Asubmarine started at 750 meters below sea level. it rose 50 meters per hour over a 4-hour period. which expression represents the new position of the submarine in relation to sea level?

Step-by-step explanation:

We want to model the position of the submarine as a function of time.

Notice that we have a constant amount. The initial depth: 750 meters

 Then we have a factor that varies over time. Every hour the submarine ascends 50 meters. therefore as t increases the depth y(t) of the submarine decreases. Then the factor is:

-50t.

Where t represents the time in hours.

Then the equation that represents the new position of the submarine in relation to the level of the sea:

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

360/29

step-by-step explanation: