# public Khan /khan-exercises

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  41183eb2 » marcia  2011-09-18 Add exercise -- distributive property 1  ed2d8a69 » marcia  2011-09-27 Add a replace the symbol distributive property problem type 2  41183eb2 » marcia  2011-09-18 Add exercise -- distributive property 3 4 5 Distributive property 6 7 8 9
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 ed2d8a69 » marcia  2011-09-27 Add a replace the symbol distributive property problem type 11
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13 randRangeUnique( 2, 10, 3 ) 14 A * ( B + C ) 15
 24ba7ea8 » Lucy Bain  2011-11-26 changed wording on question so clarify for bug 7100 16

Use distribution to solve:

 41183eb2 » marcia  2011-09-18 Add exercise -- distributive property 17

A\times(B + C)

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RESULT

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Each row has B + C = B+C rectangles, and there are A rows.

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23 init({ 24 range: [ [ 0, 1 ], [ 0, A ] ], 25 scale: [ 180, 30 ] 26 }); 27 rectchart( [ B, C ], [ "#FFA500", "#6495ED" ], A - 1 ); 28 for ( var i = 0; i < A - 1; i++ ) { 29 rectchart( [ B, C ], [ "#FFCF22", "#85B7FF" ], i); 30 } 31
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Counting by color:

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Number of orange rectangles: A \times B = A * B

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Number of blue rectangles: A \times C = A* C

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Total number of rectangles: A * B + A * C = RESULT

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Counting by row:

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Number of rows: A

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Number of rectangles in a row: B + C = B + C

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Total number of rectangles: A \times B + C = RESULT

 98d40344 » marcia  2011-09-18 Distributive tweak 41
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Both ways of counting result in the same number, and this is known as the distributive property.

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(A \times B) + (A \times C) = A\times(B + C)

 ed2d8a69 » marcia  2011-09-27 Add a replace the symbol distributive property problem type 44
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49 randRangeUnique( 2, 10, 3 ) 50 NUMS 51 A * ( B + C ) 52 binop( 1 ) 53 randRange( 0, 2 ) 54 55 (function() { 56 return jQuery.map( NUMS, function( el, i ) { 57 if ( i == SWAP_INDEX ) { 58 return SYMBOL; 59 } else { 60 return el; 61 } 62 }); 63 })() 64 65 NUMS[ SWAP_INDEX ] 66
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Using the picture below as a guide, what number could replace SYMBOL?

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A\times(B + C) = ( SYM_A \times SYM_B ) + ( SYM_A \times SYM_C )

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70 init({ 71 range: [ [ 0, 1 ], [ 0, A ] ], 72 scale: [ 180, 30 ] 73 }); 74 rectchart( [ B, C ], [ "#FFA500", "#6495ED" ], A - 1 ); 75 for ( var i = 0; i < A - 1; i++ ) { 76 rectchart( [ B, C ], [ "#FFCF22", "#85B7FF" ], i); 77 } 78
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There are A rows, and each row has B + C = B+C rectangles.

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MISSING

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Counting by color:

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Number of orange rectangles: A \times B = A * B

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Number of blue rectangles: A \times C = A* C

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Total number of rectangles: A * B + A * C = RESULT

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Counting by row:

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Number of rows: A

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Number of rectangles in a row: B + C = B + C

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Total number of rectangles: A \times B + C = RESULT

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Both ways of counting result in the same number, and this is known as the distributive property.

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(A \times B) + (A \times C) = A\times(B + C)

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Thus the symbol SYMBOL could be replaced with MISSING.

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 41183eb2 » marcia  2011-09-18 Add exercise -- distributive property 95
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