Calculate the volume of liquid that would fill the bowl of each class. Show all your calculations and explain your reasoning. (no links plz)

The volume of liquid that will fill the bowl of the glass is approximately 178.82 cm³

Step-by-step explanation:

The given dimensions of the bowl are;

The shape of the bowl containing the liquid = A cone

The base length of the triangular cross-section of the cone = The diameter of the cone, d = 6 cm

The slant height of the triangular cross-section of the cone = The slant height of the cone, l = 7 cm

Therefore, the height of the cone, h = √(l² - (d/2)²)

∴ h = √(7² - (6/2)²) = 2·√10

The height of the cone, h = 2·√10 cm

The volume of a cone, V = (1/3) × Base area, A × Height, h

V = A × h

The base area of the cone, A = π × (d/2)²

A = π×(6/2 cm)² = 9·π cm²

By plugging in the values

V = A × h

∴ V = 9·π cm² × 2·√10 cm ≈ 178.82 cm³

The volume of the cone = The volume of liquid that will (can) fill the bowl of the glass, V ≈ 178.82 cm³.

c

step-by-step explanation:

there is vertical asymptote at x = 2 and x = 0

vertical asymptote occurs at the points where the denominator is 0. so we can say that

x = 2 occurs when there is a factor (x-2) in the denominator, and

x = 0 occurs when there is a factor "x" in the denominator

thus, the denominator of the equation would have "x * (x-2)", this is option c only. that's the correct answer.