Estimating Fraction Calculator

Fractions:

A fraction is a number that can represent part of a whole. The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, one part of four, and so on. A much later development was the common or "vulgar" fractions which are still used today (½, ⅝, ¾, etc.) and which consist of a numerator and a denominator.

Calculator:

A calculator is a small (often pocket-sized), usually inexpensive electronic device used to perform the basic operations of arithmetic.(source : Wikipedia).

estimating fractions calculator

By giving input fraction we can get the output as per our operation. Let us see some problems on estimating fractions calculator.
Problems on Estimating Fractions Calculator :

1. Addition of fractions:

To add a fraction we need the common denominator.  We can add the fractions with same denominator.

Example 1:

Estimation the addition of fractions  3/4 + 6/4

Solution:

3/4 + 6/4 = (3+6) /4

= 9/4

Example 2:

Estimate the addition of fractions 5/7 + 9/6

Solution:

To add a fraction we need to make common denominator,

5/7 + 9/6 = (5 * 6) / (7 * 6) + ( 9 *7) / ( 6 * 7)

= 30/42 + 49 /42

= ( 30 + 49 ) / 42

= 79 / 42

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2.Estimating subtraction of fraction:

Example 1:

Estimating the subtraction for the following fractions 5/4 - 2/4

Solution:

Given , 5/4 - 2/4

In the above fractions , the denominator is equal.

So we can subtract the numerator of the fractions  just like integers and keep the denominator,

5/4 - 2/4 = (5-2) / 4

= 3/4

Answer: 5/4 - 2/4 = 3 / 4

Example 2:

Estimating the subtraction for the following fractions 5/4 - 9/8

Solution:

Given,  5/4 - 9/8

Both fractions are has different denominator,

So we need to find the lcd for 4 and 8.

Multiple of 8 = 8 , 16 ,24 ,.....

Multiple of 4 = 4 ,8, 12 ,16 ,....

The least common multiple of 4 and 8 is 8.

Multiply 5/4 by 2 on both numerator and denominator ,

(5*2) / ( 4 * 2) = 10 /8

Now we can add the fractions,

5/4  - 9 / 8 = (10 /8) - (9 / 8)

= (10- 9 ) /8

= 1 / 8

Answer : 5/4  - 11/8 = 1/ 8
Problems on Estimating Fractions Calculator :

3. Estimating multiplication of  fractions:

Just like integers we can multiply the fractions.

Example:

Multiply 3/8 * 24 / 27

Solution:

Given, 3/8 * 24 / 27

3/8 * 24 / 27 = (3 * 24) / ( 8 *27)

= 72 / 216

Divide by 72 on both numerator and denominator,

= 1/3

4.Estimating division of fractions:

Example :

Divide the fractions 14/18 ÷ 7/9

Solution:

Given, 14/18 ÷ 7/9

14/18 -> Dividend

7/9 -> Divisor

Take a reciprocal of the divisor,

Reciprocal of 7/9 = 9/7

Multiply the reciprocal of divisor by the dividend.

14/18 * 9/7 = (14 * 9) / ( 18 * 7)

= 126 / 126

= 1

Answer: 14/18 ÷ 7/9 = 1

Solving for a Specific Variable

Algebra includes all the concepts like variables, constants, expressions, exponents, equation and etc . Variable is one of the main terms in math. Variables do not change the meaning of the expressions. Generally algebra expression includes variables. Commonly variables can be represented using alphabets. Here is the example, 2y^2+4y+2.Here we are going to learn about solving for a specific variables.

I like to share this Different Types of Variables with you all through my article.

Simple Example Problems of Solving for a Specific Variable:

Example 1:

A= bc then solve for b .

Solution:

Step 1: divide using c on both the sides .

Step 2:So, A/c = (bc)/c . (c term will be cancelled )

Step 3:therefore, the value of b is A/c .

Example 2:

P= 2l+2w Solve for w.

Solution :

Step 1: the given question is p= 2l+2w .

Step 2: p-2l =2l-21+2w.(Subtract 2l on both the sides )

Step 3: When we simplify we get p-2l=2w.

