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Irina18 [472]
3 years ago
15

Can someone simplify these? (3x^2)3x^2 and (3x^3)^2(x^2) I'd really appreciate it.

Mathematics
2 answers:
Ira Lisetskai [31]3 years ago
7 0
<span>(3x^2)3x^2= 9x^4

</span><span>(3x^3)^2(x^2)</span>=  9x^8
MrMuchimi3 years ago
7 0
9x^8 have a nice day

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Answer:

18

Step-by-step explanation:

To find volume you have to use the formula (length x width x height.)

So by looking at it you can see that it's 3 units long, 3 units in wide, and 2 units tall. Knowing that, you then plug in those numbers to the formula. That would be 3x3x2. Any calculator will work.

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Solve for a :<br> w=7a+4b
rosijanka [135]

Answer:

w=7a+4b

or,w-4b=7a

or,(w-4b)/7=a

\frac{w - 4b}{7}  = a

3 0
3 years ago
What number is 2 hundred 15 tens and 6 one
LuckyWell [14K]
The answer is:  "2156" .
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(215)*10 + 6*(1) = 2150 + 6 = 2156 .
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7 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
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