A can of StarKist tuna has a volume LaTeX: 18\pi\:cm^318 π c m 3 and a height of 2 cm. Find the area of the StarKist label below the wraps around the entire can and does not overlap. Write your answers in terms of LaTeX: \piπ.

Area of the StarKist label around the can in terms of π = 12π cm²

Step-by-step explanation:

Given;

the volume of a can of StarKist tuna, V = 18 π cm³

height of the can of StarKist tuna, h =  2 cm

To determine the area of the StarKist label that wraps around the entire can and does not overlap, we assume the can to have a shape of a cylinder.

Volume of the can = πr²h

where;

r is radius of the can

h is height of the can

πr² x 2 = 18 π

2r² = 18

r² = 18/2

r² = 9

r = 3

Area around the can = curved surface area of the can (cylinder)

Curved surface area of the can = 2πr × h = 2πrh

Curved surface area of the can = 2πrh = 2π x 3 x 2 = 12π cm²

Area of the StarKist label around the can in terms of π = 12π cm²

answer: 12

step-by-step explanation:

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.