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AfilCa [17]
3 years ago
14

Suppose that a local trout population exhibits exponential growth by doubling every 4 years. Considering that the initial popula

tion were 375 trout. How many trout will there be after 1 year?
Mathematics
1 answer:
pshichka [43]3 years ago
7 0
<span>468.75

round up if asks for you to
</span>
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What's the answer to this question
jasenka [17]
Tan w=20/83
inverse tan 20/83=13.55 degrees
3 0
3 years ago
The sum of the ages of Berma, her mother Rinna and her father Erwin is 80. Two years from now, Rinna’s age will be 13 less than
jekas [21]

Answer: Berma is 5 years old

Rinna is 38 years old

Erwin is 37 years old

Step-by-step explanation:

Let x represent Berma's age

Let y represent Rinna's age

Let z represent Erwin's age

Since the sum of their ages is 80,

x + y + z = 80 - - - - - - -1

Two years from now, Rinna’s age will be 13 less than the sum of Erwin’s age and twice Berma’s age. This means that

y +2 = [ (z+2) + 2(x+2) ] - 13

y +2 = z + 2 + 2x + 4 - 13

2x - y + z = 13 + 2 - 4 -2

2x - y + z = 9 - - - - - - -2

Three years ago, 15 times Berma’s age was 5 less than the age of Rinna. It means that

15(x - 3) = (y - 3) - 5

15x - 45 = y - 3 - 5

15x - y = - 8 + 45

15x - y = 37 - - - - - - - -3

From equation 3, y = 15x - 37

Substituting y = 15x - 37 into equation 1 and equation 2, it becomes

x + 15x - 37 + z = 80

16x + z = 80 + 37 = 117 - - - - - - 4

2x - 15x + 37 + z = 9

-13x + 2 = -28 - - - - - - - - -5

subtracting equation 5 from equation 4,

29x = 145

x = 145/29 = 5

y = 15x - 37

y = 15×5 -37

y = 38

Substituting x= 5 and y = 38 into equation 1, it becomes

5 + 38 + z = 80

z = 80 - 43

z = 37

6 0
3 years ago
Someone please help me
solmaris [256]
37.
13 - 7 = 6
Her Mom Uses 6

38.
9 + 8 = 17
Andrew Had 17 Muffins
6 0
3 years ago
1. a) Solve the following quadratic-quadratic system of equations graphically.
anastassius [24]

Answer:

see explanation

Step-by-step explanation:

look at the photo

4 0
2 years ago
A 12-m3 oxygen tank is at 17°C and 850 kPa absolute. The valve is opened, and some oxygen is released until the pressure in the
sweet-ann [11.9K]

Answer:

Released oxygen mass: 15.92 kg

Step-by-step explanation:

ideal gas law : P*V=nRT

P:pressure

V:volume

T:temperature

n:number of moles of gas

n [mol] = m [g] /M [u]

m : masa

M: masa molar = 15,999 u (oxygen)

R: ideal gas constant = 8.314472 cm^3 *MPa/K*mol =

grados K = °C + 273.15

P1*V*M/R*T = m1

P2*V*M/R*T = m2

masa released : m1-m2 = (P1-P2) * V*M/R*T

m2-m1 = 200 * 10^-3 MPa * 12 * 10^6 cm^3 * 15.999 u / 8.314472 (cm^3 * MPa/K *mol) * 290. 15 K

m2-m1= 38 397.6 * 10^3 u*mol / 2412.44 = 15916.5 g = 15.9165 kg

3 0
3 years ago
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