The width of a rectangle is fixed at 2yd. for what lengths will the area be at least 4 yd ^2? my second question is: a rectangle has a perimeter of 8ft. the length is 5ft less than two times the width. fond the length and width.

Length of rectangle ≥ 2 yd.

Step-by-step explanation:

Suppose the length of rectangle = X yd.

Given the width of rectangle = 2 yd.

Area ≥ 4 yd²

Length * width ≥ 4 yd²

2X ≥ 4 yd²

X ≥ 2 yd

So, length of rectangle must be at least 2 yd.

length = 1 feet and width = 3 feet.

Step-by-step explanation:

Suppose the width of rectangle = X feet.

then length of rectangle = (2X - 5) feet.

Given the perimeter of rectangle = 8 feet.

The formula is Perimeter = 2 * (length + width)

8 = 2 * (2X - 5 + X)

8 = 2 * (3X - 5)

8 = 6X - 10

6X = 8 + 10 = 18

X = 18/6 = 3.

so 2X - 5 = 2(3) - 5 = 6 - 5 = 1.

Hence, length = 1 feet and width = 3 feet.

-i

Step-by-step explanation:

Given are two complex numbers as

To find quotient

We can use Demoivre theorem for products and quotients here

Quotient would be equal to

Hence quotient = -i