How to Solve Significant Figures

Introduction how to solve significant figures:

The digits, which are used to represent the accuracy of numbers are said to be significant figures. Significant figures can also be called as significant digits. The concepts of significant figures are highly linked with rounding. The term significant figures can be abbreviated as sig figs, sign figs, and sign digs. In this section, we shall discuss how to solve significant figures. In this content we have an overview on identifying and solving significant figures.

Example for How to Solve Significant Figures :

Example problem 1:

How to Identify the number of significant figures from the following.

84, 0.084, 5.8480, 2005, 8400

Solution:

84 – The number 84 is a non-zero number that has two significant figures.

0.084 – Zeroes placed before the other numbers are not significant figures and the number of significant figures in 0.084 is two.

5.8480– Zeroes placed after the other numbers but after a decimal point are significant figures and the number of significant figures in 5.8480 is five.

2005 – The number 2005 has four significant figures because the Zeroes placed between the figures are always significant.

8400:

The number 8400 may have at least two significant figures. It also has three and four significant figures.

That is:

8400 – 8.4 x 103 has two significant figures.

8400 – 8.40 x 103 has three significant figures.

8400 – 8.400 x 103 has four significant figures.

Between, if you have problem on these topics What are Irrational Numbers, please browse expert math related websites for more help on List of Prime Numbers to 100.

One more Example for How to Solve Significant Figures :

Example problem 2:

Solve the following addition 5.76 + 4.62 + 31.21 and find the number of significant figures in the solution.

Solution:

The given numbers are 5.76, 4.62, and 31.21.

The number 5.76 has three significant figures.

The number 4.62 has three significant figures.

The number 31.21 has four significant figures.

Adding the numbers, we get:

5. 7 6

4. 6 2

3 1. 2 1   +

4 0. 5 9

Therefore, the solution is 40.59.

40.59 can be rounded to 40.6

Therefore, the number of significant figures in 40.6 is three.

Point Symmetry and Line Symmetry

Introduction to point symmetry and line symmetry:
                Point of symmetry is a special center point for certain types of geometry symmetric objects. If an object or shapes can be rotated at 180 degree about a point P and end up looking an identical to the original, then the P is called as point of symmetry. A line which divides the object into their half and each figure has its mirror image is known as line symmetry.

The Explanation about Point Symmetry:

                   In geometry Point Symmetry is sometimes called as Origin Symmetry, because the "Origin" is the central point about which the figure is symmetrical. If shapes or graphs is rotated about a point by 180 degree and yet looks an identical to its original, that point is called as the Point Symmetry. If shapes have point symmetry, then its order of rotational symmetry must be 2.
Algebra is widely used in day to day activities watch out for my forthcoming posts on Multiplying Variables with Exponents and How do you Find the Degree of a Polynomial. I am sure they will be helpful.
Example of Point of Symmetry geometry:

                In the given shapes, an ellipse or rectangle is rotated 180 degree by a point. The ellipse or rectangle obtained after rotation same with the original rectangle. So, the point with which the ellipse or rectangle geometry shape rotated is called as the point symmetry.

Solving Real Analysis Problems

Solving Real Analysis Problems

Real analysis is a practice where raw record is arranging and organizing such that useful information will be extracting from it. The process of organizing about numbers and to understand what the data does and does not contain. Analysis classified into qualitative and quantitative. Qualitative analysis is involved in interpreting information which can be collected during the course of qualitative research. Quantitative analysis is involved in presenting and interpreting numerical numbers.I like to share this Anova Analysis with you all through my article.

Solving Real Analysis – Solving Example Problems

Solved real analysis example problems

Example 1: A pair of balanced dice is rolled, and what are the probabilities of getting the sum (1) 7 (2) 7 or 8 (3) 9

Solution:-

The sample space S = {(1, 1), (1, 2) … (6, 6)}

Number of possible outcomes n(S) = 36

Let A be the event of getting sum 7.

Let B be the event of getting the sum 8.

Let C be the event of getting the sum 9.

A = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)}, n(A) = 6.

B = {(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}, n (B) = 5

C = {(3, 6), (4, 5), (5, 4), (6, 3)}, n(C) = 4

(1) P (getting sum 7) = P(A) = n(A) / n(S)

= 6/36 = 1/6

Therefore P(7) = 1/6

(2) P (7 or 8) = P(A or B)

= P (A) + P (B)          (A and B are mutually exclusive i.e. AnB=f)

= 6/36 + 5/36

= 11/36

Therefore P (7 or 8) = 11/36

(3) P (getting sum 8) = P(c) = n(A) / n(S)

= 4/36 = 1/9

Therefore P (8) = 1/9.

Example 2: Find the sum of all integers, from 10 to 1000 inclusive, which are divisible by 10.

Solution:

Sequence of first few elements of integers divisible by 10 are given by 10, 20, 30, 40...

The above sequence has a first element equal to 10 and a common difference d = 10.

We need to know the rank of the term 1000.

We use the following formula for the nth term

an = a1 + (n - 1 )d

1000 = a 1 + (n - 1 )d

Substitute a1 and d by their values

1000 = 10 + 10(n - 1)

Solve for n to obtain

n = 100

1000 is the 100th term, we can use the following formula to find sum

sn = n (a1 + an) / 2

s100 = 100 (10 + 1000) / 2 = 50500.
Between, if you have problem on these topics adding subtracting multiplying and dividing rational expressions, please browse expert math related websites for more help on tutoring math online.
Solving Real Analysis – Solving Practice Problems

Solve these practice real analysis problems

Problem 1: When a pair of balanced dice is rolled, and what are the probabilities of getting the sum (1) 12 (2) 2 (3) 6 or 7.

Answer: 1) 1/36, 2) 1/36, 3) 11/36

Problem 2: Find the sum of all integers, from 11 to 1100 inclusive, which are divisible by 11.

Answer: 61105

Limit Rules Calculus

Introduction for limit rules for calculus:

In mathematics, the concepts of a “limit” are used to describe the value that a function or sequence "approaches" as the input or index approaches some value. The concepts of limit allow one to, in the complete space; define a new point from a Cauchy sequence of previously defined points. Limit is essential to calculus (and mathematical analysis in general) and is used to define continuity, derivatives and integrals.                                                                                                   (Source.Wikipedia)

Limit rules Calculus includes that differential calculus and integral calculus limits is used 

Rules for Limit Calculus:

(1) If f(x) = k for all x, then`lim_(x->c)` f(x) = k.

(2) If f(x) = x for all x, then`lim_(x->c)` f(x) = c.

(3) If f and g are two functions possessing limits and k is a constant then

(i)`lim_(x->c)` k f(x) = k`lim_(x->c)` f(x)

(ii)`lim_(x->c)` [f(x) + g(x)] =`lim_(x->c)` f(x) +`lim_(x->c)` g(x)

(iii)`lim_(x->c)`[f(x) - g(x)] =`lim_(x->c)` f(x) -`lim_(x->c)` g(x)

(iv)`lim_(x->c)` [f(x) . g(x)] =`lim_(x->c)` f(x) `lim_(x->c)` g(x)

(v)`lim_(x->c)`` f(x)/g(x)` =`lim_(x->c)` f(x) / `lim_(x->c)`g(x), g(x) ? 0

(vi) If f(x) = g(x) then `lim_(x->c)`f(x) =`lim_(x->c)` g(x).

(4) `lim_(x->a)`   xn - an/x - a = nan - 1 (a ? 0)

Examples for Limit Rules Calculus:

Example 1:

Evaluate 7`lim_(x->1)` (x3 - 1) / (x- 1)

Solution:

7`lim_(x->1)`x3 - 1 / x- 1

=7* 3(1)3 - 1                                     [limit rules:  `lim_(x->a)` `lim_(x->a)`   xn - an / x - a = nan - 1 (a ? 0)]

= 7*3(1)2

= 21

Example 2:

Find 8`lim_(x->0)`{(1 + x)4 - 1}/x

Solution:

Put 1 + x = t so that t ? 1 as x ? 0 and x= t -1

8`lim_(x->0)`(1 + x)4 - 1/ x = 8`lim_(t->1)`(t4 - 1)4/ (t - 1)           [limit rules:   `lim_(x->a)`   xn - an/x - a = nan - 1 (a ? 0)]

= 8*4(1)4-1 = 4*8=32

Example 3:

calculate the value n so that`lim_(a->2)`an - 2n/a - 2= 32

Solution: We have

`lim_(a->2)`an - 2n/a - 2 = n2n - 1                       [limit rules:   `lim_(x->a)`   xn - an/x - a = nan - 1 (a ? 0)]

? n2n - 1 = 32 = 4 × 8 = 4 × 23 = 4 × 2 4 - 1

Comparing on both sides we get n = 4

Example 4:

3`lim_(x->0)`log `(1 + x)/x ` = 1

Solution:

We know that loge (1 + x) =`x/1` - `x^2/2` +`x^3/3` - …

loge `(1 + x) / x ` = 1 - `x/2 ` +`x^2/3` - …

Therefore  3`lim_(x->0)`loge` (1 + x)/x``lim_(x->0)`loge (`(1+0)/ 0` ) 

=3 loge 0

= 3*1=3.

