Learn Area of a Rhombus

Rhombus is a quadrilateral in which all the sides are equal. Rhombus has all the properties of a parallelogram and its diagonals bisect each other at right angles. The area of a rhombus can be calculated using the conventional formula used to calculate the area of a parallelogram b*h where "b" is the base and "h" is the perpendicular distance between the base and the opposite parallel line.



ABCD is a rhombus and we shall derive the formula for finding out the area given its diagonals AC and BD


learning how to find area of a rhombus


Let the length AC be d1 and that of BD be d2. The formula for calculating the area of a triangle is 1/2 bh where 'b" is the base and "h" is height or altitude.

We shall consider the rhombus as the combinations of two triangles ABC and ADC.If we add the areas of triangles ABC and ADC we are sure to get the area of the rhombus ABCD.

The diagonals of rhombus bisect each other at right angles the height of triangles of ABC and ADC will be half of the diagonal BD which is equal to d2/2. We shall now calculate the area separately and add.


Formula to learn area of a rhombus


Area of triangle ABC =1/2  (d1* d2/ 2) =(d1 d2) /4

Area of triangle ADC=1/2  (d1* d2/ 2) =(d1 d2) /4

Area ABC +Area ADC = (d1 d2 )/4 + (d1 d2 ) /4 =d1 d2 /2

The area of the rhombus  = 1/2 product of its diagonals

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learn examples on area of a rhombus


We shall illustrate this by two examples. We are given the diagonals of a rhombus as 10 cms and 12 cms. We are asked to find the area of the rhombus. The area of a rhombus= 1/2 the product of its diagonals= 1/2 *10*12 = 60 square cm.

This is an example to find the area of the rhombus using the conventional formula. Suppose the length of the side of a rhombus is given as 12cm and the perpendicular distance between the sides is given as 10 cm the area of the rhombus =side* perpendicular distance = 10 * 12 =120square cm.

Rational Algebraic Expressions

A rational expression is an algebraic expression of the `(P)/(Q)` from ,where P and Q are simpler expressions (usually polynomials), and the denominator Q is not zero.



This rational expressions whose numerators and denominators are (or written as) polynomials. Like polynomials, rational expressions appear frequently in the Algebra II and higher mathematics. We must understand the how to perform basic operations with rational expressions and how to solve rational equations.


Two Types:-


1.Adding and Subtracting Rational Expressions

2.Multiplying and Dividing Rational Expressions



1. Adding and Subtracting Rational Expressions

Adding Subtracting Rational Expressions with the Like Denominators

Add and subtract their numerators and then write the results over the denominator. Then, simplify the numerator factor, and write the expression in lowest terms.

Adding and Subtracting Rational Expressions with the Unlike Denominators

Rational expressions can be added and subtracted if and only if they have the same denominator. Thus, we must rewrite them as expressions with a common denominator.



2.Multiplying and Dividing Rational Expressions

Multiplying Rational Expressions

To multiply two rational expressions, factor them. Then multiply their numerators and denominators, crossing out the any factors that appear in both the numerator and the denominator

Dividing Rational Expressions

To divide by rational expression,and multiply by its reciprocal.


Examples:-


Example 1:

Simplify the rational expression

4x - 2
--------
2x - 1
Solution :-

Factor both the numerator and denominator completely.

2(2x - 1)
---------
2x - 1
Cancel common factors to reduce and then simplify the given expression.

2(2x - 1)
= ------------ = 2 , with x not equal to 1
2x - 1


Example 2:

Simplify the rational expression

4x + 16
---------
x2 - 16
Solution:-

Factor both the numerator and denominator completely.

4x + 16         4(x + 4)
--------- = ----------------
x2  - 16      (x + 4)(x - 4)
Cancel common factors to simplify the expression.

4(x + 4)
= ---------------
(x - 4)(x + 4)


4
= --------    , with x not equal to -4.
(x - 4)

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Example 3:

Simplify the rational expression

x2 + x - 2
-------------
-x2 -2x + 3

Solution:-

Factor both the numerator and denominator completely.

x2 + x - 2      (x + 2)(x - 1)
------------- = ----------------
-x2 -2x + 3      (x + 3)(-x + 1)


Note that - x + 1 = - (x - 1) in the denominator.

(x + 2)(x - 1)
= -----------------
-(x + 3)(x - 1)


Cancel common factors.

(x + 2)(x - 1)
=--------------------
-(x + 3)(x - 1)


(x + 2)
= -------------   , with x not equal to 1
(x + 3)

Learning Compound Interest

Compound interest is paid on the principal and also for the interest accumulated in the past years. Compound interest emerges when interest is added to the principal, so the interest that has been added also earns interest. This adding of interest with the principal is said to be compounding. Compounding can be done for a time like yearly, quarterly, monthly, daily etc.


Learning Compound Interest Formula:


Learning The basic formula for Compound Interest is:

FV = PV (1+r)n

PV is the present value

r is the annual rate of interest (percentage)

n is the number of years the amount is deposit or borrowed for.

FV = Future Value is the amount of money accumulate after n years, including interest.

Learning Frequent Compounding of Interest:

If interest is paid more frequently, Here are a few examples of the formula:

Annually = P × (1 + r) = (annual compounding)

Quarterly = P (1 + r/4)4 = (quarterly compounding)

Monthly = P (1 + r/12)12 = (monthly compounding)

Learning Find the Present Value when know a Future Value, the Interest Rate and number of Periods.

PV = FV / (1+r)n

Learning Find the Interest Rate when know the Present Value, Future Value and number of Periods.

r = ( FV / PV )1/n – 1

Learning Find the number of Periods when know the Present Value, Future Value and Interest Rate

n = ln(FV / PV) / ln(1 + r)

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Example of compound interest


Ex 1: I have $1000.00 to invest for 3 years at rate of 6% compound interest.

Sol:  Here p=$1000, n=3, r=5

A = 1000 (1 + 0.06)3 = $1191.02.

1000.00 is worth $1191.02

Ex 2:  How many years to turns 1000into10,000 at 5% interest?

Sol :    n = ln( 10000 / 1,000 ) / ln( 1 + 0.05 ) = ln(10)/ln(1.05) = 2.3026/0.04879 = 47.19

Ex 3 :  I have $2000.00 to invest for 3 years at rate of 7% compound interest.

Here p=$2000, n=3, r=7

A = 2000 (1 + 0.07)3 = $2450.09.

2000.00 is worth $2450.09.

Angle of Elevation Learning

Introduction:

Assume that you are walking on a road. Normally your eyes will be viewing objects at a horizontal level at a distance of your height from the ground. Suppose you see a tall building in the vicinity and you want to see the top of the building. What you do is, rotate your line of sight from horizon to spot the top of thr building. The angle by which you do that is called the Angle of Elevation.

The angle of elevation and depression are a wonderful concepts in the subject of trigonometry.


Angle of elevation and depression- Definition


Let AB be an object. The angle θ that is required to see the point A from point O is called the angle of elevation of the object at point O.

The angle of elevation of an object depends on the point from where it is measured. The more you are closer to the object, the more is the angle of elevation and vice versa.

Hence, it is a must that the angle of elevation of an object should always be accompanied by the information from where it is measured.

Angle of Depression is the angle by which you tilt your eye down from horizontal line to see an object which is below you.


Angle of elevation – Practical applications

The concept of angle of elevation and depression is used very widely in estimations of heights and depths. In fact, in trigonometry, a separate topic heights and distances deals with angle of elevation and angle of depression. The angle of depression is the angle made below the horizon for objects at lower levels.



Suppose the height of the building shown in the above picture (h) is to be estimated. It may not be practicable to measure that directly. But it is always possible to measure the angle of elevation θ (equipment for such equipment is available) and also the horizontal distance between the building and the point of angle measurement (d).

The basic trigonometric identity, tan θ =`(h)/(d)`, will easily give you the estimate.


Problems based on Angle of elevation and angle of depression


Let us do a  problems relating to angle of elevation and angle of depression to understand the concept properly.

Angle of Elevation Problems :

1)  A balloon is flying high. A boy looks at it. His eyes make an angle of 30 º from the ground to the balloon.

This is angle of elevation. The distance between the boy and  the flying balloon is 10 m. Find the height of the balloon above the Ground.

Here we use sine 30º formula =  opposite side / hypotenuse   = height / hypotenuse

Sine 30º= 1/2     hypotenuse = 10 m

so height = sine 30º x 10  =    `(1)/(2)` x 10m =  5 meter

So the ballon flies at a height of 5 meter above the ground

Problems based on Angle of depression

Problem 2)From the top of a light house, the angle of depression of 2 ships on the opposite side of the light house  are

observed as  45º and 3 angle 0º. The height of the light house is 200 m. Find the distance between the ships.


Let  M be  a ship and N be another ship.   Let AB be the light house

CD is a line drawn horizontally through B.

From B  the angle of depression is 45º and 30º.

CD is parallel to MN

So angle CBM= angle BMA and angle DBN = angle BNA.

That makes angle BMA=45º and angle BNA= 30 degrees(Interior alternate angles  are equal)

In the right triangle BAN ,  tan 30º = AB/AN

Tan 30º= `(1)/(sqrt(3))`        So `(1)/(sqrt(3))`  = `(200)/(AN)`

So AN =  200√3 = 200 x 1.732 = 346.4 m

Now let us measure AM

Tan 45º = 1   Tan 45º = `(AB)/(AM)`  so AM x 1 = AB   AM= AB   AB= 200m so AM = 200 m

Distance between Ship M and the light house is 200m

Distance between ship N and the light house is 346.4 m

Add them both.

Distance between ship M and N is 200+346.4 = 546.4 meters

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Probability Problems with Dice Eighth Grade

Probability is the chance that something will happen - how likely it is that some event will happen.If a trial will produce N commonly exclusive and equally probable outcomes out of which n outcomes are positive to the amount of event A, then the probability of A is mentioned by P(A) and is defined as the ratio n/ N. Thus the probability of A is given by

                                           Event (P)   = Number of Successful outcomes
                                                                   Number of total outcomes

probability problem with dice for 8th grade:

Problem:

A fair die is thrown. Find the probabilities that the face on the die Is 

 (1) Maximum (2) Prime (3) Multiple of 3 (4) Multiple of 7

Solution:
            In the above probability problem there are 6 possible outcomes when a die is tossed. We assumed that all the 6 faces are equally likely. The classical definition of probability is to be applied here
            

The sample space is ,S={1,2,3,4,5,6}  => n(S)=6

(1) Let A be the event that the face is maximum
            Thus,
                       A=(6),        n(A)=1
            Therefore, P(A) = n(A) / n(S)

                                  = 1/6

(2) Let B be the event that the face is prime

            Thus,
                       B=(2,3,5)        n(B)=3   
            Therefore, P(B) = n(B) / n(S)

                                  = 3/6

                                  => 1/2

(3) Let C be the event that the face is multiple of 3
            Thus,          C=(3,6)    n(C)=2
          Therefore, P(C) = n(C) / n(S)

                                = 2/6

                                =>1/3

(4) Let D be the event that the face is multiple of 7
            Thus,          D  =0,           n(D)=0
           Therefore,P(D)  = n(D)/n(S)

                                  =0/6

                                  =0   (not possible)

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Dice problem for 8th grade:

Example:

Solve the problem for probability that when 2 standard 6-sided dice are rolled, the sum of the numbers on the top faces is 4.

Solution:

In the above problem, A dice contains 6 faces, so the possible out comes from the  both dices are      

                            = 6 * 6 
                            = 36 total number of possible outcomes 
Here the given condition is “Some of the numbers on the top of the face is 4”.
So the possible out comes are 
                   Dice 1             Dice 2
                      1         +         3           = 4
                      2         +         2           = 4
                      3         +         1           = 4
the total number of successful outcomes are 3

Final solution 
                          P (sum of 4)     = 3 / 36

                                                 = 1 / 12.

Volume of a Cylinder

Volume of a cylinder is a measurement of the occupied units of a cylinder. The volume of a cylinder is represented by cubic units like cubic centimeter, cubic millimeter and so on. Volume of a cylinder is the number of units used to fill a cube.

A cylinder has two parallel faces in its structure. The height of a cylinder is perpendicular to the two bases. The cylinder has a round base and a perpendicular height.



The volume of a cylinder formula depends on the area of the cylinder. The formula for volume of a cylinder can be written as,

Volume of cylinder = pi * r2 * h.


Formula for Surface Area of a Cylinder


The formula for volume of a cylinder depends on the area of the cylinder. The area of a cylinder can be written as, A = pi * r2. Here, pi is the constant value and r is the radius of the cylinder. To find the volume of the cylinder, we have to multiply the height of the cylinder to the area of the cylinder. The volume of cylinder formula can be written as,

Volume of cylinder = pi * r2 * h.

Here, pi is the constant variable, r is the radius of the cylinder and h is the height of the cylinder. The constant variable pi is equal to (22 / 7) or 3.14.


How to Find The Volume of a Cylinder


Given below are some of the examples to find the volume of a cylinder.

Example 1:

Calculate volume of a cylinder having the radius of cylinder is 3cm and height is 4cm.

Solution:

The given radius of the cylinder is 3cm and height of the cylinder is 4cm

Formula:

Volume of a cylinder = pi * r2 * Height

Volume of cylinder = 3.14 * 32 * 4 cm3

= 113.04 cm3

Answer:

The volume of cylinder is 113. 04 cm3

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Example 2:

The radius and height of the cylinder is 5 cm, 21 cm. Find the volume of cylinder.

Solution:

Given radius r = 5 cm, height = 21 cm.

Formula:

Volume of cylinder = pi * r2 * h

= 3.14 * 52 * 21 cm3

= 1648.5 cm3

Answer:

The volume of given cylinder is 1648.5 cm3

Example 3:

Find the volume of cylinder for the given base and height. base = 6 cm and height = 4 cm

Solution:

Given base = 6 cm and height = 4 cm

Formula:

Volume of cylinder = pi * r2 * h

= 3. 14 * 62 * 4 cm3

= 452.16 cm3

Answer:

The volume of cylinder is 452.16 cm3

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Probability Mass Function Examples

Introduction:

In probability, a probability mass function (pmf) is a function that gives the probability that a discrete random variable and accurately equals some value. Probability mass function examples differ from portable document format defined only for continuous random variables are not probabilities as such examples. The integral over a range of possible values [a, b] gives the probability of the random variable. Some examples for probability mass function are below. I like to share this Probability Set with you all through my article.

Example 1:

At the same time two coins are tossed, what is the probability result of getting (i) accurately one head (ii) no less than one head (iii) more or less one head.

Solution:

The sample space is S = {HH, HT, TH, TT}, n(S) = 4

Let A be the event of getting one head, B be the event of getting at least one

Head and C be the event of getting almost 1head.

∴ A = {HT, TH}, n(A) = 2

B = {HT, TH, HH}, n(B) = 3

C = {HT, TH, TT}, n(C) = 3

(i)               P(A) = n(A) / n(S) = 2/4 = 1/2

(ii)              P(B) =n(B)/n(S) = 3/4

(iii)             P(C) = n(C)/n(S) = 3/4


Example 2:


Two cards have chosen without replacement from 54 cards X measures the number of heart cards drawn y measures the number of clubs drawn. Find probability of joint probability mass function?

Solution:

A deck involves 52 cards.

Sample space = {all ordered pair (p, q)} = {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (2, 0)}

Here, we have to find out the probability at each element of the sample space, and then we have to joint probability mass function.

X (0, 0) = (26 / 52) * (25 / 51) = 650 / 2652

X  (0,1) = (26 / 52) * (13 / 51) + (13 / 52) * (26 / 51) = 676 / 2652

X  (0,2) = (13/52) * (12/51) = 156 / 2652

X (1, 0) = (13/52) * (26/51) = 676/2652

X (1, 1) = (13/52) * (13/51) + (13/52) * (13/51) = 338/2652

X (2, 0) = (13/52) * (12/51) = 156/2652

X (s, t) = 0 if (s, t) is not in the sample space.

Free Calculus Tutoring

Introduction:

British mathematician, Isaac Newton and the German man Gottfried Leibnitz, invented the calculus. Calculus has  two classifications: Differentiation calculus, Integration calculus. Calculus is the study of rates of change, area, or volume. In symbol, we require to find f(x) where, d/dx f(x) = g(x). Integration is one of most important study of calculus in mathematics. We require to find f(x) = ∫ g(x) dx.In the trigonometric calculus the derivative of a constant is zero. There is no exact value for the integral.


Examples for free calculus tutoring:

Differential free calculus tutoring:

Free calculus tutoring problem 1:

Find the first and second derivative of f(x) = 8x4 + 7x3 - 6x2 - 5x +3

Solution;

Steps to solve:

• First differentiate f(x) = 8x4 + 7x3 - 6x2 - 5x +3                                  
• we know, d / dx (x n ) = n xn-1
• when we differentiate the first term we get  (8×4) x (4-1)
• The above equation can above simplified as 32 x 3
• Like wise we can differentiate  and simplify all the terms in the equation
• Finally after the first derivative we will get  32 x 3 + 21 x2 -12x -5
• Now we will go for Second derivative
• Second derivative of  f(x) = 32 x 3 + 21 x2 -12x -5
• Differentiate all terms   (32 × 3)x2 + (21 × 2) x - (12 × 1) x0
• End of the second derivative we will get   96x2 + 42 x - 12

First derivative:

f' = df / dx

= (8 × 4) x (4-1) + (7 × 3)x(3-1) - (6 × 2)x(2-1) - (5 × 1)x (1-1)

= 32 x 3 + 21 x2 – 12x -5

Second derivative:

f '' = df ' / dx

= 32 x 3 + 21 x2 – 12x -5

= 96x2 + 42 x - 12

Answer for the given free calculus tutoring problem is:

`d^2/dx^2 ` (8x4 + 7x3 - 6x2 - 5x +3) = 96x2 + 42 x - 12



Free calculus tutoring problem  2:

Find the first and second derivative of sin 2x.

