Question

At 11:00 AM, the temperature is 17°F. If the temperature drops 3°F per hour, what will the temperature be at 2:00 PM?

Answer 1

`8^@F`

Explanation:

If at `11:00` A.M., the temperature is `17^@F`, and you want to find the temperature at `2:00` P.M., knowing that the temperature drops `3^@F` per hour, first determine the change in time between `11:00` A.M. and `2:00` P.M.

Convert `11:00` A.M. and `2:00` P.M. into `24`-hour time.

`11:00` A.M. `stackrel(color(red)("becomes"))rArrcolor(blue)(11:00)`

`2:00` P.M. `stackrel(color(red)("becomes"))rArr12` P.M. `+2color(white)(i)"hours"=color(blue)(14:00)`

Subtract `11` from `14` to determine the change in time in terms of hours.

`14-11`

`=color(darkorange)(|bar(ul(color(white)(a/a)color(black)(3color(white)(i)"hour difference")color(white)(a/a)|)))`

Since you know that the temperature drops `3^@F` per hour from `17^@F`, and you know the change in time, you can calculate the temperature at `2:00` P.M.

Determine how much the temperature dropped over `3` hours.

`DeltaT=(3^@F)/"hour"xx3color(white)(i)"hours"`

Note: `Delta` is a symbol for "change in!" So `DeltaT` means "change in temperature!"

The unit, "hour(s)," cancels out.

`DeltaT=(3^@F)/color(red)cancelcolor(black)("hour")xx3color(white)(i)color(red)cancelcolor(black)("hours")`

`DeltaT=3^@Fxx3`

`DeltaT=color(purple)(|bar(ul(color(white)(a/a)color(black)(9^@F)color(white)(a/a)|)))`

Now that you know the temperature dropped by `9^@F` since `11:00` A.M., you can finally calculate the temperature at `2:00` P.M. Subtract `9^@F` from `17^@F` to determine the temperature at `2:00` P.M.

`17^@F-9^@F`

`=color(green)(|bar(ul(color(white)(a/a)8^@Fcolor(white)(a/a)|)))`