11x2=22
22x6=132
I believe the answer is 132
Answer:
-1/4x² + 3
Step-by-step explanation:
Given that:
Path of sprinkler is modeled by the quadratic function :
w(x) = -1/4(x^2 -12 )
w(x) = height of the water, in meters, at position x
The height of ceiling as a constant or Coefficient = w(x)
Expanding w(x)
-1/4(x^2 -12 )
-1/4 * x² + 1/4 * 12
-1/4x² + 3
The height of ceiling is -1/4x² + 3
Hi hope I helped, ok so first 6×7=42 and its less than 53 so ur answer is 42
Answer:
I= 84
Step-by-step explanation:
for
![I=\int\limits^{}_{} \int\limits^{}_D {x*y} \, dA = \int\limits^{}_{} \int\limits^{}_D {x*y} \, dx*dy](https://tex.z-dn.net/?f=I%3D%5Cint%5Climits%5E%7B%7D_%7B%7D%20%5Cint%5Climits%5E%7B%7D_D%20%7Bx%2Ay%7D%20%20%5C%2C%20dA%20%3D%20%20%5Cint%5Climits%5E%7B%7D_%7B%7D%20%5Cint%5Climits%5E%7B%7D_D%20%7Bx%2Ay%7D%20%20%5C%2C%20dx%2Ady)
since D is the rectangle such that 0<x<3 , 0<y<3
![I=\int\limits^{}_{} \int\limits^{}_D {x*y} \, dA = \int\limits^{3}_{0} \int\limits^{3}_{0} {x*y} \, dx*dy = \int\limits^{3}_{0} {x} \, dx\int\limits^{3}_{0} {y} \, dy = x^{2} /2*y^{2} /2 = (3^{2} /2 - 0^{2} /2)* (3^{2} /2 - 0^{2} /2) = 3^{4} /4 = 81/4](https://tex.z-dn.net/?f=I%3D%5Cint%5Climits%5E%7B%7D_%7B%7D%20%5Cint%5Climits%5E%7B%7D_D%20%7Bx%2Ay%7D%20%20%5C%2C%20dA%20%3D%20%20%5Cint%5Climits%5E%7B3%7D_%7B0%7D%20%5Cint%5Climits%5E%7B3%7D_%7B0%7D%20%7Bx%2Ay%7D%20%20%5C%2C%20dx%2Ady%20%3D%20%20%5Cint%5Climits%5E%7B3%7D_%7B0%7D%20%7Bx%7D%20%20%5C%2C%20dx%5Cint%5Climits%5E%7B3%7D_%7B0%7D%20%7By%7D%20%20%5C%2C%20dy%20%20%3D%20x%5E%7B2%7D%20%2F2%2Ay%5E%7B2%7D%20%2F2%20%3D%20%20%283%5E%7B2%7D%20%2F2%20-%200%5E%7B2%7D%20%2F2%29%2A%20%283%5E%7B2%7D%20%2F2%20-%200%5E%7B2%7D%20%2F2%29%20%3D%203%5E%7B4%7D%20%2F4%20%3D%2081%2F4)