Formula for Volume of Hemisphere

Hemisphere is a geometrical object. It is look like a semi circle. Half of the sphere is called as hemisphere. The sphere has the radius r as well as hemisphere also have the radius r. The volume of the hemisphere formula is given below,

Formula:

Volume of hemisphere = (2 / 3) * π * r3

Here, π is constant value; r is radius of the hemisphere.

Example problems - formula for volume of hemisphere

Example 1:

Find the volume of hemisphere with the radius 12 cm.

Solution:

Given hemisphere radius r = 12 cm

Formula:

Volume of hemisphere = (2 / 3) * π * r3

Substitute the given radius value in the above formula, we get

Volume of hemisphere = (2 / 3) * π * 123

= (2 / 3) * π * 1728 cm3

= 1152π cm3

Answer:

The volume of the hemisphere is 1152π cm3

Example 2:

Find the volume of hemisphere with the radius 23 cm.

Solution:

Given hemisphere radius r = 23cm

Formula:

Volume of hemisphere = (2 / 3) * π * r3

Substitute the given radius value in the above formula, we get

Volume of hemisphere = (2 / 3) * π * 233

= (2 / 3) * π * 12167 cm3

= 8111.33π cm3

Answer:

The volume of the hemisphere is 8111.33π cm3

Example 3:

Find the volume of hemisphere with the radius 6 cm.

Solution:

Given hemisphere radius r = 6 cm

Formula:

Volume of hemisphere = (2 / 3) * π * r3

Substitute the given radius value in the above formula, we get

Volume of hemisphere = (2 / 3) * π * 63

= (2 / 3) * π * 216 cm3

= 144π cm3

Answer:

The volume of the hemisphere is 144π cm3

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Example 4:

Find the radius value of the hemisphere. The volume of the hemisphere is 244cm3

Solution:

Given volume of hemisphere is v = 244 cm3

Formula:

Volume of hemisphere = (2 / 3) * π * r3

Substitute the volume value in the above formula, we get

244 cm3 = (2 / 3) * π * r3

Rearrange the above, we get

r3 = 244 * (3 / 2 * π)

= 116.56 cm

r = 4.88 cm

Answer:

The radius of the hemisphere is 4.88 cm

Practice problems - formula for volume of hemisphere

Example 1:

Find the volume of hemisphere with the radius 8 m.

Answer:

The final answer is 341.33 m3

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Example 2:

Find the volume of hemisphere with the radius 15 cm

Answer:

The final answer is 2250 cm3

Example 3:

The area of the hemisphere is 18π cm3. Find the radius of the hemisphere.

Answer:

The radius of the hemisphere is 3 cm.

Learning about Permutation

Permutation is known as an arrangement of things. The word arrangement is used, if the order of things is measured. Example: assume we have to shape a number of consisting of three digits using the digits 6,7,8,9, to form this number the digits have to be arranged.
Tutoring:

Tutor is a personality working in the education of others, either independently or in group. A tutor gives knowledge, experience, and support. Tutors do not offer "solution," but rather help in problem solving, in getting solution.

Formula for learning about permutation tutoring:

The following permutation formula to estimate the how many possible ways possible in particular set.

P (n, r) = (n!)/ ((n-r)!) ,

Where, n offers number of stuffs, and r offers number of times. In this article we can learn about permutation problems for learning about permutation tutoring.
Example problems for learning about permutation tutoring:
Learning about permutation tutoring – Example 1:

How many 2 digits numbers can make with the digits 0 - 9?

Solution:

This problem occupies 10 digits, taken 2 at a time.

P (10, 3) = (10!) / ((10-2)!)

= (10* 9* 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)/ (8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)

= 90

Therefore 90 two digits numbers can make with the digits 0 - 9.
Learning about permutation tutoring – Example 2:

In how many ways can 4 parcels elected among 9 different colors of parcels?

Solution:

This problem involves 10 books, taken 3 at a time.

P (9, 4) = (9!)/ ((9-4)!)

= (9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)/ (5 * 4 * 3 * 2 * 1)

= 3024

There are 3024 possible ways to select 4 parcels among 9 parcels.
Learning about permutation tutoring – Example 3:

How many 3 characters word can create with the characters in the word ‘IDCARD’?

Solution:

This problem occupies 5 unrelated characters, select 3 at similar time.

P (6, 3) = (6!)/ ((6-3)!)

= (6 * 5 * 4 * 3 * 2 * 1) /(3 * 2 * 1)

= 120

Therefore 120 three letters word can make from the word 'IDCARD'.
Learning about permutation tutoring – Example 4:

How many 5 letters words can create using the letters P, Q, R, S, A, B, C and D?

Solution:

This problem involves 8 dissimilar characters, select 5 characters at a time.

P (8, 5) = (8!)/((8 - 5)!)

= (8 * 7 * 6 * 5 * 4 * 3 * 2 * 1)/(3 * 2 * 1)

= 6720

Therefore 6720 different words can make using those 5 characters.

Practice problems for learning about permutation tutoring:

Problem 1: In how many ways can 3 ships elected among 9 different colors of ships?

Problem 2: How many 2 letters word can make with the letters in the word 'LIGHT?

Problem 3: How many 4 digits numbers can make with the digits 0 - 7.

