Searches related to Elon sat on the dock with his fishing rod, when suddenly he felt the rod being pulled down! He reeled in his
catch, causing it to ascend at a constant rate of 0.10, point, 1 meters per second. When it reached the water's surface after 35 seconds, Elon found it was only an old shoe. Graph the shoe's altitude (in meters relative to the water's surface) as a function of time (in seconds).
1 answer:
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.
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.
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Answer:
Diameter of pound = 10 yd
Step-by-step explanation:
Given area of circular pond = 78.5 square yd
To find the diameter of the pond.
Solution:
The pond being circular in shape, the area of pond can be given as:
⇒
where represents radius of the pond.
Thus, we have:
⇒
Using
⇒
Solving for
Dividing both sides by 3.14
⇒
⇒
Taking square root both sides.
⇒
⇒
So, radius = 5 yd, as distances are always positive.
Diameter =
Thus, diameter of pound = 10 yd
The problem is modelled in the diagram below
Question 1:
The largest area of the pool is half of the area of the circle
Area of circle = πr², where r is the radius given by 1/2 of the diameter
Area of circle = π(60)² = 11309.73355 feet
Area of half of the circle is 11309.73355/2 = 5654.866... ≈ 5654.87 feet (2dp)
Question b:
The area of the pool is the area of the circle subtracts the area of the triangle.
The area of the circle is 11309.73²
The area of the triangle is 1/2 (60×103.92) = 3117.6 feet²
The area of the pool is 11309.73 - 3117.6 = 7922.13 feet²
The volume of the pool = 7922.13 × 4 = 31688.52 feet³
Note: there isn't any information on the fish tank part so the answer above is assumed for the whole pool.
Question 1:
The largest area of the pool is half of the area of the circle
Area of circle = πr², where r is the radius given by 1/2 of the diameter
Area of circle = π(60)² = 11309.73355 feet
Area of half of the circle is 11309.73355/2 = 5654.866... ≈ 5654.87 feet (2dp)
Question b:
The area of the pool is the area of the circle subtracts the area of the triangle.
The area of the circle is 11309.73²
The area of the triangle is 1/2 (60×103.92) = 3117.6 feet²
The area of the pool is 11309.73 - 3117.6 = 7922.13 feet²
The volume of the pool = 7922.13 × 4 = 31688.52 feet³
Note: there isn't any information on the fish tank part so the answer above is assumed for the whole pool.