I believe that the correct answer is 72
Hope this helps!<3
PEMDAS
multiply 2x2=4 and simplify the equation by looking for like terms to combine.
4+16x+y+34=0
4 and 34 are like terms so add them.
the simplified expression is 16x+y+38
S casts a "shadow" on the x-y plane given by the set
![S' = \left\{(x,y) ~:~ -1 \le x \le 1 \text{ and } x^2 \le y \le 1\right\}](https://tex.z-dn.net/?f=S%27%20%3D%20%5Cleft%5C%7B%28x%2Cy%29%20~%3A~%20-1%20%5Cle%20x%20%5Cle%201%20%5Ctext%7B%20and%20%7D%20x%5E2%20%5Cle%20y%20%5Cle%201%5Cright%5C%7D)
Each cross section is a square whose side length is determined by the vertical distance between y = 1 and y = x², which is |1 - x²|. But since -1 ≤ x ≤ 1, this distance simplifies to 1 - x².
The volume of an infinitesimally thin section is then (1 - x²)² ∆x (where ∆x represents its thickness), and so the volume of S is
![\displaystyle \int_{-1}^1 (1-x^2)^2 \, dx = \int_{-1}^1 (1 - 2x^2 + x^4) \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B-1%7D%5E1%20%281-x%5E2%29%5E2%20%5C%2C%20dx%20%3D%20%5Cint_%7B-1%7D%5E1%20%281%20-%202x%5E2%20%2B%20x%5E4%29%20%5C%2C%20dx)
The integrand is even, so this integral is equal to twice the integral over [0, 1] :
![\displaystyle \int_{-1}^1 (1-x^2)^2 \, dx = 2 \int_0^1 (1 - 2x^2 + x^4) \, dx = 2 \left(1 - \frac23 + \frac15\right) = \boxed{\frac{16}{15}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cint_%7B-1%7D%5E1%20%281-x%5E2%29%5E2%20%5C%2C%20dx%20%3D%202%20%5Cint_0%5E1%20%281%20-%202x%5E2%20%2B%20x%5E4%29%20%5C%2C%20dx%20%3D%202%20%5Cleft%281%20-%20%5Cfrac23%20%2B%20%5Cfrac15%5Cright%29%20%3D%20%5Cboxed%7B%5Cfrac%7B16%7D%7B15%7D%7D)
Answer:
addition +
Step-by-step explanation:
You always do what's in parenthesis first.
Answer:
5√5
Step-by-step explanation: