If you project S onto the (x,y)-plane, it casts a "shadow" corresponding to the trapezoidal region
T = {(x,y) : 0 ≤ x ≤ 10 - y and -4 ≤ y ≤ 4}
Let z = f(x, y) = √(16 - y²) and z = g(x, y) = -√(16 - y²), each referring to one half of the cylinder to either side of the plane z = 0.
The surface element for the "positive" half is
dS = √(1 + (∂f/∂x)² + (∂f/dy)²) dx dy
dS = √(1 + 0 + 4y²/(16 - y²)) dx dy
dS = √((16 + 3y²)/(16 - y²)) dx dy
The the surface integral along this half is
You'll find that the integral over the "negative" half has the same value, but multiplied by -1. Then the overall surface integral is 0.
Sorrysorrysorrysorrysorry
Answer:
x = 128 degrees
Step-by-step explanation:
First figure out the angle inside the triangle
1st step: add 55 and 73 together:
⇒ 55 + 73 = 128
2nd step: subtract the sum from 180:
⇒ 180 - 128 = 52
to find x, take 180 again and then subtract 52 from it:
⇒ 180 - 52 = 128
Answer:
(a) x = 260.52
(b) x = 249.24
(c) x = 265.22
(d) x = 297.65
Step-by-step explanation:
Here,
Mean = 
= 300
Standard deviation = 
= 47
(a)
Using standard normal table,
P(Z > z) = 80%
1 - P(Z < z) = 0.8
P(Z < z) = 1 - 0.8
P(Z < -0.52) = 0.2
z = -0.84
Using z-score formula,
x = z × σ + μ
x = -0.84 × 47 + 300 = 260.52
(b) Using standard normal table,
P(Z < z) = 14%
P(Z < -1.08) = 0.1
4
z = -1.08
Using z-score formula,
x = z × σ + μ
x = -1.08 × 47 + 300 = 249.24
(c)
Using standard normal table,
P(Z < z) = 23%
P(Z < -0.74) = 0.243
z = -0.714
Using z-score formula,
x = z × σ + μ
x = -0.74 × 47 + 300 = 265.22
(d) Using standard normal table,
P(Z > z) = 52%
1 - P(Z < z) = 0.52
P(Z < z) = 1 - 0.52
P(Z < -0.25) = 0.4
8
z = -0.05
Using z-score formula,
x = z × σ + μ
x = -0.05 × 47 + 300 = 297.65
Answer:
Since a triangle has 180 interior degrees, and 81+57=138, and 180-138=42, you need to find _*2+18=42. (Hint, you can subtract the 18 from the 42, and find _*2=24).
Step-by-step explanation:
