The graph is attached.
We first graph the point where his catch reached the surface, (35, 0). Since it travels upward at a constant rate, the graph will be linear. We also need to know where it starts (what depth it is at when he begins reeling it in). We can use the formula d=rt as a template for our function. d would be distance (in our case, depth), r is the rate (speed) and t is the amount of time.
To find how far the catch had to travel to reach the surface, we set up our equation as:
d = 0.1(35)
This will tell us how much distance it traveled in 35 seconds. 0.1(35)=3.5, so the catch started 3.5m under water. It then travels up at 0.1 m per second.
Answer:
Get a calculator a graphing calculator and insert each one of those in it
Answer:
15%?
Step-by-step explanation:
sorry if im wrong
The answer is x=7/3 hope this helps
Answer: The angle is in the second quandrant
Step-by-step explanation:
If sin of theta is greater than 0, the angle theta has to be in either the first or second quandrant. If tangent of theta is less than 0, it cannot be in the first or third quandrant. Therefore, we know the angle is in the second quandrant.