Answer:
Step-by-step explanation:
As per given information there are 18000 CCTV cameras in the area of 783.8 square kilometers
To find the average number of cameras you divide the number by the area.
The average number of CCTV cameras per square kilometer in NYC = 18000/783.8 = 22.9650421026 ≈ 23
It should be rounded up if taking approximate number
Yes you can have as many equal signs as you need and the approximately equal sign in the same equation.
2
it like trying to imagine re-folding back a cardboard box
The answer Is 3.482 × 10^9
Answer:
4.41 feet per second.
Step-by-step explanation:
Please find the attachment.
We have been given that a man flies a kite at a height of 16 ft. The wind is carrying the kite horizontally from the man at a rate of 5 ft./s. We are asked to find how fast must he let out the string when the kite is flying on 34 ft. of string.
We will use Pythagoras theorem to solve for the length of side x as:
![x^2+16^2=34^2](https://tex.z-dn.net/?f=x%5E2%2B16%5E2%3D34%5E2)
![x^2=34^2-16^2](https://tex.z-dn.net/?f=x%5E2%3D34%5E2-16%5E2)
![x^2=900\\\\x=30](https://tex.z-dn.net/?f=x%5E2%3D900%5C%5C%5C%5Cx%3D30)
Now, we will use Pythagorean theorem to relate x and y because we know that the vertical side (16) is always constant.
![x^2+16^2=y^2](https://tex.z-dn.net/?f=x%5E2%2B16%5E2%3Dy%5E2)
Let us find derivative of our equation with respect to time (t) using power rule and chain rule as:
![2x\cdot \frac{dx}{dt}+0=2y\cdot \frac{dy}{dt}](https://tex.z-dn.net/?f=2x%5Ccdot%20%5Cfrac%7Bdx%7D%7Bdt%7D%2B0%3D2y%5Ccdot%20%5Cfrac%7Bdy%7D%7Bdt%7D)
We have been given that
,
and
.
![2(30)\cdot 5=2(34)\cdot \frac{dy}{dt}](https://tex.z-dn.net/?f=2%2830%29%5Ccdot%205%3D2%2834%29%5Ccdot%20%5Cfrac%7Bdy%7D%7Bdt%7D)
![300=68\cdot \frac{dy}{dt}](https://tex.z-dn.net/?f=300%3D68%5Ccdot%20%5Cfrac%7Bdy%7D%7Bdt%7D)
![\frac{dy}{dt}=\frac{300}{68}](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%3D%5Cfrac%7B300%7D%7B68%7D)
![\frac{dy}{dt}=4.4117647058823529](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%3D4.4117647058823529)
![\frac{dy}{dt}\approx 4.41](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdt%7D%5Capprox%204.41)
Therefore, the man must let out the string at a rate of 4.41 feet per second.