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This module is from Fundamentals of Mathematics by Denny Burzynski and Wade Ellis, Jr. This module discusses combinations of operations with fractions. By the end of the module students should gain a further understanding of the order of operations.

Section overview

  • The Order of Operations

The order of operations

To determine the value of a quantity such as

1 2 + 5 8 2 15 size 12{ { {1} over {2} } + { {5} over {8} } cdot { {2} over {"15"} } } {}

where we have a combination of operations (more than one operation occurs), we must use the accepted order of operations.

    The order of operations:

  1. In the order (2), (3), (4) described below, perform all operations inside group­ing symbols: ( ), [ ], ( ),           . Work from the innermost set to the outermost set.
  2. Perform exponential and root operations.
  3. Perform all multiplications and divisions moving left to right.
  4. Perform all additions and subtractions moving left to right.

Sample set a

Determine the value of each of the following quantities.

1 4 + 5 8 2 15 size 12{ { {1} over {4} } + { {5} over {8} } cdot { {2} over {"15"} } } {}

  1. Multiply first.

    1 4 + 5 1 8 4 2 1 15 3 = 1 4 + 1 1 4 3 = 1 4 + 1 12 size 12{ { {1} over {4} } + { { {5} cSup { size 8{1} } } over { {8} cSub { size 8{4} } } } cdot { { {2} cSup { size 8{1} } } over { {"15"} cSub { size 8{3} } } } = { {1} over {4} } + { {1 cdot 1} over {4 cdot 3} } = { {1} over {4} } + { {1} over {"12"} } } {}

  2. Now perform this addition. Find the LCD.

    4 = 2 2 12 = 2 2 3 The LCD = 2 2 3 = 12 .

    1 4 + 1 12 = 1 3 12 + 1 12 = 3 12 + 1 12 = 3 + 1 12 = 4 12 = 1 3

    Thus, 1 4 + 5 8 2 15 = 1 3 size 12{ { {1} over {4} } + { {5} over {8} } cdot { {2} over {"15"} } = { {1} over {3} } } {}

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3 5 + 9 44 5 9 1 4 size 12{ { {3} over {5} } + { {9} over {"44"} } left ( { {5} over {9} } - { {1} over {4} } right )} {}

  1. Operate within the parentheses first, 5 9 1 4 size 12{ left ( { {5} over {9} } - { {1} over {4} } right )} {} .

    9 = 3 2 4 = 2 2 The LCD = 2 2 3 2 = 4 9 = 36 .


    5 4 36 1 9 36 = 20 36 9 36 = 20 9 36 = 11 36 size 12{ { {5 cdot 4} over {"36"} } - { {1 cdot 9} over {"36"} } = { {"20"} over {"36"} } - { {9} over {"36"} } = { {"20" - 9} over {"36"} } = { {"11"} over {"36"} } } {}


    Now we have


    3 5 + 9 44 11 36 size 12{ { {3} over {5} } + { {9} over {"44"} } left ( { {"11"} over {"36"} } right )} {}

  2. Perform the multiplication.

    3 5 + 9 1 44 4 11 1 36 4 = 3 5 + 1 1 4 4 = 3 5 + 1 16 size 12{ { {3} over {5} } + { { {9} cSup { size 8{1} } } over { {"44"} cSub { size 8{4} } } } cdot { { {"11"} cSup { size 8{1} } } over { {"36"} cSub { size 8{4} } } } = { {3} over {5} } + { {1 cdot 1} over {4 cdot 4} } = { {3} over {5} } + { {1} over {"16"} } } {}

  3. Now perform the addition. The LCD=80.

    3 5 + 1 16 = 3 16 80 + 1 5 80 = 48 80 + 5 80 = 48 + 5 80 = 53 80 size 12{ { {3} over {5} } + { {1} over {"16"} } = { {3 cdot "16"} over {"80"} } + { {1 cdot 5} over {"80"} } = { {"48"} over {"80"} } + { {5} over {"80"} } = { {"48"+5} over {"80"} } = { {"53"} over {"80"} } } {}


