Y = t*e^(-t/2)
y' = t' [e^(-t/2)] + t [e^(-t/2)]' = e^(-t/2) + t[e^(-t/2)][-1/2]=
y' = [e^(-t/2)] [1 - t/2] = (1/2)[e^(-t/2)] [2 - t] = - (1/2) [e^-t/2)] [t -2]
Answer:
The Amount initially deposited is $37046.64
Step-by-step explanation:
A = p (1+r/n)^(nt)
A= final amount= $5000
P = principal amount=
r = rate = 0.05
n = number of times compounded
= 6*12
= 48
t = years= 6
A = p (1+r/n)^(nt)
50000 = p (1+0.05/48)^(48*6)
50000= p(1.001041667)^288
50000= p (1.34965)
50000/1.34965= p
37046.64 = p
The Amount initially deposited is $37046.64
Answer:
x = - 6
Step-by-step explanation:
Given that y varies directly as x then the equation relating them is
y = kx ← k is the constant of variation
To find k use the condition y = - 9 when x = 3, thus
k = 
= 
= - 3
y = - 3x ← equation of variation
When y = 18, then
18 = - 3x ( divide both sides by - 3 )
x = - 6
Answer:constant of variation = 6/5
Step-by-step explanation:
t varies inversely as r
Let k be the constant of variation
t = k/r
t = 6 and r = 5
Substitute the value of t and r
6 = k/5
Make k the subject of the formula
k = 6/5
Answer:
BIL is the answer to that one day
