randRangeUnique( 2, 10, 3 ) A * ( B + C )

Use distribution to solve:

A\times(B + C)

RESULT

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

init({ range: [ [ 0, 1 ], [ 0, A ] ], scale: [ 180, 30 ] }); rectchart( [ B, C ], [ "#FFA500", "#6495ED" ], A - 1 ); for ( var i = 0; i < A - 1; i++ ) { rectchart( [ B, C ], [ "#FFCF22", "#85B7FF" ], i); }

Counting by color:

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

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

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

Counting by row:

Number of rows: A

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

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

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

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

randRangeUnique( 2, 10, 3 ) NUMS A * ( B + C ) binop( 1 ) randRange( 0, 2 ) (function() { return \$.map( NUMS, function( el, i ) { if ( i === SWAP_INDEX ) { return SYMBOL; } else { return el; } }); })() NUMS[ SWAP_INDEX ]

Using the picture below as a guide, what number could replace SYMBOL?

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

init({ range: [ [ 0, 1 ], [ 0, A ] ], scale: [ 180, 30 ] }); rectchart( [ B, C ], [ "#FFA500", "#6495ED" ], A - 1 ); for ( var i = 0; i < A - 1; i++ ) { rectchart( [ B, C ], [ "#FFCF22", "#85B7FF" ], i); }

There are A rows, and each row has B + C = B+C rectangles.

MISSING

Counting by color:

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

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

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

Counting by row:

Number of rows: A

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

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

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

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

Thus the symbol SYMBOL could be replaced with MISSING.