    Thus, 3 5 + 9 44 5 9 1 4 = 53 80 size 12{ { {3} over {5} } + { {9} over {"44"} } left ( { {5} over {9} } - { {1} over {4} } right )= { {"53"} over {"80"} } } {}

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8 15 426 2 1 4 15 3 1 5 + 2 1 8 size 12{8 - { {"15"} over {"426"} } left (2 - 1 { {4} over {"15"} } right ) left (3 { {1} over {5} } +2 { {1} over {8} } right )} {}

  1. Work within each set of parentheses individually.

    2 1 4 15 = 2 1 15 + 4 15 = 2 19 15 = 30 15 19 15 = 30 19 15 = 11 15 3 1 5 + 2 1 8 = 3 5 + 1 5 + 2 8 + 1 8 = 16 5 + 17 8 LCD = 40 = 16 8 40 + 17 5 40 = 128 40 + 85 40 = 128 + 85 40 = 213 40


    Now we have

    8 15 426 11 15 213 40 size 12{8 - { {"15"} over {"426"} } left ( { {"11"} over {"15"} } right ) left ( { {"213"} over {"40"} } right )} {}

  2. Now multiply.

    8 15 1 426 2 11 15 1 213 1 40 = 8 1 11 1 2 1 40 = 8 11 80 size 12{8 - { { {"15"} cSup { size 8{1} } } over { {"426"} cSub { size 8{2} } } } cdot { {"11"} over { {"15"} cSub { size 8{1} } } } cdot { { {"213"} cSup { size 8{1} } } over {"40"} } =8 - { {1 cdot "11" cdot 1} over {2 cdot 1 cdot "40"} } =8 - { {"11"} over {"80"} } } {}

  3. Now subtract.

    8 11 80 = 80 8 80 11 80 = 640 80 11 80 = 640 11 80 = 629 80 or 7 69 80 size 12{8 - { {"11"} over {"80"} } = { {"80" cdot 8} over {"80"} } - { {"11"} over {"80"} } = { {"640"} over {"80"} } - { {"11"} over {"80"} } = { {"640" - "11"} over {"80"} } = { {"629"} over {"80"} } " or "7 { {"69"} over {"80"} } } {}


    Thus, 8 - 15 426 2 - 1 4 15 3 1 5 + 2 1 8 = 7 69 80

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3 4 2 8 9 5 12 size 12{ left ( { {3} over {4} } right ) rSup { size 8{2} } cdot { {8} over {9} } - { {5} over {"12"} } } {}

  1. Square 3 4 size 12{ { {3} over {4} } } {} .

    3 4 2 = 3 4 3 4 = 3 3 4 4 = 9 16 size 12{ left ( { {3} over {4} } right ) rSup { size 8{2} } = { {3} over {4} } cdot { {3} over {4} } = { {3 cdot 3} over {4 cdot 4} } = { {9} over {"16"} } } {}

    Now we have

    9 16 8 9 5 12 size 12{ { {9} over {"16"} } cdot { {8} over {9} } - { {5} over {"12"} } } {}

  2. Perform the multiplication.

    9 1 16 2 8 1 9 1 5 12 = 1 1 2 1 5 12 = 1 2 5 12 size 12{ { { {9} cSup { size 8{1} } } over { {"16"} cSub { size 8{2} } } } cdot { { {8} cSup { size 8{1} } } over { {9} cSub { size 8{1} } } } - { {5} over {"12"} } = { {1 cdot 1} over {2 cdot 1} } - { {5} over {"12"} } = { {1} over {2} } - { {5} over {"12"} } } {}

  3. Now perform the subtraction.

    1 2 5 12 = 6 12 5 12 = 6 5 12 = 1 12 size 12{ { {1} over {2} } - { {5} over {"12"} } = { {6} over {"12"} } - { {5} over {"12"} } = { {6 - 5} over {"12"} } = { {1} over {"12"} } } {}

    Thus, 4 3 2 8 9 5 12 = 1 12 size 12{ left ( { {4} over {3} } right ) rSup { size 8{2} } cdot { {8} over {9} } - { {5} over {"12"} } = { {1} over {"12"} } } {}

