Problems for Fractions Reducing

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 were the common or "vulgar" fractions which are still used today (½, ⅝, ¾, etc.) and which consist of a numerator and a denominator.

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Word problems for fractions reducing Problems:

Word problems 1: Jack bought 6/4 liters of milk but he used only 2/4 liters of the milk. How much milk did he have left? And reducing  the fractions.

Solution:

Initially jack had 6/4 liter of milk

He used 2/4 liter of milk

Now we have to find the remaining amount

The remaining amount

=6/4 -2/4

= 6-2/4

=4/4

By reducing the fraction get 1 is the answer.

Word problems 2: My pet is 1/4 years old and my dog is 7/4 years old find the total sums of ages And reducing  the fractions

Solution:

Given pet is 1/4 years old

And my dog is 7/4 years old

Now we have to find the total sums of ages

= 1/4 +7/4

=1+7/4

= 8/4

By reducing the above fraction we get 2 is the answer.

Word problems 3: Rahul needs to drink 1/5 liters of coffee and 14/5 liters of coke every day. How much fluid do rahul have to drink? And reducing  the fractions

Solution:

Given

Rahul needs to drink 1/5 liter of coffee

And 14/5 liters of coke every day

Now we have to find how much fluid do rahul have to drink?

=  1/5 +14/5

=1 +14/5

= 15/5

By reducing the above fraction we get 3 is the answer.

Word problems for fractions reducing Problems:

Word problems 4: On the eating competition my aunty ate 3/6 pizza and my uncle ate 15/6 pizza. How much pizza did they eat jointly? And reducing  the fractions

Solution:

Given:

Eating competition my aunty ate 3/6 pizza

And my uncle ate 15/6 pizza

How much pizza did they eat jointly?

=   3/6 +15/6

=  3+15/6

= 18/6

By reducing the above fraction we get 3 is the answer.

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Word problems 5: John bought 3/11 liters of oil but he used only 25/11 liters of the oil. How much oil did he have left? And reducing  the fractions

Solution:

Initially john had 3/11 liter of oil

He used 25/11 liter of oil

Now we have to find the remaining amount

The remaining amount

=3/11 -25/11

=  3-25/11

=  22/11

By reducing the fraction get 2 is the answer.

Function Rule in Math

Function rule in math Linear input output functions are finding unknown variables from the given expression with the help of known values. In the expression variable are x and y. The expression deals with linear equation. find the linear expression of the particular function. Function notation is look f(x), p(x),… to find the x value of the functions. Function notation is f(x) = mx+b. in this equation the value of x is given we need to find the value of m and b.

Example of funciton rules in math: The funciton f(x) 2x2+4x+2, using function rules in math find the f(4),f(5) and f(6).

problems in square equation of function rules in math

Problems using the square of the equation find the function rules in math.

Problem 1; find the function rule in math of square equation

F(x) = x2 +2x +4 find the x = 2,3,4,5

Solution :

Given the function of f(x) there is x value is given find the function rules in math

f(x) = x2 +2x +4 find the f(2),f(3), f(4) and f(5)

The value of x is 2 is given

f(2) = 22 +2*2 +4

f(2) = 4 +4 +4 In this step 2 square is 4 it is calculate and 2*2 is 4 be added

(ii)f(2) = 12

f(3) = 32+2*3+4

f(3)= 9+6+4 = 19

f(3) = 19

(iii)f(4)= 42+2*4+4

f(4)= 16+8+4 = 28

f(4)=28.

(iv)f(5) = 52+2*5+4

f(5) = 25+10+4 = 39

f(5) = 39

The function rule in math  f(2) = x2 +2x +4 find the f(2),f(3),f(4) and f(5) is 12,19,28,39.

Simplify the function rule in math using the linear equation

Simplifying the function rule in math f(x) = 3x+12. The input function x=4,5,6,7.

Solution: The given equation is linear substitute the linear input function of f(4).

f(x) = 3x+12 the linear input function is 4

f(4) = 3*4+12. here multiply the 3*4 is 12 and add.

f(4) = 12+12

f(4) = 24. the linear output function is 24.

f(5) = 3*5+12 = 15+12

f(5) = 27

f(6) = 3*6+12

f(6) = 18+12

f(6) = 30

f(7) = 3*7+12

f(7) = 21+12

f(7) = 33.

The function rule in math f(x) = 3x+12. The input function is f(4),f(5),f(6) and f(7) output is 24,27,30 and 33

Solving Trigonometric Functions

In mathematics, the trigonometric functions are functions of an angle. They are used to relate the angles of a triangle to the lengths of the sides of a triangle. Trigonometric functions are important in the study of triangles.