Solution;

Steps to solve:

• First differentiate f(x) = sin 2x                                                       
• (we know, d/dx (sin x) = cos x)
• when we differentiate sin x  = cos x So, we get  sin 2x = cos 2x
• Then we differentiate 2x  So, we get  2
• The differentiation of sin 2x = 2 cos 2x
• Finally after the first derivative we will get  2 cos 2x
• Now we will go for Second derivative
• we know,  d/dx (cos x) = −sin x
• The second derivative 2 cos 2x = 2 ×2 (- sin 2x)
• So, The second derivative of  sin 2x = -4 sin 2x

First derivative:

f' = df / dx

= `d/dx` (sin 2x)

= 2 cos 2x

Second derivative:

f '' = df ' / dx

= `d /dx` (2 cos 2x)

= -4 sin 2x

Answer for the given free calculus tutoring problem is:

`d^2/dx^2 `(sin 2x) = -4 sin 2x

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Integral free calculus tutoring:


Free calculus tutoring problem 1:

Integrate  `int ` (-2x2 + 8x3 + 2x4 + x5) dx

Solution;

Steps to solve:

• First seperate each term ` int` (- 2x2 + 8x3 + 2x4 + x5) dx
• So, we get `int` (- 2x2 +8x3 + 2x4 + x5) dx =  `int ` - 2x2 dx + ` int ` 8x3 dx + `int` 2x4 dx + `int`   x5  dx
• Integrate first term ` int ` - 2x2 dx =  -2 (1/3) x3                                                                  
•  [ we know,   `int` xn dx = 1/ n+1 (xn+1)]
• Integrate all term and we get   (- 2/3) x3 + (8/4) x4 + (2/5) x5+(1/6) x6
• Finally after the integration we will get  (- 2/3) x3 + 2 x4 + (2/5) x5+(1/6) x6

` int ` (-2x2 + 8x3 + 2x4 + x5) dx =` int` -2x2 dx + `int` 8x3 dx + ` int ` 2x4 dx +` int` x5  dx

= -2 `int ` x2 dx +  8  `int` x3 dx + 2` int`x4 dx + `int` x5  dx

= -2 (1/3) x3 + 8(1/4) x4 + 2(1/5) x5+(1/6) x6

= (-2/3) x3 + (8/4) x4 + (2/5) x5+(1/6) x6

= (-2/3) x3 + 2 x4 + (2/5) x5+(1/6) x6

Answer for the given free calculus tutoring problem is:

` int` (-2x2 + 8x3 + 2x4 + x5) dx = (-2/3) x3 + 2 x4 + (2/5) x5+(1/6) x6

How to solve linear programming problems

Linear Programming is one of the operations research techniques. It is one of the best mathematical techniques for finding the limited use of resources of a concern in a best way. Complex problems can be modeled using linear functions in a presentable way by the management. The linear programming technique is used in solving a wide range of operations management problems.

Definition of linear programming problems:

Linear Programming is defined as a technique which allocates the available resources in an optimum manner for achieving the company’s objective which is for maximizing the overall profit or to minimize the overall cost under conditions of certainty.

Linear Programming can be applied to areas which are given below:

Allocation of resources to various activities of the concern, for example: man power, machine etc.
Production scheduling.
The common characteristics in the above mentioned areas are to allocate limited resources to the activities of the concern.

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How to solve Linear Programming Problems: Mathematical Formulation

Linear Programming can be used in a variety of situations. In most of the business or economic situations, the resources will be limited, the problem there will be to make use of the available resources in such a way as to maximize the production or to maximize the profit or to minimize the expenditure. This can be formulated as linear programming models.

Mathematical Formulation of the problem:

How to solve linear programming problems?? here are the steps which you need to follow:

Step 1:

Write down the decision variables of the problem.

Step 2:

Formulate the objective function to be optimized as a linear function of the decision variables.

Step 3:

Formulate the other conditions of the problem as Linear equations or In equations in terms of the decision variables.

Step 4:

Add the non negativity constraint from the consideration that negative values of the decision variables do not have any valid physical interpretation.

The objective function, the set of constraints, and the non negative constraints together form an LPP.


Steps to solve linear programming problems using Graphical Method:


When a LPP has only two variables in the objective function and constraints, it can be easily solved using the graphical method. The given information of a LPP can be plotted on the graph and the optimal solution can be obtained from the graph.

The steps to solve an Linear Programming Problem using Graphical method is given below:

Step 1:

Identify the decision variables, the objective function and the restrictions for the given Linear Programming Problem (LPP).

Step 2:

Write the Mathematical Formulation of the problem.

Step 3:

Plot the points on the graph representing all the constraints of the problem. Find the feasible region or solution space. The intersection of all the regions represented by the constraints of the problem is called the feasible region and is restricted to the first quadrant only.

Step 4:

The Feasible region obtained in the step 3 may be bounded or un bounded. Determine the Co-ordinates (x, y) values of all the corner points of the feasible region.

Step 5:

Find the value of the objective function at each corner points (solution) determined in step 3.

Step 6:

Select a point from all the corner points that optimizes (Maximizes or Minimizes) the values of the objective function. It gives the Optimum Feasible Solution.

