I need help on my assignment down below

Question

This is the link to my assignment. I need help on questions 3 and 4. They are on the last page.

http://static.k12.com/eli/bb/1211/5_68981/1_124222_8_68989/73a9e1e937f929aa0273c1ce37d80775f5e5d291/media/d707bf1f569b336d45e26860d378b0acb041bfe9/AAHS_M5S2L4_1_teacher_graded_assignment_PROJECTGUIDE.pdf

Answer 1

3).

a). Area of the rectangle (A) = length * width = l * w.

From the given data (A) = l * w = x ² + 40x + 300.

l * w = x ² + 30x + 10x  + 300

l * w = x (x + 30) + 10(x  + 30)

l * w = (x + 30)(x  + 10)

l * w = (x + 30) * (x  + 10)

Therefore, length of the rectangle (l) is (x + 30) units and width of the rectangle (w) is (x + 10) units.

b). Area of the foundation = length * width = x ² + 40x + 300.

If, x = 20 feet, then the length of the foundation (l ) = x + 30 = 20 + 30 = 50 feet.

If, x = 20 feet, then the width of the foundation (w ) = x + 10 = 20 + 10 = 30 feet.

If, x = 20 feet, then the area of the foundation:

x ² + 40x + 300

=  20 ² + (40*20) + 300

= 400 + 800 + 300

= 1500 square feet.

Answer 2

(4).(a).

The area of the rectangle is x² + 8x - 9.

To find the dimensions of the rectangle, write the trinomial x² + 8x - 9 in factors form.

x² + 8x - 9= x² + 9x - x – 9

= x(x + 9) - 1(x + 9)

= (x + 9)(x - 1)

x² + 8x – 9 = (x + 9)(x - 1).

The dimensions of the rectangle are (x + 9) and (x – 1).

Let x be any positive integer except 1, so (x + 9) represent the length of the rectangle and (x – 1) represent the width of the rectangle.

(4)(b).

Let x = 2 units the

Length = x + 9

            = 2 + 9 = 11 units

Width = x – 1  

           = 2 – 1 = 1 unit

Area = x² + 8x - 9

         = (2)² + 8(2) – 9 = 4 + 16 – 9 = 11 square units.