The area of a rectangle is 33 ft^2, and the length of the rectangle is 5 ft less than double the width. Find the dimensions of the rectangle.

Length = 6

Width = 5.5

Step-by-step explanation:

Let's use the variable x to represent the width:

x = width

The question says that the length of the rectangle is "5 ft less than double the width". This length can be represented like this:

2x - 5 = length

To find the area of a rectangle, you multiply the width and length, so we are going to write an equation where we multiply the width and length and set it equal to 33 ft^2, which is the area:

(x)(2x - 5) = 33

Now use basic algebra to solve for x:

2x^2 - 5x = 33

2x^2 - 5x - 33 = 0

2x^2 + 6x - 11x - 33 = 0

(2x^2 + 6x) + (-11x - 33) = 0

2x(x + 3) - 11(x + 3) = 0

(2x - 11) (x + 3) = 0

x = 11/2, -3

x either equals 11/2 or -3. However, remember that x represents the width, and you can't have a negative number for a width, so x = 11/2 or 5.5.

Now plug in 5.5 to the length expression and simplify:

2(5.5) - 5

11 - 5 = 6

The length is 6 and the width is 5.5.

Hope this helps (●'◡'●)

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this