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WINSTONCH [101]

3 years ago

15

I will mark you brainiest if you can answer this

1 answer:

Anettt [7]3 years ago

6 0

Answer:

slope- 40

y-intercept- 10

slope-intercept form- y=40x+10

Step-by-step explanation:

hope this helps :)

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The volume of a cylinder closed at the one end is 1056cm cube. If its height is 21cm, and the radius is 4, find the total surfac

monitta

Answer:

578 cm^2

Step-by-step explanation:

The surface area is given by one circle of radius 4 (the other is missing!), plus a "rectangle" (the lateral surface) of sides 21 and $2\pi \times 4$ (the length of the circumference at the base). Adding them up we get: $\pi \times4^2 + 21\times 2\pi\times4 = 184\pi \approx 578 cm^2$

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Arlecino [84]

Answer:

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Step-by-step explanation:

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Write Three hundred forty eight in standard form

GrogVix [38]

Answer:

348

Step-by-step explanation:

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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

What is the value of 3-(-2)?

Lostsunrise [7]

Answer:

5

Step-by-step explanation:

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