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Lorico [155]
3 years ago
15

A rocket is launched into the air. The height h(t), in yards, of the rocket is a function of time, t, in seconds, as shown in th

e following table.
What is the rocket's average rate of change from 3≤t≤6 in yards/second?
Enter your answer as a number, like this: 42

Mathematics
1 answer:
RideAnS [48]3 years ago
7 0

Answer:

From 3 to 6 seconds, the rocket is falling 11 yards per second.

Step-by-step explanation:

Normal pace of progress is exactly the same thing as the slant. Since this is explanatory, we can't track down the specific pace of progress as we could if this were a straight capacity. Be that as it may, we can utilize a similar thought. At the point when t = 3, h(t) = 33, so the organize point is (3, 33). At the point when t = 6, h(t) = 0, so the facilitate is (6, 0). Attachment those qualities into the incline recipe:  

m= 0-33/6-3 and m=-33/3 which is 11  

From 3 to 6 seconds, the rocket is falling 11 yards each second.

Brainliest?

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Answer:

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Step-by-step explanation:

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Hope this helps

4 0
3 years ago
Given the circle with the equation (x - 3)2 + y2 = 49, determine the location of each point with respect to the graph of the cir
Bas_tet [7]
To find out if a point is inside, on, or outside a circle, we need to substitute the ordered pair into the equation of the circle:
(x-xc)^2+(y-yc)^2=r^2
where (xc,yc) is the centre of the circle, and r=radius of the circle.

If the left-hand side [(x-xc)^2+(y-yc)^2] is less than r^2, then point (x,y) is INSIDE the circle.  If the left-hand side is equal to r^2, the point is ON the circle.
Finally, if the left-hand side is greater than r^2, the point is OUTSIDE the circle.

For the given problem, we have xc=3, yc=0, or centre at (3,0), r=sqrt(49)=7
(x-xc)^2+(y-yc)^2=r^2 => (x-3)^2+y^2=7^2

A. (-1,1), 
(x-3)^2+y^2=7^2 => (-1-3)^2+1^2=16+1=17 <49  [inside circle]

B. (10,0)
(x-3)^2+y^2=7^2 => (10-3)^2+0^2=49+0=49  [on circle]

C. (4,-8)
(x-3)^2+y^2=7^2 => (4-3)^2+(-8)^2=1+64=65 > 49  [outside circle]


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