advanced algebra/algebra 2 help please!?

Question

I have a question here and I need help. Please help me solve this problem and tell me the steps you took to solve it. Please and thanks in advance!

In 1980 the Wincon river was 32 feet below the bridge. Because of silt build-up in the river bottom, the river was only 20 feet below the bridge by 1986. Write an equation for the distance of the river from the bridge, "d", with t=0 representing 1980. If nothing is done about the silt, what year will the river reach the bridge?

thanks but now what year will the river reach the bridge then?!

Answer 1

*******1996*******


[1] The river rises 2ft per year: (32ft - 20ft = 12, 12ft / 6 years = 2ft/1year)

And 32 is the starting distance

so, FORMULA: d = 32 - 2t

[2] the river will reach the bridge when d (distance) = 0, so

d = 32-2t

> 0 = 32-2t

>> -32 = -2t

>>> t = 16

>>>> 1980 + 16 = 1996

Answer 2

First we'll assume a linear relationship between the distance from the bridge and time (y=mx+b or in our case d=mt+b)

our slope is (d1-d2)/(t1-t2) = (20-32)/(6-0) = -12/6=-2
so d=-2t+b

Since the distance was 32 feet in 1980 (t=0) then
32=-2*0+b or b=32

Then d = -2 *t +32

Answer 3

1980=32 ft. 1986=20 ft.
32 minus 20= 12
12 divided by 6 (number of years between 1980 and 1986) = 2
Therefore, every year it gets 2ft. closer to bridge. So...
32 divided by 2= 16
1980+16(years)=1996
1996 is the year it reaches the bridge

Answer 4

1996