Answers: 1) 504

2) 20

3) 1680

Parallel Planes

A parallel planes are defined as a surface which is line joining of any two points on it lies completely on the surface. Parallel Planes are a horizontal and two-dimensional surface. Plane has two-dimensional analogue of a line (one-dimension), a dot (zero-dimensions), and a space (three-dimensions). Planes can occur as subspace of some upper dimensional space, as with a room wall, or they may like a separate own right in their continuation, as in the location of Euclidean geometry.

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Parallel is a term in geometry and in everyday life that refers to a property in Euclidean space of two or more lines or planes, or a combination of these. The existence and properties of parallel lines are the basis of Euclid's parallel postulate. Two lines in a plane that do not intersect or meet are called parallel lines.(Source: from Wikipedia)

Parallel planes and its various areas:

The parallel planes which do not intersect are known as parallel. One more planes especially in Hessian normal form is parallel if | n1. n2 | = 1 or n1 * n2 = 0.

Parallel  planes which are not parallel any time intersects in a line.

Parallel planes
Planes in various areas of mathematics:

Besides its familiar geometric structure, with isomorphism’s that are isometric with respect to the usual inner product, the plane may be viewed at various other levels of abstraction. Each intensity of abstraction corresponds with the specific categories.

At one extreme, all geometrical and metric concepts might be dropped to leave the topological plane, which may be thought of as an idealized phonologically trivial infinite rubber sheet, which retains a notion of proximity, but has no distances.  The topological plane, or its equal of the open recording, is the basic topological neighborhood used to built surfaces (or 2-manifolds) classified in low-dimensional topology.

Example for parallel planes:

Examples:

Equation of the plane passing through a given point and parallel to two given vectors:

Let  `bara`   be the position vector of the given point A referred to the origin O.

Let `u` and `v` be the given vectors, which are parallel to the plane.

Let `p` be any point on the plane and let its position vector be `r`  (i.e.,) = `op` = `r` .

Through A, draw a lines AB and AC parallel to `u` and  `v`  lying in the planes such that `AB` =   `u`   and  `AC` = `v`

Now `AP` is coplanar with `AB` and `AC` .

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Therefore  `AP` = s `AB` +t `AC` Where s and t are scalars

=  `su` + `tv`

= `OP` + `OA` + `AP`

`=>`   `r` = `a` + `s`  `u` + `tv`

This is the vector equation of the plane (in parametric form).

Learn About Volume Learning

In General, volume is the 3d space occupied by an object or matter under study. The space occupied by an object in three dimension is said to be volume and not in two dimension and  it is measured in cubic units.. In mathematics, we study about the volume of different objects and shapes. General formulas have been derived for the calculation of volume and learning as well. The general solid shapes frequently used  on mathematics are cylinder, sphere, cone, pyramid prism, etc. Formulas and problems for learning  volume calculation are given in the following sections.



Learn about volume learning- Volume formulas:

The derived formulas to calculate the volume are given here for learning volume calculation. These formulas are very impotant for learning volume calculation,

Pyramid:
Volume of pyramid = Area of base * h * (1/3)
where,
Area of base = width of the base * lenght of the base

h = Height of the pyramid from the apex,

Triangular prism:
Volume of triangular prism: (1/2) b * h * l
where,
b = width of the base
l  = lenght of the base

h = Height of the pyramid from the apex,

Cube:
Volume of a cube = a3
where,
a = length of any side

Cylinder:
Volume of a cylinder =  Pir 2 h
where,
r =radius of the circular side.
h= height of the cylinder.
Volume of a sphere:
Volume of a sphere = (4/3) (pi) r^3
where,
r =radius

Cone:
Volume of a cone: (1/3) (pi) (r^2) h
r = base radius
h = the height of the cone from the apex.

Pectangular Prim:
Volume of the rectangular prism = l * w * h.
l= length of the prism,
w = width of the prism,
h= height of the prism.

Prism:
Volume of the prism = b* h
where,
b = l* w,
l  =  length of the prism,
w = width of the prism,

h = height of the prism.

Cuboid:
Volume of Cuboid: l * b* h
where,
l = lenght of the base,
b = width of the base,

h = height of the cuboid.



These formulas are very essential to learn about volume.

learn about volume learning-Pictorial view

Here are the pictorial view of the shapes mentioned above to help volume learning easy,

learn about volume learning

Ordinary Second Derivatives

The change of measure of function when the input of the function changes is called an ordinary derivative of a function. It is also defined as the insignificant change in the function with reference to one of its variables. An ordinary derivative for a specified input value defines the linear approximation of the function which is closer to that input value. At greater dimensions the ordinary derivative at a point is a linear transformation and is also called as the linearization.


Formulas of ordinary derivatives:

A simple ordinary derivative of a function f with respect to the variable x is indicated by df /dx.

The derivative of a function f which is given by the variable x is defined by,

f´(x) =  lim f(x+h)-f(x)    and it can also be calculated symmetrically in the form of
h->0         h

f´(x) =  lim f(x+h)-f(x-h)
h->0      2 h

When a function (say f(x, y)) includes more than a single variable then the partial derivatives are used to specify the derivatives of those functions with respect to that variables.

Some of the ordinary derivatives of a function are mostly defined on the complex plane and they are referred to as the complex derivatives.

The second derivative of a function f with its variable x is given by,

f´´(x) = lim f(x+h)-f´ (x)    and it can also be calculated symmetrically in the form of
h->0         h

f´´(x) = lim f(x+2h)-2f(x+h)+f(x)
h->0          h2

Simple derivatives of some functions:

`d/(dx)`  (xn) = nxn-1

`d/(dx)`  (ln x) = 1 / x

`d/(dx)` (sin x) = cos x

`d/(dx)`  (cos x) = -sin x

`d/(dx)`  (tan x) = sec2 x

Properties of Ordinary derivatives:

The derivatives of the sum are equal to the sum of the derivatives which is given by

[f(x) + …….+ h(x)]´ = f´(x) +…..+h´(x)

The product rule for the differentiation of the derivative is given by

`d/(dx)`  [f(x) g(x)] = f(x) g´(x) + f´(x) g(x).

Here f´ is called as the ordinary derivative of the function f with respect to its variable x.



What are second derivatives:

Second derivatives involves the process differentiating the given algebraic function twice with respect to the given variable. Generally the derivative is discussed in calculus whereas it is mainly used to find the rate of change of the given function with respect to the change in the input. The following are the solved example problems with detailed step by step solution in second derivatives to study for the test.

Examples on second derivatives:

Ex 1: Find the second derivative for the function.

f(x) = 2x6 + 2 x5 + 3 x4 + 3x

Sol:

The given equation is

f(x) = 2x6 + 2 x5 + 3 x4 + 3x

The above function is differentiated with respect to x to find the first derivative

f '(x) =  2(6x 5)  +2 (5 x4 ) +3(4 x3) + 3

By solving above terms

f '(x) =  12x 5  +  10x4  + 12 x3 – 3

The above function is again differentiated with respect to x to find second derivative

f ''(x) =  12(5x 4 ) – 10(4x3)  + 12(3x2)

f ''(x) =  60x 4 – 40x3 +36x2  is the answer.

Ex 2: Find the second derivative for the function.

f(x) = 5x 2 +5x 4  + 12

Sol:

The given function is

f(x) = 5x 2 +5x 4  + 12

The above function is differentiated with respect to x to find the first derivative

f '(x) = 5(2x  )+5(4 x 3 ) + 0

By solving above terms

f '(x) = 10x +20x3

The above function is again differentiated with respect to x to find second derivative

f ''(x) =  10(1 ) +20(3x2)

f ''(x) =  10 + 60x2 is the answer.

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Ex 3:  Find the second derivative for the function.

f(x) = 4x4 +5x 5 +6x 6  + 2x

Sol:

The given function is

f(x) = 4x4 +5x 5 +6x 6  + 2x

The above function is differentiated with respect to x to find the first derivative

f '(x) = 4(4x 3 )+5(5x 4 ) +6( 6x 5) +2

By solving above terms

f '(x) = 16x 3 +25x 4 +36 x 5 + 2

The above function is again differentiated with respect to x to find second derivative

f ''(x)= 16(3x 2) +25(4x 3) +36 (5x 4)

f ''(x)= 48x 2 +100x 3 +180x 4 is the answer

Practice problems

1) Find the second derivative for the function.

f(x) = x 3 + x 4 + x 5

Ans : f ''(x) = 6x +12x2+ 20x 3

2) Find the second derivative for the function.

f(x) = 2x 3+3x5 + 4x 6

Ans : f ''(x) = 12x + 60x3 + 120 x 4

Solving Number Problems

Number problems are an important research field of mathematics. In mathematical competitions, problems of basic number theory occur frequently. These problems use slight knowledge and have many variations. They are flexible and diverse. The number problems are much related to the human life. Number problems used to calculate the particular value by using the remaining values.

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Solved examples on number problem:

Example 1 on number problem:-

1) Twice the larger of the 2 numbers is 3 more than 5 times the smaller and the sum of four times the larger and three times the smaller is 71. What are the numbers?

The point is in the solving the problem, not in the relative reality of the problem. That said, how do you solve this number problem? The best first step is to start labeling:

The larger number:  x
the smaller number:  y

Twice the larger:  2x
three more than five times the smaller:  5y + 3
relationship between ("is"):  2x = 5y + 3

Four times the larger:  4x
three times the smaller:  3y
relationship between ("sum of"):  4x + 3y = 71

Now I have two equations in two variables:

2x = 5y + 3
4x + 3y = 71

I will solve, say, the first equation for x:

x = (5/2)y + (3/2)

Then I'll plug the right-hand side of this into the 2nd equation in place of the "x":

4[ (5/2)y + (3/2) ] + 3y = 71
10y + 6 + 3y = 71
13y + 6 = 71
13y = 65
y = 65/13 = 5

Now that I have the value for y, I can solve for x:

x = (5/2)y + (3/2)
x = (5/2)(5) + (3/2)
x = (25/2) + (3/2)
x = 28/2 = 14

The answer here is not "x = 14", but is the following sentence:

Answer: The larger number is 14, and the smaller number is 5.

Example 2 on number problem:-

2) Double the bigger of the 2 numbers is 3 more than 5 times the smaller and the sum of four times the bigger and three times the lesser is 71. What are the numbers?