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2 7 8 + 25 36 ÷ 2 1 2 1 1 3 size 12{2 { {7} over {8} } + sqrt { { {"25"} over {"36"} } } div left (2 { {1} over {2} } - 1 { {1} over {3} } right )} {}

  1. Begin by operating inside the parentheses.

    2 1 2 1 1 3 = 2 2 + 1 2 1 3 + 1 3 = 5 2 4 3 = 15 6 8 6 = 15 8 6 = 7 6

  2. Now simplify the square root.

    25 36 = 5 6 since 5 6 2 = 25 36 size 12{ sqrt { { {"25"} over {"36"} } } = { {5} over {6} } left ("since " left ( { {5} over {6} } right ) rSup { size 8{2} } = { {"25"} over {"36"} } right )} {}

    Now we have

    2 7 8 + 5 6 ÷ 7 6 size 12{2 { {7} over {8} } + { {5} over {6} } div { {7} over {6} } } {}

  3. Perform the division.

    2 7 8 + 5 6 1 6 1 7 = 2 7 8 + 5 1 1 7 = 2 7 8 + 5 7 size 12{2 { {7} over {8} } + { {5} over { {6} cSub { size 8{1} } } } cdot { { {6} cSup { size 8{1} } } over {7} } =2 { {7} over {8} } + { {5 cdot 1} over {1 cdot 7} } =2 { {7} over {8} } + { {5} over {7} } } {}

  4. Now perform the addition.

    2 7 8 + 5 7 = 2 8 + 7 8 + 5 7 = 23 8 + 5 7 LCD = 56 . = 23 7 56 + 5 8 56 = 161 56 + 40 56 = 161 + 40 56 = 201 56  or  3 33 56

    Thus, 2 7 8 + 25 36 ÷ 2 1 2 1 1 3 = 3 33 56 size 12{2 { {7} over {8} } + sqrt { { {"25"} over {"36"} } } div left (2 { {1} over {2} } - 1 { {1} over {3} } right )=3 { {"33"} over {"56"} } } {}

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Practice set a

Find the value of each of the following quantities.

5 16 1 10 1 32 size 12{ { {5} over {"16"} } cdot { {1} over {"10"} } - { {1} over {"32"} } } {}

0

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6 7 21 40 ÷ 9 10 + 5 1 3 size 12{ { {6} over {7} } cdot { {"21"} over {"40"} } div { {9} over {"10"} } +5 { {1} over {3} } } {}

35 6 size 12{ { {"35"} over {6} } } {} or 5 5 6 size 12{5 { {5} over {"6"} } } {}

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8 7 10 2 4 1 2 3 2 3 size 12{8 { {7} over {"10"} } - 2 left (4 { {1} over {2} } - 3 { {2} over {3} } right )} {}

211 30 size 12{ { {"211"} over {"30"} } } {} or 7 1 30 size 12{7 { {1} over {"30"} } } {}

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17 18 58 30 1 4 3 32 1 13 29 size 12{ { {"17"} over {"18"} } - { {"58"} over {"30"} } left ( { {1} over {4} } - { {3} over {"32"} } right ) left (1 - { {"13"} over {"29"} } right )} {}

7 9 size 12{ { {7} over {9} } } {}

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1 10 + 1 1 2 ÷ 1 4 5 1 6 25 size 12{ left ( { {1} over {"10"} } +1 { {1} over {2} } right ) div left (1 { {4} over {5} } - 1 { {6} over {"25"} } right )} {}

2 6 7 size 12{2 { {6} over {7} } } {}

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2 3 3 8 4 9 7 16 1 1 3 + 1 1 4 size 12{ { { { {2} over {3} } - { {3} over {8} } cdot { {4} over {9} } } over { { {7} over {"16"} } cdot 1 { {1} over {3} } +1 { {1} over {4} } } } } {}

3 11 size 12{ { {3} over {"11"} } } {}

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3 8 2 + 3 4 1 8 size 12{ left ( { {3} over {8} } right ) rSup { size 8{2} } + { {3} over {4} } cdot { {1} over {8} } } {}

15 64 size 12{ { {"15"} over {"64"} } } {}

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2 3 2 1 4 4 25 size 12{ { {2} over {3} } cdot 2 { {1} over {4} } - sqrt { { {4} over {"25"} } } } {}

11 10 size 12{ { {"11"} over {"10"} } } {}

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Exercises

Find each value.

4 3 1 6 1 2 size 12{ { {4} over {3} } - { {1} over {6} } cdot { {1} over {2} } } {}

5 4 size 12{ { {5} over {4} } } {}

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7 9 4 5 5 36 size 12{ { {7} over {9} } - { {4} over {5} } cdot { {5} over {"36"} } } {}

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2 2 7 + 5 8 ÷ 5 16 size 12{2 { {2} over {7} } + { {5} over {8} } div { {5} over {"16"} } } {}

4 2 7 size 12{4 { {2} over {7} } } {}

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3 16 ÷ 9 14 12 21 + 5 6 size 12{ { {3} over {"16"} } div { {9} over {"14"} } cdot { {"12"} over {"21"} } + { {5} over {6} } } {}

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4 25 ÷ 8 15 7 20 ÷ 2 1 10 size 12{ { {4} over {"25"} } div { {8} over {"15"} } - { {7} over {"20"} } div 2 { {1} over {"10"} } } {}

2 15 size 12{ { {2} over {"15"} } } {}

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2 5 1 19 + 3 38 size 12{ { {2} over {5} } cdot left ( { {1} over {"19"} } + { {3} over {"38"} } right )} {}

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3 7 3 10 1 15 size 12{ { {3} over {7} } cdot left ( { {3} over {"10"} } - { {1} over {"15"} } right )} {}

1 10 size 12{ { {1} over {"10"} } } {}

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10 11 8 9 2 5 + 3 25 5 3 + 1 4 size 12{ { {"10"} over {"11"} } cdot left ( { {8} over {9} } - { {2} over {5} } right )+ { {3} over {"25"} } cdot left ( { {5} over {3} } + { {1} over {4} } right )} {}

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2 7 6 7 3 28 + 5 1 3 1 1 4 1 8 size 12{ { {2} over {7} } cdot left ( { {6} over {7} } - { {3} over {"28"} } right )+5 { {1} over {3} } cdot left (1 { {1} over {4} } - { {1} over {8} } right )} {}

6 3 14 size 12{6 { {3} over {"14"} } } {}

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6 11 1 3 1 21 + 2 13 42 1 1 5 + 7 40 size 12{ { { left ( { {6} over {"11"} } - { {1} over {3} } right ) cdot left ( { {1} over {"21"} } +2 { {"13"} over {"42"} } right )} over {1 { {1} over {5} } + { {7} over {"40"} } } } } {}

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1 2 2 + 1 8 size 12{ left ( { {1} over {2} } right ) rSup { size 8{2} } + { {1} over {8} } } {}

3 8 size 12{ { {3} over {8} } } {}

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3 5 2 3 10 size 12{ left ( { {3} over {5} } right ) rSup { size 8{2} } - { {3} over {"10"} } } {}

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36 81 + 1 3 2 9 size 12{ sqrt { { {"36"} over {"81"} } } + { {1} over {3} } cdot { {2} over {9} } } {}

20 27 size 12{ { {"20"} over {"27"} } } {}

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49 64 9 4 size 12{ sqrt { { {"49"} over {"64"} } } - sqrt { { {9} over {4} } } } {}

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2 3 9 4 15 4 16 225 size 12{ { {2} over {3} } cdot sqrt { { {9} over {4} } } - { {"15"} over {4} } cdot sqrt { { {"16"} over {"225"} } } } {}

0

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3 4 2 + 25 16 size 12{ left ( { {3} over {4} } right ) rSup { size 8{2} } + sqrt { { {"25"} over {"16"} } } } {}

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1 3 2 81 25 + 1 40 ÷ 1 8 size 12{ left ( { {1} over {3} } right ) rSup { size 8{2} } cdot sqrt { { {"81"} over {"25"} } } + { {1} over {"40"} } div { {1} over {8} } } {}

2 5 size 12{ { {2} over {5} } } {}

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4 49 2 + 3 7 ÷ 1 3 4 size 12{ left ( sqrt { { {4} over {"49"} } } right ) rSup { size 8{2} } + { {3} over {7} } div 1 { {3} over {4} } } {}

