The US standard can is 4.83 inches high, 2.13 inches in diameter at the lid, and 2.60 inches in diameter at the widest point of
the body. Assume that the can has the largest diameter 2) One cubic inch of volume of liquid is (equivalent to) the same as 16.3891 milliliters of liquid. What will be the resulting amount of liquid that has been bottled in milliliters?
The first thing we must do for this case is to calculate the volume of the can using the largest diameter: V = pi * ((d / 2) ^ 2) * (h) Where, h: height d: diameter. Calculating we have: V = pi * ((2.60 / 2) ^ 2) * (4.83) = 25.64 in ^ 3 Then, to determine the amount of liquid in millimeters, we can use a simple rule of three: 1 in ^ 3 ---> 16.3891 milliliters 25.64 in ^ 3 ---> x Clearing x we have: x = ((25.64) / (1)) * (16.3891) = 420.22 mm answer The amount of liquid that has been bottled in milliliters is 420.22 mm
Radius is half of the diameter. So because the area of a circle is A= pi • r^2, to find the radius first, we have to change the equation to solve for r. Now it is, r= square root of A/pi. We plug in 1134.1. (Look at pic) The diameter is 38
The simplest fraction for is . Write the upper bound as a fraction with the same denominator:
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Hence the range for would be:
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If the denominator of is also , then the range for its numerator (call it ) would be . Apparently, no whole number could fit into this interval. The reason is that the interval is open, and the difference between the bounds is less than .
To solve this problem, consider scaling up the denominator. To make sure that the numerator of the bounds are still whole numbers, multiply both the numerator and the denominator by a whole number (for example, 2.)
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At this point, the difference between the numerators is now . That allows a number ( in this case) to fit between the bounds. However, can't be written as finite decimals.
Try multiplying the numerator and the denominator by a different number.
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It is important to note that some expressions for can be simplified. For example, because of the common factor .