The point is in the solving the number problem, not in the relative reality of the problem. That said, how do you solve this number problem? The best first step is to start labeling:

The larger number:  x
the smaller number:  y

Twice the larger:  2x
three more than five times the smaller:  5y + 3
relationship between ("is"):  2x = 5y + 3

Four times the larger:  4x
three times the smaller:  3y
relationship between ("sum of"):  4x + 3y = 71

Now I have two equations in two variables:

2x = 5y + 3
4x + 3y = 71

I will solving, say, the first equation for x:

x = (5/2)y + (3/2)

Then I'll plug the right-hand side of this into the 2nd equation in place of the "x":

4[ (5/2)y + (3/2) ] + 3y = 71
10y + 6 + 3y = 71
13y + 6 = 71
13y = 65
y = 65/13 = 5

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Now that I have the value for y, I can solving for x:

x = (5/2)y + (3/2)
x = (5/2)(5) + (3/2)
x = (25/2) + (3/2)

x =14

The answer here is not "x = 14", but is the following sentence:

Answer: The larger number is 14, and the smaller number is 5.

Binary Number System

The binary number system, or base-2 number system, signifies numeric values using two numbers, 0 and          1.The usual binary number system or base-2 system is a positional symbol with a radix of 2.


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For example:

102 is a binary number.

Therefore,

the first 10 numbers in binary notation 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 in decimal notation,

are 02, 12, 102, 112, 1002, 1012, 1102, 1112, 10002, and 10012.


Binary number's place value

solving decimal numbers to binary numbers:



Example 1:

Convert the number 5 to the binary number system.

Solution:

The given decimal number is 5.

We have to convert 5 into binary number system by dividing the number 2.

Divide 5 by 2 that is 5 ÷2 =1(remainder) and the quotient is 2.

Divide 2 by 2 that is 2 ÷2 =0(remainder) and the quotient is 1.


decimal to binary

Example 2:

Convert the number 9 to the binary number system.

Solution:

The given decimal number is 9.

We have to convert 9 into binary number system by dividing the number 2.

Divide 9 by 2 that is 9 ÷2 =1(remainder) and 4 (quotient).

Then divide the quotient 4 by 2 that is 4÷2=0(remainder) and 2(quotient).

Then divide the quotient 2 by 2 that is 2÷2=0(remainder) and 1(quotient)

We have to start from the remainder that is 9 = 10012.


decimal to binary

example 3:   Converts  the  binary number 10111102 into  decimal number

Solution:

=  (1 × 26) + (0 × 25) + (1 × 24) + (1 × 23) + (1 × 22) + (1 × 21) + (0 × 20) = 64 + 0 + 16 + 8 + 4 + 2 + 0                         = 94.

hence  1011110    =  94

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solving of arithematic operations on binary numbers:



Arithmetic in binary is much like arithmetic in other numeral systems. Addition, subtraction, multiplication, and           division can be performed on binary numerals.
Addition:

The simplest arithmetic operation in binary is  addition. Adding two single-digit binary numbers is relatively                     simple, using a form of carrying:

0 + 0 → 0
0 + 1 → 1
1 + 0 → 1
1 + 1 → 0, carry 1 (since 1 + 1 = 0 + 1 × 10 in binary)

For example:
two numerals are being added together: 011012 (1310) and 101112 (2310).

01101
10111
+    -------------
100000
-------------------

Subtraction:

Subtraction   works in much the same way:

0 − 0 → 0
0 − 1 → 1, borrow 1
1 − 0 → 1
1 − 1 → 0

For  example :
1010
1110
_         --------------
0100
-----------------

Division:

Binary   division is again similar to its decimal counterpart:

for  example:

the divisor is 1012, or 5 decimal, while the dividend is 110112, or 27 decimal.

The procedure is the same as that of decimal long division,  here, the divisor 1012 goes into the first three digits 1102of the dividend one time,

1
___________
1 0 1   ) 1 1 0 1 1
− 1 0 1
-----
0 1 1



Multiplication
Multification in binary is similar to its decimal counterpart.

Two numbers A and B can be multiplied by partial products:

for each digit in B, the product of that digit in A is calculated and written on a new line, shifted leftward so that its rightmost digit lines up with the digit in B that was used. The sum of all these partial products gives the final result.


For example:

the binary numbers 1011 and 1010 are multiplied as follows:



1 0 1 1   (A)
× 1 0 1 0   (B)
---------
0 0 0 0   ← Corresponds to a zero in B
+     1 0 1 1     ← Corresponds to a one in B
+   0 0 0 0
+ 1 0 1 1
---------------
= 1 1 0 1 1 1 0

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Three linear Functions

Functions are the ordered pair of things .The first member of the ordered pairs are different from the others. Whole set of members of the functions are known as Domain. And the first member of the ordered pairs is called arguments. 1st degree polynomial functions are known as linear functions.

Forms of three linear functions:

Slope intercept form
Point- slope form
General form

m of Slope intercepts form of three linear functions:

It is most probably used to express the equation of a line. To do this we need to find the Slope of a line and y-intercept of a line.

Slope intercepts formula:

Y=mx+b

Where m is the slope of the line and b is the y- intercept. Slope intercept equation for a vertical line= x=b

Slope intercept equation for a Horizontal line y=b

Point slope form of three linear functions:

It refers the method of graphing the linear equation in the x-y axis. To draw the graph of a linear equation we have to plot the x and y coordinates on the graph. It is used to find the particular equation of a line. The equation of a point slope form is  Y-y1=m(x-x1)

General form:

The general form of the linear function is Ax+by+C=0.

