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:
Please mark as brainliest.
Answer:
3
Step-by-step explanation:
It's given that ΔDEF ∼ΔXYZ . So the corresponding sides of both triangles will be proportional to each other.
DE = 4 ; XY = 4(n + 1) ; EF = 5 ; YZ = 7n - 1
Putting all these values gives ,
Answer:
27.5
Step-by-step explanation:
1.5(5)+20=27.5
:) hope its right :)
