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Ber [7]
3 years ago
14

The volume of a container varies jointly with the square of its radius, r, and its height, h. The

Mathematics
1 answer:
xenn [34]3 years ago
7 0
<h3>The volume of a container with a radius of 4 centimeters and a height of 4  centimeters is 67.2 cubic centimeter</h3>

<em><u>Solution:</u></em>

<em><u>The volume of a container varies jointly with the square of its radius, r, and its height, h </u></em>

Therefore,

v \propto r^2h\\\\v = k \times r^2h ----------- eqn\ 1

<em><u>The  container has a height of 10 centimeters, and radius of 6 centimeters, and a volume of 377 cubic  centimeters </u></em>

<em><u> Substitute v = 377 and h = 10 and r = 6 in eqn 1</u></em>

377 = k \times 6^2 \times 10\\\\377 = 360 \times k\\\\k = \frac{377}{360}\\\\k = 1.047 \approx 1.05

<em><u>What is the volume of a container with a radius of 4 centimeters and a height of 4  centimeters?</u></em>

<em><u>Substitute k = 1.05 and r = 4 and h = 4 in eqn 1</u></em>

v = 1.05 \times 4^2 \times 4\\\\v = 1.05 \times 16 \times 4\\\\v = 67.2

Thus volume of a container with a radius of 4 centimeters and a height of 4 centimeters is 67.2 cubic centimeter

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Answer:

f(x) = $25 + $75x.

Domain: positive whole numbers.

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Step-by-step explanation:

Let x denote the total number of tour member, charging $75 per tour member implies $75x.

Therefore, the tour company charges:

f(x) = $25 + $75x

The Domain of the function, which is the possible value of x, is the set of all positive whole numbers. You can't have negative numbers because no such thing as "negative member," also you can't have fractions because no such thing is "2/3 member (for instance)."

The range, which are all the values of f(x) after putting a particular value from the domain in f(x), is the numbers from 25 and above. That is f(x) ≥ 25. This is because, if there are no tour members, they pay for hiring director only which is $25, if we have one member, the money increases to $100 and so on. These values are from 25 upwards.

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Please help. I only need parts a and b.
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