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Some Exceptional Cases of Linear Programming Problem:


There may be an LPP for which no solution exists or for which the only solution obtained is an unbounded one. The exceptional cases arise in the application of graphical method are

• Alternative Optima
• Unbounded Solution
• Infeasible Solution or Non existing Solution

Alternative Optima:

When the objective function is parallel to the binding constraint, the objective function will assume the same optimal value at more than one solution point, because of this reason, they are called as Alternative Optima.

Unbounded Solution:

When the values of the decision variables may be increased in definitely without violating any of the constraints, the feasible region is unbounded. In such cases, the value of the objective function may increase (for maximisation) or decrease (for minimisation) in definitely. Thus, both the solution space and the objective function value are unbounded.

Infeasible Solution:

When the constraints are not satisfied simultaneously, the LPP has no feasible solution. This solution can never occur, if all the constraints are of less than or equal to type.




Example for some exceptional cases:


The general form of the LPP is used to develop the procedure for solving a common programming problem.

A standard LPP Some exceptional cases is of the form
Max (or min) Z = c1x1 + c2x2 + … +cnxn
x1, x2, ....xn these are called decision variable.

Ex: Show graphically that the model

Maximize Z = -5y

Subject to

x+y<span style="font-family: Serif;">?</span> 1

0.5x-5y<span style="font-family: Serif;">?</span> -10

x<span style="font-family: Serif;">?</span> 0

y<span style="font-family: Serif;">?</span> 0 has no feasible solution.

Sol:

Draw the graphs x + y = 1

- 0.5 -5y = - 10

Shade the half planes of the constraints x + y 1 …(1)

-0.5x - 5y -10 …(2)



Points are (0,1)(0,2)(1,0)(20,0)

Note that the origin (0, 0) does not satisfy the in 2nd equation hence the required region is the upper half plane.

From the graph, that the intersection of the constraints is empty. Therefore the given problem has no feasible solution. So, the some exceptional cases of given LPP has no solution.

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slope formula calculator

• If an area of surface tends evenly towards top or down, it is referred as slope.

• The slope of a line is usually denoted by m.

• In other words, slope is the ratio of change in the y coordinates to the change in the x coordinate. The slope is otherwise named as gradient. Slope is equal to rise divided by run.

• In general, mathematical calculators are used to perform mathematical operations .In this article of slope formula calculator, we are going to learn how to find the slope between points using the calculator.

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How to calculate Slope formula:


If the two points (x1, y1), (x2, y2) are given, the slope formula is given by

Slope   m = (y2-y1) / (x2-x1)

Step by step explanation:

The steps necessary for finding slope between two points using calculator are given below:

Step 1:  Enter the x1 and x2 values.

Step 2:  Enter the y1 and y2 values.

Step 3: The slope between two points will be shown in the result area.

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Example calculation using slope formula:


1) Using the slope formula, Find the slope of the line through the points (6, 10) and (9, 11)

Solution:

Given:            x1 = 6     x2 = 9

y1 = 10    y2 = 11

Slope   m = (y2-y1) / (x2-x1)

= (11 -10) / (9 - 6)

= 1/3

2) Find the slope of the line through the points (7, 11) and (12, 14)

Solution:

Given:            x1 = 7     x2 = 12

y1 = 12    y2 = 14

Slope   m = (y2-y1) / (x2-x1)

= (14 -11) / (12 - 7)

= 3/5

3) Find the slope of the line through the points (11, 14) and (16, 23)

Solution:

Given:            x1 = 11     x2 = 16

y1 = 14    y2 = 23

Slope   m = (y2-y1) / (x2-x1)

= (23 -14) / (16 - 11)

= 9/5

4) Find the slope of the line through the points (21, 25) and (31, 35)

Solution:

Given:            x1 = 21     x2 = 31

y1 = 25    y2 = 35

Slope   m = (y2-y1) / (x2-x1)

= (35 -25) / (31 - 21)

= 10/10

= 1.


Practice problems on slope formula calculator:


1) Find the slope of the line through the points (11, 10) and (12, 11)

Answer: m = 1

2) Find the slope of the line through the points (17, 15) and (23, 18)

Answer: m = 1/2

Solving Online Type of Quadrilaterals

Solving online problems is used for learning problems through online that help students to learn easy and transfer knowledge and skills to people through online. Learn online will help kids to study anywhere at anytime.Quadrilaterals are four sided polygons. They are classified by their sides and angles. an important distinction between quadrilaterals is whether or not one or more pairs of sides are parallel. One of the more familiar quadrilaterals is a parallelogram. We see that a square, a rectangle, and a rhombus are all different types of a parallelogram. The quadrilaterals are of 4 types basically but there are some other types that satisfy the properties of quadrilaterals. Let us see about solving online type of quadrilaterals.



Solving Online Type of Quadrilaterals:

Here let us see type of quadrilaterals and its properties,

Trapezoid:

A trapezoid is a quadrilateral that has one pair of parallel sides.



Parallelogram:

A parallelogram is a quadrilateral of  two pairs of parallel sides.



Additional properties:

Opposite sides parallel
Opposite sides equal in measure
Opposite angles equal in measure

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Rectangle:

A rectangle is a parallelogram with four right angles.


Additional properties:

Opposite sides parallel
Opposite sides equal in measure
All angles measure 90°
Diagonals equal in length

Square:

A square is a rectangle with all sides equal.



Additional properties:

Opposite sides parallel
All sides equal in measure
All angles measure 90°
Diagonals equal in length

Rhombus:

A rhombus is a parallelogram with all sides equal.



Additional properties:

Opposite sides parallel
All sides equal in measure
Opposite angles equal in measure

Isosceles Trapezoid:

An isosceles trapezoid is a quadrilateral.


Additional properties:

One pair of parallel sides
Nonparallel sides are equal in length
Solving Online Type of Quadrilaterals:

Practice problems for solving online type of quadrilaterals,

Example 1:

Find the base of a parallelogram if its area is 512 cm2 and altitude is 14 cm.

Solution:

Area = base × height.

512 = base × 14.

b = 512 / 14

= 512 cm.

Base = 36.5 cm.

Example 2:

Find the perimeter of square whose sides are 11 cm.

Solution:

given the side if square is 11cm

Perimeter of the square, P = 4a

= 4 × 11 cm

= 44 cm

Hence the perimeter of square is 44 cm.

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Squeeze Theorem Proof

The proof is a correct demonstration of math statement. It should be true. The proof of one statement is used in other statement proof. The limits of a function are defined by squeeze theorem. We can also refer the squeeze theorem as sandwich theorem or pinching theorem. Now we are going to see about squeeze theorem.

Explanation for Squeeze Theorem Proof

Define squeeze theorem:

In calculus, the squeeze theorem is used and we can analyze the function’s limits by using this theorem. The interval ‘I’ is including the point ‘a’. The functions f, g and h are derived by interval ‘I’. We gets the function as g(x) `lt=` f(x) `lt=` h(x) if we have unequal 'x' values present. In other words, the definition is `lim_(x->a)` g(x) = `lim_(x->a)` h(x) = L. Final result is `lim_(x->a)` f(x). The lower bound g(x) and upper bound h(x) are used to bounds the f(x). We does not including the value 'a' in the interval 'I'.

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More about Squeeze Theorem Proof

The squeeze theorem proof is derived by previous statements and the proof of this theorem is using the special case and general notations.

Using the special cased for squeeze theorem proof:

The special case is g(x) = 0 for all x and L = 0.

First prove the special case as,

Let us take the special case as `lim_(x->a)` h(x) = 0.

Use the fixed positive number ε > 0 and δ > 0 from limits of function.

if 0 < | x-a | < δ then |h(x)| < ε.

Take the terms from above interval and the terms are 0 = g(x) `<=` f(x) `<=` h(x). So | f(x) | `<=` | h(x) |.

The conclusion of theorem is if 0 < | x-a | < δ then | f(x) | `<=` | h(x) | `<` ε.

Final proof of squeeze theorem by above condition is `lim_(x->a)` f(x) = 0 = L.

Use the general notations for squeeze theorem proof:

The g and L are arbitrary and these are used for proof. We have g(x) `<=` f(x) `<=` h(x).

Subtract the g(x) from both sides as 0 `<=` f(x) - g(x) `<=` h(x) - g(x).

Let us take x--> a and g(x) and h(x) as 'L'.

Therefore, h(x) - g(x) --> L - L = 0.

The special case is used in theorem conclusion as f(x) = (f(x) - g(x)) + g(x) --> 0 + L = L.

Hence the sandwich theorem is proved.

Metric Calculator

Metric number line is one of the important topics on the metric number system in mathematics subject. The quantities used to find lengths, capacities, weights of things etc are called measures. Many Countries have their own system of measures. But Metric System of measures is very simple and easy To calculate. Hence most countries in the world use Metric System of measures.

In Metric System,

• Length’s basic unit is metre(m)
• Weight’s basic unit is gram(g)
• Capacity‘s basic unit is litre(l)

And Another name of metric system is decimal system.

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Explanations of the Metric Number Line:

• The metric system is used in all the places of the world, because of its superior basis metric units are linked to each other by factor of 10.

• So while we convert one metric unit to another, we have to move the decimal point in the unique value.

• These metric number lines give us an easy way to do these unit conversions.

Metric units Names and Abbreviations are given below:

• Centi-meter = cm
• Milli-meter = mm
• Kilo-meter = km
• Mega meter = Mm
• Decimeter = dm
• Dekameter = dam
• Hectometer = hm
• Micrometer = mcm
• Meter = m
• Liter = L
• Grams = g
• Volt = v

The above names of units are important. And those all units are different like meter is length, liter is volume, gram is mass or weight.

Metric Number Line Units:

• In number line we can forever multiply the unit by a factor of 10.
• Let us make the number line and use unit as the basic unit.
• Then we can use the metric number line for all units.
• The unit could be decigrams, deciliters, decimeters, decivolts, etc…

Example figure for metric number line:



Ex :  Convert dekagrams (dag) to centigrams (cg) :



The above figure represents the example of converting metric numbers from line, here we have to choose the start point to end with stop point. Here we note two things, there is

1. Direction and,

2. The number of points necessary to move to get the stop point.

Then we have to make the decimal changes into the original number consequently.

For the above number line, we have to move three places to the right direction.



For ex it can be 4.5 dekagram means; it could be changed into 4500 cg. Because of moved three places to right direction.

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These all are the metric number line details.

Metric Calculator - Conversion Tables and Examples:

In decimal system  to concert a higher value into lower value then we need to multiply it by powers of ten and to convert lower value into a higher value then we need to divide the number by ten.

Let us discuss about linear measure  like measures of length,weight,capacity.

Length-Conversion Table:

10mm	1cm
10cm	1dm
10dm	1m
10m	1dam
10dam	1hm
10hm	1km
100cm	1m
1000m	1km

Weight - Conversion Table:

10mg	1cg
10cg	1dg
10dg	1g
10g	1dag
10dag	1hg
10hg	1kg
1000mg	1g
1000g	1kg
100kg	1quintal(q)
1000kg	1 tonne(ton)

Capacity / Volume :- Conversion Table:

10ml	1cl
10cl	1dl
10dl	1l
10l	1dal
10dal	1hl
10hl	1kl

Ex 1 : Convert 4 km into lower units.

Sol:

4 km = 40 hm (4 × 10) = 4 × 101 hm

= 400 dam (4 × 100) = 4 × 102 dam

= 4000 m (4 × 1000) = 4 × 103 m

=40000 dm (4 × 10000) = 4 × 104 dm

= 400000 cm (4 × 100000) = 4 × 105 cm

= 4000000 mm (4 × 1000000) = 4 × 106 mm

Ex 2: Express 1267547 mm into higher units.

Sol:

12675477 mm   = 126754.7cm[1267547/10]  =1267547 x 10-1 cm

=12675.47dm[1267547/100] =1267547 x 10-2 dm

= 1267.547m[1267547/1000] =1267547 x 10-3 m

=126.7547dam[12.6754]/10000  = 1267547 x 104 dam

=12.67547hm[1267547/100000]  =1267547 x 10-5 hm

=1.267547km[1267547/1000000] =1267547 x 10-6 km

Ex 3: Convert 7m into millimeter.

Sol:

We know that ,

1m =1000mm

Therefore ,7m  = 7*1000mm

=7000mm

Ex 4: Express 10kg5dag in grams

Sol:

We know that,

1kg   =1000g

1dag =10g

Therefore, 10kg5dag = 10 *1000g +5 *10g

=10000g +50g

=10050g

Understanding very hard math problems is always challenging for me but thanks to all math help websites to help me out.

Practice problem to help with metric calculation:

1)Convert 6m into millimetre.

Ans: 6000mm

2)Express 2769 g in kilograms.

Ans: 2.769 kg

3)Convert 25 kl 37 l into litres.

Ans: 25037 l

Definition Data Table

Let us see about definition data table. Generally, the data tables are made by the number of rows and the columns. In data table, the rows and columns are separated by the number of lines or line segments. In data table, each row and column has the data. Generally the data table is premeditated by the statistical graphs. The group of data can be creating the data table. Data tables are conniving by the graphs. The statistical graphs are depends on the data table. The data table rows are positioned as horizontal and the columns are positioned as vertical.

Examples of Data Table:

Let us see the example problems of data table.

Example 1:

Define the data table.

Name of pets	Number of students
Dogs	80%
Cats	20%
Fish	64%
Parrot	75%
Peacock	64%
Dove	29%
Sparrow	34%

Solution:

The definition of data table is described below. The data table exhibits the name of pets and the number of students. The data table is defined by the bar graph.  Name of pets is represented as x-axis and the number of students is represented as y-axis.



This is the definition of data table.

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Example 2:

Define the data table.

Activity	Number of students
Visit friends	20%
Talk on phone	15%
Play sports	46%
Earn money	50%
Use computers	61%
Preparing multimedia	31%
Playing cards	21%
Doing home work	11%

Solution:

The definition of the data table is defined below. The data table exhibits the activity and the number of students. The data table is defined by line graph.  Activity is represented as x-axis and the number of students is represented as y-axis.



This is the definition of data table.

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One more Example of Data Table:

Define the sample data table.

Year	Non-Employees rate
2000	12%
2001	45%
2002	31%
2003	61%
2004	55%
2005	72%
2006	81%
2007	74%
2008	94%
2009	75%
2010	95%

Solution:

The definition of the data table is described below. The data table exhibits the year and the non-employees rate. The data table is defined by the scatter plot graph.  The year is represented as x-axis and the non-employees rate is represented as y-axis.



This is the definition of data table.

Trigonometry Xi

In this section we will see about trigonometry xi. Eleventh average trigonometry is also recognized as the division of the main dealing with trigonometry functions, angle, etc. It gives the association and angles in detail with their problems. Sine, Cosine and Tangent are the trigonometric meaning concerned in trigonometric semi position identity. We have worked problems and practice problems along with solution in below. Let us see about the topic trigonometry xi.

Solved Problems for Trigonometry Xi:

Let us see about the topic trigonometry xi,

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Solved problem 1: Try to calculate the radius of the circle in which a central angle of 45 degree intercepts an arc of length 10 cm. (use π value as `22/7` )

Solution:

Given length = 10 cm and angle = 45 degree

θ = 45 degree = `(45Pi) / 180` = π/4

r = l/ θ

r = `(10 * 4) /Pi` = `(10 * 4 *7) / 22` = 12.72 cm

Therefore, the radius of the circle = 12.72 cm.

Answer: The radius of the circle = 12.72 cm.

Solved problem 2: If cos x = `1/5` , x lies in the first quadrant. Carry out step of the values of other five trigonometric functions.

Solution:

Given cos x = `1/5` , therefore, sec x = 5

We know that,

sin2 x + cos2 x = 1, that is sin2 x = 1 – cos2 x

sin2 x = 1 - `1/5` =`4/5`

sin x = ± `2/(sqrt 5)` (take square root on both sides)

x lies in 1st quadrant, sin x is negative.

Therefore, sin x = `2/(sqrt 5)` which also gives

cosec x = `2/(sqrt 5)`

Further, we have tan x = `(sin x)/(cos x)` = `(5(sqrt 5))/2` and cot x = `(cos x) /(sin x)` = `(2)/(5(sqrt 5))`

Answer: sec x = 5, sin x = `2/(sqrt 5),` cosec x = `2/(sqrt 5)` , tan x = `(5(sqrt 5))/2` , cot x =`2/(5(sqrt 5))`



Practice Problems for Trigonometry Xi:

Let us see about the topic trigonometry xi,

Practice problem 1: Determine the value of cos (370°).

Practice problem 2: Find the value of sin 5pi/3.

Solutions for prepare for trigonometry xi:

Solution 1: The value of cos (370°) is 0.98.

Solution 2: The value of sin 5pi/3 is -0.87.

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Solving Online Calculus Optimizing Problems

Study of rate of transformation is called calculus. Study of optimization calculus or mathematical encoding is disturbed to influence the excellent element from the group of elements. Optimization is a single technique to obtain a maximum or minimum value of a function. The smaller value of the function is known as minimum. The greater value of the function is known as maximum.

Online has emerged as one of the main key source for students to increase their knowledge topic wise.

Examples to Solving Online Calculus for Optimizing Problems:

Solving online calculus for optimizing example problems 1:

`y = 3x^2 - 5x` , solving for x and y for the optimizing calculus problems.

