draw a perpendicular line from the directrix passing through the focus, this will be the line of symmetry.
The vertex(h, k) will be located on the line half way between the focus and directrix.
The distance from the focus to the vertex is called the focal length, call it a. The then equation is
(x - h)^2 = 4a(y - k)
the equation can be manipulated to
y = 1/4a(x - h)^2 + k
hope it helps
Answer:
228
Step-by-step explanation:
We are adding 3 each time so general rule is
Tn= 3n + 3
Therefore
T= 3(75)+3
= 228
think the answer is
x=6+sqrt0/2 x=6-sqrt0/2 both equaling into 3.
But I might be wrong.
Answer:
Q(t) = Q_o*e^(-0.000120968*t)
Step-by-step explanation:
Given:
- The ODE of the life of Carbon-14:
Q' = -r*Q
- The initial conditions Q(0) = Q_o
- Carbon isotope reaches its half life in t = 5730 yrs
Find:
The expression for Q(t).
Solution:
- Assuming Q(t) satisfies:
Q' = -r*Q
- Separate variables:
dQ / Q = -r .dt
- Integrate both sides:
Ln(Q) = -r*t + C
- Make the relation for Q:
Q = C*e^(-r*t)
- Using initial conditions given:
Q(0) = Q_o
Q_o = C*e^(-r*0)
C = Q_o
- The relation is:
Q(t) = Q_o*e^(-r*t)
- We are also given that the half life of carbon is t = 5730 years:
Q_o / 2 = Q_o*e^(-5730*r)
-Ln(0.5) = 5730*r
r = -Ln(0.5)/5730
r = 0.000120968
- Hence, our expression for Q(t) would be:
Q(t) = Q_o*e^(-0.000120968*t)
2t<-4 divide both sides by 2
t<-2
...
7t>49 divide both sides by 7
t>7
So t=(-oo,-2) and t=(7,+oo)
t=(-oo,-2)U(7,+oo)
