Answer:
a_n = 2^(n - 1) 3^(3 - n)
Step-by-step explanation:
9,6,4,8/3,…
a1 = 3^2
a2 = 3 * 2
a3 = 2^2
As we can see, the 3 ^x is decreasing and the 2^ y is increasing
We need to play with the exponent in terms of n
Lets look at the exponent for the base of 2
a1 = 3^2 2^0
a2 = 3^1 2^1
a3 = 3^ 0 2^2
an = 3^ 2^(n-1)
I picked n-1 because that is where it starts 0
n = 1 (1-1) =0
n=2 (2-1) =1
n=3 (3-1) =2
Now we need to figure out the exponent for the 3 base
I will pick (3-n)
n =1 (3-1) =2
n =2 (3-2) =1
n=3 (3-3) =0
Answer:
Step-by-step explanation:
we have
Solve for the variable e
That means -----> Isolate the variable e
Multiply by e both sides
tex](e)t=\frac{4u}{e}(e)[/tex]
Divide by t both sides
Answer:
x(15) = 21 lb
Step-by-step explanation:
Rate of change in volume of salt water solution = rate of volume incoming - rate of volume outgoing
dV/dt = 4 - 2 =2gal/min
So, the equation for volume at cetain time t at given conditions and values becomes,
V(t) = 2t + V
V(t) = 2t + 20 gal-------------------euqation (1)
Rate of change in amount of salt = rate of salt in - rate of salt out
dx/dt = {0.5*4} - {[x(t)/V(t)]*2}
dx/dt = 2-2[(x(t))/(2t+20)]
dx/dt = 2-[(x(t))/(v(t))] lb/min
Now, with integrating factor, we get
exp[∫(1/(1+10))dt)] = t+10
the equation becomes
(t + 10)*x' + x = 2*(t+10)
((t+10)*x') = 2*(t+10)
(t+10)*x = t² + 20t + C
As x(0) = 0,
x(t) = (t²+20t)/(t+10)
x(15) = (15²+20*15)/(15+10)
x(15) = 21 lb
Answer:
Step-by-step explanation:
A. 3.11
B.7.5
C.6.46
D.133.17
Answer:
4x3y
Step-by-step hmmm ask your teacher
