The shaded areas add to 2/5 of the area of the circle, which is given by
A = πr²
Then the shaded area is ...
(2/5)A = (2/5)π(10 in)²
shaded area = 40π in²
A = πr²
Then the shaded area is ...
(2/5)A = (2/5)π(10 in)²
shaded area = 40π in²
On a number line, the distance between teo points with coordinates -15 and -9 is:
1 answer:
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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.