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2 years ago

8

Two groups are going on a trip to the theater. The first group has 30 students in for adult chaperones. The second group has 25

students in for adult chaperones. The cost, in dollars, for each student ticket, S, and each adult ticket, A, can be determined using the system of equations below

1 answer:

Nata [24]2 years ago

7 0

Answer:

$20

Step-by-step explanation:

Two groups are going on a trip to a theater. The first group has 30 students and 4 adult chaperones. The second group has 25 students and 4 adult chaperones. The cost, in dollars, for each student ticket, s, and each adult ticket, a, can be determined using the system of equations below.

30s+4a=720

25s+4a=620

What is the cost for each student ticket?

Solution:

30s+4a=720

(1)

25s+4a=620 (2)

Subtract (2) from (1)

30s - 25s = 720 - 620

5s = 100

Divide both sides by 5

s = 100/5

= 20

s = $20

Substitute the value of s into (1)

30s + 4a = 720

30(20) + 4a = 720

600 + 4a = 720

4a = 720 - 600

4a = 120

a = 120 / 4

= 30

a = $30

Each student tickets cost $20

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The length of a rectangle is 4 cm longer than its width. If the perimeter of the rectangle is 44 cm, find its area.

adell [148]

Length: w+4
Width:w
Perimeter:44
Area: L times w
P=44
44=2(w)+2(w+4)
44=2w+2w+8
44=4w+8
36=4w
w=9
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Area should equal 117 cm squared

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3 years ago

During optimal conditions, the rate of change of the population of a certain organism is proportional to the population at time

Lana71 [14]

Answer:

The population is of 500 after 10.22 hours.

Step-by-step explanation:

The rate of change of the population of a certain organism is proportional to the population at time t, in hours.

This means that the population can be modeled by the following differential equation:

$\frac{dP}{dt} = Pr$

In which r is the growth rate.

Solving by separation of variables, then integrating both sides, we have that:

$\frac{dP}{P} = r dt$

$\int \frac{dP}{P} = \int r dt$

$\ln{P} = rt + K$

Applying the exponential to both sides:

$P(t) = Ke^{rt}$

In which K is the initial population.

At time t = 0 hours, the population is 300.

This means that K = 300. So

$P(t) = 300e^{rt}$

At time t = 24 hours, the population is 1000.

This means that P(24) = 1000. We use this to find the growth rate. So

$P(t) = 300e^{rt}$

$1000 = 300e^{24r}$

$e^{24r} = \frac{1000}{300}$

$e^{24r} = \frac{10}{3}$

$\ln{e^{24r}} = \ln{\frac{10}{3}}$

$24r = \ln{\frac{10}{3}}$

$r = \frac{\ln{\frac{10}{3}}}{24}$

$r = 0.05$

So

$P(t) = 300e^{0.05t}$

At what time t is the population 500?

This is t for which P(t) = 500. So

$P(t) = 300e^{0.05t}$

$500 = 300e^{0.05t}$

$e^{0.05t} = \frac{500}{300}$

$e^{0.05t} = \frac{5}{3}$

$\ln{e^{0.05t}} = \ln{\frac{5}{3}}$

$0.05t = \ln{\frac{5}{3}}$

$t = \frac{\ln{\frac{5}{3}}}{0.05}$

$t = 10.22$

The population is of 500 after 10.22 hours.

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pentagon [3]

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