PLEASE HELP
1 answer:
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Part A
The notation means that we're approaching x = 2 from the right hand side (aka positive side). This is known as a right hand limit.
So we could start at say x = 2.5 and get closer to 2 by getting to x = 2.4 then to x = 2.3 then 2.2, 2.1, 2.01, 2.001, etc
We don't actually arrive at x = 2 itself. We simply move closer and closer.
Since we're on the positive or right hand side of 2, this means we go with the rule involving x > 2
Therefore f(x) = (x/2) + 1
Plug in x = 2 to find that...
f(x) = (x/2) + 1
f(2) = (2/2) + 1
f(2) = 2
This shows
Then for the left hand limit , we'll involve x < 2 and we go for the first piece. So,
f(x) = 3-x
f(2) = 3-2
f(2) = 1
Therefore,
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Part B
Because this means that the limit
does not exist.
If you are a visual learner, check out the graph below of the piecewise function. Notice the gap or disconnect at x = 2. This can be thought of as two roads that are disconnected. There's no way for a car to go from one road to the other. Because of this disconnect, the limit doesn't exist at x = 2.
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Part C
You'll follow the same type of steps shown in part A.
However, keep in mind that x = 4 is above x = 2, so we'll deal with x > 2 only.
So you'd only involve the second piece f(x) = (x/2) + 1
You should find that f(4) = 3, and that both left and right hand limits equal this value. The left and right hand limits approach the same y value. The limit does exist here. There are no gaps to worry about when x = 4.
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Part D
As mentioned earlier, since , this means the limit
does exist and it's equal to 3.
As x gets closer and closer to 4, the y values are approaching 3. This applies to both directions.
(a) The projectile remains in the air for 10 seconds
(b) It travels a horizontal distance of 600 meters
Step-by-step explanation:
A projectile is fired at such an angle that
1. The vertical component of its velocity is 49 m/sec
2. The horizontal component of its velocity is 60 m/sec
We need to find:
(a) How long the projectile remains in the air
(b) The horizontal distance it travels
∵ The vertical distance y = t -
g t², where
is the vertical component of its velocity, g is the acceleration
of gravity and t is the time
∵ y = 0 ⇒ it return to the same initial height
∵ = 49 m/s
∵ g = 9.8 m/s²
- Substitute these values in the rule above
∴ 0 = 49 t - (9.8) t²
∴ 0 = 49 t - 4.9 t²
- Take t as a common factor
∴ 0 = t (49 - 4.9 t)
- Equate each term by 0
∴ t = 0 ⇒ at initial position
∴ 49 - 4.9 t = 0
- Add 4.9 t to both side
∴ 49 = 4.9 t
- Divide both sides by 4.9
∴ t = 10 seconds
(a) The projectile remains in the air for 10 seconds
∵ The horizontal distance = t, where
is the
horizontal component of its velocity
∵ = 60 m/sec
∵ t = 10 seconds
∴ x = 60 × 10 = 600 meters
(b) It travels a horizontal distance of 600 meters
Learn more:
You can learn more about the component of velocity in brainly.com/question/4464845
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