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zysi [14]
3 years ago
9

Which equation is related to..???

Mathematics
1 answer:
Trava [24]3 years ago
8 0

Answer:

that is the solution to the question

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Gre4nikov [31]

Answer:

What are you trying to figure out? Where the card goes?

Step-by-step explanation:

8 0
2 years ago
Peyton says the sum of two numbers can never be zero. Explain why Peyton's reasoning is incorrect. Use an example.
Elena-2011 [213]

Answer:

Well, there can be negative numbers added to their positive opposite.

Step-by-step explanation:

For example, -3 + 3 = 0

or you could 0 + 0 = 0

4 0
2 years ago
I need help with this
Alex787 [66]
The area of a triangle is 1/2(base)(height) and since this is a right triangle the two legs can be base and height. So the area is 1/2(2x-4)^2 which becomes 1/2(4x^2-16x+16) which is simplified to 2x^2-8x+8. B is the correct answer.
3 0
2 years ago
Read 2 more answers
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

5 0
2 years ago
What is the derivative of y = x^(lnx)? Show your solution.
aleksley [76]
Hello,

ln(y)=ln(x^(ln(x))=ln(x)*ln(x)=(ln(x))²

(ln(y))'=2ln(x)*1/x
(1/y)*y'=2ln(x) / x
y'=(2 ln(x) * x^(ln(x)) )  /x
4 0
3 years ago
Read 2 more answers
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