Question

(Need help on last part... got the rest)


A certain mountain has an elevation of 19,915. In 1906 the glacier on this peak covered 10 acres. By 2001 the glacier had melted to only 1 acre.
a) Assume that this glacier melted at a constant rate each year. Find this yearly rate.
b) Use your answer from part a to write a linear equation that gives the acreage A of this glacier t years past 1906.
The yearly rate of change is [ ] acres/year?
(Type an integer or decimal. Round to nearest thousandth)
What is the equation that gives the acreage of the glacier t years after 1906?

Answer 1

yearly rate means amount melted per year. per: divided by.
looks like from ten acres to one acre is a loss of 9 acres, and from 1906 to 2001 is 95 years. compute \(-9\div 95\) for the average rate of change

Answer 2

notice that this is identical to computing the slope of the line between the two points
\[(1906,10)\text { and } (2001,1)\] you would write
\[m=\frac{1-10}{2001-1906}=\frac{-9}{95}\]

Answer 3

so the rate of change is .09 acres/year?

Answer 4

it should be negative, i would say \(-.095\) to the nearest thousandth which is what it asks for
negative because the acreage is decreasing

Answer 5

ok.

Answer 6

you got the last part?

Answer 7

so, I would write A= 1−10
---------
2001-1906

Answer 8

yes?

Answer 9

is that the correct equation showing t years after 1906?

Answer 10

ok now we got the slope, it is \(-.095\) and we want the model (linear equation) that will give the acres in the year \(t\) after 1906

Answer 11

it goes down by \(-.095\) acres per year and when you start counting, in 1906, you have 10 acres. since it goes down at a rate of \(-.095\) per year and you start with 10, your equation is
\[y=-0.095t+10\]

Answer 12

Thanks for the stepy by step so I understand and appreciate the help.

Answer 13

yw