No because the exact opposite of -3 would be 3.
Y1 is the simplest parabola. Its vertex is at (0,0) and it passes thru (2,4). This is enough info to conclude that y1 = x^2.
y4, the lower red graph, is a bit more of a challenge. We can easily identify its vertex, which is (-4,0), and several points on the grah, such as (2,-3).
Let's try this: assume that the general equation for a parabola is
y-k = a(x-h)^2, where (h,k) is the vertex. Subst. the known values,
-3-(-4) = a(2-0)^2. Then 1 = a(2)^2, or 1 = 4a, or a = 1/4.
The equation of parabola y4 is y+4 = (1/4)x^2
Or you could elim. the fraction and write the eqn as 4y+16=x^2, or
4y = x^2-16, or y = (1/4)x - 4. Take your pick! Hope this helps you find "a" for the other parabolas.
Answer:
The dimensions of the rectangle = 60ft by 107ft
Where 60 ft = Width of the playing field
107ft = Length of the playing field
Step-by-step explanation:
A playing field is Rectangular is shape, hence,
The formula for Perimeter of a rectangle = 2(L + W)
P = 334 ft
L = 47 + W
W = W
Hence we input these values in the formula and we have:
334 = 2(47 + W + W)
334 = 2(47 + 2W)
334 = 94 + 4W
334 - 94 = 4W
240 = 4W
W = 240/4
W = 60
There fore, the width of this playing field = 60 ft
The length of this rectangle is calculated as:
47 + W
47 + 60
= 107 ft
The length of this playing field = 107ft
Therefore the dimensions of the rectangle = 60ft by 107ft