Write a Mixed Decimal

Mixed number: Mixed number is formed by associating a whole number with fraction numbers.

Decimal number: A number along decimal a point in it is called decimal number.

Mixed decimal number:  A number having a decimal point in its mixed format is mixed decimal number.

Example: 7.5 `3/6`.  Let us see how to write mixed decimal number.


Writing a mixed decimal:

Write a mixed decimal:

In the whole number part of mixed number the decimal point appears show that the number is mixed decimal.

It should be written as,

133.63 `1/5`

Conversion of mixed decimal into fraction:

Step 1: First see how many numbers present after the decimal point.

Step 2: Calculate number of 10’s equal to the counted digits after the decimal point.

Step 3: Put number of 10’s as denominator and the number is mixed decimal number.

Step 4: Convert the numerator mixed decimal number into fraction.

Step 5: Now simplify the terms by a common term.

Example problems for writing a mixed decimal:

Example: 1

Write the following mixed decimal number into fraction number.

12.5 2/3

Solution:

Given, mixed decimal number =12.5 2/3.

We have to add the 10 as the denominator and the mixed decimal number as numerator.

That is, `(125 2/3)/10`

= `(377/3)/10`

= `377/30`

=12`17/30`

Example: 2

Write the following mixed decimal number into fraction number.

5.22 1/2

Solution:

Given, mixed decimal number =5.22 1/2.

We have to add the 100 as the denominator and the mixed decimal number as numerator.

That is, `(522 1/2)/100`

= `(1045/2)/100`

= `1045/200`

=`209/20`

=1`9/20` .

Example: 3

Write the following mixed decimal number into fraction number.

15.123 1/5

Solution:

Given, mixed decimal number =15.123 1/5.

We have to add the 1000 as the denominator and the mixed decimal number as numerator.

That is, `(15123 1/5)/1000`

= `(75616/5)/1000`

= `75616/5000`

= 25 `616/5000`

= 25 `77/625` .

Answer: 25 `77/625` .

Practice problems for write a mixed decimal number:

Problem: 1

Write the following mixed decimal into mixed number.

5.9 3/4

Answer: 5`39/40`.

Problem: 2

Write the following mixed decimal into mixed number.

5.633 1/2 .

Answer: 5`1267/2000`.

Hypothesis Testing Variance

Hypothesis testing is the use of statistics  found in the probability that a specified hypothesis is correct. Hypothesis is specified as declaration which may or may not be accurate. In statistics two hypothesis testing are used. They are null hypothesis and alternative hypothesis. These two hypothesis tested are opposed to each other. In statistics the significance level is symbolized through alpha. Let us see about the probability of hypothesis testing variance.

Hypothesis testing variance

There are five constituent to either statistical test:

Null Hypothesis

Alternate Hypothesis

Test Statistic

Level of significance

Conclusion

Hypothesis testing variance

Consider the population is standard, we are able to test the variance of the method using the chi-square distribution through (n – 1) degrees of freedom.

To test a variance or standard deviation of a population to be exact normally distributed, we can utilize the χ2–test.

The χ2- test for a variance or standard deviation is not as robust as the samples for the population mean otherwise the population proportion.

Therefore it is necessary to while performing the χ2–test used for a variance that the population is usually circulated. The results can be deceptive if the population is not standard.

Examples for hypothesis testing variance

χ2- test for population variance

In a sample of size 16 drawn from a normal population standard deviation is 4 can you say that population standard deviation is 5.

Solution

Null hypothesis:

H0: The population standard deviation is 5

Test statistic:

χ2   =`(ns^2)/sigma^2` ~ χ2n-1

Level of significance:

α= 0.05 at 5% level of χ2  table values for 16 degrees of freedom is 24.996

Calculation:

n = 16, s = 4, σ = 5

χ2   =`(16xx16)/25 = 256/25`

χ2   = 10.24

Calculated value = 10.24

Table value = 24.996

Calculated value < Table value

Therefore the null hypothesis is accepted.

The population standard deviation is 5

Result

The population standard deviation is 5

Whole Set Fraction

This article is about whole set fraction. Whole set fraction is nothing but it is about the set of fractions. A fraction is a number that can be represented by an ordered pair of whole numbers `a/b` where `b!= 0`. Here a is represented as numerator and b as denominator. The tutors in tutor vista are always ready to help the students in any topics like whole set fraction. The online tutors help students in online. Tutor vista is the best tutoring website where many students follow this. Below we can see about the whole set of fraction.

whole set fractions

In set notation, the set of fractions is

F ={ `a/b` where a and b are whole numbers,  b` !=` 0 }

Two fractions that represent the same relative amount are termed to be equivalent fractions.

Proper Fraction:

When the numerator is less than the denominator then those fractions are called as Proper fraction.

Example: `2/3 `

Improper fraction:

When the numerator is greater than the denominator then this fraction is called as Improper fraction

Example: `7/5`

All the integers are simply a improper fraction

Example

3 is nothing but `3 / 1` which is an improper fraction

Mixed fraction:

Mixed fraction is a whole number with proper fraction

Example:  2 `1/3`

whole set fractions

Example 1:

Add `5/3`  + ` 8/3`

Here both the denominator is same

Add numerator alone.

`(5+8)/3`

`13/3`

Adding improper fraction with different denominator

Example 2:

Add `5/2`  + `4/3`

Here find LCM and solve to make the denominator equal

LCM of 2 and 3 is 6

`(5xx3)/ (2xx3) = 15/6`

`(4xx2)/ (3xx2) = 8/6`

So `(15+8)/6 = 23/6`

whole set fractions

Example 3: Add `3/11` +`6/11`

Solution

Here the denominators are same. So just add the numerator alone

` (3+6)/11`

`9/11` is the answer.

Example 4: Subtract `8/9` – `4/9`

Solution

Here the denominators are same. So just subtract the numerator alone.

`(8 - 4)/9`

`4/9`

Example 5: Multiply` 3/5` x `3/ 4 `

Solution

`3 / 5 ` x `3 / 4`

`(3xx3)/(5xx4) `

= `9 / 20 `

Addition Angle Formulas

Addition angle formula is based on trigonometric functions. We are having  the addition angle formulas to find the value of  the trigonometric equations and values for the trigonometric angles.
                         Cos (A+B) = Cos A Cos B - Sin A Sin B
                         Cos (A-B) = Cos A Cos B + Sin A Sin B
                         Sin (A+B) = Sin A Cos B + Cos A Sin B
                         Sin (A-B) = Sin A Cos B - Cos A Sin B
                         Tan (A+B) = `(Tan A + Tan B)/(1 - Tan A Tan B)`
                         Tan (A-B) =  `(Tan A - Tan B)/(1 + Tan A Tan B)`
                         Here we will some problems based on addition angle formulas.

Addition Angle Problems:

Problem 1:
         Solve the following trigonometric function using Addition angle formula Sin 75^o
Solution:
            Sin 75^o
                 We can write sin 75^o as Sin (45^o + 30^o)
                              We have the formula for Sin (A+B) = Sin A Cos B + Cos A Sin B
                              Where A = 45^o and  B = 30^o
                              Sin 45^o = Cos 45 = `(1)/(sqrt(2))`
                              Sin 30 = `(1)/(2)`    Cos 30 = `(sqrt(3))/(2)`
                              Sin (45^o + 30^o) = Sin 45^o Cos 30^o + Cos 45^o Sin 30^o
                                                          =  `(1)/(sqrt(2))` `(sqrt(3))/(2)` `(1)/(sqrt(2))` `(1)/(2)`
                                                          = `(sqrt(3))/(2sqrt(2))` + `(1)/(2sqrt(2))`
                                                          = `(sqrt(3)+1)/(2sqrt(2))`
Problem 2:
       Solve the following trigonometric function using Addition angle formula Cos 135^o
Solution:
               Cos 135^o
                We can write Cos 135^o as Sin (90^o + 45^o)
                              We have the formula for Cos (A+B) = Cos A Cos B - Sin A Sin B
                              Where A = 90^o and  B = 45^o
                              Sin 45^o = Cos 45 = `(1)/(sqrt(2))`
                              Sin 90^o = 1  Cos 90^o = 0
                              Cos (90^o + 45^o) = Cos 90^o Cos 45^o - Sin 90^o Sin 45^o
                                                      = 0 . `(1)/(sqrt(2))` - 1 . `(1)/(sqrt(2))`
                                                      = 0 - `(1)/(sqrt(2))`
                                                      = - `(1)/(sqrt(2))`

Problem 3:

       Solve the following trigonometric function using Addition angle formula Tan 135^o
Solution:
               Tan 135^o
               We can write Tan 135^o as Tan (180^o - 45^o)
                              We have the formula Tan (A - B) = `(Tan A - Tan B)/(1 + Tan A Tan B)`
                              Where A = 180^o and  B = 45^o
                              Tan 45^o = 1 Tan 180^o = 0
                              Tan 135^o = `(Tan 180^o - Tan 45^o)/(1 + Tan 180^o Tan 45^o)`
                              Tan 135^o = `"(0 - 1)/(1 + (0) (1)) `
                              Tan 135^o = `(- 1)/(1)`
                              Tan 135^o = -1

Compute Percentages

In mathematics, a percentage is a way of expressing a number as a fraction of 100 (per cent meaning "per hundred" in French). It is often denoted using the percent sign, "%", or the abbreviation "pct". For example, 45% (read as "forty-five percent") is equal to 45 / 100, or 0.45. Let us see how to do percentages.                                                                                                                                           - Source from Wikipedia

How to compute percentages:

STEP 1: While begin the percentage` x / 100 ` = `(is) / (of)` . Out of hundred x may be the percentage, "is" denotes as fraction, and "of" denotes as whole.
STEP 2: If we are having the questions 80 : 40 percentage then we will write it as X = 40, is = 40 ("80 is"), and then of = as unknown value. Now the value is possible to write like this `40 / 100 ` = `80 / x` .
STEP 3: Let us do the cross multiplication. Now we can get a Constant value on 1 side and then multiply it with another side. Now we will get a result likes this 40x = 8,000.
STEP 4: Now you have to find out the x value. Where, x = `8000 / 40` = 200, now the x value will be 200.

How do you solve percentages some examples here:

Problem 1:
           In a question paper there is 80 questions. Laura took that test. If she gets 75% correct, how many questions did Laura missed?
Solution:
            Therefore total correct answers are 75% of 80 or else `75 / 100` × 80
            ` 75 / 100` × 80 = 60%
            So the question paper contains 80 questions and Laura got 60 exact answers, the number of questions Laura left is 80 − 60 = 20.
            Therefore Laura missed 20 questions.
Problem 2:
            Compute this, what is 85% of 15?
Solution:
Step 1: Compute the percent.
           The percent value is 85.
            P = 85
Step 2: Find out the base.
           The base is the number next the word OF, 15
            b = 15
Step 3: Identify the quantity.
           The calculation is the unknown.
            a =?
Step 4: Enter the value in the percent proportion formula.
         `a/ 15` = `85 / 100`
Step 5: Exercises the equation for the unknown.
           The Least Common Divisible of 15 and 100 is 100
     `100 / 1` * `a / 15` = `100 / 1` * `85 / 100`
            6.6a = 85
          ` (6.6a) / (6.6)` = `85 / (6.6)`
            a = 12.8
            12.8 is 85% of 15.


Practice problem for compute percentages:
Problem 1:
            What is the percentage of 67%?
Solution:
             = 0. 67.
Problem 2:
            What is percentage of 87% of 18?
Solution:
             = 15.66
            Therefore 15.66 is 87% of 18.

Taylor Polynomial Series

In mathematics, the Taylor series is a representation of a function as an infinite sum of terms calculated from the values of its derivatives at a single point. It is named after the English mathematician Brook Taylor. If the series is centered at zero, the series is also called a Maclaurin series, named after the Scottish mathematician Colin Maclaurin. It is common practice to use a finite number of terms of the series to approximate a function. The Taylor series may be regarded as the limit of the Taylor polynomials.  In the article we shall discuss about Taylor polynomial series.(Source: Wikipedia)

Taylor polynomial series:

Find the Taylor series for sin x.

    Sin x = `sum_(0)^oo` `((-1)^k)/((2k + 1)!)` x^2k+1

The remainder term is not expressible in any simple way but can be estimated by using the Lagrange's form of the remainder. The coefficients

   `((-1)^k)/((2k + 1)!)`

are easily verified by calculating successive derivatives of f(x) = sin x and using the formulas

 a_k=`(f^(k)(0))/(k!)`

To check convergence of the series, apply Lagrange's form for R_a(x); For each x`in` R. there exists Z such that

R_n(x) = `(f^(n + 1)(z))/((n + 1)!)` xⁿ⁺¹

Now |fⁿ⁺¹(z)| equals either |cos z| or |sin z| So, in either case,|fⁿ⁺¹ (z)|`<=`1 ,and

       |R_n(x) |`<=` | x |ⁿ⁺¹ /(n+1)!

Since | x |ⁿ⁺¹ /(n+1)!`->`0 as n `|->` `oo`  for all x`in` R, we can see that the remainder term |R_n(x)|`|->` 0 as n `|->` `oo`

 for all x`in` R. Thus the series representation is completely justified for all real x.

  Observe that our estimate for |R_n (x)|,

                                                    |R_n (x)|`<=`|x|^_n+1/(n+1)!

gives also a sense of the rate of convergence of the series for fixed x, for example, for | x|`<=` 1, we find

                                                    |R_n (x)|`<=`1/(n+1)!

Thus, if we want to calculate sin x on (-1, 1) to within .01, we need take only the first five terms of the series (n = 4) to achieve that degree of accuracy.

 Had we used the integral form for R_n (x) we would have obtained a similar estimate.

Sample problem for Taylor polynomial series:

Pro:  Evaluate the definite integral `int_0^1` Sin (x) dx

Sol:  The integrand has no anti derivative expressible in the terms of familiar functions. Howebver, we know how to find its Taylor series. we know that

      Sin t = t -`(t^3)/(3!)` + `(t^5)/(5!)` - `(t^7)/(7!)` + ----

Now if we substitute t = x, we have

     Sin (x) = X - `(x^3)/(3!)` + `(x^5)/(5!)` - `(x^7)/(7!)` + ----

In spite of the fact that we cannot antidifferentiate the function, we can antidifferentiate the Taylor series:

    `int_0^1` Sin (x) dx = `int_0^1` (X - `(x^3)/(3!)` + `(x^5)/(5!)` - `(x^7)/(7!)` + ----) dx

                            =(`(x^2)/(2)` -`(x^)/(4*3!)` + `(x^6)/(6*5!)`-`(x ^ 7)/(8*7!)`+ -----) |₀¹

                            = (`(x^2)/(2)` -`(x^)/(4*3!)` + `(x^6)/(6*5!)`+`(x^7)/(8*7)` + ---)

Notice that this is an alternating series so we know that it converges. if we add up the first four terms, the pattern becomes ckear: the series converges to 0.2871

Negative Integer Exponents

The exponents are which integer is placed in the power, of base numbers. It can be easily represent as, “a small number to the right side and above of base number”. It is called as exponents. These exponents have some of important rules and laws. Power with negative integer exponents is also one of the rules. Here we are going to explain about this negative integer exponent rule.

If we are having variables, which is containing the exponents and it have equal bases means, we can do some mathematical operations with the exponents. Those operations are called as the “laws of exponents” or “rules of exponents”. In this rule based negative integer rule of exponent is defined as following ways,

Definition for negative integer exponents:

It is otherwise called as power with negative exponent rules. This negative exponent rule is defined as, if m is a positive integer and x is a non-zero rational number, then it can be denoted as,

X-m = (1/x)^m (or)

= (1/x)^m

Which is (x)^-m is the reciprocal of (x)^m

And we adopt the same rule for rational exponents also. If p/q is a positive rational number means, and x>0 is a rational number, then

 X^ - (p/q) = (1/x)^ (p/q) = (1/x)^ (p/q) .

Which is, (x)^-(p/q) is the reciprocal of (x) ^(p/q) or the number obtained by raising the reciprocal of x to the exponent p/q.

For example: 1). 3^-2

= (1/3)^2

= (1/3)^-2

= -6 .

2). (4)^-(2/3)

= (1/4)^-(2/3)

= (1/4)^ (2/3)

This kind of exponentiation used for discovers the negative integer exponents and simplify the problems.


Example problems for negative integer exponents:

1) Solve: (8)^-(2/3)

Solution:

Given: (8)^-(2/3)

= (1/8)^ (2/3)

= [(1/8)^ (1/3)]^ 2

= (1/2)^2, since (1/2)^3 = 1/8

= 1/4 .

2) Solve: (32/243)^-(4/5)

Solution:

Given: (32/243)^-(4/5)

= (243/32)^(4/5)

= [(243/32)^(1/5)]^4

= [(3^5/2^5)^(1/5)]^4

= [((3/2)^5)^(1/5)]^4

= (3/2)^4

= 81/16.

3) Evaluate, and find the following negative integer exponent value:

Evaluate: (27/125)^-(2/3)  xx    (27/125)^-(4/3)

Solution:

 (27/125)^- (2/3) xx (27/125)^-(4/3)

= (125/25)^(2/3) xx (125/27)^(4/3)

= [(5^3/3^3)^(1/3)]^ 2 xx [(5^3/3^3)^(1/3)]^4

= [((5/3)^3)^(1/3)]^2 xx [((5/3)^3)^(1/3)]^4

= (5/3)^2 xx (5/3)^4

= (5/3)^6

= 15625/729.

These all are the explanations and example problems may clear about the negative integer exponents.

Math Absolute Value Inequalities

In math, the absolute value |x| of a real number x is x's arithmetical value lacking view to its symbol. So, for example, 86 is the absolute value of both 86 and −86. Generalization of the absolute value for real numbers occurs in a extensive diversity of math settings. Consider an absolute value is also definite for the complex numbers, the quaternions, prepared rings, fields and vector spaces. The absolute value is strictly associated to the ideas of magnitude, distance, and norm in different math and physical contexts.

Properties of the Math absolute value inequalities

The absolute value fundamental properties are:

 |x| = sqrt(x^2)                                      (1) Basic

 |x| \ge 0                                            (2)     Non-negativity

 |x| = 0 \iff x = 0                            (3)     Positive-definiteness

 |xy| = |x||y|\,                                  (4)     Multiplicativeness

 |x+y| \le |x| + |y|                            (5)     Subadditivity

Another important property of the absolute value includes these are the:

 |-x| = |x|\,                                      (6)     Symmetry

|x - y| = 0 \iff x = y                     (7)     Identity of indiscernible (equivalent to positive-definiteness)

|x - y| \le |x - z| +|z - y|                 (8)     Triangle inequality (equivalent to sub additivity)

|x/y| = |x| / |y| \mbox{ (if } y \ne 0) \,                   (9)      Preservation of division (equivalent to multiplicativeness)

|x-y| \ge ||x| - |y||                           (10)     (Equivalent to sub additivity)

Math absolute value inequalities – Examples:

Math absolute value inequalities – Example 1:

|x - 5| < 3

Set up the two times inequality -3 < x -5 < 3 and then solve.

-3 < x – 5 < 3

2 < x < 8

In interval notation, the answer is (2, 8).
Math absolute value inequalities – Example 2:

|2x + 3| <=-8

This solution will involve setting up two separate inequalities and solving each.

2x + 3 <= -8

2x <=-11

x<=-11/2

Else

2x+3 >=8

2x >= 5

x>=5/2

In interval notation, the answer is (-oo, -11/2) U (5/2, oo) .
Math absolute value inequalities – Example 3:

|x - 9| < 4

Set up the two times inequality -4 < x -9 < 4 and then solve.

-4 < x – 9 < 4

5 < x < 13

In interval notation, the answer is (5, 13).

Math absolute value inequalities – Example 4:

|3x + 3| <= - 8

This solution will involve setting up two separate inequalities and solving each.

3x + 3 <= -8

 3x <=-11

x<=-11/3

Else

3x+3 >=8

 3x >= 5

x>=5/3

In interval notation, the answer is (-oo, -11/3) U (5/3, oo).

Negative Number Calculator

In this article we are discussing about subtracting negative number by using calculator. Negative number also real number. The negative number is represented as minus symbol “- “. Negative number or elements are less than zero such as -4, -7, -/3. A negative number may be parenthesized with its symbol, For example a subtracting is clearer if written (-7) + (−5) = -13 Using calculator we can the negative number.
Negative number calculator:

Let us see how to negative number using calculator.

Negative number is the similar as subtracting the corresponding negative number:

Example: (-7) + (-7) = -14

Negative number calculator – Example problems:

Example 1:

(-3) + (-4) = ?

Solution:

First enter the negative number in the calculator.

Then press the equal to button.

The solution is displayed on the calculator.

negative number calculator

The subtract solution is -7.

Example 2:

(-5) - (-4) = ?

Solution:

First enter the negative number in the calculator.

Then press the equal to button.

The solution is displayed on the calculator.

negative number calculator

The subtract solution is -1.

Example 3:

(-4) x (-4) = ?

Solution:

First enter the negative number in the calculator.

Then press the equal to button.

The solution is displayed on the calculator.

negative number calculator

The subtract solution is 16.

Example 4:

(-9) `-:` (-3) = ?

Solution:

First enter the negative number in the calculator.

Then press the equal to button.

The solution is displayed on the calculator.

negative number calculator

The subtract solution is 3.

Example 5:

(-10) x (-20) = ?

Solution:

First enter the negative number in the calculator.

Then press the equal to button.

The solution is displayed on the calculator.

negative number calculator

The subtract solution is 200.

Negative number calculator – practice problems:

Problem 1: (-5) + (-6)

Problem 2: (-4) – (-2)

Problem 3: (-9) x (-18)

Problem 4: (-5) `-:` (-20)

Negative number calculator – answer key:

Problem 1: -11

Problem 2: -2

Problem 3: 162

Problem 4: -0.25

Solving Monotonicity and Concavity

A function is said to be rising on an interval [a, b] = I if f (z1) < f (z2) whenever z1 < z2 for z1, z2 `in` I. A function is said to be decreasing on an interval [a, b] = I if f (z1) > f (z2) whenever z1 < z2 for z1, z2 `in` I. A function which is strictly increasing or strictly decreasing on an interval is said to be monotonic on that interval. A function is said to be curved in up on an interval [a, b] if f ’’ (z) > 0 on [a, b]. A function is said to be curved in down on an interval [a, b] if f ’’ (z) < 0 on [a, b]. A point (c, f(c)) on the graph of y = f (z) is called an inflection point if the concavity changes at the point. (I.E., it is curved in up on some interval (a, c) and curved in down on some interval (c, b) or vice versa.)

Solving monotonicity and concavity - Examples:

Solving monotonicity and concavity - Example 1:

Find inflection points & determine concavity for f (x)

f ‘(x) = `(x^5)/20 + (x^4)/12 - (3x^3)/3 - 10`

f ’(x) = `(x^4)/4 + (x^3)/3 - 3(x^2)`

f ‘’ (x) = x3 + x2 – 3x

= x(x2 + x - 3)

f ‘‘(x) = 0

x = 0, -2.3, 1.3

Inflection pts: x= -2.3, 0, 1.3

Curved in up: (-2.3,0), (1.3,infinity)

Curved in down: (-oo,-2.3), (0,1.3)

Solving monotonicity and concavity

Solving monotonicity and concavity - More Examples:

Solving monotonicity and concavity - Example 1:

Find inflection points & determine concavity for f (x)

f (x) = `(x^5)/20 + (x^4)/12 - (x^3)/3 +10`

f ‘ (x) = `(x^4)/4 + (x^3)/3 - x^2`

f ‘’ (x) = x3 + x2 – 2x = x(x2 + x - 2)

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

f ‘‘ (x) = 0

x = 0, -2, 1

f ’’(-5) < 0, f’’(-1) > 0, f ’’(.5) < 0, f ’’(10) > 0

Inflection pts: x=-2,0,1

Curved in up: (-2,0), (1,`oo` )

Curved in down: (`-oo` .,-2), (0,1)

Solving monotonicity and concavity

Trinomial Multiplication

In algebraic expression consisting of only one term is called a monomial, two terms is called a binomial and when there are three terms separated by an addition or a subtraction operation is called a trinomial. ‘Tri’ in the word trinomial means three and hence the name. It is also referred to as a polynomial as ‘poly’ means more than two terms. For example, a+3b-c consists of three terms a, b and c and hence a trinomial. In 3b, 3 is called the coefficient of b. In a given polynomial we come across like terms and unlike terms.

Like terms are the terms which have the same variable or literal but a different coefficient. For example, (3b, -7b);(5x2, 12x2);(-4xy, 7xy) are some of the like terms. Unlike terms as the name suggests are the terms which have different variables. For example, (7x, 8y); (4ab, -4ac); (2y2, -2x2) etc are some of the unlike terms.

It is not necessary that there would be only two like or unlike terms, it depends on the number of terms in the given polynomial. Now let us learn as to How do you multiply trinomials. There are two methods in which we can multiply trinomials, one is the horizontal method the other is the vertical method.

The steps to be followed in a Horizontal method of multiplication are as follows:
Example: Multiply (4x2-3x+5)(x2+5x-3)
Solution: First and second terms are chosen irrespective of the order
Let (4x2-3x+5) be the first term and (x2+5x-3) be the second term

Arrange the two polynomials horizontally
(4x2-3x+5) X (x2+5x-3)
Now distribute each of the terms of the first trinomial with each of the terms of the second trinomial
=4x2(x2+5x-3)- 3x(x2+5x-3) +5(x2+5x-3)
=4x4+20x3-12x2 - 3x3-15x2+9x +5x2+25x-15

Combine the like terms and simplify
=4x4+x3(20-3)+x2(-12-15+5)+ x(9+25) – 15
= 4x4+17x3-22x2+34x-15

The steps to be followed in a Vertical method of multiplication are as follows:
Example: Multiply (4x2-3x+5)(x2+5x-3)
The first and the second terms are chosen irrespective of the order
Let (4x2-3x+5) be the first term and (x2+5x-3) be the second term
Arrange the two polynomials in a vertical form
4x2-3x+5
X  x2+5x-3

Working from right towards left each term of the lower trinomial is multiplied with each of the terms of the upper trinomials. Then the products are written underneath the second trinomial in three rows in the order of the degree of each term and then the like terms simplified as shown below
4x2-3x+5
X  x2+5x-3
-12x2+ 9x – 15
+20x3-15x2 + 25x
4x4 - 3x3  + 5x2                      .
4x4+17x3-22x2+ 34x-15 is the final product!

Solve Explicit Differentiation

In calculus,Explicit is a function which the independent variable. The function f explicitly is to provide a preparation for determining the output of the given function y in terms of the input value x: y = f(x). Derivative of an explicit function is called as explicit differentiation. for example  y = x3 + 5. The process of finding the differentiation of the independent variable in an explicit function by differentiating each term separately, by expressing the derivative of the independent variable as a symbol, and by solving the resulting expression for the symbol.

Solve explicit differentiation problems:

Let us see some problems and its helps to solve an explicit differentiation.

Solve explicit differentiation problem 1:

Find the differentiation of given explicit function   y = x2 - 15x + 3.

Solution:

Given explicit function is  y = x2 - 15x + 3.

Differentiation of explicit function is   dy/dx  = d/dx ( x2 - 15x + 3)

Separate the each term, so, we get

= d/dx (x2) - d/dx (15x) + d/dx (3)

= d/dx (x2) - 15d/dx(x) + d/dx (3).

= 2 x(2-1) - 15 + 0

= 2x - 15

The differentiation of an explicit function is  2x - 15

Solve explicit differentiation problem 2:

Find the differentiation of an explicit function  x2 +  cot x = - y

Solution:

Given explicit function is  x2 +  cot x = - y

Multiply by (-1) on both sides,  y = - x2 -  cot x

Differentiation of an explicit function,

y =   - x2 -  cot x

dy/dx = d/dx -x2   - d/dx (cot x) .

=  - 2x - (-cosec2 x) .

= - 2x + cosec2x

The Differentiation of an explicit function is  - 2x + cosec2x

Solve explicit differentiation problem 3:

Find the differentiation of given explicit function  2x2 + y2 = 1

Solution:

Given explicit function is  2x2 + y2 = 1

Subtract 2x2 on both sides.we get,

2x2 + y2 - 2x2 = 1 - 2x2

y 2 = 1 - 2x2

Take square root on both sides, we get

 sqrt (y^2) =  +- sqrt (1 - 2x^2)

y = +- sqrt (1 - 2x^2) .

Differentiate the function,  Let u = 1 - 2x2            and           y = sqrt u

(du)/(dx) = - 4x                              dy/(du) = 1/(2sqrtu)

So,   dy/dx =  ((dy)/(du)) ((du)/(dx)) .

=  1/(2sqrtu) . (-4x )

=  (- 2x) / sqrtu .

Substitute u = y, So we get

= - (2x)/y   .

The differentiation of an explicit function is  - (2x)/y    .

Substitution Geometry

Geometry is a module of mathematics, which involves the study of shapes, line equation, angles problem, dimensions, relative position of figures etc.  The term ‘Geometry’ means study of properties. A point is used to represent a position in space. A plane to be a surface extending infinitely in all directions such that all points lying on the line joining any two points on the surface. Substitution geometry problems are given below.

Example problems for substitution geometry :

1. Find out the geometry equation of straight line passing through the points 2x + y = 8 and 3x - 2y + 7 = 0 and parallel to 4x+ y - 11 = 0
Solution:
Let (x1, y1) be the intersection lines
2x1 +  y1 =  8       …  (1)
3x1 - 2y1 = - 7   …   (2)


(1) × 2 ? 4x1 + 2y1 = 16      …    (3)
(2) + (3) ? x1 =9/7 `=>` y1 =38/7   (x1, y1) =( 9/7 ,38/7)


The straight line parallel to 4x + y - 11 = 0 is of the form 4x + y + k = 0
But it passes through (9/7 ,38/7)


36/7 +38/7 + k = 0 ? k = -74/7
4x + y -74/7 = 0
28x + 7y - 74 = 0 this is the equation of straight line.


2. For what values of ‘a’, the three straight lines 3x + y + 2 = 0, 2x - y + 3 = 0and x + a y - 3 = 0 are concurrent?

Solution:

Let (x1, y1) be the point of concurrency. This point satisfies the first two equations.
3x1 + y1 + 2 = 0 … (1)
2x1 - y1 + 3 = 0 … (2)

Solving (1) and (2) By using substitution method, we get (- 1, 1) as the point of intersection. Since it is a point of concurrency, it lies on x + a y - 3 =0
- 1 + a - 3 = 0
a -4 = 0

a = 4

Practice problems for substitution geometry:

1. Find the point of intersection of the straight lines 5x + 4y - 13 = 0 and 3x + y - 5 = 0

Ans: The point of intersection is (1, 2)

2. Find the geometry equation of straight- line perpendicular to the straight line 3x + 4y + 28 = 0 and passing through the  point (- 1, 4).

Ans: 4x - 3y + 16 = 0

3. Find the equation of straight line passing through the intersection of straight lines 2x + y = 8 and 3x - y = 2 and through the point (2, - 3).

Ans: x = 2

Proportionality Problems

In mathematics, proportionality indicates that two variables are related in a linear manner. If one number doubles in size, so does the other; if one of the variables diminishes to 1/10 of its former value, so does the other.The symbol for proportionality resemble a stretched-out, lowercase Greek letter alpha . When this symbol appear that two quantities or variables, it is read "is proportional to" or "varies in direct proportion with." Thus, the expression x alpha y is read " x is proportional to y " or " x varies in direct proportion with y ." In this condition, as long as x and y do not attain values of zero, the quotient x / y is always equal to the same value k , which is called the proportionality constant. (source: wiki)

Example proportionality problems:

Proportionality problem 1:

Find a if a/4= 3/2.

Proportionality  solution:

By using property 1:

a/4=3/2

(a) (2) = (4) (3)

2a = 12

Dividing both side by 2

a=6

Therefore the value of a=6

Proportionality problem 2:

Is 6: 4 = 3: 2 a proportion?

No. If this were a proportion, Property 1 would produce

(6) (2) = (4) (3)

12   = 12, this is true. It is a proportion.

Practice proportionality problems:

Practice proportionality problem 1:

Janette's car uses 9 gallons to go 200 miles.

a) How many gallons will she use to go 400 miles?

B) How many gallons will she use to go 600 miles?

c) How far can she drive with 36 gallons?

(a)18

(b) 27

(c)800 miles

Practice proportionality problem 2:

Carlos makes $25 in 4 hours.

a) How much will he make in 2 hours?

b) How much will he make in 20 hours?

c) How much will he make in 22 hours?

(a)$12.50

(b)$125

(c)$137.50


practice proportionality problem 3:

Which is a better value?

a) 15 ounces for $ 9.25 or 30 ounces for $18.00?

b) 15 ounces for $ 9.25 or 5 ounces for $3.29?

Answer:  ( a )30 ounces for $18.

( b )15 ounces for $9.25.

practice proportionality problem 4:

6 bottles cost $7.00.

a) How much will 18 bottles cost?

b) How much will 15 bottles cost?

c) How much will 27 bottles cost?

d) How many bottles can you buy with $10.50?

Answer:   ( a ) $21

( b ) $17.50

( c ) $31.50

( d ) 9

Geometry Area and Volume

Geometry” Earth-measuring" is an part of the mathematics concerned with questions of size, shape, relative position of figures, and the properties of space. Geometry is one of the oldest sciences. Initially a body of the  practical knowledge concerning lengths, areas, and volumes. And now we can see about the problems in geometry area and volume.

Geometry area and volume problem 1:

Pro 1 :Find the volume of cylinder with the radius 8 cm and the height 12 cm.

Solution:We can find the volume of  an cylinder by using the following formula:

Volume of cylinder V=πr2h

Substitute the values of r and h into the above formula. Than, we get

V=π*82*12

=3.14*64*12

=2411.52 cm3

Pro 2 :Find the volume of sphere with the radius is 12 cm.

Solution:We can find the volume of  the sphere by using the following formula

Volume of sphere V= (4/3) πr3

Substitute the value of radius into the above formula. Then we get,

V= (4/3) *3.14*123

= 1.333*3.14*144

= 602.72 cm3

Ans: 602.72 cm3

Problem 3:Find the amount of pyramid with the base 8.2 mt and height 10.2 mt.

Solution:We can find the volume of  the pyramid by using the following formula

Volume of pyramid V= (1/3) b h

Substitute the values of  the base and height into the above formula. Then we get,

V= (1/3)*8.2*10.2

=0.333*8.2*10.2

=27.8521 mt3

Ans: 27.8521 mt3

Geometry - Find the volume of shapes when area is given

Find the volume of the right prism whose area of the base is 550 cm2 and height is 38cm



Solution:Given that area of the base, A = 550 cm2 and height (h) of the prism = 38 cm

Volume of the right prism = area of the base * height cu.units

= A * h

= 20900

Volume of right prism = 20900 cm3

Find the volume of the right prism whose area of the base is 450 cm2 and height is 34cm

Solution:Given that area of the base, A = 450 cm2 and height (h) of the prism = 34 cm

Volume of the right prism = area of the base * height cu.units

= A * h

= 15300

Tutoring About Correlation Coefficient

Concept of correlation:

Correlation is a method of studying the relationship between the two variables. In statistical analysis we come across the study of two variables wherein the change in the value of one variable produces a change in the value of other variable. In that case we say that the variables are correlated or there is a correlation between the two variables.

The formula for the correlation coefficient r can be expressed in the form,

r = `(sum (X - barX) ( Y - barY))/(sqrt(sum (X - barX)^2) sqrt(sum(Y - barY)^2))`

It is conventionally taken as x = X - X and y = Y - Y and hence we write

r = `(sum xy)/(sqrt(sum x^2) sqrt(sum y^2))`

The above formula is expressed in terms of deviations of the variables from their means. Instead, if the actual values of the observations are taken then the formula can be written as,

r= `(N sum XY - sum X sum Y)/(sqrt(N sum X^2 - (sumX)^2) sqrt(NsumY^2 - (sum Y)^2))`

Instead of the deviations from their means, the deviations are measured from the value A and B for X and Y variables by taking dx = X - A, dy = Y - B, the correlation coefficient r is given by,

r = `(N sumdxdy - sumdx.sumdy)/(sqrt(Nsumdx^2- (sumdx)^2) sqrt(Nsumdy^2 - (sumdy)^2))`

Tutoring about formulas for calculating correlation coefficient:

r = `(sum xy)/(sqrt(sum x^2) sqrt(sum y^2))`

This formula is used when deviations are measured from their mean.
r= `"(N sum XY - sum X sum Y)/(sqrt(N sum X^2 - (sumX)^2)(sqrt(N sum Y^2 - (sumY)^2))) `

This formula is used if no assumed average is taken for x and y series
r = `"(N sumdxdy - sumdx.sumdy)/(sqrt(Nsumdx^2- (sumdx)^2)(sqrtNsumdy^2 - (sumdy)^2)) `

This formula is applied when deviations for x and y series are taken from some assumed values.


Tutoring on problems on correlation coefficient::

Calculate the correlation coefficient between x and y from the following data:
x 1 3 5 8 9 10
y 3 4 8 10 12 11


Solution:
x y x - `barx` y - `bary` (x - `barx` )2 (y - `bary` )2 (x-`barx` )(y - `bary` )
1 3 -5 -5 25 25 25
3 4 -3 -4 9 16 12
5 8 -1 0 1 0 0
8 10 2 2 4 4 4
9 12 3 4 9 16 12
10 11 4 3 16 9 12
36 48 0 0 64 70 65


`barx` = `(sum x)/(n)` = 36/6 = 6

`bary` = `(sum y)/(n)` = 48/6 = 8

r = `(sum(x- barx)(y - bary))/(sqrt(sum(x-barx)^2)sqrt(sum(y-bary)^2))`

= `(sum xy)/(sqrt(sumx^2)sqrt(sumy^2))`

= `(65)/(sqrt(64)sqrt(70))` = 0.97

Limit of a Function

Students can study about Limit of a Function here. Consider the function f(x). Let the independent variable x take values near a given constant a. Then f(x) takes a corresponding set of values. Suppose that when x is close to a, the values of f(x) are close to some constant. Suppose f(x) can be made to differ arbitrarily small from A by taking values of x that are sufficiently close to a but not equal to a and that is true for all such values of x. Then f(x) is said to approach limit A as x approaches a.

If the function f(x) approaches a constant A when x approaches a in whatever manner without assuming the value a, A is said to be the limit of f(x) as x approaches a. Thus we write lim_(x->a) f(x) = A

Find the Limit of a Function

Students can learn to Find the Limit of a Function if they know what Functions are and how they behave at the given limits.

A function may approach two different limits. One where the variable approaches its limit through values larger than the limit and the other where the variable approaches its limit through values smaller than the limit. In such a case the limit is not defined but the right and left hand limit exists.

The right hand limit of a function is the value of the function approaches when the variable approaches its limit from the right. Here, we write lim_(x->a^+) f(x) = A+

The left hand limit of a function is the value of the function approaches when the variable approaches its limit from the left.

here, we write lim_(x->a^-) f(x) = A-

The limit of a function exists if and only if the left hand limit = right hand limit.

In that case, lim_(x->a^+) f(x) = lim_(x->a^-) f(x) = f(x)

Properties of limit of a function

The following are some of the properties of limits which are useful in evaluating the limit of a function.

1. lim_(x->a) k = k ( k is a constant)

2. lim_(x->a) [f(x) ± g(x)] =  lim_(x->a) f(x) ± lim_(x->a) g(x)

3. lim_(x->oo) [f(x).g(x)] = lim_(x->oo) f(x) . lim_(x->oo) g(x)

4. lim_(x->a) (f(x))/(g(x)) = (lim_(x->a)f(x))/(lim_(x->a)g(x))

5. lim_(x->a) [f(x)]n =  [ lim_(x->a) f(x)]n

Standard limit theorems:

1. lim_(x->a) (x^n- a^n)/(x - a) = nan-1

2. lim_(x->0) (e^x-1)/(x) = 1

3. lim_(x->0) (sin x)/(x) = 1

4. lim_(x->0) (1 + (1)/(n))^(n) = e



Solved Examples

Example 1: Evaluate the right hand limit of the function

f(x) = {│x – 4│/x – 4, x ≠ 4, 0 x = 4

at x = 4

Sol: (RHL of f(x) at x = 4)

= lim f(x) = lim f(4 + h) = lim │4+ h – 4│4 + h – 4

x→4+      h→0             h→0

= lim │h│/h = lim h/h = lim 1 = 1

h→0            h→0      h→0

Example 2: Let f be the function given by f(x) = x2 – a2/x – a, x ≠ a

Using (in , δ) definition show that lim f(x) = 2a

x → 0

Sol: Let in > 0 be given. In order to show that

lim f(xi) = 2a

x → a

We have to show that that for any given in > 0, there exists a number δ >0 such that

│f(x) – 2a│< in whenever 0 < │x – a│< δ

If x ≠ a, then │f(x) – 2a│= │x3 – a2/x – a│

= │(x + a) – 2a│                                         [... x ≠ a]

= │x – a│

... │f(x) – 2a│< in , if │x – a│< in

Choosing a number δ such that 0 < δ < in , we have

│f(x) – 2a│< in when whenever 0 < │x – a│< δ

Hence    lim f(x) = 2a

x → 0

Standard Deviation in Statistics

Statistics is a division of applied mathematics which contracts with the particular interpolation of data. The term ‘Statistics’ has been taken from the Latin name ‘Status’ in which it defines ‘political state’. This statistics mainly used to measure the arithmetic mean, median, mode and standard deviation. This measurement gives idea about where the data points are centered. Let us discuss about standard deviation with some example problems.

Evaluation of Standard Deviation in statistics:

Mean:

In Statistics, Mean is defined as the average of the given total numbers, i.e., total number of data divided by the number of data set given.

Formula for finding mean,

barx   = (sum(x))/(n)

Standard Deviation:

In Statistics, Standard Deviation is the measure of describing squared mean difference variability and spread of the Data set in the given total numbers. It is used to take the measurement of taking square root and average of numbers in the Data set.

Formula for standard deviation,

S =  sqrt ((sum(x - barx)^2 )/ (n-1))

Here Standard Deviation is calculated by using the mean Value  barx

Example Problems to find standard deviation in statistics :

Problem 1:

Here are 4 measurements   66, 45, 67, 45, 34, 56, 78 and 57. Calculate statistics standard deviation for the given measurements

Sol:

Mean: Calculate the mean the using the formula,

barx   = (sum(x)) / n

barx   =( 66+45+67+45+34+56+78+57) / 8

= 448 / 8

barx   = 56

Standard Deviation,

S = sqrt((sum(x-barx)^2) / (n-1) )

S = sqrt((( 66-56)^2+(45-56)^2+(67-56)^2+(45-56)^2+(34-56)^2+(56-56)^2+(78-56)^2+(57-56)^2) / (8-1))

= sqrt((100+121+121+121+484+0+484+1)/7)

=  sqrt( 1432 / 7)

S = sqrt(204.571429)

Standard Deviation S =  14.3028469

Problem 2:

Find the Statistics Standard deviation of the given Data 16, 17, 18, 20 and 24

Sol:

Mean: Calculate the mean using the formula,

barx   = (sum(x)) / n

barx  = ( 16 + 17 + 18 + 20 + 24 ) / 5

barx = 95 / 5

barx = 19

X               x-barx              (x-barx )^2

6               16 - 19 = -3           9

7               17 - 19 = -2           4

8               18 - 19 = -1           1

9               20 - 19 =  1           1

10              21 - 19 =  2           4

Sum of the  (x-barx)^2

9+4+1+1+4 = 19

Standard Deviation: Calculate the standard deviation

S =   sqrt((sum(x- barx)^2 ) / (n-1))

S = sqrt( ( 9+4+1+1+4) / 4 )

S = sqrt (19 / 4 )

S = sqrt(4.75 )

S = 2.17944947

Practice Problems for Statistics Standard Deviation:

1. Find the statistics standard deviation for the following given data.37, 56, 54, 54, 26, 67, 12, 65 and 34.

Answer:                     Mean = 45

Standard Deviation = 18.714967272213

2. Calculate the statistics standard deviation for the following. 77, 56, 33, 87, 90, 23, 67, 80 and 99.

Answer:                    Mean = 68

Standard Deviation = 26.043233286211

Study Online Second Derivatives

Study online second derivatives involves the process differentiating the given polynomial function twice with respect to the given variables whereas all the process is clearly explained with the help of online. Generally the second derivative is discussed in calculus whereas it is mainly helps 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 for second derivatives study discussed in online.

Example 1:

Determine the second derivative from the polynomial.

f(b) = 5b 2 +5b 4  + 12

Solution:

The given function is

f(b) = 5b 2 +5b 4  + 12

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

f '(b) = 5(2b  )+5(4 b 3 ) + 0

By solving above terms

f '(b) = 10b +20b3

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

f ''(b) =  10(1 ) +20(3b2)

f ''(b) =  10 + 60b2 is the answer.

Example 2:

Determine the second derivative from the polynomial.

f(b) = 4b4 +5b 5 +6b 6  + 2b

Solution:

The given function is

f(b) = 4b4 +5b 5 +6b 6  + 2b

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

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

By solving above terms

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

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

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

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

Example 3:

Determine the second derivative from the polynomial.

f(b) = 2b6 + 2 b5 + 3 b4 + 3b

Solution:

The given equation is

f(b) = 2b6 + 2 b5 + 3 b4 + 3b

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

f '(b) =  2(6b 5)  +2 (5 b4 ) +3(4 b3) + 3

By solving above terms

f '(b) =  12b 5  +  10b4  + 12 b3 – 3

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

f ''(b) =  12(5b 4 ) – 10(4b3)  + 12(3b2)

f ''(b) =  60b 4 – 40b3 +36b2  is the answer.

Online second derivatives practice problems for study:

1) Determine the second derivative from the polynomial.

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

Answer: f ''(b) = 6b +12b2+ 20b 3

2) Determine the second derivative from the polynomial.

f(b) = 2b 3+3b5 + 4b 6

Answer: f ''(b) = 12b + 60b3 + 120 b 4

Solving Negative Number

Definition:

Negative number is defined as the number which indicate by negative sign or minus ('-') . Negative number is less then zero and placed left to zero.

Ex: ...-5,-4,-3,-2,-1,0,1,2,3,4...

Comparison of Positive and Negative:

For each negative number , there is a positive number that is its opposite . Here we can write the opposite of negative number with a positive of same number or plus sign used In front of the number and call these numbers are positive numbers. Ex : 1,2,3 ,.....positive numbers are grater than zero. Similarly, the opposite of any positive number is a negative number .

Ex:    1,2,3 is -1,-2,-3.

Solving examples for negative numbers:

Zero cannot be taken as a negative number or positive number.
For every positive number x, there exists a negative number y such that x + y = 0
Positive number is denoted as plus ('+')sign and negative number is denoted as minus sign('-').
Example of negative number:-2,-43,-34 and example for positive number is 2,43,34.
negative and positive number may be written as mixed numbers or fraction numbers.

The equal fraction of negative numbers are given bellow:

(-3)/7,3/(-7),-(3/7) and -3/7 .

The equal mixed numbers are given bellow.

-2/5,-(2/5)       (-4)/9, 4/-(9) and -4/9

More about solving negative numbers:

Solving Addition of Negative Numbers:

To add the negative numbers which consist of minus sign. To provide the answer of addition of negative number.

Solving examples for addition negative numbers:

-12+(-6)=?

Solution:

-12+(-6)= -18.

Here the values of  -12 and -6are 12 and 6 adding the smaller from the larger gives -12+(-6)= -18, and since the larger  value was 12, so we can give result the same sign as -12, '-' so -12+(-6)= -18.

Example:

(-6) + (-6) = ?

Here the absolute values of -6 and -6 are 6 and 6.  Adding the smaller from the larger gives -6 - 6 = 12 ,  but here both has same value . In this case sign is the matter, here 12 and -12 are the not same and then -6 and -6 are same numbers. The property of all same number sum is 12. The addition of two number up to zero are called as additive inverses.

Multiplying Negative Numbers:

Solving example for multiplying negative numbers :

Product of negative number ,here we can take the product of their values.

(-3.3) × (-5) = ?


(-3.3) × (-5) = (-3.3) × (-5)

= (-3.3) × (-5)

=  +16.5.
Dividing Negative Numbers

Solving example for dividing negative number:

To divide two negative numbers, here we can divide the value of the first by the value of the second.

(-1.6) ÷ (-4) = (-1.6) ÷ (-4)

= (-1.6) ÷ (-4)

=  -0.4

Geometry Without Common Vertices

In geometry, some figures have common vertices. Mostly the geometric figures are without common vertices. If a triangle, quadrilateral and some geometric figures in a graph are lying without common vertices are refered same. In graph there are four quadrants in that some vertices are fall on common vertices, with out common vertices points of are plotted.

Geometric figure without common vertices

Without common vertices find the distance of two points

In geometry the triangle has three vertices; the vertices are not common vertices. The common vertices are formed only if two triangles are in same point without three common vertices. The distance between two un common vertices are find out by using the coordinates of the vertices (x1,x2) and (x2,x2) of the vertices.

Distance between two vertices = √(x2-x1)2 +(y2-y1)2

If the geometry figure having the common vertices in a graph



Examples for without common vertices

Distance of a vertices are find using distance formula:

Examples for distance between two vertices:

Ex 1:   Find the distance formed by without common vertices, Vertices A(4,5), B(7,4)

Sol :            Distance AB = √(x2-x1)2 +(y2-y1)2

X1=4    X2=7    Y1=5    Y2=4

AB = √(7-4)2+(4-5)2

= √(3)2+ (-1)2

= √9+1

= √10 units

Ex 2 :  Find the distance formed by without common vertices, Vertices A(3,2), B(5,4)

Sol :            Distance AB = √(x2-x1)2 +(y2-y1)2

X1=3    X2=5    Y1=2    Y2=4

AB = √(5-3)2+(4-2)2

= √(2)2+ (2)2

= √4+4

= √8 units

Ex 3 :      Find the distance formed by without common vertices, Vertices A(6,4), B(10,8)

Sol :            Distance AB = √(x2-x1)2 +(y2-y1)2

X1=6    X2=10    Y1=4    Y2=8

AB = √(10-6)2+(8-4)2

= √(4)2+ (4)2

= √16+16

= √32 units

Ex 4:    Find the distance formed by without common vertices, Vertices A(3,3), B(4,8)

Sol :            Distance AB = √(x2-x1)2 +(y2-y1)2

X1=3    X2=4    Y1=3    Y2=8

AB = √(4-3)2+(8-3)2

= √(1)2+ (5)2

= √1+25

= √26 units

Practice problems:

Q 1   Find the distance of two vertices A(1,1) B(1,2)   Answer: √1 units

Q 2   Find the distance of two vertices A(2,2) B(1,1)   Answer: √2 units

Straight Line Equation

The equation of straight line is generally written as    y = mx + b

Graphical Representation

where, m= slope or gradient of the straight line equation

b = the y-intercept

Suppose we want to find equation of a straight line that passes through a known point and has a known slope. Let (x ,y) represent the co-ordinates of any point on the line and let (x1 ,y1) represent the co-ordinates of other point. The slope of the straight line equation is given as,

m = (y-y_1)/(x-x_1)

After finding the slope m as we are given the co-ordinates of the point (x1 ,y1) in the equation    y= mx + b Then, the constant 'b' can also be found so that finally the straight line equation is obtained.

Other Forms of straight Line Equations

Different Forms of Straight Line Equations:

There are many other forms of Straight Line Equations.

1. Straight Line Equation through two points:

The line through two different points ( x1 ,y1) and ( x2 ,y2) is given by

y-y1  = [(y_2-y_1)/(x_2-x_1)]  . (x - x1)

2. Straight Line Equation in general form:

A straight line is defined by a linear equation as

Ax+By+C=0     where A , B are not both 0

3. Straight Line Equation in Intercept-intercept form:

Let us consider that a straight line intersects x-axis at (a , 0) and y-axis at (0 , b). Then it is defined by equation

(x/a) +(y/b) = 1

4. Straight Line Equation in Point-Slope Form:

The equation of straight line through the point (a , b) with slope m is

y = m ( x - a ) + b

Solved Examples on equations of straight line

Ex:1 Find the equation of a line passing through (2 , 3) and having a slope of 3?

sol: step 1: Compare the given point(2 , 3) with the general point (x1 , y1)

Now x1=2  and  y1=3

step 2: we have the straight line equation as  y-y1=m(x-x1)

now substitute the given point in the above equation

y-3=3(x-2)

y-3=3x-6

3x-y=6-3

3x-y-3=0

The straight line equation is 3x-y-3=0

Ex:2 Find out the straight line equation of the line passing through the points (1,2) and (2,4).

Sol: Given points are compared with (x1 , y1) and (x2 , y2) and substitute the points in the equation

y-y1  = [(y_2-y_1)/(x_2-x_1) ] . (x-x1)

y-2  =  [(4-2)/(2-1)] . (x-1)

y-2 = 2(x-1)

y-2 = 2x-2

2x - y = 0

The straight line equation is 2x-y=0.

Practice Problems on equations of straight line

Pro:1 Find the equation of the line passing through the points (-3,4) and (4,-2)

Ans: we have      

m = (y_2-y_1)/(x_2-x_1)

=(-2-4)/(4+3)

=(-6/7)

let (x,y) be compared to (-3,4)

y-4 = (-6/7) . (x-(-3))

7(y-4) = -6(x+3)

7y-28 = -6x-18

7y-28 = -6x-18

7y+6x = 10

Pro:2 Write equation of line having points and slopes as follows;

P(3,3) , m=-2

P(-2,-1) , m=1/3

P1(2,2) and P2(-4,-1)

y-intercept = 2 , m=3

Ans: The answers to the above given practice problems are

y+2x=11

3y-x=-1

2y-x=2

y-3x=2

Hyperbola Axis

The locus of a point whose distance from a point rest at center shows an unaltered ratio, higher than one to its length from the fixed line is known as hyperbola. In this article, you can learn about the axis of hyperbola, general equation of the hyperbola and definitions regarding hyperbola.

Axis of Hyperbola:

Transverse Axis:

Transverse axis can be defined as the line segment, which joins the vertices of the hyperbola. 2a is calculated as the difference between two vertices. The equation of the transverse axis is y = 0. This is showing that y-coordinate is zero.

Conjugate Axis:

Conjugate axis can be defined as the line segment, which joins the y-coordinates of the hyperbola. The distance between the two coordinates are 2a (i.e., the length of the Conjugate axis is 2a.). x = 0 is considered as the equation of conjugate axis. Therefore the x-coordinates of Conjugate axis is zero.

General Equation of the Hyperbola:

Some important points to be considered:

Fixed point is represented as F.
Fixed line as l.
Eccentricity as e, then it should be greater 1.
Moving point is represented as P(x,y).

Steps:

Fixed point F is plotted and also the fixed line ‘l’ is drawn.
Perpendicular (FZ) is dropped from F to l.
As next step drop PM which is perpendicular one from P to l.
Plot the points A, A’ which divides FZ internally and externally in the ratio e : 1            
respectively.
Take AA’ = 2a and treat it as x-axis.
Draw a perpendicular bisector of AA’ and treat it as y-axis.
Consider C as the origin, then the known points are C(0,0), A(a,0) and A’(-a,0).
The general equation of the hyperbola x^2/a^2 -y^2/b^2 = 1.

Definitions Regarding Hyperbola:

Focus:

The fixed point F is known as the focus of the hyperbola.

Directrix:

The directrix is nothing but the fixed line. Then the directrix equation is given by x = a/e.

Centre:

The centre of the hyperbola is a point, at which the transverse and conjugate axes intersect and they are represented as ‘C’.

Vertices:

The vertices of the hyperbola are the points, where the curve and its transverse axis meet. The vertices are A(a,0) and A’(-a,0).

Studying Standard Deviation Examples

The standard deviation of a statistical population, a data set, or a probability distribution is the square root of its variance. Standard deviation is a widely used measure of the variability or dispersion, being algebraically more tractable though practically less robust than the expected deviation or average absolute deviation. It shows how much variation there is from the "average" (mean) (or expected/budgeted value). It helps detect tampering of data. Examples for studying standard deviation is given below.

Formula for studying standard deviation examples:

Formula for studying standard deviation examples are defined below:

Measurement of standard deviation is prepared by taking square root for the addition of mean difference with the certain data divided by the total number of values subtracted by one. The following formula for standard deviation as shows given below.

s=v?(X-M)2/n-1

Here S = Sum of values

X = Individual value

M = Mean of total all value

N = Sample size (Total number of values)

Variance:

Variance = s2

Steps for calculating Standard Deviation examples:

• Step 1: calculate the average for given n numbers using the formula this is called mean of given numbers

• Step 2: Find distance between each given numbers in the Data set from the calculated average value. This is called  "deviation" from the mean value.

• Step 3: Take the Square of each deviation value found from mean. This is squared deviation from mean.

• Step 4: Calculate the sum for all the squared standard deviations.

• Step 5: Now apply the Standard Deviation formula and find standard deviation formula. It will be the square root of variance.

Examples for Studying Standard deviation:

Examples for Studying Standard deviation are as follows:

Pro 1:   Here are 4 measurements 4, 6, 7, 9 and 10Calculate the Standard Deviation

Sol :     Mean: Calculate the average for the given values. To find the mean.
4+6+7+9+ 10
x =   ------------------------
5-1

= 36/4

= 9

Standard Deviation,
v (4-9)2 + (6-9)2 + (7-9)2 + (9-9)2 + (10-9)2
S=  -------------------------------------------------------------
5-1

= v 81 /4

= v 20.25

=  4.5

Standard Deviation  S = 2.12132

Pro 2:  Here are 4 measurements 10, 20, 30, 40 and 50 Calculate the Standard Deviation

Sol :     Mean: Calculate the average for the given values. To find the mean.
10+20+30+40+60
x = ---------------------------
5-1

= 160/4

= 40

Standard Deviation,
v (10-40)2 + (20-40)2 + (30-40)2 + (40-40)2 + (60-40)2
S=  -----------------------------------------------------------------------------
5-1

= v 1600 /4

= v 400

S = 20

Standard Deviation S = 20

Absolutely Continuous Function

DEFINITION

Let [p,q] be the close bounded interval of C. Then a function f:[p,q]→ R will be an absolutely continuous function on [p,q], if for any δ>0 there will be a ε>0 such that the certain conditions which are mentioned below holds good

If (p1q1)..............(pnqn) is a collection which is finite with disjoint open intervals in [p,q] such that

Σni=1 (qi-pi) < ε

and

Σni=1 |f(qi)-f(pi)| < δ



EQUIVALENT DEFINITIONS

The condition on a real-valued function  "  f  "  on the compact interval [ p, q ] are equivalent if

1) f is an absolutely continuous function

2) ' f ' has a derivative f1 almost everywhere which is a Lebesgue integral and

f ( x ) = f ( p ) +  ∫x a  f1  ( c ) dt

for all x on [ p , q ]

3)There exists a Lebesgue integrable function such that g on [ p , q ] such that f ( x ) = f ( p ) + ∫x p  g ( c ) dt

for all x on [ p , q ]

If these conditions are satisfied by the function the definitely g = f' almost everywhere

Properties of the absolute continuous function

PROPERTIES OF ABSOLUTELY CONTINUOUS FUNCTION

1. The sum and difference of two absolute continuous function are also absolutely continuous. The products of two absolute continuous function defined on the bounded interval will also be a absolute continuous function.

2. If an absolutely absolute continuous function is defined on a bounded closed interval and is nowhere zero then its reciprocal is also an absolutely continuous function.

3. Every absolutely continuous function is an uniformly continuous function.

4. If  f: [ p , q ] → R is absolutely continuous, then it will be a function of the bounded variation on [ p,q ]

Learning Geometric Probability

Numerical measure of the likelihood of an event to occur is called as Probability. The probability should be a range in between 0 and 1.In this case we say probability is 0. If the event is certain to occur, we say probability is likely 1. The probabilities involved in the geometric problem that also called as geometric probability. This geometry probability may be a circle or any polygon from the geometric. It involves the length, area and volume of any one of the geometric shapes.

The definition of probability of an event has shown in below that is depend on their outcomes from the possibilities



Number of successful outcomes

Probability  =        _____________________________

Total number of possible outcomes.

learning geometric probability example problem in triangle:

The figure shows a triangle divided into sectors of different colours. Find the probability of angle for blue sector and orange sector?

learning geometric probability

Solution:

We have to find the Probability for blue color:

Step 1:

Total angle of triangle is 180 degree

Step: 2

Probability finding blue color sector= The blue sector triangle angle/ The total angle of  entire triangle

The blue sector triangle angle=30 degree

=30/360

=1/12

Step 3:

Next we have to find Probability for orange color:

Probability finding orange color sector= The orange sector triangle angle/ The total angle of entire triangle

The orange sector triangle angle=45

=45/360

=9/72

learning geometric probability example in Rectangle:

Example 2:

A rectangle has four sides and that corner having each 4 ball. Find the probability of each corner having 4 balls.?

Solution:

Here the Rectangle has the five corners such as A, B, C, D

We have to find the probability here,

Probability = Number of successful outcomes / total number of possible outcome

Probability  of each corner having the balls=4/4=1

Example 3:

A triangle with area 15 cm2 is inscribed in a circle with radius 3 Cm.Find the probability that a ball thrown fall into the triangle?

Solution:

We have to find the area of the circle and area of the triangle is given. From this we can find the probability of a ball that fall into the triangle easily.

Area of the circle = `Pi` r2

Radius  = 3. So Area of the circle = 3.14 * 3 * 3 = 28.26 cm2.

Area of the triangle  = 15 cm2.

So the probability  = 15 / 28.26 = 0.53

Answer is 0.53.