anonymous
  • anonymous
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.
Mathematics
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katieb
  • katieb
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jdoe0001
  • jdoe0001
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
anonymous
  • anonymous
2005-1913 i got 92
jdoe0001
  • jdoe0001
b) is just the equation off that say it melted "Z amount" in 1year Acres = "Z amount" t

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anonymous
  • anonymous
2005-1913=92 92/6=15.33
anonymous
  • anonymous
is that right
anonymous
  • anonymous
A=-0.06+6
jdoe0001
  • jdoe0001
actually ... that was my mistake ... it should be Acres per Year melted, thus \(\bf \cfrac{acres}{years}\)
jdoe0001
  • jdoe0001
which is a very small amount
anonymous
  • anonymous
6/92=0.06
jdoe0001
  • jdoe0001
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 \(\bf \cfrac{acres}{years} \implies \cfrac{6}{92} \cdot \cfrac{261360}{92} \implies 2840.86 \frac{ft}{year}\)
jdoe0001
  • jdoe0001
\(\bf 2840.86 \frac{ft^2}{year}\)
jdoe0001
  • jdoe0001
.... actually, the B) part uses Acres too... darn... well , I gather we'll use 0.0652 then so Acreage melted = 0.0652t

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