Please answer with explanation thank you
1 answer:
9514 1404 393
Answer:
- A
- C
- B
Step-by-step explanation:
On a position-time graph, "does not move" means the position is a constant—a horizontal line.
"Positive velocity" means the position is increasing as time increases. The graph goes up to the right. The rate of change (slope) is the velocity.
"Negative velocity" means the position is decreasing as time increases. The graph goes down to the right. Again, the rate of change (slope) is the velocity.
The various sections of the graph are identified in the attachment.
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Answer:
5 sides.
Step-by-step explanation:
Unlike a triangle, which is 180°, also equal to a straight line, the length is equal to 540°.
Remark
If you don't start exactly the right way, you can get into all kinds of trouble. This is just one of those cases. I think the best way to start is to divide both terms by x^(1/2)
Step One
Divide both terms in the numerator by x^(1/2)
y= 6x^(1/2) + 3x^(5/2 - 1/2)
y =6x^(1/2) + 3x^(4/2)
y = 6x^(1/2) + 3x^2 Now differentiate that. It should be much easier.
Step Two
Differentiate the y in the last step.
y' = 6(1/2) x^(- 1/2) + 3*2 x^(2 - 1)
y' = 3x^(-1/2) + 6x I wonder if there's anything else you can do to this. If there is, I don't see it.
I suppose this is possible.
y' = 3/x^(1/2) + 6x
y' =
Frankly I like the first answer better, but you have a choice of both.
If you don't start exactly the right way, you can get into all kinds of trouble. This is just one of those cases. I think the best way to start is to divide both terms by x^(1/2)
Step One
Divide both terms in the numerator by x^(1/2)
y= 6x^(1/2) + 3x^(5/2 - 1/2)
y =6x^(1/2) + 3x^(4/2)
y = 6x^(1/2) + 3x^2 Now differentiate that. It should be much easier.
Step Two
Differentiate the y in the last step.
y' = 6(1/2) x^(- 1/2) + 3*2 x^(2 - 1)
y' = 3x^(-1/2) + 6x I wonder if there's anything else you can do to this. If there is, I don't see it.
I suppose this is possible.
y' = 3/x^(1/2) + 6x
y' =
Frankly I like the first answer better, but you have a choice of both.