 and place the result (10) below the 12 of the dividend.

2
5)125
10


Step 4:Draw  a line under the 10 and subtract it from 12 (12-10=2). 2 is smaller than divisor (5) them bring down 5 from the 125 and place it to the right of the other 2.

2
5)125
10
25

Step 5:Now Divide  25 by 5 and place that answer above the division bracket to the right of the 2 there is five 5 in 25.

25
5)125
10
25

Step 6:Multiply the 5 of the quotient by the divisor (5) to get 25and place this below the 25. Subtract 25 from 25 to give an answer of 0. The number 0 is called the remainder and indicates that there were 0 left over.
25
5)125
10
25
25
0    (remainder)
Here we estimate the quotient 25 when dividing 125 by 5

Please express your views of this topic how to calculate compound interest by commenting on blog.

Problems for Estimating Quotients:

Problem 1:
Divide 24 by 4 and estimate quotients.

Solution:
first take one digit from 24 where 2<4 br="br" so="so">take tens,24>4  we know six fours in 24 (6 x 4)

6                        quotient is 6
4)24
24
0                           Remainder is 0


Problem 2:
Divide 140 by 2 and estimate quotients.

Solution:
First take one digit from 140 where 1<2 br="br" so="so">Take tens,14>2  we know seven two's in 14 (7 x 2)
Then,take 0 down no digit to divide put 0 in quotient
70
2)140
14                                             quotients is 70
0
0
0

Problem 3:
Divide  520 by 5 and estimate quotients.

Solution:
First take one digit from 520 where 5 divide by 5 is 1  put 1 in quotient
Now bring down next two digit 21 where (4 x 5)=20 put 4 in quotient and subtract 20 from 21
we have remainder 1
14
5)521
5                                             quotients is 14  and remainder is 1
21
20
1

Problem 4:
Divide  1253 by 3 and estimate quotients.

Solution:
First take one digit from 1253 where 1<3 .so=".so" 12="12" take="take" tens="tens">3
In 12 we four three's (4x 3) and take down 5,5>3 one three's in 5
subtract 3 from 5 and bring down 3.so,23>3 in 23 we have seven three's(7x 3) and remainder is 2.
417
3)1253
12                                             quotients is 417  and remainder is 2
5
3
23
21
2

Double Exponential Function

Introduction for double exponential function:

Standard form of exponential functions is in the form of f(x) = `b^(b^x),` here b is the number greater than zero and x is the exponent. Here the value of x is given we need to find the exponential functions. For example f(x) = `2^(2^x)` when the value of x is 4 we need to substitute the value in the 4 in the x place and find the double exponential functions of f(4) = `(2^2)^4` .By using the standard form of exponential function we will discuss some important problems in the below concepts.

Problem in Double Exponential Function:

Simplify the double exponential functions.

Problem (i) Find the double exponential function of f(x)  = `4^(4^x)` here x is 3

Solution: Substitute the x values exponential function

f(x) = `4^(4^x)`

f(3) = `4^(4^3)` substitute the x value 3 in the functions.

f(3) =` 4^(4^3)` .

f(3) = 464. here 43 is 64

f(3) = 3.40282367x1038

Problem (ii) Find the double exponential function of f(x)  = `5^(5^x)` here x is 2

Solution: Substitute the x values in the standard form of exponential function

f(x) = `5^(5^x)`

f(2) = `5^(5^2) ` substitute the x value 2 in the functions.

f(2) = `5^(5^2).`

f(2) = 525. in this step 52 is 25

f(2) = 2.98023224x1017

Problem (iii) Find the double exponential function of f(x)  = `8^(8^x)` here x is 1

Solution: Substitute the x values in the standard form of exponential function

f(x) = `8^(8^1)`

f(1) = `8^(8^1)` substitute the x value 3 in the functions.

f(1) = 88 here 81 is 8.

f(1) = 88.

f(1) = 16777216. I have recently faced lot of problem while learning The Coordinate Plane, But thank to online resources of math which helped me to learn myself easily on net.

Problem Double Exponential Functions:

Simplify the standard form of exponential functions.

Problem (i) Find the double exponential function of f(x)  =` 19^(19^x)` here x is 0

Solution: Substitute the x values in the standard form of exponential function

f(x) = `19^(19^x)`

f(0) =` 19^(19^0)` substitute the x value 0 in the functions.

f(0) = 191 in this step anything power zero is 1.

f(0) = 19.

