Answer:

Step-by-step explanation:
Given that:
![\int \int _R 4xye^{x^2 \ y} \ dA, R = [0,1]\times [0,7]](https://tex.z-dn.net/?f=%5Cint%20%5Cint%20_R%204xye%5E%7Bx%5E2%20%5C%20y%7D%20%5C%20dA%2C%20R%20%3D%20%5B0%2C1%5D%5Ctimes%20%5B0%2C7%5D)
The rectangle R = [0,1] × [0,7]
R = { (x,y): x ∈ [0,1] and y ∈ [0,7] }
R = { (x,y): 0 ≤ x ≤ 1 and 0 ≤ x ≤ 7 }




![\int \int _R \ 4xy e^{x^2 \ y} \ dA = \dfrac{4}{2}[e^y -1]^7_0 \ dy](https://tex.z-dn.net/?f=%5Cint%20%5Cint%20_R%20%5C%204xy%20e%5E%7Bx%5E2%20%5C%20y%7D%20%20%5C%20dA%20%3D%20%20%5Cdfrac%7B4%7D%7B2%7D%5Be%5Ey%20-1%5D%5E7_0%20%5C%20dy)
![\int \int _R \ 4xy e^{x^2 \ y} \ dA = 2 [(e^7 -7)-(e^0 -0)]](https://tex.z-dn.net/?f=%5Cint%20%5Cint%20_R%20%5C%204xy%20e%5E%7Bx%5E2%20%5C%20y%7D%20%20%5C%20dA%20%3D%20%202%20%5B%28e%5E7%20-7%29-%28e%5E0%20-0%29%5D)
![\int \int _R \ 4xy e^{x^2 \ y} \ dA = 2 [(e^7 -7)-1]](https://tex.z-dn.net/?f=%5Cint%20%5Cint%20_R%20%5C%204xy%20e%5E%7Bx%5E2%20%5C%20y%7D%20%20%5C%20dA%20%3D%20%202%20%5B%28e%5E7%20-7%29-1%5D)

1) 3 / x = 4 / 6
2) 4x = 18
3) x = 4.5
4) 4.5 + 6 + 4.5 + 6 = 21
Answer: 21 m
Answer:
B
Step-by-step explanation:
To solve this, we use ratio.
Firstly, we need to know the number of hours traveled. The total number of hours traveled = x+y
Ratio of this used by high speed train = x/(x +y).
Total distance traveled before they meet = [x/(x + y)] × z
For low speed train = [y/(x + y)] × z.
The difference would be distance by high speed train - distance by low speed train.
= z [ (x - y)/x + y)]
Tan3A=tan(2A+A)
We know that , tan(x+y)=(tanx + tany)/(1 - (tanx)(tany))
tan(2A+A)=(tan2A+tanA)/(1 - (tan2A)(tanA))— (1)
We know that , tan2x=2tanx/(1 - tan^2x)
So, by substituting tan2A in (1),we get,
=[2tanA/(1 - tan^2A) + tanA]/1- (2tanA/(1 - tan^2A))(tanA)]
=[2tanA+tanA - tan^3A]/[1 - tan^2A - 2tan^2A]
=[3tanA - tan^3A]/[1-3tan^2A]
Therefore, tan3A= [3tanA - tan^3A]/[1-3tan^2A]