Mathematics, 06.05.2020 04:26 Squid1865

The information below is used in the last 2 questions:
The two cylinders shown below contain oatmeal. The small cylinder has a radius of 3 inches and a height of 10 inches. The large cylinder has a radius of 6 inches and a height of 10 inches (the radius is doubled). If the small container of oatmeal costs $3, what is a fair price for the large container?

$9

$12

$6

$24

The information below is used in the last 2 questions: The two cylinders shown below contain oatmeal

Answers: 2

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Answer from: countrygirlbookworm2

The fair price for the large container is $12

Step-by-step explanation:

In this question, we are tasked with calculating what would be the fair price for the large cylinder given the price for the small cylinder.

To calculate this, what we need to do firstly is to estimate the the volumes of both cylinders.

Mathematically, the volume of a cylinder can be calculated using the formula; V = π × $r^{2}$ ×h

For the small cylinder, r = 3 inches and h = 10 inches

Thus $V_{s}$ = π × $3^{2}$ × 10 = 90π $inches^{3}$

For the large cylinder, r = 6 inches and h = 10 inches

Thus $V_{l}$ = π × $6^{2}$ × 10 = 360π $inches^{3}$

Now let us find the ratio of their volumes = 360π/90π = 4

Since the ratio of their volumes is 4, it can be fair to say that the large cylinder should cost 4 times the price of the small cylinder = 4 × $3 = $12

Answer from: Quest

hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.

Answer from: Quest

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step-by-step explanation:

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