Problem (ii) Find the double exponential function of f(x)  = `25^(25^x)` here x is 1

Solution: Substitute the x values in the standard form of exponential function

f(x) = `25^(25^x)`

f(1) = `25^(25^1)` substitute the x value 1 in the functions.

f(1) = 2525.

f(1) =2525 = 8.8817842 × 1034

Statistical Data Collection

Introduction for Statistical data collection:

Data collection is one of the important topics of the data handling chapter in statistics (mathematics) subject. A data is a collection of information gathered from some one. Data handling is having so many subtopics like a recording the data, organizing the data and collecting data. Collecting data is known as “statistics data collection”. Here we have to going for brief explained about the statistics data collection.

Collection of Data:

Collection of data is defined as, the information getting in the way of first investigation or enquiry. It is known as collection of data. This collection of data’s are may not be give us to particular information related to that data. And before collecting the data, we need to know, how to use it for.

Types of Data:
There are two types in the collection of data:

Primary data,
Secondary data.


Explanatory for collection of data types:

1.     Primary data:

The primary data is the data, which is the data collected from the first hand information. It is published by the organization for some purposes. This data is mostly pure and original data’s.

Collecting methods:
It has three important methods.

1.     Data’s collecting through telephone:

The data collectors get information through the telephone. It is accurate and quick process.

2.      Collecting through investigation:

Well trained investigators are employee for collecting the data. This method, the investigators will collect the information from asking required questions from the individuals.

3.      Personal investigation method:

The information collectors conduct the survey and getting data’s. In this method we have to get more accurate and original data. But this method is very useful for small projects only. Having problem with Absolute Value Equations and Inequalities keep reading my upcoming posts, i will try to help you.

2.     Secondary data:

The secondary data is the data, which is the data collected from the second hand information. It means the given data is already collected from some one for any other purpose, and it is available for the present issues. And mostly these secondary data’s are not pure data’s.

Collecting methods:
There are two important methods for collecting the data.

1. Official:

The data collection from the ministry of finance, agriculture, industry and etc…

2.  Semi – official:

This is the method of data collection from, Board of railways, and bank, etc…

These are all the important methods in the statistical collection of data.

Plotting Cartesian Coordinates

Introduction to Cartesian coordinates: 

We are already aware of how to locate a point on a number line. In order to locate a point (ordered pair) in a plane, we consider two mutually perpendicular lines known as Coordinate axes. The system used to plot points on a plane is known as Rectangular Coordinate System or Cartesian coordinate system (named after the French Mathematician and Philosopher Rene Descartes) and the plane is known as Cartesian plane.

Consider two mutually perpendicular lines XOX’ and YOY’ intersecting each other at the point O. The horizontal line XOX’ is called X axisY axis and the point O is called the Origin. and the vertical line YOY’ is called

Let P(x,y) be a point in Cartesian plane. Then, x = the horizontal distance of point P from Y-axis is known as x-coordinate or abscissa, y = the vertical distance of Point P from X-axis is known as y-coordinate or ordinate and (x,y) = the coordinates of the point P.

Pictorial Representation of Cartesian Coordinate Plane

Cartesian Coordinates : Parts of Cartesian Plane

The X and Y axes separate the coordinate plane into four regions called quadrants, which are numbered in a counterclockwise direction as shown in the above figure.

Signs of the coordinate point in different quadrants:

Quadrant	x-coordinate	y-coordinate	Point(x, y)
I	+	+	(+, +)
II	-	+	(-, +)
III	-	-	(-, -)
IV	+	-	(+, -)

Explanation:

If both (x and y) coordinates of a point are positive, the points falls in Quadrant I. Example: (2, 3), (4, 9) etc.
If x-coordinate is negative and y-coordinate is positive, the point falls in Quadrant II. Example: (-4, 5), (-8, 3) etc.
If both (x and y) coordinates of a point are negative, the point falls in Quadrant III. Example: (-2, -4), (-7, -5) etc.
If x-coordinate is positive and y-coordinate is negative, the point falls in Quadrant IV. Example: (7, -4), (6, -6) etc. Understanding 7th grade math practice problems is always challenging for me but thanks to all math help websites to help me out.

Cartesian Coordinates : Process to Plot the Point (4, 3) and (-4, -3)

To plot the ordered pair (4, 3):

Starting from the origin O, move 4 units to the right on the X-axis and 3 units up on the Y-axis. This gives us the point (4, 3).

To plot the ordered pair (-4, -3):

Starting from the origin from O, move 4 units to the left on the X-axis and 3 units down on the Y-axis. This gives us the point(-4, -3).

