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:
9990 years
Step-by-step explanation:
The exponential function with given values filled in can be solved for the unknown using logarithms.
__
Q(t) = 12 = 36e^(-0.00011t)
1/3 = e^(-0.00011t) . . . . . . divide by 36
ln(1/3) = -0.00011t . . . . . . take natural logs
t = ln(1/3)/(-0.00011) . . . . divide by the coefficient of t
t ≈ 9990 . . . years
Step-by-step explanation:
5(tanx)^2 -2 +tanx=0
let tanx=y
5y^2 + y-2=0
y=0.5403 or y= -0.7403
tanx=0.5403
x=arctan(0.5403)
x=28.38°
or
tanx=-0.7403
x=arctan(-0.7403)
x=143.49°
Answer:
D. 4(3x-5) = 52
Step-by-step explanation:
52 is equal the sum of one side multiplied by 4
So
4(3x-5) = 52
Is the right answer
