I NEED IN THE NEXT 10 MIN PLS. GRAPH ATTACHED WILL GIVE BRAINLIEST Use the given graph to determine the limit, if it exists. A coordinate graph is shown with a horizontal line crossing the y axis at five that ends at the open point 2, 5, a closed point at 2, 1, and another horizontal line starting at the open point 2, negative 3 and continues to the right. Find limit as x approaches two from the left of f of x. and limit as x approaches two from the right of f of x..

Step-by-step explanation:

You gave very clear instructions on how to draw this graph, so that's what I did. What you need to remember in particular with limits is that you do not care in the least what happens AT the x value of 2, only what happens as it is being approached. Because we are asked the limit as x is approaching from the left and the right, this is a one-sided limit question. In order for the limit to exist as x approaches 2 (NOT from the left or the right), the limit would have to agree from the left and the right, and this one doesn't. Having said that there is "a horizontal line crossing the y-axis at 5 that ends at the open point (2, 5)..." is a limit approaching x from the left. Therefore,

Having also said there is "...another horizontal line starting at the open point (2, -3) and continues to the right..." is a limit approaching x from the right. Therefore,

The closed point at (2, 1) is where x IS, and remember that we don't care about what happens AT x. So disregard this point in limits.

what graph

step-by-step explanation:

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.