19) A submarine ascended from 56.8 meters under the sea level to 27.4 meters under the sea level. distance it moved between two positions. Use the "distance on the number line "formula.

29.4 meters

Step-by-step explanation:

Given the following :

A submarine ascended from 56.8 meters under the sea level to 27.4 meters under the sea level.

distance it moved between two positions. Use the "distance on the number line "formula

Initial Position or distance of submarine before ascension = 56.8 meters

Final position after ascension = 27.4 meters

The distance moved by the submarine between the two positions can be obtained by.

(Initial distance - Final distance)

(56.8 - 27.4) = 29.4 meters

step-by-step explanation:

in the absence of parentheses, i will have to make some assumptions:   i interpret your (3x+5/y^6)(4y^3/5)​ to mean:

3x + 5     4y^(3/5)       (3x + 5)(4)

* =

    y^6             1               y^(6-3/5)

if this is not what you expected, repost your question using parentheses to indicate whether you meant

4y^3

    or   4y^(3/5)

    5

any options?

step-by-step explanation: