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Komok [63]
3 years ago
14

I'm completely lost help

Mathematics
1 answer:
e-lub [12.9K]3 years ago
3 0
I can't see the line very well, but the initial value, also known as the y-intercept, is 1. I would tell you the rate of change but like I said I can't see the line very well, but count across from where the line touches the y axis until you can go up and hit a corner of a square on the grid and the line.. if that makes sense. I hope this helps!
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Anna35 [415]

Answer:

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Step-by-step explanation:

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If amy sends 20emalis in 5minutes what ia the init rate
Maksim231197 [3]
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Which set of points includes all of the solutions for y = 5/2x + 3/2?
Firdavs [7]

Answer:

Choice A: (x,\frac{5}{2}x+ \frac{3}{2} ). for all real numbers.

Step-by-step explanation:

We are looking for points on the x-y plane which satisfy the equation y = 5/2x + 3/2, and these point are all those that lie on the graph of the equation, which are points (x,y) = (x,\frac{5}{2}x+ \frac{3}{2} ).

Looking at the other choices we see that choice B gives only 3 points, and regardless of whether they lie on the graph of y or not, this choice cannot be correct because 3 points are not all solutions.

The same goes for choice D.

And choice C is just the whole of x-y plane, therefore this cannot be the answer because not every point on the x-y plane is a solution to  y = 5/2x + 3/2

8 0
3 years ago
Which shape is always an isosceles triangle?
Lady_Fox [76]
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3 years ago
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
3 years ago
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