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vfiekz [6]
3 years ago
6

Please help asap!

Mathematics
2 answers:
sasho [114]3 years ago
5 0
The answer is a 47 deg.
uysha [10]3 years ago
4 0

Answer:

147

Step-by-step explanation:

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Which relation represents f(x) = 2x - 5?
goblinko [34]

Answer:

{(0,-5), (1, -3), (2, -1), (3, 1)

Step-by-step explanation:

If you plug in 1 for x you get f(x)=2(1)-5 which equals 2-5=-3

You would repeat these steps just going up by one number every time

7 0
3 years ago
85,86,89,35,60,57,89,75,65​
Marysya12 [62]

Answer:

sorry but what we have to do in this question

5 0
3 years ago
Mrs. Olga’s $250 investment earns 19% interest compounded annually. What will her balance be after 3 years?
Maurinko [17]

Answer:

The answer is A. $421.29

Step-by-step explanation:

Just keep finding 19% of the numbers next, and you add that all up and you get the answer.

250 x 19%=47.5. (250+47.5)=> 297.5x19%= 56.525. (297.5+56.525) =>354.025x19%=67.26475

354.025+67.26475=> 421.28975, which can round up to 421.29.

6 0
3 years ago
Read 2 more answers
Find a positive angle less than one revolution around the unit circle that is co-terminal with the given angle: 52pi/5
Ludmilka [50]
We know that
Two angles are said to be co-terminal <span>if they have the same initial side and </span>
<span>the same terminal side. 
</span>
(52π/5)-----> 10.4π<span>
so
</span>(52π/5)-5*2π------> (2π/5)

the answer is
the positive angle less than one revolution around the unit circle that is co-terminal with angle of 52π/5 is 2π/5

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
2 years ago
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