Step 4: (p-2l)/2 =(2w)/2 (Divide using 2 on both the sides)

Step 5: Therefore the value of w is (p-21)/2 .

These are the simple examples of solving  for a specific variables.
Some more Examples of Solving for a Specific Variables.

Example 3:

Q=(c+d)/2  then solve d.

Solution :

Step 1: The given question is q= (c+d)/2  .

Step 2: Multiply 2 on both the sides. So, 2q=(c+d) /2   xx 2 .

Step 3: When we simplify we get 2q= c+d.

Step 4: Subtract c on both the sides . So 2q-c=c-c+d.

Step 5: Therefore, the value d is 2q-c

Example 4:

V= (3k)/t  then solve t .

Solution :

Step 1: Multiply t on both the sides Vt = (3k)/t  xx t

Step 2: When we simplify we get Vt = 3k.

Step 3: Divide using V on both the sides. So (Vt)/V =(3k)/V

Step 4: Therefore, the value of t = (3k)/v .

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Example 5:

Q =3a+5ac then solve a.

Solution:

Step 1: The given question is q= 3a+5ac .

Step 2: Here a is a common term. Take the common term as outside.

Step 3: Now we have q= a(3+5c). (divide using 3+5c on both the sides).

Step 4: (q)/(3+5c) = (a(3+5c))/(3+5c) .

Step 5: When we simplify we get (q)/(3+5c) = a. Therefore the answer is (q)/(3+5c) =a.

These are the example problems of solving for a specific variables.

Solve Interval Notation

The interval notation denotes the set of real datas with the property that any data that deception among in the set is also has in the set. For instance, the group of all the datas x satisfying 0 ≤ x ≤1 this is an range which has 0 and 1, as well as all numerical values among 0 and 1.

We can represent the inequality result by solving the interval notation.
Solving Interval Notation:

We can express the inequality outcomes by solving the interval notation

The symbols used in solving the interval notation are,

( - is the “not included” symbol or “open” symbol

[ - is the “included” symbol or “closed” symbol

Solving Open Interval:

(p, q)  is written as p ≤ x ≤ q  here the termination points are not included.

open interval

Solving Closed Interval:

[p, q]  is written as p < x < q  here the termination points are included.

closed interval

Solving Half-Open Interval:

(p, q]  is written p < x
half-open interval

Solving Half-Closed Interval:

[p, q) is taken as p < x
half-closed interval

Solving Non-ending Interval:

( p , ∞) is taken as x > p where p is not incorporated and infinity is always defined as being "open".

non-ending interval

Solving Non-ending Interval:

( -∞ , q ] is interpreted as x < q here q is incorporated and again, infinity is always defined as being "open".

non-ending interval1

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More about Solving Interval Notation:

If we accomplish the favored set of outcomes we can use the concoction of interval notations. For example define the interval of all values except 9.

As an inequality  x<9 or="or" x="x">9

In an interval notation ( -∞ ,9) U (9,∞).

Scalar Line Integrals

Scalar line integral is a definite integral it will be taken over a surface and integrated, so the scalar line integral is defined as the sum of all points in the surface.  Let x: [a, b] gives R3 be a path of class C and f: X subeR3 gives R be a continuous function contains the image of x.  The scalar line integral of f along x is

int_a^bf(x(t)) || x'(t) dt

Notation is usually written as int_x f ds.

Algebra is widely used in day to day activities watch out for my forthcoming posts on Definite Integrals and Anti derivative. I am sure they will be helpful.

General form of scalar line integral:

The scalar function of line integral  is

if F = F1 i + F2 j + F3 k

r = x i + y j + z k

So int_cF. dr = int_c (f_(1)dx + f_(2) dy + f_(3) dz)

Scalar line integral and parametrizations

If y is a reparametrization of x. Then

If y is orientation-preserving, then int_y F.ds =  int_x F.ds

If y is orientation-preserving, then  int_y F.ds  = -int_x F.ds


When the path is parametrized by length of arc, the natural analog of the integral has done in 1 dimension. The integral of a scalar function f along a curve r(s) is simply int f (r(s)) ds

Applications:

F is a force acting upon a the particular particle so the particle moves along a curve C in  sample space and r be the position vector of the  given particle at a point on C. Then work done by the given particle at C is F.dr and the total work is done by F along a curve C is given by the line integral  int_c F. dr



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Scalar Line Integrals don’t depend on parameterizations.
Scalar Line Integrals-example Problems

Evaluate I = int_e f(x, y, z) ds where f(x, y, z) = = x2 – y2 – 1 + z and e is the helix parametrized by c(t) = (cos t, sin t, t)  [o <= t <=pi]. 3

Solution:

I = int_alpha^beta f(c(t)) || c'(t) || dt

See that f (c (t)) = cos2t + sin2t - 1

Also  c' (t) = (-sin t, cos t, 1) =    sqrt(-sin t^(2) + cos^(2) t + 1^(2))

See that f(c (t)) = cos2t + sin2t - 1

I = int_0^(3pi)sqrt(2) t dt

= sqrt(2) [(1)/(2)  t^(2) ][]_0^(3pi)

= (sqrt(2))/(2) 9pi^(2).

The correct answer for scalar line integral is = (sqrt(2))/(2) 9pi^(2).

Practice Problem:

Evaluate int sin 6x cos 3x dx


Answer: -1/2 [(cos 9x / 9) + (cos 3x / 3)] + c.

Define Natural Logarithm

There are three mathematical quantity related to the function  ea = x ,  Here the quantity " x " is said to be natural lagarithm of the number " a ". And the quantity " e " is said to be the base of the log and last one  is x which is power of the natural logarithm .The value of natural logarithm is given by as follow:

logex =a.

We can state it as above .

To show the formula

logex =a. and  ea = x represents the same we can take some examples as .

loge 10 = 2.3025

And the      e2.3025 = 10,so both formula exists.
Graph of Natural Logarithm:


Examples on Natural Logarithms:

Addition rule –


logex + logey   = logexy

Subtraction rule –

logex + logey   = logex/y
Solved problem :

Ex 1: Solve loge2 + loge4

Sol: Assume base as e. so  log 2 +log 4 = log 8

Ex 2: Solve loge4 - loge2

Sol: Base is e than log 4 -log 2 = log 2

Practice questions:

Que 1 : Change the following from exponential form to natural log form

e2.3025 =10
e1 =e

Ans : a. is  loge10 =2.3025

b.    is   normal form  as   logee = 1

Que 2: For log 23+ log 3 = log x then x=?,where all base is e.

Ans: 69

Que 3:For log 24 – log 4 =log x ,where all base is e.What is the value of x?

Ans: 6

Simple Logarithms

In mathematics, the logarithms of a number to a given base is the power or exponent to which the base must be raised in order to produce that number. For example, the logarithm of 100000 to base 10 is 5, because 5 is the power to which ten must be raised to produce 100000: 105 = 100000, so log10100000  = 5. Only positive real number  have real number logarithms; negative and complex numbers have complex logarithms.

I like to share this Simplifying Logarithms with you all through my article.

Simple logarithms are simple step produced by the problem.
Simple Logarithms Rules:

Let us see some of the simple steps that used to solve the logarithims.

Product rule: If a, p and q are positive numbers and a ?1, then

loga(pq) = logap +logaq

Quotient rule: If p, q and a are positive numbers and a ? 1, then,

log a(p/q) = log ap –loga q

Power rule: If a and q are positive numbers, a ? 1 and m is a real number, then

logapq =qlogap

Change of base rule: If p, q and a are positive numbers and p ? 1, a ? 1, then

Log pq = logap* logqa

Reciprocal rule: If p and q is the positive numbers other than 1, then

Log pq =1 / log pq
Examples Simple Logarithms:

Example 1:

Reduce: 22log3 27 + 22log3 729      (ii)75 log5 8 +75  log5 5/1000

Solution:

(i) Since the expressions is a sum of two logarithms and the bases are equal, we can apply the product rule

(i) 22log3 27 +4log3 729 =22 [log 3 (27*729)]

=22[ log3 (33*36)]

