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Phantasy [73]
3 years ago
11

A spherical balloon has a maximum surface area of 1,500 square centimeters.

Mathematics
1 answer:
topjm [15]3 years ago
4 0

Answer:

Option (1).

Step-by-step explanation:

Given question is incomplete: here is the complete question.

The radius of a sphere, r, is given by the formula below, where s is the surface area of the sphere. A spherical balloon has a maximum surface area of 1,500 square centimetres. Use the given formula to write a function, r(s), that models the situation. Then, use the function to predict how the radius of the balloon changes as the balloon is inflated.

1). As the surface area of the balloon increases, the radius of the balloon increases until the maximum surface area is reached.

2). As the surface area of the balloon increases, the radius of the balloon increases without bound.

3). As the surface area of the balloon increases, the radius of the balloon decreases without bound.

4). As the surface area of the balloon increases, the radius of the balloon decreases until the maximum surface area is reached.

Since surface area of a sphere is represented by,

S= 4πr²

r = \sqrt{\frac{S}{4\pi }}

Function representing the relation between the radius and surface area will be,

r(s) = \sqrt{\frac{S}{4\pi } }

Maximum surface area that a balloon can get = 1500 cm²

From this function we find that radius of the balloon is directly proportional to the square root of the surface area.

Therefore, As the surface area of the balloon increases, radius of the balloon increases until the balloon achieves maximum surface area.

Option 1). will be the answer.

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a milk tank at a a dairy farm has the form of a rectangular prism. The tank is 3.5 feet wide. The width of the tank is 5/7 of it
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Answer:

17.5 cubic feet

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Step-by-step explanation:

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L = 3.5                                    Given

W = 5/7  *  3.5                        Divide by 7

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Find all of the equilibrium solutions. Enter your answer as a list of ordered pairs (R,W), where R is the number of rabbits and
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Answer:

(0,0)   (4000,0) and (500,79)

Step-by-step explanation:

Given

See attachment for complete question

Required

Determine the equilibrium solutions

We have:

\frac{dR}{dt} = 0.09R(1 - 0.00025R) - 0.001RW

\frac{dW}{dt} = -0.02W + 0.00004RW

To solve this, we first equate \frac{dR}{dt} and \frac{dW}{dt} to 0.

So, we have:

0.09R(1 - 0.00025R) - 0.001RW = 0

-0.02W + 0.00004RW = 0

Factor out R in 0.09R(1 - 0.00025R) - 0.001RW = 0

R(0.09(1 - 0.00025R) - 0.001W) = 0

Split

R = 0   or 0.09(1 - 0.00025R) - 0.001W = 0

R = 0   or  0.09 - 2.25 * 10^{-5}R - 0.001W = 0

Factor out W in -0.02W + 0.00004RW = 0

W(-0.02 + 0.00004R) = 0

Split

W = 0 or -0.02 + 0.00004R = 0

Solve for R

-0.02 + 0.00004R = 0

0.00004R = 0.02

Make R the subject

R = \frac{0.02}{0.00004}

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When R = 500, we have:

0.09 - 2.25 * 10^{-5}R - 0.001W = 0

0.09 -2.25 * 10^{-5} * 500 - 0.001W = 0

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Collect like terms

- 0.001W = -0.07875

Solve for W

W = \frac{-0.07875}{ - 0.001}

W = 78.75

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(R,W) \to (500,79)

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0.09 - 2.25 * 10^{-5}R - 0.001W = 0

0.09 - 2.25 * 10^{-5}R - 0.001*0 = 0

0.09 - 2.25 * 10^{-5}R = 0

Collect like terms

- 2.25 * 10^{-5}R = -0.09

Solve for R

R = \frac{-0.09}{- 2.25 * 10^{-5}}

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(R,W) \to (4000,0)

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-0.02W + 0.00004W*0 = 0

-0.02W + 0 = 0

-0.02W = 0

W=0

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(R,W) \to (0,0)

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