Answer:
percentage of the total capacity is 75.6%
Step-by-step explanation:
Hello! To solve this problem we follow the following steps
1. draw the complete scheme of the problem (see attached image)
2. To solve this problem we must find the area of the circular sector using the following equation.(c in the second attached image)
![A=\frac{R^2}{2} (\alpha -sin\alpha )](https://tex.z-dn.net/?f=A%3D%5Cfrac%7BR%5E2%7D%7B2%7D%20%28%5Calpha%20-sin%5Calpha%20%29)
![\alpha =2arccos(\frac{d}{R})](https://tex.z-dn.net/?f=%5Calpha%20%3D2arccos%28%5Cfrac%7Bd%7D%7BR%7D%29)
3. observing the attached images we replace the values in the equations and find the area of the circular sector, remember that you must transform the angle to radians
![\alpha =2arccos(\frac{5}{7})=88.83](https://tex.z-dn.net/?f=%5Calpha%20%3D2arccos%28%5Cfrac%7B5%7D%7B7%7D%29%3D88.83)
![A=\frac{R^2}{2} (\alpha -sin\alpha )\\A=\frac{7^2}{2} (88.83-sen88.83)*\frac{\pi rad}{180} =37.57ft^2](https://tex.z-dn.net/?f=A%3D%5Cfrac%7BR%5E2%7D%7B2%7D%20%28%5Calpha%20-sin%5Calpha%20%29%5C%5CA%3D%5Cfrac%7B7%5E2%7D%7B2%7D%20%2888.83-sen88.83%29%2A%5Cfrac%7B%5Cpi%20rad%7D%7B180%7D%20%3D37.57ft%5E2)
4.we calculate the area of the total circle (At), then subtract the area of the circular sector (Ac) to find the area occupied by water (Aw)
![At=\frac{\pi }{4} (14ft)^2=153.93ft^2](https://tex.z-dn.net/?f=At%3D%5Cfrac%7B%5Cpi%20%7D%7B4%7D%20%2814ft%29%5E2%3D153.93ft%5E2)
Aw=At-Ac=153.93-37.57=116.36ft^2
5.Finally, we calculate the percentage that represents the water in the tank by dividing the area of the water over the total area of the tank
![\frac{116.36}{153.93} *100=75.6](https://tex.z-dn.net/?f=%5Cfrac%7B116.36%7D%7B153.93%7D%20%2A100%3D75.6)
percentage of the total capacity is 75.6%
0.15 * 60 = 9 or 15/100 * 60/1 = 900/100 = 9
For this case what we must do is use the law of cosines.
We then have the following equation:
![c ^ 2 = a ^ 2 + b ^ 2 - 2 * a * b * cos (x) ](https://tex.z-dn.net/?f=c%20%5E%202%20%3D%20a%20%5E%202%20%2B%20b%20%5E%202%20-%202%20%2A%20a%20%2A%20b%20%2A%20cos%20%28x%29%0A)
Where,
a, b: sides of the triangle
x: angle between sides a and b.
Substituting values we have:
![c^2 = 90^2 + 75^2 - 2*90*75*cos(85) ](https://tex.z-dn.net/?f=c%5E2%20%3D%2090%5E2%20%2B%2075%5E2%20-%202%2A90%2A75%2Acos%2885%29%0A)
Clearing the value of c we have:
Answer:
An expression that is equivalent to how many feet the oak trees are from each other is:
Answer:
The rate at which the distance between them is changing at 2:00 p.m. is approximately 1.92 km/h
Step-by-step explanation:
At noon the location of Lan = 300 km north of Makenna
Lan's direction = South
Lan's speed = 60 km/h
Makenna's direction and speed = West at 75 km/h
The distance Lan has traveled at 2:00 PM = 2 h × 60 km/h = 120 km
The distance north between Lan and Makenna at 2:00 p.m = 300 km - 120 km = 180 km
The distance West Makenna has traveled at 2:00 p.m. = 2 h × 75 km/h = 150 km
Let 's' represent the distance between them, let 'y' represent the Lan's position north of Makenna at 2:00 p.m., and let 'x' represent Makenna's position west from Lan at 2:00 p.m.
By Pythagoras' theorem, we have;
s² = x² + y²
The distance between them at 2:00 p.m. s = √(180² + 150²) = 30·√61
ds²/dt = dx²/dt + dy²/dt
2·s·ds/dt = 2·x·dx/dt + 2·y·dy/dt
2×30·√61 × ds/dt = 2×150×75 + 2×180×(-60) = 900
ds/dt = 900/(2×30·√61) ≈ 1.92
The rate at which the distance between them is changing at 2:00 p.m. ds/dt ≈ 1.92 km/h
Answer:
b ata.....
Step-by-step explanation:
no explanation