Answer:
x(t) = -1 + 4t
y (t) = 3 - 5t
Step-by-step explanation:
The end points of the line segment are (-1, 3) and (3, -2).
Let a = (-1,3) and b = (3,-2)
Now, find the value of (b-a)
(b-a) = ((3+1), (-2-3))
b-a = (4, -5)
Therefore, the parametrization of the given line segment is
r(t) = a + (b-a)t
r(t) = (-1,3) +(4,-5)t
r(t) = (-1+4t, 3-5t)
We can rewrite this as
x(t) = -1 + 4t
y (t) = 3 - 5t
Answer:
very good
Step-by-step explanation:
yes
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Answer: (y-3)^2= 52(x+7)
The focus is (-10, -7) and the directrix is x=16. The y-coordinate of the vertex should be same as the focus(k=-7). Then the x-coordinate of the vertex would be:
p + (-10)= 16 - p
2p= 16 + 10
p=26/2= 13
The x-coordinate of the vertex would be:
h= p+ (-10)
h= 13 - 10= 3
The vertex coordinate would be: (h, k)= (3, -7)
For a vertex (h, k), the formula for equation would be
(y-k)^2=4 p(x-h)
(y-3)^2= 4*13(x--7)
(y-3)^2= 52(x+7)
Answer:
Approximately 
.
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: 
.
Apply the power rule to find the first derivative of this 
with respect to time 
.
By the power rule:
- the first derivative of (same as
- the first derivative of
(same as 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}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cfrac%7Bdy%7D%7Bd%20t%7D%20%26%3D%20%5Cfrac%7Bd%7D%7Bd%20t%7D%5Cleft%5B62%20%5C%2C%20t%20-%201.86%5C%2C%20t%5E2%5Cright%5D%20%5C%5C%20%26%3D%2062%5C%2C%5Cleft%28%5Cfrac%7Bd%7D%7Bd%20t%7D%5Cleft%5Bt%5Cright%5D%5Cright%29%20-%201.86%5C%2C%20%5Cleft%28%5Cfrac%7Bd%7D%7Bd%20t%7D%5Cleft%5Bt%5E2%5Cright%5D%5Cright%29%20%5C%5C%20%26%3D%2062%20%5Ctimes%201%20-%201.86%5Ctimes%5Cleft%282%5C%2C%20t%29%20%3D%2062%20-%203.72%5C%2C%20t%5Cend%7Baligned%7D)
.
In other words, the (vertical) velocity of this arrow at time would be 
meters per second.
Evaluate this expression for 
to find the (vertical) velocity of this arrow at that moment: 
.
The answer to your question is C