Graph of a function for three linear functions:

The straight line is also known as the graph of the linear function. To graph the linear functions can be done in the three methods.

First method is the table of values is employed that is to assign the values for x and then solve the y.
The second method is known as slope intercept method.
Third method is the Intercept- intercept method


Example to three linear functions:

Example: To identify the degree of each of the polynomials given below:

(i)    a5 – a4 + 3

(ii)  2 – r2 – r3 + 2r8

Solution:

The highest power of the variable is 5. So the degree of the polynomial 5.

The highest power of the variable is 8. So the degree of the polynomial is 8.

The only term here is 2 which can be written as 2a0. So the exponent of a is 0.

Here 0 is the degree of the polynomial.

Normal Distribution

The normal distribution refers to a family of continuos probability distributions described by the normal equation. The normal distribution is defined by the equation with the random variable Y is   Y = [ 1/σ * sqrt(2π) ] * e-(x - μ)^2/2σ^2 .where  X is a normal random variable, μ is the mean, σ is the standard deviation, π is approximately 3.14159, and e is approximately 2.71828. The random variable X in the normal equation is called the normal random variable.

The normal distribution graph depends on two factors - the mean and the standard deviation. The mean of the distribution determines the location of the centre of the graph, and the standard deviation determines the height and width of the graph. When the standard deviation is large, the curve is short and wide. When the standard deviation is small, the curve is tall and narrow. All normal distributions looks like a symmetric, bell-shaped curve, as shown below. The curve on the left is shorter and wider than the curve on the right, because the curve on the left has a bigger standard deviation.

Normal Distribution

Probability and the Normal Curve

The normal distribution is a continuous probability distribution. This has several implications for probability.

1)       The total area under the normal curve is equal to 1.

2)       The probability that a normal random variable X equals any particular value is 0.

3)       The probability that X is greater than a equals the area under the normal curve bounded by a and plus infinity (as indicated by the non-shaded area in the figure below).

4)       The probability that X is less than a equals the area under the normal curve bounded by a and minus infinity (as indicated by the shaded area in the figure below).

normal distribution graph

Additionally, every normal curve (regardless of its mean or standard deviation) conforms to the following "rule

1)       About 68% of the area under the curve falls within 1 standard deviation of the mean.

2)       About 95% of the area under the curve falls within 2 standard deviations of the mean.

3)       About 99.7% of the area under the curve falls within 3 standard deviations of the mean

Examples

Listed below are some of the normal distribution examples.

Example 1
An average light bulb manufactured by the Acme Corporation lasts 300 days with a standard deviation of 50 days. Assuming that bulb life is normally distributed, what is the probability that an Acme light bulb will last at most 365 days?

Solution:

Given a mean score of 300 days and a standard deviation of 50 days, we want to find the cumulative probability that bulb life is less than or equal to 365 days. Thus, we know the following:

1) The value of the normal random variable is 365 days.

2) The mean is equal to 300 days.

3)The standard deviation is equal to 50 days.

We enter these values into the Normal Distribution Calculator and compute the cumulative probability. The answer is: P( X < 365) = 0.90. Hence, there is a 90% chance that a light bulb will burn out within 365 days.

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Example 2:
Suppose scores on an IQ test are normally distributed. If the test has a mean of 100 and a standard deviation of 10, what is the probability that a person who takes the test will score between 90 and 110?

Solution: Here, we want to know the probability that the test score falls between 90 and 110. The "trick" to solving this problem is to realize the following:

P( 90 < X < 110 ) = P( X < 110 ) - P( X < 90 )

We use the Normal Distribution Calculator to compute both probabilities on the right side of the above equation.

1) To compute P( X < 110 ), we enter the following inputs into the calculator: The value of the normal random variable is 110, the mean is 100, and the standard deviation is 10. We find that P( X < 110 ) is 0.84.

2)  To compute P( X < 90 ), we enter the following inputs into the calculator: The value of the normal random variable is 90, the mean is 100, and the standard deviation is 10. We find that P( X < 90 ) is 0.16. We use these findings to compute our final answer as follows:

P( 90 < X < 110 ) = P( X < 110 ) - P( X < 90 )
P( 90 < X < 110 ) = 0.84 - 0.16
P( 90 < X < 110 ) = 0.68

Thus, about 68% of the test scores will fall between 90 and 110.

construct a Right Angle

The triangle is having a single angle only as a right angle which means the angle is at 90° degree. In a right angled triangle the three sides which are given some names. The side which is opposite to that of the right angle 90° is called as the hypotenuse. This is always the longest side of the triangle. The other two sides are the adjacent side and the opposite side.

Construct an right angle triangle:

From the property of the sum of the angles of a triangle ABC it is clear that the other two angles will be acute angles. The side which is opposite to that of the right angle 90° is called as the hypotenuse. The other two sides are its legs or sides. Since in the above right triangles ?C is right angle when construct a right angle. Therefore AB is its hypotenuse and AC and BC are its sides.

Example of constructing right angle triangle:

Construct a right angle triangle whose hypotenuse is of length 4cm. and its one side is of length 3cm.

