For every 7.5 miles alberto jogs,he walks 2 miles.How many miles does he walk if he jogs 60 miles?
2 answers:
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Answer:
<em>Interval notation: [0, 2.236]</em>
<em>Set Builder notation: </em>
Step-by-step explanation:
Given that:
Equation of height of the ball dropped from a height of 10 foot, as:
Where is the time since the ball was dropped.
To find:
The domain of the function in Interval and set builder notation.
Solution:
<em>Domain of a function </em>is defined as the set of valid input values that can be given to the function for which the function is defined.
Here, input is time.
We can not have negative values for time.
Therefore, starting value for time will be <em>0 seconds</em>.
And the value of height can not be lesser than that of 0 ft.
Maximum value for time can be 2.236 seconds.
Therefore the domain is:
<em>Interval notation: [0, 2.236]</em>
<em>Set Builder notation: </em>
Define
g = 9.8 32.2 ft/s², the acceleration due to gravity.
Refer to the diagram shown below.
The initial height at 123 feet above ground is the reference position. Therefore the ground is at a height of - 123 ft, measured upward.
Because the initial upward velocity is - 11 ft/s, the height at time t seconds is
h(t) = -11t - (1/2)gt²
or
h(t) = -11t - 16.1t²
When the ball hits the ground, h = -123.
Therefore
-11t - 16.1t² = -123
11t + 16.1t² = 123
16.1t² + 11t - 123 = 0
t² + 0.6832t - 7.64 = 0
Solve with the quadratic formula.
t = (1/2) [-0.6832 +/- √(0.4668 + 30.56) ] = 2.4435 or -3.1267 s
Reject the negative answer.
The ball strikes the ground after 2.44 seconds.
Answer: 2.44 s
g = 9.8 32.2 ft/s², the acceleration due to gravity.
Refer to the diagram shown below.
The initial height at 123 feet above ground is the reference position. Therefore the ground is at a height of - 123 ft, measured upward.
Because the initial upward velocity is - 11 ft/s, the height at time t seconds is
h(t) = -11t - (1/2)gt²
or
h(t) = -11t - 16.1t²
When the ball hits the ground, h = -123.
Therefore
-11t - 16.1t² = -123
11t + 16.1t² = 123
16.1t² + 11t - 123 = 0
t² + 0.6832t - 7.64 = 0
Solve with the quadratic formula.
t = (1/2) [-0.6832 +/- √(0.4668 + 30.56) ] = 2.4435 or -3.1267 s
Reject the negative answer.
The ball strikes the ground after 2.44 seconds.
Answer: 2.44 s