Option (2.)approximately uniform
So you can use substitution...which is where you have Y all alone on one side of your equation and then plug it in to the other equation...2x-5=-3x+10 and you.solve for X. So x= 3. Then plug 3 into one of the original equations... Y=2(3)+5. So Y= 11.
Answer:
the second one is the better deal
Step-by-step explanation:
Answer:
least to greatest: -5 and 1
Step-by-step explanation:
they represent the zeroes of the function and are sorted from least to greatest
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.
