Question

Help me please! LAST TWO QUESTIONS.

Answer 1

Answer:

1) $(4x)(x+2)-x(3x+5)$

2A) $w^2+3w$

2B) $28\text{ in}^2$

Step-by-step explanation:

1)

First, let's write the expressions for the area of the entire shaded region and the white area.

The formula for the area of a rectangle is given by:

$A=lw$

For the entire shaded rectangle, the length is (4x) and the width is (x+2). So, the area is:

$(4x)(x+2)$

For the white rectangle, the length is (3x+5) and the width is (x). So, the area is:

$x(3x+5)$

The shaded area is the entire shaded rectangle minus the area of the white rectangle. Therefore, our expression would be:

$(4x)(x+2)-x(3x+5)$

2)

Part A)

So, we are given that the length of the rectangle is 3 inches greater than the width.

The area of a rectangle is given by:

$A=lw$

So, let w be width and let l be the length.

Since the length is 3 inches greater than the width, this means that the l is (w+3).

Thus, substitute. Our expression will therefore be:

$lw\\w(w+3)$

Since we want a polynomial, let's expand:

$=w^2+3w$

Part B)

To find the area when the width is 4, substitute 4 for w:

$w^2+3w\\=(4)^2+3(4)$

Square and multiply:

$=16+12$

Add:

$=28\text{ in}^2$

So the area is 28 square inches.