Answer:
Alice reaches 11 metres below the centre as lowest height.
Step-by-step explanation:
Cosine is a bounded function between -1 and 1, so that the lowest height that Alice achieves in Ferris wheel is:
Where:
- Time, measured in seconds.
- Height with respect to centre, measured in metres.
If 
, then:
Alice reaches 11 metres below the centre as lowest height.
Refer to the figure shown below.
Because the maximum height of the parabola is 50 m, its equation is of the form
y = ax² + 50
This equation places the vertex at (0,50). The constant a should be negative for the vertex to be the maximum of y.
The base of the parabola is 10 m wide. Therefore the x-intercepts are (5,0) and (-5,0).
Set x=5 and y=0 to obtain
a(5²) + 50 = 0
25a = -50
a = -2
The equation of the parabola is
y = - 2x² + 50
At 2 m from the edge of the tunnel, x = 5 - 2 = 3 m.
Therefore the height of the tunnel (vertical clearance) at x = 3 m is
h = y(3)
= -2(3²) + 50
= - 18 + 50
= 32 m
Answer: 32 m
Solve for x over the real numbers:
x^2 - 4 x = 5
Subtract 5 from both sides:
x^2 - 4 x - 5 = 0
x = (4 ± sqrt((-4)^2 - 4 (-5)))/2 = (4 ± sqrt(16 + 20))/2 = (4 ± sqrt(36))/2:
x = (4 + sqrt(36))/2 or x = (4 - sqrt(36))/2
sqrt(36) = sqrt(4×9) = sqrt(2^2×3^2) = 2×3 = 6:
x = (4 + 6)/2 or x = (4 - 6)/2
(4 + 6)/2 = 10/2 = 5:
x = 5 or x = (4 - 6)/2
(4 - 6)/2 = -2/2 = -1:
Answer: x = 5 or x = -1
Draw the picture and label the
width = w
The length of the monitor is six times the quantity of five less than half its width:
length = 6(w/2-5)
length = 3w-30
Area = (length)*(width)
384=(3w-30)*(w)
384=3w^2-30w
Answer:
the dimensions of the monitor in terms of its width is:
384=3w^2-30w
Answer:
Whats the question
Step-by-step explanation:
