Aconical oil cup with a radius of 5.6cm must be design to hold 69 cu cm of oil. what should be the altitude of the cup

The height of the conical oil cup is 2.102 cm

Step-by-step explanation:

Given as for a conical cup :

The radius of the conical oil cup =  r = 5.6 cm

The volume of the conical oil cup = v = 69 cubic cm

Or, v = 69 cm³

Now, ∵ The volume of cone = ××r²×h

Where r is the radius of the cone

And h is the height of the cone

The value of = 3.14

So, The volume of conical oil cup =  The volume of cone = ××r²×h

Or, The volume of conical oil cup =   × 3. 14 × (5.6)² × h

Or, 69 cm³ =   × 98.47 × h

Or, 69 cm³ = 32.823 cm² × h

∴ h =  

I,e h = 2.102 cm

So, height of cup is 2.102 cm

Hence The height of the conical oil cup is 2.102 cm , Answer

there are lots of ways.   pick's theorem tells us if the vertices are lattice points the area a is

a = b/2 + i - 1

where b is the number of boundary lattice points and i is the number of interior lattice points.

in the figure i count b=16, i=50 so

a = 8 + 50 - 1 = 57

let's try the shoelace formula.   the area is half the sum of the cross products of the sides.  

a(-2,6)   -2(2)-6(-5) = 26

b(-5,2)   -5(-3) - 2(-2) = 19

c(-2,-3) -2(-3) - (-3)(4) = 18

d(4,-3)   4(3) - (-3)(5) = 27

e(5,3)   5(3) - 3(1) = 12

f(1,3)     1(6) - 3(-2) = 12

a(-2,6)

each of the cross products is twice the area of the triangle with the associated side and one vertex the origin. so the area of abo=26/2=13.

area = (1/2) (26 + 19 +   18 + 27 + 12 + 12) = 57

that checks!

d? maybe

step-by-step explanation: