Answer:
Multiply by ∛2 and translate the graph to left by 4 units.
Step-by-step explanation:
The initial function given is:
y = -∛(x - 4)
The transformed function is:
y = -∛(2x - 4)
Consider the initial function.
y = -∛(x - 4)
(Represented by Black line in the graph)
Multiply the function by ∛2. The function becomes:
y = -∛(x - 4) × ∛2
y = -∛(2)(x-4)
y = -∛(2x-8)
(Represented by Red line in the graph represents this function)
Translate the graph 4 units to the left by adding 4 to the x component:
y = -∛(2x-8+4)
y= -∛(2x - 4)
(Represented by Blue line in the graph)
Answer:
Text me back for an answer.
Step-by-step explanation:
Make sure you text back.
Answer:
$95.78
Step-by-step explanation:
f(t) = 300t / (2t² + 8)
t = 0 corresponds to the beginning of August. t = 1 corresponds to the end of August. t = 2 corresponds to the end of September. So on and so forth. So the second semester is from t = 5 to t = 10.
$T₂ = ∫₅¹⁰ 300t / (2t² + 8) dt
$T₂ = ∫₅¹⁰ 150t / (t² + 4) dt
$T₂ = 75 ∫₅¹⁰ 2t / (t² + 4) dt
$T₂ = 75 ln(t² + 4) |₅¹⁰
$T₂ = 75 ln(104) − 75 ln(29)
$T₂ ≈ 95.78
21,600 v^3 been a while so not sure wether you need the sq symbol afterword let me know
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