The most familiar trigonometric functions are the sine, cosine, and tangent. Trigonometric functions can also be called as circular functions.

A few example problems are given below to learn solving trigonometric functions.

(Source: Wikipedia)

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Example problems for solving trigonometric functions:

Example 1:

Prove that  (cos6x + cos 4x)/(sin 6x - sin4x) = cot x

Solution:

To prove the above equation,  take the left hand side (LHS) of the equation and simplify it step by step to get right hand side of the equation (RHS).

Step 1: Given

(cos6x + cos 4x)/(sin 6x - sin4x) = cot x

Step 2: Take left hand side of the equation and prove it

LHS = (cos6x + cos 4x)/(sin 6x - sin4x)

= (2cos((6x+4x)/2)cos((6x-4x)/2))/(2cos((6x+4x)/2)sin((6x-4x)/2))       [Using trigonometric formulas]

Solving above function, we get

= (2cos5xcosx)/(2cos5xsinx)

=  cosx/sinx

= cot x

= RHS

Hence proved

Example 2:

Prove that  (sin3x - sinx)/(cosx - cos3x) = cot 2x

Solution:

To prove the above equation,  take the left hand side (LHS) of the equation and simplify it step by step to get right hand side of the equation (RHS).

Step 1: Given

 (sin3x - sinx)/(cosx - cos3x)  = cot 2x

Step 2: Take left hand side of the equation and prove it

LHS = (sin3x - sinx)/(cosx - cos3x)

= (2cos((3x+x)/2)sin((3x-x)/2))/(-2sin((x+3x)/2)sin((x-3x)/2))         [Using trigonometric formulas]

Solving above function, we get

= (cos2xsinx)/(-sin2xsin(-x))

= (cos2xsinx)/(sin2xsinx)

= cot 2x

= RHS

Hence proved

Example 3:

Solve the function (sin3x-sinx)/(cos2x)

Solution:

Given:

(sin3x-sinx)/(cos2x)

Solving above function using trigonometric formula sin C - sin D = 2cos(C+D)/2 sin(C - D)/2 , we get

(sin3x-sinx)/(cos2x) = (2cos((3x+x)/2)sin((3x-x)/2))/(cos2x)

= (2cos2xsinx)/(cos2x)

= 2sin x

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Example 4:

Solve (sinx - siny)/(cosx + cosy)

Solution:

We can write (sinx - siny)/(cosx + cosy) as follows using trigonometric identities,

(sinx - siny)/(cosx + cosy) = (2cos((x+y)/2)sin((x-y)/2))/(2cos((x+y)/2)cos((x-y)/2))

= tan((x-y)/2)

Practice problems for solving trigonometric functions:

1) Prove that sin x + sin 3x + sin 5x + sin 7x = 4sin 4x cos 2x cos x

2) Prove that (sin5x - 2sin3x + sinx)/(cos5x - cosx) = cosec 2x - cot 2x

3) Simplify (sinx + sin3x)/(cosx - cos 3x)

Ans: cot x

Geometric Progression

In mathematics, a geometric progression, are called as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed non-zero number called the common ratio. For example, the sequence 2, 6, 18, 54 ... is a geometric progression with common ratio 3. Similarly 10, 5, 2.5, 1.25, ... is a geometric sequence with common ratio 1/2. The sum of the conditions of a geometric progression

Source: Wikipedia

Concepts of geometric progression:

A geometric progression is a sequence of number each of which, after the first, is obtained by multiplying the preceding number by a constant r. This constant r is called the common ration of the geometric progression.

In other words, a series is said in geometric progression if the ratio of any term to the preceding term is constant throughout. It is briefly written as geometric progression.

The constant ratio is called the common ratio. It is denoted by “r”.

Thus, the general form of a geometric sequence is

a,ar,ar2,ar3,ar4,…

And that of a geometric series is

a+ar+ar2+ar3+ar4+…

Where r ≠ 0 is the common ratio and a is a scale factor, equal to the sequence's start value.

Formula for geometric progression:

For the geometric progression represented by

a,ar,ar2,ar3,…,arn-1

With common ratio r, the sum of the first n terms, denoted by Sn is

Sn=a+ar+ar2+ar3+…+arn-1

Multiplying both sides by r, we get

rSn=ar+ar2+ar3+…+arn-1+arn

Subtracting, we have

Sn-rSn=a-arn

Sn(1-r)=a(1-rn)

So that Sn= a(1-rn)/(1-r)       if r≠1.

