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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

what is variations in raman spectra for nanomaterials
Jyoti Reply
I only see partial conversation and what's the question here!
Crow Reply
what about nanotechnology for water purification
RAW Reply
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
what is the stm
Brian Reply
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
LITNING Reply
what is a peer
LITNING Reply
What is meant by 'nano scale'?
LITNING Reply
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
what is Nano technology ?
Bob Reply
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
Damian Reply
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
Stoney Reply
why we need to study biomolecules, molecular biology in nanotechnology?
Adin Reply
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
Adin
why?
Adin
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
anyone know any internet site where one can find nanotechnology papers?
Damian Reply
research.net
kanaga
sciencedirect big data base
Ernesto
Introduction about quantum dots in nanotechnology
Praveena Reply
what does nano mean?
Anassong Reply
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
Damian Reply
absolutely yes
Daniel
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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