Answer:
3rd option
Step-by-step explanation:
Answer:
AC and DF
Step-by-step explanation:
We can see that the triangles have a scale factor of 3.
1 * 3 = 3 (AB and DE)
2 * 3 = 6 (BC and EF)
However...
2 * 3 = 6 not 7 (AC and DF)
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:
Yes
Step-by-step explanation:
Triangle inequality theorem states that a triangle can exist if
a + b > c; a + c > b; b + c > a
a = 2; b = 10; c = 11
2 + 10 = 12; 12 > 11.
2 + 11 = 13; 13 > 10.
10 + 11 = 21; 21 > 2.
