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Andru [333]
3 years ago
6

Two researchers are studying the decline of orangutan populations. In one study, a population of 784 orangutans is expected to d

ecrease at a rate of 25 orangutans per year. In a second study, the population of a group of 817 orangutans is expected to decrease at a rate of 36 per year. After how many years will the two populations be the same?
Mathematics
2 answers:
zhuklara [117]3 years ago
5 0
Step 1:
Set Variables (We will use x & y)

x = years
y = total orangutan population

Step 2:
Set up Equations

784 - 25x = y
817 - 36x = y

Step 3:
Set equations equal to each other & solve

784 - 25x = 817 - 36x
784 = 817 - 11x
-33 = -11x
3 years = x
earnstyle [38]3 years ago
5 0

Answer:

The answer is 3 years.

Step-by-step explanation:

Let the years be denoted by 't' ,when both populations will be same.

1st study says a population of 784 orangutans is expected to decrease at a rate of 25 orangutans per year.

Equation becomes:

y=784-25t

In a second study, the population of a group of 817 orangutans is expected to decrease at a rate of 36 per year.

Equation becomes:

y=817-36t

Now to solve for 't' we will equal both the equations.

784-25t=817-36t

36t-25t=817-784

11t=33

So, t = 3 years.

So, the answer is 3 years.

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A cylindrical can without a top is made to contain 25 3 cm of liquid. What are the dimensions of the can that will minimize the
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Answer:

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

Step-by-step explanation:

Given that, the volume of cylindrical can with out top is 25 cm³.

Consider the height of the can be h and radius be r.

The volume of the can is V= \pi r^2h

According to the problem,

\pi r^2 h=25

\Rightarrow h=\frac{25}{\pi r^2}

The surface area of the base of the can is = \pi r^2

The metal for the bottom will cost $2.00 per cm²

The metal cost for the base is =$(2.00× \pi r^2)

The lateral surface area of the can is = 2\pi rh

The metal for the side will cost $1.25 per cm²

The metal cost for the base is =$(1.25× 2\pi rh)

                                                 =\$2.5 \pi r h

Total cost of metal is C= 2.00 \pi r^2+2.5 \pi r h

Putting h=\frac{25}{\pi r^2}

\therefore C=2\pi r^2+2.5 \pi r \times \frac{25}{\pi r^2}

\Rightarrow C=2\pi r^2+ \frac{62.5}{ r}

Differentiating with respect to r

C'=4\pi r- \frac{62.5}{ r^2}

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C''=4\pi + \frac{125}{ r^3}

To find the minimize cost, we set C'=0

4\pi r- \frac{62.5}{ r^2}=0

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\left C''\right|_{x=1.71}=4\pi +\frac{125}{1.71^3}>0

When r=1.71 cm, the metal cost will be minimum.

Therefore,

h=\frac{25}{\pi\times 1.71^2}

⇒h=2.72 cm

Therefore the radius of the can is 1.71 cm and height of the can is 2.72 cm.

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