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3 years ago

Determine the initial value

Mathematics

1 answer:

Ghella [55]3 years ago

6 0

Hey there I am just telling you , it seems that your questions is incomplete . Can you kindly post the whole question then I am able to solve it.

please don't mind.

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What numbers do you multiply to find 900 this partical product

Bad White [126]

You mean square root of 900?

I gotcha covered

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2 years ago

Jim read book for 3/5 hours.He done at 1018. <br>At what time did he start reading a book?

Andrei [34K]

Step-by-step explanation:

Well, 3/5 hours equals 30 secs.This means that he started 3:30 hours ago .To find out what time was it you have to subtract 3:30 from the finishing time. So, 10:18-3:30=6:48.If you have any further questions please contact me.

Your sincerely,

Manos

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3 years ago

-2(x+1)=-3 using <br> distributive property

Natalija [7]

Answer:

X=1/2

Step-by-step explanation:

-2x1/2=-1

-2x1=-2

-1+-2=-3

7 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

If s(x) = x-7 and f(x) = 4x? – X+3, which expression is equivalent to (tos)(x)?

lesya692 [45]

Answer:

4(x-7)^2-(x-7)+3 (Assuming t is f)

Step-by-step explanation:

Let s(x)=x-7 and t(x)=4x^2-x+3 .

(t o s)(x)=t(s(x))=t(x-7)

Before I continue this means replace the orginal x in t with x-7.

This will then give you

4(x-7)^2-(x-7)+3

4 0

3 years ago