Question

A certain mountain has elevation of 19,710 feet. in 1913, the glacier on this peak covered 7 acres. By 2005, this glacier had melted to only 1 acre. Assume that this glacier melted at a constant rate each year. Find this yearly rate.
b) use the answer from part a to write a linear equation that gives the acreage A of this glacier t years 1913.

Answer 1

hmmm
so you have a mountain covered with snow at the top
back in 1913, the snow was covering 7 acres at the top of the mountain

and then in 2005, it only covered 1 acre
so it melted 6 acres since 1913

a) is the yearly rate, that is, 2005-1913 = 94 years
so in 94 years it melted 6 acres, so \(\bf \cfrac{years}{acres} \) will be the yearly rate

Answer 2

2005-1913 i got 92

Answer 3

b) is just the equation off that
say it melted "Z amount" in 1year

Acres = "Z amount" t

Answer 4

2005-1913=92
92/6=15.33

Answer 5

is that right

Answer 6

A=-0.06+6

Answer 7

actually ... that was my mistake ... it should be Acres per Year melted, thus \(\bf \cfrac{acres}{years}\)

Answer 8

which is a very small amount

Answer 9

6/92=0.06

Answer 10

yes, then again, one can say, that there are 43560 square feet in 1 acre, thus in 6 acres it'll be 261360 square feet, and if use that to get a rational value, we'd end up with \(\bf \cfrac{acres}{years} \implies \cfrac{6}{92} \cdot \cfrac{261360}{92} \implies 2840.86 \frac{ft}{year}\)

Answer 11

\(\bf 2840.86 \frac{ft^2}{year}\)

Answer 12

.... actually, the B) part uses Acres too... darn... well , I gather we'll use 0.0652 then


so
Acreage melted = 0.0652t