Q5. 23
Q6. 2.4
Q7. 4.7
Q8. $19.20
Q9. 4.77
Answer:
The approximated length of the cables that stretch between the tops of the two towers is 1245.25 meters.
Step-by-step explanation:
The equation of the parabola is:

Compute the first order derivative of <em>y</em> as follows:

![\frac{\text{d}y}{\text{dx}}=\frac{\text{d}}{\text{dx}}[0.00035x^{2}]](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7Bd%7Dy%7D%7B%5Ctext%7Bdx%7D%7D%3D%5Cfrac%7B%5Ctext%7Bd%7D%7D%7B%5Ctext%7Bdx%7D%7D%5B0.00035x%5E%7B2%7D%5D)

Now, it is provided that |<em>x </em>| ≤ 605.
⇒ -605 ≤ <em>x</em> ≤ 605
Compute the arc length as follows:


Now, let



Plug in the solved integrals in Arc Length and solve as follows:


Thus, the approximated length of the cables that stretch between the tops of the two towers is 1245.25 meters.
Answer:
n - 13
Step-by-step explanation:
n-4-9
We can combine like terms.
n - (4+9)
n - 13
The upper quartile for the data set given below is. 14, 8, 23, 9, 11, 27, 22, 3, 17, 12, 29
LekaFEV [45]
Ok, first we need to organize.
<span>14, 8, 23, 9, 11, 27, 22, 3, 17, 12, 29
</span>3, 8, 9, 11, 12, 14, 17, 22, 23, 27, 29.
First, we need to find the median, the or the middle, which is 14.
Now that we know the median is fourteen, we take away all numbers to the left of 14, including 14: 17, 22, 23, 27, 29. Now just find the median in this new set of numbers. This will give us our upper quartile, which is 23. Hope this helped, and don't forget to drop a like.