The volumes of two cylindrical cans of the same shape vary directly as the cubes of their radii. If a can with a six-inch radius holds 1½ pints, how many gallons will a similar can with a 24-inch radius hold?

The second similar can with a radius of 24-inch radius will hold 12 gallons

Step-by-step explanation:

The given parameters are;

The volumes of the cylindrical can ∝ (The radius of the cans)³

The volume of first can, V₁ = 1¹/₂ pints

The radius of the first can, r₁ = 6-inch

The radius of the second can, r₂ = 24-inch

Therefore, we have;

V ∝ r³

V = k × r³

∴ 1¹/₂ pints ∝ (6 in.)³

1¹/₂ pints = 43.3125 in³

∴ 43.3125 in³ = k × 216 in.³

The constant of proportionality, k = 43.3125/216 = 0.2005208333

Therefore, we have for the second can, we have;

V₂ = k × r₂³ = 43.3125/216 × (24 in.)³ = 2772 in.³

The volume of the second can = 2772 in.³

1 in.³ = 0.004329 gallons

∴ 2772 in.³ = 2772 × 0.004329 gallons = 12 gallons

Therefore, the volume the second similar can with a 24-inch radius will hold = 12 gallons.

option: d is correct

step-by-step explanation:

the sum of the lengths of any two sides of a triangle is greater than the length of the third side. if you take the three sides of a triangle and add them in pairs, the sum is greater than (not equal to) the third side. if that is not true, then it is not possible to construct a triangle with the given side lengths.

step-by-step explanation: