0.78 strikeouts per minute
Given that plane P is parallel to the planes containing the base faces of the prism; then, if the plane meets the prism between the planes containing the hexagonal bases, then P meets the prism in a hexagonal region that is congruent (with the same size) to the bases of the prism.
Answer:
160π cm²
Step-by-step explanation:
The area of the shaded region is
area of outer circle - area of inner circle
A = πr₁² - πr₂²
r₁ is the radius of the outer circle = 10 + 3 = 13 cm
r₂ is the radius of the inner circle = 3 cm
A = π × 13² - π × 3²
= π(169 - 9) = 160π cm² ≈ 502.65 cm² ( to 2 dec. places )
Answer:
22.96
Step-by-step explanation:
Answer:
<h2>(4, -6)</h2>
Step-by-step explanation:
The formula of a midpoint of the segment AB:
We have A(7, -2) and B(1, -10). Substitute:
