PLS HALP A submersible is a type of underwater vessel. Suppose a submersible starts at a depth of 40 feet below sea level, represented by –40 feet. The table shows the submersible's change in depth from one minute to the next.

Time and Depth of Submersible
2:15 p. m. 2:16 p. m. 2:17 p. m. 2:18 p. m.
–40 feet 180 feet lower 175 feet lower 185 feet lower
Answer the questions to find the depth of the submersible after 3 minutes.

1. Write and evaluate a subtraction expression to find the depth of the submersible at 2:16 p. m.

Write your answer in the space below.

2. Write and evaluate a subtraction expression to find the depth of the submersible at 2:17 p. m.

Write your answer in the space below.

3. Write and evaluate a subtraction expression to find the depth of the submersible at 2:18 p. m.

Write your answer in the space below.

4. If the submersible dove straight down, what is the distance between the submersible's starting position at 2:15 and its position at 2:18? Show your work.

Write your answer in the space below.

Previous question Rational numbers prior question delete your full name you accidently wrote in your question for rational numbers by pressing edit icon answered here so your edit is possible ok!   Answer is -2t +7t = 5t       Step-by-step explanation: question asks for addition only (4t-6t= -2t)+(8t-1t=7t)   This question on page = 1) At 2:16 p.m.= -180ft - 40ft = -220ft   2) At 2:17 p.m = -175ft-40ft= -115ft  3) At 2:18 p.m. = -185ft - 40ft = -225ft  4) The difference is show your work by reversal 2:18-2:15 = 3 minutes so the distance is also reversed at (-225) - (-40 )= -225 + 40 = -175

Step-by-step explanation: This is minus - number minus- number = minus number counting bkwards  but when when we subtract two minus number with another minus it becomes a  plus +  just between the brackets see question 4 brackets use. The work is using brackets for question 4.

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

put in or use mine