The rate of change of the distance from the particle to the origin at this instant is 3 units per second.
<h3>What is the rate of change?</h3>
The instantaneous rate of change is the rate of change at a particular instant.
A particle is moving along the curve

The rate of change of y is given as:
dy / dx = 2
by differentiating both sides,

From the question, we have:
(x, y) = (5,24)
Substitute 5 for x and dy / dx = 2

Hence, the rate of change of the distance from the particle to the origin at this instant is 3 units per second.
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Answer:
7
Step-by-step explanation:
Answer:
-8b^3ax^2
Step-by-step explanation:
The total weekly pay of the company is $754.98
What is total hours worked for the week?
The overtime is the excess of total hours, which is 47 hours, over the regular hours of 40 hours of work a week, in essence, overtime hours is 7(47-40)
Overtime rate=regular hour pay*1.5
regular hour pay=$14.95
overtime rate=$14.95*1.5
overtime rate=$22.425
overtime pay for 7 hours=7*$22.425
overtime pay for 7 hours=$156.98
The regular pay is 40 hours of regular time multiplied regular pay per hour
regular pay=$14.95*40
regular pay=$598.00
total pay=regular pay+overtime
total pay=$598.00+$156.98
total pay=$754.98
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Find Radius :
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





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Find volume of the cylinder :
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


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Find volume of the gap :
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

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
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