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Leto [7]
3 years ago
5

PLEASE HELP!!! Is my answer correct

Mathematics
1 answer:
alexandr1967 [171]3 years ago
6 0

Answer:

Yes You Are Correct :)

Step-by-step explanation:

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r = 12.5
2 * \pi * 12.5 ~ 78.540
The answer is about 78.540
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If an arrow is shot upward on Mars with a speed of 62 m/s, its height in meters t seconds later is given by y = 62t − 1.86t². (R
Soloha48 [4]

Answer:

Approximately 58.28\; \rm m \cdot s^{-1}.

Step-by-step explanation:

The velocity of an object is the rate at which its position changes. In other words, the velocity of an object is equal to the first derivative of its position, with respect to time.

Note that the arrow here is launched upwards. (Assume that the effect of wind on Mars is negligible.) There would be motion in the horizontal direction. The horizontal position of this arrow will stays the same. On the other hand, the vertical position of this arrow is the same as its height: y = 62\, t - 1.86\, t^2.

Apply the power rule to find the first derivative of this y with respect to time t.

By the power rule:

  • the first derivative of t (same as
  • the first derivative of t^2 (same as t to the second power) with respect to

Therefore:

\begin{aligned}\frac{dy}{d t} &= \frac{d}{d t}\left[62 \, t - 1.86\, t^2\right] \\ &= 62\,\left(\frac{d}{d t}\left[t\right]\right) - 1.86\, \left(\frac{d}{d t}\left[t^2\right]\right) \\ &= 62 \times 1 - 1.86\times\left(2\, t) = 62 - 3.72\, t\end{aligned}.

In other words, the (vertical) velocity of this arrow at time t would be (62 - 3.72\, t) meters per second.

Evaluate this expression for t = 1 to find the (vertical) velocity of this arrow at that moment: 62 - 3.72 \times 1 =58.28.

6 0
3 years ago
Read 2 more answers
Which are the like terms in 2y3 – 4y2 + y + y3?
stealth61 [152]
C is the correct answer.  A, B, and C have terms with different powers of y, and as a result, these terms are not like.  C, however, has only multiples of y cubed, so it is a set of like terms.
8 0
4 years ago
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