Answer:
Step-by-step explanation:
Given a sample M(t)
M(t) = 120 • ( 81 / 625)^t
When is the fraction of the mass decay to 3/5 of it's mass
Generally
M(t) = Mo•(k^t)
The original mass is 120
Mo = 120
So, we want to find time when it decay to 3/5 of it's original mas
M = 3/5 × 120
M = 72
Then,
M(t) = 120 • ( 81 / 625)^t
72 = 120 • ( 81 / 625)^t
72 / 120 = ( 81 / 625)^t
0.6 = ( 81 / 625)^t
Take natural logarithmic of both sides
In(0.6) = In(81/625)^t
In(0.6) = t•In(81/625)
t = In(0.6) / In(81/625)
t = In(0.6) / In(0.1296)
t = 0.25 monthly
t = ¼ monthly
Answer:
5 zeros maximum
Step-by-step explanation:
Recall that a polynomial has as many roots (zeros) in the Complex number system, as its degree. So in the given case, where the leading term of the polynomial carries order 5, the maximum number of zeros for the given function cannot be larger than 5.
Answer: I think it’s 120 for TPS
Step-by-step explanation:
Something funny is that the x value of the vertex lies directl in the middle of the x intercepts
so
we see the x intercepts or 0's at x=8 and 2
the average is x=5
so find f(5) to find the y value of the vertex
f(5)=(5-8)(5-2)
f(5)=(-3)(3)
f(5)=-9
vertex is at (5,-9)
the actual way the teacher wants is to expand then compltete the square to get into the form f(x)=a(x-h)^2+k where the vertex is (h,k)
but whatever
verrtex is at (5,-9)