1. Suppose the height, h, in feet, of a trampolinist above the ground during one bounce is modelled by the quadratic function h(t) = -16t2 + 42t + 3.75 . For what period of time is the trampolinist at least 22 ft above the ground? Round your answers to the nearest hundredth. (2 marks)

2. The length of a rectangle is 5 cm more than its width. What are the possible dimensions of the rectangle if the area must be at least 84 cm2? (2 marks)
3. A square storage area measures 10 m on a side. By how much must each side be shortened to decrease this area to less than half the original area? (3 marks - show your work)

parabola opens down (sheds water)

when is it between the vertex and 22 feet?

16 t^2 - 42 t - 3.75 = -22

16 t^2 - 42 t + 18.25 = 0

t = [ 42 +/- sqrt (1764 - 1168) ] / 32

t = [ 42 +/- 24.4 ] /32

t = .55 to 2.07 seconds

w(w+5) >/= 84

w^2 + 5 w - 84 >/= 0
solve for zero, anything bigger works

w = [ -5 +/- sqrt(25 + 336) ]/ 2

w = [ -5 + 19 ]/2 = 7
then L = 12
so anything at least 7 by 12

divide length dimension by sqrt 2 to get half the area

1. To find the period of time during which the trampolinist is at least 22 ft above the ground, we need to solve the quadratic equation h(t) = -16t^2 + 42t + 3.75 ≥ 22.

First, set the equation equal to 22:
-16t^2 + 42t + 3.75 = 22

Rearranging the equation to standard form:
-16t^2 + 42t - 18.25 = 0

To solve this quadratic equation, we can either factor it or use the quadratic formula. In this case, factoring is difficult, so let's use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / 2a

For our equation, a = -16, b = 42, and c = -18.25. Plugging these values into the formula:

t = (-42 ± √(42^2 - 4(-16)(-18.25))) / 2(-16)

Simplifying the equation:

t = (-42 ± √(1764 - 1168)) / (-32)

t = (-42 ± √596) / (-32)

Now, let's calculate the values of t:

t ≈ (-42 + √596) / (-32) ≈ 0.58

t ≈ (-42 - √596) / (-32) ≈ 2.92

So the trampolinist is at least 22 ft above the ground for the period of time between approximately 0.58 seconds and 2.92 seconds.

2. Let's represent the width of the rectangle as "w". According to the problem, the length is 5 cm more than the width, so the length is "w + 5".

To find the area of the rectangle, we use the formula: Area = Length × Width.

So, the area of the rectangle is: Area = (w + 5) × w.

We have the condition that the area must be at least 84 cm^2. Therefore, we can set up the inequality:

(w + 5) × w ≥ 84

Expanding the above expression gives:

w^2 + 5w ≥ 84

Rearranging the equation:

w^2 + 5w - 84 ≥ 0

To solve the inequality, we can factor the left side of the inequality:

(w - 7)(w + 12) ≥ 0

Now let's consider the two factors separately:

w - 7 ≥ 0 --> w ≥ 7

and

w + 12 ≥ 0 --> w ≥ -12

Since width cannot be negative, we can ignore the second inequality.

So, the possible dimensions of the rectangle are widths greater than or equal to 7 cm and lengths equal to (w + 5).

3. The original area of the square storage area is 10 m * 10 m = 100 m^2.

To decrease the area to less than half the original area, we need to find the area that is less than 100 m^2 / 2 = 50 m^2.

Let's represent the amount by which each side needs to be shortened as "x".

The new side length will then be (10 - x).

The new area will be: (10 - x) * (10 - x) = (10 - x)^2.

Setting up the inequality:

(10 - x)^2 < 50

Taking the square root of both sides:

10 - x < √50

Simplifying the equation:

10 - x < √(25 * 2)

10 - x < 5√2

Subtracting 10 from both sides:

- x < 5√2 - 10

Multiplying both sides by -1 (which reverses the inequality):

x > 10 - 5√2

Therefore, each side must be shortened by more than 10 - 5√2 meters to decrease the area to less than half the original area.