Answer:
![\mathbf{\dfrac{\pi}{6}[5 \sqrt{5}-1]}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Cdfrac%7B%5Cpi%7D%7B6%7D%5B5%20%5Csqrt%7B5%7D-1%5D%7D)
Step-by-step explanation:
Given that:
The surface area (S.A) 
Hence the S.A is of form z = f(x,y)
Then the S.A can be represented with the equation

where :
D = cylinder 
In polar co-ordinates:
D = {(r, θ): 0≤ r ≤ 1, 0 ≤ θ ≤ 2π)
Similarly,
and 
Therefore;



![= [\theta]^{2 \pi}_{0} \dfrac{1}{8}\times \dfrac{2}{3}\begin {bmatrix} (1+4r^2)^{\dfrac{3}{2}}\end {bmatrix}^1_0](https://tex.z-dn.net/?f=%3D%20%5B%5Ctheta%5D%5E%7B2%20%5Cpi%7D_%7B0%7D%20%5Cdfrac%7B1%7D%7B8%7D%5Ctimes%20%5Cdfrac%7B2%7D%7B3%7D%5Cbegin%20%7Bbmatrix%7D%20%281%2B4r%5E2%29%5E%7B%5Cdfrac%7B3%7D%7B2%7D%7D%5Cend%20%7Bbmatrix%7D%5E1_0)
![= 2 \pi \times \dfrac{1}{12}[5^{\dfrac{3}{2}} - 1]](https://tex.z-dn.net/?f=%3D%202%20%5Cpi%20%5Ctimes%20%5Cdfrac%7B1%7D%7B12%7D%5B5%5E%7B%5Cdfrac%7B3%7D%7B2%7D%7D%20-%201%5D)
![\mathbf{=\dfrac{\pi}{6}[5 \sqrt{5}-1]}](https://tex.z-dn.net/?f=%5Cmathbf%7B%3D%5Cdfrac%7B%5Cpi%7D%7B6%7D%5B5%20%5Csqrt%7B5%7D-1%5D%7D)
Let the two numbers be x and y.
x = 4y
x - y = 342
3y = 342
y = 114
xy = 4y * y = 456 * 114 = 51984
Answer:
The speed of the biker is 15
Step-by-step explanation:
First, you call the speed of the walker x, and the speed of the biker 2.5x.
Then you know that speed times time equals distance, so you set up and equation. You do x times 2, which is 2x, and then 2.5x times 2, which is 5x. Then, since the distance between them is 18 miles, the equation would be 5x-2x=18. You would get 3x=18, and x is 6. So 6 is the speed of the walker, and 6 times 2.5 = 15, so the speed of the biker is 15.
Answer:
38°
Step-by-step explanation:
The angle 1 is 63°
The angle 2 is the same of angle 1
so :
63°+79°=142°
then :
180° - 142° = 38°.