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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 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 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
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
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.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
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
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
Sanket Reply
what's the easiest and fastest way to the synthesize AgNP?
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Cied
types of nano material
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I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
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what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
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what is nano technology
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what is system testing?
AMJAD
preparation of nanomaterial
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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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