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Alecsey [184]
3 years ago
8

Fish enter a lake at a rate modeled by the function E given by ( ) 20 15sin 6 t E t          . Fish leave the lake at a

rate modeled by the function L given by 2 0.1 L( ) 4 2 t t   . Both E(t) and L(t) are measured in fish per hour, and t is measured in hours since midnight (t = 0). a) How many fish enter the lake over the 5-hour period from midnight (t = 0) to 5 A.M. (t = 5)? Give your answer to the nearest whole number.

Mathematics
1 answer:
liraira [26]3 years ago
5 0

Answer: The number of fish that enter the lake from midnight (t = 0) to 5 A.M. (t = 5) is 101 fish

Step-by-step explanation:

Given that

Fish enter a lake at a rate modeled by the function E given by E(t) = 20 + 15 sin(πt/6). Fish leave the lake at a rate modeled by the function L given by L(t) = 4 + 2^0.1t2. Both E(t) and L(t) are measured in fish per hour, and t is measured in hours since midnight (t = 0).

(a) How many fish enter the lake over the 5-hour period from midnight (t = 0) to 5 A.M. (t = 5)?

Please find the attached file for the solution

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Am i correct?? thanks for checking my answer :)
UNO [17]

Answer:

no

Step-by-step explanation:

since the y=mx+b format is used here, your b is -5. for this to be correct, your y-int needs to be on -5, not -2 as you have it. Other than that, i believe you are correct


3 0
3 years ago
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Find the sum of the arithmetic series given ai = 45, an = 85, and n = 5.
seraphim [82]

Answer:

C. 325.

Step-by-step explanation:

The last term a5 = 85

a1 = 45

Sum of n terms = n/2 (a1 + l)

So here we have n = 5, a1 = 45 and the last term l = 85

= (5/2)(45 + 85)

= 5/2 * 130

= 325.

5 0
3 years ago
In the circle below, suppose m FEH=272º and m EFG=116º. Find the following.​
adoni [48]

Answer:

m∠FEH = 44°

m∠EHG =  64°

Step-by-step explanation:

1) The given information are;

The angle of arc m∠FEH = 272°, the measured angle of ∠EFG = 116°

Given that m∠FEH = 272°, therefore, arc ∠HGF = 360 - 272 = 88°

Therefore, angle subtended by arc ∠HGF at the center = 88°

The angle subtended by arc ∠HGF at the circumference = m∠FEH

∴ m∠FEH = 88°/2 = 44° (Angle subtended at the center = 2×angle subtended at the circumference)

m∠FEH = 44°

2) Similarly, m∠HGF is subtended by arc m FEH, therefore, m∠HGF = (arc m FEH)/2 = 272°/2 = 136°

The sum of angles in a quadrilateral = 360°

Therefore;

m∠FEH + m∠HGF + m∠EFG + m∠EHG = 360° (The sum of angles in a quadrilateral EFGH)

m∠EHG = 360° - (m∠FEH + m∠HGF + m∠EFG) = 360 - (44 + 136 + 116) = 64°

m∠EHG =  64°.

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3 years ago
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Alja [10]

Answer:

It is Acceleration have a good day!

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A certain spinner is divided into 6 sectors of equal size, and the spinner is equally likely to land in any sector. Four of the
ICE Princess25 [194]

Answer:

4/6

Step-by-step explanation:

If there's 6 sectors, and each sector is equal, the probability of the spin landing into any sector has to be the same. Therefore, each sector should be equal to 1, or 1/6. If 4/6 are shaded, there's a 4/6 probability of the spin landing in a shaded sector.

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