Ask question

Ask question

All categories

4 years ago

14

Limestone is dissolved by chemical weathering at a rate of 0.2 cm every 100 years. At this rate, after how many years souldva la

yer of limestone 15m thick completely dissolve

1 answer:

Zinaida [17]4 years ago

4 0

At this rate, it should take 7,500 years to dissolve 15m thiccccc

You might be interested in

Help plz i need this

kicyunya [14]

Answer:

B

Step-by-step explanation:

Plug in:

y = 2

x = 0

2 x 0 = 0 - 1 = -1

2 is greater than / equal to -1

4 0

3 years ago

Read 2 more answers

Each face of a small cube has a surface area of 0.25 square meters. A larger cube has edges that are six times as long as the ed

Natalija [7]

Answer:

1534

Step-by-step explanation:

34 35 baby you might need a seatbelet

4 0

3 years ago

A bulletin board is 23 square feet. What is the approximate perimeter of the bulletin board? And wants to have an area of 32 squ

miskamm [114]

Answer: The answer to the first question is 20

Step-by-step explanation:

5 0

3 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

3 years ago

Write a power that has a value greater than 190 and less than 200

Julli [10]

199 is greater than 190, and less than 200.

3 0

3 years ago

Read 2 more answers