A submarine after traveling 100 km from the shore descended 1 1/10 below the sea level.Then it ascended about 3/5 km.The next day,it again descended by 9/10 km.what is its current position with respect to the sea ? what the horizontal distance of the submarine from the sea shore?
Answers
Answered by
14
1 ¹/₁₀ = (10 * 1 + 1)/10 = (10 +1)/10 = 11/10
Therefore,
- (11/10) + (6/10) - (9/10) {since, 3/5 = 6/10}
= - (14/10)
Therefore , the current position with respect to sea is 14/10 km .
Therefore,
- (11/10) + (6/10) - (9/10) {since, 3/5 = 6/10}
= - (14/10)
Therefore , the current position with respect to sea is 14/10 km .
eahu:
yes
Ok... Yes, I will.
tnx
I have updated my answer, please have a look.
tnx dear
was that answer helpful..? If yes, then please give me ratings.
Happy studying. Thanks
can u explain any simple way..I can't understand it
Ok.I understood..tnx
ur welcome
Answered by
9
the submarine descended by 1.1 km
then it ascended by 0.6 km
then again it descended by 0.9 km
so the position of the submarine-
=0 -1.1 + 0.6 - 0.9
=-1.4km
so the submarine is 1.4 km below the sea level
then it ascended by 0.6 km
then again it descended by 0.9 km
so the position of the submarine-
=0 -1.1 + 0.6 - 0.9
=-1.4km
so the submarine is 1.4 km below the sea level
Similar questions