Find the solution set for y = 2x + 1, given the replacement set {(-2, -3), (-1, -1), (0, 2), (1, 3), (2, 6)} ...?
1 answer:
You might be interested in
Answer:
See the attached figure.
Step-by-step explanation:
Modeling each part as a function.
Let x is hours of the working day
The company packs 80 boxes of toys every hour for the first 3 hours of the day. ⇒ y₁ = 80 x , x ∈ [0,3]
At the end of the 3 hours The company packs = 80 * 3 = 240
They stop packaging for the next 2 hours in order to carry out a training session. ⇒ y₂ = 240 , x ∈ [3,5]
Then, for the next 4 hours, they pack 20 boxes of toys every hour
y₃ = 20x + c , x ∈ [5,9]
To find c, y₂ = y₃ at x =5
∴ 240 = 20 * 5 + c
∴ c = 240 - 20 * 5 =240 - 100 = 140
∴ y₃ = 20x + 140 , x ∈ [5,9]
So,
y₁ = 80 x , x ∈ [0,3]
y₂ = 240 , x ∈ [3,5]
y₃ = 20x + 140 , x ∈ [5,9]
The graph models the piecewise function for the given situation is as shown in the attached figure.
9514 1404 393
Answer:
- maximum: 15∛5 ≈ 25.6496392002
- minimum: 0
Step-by-step explanation:
The minimum will be found at the ends of the interval, where f(t) = 0.
The maximum is found in the middle of the interval, where f'(t) = 0.
This derivative is zero when the numerator is zero, at t=5. The function is a maximum at that point. The value there is ...
f(5) = (∛5)(20-5) = 15∛5
The absolute maximum on the interval is 15∛5 at t=5.
Answer:
I sorted them by colors: red goes with red, blue with blue, purple with purple and brown with brown :))
Step-by-step explanation:
