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koban [17]
3 years ago
15

Help me plz I don't understand math today?

Mathematics
2 answers:
Katarina [22]3 years ago
6 0

Answer:

C-42

B-156

A-66

Step-by-step explanation:

im pretty sure

Natasha2012 [34]3 years ago
4 0

Answer:

C-42

B-156

A-66

Step-by-step explanation:

Now I'm not a 100% sure on this but I'm pretty confident.

The easiest on to find was C because on the opposite side was 24 so the side we are working with must equal 24. In order to find the answer I did 24 + 18 and got 42. That means C = 42.

The next one I found was B. A straight line is 180 degrees so I subtracted 24 from 180 and got 156.

Finally I found A. This was considerably the quickest on to find for me because of all the other measurements I found and the 90 degrees beside the A. I subtracted 90 from the already found 156. I got 156 because if you take out the 90 degree line it would be a mirror image of B. Once I subtracted 90 from 156 I got 66 as my answer.

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Find the slope of the line represented by the table of values. A) 1/2 B) 1 C) 2 D) 4
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3 years ago
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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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adoni [48]

Answer:

0.00109

Step-by-step explanation:

1.09(10^−3)

=1.09*10^−3

=1.09* 1 /10*10*10

=0.00109

5 0
3 years ago
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