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The scuba diver's position relative to sea level after the 4.6 minutes is 5.1 ft.
Given:
A scuba diver descends in the water at a rate of 23 1/2 feet per minute for 2.6 minutes.
A scuba diver ascends at a rate of 28 feet per minute for 2 minutes after seeing a big fish.
To find :
The scuba diver's position relative to sea level after the 4.6 minutes.
Solution:
Rate of a scuba diver descending in the water for 2.6 minutes = 

Distance covered by a scuba diver in 2.6 minutes = 

Rate of a scuba diver ascending for 2 minutes = 

Distance covered by a scuba diver in 2 minutes = 

The position of scuba diver after 4.6 minutes relative to sea level:

The scuba diver's position relative to sea level after the 4.6 minutes is 5.1ft.
Learn more about ascending rate and descending rate here:
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Answer:
no diagram
Step-by-step explanation:
Answer: D) 13y^25 and 2y^25
Like terms involve the same variables, and each of those variables must have the same exponents.
Another example of a pair of like terms would be 5x^3y^2 and 7x^3y^2. Both involve the variable portion "x^3y^2" which we can replace with another variable, say the variable z. That means 5x^3y^2 becomes 5z and 7x^3y^2 becomes 7z. After getting to 5z and 7z, it becomes more clear we have like terms.