Steps of construction:

Draw line segment BC = 3cm

Draw ?BCX=90°

With centre B and a radius of AB = 4 cm (hypotenuse) draw an arc at A.

Join points B and A.

? ABC is the required right angle triangle.


Problem for a right angle triangle:

Find the height of the tower where the length of one side is 500m and the angle is 60°.

Solution: Adjacent side is nothing but the side which is next to the angle, the Opposite side is opposite to the angle and the longest side is the Hypotenuse.

one of the side we know is the Hypotenuse and the one we going to find is Adjacent to the angle.

In this we have to use the cosine formula to find its height,

cos 60° = Adjacent / Hypotenuse = h / 500

cos 60° = 1/2

h / 500 = 1/2

h = (1/2) x 500

= 250.

I am planning to write more post on how do you evaluate an algebraic expression and sample papers for class 12. Keep checking my blog.

Mixed Fractions

In this page we are going to discuss about mixed fractions. Before that a small introduction to fractions. A fraction is defined as the ratio of a whole number. For example 7/4 is a fraction.  What does “4” stand for? It is the number parts into which the whole number division. What does “7” stand for? It is the number of equal parts which have been taken out. Here 5 are called the numerator and 2 are called the denominator.

A fraction is a number that can represent part of a whole. The fractions are reciprocals of integers, symbols representing one half, one third, one quarter, and so on. A much after development were the common or "vulgar" fractions which are still used today, and which consist of a numerator and a denominator, the numerator representing a number of equal parts and the denominator telling how many of those parts make up a whole.


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Below are the Forms of Fractions:

Proper fraction: A Proper fraction is a rational number in which the numerator is less than the denominator. Examples: 1/3, 3/4, 2/7

Improper fraction: An Improper fraction has a top number larger than the bottom number, It is "top-heavy". Improper fractions can be written as a number plus a fraction. Improper fraction convert to a mixed number is to divide the fraction's numerator by its denominator. One way Improper fraction convert to a mixed number is to divide the fraction's numerator by its denominator. The integer part of the answer of division is the integer part of the mixed number. The denominator is the same as the original denominator. Another way Divide the numerator of fractions by its denominator and then convert the decimal portion of the answer to a fraction. Examples: 4/3, 11/4, 7/7

Mixed Fraction: A mixed fraction is defined as the whole number and a fraction number combined into one number known as “mixed number”. The combination of whole number and fraction is known as mixed fractions. Examples: 1 1/3, 2 1/4, 16 2/5

Converting Mixed Fractions to Improper Fractions:

To convert mixed fraction to an improper fraction.

Steps to be followed:

Multiply the whole number part by fraction’s denominator.
Sum up to the numerator.
Then note down the solution on top of the denominator.

Example: Convert 4 2/5   to an improper fraction.

Multiply the whole number by the denominator: 4 × 5 = 20

Sum up the numerator to that: 20 + 2 = 22.

Then note down that above number as denominator, like this: 22/5 .

Converting Improper Fractions to Mixed Fractions:

To convert an improper fraction to mixed fraction, follow these steps:

By Dividing the numerator with the denominator and then
Write down the whole number answer

Write down any remainder which we got above the denominator.

Example: Convert 11/4 to a mixed fraction.

Divide: 11 ÷ 4 = 2 with a remainder of 3

Take down the 2 and then write down the remainder (three) above the denominator (four), like this:2 3/4
When to Use Improper Fractions or Mixed Fractions

For everyday use, people understand mixed fractions better:

Example: It is comfortable to say "I had 21/4 cup of milk", than "I had 9/4 cup of milk"

Solved Examples

Below are the problems based on Mixed fractions:
Example 1: Convert  13/3   into a mixed fraction.
Solution: First Divide 13 ÷ 3 = 4 with a remainder of 1
Write down the 4 and then write down the remainder (1) above the denominator (3)
= 4 1/3
Example 2:  Convert 12/5   into a mixed fraction.

Solution: First Divide 12 ÷ 5 = 2 with a remainder 2
Write down the 2 and then write down the remainder (2) above the denominator (5)
=     2 2/5  this  is improper fraction
Example 3: Convert   21/4   into a mixed fraction.

Solution: First Divide 21 ÷ 4 = 5 with a remainder 1
Write down the 5 and then write down the remainder (1) above the denominator (4)
=     5 1/4  this  is improper fraction
Example 4: Convert 1 3/5 to an improper fraction.
Solution: Multiply the whole number by the denominator: 1 × 5 = 5
Add the numerator to that: 5 + 3 = 8
Then write that down above the denominator, like this:
= 8/5

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Example 5: Convert 3  2/5 to an improper fraction.
Solution: Multiply the whole number by the denominator: 3 × 5 = 15
Add the numerator to that: 15 + 2 = 17
Then write that down above the denominator, like this:
= 17/5
Example 6: Convert  5  6/7 to an improper fraction.
Solution: Multiply the whole number by the denominator: 5 × 7 = 35
Add the numerator to that: 35 + 6 = 41
Then write that down above the denominator, like this:
= 41/7

Linear Measurement Definition

The Word Linear is taken from the Latin word Linearis. Linear literally means created by lines. However Linear on the other hand can be mathematically defined as something which is in a straight line.

The word Measurement is derived from the greek word Metron meaning limited proportion. Measurement on the other hand is scientifically defined as the process of identifying the magnitude of a quantity.