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100 121 2 + 21 11 2 size 12{ left ( sqrt { { {"100"} over {"121"} } } right ) rSup { size 8{2} } + { {"21"} over { left ("11" right ) rSup { size 8{2} } } } } {}

1

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3 8 + 1 64 1 2 ÷ 1 1 3 size 12{ sqrt { { {3} over {8} } + { {1} over {"64"} } } - { {1} over {2} } div 1 { {1} over {3} } } {}

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1 4 5 6 2 + 9 14 2 1 3 1 81 size 12{ sqrt { { {1} over {4} } } cdot left ( { {5} over {6} } right ) rSup { size 8{2} } + { {9} over {"14"} } cdot 2 { {1} over {3} } - sqrt { { {1} over {"81"} } } } {}

125 72 size 12{ { {"125"} over {"72"} } } {}

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1 9 6 3 8 + 2 5 8 16 + 7 7 10 size 12{ sqrt { { {1} over {9} } } cdot sqrt { { {6 { {3} over {8} } +2 { {5} over {8} } } over {"16"} } } +7 { {7} over {"10"} } } {}

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3 3 4 + 4 5 1 2 3 67 240 + 1 3 4 9 10 size 12{ { {3 { {3} over {4} } + { {4} over {5} } cdot left ( { {1} over {2} } right ) rSup { size 8{3} } } over { { {"67"} over {"240"} } + left ( { {1} over {3} } right ) rSup { size 8{4} } cdot left ( { {9} over {"10"} } right )} } } {}

252 19 size 12{ { {"252"} over {"19"} } } {}

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16 81 + 1 4 6 size 12{ sqrt { sqrt { { {"16"} over {"81"} } } } + { {1} over {4} } cdot 6} {}

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81 256 3 32 1 1 8 size 12{ sqrt { sqrt { { {"81"} over {"256"} } } } - { {3} over {"32"} } cdot 1 { {1} over {8} } } {}

165 256 size 12{ { {"165"} over {"256"} } } {}

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Exercises for review

( [link] ) True or false: Our number system, the Hindu-Arabic number system, is a positional number system with base ten.

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( [link] ) The fact that 1 times any whole number = that particular whole number illustrates which property of multiplication?

multiplicative identity

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( [link] ) Convert 8 6 7 size 12{8 { {6} over {7} } } {} to an improper fraction.

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( [link] ) Find the sum. 3 8 + 4 5 + 5 6 size 12{ { {3} over {8} } + { {4} over {5} } + { {5} over {6} } } {} .

241 120 size 12{ { {"241"} over {"120"} } } {} or 2 1 120 size 12{2 { {1} over {"120"} } } {}

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( [link] ) Simplify 6 + 1 8 6 1 8 size 12{ { {6+ { {1} over {8} } } over {6 - { {1} over {8} } } } } {} .

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Questions & Answers

how to know photocatalytic properties of tio2 nanoparticles...what to do now
Akash Reply
it is a goid question and i want to know the answer as well
Maciej
Do somebody tell me a best nano engineering book for beginners?
s. Reply
what is fullerene does it is used to make bukky balls
Devang Reply
are you nano engineer ?
s.
what is the Synthesis, properties,and applications of carbon nano chemistry
Abhijith Reply
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
s. Reply
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
SUYASH Reply
for screen printed electrodes ?
SUYASH
What is lattice structure?
s. Reply
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
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what's the easiest and fastest way to the synthesize AgNP?
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Cied
types of nano material
abeetha Reply
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
Ramkumar Reply
what is nano technology
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what is system testing?
AMJAD
preparation of nanomaterial
Victor Reply
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
Himanshu Reply
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AMJAD
what is system testing
AMJAD
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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anybody can imagine what will be happen after 100 years from now in nano tech world
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after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
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Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
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how did you get the value of 2000N.What calculations are needed to arrive at it
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Source:  OpenStax, Fundamentals of mathematics. OpenStax CNX. Aug 18, 2010 Download for free at http://cnx.org/content/col10615/1.4
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