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mylen [45]
2 years ago
9

Please help if you know the answer

Mathematics
2 answers:
Natali5045456 [20]2 years ago
5 0
The correct answer of this problem is C
worty [1.4K]2 years ago
3 0
It's C. When you're dividing g(x) by f(x), you get the equations in both B. and C., but C is more correct because you can't have x=-1/3 because the function is undefined at that number (you can't have the denominator equal 0)
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in two or more complete sentences identify the parent function and describe the transformations that were applied to obtain the
erastovalidia [21]
The parent function is f(x)=√x.

This graph has been transformed by a translation left 6 units, a translation up 2 units, and a stretch by a factor of 2.

Adding a number at the end of a function results in a vertical translation.
Adding a number inside of a function, in this case under the square root, translates the graph horizontally.
Multiplying the variable by a number before another operation results in a stretch or shrink of the graph.
6 0
3 years ago
SOMEONE HELP ME PLSSSSSSSSSSSSSSSSSSS
fomenos

Answer: drag the first one to the fourth one, drag the second one to the first one,drag the third one to the second one,drag the fourth one to the third one.

Step-by-step explanation:if its wrong then im sorry

3 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
2 years ago
a standard deck of 52 playing cards contains 13 cards in each of 4 suits. what is the appropriate probability that exactly three
kvv77 [185]

Answer:

2%

Step-by-step explanation:

After 4 cards are drawn at random, that leaves 48 cards. 1 divided by 48 is .0208.

Hence, 2%

7 0
3 years ago
under what circumstances is the ERR more appropriate method than IRR to evaluate a project ANSWER is C WHEN THE LENGTH OF THE PR
Vilka [71]

Answer:

yes

Step-by-step explanation:

3 0
2 years ago
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