Answer:
2p+2q
Step-by-step explanation:
With the given order of integration, the interval over D is
![\displaystyle \iiint_D f(x,y,z) \, dV = \int_{-3}^{3} \int_{-\sqrt{9-x^2}}^{\sqrt{9-x^2}} \int_{-\sqrt{9-x^2-z^2}}^{\sqrt{9-x^2-z^2}} f(x,y,z) \, dy \, dz \, dx](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ciiint_D%20f%28x%2Cy%2Cz%29%20%5C%2C%20dV%20%3D%20%5Cint_%7B-3%7D%5E%7B3%7D%20%5Cint_%7B-%5Csqrt%7B9-x%5E2%7D%7D%5E%7B%5Csqrt%7B9-x%5E2%7D%7D%20%5Cint_%7B-%5Csqrt%7B9-x%5E2-z%5E2%7D%7D%5E%7B%5Csqrt%7B9-x%5E2-z%5E2%7D%7D%20f%28x%2Cy%2Cz%29%20%5C%2C%20dy%20%5C%2C%20dz%20%5C%2C%20dx)
It's kinda blurry. I'm sorry!!
Answer:
Distance= 6.6 miles
Bearing= N 62.854°W
Step-by-step explanation:
Let's determine angle b first
Angle b=20° (alternate angles)
Using cosine rule
Let the distance between the liner and the port be x
X² =8.8²+2.4²-2(8.8)(2.4)cos20
X²= 77.44 + 5.76-(39.69)
X²= 43.51
X= √43.51
X= 6.596
X= 6.6 miles
Let's determine the angles within the triangle using sine rule
2.4/sin b = 6.6/sin20
(2.4*sin20)/6.6= sin b
0.1244 = sin b
7.146= b°
Angle c= 180-20-7.146
Angle c= 152.854°
For the bearing
110+7.146= 117.146
180-117.146= 62.854°
Bearing= N 62.854°W
QUESTION:- EVALUATE THE EXPRESSION
EXPRESSION:-
![{( \sqrt[6]{27 {a}^{3} {b}^{4} } )}^{2}](https://tex.z-dn.net/?f=%7B%28%20%5Csqrt%5B6%5D%7B27%20%7Ba%7D%5E%7B3%7D%20%20%7Bb%7D%5E%7B4%7D%20%7D%20%29%7D%5E%7B2%7D%20)
<h3>ANSWER-></h3>
![\\ ANSWER->3ab \sqrt[3]{b} \: \: \: \: \: or \: \: 3a {b}^{ \frac{4}{3} }](https://tex.z-dn.net/?f=%20%5C%5C%20ANSWER-%3E3ab%20%5Csqrt%5B3%5D%7Bb%7D%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20or%20%5C%3A%20%5C%3A%203a%20%7Bb%7D%5E%7B%20%5Cfrac%7B4%7D%7B3%7D%20%7D%20)