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Aloiza [94]
3 years ago
10

1 + cot^2 x is equivalent to

Mathematics
1 answer:
zysi [14]3 years ago
6 0

Answer:

csc^2x

Step-by-step explanation:

It's a trig identity where 1+cot^2x = csc^2x

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Animal populations are not capable of unrestricted growth because of limited habitat and food supplies. Under such conditions th
trasher [3.6K]

Answer:

(a) 100 fishes

(b) t = 10: 483 fishes

    t = 20: 999 fishes

    t = 30: 1168 fishes

(c)

P(\infty) = 1200

Step-by-step explanation:

Given

P(t) =\frac{d}{1+ke^-{ct}}

d = 1200\\k = 11\\c=0.2

Solving (a): Fishes at t = 0

This gives:

P(0) =\frac{1200}{1+11*e^-{0.2*0}}

P(0) =\frac{1200}{1+11*e^-{0}}

P(0) =\frac{1200}{1+11*1}

P(0) =\frac{1200}{1+11}

P(0) =\frac{1200}{12}

P(0) = 100

Solving (a): Fishes at t = 10, 20, 30

t = 10

P(10) =\frac{1200}{1+11*e^-{0.2*10}} =\frac{1200}{1+11*e^-{2}}\\\\P(10) =\frac{1200}{1+11*0.135}=\frac{1200}{2.485}\\\\P(10) =483

t = 20

P(20) =\frac{1200}{1+11*e^-{0.2*20}} =\frac{1200}{1+11*e^-{4}}\\\\P(20) =\frac{1200}{1+11*0.0183}=\frac{1200}{1.2013}\\\\P(20) =999

t = 30

P(30) =\frac{1200}{1+11*e^-{0.2*30}} =\frac{1200}{1+11*e^-{6}}\\\\P(30) =\frac{1200}{1+11*0.00247}=\frac{1200}{1.0273}\\\\P(30) =1168

Solving (c): \lim_{t \to \infty} P(t)

In (b) above.

Notice that as t increases from 10 to 20 to 30, the values of e^{-ct} decreases

This implies that:

{t \to \infty} = {e^{-ct} \to 0}

So:

The value of P(t) for large values is:

P(\infty) = \frac{1200}{1 + 11 * 0}

P(\infty) = \frac{1200}{1 + 0}

P(\infty) = \frac{1200}{1}

P(\infty) = 1200

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3 years ago
at a lunch stand, each hamburger has 50 more calories than each order of fries. if 2 hamburgers and 3 orders of fries have a tot
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The correct answer is 370. A system of equations can be used, where h represents the number of calories in a hamburger and f represents the number of calories in an order of fries. The equation <span> 2</span> h plus 3 f equals 1,700 represents the fact that 2 hamburgers and 3 orders of fries contain a total of 1700 calories, and the equation  h<span> equals f plus 50</span> represents the fact that one hamburger contains 50 more calories than an order of fries. Substituting  f plus 50 for h in  2 h plus 3 f equals 1,700 gives  <span>2, parenthesis, f plus 50, close parenthesis, plus 3</span> f, equals 1,700. This equation can be solved as follows:

 

 2 f plus 100 plus 3 f equals 1,700

 5 f plus 100 equals 1,700

 5 f equals 1,600

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<span>The number of calories in an order of fries is 320, so the number of calories in a hamburger is 50 more than 320, or 370.</span>

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