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:
8/3
Step-by-step explanation:
3 parts yellow to 2 parts red
4 parts yellow to x parts red
x=4*2/3 =8/3 parts red.
If it is right then there you go
The Intersecting Secant Theorem says
EC × ED = EB × EA
(x+4)(x+4+1) = (x+1)(x+1+11)
(x+4)(x+5)=(x+1)(x+12)
x^2 + 9x + 20 = x^2 + 13 x + 12
8 = 4x
x = 2
Check:
6(7)=42
3(14)=42, good
Answer: 2
Answer:
17.85
Step-by-step explanation:
just times 5.95 by 3
