Answer:
![A=486\ \text{inches}^2](https://tex.z-dn.net/?f=A%3D486%5C%20%5Ctext%7Binches%7D%5E2)
Step-by-step explanation:
The area of a cube shaped box is given by :
![A=6s^2](https://tex.z-dn.net/?f=A%3D6s%5E2)
Where
s is the side length and A is the surface area
We have, s = 9 inches
So,
![A=6\times (9)^2\\\\A=486\ \text{inches}^2](https://tex.z-dn.net/?f=A%3D6%5Ctimes%20%289%29%5E2%5C%5C%5C%5CA%3D486%5C%20%5Ctext%7Binches%7D%5E2)
So, the required area is
.
Answer:
b. 45
Step-by-step explanation:
895 ÷ 19 = 47.1
47.1 rounded is 50, but since there is no 50 option, we'll just round it down to 45.
--
another way:
round 895 (900) and 19 (20)
divide:
895 ÷ 20 = 45
you'll get 45 either way :)
If
![\nabla f=(ye^x+\sin y)\,\vec\imath+(e^x+x\cos y)\,\vec\jmath](https://tex.z-dn.net/?f=%5Cnabla%20f%3D%28ye%5Ex%2B%5Csin%20y%29%5C%2C%5Cvec%5Cimath%2B%28e%5Ex%2Bx%5Ccos%20y%29%5C%2C%5Cvec%5Cjmath)
then
![\dfrac{\partial f}{\partial x}=ye^x+\sin y\implies f(x,y)=ye^x+x\sin y+g(y)](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20x%7D%3Dye%5Ex%2B%5Csin%20y%5Cimplies%20f%28x%2Cy%29%3Dye%5Ex%2Bx%5Csin%20y%2Bg%28y%29)
Differentiating with respec to
gives
![\dfrac{\partial f}{\partial y}=e^x+x\cos y=e^x+x\cos y+g'(y)](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%20f%7D%7B%5Cpartial%20y%7D%3De%5Ex%2Bx%5Ccos%20y%3De%5Ex%2Bx%5Ccos%20y%2Bg%27%28y%29)
![\implies g'(y)=0\implies g(y)=C](https://tex.z-dn.net/?f=%5Cimplies%20g%27%28y%29%3D0%5Cimplies%20g%28y%29%3DC)
So
is indeed conservative, and
![f(x,y)=ye^x+x\sin y+C](https://tex.z-dn.net/?f=f%28x%2Cy%29%3Dye%5Ex%2Bx%5Csin%20y%2BC)