When r=1, the progression is merely a,a,…a so that Sa in this case is simply na. Note that Sn can also be written as

Sn= (r_(n-1))/(r-1)                    r≠1

Example 1:

Find the 12th term of the series:

3,6,12,24,48…

Solution:

The given series is a geometric progression with a=3,r=6/2 =2

T12=ar11=3 x 211=6144

Practice problem for geometric progression:

The sum of the first 8 terms of the geometric progression with the first term a=25 and the common ration r=- 1/5 is

Answer: S8=25 (1-(1/(5_8) )) / (6/5 )

Multiply Rational Expressions

How to Multiply Rational Expressions

Multiplication of rational expressions is similar to the multiplication of fractions of rational numbers. Let us consider an example of multiplying fractions, 3/4 x 7/2. Here the method to be followed is, first multiply the terms in the numerator and then multiply the terms in the denominator which gives 3x7/4x2 = 21/8.

Then any further simplification possible is carried out. In this example there was no possibility of cancelations as there were no common factors and hence the fraction got by multiplying the numbers in the numerator and the numbers in the denominators would be the final answer. Similar set of steps are used while multiplying rational expressions.

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A rational expression  consist of coefficients which are constants for a given term and hence first the coefficient parts are simplified then the product of the variable parts in the numerator is found, then the product of the variable part in the denominator is found and finally the products are written in the simplified form. In this process we use exponential laws and common factors for simplification.

Let us consider an example for a better understanding. Multiply Rational Expressions 4x2y2/3x and 3x/4y.  The product of the given expressions would be,  4x2y2/3x . 3x/4y Here first the coefficients are simplified and any cancelations possible are carried out. 4 and 3 get canceled and we get x2y2/x . x/y, this can be written as x.x.y.y/x. x/y; x and y are the common factors which can be canceled and we get x2y which is the final answer. Here we can use the exponential rule in simplification.

To multiply rational expressions involving polynomials the steps to be followed are:
First factor the terms in the numerator and the denominator
Reduce all the common factors
Further simplify the terms by either multiplying the numerators and denominators or leave the product in the fraction form

Example: Multiply (x2-9)/(x2+6x+9) and (3x+9)/(3x-9)
Solution: First factor the terms in the numerator and denominator
(x2- 9)= (x+3)(x-3) [special products]
(x2+6x+9)= (x+3)(x+3) [(a+b)2=a2+2ab+b2]
(3x+9)=3(x+3)                               [taking out common factor]
(3x-9) = 3(x-3)                               [taking out common factor]

The product of the expressions would be in the form, (x2-9)/(x2+6x+9) . (3x+9)/(3x-9)
Re-writing the given expressions as factors gives,
(x+3)(x-3)/(x+3)(x+3)  .  3(x+3)/(x-3)
Reducing the common factors gives,
(x+3)(x-3)/(x+3)(x+3)  .  3(x+3)/3(x-3)
Canceling the common terms gives 3/3 = 1, the final answer!

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Multiplying Rational Expressions Solver performs the multiplication and division of rational numbers when the expressions are entered in the given fields. This helps to check the answers.

Exam for Total Differentiation

The total differentiation of a function, f of several variables. For example, t, x, y etc., with respect to t is different from the partial derivative. The total differentiation of f with respect to t does not assume and other variables are constant while t varies. The  total differentiation adds in these indirect dependencies to find the overall dependency of f on t. For example, the total differentiation of f(t,x,y) with respect to t is

(df)/(dt) = (df)/(dt) +  ((df)/(dx)) ((dx)/(dt)) + ((df)/(dy))  ((dy)/(dt)) .

Exam for total differentiation problems:

Let us see some example problems for total differentiation and its helps to exam preparation.

Exam for total differentiation problem 1:

Find the total differentiation of trigonometric term (cot x2) with respect to x.

Solution:

Given trigonometric term is  (cot x2)

Let      y = (cot x2)

Putting x2 = u and  cot x2 = cot u = y

y = cot u,        and       u = x2

dy/(du) = - cosec2 u, and  (du)/(dx) = 2x

The total differentiation is  dy/dx

dy / dx = [  dy/(du) ×  (du)/(dx)]

= [( - cosec2 u) × 2x]

=  2x ( - cosec2 u)                                                         (u = x2)

= - 2x cosec2 (x2).

dy/dx = - 2x cosec2 (x2).