Linear Measurement: Thus we combine the definition of Linear and measurement to arrive at the definition of Linear Measurement. In simple words, Linear Measurement can be defined as the measurement of Length.

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Examples of Linear Measurement:

1. Diameter of a Circle.

2. Length, breadth or height of any two dimensional or three dimensional object.

3. Distance between any two points or places.
Units of Linear Measurement:

The basic unit of Linear measurement  is metre. one metre is equal to the bar of metal called as the International Metre bar and kept at the National Burea of Standards, Washington D.C. Other units are millimetre, centrimetre, kilometre

However the British or the English have known to be measuring length using Inches, Feets, Yards, Furlongs and miles. But since the advent of metrificaiton of units world over, the accepted norm is to use Metre as the basic unit of measurement.

The basic difference between the Metric system and the British system is that the Metric system has a tremendous ease of conversion between different units which is comparitively difficult in the British system.
Examples of Units of Measurement

Different parameters are measured in different units. We will be looking into each of these parameters separately in the below paragraphs.

At sea, we measure the depth of the sea in Fathoms and the distance in nautical miles. One fathom is equal to six Feet or 182 centimetres and one nautical mile equals to 6076 feet or 1852 metres.


Chest, Waist, Collar and other measurements concerned with stitching are measure in Inches. on the other hand length of foot while purchasing footwear also uses Inches mostly.

At Land or in air, Distance between 2 places is mostly measured in Kilometres or Miles. However in outer space, the distance measurement is carried in Light Years. One Light Year is the distance travelled by Light in 1 year.

Distance travelled by sound is measure in MAch which is nothing but distance travelled by sound in air.

Natural Squares Calculator

The natural squares are the process of squaring the given natural numbers. The natural numbers are the countable numbers in the number system. The natural numbers are used for the ordering and also the counting. The squares for the numbers are processes in which the given number is multiplied by the same number. This article has the information about the natural squares calculator.

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Examples for the Natural Squares Calculator:

Example 1 for the natural squares calculator:

Find the squares for the natural number 16.

Solution:

The given natural number is 16.

We have to find the square for the given natural number.

Multiply the given number two times and then we get the squared value for the given number.

162 =16 x 16

162 = 256

natural squares calculator

The squares value for the natural number 16 is 256.

Example 2 for the natural squares calculator:

Find the squares for the natural number 25.

Solution:

The given natural number is 25.

We have to find the square for the given natural number.

Multiply the given number two times and then we get the squared value for the given number.

252 =25 x 25

252 = 625

natural squares calculator

The squares value for the natural number 25 is 625.

Example 3 for the natural squares calculator:

Find the squares for the natural number 47.

Solution:

The given natural number is 47.

We have to find the square for the given natural number.

Multiply the given number two times and then we get the squared value for the given number.

472 =47 x 47

472 = 2209

natural squares calculator

The squares value for the natural number 47 is 2209.

Example 4 for the natural squares calculator:

Find the squares for the natural number 53.

Solution:

The given natural number is 53.

We have to find the square for the given natural number.

Multiply the given number two times and then we get the squared value for the given number.

532 =53 x 53

532 = 2809

natural squares calculator

The squares value for the natural number 58 is 2809.

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Practice Problems for the Natural Squares Calculator:

Find the squares for the natural numbers 40.

Answer: 1600.

Find the squares for the natural numbers 67.

Answer: 4489.

Find the squares for the natural numbers 80.

Answer: 6400.

Subtracting Fractions

Fraction is one of the most important topics in mathematics taught in middle school. When a number is divided into two parts, each part is called as a fraction. For example: 89/100 of parents said good things about construction toys . Here, 89 is the numerator and 100 is the denominator. We can perform all four basic mathematical operations with fractions. Let’s have a look at the concept of subtracting fractions. Subtracting fractions varies based on its denominators. The concept of subtracting fractions with same and different denominators is elaborated with relevant examples in the below :
Subtracting Fractions with Same Denominators:
While subtracting fractions with same denominators, first the denominators should be taken as common denominators and then the numerators should be subtracted. If the result is a greater value, it can be further reduced.
1. Example: The shop keeper wanted to sell at least 5/7 of construction toys but Kiran bought only 3/7? How many toys remaining?
5/7 – 3/7
(5-3)/7
2/7
2. Example: 78/100 parents voted for soft baby toys and 20/100 parents voted for electronic toys for kids. What is the difference?
78/100 – 20/100
(78-20)/100
58/100
29/50
Subtracting Fractions with Different Denominators:
While subtracting fractions with different denominators, first the LCM that is the lowest common multiple of the denominators is calculated and then the numerators are subtracted. If the result is a greater value, it can be further reduced.
3. Example:  Mahesh bought 4/3 parts of electronic toys for kids and Hari bought 2/4 of the soft baby toys collection. What is the difference?
4/3 – 2/4
Finding LCM, [(4*3) – (2*4)] / 12
(12 – 8)/12
4/12
1/3
4. Example: He has 55/60 part of the property and Niketan has 40/55. Find the difference:
55/60 – 40/55
Finding LCM, [(55*11) – (40*12)]/132
125/132
These are some of the basics on subtracting fractions along with solved examples for better insights.

