Will give brainliest pls help 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.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

the diver is 33 feet below the surface.

step-by-step explanation:

hope this