who is smart at math???

Question

The sand on a beach is being washed away by slowly by erosion. In 1870, a lighthouse was 1500
feet from the shoreline. In 1990, the lighthouse was only 150 feet from the shoreline. If the
erosion continues at this rate, how many more years will it be before the lighthouse is at the
shoreline? Write an equation that will tell you the distance of the lighthouse from the shore at
any year, where t is the year.

Answer 1

t = years
d = distance from shoreline

when t = 1870, d = 1500
when t = 1990, d = 150

So, you have two points (1870, 1500) and (1990, 150).

Write a linear equation to solve.
slope is
m = (1500 - 150) / (1870 - 1990) = 1350 / -120 = -11.25

Point-slope form:
d - 1500 = (-11.25)(t - 1870)
d - 1500 = -11.25t + 21037.5
d = -11.25t + 22537.5

If the lighthouse is at the shoreline, then d = 0.

0 = -11.25 t + 22537.5
11.25t = 22537.5
t = 2003.3333

So it will happen in 2003.

Answer 2

First, we need to find the rate of erosion which we'll call r. In 120 years, 1350 feet were removed (it went from 1500 to 150) which can be written line this:

120 * r = 1350

Therefore, r = 11.25 which is to say, 11.25 feet per year.
Now, let's create an equation... What we want is something like this:
The distance to the shoreline is the number of years since 1870 multiplied by our rate of erosion subtracted from 1500 which is written like this:

1500 - ((t - 1870) * r) = d

We know that r is 11.25 so plug that in and simplify:

1500 - ((t - 1870) * 11.25) = d
1500 - (11.25t - 21037.5) = d
1500 - 11.25t + 21037.5 = d

22537.5 - 11.25t = d

That's our equation. To solve for the year in which the lighthouse is at the shorline, solve for t using d = 0.

22537.5 - 11.25t = 0
-11.25t = -22537.5
t = 2003.3

So, about a third of the way into 2003, the lighthouse was at the shoreline and the equation to solve for any year is:

22537.5 - 11.25t = d

Answer 3

think of it as a coordinate system, one point being (1870, 1500) and (1990, 150) where x = year and y = distance

get the slope y1-y0/x1-x0 = - 45/4

get and eq = (y - y0) = m(x - x0) =
y - 1500 = -45/4 (x - 1870) =
y = -45/4x + 22537.5 (where x = year and y = distance)

solve for y = 0 to get the year where the lighthouse will have no shoreline

you get 2003.3333333.....

so the shoreline will become 0 at approximately the year 2003.

The equation will be (in terms of t) = y = -45/4t + 22537.5 where t is in years.

Answer 4

ever 120 years it erodes by 1350 feet. every 12 years, 135 feet, therefore 11.25 feet per year.

with 150 feet left, you divide it by 11.25. the answer is 13 years and 4 months.

(t-1870)*11.25=1/distance from the lighthouse

Answer 5

you divide the amount of feet eroded by the amount of years to reach the amount eroded per year. use the guess and check method to multiple the amount per year and subtract it from 150 to find the amount of years till the lighthouse is at the shoreline.

Answer 6

make a table

y= how many feets left
x= how many year

y=1500-11.25x

120 years passed between the two dates.
1350 feets are gone.

1350/120= 11.25
150/11.25=13.3333...

it'll take 13 years so it should be in 2003 (i think)

Answer 7

I can do that but my mind's still asleep.

I recently learned though that I'm NOT smarter than a fifth grader.

Answer 8

Easy
1500-150 = 1350 / ( number of years ) 120 = 11.25' eroding each year

150'/11.25' = 13.33333 years till its onm the shore
So it probably fell in sometime during 2003

please answer my last question!!thanks

Answer 9

y does that matter? you dont need to do math in the summmer!! lol

Answer 10

why dont you just figure it out your self