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Ede4ka [16]
3 years ago
9

A rabbit population of 500 was introduced in a certain county of Eagleville. Biologists observe that the population increases 25

% each year. Complete the following:
a. Identify a and b.
b. Write an exponential function that models the situation if the trend continues.
c. How many rabbits will be in Eagleville after 10 years?.
Mathematics
2 answers:
Nostrana [21]3 years ago
7 0

Answer:

c. How many rabbits will be in Eagleville after 10 years?.

Step-by-step explanation:

Answer.

Ipatiy [6.2K]3 years ago
5 0

Answer:

Step-by-step explanation:

The exponential growth equation that represents the rabbit population y) after x years is y = 500(1.25)ˣ

Exponential growth is given by:

y = abˣ

where a is the initial value of y, b is the multiplier and b>1, y, x are variables.

Let y represent the total rabbit population after x years.

Since there were 500, introduced hence a = 500. Also the population increases 25% each year, hence:

b = 100% + 25% = 1 + 0.25 = 1.25

y = 500(1.25)ˣ

The exponential growth equation becomes y = 500(1.25)ˣ

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OLEGan [10]

Answer:

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4 0
3 years ago
BRAINLIEST!!! PLEASE HELP!! 6 points are placed on the line a 4 points are placed on the line b How many triangles is it possibl
amid [387]

Answer:

all vertices on the two lines=60+36=96

Step-by-step explanation:

line a has 6 points and line b has 4 points

now each triangle drawn has on vertices on one line and two vertices on another line:

the line with six points :C(6,2) six points , has two vertices

C(6,2)=6*5/2*1=30/2=15 pairs of points

each pair connected to one point on the other line (4 points)

4*15=60 vertices

now let us consider the two lines on the other line with 4 points:

6c(4,2)=6*((4*3(/2))=6*6=36 vertices

all vertices on the two lines=60+36=96

8 0
3 years ago
George is comparing two different phones. The screen of phone A has a width of 5.3 cm. The width of the screen on phone B is 5 a
BaLLatris [955]

Answer:

The answer is "Screen B is wider , because 5.3 is greater than 5 and one-third".

Step-by-step explanation:

The mobile display unit must be made similar. Phone A uses a decimal device, and telephone B uses a fraction in this case.

It's easier for you to use. Attempt using the decimal. The width of the phone B fraction could then be translated to decimal width. The computing is:

= 5 cm + \frac{1}{3} cm

= 5 cm + 0.333cm

= 5.333 cm

It is evident from here that the wider screen of Phone B is than that of Phone A.

8 0
3 years ago
Read 2 more answers
Refer to the previous question.
Alexxandr [17]

Answer:

|AC| = 4.47 units (question attached as image)

Step-by-step explanation:

The question lacks sufficient details. We wil however consider a similar example (image attached) to understand how to solve questions using Pythagoras theorem.

We will calculate for |AC| using Pythagoras theorem, we have:

|AC|² = |AB|² + |BC|²

|AC| = ?, |AB| = 4 units, |BC| = 2 units

|AC|² = 4² + 2² = 16 + 4

|AC|² = 20 ⇒ |AC| = \sqrt{20}

|AC| = 4.472135955 ≈ 4.47

|AC| = <u>4.47</u> units

5 0
3 years ago
A large storage tank, open to the atmosphere at the top and fi lled with water, develops a small hole in its side at a point 16.
Arada [10]

Answer:

The speed at which the water leaves the hole V_{2} = 4.21 \frac{m}{s}

The value of  diameter of the hole d = 0.112 m = 11.2 cm

Step-by-step explanation:

Given data

Flow rate = 2.5 × 10^{-3} \frac{m^{3} }{min} = 0.0416 × 10^{-3} \frac{m^{3} }{sec}

Height (h) = 16 m

(a) The speed at which the water leaves the hole :-

Apply bernouli equation for the water tank at point 1 & 2

\frac{P_{1} }{\rho g} + \frac{V_{1} ^{2} }{2g}  + Z_{1} = \frac{P_{2} }{\rho g} + \frac{V_{2} ^{2} }{2g}  + Z_{2} ------ (1)

Since P_{1} = P_{2} , V_{1}  =  0 & Z_{1} - Z_{2} = h

Equation (1) becomes

\frac{V_{2}^{2}  }{2 g} = h

V_{2} = \sqrt{2 g h}

This is the speed at which the water leaves the hole.

Put the values of g & h in the above formula

⇒ 2 g h = 2 × 9.81 × 16 = 17.71

V_{2} = \sqrt{2 g h} = \sqrt{17.71}

V_{2} = 4.21 \frac{m}{s}

This is the speed at which the water leaves the hole.  

(b)Diameter of the hole :-

We know that flow rate Q = A × V

A = \frac{Q}{V}

Put the values of Q & V in the above formula we get

A =  \frac{0.0416}{4.21} × 10^{-3}

A = 9.88 × 10^{-6}

We know that area A = \frac{\pi}{4} d^{2}

⇒ d^{2} = \frac{4}{\pi} × 9.88 × 10^{-6}

⇒ d^{2} = 0.01258

⇒ d = 0.112 m = 11.2 cm

this is the value of  diameter of the hole.

7 0
3 years ago
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