Solving Exponential Equations Calculator

Introduction of solving exponential equations calculator:-

In math exponential is the function ex, where e is the digit such to the significance ex the same its contain consequent.
For example,

`e^x` is an exponential function
Put the value x=0 and get `e^0 = 1` . This is a basic for how to write exponential function. Solving exponential equations is the new way for the students. Students easily to get a values in calculator and also easily to solving the problems.

Basic Solving Exponential Equations Calculator:-

In the following figure shows the solving exponential equations calculator:-

Enter the exponential equations then click the answer button. Perform the exponential operation and get the answer.

Check the answer in manual calculation `7^(2x+1) 7^(3x-2) `

`7^(2x+1) = 7^(3x-2)`

2x+1 = 3x-2

2x = 3x -2-1

2x = 3x -3

3 = 3x-2x

3 = x   (or) x = 3

Verify the answer in calculator the above answer is correct.I like to share this online math tutor with you all through my article.

Example Problems for Solving Exponential Equations Calculator:-

Problem 1:-

Find whether the point (0, 1) lies on the graph of the function y = 3(2)x.

Solution:-

Substitute x = 0 in the function y = 3(2)x.

y = 3(2)0

= 3(1)

= 3

The y–coordinate of the point is 1, which does not match with the obtained value y = 3.

So, the graph of the function y = 3(2)x does not contain the point (0, 1).

Problem 2:-

Solving the exponential equations  `3^(2x-1)` .

Solution:-

`3^(2x-1) = (3^3)^x`

`= 3^3x`

If the base value is same equal to one another

2x-1  = 3x

-1 = 3x-2x


-1= x


Verify the answer in calculator the above answer is correct.

Problem 3:-

Solving add the exponential equation `4^2x +4^3x`

Solution:

Given: `4^2x +4^3x`

Take the common term 4.

`= 4^(2x+3x)`

`= 4^(5x)`

Adding the both values and get 5x.

Finally we get an answer as `4^5x`

Problem 4:-

Solving the exponential equation `5^(3x-8)` .

Solution:-

`5^(3x-8) = (5^2)^2x`

`= 5^4x`

If the base value is same equal to one another

`3x-8 = 4x `

-8 = 4x-3x

-8 = x

Verify the answer in calculator the above answer is correct.

Word Problem with Solution

Introduction to the word problems with solution:

In mathematics the word problem for a recursively presented group G is the algorithmic problem of deciding whether two words represent the same element. Although it is common to speak of the word problem for the group G strictly speaking it is a presentation of the group that does or does not have solvable word problem.


Steps for solving the word problems:

Convert the phrasing into a numeric equation

that combines smaller "expressions"

Solve the equation.

Example for Word Problems with Solutions

Area/volume/perimeter problems:

A square contains area of 16 square centimeters. What is the length of every side?

Sol:

A of a square with side-length s is the formula for area:

A = s2

Here area has given, so we can insert this value into the area formula, and distinguish what it leads:

16 = s2
4 = s

Therefore, the length of every side is 4 centimeters.

Percent problem:

Jason got 30 marks in his term test. He wants to know the percentage of marks out of 40?

Sol:

Jason had the actual mark (30) and the relative mark (40). The unfamiliar in this problem is the rate or percentage. Because the report is (40) is (some percentage) of (30), in that case the variable stands for the percentage, and the equation is:

40 = (x) (30)

40 ÷ 30 = x = 1.33

Because x situate for a percentage, we want to consider translating this decimal back into a percentage:

1.33 = 133%

30 is 133% of 20.

Jason percentage is 133%.

Investment problem:

We deposit 1000 dollars into a savings yielding 6% yearly interest; we leave that money for 2 years. How much interest achieves you get at that end of the 2 years?

Sol:

In this situation, P = 1000 dollars, r = 0.06 (because we have to transfer the percent to decimal form), then the time is t = 2. From this we can get:

I = (1000) (0.06) (2) = 120

Therefore the interest is 120 dollars.

Time and distance:

John travelling in his bike with a speed of 25 m / sec. Help John to convert  his speed in km / hr.

Sol:

John's Speed = 25 m / sec

To convert the speed in km/hr multiply 25 by 18/5

Therefore Speed in km/hr =  25 * 18 / 5

John's speed is 90 km / hr.

Practice Problems on Word Problems with Solutions:

1.)     A tunnel is 160 m long. A car crosses it in 10 seconds. Find its speed?

Solution:

16 m / sec

2.) Find the volume of brick whose dimensions are 2.8 cm * 0.5 dm * 0.7 dm?

Solution:

98 CubicCentimeters.