Between, if you have problem on these topics Converting Decimals to Mixed Numbers, please browse expert math related websites for more help on How to Find the Percentage of something.

Calculus Integration Problems

Introduction to calculus integration problems: 

Integration calculus is the definition, properties, and applications of two related concepts, definite integral and indefinite integral. Finding the value of an integral process is called integration Methods. Integral calculus is two related linear operators.Indefinite integral is antiderivative, the inverse operation to the derivative.Definite integral inputs a function and outputs a number, gives the area between graph of input and x-axis. Limit of a sum of areas of rectangles is known as Riemann sum.

Calculus Integration Problems - Importance

Ancient period introduced ideas of integral calculus, have developed these ideas in a rigorous or systematic way. Manipulating very small quantities are usually developed Calculus.Infinitesimals are the first method of the calculus. dx is infinitesimal number could be greater than 0, but less than in the sequence 1, 1/2, 1/3, .... Infinitely small with the Integer multiple of an infinitesimal.    

Provided solid foundations for the manipulation of infinitesimals is introduce of non-standard analysis and smooth infinitesimal analysis.

Objective of integration calculus problems is settle on rate of change.

Main wing of integration calculus problems is

Integral calculus.
Integration calculus problems is decide function of  the rate of change.

Between, if you have problem on these topics find the value for the correlation coefficient r., please browse expert math related websites for more help on is the square root of 5 a rational number.

Calculus Integration Problems - Sample Problems

Problem 1:
Find the integral of the given equation

2x + exdx

Solution:

?2x + ex dx=?2x dx + ?ex dx

Integrating the above equation.

We get

2x2 / 2 + eX

x^2 + ex

Problem 2:

Find the integral of the given equation

x2+2x+3 dx

Solution:

?x2+2x+3 dx  = ?x2 dx + ?2x dx +?3 dx

Integrating the above equation

We get

= x3/3 + 2x2/2 + 3x

= x3/3 + x2 + 3x

Problem 3

Find the integral of the given equation

X4+2x2+3x dx

Solution:

?x4+2x2+3x dx  = ?x4 dx + ?2x2 dx +?3x dx
Integrating the above equation

We get

= x5/5+ 2x3/3 + 3x2/2

= x5/5 + 2x3/3+ 3x2/2

Volume of a Sphere Units

Introduction about volume sphere:

A sphere is a perfectly round geometrical object in three-dimensional space, such as the shape of a round ball. Like a circle in two dimensions, a perfect sphere is completely symmetrical around its center, with all points on the surface laying the same distance r from the center point. Let us see how to calculate the volume of sphere.

(Source – Wikipedia)

Examples Problems in Volume of Sphere Calculator:

Volume of the sphere (V) = `4/3` p r ^ 3 cubic unit.

r = radius.

Volume of Sphere Units - Problems:

1. The sphere has the radius 20.3 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 20.3 cm

Formula:

Volume of the sphere (V) = `4/3` p r ^ 3 cubic units

= `4/3` * 3.14 * (20.3) 3

=`4/3` * 3.14 * 8365.427

Volume of the sphere (V) = 35023.25

2. The sphere has the radius 15.2 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 15.2 cm

Formula:

Volume of the sphere (V) = `4/3` p r ^ 3 cubic units

= `4/3` * 3.14 * (15.2) 3

=`4/3` * 3.14 * 3511.80

Volume of the sphere (V) = 14702.76 cm3

3. The sphere has the radius 11.25 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 11.25 cm

Formula:

Volume of the sphere (V) = `4/3` p r ^ 3 cubic units

= `4/3` * 3.14 * (11.25) 3

=`4/3` * 3.14 * 1423.82

Volume of the sphere (V) = 5961.09 cm3

4. The sphere has the radius 3.4 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 3.4 cm

Formula:

Volume of the sphere (V) = 4/3 p r ^ 3 cubic units

= 4/3 * 3.14 * (39.304) 3

=4/3 * 3.14 * 164.552

Volume of the sphere (V) = 164.552 cm3

5. The sphere has the radius 5.6 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 5.6 cm

Formula:

Volume of the sphere (V) = `4/3 ` p r ^ 3 cubic units

= `4/3` * 3.14 * (5.6) 3

= `4/3` * 3.14 * 175.61

Volume of the sphere (V) = 735.22 cm3

I am planning to write more post on geometry help online, mathematical induction. Keep checking my blog.

