Answer:
a) Average velocity at 0.1 s is 696 ft/s.
b) Average velocity at 0.01 s is 7536 ft/s.
c) Average velocity at 0.001 s is 75936 ft/s.
Step-by-step explanation:
Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by
.
To find : The average velocity for the time period beginning when t = 2 and lasting. a. 0.1 s.
, b. 0.01 s.
, c. 0.001 s.
Solution :
a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.





b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.





c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.




Answer:
-1.375
Step-by-step explanation:
Like all problems that involve images within the question, we should definitely try to draw this out. In the picture above, I have done this.
Now, we can see that this is just a simple proportion problem. For every 2.5 cm of height of the flower, we are 2 cm from the opening, or aperture. For every 20 cm of height, how far are we? We can set up the problem like this:
20 ............2.5
-------- = ---------
...x ............. 2
where x is the unknown distance to the aperture from the flower. Now, we just need to get x by itself. A typical way of solving something like this is by doing "butterfly multiplication" which is really just a shortcut haha. Anyway, I can rewrite that equation ^ as:
20×2 = 2.5 × x
Then, to solve for x, we would divide both sides by 2.5. (If you don't know why that is, please let me know and I'll elaborate).
We would then have:
20×2
------- = x
2.5
Which then simplifies to:
x = 16
Try using the same logic for your second question, and if you get stuck, I'd be happy to help! please let me know if any of this doesn't make sense. :)
I’m pretty sure that it’s the third answer (q^12)! :)
Answer:
Step-by-step explanation:
It will be 7x+1