SOLUTION: Due to melting, an ice sculpture loses one-half its weight every hour. After 8 hours, it weighs 5/16 of a pound. How much did it weigh in the beginning. I did 5/16+8 and put the

Algebra ->  Coordinate Systems and Linear Equations  -> Linear Equations and Systems Word Problems -> SOLUTION: Due to melting, an ice sculpture loses one-half its weight every hour. After 8 hours, it weighs 5/16 of a pound. How much did it weigh in the beginning. I did 5/16+8 and put the      Log On


   



Question 362889: Due to melting, an ice sculpture loses one-half its weight every hour. After 8 hours, it weighs 5/16 of a pound. How much did it weigh in the beginning.
I did 5/16+8 and put the fraction into decimal form then added the two together and got 8.3125. Then its says 1/2 of the that is lost in weight. Now I multiplied the 8.3125 to .05(The decimal of 1/2) then got 4.15625. I dont know if that is correct or what and i used samdep instead of pemdas its pemdas backwards because it says to solve backwards using asmdep.

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
Let's say it weighed W to start.
After 1 hour, it weighs W%2F2
After 2 hours, it weighs W%2F4
After 3 hours, it weighs W%2F8
After 4 hours, it weighs W%2F16
After 5 hours, it weighs W%2F32
After 6 hours, it weighs W%2F64
After 7 hours, it weighs W%2F128
After 8 hours, it weighs W%2F256
But you know that
W%2F256=5%2F16
So then,
W=%285%2F16%29%28256%29
highlight%28W=80%29lbs
.
.
.
You could also work it backwards,
After 8 hours it weighs 5%2F16lb.
After 7 hours it weighed 2%285%2F16%29=5%2F8lb.
After 6 hours it weighed 2%285%2F8%29=5%2F4lb.
After 5 hours it weighed 2%285%2F4%29=5%2F2lb.
After 4 hours it weighed 2%285%2F2%29=5lb.
After 3 hours it weighed 2%285%29=10lb.
After 2 hours it weighed 2%2810%29=20lb.
After 1 hours it weighed 2%2820%29=40lb.
After 0 hours it weighed 2%2840%29=highlight%2880%29lb.