Use distribution to solve:

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

`RESULT`

Each row has

rectangles, and there are `B` + `C` = `B+C`

rows.`A`

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`)

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

rows, and each row has `A`

rectangles.`B` + `C` = `B+C`

`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

could be replaced with `SYMBOL`

.`MISSING`