Solve Arc Lengths

Introduction to solve arc lengths:
The arc length is the term which we use in the geometry circles in mathematics. The arc length is nothing but the curve where it is the part of the circumference of the circle. The arc lengths in the circles are determined with the help of the example problems. Let us see about solve for arc lengths with the help of formula.

Formula and Procedure to Solve Arc Lengths:
Now we see about the formula and the procedure to solve for the arc lengths with the help of the example problems.

Formula to solve arc lengths:

The arc lengths of the circle are determined with the help of the angle values and the radius value of the circle. The arc length of the circle which can be solved as follows,

Arc length = `(theta/360)` (2πr)

Where `theta` = central angle of the circle

r = radius of the circle.

Procedure to solve for arc lengths:

The procedure with steps are given below to solve for the arc lengths

The given data of the radius and the central angle measurements are taken first.

Then, the values are substituted in the arc lengths formula.

First divide the given angle by 360 degrees and the other value of the circumference of the circle with radius value is multiplied.

Then, multiply both the values we get the result of the arc length of the circle.

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Problem to Solve for Arc Lengths:

Example:

Solve for the measure of the arc length, where measure of the radius is given as 6 cm and the central angle measurement is about 200 degrees?

Solution:

The arc lengths are calculated as follows,

Arc length = `(theta/360)` (2πr)

Arc length = `(200/360)` (2 × 3.14 × 6)

We have to determine the fraction of an angle by using the `theta/360`

Now we substitute the value in the formula we can get the length as follows,

Arc length = 0.555 × 2 × 3.14 × 6

Arc length = 0.555 × 37.68

Arc length = 20.91 cm.

Multiplying Decimals by Powers of 10

Introduction to Multiplying Decimals by Powers of 10:

              The decimals are the number which represents a real number and every decimal place after the decimal point is called a multiple of a negative power of 10. The powers of 10 are 1, 10, 100, 1000, and 10000 etc. these powers can be written using the exponents such as 1 = 100, 10 = 101, 100 = 102, 1000 = 103, etc. Let us see about multiplying decimals by powers of 10 in this article.

Exponents Used for Multiplying Decimals Powers of 10

Definition of Exponents:

The exponent is said to be how many times the number multiplied by it.

For example:

102 = 10 × 10

103 = 10 × 10 × 10

Exponents for Decimals in Powers of 10:

The fractions 1/10, 1/100, 1/1000, 1/10000, and etc is called powers of 10.

The above fractions can be written in terms of decimals.

0.1, 0.01, 0.001, 0.0001, and etc.

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Rules for Multiplying Decimals by Powers of 10

For multiplying decimals we need to move only the decimal place.
Multiply the number by the power of 10 like 10, 100, 1000, and etc, we have to move the decimal point right to appropriate places in a real number.
To multiply the number by 10 we have to move right to one place.
To multiply the number by 100 we have to move right to two places.
To multiply the number by 1000 we have to move right to three places.

Example Problems to Multiplying Decimals by Powers of 10

Example 1:

4.78 × 10

Solution:

For multiplying 4.78 by 10 we have to move the decimal place to right for one  appropriate place, we get,

4.78 × 10 = 47.8

The solution for multiplying 10 to the decimal 4.78 is 47.8.

Example 2:

3.86 ×100

Solution:

For multiplying 3.86 by 100 we have to move the decimal place to right for two appropriate places, we get,

3.86 × 100 = 386

The solution for multiplying 100 to the decimal 3.86 is 386.

Problems to Practice for Multiplying Decimals by Powers of 10

1. Solve 6.7 × 10

Key: 67

2. Solve 7.8 × 103

Key: 7800

3. Solve 12.47 × 104

Key; 124700

4. Solve 87.2356 × 106

Key: 87235600

Introduction for adding decimals calculator

Introduction for adding decimals calculator:

In adding decimal calculator we have to arrange decimal point in order in given decimal number. When we adding decimals numbers like ordinary addition from left to right and then keep decimal point in the added value. In decimals calculator we have to add decimal numbers with the same length and take added value into decimal number.In decimals calculator we can adding many decimals  numbers.decimals number was indicated by point representation.

Addition of Decimals Calculator

Place value for decimal number system:
A decimal numbers are like a fractions are same whose denominators are 10, 100 etc. A decimal number addition may be contains a whole number and a decimal number. In a decimal number system, when a whole number part is not given, we can be indicate decimal number by writing a zero on the left of the decimal point. In decimal number we have some simple notation for addition. Decimal numbers has point representation for accurate measurements.