Solution:

Step 1: Equation is` y = 3x^2- 5x`

Step 2: Differentiate with respect to x

`dy / dx` = 6x - 5

Equate `dy / dx ` to 0.

`dy / d` x = 6x - 5 = 0

6x = 5

x = `(5) / 6` or   0.83

Step 3: Plug x = 0.83 in the given equation

y = `3 (0.83) ^2- 5(0.83)`

= 3(0.6889) - (4.15)

= 2.06 - 4.15

= -2.09

Therefore, x = 0.83 and y = -2.09

Step 4: From the given equation plot the graph and mark out the points in the graph.





Solving Online Calculus for Optimizing Example Problems 2:

`y = 5x^2 - 19x,` solving for x and y for the optimizing calculus problems.

Solution:

Step 1: Equation is `y = 5x^2- 19x`

Step 2: Differentiate with respect to x

`dy / dx ` = 5x - 19

Equate `dy / dx` to 0.

` dy / dx` = 10x - 19 = 0

10x = 19

x = `(19)/10`  or 1.9

Step 3: Plug x = 1.9 in the equation

y = `5(1.9) ^2- 19(1.9)`

= 5(3.61) - 36.1

= 18.05 - 18.05

= -18.05

Therefore, x = 1.9 and y = -18.05

Step 4: From the given equation plot the graph and mark out the points in the graph.



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Solving Online Calculus for Optimizing Example Problems 3:

`y = 4x^2- 7` , solving for x and y for the optimizing calculus problems.

Solution:

Step 1: The given equation is `y = 4x^2- 7`

Step 2: Differentiate with respect to x

`dy / dx ` = 8x

Step 3: Equate `dy / dx` = 0

8x = 0

x = 0

Step 4: Thus, `y = 4(0) ^2-7`

y = -7

So, x = 0 and y = -7.

Step 4: From the given equation plot the graph and mark out the points in the graph.

Same Perimeter Different Area

Perimeter is nothing but the path around the shape. And area is nothing but the space occupied by the 2 dimensional object. Here we are going to deal with the same perimeter and different area of the shape. Every shape is having different formulas for area. And for all shape if we want to find the perimeter we have to add the length of all sides. We will see some example problems for same perimeter and different area.

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Example Problems for same Perimeter and Different Area:

Example 1 for same perimeter and different area:



Find the area and perimeter of the following shapes and compare the area and perimeter.

Solution:

Shape 1:

The first shape is a rectangle. We know the area of the rectangle is l X w

Here length l = 3 cm and the width w = 4 cm

So the area of the rectangle = 3 X 4 = 12 cm2

Perimeter of the rectangle is = 2 (l + w) = 2 (3 + 4) = 2 X 7 =14 cm.

Shape 2:

The shape 2 is a triangle. To find the area of the triangle we have to use the following formula.

Area of the triangle = (`1 / 2` ) bh.

From the above b = 7 cm and h = 2 cm.

So the area =` (1 / 2) xx 7 xx 2 ` = 7 cm2

Perimeter of the triangle   = sum of all sides = 7 + 4 +3 = 14 cm.

From the above bath shape is having the same perimeter and different area.

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Example 2 for same Perimeter and Different Area:

Find the area and perimeter of the following shapes and compare the area and perimeter.



Solution:

Shape 1:

The first shape is a trapezoid. We know the area of the trapezoid is `(1/2) h (a + b)`

Here a and b are the base lengths. A = 6 cm. b = 4 cm and the height is 5 cm.

So the area of the rectangle = `(1/2) 5 (6 + 4) ` = 25 cm2

Perimeter of the rectangle is = sum of all sides = 6 + 5 + 4 + 3 = 18 cm.

Shape 2:

The second shape is a rectangle. We know the area of the rectangle is l X w

Here length l = 6 cm and the width w = 3 cm

So the area of the rectangle = 6 `xx` 3 = 18 cm2

Perimeter of the rectangle is = 2 (l + w) = 2 (6 + 3) = 2 X 9 =18 cm.

From the above bath shape is having the same perimeter and different area.

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Pictures of Division Sets

In this article we shall discuss  pictures of division sets. Here, division is also meant by fraction of a whole. A division can be creation over to a decimal through dividing the upper digit, or numerator, during the lower digit, or denominator. Division is instead of as ratios, and significance for fraction which is one of the important math processes. Thus the division `3/5` is also used to point out the ratio 3:5 and the division 3 ÷ 5 as well.

How to do Pictures of Division Sets:

The pictures of division sets are shown given below that,



 
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Example Problems Based on Learn Pictures of Division Sets:

The example problems based on learn about pictures of division sets are given below that,

Example 1:

How to learn about pictures of division for 277 divide by 8 sets?

Solution:

Step 1:

The given value is 277 divide by 8 sets.

Step 2:

Here, 277 divide by 8 sets is also denoted as 277/8.

Step 3:

Now, 277 divide by 8 sets is explain about using long division procedure pictures.

Here, using long division procedure pictures are shown given below that,



Step 4:

The final answer for pictures of division sets is 34.625.

Example 2:

How to learn about pictures of division for 197 divide by 6 sets?

Solution:

Step 1:

The given value is 197 divide by 6 sets.

Step 2:

Here, 197 divide by 6 sets is also denoted as 197/6.

Step 3:

Now, 197 divide by 6 sets is explain about using long division procedure pictures.

Here, using long division procedure pictures are shown given below that,



Step 4:

The final answer for pictures of division sets is 32.833







Understanding long division with decimals is always challenging for me but thanks to all math help websites to help me out.

Practice problems based on learn about pictures of division sets:

The practice problems based on learn about pictures of division sets are given below that,

Problem 1:

How to learn about pictures of division for 163 divide by 4 sets?

Answer: The final answer for pictures of division sets is 40.75

Problem 2:

How to learn about pictures of division for 52 divide by 4 sets?

Answer: The final answer for pictures of division sets is 13.