Log Base Formula

The concept  of logarithms arrived  from John Napier's work in the early 17th century. Since then, logarithm tables are used. logarithms were an important key to simplifying scientific calculations. Today's there are numerous applications of logarithms.

The logarithms of a number 'N' to the base 'a' (a>0, a`!=` 1)  is defined as the exponent (or power) to which the base (a) is raised to produce the number (N).

i.e;      logarithmbase( Positive number) = any real number

Mathematically , if  ax = N  a>0, a`!=` ,  N>0

then  loga N = x

Conversely, loga N = x

=>  ax = N

In the above context, it is to be noted that logarithmic function can be defined in real number system only if following conditions are satisfied at a time.

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base is positive.
base is not equal to 1.
number ( whose logarithm is to be found out ) is positive.

Important Log Base Formula:

1) loga(1) =0 ,

where  'a' is any base.   The log of any base is zero.

2) logN N =1

The log of a number to the same base is one.

3) loga N =   1/ logN a

Here base and number are interchanged.  =>  loga N *logN a =1

4) a loga N = N

5) loga b * logbc * logc a =1

Circle of log base

Here a,b and c are arranged in such a way that each number moves to the base once alternatively.

In such case, the continued product of the logarithm of the numbers is equal to 1. It is also true for more than 3 numbers.

log base formula

7 ):  ( a). loga N > 0, if a>1, N>1

or      a<1 n="" p="">
(b). loga N < 0, if a>1, N<1 p="">
or   a<1 n="">1.

8) (a). If  a (=base) > 1, and loga N1 > logaN2, then N1 > N2

(b)   If  a (=base) < 1, and loga N1 > logaN2, then  N1 Examples on Log Base:

Q:1 Evaluate log2log2 log2 16

Sol:  log2log2 log2 16

= log2log2 log2 24

= log2  log2 4              ( since log2 2=1 )

=log2 log222  =  log2 2  =1

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Q:2 Evaluate '(logm  (logk m))/ logk (logm k)'

Sol:            '(logm  (logk m))/ logk (logm k)'

let  logk m =x    so we can write as logm k = 1/x

now we can write log mx  / logk  (1/x)  =  logm x/ -logk x

so our answer =   -logm k

Triangles Activity for Middle School

Triangles is one of the important geometrical shapes, it is a closed shape with three sides and angles. The triangles activity is divided into two types they are based on sides and angles. In triangle activity the sides are consist of three types they are equilateral triangle, scalene triangle, and isosceles triangle. In triangle activity the angles are consist of  three types they are right angle triangle, acute angle and obtuse angle.  In this article, we will discuss about triangles activity for middle school with suitable example problem.
Triangles Activity in Sides

Triangles activity in sides are classified into three types.They are,

Equilateral triangle
Scalene triangle and
Isosceles triangle

Equilateral triangle:

equaliteral triangle

All the three sides are equal then it is said to be equilateral triangle.

Scalene triangle:

scalene triangle

All the three sides are different then it is said to be scalene triangle

Isosceles triangle:

isosceles triangles

If the two sides are equal then it is said to be scalene triangle.

Problems in triangle activity for middle school in sides:

Problem (i): Give the statement about which type of the triangle the sides are 10cm, 10cm and 10cm ?

Solution:

Give the statement about which type of the triangle the sides are 10cm, 10cm and 10cm

Step1: Here all the sides are different

10 cm, 10cm, and 10cm.

Step2: All the three sides are equal then it is said to be equilateral triangle.

So, this is the equilateral triangle.

Problem (ii): The sides of the equilateral triangle is 6cm, find the area of equilateral triangle and perimeter

Solution

The given problem sides of the equilateral triangle is 6cm, find the area of equilateral triangle

Step 1: Using the formula Area of the equilateral triangle is sqrt(3)/4 xx s2, here s is the given sides

Step 2: The sides in the equilateral triangle is 6cm

Step 3: substitute the  sides in the formula

Area of the equilateral triangle is sqrt(3)/4 xx s2

= sqrt(3)/4 xx 62

=sqrt(3)/4 xx 36

Step 4: The value of sqrt(3)  is 1.732.

Area of the equilateral triangle is  (1.732xx36)/4

= 62.352/4

= 15.588cm2

Area of the equilateral triangle is  15.588cm2

Step 5: The  perimeter of equilateral triangle is P = A+B+C

Perimeter  = 6+6+6

Perimeter of the equilateral triangle is 18cm.
Triangles Activity in Angles

Triangles activity in angles are classified into three types.They are,

Right angle triangle
Acute angle triangle
Obtuse angle triangle

Right angle triangle:

right angle triangle

If one of the angle is 900 then it is called right angle triangle the total angles is 1800

Acute angle triangle:

acute angle triangle

If one of the angle is less than 90o then it is said to be acute angle. Triangles'  total angle is 180o.

Obtuse angle triangle:

obtuse angle triangle

if any one of the angle is greater than 90o then it is said to be obtuse angle triangle the total angle is 180o.

Problems in triangle activity for middle school in sides:

Problem (i): Give the statement about which type of the triangle, the angles are 110o, 40o, and 30o ?

Solution:

The angle of the triangle are given  110o, 40o and 30o.

Step 1: Here the angle is 110o

110o, 40o and 30o

Step 2: one of the angle is greater than 90o. so it is called obtuse angle triangle

So  this is called obtuse angle triangle.