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Example Problems in Decimal Addition:

Example 1:
         Add 0.256 and 1.31

Solution:
        Here 1.31 written into 1.310
          0.256                  
          1.310
          1.566

Example 2:
         Add 0.1 and 0.006

Solution:
        Here 0.1 written as 0.100
          0.100
          0.006
          0.106

Example 3:
       Add 3 and 4.22

Solution:
     Here we represent 3 into 3.00 and we add
       3.00
       4.22
       7.22

Example 4:
        Add 0.870 and 2.36

Solution:
     Here we represent 2.36 into 2.360 and we add
    0.870
    2.360
    3.230


Example 5:
        Add 2.1 and 31.56

Solution:
     Here we represent 2.1 into 2.10 and we add
    2.100
  31.560
  33.660

Introduction to word problems in rates and ratios

Introduction to word problems in rates and ratios:

           Math word problems we solve the problems by replacing the data’s given in the problem by alphabets (text manner) rather than using numerical notation to get the solution of the problem. We can solve the mathematical problems in two steps:

1) Prepare equations with the help of data’s given in the problem.

2) Solve the equations which you have prepared.

Here are some example word problems in rates and ratios:

Example Word Problems in Rates:

          In rates we see problems in related to money and work.

Example 1 [rates]:

          Last month Craig earned $330 by working in a shoe company as part time job for 30 hours. Find

a)    Find his rate of pay

b)    Find how much he would have earned if he had worked for 40 hours.

Given data:

                   Craig earned by working: $310,

                   Craig worked for: 30 [hours].

Solution:

a)    To find his rate of pay we use:

    Rate of pay=money earned/hours worked

                       =`330/30`

                        =11 [per hour]

b)To find the earnings if he had worked 40 hours:

                            = rate xx hours he would work

                            =`11 xx 40`

                            =440 [dollars]

Therefore, Craig would have eabrown $440 if he had worked for 40 hours.

Example 2 [rates]:                                                    

          Two students can wash the car 6 cars in 1 hour. How much would 8 students if they are charged $10 to wash a car?

Given data:

          Number of cars two students can wash in an hour: 6 [cars],

          Washing one car is charged about: $10.

Solution:

Step 1:

          Number of cars which can be washed by 8 students let it be x,

          Therefore,

                             `2/8` =`6/x`

                             2x=`8 xx 6` ,

                             2x=48,

                             x=24.

Step 2:

          Total money earned if they charged $10 a car:

                                      =`24 xx 10`

                                      =240 [dollars]

          Therefore, they earn $240 by washing cars.

Example Problems in Ratios:

Example 3 [ratios]:

Fred and George have pens at the ratio 6:1.fred gave half of his pens to George. Find the ratio of pen’s that fred and George have at the end.

Solution:

        In the question it’s given that Fred has given half of his pens to George.

        Therefore, the unit in which Fred has right now,

                                                =62

                                                =3 [units]

Now, George gets 3 units of pens,

Therefore, the unit that George has now=1+3

                                                                =4 [units]

Now, the ratio of pens that Fred and George have now is 3:4.

Example 4 [ratios]:

                Ricky arranges some brown and green covered book in his self. The ratio of the number of brown covered book to the number green covered books is 2:1.He arranges 9 more green covered books in his self and the ratio becomes 4:5.

        Now find the following questions

a)   How many Brown covered book are in the box?

b)   How many green covered books does Ricky have in the end?

Solution:

                The ratio of the number of brown covered books to the number of green covered book is 2:1

Therefore,                                           b=2g ----------- (1)                   

                          Now she adds 12 more green books in the shelf and the ratio becomes 4:5

Therefore,             

                                                              5b=4(g+9)

                                                              5b=4g+36 ----- (2)

Now substituting 1 in 2,

We get,

                                 5(2g) =4g+36

                                10g=4g+36 

             [Now by moving 4r to the other side, as it is positive sign it becomes negative when placed in the other side.]

                                10g-4g=36

                                   6g=36

  [Now by moving 6 on the other side]

                                  g=`36/6`

                                   g=6.                             

a)   To find the no of brown covered books in the box, we use equation 1:

                      b=2g

                      b=2×6        [where b=6]

                      b=12 [books]

Therefore, the no of brown covered books that Ricky has arranged is 12 books.

b)   To find the no of green covered books that Ricky has to arrange at the end, we use the equation,

                            g=b+12

                             g=12+12     [where b=12]

                             g=24.

Therefore, the number of green covered books does Ricky at the end is 24 books

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Introduction to largest common factor

Introduction to largest common factor: 
    
               In mathematics, the largest common factor, also known as the greatest common denominator, greatest common factor (gcf), or highest common factor (hcf), of two or more non-zero integers, is the largest positive integer that divides the numbers without a remainder.
              The example and practice problems are given below which will help you to learn finding largest common factor.

Example Problems on Largest Common Factor:

Example 1:

Find the largest common factor of 8 and 14.

Solution:

      Step 1: Find the factors of 8

                        We can divide 8 by 1 or 2 or 4 or 8, so

                                    Factors of 8 = 1, 2, 4, 8

      Step 2: Find the factors of 14

                        We can divide 14 by 1 or 2 or 7 or 14, so

                                    Factors of 14 = 1, 2, 7,14

      Step 3: Find the common factors of 8 and 14

                              Common factors of 8 and 14 = 1, 2

                    Here, 2 is greater than 1. So, 2 is the greatest common factor

      Step 4: Solution

                          Largest common factor of 8 and 14 is 2

Example 2:

Find the largest common factor of 9 and 27.

