Answer:
E
Step-by-step explanation:
Answer:
The time interval when
is at 
The distance is 106.109 m
Step-by-step explanation:
The velocity of the second particle Q moving along the x-axis is :

So ; the objective here is to find the time interval and the distance traveled by particle Q during the time interval.
We are also to that :
between 
The schematic free body graphical representation of the above illustration was attached in the file below and the point when
is at 4 is obtained in the parabolic curve.
So,
is at 
Taking the integral of the time interval in order to determine the distance; we have:
distance = 
= 
= By using the Scientific calculator notation;
distance = 106.109 m
When you divide by a fraction all you have to is multiply by the reciprocal. The reciprocal of 3/8 is 8/3.
<span>The real problem is 2/3 TIMES 8/3. Multiply the numerators across 2 X 8 = 16. Then multiply the denominators across 3 X 3 = 9. The answer is 16/9. If you need to simplify then the answer is 1 7/9.</span>
Answer:
<u>32m^2</u>
Step-by-step explanation:
32m^2
8*4 = <u>32</u>
Have a nice day! :-)
(7x-3y)^2 hope this helps