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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

a perfect square v²+2v+_
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kkk nice
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
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or infinite solutions?
Kim
y=10×
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if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
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rolling four fair dice and getting an even number an all four dice
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Kristine 2*2*2=8
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Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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is it 3×y ?
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J, combine like terms 7x-4y
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f(x)= 2|x+5| find f(-6)
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f(n)= 2n + 1
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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preparation of nanomaterial
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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7hours 36 min - 4hours 50 min
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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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