If a stack of 45 coins fits snugly in the coin wrapper shown, how thick is 1 coin? coin
wrapper
d=21.17 mm V=28,470 mm

1.797 mm

Step-by-step explanation:

The key to solving this problem is finding the length of the coin wrapper, or "height" of the cylinder.

The volume of a cylinder equation is:

V = hπr^2

- We want to find h

- We are given diameter and we will want the radius so we divide

21.17/2 = 10.585

r = 10.585

So to find h we manipulate the equation by dividing both sides by pi and r^2

resulting in the following:

h =

now substitute given values and we get

h = = 80.883

Now to find the thickness of a single coin we divide the length of the sleeve (h) by the number of coins in the sleeve (45).

Thickness = 80.883/45 = 1.797 mm

is should be y = 5x

step-by-step explanation:

correct me if im wrong

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)