A submarine hovers at 66 and 2/3 yards below sea level. If it ascends 24 and 1/8 yards and then descends again 78 and 3/4 yards, what is the submarine’s new position, in yards, below sea level?
Answer:
The submarine's new position is [tex]-121\frac{7}{24}\text{ yards}[/tex] below sea level.
Solution:
Given that a submarine hovers at [tex]66\frac{2}{3}[/tex] yards below sea level. This means that the position of submarine is negative
Therefore, we get [tex]-66\frac{2}{3}[/tex]
Also given that it ascend [tex]24\frac{1}{8}[/tex] yards
So we have to add [tex]-66\frac{2}{3}[/tex] and [tex]24\frac{1}{8}[/tex]
[tex]\rightarrow -66\frac{2}{3} + 24\frac{1}{8}\\\\\text{Convert mixed fractions to improper fractions }\\\\\rightarrow -\frac{66 \times 3+2}{3} + \frac{24 \times 8+1}{8} = \frac{-200}{3} + \frac{193}{8}\\\\\rightarrow \frac{-200 \times 8}{3 \times 8} + \frac{193 \times 3}{8 \times 3}\\\\\text{Simplify the above expression }\\\\\rightarrow \frac{-1600}{24} + \frac{579}{24} = \frac{-1021}{24}[/tex]
The submarine descends again 78 and 3/4 yards
This means we need to subtract [tex]-78\frac{3}{4}\text{ yards from } -\frac{1021}{24}\text{ yards}[/tex]
[tex]\rightarrow \frac{-1021}{24} -(78\frac{3}{4})\\\\\rightarrow \frac{-1021}{24} - \frac{78 \times 4 + 3}{4}\\\\\rightarrow \frac{-1021}{24} - \frac{315}{4} = \frac{-1021}{24} - \frac{315 \times 6}{4 \times 6}\\\\\rightarrow \frac{-1021 -1890}{24} = \frac{-2911}{24}\\\\\text{Simplify the expression }\\\\\rightarrow \frac{-2911}{24} = -121\frac{7}{24}[/tex]
Therefore, the submarine's new position is [tex]-121\frac{7}{24}\text{ yards}[/tex] below sea level.
Answer:
The answer-121 and 7/24
Hope this helps
:D
If I'm wrong sorry
And sorry that I'm super late
