Hello,
I am going to remember:
y'+3y=0==>y=C*e^(-3t)
y'=C'*e^(-3t)-3C*e^(-3t)
y'+3y=C'*e^(-3t)-3Ce^(-3t)+3C*e^(-3t)=C'*e^(-3t) = t+e^(-2t)
==>C'=(t+e^(-2t))/e^(-3t)=t*e^(3t)+e^t
==>C=e^t+t*e^(3t) /3-e^(3t)/9
==>y= (e^t+t*e^(3t)/3-e^(3t)/9)*e^(-3t)+D
==>y=e^(-2t)+t/3-1/9+D
==>y=e^(-2t)+t/3+k
Answer:
Step-by-step explanation:
__
The applicable rules of logarithms are ...
log(ab) = log(a) +log(b)
log(a/b) = log(a) -log(b)
log(a^b) = b·log(a)
Answer:
answer = $144.40
Step-by-step explanation:
34/100 × $424.71
Answer:
Down below.
Step-by-step explanation:
Use Pythagorean’s Theron
9 + 16 = c squared
25 = c squared
c = 5