Solution:

      Step 1: Find the factors of 9

                        We can divide 9 by 1 or 3 or 9, so

                                    Factors of 9 = 1, 3, 9

      Step 2: Find the factors of 27

                        We can divide 27 by 1 or 3 or 9 or 27, so

                                    Factors of 27 = 1, 3, 9, 27

      Step 3: Find the common factors of 9 and 27

                              Common factors of 9 and 27 = 1, 3, 9

                    Here, 9 is greater than 1 and 3. So, 9 is the greatest common factor

      Step 4: Solution

                          Largest common factor of 9 and 27 is 9

Example 3:

Find the largest common factor of 15 and 35.

Solution:

      Step 1: Find the factors of 15

                        We can divide 15 by 1 or 5 or 15, so

                                    Factors of 15 = 1, 5, 15

      Step 2: Find the factors of 35

                        We can divide 35 by 1 or 5 or 7 or 35, so

                                    Factors of 35 = 1, 5, 7, 35

      Step 3: Find the common factors of 15 and 35

                              Common factors of 15 and 35 = 1, 5

                    Here, 5 is greater than 1. So, 5 is the greatest common factor

      Step 4: Solution

                          Largest common factor of 15 and 35 is 5

Practice Problems on Largest Common Factor:

1) Find the largest common factor of 6 and 16.

2) Find the largest common factor of 18 and 24.

3) Find the largest common factor of 12 and 56.

Solutions:

1) Largest common factor of 6 and 16 is 2

2) Largest common factor of 18 and 24 is 6

3) Largest common factor of 12 and 56 is 4.

Introduction to multiplying rational expressions calculator

 Introduction to multiplying rational expressions calculator:  

     In the algebraic expression the  variable does not occur in the fraction  or negative index. While using the rational expression calculator the variable that occur in the fractional notation. The calculator show the error mistake. while in  the multiply algebraic expression the calculator will be mention in the integer it may not occur any fraction or negative variable.

  
For example  5x2- 3x +2 this is the algebraic expression
      An rational expression of the form A(x) * B(x) where A(x) and B(x) are two polynomials over the set of real numbers and QA(x) ? 0 is called a rational expression.
For example 2/x^2 , ((x^4+x^3+x+1))/((x+5)), are rational expressions.

Rational Expression on Multiplying Rational Expressions Calculator

Problem in rational expression calculator In this expression the variable only in the integer not in the fraction form.
1. Simplify: (x2-x-6)/(x2+5x+6)
= ((x^2-x-6))/((x^2+5x+6))
= ((x-3)(x+2))/((x+2)(x+3))
= ((x-3))/((x+3))
Multiplication of rational expressions
The  product of rational expression in the form. The resulting expression is then reduced to its lowest form. If p(x)*g(x)
= (p(x))/(q(x)) + (g(x))/(h(x))
= (g(x))/(h(x)) *  (g(x))/(h(x))
     In this expression the multiplication in the rational expression are in the status of variable must be in the integers not in the fraction. The multiplication  rational expression  calculator is  reduced to its lowest form.

Problems on Multiplying Rational Expressions Calculator

A rational expression A(x) * B(x) can be reduced to its lowest term by multiplying A(x) and B(x) using calculator method.multiplying the  rational term A(x) the  expression  B(x)  in the calculator method  the variable x is only in the integer not in the fraction  term.
Multiply Rational expression calculation:
2. (x2-2x+ 1) / (x2-3x+2) * (3x-6) / (6x-6)
Solution :
x2 – 2x + 1 = (x–1)2
x2– 3x + 2 = (x – 2) (x – 1);
6x – 6 = 6(x –1)
3x – 6 = 3( x – 2)
((x-1)(x-1))/((x-1)(x-2))  * (6(x-1))/(3(x-2))
2(x-1)^2/((x-2))
3. Simplify (5ab)/(15cd) * (4cb)/(32ad) * (16ac)/(2bc)  in the rational expression calculator
Solution (5ab)/(15cd) * (4cb)/(32ad)  * (16ac)/(2bc)
= (5*a*b*4*c*b*16*a*c) /(5*3*c*d*2*16*a*d*2*b*c)
= (ab)/(3d^2).
Solution (a^3+b^3)/( a^2+2ab+b^2) *( a2-b2) / (a – b)
=(a+b)(a2-ab+b2) *` ((a+b)(a-b))/((a+b)(a+b)(a-b))
=a2-ab+b2
3. Simplify (x^2-x-6)/(x^2+5x+6)
Solution:
=(x^2-x-6)/(x^2+5x+6)
=(x-3)(x+2) / (x+2)(x+3)
=(x-3)/(x+3)