Answer:
[tex]-47\frac{3}{4} \text{ feet}[/tex]
Step-by-step explanation:
We have been given that the position of a submarine relative to the water surface was [tex]-32\frac{1}{4} \text{ feet}[/tex]. A downward navigational maneuver increased its depth by [tex]15\frac{1}{2} \text{ feet}[/tex].
To find the final position of submarine relative to the water surface we will add both elevation of submarine.
[tex]\text{Final position of submarine relative to water surface}=-32\frac{1}{4} +(-15\frac{1}{2} )[/tex]
We are adding negative [tex]15\frac{1}{2} \text{ feet}[/tex] as it is downward increase in depth which will make the position of submarine more negative.
[tex]\text{Final position of submarine relative to water surface}=(-32-15)+(-\frac{1}{4} -\frac{1}{2} )[/tex]
[tex]\text{Final position of submarine relative to water surface}=(-47)+(-\frac{1}{4} -\frac{2}{4} )[/tex]
[tex]\text{Final position of submarine relative to water surface}=(-47)+(-\frac{3}{4})[/tex]
[tex]\text{Final position of submarine relative to water surface}=-47\frac{3}{4}[/tex]
Therefore, final position of the submarine relative to the water surface will be [tex]-47\frac{3}{4} \text{ feet}[/tex].
