How many times will a 3 and a half inch circumference of a tire make full rotations every 10,000 miles. What would happen in the number of rotations if the circumference is both increased and or decreased by 20%?
no
I need the equation and help.
first thing I would do is convert `10,000 miles` to inches
do you know how to do that @AnnaLee607 ?
63,360 inches is = 1 mile right?
yes
Would you multiple 10,000 by 63,360?
correct
633,600,000?
yep
633,600,000 inches = 10,000 miles
the tire has a `3 and a half inch circumference` so how many full rotations will it do to cover 10,000 mi? equivalently, this is the same as asking so how many full rotations will it do to cover 633,600,000 inches?
Would I divide 3.5 inches by 633,600,000
example: if I asked you "how many full rotations would the tire have to roll if the tire covers 7 inches?" then you would do 7/3.5 = 2 so the tire has to roll a full 2 rotations to cover 7 inches
633,600,000/3.5=181,028,571.43
good, so it has to do 181,028,571.43 rotations if the tire is 3.5 inches in circumference
now increase the circumference by 20% and recompute the number of rotations
Would I multiply it by .20?
100% is the initial amount increase by 20% 100% + 20% = 120% = 1.2 so you multiply by 1.2
if you decrease by 20%, then 100% - 20% = 80% = 0.80 so you'd multiply by 0.80
Increase 217,234,285.72 Decrease 144,822,857.14
let me check
Is that all?
no you should multiply 3.5 by 1.2 NOT the number 181,028,571.43
step 1) multiply 3.5 by 1.2 step 2) divide 633,600,000 over the result from step 1
Increase 150,857,142.86 Decrease 226,285,714.29
very good
if you increase the circumference by 20% then you go from `181,028,571.43 rotations` to `150,857,142.86 rotations`
if you decrease the circumference by 20% then you go from `181,028,571.43 rotations` to `226,285,714.29 rotations`
Thank you so much.
no problem
Join our real-time social learning platform and learn together with your friends!