= 22log3 39 =22* 9 log33 = 22*9=198

(ii)               75 log58+75log5(`5/1000` )  =75 log5 (`(8*5)/1000` )

=75 log 5(`1/25` )

=75log 5(`1/ 5^2` )

=75 log5(5-2)

= -2*75 log55 = -150*1= -150



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Example 2:

solve: 89log7 98-89log714

Solution:

89log7 98-89log714 =89log 7(`98/14` )

=89 log77 =89*1= 89

Example 3:

Solve .log650x – log6(4x+1)

Solution:

Using quotient law, we can write the equations as log6 (50x / 4x+1) Changing into exponential form, we get

(50x /4x+1) = 60

50x = 4x + 1

46x=1

x=`1/46`

polynomials

In mathematics, a polynomial is an expression of finite length constructed from variables (also known as indeterminates) and constants, using only the operations of addition, subtraction, multiplication, and non-negative, whole-number exponents.

Rational function:

In mathematics, a rational function is any function which can be written as the ratio of two polynomial functions. It can be  written as((x - 2) / (x + 3))

(Source: Wikipedia)
Example Problems for Polynomials Rational Expressions

Polynomials rational expressions example problem 1:

Simplifying the given rational expressions ((5x + 15) / (10x + 40))

Solution:

Given rational expression is ((5x + 15) / (10x + 40))

Take 5 as common in the numerator value, we get

= ((5(x + 3)) / (10x + 40))

Take 10 as common in the denominator value, we get

= ((5(x + 3)) / (10(x + 4)))

Divide the both numerator and denominator value by 5, we get

= ((x + 3) / (2 (x + 4)))          

Answer:

The final answer is ((x + 3) / (2(x + 4)))

Polynomials rational expressions example problem 2:

Simplifying the given rational expressions ((4x + 12) / (2x + 84))

Solution:

Given rational expression is ((4x + 12) / (2x + 84))

Take 4 as common in the numerator value, we get

= ((4(x + 3)) / (2x + 84))

Take 2 as common in the denominator value, we get

= ((4(x + 3)) / (2(x + 42)))

Divide the both numerator and denominator value by 2 , we get

= ((2(x + 3)) / (x + 42))          

Answer:

The final answer is ((2(x + 3)) / (x + 42))

Algebra is widely used in day to day activities watch out for my forthcoming posts on Sphere Definition and Hemisphere Definition. I am sure they will be helpful.

Polynomials rational expressions example problem 3:

Simplifying the given rational expressions ((3x + 12) / (12x + 84))

Solution:

Given rational expression is ((3x + 12) / (12x + 84))

Take 3 as common in the numerator value, we get

= ((3(x + 4)) / (12x + 84))

Take 12 as common in the denominator value, we get

= ((3(x + 4)) / (12(x + 7)))

Divide the both numerator and denominator value by 3, we get

= ((x + 4) / (4 (x + 7)))          

Answer:

The final answer is ((x + 4) / (4(x + 7)))

Polynomials rational expressions example problem 4:

Simplifying the given rational expressions ((13x + 13) / (13x + 26))

Solution:

Given rational expression is ((13x + 13) / (13x + 26))

Take 13 as common in the numerator value, we get

= ((13(x + 1)) / (13x + 26))

Take 13 as common in the denominator value, we get

= ((13(x + 1)) / (13(x + 2)))

Divide the both numerator and denominator value by 13, we get

= ((x + 1) / (x + 2))          

Answer:

The final answer is ((x + 1) / (x + 2))
Practice Problems for Polynomials Rational Expressions

Polynomials rational expressions practice problem 1:

Simplifying the given rational expressions ((5x + 75) / (15x + 45))

Answer:

The final answer is ((x + 15) / (3(x + 3)))

Polynomials rational expressions practice problem 2:

Simplifying the given rational expressions ((3x + 30) / (6x + 54))

Answer:

Given rational expression is ((x + 10) / (2(x + 9)))

Polynomials rational expressions practice problem 3:

Simplifying the given rational expressions ((2x + 30) / (4x + 56))

Answer:

Given rational expression is ((x + 15) / (2(x + 14)))

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