Answer:    dy/dx =  - 2x cosec2 (x2)

Exam for total differentiation problem 2:

Find the total differentiation of  function  P = xe^(y + z) ,  x = uv,  y = u - v , z = u + v

Find     (dP)/(dv)      when u = 1  and v = -1

Solution:

Given function is    P =  xe^(y+ z),

x = uv,        y = u - v        and     z = u + v

((dx)/(dv))= u = 1          ((dy)/(dv))  = -1     and    ((dz)/(du)) = 1

We know the chain rule of partial differentiation. it is the Total differentiation

 (dP)/(du) =  ((dP)/(dx)) ((dx)/(du)) +  ((dP)/(dy))   ((dy)/(du))  + ((dP)/(dz))   ((dz)/(du)) .

Substitute the u and v value we get   x = -1,    y = 2   and z = 0

P = xe^(y + z),

Differentiate P with respect to x .       (dP)/(dx)   = e^(y + z).

Differentiate P with respect to y .        (dP)/(dy) =  xe^(y + z)

Differentiate P with respect to z.       (dP)/(dz)   = xe^(y + z) .

Total differentiation is       (dP)/(dv) =  ((dP)/(dx)) ((dx)/(dv)) +  ((dP)/(dy))   ((dy)/(dv))  + ((dP)/(dz))   ((dz)/(dv)).

= e^(y + z).(1) +  xe^(y + z) .(-1) +  xe^(y + z).1

= e^(2 + 0).(1) +  (-1)e^(2 + 0) .(-1) +  (-1)e^(2 + 0).1

= e2 + e2 - e2

= e2

Answer:   e2

Addition of Decimals Methods

In this article we are going to learn about adding decimals concept.Decimal is base 10 number system. A number consists of decimal part and whole number part is known as decimal number. The point which splits the whole number part and decimal part is known as decimal point. The digit after a decimal point is decimal part and the left to the point is whole number part.

Addition of Decimals Methods

Th following are the steps for addition with decimals:

Step 1: Checking the decimal part of given numbers and making them equal in number of digits by adding zeros at the end.

Step 2: Write down the numbers vertically one under the other with the aligned decimal point

Step 3: Add the numbers as normal addition and place the decimal point in the result

Examples

Below are the examples based on Adding decimals:

Example 1:

1)      Add 12.43 with 5

Sol:

Write 5 as 5.00 (as 12.43 has 2 digits after decimal point)

Add 12.43+5.00

12.43

5.00

---------

17.43

Hence the sum is, 12.43 + 5 = 17.43

Example 2:

Add 1.34 + 6 + 12.7.

Sol:

Making decimal part equal ,

6= 6.00, 12.7 = 12.70

1.34

6.00

12.70

--------

20.04

Hence the sum is, 1.34 + 6 + 12.7 = 20.04

Example 3:

Add 0.803+ 3.1+ 12

Sol:

Making decimal part equal,

3.1 = 3.100, 12 = 12.000

0.803

3.100

12.000

------------

15.903

Hence the sum is 0.803+ 3.1+ 12 = 15.903

Example 4:

Add   7.09 +9.20 + 0.36

Sol:

7.09

9.20

0.36

---------

16.65

Hence the result is 7.09 +9.20 + 0.36 = 16.65

Example 5:

Add 45. 56 + 6.7+ 2

Sol:

Making decimal part equal,

6.7= 6.70, 2=2.00

45.56

6.70

2.00

-------------

54.26

Hence the sum is 45. 56 + 6.7+ 2 = 54.26

Addition with decimals Ex 5:

Add   0.5795 + 2.5301

Sol:

0.5795

2.5301

-----------

3.1096

Hence the sum is 0.5795 + 2.5301= 3.1096

Addition with decimals Ex 6:

Add 82.543+322.916

Sol:

82.543

322.916

----------------

405.459

Hence the sum is 82.543+322.916 = 405.459

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Addition with decimals Ex 7:

Add 369.2165 with 100

Making decimal part equal,

100 = 100 .000

369.2165

100 .0000

-------------

469.2165

Hence the sum is 369.2165+100 = 469.2165.

Addition with decimals Ex 8 :

Add 0.00013+3.902+56.7

Sol:

Making decimal part equal ,

3.902 = 3.90200 , 56.7 = 56.70000

0.00013

3.90200

56.70000

------------------------

60.60213

Hence the sum is 0.00013+3.902+56.7 =   60.60213.

Practice problems

Below are the practice problems on adding decimals:

Add 32+4.5+7.03

Answer: 43.53

Add 2.34+5.6+0.007

Answer: 7.947

Add 7.985 with 9.71485

Answer: 17.69985

Add 0.009+7.89+6.0

Answer: 13.899

Add 1.111+4.67+17

Answer: 31.899