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Nadusha1986 [10]

3 years ago

12

Hey, people I'm looking for some new friends.

2 answers:

eimsori [14]3 years ago

7 0

:> ok.....this isn’t a friends app this is a hw site....

uysha [10]3 years ago

4 0

Hey we can be friends if you would like !!

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Simplify;<br> 26 ⋅ 26^8<br><br> Please also explain how you got the answer.

antoniya [11.8K]

26 ⋅ 26^8

Multiply by adding the exponents:

Raise 26 to the power of 1 to get:

26^1 * 26^8

Now when you multiply the same number raised to different powers, you add the powers together:

26^1 * 26^8 = 26^(8+1) = 26^9

So 26 * 26^8 simplifies is 26^9

6 0

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

Pls help I’ll give brainliest

alexandr402 [8]

What grade is this? are these meant to be fractions?

7 0

3 years ago

A kite with a 100 foot-long string is caught in a tree. When the full length of the string is stretched in a straight line to th

spayn [35]

Hyp=100,adj side=30

angle= $cos^{-1}(30/100)$

6 0

4 years ago

Read 2 more answers

Number 1 write in standard form correct answer will get brainliest

Fudgin [204]

Answer:

Step-by-step explanation:

5 0

3 years ago