Your friend is wrapping 1 meter of twine around a spool with a 2-centimeter diameter. The spool is thin and accommodates only one wrap of twine before the twine stacks on top of itself. The twine has a diameter of 1 2 centimeter, which increases the diameter of the spool by 1 centimeter with each wrap. a. Find how many complete times your friend will wrap the twine around the spool. b. Find the percentage of a complete circle that the last wrapping of the twine will make. Round your answer to the nearest tenth.

a) 6 complete times

b) 19.3%

Step-by-step explanation:

In every turn, the diameter of the twine increases 1 cm.

After the first turn, there is a total wrapped of 3.14*2=6.3 cm.

After the second turn, we have wrapped 6.3+3.14*(2+1)=15.7 cm.

If we continue with these algorithm, we have that after the 6th turn we have 84.8 cm wrapped. This is the last complete turn.

T = 0D = 2 cm.Wrapped = 6.3 cm.Total wrapped = 6.3 cm.

T = 1         D = 3 cm.Wrapped = 9.4 cm.Total wrapped = 15.7 cm.

T = 2D = 4 cm.Wrapped = 12.6 cm.Total wrapped = 28.3 cm.

T = 3D = 5 cm.Wrapped = 15.7 cm.Total wrapped = 44 cm.

T = 4D = 6 cm.Wrapped = 18.8 cm.Total wrapped = 62.8 cm.

T = 5D = 7 cm.Wrapped = 22.0 cm.Total wrapped = 84.8 cm.

T = 6D = 8 cm.Wrapped = 25.1 cm.Total wrapped = 110 cm.

There are left (100-84.8) = 15.2 cm to wrap around the twine, which has a 25.1 cm diameter by now.

The perimeter of the twine is 3.14*25.1=78.8 cm.

The percentage of a complete circle that the last wrapping of the twine will make is:

P = 15.2/78.8 = 0.193 = 19.3%

see explanation

step-by-step explanation:

a

(6r + 7) + (13 + 7r) ( remove brackets and collect like terms )

= 6r + 7 + 13 + 7r

= 13r + 20

b

remove brackets and collect like terms

13 - r - 1 + r

= 12 - r

c

remove brackets and collect like terms

= - 8 - r + 2r - 4

= - 12 + r

d

remove brackets and collect like terms

= 7r - - - 6r

[-   - = - - = - ]

= r -

mathpapa

step-by-step explanation: