Help please pls help help help

A

Step-by-step explanation:

The solution we're looking at here is a set of all the points between two parallel lines. Let's see what we can find out about those lines from the inequality, and use that the pick the correct option.

We can start by splitting the inequality at the top into two seperate inequalities:

From here, we can put each inequality into slope-intercept form:

Our slope for both lines is positive, so we can immediately eliminate B and D. A and C both display lines with slopes of 3 and y-intercepts of 7 and 11; the only difference is in which line is dashed and which line is solid.

As a reminder, a dashed line means that the line itself is not included in the solution, a solid line means that it is. Dotted lines always pop up when you see the symbols < and >, while solid lines pop up when you see ≤ and ≥.

(Solid)

(Dashed)

This matches up very nicely with the graph in A, our solution.

answer: ratio of volume of cone (with 3 times the height and equal diameter) to volume of cylinder is 2 to 1. the value of π is not used.

step-by-step explanation:

radius is diameter/2 = 4.

area of circle is π r^2.

volume of cylinder is height × area of base = hπr^2 = 3(4^2) π = 48π cubic inches.

not knowing the formula for volume of cone, i calculate it using calculus (which is for calculating things).

volume of cone is integral 0 to h area(y) dy. area(y) is π r(y)^2. r(y) runs r to zero as y runs zero to h, r(0) = r, r(h) = 0.

r(y) = r - r/h y

area(y) = π (r - r/h y)^2 = π (r^2 - 2r^2y/h + r^2y^2/h^2) = π r^2(1-2y/h+y^2/h^2)

antiderivative area(y) = π r^2 (y-y^2/h+y^3/(3h^2))

integral 0 to h = π r^2(h-h^2/h+h^3/(3h^2) - (0)) = π r^2(h - h + h/3) = π r^2(h/3) = π/3 r^2 h

volume of this cone is π/3(4^2)18 cubic inches.

= 6 16 π = 96π cubic inches.

cone has twice the volume, 96/48.

cone height is 6 times the cylinder.

ratio of volume formula: π r^2 cancels.

cone over cylinder ratio is 6h/3 to h, 2 to 1.

answer: what are the answer choices?

step-by-step explanation: