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Vera_Pavlovna [14]
3 years ago
5

Identify the middle line for the function.

Mathematics
2 answers:
Ierofanga [76]3 years ago
7 0
The solution to the problem is as follows:

<span>'(t) = 0 gives:
 
-24t + 60 = 0
 
t = 2.5
 
For this t, the second derivative s"(t) = -24 is negative.

And so, s(t) is maximum for t = 2.5.
 
Maximum height is:
 
S = -12(2.5)^2 + 60(2.5) +8 = 233ft
</span>
I hope my answer has come to your help. God bless and have a nice day ahead!
soldi70 [24.7K]3 years ago
6 0
Hello there.

<span>Identify the middle line for the function.

Identify the middle line for the function: f (t) = 5sec(3t − π) + 1.

</span><span>y = 2
</span>
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3^x grows faster than x^3. Explain why.
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Because 3 is a number and the x is unknow so its just an exponent the x ^ 3 grows three each time

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3 years ago
A cable television compan is laying cable in an area with underground utilities. Two subdivisions are located on opposite sides
Schach [20]

Answer:

Step-by-step explanation:

Given:

Point Q = Q is on the north bank 1200 m east of P $40/m to lay cable underground and $80/m way to lay the cable.

Lets denote T be a point on north bank, therefore, point T is t m east and 100m north of P.

Note that 0 < t < 1200

Distance from P to T using Pythagoras theorem:

PT^2 = PQ^2 + QT^2

PT = √(t²+100²) = √(t²+10,000)

Distance from T to Q:

TQ = 1200 - t

From above, Cable from P to T is underwater, and costs $80/m to lay.

Cable from T to Q is underground, and costs $40/m to lay.

Total cost of the cable:

C(t) = 80√(t²+10,000) + 40(1200 - t)

Cost is at its minimum: when dC/dt = C'(t) = 0 and d^2C/dt^2 = C''(t) > 0

dC/dt = C'(t)

= 80 * 1/2 (t² + 10,000)^(-1/2) × 2t + 40 × (-1)

dC/dt = C'(t)

= 80t/√(t² + 10,000) - 40 = 0

80t/√(t² + 10,000) = 40

80t = 40√(t² + 10,000)

2t = √(t² + 10,000)

square both sides:

4t² = t² + 10,000

3t² = 10,000

t² = 10,000/3

t = 100/√3

≈ 57.735

d^2C/dt^2 =

C''(t) = [80√(t² + 10,000) - 80t × t/√(t²+10,000)] /√(t²+10,000)

C''(t) = [(80(t²+10,000) - 80t²) / √(t²+10,000)] /√(t²+10,000)

C''(t) = 800,000 / (t² + 10,000)^(3/2)

C''(t) > 0 for all t

Therefore C(t) is minimized at t = 100/√3

C(100/√3) = 80√(10,000/3 + 10,000) + 40 × (1200 - 100/√3)

= 80 √(40,000/3) + 48,000 - 4000/√3

= 16000/√3 + 48,000 - 4000/√3

= 48,000 + 12,000/√3

= 48,000 + 4,000√3

= 4000 (12 + √3)

≈ 54,928.20

Minimum cost is $54,928.20

This occurs when cable is laid underwater from point P to point T(57.735 m east and 100m north of P) and then underground from point T to point Q (1200 - 57.735)

= 1142.265 m

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Answer:

Since it took Juan 16 hours to wrap 2 presents, it will take Juan 32 hours.

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