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.
It’s 90 after you multiply the length, width, and height
3 divided by 2 is 1.5 so c
The line is horizontal, therefore the slope is 0.
Answer=0
5x + 14 = 26
subtract 14 on both sides
5x = 26 - 14
5x = 12
Divide 5 on both sides
x = 12/5
x = 2.4
That's only to the nearest tenth. To make it to the hundreth, just add another 0 at the end. It doesn't change the value of it.
x = 2.40 is your final answer.