Practice problems in volume of sphere:

1. The sphere has the radius 8.4 cm. Find the volume of the sphere.

Answer: 2481.45408 cm3

2. The sphere has the radius 6.6 cm. Find the volume of the sphere.

Answer: 1203.64992 cm3

Solving Linear Equations Substitution

Introduction for solving linear equations using substitution:

Linear equations substitution is nothing but a process of exchanging a variable in a linear expression with its actual value. In linear algebra the substitution process plays a major role to solve the system of linear equation by substituting the value for the given variable, linear substitution is mainly used to identify a variable in an expression and to find linear relationships between the equations. Here we use the linear substitution method to solve the different types of linear equations.

Having problem with Solving Systems of Linear Inequalities keep reading my upcoming posts, i will try to help you.

Solved Examples on Linear Equation Substitution

Ex 1:

Solve the linear equation by substitution method.

1 / (y - 1) 2 - 4 / (y - 1) + 4 = 0

Solution:

Let s = `1 / (y - 1)` and substitute this term in the given equation.

s 2 – 4s + 4 = 0

By solving the above quadratic equation, we get:
s = 2 and s=2.

Now substitute s by` 1 / (y - 1)` and solve for y
`1 / (y - 1)` = 2

1 = 2(y - 1)

1=2y - 2

3= 2y

Y = `3/2` is the solution for above equation.

Ex 2:

Solve the linear equation by substitution method.

z - 5 `sqrt (z)` = - 6

Solution:

Let s = `sqrt (z)` so that s 2 = z. Substitute z by s and `sqrt (z)` by s 2 respectively to obtain an equation in s.
s 2 - 5 s = - 6

The above equation looks like quadratic form, so rewrite the above term

s 2 - 5 s + 6 = 0

By solving the above equation we get
s = 2 or s = 3

We now substitute s by `sqrt (z)` and solve for z
`sqrt (z)` = 2 or `sqrt ( z )` = 3

z = 4 or z = 9  Is the solution for above equation.

My forthcoming post is on example of a algebraic expression, how to write an algebraic expression in words will give you more understanding about Algebra.

Practice Problems on Linear Equations Substitution for Solving:

1) Solve the linear equation by substitution method.

1 - 2 / (a - c) = 8 at c=1.

Answer:    a = 5/7.

2) Solve the linear equation by substitution method.

1 - 1 / (x - z) = -8 / (x 2 - z 2) at z=4

Answer:    x = -3.

Horizontal Line Segments

Introduction to horizontal line segments:
The line segments are distinct as the distance between two points. The line segments are make clear in another method is, the point that is joined the points of both directions. We can define a line PQ as `bar(PQ)` . Here we are going discuss about the horizontal line segment. Also we shall solve an example problem based on horizontal line segments.

Special Cases of Horizontal Line Segments:
Horizontal line segments have some special properties when compared to normal line segments,

Horizontal and vertical line segments are identified easily, that they have either x- value as similar or y value as same from both the given points of (x1, y1) (x2, y2).

In horizontal line segments, if y-value is similar then the line segment is easily identified as horizontal line segments.

I am planning to write more post on how to do long division with decimals step by step, how to simplify large fractions. Keep checking my blog.

Example Problem for Horizontal Line Segments:

Plot the given horizontal line segment on the graph (8, 5) (-8, 5).

Solution:

Given: Two points to plot line segments are (8, 5) (-8, 5).

Two pairs of points are (x1, y1) (x2, y2) needed to plot the horizontal line segments in graph.

Here, from the given points x1 is 8 and y1is 5.

From the given points x2 is -8 and y2 is 5.

Now, from the given values of x co-ordinates and the values of the y co-ordinates, we can say that the values of y-coordinates are similar, so the line is horizontal line segments.

According, to the values of the x co-ordinate we have to plot the corresponding x value of the line segment and also on the y co-ordinate, plot the corresponding y value of the horizontal line segments.

These mentioned above, process is done by using graph is shown.

Thus, horizontal line segment is explained clearly and through diagrammatically is explained successfully.