Estamate the value of y when x = 2.5
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The data for linear pair are;
The domain are the values (input) on the x-axis which is the time
The range are the values input on the y-axis which is the height reached by the balloon
Part A
The interval of the domain during which the water balloon height is increasing is 0 ≤ x ≤ 2
Part B
The intervals of the domain the water balloon’s height stays the same are;
2 ≤ x ≤ 3 and 6 ≤ x ≤ 8
Part C
The water balloon height is decreasing at the following intervals;
At the interval 3 ≤ x ≤ 4
The rate of decrease = (20 ft. – 80 ft.)/(4 s – 3 s) = -20 ft./s.
At the interval 4 ≤ x ≤ 6
The rate of decrease = (0 ft. – 20 ft.)/(6 s – 4 s) = -10 ft./s
Therefore, the interval of the domain that the balloon’s height is decreasing the fastest is 3 ≤ x ≤ 4
Part D
According to Newton’s law of motion, provided that the no additional force is applied to the the balloon, at 10 seconds, the height of the water balloon is 0 ft. given that the height of the balloon is constantly decreasing from 3 seconds after being thrown off the roof, reaching a height of 0ft. at 6 seconds and maintaining that height up until 8 seconds.
By extending the graph further, the height of 0 ft. is obtained at 10 seconds after the balloon is thrown
To solve this we are going to use the formula for speed: 
where
is the speed
is the distance
is the time
Let
be the speed of the boat in the lake, 
the speed of the boat in the river, 
the time of the boat in the lake, and 
the time of the boat in the river.
We know for our problem that <span>the current of the river is 2 km/hour, so the speed of the boat in the river will be the speed of the boat in the lake minus 2km/hour:
</span>
We also know that in the lake the boat<span> sailed for 1 hour longer than it sailed in the river, so:
</span>
<span>
Now, we can set up our equations.
Speed of the boat traveling in the river:
</span>
But we know that, so:
equation (1)
Speed of the boat traveling in the lake:
But we know that, so:
equation (2)
Solving for in equation (1):
equation (3)
Solving for in equation (2):
equation (4)
Replacing equation (4) in equation (3):
Solving for:
or 
We can conclude that the speed of the boat traveling in the lake was either 6 km/hour or 5 km/hour.
where
Let
We know for our problem that <span>the current of the river is 2 km/hour, so the speed of the boat in the river will be the speed of the boat in the lake minus 2km/hour:
</span>
We also know that in the lake the boat<span> sailed for 1 hour longer than it sailed in the river, so:
</span>
<span>
Now, we can set up our equations.
Speed of the boat traveling in the river:
</span>
But we know that
Speed of the boat traveling in the lake:
But we know that
Solving for
Solving for
Replacing equation (4) in equation (3):
Solving for
We can conclude that the speed of the boat traveling in the lake was either 6 km/hour or 5 km/hour.
Answer:
12.4967 m
≈ 12.5 m
Step-by-step explanation:
From the solution diagram attached, to find /AB/, we apply sine rule:
AB/Sin 140° =AD/Sin 15°
AB = 10/Sin 15° x Sin 140°
= 24.8372 m
To calculate the height of the pole we will calculate BC first. Using the right angle triangle ∠ABC, we have:
Sin ∅ = BC/AB
Sin 25° = BC/ 24.8372
BC = 24.8372 Sin 25°
= 10.4967 m
The height of the pole = BC + 2.0
= 10.4967 + 2.0
= 12.4967
≈ 12.5 m
